Instructor Manual and Resource for Calculus with Applications, 12th Edition by Ace Test Banks

Instructor Manual for Calculus with Applications, 12th Edition.

richard@qwconsultancy.com

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CONTENTS PREFACE...................................................................................................................................................................

HINTS FOR TEACHING CALCULUS WITH APPLICATIONS ............................................................................

PRETESTS ................................................................................................................................................................. xix ANSWERS TO PRETESTS ....................................................................................................................................... xxvii FINAL EXAMINATIONS ......................................................................................................................................... xxix ANSWERS TO FINAL EXAMINATIONS............................................................................................................... xliii

SOLUTIONS TO ALL EXERCISES CHAPTER R ALGEBRA REFERENCE R.1 Polynomials ..................................................................................................................................................

R.2 Factoring .......................................................................................................................................................

R.3 Rational Expressions ....................................................................................................................................

R.4 Equations ......................................................................................................................................................

R.5 Inequalities ...................................................................................................................................................

R.6 Exponents .....................................................................................................................................................

R.7 Radicals ........................................................................................................................................................

CHAPTER 1 LINEAR FUNCTIONS 1.1 Slopes and Equations of Lines ......................................................................................................................

1.2 Linear Functions and Applications................................................................................................................

1.3 The Least Squares Line .................................................................................................................................

Chapter 1 Review Exercises..................................................................................................................................

Extended Application: Predicting Life Expectancy ..............................................................................................

iii

CHAPTER 2 NONLINEAR FUNCTIONS 2.1 Properties of Functions..................................................................................................................................

2.2 Quadratic Functions; Translation and Reflection ..........................................................................................

2.3 Polynomial and Rational Functions .............................................................................................................. 102 2.4 Exponential Functions ................................................................................................................................... 115 2.5 Logarithmic Functions .................................................................................................................................. 125 2.6 Applications: Growth and Decay; Mathematics of Finance .......................................................................... 137 Chapter 2 Review Exercises.................................................................................................................................. 146 Extended Application: Power Functions ............................................................................................................... 162

CHAPTER 3 THE DERIVATIVE 3.1 Limits ............................................................................................................................................................ 163 3.2 Continuity...................................................................................................................................................... 177 3.3 Rates of Change ............................................................................................................................................ 184 3.4 Definition of the Derivative .......................................................................................................................... 195 3.5 Graphical Differentiation .............................................................................................................................. 219 Chapter 3 Review Exercises.................................................................................................................................. 223 Extended Application: A Model for Drugs Administered Intravenously .............................................................. 235

CHAPTER 4 CALCULATING THE DERIVATIVE 4.1 Techniques for Finding Derivatives .............................................................................................................. 237 4.2 Derivatives of Products and Quotients .......................................................................................................... 247 4.3 The Chain Rule ............................................................................................................................................. 258 4.4 Derivatives of Exponential Functions ........................................................................................................... 269 4.5 Derivatives of Logarithmic Functions ........................................................................................................... 283 Chapter 4 Review Exercises.................................................................................................................................. 296 Extended Application: Electric Potential and Electric Field ................................................................................. 308

CONTENTS

CHAPTER 5 GRAPHS AND THE DERIVATIVE 5.1 Increasing and Decreasing Functions ............................................................................................................ 309 5.2 Relative Extrema ........................................................................................................................................... 324 5.3 Higher Derivatives, Concavity, and the Second Derivative Test .................................................................. 339 5.4 Curve Sketching

............................................................................................................................... 362

Chapter 5 Review Exercises.................................................................................................................................. 383 Extended Application: A Drug Concentration Model for Orally Administered Medications ............................... 400

CHAPTER 6 APPLICATIONS OF THE DERIVATIVE 6.1 Absolute Extrema .......................................................................................................................................... 401 6.2 Applications of Extrema................................................................................................................................ 411 6.3 Further Business Applications....................................................................................................................... 432 6.4 Implicit Differentiation ................................................................................................................................. 440 6.5 Related Rates ................................................................................................................................................. 457 6.6 Differentials: Linear Approximation ............................................................................................................. 466 Chapter 6 Review Exercises.................................................................................................................................. 472 Extended Application: A Total Cost Model for a Training Program .................................................................... 483

CHAPTER 7 INTEGRATION 7.1 Antiderivatives .............................................................................................................................................. 485 7.2 Substitution ................................................................................................................................................... 497 7.3 Area and the Definite Integral ....................................................................................................................... 507 7.4 The Fundamental Theorem of Calculus ........................................................................................................ 521 7.5 The Area Between Two Curves .................................................................................................................... 538 7.6 Numerical Integration ................................................................................................................................... 556 Chapter 7 Review Exercises.................................................................................................................................. 570 Extended Application: Estimating Depletion Dates for Minerals ......................................................................... 588

CONTENTS

CHAPTER 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION 8.1 Integration by Parts ....................................................................................................................................... 589 8.2 Volume and Average Value .......................................................................................................................... 601 8.3 Continuous Money Flow ............................................................................................................................... 612 8.4 Improper Integrals ......................................................................................................................................... 619 Chapter 8 Review Exercises.................................................................................................................................. 633 Extended Application: Estimating Learning Curves in Manufacturing with Integrals ......................................... 643

CHAPTER 9 MULTIVARIABLE CALCULUS 9.1 Functions of Several Variables...................................................................................................................... 645 9.2 Partial Derivatives ........................................................................................................................................ 655 9.3 Maxima and Minima ..................................................................................................................................... 671 9.4 Lagrange Multipliers ..................................................................................................................................... 686 9.5 Total Differentials and Approximations ........................................................................................................ 702 9.6 Double Integrals ............................................................................................................................................ 711 Chapter 9 Review Exercises.................................................................................................................................. 730 Extended Application: Using Multivariable Fitting to Create a Response Surface Design .................................. 749

CHAPTER 10 DIFFERENTIAL EQUATIONS 10.1 Solutions of Elementary and Separable Differential Equations .................................................................. 751 10.2 Linear First-Order Differential Equations ................................................................................................... 769 10.3 Euler’s Method ............................................................................................................................................ 778 10.4 Applications of Differential Equations........................................................................................................ 788 Chapter 10 Review Exercises ............................................................................................................................... 800 Extended Application: Pollution of a Lake ........................................................................................................... 813

CHAPTER 11 PROBABILITY AND CALCULUS 11.1 Continuous Probability Models ................................................................................................................... 815 11.2 Expected Value and Variance of Continuous Random Variables ............................................................... 828 11.3 Special Probability Density Functions ........................................................................................................ 845 Chapter 11 Review Exercises ................................................................................................................................ 857 Extended Application: Exponential Waiting Times .............................................................................................. 871 vi

CONTENTS

CHAPTER 12 SEQUENCES AND SERIES 12.1 Geometric Sequences ................................................................................................................................. 873 12.2 Annuities: An Application of Sequences .................................................................................................... 882 12.3 Taylor Polynomials at 0 .............................................................................................................................. 896 12.4 Infinite Series .............................................................................................................................................. 914 12.5 Taylor Series at 0 ........................................................................................................................................ 923 12.6 Newton’s Method ........................................................................................................................................ 940 12.7 L’Hospital’s Rule ........................................................................................................................................ 951 Chapter 12 Review Exercises ................................................................................................................................ 962 Extended Application: Living Assistance and Subsidized Housing ...................................................................... 979

CHAPTER 13 THE TRIGONOMETRIC FUNCTIONS 13.1 Definitions of the Trigonometric Functions ................................................................................................ 981 13.2 Derivatives of Trigonometric Functions ..................................................................................................... 993 13.3 Integrals of Trigonometric Functions .......................................................................................................... 1010 Chapter 13 Review Exercises ................................................................................................................................ 1019 Extended Application: The Shortest Time and the Cheapest Path ........................................................................ 1031

CONTENTS

vii

PREFACE This book provides several resources for instructors using Calculus with Applications, Twelfth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey, with Katherine Ritchey, Sarah Ritchey Patterson, and Blain Patterson. ·

Hints for teaching Calculus with Applications are provided as a resource for faculty.

One open-response form and one multiple-choice form of a pretest are provided. These tests are an aid to instructors in identifying students who may need assistance.

One open-response form and one multiple-choice form of a final examination are provided.

Solutions for nearly all of the exercises in the textbook are included. Solutions are usually not provided for exercises with open-response answers.

PREFACE

HINTS FOR TEACHING CALCULUS WITH APPLICATIONS Algebra Reference Some instructors obtain best results by going through this chapter carefully at the beginning of the semester. Others find it better to refer to it as needed throughout the course. Use whichever method works best for your students. As in the previous edition, we refer to the chapter as a “Reference” rather than a “Review,” and the regular page numbers don’t begin until Chapter 1. We hope this will make your students less anxious if you don’t cover this material.

Chapter 1 Instructors sometimes go to either of two extremes in this chapter and the next. Some feel that their students have already covered enough precalculus in high school or in previous courses, and consequently begin with Chapter 3. Unfortunately, if they are wrong, their students may do poorly. Other instructors spend at least half a semester on Chapters 1 and 2 and the algebra reference chapter, and subsequently have little time for calculus. Such a course should not be labeled as calculus. We recommend trying to strike a balance, which may still not make all your students happy. A few may complain that the review of algebra, functions, and graphs is too quick; such students should be sent to a more basic course. Those students who are familiar with this material may become lazy and develop habits that will hurt them later in the course. You may wish to assign a few challenging exercises to keep these students on their toes. Chapter 1 of Calculus with Applications is identical to Chapter 1 of Finite Mathematics and may be skipped by students who have already taken a course using that text. In this edition, we have streamlined the chapter from four to three sections, allowing instructors to reach the calculus material more quickly.

Section 1.1 This section and the next may seem fairly basic to students who covered linear functions in high school. Nevertheless, many students who have graphed hundreds of lines in their lifetime still lack a thorough understanding of slope, which hampers their understanding of the derivative. Such students could benefit from doing dozens of exercises similar to 43–46. Perpendicular lines are not used in future chapters and could be skipped if you are in a hurry.

Section 1.2 Much recent research has been devoted to students’ misunderstandings of the function concept. Such misunderstandings are among the major impediments to learning calculus. One way to help students is to study a simple class of functions first, as we do in this section. In this edition, even more of the general material on functions, including domain and range, is postponed until Chapter 2. Supply and demand provides the students’ first experience with a mathematical model. Spend time developing both the economics and the mathematics involved. Stress that for cost, revenue, and profit functions, x represents the number of units. For supply and demand functions, we use the economists’ notation of q to represent the number of units. Emphasize the difference between the profit earned on 100 units sold as opposed to the number of units that must be sold to produce a profit of $100.

Section 1.3 The statistical functions on a calculator can greatly simplify these calculations, allowing more time for discussion and further examples. In this edition, we use “parallel presentation” to allow the instructor a choice on the extent technology is used. This section may be skipped if you are in a hurry, but your students can benefit from the realistic model and the additional work with equations of lines.

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Hints for Teaching Calculus with Applications

Section 2.1 After learning about linear functions in the previous chapter, students now learn about functions in general. This concept is critical for success in calculus. Unless sufficient time is devoted to this section, the results will become apparent later when students don’t understand the derivative. One device that helps students distinguish f ( x + h) from f ( x) + h is to use a box in place of the letter x, as we do in this section before Example 6.

Section 2.2 This section combines the topics of quadratic functions and translation and reflection, with a minimal amount of material on completing the square. Our experience is that students graph quadratics most easily by first finding the y-intercept, then finding the x-intercepts when they exist (using factoring or the quadratic formula), and finding the vertex last by locating the point midway between the x-intercepts or, if the quadratic formula was used, by letting x = -b /(2a). Quadratics are among a small group of functions that can be analyzed completely with ease, so they are used throughout the text. On the other hand, the advent of graphing calculators has made ease of graphing less important, so we rely on quadratics less than in previous editions. Some instructors pressed for time may choose to skip translations and reflections. But we have found that students who understand that the graph of f ( x) = 5 - 4 - x is essentially the same as the graph of f ( x) = x , just shifted and reflected, will have an easier time when using the derivative to graph functions. Since students are familiar with very few classes of functions at this point, it helps to work with functions defined solely by their graphs, such as Exercises 35–38. Exercises 43–50 cover stretching and shrinking of graphs in the vertical and horizontal directions. Covering these exercises carefully will not only give students a better grasp of functions, but will help them later to interpret the chain rule.

Section 2.3 Graphing calculators have made point plotting of functions less important than before. Plotting points by hand should not be entirely neglected, however, because a small amount is helpful when using the derivative to graph functions. The two main goals of this section are to have an understanding of what an n-th degree polynomial looks like, and to be able to find the asymptotes of a rational function. Students who master these ideas will be better prepared for the chapter on curve sketching. Exercise 59 is the first of several in this chapter asking students to find what type of function best fits a set of data. (See also Section 2.4, Exercises 57, 58 and 59, and Review Exercise 111.) The class can easily get bogged down in these exercises, particularly if you decide to explore the regression features in a calculator such as the TI-84 Plus C. But there is a powerful payoff in terms of mastery of functions for the student who succeeds at these exercises.

Section 2.4 Some instructors may prefer at this point to continue with Chapter 3 and to postpone discussion of the exponential and logarithmic functions until later. The overwhelming preference of instructors we surveyed, however, was to cover exponential and logarithmic functions early and then to use these functions throughout the rest of the course. Instructors who wish to postpone this material will also need to omit for now those examples and exercises in Sections 3.1–4.3 that refer to exponential and logarithmic functions. Students typically have no problem with f ( x) = 2 x , but the number e often remains a mystery. Like π, the number e is a transcendental number, but students have had years of schooling to get used to π. Have your students approximate e with a calculator, as the textbook does before the definition of e. Notice how we use compound interest to help students get a handle on this number.

Hints for Teaching Calculus with Applications

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Section 2.5 Logarithms are a very difficult topic for many students. It’s easy to say that a logarithm is just an exponent, but the fact that it is the exponent to which one must raise the base to get the number whose logarithm we are calculating is a rather obtuse concept. Therefore, spend lots of time going over examples that can be done without a calculator, such as log 2 8. Students will also tend to come up with many incorrect pseudoproperties of logarithms, similar in form to the properties of logarithms given in this section. Take as much time and patience as necessary in gently correcting the many errors students inevitably will make at first. Even after receiving a thorough treatment of logarithms, some students will still be stumped when solving a problem such as Example 8. Some of these students can get the correct answer using trial-and-error. The instructor should take consolation in the fact that at least such a student understands exponentials better than the one who uses logarithms incorrectly to solve Example 8 and comes up with the nonsensical answer t = -7.51 without questioning whether this makes sense. Be sure to teach your students to question the reasonableness of their answers; this will help them catch their errors.

Section 2.6 This section gives students much needed practice with exponentials and logarithms, and the applications keep students interested and motivated. Instructors should keep this in mind and not worry about having students memorize formulas. We have removed the formulas for present value in this edition, having decided that it’s better for students to just solve the compound amount formula for P. This reduces by two the number of formulas that students need to remember. There is a summary of graphs of basic functions in the end-of-chapter review.

Section 3.1 This is the first section on calculus, and perhaps the most important, since limits are what really distinguish calculus from algebra. Students will have the best understanding of limits if they have studied them graphically (as in Exercises 11–18), numerically (as in Exercises 21–26), and analytically (as in Exercises 35–62). The graphing calculator is a powerful tool for studying limits. Notice in Example 12 (c) and (d) that we have modified the method of finding limits at infinity by dividing by the highest power of x in the denominator, which avoid the problem of division by 0.

Section 3.2 The section on continuity should be straightforward if students have mastered limits from the previous section.

Section 3.3 This section introduces the derivative, even though that term doesn’t appear until the next section. In a class full of business and social science majors, an instructor may wish to place less emphasis on velocity, an approach more suited to physics majors. But we have found velocity to be the manifestation of the derivative that is most intuitive to all students, regardless of their major. Instructors in a hurry can skip the material on estimating the instantaneous rate of change from a set of data, but it helps solidify students’ understanding of the derivative by giving them one more point of view.

Section 3.4 Students who have learned differentiation formulas in high school usually want (and deserve) some explanation of why they need to learn to take derivatives using the definition. You might try explaining to your students that getting the right formula is not the only goal; graphing calculators can give derivatives numerically. The most important thing for students to learn is the concept of the derivative, which they don’t learn if they only memorize differentiation formulas. Zooming in on a function with a graphing calculator until the graph appears to be a straight line gives students a very concrete image of what the derivative means. After students have learned the differentiation formulas, they may forget about the definition of derivative. We have found that if we want them to use the definition on a test, it is important to say so clearly and emphatically, or they will simply use the shortcut formulas.

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Section 3.5 One way to get students to focus on the concept of the derivative, rather than the mechanics, is to emphasize graphical differentiation. We have therefore devoted an entire section to this topic. Graphical differentiation is difficult for many students because there are no formulas to rely on. One must thoroughly understand what’s going on to do anything. On the other hand, we have seen students who are weak in algebra but who possess a good intuitive grasp of geometry find this topic quite simple.

Section 4.1 Students tend to learn these differentiation formulas fairly quickly. These and the formulas in the next few sections are included in a summary at the end of the chapter.

Section 4.2 The product and quotient rules are more difficult for students to keep straight than those of the previous section. People seem to remember these rules better if they use an incantation such as “The first times the derivative of the second, plus the second times the derivative of the first.” Some instructors have argued that this formulation of the product rule doesn’t generalize well to products of three or more functions, but that’s not important at this level. Some instructors allow their students to bring cards with formulas to the tests. This does not eliminate the need for students to understand the use of the formulas, but it does eliminate the anxiety students may have about forgetting a key formula under the pressure of an exam.

Section 4.3 No matter how many times an instructor cries out to his or her students, “Remember the chain rule!”, many will still forget this rule at some time later in the course. But if a few more students remember the rule because the instructor reminds them so often, such reminders are worthwhile.

Section 4.4 and 4.5 In going through these sections, you may need to frequently refer to the rules of differentiation in the previous sections. You may also need to review the last three sections of Chapter 2.

Section 5.1 and 5.2 If students have understood Chapter 3, then the connection between the derivative, increasing and decreasing functions, and relative extrema should be obvious, and these sections should go quickly and smoothly.

Section 5.3 Students often confuse concave downward and upward with increasing and decreasing; carefully go over Figure 31 or the equivalent with your class.

Section 5.4 Graphing calculators have made curve sketching techniques less essential, but curve sketching is still one of the best ways to unify the various concepts introduced in this and the previous two chapters. Students should use graphing calculators to check their work. Because this section is the culmination of many ideas, students often find it difficult and start to forget things they previously knew. For example, a student might state that a function is increasing on an interval and then draw it decreasing. The best solution seems to be lots of practice with immediate help and feedback from the instructor. Students sometimes stumble over this topic because they use the rules for differentiation incorrectly, or because they make mistakes in algebra when simplifying. Exercises such as 29–34 are excellent for testing whether students really grasp the concepts.

Section 6.1 This section should not be conceptually difficult, but students need constant reminders to check the endpoints of an interval when finding the absolute maxima and minima.

Hints for Teaching Calculus with Applications

Section 6.2 This section is one of the high points of the course. Some of the best applications of calculus involve maxima and minima. Notice that some exercises have a maximum or minimum at the endpoint of an interval, so students cannot ignore checking endpoints. Almost everyone finds this material difficult because most people are not skilled at word problems. Remind your students that if they ever wonder whether mathematics is of any use, this section will show them. Why are word problems so difficult? One theory is that word problems require the use of two different modes of thinking, which students are using simultaneously for the first time. People use words in daily life without difficulty, but when they study mathematics, they often turn off that part of their brain and begin thinking in a very formal, mechanical way. In word problems, both modes of thinking must be active. If and when the NCTM Standards become widely accepted in the schools, children will get more practice at an early age in such ways of thinking. Meanwhile, the steps for solving applied problems given in this book might make the process a little more straightforward, and hence achievable by the average student.

Section 6.3 This section continues the ideas of the previous one. The point of studying economic lot size should not be to apply Equation (3), but to learn how to apply calculus to solve such problems. We therefore urge you to cover Exercises 17–20, in which we vary the assumptions, so Equation (3) does not necessarily apply. In this edition, we have changed the presentation to be consistent with that of business texts. The material on elasticity can be skipped, but it is an important application that should interest students who have studied even a little economics.

Section 6.4 There are two main reasons for covering implicit differentiation. First, it reinforces the chain rule. Second, it is needed for doing related rate problems. If you skip related rates due to lack of time, it is not essential to cover implicit differentiation, either.

Section 6.5 Related rate applications are less important than applied extrema problems, but they use some of the same skills in setting up word problems, and for that reason are worth covering. The best application exercises are under the heading “Physical Sciences,” because those are the exercises in which no formula is given to the student; the student must construct a formula from the words. The geometrical formulas needed are kept to a minimum: the Pythagorean theorem, the area of a circle, the volume of a sphere, the volume of a cone, and the volume of a cylinder with a triangular cross section. Some instructors allow their students to use a card with such formulas on the exam. These formulas are summarized in a table in the back of the book.

Section 6.6 Differentials may be skipped by instructors in a hurry; you need not fear that this omission will hamper your students in the chapter on integration. The differentials used there are not the same as those used here, and the required techniques are easily picked up when integration by substitution is covered. The exception is for instructors who intend to cover Section 10.3 on Euler’s Method, since differentials are used to motivate and derive that method. As in the previous edition, our presentation of differentials emphasizes linear approximation, a topic of considerable importance in mathematics.

Section 7.1 Students sometimes start to get differentiation and antidifferentiation confused when they reach this section. Some believe the antiderivative of x-2 is (-1/3) x-3; after all, if n is −2, isn’t n + 1 = -3? Carefully clarify this point.

Section 7.2 The main difficulty here is teaching students what to choose as u. The advice given before Example 3 should be helpful.

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Section 7.3 Some instructors who are pressed for time go lightly over the topic of the area as the limit of the sum of rectangular areas. This is possible, but care should be taken that students don’t lose track of what the definite integral represents. Also, a light treatment here lessens the excitement of the Fundamental Theorem of Calculus. We have continued the trend of the previous edition in covering three ways of approximating a definite integral with rectangles: the right endpoint, the left endpoint, and the midpoint. The trapezoidal rule is briefly introduced here as the average of the left sum and the right sum.

Section 7.4 The Fundamental Theorem of Calculus should be one of the high points of the course. Make a big deal about how the theorem unifies these two separate topics of area as a limit of sums and the antiderivative. When using substitution on a definite integral, the text recommends changing the limits and the variable of integration. (See Example 4 and the Caution which follows the example.) Some instructors prefer instead to have their students solve an indefinite integral, and then to evaluate the integral using the limits on x. One advantage of this method is that students don’t have to remember to change the limits. This method also has two disadvantages. The first is that it takes slightly longer, since it requires changing the integral to u and then back to x. Second, it prevents students at this stage from solving problems such as ò

1/2

that the integral ò

x 1 - 16 x 4 dx, which can be solved using the substitution u = 4 x 2 and the fact

1 - u 2 du represents the area of a quarter circle. This is one section in which we deliberately did not

use more than one method of presentation, because this would inevitably lead to confusion in the minds of some students.

Section 7.5 This section gives more motivation to the topic of integration. Consumers’ and producers’ surplus are important, realistic applications. We have downplayed sketching the curves that bound the area under consideration. Such sketches take time and are not necessary in solving these problems. But they clarify what is happening and make it possible to avoid memorizing formulas. Using a graphing calculator to sketch the curves can be helpful.

Section 7.6 The ubiquity of computers and graphing calculators has made numerical integration more important. Rather than computing a definite integral by an integration technique, one can just as easily enter the function into a calculator and press the integration key. Point out to students that this is valuable when the function to be integrated is complicated. On the other hand, using the antiderivative makes it easier to see the effect of changing the upper limit, say, from 2 to 3, or for working with the definite integral when one or both limits are parameters, such as a and b, rather than numbers. Simpson’s rule is the most accurate of the simple integration formulas. To achieve greater accuracy, a more complicated method must be used. This is why, unlike the trapezoidal rule, Simpson’s rule is actually used by mathematicians and engineers. You may wish to give your students the programs for the trapezoidal rule and Simpson’s rule in The Graphing Calculator Manual that is available with the text.

Section 8.1 Students usually find column integration simpler than traditional integration by parts. We show both methods to give instructors a choice, and also to emphasize that the two methods are equivalent. Column integration makes organizing the details simpler, but is no more mechanical than the traditional method, as some have mistakenly claimed. At Hofstra University, students even use this method when neither the instructor nor the book discuss it. They find out about it from other students, and so it has become an underground method. Some instructors feel that students will lose any theoretical understanding of what they are doing if they use this method. Our experience is that almost no students at this level have a theoretical understanding of what integration by parts is about, but the better students can at least master the mechanics. With column integration, almost all of the students master the mechanics.

Section 8.2 and 8.3 These two sections give more applications of integration. Coverage of either section is optional.

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Section 8.4 Improper integration is not really an application of integration, but it makes further applications of integration possible. Many mathematicians use shorthand notation such as the following: ò

¥ -x

dx = -e-x

¥ 0

= 0 - (-1) = 1.

For students at this level, it may be best to avoid the shorthand notation.

Section 9.1 The major difficulty students have with this section, and indeed with this entire chapter, is that they cannot visualize surfaces in 3-dimensional space, even though they live there. Fortunately, such visualization is not really necessary for doing the exercises in this chapter. A student who wants to explore what various surfaces look like can use any of the commercial or public domain computer programs available.

Section 9.2 Students who have mastered the differentiation techniques should have no difficulty with this section.

Section 9.3 This section corresponds to the section on applied extrema problems in the chapter on applications of the derivative, but with less emphasis on word problems. In Exercise 36, we revisit the topic of the least squares line, first covered in Section 1.3.

Section 9.4 Lagrange multipliers are an important application of calculus to economics. At some colleges, the business school is very insistent that the mathematics department cover this material.

Section 9.5 This section corresponds to the section on differentials in the chapter on applications of the derivative.

Section 9.6 Students who have trouble visualizing surfaces in 3-dimensional space are sometimes bothered by double integrals over variable regions. Instructors should assure such students that all they need to do is draw a good sketch of the region in the xy-plane, and not try to draw the volume in three dimensions.

Section 10.1 Differential equations of the form dy /dx = f ( x) are treated lightly in this section because they were already covered in the chapter on integration, although the terminology and notation were different then. Remind students that solving such differential equations is the same as antidifferentiation. The rest of the section is on separable differential equations. Students sometimes have trouble with this section because they have forgotten how to find an antiderivative, particularly one requiring substitution.

Section 10.2 If you get this far, your students have covered most of the techniques for solving first-order differential equations. You can find further techniques in differential equations texts, but most first-order equations that come up in real applications are separable, linear, or not solvable by any exact method.

Section 10.3 This is one of the most calculator/computer-intensive sections of the book. In practice, more accurate methods than Euler’s method are almost always used, but Euler’s method introduces students to a way of solving problems that would otherwise be beyond their grasp. You may wish to give your students the program for Euler’s method in The Graphing Calculator Manual that is available with the text.

Section 10.4 This is a fun section of assorted applications, showing students that the techniques they have learned were not in vain. You can pick and choose those applications of greatest interest to yourself and your students. You can also supplement

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the text with applications from other sources, such as those published by the Consortium for Mathematics and Its Applications Project (COMAP).

Chapter 11 Probability is one of the best applications of calculus around. In fact, statistics instructors sometimes feel the temptation to start discussing the definite integral even when their students know no calculus. This chapter is just a brief introduction, but it covers some of the most important concepts, such as mean, variance, standard deviation, expected value, and probability as the area under the curve. The third section covers three of the most important continuous probability distributions: uniform, exponential, and normal.

Chapter 12 Except for geometric series, most of the material in this chapter is of greater interest to the professional mathematician or engineer than to the student in business, management, economics, life sciences or social sciences, so many instructors may choose to skip this chapter. Many students confuse sequences with series, and often have trouble with the various tests for convergence. To avoid these sources of confusion, we emphasize summation as a unifying concept, whether of a few terms of a sequence or of an infinite series. Except for convergence of geometric series and a few words on the interval of convergence, we devote little coverage to the topic of convergence. We believe this approach is appropriate for a course at this level.

Section 12.1 Geometric sequences are the most important type of sequences for applications. Even instructors who wish to skip most of this chapter may want to cover the first two sections.

Section 12.2 This material is similar to the material in the Mathematics of Finance chapter in our textbook Finite Mathematics. The section shows how geometric sequences are critical for an understanding of annuities, mortgages, and amortization. Don’t be put off by the strange notation for the amount or the present value of an annuity; this notation is not at all strange in the world of finance.

Section 12.3 Taylor polynomials are introduced here as an approximation method, with no hint of infinite series.

Section 12.4 The emphasis here is on geometric series, the most important and simplest type of infinite series to understand.

Section 12.5 Some students find Taylor series a strange and abstract concept. To help make this concept more concrete, cover Example 4 on the normal curve, as well as the derivation of the rule of 70 and the rule of 72, introduced without proof in Chapter 2.

Section 12.6 Students with a “zero” feature on their calculator may lack motivation to learn Newton’s method unless you can interest them in how one might develop techniques for finding a zero. Newton’s method is not, however, the method typically used by calculators; calculator manufacturers are usually reluctant to discuss the actual algorithms.

Section 12.7 We have no applications for this sections, but students and instructors who enjoy symbol manipulation may still find this section satisfying.

Chapter 13 This chapter is a brief introduction to trigonometry and its uses in calculus. Students who need a more thorough treatment of this subject would be better served by a calculus book designed for mathematics majors. The presentation is brief, with a limited number of examples. As a result, students may find some of these exercises challenging. Therefore, tread carefully through this chapter.

PRETESTS AND ANSWERS

CALCULUS WITH APPLICATIONS

Name: Pretest, Form A

Evaluate the expression -x 2 + 3 y - z 3 when x = -2, y = 3, and z = -1.

1. ____________________

Perform the indicated operations. 2.

(2 x 2 - 3x + 7) - (5 x 2 - 8x - 9)

2. ____________________

( y - 4) 2

3. ____________________

(2 x - 1)( x 2 + 3x - 4)

4. ____________________

(5a + 2b)(5a - 2b)

5. ____________________

Factor each polynomial completely. 6.

3x 2 y3 - 6 x3 y 2 + 15 x 2 y 2

6. ____________________

2ac - 3ad + 8bc - 12bd

7. ____________________

Solve each equation. 8.

6( x - 1) - 4(2 x + 8) = 2 - 5( x + 2)

8. ____________________

1 2x 1 x-7 = + 4 3 2

9. ____________________

10.

x -1 3 - 2x = 5 4

10. ____________________

11.

( x - 2)( x + 1) = 4

11. ____________________

12.

Solve the inequality 6 x - 8( x + 2) £ 5 - ( x + 14). Write your answer in interval notation.

12. ____________________

13.

Solve the equation 2 x + 4 y = ay - bx for y.

13. ____________________

14.

Solve the equation a5 + 2cb = 37a for a.

14. ____________________

PRETEST, FORM A

xxi

xxii 15.

PRETEST, FORM A Solve the following system of equations. 3x - 2 y = 12 -4 x + 5 y = -23

15. ____________________

Perform the indicated operation. Write answers in lowest terms. 16.

18a 2b 2

10 xy 2

27 xy

8ab 2

⋅ 3

17.

2x + 8 x 2 + 7 x + 12 ¸ 2 3 y - 15 y - 10 y + 25

18.

2 + 2 x -1 x + 3x + 2

16. ____________________

17. ____________________

18. ____________________

Simplify each expression. Write answers with only positive exponents. 19.

20.

(2 x 2 y-3 )-3

19. ____________________

6 x-2 y 5

20. ____________________

15 x 4 y-2

Simplify each radical. Assume all variables represent nonnegative real numbers. 21.

22.

72x3 y 4

21. ____________________

72x3 y 4

22. ____________________

Write an equation in the form ax + by = c for each line. 23.

The line through the points with coordinates (-3, -14) and (5, 2)

23. ____________________

24.

The line through the points with coordinates (3, - 7) and (3, 5)

24. ____________________

25.

Graph the solution of the inequality 2 x - 3 y £ 12.

25.

CALCULUS WITH APPLICATIONS

Name: Pretest, Form B

Choose the best answer. 1.

Evaluate the expression (a)

12 5

(b) 4

y2 - z when x = -2, y = 4, and z = -8. -x 2 + 6

(d)

1. ____________________

4 5

Perform the indicated operations. 2.

( x 2 - 3x + 7) + (4 x3 - 6 x 2 - 5x + 4)

(a) 4 x3 - 6 x 4 - 8 x 2 + 11

(b) 4 x3 - 5 x 2 - 8 x + 11

(d) 4 x3 - 7 x 2 - 8 x + 11

(5 x - 4w)2

2. ____________________

3. ____________________

(a) 25 x 2 - 16w2

(b) 25 x 2 + 16w2

(d) 25 x 2 - 20wx + 16w2

( x - 3) ( x 2 + x - 2)

4. ____________________

(a) x3 - 2 x 2 - 5x + 6

(b) x3 - 3x 2 - 5 x + 6

(d) x3 - 2 x 2 + 5x + 6

(4 x + 9) (4 x - 9)

5. ____________________

(a) 16 x 2 + 72 x - 81y 2

(b) 16 x 2 - 72 xy - 81y 2

(d) 16 x 2 - 72 xy + 81y 2

Factor each polynomial completely. 6.

14a 2b3 - 7a 2b 2 + 35ab 4

6. ____________________

(a) 7ab(2ab 2 - ab + 5b3 )

(b) a 2b3 (14 - 7b + 35b 2 )

(d) 7ab 2 (2ab - a + 5b 2 )

6 xt + 8x - 3 yt - 4 y

7. ____________________

(a) (2 x - y) (3t + 4)

(b) (2 x + y) (3t - 4)

(d) (2 x + y) (3t + 4)

PRETEST, FORM B

xxiii

xxiv

PRETEST, FORM B

Solve each equation. 8.

4m - (7m - 6) = -m

(a)

3 2

(b) - 32

12.

43 7

(b)

(d) -6

27 7

10. ____________________ (c)

27 17

43 17

(d)

11. ____________________

(a) -4, 3

(b) 4, - 3

(d) 4, - 72

Solve the inequality -6 y + 2 ³ 4 y - 7. Write your answer in interval notation.

( -¥, -109 ùúû

9 ,¥ (c) éê 10 ë

14.

23 11

9. ____________________

2 z 2 + z = 28

(a)

13.

(d) 3

3y + 2 7- y = 5 4

(a) 11.

5 2 1 + = 3 r -1 6

(a) - 13 (b) 10.

8. ____________________

)

Solve the equation x =

(b) éê 12 , ¥ ë (d) 4( y - z ) 3k

12. ____________________

)

( -¥, 109 ùúû for y.

13. ____________________

(a)

y = 3kx 4+ 4 z

(b)

y = kx 4- z

(c)

y = - 3kxz 4

(d)

y =

3x + 4 z 4

Solve the equation 4x + ay = 23 for x.

14. ____________________

(a) x =

2 xy - 12 y 3s

(b) x = 2 y - 3a

12 y + 3ax 2y

(d) x =

12 y

3a - 2 y 12 y

PRETEST, FORM B 15.

xxv

Solve the following system of equations and then determine the value of x + y for the solution.

15. ____________________

3x + 2 y = 0 x - 5 y = 17

(a) -1

(b) 1

(d) -3

Perform the indicated operation. Write answer in lowest terms. 16.

17.

6 yz 3

15a3b3 ⋅ 30a 2b 4 12 y 2 z 3

(a)

3a 2 12aby

(b)

(c)

a 4by

(d)

x2 - y 2 4a5b7 (a) (c)

18.

16. ____________________

4 y3 z 6 25a5b7 b 4ay

x 2 + 3xy + 2 y 2

17. ____________________

12a3b12

3b5 2a 2

(b)

3b5 ( x - y )

(d)

a ( x + 2 y)

3b5 ( x + y ) a2 ( x - 2 y)

3b3 ( x + y ) ( x - 2 y)

1 2 5 + 2x 3y 6 xy

(a)

18. ____________________

2 3xy

(b)

3y - 4x + 5 6 xy

(d)

4x - 3y + 5 6 xy

Simplify each expression. Write answers with only positive exponents. 19.

(3a-2b3 )-4

19. ____________________ 8

(b) - 1212a

(a) - 12 6

a b

(c)

20.

a8 12b12

(d)

a8 81b12

æ r 2m-1 ö÷-2 çç ÷ çç 3 2 ÷÷ è r m ø÷

(a) r 2m2

20. ____________________ (b)

1 r 2m2

(d)

1 r 2 m6

xxvi

PRETEST, FORM B

Simplify each radical. Assume all variables represent nonnegative real numbers. 108a 4b3

21.

22.

21. ____________________

(a) 3a 2b 12b

(b) 6a 2b 3b

(d) 54a 2b b

108a 4b3

22. ____________________

(a) 3ab 3 4a

(b) 6a 2b 3 3b

(d) 9a 2b 3 12a

Write an equation in the form ax + by = c for each line. 23.

24.

25.

The line through the points with coordinates (1, -1) and (-1, -2) (a) x - 2 y = 3

(b) x + 2 y = 3

(d) x - 2 y = -3

The line through the points with coordinates (-5, - 2) and (-3, - 2) (a) x + y = -2

(b) x = -2

(d)

23. ____________________

24. ____________________

y = -2

Graph the solution of the inequality 5 x - 3 y ³ 15.

25. ____________________

ANSWERS TO PRETESTS

xxvii

ANSWERS TO PRETESTS PRETEST, FORM A 2 y7

11.

-2, 3

20.

-3x 2 + 5 x + 16

12.

[-7, ¥)

21.

6 xy 2 2 x

y 2 - 8 y + 16

13.

y =

22.

2 xy 3 9 y

2 x3 + 5x 2 - 11x + 4

14.

b a = - 16 3c

23.

2x - y = 8

25a 2 - 4b 2

15.

(2, - 3)

24.

x =3

3x 2 y 2 ( y - 2 x + 5)

16.

5a 6y

25.

(b + 2) x a-4

5 x6

PRETEST, FORM B 1. (c)

6. (d)

11. (c)

16. (c)

21. (b)

2. (b)

7. (a)

12. (d)

17. (c)

22. (a)

3. (c)

8. (d)

13. (d)

18. (b)

23. (a)

4. (a)

9. (b)

14. (b)

19. (d)

24. (d)

5. (c)

10. (c)

15. (a)

20. (c)

25. (b)

FINAL EXAMINATIONS AND ANSWERS

CALCULUS WITH APPLICATIONS

Name: Final Examination, Form A

1. Find the coordinates of the vertex of the graph of f ( x) = -2 x 2 + 30 x + 45.

1. ________________________

2. Write the equation 2a = d using logarithms.

2. ________________________

3. Graph the function y = log 2 ( x - 3) + 4.

4. Evaluate lim

x4

x-4 . x -2

4. ________________________

x-2

5. Find all values of x at which g ( x) = x - 5 is not continuous.

5. ________________________

6. Find the average rate of change of y = between x = 4 and x = 12.

6. ________________________

2x + 1

7. Find the derivative of y = 7 x 2e3x .

7. ________________________

ln (3 x ) . x +1

8. ________________________

8. Find the derivative of y =

9. Find the instantaneous rate of change of s(t ) = 3t 2 -

8 at t = 2. t

9. ________________________

10. Find an equation of the tangent line to the graph of y = ( x 2 + 3x + 3)8 at the point (-1, 1). 11. Find h¢(2) if h( x) =

6 . 4 x +17

10. ________________________ 11. ________________________

12. Find the largest open intervals where f is increasing or decreasing if f ( x) = 2 x3 - 24 x 2 + 72 x.

FINAL EXAM, FORM A

12. ________________________

xxxi

xxxii

FINAL EXAM, FORM A

13. Find the locations and values of all relative extrema x2 + 5x + 3 of g if g ( x) = . x -1 14.

13. ________________________

For the function f ( x) = x 4 - 4 x3 - 5, find (a) all intervals where f is concave upward;

14. (a) _____________________

(b) all intervals where f is concave downward;

(b) _____________________

15. Find the fourth derivative of h( x) = xe x .

15. ________________________

16. Find the locations and values of all absolute extrema of f ( x) = x3 - 6 x 2 on the interval [-1, 2]. 17. For a particular commodity, its price per unit in dollars x2 , is given by P( x) = 120 10 where x, measured in thousands, is the number of units sold. This function is valid on the interval [0, 34]. How many units must be sold to maximize revenue? dy

18. Find dx if 3 x - 5 y 3 = 7 xy.

16. ________________________

17. ________________________ 18. ________________________

19. If xy = x - 9, find dt if dx = 12, x = -3, and y = 4. dt

19. ________________________

20. If y = -3(2 + x 2 )3 , find dy.

20. ________________________

21. Using differentials, approximate the volume of coating on a sphere of radius 5 inches, if the coating is 0.03 inch thick.

21. ________________________

22. Find

ò (3x - 7 x + 2) dx. 2

22. ________________________

23. Find

8 dx. x+9

23. ________________________

24. Evaluate

ò0 x 3x2 + 4 dx.

24. ________________________

25. Find the area of the region between the x-axis and the 2

graph of f ( x) = xe x on the interval [0, 2].

25. ________________________

FINAL EXAM, FORM A

xxxiii

26. Find the area of the region enclosed by the graphs of f ( x) = x 2 - 4 and g (x) = 1 - x 2.

27. Find

26. ________________________

ò x e dx.

28. Evaluate

2 x

27. ________________________

ò 3x In x dx.

28. ________________________

29. Find the average value of f ( x) = x( x 2 + 1)3 over the interval [0, 2].

29. ________________________

30. Find the volume of the solid of revolution formed by rotating the region bounded by the graphs of f ( x) = y = 0, and x = 12 about the x-axis.

30. ________________________

31. Determine whether the improper integral

-1

-¥

x + 4,

x-2 dx

converges or diverges. If it converges, find its value.

31. ________________________

32. If f ( x, y) = 3x 2 - 4 xy + y3, find f xy (1, -3). 33.

32. ________________________

For f ( x, y) = 2 x 2 - 2 xy + 2 y 2 + 4 x + 4 y + 5, find all of the following: (a) relative maxima;

33. (a) _____________________

(b) relative minima;

(b) _____________________

34. Maximize f ( x, y ) = x 2 y, subject to the constraint y + 4 x = 84.

34. ________________________

35. Find dz, given z = 3x 2 - 4 xy + y 2 , where x = 0, y = 2, dx = 0.02, and dy = 0.01.

35. ________________________

36. Evaluate

ò1 ò0 e2 x dy dx.

36. ________________________

37. Find the general solution of the differential equation dx =

1 + ex . 2y

37. ________________________

38. Find the particular solution of the differential dy

equation dx = 3x 2 - 7 x + 2; y = 4 when x = 0.

38. ________________________

xxxiv

FINAL EXAM, FORM A

39. The marginal cost function for a particular commodity dy

is dx = 8 x - 9 x 2. The fixed cost is $20. Find the cost function.

39. ________________________

40. The probability density function of a random variable is defined by f ( x) = 14 for x in the interval [12, 16]. Find P( X £ 14). 41.

40. ________________________

For the probability density function f ( x) = 3x-4 on [1, ¥), find (a) the expected value;

41. (a) _____________________

(b) the variance;

(b) _____________________

42. The average height of a particular type of tree is 20 feet, with a standard deviation of 2 feet. Assuming a normal distribution, what is the probability that a tree of this kind will be shorter than 17 feet?

42. ________________________

43. For the function f ( x) = e-3x , find (a) the Taylor polynomial of degree 4 at x = 0;

43. (a) _____________________

(b) an approximation, rounded to four decimal places, of e-0.03 using the Taylor polynomial from part (a). 44. Find the sum of the convergent geometric series 5 +

5 5 5 + + + . 3 9 27

(b) _____________________ 44. ________________________

45. Use Newton’s method to find a solution to the nearest hundredth of 3x3 - 2 x 2 + x - 1 = 0 in the interval [0, 1].

45. ________________________

46. Use l’Hospital’s rule to find lim

3e2 x - 3 . x x ¥

46. ________________________

47. Find the derivative of f ( x) = tan ( x 2 + 1).

47. ________________________

48. Find the derivative of y = x3 sin 2 x.

48. ________________________

ò sec2 (3x + 1) dx.

49. ________________________

49. Find

50. Evaluate

 /6

ò0 sin3 x cos x dx.

50. ________________________

CALCULUS WITH APPLICATIONS

Name: Final Examination, Form B

Choose the best answer. 1. Find the coordinates of the vertex of the graph of f ( x) = 2 x 2 + 10 x - 17. æ 5 ö çç - , - 59 ÷÷ (b) çè 2 2 ÷ø

(a)

æ5 ö çç , - 59 ÷÷ çè 2 2 ÷ø

(-5, -42) (d) (5, -42)

(c)

2. Write the equation m P = q using logarithms. (a) log m p = q

(b) log q p = m

(d) log m q = p

3. Solve

( 1e )

2. _______________

= 14. (Round to nearest thousandth.)

(a) -1.320 4. Evaluate lim

x9

3. _______________

(b) 2.639

(d) 1.320

x-9 . x-3

(a) 0

4. _______________

(b) 3

5. Find all values of x at which f ( x) = (a) 3

1. _______________

(b) 4

x2 - 5x + 6 x 2 - 7 x + 12

(d) The limit does not exist.

is not continuous.

5. _______________

(d) The function is continuous everywhere. 6. Find the average rate of change of y = x 2 + x between x = 4 (a) 9

(b) 10

(d) 25

7. Find the derivative of y = -3x3e 2 x .

7. _______________

(a) -27 x 2e2 x - 6 x3e2 x

(b) -18x 2e3x

(d) -6 x 2e2 x - 6 x3e2 x 2

8. Find the derivative of y = Inx 3x . (a) (c)

2x2 (ln 3x )

2 x ln 3x - x (ln 3x )

8. _______________ (b)

FINAL EXAM, FORM B

6. _______________

(d)

6 x ln 3x - x 3(ln 3x )2 6 x ln 3x + x 3(ln 3x )2

xxxv

xxxvi

FINAL EXAM, FORM B

9. Find the instantaneous rate of change of s(t ) =

(a)

3 2

(b)

5 2

(c)

t2 2 - 2 at t = 2. 2 t

7 2

(d)

9. _______________

9 2

(

)

1 10. Find an equation of the tangent line to the curve f ( x) = - x + at the point 2, - 14 . 2

(a) x - 16 y = 6 (c)

f ¢( x ) =

(b) x - 4 y = 3

1 16

(d)

y =

1 ( x + 2)2

11. Find h¢(3) if h( x) = - 400 .

11. _______________

3x + 7

(a) 100

10. _______________

(b)

25 8

(c)

75 8

(d) 200

12. Find the largest open interval(s) over which the function g ( x) = x3 - 12 x 2 + 36 x + 1 is increasing.

(a) (2, 6)

(b) (-¥, 2)

12. _______________

13. Find the location and value of all relative extrema for x2 + 5x + 3 . the function f ( x) = x -1

13. _______________

(a) Relative minimum of 1 at -2; relative maximum of 13 at 4 (b) Relative maximum of 1 at -2; relative minimum of 13 at 4 (c) Relative minimum of 1 at 4; relative minimum of 13 at -2 (d) No relative extrema 14. Find the coordinates of all points of inflection of the function g ( x) = x3 - 6 x 2.

(a) (2, -16)

(b) (2, -12)

15. Find the third derivative of f ( x) =

(a)

3 (2 x + 1)-5/2 8

14. _______________

(d) (-2, 16)

2 x + 1.

(b) 3(2 x + 1)-5/2 1 (d) - (2 x + 1)-3/2 4

15. _______________

FINAL EXAM, FORM B

xxxvii

16. Find the absolute minimum of f ( x) = x 4 - 4 x3 - 5 on the interval [-1, 2].

(a) -30

(b) 0

16. _______________

(d) No absolute minimum

17. Botanists, Inc., a consulting firm, monitors the monthly growth of an unusual plant. They determine that the growth (in inches) is given by g ( x) = 4 x - x 2 ,

where x represents the average daily number of ounces of water the plant receives. Find the maximum monthly growth of the plant. (a) 2 inches

(b) 8 inches

(d) 6 inches

18. Find dx , given 4 x 2 - 7 = 3 y + 23 .

18. _______________

(a)

(c)

4 y (4 x5 - 3)

(b)

3x 4 4 y (4 x3 - 3)

(d)

3x 2

4 y (4 x5 + 3) 3x 4 4 y (4 x3 + 3) 3x 2

19. If xy = y + 6, find dt if dx = -3, x = 4, and y = 2. dt

(a) 2

(b) 0

19. _______________ (d) -4

20. Evaluate dy, given y = 45 - 2 x - 3x 2 , with x = 3 and  x = 0.02.

(a) -0.4

(b) 0.4

17. _______________

(d) -0.32

21. Using differentials, approximate the volume of coating on a cube with 3-cm sides, if the coating is 0.02 cm thick.

(a) 0.54 cm3

(b) 27.54 cm3

(d) 81.54 cm3

22. Find

ò (4x + 3x - 5) dx.

(a)

4 3 3 x + x2 - 5 + C 3 2

(b)

(c)

3 3 3 x + x2 - 5 + C 4 2

(d) 8x + 3

20. _______________

21. _______________

22. _______________ 4 3 3 x + x 2 - 5x 3 2

xxxviii 23. Find

FINAL EXAM, FORM B

ò x

-3 dx. (2 + 5)

23. _______________

3 (a) - ln |2x + 5| + C 2

(b) -3 ln |2x + 5| + C

1 (c) - ln |2 x + 5| + C 6

(d)

24. Evaluate

1 ln |2x + 5| + C 2

ò 2x 4x + 5 dx. 2

24. _______________

(a)

3 8

(b)

27 - 5 5 6

(c)

27 - 5 5 8

(c)

27 - 5 5 4

25. Find the area of the region between the x-axis and the graph of f ( x) = x + 1 on the interval [0,3].

(a)

14 3

(b) 7

(c)

1 4

25. _______________ (d)

21 2

26. Find the area of the region enclosed by the

graphs of f ( x) = 9 - x 2 and g(x) = x 2 - 9. (a) -36 27. Find

(b) 36

26. _______________

(d) 72

ò 3x e x dx. 2 2

27. _______________ 3 2 2x 3 3 x e - xe2 x + e 2 x + C 2 2 4

(a) 3e x + C

(b)

(d) 3x 2e 2 x - 6 xe2 x + 6e 2 x + C

28. Evaluate

ò 9x ln x dx. 2

28. _______________

(a) 192 ln 4 - 45

(b) 192 ln 4 - 65

(d) 192 ln 4 - 63

29. Find the average value of the function f ( x) = interval [0, 8].

(a)

31 3

(b)

248 9

(c)

31 9

3x + 1 over the

29. _______________ (d)

248 3

FINAL EXAM, FORM B

xxxix

30. Find the volume of the solid of revolution formed by rotating the region bounded by f ( x) = x - 1, y = 0, and x = 10 about the x-axis.

(a)

40.5

31. Evaluate

(a) -

(b)

-2

-¥ x

81 2

(d) 40

31. _______________

dx.

1 8

(b) Diverges

(c)

1 64

(d) -

1 64

32. Given z = f ( x, y ) = x 2 - 3xy + 2 y 3, find f yx (0, -2).

(a) -16

30. _______________

(b) 2

32. _______________ (d) -3

33. Which of the following applies to the

function f ( x, y ) = 2 x 2 + 8 y 3 - 12 xy + 7 ? (a) f has a relative maximum at

33. _______________

( 92 , 23 )

(b) f has a relative minimum at (0, 0) (c) f has a relative minimum at (d) f has a saddle point at

( 92 , 23 )

34. Maximize f ( x, y) = x 2 y, subject to the constraint y + 4 x = 84.

(a) 0

(b) 14

(d) 3642

x 2 + y 2 , with x = 3, y = 4, dx = -0.01,

35. Find dz, given z =

and dy = 0.02.

35. _______________

(a) 0.022

36. Evaluate

(b) 0.01 2

(a)

112 3

(d) 0.05

ò ò x y dy dx. 0

34. _______________

36. _______________

(b) 16

(d) 12

FINAL EXAM, FORM B dy

37. Find the general solution of the differential equation dx =

(a)

æ3 ö1/3 y = çç e 2 x - 12 x + C ÷÷÷ çè 2 ø

(b)

y =

(c)

æ1 ö1/3 y = çç e 2 x - 4 x + C ÷÷÷ çè 2 ø

(d)

y 2 = e2 x - 4

e2 x - 4 y2

37. _______________

1 2x e - 4x + C 2

38. Find the particular solution of the differential equation dy = 7 - 2 x + 3x 2 ; y = -2 when x = 0. dx

38. _______________

(a)

y = 7 x - x 2 + x3 + C

(b)

y = 7 x - x 2 + x3 - 2

(c)

y = -x 2 + x 3 + C

(d)

y = 7 x - x 2 + x3 + 2 + C dy

39. The marginal cost function for a particular commodity is dx = 10 x - 7 x 2.

The fixed cost is $200. Find the cost function. (a)

y = 200 + 5 x 2 - 14 x

(b)

y = 5x 2 - 14 x

(c)

y = 5x2 -

7 x3 + 200 3

(c)

y = 10 - 14 x

39. _______________

40. The probability density function of a random variable is defined by f ( x) = 15 for [20, 25]. Find P( X £ 22).

(a) 0.2

(b) 0.8

40. _______________ (d) 0.6

41. Find the standard deviation for the probability

density function f ( x ) = 3x-4 on [1, ¥). (a) 1.5

(b) 0.866

41. _______________ (d) 0.67

(a) 0.0668

(b) 0.9332

(d) 0.1977

43. Find the monthly house payment necessary to amortize a $150,000 loan at 7.2% for 30 years.

(a) $1018.18

(b) $975.34

42. _______________

(d) $5478.44

43. _______________

FINAL EXAM, FORM B

xli

44. Find the sum of the convergent geometric series 2 +

(a) 2

(b) 8

1 1 1 + + + . 2 8 32

(d)

44. _______________

8 3

45. The nth term of the Taylor series expansion of ln(1 + 2x ) is

given by which of the following? (a)

(-1)n x n +1 (n + 1) ⋅ 2n +1

(b)

45. _______________

x n +1 n +1

(c)

x n +1 (n + 1) ⋅ 2n +1

(d)

(-1) n x n +1 n +1

-2e 4 x + 2 . x x 0

46. Evaluate lim

(a) -2

46. _______________

(b) -8

(d) The limit does not exist.

47. Find the derivative of f ( x) = tan 2 x + 1.

(a)

sec 2 2 x + 1 2x + 1

(b)

sec 2 2 x + 1 2

(d) 2 x sec2 2 x + 1

48. Find the derivative of y = x 2 cos3 x.

(a) x cos2 x (2 cos x + 3x)

(b) x cos2 x (2 cos x - 3x sin x)

(d) x cos2 x (2 cos x + 3x sin x)

49. Find

ò tan (2x) sec (2x) dx. 2

1 tan 2 (2 x) + C 4

(b) tan 2 (2 x) + C

(c)

1 sec (2 x) + C 4

(d) sec (2 x) + C

 /2

48. _______________

49. _______________

(a)

50. Evaluate

47. _______________

(10 - 10sin 2 x cos x) dx.

(a)

20 3

(b) 5 - 10

(c)

30 - 10 3

(d)

15 - 10 3

50. _______________

ANSWERS TO FINAL EXAMINATIONS

xliii

ANSWERS TO FINAL EXAMINATIONS FINAL EXAMINATION, FORM A 1.

(7.5,157.5)

log 2 d = a

15.

( x + 4)e x

16.

Absolute maximum of 0 at 0; absolute minimum of -16 at 2

17.

20, 000

18.

3 - 14 x1/2 y dy = dx 14 x 3/2 + 30 x1/2 y 2

19.

20.

dy = -18 x(2 + x 2 ) 2 dx

21.

3 in.3

1 4

22.

x3 - 72 x 2 + 2 x + C

y ¢ = 7 xe3x (2 + 3x)

23.

1 ln | x + 9 | + C 8

y¢ =

24.

56 9

25.

1 (e4 - 1) 2

10.

y = 8x + 9 26.

10 10 3

27.

e x ( x 2 - 2 x + 2) + C

on (2, 6)

28.

3 4

13.

Relative maximum of 1 at -2; relative minimum of 13 at 4

29.

14.

(a) (-¥, 0) and (2, ¥)

30.

120

11. 12.

x + 1 - x ln (3 x ) x ( x + 1) 2

12 - 125

» 10.54

Increasing on (-¥, 2) and (6, ¥); decreasing

(b) (0, 2)

( 3e4 + 1) » 123.60

xliv

ANSWERS TO FINAL EXAMINATIONS

31.

Converges; 1

42.

0.0668

32.

-4

43.

(a) 1 - 3x + 92x - 92x + 278x

33.

(a) None

34.

5488 when x = 14 and y = 28

35.

-0.12

36.

(b) -3 at ( - 2, - 2)

(b) 0.9704 44.

15 2

45.

0.78

3 e4 - 3 e2 » 70.81 2 2

46.

37.

y2 = x + ex + C

47.

2 x sec2 x 2 + 1

38.

y = x3 - 72 x 2 + 2 x + 4

48.

x 2 sin x (3sin x + 2 x cos x)

39.

y = 4 x 2 - 3x3 + 20

49.

1 tan (3x + 1) + C 3

40.

0.5

50.

1 64

41.

(a)

3 2

(b) 0.75

(

)

FINAL EXAMINATION, FORM B 1.

(a)

11.

(c)

21.

(a)

31.

(a)

41.

(b)

(d)

12.

(d)

22.

(c)

32.

(d)

42.

(b)

(a)

13.

(b)

23.

(a)

33.

(c)

43.

(a)

(c)

14.

(a)

24.

(b)

34.

(c)

44.

(d)

(b)

15.

(b)

25.

(a)

35.

(b)

45.

(a)

(c)

16.

(c)

26.

(d)

36.

(b)

46.

(b)

(c)

17.

(c)

27.

(b)

37.

(a)

47.

(a)

(c)

18.

(b)

28.

(d)

38.

(b)

48.

(b)

19.

(a)

29.

(c)

39.

(c)

49.

(a)

10.

(a)

20.

(a)

30.

(b)

40.

(c)

50.

(d)

SOLUTIONS TO ALL EXERCISES

Chapter R

ALGEBRA REFERENCE R.1 Polynomials

(-4 y 2 - 3 y + 8) - (2 y 2 - 6 y - 2)

Your Turn 1

= (-4 y 2 - 3 y + 8) + (-2 y 2 + 6 y + 2)

3( x 2 - 4 x - 5) - 4(3x 2 - 5x - 7)

= -4 y 2 - 3 y + 8 - 2 y 2 + 6 y + 2

= 3x 2 - 12 x - 15 - 12 x 2 + 20 x + 28

= (-4 y 2 - 2 y 2 ) + (-3 y + 6 y ) + (8 + 2)

= -9 x 2 + 8 x + 13

= -6 y 2 + 3 y + 10

Your Turn 2

(2 x + 7)(3x - 1)

æ1 ö æ1 1 ö æ2 4ö = ççç z 2 + z 2 ÷÷÷ + ççç z + z ÷÷÷ + ççç + ÷÷÷ è2 ø è3 ø è 2 3 3ø

= (2 x )(3x ) + (2 x )(-1) + (7)(3x ) + (7)(-1) = 6 x 2 - 2 x + 21x - 7

= 6 x 2 + 19 x - 7

Your Turn 3

(3 y + 2)(4 y - 2 y - 5) = (3 y )(4 y 2 - 2 y - 5) + (2)(4 y 2 - 2 y - 5) = 12 y 3 - 6 y 2 - 15 y + 8 y 2 - 4 y - 10

Your Turn 4

æ ö æ ö çç 2 t 2 + 3 t + 5 ÷÷ - çç 1 t 2 + t + 1 ÷÷ ÷ çè 3 ç 2 6ø è6 6 ø÷ æ2 3 5ö æ 1 1ö = ççç t 2 + t + ÷÷÷ + ççç - t 2 - t - ÷÷÷ è3 ø è 2 6 6 6ø

(3x + 2 y )3

= (3x + 2 y )(3x + 2 y )(3x + 2 y )

- 6(2q 2 + 4q - 3) + 4(-q 2 + 7q - 3) + (-4q 2 + 28q - 12)

= (9 x 2 + 12 xy + 4 y 2 )(3x + 2 y )

= (-12q 2 - 4q 2 ) + (-24q + 28q) + (18 - 12)

= 9 x 2 (3x + 2 y ) + 12 xy (3x + 2 y ) + 4 y 2 (3x + 2 y ) = 27 x 3 + 18 x 2 y + 36 x 2 y + 24 xy 2 + 12 xy 2 + 8 y 3

= -16q 2 + 4q + 6

= 27 x 3 + 54 x 2 y + 36 xy 2 + 8 y 3

R.1 Exercises

2(3r 2 + 4r + 2) - 3(-r 2 + 4r - 5) = (6r 2 + 8r + 4) + (3r 2 - 12r + 15)

(2 x - 6 x + 11) + (-3x + 7 x - 2)

= (6r 2 + 3r 2 ) + (8r - 12r ) + (4 + 15)

= 2 x 2 - 6 x + 11 - 3x 2 + 7 x - 2

= 9r 2 - 4r + 19

= (2 - 3) x + (7 - 6) x + (11 - 2) = -x 2 + x + 9

1 2 1 2 t + t+ 2 2 3

= (-12q 2 - 24q + 18)

= (9 x 2 + 6 xy + 6 xy + 4 y 2 )(3x + 2 y )

3 2 5 z + z+2 2 6

æ2 ö æ5 1 ö æ3 1ö = ççç t 2 - t 2 ÷÷÷ + ççç t - t ÷÷÷ + ççç - ÷÷÷ è ø è3 ø è 6 2 6 6ø

= 12 y 3 + 2 y 2 - 19 y - 10

æ1 2 ö æ ö çç z + 1 z + 2 ÷÷ + çç z 2 + 1 z + 4 ÷÷ ÷ ç èç 2 ø è 3 3 2 3 ø÷

(0.613x 2 - 4.215x + 0.892) - 0.47(2 x 2 - 3x + 5) = 0.613x 2 - 4.215x + 0.892 - 0.94 x 2 + 1.41x - 2.35 = -0.327 x 2 - 2.805x - 1.458

2 8.

Chapter R ALGEBRA REFERENCE 0.5(5r 2 + 3.2r - 6) - (1.7r 2 - 2r - 1.5)

16.

= (2.5r 2 + 1.6r - 3) + (-1.7r 2 + 2r + 1.5)

æ 3 öæ 5 ö æ 3 öæ 1 ö æ 2 öæ 5 ö = çç r ÷÷÷çç r ÷÷÷ + çç r ÷÷÷çç s ÷÷÷ + çç - s ÷÷÷çç r ÷÷÷ èç 4 øèç 4 ø èç 4 øèç 3 ø èç 3 øèç 4 ø

= (2.5r 2 - 1.7r 2 ) + (1.6r + 2r ) + (-3 + 1.5) = 0.8r 2 + 3.6r - 1.5

æ 2 öæ 1 ö + çç - s ÷÷÷çç s ÷÷÷ çè 3 øèç 3 ø 15 2 1 5 2 = r + rs - rs - s 2 16 4 6 9 15 2 7 2 2 = r rs - s 16 12 9

-9m(2m 2 + 3m - 1) = -9m(2m2 ) - 9m(3m) - 9m(-1) = -18m3 - 27m 2 + 9m

10.

6 x(-2 x3 + 5x + 6) = -12 x 4 + 30 x 2 + 36 x

11.

(3t - 2 y )(3t + 5 y )

17.

= 3 p(9 p 2 ) - 1(9 p 2 ) + 3 p(3 p )

= 9t 2 + 15ty - 6ty - 10 y 2

- 1(3 p) + 3 p(1) - 1(1)

= 9t 2 + 9ty - 10 y 2

= 27 p 3 - 9 p 2 + 9 p 2 - 3 p + 3 p - 1

(9k + q)(2k - q)

= 27 p 3 - 1

= (9k )(2k ) + (9k )(-q) + (q)(2k ) + (q)(-q)

18.

= 18k 2 - 9kq + 2kq - q2

+ (2)(5 p 2 ) + (2)( p ) + (2)(-4)

(2 - 3x )(2 + 3x )

= 15 p 3 + 3 p 2 - 12 p + 10 p 2 + 2 p - 8

= (2)(2) + (2)(3x ) + (-3x )(2) + (-3x )(3x ) = 4 + 6x - 6x - 9x

= 15 p 3 + 13 p 2 - 10 p - 8

19.

= 4 - 9x2

14.

(3 p + 2)(5 p 2 + p - 4) = (3 p )(5 p 2 ) + (3 p )( p) + (3 p )(-4)

= 18k 2 - 7kq - q2

13.

(3 p - 1)(9 p 2 + 3 p + 1) = (3 p - 1)(9 p 2 ) + (3 p - 1)(3 p) + (3 p - 1)(1)

= (3t )(3t ) + (3t )(5 y ) + (-2 y )(3t ) + (-2 y )(5 y )

12.

æ3 öæ ö çç r - 2 s ÷÷çç 5 r + 1 s ÷÷ ÷ ÷ çè 4 ç 3 øè 4 3 ø

(2m + 1) (4m2 - 2m + 1) = 2m(4m 2 - 2m + 1) + 1(4m2 - 2m + 1) = 8m3 - 4m 2 + 2m + 4m2 - 2m + 1

(6m + 5)(6m - 5) = (6m)6m) + (6m)(-5) + (5)(6m) + (5)(-5)

= 8m3 + 1

= 36m - 30m + 30m - 25

20.

= 36m2 - 25

15.

(k + 2)(12k 3 - 3k 2 + k + 1) = k (12k 3 ) + k (-3k 2 ) + k (k ) + k (1) + 2(12k 3 ) + 2(-3k 2 ) + 2(k ) + 2(1)

æ2 öæ ö çç y + 1 z ÷÷çç 3 y + 1 z ÷÷ ÷ çè 5 ç 8 øè 5 2 ÷ø

= 12k 4 - 3k 3 + k 2 + k + 24k 3 - 6k 2 + 2k + 2

æ 2 öæ 3 ö æ 2 öæ 1 ö æ 1 öæ 3 ö = çç y ÷÷÷çç y ÷÷÷ + çç y ÷÷÷çç z ÷÷÷ + çç z ÷÷÷çç y ÷÷÷ èç 5 øèç 5 ø èç 5 øèç 2 ø èç 8 øèç 5 ø æ 1 öæ 1 ö + çç z ÷÷÷çç z ÷÷÷ çè 8 øèç 2 ø

6 2 1 3 1 2 y + yz + yz + z 25 5 40 16 6 2 æç 8 3 ö÷ 1 2 = y +ç + z ÷÷ yz + ç è 40 25 40 ø 16 6 2 11 1 2 = y + yz + z 25 40 16

= 12k 4 + 21k 3 - 5k 2 + 3k + 2

21.

( x + y + z )(3x - 2 y - z ) = x(3x ) + x(-2 y ) + x(-z ) + y(3x ) + y(-2 y )

+ y(-z ) + z(3x ) + z(-2 y ) + z(-z ) = 3x 2 - 2 xy - xz + 3xy - 2 y 2 - yz + 3xz - 2 yz - z 2 = 3x 2 + xy + 2 xz - 2 y 2 - 3 yz - z 2

Section R.2 22.

( r + 2s - 3t )(2r - 2s + t ) = r(2r ) + r(-2s ) + r(t ) + 2s(2r ) + 2s(-2s )

28.

(3x + y )3 = (3x + y )(3x + y )2 = (3x + y )(9 x 2 + 6 xy + y 2 )

+ 2s(t ) - 3t (2r ) - 3t (-2s ) - 3t (t )

= 27 x 3 + 18 x 2 y + 3xy 2 + 9 x 2 y

= 2r - 2rs + rt + 4rs + 2st - 6rt + 6st 2

- 3t + 2rt - st + t

+ 6 xy 2 + y 3

= 27 x 3 + 27 x 2 y + 9 xy 2 + y 3

= 2r + 2rs - 5rt - 4s 2 + 8st - 3t 2

23.

R.2 Factoring

( x + 1)( x + 2)( x + 3) = [ x( x + 2) + 1( x + 2)]( x + 3)

Your Turn 1

= [ x 2 + 2 x + x + 2]( x + 3)

Factor 4 z 4 + 4 z 3 + 18z 2 .

= [ x 2 + 3x + 2]( x + 3)

4 z 4 + 4 z 3 + 18z 2

= x 2 ( x + 3) + 3x( x + 3) + 2( x + 3)

= (2 z 2 ) ⋅ 2 z 2 + (2 z 2 ) ⋅ 2 z + (2 z 2 ) ⋅ 9

= x 3 + 3x 2 + 3x 2 + 9 x + 2 x + 6

= (2 z 2 )(2 z 2 + 2 z + 9)

= x 3 + 6 x 2 + 11x + 6

24.

Your Turn 2

( x - 1)( x + 2)( x - 3) = [ x( x + 2) + (-1)( x + 2)]( x - 3) = ( x 2 + 2 x - x - 2)( x - 3)

Your Turn 3

= ( x 2 + x - 2)( x - 3) = x 2 ( x - 3) + x( x - 3) + (-2)( x - 3) = x 3 - 3x 2 + x 2 - 3x - 2 x + 6 = x 3 - 2 x 2 - 5x + 6

25.

x 2 - 3x - 10 = ( x + 2)( x - 5) since (2)(-5) = -10 and -5 + 2 = -3.

Factor 6a 2 + 5ab - 4b2 . 6a 2 + 5ab - 4b2 = (2a - b)(3a + 4b)

R.2 Exercises 1.

( x + 2) 2 = ( x + 2)( x + 2) = x( x + 2) + 2( x + 2)

= 7a 2 (a + 2)

= x2 + 2x + 2x + 4

7a3 + 14a 2 = 7a 2 ⋅ a + 7a 2 ⋅ 2

3 y 3 + 24 y 2 + 9 y = 3 y ⋅ y 2 + 3 y ⋅ 8 y + 3 y ⋅ 3 = 3 y( y 2 + 8 y + 3)

= x2 + 4x + 4

26.

(2a - 4b) 2 = (2a - 4b)(2a - 4b) = 2a(2a - 4b) - 4b(2a - 4b)

= 13 p 2q ⋅ p 2q - 13 p 2q ⋅ 3 p + 13 p 2q ⋅ 2q = 13 p 2q( p 2q - 3 p + 2q)

= 4a 2 - 8ab - 8ab + 16b 2 = 4a 2 - 16ab + 16b 2

27.

( x - 2 y )3 = [( x - 2 y )( x - 2 y )]( x - 2 y ) 2

60m4 - 120m3n + 50m2n 2 = 10m2 ⋅ 6m2 - 10m2 ⋅ 12mn + 10m2 ⋅ 5n 2 = 10m2 (6m2 - 12mn + 5n 2 )

= ( x - 2 xy - 2 xy + 4 y )( x - 2 y ) = ( x 2 - 4 xy + 4 y 2 )( x - 2 y )

m2 - 5m - 14 = (m - 7)(m + 2) since (-7)(2) = -14 and -7 + 2 = -5.

x 2 + 4 x - 5 = ( x + 5)( x - 1) since 5(-1) = -5 and -1 + 5 = 4.

z 2 + 9 z + 20 = ( z + 4)( z + 5) since 4 ⋅ 5 = 20 and 4 + 5 = 9.

= ( x 2 - 4 xy + 4 y 2 ) x + ( x 2 - 4 xy + 4 y 2 )(-2 y ) = x 3 - 4 x 2 y + 4 xy 2 - 2 x 2 y + 8 xy 2 - 8 y 3 = x 3 - 6 x 2 y + 12 xy 2 - 8 y 3

13 p 4q 2 - 39 p3q + 26 p 2q 2

4 8.

Chapter R ALGEBRA REFERENCE b 2 - 8b + 7 = (b - 7)(b - 1) since (-7)(-1) = 7 and -7 + (-1) = -8.

s 2 + 2st - 35t 2 = (s - 5t )(s + 7t )

since (-5t )(7t ) = -35t 2 and 7t + (-5t ) = 2t. 11.

y 2 - 4 yz - 21z 2 = ( y + 3z )( y - 7 z )

since (3z )(-7 z ) = -21z 2 and 3z + (-7 z ) = -4 z. 12.

13.

= 10( x 2 - 42 ) = 10( x + 4)( x - 4)

a 2 - 6ab + 5b 2 = (a - b)(a - 5b)

since (-b)(-5b) = 5b 2 and -b + (-5b) = -6b. 10.

23. 10 x 2 - 160 = 10( x 2 - 16)

3x 2 + 4 x - 7

24.

9 x 2 + 64 is the sum of two perfect squares. It cannot be factored. It is prime.

25.

z 2 + 14 zy + 49 y 2 = z 2 + 2 ⋅ 7 zy + 72 y 2 = ( z + 7 y )2

s 2 - 10st + 25t 2 = s 2 - 2 ⋅ 5st + (5t )2

26.

= ( s - 5t )2

The possible factors of 3x2 are 3x and x and the possible factors of –7 are –7 and 1, or 7 and –1. Try various combinations until one works.

27.

3x 2 + 4 x - 7 = (3x + 7)( x - 1)

28.

9 p 2 - 24 p + 16 = (3 p)2 - 2 ⋅ 3 p ⋅ 4 + 42 = (3 p - 4)2

a 3 - 216 = a 3 - 63 = (a - 6)[(a)2 + (a)(6) + (6)2 ]

3a + 10a + 7

The possible factors of 3a2 are 3a and a and the possible factors of 7 are 7 and 1. Try various combinations until one works.

= (a - 6)(a 2 + 6a + 36)

27r 3 - 64s 3 = (3r )3 - (4s)3

29.

= (3r - 4s )(9r 2 + 12rs + 16s 2 )

3a + 10a + 7 = (a + 1)(3a + 7) 2

14. 15 y + y - 2 = (5 y + 2)(3 y - 1) 2

21m + 13mn + 2n = (7m + 2n)(3m + n)

16.

6a 2 - 48a - 120 = 6(a 2 - 8a - 20) = 6(a - 10)(a + 2)

18.

19.

3m3 + 12m 2 + 9m = 3m(m 2 + 4m + 3) = 3m(m + 1)(m + 3)

= 3(m + 5)(m2 - 5m + 25)

x 4 - y 4 = ( x 2 )2 - ( y 2 )2

31.

= ( x 2 + y 2 )( x 2 - y 2 ) = ( x 2 + y 2 )( x + y )( x - y )

32. 16a 4 - 81b4 = (4a 2 )2 - (9b2 )2

4a 2 + 10a + 6 = 2(2a 2 + 5a + 3) = 2(2a + 3)(a + 1)

= (4a 2 + 9b2 )(4a 2 - 9b2 )

24a 4 + 10a3b - 4a 2b 2

= (4a 2 + 9b2 )(2a + 3b)(2a - 3b)

= 2a 2 (12a 2 + 5ab - 2b 2 ) = 2a 2 (4a - b)(3a + 2b)

20.

= 3(m3 + 53 )

15.

17.

3m3 + 375 = 3(m3 + 125)

30.

R.3 Rational Expressions Your Turn 1

2 2

24 x + 36 x y - 60 x y 2

= (4a 2 + 9b2 )[(2a )2 - (3b)2 ]

= 12 x (2 x + 3xy - 5 y ) = 12 x 2 ( x - y)(2 x + 5 y)

21.

x 2 - 64 = x 2 - 82 = ( x + 8)( x - 8)

22.

9m2 - 25 = (3m)2 - (5)2 = (3m + 5)(3m - 5)

Write in lowest terms z 2 + 5z + 6

z 2 + 5z + 6 2z 2 + 7z + 3

( z + 3)( z + 2) ( z + 3)(2 z + 1) 2z + 7z + 3 z+2 = 2z + 1 2

Section R.3

Your Turn 2 Perform each of the following operations. (a)

2z 2 - z - 1 ⋅ 2 z 2 - 5z - 3 z 2 + 2 z - 3 (z + 2)( z + 3) (2 z + 1)( z - 1) = ⋅ (2 z + 1)( z - 3) ( z + 3)( z - 1)

z 2 + 5z + 6

( z + 2) ( z + 3) (2 z + 1) ( z - 1) (2 z + 1) ( z - 3) ( z + 3) ( z - 1)

z+2 z-3 a-3

(b)

5a + 2 a + 3a + 2 a -4 5a a-3 = + (a + 2)(a + 1) (a - 2)(a + 2) (a - 2) a-3 = ⋅ (a + 2)(a + 1) (a - 2) 5a (a + 1) + ⋅ (a - 2)(a + 2) (a + 1)

(a 2 - 5a + 6) + (5a 2 + 5a) (a - 2)(a + 2)(a + 1)

6a 2 + 6 (a - 2)(a + 2)(a + 1)

6(a 2 + 1) (a - 2)(a + 2)(a + 1)

10.

11.

m 2 - 4m + 4

(m - 2)(m - 2) (m - 2)(m + 3) m +m-6 m-2 = m+3 2

r2 - r - 6

( r - 3)( r + 2) ( r + 4)( r - 3) r + r - 12 r+2 = r+4 =

3x 2 + 3x - 6 2

x -4

z 2 - 5z + 6 2

z -4 m 4 - 16 4m2 - 16

( z - 3)( z - 2) z-3 = ( z + 2)( z - 2) z+2

(m2 + 4)(m + 2)(m - 2) 4(m + 2)(m - 2)

m2 + 4 4

12.

6 y 2 + 11 y + 4 (3 y + 4)(2 y + 1) 2 y +1 = = 2 3y + 7 y + 4 (3 y + 4)( y + 1) y +1

13.

9k 2 5 3 ⋅ 3 ⋅ 5k 2 3k 2 3k ⋅ = = = 25 3k 5 ⋅ 5 ⋅ 3k 5k 5

14.

R.3 Exercises

15 p3

10 p

¸ 2

= 2

5v 2 5⋅v⋅v v = = 35v 5⋅7⋅v 7

25 p 3

5⋅5⋅ p⋅ p⋅ p 5p = = 2 2⋅5⋅ p⋅ p 2 10 p

8k + 16 8(k + 2) 8 = = 9k + 18 9(k + 2) 9

2(t - 15) 2 = (t - 15)(t + 2) t+2

4 x3 - 8 x 2 4x2

4 x 2 ( x - 2) 4x2

15.

= x-2

36 y 2 + 72 y 36 y( y + 2) = 9y 9y 9 ⋅ 4 ⋅ y( y + 2) = 9⋅ y = 4( y + 2)

3( x + 2)( x - 1) 3( x - 1) = ( x + 2)( x - 2) x-2

16.

15 p3 10 p 2 ⋅ 6p 9 p2 150 p5 54 p3 25 ⋅ 6 p5 9 ⋅ 6 p3 25 p 2 9

3a + 3b 12 3(a + b) 3⋅ 4 ⋅ = ⋅ 4c 5(a + b) 4c 5(a + b) 3⋅3 = c⋅5 9 = 5c a-3 a-3 a-3 32 ¸ = ⋅ 16 32 16 a-3 a - 3 16 ⋅ 2 = ⋅ 16 a-3 2 = = 2 1

6 17.

18.

19.

20.

Chapter R ALGEBRA REFERENCE 2k - 16 4k - 32 2k - 16 3 ¸ = ⋅ 6 3 6 4k - 32 2(k - 8) 3 = ⋅ 6 4(k - 8) 1 = 4

4a + 12 a2 - 9 ¸ 2 2a - 10 a - a - 20 4(a + 3) (a - 5)(a + 4) = ⋅ 2(a - 5) (a - 3)(a + 3) 2(a + 4) = a-3

25.

6r - 18

12r - 16 9r + 6r - 24 4r - 12 6(r - 3) 4(3r - 4) = ⋅ 2 3(3r + 2r - 8) 4(r - 3) ⋅

6(r - 3) 4(3r - 4) ⋅ 3(3r - 4)(r + 2) 4(r - 3) 6 2 = = 3(r + 2) r+2 2

22.

k + 4k - 12 2

⋅

m 2 + 3m + 2 m 2 + 5m + 4 =

26.

27. 28.

m2 + 10m + 24 m 2 + 10m + 24

m 2 + 5m + 4 m2 + 5m + 6 (m + 1)(m + 2) (m + 6)(m + 4) = ⋅ (m + 4)(m + 1) (m + 3)(m + 2) m+6 = m+3

2m2 - 5m - 12

⋅

m2 - 9m + 18

4n 2 + 4n - 3

⋅

8n 2 + 32n + 30

6n 2 - n - 15 4n 2 + 16n + 15 (2n + 3)(2n - 1) 2(2n + 3)(2n + 5) = ⋅ (2n + 3)(3n - 5) (2n + 3)(2n + 5) 2(2n - 1) = 3n - 5 (a + 1) - (a - 1) a +1 a -1 = 2 2 2 a +1- a +1 = 2 2 = =1 2

3 1 + p 2 2⋅3 p ⋅1 6 p 6+ p + = + = 2⋅ p p⋅2 2p 2p 2p

m2 + 5m + 6

⋅

m2 - 9m + 18

k + k - 12

m2 + 3m + 2

4m2 - 9

Multiply the first term by 22 and the second by p .

k + 10k + 24 k2 - 9 (k + 6)(k - 2) (k + 4)(k - 3) = ⋅ (k + 6)(k + 4) (k + 3)(k - 3) k-2 = k +3

m2 - 10m + 24

m2 - 10m + 24 4m 2 - 9 (2m + 3)(m - 4)(m - 6)(m - 3) = (m - 6)(m - 4)(2m - 3)(2m + 3) m-3 = 2m - 3

24.

2m2 - 5m - 12

9 y - 18 3 y + 6 9( y - 2) 3( y + 2) ⋅ = ⋅ 6 y + 12 15 y - 30 6( y + 2) 15( y - 2) 27 3⋅3 3 = = = 90 10 ⋅ 3 10

21.

23.

29.

6 3 6⋅2 3⋅5y 12 -15 y - = = 5y 2 5y ⋅ 2 2⋅5y 10 y 1 2 4 5 ⋅1 6⋅2 30 ⋅ 4 + + = + + 6m 5m 5 ⋅ 6m 6 ⋅ 5m 30 ⋅ m m 5 12 120 = + + 30m 30m 30m 5 + 12 + 120 = 30m 137 = 30m 1 2 m æ 1 ö÷ m -1 çæ 2 ÷ö + = çç ÷+ ç ÷ m -1 m m èç m -1ø÷ m -1 çè m ÷ø m + 2m - 2 = m( m -1) 3m - 2 = m( m -1)

Section R.3 30.

31.

32.

33.

5 2 5r 2(2r + 3) - = 2r + 3 r r (2r + 3) r (2r + 3) 5r - 2(2r + 3) 5r - 4r - 6 = = r (2r + 3) r (2r + 3) r -6 = r (2r + 3) 8 2 8 3 æ 2 ö÷ + = + çç ÷ 3( a -1) 3( a -1) 3 çè a -1ø÷ a -1 8+6 = 3( a -1) 14 = 3( a -1)

4 2

4( x - 2) ( x - 2)( x + 3)( x + 1)

3( x + 3) ( x - 2)( x + 3)( x + 1) 4( x - 2) + 3( x + 3) = ( x - 2)( x + 3)( x + 1) 4 x - 8 + 3x + 9 = ( x - 2)( x + 3)( x + 1) 7x + 1 = ( x - 2)( x + 3)( x + 1) +

y2 + y - y + 1 ( y + 3)( y + 1)( y - 1)

y2 + 1 ( y + 3)( y + 1)( y - 1) 3k

1 2

y + 2y - 3 y + 4y + 3 y 1 = ( y + 3)( y - 1) ( y + 3)( y + 1) y( y + 1) = ( y + 3)( y + 1)( y - 1) 1( y - 1) ( y + 3)( y + 1)( y - 1)

2k 2

2k + 3k - 2 2k - 7k + 3 3k 2k = (2k - 1)(k + 2) (2k - 1)(k - 3) æ k - 3 ÷ö 3k = çç çè k - 3 ÷÷ø (2k - 1)(k + 2)

(3k 2 - 9k ) - (2k 2 + 4k ) (2k - 1)(k + 2)(k - 3)

k 2 - 13k (2k - 1)(k + 2)(k - 3) k (k - 13) = (2k - 1)(k + 2)(k - 3) =

3 2

x + 4x + 3 x -x-2 4 3 = + ( x + 3)( x + 1) ( x - 2)( x + 1)

y( y + 1) - ( y - 1) ( y + 3)( y + 1)( y - 1)

æ k + 2 ö÷ 2k - çç çè k + 2 ÷÷ø (2k - 1)(k - 3)

2 3 4⋅2 5⋅ 3 + = + 5( k - 2) 4( k - 2) 4 ⋅ 5( k - 2) 5 ⋅ 4( k - 2) 8 15 = + 20( k - 2) 20( k - 2) 8 + 15 = 20( k - 2) 23 = 20( k - 2)

34.

35.

36.

4m 2

m 2

3m + 7m - 6 3m - 14m + 8 m 4m = (3m - 2)(m + 3) (3m - 2)(m - 4) 4m(m - 4) = (3m - 2)(m + 3)(m - 4) m(m + 3) (3m - 2)(m - 4)(m + 3) 4m(m - 4) - m(m + 3) = (3m - 2)(m - 4)(m + 3) =

4m2 - 16m - m2 - 3m (3m - 2)(m + 3)(m - 4)

3m 2 - 19m (3m - 2)(m + 3)(m - 4) m(3m - 19) = (3m - 2)(m + 3)(m - 4) =

Chapter R ALGEBRA REFERENCE

37.

a -1 2 1 + + 2 a+2 a a + 2a a -1 2 1 = + + a+2 a a(a + 2)

Your Turn 3

Solve z 2 + 6 = 8z. z 2 - 8z + 6 = 0

æaö 2 æ a + 2 ö÷ 1 a -1 = çç ÷÷÷ + çç + çè a ø a + 2 çè a + 2 ø÷÷ a a(a + 2)

2a + a + 2 + a - 1 a(a + 2) 4a + 1 = a(a + 2) =

38.

5x + 2 2

x -1

Use the quadratic formula with a = 1, b = -8, and c = 6. -(-8)  (-8)2 - 4(1)(6) 2(1)

8  64 - 24 2 8  40 = 2 =

8  4 ⋅ 10 2 8  2 10 = 2 = 4  10

x +x x -x 5x + 2 3 1 = + x( x + 1) x( x - 1) ( x + 1)( x - 1) ö ö÷ æ x ö÷æç æ 5x + 2 3 ÷÷ + ç x - 1 ö÷÷æçç ÷ = çç ÷÷ç ç ÷ ÷ çè x øçè ( x + 1)( x - 1) ø÷ çè x - 1 øçè x( x + 1) ÷ø÷ ö÷ æ x + 1 ö÷æç 1 ÷ - çç ÷ç ÷ çè x + 1 øçè x( x - 1) ÷÷ø =

x(5 x + 2) + ( x - 1)(3) - ( x + 1)(1) x( x + 1)( x - 1)

5 x 2 + 2 x + 3x - 3 - x - 1 x( x + 1)( x - 1)

5x 2 + 4 x - 4 x( x + 1)( x - 1)

Your Turn 4 1 2 1 Solve 2 + = . x-2 x x -4

1 2 1 + = x-2 x ( x - 2)( x + 2) 1 ( x - 2)( x + 2)( x ) ⋅ ( x - 2)( x + 2) 2 + ( x - 2)( x + 2)( x ) ⋅ x-2 = ( x - 2)( x + 2)( x ) ⋅

R.4 Equations

x + 2x2 + 4x = x 2 - 4

Your Turn 1 Solve 3x - 7 = 4(5x + 2) - 7 x.

x 2 + 5x + 4 = 0 ( x + 1)( x + 4) = 0

3x - 7 = 20 x + 8 - 7 x 3x - 7 = 13x + 8

x = -1 or x = -4

-10 x = 15 15 x=10 3 x=2

Neither of these values makes a denominator equal to zero, so both are solutions.

R.4 Exercises 1.

x = -12

Your Turn 2

The solution is –12.

Solve 2m2 + 7m = 15.

2m - 3 = 0

2x + 8 = x - 4 x + 8 = -4

2m2 + 7m - 15 = 0 (2m - 3)(m + 5) = 0 3 m= or 2

5 x + 2 = 8 - 3x 8x + 2 = 8

or m + 5 = 0 m = -5

1 x

8x = 6 x =

3 4

Section R.4 3.

0.2m - 0.5 = 0.1m + 0.7

10(0.2m - 0.5) = 10(0.1m + 0.7)

4[2 p - (3 - p) + 5] = -7 p - 2 4[2 p - 3 + p + 5] = -7 p - 2

2m - 5 = m + 7

4[3 p + 2] = -7 p - 2

m-5 = 7

12 p + 8 = -7 p - 2

m = 12

19 p + 8 = -2

The solution is 12.

19 p = -10

2 3 1 k-k + = 3 8 2

p =-

Multiply both sides of the equation by 24. æ2 ö æ3ö æ1ö 24 çç k ÷÷ - 24(k ) + 24 çç ÷÷ = 24 çç ÷÷ çè 2 ÷ø èç 3 ø÷ èç 8 ÷ø 16k - 24k + 9 = 12 -8k + 9 = 12 -8k = 3 3 k =8

The solution is - 10 . 19 9.

x+3= 0

10.

x - 3 = 0 or

-7 = 8r

11.

7 = r 8

(m - 7)2 = 0 m-7 = 0 m=7 The solution is 7.

5a + 15 + 4a - 5 = - 2a + 4 9a + 10 = - 2a + 4 11a + 10 = 4

12.

11a = -6 6 a =11 6. The solution is - 11

2[3m - 2(3 - m) - 4] = 6m - 4 2[3m - 6 + 2m - 4] = 6m - 4 2[5m - 10] = 6m - 4 10m - 20 = 6m - 4

m= 4

m2 = 14m - 49 (m) 2 - 2(7m) + (7)2 = 0

5(a + 3) + 4a - 5 = - (2a - 4)

4m = 16

x = -1

m 2 - 14m + 49 = 0

The solution is - 78 .

4m - 20 = -4

x +1 = 0

x = 3 or The solutions are 3 and -1.

-3 = 8r + 4

The solution is 4.

x2 = 3 + 2x ( x - 3)( x + 1) = 0

-3 - 2r = 6r + 4

x = -2

x2 - 2x - 3 = 0

3r + 2 - 5r - 5 = 6r + 4

x+2 =0

x = -3 or The solutions are -3 and -2.

3r + 2 - 5(r + 1) = 6r + 4

x 2 + 5x + 6 = 0 ( x + 3)( x + 2) = 0

The solution is - 83 . 5.

10 19

2k 2 - k = 10 2k 2 - k - 10 = 0 (2k - 5)(k + 2) = 0 2k - 5 = 0 or k + 2 = 0 5 k = k = -2 or 2 The solutions are 52 and -2.

13. 12 x 2 - 5 x = 2 12 x 2 - 5 x - 2 = 0 (4 x + 1)(3x - 2) = 0 4x + 1 = 0 or 3x - 2 = 0 4 x = -1 or 3x = 2 1 2 x = - or x = 4 3 1 2 The solutions are - 4 and 3 .

10 14.

Chapter R ALGEBRA REFERENCE m(m - 7) = -10

19.

m 2 - 7m + 10 = 0 (m - 5)(m - 2) = 0 m - 5 = 0 or m - 2 = 0 m = 5 or m= 2 The solutions are 5 and 2.

15.

2m 2 - 4m = 3 2m 2 - 4m - 3 = 0 m= =

4 x 2 - 36 = 0 x2 - 9 = 0 ( x + 3)( x - 3) = 0 x+3=0 or x - 3 = 0 x = -3 or x =3 The solutions are -3 and 3. z (2 z + 7) = 4 2

2z + 7 z - 4 = 0 (2 z - 1)( z + 4) = 0 2 z - 1 = 0 or z + 4 = 0 1 z = z = -4 or 2 The solutions are 12 and -4.

The solutions are 2 - 10 2

20.

18.

y-4=0 y = 4

Use the quadratic formula.

5

25 - 12 6 5 + 13 5 - 13 x = or x = 6 6 » 1.4343 » 0.2324 5 + 13 » 1.4343 and 6

5 - 13 » 0.2324. 6

4 ⋅ 10 4

4

2 + 10 2

» 2.5811 and

-1  12 - 4(1)(-1) 2(1) -1  5 2

The solutions are

-1 + 5 » 0.6180 and 2

21.

k 2 - 10k = -20 k 2 - 10k + 20 = 0 - (-10) 

(-10)2 - 4(1)(20) 2(1)

10  100 - 80 2 10  20 k = 2 10  4 5 k = 2 10  2 5 k = 2 2(5  5) k = 2 k =5 5 k =

-(-5)  (-5) 2 - 4(3)(1) 2(3)

The solutions are

4

» -0.5811.

k =

3x - 5 x + 1 = 0

-1 - 5 » -1.6180. 2

x =

40 4

p2 + p - 1 = 0

p =

17. 12 y - 48 y = 0 12 y( y) - 12 y(4) = 0 12 y( y - 4) = 0 12 y = 0 or y = 0 or The solutions are 0 and 4.

4

4 10 4 4  2 10 2  10 = = 4 2 =

Divide both sides of the equation by 4.

16.

(-4) 2 - 4(2)(-3) 2(2)

-(-4) 

The solutions are 5 + 5 » 7.2361 and 5 - 5 » 2.7639.

Section R.4 22.

5x 2 - 8x + 2 = 0 x = = = = =

-(-8)  (-8)2 - 4(5)(2) 2(5)

23.

5(3x - 2) = 7( x + 2) 15x - 10 = 7 x + 14

8

4⋅6 10

8

4 6 82 6 = 10 10

8 x = 24

4 6 5 4+ 6 5

3x - 2 x+2 = 7 5 æ 3x - 2 ö÷ æ x + 2 ÷ö 35çç ÷ = 35ççç ÷ èç 7 ÷ø è 2 ÷ø

8  24 10

The solutions are 4- 6 5

27.

x =3 The solution is x = 3.

28.

x 3x -7 = 63 4

Multiply both sides by 12, the least common denominator of 3 and 4. æx ö æ 3x ö÷ 12 çç - 7 ÷÷ = 12 çç 6 ÷ ÷ø çè 3 çè 4 ø÷ æxö æ 3x ö 12 çç ÷÷÷ - (12)(7) = (12)(6) - 12 çç ÷÷÷ èç 3 ø èç 4 ø

» 1.2899 and

» 0.3101.

2r 2 - 7r + 5 = 0 (2r - 5)( r - 1) = 0 2r - 5 = 0 or r - 1 = 0

4 x - 84 = 72 - 9 x

2r = 5 5 or r = 2 The solutions are 52 and 1.

24.

13x - 84 = 72 x = 12

The solution is 12.

2 x 2 - 7 x + 30 = 0

29.

(-7)2 - 4(2)(30) 2(2) 7  49 - 240 x = 4 7  -191 x = 4 Since there is a negative number under the radical sign, -191 is not a real number. Thus, there are no real number solutions. x =

25.

- (-7) 

3k 2 + k = 6

4 8 3 + =0 x-3 2x + 5 x-3 4 3 8 + =0 x-3 x-3 2x + 5 7 8 =0 x-3 2x + 5

Multiply both sides by ( x - 3)(2 x + 5). Note that x ¹ 3 and x ¹ 52 . æ 7 8 ö÷ ( x - 3)(2 x + 5) çç ÷ çè x - 3 2 x + 5 ÷ø = ( x - 3)(2 x + 5)(0)

3k 2 + k - 6 = 0 k =

13x = 156

r =1

7(2 x + 5) - 8( x - 3) = 0

-1  1 - 4(3)(-6) 2(3)

14 x + 35 - 8x + 24 = 0 6 x + 59 = 0

-1  73 = 6

6 x = -59

-1 + 73 The solutions are » 1.2573 and 6 -1 - 73 » -1.5907. 6

26.

5m 2 + 5m = 0 5m(m + 1) = 0

x =-

59 6

Note: It is especially important to check solutions of equations that involve rational expressions. Here, a check shows that - 59 is a solution. 6

5m = 0 or m + 1 = 0 m = 0 or m = -1 The solutions are 0 and –1.

12 30.

Chapter R ALGEBRA REFERENCE 5 7 12 = 2 p-2 p+2 p -4

32.

2y 5 10 - 8 y = + 2 y -1 y y -y

5 7 12 = p-2 p+2 ( p - 2)( p + 2)

2y 5 10 - 8 y = + y -1 y y( y - 1)

Multiply both sides by ( p - 2)( p + 2). Note that p ¹ 2 and p ¹ -2.

Multiply both sides by y( y - 1). Note that y ¹ 0 and y ¹ 1.

æ 5 7 ö÷ ÷ ( p - 2)( p + 2) çç çè p - 2 p + 2 ÷÷ø æ ö÷ 12 ÷ = ( p - 2)( p + 2) çç ÷ çè ( p - 2)( p + 2) ÷ø

æ 2 y ö÷ é5 10 - 8 y úù ÷ = y( y - 1) ê + y( y - 1) çç êy çè y - 1 ÷÷ø y( y - 1) úû ë æ 2 y ÷ö æ5ö ÷ = y( y - 1) çç ÷÷÷ y( y - 1) çç çè y - 1 ÷÷ø çè y ÷ø

æ 5 ö÷ æ 7 ÷ö ÷÷ - ( p - 2)( p + 2) çç ÷ ( p - 2)( p + 2) çç çè p + 2 ÷÷ø èç p - 2 ø÷

é 10 - 8 y ù ú + y( y - 1) ê ê y( y - 1) ú ë û y(2 y) = ( y - 1)(5) + (10 - 8 y)

æ ö÷ 12 ÷ = ( p - 2)( p + 2) çç ÷ èç ( p - 2)( p + 2) ÷ø

2 y 2 = 5 y - 5 + 10 - 8 y

( p + 2)(5) - ( p - 2)(7) = 12 5 p + 10 - 7 p + 14 = 12 -2 p + 24 = 12 -2 p = -12 p =6

2 y2 = 5 - 3y 2 y2 + 3y - 5 = 0 (2 y + 5)( y - 1) = 0 2y + 5 = 0 5 y =2

The solution is 6. 31.

2m 6 12 = 2 m-2 m m - 2m 2m 6 12 = m-2 m m(m - 2)

y -1 = 0

y =1

Since y ¹ 1, 1 is not a solution. The solution is - 52 .

Multiply both sides by m(m - 2). Note that m ¹ 0 and m ¹ 2. æ 2m 6ö m(m - 2) çç - ÷÷÷ çè m - 2 mø æ ö÷ 12 ÷ = m(m - 2) çç çè m(m - 2) ÷÷ø

33.

1 3x 2x + 1 = 2 x-2 x -1 x - 3x + 2 1 3x 2x + 1 = x-2 x -1 ( x - 2)( x - 1)

Multiply both sides by ( x - 2)( x - 1). Note that x ¹ 2 and x ¹ 1. æ 1 3x ö÷ ( x - 2)( x - 1) çç ÷ çè x - 2 x - 1 ÷ø

m(2m) - 6(m - 2) = 12 2m2 - 6m + 12 = 12 2m 2 - 6m = 0 2m(m - 3) = 0 2m = 0 or m - 3 = 0 m = 0 or

m=3

Since m ¹ 0, 0 is not a solution. The solution is 3.

é ù 2x + 1 ú = ( x - 2)( x - 1) ⋅ ê ê ( x - 2)( x - 1) ú ë û æ 1 ö÷ æ 3x ÷ö ( x - 2)( x - 1) çç ÷ - ( x - 2)( x - 1) ⋅ ççç ÷ èç x - 2 ø÷ è x - 1 ÷ø ( x - 2)( x - 1)(2 x + 1) = ( x - 2)( x - 1)

Section R.4

( x - 1) - ( x - 2)(3x) = 2 x + 1

5b(b + 2) - 4(b + 5) = 6b

x - 1 - 3x + 6 x = 2 x + 1

5b 2 + 10b - 4b - 20 = 6b

-3 x 2 + 7 x - 1 = 2 x + 1

5b 2 - 20 = 0 b2 - 4 = 0

-3 x 2 + 5 x - 2 = 0

(b + 2)(b - 2) = 0 b+2= 0 or b-2= 0

3x 2 - 5 x + 2 = 0 (3x - 2)( x - 1) = 0 3x - 2 = 0 or x =

2 3

b = -2

x -1 = 0

Since b ¹ -2, - 2 is not a solution. The solution is 2.

x =1

1 is not a solution since x ¹ 1. The solution is 23 . 36.

34.

5 a - 2a + 4 -7 + = a a +1 a2 + a æ5 -7 ÷ö a(a + 1) çç + ÷ çè a a + 1 ø÷ æ a 2 - 2a + 4 ö÷ ç ÷÷ = a(a + 1) çç çè a 2 + a ÷ø÷

Note that a ¹ 0 and a ¹ -1. 5a + 5 - 7a = a 2 - 2a + 4

5 + 2 x - 2x - 3 x -x-6 1 = 2 x + 3x + 2 2 5 + ( x - 3)( x + 1) ( x - 3)( x + 2) 1 = ( x + 2)( x + 1)

æ ö÷ 2 ÷ ( x - 3)( x + 1)( x + 2) çç çè ( x - 3)( x + 1) ÷ø÷

5 - 2a = a - 2a + 4

æ ö÷ 5 ÷ + ( x - 3)( x + 1)( x + 2) çç çè ( x - 3)( x + 2) ÷÷ø

5 = a2 + 4 0 = a2 - 1

æ 1 ÷÷ö = ( x - 3)( x + 1)( x + 2) çç çè ( x + 2)( x + 1) ÷÷ø 2( x + 2) + 5( x + 1) = x - 3

0 = (a + 1)(a - 1) a = -1

Multiply both sides by ( x - 3)( x + 1)( x + 2). Note that x ¹ 3, x ¹ -1, and x ¹ -2.

5(a + 1) + (-7)(a) = a 2 - 2a + 4

a +1= 0

a -1 = 0

a =1

2 x + 4 + 5x + 5 = x - 3

Since 1 would make two denominators zero, 1 is the only solution. 35.

b= 2

5 4 6 - 2 = 2 b+5 b + 2b b + 7b + 10 5 4 6 = b+5 b(b + 2) (b + 5)(b + 2)

Multiply both sides by b(b  5)(b  2). Note that b  0, b  5, and b  2. æ 5 ö÷ 4 ÷ b(b + 5)(b + 2) çç çè b + 5 b(b + 2) ø÷÷ æ ö÷ 6 ÷ = b(b + 5)(b + 2) çç çè (b + 5)(b + 2) ÷÷ø

7x + 9 = x - 3 6 x + 9 = -3 6 x = -12 x = -2 However, x ¹ -2. Therefore there is no solution.

37.

4 2

2 x + 3x - 9

2 2

2x - x - 3 =

3 2

x + 4x + 3 4 2 + (2 x - 3)( x + 3) (2 x - 3)( x + 1) 3 = ( x + 3)( x + 1)

Multiply both sides by (2 x - 3)( x + 3)( x + 1). Note that x ¹ 32 , x ¹ -3, and x ¹ -1.

Chapter R ALGEBRA REFERENCE (2 x - 3)( x + 3)( x + 1) æ 4 2 ÷÷ö ⋅ çç + çè (2 x - 3)( x + 3) (2 x - 3)( x + 1) ÷÷ø æ ö÷ 3 ÷ = (2 x - 3)( x + 3)( x + 1) çç çè ( x + 3)( x + 1) ø÷÷ 4( x + 1) + 2( x + 3) = 3(2 x - 3)

The numbers in interval B satisfy the inequality, and since the sign was less than or equal to, the boundary points of interval B are also part of the solution. The solution is [-2/3, 6]. Your Turn 3

Solve

4x + 4 + 2x + 6 = 6x - 9 6 x + 10 = 6 x - 9

k 2 - 35 ³ 2. k

First solve the corresponding equation

10 = -9

k 2 - 35 =2 k

This is a false statement. Therefore, there is no solution.

k 2 - 35 = 2k k 2 - 2k - 35 = 0 (k - 7)(k + 5) = 0

R.5 Inequalities

k = 7 or k = -5

Your Turn 1 Solve 3z - 2 > 5z + 7. 3z - 2 > 5z + 7

The denominator is 0 when k = 0, so there are four intervals to consider: (-¥, -5), (-5, 0), (0, 7), and (7, ¥).

3z - 2 + 2 > 5z + 7 + 2 3z > 5z + 9

Choose a test point in each interval.

3z - 5z > 5z - 5z + 9 -2 z > 9 -2 z 9 < -2 -2 9 z<2

k = -8;

(-8)2 - 35 <2 -8

k = -1;

(-1)2 - 35 >2 -1

k = 5;

(-5)2 - 35 <2 5

k = 10;

(10)2 - 35 >2 10

Your Turn 2

Solve 3 y 2 £ 16 y + 12. 3 y 2 £ 16 y + 12 2

3 y - 16 y - 12 £ 0

First solve the equation 3 y 2 - 16 y - 12 = 0. 3 y 2 - 16 y - 12 = 0 (3 y + 2)( y - 6) = 0 3y + 2 = 0 y =-

The second and fourth intervals are part of the solution. Since the inequality is greater than or equal to, we can include the endpoints -5 and 7 but not the endpoint 0, which makes the denominator 0. The solution is [-5, 0) È [7, ¥).

R.5 Exercises 1. x  4

or y - 6 = 0 2 3

k 2 - 35 = 2. k

y =6

Because the inequality symbol means “less than,” the endpoint at 4 is not included. This inequality is written in interval notation as (-¥, 4). To graph this interval on a number line, place an open circle at 4 and draw a heavy arrow pointing to the left.

Determine three intervals on the number line and choose a test point in each interval. B

A 2 3

C 6

Choose -1 from interval A: 3(-1)2 - 16(-1) - 12 > 0 Choose 0 from interval B: 3(0)2 - 16(0) - 12 < 0 Choose 7 from interval C: 3(7)2 - 16(7) - 12 > 0

x ³ -3

Because the inequality sign means “greater than or equal to,” the endpoint at -3 is included. This inequality is written in interval notation as [-3, ¥).

Section R.5

To graph this interval on a number line, place a closed circle at -3 and draw a heavy arrow pointing to the right. –3

1£ x < 2

The endpoints at -2 and 3 are both included. This inequality is written in interval notation as [-2, 3]. To graph this interval, place an open circle at -2 and another at 3 and draw a heavy line segment between them.

11. Notice that the endpoint -2 is included, but 6 is not. The interval shown in the graph can be written as the inequality -2 £ x < 6. 12. Notice that neither endpoint is included. The interval shown in the graph can be written as 0 < x < 8. 13. Notice that both endpoints are included. The interval shown in the graph can be written as x £ -4 or x ³ 4.

-2 £ x £ 3

Ð2

14. Notice that the endpoint 0 is not included, but 3 is included. The interval shown in the graph can be written as x < 0 or x ³ 3. 15.

p£2

This inequality may be rewritten as x < -9, and is written in interval notation as (-¥, -9). Note that the endpoint at -9 is not included. To graph this interval, place an open circle at -9 and draw a heavy arrow pointing to the left.

The solution in interval notation is (-¥, 2].

16. 6.

6k < 3k + 3 3k < 3

This inequality may be written as x ³ 6, and is written in interval notation as [6, ¥). Note that the endpoint at 6 is included. To graph this interval, place a closed circle at 6 and draw a heavy arrow pointing to the right.

k <1

m - (3m - 2) + 6 < 7m - 19 m - 3m + 2 + 6 < 7m - 19 -2m + 8 < 7m - 19

[-7, -3]

[4, 10)

This represents all the numbers between 4 and 10, including 4 but not including 10. This interval can be written as the inequality 4 £ x < 10. 9.

The solution in interval notation is (-¥,1).

17.

This represents all the numbers between -7 and -3, including both endpoints. This interval can be written as the inequality -7 £ x £ -3. 8.

6k - 4 < 3k - 1

6£ x

6 p + 7 £ 19 6 p £ 12 æ 1 ö÷ æ ö çç ÷ (6 p) £ çç 1 ÷÷ (12) çè 6 ÷ø èç 6 ÷ø

-9 > x

–9

(3, ¥)

This represents all the numbers to the right of 3, and does not include the endpoint. This interval can be written as the inequality x > 3.

The endpoint at 1 is included, but the endpoint at 2 is not. This inequality is written in interval notation as [1, 2). To graph this interval, place a closed circle at 1 and an open circle at 2; then draw a heavy line segment between them.

10.

-9m + 8 < -19 -9m < -27

1 1 - (-9m) > - (-27) 9 9 m>3

The solution is (3, ¥).

(-¥, -1]

This represents all the numbers to the left of -1 on the number line and includes the endpoint. This interval can be written as the inequality x £ -1.

16 18.

Chapter R ALGEBRA REFERENCE 22.

-2(3 y - 8) ³ 5(4 y - 2)

8 £ 3r + 1 £ 13 8 + (-1) £ 3r + 1 + (-1) £ 13 + (-1)

-6 y + 16 ³ 20 y - 10

7 £ 3r £ 12

-6 y + 16 + (-16) ³ 20 y - 10 + (-16)

1 1 1 (7) £ (3r ) £ (12) 3 3 3 7 £r £4 3 The solution is éê 73 , 4 ùú . ë û

-6 y ³ 20 y - 26 -6 y + (-20 y) ³ 20 y + (-20 y) - 26 -26 y ³ -26 -

1 1 (-26) y £ - (-26) 26 26 y £1

19.

23.

3 p - 1 < 6 p + 2( p - 1) 3p - 1 < 6 p + 2 p - 2

1 1 1 - (-9) > - (-3k ) ³ - (15) 3 3 3 Rewrite the inequalities in the proper order. -5 £ k < 3

1 1 - (-5 p) > - (-1) 5 5 1 p> 5

The solution is [ -5, 3 ).

( 15 , ¥ ). 01

Ð5

24.

x + 5( x + 1) > 4 (2 - x) + x 6 x + 5 > 8 - 3x

5y + 2 £4 3

-5 £ 5 y £ 10

9x > 3

-1 £ y £ 2

1 x> 3

The solution is [ -1, 2 ].

( 13 , ¥ ). 0 1

–1

25.

-11 < y - 7 < -1 -11 + 7 < y - 7 + 7 < -1 + 7 -4 < y < 6

3 1 (2 p + 3) ³ (5 p + 1) 5 10

æ3ö æ1 ö 10 çç ÷÷ (2 p + 3) ³ 10 çç ÷÷ (5 p + 1) ÷ èç 5 ø èç 10 ÷ø 6(2 p + 3) ³ 5 p + 1 12 p + 18 ³ 5 p + 1

The solution is (-4, 6). Ð4

-3 £ 5 y + 2 £ 12

6 x > 3 - 3x

21.

-1 £

æ 5 y + 2 ÷ö 3(-1) £ 3çç £ 3(4) çè 3 ÷÷ø

x + 5x + 5 > 8 - 4 x + x

The solution is

1 - 3k £4 4 æ 1 - 3k ö÷ 4(-2) < 4 çç £ 4(4) çè 4 ÷÷ø

-2 <

-9 < -3k £ 15

- 5 p - 1 < -2 -5 p < -1

20.

-8 < 1 - 3k £ 16

3p - 1 < 8p - 2

The solution is

7 3

The solution is (-¥, 1].

7 p ³ -17 17 p³7

The solution is éê - 17 , ¥ ). ë 7 – 17 7

Section R.5 26.

8 2 ( z - 4) £ (3z + 2) 3 9 8 2 (9) ( z - 4) £ (9) (3z + 2) 3 9 24( z - 4) £ 2(3z + 2)

29.

y2 - 3y + 2 < 0 ( y - 2)( y - 1) < 0

Solve ( y - 2)( y - 1) = 0. y = 2

24 z - 96 £ 6 z + 4

Intervals: (-¥, 1), (1, 2), (2, ¥) For (-¥, 1), choose y = 0.

24 z £ 6 z + 100 18z £ 100

02 - 3(0) + 2 = 2 < / 0

100 8 50 z £ 9

For (1, 2), choose y = 32 .

z £

æ 3 ÷ö2 æ ö çç ÷ - 3çç 3 ÷÷ + 2 = 9 - 9 + 2 çè 2 ÷ø çè 2 ÷ø 4 2

(

ù. The solution is -¥, 50 9 úû

9 - 18 + 8 4 1 =- <0 4 =

50 9

27.

(m - 3)(m + 5) < 0

For (2, ¥), choose 3.

Solve (m - 3)(m + 5) = 0. (m - 3)(m + 5) = 0 m=3 or m = -5

32 - 3(3) + 2 = 2 < / 0

The solution is (1, 2).

Intervals: (-¥, - 5), (-5, 3), (3, ¥) For (-¥, - 5), choose -6 to test for m. (-6 - 3)(-6 + 5) = -9(-1) = 9 < / 0 For (-5, 3), choose 0. (0 - 3)(0 + 5) = -3(5) = -15 < 0 For (3, ¥), choose 4. (4 - 3)(4 + 5) = 1(9) = 9 < / 0

30.

28.

2k 2 + 7k - 4 > 0 Solve 2k 2 + 7k - 4 = 0. 2k 2 + 7 k - 4 = 0 (2k - 1)(k + 4) = 0 1 k = or k = -4 2

The solution is (-5, 3). Ð5

y =1

Intervals: (-¥, -4), ( -4, 12 ),( 12 , ¥ )

For (-¥, -4), choose -5.

(t + 6)(t - 1) ³ 0

2(-5)2 + 7(-5) - 4 = 11 > 0

Solve (t + 6)(t - 1) = 0. (t + 6)(t - 1) = 0 t = -6 or t =1

For -4, 12 , choose 0.

Intervals: (-¥, -6), (-6, 1), (1, ¥) For (-¥, -6), choose -7 to test for t. (-7 + 6)(-7 - 1) = (-1)(-8) = 8 ³ 0 For (-6, 1), choose 0. (0 + 6)(0 - 1) = (6)(-1) = -6 ³ / 0 For (1, ¥), choose 2. (2 + 6)(2 - 1) = (8)(1) = 8 ³ 0 Because the symbol ³ is used, the endpoints -6 and 1 are included in the solution. The solution is (-¥, -6] È [1, ¥).

For

–6

(

)

2(0)2 + 7(0) - 4 = -4 > / 0

0 1

( 12 , ¥ ), choose 1.

2(1) 2 + 7(1) - 4 = 5 > 0

The solution is (-¥, -4) È ( 12 , ¥ ). –4

01 2

31.

x 2 - 16 > 0

Solve x 2 - 16 = 0. x 2 - 16 = 0 ( x + 4)( x - 4) = 0 x = -4 or x = 4

Chapter R ALGEBRA REFERENCE Intervals: (-¥, -4), (-4, 4), (4, ¥)

For (5, ¥), choose 6.

For (-¥, -4), choose - 5.

(6)2 - 4(6) = 12 ³ 5 The solution is (-¥, -1] È [5, ¥).

(-5) 2 - 16 = 9 > 0

For (-4, 4), choose 0. 02 - 16 = -16 > / 0

Ð1 0

34. 10r + r £ 2

For (4, ¥), choose 5.

Solve 10r 2 + r = 2.

5 - 16 = 9 > 0

10r 2 + r = 2

The solution is (-¥, -4) È (4, ¥).

10r 2 + r - 2 = 0

Ð4

32.

(5r - 2)(2r + 1) = 0 2 1 or r = r = 5 2

2k 2 - 7k - 15 £ 0

(

(2k + 3)(k - 5) = 0 3 2

(

)

For - 12 , 52 , choose 0. 10(0)2 + 0 = 0 £ 2

Intervals: ( -¥, - 23 ), ( - 23 , 5 ), (5, ¥)

For

For ( -¥, - 23 ), choose -2.

)

For - 32 , 5 , choose 0.

–1

2(0)2 - 7(0) - 15 = -15 £ 0

2 5

35.

For (5, ¥), choose 6.

( 52 , ¥ ), choose 1.

10(1) 2 + 1 = 11 £ / 2 The solution is éê - 12 , 52 ùú . ë û

2(-2) 2 - 7(-2) - 15 = 7 £ / 0

(

)

10(-1) 2 + (-1) = 9 £ / 2

k =5

3x 2 + 2 x > 1

Solve 3x 2 + 2 x = 1.

2(6)2 - 7(6) - 15 £ / 0

3x 2 + 2 x = 1

The solution is éê - 23 , 5 ùú . ë û

3x 2 + 2 x - 1 = 0

x2 - 4x ³ 5

(3x - 1)( x + 1) = 0 1 or x = -1 x = 3

Solve x 2 - 4 x = 5.

Intervals: (-¥, -1), -1, 13 ,

–3

(

3(-2)2 + 2(-2) = 8 > 1

x2 - 4x - 5 = 0

(

)

For -1, 13 , choose 0.

( x + 1)( x - 5) = 0 or

) ( 13 , ¥ )

For (-¥, -1), choose -2.

x2 - 4x = 5

x +1= 0

) ( 52 , ¥ )

For -¥, - 12 , choose -1.

2k 2 - 7k - 15 = 0

k =-

)(

(

Intervals: -¥, - 12 , - 12 , 52 ,

Solve 2k 2 - 7k - 15 = 0.

33.

3(0)2 + 2(0) = 0 > / 1

x-5 = 0

x = -1 or x =5 Intervals: (-¥, -1), (-1, 5), (5, ¥) For (-¥, -1), choose -2. (-2)2 - 4(-2) = 12 ³ 5

For

( 13 , ¥ ), choose 1. 3(1)2 + 2(1) = 5 > 1

The solution is (-¥, -1) È ( 13 , ¥ ).

For (-1, 5), choose 0.

–1

02 - 4(0) = 0 ³ / 5

01 3

Section R.5 36.

19 For (-¥, 0), choose -1.

3a 2 + a > 10

(-1)2 - 16(-1) = 17 > 0

Solve 3a 2 + a = 10. 3a 2 + a = 10

For (0, 16), choose 1.

3a 2 + a - 10 = 0

(1)2 - 16(1) = -15 > / 0

(3a - 5)(a + 2) = 0 5 or a = -2 a = 3

For (16, ¥), choose 17. (17)2 - 16(17) = 17 > 0

The solution is (-¥, 0) È (16, ¥).

æ5 ö Intervals: (-¥, -2), ( -2, 53 ), çç , ¥ ÷÷÷ çè 3 ø For (-¥, -2), choose -3. 3(-3)2 + (-3) = 24 > 10

(

39.

x3 - 4x = 0

3(0) 2 + 0 = 0 ³ / 10

x( x 2 - 4) = 0 x( x + 2)( x - 2) = 0 x = 0, or x = -2, or

For ( 53 , ¥ ), choose 2. 3(2) 2 + 2 = 14 > 10

The solution is (-¥, - 2) È ( 53 , ¥ ).

37.

x3 - 4 x ³ 0

Solve x3 - 4 x = 0 .

)

For -2, 53 , choose 0.

–2

x = 2

Intervals: (-¥, -2), (-2, 0), (0, 2), (2, ¥) For (-¥, -2), choose -3. (-3)3 - 4(-3) = -15 ³ / 0

5 3

For (-2, 0), choose 1.

9 - x2 £ 0

(-1)3 - 4(-1) = 3 ³ 0

Solve 9 - x 2 = 0.

For (0, 2), choose 1.

9 - x2 = 0

(1)3 - 4(1) = -3 ³ / 0

(3 + x )(3 - x ) = 0 x = -3 or x = 3 Intervals: (-¥, -3), (-3, 3), (3, ¥) For (-¥, -3), choose -4.

For (2, ¥), choose 3. (3)3 - 4(3) = 15 ³ 0

The solution is [-2, 0] È [2, ¥).

9 - (-4) 2 = -7 £ 0

For (-3, 3), choose 0.

2

9 - (0)2 = 9 £ / 0

For (3, ¥), choose 4. 9 - (4)2 = -7 £ 0

40.

Ð3

x3 + 7 x 2 + 12 x £ 0

Solve x 3 + 7 x 2 + 12 x = 0. x3 + 7 x 2 + 12 x = 0

The solution is (-¥, -3] È [3, ¥).

38.

x( x 2 + 7 x + 12) = 0 x( x + 3)( x + 4) = 0 or x = -3, or x = -4

p 2 - 16 p > 0

x = 0,

Solve p 2 - 16 p = 0.

Intervals: (-¥, -4), (-4, -3), (-3, 0), (0, ¥) For (-¥, -4), choose -5.

p 2 - 16 p = 0

(-5)3 + 7(-5) 2 + 12(-5) = -10 £ 0

p( p - 16) = 0 p = 0 or p = 16

For (- 4, - 3), choose - 72 . 3

Intervals: (-¥, 0), (0, 16), (16, ¥)

( - 72 ) + 7 ( - 72 ) + 12( - 72 ) = 78 £/ 0 For (-3, 0), choose -1.

Chapter R ALGEBRA REFERENCE For (0, 4), choose 1.

(-1)3 + 7(-1)2 + 12(-1) = -6 £ 0

3(1)3 - 9(1)2 - 12(1) = -18 > / 0

For (0, ¥), choose 1.

For (4, ¥), choose 5.

(1)3 + 7(1)2 + 12(1) = 20 £ / 0

3(5)3 - 9(5) 2 - 12(5) = 90 > 0

The solution is (-¥, -4] È [-3, 0]. −4 −3

41.

The solution is (-1, 0) È (4, ¥).

 0

2 x3 - 14 x 2 + 12 x < 0 Solve 2 x3 - 14 x 2 + 12 x = 0 .

43.

2 x3 - 14 x 2 + 12 x = 0

(m + 5)

x =6

Intervals: (-¥, 0), (0, 1), (1, 6), (6, ¥) For (-¥, 0), choose -1.

For (0, 1), choose 12 . æ 1 ö3 æ 1 ö2 æ1ö 11 2 çç ÷÷ - 14 çç ÷÷ + 12 çç ÷÷ = < / 0 çè 2 ÷ø çè 2 ø÷ çè 2 ø÷ 4 2(2)3 - 14(2)2 + 12(2) = -16 < 0

For (-5, 3), choose 0.

For (6, ¥), choose 7.

0-3 3 =- £0 0+5 5

2(7)3 - 14(7)2 + 12(7) = 84 < / 0

For (3, ¥), choose 4.

The solution is (-¥, 0) È (1, 6).

42.

4-3 1 = £ / 0 4+5 9 Although the £ symbol is used, including -5 in the solution would cause the denominator to be zero.

3x3 - 9 x 2 - 12 x > 0 Solve 3x3 - 9 x 2 - 12 x = 0 .

The solution is (-5, 3].

3x3 - 9 x 2 - 12 x = 0 3x( x 2 - 3x - 4) = 0 3x( x - 4)( x + 1) = 0 x = 0, or x = 4, or x = -1

Intervals: (-¥, -1), (-1, 0), (0, 4), (4, ¥) For (-¥, -1), choose -2. 3(-2)3 - 9(-2)2 - 12(-2) = -36 > / 0

For (-1, 0), choose - 12 . 3

( ) - 9( - 12 ) - 12 ( - 12 )

3 - 12

m-3 = (m + 5)(0) m+5 m-3= 0 m=3

-6 - 3 =9£ / 0 -6 + 5

For (1, 6), choose 2.

m-3 = 0. m+5

Set the denominator equal to 0 and solve. m+5=0 m = -5 Intervals: (-¥, -5), (-5, 3), (3, ¥) For (-¥, -5), choose - 6.

2(-1)3 - 14(-1) 2 + 12(-1) = -28 < 0

m-3 £0 m+5

Solve

2 x( x 2 - 7 x + 6) = 0 2 x( x - 1)( x - 6) = 0 x = 0, or x = 1, or

44.

r +1 >0 r -1

Solve the equation

r +1 = 0. r -1

r +1 =0 r -1 r +1 (r - 1) = (r - 1)(0) r -1 r + 1 = 0 r = -1

Find the value for which the denominator equals zero. 27 = >0 8

r -1 = 0 r =1

Intervals: (-¥, -1), (-1, 1), (1, ¥)

Section R.5

(

For (-¥, -1), choose -2.

)

For -2, 32 , choose 0.

-2 + 1 -1 1 = = >0 -2 - 1 -3 3

0-5 -5 5 = = - < -1 0+2 2 2

For (-1, 1), choose 0.

For

0 +1 1 = = -1 > / 0 0 -1 -1

( 32 , ¥ ), choose 2. 2-5 -3 3 = =- < / -1 2+2 4 4

For (1, ¥), choose 2. 2 +1 3 = =3>0 2 -1 1 The solution is (-¥, -1) È (1, ¥).

45.

(

47.

k -1 >1 k +2

Solve

2y + 3 £1 y-5

Solve

k -1 = 1. k +2

Solve y - 5 = 0. y =5

Intervals: (-¥, - 8), (-8, 5), (5, ¥) For (-¥, -8), choose y = -10. 2(-10) + 3 17 = £ / 1 -10 - 5 15

-3 - 1 = 4>1 -3 + 2

For (-8, 5), choose y = 0.

For (-2, ¥), choose 0.

2(0) + 3 3 =- £1 0-5 5

0 -1 1 =- > / 1 0+2 2 The solution is (-¥, -2).

For (5, ¥), choose y = 6. 2(6) + 3 15 = £ / 1 6-5 1

a-5 < -1 a+2

Solve the equation

2y + 3 = 1. y-5 2y + 3 = y - 5 y = -8

k -1 = k + 2 -1 ¹ 2 The equation has no solution. Solve k + 2 = 0. k = -2 Intervals: (-¥, -2), (-2, ¥) For (-¥, -2), choose -3.

46.

)

The solution is -2, 32 .

The solution is [-8, 5). a-5 = -1. a+2

48.

a-5 = -1 a+2 a - 5 = -1(a + 2) a - 5 = -a - 2 2a = 3 3 a = 2 Set the denominator equal to zero and solve for a. a+2=0 a = -2

Intervals: (-¥, -2), ( -2, 23 ), ( 23 , ¥ )

a+2 £5 3 + 2a a+2

Solve the equation 3 + 2a = 5. a+2 =5 3 + 2a a + 2 = 5(3 + 2a) a + 2 = 15 + 10a -9a = 13 13 a =9 Set the denominator equal to zero and solve for a.

For (-¥, -2), choose -3. -3 - 5 -8 = =8< / -1 -3 + 2 -1

3 + 2a = 0 2a = -3 3 a =2

Chapter R ALGEBRA REFERENCE Intervals: ( -¥, - 23 ), ( - 23 , - 13 , - 13 , ¥ ) 9 ) ( 9

(

For (3, ¥), choose 4. 2(4) = 8 and 4-3 2(4) 4 . £ / 4-3 4-3

)

For -¥, - 32 , choose -2. -2 + 2 0 = =0£5 3 + 2(-2) -1

(

)

The solution is [2, 3).

, choose -1.46. For - 32 , - 13 9 -1.46 + 2 0.54 = = 6.75 £ / 5 3 + 2(-1.46) 0.08

(

50.

)

, ¥ , choose 0. For - 13 9

5 12 > p +1 p +1

Solve the equation

0+2 2 = £5 3 + 2(0) 3

since it would make the denominator zero. The ,¥ . solution is -¥, - 32 È éê - 13 ë 9 49.

)

The equation has no solution. Set the denominator equal to zero and solve for p.

)

p +1= 0 p = -1

2k 4 £ k -3 k -3 2k 4 Solve = . k -3 k -3

Intervals: (-¥, -1), (-1, ¥) For (-¥, -1), choose -2. 5 = -5 and -2 + 1 5 12 > . -2 + 1 -2 + 1

2k 4 = k -3 k -3 2k 4 =0 k -3 k -3 2k - 4 =0 k -3 2k - 4 = 0 k = 2

For (-1, ¥), choose 0. 5 = 5 and 0 +1 5 12 > / . 0 +1 0 +1

Set the denominator equal to 0 and solve for k. k -3= 0 k =3

51.

For (-¥, 2), choose 0. 4 4 = - , so 0-3 3

For (2, 3), choose 52 .

( ) = 5 = -10

2 52

5 -3 2

and

4 5 -3 2

2 ( 52 ) 5 -3 2

- 12

4 = -8, so - 12 4 . 3 2

£ 5

12 = -12, so -2 + 1

12 = 12, so 0 +1

The solution is (-¥, -1).

Intervals: (-¥, 2), (2, 3), (3, ¥)

2(0) = 0 and 0-3 2(0) 4 £ / . 0-3 0-3

5 12 = . p +1 p +1

5 12 = p +1 p +1 5 = 12

The value - 32 cannot be included in the solution

(

4 = 4, so 4-3

2x 2

x -x-6

Solve

³0

2x 2

x -x-6

= 0.

=0 x -x-6 2x = 0 x =0 2

Set the denominator equal to 0 and solve for x. x2 - x - 6 = 0 ( x + 2)( x - 3) = 0 x+2 = 0 or x - 3 = 0 x = -2 or x =3

Intervals: (-¥, -2), (-2, 0), (0, 3), (3, ¥)

Section R.5

For (-¥, -2), choose -3.

For (2, ¥), choose 3.

2(-3)

= -1 ³ / 0

(-3) - (-3) - 6

(3) + (2)(3)

2(-1) (-1) - (-1) - 6

1 ³0 2

53.

For (0, 3), choose 2. 2(2) 2

2 -2-6 For (3, ¥), choose 4. 2(4)

8 2

p + 2p

z2 - 1

8 2

p + 2p

= 3.

z 2 + z = 3z 2 - 3 4 ³0 3

-2 z 2 + z + 3 = 0 -1(2 z 2 - z - 3) = 0 -1( z + 1)(2 z - 3) = 0 3 z = -1 or z = 2

Solve the equation

³3

z2 - 1 Solve

z2 + z

4 -4-6 The solution is (-2, 0] È (3, ¥).

52.

z2 + z

= -1 ³ / 0

Set z 2 - 1 = 0.

= 1.

p + 2p

z2 = 1 z = -1 or z = 1

Intervals: (-¥, -1), (-1, 1), (1, 23 ),( 23 , ¥ )

8 = p2 + 2 p

For (-¥, -1), choose x = -2.

0 = p2 + 2 p - 8 0 = ( p + 4)( p - 2) p+4 =0 or p - 2 = 0 p = -4 or p = 2

For (-1, 1), choose x = 0.

Set the denominator equal to zero and solve for p.

02 + 3

p2 + 2 p = 0 p( p + 2) = 0 p = 0 or p + 2 = 0 p = -2 Intervals: (-¥, -4), (-4, -2), (-2, 0), (0, 2), (2, ¥) For (-¥, -4), choose -5.

02 - 1

8 2

(-5) + 2(-5)

(-2)2 + 3 2

(-2) - 1

22 + 3 2 -1

8 8 = = >1 2 9-6 3 (-3) + 2(-3) 8 = = -8 > / 1 2 -1 (-1) + 2(-1)

7 ³ / 3 3

The solution is 1, 32 ùú . û 54.

a 2 + 2a a2 - 4

£2 2

Solve the equation a 2+ 2a = 2. a -4

For (0, 2), choose 1. (1) + 2(1)

(

= -3 ³ / 3

( 32 ) + 3 = 21 ³ 3 2 ( 32 ) - 1 5 For ( 32 , ¥ ) , choose x = 2. 2

7 ³ / 3 3

( )

8 > / 1 15

For (-2, 0), choose -1.

For 1, 32 , choose x = 32 .

For (-4, -2), choose -3.

8 > / 1 15

The solution is (-4, -2) È (0, 2).

For (-2, 0), choose -1. 2

8 >1 3

Chapter R ALGEBRA REFERENCE a 2 + 2a a2 - 4

= 2

a 2 + 2a = 2(a 2 - 4)

-2

( (

) )

æ y 2 z-4 ö÷-2 y 2 z -4 y (2)(-2) z (-4)(-2) çç ÷÷ = = çç -3 4 ÷÷ -2 y (-3)(-2) z (4)(-2) èy z ø y -3 z 4

a 2 + 2a = 2a 2 - 8 0 = a 2 - 2a - 8 0 = (a - 4)(a + 2) a - 4 = 0 or a + 2 = 0

a = 4 or

a = -2

But -2 is not a possible solution. Set the denominator equal to zero and solve for a. a2 - 4 = 0 (a + 2)(a - 2) = 0 a + 2 = 0 or a - 2 = 0 a = -2 or a = 2

Intervals: (-¥, -2), (-2, 2), (2, 4), (4, ¥) For (-¥, -2), choose -3. (-3) 2 + 2(-3) 2

(-3) - 4

9-6 3 = £2 9-4 5

For (-2, 2), choose 0. (0) 2 + 2(0) 0 = = 0£2 -4 0-4

(3)2 - 4

9+6 15 = = £ / 2 9-5 4

For (4, ¥), choose 5. (5)2 + 2(5) (5)2 - 4

25 + 10 35 = = £2 25 - 4 21

The value 4 will satisfy the original inequality, but the values -2 and 2 will not since they make the denominator zero. The solution is (-¥, -2) È (-2, 2) È [4, ¥).

R.6 Exponents Your Turn 1

æ 2 ö÷-3 çç ÷ = çè 3 ø÷

1 27 = 8 = 3 8 2 27

1251/ 3 = 5, since 53 = 125.

Your Turn 4

16-3/ 4 = (161/ 4 )-3 = 2-3 =

1 8

æ x1/ 2 x 4 ö÷1/ 3 æ (1/ 2) + 4 ÷ö1/ 3 æ 9 / 2 ÷ö1/ 3 çç ÷÷ = çç x ÷÷ = çç x ÷ çç 3/ 2 ÷÷ çç çç 3/ 2 ÷÷÷ 3/ 2 ÷÷ è x ø è x ø èx ø

(

= x (9 / 2) -(3/ 2)

1/ 3

)

1/ 3

( )

= x3

= x 3(1/ 3) = x1 = x

Your Turn 6

Factor 5z1/3 + 4 z-2/3.

(

5z1/3 + 4 z-2/3 = z-2/3 5z (1/3)+(2/3) + 4 z (-2/3)+(2/3) = z-2/3(5z + 4)

R.6 Exercises 1.

1 1 8-2 = 2 = 64 8

1 1 3-4 = 4 = 81 3

50 = 1, by definition.

æ 3 ö÷0 çç - ÷ = 1, by definition. çè 4 ÷ø

-(-3)-2 = -

æ 1 ö æ 1ö 1 -(-3-2 ) = -çç - 2 ÷÷ = -çç - ÷÷÷ = çè 9 ø çè 3 ÷ø 9

æ 1 ö-2 ççç ÷÷÷ = è6ø

Your Turn 2 æ y 2 z-4 ö÷-2 Simplify ççç -3 4 ÷÷÷ . çè y z ÷ø

Your Turn 5

(3)

z8-(-8) z16 = = y 6 z -8 y 6-(-4) y10

Your Turn 3

For (2, 4), choose 3. (3) 2 + 2(3)

y -4 z 8

1 (-3)

1 2

( 16 )

1 9

1 = 1 = 36 36

)

Section R.6 8.

æ 4 ö÷-3 çç ÷ = çè 3 ÷ø

11.

1 27 = 64 = 3 64 4

(3)

18.

-7

8 ⋅8

5-2 m 2 y-2 = ⋅ ⋅ 52 m-1 y-2 52 m-1 y-2 = 5-2- 2 m2-(-1) y-2-(-2) = 5-2- 2 m2 +1 y-2 + 2 = 5-4 m3 y 0 =

= 89 + (-7)-(-3) = 89-7 +3 = 85

8-3

104 ⋅ 102 108+ (-10) -2 - 6

= 10 =

10-2

104 + 2

19.

106

= 10

1 108

20. -1

14.

15.

x 4 ⋅ x3 x5

y10 ⋅ y-4 y6

(4k -1)2 2k -5

x 4 +3

(-1)(-1)

c (3)(-2) c-6 = -2 (-2)(-2) = -2 4 7 d 7 d 72 49 = 6 4 = 6 4 c d c d

=7 =7

x7 x

= x 7 -5 = x 2 5

21.

1 1 + a b æ b öæ ö æ a öæ 1 ö 1 = çç ÷÷çç ÷÷ + çç ÷÷çç ÷÷ ÷ç b ÷ø çè b ÷øèç a ø÷ çè a øè

a-1 + b-1 =

= y10- 4-6 = y 0 = 1

b a + ab ab b+a = ab a+b = ab =

4 2 k -2 2 k -5

16k -2 - (-5) 2

= 8k -2 + 5 = 8k 3

16.

(3z 2 )-1 z5

= =

3-1( z 2 )-1

z5 3-1 z-2 z5

3-1 ⋅ x ⋅ y 2 x-4 ⋅ y 5

-1

3-1 z 2(-1) z5

b-2 - a =

1 1 1 ⋅ 7 = 7 3 z 3z 1-(-4)

⋅x

⋅ y

= 3-1 ⋅ x1+ 4 ⋅ y-3 =

22.

-1 -2 -5

= 3-1 z-7 =

17.

æ c3 ö÷-2 (c 3 )-2 çç ÷÷ = çç -2 ÷÷ 7-2 (d -2 )-2 è 7d ø

= (7-12 +3+8 )-1 = (7-1)-1

13.

⋅ m3 ⋅ 1

æ a-1 ö÷-3 (a-1)-3 a (-1)(-3) çç ÷ = çç 2 ÷÷÷ = 2 -3 (b ) b2(-3) è b ø

= (7-12 +3-(-8) )-1

a3 = -6 = a 3b6 b

-8

æ 7-12 ⋅ 73 ö÷ çç ÷÷ çç ÷÷ -8 7 è ø

108 ⋅ 10-10

12.

5-2 m2 y-2

4-2 1 1 = 4-2-1 = 4-3 = 3 = 4 64 4 9

10.

= =

2-5

1 5 1 ⋅x ⋅ 3 3 y x5 3 y3

1 b2

-a

æ b 2 ö÷ çç ÷ a çç 2 ÷÷ 2 b è b ø÷ 1 1 b

ab 2 b2

1 - ab 2 b2

26 23.

Chapter R ALGEBRA REFERENCE 2

2n-1 - 2m-1

æ xy - 1 ö÷-2 ÷÷ = ççç çè y 2 ÷÷ø

- m = n m + n2

m + n2

2 ⋅ m - 2 ⋅ n m m n 2

= n =

or -1

24.

æ m ö÷ çç ÷ èç 3 ø÷

-2

(m + n ) 2m - 2n

mn(m + n 2 ) 2(m - n)

mn(m + n 2 ) 1

ænö + çç ÷÷ èç 2 ø÷

æ y 2 ö÷2 ç ÷÷ = çç çè xy - 1 ÷÷ø

= 2

æ3ö æ2ö = çç ÷÷ + çç ÷÷ èç m ø÷ èç n ø÷

3 4 + 2 m n æ æ 3 öç n 2 ö÷ æ 4 öæ m ö = çç ÷÷çç 2 ÷÷÷ + çç 2 ÷÷çç ÷÷ èç m ø÷çè n ø÷ çè n ÷øçè m ø÷ 3n

4m mn 2

3n 2 + 4m

25.

29.

322/5 = (321/5 )2 = 22 = 4

30.

-1252 / 3 = -(1251/ 3 )2 = -52 = -25

31.

æ 36 ö÷1/ 2 361/ 2 6 1 çç = = = ÷ 1/ 2 çè 144 ÷ø 12 2 144

mn 2

This can also be solved by reducing the fraction first. æ 36 ö÷1/2 æ 1 ö1/ 2 11/2 1 çç = çç ÷÷ = 1/2 = ÷ çè 144 ÷ø çè 4 ÷ø 2 4

1 1 ⋅ y - 1 ⋅ x x y y x

x - xy xy

1 = y- x

32.

1/3 æ 64 ö÷1/3 çç ÷ = 64 = 4 ÷ çè 27 ø 3 271/3

33.

8-4 / 3 = (81/3 )-4 = 2-4 =

34.

625-1/4 =

35.

æ 27 ö÷-1/3 27-1/3 641/3 4 çç ÷ = = = çè 64 ÷ø 3 64-1/3 271/3

26.

( xy - 1) 2

271/ 3 = 3 27 = 3

1 - 1y x

= y

28.

( x-1 - y-1)-1 = 1 =

( xy - 1) 2

27. 1211/2 = (112 )1/2 = 112(1/2) = 111 = 11

( y 2 )2

xy y-x

-2 æx 1 ö÷ ( x ⋅ y-1 - y-2 )-2 = ççç - 2 ÷÷ çè y y ÷÷ø

36.

-2 é æ x öæ y ö 1 ùú ÷ ÷ ç ç ê = ê ç ÷÷ç ÷÷ - 2 ú y úû êë çè y ÷øèç y ÷ø -2 æ xy 1 ö÷ = ççç 2 - 2 ÷÷ çè y y ÷÷ø

1/4

625

1 2

1 16

1 5

æ 121 ö÷-3/2 1 1 çç = = 3/2 çè 100 ÷÷ø 121 é 121 1/2 ù 3 ê 100 ú 100 ëê ûú =

( )

1 1000 = 1331 = 1331 1000

( ) 11 10

37.

32/3 ⋅ 34 / 3 = 3(2/3) + (4/3) = 36/3 = 32 = 9

38.

272/3 ⋅ 27-1/3 = 27(2/3) + (-1/3) = 272 / 3-1/ 3 = 271/ 3 = 3

Section R.6 39.

49/4 ⋅ 4-7/4

= 49/4-7/4-(-10/4)

-10/4

45.

= 412/4 = 43 = 64

40.

3-5/2 ⋅ 33/2 37/2 ⋅ 3-9/2

41/2 ⋅ k 7/2 =

= 3(-5/2) + (3/ 2)-(7/2)-(-9/2)

= 30 = 1

æ x6 y-3 ö÷1/ 2 çç ÷ = ( x6-(-2) y-3-5 )1/ 2 çç -2 5 ÷÷ è x y ø÷

3k 2 ⋅ 4-1k 3 2 ⋅ k 7/2

= 3 ⋅ 2-1 ⋅ 4-1k 2 + 3-(7/2) 3 = ⋅ k 3/2 8

= 3-5/2 + 3/2-7/2 + 9 / 2

41.

3k 2 ⋅ (4k -3 )-1

46.

3k 3/2 8

8 p-3 (4 p 2 )-2 p-5

= ( x8 y-8 )1/ 2 = ( x8 )1/ 2 ( y-8 )1/ 2

= x 4 y -4 = æ a-7b-1 ö÷ çç ÷ çç -4 2 ÷÷÷ è b a ø

= (a-7- 2b-1-(-4) )1/ 3

= a-3b1 b = 3 a

47.

7-1/3 ⋅ 7r -3

7-1/3+1r -3 = 2/3 -4 2/3 -2 2 7 ⋅ (r ) 7 ⋅r = 70 r -3+ 4 = 1 ⋅ r1 = r

48.

128/4 ⋅ y-2

12-1 y 6

= 122-(-1) ⋅ y-2-6 = 123 y-8 =

⋅

b1/2 -3/2

1 2 p2

= a 4/3- 2/3b1/2-(-3/2)

x3/2 ⋅ y 4/5 ⋅ z-3/4 x5/3 ⋅ y-6/5 ⋅ z1/2

12-1 ⋅ y 6

122 ⋅ y-2

123

2/3

16 p

= x-1/6 ⋅ y 2 ⋅ z-5/4

12-1 ⋅ ( y-3 )-2 12-1 ⋅ y (-3)(-2)

a 4/3

= x3/2-(5/3) ⋅ y 4/5-(-6/5) ⋅ z-3/4-(1/2)

123/4 ⋅ 125/4 ⋅ y-2

p-5

= a 2/3b 2

= 7-1/3+3/3- 2/3 r -3-(-4)

123/4 + 5/4 ⋅ y-2

8 p-3 4-2 p-4

= 8 ⋅ 4-2 p-2 1 1 = 8⋅ 2 ⋅ 2 p 4 1 1 = 8⋅ ⋅ 2 16 p

= (a-9 )1/ 3 (b3 )1/ 3

44.

p -5

= 8 ⋅ 4-2 p-3-4 + 5

= (a-9b3 )1/ 3

43.

8 p-3 ⋅ 4-2 p (2) (-2)

= 8 ⋅ 4-2 p (-3) + (-4)-(-5)

1/ 3

42.

49.

y2 x1/6 z 5/4

k -3/5 ⋅ h-1/3 ⋅ t 2/5 k -1/5 ⋅ h-2/3 ⋅ t1/5 = k -3/5-(-1/5) h-1/3-(-2/3)t 2/5-1/5 = k -3/5+1/5h-1/3+ 2/3t 2/5-1/5 = k -2/5h1/3t1/5 =

h1/3t1/5 k 2/5

28 50.

Chapter R ALGEBRA REFERENCE m7/3 ⋅ n-2/5 ⋅ p3/8

56.

(4 x 2 + 1)2 (2 x - 1)-1/2

m-2/3 ⋅ n3/5 ⋅ p-5/8

+ 16 x(4 x 2 + 1)(2 x - 1)1/ 2

= m7/3-(-2/3) n-2/5-(3/5) p3/8-(-5/8)

= (4 x 2 + 1)(4 x 2 + 1)(2 x - 1)-1/2

= m7/3+ 2/3n-2/5-3/5 p3/8+5/8

+ 16 x(4 x 2 + 1)(2 x - 1)-1/2 (2 x - 1)

= m9/3n-5/5 p8/8

= (4 x 2 + 1)(2 x - 1)-1/2

= m3n-1 p1 =

51.

⋅ [(4 x 2 + 1) + 16 x(2 x - 1)]

m3 p n

= (4 x 2 + 1)(2 x - 1)-1/2

⋅ (4 x 2 + 1 + 32 x 2 - 16 x)

3x 3 ( x 2 + 3x )2 - 15x( x 2 + 3x )2

= (4 x 2 + 1)(2 x - 1)-1/2 (36 x 2 - 16 x + 1)

= 3x ⋅ x 2 ( x 2 + 3x )2 - 3x ⋅ 5( x 2 + 3x )2 = 3x( x 2 + 3x )2 ( x 2 - 5)

52.

6 x( x 3 + 7)2 - 6 x 2 (3x 2 + 5)( x 3 + 7) = 6 x( x 3 + 7)( x 3 + 7) - 6 x( x )(3x 2 + 5)( x 3 + 7) = 6 x( x 3 + 7)[( x 3 + 7) - x(3x 2 + 5)] = 6 x( x 3 + 7)( x 3 + 7 - 3x 3 - 5x )

R.7 Radicals Your Turn 1

28 x 9 y 5 .

Simplify

28 x 9 y 5 =

= 2 x 4 y 2 7 xy

= 6 x( x 3 + 7)(-2 x 3 - 5x + 7)

53. 10 x 3 ( x 2 - 1)-1/2 - 5x( x 2 - 1)1/ 2 = 5x ⋅ 2 x 2 ( x 2 - 1)-1/2 - 5x( x 2 - 1)-1/2 ( x 2 - 1)1 2

-1/2

= 5x( x - 1)

Your Turn 2

Rationalize the denominator in

[2 x - ( x - 1)]

= 5x( x 2 - 1)-1/ 2 ( x 2 + 1)

54.

4 ⋅ x8 ⋅ y 4 ⋅ 7 xy

5 x-

9(6 x + 2)1/2 + 3(9 x - 1)(6 x + 2)-1/2

5 x-

= 3(6 x + 2)-1/2[3(6 x + 2) + (9 x - 1)] = 3(6 x + 2)-1/2 (18 x + 6 + 9 x - 1) = 3(6 x + 2)-1/2 (27 x + 5)

55.

x(2 x + 5)2 ( x 2 - 4)-1/2 + 2( x 2 - 4)1/2 (2 x + 5) = (2 x + 5)2 ( x 2 - 4)-1/2 ( x )

y y

Your Turn 3

4+ x 4+ x 4- x = ⋅ 16 - x 16 - x 4 - x 16 - x 1 = = (16 - x)(4 - x ) 4- x

R.7 Exercises 125 = 5 because 53 = 125.

= (2 x + 5)( x 2 - 4)-1/2

4 1296 = 4 64 = 6

⋅ [(2 x + 5)( x ) + ( x 2 - 4)(2)] = (2 x + 5)( x 2 - 4)-1/2 ⋅ (2 x 2 + 5x + 2 x 2 - 8)

= (2 x + 5)( x 2 - 4)-1/ 2 (4 x 2 + 5x - 8)

x-y

+ ( x 2 - 4)1( x 2 - 4)-1/2 (2)(2 x + 5)

( x + y)

= 3 ⋅ 3(6 x + 2)-1/2 (6 x + 2)1 + 3(9 x - 1)(6 x + 2)-1/ 2

x+ x+

⋅

5 x-

-3125 = -5 because (- 5)5 = -3125.

50 =

25 ⋅ 2 =

2000 =

4 ⋅ 100 ⋅ 5

= 2 ⋅ 10 5 = 20 5

25 2 = 5 2

Section R.7

32 y 5 =

(16 y 4 )(2 y)

14.

23 5 - 43 40 + 33 135 = 23 5 - 43 8 ⋅ 5 + 33 27 ⋅ 5

= 16 y 4 2 y

= 23 5 - 4(2) 3 5 + 3(3) 3 5

= 4 y2 2 y

= 2 3 5 - 83 5 + 9 3 5

27 ⋅ 3 =

81 = 9

2 ⋅ 32 =

64 = 8

7 2 - 8 18 + 4 72 = 7 2 - 8 9 ⋅ 2 + 4 36 ⋅ 2

= 33 5

15.

2 x3 y 2 z 4 =

16.

160r 7 s 9t12

= 7 2 - 8(3) 2 + 4(6) 2

(16 ⋅ 10)( r 6 ⋅ r )( s8 ⋅ s )(t12 )

= 7 2 - 24 2 + 24 2

(16r 6 s8t12 )(10rs)

16r 6 s8t12 10rs

=7 2

10.

= 4r 3s 4t 6 10rs

4 3 - 5 12 + 3 75 = 4 3 - 5( 4 3) + 3( 25 3) = 4 3 - 5(2 3) + 3(5 3)

11.

17.

128 x3 y8 z 9 = 3 64 x3 y 6 z 9 ⋅ 2 y 2

= 4 3 - 10 3 + 15 3

= 3 64 x3 y 6 z 9 3 2 y 2

= (4 - 10 + 15) 3 = 9 3

= 4 xy 2 z 3 3 2 y 2

4 7 - 28 + 343 = 4 7 - 4 7 +

18.

4 x8 y 7 z11 = 4 ( x8 )( y 4 ⋅ y 3 )( z 8 z 3 )

49 7

= 4 ( x8 y 4 z 8 )( y3 z 3 )

= 4 7 -2 7 +7 7

12.

= (4 - 2 + 7) 7

= 4 x8 y 4 z 8 4 y 3 z 3

=9 7

= x 2 yz 2 4 y 3 z 3

3 28 - 4 63 + 112

19.

= 3( 4 7) - 4( 9 7) + ( 16 7)

a 2b 4ab - 2 a 6b 2ab +

a 2b8ab

= (ab 2 - 2a3b + ab 4 ) ab

= (6 - 12 + 4) 7

= ab ab (b - 2a 2 + b3 )

= -2 7 2 - 3 16 + 23 54

a3b9

= ab 2 ab - 2a3b ab + ab 4 ab

= 6 7 - 12 7 + 4 7

a3b5 - 2 a 7b3 + =

= 3(2 7) - 4(3 7) + (4 7)

13.

x 2 y 2 z 4 ⋅ 2 x = xyz 2 2 x

20.

p 7 q3 -

p5q 9 +

p9 q

= 3 2 - ( 3 8 ⋅ 2) + 2( 3 27 ⋅ 2)

( p 6 p)(q 2q) - ( p 4 p)(q8q) + ( p8 p)q

= 3 2 - 3 8 3 2 + 2( 3 27 3 2)

( p 6q 2 )( pq) - ( p 4q8 ) ( pq) + ( p8 ) pq

= 3 2 - 23 2 + 2(33 2)

p 6q 2 pq -

= 3 2 - 23 2 + 63 2

= p3q pq - p 2q 4 pq + p 4 pq

= 53 2

= p 2 pq pq - p 2q 4 pq + p 2 p 2 pq

p 4q8 pq +

= p 2 pq ( pq - q 4 + p 2 )

21.

a ⋅ 3 a = a1/2 ⋅ a1/3

= a1/2 + (1/3) = a5/6 =

6 5

p8 pq

30 22.

Chapter R ALGEBRA REFERENCE b3 ⋅ 4 b3 = b3/2 ⋅ b3/4

32.

= b3/2 + (3/4) = b9/4 = 4 b9 = 4 b8 ⋅ b

( )

5(2 + 6) 4 + 2 6 - 2 6 - 36

5(2 + 6) 4 - 36

= 4 b8 4 b = b2 4 b

23.

16 - 8 x + x 2 =

(4 - x)

5(2 + 6) 4-6 5(2 + 6) = -2 5(2 + 6) =2 =

=| 4 - x|

24.

9 y 2 + 30 y + 25 =

25.

4 - 25z 2 =

(3 y + 5) 2 = |3 y + 5|

(2 + 5z )(2 - 5z )

This factorization does not produce a perfect square, so the expression 26.

4 - 25 z

cannot be simplified.

33.

and cannot be factored. Therefore, cannot be simplified. 27.

5 5 = ⋅ 7 7

28.

5 = 10

6 6 2- 2 = ⋅ 2+ 2 2+ 2 2- 2 6(2 - 2) 4-2 2 +2 2 - 4

9k 2 + h 2 The expression 9k 2 + h 2 is the sum of two squares

29.

5 5 2+ 6 = ⋅ 2- 6 2- 6 2+ 6

9k + h

7 5 7 = 7 7

6(2 - 2) 6(2 - 2) = 4-2 2 = 3(2 - 2) =

34.

5 = 5 + 2

5 10 5 10 ⋅ = 10 10 100

5 10 5 10 = 10 100

10 2

-3 = 12

-3 4⋅3

-3 ⋅ 2 3

3 -3 3 3 = =6 2 3

4 4 = ⋅ 8 8

31.

3 3 1+ = ⋅ 1- 2 1- 2 1+

5- 2 5- 2

5( 5 - 2) 25 - 10 + 10 - 4

5 - 10 5-2 5 - 10 = 3

35.

30.

5 ⋅ 5 + 2

2 4 2 4 2 = = = 4 2 16

1 = r - 3 =

36.

5 = m- 5 =

2 2

3(1 + 2) 1- 2 = -3(1 + 2) =

1 ⋅ r - 3

r + 3 r + 3

r + 3 r-3

5 ⋅ m- 5

m + 5 m + 5

5( m + 5) 2

( m ) + 5m - 5m - 25 5( m + 5) 2

( m ) - 25 5( m + 5) m-5

Section R.7 37.

y-5 = y - 5

42.

3- 3 3- 3 3+ 3 = ⋅ 6 6 3+ 3

( y - 5)( y + 5) y-5

38.

y + 5 y + 5

y-5 ⋅ y - 5

y + 5

z -1 = z - 5

z -1 ⋅ z - 5

= =

z + 5 z + 5

( z ) + 5z - z - 5 = 2 ( z ) + 5z - 5z - 25 z + 5z - z - 5 z-5

39.

x + x -

x +1 = x +1

x + x -

x +1 ⋅ x +1

43. x + x +

x +1 x +1

2 x + 2 x( x + 1) + 1 -1 = -2 x - 2 x( x + 1) - 1 p +

p2 - 1

p -

p2 - 1

44.

p2 - 1

p -

p2 - 1

⋅

p +

p2 - 1

p +

p2 - 1

( p) + 2

= =

41.

p -1 -

p + 2 p p 2 - 1 + ( p 2 - 1)

p -

(

p -1

)

= =

p 2 + p + 2 p( p 2 -1) - 1

- p2 + p + 1

(

)(

(

1- 2

2 1- 2

(

)

2 1- 2

)

1 3+ 3

p-2 ⋅ p p

)

p + p +

p-2 p-2

p -2 -

( p) +

p -2

p - ( p - 2) p + p( p - 2)

p - ( p 2 - 1)

1+ 2 1- 2 1+ 2 = 2 2 1- 2

(

6 3+ 3

p-2 p

)

x +1 x +1

6 3+ 3

( p) + =

( p ) + 2 p p - 1 + ( p - 1) 2

p -

p +

(

9-3

)

x +1 x - x +1 ⋅ x +1 x - x +1 x - ( x + 1) = x - 2 x ⋅ x + 1 + ( x + 1) -1 = 2 x - 2 x( x + 1) + 1

40.

(

6 3+ 3

x + x -

x + 2 x( x + 1) + ( x + 1) x - ( x + 1)

x + x -

9+3 3-3 3- 9

2 p( p - 2)

( p - 2)

Chapter 1

LINEAR FUNCTIONS 1.1 Slopes and Equations of Lines Your Turn 1 Find the slope of the line through (1, 5) and (4, 6). Let ( x1, y1) = (1, 5) and ( x2 , y2 ) = (4, 6). m=

6-5 1 = 4 -1 3

First find the slope of the line 3x - 6 y = 7 by solving this equation for y. 3x - 6 y = 7

Your Turn 2 Find the equation of the line with x-intercept -4 and y-intercept 6. We know that b = 6 and that the line crosses the axes at (-4, 0) and (0, 6). Use these two intercepts to find the slope m. m=

Your Turn 5 Find (in slope-intercept form) the equation of the line that passes through the point (4, 5) and is parallel to the line 3x - 6 y = 7.

6-0 6 3 = = 0 - (-4) 4 2

6 y = 3x - 7 y = 3x- 7 6 6 1 y = x-7 2 6 Since the line we are to find is parallel to this line, it will also have slope 1/2. Use the point-slope form with ( x1, y1) = (4, 5).

y - y1 = m( x - x1)

Thus the equation for the line in slope-intercept form is 3 y = x + 6. 2

y - 5 = 1 ( x - 4) 2 y-5 = 1x-2 2 1 y = x+3 2

Your Turn 3 Find the slope of the line whose equation is 8 x + 3 y = 5. Solve the equation for y.

Your Turn 6 Find (in slope-intercept form) the equation of the line that passes through the point (3, 2) and is perpendicular to the line 2 x + 3 y = 4.

8x + 3 y = 5 3 y = -8 x + 5 8 5 y =- x+ 3 3

First find the slope of the line 2 x + 3 y = 4 by solving this equation for y.

The slope is -8/3. Your Turn 4 Find the equation (in slope-intercept form) of the line through (2, 9) and (5, 3). First find the slope. 3-9 -6 = = -2 5-2 3 Now use the point-slope form, with ( x1, y1) = (5, 3).

y - y1 = m( x - x1) y - 3 = -2( x - 5) y - 3 = -2 x + 10 y = -2 x + 13

2x + 3 y = 4 3 y = -2 x + 4 y = -2 x + 4 3 3 Since the line we are to find is perpendicular to a line with slope –2/3, it will have slope 3/2. (Note that (-2/3)(3/2) = -1.)

Use the point-slope form with ( x1, y1) = (3, 2). y - y1 = m( x - x1) y - 2 = 3 ( x - 3) 2 y-2= 3x-9 2 2 3 5 y = x2 2

Chapter 1 LINEAR FUNCTIONS

1.1 Warmup Exercises W1.

15 - (-3) 18 = = -3 -2 - 4 -6

W2. y - (-3) = -2( x + 5) y + 3 = -2 x - 10 y = -2 x - 13

1 2æ 1ö = çç x + ÷÷÷ ç 2 5è 3ø 1 2 2 y- = x+ 2 5 15 2 4 15 + y = x+ 5 30 30 2 19 y = x+ 5 30

10.

11.

y = 3x - 2

5x - 9 y = 11

Rewrite the equation in slope-intercept form. 9 y = 5x - 11 5 11 y = x9 9

-3 y = -2 x + 7 2 7 y = x3 3

The slope is 95 . 12.

1.1

Exercises

False. Any nonvertical line can be represented in the form y = mx + b.

True

False. A vertical line has an undefined slope. A horizontal line has a slope of 0.

False. If two lines are perpendicular, the product of their slopes is –1.

Find the slope of the line through (4, 5) and (-1, 2).

4x + 7 y = 1

Rewrite the equation in slope-intercept form. 7 y = 1 - 4x 1 1 1 (7 y ) = (1) - (4 x ) 7 7 7 1 4 y = - x 7 7 4 1 y =- x+ 7 7 The slope is - 74 .

5-2 4 - (-1) 3 = 5

13.

Find the slope of the line through (5, -4) and

15.

x 5

This is a vertical line. The slope is undefined. 14. The x-axis is the horizontal line y = 0. Horizontal lines have a slope of 0.

(1, 3).

y =8

This is a horizontal line, which has a slope of 0.

3 - (-4) m= 1- 5 3+4 = -4 7 =4

16.

y = -6

By rewriting this equation in the slope-intercept form, y = mx + b, we get y = 0 x - 6, with the slope, m, being 0. 17. Find the slope of a line parallel to 6 x - 3 y = 12.

Find the slope of the line through (8, 4) and (8, -7). 4 - (-7) 11 = 8-8 0 The slope is undefined; the line is vertical. m=

y = x

This equation is in slope-intercept form, y = mx + b. Thus, the coefficient of the x-term, 3, is the slope.

W4. 2 x - 3 y = 7

5-5 0 = =0 -2 - 1 -3

Using the slope-intercept form, y = mx + b, we see that the slope is 1.

W3. y -

Find the slope of the line through (1, 5) and (-2, 5).

Rewrite the equation in slope-intercept form. -3 y = -6 x + 12 y = 2x - 4 The slope is 2, so a parallel line will also have slope 2.

Section 1.1

18. Find the slope of a line perpendicular to 8 x = 2 y - 5.

First, rewrite the given equation in slope-intercept form. 8x = 2 y - 5 8x + 5 = 2 y 5 5 4 x + = y or y = 4 x + 2 2 Let m be the slope of any line perpendicular to the given line. Then. 4 ⋅ m = -1 1 m =- . 4 19. The line goes through (1, 3), with slope m = -2. Use point-slope form. y - 3 = -2( x - 1) y = -2 x + 2 + 3 y = -2 x + 5

20. The line goes through (2, 4), with slope m = -1. Use point-slope form.

24. The line goes through (8, -1) and (4, 3). Find the slope, then use point-slope form with either of the two given points. 3 - (-1) 4-8 3+1 = -4 4 = = -1 -4 y - (-1) = -1( x - 8) m=

y + 1 = -x + 8 y = -x + 7

25. The line goes through m =

3 2 3 y = 6x - - 2 2 3 4 y = 6x - 2 2 7 y = 6x 2

(

)

26. The line goes through -2, 34 and

( 23 , 52 ).

5 - 3 10 - 3 4 4 m = 22 = 24 6 ( 2) + 3 3 3

22. The line goes through (-8, 1), with undefined slope. Since the slope is undefined, the line is vertical. The equation of the vertical line passing through (-8, 1) is x = -8.

3-2 1 =1- 4 3 1 y - 3 = - ( x - 1) 3 1 1 y =- x+ +3 3 3 1 10 y =- x+ 3 3

- 42 - 12

y + 2 = 6x -

y = -x + 6

æ 1ö y - (-2) = 6 çç x - ÷÷÷ çè 4ø

y - 4 = -x + 2

23. The line goes through (4, 2) and (1, 3). Find the slope, then use point-slope form with either of the two given points.

-2 - 12

3 - 8 1 - 2 4 3 12 12 -5 60 m = 25 = =6 - 12 10

y - 4 = -1( x - 2)

21. The line goes through (-5, - 7) with slope m = 0. Use point-slope form. y - (-7) = 0[ x - (-5)] y+7 = 0 y = -7

( 23 , 12 ) and ( 14 , -2 ).

= 84 = 3

21 32

3 21 = [ x - (-2)] 4 32 3 21 42 y- = x+ 4 32 32 21 42 3 y = x+ + 32 32 4 21 21 12 y = x+ + 32 16 16 21 33 y = x+ 32 16 y-

Chapter 1 LINEAR FUNCTIONS

27. The line goes through (-8, 4) and (-8, 6). m=

4-6 -2 = ; -8 - (-8) 0

Use m = - 32 and the point (-4, 6) in the pointslope form. 3 y - 6 = - [ x - (-4)] 2 3 y = - ( x + 4) + 6 2 3 y =- x-6+6 2 3 y =- x 2 2 y = -3 x

which is undefined. This is a vertical line; the value of x is always -8. The equation of this line is x = -8. 28. The line goes through (-1, 3) and (0, 3). m=

3-3 0 = =0 -1 - 0 -1

This is a horizontal line; the value of y is always 3. The equation of this line is y = 3. 29. The line has x-intercept -6 and y-intercept -3. Two points on the line are (-6, 0) and (0, -3). Find the slope; then use slope-intercept form. 1 -3 - 0 -3 = =0 - (-6) 6 2 b = -3 1 y = - x-3 2 2 y = -x - 6

m =

x + 2 y = -6

30. The line has x-intercept –2 and y-intercept 4. Two points on the line are (-2, 0) and (0, 4). Find the slope; then use slope-intercept form. 4-0 4 m= = = 2 0 - (-2) 2 y = mx + b y = 2x + 4 2 x - y = -4 31. The vertical line through (-6, 5) goes through the point (-6, 0), so the equation is x = -6. 32. The line is horizontal, through (8, 7). The line has an equation of the form y = k where k is the y-coordinate of the point. In this case, k = 7, so the equation is y = 7. 33. Write an equation of the line through (-4, 6), parallel to 3x + 2 y = 13.

Rewrite the equation of the given line in slopeintercept form. 3x + 2 y = 13 2 y = -3x + 13 3 13 y =- x+ 2 2 The slope is - 32 .

3x + 2 y = 0

34. Write the equation of the line through (2, -5), parallel to y - 4 = 2 x. Rewrite the equation in slope-intercept form. y - 4 = 2x y = 2x + 4

The slope of this line is 2. Use m = 2 and the point (2, -5) in the pointslope form. y - (-5) = 2( x - 2) y + 5 = 2x - 4 y = 2x - 9 2x - y = 9 35. Write an equation of the line through (3, -4), perpendicular to x + y = 4.

Rewrite the equation of the given line as y = -x + 4. The slope of this line is -1. To find the slope of a perpendicular line, solve -1m = -1. m=1 Use m = 1 and (3, -4) in the point-slope form. y - (-4) = 1( x - 3) y = x-3-4 y = x-7 x- y =7

36. Write the equation of the line through (-2, 6), perpendicular to 2 x - 3 y = 5.

Rewrite the equation in slope-intercept form. 2x - 3 y = 5 -3 y = -2 x + 5 2 5 y = x3 3

Section 1.1

37 æ 2 öù 1é y - 0 = - ê x - çç - ÷÷÷ ú èç 3 ø úû 2 êë 1æ 2ö y = - çç x + ÷÷÷ ç 2è 3ø

The slope of this line is 23 . To find the slope of a perpendicular line, solve 2 m = -1. 3 3 m=2

1 1 y =- x2 3 6 y = -3 x - 2 3 x + 6 y = -2

Use m = - 32 and (-2, 6) in the point-slope form. 3 y - 6 = - [ x - (-2)] 2 3 y - 6 = - ( x + 2) 2 3 y-6 = - x-3 2 3 y =- x+3 2 2 y = -3 x + 6 3x + 2 y = 6

37. Write an equation of the line with y-intercept 4, perpendicular to x + 5 y = 7.

39. Do the points (4, 3), (2, 0), and (-18, -12) lie on the same line? Find the slope between (4, 3) and (2, 0). m =

Find the slope between (4, 3) and (-18, -12). -12 - 3 -15 15 = = -18 - 4 -22 22 Since these slopes are not the same, the points do not lie on the same line. m =

40. (a) Write the given line in slope-intercept form.

Find the slope of the given line. x + 5y = 7 5 y = -x + 7 1 7 y =- x+ 5 5

2x + 3 y = 6 3 y = -2 x + 6 2 y =- x+2 3

This line has a slope of - 23 . The desired line

The slope is - 15 , so the slope of the perpendicular

has a slope of - 23 since it is parallel to the

line will be 5. If the y-intercept is 4, then using the slope-intercept form we have y = mx + b y = 5 x + 4, or 5x - y = -4

given line. Use the definition of slope. m =

2 2 - (-1) = k-4 3 2 3 - = k-4 3 -2(k - 4) = (3)(3)

Find the slope of the given line. 2x - y = 4 2x - 4 = y

-2k + 8 = 9 -2 k = 1

The slope of this line is 2. Since the lines are perpendicular, the slope of the needed line is - 12 . The line also has an x-intercept of - 23 . Thus, it

)

passes through the point - 23 , 0 .

y2 - y1 x2 - x1

38. Write the equation of the line with x-intercept - 23 , perpendicular to 2 x - y = 4.

(

0-3 -3 3 = = 2-4 -2 2

1 2 Write the given line in slope-intercept form. 5 x - 2 y = -1 k =-

(b)

2 y = 5x + 1

Using the point-slope form, we have

y =

5 1 x+ 2 2

This line has a slope of 52 . The desired line has a slope of - 52 since it is perpendicular to the given line. Use the definition of slope.

Chapter 1 LINEAR FUNCTIONS m =

y2 - y1 x2 - x1

2 - (-1) k-4 2 2 +1 - = k-4 5 3 -2 = k-4 5 -2(k - 4) = (3)(5)

The product of the slopes is (1)(-1) = -1, so the diagonals are perpendicular.

43. The line goes through (0, 2) and (-2, 0) m=

41. A parallelogram has 4 sides, with opposite sides parallel. The slope of the line through (1, 3) and (2, 1) is 3-1 m= 1- 2 2 = -1 = -2.

)

(

The slope of the line through - 52 , 2 and - 72 , 4 is m=

2-4

- 52 -

( ) - 72

-2 = -2. 1

Since these slopes are equal, these two sides are parallel.

(

)

The slope of the line through - 72 , 4 and (1, 3) is

)

44. The line goes through (1, 3) and (2, 0). 3-0 3 m= = = -3 1- 2 -1 The correct choice is (f). 45. The line appears to go through (0, 0) and (-1, 4). m=

Slope of the line through m =

(

) and (2, 1) is

2 -1 1 2 = 9 =- . - 52 - 2 -2 9

Since these slopes are equal, these two sides are parallel. Since both pairs of opposite sides are parallel, the quadrilateral is a parallelogram. 42. Two lines are perpendicular if the product of their slopes is -1.

The slope of the diagonal containing (4, 5) and (-2, -1) is 5 - (-1) 6 = = 1. m= 4 - (-2) 6

The slope of the diagonal containing (-2, 5) and (4, -1) is

4-0 4 = = -4 -1 - 0 -1

46. The line goes through (-2, 0) and (0, 1). 1- 0 1 m= = 0 - (-2) 2 47. (a) See the figure in the textbook. Segment MN is drawn perpendicular to segment PQ. Recall that MQ is the length of segment MQ.

m1 =

4-3 1 2 m= 7 = 9 =- . 9 -2 -1 -2 - 52 , 2

2-0 2 = =1 0 - (-2) 2

The correct choice is (a).

-2k + 8 = 15 -2k = 7 7 k =2

(

5 - (-1) 6 = = -1. -2 - 4 -6

y MQ = x PQ

From the diagram, we know that PQ = 1. MQ

Thus, m1 = 1 , so MQ has length m1. y -QN -QN (b) m2 = = = PQ x 1 QN = -m2 (c)

Triangles MPQ, PNQ, and MNP are right triangles by construction. In triangles MPQ and MNP, angle M = angle M , and in the right triangles PNQ and MNP, angle N = angle N .

Since all right angles are equal, and since triangles with two equal angles are similar, triangle MPQ is similar to triangle MNP and triangle PNQ is similar to triangle MNP. Therefore, triangles MNQ and PNQ are similar to each other.

Section 1.1

(d) Since corresponding sides in similar triangles are proportional, MQ = k ⋅ PQ and

x y + =1 a b y 0+ =1 b y =1 b y = b

PQ = k ⋅ QN .

MQ k ⋅ PQ = PQ k ⋅ QN MQ PQ = PQ QN

From the diagram, we know that PQ = 1. MQ =

(c)

1 QN

From (a) and (b), m1 = MQ and -m2 = QN .

49.

Substituting, we get

The y-intercept is b. If the equation of a line is written as x y + = 1 , we immediately know the a b intercepts of the line, which are a and b.

y = x -1

Three ordered pairs that satisfy this equation are (0, -1), (1, 0), and (4, 3). Plot these points and draw a line through them.

1 m1 = . -m2

Multiplying both sides by m2 , we have m1m2 = -1. 48. (a) Multiplying both sides of the equation x y + = 1 by ab, we have a b æxö æ yö ab çç ÷÷÷ + ab çç ÷÷÷ = ab (1) èç a ø èç b ø

50.

bx + ay = ab.

Three ordered pairs that satisfy this equation are (-2, -3), (-1, 1), and (0, 5). Plot these points and draw a line through them.

Solve this equation for y. bx + ay = ab ay = ab - bx ab - bx y = a b y =- x+b a b If we let m = - , then the equation a becomes y = mx + b.

(b)

Let y = 0 .

y = 4x + 5

y y = 4x + 5

1 –1

51.

y = -4 x + 9

Three ordered pairs that satisfy this equation are (0, 9), (1, 5), and (2, 1). Plot these points and draw a line through them.

x y + =1 a b x +0 =1 a x =1 a x = a The x-intercept is a. Let x = 0 .

40 52.

Chapter 1 LINEAR FUNCTIONS y = -6 x + 12

Plot the ordered pairs (-3, 0) and (0, 9) and draw a line through these points. (A third point may be used as a check.)

There ordered pairs that satisfy this equation are (0, 12), (1, 6), and (2, 0). Plot these points and draw a line through them.

y 3x – y = –9

y = –6x + 12

55. 53.

2 x - 3 y = 12

3 y - 7 x = -21

Find the intercepts. If y = 0, then

3(0) + 7 x = -21 -7 x = -21 x =3

2 x - 3(0) = 12 2 x = 12 x =6

so the x-intercept is 3. If x = 0, then

so the x-intercept is 6. If x = 0, then

3 y - 7(0) = -21 3 y = -21 y = -7

2(0) - 3 y = 12 -3 y = 12 y = -4

So the y-intercepts is -7.

so the y-intercept is -4.

Plot the ordered pairs (3, 0) and (0, -7) and draw a line through these points. (A third point may be used as a check.)

Plot the ordered pairs (6, 0) and (0, -4) and draw a line through these points. (A third point may be used as a check.)

3y – 7x = –21 –7

54.

56.

3 x - y = -9

Find the intercepts. If y = 0, then 3 x - 0 = -9 3 x = -9 x = -3

If x = 0, then 3(0) - y = -9 - y = -9 y =9

5 y + 6 x = 11

Find the intercepts. If y = 0, then 5(0) + 6 x = 11 6 x = 11 11 x = 6 11 so the x-intercept is 6 .

If x = 0, then 5 y + 6(0) = 11 5 y = 11 11 y = 5

so the y-intercept is 9.

Section 1.1

( 116 , 0 ) and ( 0, 115 ) and

Plot the ordered pairs

for any value of x. The graph is the horizontal line with y-intercept -8.

draw a line through these points. (A third point may be used as a check.)

y x

5y + 6x = 11

61. 57.

y = 2x

Three ordered pairs that satisfy this equation are (0, 0), (-2, -4), and (2, 4). Use these points to draw the graph.

y = -2

The equation y = -2, or, equivalently, y = 0 x - 2, always gives the same y-value, -2, for any value of x. The graph of this equation is the horizontal line with y-intercept -2.

62.

58.

y = -5 x

Three ordered pairs that satisfy this equation are (0, 0), (-1, 5), and (1, -5). Use these points to draw the graph.

x = 4

For any value of y, the x-value is 4. Because all ordered pairs that satisfy this equation have the same first number, this equation does not represent a function. The graph is the vertical line with x-intercept 4.

y y = –5x 1 0

63.

x=4

59.

–5

y+8=0

–8

4 x

x+5=0

This equation may be rewritten as x = -5. For any value of y, the x-value is -5. Because all ordered pairs that satisfy this equation have the same first number, this equation does not represent a function. The graph is the vertical line with x-intercept -5.

x + 4y = 0

If y = 0, then x = 0, so the x-intercept is 0. If x = 0, then y = 0, so the y-intercept is 0. Both intercepts give the same ordered pair, (0, 0). To get a second point, choose some other value of x (or y ). For example if x = 4, then x + 4y = 0 4 + 4y = 0 4 y = -4 y = -1, giving the ordered pair (4, -1). Graph the line through (0, 0) and (4, -1).

60.

y+8= 0

42 64.

Chapter 1 LINEAR FUNCTIONS 3x - 5 y = 0

(c)

If y = 0, then x = 0, so the x-intercept is 0. If x = 0, then y = 0, so the y-intercept is 0. Both intercepts give the same ordered pair (0, 0). To get a second point, choose some other value of x (or y ). For example, if x = 5, then

y = 0.9 x + 36 y = 0.9(180) + 36 = 198 180 gourmet cupcakes would cost $198.

67. (a)

3x - 5 y = 0 3(5) - 5 y = 0 15 - 5 y = 0 -5 y = -15 y =3

giving the ordered pair (5, 3). Graph the line through (0, 0) and (5, 3).

(b)

Yes, the data appear to lie roughly along a straight line. The line goes through (10, 31,500) and (19, 36,880).

m= 5 x

36,880 - 31,500 » 597.8 19 - 10

Use (10, 31,500) and the point-slope form. y - y1 = m(t - t1)

3x – 5y = 0

y - 31,500 = 597.8(t - 10)

65. (a) The line goes through (2, 27,000) and (5, 63,000). 63, 000 - 27, 000 5-2 = 12, 000

y = 597.8t - 25,522

The slope 597.8 indicates that tuition and fees have increased approximately $598 per year.

(c)

y - 27, 000 = 12, 000( x - 2) y - 27, 000 = 12, 000 x - 24, 000 y = 12, 000 x + 3000

(b) Let y = 100, 000; find x.

The year 2035 is too far in the future to rely on this equation to predict costs; too many other factors may influence these costs by then.

68. (a)

100, 000 = 12, 000 x + 3000 97, 000 = 12, 000 x 8.08 = x

Sales would surpass $100,000 after 8 years, 1 month. 66. (a) The line goes through (100, 126) and (120, 144) . 144 - 126 m = 120 - 100 = 0.9 The average cost of producing gourmet cupcakes increases by $0.90 per cupcake. (b) Use the point slope form with the given points. y - 126 = 0.9( x - 100) y = 0.9 x - 90 + 126 y = 0.9 x + 36

(b)

The line goes through (6, 233.04) and (18, 421.79). 421.79 - 233.04 m= » 15.729 18 - 6 Use (10, 31,500) and the point-slope form. y - y1 = m(t - t1) y - 233.04 = 15.729(t - 6)

y = 15.729t + 138.67

Section 1.1 (c)

The line goes through (10, 296.29) and (18, 421.79). m = 421.79 - 296.29 » 15.688 18 - 10 Use (10, 296.29) and the point-slope form. y - 296.29 = 15.688(t - 10) y = 15.688t + 139.41

(d)

(e)

The data is approximately linear because all the data points do not fall on a straight line. So the lines between different pairs of points have different slopes that are close in value. y = 15.729t + 138.67 y = 15.729(16) + 138.67 y » 390.33 million subscribers y = 15.688t + 139.41

71. (a) Let x = age. u = 0.85(220 - x ) = 187 - 0.85x l = 0.7(200 - x ) = 154 - 0.7 x

(b)

u = 187 - 0.85(20) = 170 l = 154 - 0.7(20) = 140

The target heart rate zone is 140 to 170 beats per minute. (c)

u = 187 - 0.85(40) = 153 l = 154 - 0.7(40) = 126

The target heart rate zone is 126 to 153 beats per minute. (d)

154 - 0.7 x = 187 - 0.85( x + 36) 154 - 0.7 x = 187 - 0.85x - 30.6 154 - 0.7 x = 156.4 - 0.85 x

y = 15.688(16) + 139.41

0.15 x = 2.4 x = 16

y » 390.42 million subscribers

Both the estimated values are slightly less than the actual number of subscribers of 395.88 million.

The younger woman is 16; the older woman is 16 + 36 = 52. l = 0.7(220 - 16) » 143 beats per minute.

69. (a) The line goes through (3, 100) and (35, 251.1). m = 251.1 - 100 » 4.722 35 - 3 Use the point (3, 100) and the point-slope form. y - 100 = 4.722(t - 3) y = 4.722t - 14.166 + 100 y = 4.722t + 85.834

72. Let x represent the force and y represent the speed. The linear function contains the points (0.75, 2) and (0.93, 3).

(b)

(c)

The year 2000 corresponds to t = 2000 – 1980 = 20. y = 4.722(20) + 85.834 » 180.274 The predicted value is slightly more than the actual CPI of 172.2. The annual CPI is increasing at a rate of approximately 4.7 units per year.

70. (a) The line goes through (4, 0.17) and (7, 0.33). m = 0.33 - 0.17 » 0.053 7-4 y - 0.33 = 0.16 (t - 7) 3 y - 0.33 = 0.053t - 0.373 y » 0.053t - 0.043

(b)

Let y = 0.5; solve for t. 0.5 = 0.053t - 0.043 0.543 = 0.053t 10.2 = t

3-2 1 = 0.93 - 0.75 0.18 1 100 50 = 18 = = 18 9 100

Use point-slope form to write the equation. 50 ( x - 0.75) y-2= 9 50 50 (0.75) y-2= x9 9 50 75 +2 y = x9 18 50 13 y = x9 6 Now determine y, the speed, when x, the force, is 1.16. 50 13 y = (1.16) 9 6 58 13 77 = = » 4.3 9 6 18 The pony switches from a trot to a gallop at approximately 4.3 meters per second.

Chapter 1 LINEAR FUNCTIONS

73. Let x = 0 correspond to 1900. Then the “life expectancy from birth” line contains the points (0, 46) and (110, 78.7). 78.7 - 46 32.7 m = = » 0.297 110 - 0 110

Since (0, 46) is one of the points, the line is given by the equation. y = 0.297 x + 46. The “life expectancy from age 65” line contains the points (0, 76) and (110, 84.1). 84.1 - 76 8.1 m = = » 0.074 110 - 0 110 Since (0, 76) is one of the points, the line is given by the equation y = 0.074 x + 76. Set the two expressions for y equal to determine where the lines intersect. At this point, life expectancy should increase no further. 0.297 x + 46 = 0.074 x + 76 0.223x = 30 x » 135 Determine the y-value when x = 129. Use the first equation. y = 0.297(135) + 46 = 40.095 + 46 = 86.095

75. (a) The line goes through (50, 249,187) and (117, 1,127,167). 1,127,167 - 249,187 m= » 13,104.18 117 - 50

Use the point (50, 249,187) and the pointslope form. y - 249,187 = 13,104.18(t - 50) y = 13,104.18t - 655, 209 + 249,187 y = 13,104.18t - 406, 022

(b)

(c)

Use the point (9, 26.2) and the point-slope form. y - 26.2 = 0.12(t - 9) y = 0.12t + 25.12 (b)

y = -1.93t + 266.7

(c)

Since 0.14 > 0.12 , women seem to have the faster increase in median age at first marriage.

(d)

Let y = 32. 32 = 0.12t + 25.12 6.88 = 0.12t 57.33 » t

25 = -1.93t + 266.7 -241.7 = -1.93t 125.2 » t

If the trend continues, the goal of the mortality rate for children under 5 years of age being 25 per 1000 live births would be reached in 2026.

The line (for the data for women) goes through (9, 23.8) and (39, 28.0). 28.0 - 23.8 m= = 0.14 39 - 9 Use the point (9, 23.8) and the point-slope form. y - 23.8 = 0.14(t - 9) y = 0.14t + 22.54

y - 93 = -1.93(t - 90) y = -1.93t + 266.7

(b) Let y = 25 and solve for t.

The number of immigrants admitted to the United States in 2020 will be about 1,166,480. The equation y = 13,104.18t - 406, 022 has –406,022 for the y-intercept, indicating that the number of immigrants admitted in the year 1900 was –406,022. Realistically, the number of immigrants cannot be a negative value, so the equation cannot be used for valid predicted values.

76. (a) The line (for the data for men) goes through (9, 26.2) and (39, 29.8). 29.8 - 26.2 m= = 0.12 39 - 9

Thus, the maximum life expectancy for humans is about 86 years. 74. (a) Let t = 0 correspond to 1900. Then the “mortality rate for children under 5 years of age” line contains the points (90, 93) and (118, 39). 39 - 93 m= » -1.93 118 - 90 Use the point (90, 93) and the point-slope form.

The year 2020 corresponds to t = 120. y = 13,104.18(120) - 406, 022 y » 1,166, 480

The median age at first marriage for men will reach 32 in the year 1980 + 58 = 2038.

Section 1.1 (e)

45 So, 1970 + 133 = 2103.

Let t = 57.33. y = 0.14(57.33) + 22.54 y » 30.6

The temperature will rise to 19°C in about the year 2103.

The median age at first marriage for women will reach be 30.6 when the median age for men is 30. 77. (a) Plot the points (15, 1600), (200,15, 000), (290, 24, 000), and (520, 40, 000).

(c)

From part (a) If the temperature rises 0.2C° per decade, it rises 0.02C° per year. m = 0.02 b = 15, since a point is (0, 15). T = 0.02t + 15

Velocity

5 10

2.5 10

From part (b) Let T = 19; find t. 19 = 0.02t + 15

4 = 0.02t t = 200 0 100 200 300 400 500 600 Distance

The points lie approximately on a line, so there appears to be a linear relationship between distance and time. (b) The graph of any equation of the form y = mx goes through the origin, so the line goes through (520, 40, 000) and (0, 0). 40, 000 - 0 » 76.9 520 - 0 b=0

So, 1970 + 200 = 2170 The temperature will rise to 19°C in about the year 2170. 79. (a) Let t = 0 correspond to 2000. Then the line representing the percentage of Americans 12 or older who had listened to online radio contains the points (10, 27) and (18, 64). m=

Use the point (10, 27) and the point-slope form.

y = 76.9 x + 0

yo - 27 = 4.625(t - 10)

y = 76.9 x

(c)

Let y = 60, 000; solve for x. 60, 000 = 76.9 x 780.23 » x Hydra is about 780 megaparsecs from earth.

(d)

yo = 4.625t - 19.25

(b) The line representing the percent of U.S. cellphone users who ever listened to online radio in a car using a phone contains the points (10, 6) and (18, 44).

9.5 ´ 1011 , m = 76.9 m

9.5 ´ 1011 76.9 = 12.4 billion years

44 - 6 = 4.75 18 - 10

Use the point (10, 6) and the point-slope form.

yc  6  4.75(t  10)

78. (a) If the temperature rises 0.3C° per decade, it rises 0.03C° per year. m = 0.03 b = 15, since a point is (0,15).

T = 0.03t + 15

(b)

64 - 27 = 4.625 18 - 10

yc  4.75t  41.5

(c) The percent of Americans who had listened to online radio in the previous month increased by 4.625% per year, while the percent of U.S. cellphone users who had ever listened to online radio in the car using a phone increased by 4.75% per year.

Let T = 19; find t. 19 = 0.03t + 15 4 = 0.03t t = 133.3 » 133

Chapter 1 LINEAR FUNCTIONS

1.2 Linear Functions and Applications Your Turn 1 For g ( x ) = -4 x + 5, calculate g (-5).

g ( x ) = -4 x + 5 g (-5) = -4(-5) + 5 = 20 + 5 = 25 Your Turn 2 For the demand and supply functions given in Example 2, find the quantity of watermelon demanded and supplied at a price of $3.30 per watermelon. p = D(q) = 9 - 0.75q 3.30 = 9 - 0.75q 0.75q = 5.7 5.7 q= = 7.6 0.75

Since the quantity is in thousands, 7600 watermelon are demanded at a price of $3.30. p = S (q) = 0.75q 3.30 = 0.75q

Your Turn 5 The fixed cost is b, this function has the form C ( x)  mx  7145. To find m, use the fact that producing 100 items costs $7965. C ( x) = mx + 7145 C (100) = 100m + 7145 7965 = 100m + 7145 820 = 100m m = 8.2

Thus the cost function is C ( x)  8.2 x  7145. Your Turn 6 The cost function is C ( x ) = 35x + 250 and the revenue function is R( x) = 59 x. Thus the profit function is P ( x) = R ( x ) - C ( x ) = 59 x - (35x + 250) = 24 x - 250

The profit is to be $8030. P( x) = 24 x - 250

3.3 q= = 4.4 0.75

8030 = 24 x - 250 24 x = 8280

Since the quantity is in thousands, 4400 watermelon are supplied at a price of $3.30. Your Turn 3 Set the two price expressions equal and solve for the equilibrium quantity q. 10 - 0.85q = 0.4q 10 = 1.25q 10 =8 q= 1.25

8280 = 345 24

Sale of 345 units will produce $8030 profit.

1.2 Warmup Exercises W1. 3( x - 2)2 + 6( x + 4) - 5x + 4 3(5 - 2)2 + 6(5 + 4) - 5(5) + 4 = 3(3)2 + 6(9) - 5(5) + 4 = 27 + 54 - 25 + 4 = 60

The equilibrium quantity is 8000 watermelon. Use either price expression to find the equilibrium price p. W2.

p = 0.4q p = 0.4(8) = 3.2 The equilibrium price is $3.20 per watermelon. Your Turn 4 The marginal cost is the slope of the cost function C ( x ), so this function has the form C ( x ) = 15x + b. To find b, use the fact that producing 80 batches costs $1930.

C ( x ) = 15x + b C (80) = 15(80) + b 1930 = 1200 + b b = 730

p 8 6 4 2 0

y = 7 – 2.5x

4 q

1.2

Exercises

This statement is true. When we solve y = f ( x) = 0, we are finding the value of x when y = 0, which is the x-intercept. When we evaluate f (0), we are finding the value of y when x = 0, which is the y-intercept.

Thus the cost function is C ( x ) = 15x + 730.

This statement is false. The graph of f ( x) = -5 is a horizontal line.

Section 1.2

This statement is true. Only a vertical line has an undefined slope, but a vertical line is not the graph of a function. Therefore, the slope of a linear function cannot be undefined.

This statement is true. For any value of a, f (0) = a ⋅ 0 = 0,

so the point (0, 0), which is the origin, lies on the line.

24. $10 is the fixed cost and $0.99 is the cost per downloaded song—the marginal cost. Let x = the number of downloaded songs and C ( x) = cost of downloading x songs. Then, C ( x ) = (marginal cost) ⋅ (number of downloaded songs) + fixed cost C ( x ) = 0.99 x + 10.

25. $2 is the fixed cost and $0.75 is the cost per halfhour.

Let x = the number of half-hours;

False. The variable x in the linear function f ( x) = 3x + 4 is the independent variable.

True

f (2) = 7 - 5(2) = 7 - 10 = -3

10.

f (4) = 7 - 5(4) = 7 - 20 = -13

11.

f (-3) = 7 - 5(-3) = 7 + 15 = 22

12.

f (-1) = 7 - 5(-1) = 7 + 5 = 12

13.

g (1.5) = 2(1.5) - 3 = 3 - 3 = 0

14.

g (2.5) = (2.5) - 3 = 5 - 3 = 2

15.

æ 1ö æ 1ö g çç - ÷÷ = 2 çç - ÷÷ - 3 = -1 - 3 = -4 çè 2 ÷ø çè 2 ø÷

C ( x) = mx + b, where b is the fixed cost.

16.

æ 3ö æ 3ö 3 9 g çç - ÷÷ = 2 çç - ÷÷ - 3 = - - 3 = èç 4 ø÷ èç 4 ø÷ 2 2

Now,

17.

f (t ) = 7 - 5(t ) = 7 - 5t

18.

g (k 2 ) = 2(k 2 ) - 3 = 2k 2 - 3

C ( x) = the cost of parking a car for x half-hours.

Thus,

C ( x) = 2 + 0.75x = 0.75x + 2

26. $44 is the fixed cost and $0.28 is the cost per mile. Let x = the number of miles; R( x) = the cost of renting for x miles.

Thus, R( x ) = fixed cost + (cost per mile) ⋅ (number of miles) R( x ) = 44 + 0.28 x. 27. Fixed cost, $100; 50 items cost $1600 to produce.

Let C ( x) = cost of producing x items. C ( x) = mx + 100 C ( x) = 1600 when x = 50, so 1600 = m(50) + 100 1500 = 50m 30 = m.

19. The fixed cost is constant for a particular product and does not change as more items are made. The marginal cost is the rate of change of cost at a specific level of production and is equal to the slope of the cost function at that specific value; it approximates the cost of producing one additional item. 23. $10 is the fixed cost and $2.25 is the cost per hour. Let x = number of hours;

Thus, C ( x) = 30 x + 100. 28. Fixed cost: $35; 8 items cost $395.

Let C ( x) = cost of x items C ( x) = mx + b, where b is the fixed cost C ( x) = mx + 35

Now, C ( x) = 395 when x = 8, so 395 = m(8) + 35 360 = 8m 45 = m.

R( x) = cost of renting a snowboard for x hours.

Thus, R( x ) = fixed cost + (cost per hour) ⋅ (number of hours) R( x ) = 10 + (2.25)( x ) = 2.25x + 10

Thus, C ( x) = 45x + 35.

Chapter 1 LINEAR FUNCTIONS

29. Marginal cost: $75; 50 items cost $4300.

(g)

C ( x) = 75 x + b

Now, C ( x) = 4300 when x = 50. 4300 = 75(50) + b 4300 = 3750 + b 550 = b

Thus, C ( x) = 75 x + 550.

(h)

p = 16 – 1.25q

2 4 6 8 10 12 14 q

S (q) = 0.75q

Let S (q) = 0. Find q.

30. Marginal cost, $120; 700 items cost $96,500 to produce. C ( x) = 120 x + b Now, C ( x) = 96,500 when x = 700. 96,500 = 120(700) + b 96,500 = 84, 000 + b 12,500 = b Thus, C ( x ) = 120 x + 12,500. 31.

p 16 14 12 10 8 6 4 2

0 = 0.75q 0 = q

When the price is $0, the number of watches supplied is 0. (i)

Let S (q) = 10. Find q. 10 = 0.75q 40 = q 3 q = 13.3

D(q) = 16 - 1.25q (a) D(0) = 16 - 1.25(0) = 16 - 0 = 16

When 0 watches are demanded, the price is $16. (b)

(c)

When the price is $10, The number of watches supplied is about 1333.

D(4) = 16 - 1.25(4) = 16 - 5 = 11 When 400 watches are demanded, the price is $11.

(j) Let S (q) = 20. Find q. 20 = 0.75q 80 = q 3 q = 26.6

D(8) = 16 - 1.25(8) = 16 - 10 = 6 When 800 watches are demanded, the price is $6.

When the price is $20, the number of watches demanded is about 2667.

(d) Let D(q) = 8. Find q. 8 = 16 - 1.25q 5 q =8 4 q = 6.4

(k)

When the price is $8, the number of watches demanded is 640. (e)

Let D(q) = 10. Find q. 10 = 16 - 1.25q 5 q =6 4 q = 4.8

(l)

When the price is $10, the number of watches demanded is 480. (f)

Let D(q) = 12. Find q. 12 = 16 - 1.25q 5 q = 4 4 q = 3.2 When the price is $12, the number of watches demanded is 320.

p 16 14 12 10 8 6 4 2

p = 16 – 1.25q

(8, 6) p = 0.75q 2 4 6 8 10 12 14 q

D( q ) = S ( q ) 16 - 1.25q = 0.75q 16 = 2q 8= q S (8) = 0.75(8) = 6

The equilibrium quantity is 800 watches, and the equilibrium price is $6. 32.

D(q) = 6 - 0.25q

(a)

D(0) = 6 - 0.25(0) = 6 - 0 = 6

When 0 quarts are demanded, the price is $6.

Section 1.2 (b)

49 When the price is $6.30, 1800 quarts are supplied.

D(4) = 6 - 0.25(4) = 6 - 1 = 5

When 400 quarts are demanded, the price is $5. (c)

(k)

D(8.4) = 6 - 0.25(8.4) = 6 - 2.1 = 3.9

When 840 quarts are demanded, the price is $3.90. (d) Let D(q) = 4.5. Find q. 4.5 = 6 - 0.25q 0.25q = 1.5 q =6

(l)

When the price is $4.50, 600 quarts are demanded. (e)

Let D(q) = 3.25. Find q.

S (10) = 0.35(10) = 3.5

3.25 = 6 - 0.25q 0.25q = 2.75 q = 11

When the price is $3.25, 1100 quarts are demanded. (f)

D( q ) = S ( q) 6 - 0.25q = 0.35q 6 = 0.6q 10 = q

33.

The equilibrium quantity is 1000 quarts and the equilibrium price is $3.50. 2 4 p = S (q) = q; p = D(q) = 120 - q 5 5 (a)

Let D(q) = 2.4. Find q. 2.4 = 6 - 0.25q 0.25q = 6.6 q = 14.4

When the price is $2.40, 1440 quarts are demanded. (g)

(h)

(b)

S (q) = 0.35q

The equilibrium quantity is 100, the equilibrium price is $40.

Let S (q) = 0. Find q. 0 = 0.35q q =0

34. (a)

Let S (q) = 2.80. Find q.

p = 1.4q – 0.6

1 0

When the price is $2.80, 800 quarts are supplied. Let S (q) = 6.30. Find q.

p = –2q + 3.2

2.80 = 0.35q q =8

(j)

p 3

When the price is $0, 0 quarts are supplied. (i)

S (q ) = D(q ) 2 4 q = 120 - q 5 5 6 q = 120 5 q = 100 2 S (100) = (100) = 40 5

(b)

S (q) = p = 1.4q - 0.6 D(q) = p = -2q + 3.2

Set supply equal to demand and solve for q.

6.30 = 0.35q q = 18

Chapter 1 LINEAR FUNCTIONS 1.4q - 0.6 = -2q + 3.2

(b)

1.4q + 2q = 0.6 + 3.2

P( x ) = 15x - (5x + 20)

3.4q = 3.8 3.8 q = » 1.12 3.4 S (1.12) = 1.4(1.12) - 0.6 = 0.968

P(100) = 15(100) - (5 ⋅ 100 + 20) = 1500 - 520 = 980

The equilibrium quantity is about 1120 pounds; the equilibrium price is about $0.96 35. Use the supply function to find the equilibrium quantity that corresponds to the given equilibrium price of $4.50. S (q) = p = 0.3q + 2.7 4.50 = 0.3q + 2.7 1.8 = 0.3q 6= q The line that represents the demand function goes through the given point (2, 6.10) and the equilibrium point (6, 4.50). 4.50 - 6.10 m= = -0.4 6-2

Use point-slope form and the point (2, 6.10) . D(q) - 6.10 = -0.4(q - 2) D(q) = -0.4q + 0.8 + 6.10 D(q) = -0.4q + 6.9

36. Use the supply function to find the equilibrium quantity that corresponds to the given equilibrium price of $5.85. p = S ( q) 5.85 = 0.25q + 3.6 2.25 = 0.25q 9 = q

(c)

5.85 - 7.60 9-4 = -0.35 Use point-slope form and the point (4, 7.60) .

The profit from 100 units is $980. P( x ) = 500 15x - (5x + 20) = 500 10 x - 20 = 500 10 x = 520 x = 52 For a profit of $500, 52 units must be produced.

38.

C ( x) = 12 x + 39; R( x) = 25 x

(a)

C ( x ) = R( x ) 12 x + 39 = 25x 39 = 13x 3= x

The break-even quantity is 3 units. (b) P( x ) = R( x ) P( x ) = 25x - (12 x + 39) P( x ) = 13x - 39 P(250) = 13(250) - 39 = 3250 - 39 = 3211 The profit from 250 units is $3211. (c)

P( x) = $130; find x. 130 = 13x - 39 169 = 13x 13 = x

The line that represents the demand function goes through the given point (4, 7.60) and the equilibrium point (9, 5.85) . m=

P( x ) = R( x ) - C ( x )

For a profit of $130, 13 units must be produced. 39. (a)

C ( x) = mx + b ; m = 3.50; C (60) = 300 C ( x) = 3.50 x + b

Find b.

D(q) - 7.60 = -0.35(q - 4)

300 = 210 + b

D(q) = -0.35q + 1.4 + 7.60

90 = b C ( x ) = 3.50 x + 90

D(q) = -0.35q + 9

37.

C ( x) = 5x + 20; R( x) = 15x

(a)

300 = 3.50(60) + b

(b)

R( x) = 9 x C ( x) = R( x)

C ( x ) = R( x ) 5x + 20 = 15x

3.50 x + 90 = 9 x

20 = 10 x

90 = 5.5x

2 = x

16.36 = x

The break-even quantity is 2 units.

Joanne must produce and sell 17 shirts.

Section 1.2 (c)

51 P( x) = R( x) - C ( x); P( x) = 500

(c)

500 = 9 x - (3.50 x + 90)

= 97 + 1.32

500 = 5.5x - 90

= 98.32

The total cost of producing 1000 cups is $98.32.

590 = 5.5x 107.27 = x

(d)

To make a profit of $500, Joanne must produce and sell 108 shirts. 40. (a)

C (1000) = 2675; b = 525

(e)

2675 = m(1000) + 525 2150 = 1000m

(f)

2.15 = m C ( x ) = 2.15x + 525

(b)

C (1001) = 0.097(1001) + 1.32 = 97.097 + 1.32 = 98.417

C ( x) = mx + b

Find m.

C (1000) = 0.097(1000) + 1.32

R( x) = 4.95x

42.

C ( x) = R( x)

The total cost of producing 10001 cups is $98.42. Marginal cost = 98.417 - 98.32 = $0.097 or 9.7¢ The marginal cost for any cup is the slope, $0.097 or 9.7¢. This means the cost of producing one additional cup of coffee would be 9.7¢.

C (10, 000) = 547,500; C (50, 000) = 737,500

(a)

C ( x) = mx + b

2.15 x + 525 = 4.95 x

737,500 - 547,500 50, 000 - 10, 000 190, 000 = 40, 000 = 4.75

525 = 2.80 x 187.5 = x

In order to break even, he must produce and sell 188 books. (c)

P( x) = R( x) - C ( x); P( x) = 1000

y - 547,500 = 4.75( x - 10, 000) y - 547,500 = 4.75 x - 47,500 y = 4.75 x + 500, 000 C ( x) = 4.75 x + 500, 000

1000 = 4.95 x - (2.15 x + 525) 1000 = 4.95 x - 2.15 x - 525 1000 = 2.80 x - 525 1525 = 2.80 x

(b)

The fixed cost is $500,000.

544.6 = x

(c)

C (100, 000) = 4.75(100, 000) + 500, 000 = 475, 000 + 500, 000 = 975, 000

In order to make a profit of $1000, he must produce and sell 545 books. 41. (a) Using the points (100, 11.02) and (400, 40.12), 40.12 - 11.02 m = 400 - 100 29.1 = = 0.097. 300 y - 11.02 = 0.097( x - 100) y - 11.02 = 0.097 x - 9.7 y = 0.097 x + 1.32 C ( x ) = 0.097 x + 1.32

(b) The fixed cost is given by the constant in C ( x). It is $1.32.

(d)

43.

The total cost to produce 100,000 items is $975,000. Since the slope of the cost function is 4.75, the marginal cost is $4.75. This means that the cost of producing one additional item at this production level is $4.75.

C ( x) = 85 x + 900 R( x) = 105x

Set C ( x) = R( x) to find the break-even quantity. 85 x + 900 = 105x 900 = 20 x 45 = x The break-even quantity is 45 units. You should decide not to produce since no more than 38 units can be sold.

Chapter 1 LINEAR FUNCTIONS P( x) = R( x) - C ( x) = 105 x - (85 x + 900) = 20 x - 900

105 x + 6000 = 250 x 6000 = 145 x 41.38 » x

The profit P( x) = R( x) - C ( x) is 0 at the given break-even quantity of 80. P( x ) = px - (mx + 400) P( x ) = px - mx - 400 P( x ) = Mx - 400 (Let M = p - m.) P(80) = M ⋅ 80 - 400 0 = 80 M - 400 400 = 80 M 5= M

The break-even quantity is about 41 units, so you should decide to produce.

So, the linear profit function is P( x) = 5 x - 400, and the marginal profit is 5.

The profit function is P( x) = 20 x - 900. 44.

C ( x) = 105 x + 6000 R( x) = 250 x

Set C ( x) = R( x) to find the break-even quantity.

P( x) = R( x) - C ( x) = 250 x - (105 x + 6000) = 145x - 6000

The profit function is P( x) = 145x - 6000. 45.

C ( x) = 70 x + 500 R( x) = 60 x 70 x + 500 = 60 x 10 x = -500 x = -50

This represents a break-even quantity of -50 units. It is impossible to make a profit when the break-even quantity is negative. Cost will always be greater than revenue. P( x) = R( x) - C ( x) = 60 x - (70 x + 500) = -10 x - 500

The profit function is P( x) = -10 x - 500. 46.

The revenue function is R( x) = px, where p is the price per unit.

C ( x) = 1000 x + 5000 R( x) = 900 x 900 x = 1000 x + 5000 -5000 = 100 x -50 = x

It is impossible to make a profit when the breakeven quantity is negative. Cost will always be greater than revenue. P ( x ) = R ( x ) - C ( x) = 900 x - (1000 x + 5000) = -100 x - 5000

The profit function is P( x) = -100 x - 5000 (always a loss). 47. Since the fixed cost is $400, the cost function is C ( x) = mx + 100 , where m is the cost per unit.

48. Since the fixed cost is $650, the cost function is C ( x) = mx + 650 , where m is the cost per unit. The revenue function is R( x) = px, where p is the price per unit.

The profit P( x) = R( x) - C ( x) is 0 at the given break-even quantity of 25. P( x ) = px - (mx + 650) P( x ) = px - mx - 650 P( x ) = Mx - 650 (Let M = p - m.) P(25) = M ⋅ 25 - 650 0 = 25M - 650 650 = 25M 26 = M So, the linear profit function is P( x) = 26 x - 650, and the marginal profit is 26. 49. (a)

f ( x) = 34 x + 230 1000 = 34 x + 230 770 = 34 x x = 22.647

Approximately 23 acorns per square meter would produce 1000 deer tick larvae per 400 square meters. (b) The slope is 34, which indicates that the number of deer tick larvae per 400 square meters in the spring will increase by 34 for each additional acorn per square meter in the fall. 50. (a) Let t correspond to the number of years since 1990. Then for the cause of death due to tobacco, we have at t  0, m  2.8 since the quantity was rising at the rate of 28 million years per decade and b  35 since 35 million years of healthy life were lost. The linear function is f1 (t )  2.8t  35.

Section 1.2

(b) For the cause of death due to diarrhea, we have at t = 0, m = -2.2 since the quantity was falling at the rate of 22 million years per decade and b = 100 since 100 million years of healthy life were lost. The linear function is f d (t ) = -2.2t + 100.

2.8t + 35 = -2.2t + 100 5.0t = 65 t = 13

51. Use the formula derived in Example 8 in this section of the textbook. 9 5 C + 32 or C = ( F - 32) 5 9 F = 58; find C.

(a)

5 (58 - 32) 9 5 C = (26) = 14.4 9 The temperature is 14.4°C.

9 (37) + 32 5 333 F = + 32 = 98.6 5 The Fahrenheit equivalent of 37°C is 98.6°F. C = 36.5; find F. 9 F = (36.5) + 32 5 F = 65.7 + 32 = 97.7

C = 37.5; find F. 9 F = (37.5) + 32 5 = 67.5 + 32 = 99.5

The amount of healthy life lost to tobacco was expected to equal that lost to diarrhea in 2003.

F =

C = 37; find F. F =

(b)

f t (t ) = f d ( t )

(c)

(a)

The range is between 97.7°F and 99.5°F. 53. If the temperatures are numerically equal, then F = C. 9 C + 32 5 9 C = C + 32 5 F =

C =

(b)

F = -20; find C. 5 ( F - 32) 9 5 C = (-20 - 32) 9 5 C = (-52) = -28.9 9 C =

4 - C = 32 5 C = -40 The Celsius and Fahrenheit temperatures are numerically equal at -40.

54. (a) m = 7.17 b = 3.308 ´ 109 C ( x) = mx + b C ( x) = 7.17 x + 3.308 ´ 109

The temperature is -28.9C. (c)

(b) C (120) = 7.17 ⋅ 120, 000, 000 + 3.308 ´ 109 = 4,168, 400, 000

C = 50; find F.

9 C + 32 5 9 F = (50) + 32 5 F = 90 + 32 = 122

The total cost to generate 120 million MWh will be about $4.168 billion.

F =

The temperature is 122°F.

(c)

Let C ( x) = 4,500, 000, 000 . 4.5 ´ 109 = 7.17 x + 3.308 ´ 109 1.192 ´ 109 = 7.17 x

52. Use the formula derived in Example 8 in this section of the textbook. 9 5 F = C + 32 or C = ( F - 32) 5 9

166, 248, 256.6 = x

Spending $4.5 billion would result in generating about 166 million MWh of energy.

Chapter 1 LINEAR FUNCTIONS

1.3

The Least Squares Line

Your Turn 1

True

(a)

y 6 5

2 3

0.5 1

1 3

4 9

0.25 1

4 5

2 2.5

8 12.5

16 25

4 6.25

6 7

3 3

18 21

36 49

9 9

8 9

4 4.5

32 40.5

64 81

16 20.25

10 55

5 25.5

50 186

100 385

25 90.75

3 2 1

The number of data points is n  6. Putting the column totals into the formula for the slope m, we get

n ( å xy ) - ( å x )( å y )

(

)

n å x2 - (å x) 6(132) - (21)(33)

6(91) - (21) m » 0.9429

r =

 y - m ( x ) n 33 - (0.9429)(21) = » 2.2 6

x = 21 y = 33 xy = 132 x 2 = 91 y 2 = 199

(b)

(

(c)

n(å x ) - (å x )2 ⋅ n(å y 2 ) - (å y )2 10(186) - (55)(25.5) 10(385) - (55)2 ⋅ 10(90.75) - (25.5)2

The least squares line is of the form Y = mx + b. First solve for m. m =

)

n å x 2 - (å x) ⋅ n å y 2 - (å y ) 6(132) - (21)(33) 6(91) - (21)2 ⋅ 6(199) - (33)2

» 0.9429

n(å xy ) - (å x )(å y ) n(å x 2 ) - (å x )2 10(186) - (55)(25.5)

10(385) - (55)2 = 0.5545454545 » 0.555

Now find b. å y - m(å x) b= n 25.5 - 0.5545454545(55) = 10 = -0.5

1.3

Exercises

True

y 6

False. Additionally, å y 2 must be known to

calculate the correlation coefficient.

n(å xy ) - (å x )(å y )

n ( å xy ) - ( å x )( å y )

)

Your Turn 2 Put the column totals computed in Your Turn 1 into the formula for the correlation r.

(

10 x

» 0.993

The least square line is Y  0.9429 x  2.2.

Thus, Y = 0.555x - 0.5.

False. A positive correlation coefficient between two variables means that the line has a positive slope.

4 2 1

10 x

Section 1.3 (d)

Let x = 11. Find Y.

(b)

Y = 0.55(11) - 0.5 = 5.6

y2 0.64 1.44

5.44 8.4

x2 46.24 49.0

0.9 0.9 1.5

6.39 6.48 11.1

50.41 51.84 54.76

0.81 0.81 2.25

5.3

37.81 252.25

5.95

6.8 7.0

0.8 1.2

7.1 7.2 7.4 35.5

n= 4 n ( å xy ) - ( å x )( å y )

m =

(

5(37.81) - (35.5)(5.3)

r =

4(10) - (6)2

» 0.6985

r = (0.6985)2 » 0.5

The answer is choice (c). 9.

å y - m( å x ) 6 - (0)(6) = = 1.5 n 4

b =

4(9) - (6)(6)

5(252.25) - (35.5)2 ⋅ 5(5.95) - (5.3)2

)

n å x2 - ( å x )

Thus, Y = 0 x + 1.5, or Y = 1.5 .

n ( å xy ) - ( å x )( å y )

r =

(

)

n å x2 - ( å x )

(

)

⋅ n å y 2 - ( å y )2

4(9) - (6)(6) 4(10) - (6)2 ⋅ 4(10) - (6)2

(c)

y 10 8 6

n=5

(a)

m = =

4 2

n ( å xy ) - ( å x )( å y )

(

n åx

) - (å x)

5(90) - (15)(15)

10.

å y - m( å x ) n 15 - (0.9782608)(15) = » 0.0652 5 Thus, Y = 0.9783x + 0.0652 . b =

n ( å xy ) - ( å x )( å y )

(

)

n å x2 - ( å x )

(

10 x

The point (9, 9) is an outlier that has a strong effect on the least squares line and the correlation coefficient.

5(91) - (15)2 = 0.9782608 » 0.9783

r =

)

⋅ n å y 2 - ( å y )2

-20

-180

400

-10

-150

111

430

5(90) - (15)(15) » 0.9783 2 2 5(91) - (15) ⋅ 5(91) - (15)

Chapter 1 LINEAR FUNCTIONS (a)

n=5 m= =

n ( å xy ) - ( å x )( å y )

(

n åx

) - (å x)

(

)

n å x2 - ( å x )

(

)

⋅ n å y 2 - ( å y )2

4(30) - (10)(10)

4(30) - (10)2 ⋅ 4(30) - 102

5(-150) - (19)(-10)

5(111) - (19)2 = -2.886597 » -2.887 b=

n ( å xy ) - ( å x )( å y )

r =

(c)

å y - m( å x )

y 10 5 0 –5 –10 –15 –20 –25

n -10 - (-2.886597)(19) = 5 = 8.969069 » 8.969

10 x

Thus, Y = -2.887 x + 8.969 . The point (9, -20) is an outlier that has a strong effect on the least squares line and the correlation coefficient.

n (å xy ) - (å x)(å y )

r =

(

n åx

) - (å x)2 ⋅ n (å y 2 ) - (å y)2

5(-150) - (19)(-10)

5(111) - (19)2 ⋅ 5(430) - (-10)2

11.

= -0.887994 » -0.8880

(b)

1.1

4.4

1.21

4.1

10.4

4.21

(a)

n=4

(

n ( å xy ) - ( å x )( å y )

(

n åx

) - (å x)

4(30) - (10)(10) 4(30) - (10)2

)

n å x 2 - (å x )

(

)

⋅ n å y 2 - (å y )2

4(10.4) - (10)(4.1)

n=4 m=

n (å xy ) - (å x )(å y )

4(30) - (10)2 ⋅ 4(4.21) - (4.1)2

= 0.7745966 » 0.7746

(b)

y 2

=1 1

å y - m( å x ) n 10 - (1)(10) = =0 4

b =

(c)

Thus, Y = 1x + 0, or Y = x .

Yes; because the data points are either on or very close to the horizontal line y = 1 , it seems that the data should have a strong linear relationship. The correlation coefficient does not describe well a linear relationship if the data points fit a horizontal line.

Section 1.3 12.

(å x)[(å y) - m(å x)] + nm(å x 2 ) = n(å xy) (å x)(å y) - m(å x) 2 + nm(å x 2 ) = n(å xy)

nm(å x 2 ) - m(å x )2 = n(å xy ) - (å x )(å y ) m[n(å x 2 ) - (å x )2 ] = n(å xy ) - (å x )(å y ) n(å xy ) - (å x )(å y ) m= n(å x 2 ) - (å x )2

(a) n = 5 m = =

æ å y - m(å x) ö÷ 2 (å x) çç ÷÷ + (å x )m = å xy çè ø n

14. (a)

(

)

n å x2 - ( å x ) 5(20) - (10)(10) 5(30) - (10)2

10(1645) - (125) 2 = 0.06814848 » 0.06815

(b)

Thus, Y  0.06815 x  1.844. .

n ( å xy ) - ( å x )( å y )

(

)

n å x2 - ( å x )

(b)

⋅ n ( å y 2 ) - ( å y )2

5(20) - (10)(10) =0 5(30) - (10)2 ⋅ 5(34) - (10)2

(c)

Let Y  6 and find x. 6  0.06815 x  1.844 4.156  0.06815 x 60.98  x The total value of consumer durable goods will reach at least $6 trillion in the year 2000 + 61 = 2061.

(d)

y 4 3 2

(c)

13.

No; a correlation coefficient of 0 means that there isn’t a linear relationship between the x and y values. A parabola (a quadratic relationship) seems to fit the given data points. nb + (å x)m = å y

The year 2025 corresponds to x = 25. Y  0.06815(25)  1.844  3.54775 The total value of consumer durable goods in 2025 will be about $3.548 trillion.

1 0

å y - m (å x)

n 26.9571 - (0.06815)(125) = 10 = 1.843835 » 1.844

Thus, Y = 0 x + 2, or Y = 2 .

n(å x 2 ) - (å x)2 10(342.586) - (125)(26.9571)

n ( å xy ) - ( å x )( å y )

å y - m( å x ) b = n 10 - (0)(10) = = 2 5

r =

n (å xy ) - (å x )( å y )

n ( å xy ) - ( å x)( å y ) n(å x 2 ) - (å x)2 ⋅ n(å y 2 ) - (å y)2 10(342.586) - (125)(26.9571) 10(1645) - 1252 ⋅ 10(73.2813) - 26.95712

» 0.7907

(å x)b + (å x 2 )m = å xy nb + (å x)m = å y nb = (å y) - (å x)m å y - m(å x) b = n

Since r is very close to 1, the data has a strong linear relationship, and the least squares line fits the data very well.

Chapter 1 LINEAR FUNCTIONS m=

15. (a)

n (å xy ) - ( å x )(å y ) 2

n 154 - 0.815(70) = 5 = 19.39 » 19.4

n(å x ) - (å x) 10(753.387) - (135)(57.232)

10(1905) - (135)2 = -0.2332727 » -0.2333 å y - m (å x)

Thus, Y = 0.815 x + 19.4 .

n 57.232 - (-0.2333)(135) = 10 = 8.87275 » 8.873 Thus, Y  0.2333 x  8.873.

(b)

(b) The number of subscriptions is growing at a rate of about 0.815 per year. (c) The year 2026 corresponds to x = 26. Y = 0.815(26) + 19.4 = 40.59 » 40.6

The year 2025 corresponds to x = 25.

If the trend continues linearly, the percent of subscriptions will be about 40.6 in 2026.

Y  0.2333(25)  8.873  3.0405

If the trend continues linearly, there will be about 3041 banks in 2025. (c)

å y - m( å x )

(d)

Let Y = 45 and find x. 45 = 0.815x + 19.4 25.6 = 0.815x

Let Y = 4 (since Y is the number of banks in thousands) and find x.

31.4 » x

1.800  0.2333 x  8.873

The number of subscriptions will exceed 45 in the year 2000 + 32 = 2032.

7.073  0.2333 x

30.32  x

The number of U.S. banks will drop below 1800 in the year 2000 + 31 = 2031. (d)

(e)

n ( å xy ) - ( å x )( å y )

n ( å x 2 ) - ( å x ) 2 ⋅ n( å y 2 ) - ( å y ) 2

(a)

27.3

273

100

745.29

29.4

352.8

144

864.36

30.8

431.2

196

948.64

32.7

523.2

256

1069.29

33.8

608.4

324

1142.44

154

2188.6

1020

4770.02

m= =

n (å xy ) - (å x )( å y ) 2

⋅ n (å y 2 ) - (å y)2

5(2188.6) - (70)(154) 5(1020) - 702 ⋅ 5(4770.02) - 1542

17.

(a)

58.1

755.3

169

3375.61

53.3

746.2

196

2840.89

49.3

739.5

225

2430.49

47.5

760

256

2256.25

43.8

744.6

289

1918.44

41.7

750.6

324

1738.89

293.7

4496.2

1459

14560.57

n(å x ) - (å x) 5(2188.6) - (70)(154)

5(1020) - (70) = 0.815

n ( å x 2 ) - ( å x)

This strong positive correlation means that the least squares line fits the data points extremely well.

» -0.9991

Since r is very close to –1, the data has a strong linear relationship, and the least squares line fits the data very well.

n (å xy ) - (å x)(å y)

» 0.9953

10(753.387) - (135)(57.232) 10(1905) - 1352 ⋅ 10(332.048) - 57.2322

16.

r =

n ( å xy ) - ( å x )( å y ) n( å x 2 ) - ( å x ) 2 6(4496.2) - (93)(293.7)

6(1459) - (93)2 » -3.20857 » -3.21

Section 1.3

59 b=

å y - m( å x )

n 293.7 - (-3.21)(93) = 6 » 98.7 Thus, Y = -3.21x + 98.7 .

å y - m( å x )

n 21,319.8 - 176.84(93) = 6 = 812.28

Thus, Y = 176.84 x + 812.28 .

(b)

The percent of households with landlines is decreasing at a rate of about 3.21% per year.

(b)

The consumer credit is growing at a rate of about $176.84 billion per year.

(c)

The year 2024 corresponds to x = 24.

(c)

The year 2030 corresponds to x = 30.

Y = -3.21(24) + 98.7 = 21.66 » 21.7 If the trend continues linearly, the percent of households will be about 21.7% in 2024.

Let Y = 20 and find x.

(d)

-78.7 = -3.21x 24.5 » x

The percent of households with landlines will dip below 20% in the year 2000 + 25 = 2025. r =

n (å xy) - (å x)(å y) n ( å x ) - ( å x)

⋅ n (å y ) - (å y ) 2

6(4496.2) - (93)(293.7)

6(1459) - 93

If the trend continues linearly, the consumer credit will be about $6117 in 2030. (d) Let Y  7000 and find x. 7000 = 176.84 x + 812.28 6187.72 = 176.84 x 35 » x

20 = -3.21x + 98.7

(e)

Y = 176.84(30) + 812.28 = 6117.48

⋅ 6(14,560.57) - 293.7

» -0.9896

The total debt will first exceed $7000 billion in the year 2000 + 35 = 2035. (e) r =

n (å xy) - (å x)(å y) 2

n( å x ) - ( å x )

⋅ n(å y 2 ) - (å y)2

6(333,551.6) - (93)(21,319.8) 6(1459) - 932 ⋅ 6(76,308,191.92) - 21,319.82

» 0.9952

This strong negative correlation means that the least squares line fits the data points extremely well. 18.

The strong positive correlation means that the least squares line fits the data points extremely well. 19. (a)

3078.7

40,023.1

169

9,478,393.69

3287.7

46,027.8

196

10,808,971.29

3527.0

52,905

225

12,439,729

3632.3

58,116.8

256

13,193,603.29

3814.9

64,853.3

289

14,553,462.01

3979.2

71,625.6

324

15,834,032.64

21,319.8

333,551.6

1459

76,308,191.92

(a)

n ( å xy ) - ( å x )( å y ) 2

Yes, the data points lie in a linear pattern. (b)

r  0.9516; there is a strong positive correlation among the data. (c)

The mean earnings of high school graduates are growing by about $615.16 per year.

(d)

Y  1071.91x  48,383.

r  0.9568; there is a strong positive correlation among the data.

n( å x ) - ( å x ) 6(333,551.6) - 93(21,319.8)

6(1459) - (93) 2 = 176.84

Y  615.16 x  26,104.

(e)

The mean earning of workers with a bachelor’s degree are growing by about $1072 per year.

Chapter 1 LINEAR FUNCTIONS 80, 000  615.16 x  26,104

(f)

53,896  615.16 x

å y - m( å x )

n 20.468 - 0.0721(126) = 9 » 1.2648 Thus, Y = -0.0721x + 1.2648 .

87.6  x The mean earnings for high school graduates will exceed $80,000 in the year 2000 + 88 = 2088. 80, 000  1071.91x  48,383

(b)

31, 617  1071.91x

n (å xy) - (å x)(å y)

r =

29.5  x The mean earning for workers with a bachelor’s degree will exceed $80,000 in the year 2000 + 30 = 2030.

n ( å x ) - ( å x)

⋅ n (å y 2 ) - (å y)2

9(290.878) - (126)(20.468)

9(1824) - 1262 ⋅ 9(48.237098) - 20.4682

» 0.4298

20. (a)

(c)

No, the data points do not lie in a linear pattern. r  0.2050; there is a negative correlation between price and distance. The cost of a ticket tends to decrease as the distance flown increases.

(b)

(c)

The outlier in the scatterplot is Philadelphia. Removing that data point results in r  0.0006244;

Y  0.00003964 x  191.16; the marginal cost per mile is about $0.00003964 per mile. (Much less than a penny.)

(d)

21.

(a)

22.

1.251

12.51

100

1.565001

2.387

26.257

121

5.697769

2.241

26.892

144

5.022081

2.369

30.797

169

5.612161

2.879

40.306

196

8.288641

2.574

38.61

225

6.625476

2.116

33.856

256

4.477456

2.068

35.156

289

4.276624

2.583

46.494

324

6.671889

126

20.468

290.878

1824

48.237098

n ( å xy ) - ( å x )( å y ) n( å x 2 ) - ( å x ) 2 9(290.878) - (126)(20.468)

(a)

9.8

100

96.04

9.1

100.1

121

82.81

8.3

99.6

144

68.89

8.0

104

169

6.6

92.4

196

43.56

5.7

85.5

225

32.49

4.9

78.4

256

24.01

4.8

81.6

289

23.04

4.1

73.8

324

16.81

126

61.3

813.4

1824

451.65

n ( å xy ) - ( å x )( å y )

n( å x 2 ) - ( å x ) 2 9(813.4) - (126)(61.3)

9(1824) - (126) 2 » -0.74667 » -0.7467

å y - m( å x )

n 61.3 - (-0.74667)(126) = 9 » 17.264 Thus, Y = -0.7467 x + 17.264 .

9(1824) - (126)2 = 0.0721

Section 1.3

(b) r =

n ( å x 2 ) - ( å x)

5(77.36) - (13.9)(19.6)

r =

(d)

n (å xy) - (å x)(å y)

⋅ n (å y 2 ) - (å y)2

5(53.07) - 13.92 ⋅ 5(113.18) - 19.62

» 0.999

9(813.4) - (126)(61.3) 9(1824) - 1262 ⋅ 9(451.65) - 61.32

24. (a)

» -0.9900

(c)

16 10

Yes, the points lie in a linear pattern. Using a calculator’s STAT feature, the correlation coefficient is found to be r » 0.959.

(b)

23. (a)

This indicates that the percentage of successful hunts does trend to increase with the size of the hunting party.

Length (cm)

y 8

Y = 3.98 x + 22.7

(c)

4 2 0

Yes, the data appear to be linear. 2

5.8

8.6

49.88

33.64

73.96

1.5

1.9

2.85

2.25

3.61

88.6

20.0

1772

7849.96

400.0

2.3

3.1

7.13

5.29

9.61

71.6

16.0

1145.6

5126.56

256.0

1.0

93.3

19.8

1847.34

8704.89

392.04

3.3

5.0

16.5

10.89

25.0

84.3

18.4

1551.12

7106.49

338.56

13.9

19.6

77.36

53.07

113.18

80.6

17.1

1378.26

6496.36

292.41

75.2

15.5

1165.6

5655.04

240.25

69.7

14.7

1024.59

4858.09

216.09

82.0

17.1

1402.2

6724

292.41

69.4

15.4

1068.76

4816.36

237.16

83.3

16.2

1349.46

3938.89

262.44

79.6

15.0

1194

6336.16

225

82.6

17.2

1420.72

6822.76

295.84

80.6

16.0

1289.6

6496.36

256.0

83.5

17.0

1419.5

6972.25

289.0

76.3

14.4

1098.72

5821.69

207.36

1200.6

249.8

20,127.47

96,725.86

4200.56

(b)

n(å xy) - (å x)(å y) 2

n(å x ) - (å x) 5(77.36) - (13.9)(19.6)

5(53.07) - 13.92 = 1.585250901 » 1.585 å y - m(å x ) b = n 19.6 - 1.585250901(13.9) = » -0.487 5 Y = 1.585x - 0.487 Length (cm)

y 8 6

25. (a)

m =

2 0

(c)

16 10

No, it gives negative values for small widths.

n(å xy) - (å x)(å y) n(å x 2 ) - (å x)2 15(20,127.47) - (1200.6)(249.8)

15(96,725.86) - 1200.62 = 0.211925009 » 0.212

Chapter 1 LINEAR FUNCTIONS å y - m(å x ) n 249.8 - 0.211925(1200.6) = » -0.309 15

b =

Y = 0.212 x - 0.309

Y = 0.212(73) - 0.309 » 15.2

If the temperature were 73° F, you would expect to hear 15.2 chirps per second. Let Y = 18; find x.

(c)

27.

86.4 » x

When the crickets are chirping 18 times per second, the temperature is 86.4°F. (d) r =

15(20,127) - (1200.6)(249.8) 15(96,725.86) - (1200.6)2 ⋅ 15(4200.56) - (249.8)2

= 0.835

26.

17.8

316.84

16.6

149.4

275.56

16.4

213.2

169

268.96

15.7

266.9

289

246.49

16.3

342.3

441

265.69

16.2

405

625

262.44

90 (a)

1465.8

1630

1635.98

6(1630) - (90)2 ⋅ 6(1635.98) - (99)2

13,359

178,462,881

15,569

77,845

24,2393,761

17,604

176,040

100

309,900,816

19,971

229,565

225

398,840,841

22,315

446,300

400

497,959,225

24,257

606,425

625

588,402,049

25,701

719,628

784

660,541,401

103

138,776

2,325,803

2159

2,876,500,974

(a)

Yes, the data appear to lie along a straight line. (b) r =

n(å xy ) - (å x)(å y ) 2 2 2 n( å x ) - ( å x ) ⋅ n( å y ) - ( å y ) 2

7(2, 325, 803) - (103)(138, 776) 2 2 7(2159) - 103 ⋅ 7(2, 876, 500, 974) - 138, 776

n ( å xy ) - ( å x )( å y )

n( å x 2 ) - ( å x ) 2 6(1465.8) - (90)(99)

= 0.9996

Yes, the value indicates a strong linear correlation.

6(1630) - (90) 2 » -0.06857

6(1465.8) - (90)(99)

18 = 0.212 x - 0.309 18.309 = 0.212 x

n( å x 2 ) - ( å x ) 2 ⋅ n( å y 2 ) - ( å y )

» -0.7286 The value indicates a strong negative linear correlation.

Let x = 73; find Y.

(b)

n ( å xy ) - ( å x )( å y )

å y - m( å x )

n 99 - (-0.06857)(90) = 6 » 17.53 Thus, Y  0.06857 x  17.53.

n ( å xy ) - ( å x )( å y ) n( å x 2 ) - ( å x ) 2 7(2,325,803) - (103)(138, 776)

7(2159) - (103) 2

» 441.1

(b) The year 2025 corresponds to x = 35. Y  0.06857(35)  17.53  15.1 If the trend continues linearly, the pupil-teacher ratio will be about 15.1 in 2035.

å y - m(å x) n 138, 776 - (441.1)(103) = » 13,335 7

Thus, Y = 441.1x + 13,335.

Section 1.3

The year 2030 corresponds to x = 40.

(d)

Y = 441.1(40) + 13,335 = 30,979

The predicted poverty level in the year 2030 is about $13,335.

3894

3481

4356

4402

3844

5041

4752

4356

5184

4964

4624

5329

5325

5041

5625

4221

4489

3969

4410

4900

3969

4757

5041

4489

4818

5329

4356

4950

5625

4356

682

46,493

46,730

46,674

28.

m =

(a)

)

(

n åx =

n ( å xy ) - ( å x )( å y )

(

) - ( å x )2 ⋅ n ( å y 2 ) - ( å y )2

» 0.9459

Data for male students:

» -0.08915

4221

4489

3969

4419

4900

3969

4757

5041

4489

4818

5329

4356

4950

5625

4356

356

325

23,156

25,384

21,139

m =

n ( å xy ) - ( å x )( å y )

(

)

n å x2 - ( å x )

5(23,156) - (356)(325)

5(23, 337) - (326)(357) 2 2 5(21, 346) - (326) ⋅ 5(25,535) - (357)

) - ( å x) ⋅ n ( å y ) - ( å y ) 2

» 0.6674

5(21,346) - (326)2

n åx

n (å xy) - (å x)(å y) 2

5(23,337) - (326)(357)

r =

å y - m( å x ) b = n 682 - (-0.08915)(682) = » 74.28 10 Thus, Y = -0.08915x + 74.28 . r =

å y - m( å x ) n 357 - (0.6674)(326) = » 27.89 5 Thus, Y = 0.6674 x + 27.89 .

10(46,730) - (682)2

)

b =

n å x2 - ( å x )

10(46,493) - (682)(682)

(

n å x2 - ( å x )

n ( å xy ) - ( å x )( å y )

(

n ( å xy ) - ( å x )( å y )

m =

5(25,384) - (356)2

» 0.4348

å y - m( å x ) n 325 - (0.4348)(356) = » 34.04 5 Thus, Y = 0.4348 x + 34.04 . b =

10(46, 493) - (682)(682) 10(46,730) - (682)2 ⋅ 10(46,674) - (682)2

» -0.1035

The taller the student, the shorter the ideal partner’s height is. (b) Data for female students:

3894

3481

4356

4402

3844

5041

4752

4356

5184

4964

4624

5329

5325

5041

5625

326

357

23,337

21,346

25,535

n ( å xy ) - ( å x )( å y )

r =

(

n åx

) - ( å x )2 ⋅ n ( å y 2 ) - ( å y )2

5(23,156) - (356)(325) 2 2 5(25, 384) - (356) ⋅ 5(21,139) - (325)

» 0.7049

(c)

80 60

Chapter 1 LINEAR FUNCTIONS There is no linear relationship among all 10 data pairs. However, there is a linear relationship among the first five data pairs (female students) and a separate linear relationship among the second five data pairs (male students).

(c)

Let x = 620; find Y. Y = -0.0067(620) + 14.75 = 10.596 » 11

(d) 19(115, 400) - (10, 490)(210) 2 2 19(5,872,500) - (10, 490) ⋅ 19(2522) - 210

r =

29.

(a)

540

10,800

291,600

400

510

8160

260,100

256

490

4900

240,100

100

560

4480

313,600

470

5640

220,900

144

600

6600

360,000

121

540

5400

291,600

100

580

4640

336,400

680

10,200

462,400

560

4480

560

500

» -0.13

30. (a)

y 3 2 1 0 1

(b)

225

1.0

1.11

1.2321

313,600

1.5

1.36

2.04

2.25

1.8496

7280

313,600

169

2.0

1.57

3.14

2.4649

7000

250,000

196

2.5

1.76

4.4

6.25

3.0976

470

4700

220,900

100

3.0

1.92

5.76

3.6864

440

4400

193,600

100

3.5

2.08

7.28

12.25

4.3264

520

5720

270,400

121

4.0

2.22

8.88

4.9844

620

6820

384,400

121

17.5

12.02

32.61

50.75

21.5854

680

5440

462,400

550

4400

302,500

620

4340

384,400

10,490

210

115,400

5,872,500

2522

m= m=

n(å xy) - (å x)(å y) n(å x 2 ) - (å x)2 19(115, 400) - (10, 490)(210)

19(5,872,500) - 10, 4902 m = -0.0066996227 » -0.0067 å y - m(å x ) n 210 - (-0.0066996227)(10, 490) b = » 14.75 19 b =

Y = -0.0067 x + 14.75

(b)

(e) There is no linear relationship between a student’s math SAT and mathematics placement test scores.

m = m =

n(å xy ) - (å x )(å y ) n(å x 2 ) - (å x )2 7(32.61) - (17.5)(12.02)

7(50.75) - 17.52 m = 0.3657142857 » 0.366 å T - m(å L) n 12.02 - 0.3657142857(17.5) b = 7 » 0.803 b =

Y = 0.366 x + 0.803

The line seems to fit the data. y

Let x = 420; find Y.

Y = -0.0067(420) + 14.75 = 11.936 » 12

2 1 0 1

Section 1.3

(c)

r =

Let x = 230; find Y. Y = 14.9(230) + 2820 » 6250

= 0.995,

Adam would need to buy a 6500 BTU air conditioner.

which is a good fit and confirms the conclusion in part (b). 32. (a) 31. (a) x2

750,000

22,500

25,000,000

5500

962,500

30,625

30,250,000

215

6000

1,290,000

46,225

36,000,000

250

6500

1,625,000

62,500

42,250,000

280

7000

1,960,000

78,400

49,000,000

310

7500

2,325,000

96,100

56,250,000

350

8000

2,800,000

122,500

64,000,000

370

8500

3,145,000

136,900

72,250,000

420

9000

3,780,000

176,400

81,000,000

450

9500

4,275,000

202,500

90,250,000

2970

72,500

22,912,500

974,650

546,250,000

150

5000

175

m = m =

r =

= -0.995

Yes, there does appear to be a linear correlation. (b)

å y - m(å x ) n 72,500 - 14.9(2970) b = 10 b » 2820 b =

å y - m(å x ) n 20.668 - (-0.0768775758)(500) b = 10 » 5.91 Y = -0.0769 x + 5.91

33. (a) Use a calculator’s statistical features to obtain the least squares line. Y = -0.1178 x + 113.10 (b)

Y = -0.3441x + 147.89

(c)

Set the two expressions for Y equal and solve for x. -0.1178x + 113.10 = -0.3441x + 147.89 0.2263x = 34.79 x » 154 The women’s record will catch up with the men’s record in 1900 + 154, or in the year 2054.

(b) Let x = 150; find Y. (d)

rmen » -0.9608 rwomen » -0.9208 Both sets of data points closely fit a line with negative slope.

Let x = 280; Find Y. Y = 14.9(280) + 2820 » 6990, compared to actual 7000

Let x = 50 Y = -0.0769(50) + 5.91 » 2.07 The predicted number of points expected when a team is at the 50 yard line is 2.07 points.

Y = 14.9 x + 2820

Y = 14.9(150) + 2820 Y » 5060, compared to actual 5000

n( å x 2 ) - ( å x ) 2 10(399.16) - (500)(20.668)

b =

» 14.9

n(å xy ) - (å x )(å y )

10(33, 250) - 5002 m = -0.0768775758 » -0.0769

n(å x 2 ) - (å x )2 10(22,912,500) - (2970)(72,500)

10(974,650) - 2970 m = 14.90924806

m = m =

(c)

n(å xy ) - (å x )(å y )

10(399.16) - (500)(20.668) 10(33, 250) - 5002 ⋅ 10(91.927042) - (20.668)2

(e)

Let x = 420; find Y. Y = 14.9(420) + 2820 » 9080, compared to actual 9000

Chapter 1 LINEAR FUNCTIONS

34. (a) Use a calculator’s statistical features to obtain the least squares line. Y  0.009498 x  10.79 (b) Y  0.01488 x  12.27

36. (a) Use a calculator’s statistical features to obtain the correlation coefficient between the number of hits and the number of triples. r » 0.5259 (b) Use a calculator’s statistical features to obtain the correlation coefficient between the number of hits and the number of home runs. r » 0.2926

(c) Set the two expressions for Y equal and solve for x. 0.009498 x  10.79  0.01488 x  12.27 0.005382 x  1.48 x  274.99

(c)

The women’s record will catch up with the men’s record in 1900 + 275, or in the year 2175. (d)

Use a calculator’s statistical features to obtain the correlation coefficient between the number of triples and the number of home runs. r » -0.0956

rmen » -0.8834 rwomen » -0.8915

Both sets of data points closely fit a line with negative slope. (e)

35. (a) Skaggs’ average speed was 100.5 / 26.339 » 3.816 miles per hour. (b)

Chapter 1 Review Exercises 1.

False; a line can have only one slant, so its slope is unique.

False; the equation y = 3x + 4 has slope 3.

True; the point (3, -1) is on the line because -1 = -2(3) + 5 is a true statement.

False; the points (2, 3) and (2, 5) do not have the same y-coordinate.

True; the points (4, 6) and (5, 6) do have the same y-coordinate.

False; the x-intercept of the line y = 8x + 9 is - 98 .

The data appear to lie approximately on a straight line. (c)

Using a graphing calculator, Y = 3.714 x + 3.809.

(d)

Using a graphing calculator, r » 0.9969 Yes, the least squares line is a very good fit to the data.

(e)

A good value for Skaggs’ average speed would be the slope of the least squares line, or m = 3.714 miles per hour. This value is faster than the average speed found in part (a). The value 3.714 miles per hour is most likely the better value because it takes into account all 14 data pairs.

True; f ( x) =  x + 4 is a linear function because it is in the form y = mx + b, where m and b are real numbers.

False; f ( x) = 2 x 2 + 3 is not linear function because it isn’t in the form y = mx + b, and it is a second-degree equation.

False; the line y = 3x + 17 has slope 3, and the line y = -3x + 8 has slope -3 . Since 3 ⋅ -3 = / -1 , the lines cannot be perpendicular.

10. False; the line 4 x + 3 y = 8 has slope - 43 , and

the line 4 x + y = 5 has slope -4 . Since the slopes are not equal, the lines cannot be parallel. 11. False; a correlation coefficient of zero indicates that there is no linear relationship among the data. 12. True; a correlation coefficient always will be a value between -1 and 1.

Chapter 1 Review

13. Marginal cost is the rate of change of the cost function; the fixed cost is the initial expenses before production begins. 14. To compute the coefficient of correlation, you need the following quantities: å x, å y, å xy, å x 2 , å y 2 , and n.

15. Through (-3, 7) and (2, 12) 12 - 7 5 m= = =1 2 - (-3) 5 16. Through (4, -1) and (3, -3). -3 - (-1) m= 3-4 -3 + 1 = -1 -2 = = 2 -1 17. Through the origin and (11, -2) -2 - 0 2 m= =11 - 0 11

4x + 3 y = 6 3 y = -4 x + 6 4 y =- x+2 3 Therefore, the slope is m = - 43 .

y+4=9 y =5 y = 0x + 5 m=0

22.

3 y - 1 = 14 3 y = 14 + 1 3 y = 15 y =5

x = 5y 1 x = y 5 1 m= 5

25. Through (5, - 1); slope 23

Use point-slope form. 2 y - (-1) = ( x - 5) 3 2 y + 1 = ( x - 5) 3 3( y + 1) = 2( x - 5) 3 y + 3 = 2 x - 10 3 y = 2 x - 13 2 13 y = x3 3

27. Through (-6, 3) and (2, -5) -5 - 3 -8 = = -1 m= 2 - (-6) 8

Use point-slope form. y - 3 = -1[ x - (-6)] y - 3 = -x - 6 y = -x - 3

4x - y = 7 - y = -4 x + 7 y = 4x - 7 m= 4

21.

24.

Use point-slope form. 1 y - 0 = - ( x - 8) 4 1 y =- x+2 4

The slope of the line is undefined.

20.

y = 5x + 4 m=5

26. Through (8, 0), with slope - 14

18. Through the origin and (0, 7) 7-0 7 m= = 0-0 0 19.

23.

This is a horizontal line. The slope of a horizontal line is 0.

28. Through (2, -3) and (-3, 4) 4 - (-3) 7 m= =-3 - 2 5 Use point-slope form. 7 y - (-3) = - ( x - 2) 5 7 14 y+3=- x+ 5 5 7 14 -3 y =- x+ 5 5 7 14 15 y =- x+ 5 5 5 7 1 y =- x5 5

Chapter 1 LINEAR FUNCTIONS

29. Through (2, -10), perpendicular to a line with undefined slope A line with undefined slope is a vertical line. A line perpendicular to a vertical line is a horizontal line with equation of the form y = k. The desired line passed through (2, -10), so k = -10. Thus, an equation of the desired line is y = -10.

The equation of the vertical line through (-1, 4) is x = -1. 34. Through (7, -6) , parallel to a line with undefined slope. A line with undefined slope has the form x = a (a vertical line). The vertical line that goes through (7, -6) is the line x = 7.

30. Through (-2, 5), with slope 0

Horizontal lines have 0 slope and an equation of the form y = k. The line passes through (-2, 5) so k = 5. An equation of the line is y = 5.

35. Through (3, -5), parallel to y = 4 Find the slope of the given line. y = 0 x + 4, so m = 0, and the required line will also have slope 0. Use the point-slope from.

31. Through (3, -4) parallel to 4 x - 2 y = 9

Solve 4 x - 2 y = 9 for y. - 2 y = -4 x + 9 9 y = 2x 2 m= 2 The desired line has the same slope. Use the pointslope form. y - (-4) = 2( x - 3) y + 4 = 2x - 6 y = 2 x - 10 Rearrange. 2 x - y = 10

y - (-5) = 0( x - 3) y+5=0 y = -5

36. Through (-3, 5), perpendicular to y = -2 The given line, y = -2, is a horizontal line. A line perpendicular to a horizontal line is a vertical line with equation of the form x = h.

The desired line passes through (-3, 5), so h = -3. Thus, an equation of the desired line is x = -3. 37.

y = 4x + 3

Let x = 0:

y = 4(0) + 3 y =3

Let y = 0:

0 = 4x + 3

32. Through (0, 5), perpendicular to 8 x + 5 y = 3

Find the slope of the given line first. 8x + 5 y = 3 5 y = -8 x + 3 -8 3 y = x+ 5 5 8 m=5

- 3 = 4x -

3 = x 4

(

The perpendicular line has m = 85 . Use point-slope form. 5 ( x - 0) 8 5 y = x+5 8

y-5 =

Rearrange.

38.

8 y = 5x + 40 5x - 8 y = -40

y = 6 - 2x

Find the intercepts. Let x = 0. y = 6 - 2(0) = 6

33. Through (-1, 4); undefined slope

)

Draw the line through (0, 3) and - 34 , 0 .

The y-intercept is 6.

Chapter 1 Review

Let y = 0.

42.

0 = 6 - 2x

y =1

This is the horizontal line passing through (0, 1).

2x = 6

x =3

The x-intercept is 3.

y=1

Draw the line through (0, 6) and (3, 0). y 6 y = 6 – 2x

43.

y = 2x

When x = 0, y = 0.

When x = 1, y = 2. 39.

Draw the line through (0, 0) and (1, 2).

3x - 5 y = 15 -5 y = -3x + 15

y =

3 x-3 5

When x = 0, y = -3. When y = 0, x = 5. Draw the line through (0, -3) and (5, 0).

44.

x + 3y = 0

When x = 0, y = 0. When x = 3, y = -1. Draw the line through (0, 0) and (3, -1). y x + 3y = 0 1

40.

–3

4 x + 6 y = 12

Find the intercepts. When x = 0, y = 2, so the y-intercept is 2. When y = 0, x = 3, so the x-intercept is 3. Draw the line through (0, 2) and (3, 0). y 3 4x + 6y = 12

(a) Let t = 0 represent the year 2000. The line goes through (0, 100) and (18, 540). 540 - 100 m= » 24.4 18 - 0 y - 100 = 24.4(t - 0) y = 24.4t + 100 (b) The imports from China are increasing by about $24.4 billion per year.

x 3

41.

45.

x-3= 0 x =3

This is the vertical line through (3, 0). y x–3=0 3 0

y = 24.4t + 100 y = 24.4(20) + 100 y = 588

The amount of imports from China in 2020 will be about $588 billion.

Chapter 1 LINEAR FUNCTIONS (d) Let y = 750; solve for t. 750 = 24.4t + 100 650 = 24.4t 26.6 » t

p  S (q)  4q  10; p  D(q )  40  4q

48.

The imports from China would be at least $750 billion in the year 2000 + 27 = 2027. 46. (a) Let t = 0 represent the year 2000. The line goes through (0, 16) and (18, 120). 120 - 16 » 5.78 18 - 0 y - 16 = 5.78(t - 0)

y = 5.78t + 16

(b) The exports to China are increasing by about $5.78 billion per year. (c) Let t = 20. y = 5.78t + 16 y = 5.78(20) + 16 y = 131.6

The amount of exports to China in 2020 will be about $132 billion. (d) Let y = 250; solve for t. 250 = 5.78t + 16 234 = 5.78t 40.48 » t

(a) 20 = 4q + 10 10 = 4q 5 = q (supply) 2

20 = 40 - 4q -20 = -4q 5 = q (demand)

When the price is $20 per pound, the supply is 2.5 pounds per day, and the demand is 5 pounds per day. (b) 24 = 4q + 10 24 = 40 - 4q 14 = 4q -16 = -4q 7 = q (supply) 4 = q (demand) 2 When the price is $24 per pound, the supply is 3.5 pounds per day, and the demand is 4 pounds per day. (c) 32 = 4q + 10 22 = 4q 11 = q (supply) 2

32 = 40 - 4q -8 = -4q 2 = q (demand)

When the price is $32 per pound, the supply is 5.5 pounds per day, and the demand is 2 pounds per day. (d)

The exports to China would be at least $250 billion in the year 2000 + 41 = 2041. 47. (a) Let t = 0 represent the year 1900. The line goes through (90, 29,943) and (218, 63,179). 63,179 - 29,943 = 1187 118 - 90 y - 29,943 = 1187(t - 90)

y = 1187t - 76,887

(b) The median income for all U.S. households is increasing by about $1187 per year. (c) Let t = 125. y = 1187t - 76,887 y = 1187(125) - 76,887 y = 71, 488

The median income for all U.S. households in 2025 will be about $71,488.

(e) The graph shows that the lines representing the supply and demand functions intersect at the point (3.75, 25). The y-coordinate of this point gives the equilibrium price. Thus, the equilibrium price is $25 per pound. (f) The x-coordinate of the intersection point gives the equilibrium quantity. Thus, the equilibrium quantity is 3.75, representing 3.75 pounds of crabmeat per day. 49. (a) The line that represents the supply function goes through the points (60, 40) and (100, 60). 60 - 40 m = » 0.5 100 - 60

Use (60, 40) and the point-slope form. p - 40 = 0.5(q - 60) p = 0.5q - 30 + 40

(d) Let y = 80,000; solve for x. 80, 000 = 1187t - 76,887 156,887 = 1187t t » 132.17

p = S ( q) = 0.5q + 10

Chapter 1 Review

(b) The line that represents the demand function goes through the points (50, 47.50) and (80, 32.50). 32.50 - 47.50 m = » -0.5 80 - 50 Use (50, 47.50) and the point-slope form. p - 47.50 = -0.5( q - 50) p = 0.5q + 25 + 47.50 p = D( q) = -0.5q + 72.50 (c) Set supply equal to demand and solve for q. 0.5q + 10 = -0.5q + 72.50 1q = 62.50 q = 62.5

53. Thirty units cost $1500; 120 units cost $5640. Two points on the line are (30, 1500), (120, 5640), so m=

5640 - 1500 4140 = = 46. 120 - 30 90

Use point-slope form. y - 1500 = 46( x - 30) y = 46 x - 1380 + 1500 y = 46 x + 120 C ( x) = 46 x + 120 54.

C ( x) = 200 x + 1000 R( x) = 400 x

(a) C ( x) = R( x)

S (62.5) = 0.5(62.5) + 10 = 41.25

200 x + 1000 = 400 x

The equilibrium quantity is about 62.5 dietary supplement pills, and the equilibrium price is about $41.25.

1000 = 200 x

50. Eight units cost $300; fixed cost is $60. The fixed cost is the cost if zero units are made. (8, 300) and (0, 60) are points on the line. 60 - 300 m= = 30 0-8

5= x

The break-even quantity is 5 cartons. (b)

The revenue from 5 cartons of CD’s is $2000. 55. (a)

Use slope-intercept form. y = 30 x + 60 C ( x) = 30 x + 60 51. Fixed cost is $2000; 36 units cost $8480. Two points on the line are (0, 2000) and (36, 8480), so 8480 - 2000 6480 m= = = 180. 36 - 0 36

Use point-slope form. y = 180 x + 2000 C ( x) = 180 x + 2000 52. Twelve units cost $445; 50 units cost $1585. Points on the line are (12, 445) and (50, 1585). 1585 - 445 m= = 30 50 - 12

Use point-slope form. y - 445 = 30( x - 12) y - 445 = 30 x - 360 y = 30 x + 85 C ( x) = 30 x + 85

R(5) = 400(5) = 2000

C ( x) = 3x + 160; R( x) = 7 x C ( x ) = R ( x) 3x + 160 = 7 x 160 = 4 x 40 x = x The break-even quantity is 40 pounds.

(b)

R(40) = 7 ⋅ 40 = $280

The revenue for 40 pounds is $280. 56. (a) E ( x) = 42 x + 352 (where x is in thousands) (b) R( x) = 130 x (where x is in thousands) (c) R( x) > E ( x) 130 x > 42 x + 352 88x > 352 x > 4 For a profit to be made, more than 4000 chips must be sold. 57. (a) Use the points (12, 30,581) and (18, 35,249) to find the slope. m=

35, 249 - 30,581 = 778 18 - 12

y - 30,581 = 778(t - 12) y = 778t + 21, 245

Chapter 1 LINEAR FUNCTIONS (b) Use the points (14, 32,362) and (18, 35,249) to find the slope. m=

35, 249 - 32,362 = 721.75 18 - 14

y - 32,362 = 721.75(t - 14) y = 721.75t + 22, 257.5

y - 54.2 = 0.556( x - 0) y = 0.556 x + 54.2 Thus c(t ) = 0.556t + 54.2

(b) Beef is decreasing by about 0.725 lb/yr; pork is decreasing by about 0.075 lb/yr; chicken is increasing by about 0.556 lb/yr. (c) -0.725t + 64.5 = 0.556t + 54.2 -1.281t = -10.3 t » 8.04

The consumption of chicken surpassed the consumption of beef in the year 2000 + 9 = 2009.

(d)

(d) (f) r = 0.9888

b(t ) = -0.725t + 64.5 b(20) = -0.725(20) + 64.5 = 50.0 p(t ) = -0.075t + 47.8 p(20) = -0.075(20) + 47.8 = 46.3

58. (a) Y = 13.981x - 288.43

c(t ) = 0.556t + 54.2 c(20) = 0.556(20) + 54.2 = 65.32

(b) Y = 13.981(125) - 288.43 Y » $1459.20

The consumption of beef will be 50.0 lb, the consumption of pork will be 46.3 lb, and the consumption of chicken will be about 65.3 lb in 2020.

(c)

The data points lie in a linear pattern. (d) r = 0.9928 ; There is a strong positive correlation among the data.

60. (a) Using a graphing calculator, r » 0.8988 . Yes, the data seem to fit a straight line. (b)

59. (a) For b(t): Use the points (0, 64.5) and (16, 52.9) to find the slope. 52.9 - 64.5 = -0.725 16 - 0 y - 64.5 = -0.725( x - 0) y = -0.725 x + 64.5 Thus b(t ) = -0.725t + 64.5 For p(t): Use the points (0, 47.8) and (16, 46.6) to find the slope. 46.6 - 47.8 m= = -0.075 16 - 0 y - 47.8 = -0.075( x - 0) y = -0.75x + 47.8 Thus p(t ) = -0.075t + 47.8 For c(t): Use the points (0, 54.2) and (16, 63.1) to find the slope. 63.1 - 54.2 m= = 0.55625 » 0.556 16 - 0 m=

(c)

(d)

Using a graphing calculator, Y = 0.01315x + 35.92 . The data somewhat fit a straight, but a curve would fit the data better. Let x = 3494 . Find Y. Y = 0.01315(3494) + 35.92 » 81.87 The predicted life expectancy in Canada, with a daily calorie supply of 3494, is about 81.87 years. This agrees with the actual value of 81.77 years.

(e)

The higher daily calorie supply most likely contains more healthy nutrients, which might result in a longer life expectancy.

Chapter 1 Review 61.

(a) 2

130

170

22,100

16,900

28,900

138

160

22,080

19,044

25,600

142

173

24,566

20,164

29,929

159

181

28,779

25,281

32,761

165

201

33,165

27,225

40,401

200

192

38,400

40,000

36,864

210

240

50,400

44,100

57,600

250

290

72,500

62,500

84,100

1394

1607

291,990

255,214

336,155

m= m=

(c) For f (t ) : Use the points (8, 24.7) and (28, 29.5) to find the slope. 29.5 - 24.7 m= = 0.24 28 - 8 y - 24.7 = 0.24( x - 8) y = 0.24 x + 22.78 Thus, f (t ) = 0.24t + 22.78 (d) The percent of never-married females is increasing by about 0.24 percent per year. (e) m(t ) = 0.2t + 29.6 m(20) = 0.2(30) + 29.6 = 35.6 f (t ) = 0.24t + 22.78 f (30) = 0.24(30) + 22.78 = 29.98 » 30.0

n(å xy) - (å x)(å y)

The percent of never-married males will be about 35.6%, and the percent of never-married females will be about 30.3% in 2020.

n(å x 2 ) - (å x) 2 8(291,990) - (1394)(1607)

8(225, 214) - 13942 m = 0.9724399854 » 0.9724

å y - m(å x ) n 1607 - 0.9724(1394) b = » 31.43 8 b =

63. (a) Use the points (40, 91.5) and (84, 50.3) to find the slope. 50.3 - 91.5 m= » -0.9364 84 - 40 y - 91.5 = -0.9364(t - 40) y = -0.9364t + 128.956

Y = 0.9724 x + 31.43

(b)

y = -0.9364t + 129.0

Let x = 190; find Y.

The linear equation for mobility in terms of t, the number of years since 1900, is y = -0.9364t + 129.0 .

Y = 0.9724(190) + 31.43 Y = 216.19 » 216

(b)

The cholesterol level for a person whose blood sugar level is 190 would be about 216. (c)

r =

8(291, 990) - (1394)(1607) 2 2 8(255, 214)-1394 ⋅ 8(336,155)-1607

Use the points (60, 62.3) and (84, 50.3) to find the slope. 50.3 - 62.3 m= = -0.5 84 - 60 y - 62.3 = -0.5(t - 60) y = -0.5t + 92.3

= 0.933814 » 0.93

62. (a)

The linear equation for mobility in terms of t, the number of years since 1900, is y = -0.9364t + 129.0 .

For m(t): Use the points (8, 31.2) and (28, 35.2) to find the slope. m=

(c)

35.2 - 31.2 = 0.2 28 - 8

Using a graphing calculator, the least squares line is Y = -0.9256t + 124.5 .

y - 31.2 = 0.2( x - 8) y = 0.2 x + 29.6

Thus, m(t ) = 0.2t + 29.6

(b) The percent of never-married males is increasing by about 0.2 percent per year. (e)

(d) The least squares line best describes the data. Since the data seems to fit a straight line, a linear model describes the data well. Using a graphing calculator, r » -0.9614 .

Chapter 1 LINEAR FUNCTIONS

64. (a) Using a graphing calculator, r = 0.6683. The data seem to fit a line but the fit is not very good.

(b)

n(å x 2 ) - (å x) 2 8(1, 260, 055.5) - 15,965(631.3)

» 0.134066 8(31,861, 775) - 15,9652 å y - m( å x ) b= n 631.3 - 0.134066(15,965) b= » -188.633 8

Using a graphing calculator, Y = 4.159 x + 112.78

(c)

n(å xy) - (å x)(å y)

Y = 0.1341x - 188.63

(d) The slope is 4.159 thousand (or 4159). On average, the governor’s salary increases $4159 for each additional million in population. 65. Use a graphing calculator to find these correlations. (a) Correlation between years since 2000 and length: r = 0.4529 (b)

Correlation between length and rating: r = 0.3955

(c)

Correlation between years since 2000 and rating: r = -0.4768

Let x = 1900. Find Y. Y = 0.134066(1900) - 188.633 » 66.09 From the equation, the guess for the life expectancy of females born in 1900 is 66.09 years.

(d)

This calculator graph plots year (on the horizontal axis) versus rating (on the vertical axis). Squares represent movies with lengths no more than 110 minutes, and plus signs represent movies with lengths 115 minutes or more.

The poor prediction isn’t surprising, since we were extrapolating far beyond the range of the original data. x

Predicted value

Residual

1970

74.7

75.48

–0.78

1980

77.4

76.82

0.58

1990

78.8

78.16

0.64

1995

78.9

78.83

0.07

2000

79.3

79.50

– 0.20

2005

80.1

80.17

– 0.07

2010

81.0

80.84

0.16

2015

81.1

81.51

– 0.41

Extended Application: Using Extrapolation to Predict Life Expectancy 1. x

1970

74.7

147,159.0

3,880,900

5580.09

1980

77.4

153,252.0

3,920,400

5990.76

1990

78.8

156,812.0

3,960,100

6209.44

1995

78.9

157,405.5

3,980,025

6225.21

2000

79.3

158,600.0

4,000,000

6288.49

2005

80.1

160,600.5

4,020,025

6416.01

2010

81.0

162,810.0

4,040,100

6561.0

2015

81.1

163,416.5

4,060,225

6577.21

1,260,055.5

31,861,775

49,848.21

15,965 631.3

6. 7.

You’ll get 0 slope and 0 intercept, because you’ve already subtracted out the linear component of the data. A cubic would fit the data; Y = 0.00001052 x3 - 0.6303x 2 + 12.5936 x - 838, 720.8

Chapter 2

NONLINEAR FUNCTIONS 2.1 Properties of Functions

W3. 16  x 2  0

Your Turn 1 y =

x 2  16

4  x  4

16  x 2  0 on the interval [ 4, 4].

x -4

Since the denominator cannot be zero and the radicand cannot be negative, the domain includes only those values of x satisfying x 2 - 4 > 0. Using the methods for solving a quadratic inequality, we get the domain (-¥, -2) È (2, ¥). Since the denominator can never be negative, y cannot be negative or zero. So, the range is (0, ¥).

W4. Solve 21  4 x  x 2  0. (7  x)(3  x)  0 x  7 or x  3 These roots divide the real line into three intervals: ( , 3)

Your Turn 2

Test a point in each interval.

f ( x) = 2 x 2 - 3 x - 4

(a)

At x  5, 21  4 x  x 2  24  0

f ( x + h) = 2( x + h) - 3( x + h) - 4

At x  0, 21  4 x  x 2  21  0

= 2( x 2 + 2 xh + h 2 ) - 3( x + h) - 4

(b)

 3, 7  7,  

= 2 x 2 + 4 xh + 2h 2 - 3x - 3h - 4

At x  10, 21  4 x  x 2  39  0

f ( x) = -5

So, 21  4 x  x 2  0 on (, 4]  [7, ).

2 x - 3x - 4 = -5 2 x 2 - 3x + 1 = 0 (2 x - 1)( x - 1) = 0 x =

1 or x = 1 2

2.1 Warmup Exercises W1. 4 x 2  9  0 (2 x  3)(2 x  3)  0 3 3 x  or x   2 2 W2. 2 x 2  9 x  5  0 (2 x  1)( x  5)  0 1 x  or x  5 2

2.1 Exercises 1.

False. A function is a rule that assigns to each element from one set exactly one element from another set.

False. Each x-value corresponds to exactly one y-value.

False. The range of f ( x)  1/ x is (, 0)  (0, ).

True

The x-value of 82 corresponds to two y-values, 93 and 14. In a function, each value of x must correspond to exactly one value of y. The rule is not a function.

Each x-value corresponds to exactly one y-value. The rule is a function.

Chapter 2 NONLINEAR FUNCTIONS

9 corresponds to 3 and -3, 4 corresponds to 2 and -2, and 1 corresponds to -1 and 1. The rule is not a function.

y = x3 + 2

Each x-value corresponds to exactly one y-value. The rule is a function. 10.

11.

12.

y =

15.

2y - x = 5 2y = 5 + x

y =

1 5 x+ 2 2

Each x-value corresponds to exactly one y-value. The rule is a function.

x -2 -1 0 1 2 3

x = | y|

3 2

5 2

7 2

(

)

Each value of x (except 0) corresponds to two y-values. The rule is not a function.

Pairs: -2, 32 , (-1, 2), 0, 52 , (1, 3),

x = y2 + 4

Range:

( ) ( 2, 72 ), (3, 4).

{ 32 , 2, 52 ,3, 72 , 4 }

Solve the rule for y. y 2 = x - 4 or y =  x - 4

Each value of x (greater than 4) corresponds to two y-values y =

x - 4 and y = - x - 4.

The rule is not a function. 13.

16.

y = 2x + 3

6 x - y = -1 - y = -6 x - 1 y = 6x + 1

x -2 - 1 0 1 2 3 y -1

1 3 5 7 9

-2 -1 0

y -11 -5 1 7 13 19

Pairs: (-2, -1), (-1, 1), (0, 3), (1, 5), (2, 7), (3, 9)

Pairs: (-2, -11), (-1, -5), (0, 1), (1, 7), (2, 13), (3, 19)

Range: {-1, 1, 3, 5, 7, 9}

Range: {-11, -5, 1, 7, 13, 19}

14.

y = -3 x + 9

x -2 - 1 0 y

1 2 3

12 9 6 3 0

Pairs: ( -2, 15), ( -1 , 12), (0, 9), (1, 6), (2, 3), (3, 0) Range: {0, 3, 6, 9, 12, 15}

17.

y = x( x + 2) x -2 - 1 0 1 2 y

0 -1 0 3 8 15

Section 2.1

Range: {-1, 0, 3, 8, 15}

21.

f ( x) = 2 x

x can take on any value, so the domain is the set of real numbers, (-¥, ¥). 22.

f ( x) = 2 x + 3

x can take on any value, so the domain is the set of real numbers, which is written (-¥, ¥). 18.

y = ( x - 2)( x + 2) x -2 -1

23. 0

1 2 3

x can take on any value, so the domain is the set of real numbers, (-¥, ¥).

0 - 3 -4 -3 0 5

Pairs: (-2, 0), (-1, -3), (0, -4), (1, -3), (2, 0), (3, 5)

f ( x) = x 4

24.

f ( x) = ( x + 3)2

x can take on any value, so the domain is the set of real numbers, (-¥, ¥).

Range: {-4, -3, 0, 5}

25.

f ( x) =

4 - x2

For f (x) to be a real number, 4 - x 2 ³ 0. Solve 4 - x 2 = 0. (2 - x)(2 + x) = 0 x = 2 or x = -2 19.

y = x2

The numbers form the intervals (-¥, -2), (-2, 2), and (2, ¥). Values in the interval (-2, 2) satisfy the inequality; x = 2 and x = -2 also satisfy the inequality. The domain is [-2, 2].

x -2 -1 0 1 2 3 y

1 0 1 4 9

Pairs: (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (3, 9) Range: {0, 1, 4, 9}

26.

f ( x) = |3x - 6|

x can take on any value, so the domain is the set of real numbers, (-¥, ¥). 27.

20.

-2

-1 0

The domain is [3, ¥).

y -16 -4 0 -4 -16 -36

Pairs: (–2,–16), (–1,–4), (0, 0), (1, -4), (2, -16 ), (3, -36 ) Range: {–36,–16,–4, 0} y 3 (–2, –16)

–20

–35

x-3

For f (x) to be a real number, x-3³ 0 x ³ 3.

y = -4 x 2 x

f ( x) = ( x - 3)1/2 =

(3, –36)

28.

f ( x) = (3x + 5)1/2 =

3x + 5

For f (x) to be a real number, 3x + 5 ³ 0 3 x ³ -5 1 (3x) ³ 1 (-5) 3 3 x ³ -5. 3

)

78 29.

Chapter 2 NONLINEAR FUNCTIONS f ( x) =

2 2 = 2 x + x) (1 )(1 1- x

33.

x =1

when x ³ 5 and when x £ -1. The domain is (-¥, -1] È [5, ¥).

1+ x = 0 x = -1

Thus, x can be any real number except 1. The domain is

34.

f ( x) =

f ( x) = 15 x 2 + x - 2

The expression under the radical must be nonnegative.

(-¥, -1) È (-1, 1) È (1, ¥).

30.

15 x 2 + x - 2 ³ 0

-8 x - 36 2

(5 x + 2)(3x - 1) ³ 0

In order for f ( x) to be a real number, x 2 - 36 cannot be equal to 0.

Solve (5 x + 2)(3x - 1) = 0.

When x 2 - 36 = 0,

5x + 2 = 0

x = 6 or x = -6. Thus, the domain is any real number except 6 or -6. In interval notation, the domain is

x =-

(

x = 4 or

x+4= 0 x = -4

Use the values –4 and 4 to divide the number line into 3 intervals, (-¥, -4), (-4, 4) and (4, ¥). Only the values in the intervals (-¥, -4) and (4, ¥) satisfy the inequality. The domain is (-¥, -4) È (4, ¥).

x =

) (

1 3

)

(

Only values in the intervals -¥, - 52

2 f ( x) = - 2 =. ( x - 4)( x + 4) x - 16

Solve ( x - 4) ⋅ ( x + 4) = 0.

2 5

intervals, -¥, - 52 , - 52 , 13 , and ( 13 , ¥ ).

( x - 4) ⋅ ( x + 4) > 0, since ( x - 4) ⋅ ( x + 4) < 0 would produce a negative radicand and ( x - 4) ⋅ ( x + 4) = 0 would lead to division by zero.

3x = 1

Use these numbers to divide the number line into 3

(-¥, -6) È (-6, 6) È (6, ¥).

x - 4 = 0 or

or 3x - 1 = 0

5 x = -2

x 2 = 36

31.

( x - 5)( x + 1)

( x - 5)( x + 1) ³ 0

When (1 - x)(1 + x) = 0, or

x2 - 4x - 5 =

See the method used in Exercise 25.

Since division by zero is not defined, (1 - x) ⋅ (1 + x) ¹ 0. 1- x = 0

f ( x) =

) and

( 13 , ¥ ) satisfy the inequality. The domain is ( -¥, - 52 ùúûú È éêëê 13 , ¥ ). 35.

f ( x) =

1 3x 2 + 2 x - 1

( 3x - 1)( x + 1)

(3 - 2)( x + 1) > 0, since the radicand cannot be negative and the denominator of the function cannot be zero.

Solve (3 - 1) ( x + 1) = 0. 3 - 1 = 0 or x = 13

x +1= 0 x = -1

Use the values -1 and 13 to divide the number line 32.

f ( x) = -

into 3 intervals, (-¥, -1), (-1, 4) and ( 13 , ¥ ).

5 2

x + 36

Only the values in the intervals ( -¥, -1 ) and

x can take on any value. No choice for x will produce a zero in the denominator. Also, no choice for x will produce a negative number under the radical. The domain is (-¥, ¥).

( 13 , ¥ ) satisfy the inequality. The domain is (-¥, -1) È ( 13 , ¥ ).

Section 2.1 36.

f ( x) =

79 43. The domain is all real numbers between the endpoints of the curve, or [-2, 4].

x2 3- x

The range is all real numbers between the minimum and maximum values of the function or [-3, 2].

For f (x) to be a real number, x2 ³ 0 and 3 - x > 0 -¥ < x < ¥ and x<3 x < 3.

(a) f (-2) = -3 (b) f (0) = -2 æ1ö (c) f çç ÷÷÷ = -1 çè 2 ø

The domain is (-¥, 3). 37. By reading the graph, the domain is all numbers greater than or equal to -5 and less than 4. The range is all numbers greater than or equal to -2 and less than or equal to 6. Domain: [-5, 4); range: [-2, 6]

(d) From the graph, f ( x) = 1 when x = 2.5. 44. The domain is all real numbers between the end points of the curve, or [-2, 4].

The range is all real numbers between the minimum and maximum values of the function, or in this case, {3}.

38. By reading the graph, the domain is all numbers greater than or equal to -5. The range is all numbers greater than or equal to 0.

(a) f (-2) = 3

Domain: [-5, ¥) range: [0, ¥) 39.

(b) f (0) = 3

By reading the graph, x can take on any value, but y is less than or equal to 12.

æ1ö (c) f çç ÷÷÷ = 3 çè 2 ø

Domain: (-¥, ¥); range: (-¥, 12] 40. By reading the graph, both x and y can take on any values.

Domain: (-¥, ¥); range: (-¥, ¥) 41. The domain is all real numbers between the end points of the curve, or [-2; 4]. The range is all real numbers between the minimum and maximum values of the function or [0, 4]. (a) f (-2) = 0

(d) From the graph, f (x) is 1 nowhere. 45.

f ( x) = 3 x 2 - 4 x + 1

(a) f (4) = 3(4) 2 - 4(4) + 1 = 48 - 16 + 1 = 33 æ 1ö æ 1 ö2 æ 1ö (b) f çç - ÷÷ = 3çç - ÷÷ - 4 çç - ÷÷ + 1 çè 2 ÷ø çè 2 ø÷ çè 2 ø÷ 3 + 2 +1 4 15 = 4 =

(b) f (0) = 4 æ1ö (c) f çç ÷÷÷ = 3 çè 2 ø

(d) From the graph, f ( x) = 1 when x = -1.5, 1.5, or 2.5. 42. The domain is all real numbers between the end points of the curve, or [-2, 4]. The range is all real numbers between the minimum and maximum values of the function or [0, 5].

(a) f (-2) = 5 (b) f (0) = 0

12 m

æ1ö (c) f çç ÷÷÷ = 1 çè 2 ø

(d) From the graph, if f ( x) = 1, x = -0.2, 0.5, 1.2, or 2.8.

8 +1 m

12 - 8m + m 2 m2

Chapter 2 NONLINEAR FUNCTIONS 2a + 1 if a ¹ 4 a-4 f (a) = 7 if a = 4

f ( x) = 1

(e)

3x 2 - 4 x + 1 = 1 3x 2 - 4 x = 0 x(3x - 4) = 0 4 x = 0 or x = 3

( )

4 +1 2 2 +1 æ2ö (d) f çç ÷÷÷ = 2m = m 2 -4 çè m ø -4

46.

f ( x) = ( x + 3)( x - 4)

æ 1ö æ 1 öæ 1 ö (b) f çç - ÷÷ = çç - + 3 ÷÷çç - - 4 ÷÷ ÷øèç 2 çè 2 ÷ø èç 2 ø÷ æ 5 öæ 9 ö 45 = çç ÷÷çç - ÷÷ = ÷ç 2 ø÷ èç 2 øè 4

(e)

48.

ì x-4 1 ï ï if x ¹ ï ï 2x + 1 2 f ( x) = ï í ï 1 ï 10 if x = ï ï 2 ï î

1 a-4 if a ¹ 2a + 1 2 1 f (a ) = 10 if a = 2

( x + 3)( x - 4) = 1 x 2 - x - 12 = 1

2 - 4m 2 - 4 æ2ö m = 4+ (d) f çç ÷÷÷ = m m çè m ø 2 2 +1

x 2 - x - 13 = 0

(m)

1  1 + 52 x = 2 1  53 x = 2 x » -3.140 or x » 4.140

ì 2x + 1 f ( x) = ïïï x - 4 if x ¹ 4 í ïï if x = 4 ïî 7

2 - 4m 4 + m

if m ¹ -4

æ2ö f çç ÷÷÷ = 10 if m = -4 çè m ø

(e)

47.

4-4 0 = =0 2(4) + 1 9

æ 1ö (b) f çç - ÷÷÷ = 10 çè 2 ø

m2 2(2 + 3m)(1 - 2m)

f ( x) = 1

(e)

4+m 1 if m ¹ 2 - 4m 2

2x + 1 =1 x-4 2x + 1 = x - 4

(a) f (4) =

(2 + 3m)(2 - 4m) or

x = -5

æ2ö 1 f çç ÷÷÷ = 7 if m = çè m ø 2

(a) f (4) = (4 + 3)(4 - 4) = (7)(0) = 0

æ2ö æ2 öæ 2 ö (d) f çç ÷÷ = çç + 3 ÷÷çç - 4 ÷÷ ÷øèç m çè m ÷ø èç m ø÷ æ 2 + 3m öæ ÷ç 2 - 4m ö÷ = çç çè m ÷÷øèçç m ÷÷ø

m 4+ m m 2- 4m m

49.

x-4 =1 2x + 1 x - 4 = 2x + 1 x = -5

f ( x) = 6 x 2 - 2 f (t + 1) = 6(t + 1) 2 - 2

(a) f (4) = 7

= 6(t 2 + 2t + 1) - 2

( ) ( 2)

2 - 12 + 1 æ 1ö 0 (b) f çç - ÷÷÷ = = 9 =0 çè 2 ø 1 - -4 2

= 6t 2 + 12t + 6 - 2 = 6t 2 + 12t + 4

Section 2.1 50.

f ( x) = 6 x 2 - 2

56.

f (2 - r ) = 6(2 - r ) 2 - 2

f ( x) = x 2 - 3

(a) f ( x + h) = ( x + h) 2 - 3

= 6(4 - 4r + r ) - 2

= ( x 2 + 2 xh + h 2 ) - 3

= 24 - 24r + 6r - 2

= x 2 + 2 xh + h 2 - 3

= 6r - 24r + 22

51.

52.

(b) f ( x + h) - f ( x) 2

= (r + h)2 - 2(r + h) + 5

= x + 2 xh + h - 3 - x + 3

= r 2 + 2hr + h 2 - 2r - 2h + 5

= 2 xh + h

(c)

g ( z - p) = ( z - p)2 - 2( z - p) + 5 = z 2 - 2 zp + p 2 - 2 z + 2 p + 5

53.

= ( x + 2 xh + h - 3) - ( x - 3)

g ( r + h)

æ 3 ÷ö æ 3 ö÷2 æ3ö g çç ÷÷ = çç ÷÷ - 2 çç ÷÷÷ + 5 çè q ÷ø èç q ø÷ èç q ø÷

57.

f ( x + h) - f ( x ) 2 xh + h 2 = h h h(2 x + h) = h = 2x + h

f ( x) = 2 x 2 - 4 x - 5

(a) f ( x + h)

6 = 2 - +5 q q

= 2( x + h) 2 - 4( x + h) - 5

9 - 6q + 5q

= 2( x 2 + 2hx + h 2 ) - 4 x - 4h - 5

= 2 x 2 + 4hx + 2h 2 - 4 x - 4h - 5

(b) f ( x + h) - f ( x) 54.

= = =

55.

= 2 x 2 + 4hx + 2h 2 - 4 x - 4h - 5

æ 5 ö æ 5 ö2 æ 5ö g çç - ÷÷ = çç - ÷÷ - 2 çç - ÷÷ + 5 çè z ÷ø çè z ÷ø çè z ÷ø 25 z

25 z2

+ +

- (2 x 2 - 4 x - 5)

10 +5 z 10 z z2

= 2 x 2 + 4hx + 2h 2 - 4 x - 4h - 5 - 2x 2 + 4x + 5

5z 2

= 4hx + 2h 2 - 4h

25 + 10 z + 5z 2 z

(c)

4hx + 2h 2 - 4h h h(4 x + 2h - 4) = h = 4 x + 2h - 4 =

f ( x) = 2 x + 1

(a) f ( x + h) = 2( x + h) + 1 = 2 x + 2h + 1 (b) f ( x + h) - f ( x) = 2 x + 2h + 1 - 2 x - 1 = 2h f ( x + h) - f ( x ) (c) h 2h = h = 2

f ( x + h) - f ( x ) h

58.

f ( x ) = -4 x 2 + 3 x + 2

(a)

f ( x + h) = -4( x + h)2 + 3( x + h) + 2 = -4( x 2 + 2hx + h 2 ) + 3x + 3h + 2 = -4 x 2 - 8hx - 4h 2 + 3x + 3h + 2

Chapter 2 NONLINEAR FUNCTIONS (b) f ( x + h) - f ( x)

1 1 =- 2 + 2 2 x + 2 xh + h x

= -4 x 2 - 8hx - 4h 2 + 3x + 3h + 2

x2 =- 2 2 x ( x + 2 xh + h 2 )

- (-4 x 2 + 3x + 2) = -4 x 2 - 8hx - 4h 2 + 3x + 3h + 2

( x 2 + 2 xh + h 2 ) + 2 2 x ( x + 2 xh + h 2 )

+ 4 x 2 - 3x - 2 = -8hx - 4h 2 + 3h =

(c)

59.

f ( x + h) - f ( x) -8hx - 4h + 3h = h h h(-8 x - 4h + 3) = h = -8 x - 4 h + 3

f ( x) =

1 x+h

(b) f ( x + h) - f ( x) 1 1 x+h x æxö 1 1 æ x + h ö÷ = çç ÷÷ - çç ÷ èç x ÷ø x + h x èç x + h ÷ø x - ( x + h) = x( x + h) -h = x( x + h) =

f ( x + h) - f ( x) h =

f ( x) = -

ù 1 éê -h ú h êë x( x + h) úû

x 2 ( x 2 + 2 xh + h 2 ) h(2 x + h)

61. A vertical line drawn anywhere through the graph will intersect the graph in only one place. The graph represents a function. 62. A vertical line drawn anywhere through the graph will intersect the graph in only one place. The graph represents a function. 63. A vertical line drawn through the graph may intersect the graph in two places. The graph does not represent a function.

65. A vertical line drawn anywhere through the graph will intersect the graph in only one place. The graph represents a function. 66. A vertical line is not the graph of a function since the one x-value in the domain corresponds to more than one, in fact, infinitely many y-values. The graph does not represent a function.

(a) f ( x + h) = -

2 xh + h 2

64. A vertical line drawn through the graph may intersect the graph in two or more places. The graph does not represent a function.

-1 = x( x + h)

60.

x 2 ( x 2 + 2 xh + h 2 )

f ( x + h) - f ( x ) x 2 ( x 2 + 2 xh + h2 ) = (c) h h 2x + h = 2 2 x ( x + 2 xh + h 2 )

1 x

(a) f ( x + h) =

(c)

-x 2 + x 2 + 2 xh + h 2

1 ( x + h) 2 1

67.

=- 2 x + 2 xh + h 2

(b) f ( x + h) - f ( x)

æ 1 ö 1 =- 2 - çç - 2 ÷÷÷ 2 çè x ø x + 2 xh + h

f ( x) = 3 x f (-x) = 3(-x) = -(3x) = - f ( x)

The function is odd.

Section 2.1 68.

f ( x) = 5 x f (-x) = 5(-x) = -(5x) = - f ( x)

The function is odd. 69.

f ( x) = | x - 2| f (-x) = | - x - 2| = | -( x + 2)| = | x + 2| f (-x) does not equal either f ( x) or f (-x). The function is neither even nor odd.

f ( x) = 2 x 2 f (-x) = 2(-x) 2 = 2x2 = f ( x)

The function is even. 70.

74.

f ( x) = x 2 - 3

75. (a) No (b) Year, t (c) Average price of silver, S(t) (d) [2000, 2019] (e) [4, 35] (f) $15 (g) 2011

f (-x) = (-x)2 - 3 = x2 - 3 = f ( x)

The function is even. 71.

f ( x) = f (-x) = =

1 x2 + 4 1 (-x) 2 + 4 1 2

x +4 = f ( x)

The function is even. 72.

f ( x) = x 3 + x f (-x) = (-x)3 + (-x) = -x 3 - x = -( x3 + x) = - f ( x)

The function is odd. x f ( x) = 2 73. x -9 -x f (-x) = (-x)2 - 9 x =- 2 x -9 = - f ( x) The function is odd.

76. (a) Year; energy consumption (b) [2005, 2040]; [17, 55] (c) No (d) United States: about 100 quadrillion Btu; China: about 155 quadrillion Btu; India: about 35 quadrillion Btu (e) About 2028 (f) About 2009 77. If x is a whole number of days, the cost of renting a saw in dollars is S ( x) = 28x + 8. For x in whole days and a fraction of a day, substitute the next whole number for x in 28 x + 8, because a fraction of a day is charged as a whole day. æ1ö (a) S çç ÷÷÷ = S (1) = 28(1) + 8 = 36 çè 2 ø

The cost is $36. (b) S (1) = 28(1) + 8 = 36

The cost is $36. æ 1ö (c) S çç1 ÷÷÷ = S (2) = 28(2) + 8 çè 4 ø = 56 + 8 = 64

The cost is $64. æ 1ö (d) S çç 3 ÷÷÷ = S (4) = 28(4) + 8 çè 2 ø = 112 + 8 = 120

The cost is $120. (e) S (4) = 28(4) + 8 = 112 + 8 = 120

The cost is $120.

Chapter 2 NONLINEAR FUNCTIONS æ 1ö (f ) S çç 4 ÷÷÷ = S (5) = 28(5) + 8 çè 10 ø

(g) Yes, C is a function. (h) No, C is not a linear function.

= 140 + 8 = 148

(i) The independent variable is x, the number of full or partial days.

The cost is $148.

(j) The dependent variable is C, the cost of renting the car.

æ 9ö (g) S çç 4 ÷÷÷ = S (5) = 28(5) + 8 çè 10 ø = 140 + 8 = 148

The cost is $148. (h) To continue the graph, continue the horizontal bars up and to the right. (i) The independent variable is x, the number of full and partial days.

79. (a) f (250, 000) = 0.40(150, 000) + 0.333(100, 000) = 93,300

The maximum amount that an attorney can receive for a $250,000 jury award is $93,300. (b) f (350, 000) = 0.40(150, 000)

(j) The dependent variable is S, the cost of renting a saw.

+ 0.333(150, 000) + 0.30(50, 000) = 124,950

(k) S is not a linear function. Its graph is not a continuous straight line. S is a step function. 78. If x is a whole number of days, the cost of renting a car is given by C ( x) = 54 x + 44. For x in whole days plus a fraction of a day, substitute the next whole number for x in 54 x + 44 because a fraction of a day is charged as a whole day. æ3ö (a) C çç ÷÷÷ = C (1) çè 4 ø = 54(1) + 44 = $98

The maximum amount that an attorney can receive for a $350,000 jury award is $124,500. (c) f (550, 000) = 0.40(150, 000) + 0.333(150, 000) + 0.30(200, 000) + 0.24(50, 000) = 181,950

The maximum amount that an attorney can receive for a $550,000 jury award is $181,950. (d)

æ 9ö (b) C çç ÷÷÷ = C (1) çè 10 ø = $98

80. (a) f (10, 000)  (0.04)(8500)  (0.045)(10, 000  8500)  $407.50

(b)

f (12, 000)  (0.04)(8500)  (0.045)(3200)

= 54(2) + 44 = $152

(e) C (2.4) = C (3) = 54(3) + 44 = $206 (f)

 (0.0525)[12, 000  (8500  3200)]  $499.75

(d) f (50, 000)  (0.04)(8500)  (0.045)(3200)  (0.0525)(2200)  (0.059)(7500)  0.0633[50, 000  (8500  3200  2200  7500)]  $2852.38

Section 2.1

(e) [0, );[0, )

f (800) = 0.01(800)0.88

» 3.6, or approximately

(f) To graph f, compute the tax at each of the bracket transition points, up to $100,000 and at one point beyond,

3.6 kcal/km when swimming at the surface of the water.

p0  0

(ii) A sea otter weighing 20,000 g expends

p1  8500

f (20, 000) = 0.01(20, 000)0.88

p2  p1  3300  11,800 p3  p2  2200  14, 000

» 61, or approximately

p4  p3  7500  21,500

61 kcal/km when swimming at the surface of the water. (b) If z is the number of kilograms of an animal’s weight, then x = g ( z ) = 1000 z is the number of grams since 1 kilogram equals 1000 grams.

p5  p4  59, 250  80, 750 p6  p5  134, 750  215,500

The corresponding taxes are $0, $340, $484, $599.5, $1042, and $4793. Plot the points (0, 0), and so on, and truncate the graph at x  100, 000.

84. (a) Let w = the width of the field; l = the length. The perimeter of the field is 6000 ft, so 81. (a) The curve in the graph crosses the point with x-coordinate 17:37 and y-coordinate of approximately 140. So, at time 17 hours, 37 minutes the whale reaches a depth of about 140 m. (b) The curve in the graph crosses the point with x-coordinate 17:39 and y-coordinate of approximately 240. So, at time 17 hours, 39 minutes the whale reaches a depth of about 250 m.

2l + 2w = 6000 l + w = 3000 l = 3000 - w.

Thus, the area of the field is given by A = lw A = (3000 - w)w.

(b) Since l = 3000 - w and w cannot be negative, 0 £ w £ 3000.

The domain of A is 0 £ w £ 3000.

82. (a) (i) y = f (5) = 19.7(5)0.753 » 66 kcal/day

(c)

(ii) y = f (25) = 19.7(25)0.753 » 222 kcal/day (b) Since 1 pound equals 0.454 kg, then x = g ( z ) = 0.454 z is the number of kilograms equal in weight to z pounds. (c) f ( g ( z )) = f (0.454 z ) = 19.7(0.454 z )0.753 = 19.7(0.4540.753 ) z 0.753

85. (a) P = 2l + 2w

However,

500 . w

æ 500 ö÷ P( w) = 2 çç + 2w çè w ÷÷ø

» 10.9 z 0.753

83.

lw = 500, so l =

(a) (i) By the given function f, a muskrat weighing 800 g expends

P( w) =

1000 + 2w w

Chapter 2 NONLINEAR FUNCTIONS (b) Since l = 500 , w ¹ 0 but w could be any w

(

(d) Since y = 2 x - 32

positive value. Therefore, the domain of P is 0 < w < ¥, or (0, ¥).

) - 112 is in the form 3

y = a( x - h)2 + k , we get h = 2 and 11

(c)

k = - 2 . So the vertex is

( , - ). 3 2

11 2

(e)

2.2 Quadratic Functions; Translation and Reflection

Your Turn 2 f ( x)  x 2  4 x  5

Your Turn 1

(a) The y-intercept is at (0, 5).

f ( x) = 2 x 2 - 6 x - 1

(a)

(b) To find the x-intercepts, solve f ( x)  0.

y = 2x2 - 6x - 1

x2  4 x  5  0

= 2( x - 3x) - 1 æ æ9ö 9ö = 2 çç x 2 - 3x + ÷÷ - 1 - 2 çç ÷÷ èç èç 4 ø÷ 4 ø÷

( x  5)( x  1)  0

2 æ 3ö 11 = 2 çç x - ÷÷÷ çè 2ø 2

(b)

x5  0

x 1  0

x5

x  1

The x-intercepts are ( 1, 0) and (5, 0).

y = 2(0)2 - 6(0) - 1 = -1

The y-intercept is (0, -1).

b 4   2. 2a 2

The corresponding y-coordinate is f (2)  9.

The vertex is at (2, 9). (d) The axis of symmetry is the vertical line through the vertex, which has the equation x  2. (e)

y –1

æ 11 3ö = çç x - ÷÷÷ çè 4 2ø

f(x) = x 2 – 4x – 5 –5

11 3 = x2 2 3  11 = x 2 

–9

The x-intercepts are

 (3  11) / 2, 0 and  (3  11) / 2, 0 .

5 x

Section 2.2

Your Turn 3

Your Turn 5

Let x be the number of $40 decreases in the price.

(a)

y 4

Price per person = 1650 - 40 x Number of people = 900 + 80 x

f(x) =

4 x

–1 –1

R( x) = (1650 - 40 x)(900 + 80 x). = 1, 485, 000 + 96, 000 x - 3200 x 2

x–2+1

y 1

(b)

-b -96, 000 -96, 000 x = = = = 15 2a 2(-3200) -6400

–1

4 x

–4

g(x) = – x – 1

The y-coordinate is y = 1, 485, 000 + 96, 000(15) - 3200(15)2

= 2, 205, 000

2.2 Warmup Exercises

The maximum revenue is $2,205,000 when 1650 - 40(15) = $1050 per person is charged.

W1. The expression is a perfect square.

Your Turn 4

W2. The expression is a perfect square.

4 x 2  36 x  81  (2 x) 2  2(2 x)( 9)  ( 9) 2  (2 x  9) 2

R( x) = -x + 40 x and C ( x) = 8x + 192 R ( x) = C ( x)

(a) 2

-x + 40 x = 8 x + 192 0 = x 2 - 32 x + 192 0 = ( x - 24)( x - 8)

The two graphs cross when x = 8 or x = 20. The minimum break-even quantity is x = 8, so the deli owner must sell 8 lb of cream cheese to break even.

x2 

W3. 3x 2  x  10  0 (3x  5)( x  2)  0 3x  5  0 or x  2  0 5 x or x  2 3 W4.

2 x2  5x  4  0 Use the quadratic formula with a  2, b  5, and c  4.

(b) The maximum revenue for R( x) = -x + 40 x is at x =

5 25 5 5 5  ( x)2  2( x)        x   x 8 8  4 64 8

-b -40 = = 20. 2a 2(-1)

x

5  25  4(2)( 4) 2(2)



5  57 4

The maximum revenue is R(20) = -202 + 40(20) = 400, or $400.

(c)

2.2

Exercises

True

= -x 2 + 40 x - 8x - 192

True

= -x 2 + 32 x - 192

True

False. The graph of y = f (-x) is a horizontal reflection of the graph of y = f ( x).

The graph of y = x 2 - 3 is the graph of y = x 2 translated 3 units downward. This is graph D.

The graph of y = ( x - 3)2 is the graph of y = x 2 translated 3 units to the right.

P ( x ) = R ( x) - C ( x)

The maximum point of P(x) is at x =

-b -32 = = 16. 2a 2(-1)

P(16) = -(16)2 + 32(16) - 192 = 64

So, the maximum profit is $64.

This is graph F.

88 9.

Chapter 2 NONLINEAR FUNCTIONS The graph of y = ( x - 3)2 + 2 is the graph of y = x 2 translated 3 units to the right and 2 units upward. This is graph A.

16.

æ ö æ ö = -5 çç x 2 + 8 x + 16 ÷÷÷ + 3 - (-5) çç 16 ÷÷÷ è è 25 ø 5 25 ø

10. The graph of y = ( x + 3) 2 + 2 is the graph of y = x 2 translated 3 units to the left and 2 units upward. This is graph B. 11. The graph of y = -(3 - x)2 + 2 is the same as the graph of y = -( x - 3)2 + 2 . This is the graph of y = x 2 reflected in the x-axis, translated 3 units to the right, and translated 2 units upward. This is graph C.

æ ö2 = -5 çç x + 4 ÷÷÷ + 31 è 5ø 5 æ ö The vertex is çç - 4 , 31 ÷÷÷. è 5 5ø

17.

Set y = 0 to find the x-intercepts. 0 = ( x + 3)( x + 2) x = -3, x = -2 The x-intercepts are –3and –2. Set x = 0 to find the y-intercept. y = 02 + 5(0) + 6 y = 6

The y-intercept is 6. The x-coordinate of the vertex is -b -5 5 = =- . x = 2a 2 2

y = 3x 2 + 9 x + 5

= 3( x 2 + 3x) + 5 æ æ9ö 9ö = 3çç x 2 + 3x + ÷÷÷ + 5 - 3çç ÷÷÷ èç èç 4 ø 4ø 2

æ 3ö 7 = 3çç x + ÷÷÷ çè ø 2 4

(

Substitute to find the y-coordinate. æ ö2 æ ö y = çç - 5 ÷÷÷ + 5çç - 5 ÷÷÷ + 6 = 25 - 25 + 6 = - 1 è 2ø è 2ø 4 2 4

)

æ ö The vertex is çç - 5 , - 1 ÷÷ è 2 4ø

The vertex is - 2 , - 4 . 14.

y = x 2 + 5x + 6 y = ( x + 3)( x + 2)

12. The graph of y = -( x + 3)2 + 2 is the graph of y = x 2 reflected in the x-axis, translated three units to the left, and two units upward. This is graph E. 13.

y = -5 x 2 - 8 x + 3 æ ö = -5 çç x 2 + 8 x ÷÷÷ + 3 è 5 ø

The axis is x = - 52 , the vertical line through the vertex.

y = 4 x - 20 x - 7

= 4( x 2 - 5x ) - 7 æ æ 25 ö 25 ö÷ = 4 çç x 2 - 5x + ÷÷ - 7 - 4 ççç ÷÷÷ çè è 4 ø 4 ø 2 æ 5ö = 4 çç x - ÷÷÷ - 32 çè 2ø

The vertex is

15.

( , -32 ).

18.

5 2

= ( x + 5)( x - 1)

y = -2 x 2 + 8 x - 9

= -2( x 2 - 4 x) - 9 = -2( x 2 - 4 x + 4) - 9 - (-2)(4) = -2( x - 2) 2 - 1

The vertex is (2, - 1).

y = x2 + 4x - 5

Set y = 0 to find the x-intercepts. 0 = ( x + 5)( x - 1) x = -5 or x = 1 The x-intercepts are –5 and 1. Set x = 0 to find the y-intercept. y = 02 + 4(0) - 5 y = -5 The y-intercept is –5.

Section 2.2

The x-coordinate of the vertex is -b -4 = = -2. x = 2a 2

20.

y = -3 x 2 - 6 x + 4 Let y = 0.

The equation

Substitute to find the y-coordinate. y = (-2)2 + 4(-2) - 5 = 4-8-5 = -9

0 = -3 x 2 - 6 x + 4

cannot be solved by factoring. Use the quadratic formula.

The vertex is (–2,–9). The axis is x = -2, the vertical line through the vertex.

0 = -3 x 2 - 6 x + 4 x =

6  6 2 - 4(-3)(4) 2(-3)

6  36 + 48 -6

6  84 6  2 21 = -6 -6

= -1 

19.

y = -2 x 2 - 12 x - 16

21 3

The x-intercepts are -1 +

= -2( x + 6 x + 8) = -2( x + 4)( x + 2)

21 » 0.53 and 3

-1 - 321 » -2.53

Let x = 0.

Let y = 0.

y = -3(0) 2 - 6(0) + 4

0 = -2( x + 4)( x + 2) x = -4, x = -2

y = 4

-4 and -2 are the x-intercepts. Let x = 0. y = -2(0) 2 + 12(0) - 16

4 is the y-intercept. Vertex: x =

-16 is the y-intercept.

6 6 -b = = = -1 2a 2(-3) -6

y = -3(-1)2 - 6(-1) + 4

-b 12 Vertex: x = = = -3 2a -4

= -3 + 6 + 4

y = -2(-3) - 12(-3) - 16 = -18 + 36 - 16 = 2

The vertex is (-1, 7). The axis is x = -1, the vertical line through the vertex.

The vertex is (-3, 2). The axis is x = -3, the vertical line through the vertex.

90 21.

Chapter 2 NONLINEAR FUNCTIONS Let x = 0.

y = 2 x2 + 8x - 8

y = -02 + 6(0) - 6

Let y = 0. 2 x2 + 8x - 8 = 0

-6 is the y-intercept.

x2 + 4x - 4 = 0

Vertex: x =

x =

-4 

42 - 4(1)(-4) 2(1)

-b -6 -6 = = =3 2a 2(-1) -2 y = -32 + 6(3) - 6 = -9 + 18 - 6

-4  32 -4  4 2 = 2 2 = -2  2 2 =

The vertex is (3, 3).

The x-intercepts are -2  2 2 » 0.83 and -4.83.

The axis is x = 3.

Let x = 0. y = 2(0)2 + 8(0) - 8 = -8

The y-intercept is –8. The x-coordinate of the vertex is -b 8 x = = - = -2. 2a 4

23.

f ( x) = 2 x 2 - 4 x + 5

Let f ( x) = 0.

If x = -2,

y = 2(-2)2 + 8(-2) - 8 = 8 - 16 - 8 = -16. The vertex is (–2,–16). The axis is x = -2.

0 = 2x2 - 4x + 5 x =

-(-4)  (-4)2 - 4(2)(5) 2(2)

4  16 - 40 4 4  -24 = 4 =

Since the radicand is negative, there are no x-intercepts. Let x = 0. 22.

y = 2(0)2 - 4(0) + 5

f ( x ) = -x + 6 x - 6

y =5 5 is the y-intercept.

Let f ( x) = 0. 0 = -x 2 + 6 x - 6 x =

Vertex: x =

-6  62 - 4(-1)(-6) 2(-1)

-b -(-4) 4 = = =1 2a 2(2) 4

y = 2(1)2 - 4(1) + 5 = 2 - 4 + 5 = 3

-6  36 - 24 -2 -6  12 = -2 -6  2 3 = -2 = 3 3

The vertex is (1, 3).

The axis is x = 1.

Section 2.2 24.

f ( x) =

1 2 x + 6 x + 24 2

x =

Let f ( x) = 0. 0=

-16  162 - 4(-2)(-21) 2(-2)

-16  88 -4 -16  2 22 = -4 22 = 4 2 The x-intercepts are =

1 2 x + 6 x + 24 2

Multiply by 2 to clear fractions. 0 = x 2 + 12 x + 48 Use the quadratic formula.

-12  122 - 4(1)(48) x = 2(1)

22 2

» 6.35 and 4 -

22 2

» 1.65.

Let x = 0. y = -2(0)2 + 16(0) - 21 y = -21

-12  144 - 192 2 -12  -48 = 2 =

–21 is the y-intercept. Vertex: x =

Since the radicand is negative, there are no x-intercepts.

-b -16 -16 = = = 4 2a 2(-2) -4 y = -2(4) 2 + 16(4) - 21 = -32 + 64 - 21 = 11

Let x = 0. 1 2 (0) + 6(0) + 24 2 y = 24

The vertex is (4, 11).

y =

The axis is x = 4.

24 is the y-intercept. Vertex: x =

-b -6 -6 = = = -6 1 2a 1 2( ) 2

1 y = (-6)2 + 6(-6) + 24 2 = 18 - 36 + 24 = 6

The vertex is ( -6 , 6). The axis x = -6.

26.

f ( x) =

3 2 x -x-4 2

Let f ( x) = 0. 3 0 = x2 - x - 4 2 Multiply by 2 to clear fractions, then solve. 0 = 3x 2 - 2 x - 8 = (3x + 4)( x - 2) 4 x = - ,2 3

The x-intercepts are - 43 and 2. 25.

f ( x) = -2 x 2 + 16 x - 21

Let x = 0. 3 2 (0) - 0 - 4 2 y = -4 y =

Let f ( x) = 0 Use the quadratic formula.

The y-intercept is –4.

Chapter 2 NONLINEAR FUNCTIONS x =

Vertex:

y =

The axis is x = 4.

-b -(-1) 1 = = 3 2a 3 2

(2)

2 3 æç 1 ö÷ 1 1 2 24 25 - -4= - =÷ ç 2 çè 3 ÷ø 3 6 6 6 6

æ1 25 ö The vertex is çç , - ÷÷÷. çè 3 6 ø

The axis x =

1 . 3

28.

1 7 y = - x2 - x 2 2

Let y = 0. 1 7 0 = - x2 - x 2 2

Multiply by -2 to clear fractions.

27.

0 = x2 + 2x + 7

1 8 1 y = x2 - x + 3 3 3

Use the quadratic formula. 4 - 4(1)(7) 2 -2  -24 = 2

Let y = 0. 0=

x =

1 2 8 1 x - x+ 3 3 3

Multiply by 3.

Since the radicand is negative, there are no x-intercepts.

0 = x 2 - 8x + 1 x =

-(-8)  (-8)2 - 4(1)(1) 2(1)

Let

x = 0. 1 7 7 y = - (0)2 - 0 - = 2 2 2

8  64 - 4 8  60 = = 2 2 8  2 15 = = 4  15 2

- 72 is the y-intercept.

Vertex:

The x-intercepts are 4 + 15 » 7.87 and 4 - 15 » 0.13. Let x = 0. y =

-2 

1 2 8 1 (0) - (0) + 3 3 3

1 is the y-intercept. 3

-b -(-1) 1 = æ ö = = -1 1 2a -1 2çç - ÷÷÷ çè 2 ø÷

1 7 y = - (-1)2 - (-1) 2 2 1 7 = - +12 2 = -3

The vertex is (-1, -3). æç 8 ö÷

çè 3 ø÷

-ç - ÷÷ çè 3 ø -b = æ ö = 32 = 4 Vertex: x = 1 2a 2çç ÷÷÷

The axis is x = -1 .

1 2 8 1 (4) - (4) + 3 3 3 16 32 1 15 = + == -5 3 3 3 3

y =

The vertex is (4, -5).

Section 2.2

29. The graph of y = x + 2 - 4 is the graph of y = x translated 2 units to the left and 4 units downward. This is graph D. 30. The graph of y = x - 2 - 4 is the graph of y = x translated 2 units to the right and 4 units downward. This is graph A.

37. The graph of y = f (-x) is the graph of y = f ( x) reflected in the y-axis.

31. The graph of y = -x + 2 - 4 is the graph of y = -( x - 2) - 4, which is the graph of y = x reflected in the y-axis, translated 2 units to the right, and translated 4 units downward.

This is graph C. 32. The graph of y =

-x - 2 - 4 is the same as

the graph of y =

-( x + 2) - 4. This is the

38.

graph of y = x reflected in the y-axis, translated 2 units to the left, and translated 4 units downward.

y = f (2 - x) + 2 y = f [-( x - 2)] + 2

This is the graph of y = f ( x) reflected in the y-axis, translated 2 units to the right, and translated 2 units upward.

This is graph B. 33. The graph of y = - x + 2 - 4 is the graph of y = x reflected in the x-axis, translated 2 units to the left, and translated 4 units downward. This is graph E.

39. 34. The graph of y = - x - 2 - 4 is the graph of

f ( x) =

x-2 +2

Translate the graph of f ( x) = and 2 units up.

y = x reflected in the x-axis, translated 2 units to the right, and translated 4 units downward.

x 2 units right

This is graph F. 35. The graph of y = - f ( x) is the graph of y = f ( x) reflected in the x-axis. 40.

f ( x) =

x+2 -3

Translate the graph of f ( x) = and 3 units down.

36. The graph of y = f ( x - 2) + 2 is the graph

of y = f ( x) translated 2 units to the right and 2 units upward.

x 2 units left

94 41.

Chapter 2 NONLINEAR FUNCTIONS f ( x) = - 2 - x - 2 = - -( x - 2) - 2

Reflect the graph of f (x) vertically and horizontally. Translate the graph 2 units right and 2 units down.

42.

f ( x) = - 2 - x + 2 = - -( x - 2) + 2

47. If 0 < a < 1, the graph of a f ( x ) will be flatter than the graph of f (x). Each y-value is only a fraction of the height of the original y-values.

48. If 1 < a , the graph of a f (x) will be taller than the graph of f (x). The absolute value of the y-value will be larger than the original y-values, while the x-values will remain the same.

Reflect the graph of f (x) vertically and horizontally. Translate the graph 2 units right and 2 units up.

43. If 0 < a < 1, the graph of f (ax) will be the graph of f (x) stretched horizontally.

44. If 1 < a, the graph of f (ax) will be the graph of f (x) compressed horizontally.

49. If -1 < a < 0, the graph will be reflected vertically, since a will be negative. Also, because a is a fraction, the graph will be flatter because each y-value will only be a fraction of its original height.

50. If a < -1, the graph of a f (x) will be reflected vertically. It will also be taller than the graph of f (x) since the absolute value of each y-value will be larger than the original y-values, while the x-values stay the same.

45. If -1 < a < 0, the graph of f (ax) will be reflected horizontally, since a is negative. It will be stretched horizontally. 51. (a) Since the graph of y = f ( x) is reflected

vertically to obtain the graph of y = - f ( x), the x-intercept is unchanged. The x-intercept of the graph of y = f ( x) is r. 46. If a < -1, the graph of f (ax) will be reflected horizontally, since a is negative. It will also be compressed horizontally.

(b) Since the graph of y = f ( x) is reflected horizontally to obtain the graph of y = f (-x), the x-intercept of the graph of y = f (-x) is -r. (c) Since the graph of y = f ( x) is reflected both horizontally and vertically to obtain the graph of y = - f (-x), the x-intercept of the graph of y = - f (-x) is -r.

Section 2.2

95 is a quadratic function, we can find the maximum profit by finding the vertex of P. This occurs at -6 b = = 3. x =2a 2(-1)

52. (a) Since the graph of y = f ( x) is reflected vertically to obtain the graph of y = - f ( x), the y-intercept of the graph of y = - f ( x) is -b. (b) Since the graph of y = f ( x) is reflected horizontally to obtain the graph of y = f (-x), the y-intercept is unchanged. The y-intercept of the graph of y = f (-x) is b. (c) Since the graph of y = f ( x) is reflected both horizontally and vertically to obtain the graph of y = - f (-x), the y-intercept of the graph of y = - f (-x) is -b. 53. (a)

Therefore, the maximum profit is P(3) = -(3) 2 + 6(3) - 5 = 4, or $4000.

54. (a)

(b) Break-even quantities are values of x = batches of widgets for which revenue and cost are equal. Set R( x) = C ( x) and solve for x. 2 - x + 5x = 3 x + 3 2 2

(b) Break-even quantities are values of x = batches of widgets for which revenue and cost are equal.

Set R( x) = C ( x) and solve for x. -x 2 + 8 x = 2 x + 5 ( x - 5)( x - 1) = 0 x - 5 = 0 or x - 1 = 0 x =1

So, the break-even quantities are 1 and 5. The minimum break-even quantity is 1 batch of widgets. (c) The maximum revenue occurs at the vertex 2

of R. Since R( x) = -x + 8x, then the x-coordinate of the vertex is b 8 x === 4. 2a 2(-1) So, the maximum revenue is R(4) = -42 + 8(4) = 16, or $16,000.

(d) The maximum profit is the maximum difference R( x) - C ( x). Since P ( x ) = R ( x) - C ( x) = -x 2 + 8 x - (2 x + 5) = -x 2 + 6 x - 5

x2 - 7 x + 6 = 0 ( x - 1)( x - 6) = 0 x - 1 = 0 or x - 6 = 0 x = 1 or x =6

So, the break-even quantities are 1 and 6. The minimum break-even quantity is 1 batch of widgets.

x2 - 6x + 5 = 0

x = 5 or

-x 2 + 10 x = 3x + 6

of R. Since R( x) = - x2 + 5 x, then the

x-coordinate of the vertex is -b -5 = æ ö = 5. x = 2a 2çç - 1 ÷÷÷ çè 2 ø÷

So, the maximum revenue is 52 R(8) = + 5(5) = 12.5, or $12,500. 2 (d) The maximum profit is the maximum differrence R( x) - C ( x). Since

(

2 P ( x ) = R ( x) - C ( x) = - x + 5 x - 3 x + 3 2 2

)

2 = -x + 7x -3 2 2 is a quadratic function, we can find the maximum profit by finding the vertex of P. This occurs at 7

b 7 = æ 2 ö = . x =1 2a 2 2 çç - ÷÷÷ çè 2 ÷ø

Chapter 2 NONLINEAR FUNCTIONS Therefore, the maximum profit is æ7ö P çç ÷÷ = èç 2 ø÷

æ 7 ö2 -çç ÷÷÷ çè 2 ø

Therefore, the maximum profit is 4 P(5) = - 52 + 8(5) - 15 = 5, or $5000. 5

7 çæ 7 ÷ö ç ÷ - 3 = 3.125, or $3125. 2 çè 2 ÷ø

56. (a) 55. (a)

(b) Break-even quantities are values of x = batches of widgets for which revenue equals cost.

Set R( x) = C ( x) and solve for x. 4 - x 2 + 10 x = 2 x + 15 5 4 2 x - 8 x + 15 = 0 5 4 x 2 - 40 x + 75 = 0 4 x 2 - 10 x - 30 x + 75 = 0 2 x(2 x - 5) - 15(2 x - 5) = 0 (2 x - 5)(2 x - 15) = 0 2 x - 5 = 0 or 2 x - 15 = 0 x = 2.5 or x = 7.5

So, the break-even quantities are 2.5 and 7.5 with the minimum break-even quantity being 2.5 batches of widgets. (c) The maximum revenue occurs at the vertex of

R. Since R( x) = - 54 x 2 + 10 x, then the x-coordinate of the vertex is b 10 x == - æ 4 ö = 6.25. 2a 2çç - ÷÷÷ çè 5 ÷ø

So, the maximum revenue is R(6.25) = 31.25, or $31,250. (d) The maximum profit is the maximum difference R( x) - C ( x). Since P ( x ) = R ( x) - C ( x ) 4 = - x 2 + 10 x - (2 x + 15) 5 4 = - x 2 + 8x - 15 5

is a quadratic function, we can find the maximum profit by finding the vertex of P. This occurs at b 8 x == - æ 4 ö = 5. 2a 2 çç - ÷÷÷

(b) Break-even quantities are values of x for which revenue equals cost.

Set R( x) = C ( x) and solve for x. -4 x 2 + 36 x = 16 x + 24 4 x 2 - 20 x + 24 = 0 4( x - 2)( x - 3) = 0 x - 2 = 0 or x - 3 = 0 x = 2 or x =3

So, the break-even quantities are 2 and 3. The minimum break-even quantity is 2 batches of widgets. (c) The maximum revenue occurs at the vertex

of R. Since R( x) = -4 x 2 + 36 x, then x-coordinate of the vertex is -b -36 9 x = = = . 2a 2(-4) 2 So, the maximum revenue is æ 9 ö2 æ9ö R(5) = -4 çç ÷÷ + 36 çç ÷÷ = 81, or $81,000. èç 2 ø÷ èç 2 ø÷

(d) The maximum profit is the maximum differrence R( x) - C ( x). Since P( x) = R( x) - C ( x) = -4 x 2 + 36 x - (16 x + 24) = -4 x 2 + 20 x - 24

is a quadratic function, we can find the maximum profit by finding the vertex of P. This occurs at b 20 5 x === = 2.5. 2a 2(-4) 2 Therefore, the maximum profit is æ5ö æ 5 ö2 æ5ö P çç ÷÷ = -4 çç ÷÷ + 20 çç ÷÷ - 24 = 1, or $1000. çè 2 ÷ø çè 2 ÷ø çè 2 ÷ø

çè 5 ÷ø

Section 2.2 57.

R( x) = 8000 + 70 x - x 2 = -x 2 + 70 x + 8000

The maximum revenue occurs at the vertex.

59. Let x = the number of $30 increases. (a) Rent per apartment: 960 + 30x (b) Number of apartments rented: 80 - x (c) Revenue: R(x) = (number of apartments rented) × (rent per apartment) = (80 - x)(960 + 30 x)

-b -70 = = 35 x = 2a 2(-1) y = 8000 + 70(35) - (35)2 = 8000 + 2450 - 1225 = 9225

= -30 x 2 + 1440 x + 76,800 (d) Find the vertex:

The vertex is (35, 9225). The maximum revenue of $9225 is realized when 35 seats are left unsold.

x =

y = -30(24)2 + 1440(24) + 76,800 = 94, 080

58. (a) The revenue is R(x) = (Price per ticket) ⋅ (Number of people flying). Number of people flying = 100 - x

The vertex is (24, 94,080). The maximum revenue occurs when x = 24. (e) The maximum revenue is the y-coordinate of the vertex, or $94,080.

Price per ticket = 200 + 4x R( x) = (200 + 4 x)(100 - x) = 20, 000 + 200 x - 4 x 2

-b -1440 = = 24 2a 2(-30)

60. Let x = the number of weeks to wait. (a) Income per pound (in cents):

(b) R(x) = -4 x + 200 x + 20, 000

80 - 4x

x-intercepts:

(b) Yield in pounds per tree: 100 + 5x

0 = (200 + 4 x)(100 - x) x = -50 or x = 100

y-intercept:

R( x) = 8000 - 20 x 2

y = -4(0) 2 + 200(0) + 20, 000 = 20, 000

Vertex: x =

(d) Find the vertex.

-b -200 = = 25 2a -8

x =

-b 0 = =0 2a -20

y = 8000 - 20(0) 2 = 8000

y = -4(25)2 + 200(25) + 20, 000 = 22,500

The vertex is (0, 8000). To produce maximum revenue, wait 0 weeks. Pick the peaches now.

This is a parabola which opens downward. The vertex is at (25, 22,500).

(e) R(0) = 8000 - 20(0)2 = 8000

or the maximum revenue is 8000 cents per tree or $80.00 per tree.

(c) The maximum revenue occurs at the vertex, (25, 22,500). This will happen when x = 25, or there are 25 unsold seats. (d) The maximum revenue is $22,500, as seen from the graph.

61.

p = 500 - x

(a) The revenue is

R( x) = px = (500 - x)( x) = 500 x - x 2.

Chapter 2 NONLINEAR FUNCTIONS (b)

63. (a)

(b) The vertex is at (6, 1027), so f has the form (c) From the graph, the vertex is halfway between x = 0 and x = 500, so x = 250 units corresponds to maximum revenue. Then the price is p = 500 - x = 500 - 250 = $250.

Note that price, p, cannot be read directly from the graph of R( x) = 500 x - x 2.

a(t - 6)2 + 1027

Using the point (18, 637) we have a(18 - 6)2 + 1027 = 637 a =

637 - 1027 » -2.71 144

Thus f (t ) = -2.71(t - 6)2 + 1027 (c) The calculator regression produces the

function f (t )  2.67t 2  32.1t  926. The graph shows this function together with the data points in the window [0, 20] by [0, 1100].

(d) R( x) = 500 x - x 2 = -x 2 + 500 x

Find the vertex. -b -500 = = 250 x = 2a 2(-1) y = -(250)2 + 500(250) = 62,500

The vertex is (250, 62,500). The maximum revenue is $62,500. 62.

1 S ( x) = - ( x - 10)2 + 40 for 0 £ x £ 10 4 1 (a) S (0) = - (0 - 10) 2 + 40 4 = -25 + 40 = 15

The increase in sales is $15,000. (b) For $10,000, x = 10. 1 S (10) = - (10 - 10)2 + 40 4 S (10) = 40

The increase in sales is $40,000. (c) The graph of S ( x) = - 14 ( x - 10)2 + 40 is

the graph of y = - 14 x 2 translated 10 units to the right and 40 units upward.

(d) The graph shows the functions found in (b) and (c) in the window [0, 20] by [0, 1100].

No, the calculator fit is better. The fit from (b) gives too large an estimate at the last two time values. (e) From part (b), f (25) = -2.71(25 - 6)2 + 1027 = 48.69 » 49 thousand short tons From part (c),

f (25)  2.67(25) 2  32.1(25)  926  59.75  60 thousand short tons 64. (a) When t = 14, L = 2.024, so the length of the crown is 2.024 mm at 14 weeks. When t = 24, L = 6.104, so the length is 6.104 mm at 24 weeks.

Section 2.2

(b) The vertex occurs at t =

The vertex of the quadratic function

66.

y  f (t )  0.0723t 2  0.723t  52.0 is at -b -0.723 = = 5. t = 2a 2(-0.0723)

-b -0.788 = » 39.4. 2a 2(-0.01)

When t = 39.4, L » 8.48, so the maximum length is 8.48 mm and occurs 39.4 weeks after conception. 65.

Since the coefficient of the leading term is negative, the graph of the function opens downward, so the maximum is reached when t = 5, or about 2000  5  2005.

S ( x) = 1 - 0.058 x - 0.076 x 2

(a) 0.50 = 1 - 0.058 x - 0.076 x

Since the graph opens downward, the function values increase to the left of the maximum point and decrease to the right of the maximum point. So, the incidence rate was increasing during the years 2002–2005 and was decreasing during 2005–2016.

0.076 x 2 + 0.058 x - 0.50 = 0 76 x 2 + 58x - 500 = 0 38x 2 + 29 x - 250 = 0

67. (a) The vertex of the quadratic function y = 0.057 x - 0.001x 2 is at

-29  (29) - 4(38) (-250) x = 2(38) -29  38,841 = 76 -29 - 38,841 » -2.97 76

and

-29 + 38,841 » 2.21 76

We ignore the negative value. The value x = 2.2 represents 2.2 decades or 22 years, and 22 years after 65 is 87. The median length of life is 87 years.

x =-

Since the coefficient of the leading term, -0.001, is negative, then the graph of the function opens downward, so a maximum is reached at 28.5 weeks of gestation. (b) The maximum splenic artery resistance reached at the vertex is y = 0.057(28.5) - 0.001(28.5) 2 » 0.81.

0.057 x - 0.001x 2 = 0 Substitute in the expression in x for y

(b) If nobody lives, S ( x) = 0. 1 - 0.058x - 0.076 x 2 = 0 76 x 2 + 58x - 1000 = 0

x(0.057 - 0.001x) = 0 Factor

38x + 29 x - 500 = 0 x = =

-29  (29)2 - 4(38)(-500) 2(38)

x = 0 or 0.057 - 0.001x = 0 Set each factor equal to 0. x =

-29  76,841 76

-29 + 76,841 » 3.27 76

We ignore the negative value. The value x = 3.3 represents 3.3 decades or 33 years, and 33 years after 65 is 98. Virtually nobody lives beyond 98 years.

0.057 = 57 0.001

So, the splenic artery resistance equals 0 at 0 weeks or 57 weeks of gestation. No, this is not reasonable because at x = 0 or 57 weeks, the fetus does not exist.

-29 - 76,841 » -4.03 76

and

b 0.057 == 28.5. 2a 2(-0.001)

68. (a)

(b) Quadratic (c) f (t ) = 0.01375t 2 - 0.4309t + 65.92

100

Chapter 2 NONLINEAR FUNCTIONS 27.8 = a(118 - 60) 2 + 20.3 7.5 = 3364a a = 0.00223

A quadratic function that models the data is f (t ) = 0.00223(t - 60) 2 + 20.3.

(d) Given that (h, k) = (10, 62.04), the equation has the form f(t) = a(t - 10)2 + 62.04. The point (48, 76.08) is also on the curve.

(e)

76.08 = a(48 - 10) + 62.04 14.04 = 1444a a » 0.009723

A quadratic function that models the data The two graphs are very close.

is f (t ) = 0.009723(t - 10) 2 + 62.04.

(f) From part (c), f (125) = 0.00111(125) 2 - 0.0712(125) + 20.8

(e)

» 29.2 From part (d),

f (125) = 0.00223(125 - 60)2 + 20.3. » 29.7

No, the two graphs do not differ by much. 70.

(f) From part (c), f (45) = 0.01375(45) 2 - 0.4309(45) + 65.92 » 74.37 From part (d),

f ( x) = 60.0 - 2.28x + 0.0232 x 2 -b -2.28 =» 49.1 2a 2(0.0232)

The minimum value occurs when x » 49.1. The age at which the accident rate is a minimum is 49 years. The minimum rate is

f (45) = 0.009723(45 - 10)2 + 62.04 » 73.95

f (49.1) = 60.0 - 2.28(49.1) + 0.0232(49.1) 2

The function from (c) gives a prediction slightly closer to the actual value of 74.55.

= 60.0 - 111.948 + 55.930792 » 3.98.

69. (a) 71.

h(t ) = 32t - 16t 2 = -16t 2 + 32t

(b) Quadratic (c) f (t ) = 0.00111t 2 - 0.0712t + 20.8

(a) Find the vertex. x = -b = -32 = 1 2a -32 y = -16(1) 2 + 32(1) = 16

The vertex is (1, 16), so the maximum height is 16 ft. (b) When the object hits the ground, h = 0, so 32t - 16t 2 = 0

(d) Given that (h, k ) = (60, 20.3), the equation has the form 2

f (t ) = a(t - 60) + 20.3.

Since (118, 27.8) is also on the curve,

16t (2 - t ) = 0 t = 0 or t = 2. When t = 0, the object is thrown upward. When t = 2, the object hits the ground; that is, after 2 sec.

Section 2.2 72.

101 A = x(190 - x)

y = 0.056057 x 2 + 1.06657 x

= 190 x - x 2

(a) If x = 25 mph,

= -x 2 + 190 x

y = 0.056057(25)2 + 1.06657(25) y » 61.7.

Find the vertex:

At 25 mph, the stopping distance is approximately 61.7 ft. (b)

0.056057 x + 1.06657 x - 150 = 0 x =

y = -(95) 2 + 190(95)

0.056057 x 2 + 1.06657 x = 150 2

-1.06657  (1.06657) 2 - 4(0.056057)(-150) 2(0.056057) x » 43.08 or x » -62.11

We ignore the negative value. To stop within 150 ft, the fastest speed you can drive is 43.08 mph or about 43 mph.

-b -190 = = 95 2a -2

= 9025

This is a parabola with vertex (95,9025) that opens downward. The maximum area is the value of A at the vertex, or 9025 sq ft. 75. Draw a sketch of the arch with the vertex at the origin.

73. Let x = the width.

Then 380 - 2x = the length. Since the arch is a parabola that opens downward, the equation of the parabola is the form y = a( x - h)2 + k , where the vertex (h, k ) = (0, 0) and a < 0. That is, the equation is of the form y = ax 2.

Area = x(380 - 2 x) = -2 x 2 + 380 x Find the vertex:

Since the arch is 30 meters wide at the base and 15 meters high, the points (15, -15) and (-15, -15) are on the parabola. Use (15, -15 ) as one point on the parabola. -15 = a(15)2

-b -380 x = = = 95 2a -4

a =

y = -2(95) + 380(95) = 18, 050

The graph of the area function is a parabola with vertex (95,18,050). The maximum area of 18,050 sq ft occurs when the width is 95 ft and the length is 380 - 2 x = 380 - 2(95) = 190 ft.

74. Let x = the length of the lot and y = the width of the lot.

-15 15

1 15

So, the equation is y =-

1 2 x . 15

Ten feet from the ground (the base) is at y = -5. Substitute -5 for y and solve for x. -5 = -

1 2 x 15

x 2 = -5(-15) = 75

The perimeter is given by

x =  75 = 5 3

P = 2x + 2 y 380 = 2 x + 2 y

The width of the arch ten feet from the ground is then

190 = x + y 190 - x = y.

5 3 - (-5 3) = 10 3 meters

» 17.32 meters.

102

Chapter 2 NONLINEAR FUNCTIONS

76. Sketch the culvert on the xy-axes as a parabola that opens upward with vertex at (0, 0).

2.3 Polynomial and Rational Functions Your Turn 1

Graph f ( x) = 64 - x 6.

The equation is of the form y = ax 2 , Since the culvert is 18 ft wide at 12 ft from its vertex, the points (9, 12) and ( -9, 12 ) are on the parabola. Use (9, 12) as one point on the parabola. 12 = a(9) 2

Graph y =

4x - 6 . x-3

The value x = 3 makes the denominator 0, so 3 is not in the domain and the line x = 3 is a vertical asymptote. Let x get larger and larger. Then 4xx--36 » 4xx = 4 as x gets larger and larger.

12 = 81a 12 = a 81 4 = a 27

Thus, y =

Your Turn 2

So, the line y = 4 is a horizontal asymptote.

4 2 x . 27

To find the width 8 feet from the top, find the points with y-value = 12 - 8 = 4. Thus, 4=

4 2 x 27

2.3

108 = 4 x 2 27 = x

W1. The graph of f ( x ) shifted 2 units to the left and 3 units down.

x 2 = 27

W2. The graph of f ( x ) reflected across the x-axis and shifted 3 units to the right.

x =  27 x = 3 3.

(

)

The width of the culvert is 2 3 3 = 6 3 ft » 10.39 ft.

Warmup Exercises

W3. The graph of f ( x ) reflected across the y-axis and shifted 2 units up. W4. The graph of f ( x ) reflected across the y-axis and shifted 2 units to the right.

2.3 Exercises 1.

True

False.

False. The f ( x)  x 4  x5  x6 is degree 6.

Section 2.3 4.

103

x -1 False. The function f ( x) = is not a x +1 rational function since x is inside the radical.

10. The graph of f ( x) = -( x - 1)4 + 2 is the graph of y = x 4 reflected in the x-axis, and translated 1 unit to the right and 2 units upward.

The graph of f ( x) = ( x - 2)3 + 3 is the graph of y = x3 translated 2 units to the right and 3 units upward.

0 = -( x - 1) 4 + 2 x - 1 = 4 2 x = 1 42

0 = ( x - 2) + 3

x = 1 - 4 2, 1 + 4 2

x - 2 = -3 3 x = 2 - 33

The graph of f ( x) = ( x + 1)3 - 2 is the graph of y = x3 translated 1 unit to the left and 2 units downward.

11. The graph of y = x3 - 7 x - 9 has the right end up, the left end down, at most two turning points, and a y-intercept of -9.

This is graph D. 12. The graph of y = -x3 + 4 x 2 + 3x - 8 has the right end down, the left end up, at most two turning points, and a y-intercept of -8.

0 = ( x + 1)3 - 2

This is graph C.

x +1 = 32 x = -1 + 3 2

The graph of f ( x) = -( x + 3) 4 + 1 is the graph of y = x 4 reflected horizontally, translated 3 units to the left, and translated 1 unit upward. f(x) = – (x + 3)4 + 1

(–4, 0)

(–2, 0)

This is graph E.

–6

13. The graph of y = -x3 - 4 x 2 + x + 6 has the right end down, the left end up, at most two turning points, and a y-intercept of 6.

–2

0 = -( x + 3)4 + 1 x + 3 = 1 x = -3  1 x = -4, x = -2

14. The graph of y = 2 x3 + 4 x + 5 has the right end up, the left end down, at most two turning points, and a y-intercept of 5.

This is graph B. 15. The graph of y = x 4 - 5 x 2 + 7 has both ends up, at most three turning points, and a y-intercept of 7.

104

Chapter 2 NONLINEAR FUNCTIONS

16. The graph of y = x 4 + 4 x3 - 20 has both ends up, at most three turning points, and a y-intercept of -20. This is graph F.

25. The right end is up and the left end is up. There are three turning points.

The degree is an even integer equal to 4 or more. The xn term has a + sign.

17. The graph of y = -x 4 + 2 x3 + 10 x + 15 has both ends down, at most three turning points, and a y-intercept of 15. This is graph G.

26. The graph has the right end up, the left end down, and four turning points. The degree is an odd integer equal to 5 or more. The xn term has a + sign.

18. The graph of y = 0.7 x5 - 2.5 x 4 - x3 + 8 x 2 + x + 2 has the right end up, the left end down, at most four turning points, and a y-intercept of 2. This is graph H.

27. The right end is up and the left end is down. There are four turning points. The degree is an odd integer equal to 5 or more. The xn term has a + sign.

19. The graph of y = -x5 + 4 x 4 + x3 - 16 x 2 + 12 x + 5 has the right end down, the left end up, at most four turning points, and a y-intercept of 5. This is graph A.

28. Both ends are down, and there are five turning points. The degree is an even integer equal to 6 or more. The xn term has a - sign.

20. The graph of y =

2 x2 + 3

x2 - 1

has the lines

with equations x = 1 and x = -1 as vertical asymptotes, the line with equation y = 2 as a horizontal asymptote, and a y-intercept of –3. This is graph B. 21. The graph of y =

2 x2 + 3

x2 + 1

has no vertical

asymptote, the line with equation y = 2 as a horizontal asymptote, and a y-intercept of 3. This is graph D. 22. The graph of y =

-2 x 2 - 3 x2 - 1

has the lines with

equations x = 1 and x = -1 as vertical asymptotes, the line with equation y = –2 as a horizontal asymptote, and a y-intercept of 3. This is graph A. 23. The graph y =

-2 x 2 - 3 x2 + 1

has no vertical

asymptote, the line with equation y = –2 as a horizontal asymptote, and a y-intercept of –3.

29. The right end is down and the left end is up. There are six turning points. The degree is an odd integer equal to 7 or more. The xn term has a - sign. 30. The right end is up, the left end is down, and there are six turning points. The degree is an odd integer equal to 7 or more. The xn term has a + sign. -4 x+2 The function is undefined for x = -2, so the line x = -2 is a vertical asymptote. y =

31.

-102 -12 -7 -5 -3 -1

1 10 100 x + 2 -100 -10 -5 -3 -1 0.04 0.4 0.8 1.3 4 -4 -0.4 -0.04 y

The graph approaches y = 0, so the line y = 0 (the x-axis) is a horizontal asymptote. Asymptotes: y = 0, x = –2 x-intercept: none y-intercept: –2, the value when x = 0

This is graph E. 24. The graph of y =

2 x2 + 3 has the line with x3 - 1

Section 2.3 32.

y =

105 -1 x+3

34.

A vertical asymptote occurs when x + 3 = 0 or when x = -3, since this value makes the denominator 0. x

-6 -5 - 4 -2 -1

x + 3 -3 -2 - 1 y

1 3

1 2

1 -1

1 2

8 5 - 3x

Undefined for 5 - 3x = 0 3x = 5

2 -

y =

x =

3 -

1 3

5 3

Since x = 53 causes the denominator to equal 0, x = 53 is a vertical asymptote.

As | x | gets larger, x-+13 approaches 0, so y = 0 is a horizontal asymptote.

-1

5 - 3x 8 5 2 -1 - 4 -7 0.5 0.8 2 -4 -1 -0.571 y

Asymptotes: y = 0, x = –3

x-intercept: none

The graph approaches y = 0, so the line y = 0 is a horizontal asymptote.

y-intercept: - 13 , the value when x = 0

Asymptotes: y = 0, x = 53

x-intercept: none y-intercept: 85 , the value when x = 0

33.

y =

2 3 + 2x

3 + 2 x = 0 when 2 x = -3 or x = - 32 ,

so the line x = - 32 is a vertical asymptote. x

-51.5 -6.5 -2 -1 3.5 48.5

3 + 2 x -100 -10 -1 y -0.02 -0.2 -2

1 10 100 2 0.2 0.02

The graph approaches y = 0, so the line y = 0 (the x-axis) is a horizontal asymptote. Asymptote: y = 0, x = - 32

35.

2x x-3 x - 3 = 0 when x = 3, so the line x = 3 is a vertical asymptote. y =

-97

-7 -1

2.5

-194 -14 -2

x - 3 -100 -10 -4 -2 -1 -0.5 1.94

x-intercept: none y-intercept: 23 , the value when x = 0

x 2x

3.5 4

103

7 8 10

206

100

14 8

5 3.5 2.75 2.06

As x gets larger, 2 x » 2 x = 2. x-3 x

Thus, y = 2 is a horizontal asymptote. Asymptotes: y = 2, x = 3

-10

x - 3 0.5 y

1.4 0.5 -1 -4

106

Chapter 2 NONLINEAR FUNCTIONS

y-intercept: 0, the value when x = 0

3.5 4.5 5

104

x +1

4.5 5.5 6

105

x - 4 -0.5 0.5 1 10

100

-9

11 6 1.5 1.05

As x gets larger, x +1 x » = 1. x-4 x 36.

y =

4x 3 - 2x

Thus, y = 1 is a horizontal asymptote. Asymptotes: y = 1, x = 4

Since x = 32 causes the denominator to

x-intercept: -1, the value when y = 0

equal 0, x = 32 is a vertical asymptote.

y-intercept: - 14 , the value when x = 0

-3

-2

-1 0

-12

-8

-4 0 4

3 - 2x

5 3 1

-1.33 -1.14 -0.8 0 4

3 - 2 x -1 -3

-5

38.

-8 -4 -3.2

y =

x-4 x +1

Since x = -1 causes the denominator to equal 0, x = -1 is a vertical asymptote.

As x gets larger,

-4 -3 -2

4 x » 4 x = -2. 3 - 2x -2 x

x-4

Thus, the line y = –2 is a horizontal asymptote.

x +1

x-intercept: 0, the value when y = 0 y-intercept: 0, the value when x = 0

-8 -7 -6 -4

-3

-3 -2 -1

3 4

x-4

-2

-1 0

x +1

4 5

6 -4 -1.5

2.67 3.5

Asymptotes: y = -2, x = 32

-0.67 -0.25 0

As x gets larger, x-4 x » = 1. x +1 x Thus, the line y = 1 is a horizontal asymptote. 37.

y =

x +1 x-4

Asymptotes: y = 1, x = -1

x - 4 = 0 when x = 4, so x = 4 is a vertical asymptote. x

-96

-6 -1

x +1

-95

-5

x - 4 -100 -10 -5 y

0.95

0.5

x-intercept: 4, the value when y = 0 y-intercept: -4, the value when x = 0

-4 - 1

0 -0.25 -4

Section 2.3 39.

107

y = 3 - 2x 4 x + 20 4 x + 20 = 0 when 4 x = -20 or x = -5, so the line x = –5 is a vertical asymptote. -8

3x + 6 = 0 when 3x = -6 or x = -2, so the line x = -2 is a vertical asymptote.

-6

-4

-3

-2

-14

-11

-8

-5 -4

-3 -1

-x - 4

- 1 -3

-4

-5

3x + 6

-9 -6

4 x + 20 -12

-8

-4

5 -3.5 -1.38 -0.67

2.17 2.88

-0.11

As x gets larger, 3 - 2 x » -2 x = - 1 . 4 x + 20 4x 2

-3

0 0.33 -1 -0.67 -0.56

As x gets larger, -x - 4 » - x = - 1 . 3x + 6 3x 3

Thus, the line y = - 12 is a horizontal asymptote.

The line y = - 13 is a horizontal asymptote.

Asymptotes: x = -5, y = - 12

40.

y = -x - 4 3x + 6

3 - 2 x -26 -23 -20

-7

41.

Asymptotes: y = - 13 , x = -2

x-intercept: 32 , the value when y = 0

x-intercept: -4, the value when y = 0

3 , the value when x = 0 y-intercept: 20

y-intercept: - 23 , the value when x = 0

y =

42.

6 - 3x 4 x + 12

4 x + 12 = 0 when 4 x = -12 or x = -3 , so x = -3 is a vertical asymptote. x

-6

-5

-4 -2

6 - 3x

4 x + 12 -12

-8

-4

-1

-2 -2.625 -4.5

3 1.125 0.5

As x gets larger, 6 - 3 x » -3 x = - 3 . 4 x + 12 4x 4

y =

-2 x + 5 x+3

x + 3 = 0 when x = -3, so x = -3 is a vertical asymptote. x -2 x + 5 x+3 y

-6

-5

- 4 -2 -1

-3

-2

-1

-5.67 -7.5 -13

9 3.5 1.67

As x gets larger, -2 x + 5 -2 x » = -2. x+3 x

The line y = - 34 is a horizontal asymptote.

The line y = -2 is a horizontal asymptote.

Asymptotes: y = - 34 , x = -3

Asymptotes: y = -2, x = -3

x-intercept: 2, the value when y = 0

x-intercept: 52 , the value when y = 0

y-intercept: 12 , the value when x = 0

y-intercept: 53 , the value when x = 0

108 43.

Chapter 2 NONLINEAR FUNCTIONS 47.

2 y = x + 7 x + 12 x+4 ( x + 3)( x + 4) = x+4 = x + 3, x ¹ -4

f ( x) = ( x - 1)( x - 2)( x + 3), g ( x) = x3 + 2 x 2 - x - 2, h( x) = 3x3 + 6 x 2 - 3x - 6

(a)

There are no asymptotes, but there is a hole at x = –4. x-intercept: –3, the value when y = 0. y-intercept: 3, the value when x = 0.

f (1) = (0)(-1)(4) = 0

(b) f (x) is zero when x = 2 and when x = -3. g (-1) = (-1)3 + 2(-1) 2 - (-1) - 2 = -1 + 2 + 1 - 2 = 0

(c)

g (1) = (1)3 + 2(1)2 - (1) - 2 = 1+ 2 -1- 2 =0 g (-2) = (-2)3 + 2(-2)2 - (-2) - 2 = -8 + 8 + 2 - 2 =0

44.

(d) g ( x) = [ x - (-1)]( x - 1)[ x - (-2)] g ( x) = ( x + 1)( x - 1)( x + 2)

9 - 6x + x2 3- x (3 - x) (3 - x) = 3- x = 3 - x, x ¹ 3

y =

(e) h( x) = 3g ( x) = 3( x + 1)( x - 1)( x + 2)

There are no asymptotes, but there is a hole at x = 3. There is no x-intercept, since 3 – x = 0 implies x = 3, but there is a hole at x = 3. y-intercept: 3, the value when x = 0.

45. For a vertical asymptote at x = 1, put x - 1 in the denominator. For a horizontal asymptote at y = 2, the degree of the numerator must equal the degree of the denominator and the quotient of their leading terms must equal 2. So, 2x in the numerator would cause y to approach 2 as x gets larger. 2x . So, one possible answer is y = x -1

(f)

If f is a polynomial and f (a) = 0 for some number a, then one factor of the polynomial is x - a.

48. Graph 7 5 4 3 2 f ( x) = x - 4 x - 3x +7 4 x + 12 x - 12 x using a graphing calculator with the indicated viewing windows. (a) There appear to be two x-intercepts, one at x = -1.4 and one at x = 1.4.

(b) There appear to be three x-intercepts, one at x = -1.414, one at x = 1.414, and one at x = 1.442.

46. For a vertical asymptote at x = -2, put x + 2 in the denominator. For a horizontal asymptote at y = 0, the only condition is that the degree of the numerator is less than the degree of the denominator. If the degree of the denominator is 1, then put a constant in the numerator to make y approach 0 as x gets larger. So, one possible 3 . solution is y = x + 2

Section 2.3 49.

f ( x) =

109 1 5

C (1000) =

x - 2 x - 3x + 6

(a) Two vertical asymptotes appear, one at x = -1.4 and one at x = 1.4.

90 + 0.68(1000) = 0.77 1000

C (10, 000) =

90 + 0.68(10, 000) » 0.69 10, 000

C (100, 000) =

90 + 0.68(100, 000) » 0.68 100, 000

As x, the number of mugs produced, get increasingly large, the average cost per mug approaches $0.68. (b) Three vertical asymptotes appear, one at x = -1.414, one at x = 1.414, and one at x = 1.442.

(c)

52.

C ( x) =

600 x + 20

(a) C (10) = 600 = 600 = $20 10 + 20 30  C (20) =

600 600 = = $15 20 + 20 40

C (50) =

600 600 = » $8.57 50 + 20 70

200 + 0.85(100) = 2.85 100

C (75) =

600 600 = » $6.32 75 + 20 95

200 + 0.85(1000) = 1.05 1000

C (100) =

600 600 = = $5 100 + 20 120

50. (a) C ( x) = 200 + 0.85 x x (b) C (100) = C (1000) =

C (10, 000) =

200 + 0.85(10, 000) = 0.87 10, 000

C (100, 000) =

200 + 0.85(100, 000) » 0.85 100, 000

As x, the number of mugs produced, get increasingly large, the average cost per mug approaches $0.85. (c)

(b) (0, ¥) would be a more reasonable domain for average cost than [0, ¥). If zero were included in the domain, there would be no units produced. It is not reasonable to discuss the average cost per unit of zero units. (c) The graph has a vertical asymptote at x = -20, a horizontal asymptote at y = 0 (the x-axis), and y-intercept 600 = 30. 20

(d)

51. (a) C ( x) = 90 + 0.68 x x (b) C (100) =

90 + 0.68(100) = 1.58 100

110 53.

Chapter 2 NONLINEAR FUNCTIONS C ( x) =

220, 000 x + 475

54.

y = x(100 - x)( x 2 + 500),

x = tax rate;

(a) If x = 25, 220, 000 220, 000 C (25) = = = $440. 25 + 475 500

y = tax revenue in hundreds of thousands of dollars. (a) x = 10 y = 10(100 - 10)(102 + 500) = 10(90)(600) = $54 billion

If x = 50, 220, 000 220, 000 = » $419. 50 + 475 525

C (50) =

If x = 100, 220, 000 220, 000 C (100) = = » $383. 100 + 475 575

(b) x = 40 y = 40(100 - 40)(402 + 500) = 40(60)(2100) = $504 billion

If x = 200, 220, 000 220, 000 = » $326. 200 + 475 675

C (200) =

If x = 300, C (300) =

y = 50(100 - 50)(502 + 500) = 50(50)(3000) = $750 billion

220, 000 220, 000 = » $284. 300 + 475 775

If x = 400,

(d) x = 80 y = 80(100 - 80)(802 + 500) = 80(20)(6900) = $1104 billion

220,000 220,000 C (400) = = » $251. 400 + 475 875

(b) A vertical asymptote occurs when the denominator is 0. x + 475 = 0 x = -475

(e)

A horizontal asymptote occurs when C ( x) approaches a value as x gets larger. In this case, C ( x) approaches 0. The asymptotes are x = -475 and y = 0. (c) x-intercepts: 0=

220, 000 ; no such x, so no x-intercepts x + 475

f1( x) has a maximum of 100, which occurs at the vertex. The x-coordinate of the vertex lies between the two roots.

y-intercepts: C (0) =

55. Quadratic functions with roots at x = 0 and x = 100 are of the form f ( x) = ax(100 - x).

220, 000 » 463.2 0 + 475

The vertex is (50, 100).

(d) Use the following ordered pairs: (25,440), (50,419), (100,383), (200,326), (300,284), (400,251).

100 = a(50)(100 - 50) 100 = a(50)(50) 100 = a 2500 1 = a 25 1 x(100 - x) f1( x) = x(100 - x) or 25 25

Section 2.3

111 y =

f 2 ( x) has a maximum of 250, occurring at (50, 250). 250 = a(50)(100 - 50) 250 = a(50)(50)

6.7(80) = 26.8 100 - 80

The cost is $26,800. x = 90

250 = a 2500 1 = a 10

y =

6.7(90) = 60.3 100 - 90

The cost is $60,300. x = 95

1 x(100 - x ) f2 ( x) = x(100 - x ) or 10 10 é x(100 - x ) ù é x(100 - x ) ù ú⋅ê ú f1( x) ⋅ f 2 ( x) = ê êë úû êë úû 25 10

y =

6.7(95) 100 - 95

The cost is $127,300. x = 98

x 2 (100 - x)2 = 250

y =

6.7(98) = 328.3 100 - 98

The cost is $328,300. x = 99 y =

6.7(99) = 668.3 100 - 99

The cost is $663,300. (b) No, because x = 100 makes the denominator zero, so x = 100 is a vertical asymptote.

56. (a)

(c)

(b) The tax rate for maximum revenue is 29.0%. The maximum revenue is $25.2 million. 57.

y =

6.7 x , 100 - x

58.

Let x = percent of pollutant; y = cost in thousands.

x = cost in thousands of dollars. (a) x = 0

6.7(50) = 6.7 100 - 50

y =

The cost is $6700. x = 70 y =

6.5(0) 0 = = $0 102 - 0 102

x = 50

6.7(70) » 15.6 100 - 70

The cost is $15,600. x = 80

6.5 x 102 - x

y = percent of pollutant;

(a) x = 50 y =

y =

6.5(50) 325 = = 6.25 102 - 50 52

6.25(1000) = $6250 x = 80

112

Chapter 2 NONLINEAR FUNCTIONS y =

6.5(80) 520 = = 23.636 102 - 80 22

(e)

(23.636)(1000) = $23, 636 » $23,600

x = 90 6.5(90) 585 = = 48.75 102 - 90 12 (48.75)(1000) = $48, 750

y =

g(t) 5 20.006t 4 1 0.140t 3 2 0.053t 2 1 1.79t

60. (a)

» $48,800

x = 95 0

6.5(95) 617.5 = = 88.214 102 - 95 7 (88.214)(1000) = 88, 214

y =

(b) Because the leading coefficient is negative and the degree of the polynomial is even, the graph will have right end down, so it cannot keep increasing forever.

» $88, 200

x = 99 y =

6.5(99) 643.5 = = 214.5 102 - 99 3

61.

A( x) = 0.003631x - 0.03746 x + 0.1012 x + 0.009

(a)

(214.500)(1000) = 214,500 » $214,500

x = 100 y =

A( x)

0.009

0.076

0.091 0.073

0.047

0.032

6.5(100) 650 = = 325 102 - 100 2 (325)(1000) = $325, 000

(b)

(b) The peak of the curve comes at about x = 2 hours. (c) The curve rises to a y-value of 0.08 at about x = 1.1 hours and stays at or above that level until about x = 2.7 hours. 62. (a)

59. (a)

(b) y  0.1743t 2  19.95t  336.0 (c)

A(t )  0.1953t 4  13.70t 3  282.9t 2  836.7t  35,842

(b) At approximately, t  2, that is, in 2002. (c) At approximately, t  8, that is, in 2008. (d) y  0.01792t 3  1.652t 2  26.35t  504.8

Section 2.3 63.

f ( x) =

113

x 1 + (ax)

As x gets larger, Kx Kx » = K. A+x x

(a) A reasonable domain for the function is [0, ¥). Populations are not measured using negative numbers and they may get extremely large. (b) If  = a = b = 1, the function becomes f ( x) =

x 1 + x2

Kx A+x

Thus, y = K will always be a horizontal asymptote for this function. (d) K represents the maximum growth rate. The function approaches this value asymptotically, showing that although the growth rate can get very close to K, it can never reach the maximum, K. Kx (e) f ( x) = A+x Let A = x, the quantity of food present. KA KA K = = f ( x) = 2A 2 A+ A K is the maximum growth rate, so K2 is half the maximum. Thus, A represents the quantity of food for which the growth rate is half of its maximum.

x . 1+ x

30 - 1500 = 220, 65. (a) When c = 30, w = 100 30

so the brain weights 220 g when its circumference measures 30 cm. When 3

40 - 1500 = 602.5, so the c = 40, w = 100 40

(d) As seen from the graphs, when b increases, the population of the next generation, f (x), gets smaller when the current generation, x, is larger. 64. (a) A reasonable domain for the function is [0, ¥). Populations are not measured using negative numbers, and they may get extremely large. (b) f ( x) =

Kx A+x

When K = 5 and A = 2, f ( x) =

5x . 2+x

brain weighs 602.5 g when its circumference 3

50 - 1500 is 40 cm. When c = 50, w = 100 50

= 1220, so the brain weighs 1220 g when its circumference is 50 cm.

(b) Set the window of a graphing calculator so you can trace to the positive x-intercept of the function. Using a “root” or “zero” program, this x-intercept is found to be approximately 19.68. Notice in the graph that positive c values less than 19.68 correspond to negative w values. Therefore, the answer is c < 19.68. (c)

The graph has a horizontal asymptote at y = 5 since 5x 5x » =5 2+x x

as x gets larger. (d) One method is to graph the line y = 700 on the graph found in part (c) and use an “intercept” program to find the point of intersection of the two graphs. This point has the approximate coordinates (41.9, 700). Therefore, an infant has a brain weighing 700 g when the circumference measures 41.9 cm.

114

Chapter 2 NONLINEAR FUNCTIONS

66. (a) a =

(d) From part (b),

k d

y = -9.5037(55) 2 + 12, 771(55) + 383, 087 » 1, 057, 000

k = ad d a k = ad 36.000 9.37 337.32 36.125 9.34 337.4075 36.250 9.31 337.4875 36.375 9.27 337.19625 36.500 9.24 337.26 36.625 9.21 337.31625 36.750 9.18 337.365 36.875 9.15 337.40625 37.000 9.12 337.44 We find the average of the nine values of k by adding them and dividing by 9. This gives 337.35, or, rounding to the nearest integer, k = 337. Therefore, 337 a = . d

From part (c), y = -30.486(55)3 + 2224.2(55) 2 - 28,111(55) + 472,959 » 583, 000

(e) The cubic seems to be a better fit. 68. (a) Using a graphing calculator with the given data, for the 4.0-foot pendulum and for n = 1, 4.0 = k (2.22)1 k = 1.80;

n = 2, 4.0 = k (2.22)2

(b) When d = 40.50, 337 a = » 8.32. 40.50

k = 0.812; n=3

The strength for 40.50 diopter lenses is 8.32 mm of arc.

4.0 = k (2.22)3 k = 0.366.

(b)

67. (a)

(b) y = -9.5037t 2 + 12, 771t + 383, 087

From the graphs, L = 0.812T 2 is the best fit.

The period will be 2.48 seconds. (d) L = 0.812T 2

If L doubles, T 2 doubles, and T increases by factor of 2. (e) L » 0.822T 2 which is very close to L = 0.812T 2.

Section 2.4

115

2.4 Exponential Functions

True

Your Turn 1

The graph of y = 3x is the graph of an exponential

x /2

= 125

function y = a x with a > 1.

x +3

This is graph E.

= (53 ) x +3

2 x /2

(5 )

5 x = 53x +9 x = 3x + 9 2 x = -9

The graph of y = 3-x is the graph of y = 3x reflected in the y-axis. This is graph D.

x =-2

Your Turn 2

(

)

(

1- x

( 13 )

is the graph

of y = (3-1)1- x or y = 3x-1. This is

r tm A = P 1+ m

the graph of y = 3x translated 1 unit to the right.

)

0.0325 5(4) 4

= 4400 1 +

The graph of y =

This is graph C.

= 5172.97

The interest amounts to 5172.97 - 4400 = $772.97.

Your Turn 3

The graph of y = 3x +1 is the graph of y = 3x translated 1 unit to the left. This is graph F.

A = Pert

= 800e4(0.0315) = 907.43

The amount after 4 years will be $907.43.

The graph of y = 3(3) x is the same as the graph of y = 3x +1. This is the graph of y = 3x translated 1 unit to the left. This is graph F.

2.4 Warmup Exercises

10. The graph of y =

W1.

2 

2

W2.

3 

 3 (5)(7 x 3)  335 x 15

3 2 x

5 7 x 3

 (3)( 2 x )

2

( 13 ) is the graph

of y = (3-1) x = 3-x. This is the graph

6 x

of y = 3x reflected in the y-axis. This is graph D. 11. The graph of y = 2 - 3-x is the same as the

2 x 1

W3. 5  5 x

 5x 2 x 1  53 x 1

graph of y = -3-x + 2. This is the graph of y = 3x reflected in the x-axis, reflected in the y-axis, and translated up 2 units.

1 x a  b  x a  x  b  x a (  b )  x a b W4. x

This is graph A. 12. The graph of y = -2 + 3-x is the same as

2.4 Exercises 1.

False. A function cannot be both an exponential function and a polynomial function.

True

False. The base for an exponential function must be positive.

the graph of y = 3-x - 2. This is the graph of y = 3x reflected in the y-axis and translated 2 units downward. This is graph B.

116

Chapter 2 NONLINEAR FUNCTIONS

13. The graph of y = 3x -1 is the graph of y = 3x translated 1 unit to the right.

22.

(52 ) x = (53 ) x + 2

This is graph C.

52 x = 53x + 6 2 x = 3x + 6 -6 = x

15.

number of folds layers of paper

5 ...

10 ...

32 ...

1024 ...

250

23.

24 x +12 = 212 x-30 4 x + 12 = 12 x - 30

16. 500 sheets are 2 inches high

42 = 8 x 21 = x 4

500 250 = 2 in. x in. x =

2 ⋅ 250 500 12

= 4.503599627 ´ 10 = 71, 079,539.57 mi

in.

-6 x = -x + 5 -5 x = 5 x = -1

25.

3x = 1 81 x 3 = 14 3

26.

ex =

1 e5

e x = e-5 x = -5

21.

4 x = 8 x +1

2| x| = 8 2| x| = 23 | x| = 3 x = 3 or x = -3

3x = 3-4 x = -4

20.

e-x = (e4 ) x + 3 e-x = e4 x +12 -x = 4 x + 12 -5x = 12 x = - 12 5

4 x = 64 4 x = 43 x =3

19.

(e3 )-2 x = e-x + 5 e-6 x = e-x + 5

2 = 2 x =5

18.

24.

2 x = 32 x

16 x +3 = 642 x -5 (24 ) x +3 = (26 ) 2 x-5

250 = 1.125899907 ´ 1015

17.

25 x = 125 x + 2

27.

5-| x | = 1 25 5-| x | = 5-2 | x| = 2 x = 2 or x = -2

(22 ) x = (23 ) x +1 22 x = 23x +3 2 x = 3x + 3 -x = 3 x = -3

Section 2.4

117 æ öx - 4 2 x 2-4 x = çç 1 ÷÷÷ è 16 ø

28.

33. Graph of y = 5e x + 2

2 x 2-4 x = (2-4 ) x- 4 2 x 2-4 x = 2-4 x +16 x 2 - 4 x = -4 x + 16 x 2 - 16 = 0 ( x + 4)( x - 4) = 0 x = -4 or

29.

x = 4

34. Graph of y = -2e x - 3

5x + x = 1 2

5x + x = 50 x2 + x = 0 x( x + 1) = 0 x = 0 or x + 1 = 0 x = 0 or x = -1

30.

35. Graph of y = -3e-2 x + 2

8x2 = 2 x + 4 2

(23 ) x = 2 x + 4 2

23x = 2 x + 4

36. Graph of y = 4e-x / 2 - 1

3x 2 = x + 4 3x 2 - x - 4 = 0 (3x - 4)( x + 1) = 0 x =

31.

4 3

x = -1

27 x = 9 x

2+ x

(33 ) x = (32 ) x

38. 4 and 6 cannot be easily written as powers of the

2+ x

same base, so the equation 4 x = 6 cannot be solved using this approach.

33x = 32 x + 2 x 3x = 2 x 2 + 2 x

40.

0 = 2x2 - x 0 = x(2 x - 1) x = 0 or 2 x - 1 = 0 1 x = 0 or x = 2

f ( x) approaches e » 2.71828.

32.

e x +5 x + 6 = 1 2

e x + 5 x + 6 = e0 x2 + 5 x + 6 = 0 ( x + 3)( x + 2) = 0 x + 3 = 0 or x + 2 = 0 x = -3 or x = -2

118 41.

Chapter 2 NONLINEAR FUNCTIONS

(

A = P 1 + mr

) , P = 10, 000, r = 0.04,

(b) semiannually, m = 2 4(2) æ 0.06 ö÷ A = 26, 000 çç1 + ÷ çè 2 ø÷

t =5

(a) annually, m = 1

= 26, 000(1.03)8

æ ö5(1) A = 10, 000 çç1 + 0.04 ÷÷ è 1 ø

= $32.936.02 Interest = $32.936.02 - $26, 000 = $6936.02

= 10, 000(1.04) = $12,166.53 Interest = $12,166.53 - $10.000 = $21, 66.53

(b) semiannually, m = 2 æ ö5(2) A = 10, 000 çç1 + 0.04 ÷÷ è 2 ø

= 26, 000(1.015)16 = $32,993.62

= 10, 000(1.02)10

Interest = $32,993.62 - $26, 000 = $6993.62

= $12,189.94 Interest = $12,189.94 - $10, 000

(d) monthly, m = 12

= $2189.94

æ 0.06 ö÷ A = 26, 000 çç1 + ÷ çè 12 ø÷

æ ö5(4) A = 10, 000 çç1 + 0.04 ÷÷ è 4 ø = 10, 000(1.01)

= 26, 000(1.005)48

= $33, 032.72

= $12, 201.90

Interest = $33, 032.72 - $26, 000

Interest = $12, 201.90 - $10, 000 = $2201.90

= $7032.72

(e)

(d) monthly, m = 12 æ ö5(12) A = 10, 000 çç1 + 0.04 ÷÷÷ è 12 ø

A = 26, 000e(0.06)(4) = $33, 052.48 Interest = $33, 052.48 - $26, 000 = $7052.48

= 10, 000(1.003)60 = $12, 209.97

43. For 6% compounded annually for 2 years,

Interest = $12, 209.97 - $10, 000 = $2209.97

(e)

42.

A = 10, 000e(0.04)(5) = $12, 214.03 Interest = $12, 214.03 - $10, 000 = $2214.03

(

A = P 1 + mr

4(12)

) , P = 26, 000, r = 0.06,

A = 18, 000(1 + 0.06)2 = 18, 000(1.06) 2 = 20, 224.80

For 5.9% compounded monthly for 2 years, 12(2) æ 0.059 ö÷ A = 18, 000 çç1 + ÷ çè 12 ÷ø

t = 4

æ 12.059 ÷ö24 = 18, 000 çç çè 12 ÷÷ø

(a) annually, m = 1 æ ö4(1) A = 26, 000 çç1 + 0.06 ÷÷÷ è 1 ø = 26, 000(1.06)4 = $32,824.40 Interest = $32,824.40 - $26, 000 = $6824.40

= 20, 248.54

The 5.9% investment is better. The additional interest is $20, 248.54 - $20, 224.80 = $23.74.

Section 2.4 44.

119

æ ötm A = P çç1 + r ÷÷÷ , P = 5000, A = $6100, è mø t =5

(a) m = 1 5(1) æ rö 6100 = 5000 çç1 + ÷÷÷ çè 1ø

1.22 = (1 + r )5 1/ 5

(1.22 )

56 æ 1240 rö = çç1 + ÷÷÷ çè 600 4ø 56 æ 31 rö = çç1 + ÷÷÷ çè 15 4ø

-1 = r

0.0406 » r

The interest rate is about 4.06%. (b) m = 4 æ ö5(4) 6100 = 5000 çç1 + r ÷÷ è 4ø æ ö20 1.22 = çç1 + r ÷÷÷ è 4ø (1.22)1/ 20 - 1 = r 4 1/ 20 é ù 4 êë (1.22) - 1úû » r 0.0400 = r The interest rate is about 4.00%.

45.

(14)(4) æ rö 47. 1240 = 600 ççç1 + ÷÷÷ è 4ø

æ 31 ö1/ 56 r = çç ÷÷÷ çè 15 ø 4

æ 31 ö1/ 56 4 + r = 4 çç ÷÷÷ çè 15 ø æ 31 ö1/ 56 r = 4 çç ÷÷÷ -4 çè 15 ø r » 0.0522 The required interest rate is 5.22%. (4)(12) æ rö 48. (a) 35,000 = 10,500 çç1 + ÷÷÷ çè 4ø 48 æ rö 10 = çç 1 + ÷÷÷ çè 3 4ø

æ 10 ö1/ 48 4 + r = 4 çç ÷÷÷ çè 3 ø

A = Pert

(a) r = 3%

æ 10 ö1/ 48 -4 r = 4 çç ÷÷÷ çè 3 ø

A = 10e0.03(3) = $10.94

(b) r = 4% A = 10e

(c)

46.

r » 0.1016 0.04(3)

= $11.27

The required interest rate is 10.16%.

r = 5% A = 10e

0.05(3)

(12)(12) æ r ö÷ (b) 35,000 = 10500 çç1 + ÷ çè 12 ÷ø

= $11.62

P = $25, 000, r = 5.5% Use the formula for continuous compounding,

A = Pert . (a)

t =1 A = 25, 000e0.055(1) = $26, 413.52

(b)

t =5 A = 25, 000e

(12)(12) æ 10 r ö÷ = çç1 + ÷ çè 3 12 ø÷ é æ ö1/144 ù 10 r = 12 êê çç ÷÷÷ - 1 úú » 0.1008 êë çè 3 ø úû

The required interest rate is 10.08%. 49. (a) The cost in 10 years will be

0.055(5)

= $32,913.27

(c)

æ 10 ö1/ 48 r = çç ÷÷÷ çè 3 ø 4

t = 10

(165, 000)(1.024)10  209,162.35 or $209,162.

(b) The cost in 5 years will be (50)(1.024)5  56.294995 or $56.29.

A = 25, 000e0.055(10) = $43,331.33

120 50.

Chapter 2 NONLINEAR FUNCTIONS

5(2) 10 æ æ jö jö A = 1000 çç1 + ÷÷÷ = 1000 çç1 + ÷÷÷ çè èç 2ø 2ø

j = 1.04 2 j = 0.04 2 j = 0.08 or 8%

This represents the amount in Bank X on January 1, 2018.

0.1 = 1.25, is choice (a). The ratio kj = 0.08

tm æ rö A = P çç1 + ÷÷÷ çè mø

æ ötm A = P çç1 + r ÷÷÷ è mø 10 ù é æ ö3(4) jö æ = êê 1000 ççç1 + ÷÷÷ úú çç1 + k ÷÷÷ è 2 ø ûú è 4ø ëê

51. (a)

10 æ ö12 jö æ = 1000 ççç1 + ÷÷÷ çç1 + k ÷÷÷ è 2ø è 4ø

The GDP appears to grow exponentially.

This represents the amount in Bank Y on January 1, 2021, $1990.76. tm

æ rö A = P çç1 + ÷÷ èç m ø÷

8.4

æ kö = 1000 çç1 + ÷÷ èç 4 ÷ø

(b) f (t ) = f 0at f (0) = f0a 0 = 103, so f 0 = 103 The year 2019 corresponds to t = 79.

32 æ kö = 1000 çç1 + ÷÷÷ çè 4ø

f (79) = 103a 79 = 21, 429 æ 21, 429 ö÷1/ 79 » 1.0699 a = ççç ÷ è 103 ø÷

This represents the amount he could have had from January 1, 2013, to January 1, 2021, at a rate of k per annum compounded quarterly, $2203.76. So,

Thus f (t ) = 103(1.0699)t .

10 æ ö12 jö æ 1000 ççç1 + ÷÷÷ çç1 + k ÷÷÷ = 1990.76 è 2ø è 4ø

(d)

æ ö32 and 1000 çç1 + k ÷÷÷ = 2203.76. è 4ø 32 æ kö çèç1 + 4 ÷ø÷÷ = 2.20376

1 + k = ( 2.20376 ) 4 k 1 + = 1.025 4 k = 0.025 4 k = 0.1 or 10% 1/32

Substituting, we have 10 12 æ jö æ 1.0 ö÷ = 1990.76 1000 çç1 + ÷÷ çç1 + ÷ èç 2 ø÷ èç 4 ø÷ 10 æ jö 1000 çç1 + ÷÷÷ (1.025)12 = 1990.76 çè 2ø

æ ö10 çç1 + j ÷÷ = 1.480 çè 2 ø÷ 1+

The graph shows that the GDP will be approximately equal to (2)(21,429) = 42,858 when t is about 89, that is, in 2029. 52. (a) y(t ) = y0 2t / k y(0) = y0 20/ k = 10, 000, so y0 = 10, 000 For t = 5, f (5) = 10, 000(2)5/ k = 20, 000 25/ k = 2 ln 25/ k = ln 2 5 =1 k k =5

Thus, y(t ) = 10, 000(2)t /5.

j = (1.480)1/10 2

Section 2.4

121

(b) From part (a), y(t ) = 10, 000(2)t /5

f (t ) = C ⋅ at

2t /5 = (21/5 )t » 1.149t

f (7) = C ⋅ a 7

f (75) = C ⋅ a 75

Thus, y(t ) = 10, 000(1.149)t

1.1 = C ⋅ a 7 9.5 = C ⋅ a 75 C = 1.1 C = 9.5 a7 a 75 1.1 = 9.5 a7 a 75

(c) The bacteria double every 5 hours, so it will take 10 hours for 10,000 bacteria to grow to 40,000. (d) From part (a),

(b)

a 75 = 9.5 1.1 a7 a 68 = 9.5 1.1

y(24) = 10, 000(2) 24/5 » 279, 000 From part (b), y(24) = 10, 000(1.149)24 » 280, 000

53.

A(t ) = 3184e

æ ö1/ 68 a = çç 9.5 ÷÷÷ » 1.0322 è 1.1 ø

0.0158t

Then,

(a) 1970: t = 10

C = 1.1 = 1.1 7 » 0.881 a7 1.0322

A(10) = 3184e0.0158(10) = 3184e0.158 » 3729 The function gives a population of about 3729 million in 1970. This is very close to the actual population of about 3701 million.

(b) 2015: t = 55 A(55) = 3184e0.0158(55) » 7592 The function gives a population of 7592 million in 2015. This is close to the actual population of about 7383 million.

Thus, f (t ) = 0.881(1.032)t . (c) f (22) = 0.881(1.032)22 » 1.76 f (45) = 0.881(1.032)45 » 3.64

For age 22, the function gives an estimate of 1.76, and the actual value is 1.4. For age 45, the function gives an estimate of 3.64, and the actual value is 2.9. These estimates are fairly close the the values in the data. (d) The exponential regression function is Y = 0.723(1.034)t . This is close to the function found in part (b).

54. (a)

55. (a) Hispanic population: h(t ) = 46.82(1.0151)t h(16) = 46.82(1.0151)16 » 59.51 The projected Hispanic population in 2016 is 59.51 million, which is slightly more than the actual value of 57.47 million.

(b) Asian population: h(t ) = 14.83(1.0159)t h(16) = 14.83(1.0159)16

Yes, fitting an exponential curve to the data seems reasonable.

» 19.09 The projected Asian population in 2016 is 19.09 million, which is close to the actual value of 18.32 million.

122

Chapter 2 NONLINEAR FUNCTIONS Annual Asian percent increase: 1.0159 - 1 = 0.0159 = 1.59% The Asian population is growing at a slightly faster rate. (d) Black population: b(t ) = 0.3985t + 36.87 b(16) = 0.3985(16) + 36.87 » 43.25 The projected Black population in 2016 is 43.25 million, which is extremely close to the actual value of 43.00 million. (e) Hispanic population: Double the actual 2016 value is 2(57.47) = 114.94 million.

56.

A(t ) = 1.2(10)-0.0788t

(a) A(10) = 1.2(10)-0.0788(10) » 0.196 The amount present in 10 days will be 0.196 gram. (b) 0.012 = 1.2(10)-0.0788t 0.012 = 10-0.0788t 1.2 10-2 = 10-0.0788t -2 = -0.0788t t » 25.4 It will take 25.4 days to reduce the substance to 0.012 gram.

57. (a) When x = 0, P = 1013. When x = 10,000, P = 265.

First we fit P = a ⋅ b x . 1013 = a ⋅ b0 a = 1013

The doubling point is reached when t » 60, or in the year 2060. Asian population: Double the actual 2016 value is 2(18.32) = 36.64 million.

P = 1013(b) x 265 = 1013(b)10,000 265 = b10,000 1013 æ ö1/10,000 b = çç 265 ÷÷÷ è 1013 ø b » 0.999866

Therefore P = 1013(0.999866) x . Next we fit P = mx + b. The doubling point is reached when t » 57, or in the 2057. Black population: Double the actual 2016 value is 2(43.00) = 86 million.

We use the points (0, 1013) and (10,000, 265). 265 - 1013 m= = -0.0748 10, 000 - 0 b = 1013 Therefore P = -0.0748 x + 1013. Finally, we fit P =

1 a(0) + b 1 » 9.87 ´ 10-4 b = 1013 1 P = 1 ax + 1013

1013 =

The doubling point is reached when t » 123, or in the year 2123.

1 . ax + b

Section 2.4

123 265 =

For (49, 36,341,912),

1 1 10, 000a + 1013

y = at 2 + b 36,341,912 = a(49)2 + b b = 36,341,912 - 2401a

1 1 = 10,000a + 265 1013 1 1 10,000a = 265 1013 1 - 1 a = 265 1013 » 2.79 ´ 10-7 10, 000 Therefore, 1 P = . -7 (2.79 ´ 10 ) x + (9.87 ´ 10-4 )

Then 300 - 16a = 36,341,912 - 2401a 2385a = 36,341, 612 a =

36,341, 612 » 15, 237.6 2385

And b = 300 - 16a = 300 - 16(15, 237.6) » -243,501.6

(b)

Therefore, y = 15, 237.6t 2 - 243,501.6. Finally we fit y = a ⋅ bt . y = a ⋅ bt

y = a ⋅ bt

300 = a ⋅ b 4

36,341,912 = a ⋅ b 49

36,341,912 a = 300 a = 4 b b 49 300 = 36,341,912 b4 b 49

P = 1013(0.999866) x is the best fit.

b 49 = 36,341,912 300 b4 36,341,912 45 b = 300

We predict that the pressure at 1500 meters will be 829 millibars, and at 11,000 meters will be 232 millibars. (d) Using exponential regression, we obtain

æ 36,341,912 ö÷1/ 45 » 1.29707 b = çç ÷÷ è ø 300

P = 1038(0.99998661) x which is very close to the function found in part (b).

Then, 300 a = 300 = » 105.9907 4 b 1.29707 4

58. (a) First we fit y = mt + b.

Use the points (4, 300) and (49, 36,341,912) to find m. 36,341,912 - 300 m= » 807,591.4 49 - 4 y - 300 = 807,591.4( x - 4) y = 807,591.4 x - 3, 230, 066

Therefore, y = 105.9907(1.29707)t . (b)

Therefore, y = 807,591.4t - 3, 230, 066. Next we fit y = at 2 + b. Use the points (4, 300) and (49, 36,341,912) to find a. For (4, 300), y = at 2 + b

The function y = 105.9907(1.29707)t is the best fit. (c) t = 55 corresponds to 2025. y = 105.9907(1.29707)55

300 = a(4) 2 + b

= 173, 083, 458.7 » 173, 083, 000

b = 300 - 16a

124

Chapter 2 NONLINEAR FUNCTIONS (d) The regression function is

197, 026 596,556 = 10 b b18

Y = 53.7179(1.31261) . This is close to the function in part (b).

b18 = 596,556 197, 026 b10 596,556 b8 = 197, 026

59. (a) First we fit C = mt + b. Use the points (10, 197,026) and (18, 596,556) to find m. 596,556 - 197, 026 m= = 49,941.25 18 - 10

æ 596,556 ö÷1/8 » 1.148526 b = çç çè 197, 026 ÷÷ø

Use the point (10, 197,026) to find b.

Then,

y = 49,941.25t + b

a =

197, 026 = 49,941.25(10) + b b = -302,386.5

197, 026 197, 026 = » 49,330.5 10 b 1.14852610

Therefore, C = 49,330.5(1.148526)t .

Therefore, C = 49,941.25t - 302,386.5 (b)

Next we fit C = at 2 + b . Use the points (10, 197,026) and (18, 596,556) to find a and b. For (10, 197,026), y = at 2 + b

The quadratic function is the best fit.

197, 026 = a(10) 2 + b b = 197, 026 - 100a

For (18, 596,556), y = at 2 + b

C  51,815.6(1.14906)t .

596,556 = a(18)2 + b

(d) t  19 corresponds to 2019. C = 49,941.25t - 302,386.5 = 49,941.25(19) - 302,386.5 » 646, 497

b = 596,556 - 324a

Then 197, 026 - 100a = 596,556 - 324a 224a = 399,530 a =

399,530 » 1783.616 224

And b = 197, 026 - 100a = 197, 026 - 100(1783.616) » 18, 664.4

C = 49,330.5(1.148526)t = 49,330.5(1.148526)19 » 685,159 C = 1402.75t 2 + 11,578.1t - 60,590

Finally we fit C = a ⋅ bt . y = a ⋅ bt

y = a ⋅ bt

197, 026 = a ⋅ b10

596,556 = a ⋅ b18

197, 026 b10

= 1783.616(19)2 + 18, 664.4 » 662,550

Therefore, C = 1783.616t 2 + 18, 664.4.

a =

C = 1783.616t 2 + 18, 664.4

a =

596,556 b18

= 1402.75(19)2 + 11,578.1(19) - 60,590 » 665, 787 C = 51,815.6(1.14906)t = 51,815.6(1.14906)19 » 726, 059

Section 2.5

125

2.5 Logarithmic Functions

60. (a)

Your Turn 1

æ ö 5-2 = 1 means log5 çç 1 ÷÷÷ = -2 è 25 ø 25 The emissions appear to grow exponentially. (b)

f (t )  f0 a

f 0  316.19

Use the point (59, 411.76) to find a. 411.76  316.19a59 a59 

æ ö log3 çç 1 ÷÷÷ è 81 ø We seek a number x such that 3x = 1 81 3x = (3)-4

411.76 316.19

a  59

Your Turn 2

x = -4

411.76  1.0044863 316.19

f (t )  316.19(1.0044863)t

Your Turn 3 æ x 2 ö÷ ç log a çç 3 ÷÷÷ = log a x 2 - log a y 3 çè y ø÷ = 2 log a x - 3log a y

Your Turn 4 log3 50 = ln 50 » 3.561 ln 3

Your Turn 5

The doubling point is reached when x  154.8. The first year in which emissions equal or exceed that threshold is 1960 + 155, or 2115.

log 2 x + log 2 ( x + 2) = 3 log 2 x( x + 2) = 3 x( x + 2) = 23 x 2 + 2x = 8

61. Use the points (1, 1000) and (2, 10,000). x

We fit y = a ⋅ b . y = a ⋅ bt

y = a ⋅ bt

1000 = a ⋅ b1

10, 000 = a ⋅ b 2

10, 000 a = 1000 a = b b2 1000 = 10, 000 b b2 b 2 = 10, 000 b 1000 b = 10

x2 + 2x - 8 = 0 x = -4 or x = 2 Since the logarithm of a negative number is undefined, the only solution is x = 2.

Your Turn 6 2 x +1 = 3x ln 2 x +1 = ln 3x ( x + 1) ln 2 = x ln 3 x ln 2 + ln 2 = x ln 3 x ln 3 - x ln 2 = ln 2

Then, a = 1000 = 1000 = 100 b 10 x

Therefore, y = 100(10) .

x(ln 3 - ln 2) = ln 2

( )

x ln 32 = ln 2 x = ln 2 » 1.7095 ln1.5

126

Chapter 2 NONLINEAR FUNCTIONS

Your Turn 7

e0.025 x = (e0.025 ) x » 1.0253x

53 = 125 Since a y = x means y = log a x, the equation in logarithmic form is

2.5 Warmup Exercises W1.

log5 125 = 3.

x2  x  3  9

x2  x  6  0

Since a y = x means y = log a x, the equation in logarithmic form is

( x  3)( x  2)  0 x3 0

x2 0

x3

x  2

W2. Assume that x is not equal to 5. x  4 x  20 3x  20

log 7 49 = 2.

x  4 implies x5

34 = 81 The equation in logarithmic form is log3 81 = 4.

20 x 3

W3. 2  8 x 1   23 

7 2 = 49

x 1

8.  23 x  3

27 = 128

Therefore,

Since a y = x means y = log a x, the equation in logarithmic form is

21  23 x  3

log 2 128 = 7.

1  3x  3 2  3 x

x

2 3

3-2 =

1 9

The equation in logarithmic form is log3

W4. e x  4  1  e0

Therefore, x4 0

10.

x4

True

False. For a logarithm, y = log a x, x > 0, a > 0, and a ¹ 1.

æ 5 ö÷-2 çç ÷ = 16 çè 4 ø÷ 25

Since a y = x means y = log a x, the equation in logarithmic form is

2.5 Exercises

1 = - 2. 9

log5/4

11.

False. log a ( x + y) cannot be simplified. (log a x)(log a y ) cannot be simplified. Also, log a ( x + y) ¹ (log a x)(log a y) . The property states log a xy = log a x + log a y . False. (log a x) r = (log a x)(log a x) (log a x)  r factors The property states log a x r = r log a x .

16 = -2. 25

log 2 32 = 5

Since y = log a x means a y = x , the equation in exponential form is 25 = 32.

12.

log3 81 = 4

Since y = log a x means a y = x , the equation in exponential form is 34 = 81.

Section 2.5 13.

127

1 = -1 e

21.

log 2

The equation in exponential form is e-1 =

14.

log 2

1 . e

2 x = 2-4

x = -4

1 = -3 8

22.

The equation in exponential form is 2-3 =

15.

1 = x 16 1 2x = 16

log3

1 = x 81

3x =

1 . 8

1 81

3x = 3-4 x = -4

log 100, 000 = 5 log10 100, 000 = 5 105 = 100, 000

23.

log 2 3

When no base is written, log10 is understood. 16.

1 = x 4

æ ö1/3 2 x = çç 1 ÷÷ è4ø

log 0.001 = -3

æ ö1/3 2 x = çç 12 ÷÷ çè 2 ÷ø

log10 0.001 = -3 10-3 = 0.001

2 x = 2-2/3 x = -2 3

When no base is written, log10 is understood. 17. Let log8 64 = x. Then, 8 x = 64 8 x = 82

x = 2. Thus, log8 64 = 2.

18.

Let log9 81 = x. Then, 9 x = 81 9 x = 92 x = 2.

Thus log9 81 = 2. 19.

log 4 64 = x

24.

log8 4

1 = x 2 8x = 4

(23 ) x = 2-1/4 1 3x = 4 1 x =12

25. ln e = x

Recall that ln means log e . ex = e x =1

4 = 64 4 x = 43 x =3

20.

log3 27 = x

26. ln e3 = x

Recall that ln means log e .

3 = 27 x

æ 1 ö1/4 1 = çç ÷÷÷ çè 2 ø 2

e x = e3

3 =3 x =3

x =3

128 27.

Chapter 2 NONLINEAR FUNCTIONS ln e5/3 = x

37.

logb 32 = logb 25 = 5logb 2 = 5a

log e e5/3 = x x = 5 3

38. 28. ln ln1 = x

= logb (2 ⋅ 9)

ex = 1 x

e =e

= logb (2 ⋅ 32 )

= logb 2 + logb 32

x =0

= logb 2 + 2 logb 3

29. The logarithm to the base 3 of 4 is written log3 4. The subscript denotes the base. 31.

log5 (3k ) = log5 3 + log5 k

32.

log9 (4m) = log9 4 + log9 m

33.

3p log3 5k

= a + 2c

39.

logb 72b = logb 72 + logb b = logb 72 + 1 = logb 23 ⋅ 33 + 1 = logb 23 + logb 32 + 1 = 3 logb 2 + 2 logb 3 + 1 = 3a + 2c + 1

= log3 3 p - log3 5k = (log3 3 + log3 p) - (log3 5 + log3 k ) = 1 + log3 p - log3 5 - log3 k

34.

logb 18

40.

logb (9b 2 ) = logb 9 + logb b 2 = logb 32 + logb b 2

15 p log 7 7y

= 2 logb 3 + 2 logb b = 2c + 2(1)

= log7 15 p - log7 7 y

= 2c + 2

= (log7 15 + log7 p) - (log7 7 + log7 y ) = log7 15 + log7 p - log7 7 - log7 y = log7 15 + log7 p - 1 - log7 y

35.

41.

3 5 ln 3 6 = ln 3 ⋅ 51/2 - ln 61/3 = ln 3 + ln 5

1/3

3.4012 1.6094

- ln 6

93 5

ln 4 3

ln 210 ln 12 » 2.152

42.

log12 210 =

43.

log1.2 0.95 =

44.

log 2.8 0.12 =

= ln 3 + 1 ln 5 - 1 ln 6 2 3

36.

ln 30 ln 5

» 2.113

= ln 3 5 - ln 3 6 1/2

log5 30 =

ln 0.95 ln 1.2 » -0.281

= ln 93 5 - ln 4 3 = ln 9⋅ 51/3 - ln 31/4 = ln 9 + ln 51/3 - ln 31/4 = ln 9 + 1 ln 5 - 1 ln 3 3 4

ln 0.12 ln 2.8 » -2.059

Section 2.5

129

log x 36 = -2

45.

51.

log5 (9 x - 4) = 1

x-2 = 36

51 = 9 x - 4 9 = 9x

( x-2 )-1/ 2 = 36-1/2

1= x

1 6

x =

52. 46.

log9 27 = m

log 4 x - log 4 ( x + 3) = -1 x = -1 x+3 x 4-1 = x+3 x 1 = x+3 4 4x = x + 3

log 4

9m = 27 (32 )m = 33 32m = 33 2m = 3 m=

3 2

3x = 3 x =1

47.

log8 16 = z

53.

8 z = 16

log9 m - log9 (m - 4) = -2

(23 ) z = 24

log9

23 z = 24 3z = 4 z =

4 3

3 4

log y 8 =

48.

y 3/4 = 8 (y

3/4 4/3

)

This value is not possible since log9 (-0.05) does not exist. Thus, there is no solution to the original equation.

4/3

y = (81/3)4 y = 24 = 16

49.

log r 5 =

1 2

m = -2 m-4 m 9-2 = m-4 m 1 = m-4 81 m - 4 = 81m -4 = 80m -0.05 = m

54.

log ( x + 5) + log ( x + 2) = 1 log [( x + 5)( x + 2)] = 1 ( x + 5)( x + 2) = 101 x 2 + 7 x + 10 = 10

r1/2 = 5

x2 + 7 x = 0 x( x + 7) = 0

(r1/2 )2 = 52 r = 25

x = 0 or x = -7

50.

log 4 (5 x + 1) = 2 42 = 5x + 1 16 = 5x + 1 5x = 15

x = -7 is not a solution of the original equation because if x = -7, x + 5 and x + 2 would be negative, and the domain of y = log x is (0, ¥). Therefore, x = 0.

x =3

130 55.

Chapter 2 NONLINEAR FUNCTIONS log3 ( x - 2) + log3 ( x + 6) = 2

60.

log3 [( x - 2)( x + 6)] = 2

ln( x + 1) - ln x = 1 x +1 =1 x x +1 = e1 x x + 1 = ex

( x - 2)( x + 6) = 32 x 2 + 4 x - 12 = 9 x 2 + 4 x - 21 = 0 ( x + 7)( x - 3) = 0

ex - x = 1

x = -7 or x = 3

x(e - 1) = 1

x = –7 does not check in the original equation. The only solution is 3. 56.

log3 ( x 2 + 17) - log3 ( x + 5) = 1

log3

x =

61.

2x = 6 ln 2 x = ln 6

x 2 + 17 =1 x+5 31 =

1 » 0.5820 e -1

x ln 2 = ln 6 2

x + 17 x+5

ln 6 » 2.5850 ln 2

x =

3x + 15 = x 2 + 17 0 = x 2 - 3x + 2 0 = ( x - 1)( x - 2) x = 1 or x = 2

57.

62.

x log 5 = log12 log12 log 5 » 1.5440

x =

log 2 ( x 2 - 1) - log 2 ( x + 1) = 2 log 2

x2 - 1 = 2 x +1

63.

(k - 1) ln e = ln 6 k -1 =

ln 6 ln e

ln 6 1 k = 1 + ln 6

k -1 =

» 2.7918

ln(5x + 4) = 2 5x + 4 = e2

64.

5 x = e2 - 4

2 y ln e = ln 15 2 y (1) = ln 15 ln15 2 » 1.3540

ln x + ln 3x = -1

y =

ln 3x 2 = -1 3x 2 = e-1 x2 =

e-1 3

x =

e-1 = 3

e2 y = 15 ln e2 y = ln 15

e2 - 4 » 0.6778 x = 5

59.

e k -1 = 6 ln e k -1 = ln 6

x2 - 1 2 = x +1 ( x - 1)( x + 1) 4= x +1 4 = x -1 x =5 2

58.

5x = 12

1 » 0.3502 3e

Section 2.5

131 3x +1 = 5 x

65.

72.

e-4 x = (e-4 ) x » 0.0183x

73.

f ( x) = log (5 - x)

ln 3x +1 = ln 5 x ( x + 1) ln 3 = x ln 5 x ln 3 + ln 3 = x ln 5 x ln 5 - x ln 3 = ln 3

5- x > 0

x(ln 5 - ln 3) = ln 3

x<5

-x > - 5

ln 3 x = » 2.1507 ln(5/3)

66.

x +1

The domain of f is x < 5. 74.

x -1

f ( x) = ln ( x 2 - 9)

Since the domain of f ( x) = ln x is (0, ¥),

ln 2 x +1 = ln 6 x-1

the domain of f ( x) = ln ( x 2 - 9) is the set of all real numbers x for which

( x + 1) ln 2 = ( x - 1) ln 6 x ln 2 + ln 2 = x ln 6 - ln 6 x ln 6 - x ln 2 = ln 2 + ln 6

x 2 - 9 > 0.

x(ln 6 - ln 2) = ln 2 + ln 6

To solve this quadratic inequality, first solve the corresponding quadratic equation.

x ln 3 = ln12 x =

x2 - 9 = 0

ln12 » 2.2619 ln 3

( x + 3)( x - 3) = 0 x + 3 = 0 or x - 3 = 0

5(0.10) = 4(0.12)

67.

x = -3 or

ln [5(0.10) ] = ln [4 (0.12) ]

These two solutions determine three intervals on the number line: (-¥, -3), (-3, 3), and (3, ¥).

ln 5 + x ln 0.10 = ln 4 + x ln 0.12 x (ln 0.12 - ln 0.10) = ln 5 - ln 4

If x = -4, (-4 + 2) (-4 - 2) > 0.

ln 5 - ln 4 x = ln 0.12 - ln 0.10 ln1.25 = ln1, 2 » 1.2239 1.5 (1.05) x = 2 (1.01) x

68.

ln [1.5 (1.05) x ] = ln [2 (1.01) x ] ln 1.5 + x ln 1.05 = ln 2 + x ln 1.01

If x = 0, (0 + 2)(0 - 2) > 0. If x = 4, (4 + 2) (4 - 2) > 0. The domain is x < -3 or x > 3, which is written in interval notation as (-¥, -3) È (3, ¥). 75.

log A - log B = 0 log

x(ln 1.01 - ln 1.05) = ln 1.5 - ln 2 x =

ln 0.75 ln (1.01 / 1.05)

» 7.4069

69. 10 x +1 = e(ln10)( x +1) 2

70. 10 x = e(ln10) x 71.

x =3

A =0 B A = 100 = 1 B A= B

A-B = 0

Thus, solving log A - log B = 0 is equivalent to solving A - B = 0

e3x = (e3 ) x » 20.09 x

132 76.

Chapter 2 NONLINEAR FUNCTIONS (c) Let r = 0.08.

(log( x + 2)) 2 ¹ 2 log( x + 2)

t =

(log( x + 2)) 2 = (log( x + 2))(log( x + 2)) 21og( x + 2) ¹ 2(log x + log 2)

= 9.0 years

21og( x + 2) = log( x + 2)2

(d) Since 0.001 £ 0.03 £ 0.05, for r = 0.03, we use the rule of 70. 70 70 = = 23.3 years 100r 100(0.03)

log 2 ¹ 100 log 2 » 0.30103 because 100.30103 = 2

77.

Since 0.05 £ 0.06 £ 0.12, for r = 0.06, we use the rule of 72. 72 72 = = 12 years 100r 100(0.06)

Let m = log a xy , n = log a x, and p = log a y. Then a m = xy , a n = x, and a p = y.

For r = 0.08, we use the rule of 72. 72 = 9 years 100(0.08)

Substituting gives am =

x an = p = an - p. y a

80. (a) t =

So m = n - p. Therefore, log a

t =

x = log a x - log a y. y

Then, a

= x

(b) t =

and a = x.

It will take 17 years for the compound amount to at least triple. (c) The rule of 72 gives 72 = 10.29 100(0.07)

a m = x r = (a n )r = a nr .

Therefore, m = nr, or log a x r = r log a x.

years as the doubling time, which is close to the answer found in part (a).

79. From Example 8, the doubling time t in years when m = 1 is given by ln 2 . ln (1 + r )

(a) Let r = 0.03. ln 2 ln (1.03)

= 23.4 years

(b) Let r = 0.06. t =

ln 3 ln (1 + 0.07)

t » 16.24

Substituting gives

t =

ln 2 ln (1 + 0.07)

It will take 11 years for the compound amount to be at least double.

t =

ln 2 ln (1 + r )

t » 10.24

78. Let m = log a x r and n = log a x. m

ln 2 ln 1.08

ln 2 ln 1.06

= 11.9 years

81.

A = Pert 1240 = 600e r ⋅14 31 = e14r 15 æ 31 ö ln ççç ÷÷÷ = ln e14r è 15 ø

æ 31 ö ln ççç ÷÷÷ = 14r è 15 ø æ 31 ö ln ççç ÷÷÷ è 15 ø = r 14 0.0519 » r The interest rate should be 5.19%.

Section 2.5

133

82.

r (sec)

0.001

0.02

0.05

0.08

0.12

ln 2 ln(1 + r )

693.5

14.2

9.01

6.12

70 100r

700

8.75

5.83

72 100r

720

14.4

For 0.001 £ r < 0.05, the Rule of 70 is more accurate. For 0.05 < r £ 0.12, the Rule of 72 is more accurate. At r = 0.05, the two are equally accurate. 83.

f (t )  103(1.0702)t where t = 0 corresponds to 1940. 103(1.0702)t  45, 053 45, 053 e(t ) ln(1.0702)  103  45, 053  t ln(1.0702)  ln   103  1  45, 053  t  ln    89.628 ln(1.0702)  103  t  90 corresponds to 2030.

84. After x years at Humongous Enterprises, your salary

85. (a) y(t ) = y0ekt implies that ekt =

1 æ y(t ) ÷ö y(t ) ÷÷ . and thus, k = ln ççç t çè y0 ø÷ y0

(b) Using the data given in Exercise 54 in Section 2.4, and with age 7 corresponding to t  0, y0 = 1.1. The six values of k are: For age 22, t = 22 – 7 = 15 1 æç 1.4 ö÷ ln ç ÷ » 0.0161 15 çè 1.1 ÷ø

For age 35, t = 35 – 7 = 28 1 æç 2.0 ö÷ ln ç ÷ » 0.0214 28 çè 1.1 ÷ø For age 45, t = 45 – 7 = 38 1 æç 2.9 ö÷ ln ç ÷ » 0.0255 38 çè 1.1 ø÷ For age 55, t = 55 – 7 = 48 1 æç 4.6 ö÷ ln ç ÷ » 0.0298 48 çè 1.1 ÷ø For age 65, t = 65 – 7 = 58 1 æç 6.9 ö÷ ln ç ÷ » 0.0317 58 çè 1.1 ÷ø For age 75, t = 75 – 7 = 68 1 æç 9.5 ö÷ ln ç ÷ » 0.0317 68 çè 1.1 ÷ø 86. If the number N is proportional to m-0.6 , where m

would be 45,000 (1 + 0.04) x or 45,000 (1.04) x . After x years at Crabapple Inc., your salary would

is the mass, then N = km-0.6 , for some constant of proportionality k.

be 30,000 (1 + 0.06) x or 30,000 (1.06) x .

Taking the common log of both sides, we have

First we find when the salaries would be equal.

log N = log(km-0.6 ) = log k + log m-0.6 = log k - 0.6 log m.

45, 000(1.04) x = 30, 000(1.06) x (1.04) x (1.06)

30, 000 45, 000

This is a linear equation in log m. Its graph is a straight line with slope -0.6 and vertical intercept log k.

æ 1.04 ö÷x 2 ç çè 1.06 ø÷÷ = 3 æ öx æ ö log çç 1.04 ÷÷÷ = log çç 2 ÷÷÷ è 1.06 ø è3ø æ ö æ ö x log çç 1.04 ÷÷÷ = log çç 2 ÷÷÷ è 1.06 ø è3ø x =

log log

(

( 23 ) 1.04 1.06

87. (a) The total number of individuals in the community is 50 + 50, or 100.

Let P1 =

50 = 0.5, P2 = 0.5. 100

H = -1[ P1 ln P1 + P2 ln P2 ]

)

x » 21.29

= -1[0.5ln 0.5 + 0.5ln 0.5] » 0.693

(b) For 2 species, the maximum diversity is ln 2. (c) Yes, ln 2 » 0.693.

2021 + 21.29 = 2042.29 Therefore, on July 1, 2043, the job at Crabapple, Inc., will pay more.

134 88.

Chapter 2 NONLINEAR FUNCTIONS H = -[P1 ln P1 + P2 ln P2

1 = 2e-3k 1 = e-3k 2 -3k = ln 1 = ln 2-1 = - ln 2 2 k = ln 2 3 C 1 T = ln 2 k C1

+ P3 ln P3 + P4 ln P4 ] H = -[0.521 ln 0.521 + 0.324 ln 0.324 + 0.081 ln 0.081 + 0.074 ln 0.074] H = 1.101

89. (a) 3 species, 13 each: 1 3 H = -( P1 ln P1 + P2 ln P2 + P3 ln P3 ) P1 = P2 = P3 =

5C T = ln12 ln 1 C1 3

æ1 1ö = -3 çç ln ÷÷÷ çè 3 3 ø

3 ln 5 ln 2 T » 7.0 T =

1 3 » 1.099

= - ln

The drug should be given about every 7 hours.

(b) 4 species, 14 each:

92. (a) From the given graph, when x = 10 g,

y » 1.3 cm3 /g/hr, and when x = 1000 g,

P1 = P2 = P3 = P4 = 1 4 H = -( P1 ln P1 + P2 ln P2 + P3 ln P3 + P4 ln P4 ) æ ö = -4 çç 1 ln 1 ÷÷÷ è4 4ø

y » 0.41 cm3 /g/hr. (b) If y = axb , then

ln y = ln (axb ) = ln a + b ln x.

= - ln 1 4 » 1.386

Thus, there is a linear relationship between ln y and ln x. 1.3 = a(10)b

(c)

0.41 = a(1000)b b 1.3 = a(10) 0.41 a(1000)b

and 1 = ln (4-1)-1 = ln 4 » 1.386 4 by Property c of logarithms, so the populations are at a maximum index of diversity. - ln

90.

mX + N = m logb x + logb n = logb x m + logb n = logb nx m

b 1.3 = æç 10 ö÷ ÷ ÷ ç è 1000 ø 0.41 æ ö b ln çç 1.3 ÷÷÷ = ln ( 0.01) è 0.41 ø æ ö ln çç 1.3 ÷÷÷ = b ln(0.01) è 0.41 ø

= logb y

Thus, Y = mX + N.

Substituting this value into 1.3 = a(10)b , 1.3 = a(10)-0.25 1.3 a = » 2.3. (10)-0.25

C (t ) = C0e-kt

When t = 0, C(t) = 2, and when t = 3, C(t) = 1. 2 = C0e-k (0)

1.3 ( 0.41 )

ln (0.01) b » -0.25

91.

Therefore, y = 2.3x-0.25.

C0 = 2

Section 2.5

135

(d) If x = 100,

(d) -0.25

y = 2.3(100) » 0.73. We predict that the basal metabolism for a marsupial carnivore with a body mass of

x = e93.911/8.9603 x » 35, 624 The per capita GDP is about $35,600.

100 g will be about 0.73 cm /g/hr. 93. (a) h(t ) = 46.82(1.0151)t

(e)

Double the 2016 population is 2(57.47) = 114.94 million 114.94 = 46.82(1.0151)t 114.94 = (1.0151)t 46.82 æ 114.94 ö÷ log1.0151 çç =t çè 46.82 ÷÷ø t =

(

114.94 46.82

(f)

ln 1.0151 » 60

(b) a(t ) = 14.83(1.0159)t

t =

36.64 ( 14.83 )

ln 1.0159 » 57.34 The Asian population is estimated to double their 2016 population in 2057.

73 = 8.9603ln x - 13.911 86.911 = 8.9603ln x 86.911 = ln x 8.9603 x = e86.911/8.9603 x » 16,310 The per capita GDP is about $16,310.

The Hispanic population is estimated to double their 2016 population in 2060.

36.64 = 14.83(1.0159)t 36.64 = (1.0159)t 14.83 æ 36.64 ö÷ log1.0159 çç =t çè 14.83 ÷÷ø

40 = 8.9603ln x - 13.911 53.911 = 8.9603ln x 53.911 = ln x 8.9603 x = e53.911/8.9603 x » 410 The per capita GDP is about $410.

)

Double the 2016 population is 2(18.32) = 36.64 million

80 = 8.9603ln x - 13.911 93.911 = 8.9603ln x 93.911 = ln x 8.9603

95.

æ ö C = B log 2 çç s + 1÷÷÷ èn ø C = log æç s + 1÷ö ÷÷ 2 çè ø B n 2C /B = s + 1 n s = 2C /B - 1 n

96. Decibel rating: 10 log II

(a) Intensity, I = 115I 0 æ 115I ö÷ 0÷ 10 log ççç çè I ÷÷ø 0

= 10 log115 » 21

94. U = 8.9603ln x - 13.911 (a) U (59,531.66) = 8.9603ln(59,531.66) - 13.911 » 84.6 (b) U (1547.85) = 8.9603ln(1547.85) - 13.911 » 51.9 (c) U (2342.24) = 8.9603ln(2342.24) - 13.911 » 55.6

(b) I = 9,500,000I 0 æ 9.5 ´ 106 I ö÷ ç 0 ÷ = 10 log 9.5 ´ 106 10 log çç ÷÷ çè I0 ø÷ » 70 I = 1,200,000,000I 0

(c)

æ 1.2 ´ 109 I ö÷ ç 0 ÷ = 10 log 1.2 ´ 109 10 log çç ÷÷ çè I0 ø÷

» 91

136

Chapter 2 NONLINEAR FUNCTIONS (d) I = 895,000,000,000I 0

98.

æ 8.95 ´ 1011 I ÷ö ç 0 ÷ = 10 log 8.95 ´ 1011 10 log çç ÷÷ çè I0 ÷ø » 120

(e) I = 109,000,000,000,000I 0 14 ö æ ç 1.09 ´ 10 I 0 ÷÷ 14 10 log çç ÷÷ = 10 log 1.09 ´ 10 I0 ÷ø çè

» 140

(f) I 0 = 0.0002 microbars 1, 200, 000, 000 I 0 = 1, 200, 000, 000(0.0002) = 240, 000 microbars 895, 000, 000, 000 I 0 = 895, 000, 000, 000(0.0002) = 179, 000, 000 microbars

97. Let I1 be the intensity of the sound whose decibel rating is 85. I (a) 10 log 1 = 85 I0 I log 1 = 8.5 I0 log I1 - log I 0 = 8.5 log I1 = 8.5 + log I 0

Let I2 be the intensity of the sound whose decibel rating is 75. I 10 log 2 = 75 I0 I log 2 = 7.5 I0 log I 2 - log I 0 = 7.5 log I 0 = log I 2 - 7.5

Substitute for I0 in the equation for log I1. log I1 = 8.5 + log I 0 = 8.5 + log I 2 - 7.5 = 1 + log I 2 log I1 - log I 2 = 1

R( I ) = log I I0

(a) R(1, 000, 000 I 0 ) 1, 000, 000 I 0 = log I0 = log 1, 000, 000 = 6 (b) R(1, 000, 000, 000 I 0 ) 1, 000, 000, 000 I 0 = log I0 = log 100, 000, 000 = 8 (c) R( I ) = log I I0 7.1 = log I I0 107.1 = I I0 I » 12, 600, 000I 0 (d) R( I ) = log I I0 6.9 = log I I0 106.9 = I I0 I » 7,940, 000I 0 (e)

12, 600, 000I 0 2017 quake = » 1.6 2018 quake 7,940, 000I 0

The 2017 earthquake had amplitude of more than 1.6 times that of the 2018 earthquake. (f ) R( E ) = 2 log E 3 E0 For the 2017 earthquake 7.1 = 2 log E 3 E0 10.65 = log E E0 E = 1010.65 E0 E = 1010.65 E0

I log 1 = 1 I2

For the 2018 earthquake,

1 I . This means the Then I 1 = 10, so I 2 = 10 1 2

intensity of the sound that had a rating of 75 1 as intense as the sound that had a decibels is 10 rating of 85 decibels.

6.9 = 2 log E 3 E0 10.35 = log E E0 E = 1010.35 E0 E = 1010.35 E0

Section 2.6

137 For black coffee:

The ratio of their energies is 1010.65 E0 1010.35 E0

5 = - log[H + ]

= 100.3 » 2

10-5 = [H + ] 10-5 = 106 = 1, 000, 000 10-11 The coffee has a hydrogen ion concentration 1,000,000 times greater than the laundry mixture.

The 2017 earthquake had an energy about twice that of the 2018 earthquake. (g) Find the energy of a magnitude 6.7 earthquake. Using the formula from part f, 6.7 =

2 E log 3 E0

100. For concert pitch A, f = 440. P = 69 + 12 log 2 (440/440) = 69 + 12 log 2 (1)

E log = 10.05 E0

= 69 + 12 ⋅ 0 = 69

For one octave higher than concert pitch A, f = 880.

E = 1010.05 E0

P = 69 + 12 log 2 (880/440) = 69 + 12 log 2 (2)

E = E01010.05

= 69 + 12 ⋅ 1 = 69 + 12 = 81

For an earthquake that releases 15 times this much energy, E = E0 (15)1010.05. R( E0 (15)1010.05 ) =

æ E (15)1010.05 ö÷ 2 ç ÷÷ log çç 0 ÷÷ 3 E0 çè ø

2 log (15 ⋅ 1010.05 ) 3 » 7.5 So, it’s true that a magnitude 7.5 earthquake releases 15 times more energy than one of magnitude 6.7. =

99.

-7 = log[H + ]

e16k = ln(18/5)

Let A(t ) = 10 A0 and k = -(ln 2/5600)

For acid rain:

A(t ) = A0ekt

4 = - log[H + ]

1 A = A e-(ln 2/5600)t 0 0 10 1 = e-(ln 2/5600)t 10 æ ö ln çç 1 ÷÷÷ = ln e-(ln 2/5600)t è 10 ø

-4 = log[H + ] 10-4 = [H + ]

= 103 = 1000

The acid rain has a hydrogen ion concentration 1000 times greater than pure water. (b) For laundry solution: +

11 = - log[H ] -11

= [H + ]

ln(18/5) » 0.08 16

Your Turn 2

10-7 = [H + ]

18 = 5e k (16)

y = 5e0.08t

7 = - log[H + ]

y = y0ekt

k =

(a) For pure water:

-7

Your Turn 1

16k = ln(18/5)

pH = - log[H+ ]

10-4

2.6 Applications: Growth and Decay; Mathematics of Finance

æ ö ln çç 1 ÷÷÷ = -(ln 2/5600)t è 10 ø æ ö t = - 5600 ln çç 1 ÷÷÷ è 10 ø ln 2 » 18602.80

138

Chapter 2 NONLINEAR FUNCTIONS

Your Turn 3

W3. 10e3t = 45

(a) 4.25% compounded monthly

e3t = 4.5 3t = ln(4.5)

æ ö12 çç1 + 0.0425 ÷÷ - 1 = (1.0035417)12 - 1 çè 12 ÷ø

t =

» 0.0433 The effective rate is 4.33%.

ln(4.5) » 0.5014 3

2.6 Exercises

(b) 3.75% compounded continuously

True

Your Turn 4

True

A = 50, 000, r = 0.315, and m = 4.

y0 represents the initial quantity; k represents the rate of growth or decay.

The half-life of a quantity is the time period for the quantity to decay to one-half of the initial amount.

Assume that y = y0ekt represents the amount remaining of a radioactive substance decaying

e0.0375 - 1 » 0.0382 The effective rate is 3.82%.

(

50, 000 = 30, 000 1 + 5 3

)

0.0315 4t 4

= (1.007875)4t

t =

ln(5/3) » 16.28 4 ln(1.007875)

We need to round up to the nearest quarter, so $30,000 will grow to $50,000 in 16.5 years.

with a half-life of T. Since y = y0 is the amount of the substance at time t = 0, then y =

Your Turn 5

y0 2

= y0e kT , and solving for k yields

1 = e kT 2 æ1ö ln çç ÷÷÷ = kT çè 2 ø

1.40625 = e7r ln1.40625 = ln e7r 7r = ln1.40625

r =

the amount at time t = T . Therefore,

A = Pert 4500 = 3200e

y0 2

ln1.40625 » 0.0487 7

k =

The interest rate needed is 4.87%.

( 12 ) T

ln (2-1) = T ln 2 =. T

2.6 Warmup Exercises W1. e2t = 5 2t = ln(5) ln(5) t = » 0.8047 2

W2.

ek + 2 = 3 k + 2 = ln(3)

10. Assume that y = y0ekt is the amount left of a radioactive substance decaying with a half-life

of T. From Exercise 9, we know k =

-In 2 , so T -t /T

y = y0e(-ln 2/T )t = y0e-(t /T )ln 2 = y0e(-ln 2 -t /T

= y0 2

k = ln(3) - 2 » -0.9014

é æ ö-1 ù-t /T æ 1 öt /T 1 = y0 êê çç ÷÷ úú = y0 çç ÷÷ çè 2 ø÷ ú èç 2 ø÷ ëê û

)

Section 2.6 11.

139

r = 4% compounded quarterly,

17.

m= 4

A = $7300, r = 5% compounded continuously, t =3

A = Pe rt

m æ rö rE = çç1 + ÷÷÷ - 1 çè mø 4 æ 0.04 ö÷ = çç1 + -1 ÷ çè 4 ø÷

18.

» 4.06%

A = $25, 000, r = 4.6% compounded continuously, t = 8 A = Pe rt

12. r = 6% compounded monthly, m = 12 12 æ 0.06 ö÷ rE = çç 1 + ÷ -1 çè 12 ø÷

P =

r = 8% compounded continuously

19.

rE = e r - 1

r = 9% compounded semiannually

» 0.0833 = 8.33%

» 0.0920 = 9.20%

20.

r = 5% compounded continuously

r = 6.2%, m = 4 4 æ 0.062 ö÷ -1 rE = çç1 + ÷ çè 4 ÷ø

rE = e r - 1 = e0.05 - 1

» 0.0635 = 6.35%

» 0.0513 = 5.13%

21. A = $10, 000, r = 6%, m = 4, t = 8

r = 6% compounded monthly 12 æ 0.06 ö÷ -1 rE = çç1 + ÷ çè 12 ÷ø

-tm æ rö P = A çç1 + ÷÷÷ çè mø

» 0.0617

-8(4) æ 0.06 ö÷ = 10, 000 çç1 + ÷ çè 4 ø÷

» $6209.93

16.

e rt 25,000

2 æ 0.09 ö÷ rE = çç 1 + ÷ -1 çè 2 ÷ø

= e0.08 - 1

15.

= 0.046(8) » $17,302.93 e

» 0.0617 = 6.17%

14.

e rt 7300

= 0.05(3) » $6283.17 e

» 0.0406

13.

P =

A = $45,678.93, r = 7.2%, m = 12, t = 11 months -tm æ rö P = A çç1 + ÷÷÷ çè mø

» 6.17%

22.

A = $20, 000, t = 4, r = 6.5%, m = 12 mt æ rö A = P çç 1 + ÷÷÷ çè mø

12(4) æ 0.065 ö÷ 20,000 = P çç 1 + ÷ çè 12 ÷ø

20, 000

-(11/12)(12)

æ 0.072 ö÷ = 45, 678.93çç1 + ÷ çè 12 ÷ø

= P (1.05416)48 $15, 431.86 = P

» $42, 769.89

140

Chapter 2 NONLINEAR FUNCTIONS

23. (a) A = $307,000, t = 3, r = 6%, m = 2

(d)

æ ömt A = P çç1 + r ÷÷÷ è mø

40, 000 = 20, 000e5r 2 = e5r

3(2)

æ ö 307, 000 = P çç1 + 0.06 ÷÷÷ è 2 ø

ln 2 = ln e5r 5r = ln 2

307, 000 = P(1.03)6

r =

307, 000 = P (1.03)6 $257,107.67 = P

25.

(b) Interest = 307, 000 - 257,107.67 = $49,892.33 (c)

= 238,810.46

The additional amount needed is 307, 000 - 238,810.46 = $68,189.54. A = Pert 307, 000 = 200, 000e 1.535 = e

The interest rate needed is 13.86%. P = $60, 000 (a) r = 8% compounded quarterly:

5(4) æ 0.08 ö÷ = 60, 000 çç1 + ÷ çè 47 ÷ø

» $89,156.84 r = 7.75% compounded continuously A = Pert = 60, 000e0.775(5)

» $88,397.58

Mina will earn more money at 8% compoundded quarterly.

ln1.535 = ln e3r 3r = ln 1.535

(b) She will earn $759.26 more.

r = ln1.535 » 0.1428 3 The interest rate needed is 14.28%.

24.

ln 2 » 0.1386 5

tm æ rö A = P çç1 + ÷÷÷ çè mø

P = $200, 000 A = 200, 000(1.03)6

(d)

A = Pert

A = $40, 000, t = 5

(a) r = 0.064, m = 4

4 æ 0.08 ö÷ = çç1 + -1 ÷ çè 4 ÷ø

æ rö A = P çç 1 + ÷÷÷ çè mø

5(4) æ 0.064 ö÷ 40,000 = P çç 1 + ÷ çè 1 ÷ø

40,000

» 0.0824 = 8.24% r = 7.75% compounded continuously:

= P (1.016)20 $29,119.63 = P

rE = e r - 1 = e0.0775 - 1

(b) Interest = $40, 000 - $29,119.63 = $10,880.37 (c) P = $20, 000 æ ö5(4) A = 20, 000 çç1 + 0.064 ÷÷ è 4 ø = $27, 472.88 The amount needed will be $40,000 - $27,472.88 = $12,527.12.

» 0.0806 = 8.06%

Section 2.6

141 (b) 11, 000 = 5000e0.063t 11 = e0.063t 5 æ 11 ö 0.063t = ln çç ÷÷÷ çè 5 ø

(d) A = $80, 000 A = Pert 80, 000 = 60, 000e0.0775t 4 = e0.0775t 3 4 ln = ln e0.0775t 3 ln 4 - ln 3 = 0.0775t ln 4 - ln 3 =t 0.0775 3.71 = t

$60,000 will grow to $80,000 in about 3.71 years.

t =

It will take about 12.52 years. 27.

(b)

80, 000 (1.02) ³ 60, 000 4 (1.02)4 x ³ 3 æ4ö log (1.02)4 x ³ log çç ÷÷÷ çè 3 ø

4 log (1.02)

-500 = -800e-x 5 = e-x 8 ln 5 = ln e-x 8 - ln 5 = x 8 0.47 » x Sales reach 500 in about 12 year.

» 3.63

We need to round up to the nearest quarter, so it will take 3.75 years.

(d) Graphing the function y = S ( x) on a graphing calculator will show that there is a horizontal asymptote at y = 1000. This indicates that the limit on sales is 1000 units.

26. (a)

æ 0.063 ö÷ 11, 000 = 5000 çç1 + ÷ çè 4 ÷ø 11 = (1.01575)4t 5 æ 11 ö ln çç ÷÷÷ = 4t ln 1.01575 çè 5 ø t =

S ( x) = 500 500 = 1000 - 800e-x

æ4ö 4 x log (1.02) ³ log çç ÷÷÷ çè 3 ø

( )

S ( x) = 1000 - 800e-x

(a) S (0) = 1000 - 800e0 = 1000 - 800 = 200

x³

( 115 )

0.063 t » 12.52

4x æ 0.08 ö÷ ³ 80, 000 (e) 60, 000 çç1 + ÷ çè 4 ÷ø

log 43

28.

S ( x) = 5000 - 4000e-x

(a) S (0) = 5000 - 4000e0 = 1000

( )

ln 11 5

Since S(x) represents sales in thousands, in year 0 the sales are $1,000,000.

4 ln 1.01575

t » 12.61

Since the interest is only added at the end of the quarter, it will take 12.75 years.

S ( x) = 4500.

(b) Let

4500 = 5000 - 4000e-x -500 = -4000e-x 0.125 = e-x - ln 0.125 = x x » 2

142

Chapter 2 NONLINEAR FUNCTIONS (c) Graphing the function y = S ( x) on a graphing calculator will show that there is a horizontal asymptote at y = 5000. Since this represents $5000 thousand or $5,000,000, the limit on sales is $5,000,000.

(b) y = y0e(ln 2/12)t = y0e(ln 2)(t /12) = y0[eln 2 ]t /12 = y0 2t /12 since eln 2 = 2

29. (a) P = P0ekt

When t = 1600, P = 500. When t = 2020, P = 7795.

= 220

500 = P0e1600k 7795 = P0e

= 1, 048,576

2020k

For 15 days, t = 15 ⋅ 24 or 360.

2020k

y = (1)2360/12

7795 = P0e 500 P0e1600k

= 230

7795 = e 420k 500 æ ö 420k = ln çç 7795 ÷÷÷ è 500 ø k =

(

ln 7795 500

= 1, 073, 741,824

31.

y = y0 2t / k y = 40, 000, y0 = 20, 000, t = 17

)

(a) 40, 000 = 20, 000(2)17 / k

420 k = 0.006540

21 = 217 / k

Substitute this value into 500 = P0e1600k to find P0. 500 = P0e1600(0.006540) 500 P0 = 1600(0.006540) e P0 = 0.01427

1 = 17 k k = 17

The equation is y = 20, 000(2)t /17 . (b) y = 20, 000(2)t /17

Therefore, P(t ) = 0.01427e0.006540t .

y = 20, 000(21/17 )t y = 20, 000(1.0416)t

0.006540

(b) P(1) = 0.01427e » 0.014364 million, or 14,364. The exponential equation gives a world population of only 14,364 in the year 1. (c) No, the answer in part (b) is too small. Exponential growth does not accurately describe population growth for the world over a long period of time. 30. (a) y = 2 y0 after 12 hours. y = y0ekt 2 y0 = y0e12k 12k

2=e

ln 2 = ln e12k 12k = ln 2 k = ln 2 » 0.05776 12 y = y0e0.05776t

y = 20, 000(eln 2

y = 20, 000e0.0408t

(d) y = 20, 000(2)45/17 y » 125, 278 There will be about 125,000 bacteria after 45 minutes. (e) 200, 000 = 20, 000(2)t /17 10 = 2t /17 log10 = log 2t /17 1 = t log 2 17 t = 17 » 56 log 2 There will be 200,000 bacteria in about 56 minutes.

Section 2.6 32.

143 (b) If t = 0 corresponds to January 1, the date January 17 should be placed on the product. January 18 would be more than 17.9 days.

y = y0ekt

(a) For 2011, t = 1, y = 2666 For 2016, t = 6, y = 19,413 2666 = y0e1k 19, 413 = y0e

34. Use y = y0e-kt .

When t = 5, y = 0.37 y0.

y e6k 19, 413 = 0 k 2666 y0e

0.37 y0 = y0e-5k 0.37 = e-5k

19, 413 = e 5k 2666 æ 19, 413 ö÷ 5k = ln ççç ÷ è 2666 ø÷ k =

-5k = ln(0.37) ln(0.37) -5 k » 0.1989

k =

( 2666 )

ln 19,413

5 k » 0.3971 Substitute this value into 2666 = y0e1k to find y0. 2666 = y0e0.3971 y0 = 2666 e0.3971 y0 » 1792

35. (a) From the graph, the risks of chromosomal abnormality per 1000 at ages 20, 35, 42, and 49 are 2, 5, 24, and 125, respectively.

(Note: It is difficult to read the graph accurately. If you read different values from the graph, your answers to parts (b)-(e) may differ from those given here.) (b) y = Cekt When t = 20, y = 2, and when t = 35, y = 5.

Therefore, y = 1792e0.3971t . (b) Using a graphing calculator,

2 = Ce20k

y = 1176(1.5478)t .

5 = Ce35k

(c)

5 Ce35k = 2 Ce20k 2.5 = e15k 15k = ln 2.5 ln 2.5 k = k = 0.061 15

(e) From part (a), t = 15 y = 1792e0.3971(15) » 692, 000

From part (b), t = 15 y = 1176(1.5478)15 » 824, 000

Neither prediction is realistic.

When t = 42, y = 29, and when t = 49, y = 125. 24 = Ce 42k 125 = Ce 49k

33.

f (t ) = 500 e0.1t

(a)

f (t ) = 3000 3000 = 500e0.1t 6 = e0.1t ln 6 = 0.1t

125 Ce 49k = 24 Ce 42k 125 = e7k 24 æ 125 ö÷ 7k = ln çç çè 24 ÷÷ø

17.9 » t k =

ln ( 125 24 ) 7

k » 0.24

144

Chapter 2 NONLINEAR FUNCTIONS (d) Since the values of k are different, we cannot kt

assume the graph is of the form y = Ce .

39. (a)

æ ö100/14.4 A(100) = 4.0 çç 1 ÷÷÷ è2ø

(e) The results are summarized in the following table.

Value of k for [20, 35] 0.00093

Value of k for [42, 49] 0.0017

2.3 ´ 10-5

2.5 ´ 10-5

6.3 ´ 10-7

4.1 ´ 10-7

A(100) » 0.03248

After 100 years, about 0.0325 gram will remain. (b)

æ öt /14.4 0.1 = 4.0 çç 1 ÷÷÷ è2ø t /14.4 0.1 = æç 1 ö÷ ÷ çè 2 ÷ø 4.0 æ ö ln 0.025 = t ln çç 1 ÷÷÷ 14.4 è 2 ø 14.4 ln 0.025 t = ln 12

The value of n should be somewhere between 3 and 4. 36.

æ öt /14.4 A(t ) = A0 çç 1 ÷÷÷ è2ø

A(t ) = A0ekt

( )

0.60 A0 = A0e(- ln 2/5600)t

t » 76.6 It will take about 76.6 years.

0.60 = e(- ln 2/5600)t ln 2 ln 0.60 = t 5600 5600(ln 0.60) =t - ln 2 4127 » t

40. (a)

æ 1 öt /1620 A(t ) = A0 çç ÷÷÷ çè 2 ø æ 1 ö100/1620 A(100) = 4.0 çç ÷÷÷ çè 2 ø

The sample was about 4100 years old.

A(100) = 3.8

37.

1 A0 = A0e-0.131t 2 1 = e-0.131t 2 1 ln = -0.131t 2 - ln 2 = -0.131t ln 2 t = 0.131 t » 5.291

After 100 years, about 3.8 grams will remain. (b)

The half-life of cobalt-50 is about 5.29 years. 38.

1 A = A e-0.0644t 0 0 2 1 = e-0.0644t 2 1 ln = -0.0644t 2 - ln 2 = -0.0644t t = ln 2 0.0644 t » 10.76

æ 1 öt /1620 0.1 = 4.0 çç ÷÷÷ çè 2 ø æ 1 öt /1620 0.1 = çç ÷÷÷ çè 2 ø 4 1 t ln 0.025 = ln 1620 2 1600 ln 0.025 t = ln 12

( )

t = 8600

The half-life is about 8600 years.

The half-life of krypton-85 is about 10.8 years.

Section 2.6

145 (b) From part (a), we have

41. (a) y = y0ekt When t = 0, y = 500,so y0 = 500. When t = 3, y = 386. 386 = 500e 386 = e3k 500

k =

= 25.0eln(19.5/25.0) ⋅(t /50) = 25.0 [eln(19.5/25.0) ]t / 50 = 25.0(19.5/25.0)t /50 = 25(0.78)t /50

(c)

(b) From part (a), we have

( ).

ln 386 500 3

y = 500ekt

1 y0 = y0e-0.00497t 2 1 = e-0.00497t 2 æ1ö -0.00497t = ln çç ÷÷÷ çè 2 ø

= 500e[ln(386/500)/3]t

t =

= 500 [eln(386/500) ]t /3

43.

= 500(0.722)t /3

(b)

( ) 1 2

42. (a) y = y0ekt

When t = 0, y = 25.0, so y0 = 25.0

It will take about 173 days.

When t = 50, y = 19.5

19.5 = 25.0e50k

50 k = -0.00497 -0.00497t

y = 25.0e

20 = 20e-0.0004t 1 = e-0.004t 2 1 ln = -0.004t 2 ln 2 = t 0.004 173 » t

-0.0863 t » 8.0 The half-life is about 8.0 days.

( 19.5 25.0 )

t = 180 y = 40e-0.004(180) = 40e-0.72 » 19.5 watts

ln 1 = -0.0863t 2

y = 40e-0.004t

(a)

19.5 = e50k 25.0 æ 19.5 ÷ö 50k = ln çç çè 25.0 ÷÷ø

( 12 )

The half-life is about 139 days.

= 500(386/500)t /3

-0.00497 t » 139

= 500eln (386/500)⋅(t /3)

k =

= 25.0e[(ln19.5/25.0)/50]t

y = 500e-0.0863t

t =

( 386 500 ) .

y = 25.0ekt

e3k = 0.772 3k = ln 0.772 k = ln 0.772 3 k » -0.0863

k =

44.

æ 1 öt /5600 A(t ) = A0 çç ÷÷÷ çè 2 ø

æ 1 ö43,000/5600 A(43, 000) = A0 çç ÷÷÷ çè 2 ø » 0.005 A0

146 45.

Chapter 2 NONLINEAR FUNCTIONS P(t ) = 100e-0.1t

47.

f (t ) = T0 + Ce-kt

(a) P(4) = 100e-0.1(4) » 67% -0.1(10)

(b) P(10) = 100e (c)

f (t ) = 18 + 5e-0.6(9)

» 37%

= 18 + 5e-5.4

10 = 100e-0.1t

» 18.02 The temperature is about 18.02°.

0.1 = e-0.1t ln (0.1) = -0.1t - ln (0.1) =t 0.1 23 » t

48.

f (t ) = T0 + Ce-kt 25 = 20 + 100e-0.1t 5 = 100e-0.1t

It would take about 23 days. (d)

t = 9, T0 = 18, C = 5, k = .6

e-0.1t = 0.05 -0.1t = ln 0.05 ln 0.05 t = -0.1 » 30 It will take about 30 min.

1 = 100e-0.1t 0.01 = e-0.1t ln (0.01) = -0.1t - ln (0.01) =t 0.1 46 » t

49.

It would take about 46 days.

C = -14.6, k = 0.6, T0 = 18 f (t ) = 10 f t  = T0 + Ce-kt

46. (a) Let t = the number of degrees Celsius.

y = y0 ⋅ ekt

f (t ) = 18 + (-14.6)e-0.6t

y0 = 10 When t = 0.

-8 = -14.6e-0.6t 0.5479 = e-0.6t ln 0.5479 = -0.6t - ln 0.5479 =t 0.6 1»t It would take about 1 hour for the pizza to thaw.

To find k, let y = 11 when t = 10. 11 = 10e10k 11 e10k = 10 10k = ln 1.1 ln 1.1 k = 10 » 0.0095

Chapter 2 Review Exercises

The equation is y = 10e0.0095t .

True

x is a rational False; for example f ( x) = x + 1

(b) Let y = 1.5; solve for t.

function but not an exponential function.

15 = 10e0.0095t ln 1.5 = 0.0095t ln 1.5 t = 0.0095t » 42.7

True

False; an exponential function has the form f ( x) = a x .

15 grams will dissolve at 42.7°C. 6.

False; the vertical asymptote is at x = 6.

Chapter 2 Review 7. 8.

147

True

24.

False; the domain includes all numbers except x = 2 and x = -2.

(

False; the amount is A = 2000 1 + 0.04 12

) .

y =

x 2

x +1

-3

-2

-1 0 1

3 -2 -1 y - 10 5 2

1 2

2 5

3 10

10. False; the logarithmic function f ( x) = log a x is not defined for a = 1.

æ 3ö æ 2ö æ 1ö Pairs: çç -3, - ÷÷ , çç -2, - ÷÷ , çç -1, - ÷÷ , ÷ ÷ çè ç ç 10 ø è 5ø è 2 ø÷ æ 1ö æ 2ö æ 3 ö (0, 0), çç 1, ÷÷÷ , çç 2, ÷÷÷ , çç 3, ÷÷÷ èç 2 ø çè 5 ø çè 10 ø

11. False; ln(5 + 7) = ln 12 = / ln 5 + ln 7

3 , 0, 3 , 2 , 1 Range: - 12 , - 25 , -10 10 5 2

{

}

12. False; (ln 3) 4 = / 4 ln 3 since (ln 3)4 means (ln 3)(ln 3)(ln 3)(ln 3). 13. False; log10 0 is undefined since 10 x = 0 has no solution. 14. True 25. f ( x) = 5 x 2 - 3 and g ( x) = -x 2 + 4 x + 1

15. False; ln (-2) is undefined.

(a) 16.

4 = 0.6667 and False; ln ln 8

f (-2) = 5(-2) 2 - 3 = 17

(b) g (3) = -(3)2 + 4(3) + 1 = 4

ln 4 - ln 8 = ln(1/2) » -0.6931.

(c)

f (-k ) = 5(-k ) 2 - 3 = 5k 2 - 3

(d) g (3m) = -(3m) 2 + 4(3m) + 1

17. True

= -9m 2 + 12m + 1

18. True 23.

(e)

f ( x + h) = 5( x + h) 2 - 3

y = (2 x - 1)( x + 1)

= 5( x 2 + 2 xh + h 2 ) - 3

= 2x2 + x - 1

= 5x 2 + 10 xh + 5h 2 - 3

x -3 -2 -1 y

0 1 2

(f )

g ( x + h) = -( x + h)2 + 4( x + h) + 1

0 -1 1 9 20

= -( x 2 + 2 xh + h 2 ) + 4 x + 4h + 1

Pairs: (-3, 14), (-2, 5), (-1, 0), (0, -1), (1, 2), (2, 9), (3, 20) Range: {-1, 0, 2, 5, 9, 14, 20}

= -x 2 - 2 xh - h 2 + 4 x + 4h + 1

(g)

f ( x + h) - f ( x) h =

5( x + h)2 - 3 - (5x 2 - 3) h

5( x 2 + 2hx + h 2 ) - 3 - 5x 2 + 3 h

5x 2 + 10hx + 5h 2 - 5 x 2 h

10hx + 5h 2 = 10 x + 5h h

148

Chapter 2 NONLINEAR FUNCTIONS (h) g ( x + h) - g ( x ) h

(h) g ( x + h) - g ( x ) h =

-( x + h) 2 + 4( x + h) + 1 - (-x 2 + 4 x + 1) h

3( x + h)2 + 4( x + h) - 1 - (3x 2 + 4 x - 1) h

-( x 2 + 2 xh + h 2 ) + 4 x + 4h + 1 + x 2 - 4 x - 1 h

3( x 2 + 2 xh + h 2 ) + 4 x + 4h - 1 - 3x 2 - 4 x + 1 h

-x 2 - 2 xh - h 2 + 4h + x 2 h

3x 2 + 6 xh + 3h 2 + 4 x + 4h - 1 - 3x 2 - 4 x + 1 h

-2 xh - h 2 + 4h h = -2 x - h + 4

6 xh + 3h 2 + 4h h = 6 x + 3h + 4

26.

f ( x) = 2 x 2 + 5 and g ( x) = 3x 2 + 4 x - 1

(a)

27.

f (-3) = 2(-3) + 5 = 23

(b) g (2) = 3(2)2 + 4(2) - 1 = 19 (c)

Domain: (-¥, 0) È (0, ¥)

f (3m) = 2(3m) + 5 = 18m + 5

(d) g (-k ) = 3(-k )2 + 4(-k ) - 1

28.

= 3k 2 - 4k - 1 (e)

3x - 4 x x= / 0 y =

y =

x-2 2x + 3

x - 2 ³ 0 and x³2

f ( x + h) = 2( x + h) + 5 = 2( x 2 + 2 xh + h 2 ) + 5

2x + 3 = / 0 2x = / -3 / x =

= 2 x 2 + 4 xh + 2h 2 + 5

3 2

Domain: [2, ¥)

(f ) g ( x + h) = 3( x + h)2 + 4( x + h) - 1

29.

y = ln( x + 7) x+7>0

= 3( x 2 + 2 xh + h 2 ) + 4 x + 4h - 1

x > -7

= 3x 2 + 6 xh + 3h 2 + 4 x + 4h - 1

(g)

f ( x + h) - f ( x) h 2( x + h) 2 + 5 - (2 x 2 + 5) = h 2( x 2 + 2 xh + h 2 ) + 5 - 2 x 2 - 5 h 2 x 2 + 4 xh + 2h 2 + 5 - 2 x 2 + 5 = h

Domain: (-7, ¥). 30.

y = ln( x 2 - 16) x 2 - 16 > 0 x 2 > 16

x > 4 or x < -4

Domain: (-¥, -4) È (4, ¥)

4 xh + 2h 2 h = 4 x + 2h =

Chapter 2 Review

149

y = 2 x 2 + 3x - 1

31.

x =

The graph is a parabola. Let y = 0.

0 = 2 x 2 + 3x - 1

-4  48 = 22 3 -2

2 - 2 3 » -1.46. Let x = 0.

-3  9 + 8 4 -3  17 = 4 =

-3 - 17 4

42 - 4(-1)(8) 2(-1)

The x-intercepts are 2 + 2 3 » 5.46 and

-3  32 - 4(2)(-1) x = 2(2)

The x-intercepts are

-4 

1 y = - (0) 2 + 0 + 2 4 y = 2 is the y-intercept. -3 + 17 4

» 0.28 and

Vertex: x =

» -1.48.

-b -1 = 2a 2 -1

( 4)

= 2

1 y = - (2)2 + 2 + 2 4 = -1 + 4 =3

Let x = 0. y = 2(0)2 + 3(0) - 1 -1 is the y-intercept.

The vertex is (2, 3).

-b -3 3 Vertex: x = = =2a 2(2) 4 æ 3 ö2 æ 3ö y = 2 çç - ÷÷ + 3çç - ÷÷ - 1 çè 4 ÷ø çè 4 ÷ø 9 9 - -1 8 4 17 =8 =

(

33.

)

. The vertex is - 34 , - 17 8

y = -x 2 + 4 x + 2

Let y = 0. 0 = -x 2 + 4 x + 2 x =

-4 

42 - 4(-1)(2) 2(-1)

-4  24 -2 = 2 6 =

32.

1 y = - x2 + x + 2 4 The graph is a parabola. Let y = 0. 1 0 = - x2 + x + 2 4 Multiply by 4. 0 = -x 2 + 4 x + 8

The x-intercepts are 2 + 6 » 4.45 and 2 - 6 » -0.45. Let x = 0. y = -02 + 4(0) + 2

2 is the y-intercept. -b -4 -4 Vertex: x = = = = 2 2a 2(-1) -2 y = -22 + 4(2) + 2 = 6

The vertex is (2, 6).

150

34.

Chapter 2 NONLINEAR FUNCTIONS

y = 3x 2 - 9 x + 2

38.

Translate the graph of y = x3 2 units to the left and reflect vertically. Translate 2 units downward.

x-intercepts: 0.24 and 2.76 y-intercept: 2 Vertex:

( 32 , -194 )

39. 35.

y = -( x + 2)3 - 2

f ( x) = x 3 - 3

f ( x) =

8 x

Vertical asymptote: x = 0 3

Horizontal asymptote:

Translate the graph of f ( x) = x 3 units down.

8 x

approaches zero as x gets larger.

y = 0 is an asymptote.

x -4 y

36.

-3

-2 -1 1 2

-2 -2.7 -4 -8 8 4 2.7

f ( x) = 1 - x 4

= -x 4 + 1 Reflect the graph of y = x 4 vertically then translate 1 unit upward. 40.

f ( x) =

2 3x - 6

Vertical asymptote: 3x - 6 = 0 or x = 2 2 Horizontal asymptote: y = 0, since 3x6

approaches zero as x gets larger. 37.

y = -( x - 1)4 + 4

- 13

- 23

2 3

1 3

2 9

Translate the graph of y = x 4 1 unit to the right and reflect vertically. Translate 4 units upward.

Chapter 2 Review 41.

f ( x) =

151

4x - 2 3x + 1

44.

Vertical asymptote: 3x + 1 = 0 x =-

y = 4-x + 3

x -2 -1 0

13 4

49 16

1 3

Horizontal asymptote: As x gets larger, 4x - 2 4x 4 » = . 3x - 1 3x 3

y = 43 is an asymptote. x - 3 -2 -1 0 1 2 3 y 1.75 2 3 -2 0.5 0.86 1

45.

æ 1 ö2 x -3 y = çç ÷÷÷ çè 5 ø

1 2

y 125 5

42.

f ( x) =

6x x+2

Vertical asymptote: x = -2

46.

Horizontal asymptote: y = 6

y = 4x

47.

y = log 2 ( x - 1)

2y = x - 1

1 4 16

x = 1 + 2y

-2 -1 0 1

1 16

1 4

æ 1 öx -1 y = çç ÷÷÷ çè 2 ø x -2 -1 0 1 2 y 8 4 2 1 1

x -5 -4 -3 -1 0 1 2 y 10 12 18 -6 0 2 3

43.

1 5

x 2 3 5 9 y 0 1 2 3

152

Chapter 2 NONLINEAR FUNCTIONS y = 1 + log3 x

48.

y - 1 = log3 x

52.

3 y -1 = x x

1 9

1 3

æ 9 ö÷x çç ÷ = 3 çè 16 ÷ø 4 æ 3 ö÷2 x æ ö1 çç ÷ = çç 3 ÷÷ çè 4 ÷ø èç 4 ÷ø

1 3 9

2x = 1

y -1 0 1 2 3

1 2

x =

53.

92 y +3 = 27 y (32 ) 2 y +3 = (33 ) y 34 y + 6 = 33 y

49.

4 y + 6 = 3y

y = - ln ( x + 3) - y = ln ( x + 3)

y = -6

e- y = x + 3 e- y - 3 = x

54.

x -2.63 -2 -0.28 4.39 y 1 0 -1 -2

æ b ö1/4 1 = çç ÷÷÷ çè 4 ø 2 æ 1 ÷ö4 çç ÷ = b çè 2 ÷ø 4 æ 1 ö4 4 çç ÷÷÷ = b çè 2 ø æ1 ö 4 çç ÷÷÷ = b çè 16 ø

50.

1 =b 4

y = 2 - ln x 2 x -4 - 3 -2 -1 y -0.8 -0.2 0.6 2

55.

35 = 243

The equation in logarithmic form is

x 1 2 3 4 y 2 0.6 -0.2 -0.8

log3 243 = 5.

56.

51/2 =

The equation in logarithmic form is log5 5 =

57. 51.

2x+2 = 2x+2 =

1 8 1 23

2 x + 2 = 2-3 x + 2 = -3

1 . 2

e0.8 = 2.22554

The equation in logarithmic form is ln 2.22554 » 0.8.

58. 101.07918 = 12

The equation in logarithmic form is

x = -5

log12 » 1.07918.

Chapter 2 Review 59.

153

log 2 32 = 5

67.

The equation in exponential form is

= log5 3k (7k 3 )

25 = 32.

60.

= log5 (21k 4 )

1 2 The equation in exponential form is log9 3 =

68.

e 62.

4.41763

69.

» 82.9.

æ y 4 ö÷ ç = log3 çç 2 ÷÷÷ çè x ø÷

The equation in exponential form is

Recall that log x means log10 x. 63.

70.

æ r 6 ö÷ ç = log 4 çç 2 ÷÷÷ çè r ø÷

log3 81 = x

= log 4 (r 4 )

3x = 34 x = 4

71. log32 16 = x 25 x = 24 5x = 4 4 x = 5

72. log 4 8 = x x

4 =8 (22 ) x = 23 2x = 3 3 x = 2

66.

6 p = 17 ln 6 p = ln 17 p ln 6 = ln 17 ln 17 p = ln 6 » 1.581

32 x = 16

65.

3log 4 r 2 - 2 log 4 r = log 4 (r 2 )3 - log 4 r 2

3x = 81

64.

4 log3 y - 2 log3 x = log3 y 4 - log3 x 2

log 3.21 = 0.50651 100.50651 » 3.21.

2 y3

8 y2 y = log3 4

ln 82.9 = 4.41763

The equation in exponential form is

log3 2 y 3 - log3 8 y 2 = log3

91/2 = 3. 61.

log5 3k + log5. 7k 3

3z -2 = 17 ln 3z -2 = ln 11 ( z - 2) ln 3 = ln 11 z-2= z =

ln 11 ln 3 ln 11 +2 ln 3

» 4.183

log100 1000 = x 100 x = 1000 (102 ) x = 103 2x = 3 3 x = 2

154

Chapter 2 NONLINEAR FUNCTIONS 21- m = 7

73.

ln 21- m = ln 7

78.

(1 - m) ln 2 = ln 7

ln 7 1- m = ln 2

2 p = -5  3 -5  5 3 2 -5 + 5 3 » 1.830 p = 2 -5 - 5 3 » -6.830 or p = 2 p =

ln 7 m = 1ln 2 » -1.807 12-k = 9 ln 12-k = ln 9 -k ln 12 = ln 9 k =-

ln 9 ln 12

2p = 3 5

5 + 2 p = 5 3

ln 7 -m = -1 ln 2

74.

æ ö2 çç1 + 2 p ÷÷ = 3 çè 5 ÷ø

79.

log k 64 = 6 k 6 = 64

» -0.884

k 6 = 26 k = 2

75.

e-5- 2 x = 5 ln e-5- 2 x = ln 5

80.

35 = 2 x + 5 243 = 2 x + 5 238 = 2 x x = 119

-5 - 2 x = ln 5 -2 x = ln 5 + 5 ln 5 + 5 -2 » -3.305

x =

81. 76.

e ln (e

3 x -1

3x -1

log3 (2 x + 5) = 5

= 14

log (4 p + 1) + log p = log 3 log[ p(4 p + 1)] = log 3 log (4 p 2 + p) = log 3

) = ln 14

3x - 1 = ln 14 3x = 1 + ln 14

4 p2 + p = 3

1 + ln 14 3 » 1.213

4 p2 + p - 3 = 0 (4 p - 3)( p + 1) = 0 4 p - 3 = 0 or p + 1 = 0 3 p = p = -1 4

x =

77.

æ ö5 çç1 + m ÷÷ = 15 çè 3 ø÷ 1/5 5ù éæ ê ç1 + m ö÷ ú = 151/5 ÷ ú êç 3 ø÷ ú êë çè û m = 151/5 1+ 3 m = 151/5 - 1 3

p cannot be negative, so p = 34 . 82.

log 2 (5m - 2) - log 2 (m + 3) = 2 log 2

m = 3(151/5 - 1)

5m - 2 = 2 m+3 5m - 2 = 22 m+3 5m - 2 = 4(m + 3) 5m - 2 = 4m + 12

» 2.156

m = 14

Chapter 2 Review 83.

155 æ 1ö C çç 2 ÷÷÷ = $40 + $60(3) çè 9 ø

f ( x) = a x ; a > 0, a = / 1

(a) The domain is (-¥, ¥).

= $40 + $180 = $220.

(b) The range is (0, ¥). (f )

(c) The y-intercept is 1. (d) The x-axis, y = 0, is a horizontal asymptote. (e) The function is increasing if a > 1. (f) The function is decreasing if 0 < a < 1. 84.

f ( x) = log a x; a > 0, a = / 1

(a) The domain is (0, ¥).

(g) The independent variable is the number of days, or x.

(b) The range is (-¥, ¥).

(h) The dependent variable is the cost, or C(x).

87.

(e) f is increasing if a > 1.

y =

7x 100 - x

(a) y =

(f) f is decreasing if 0 < a < 1.

7(80) 560 = = 28 100 - 80 20

The cost is $28,000.

86. (a) For x in the interval 0 < x £ 1, the renter is charged the fixed cost of $60 and 1 day’s rent of $60 so

(b) y =

æ3ö C çç ÷÷÷ = $40 + $60(1) çè 4 ø

7(50) 350 = =7 100 - 50 50

The cost is $7000. (c)

= $40 + $60 = $100.

7(90) 630 = = 63 100 - 90 10

The cost is $63,000.

æ9 ö (b) C çç ÷÷÷ = $40 + $60(1) çè 10 ø

(d) Plot the points (80, 28), (50, 7), and (90, 63).

= $40 + $60 = $100.

(c) C (1) = $40 + $60(1) = $40 + $60 = $100. (d) For x in the interval 1 < x £ 2, the renter is charged the fixed cost of $40 and 2 days rent of $120, so æ 5ö C çç1 ÷÷÷ = $40 + $60(2) çè 8 ø

(e) No, because all of the pollutant would be removed when x = 100, at which point the denominator of the function would be zero. 88.

p = $6902, r = 6%, t = 8, m = 2 tm æ rö A = P çç1 + ÷÷÷ çè mø

= $40 + $120 = $160.

(e) For x in the interval 2 < x £ 3 the renter is charged the fixed cost of $40 and 3 days rent of $180. So

8(2) æ 0.06 ö÷ A = 6902 çç1 + ÷ çè 2 ÷ø

= 6902(1.03)16 = $11, 075.68

156

Chapter 2 NONLINEAR FUNCTIONS Interest = A - P = $11, 075.68 - $6902 = $4173.68

89.

To double: t ⋅2 æ 0.06 ö÷ 2(1000) = 1000 çç1 + ÷ çè 2 ø÷

2 = 1.032t ln 2 = 2t ln 1.03 ln 2 t = 2 ln1.03 » 12 years

P = $2781.36, r = 4.8%, t = 6, m = 4 tm æ rö A = P çç1 + ÷÷÷ çè mø

(6)(4) æ 0.048 ÷ö A = 2781.36 çç1 + ÷ çè 4 ÷ø

= 2781.36(1.012) = $3703.31

To triple:

t ⋅2 æ 0.06 ö÷ 3(1000) = 1000 çç1 + ÷ çè 2 ÷ø

Interest = $3703.31 - $2781.36 = $921.95

90.

3 = 1.032t ln 3 = 2t ln 1.03 ln 3 t = 2 ln 1.03 » 19 years

P = $12,104, r = 6.2%, t = 2 A = Pert = 12,104e0.062(2) = $13, 701.92

91.

95. $2100 deposited at 4% compounded quarterly. tm æ rö A = P çç1 + ÷÷÷ çè mø

P = $12,104, r = 6.2%, t = 4

To double:

A = Pert

t ⋅4 æ 0.04 ö÷ 2(2100) = 2100 çç1 + ÷ çè 4 ÷ø

A = 12,104e0.062(4) = 12,104e0.248 = $15,510.79

92.

2 = 1.014t ln 2 = 4t ln 1.01 ln 2 t = 4 ln 1.01 » 17.4 Because interest is compounded quarterly, round the result up to the nearest quarter, which is 17.5 years or 70 quarters.

A = $1500, r = 0.06, t = 9 A = Pert = 1500e0.06(9) = 1500e0.54 = $2574.01

93.

P = $12, 000, r = 0.05, t = 8 A = 12, 000e

To triple: t ⋅4 æ 0.04 ö÷ 3(2100) = 2100 çç1 + ÷ çè 4 ÷ø

0.05(8)

= 12, 000e0.40 = $17,901.90

94. $1000 deposited at 6% compounded semiannually. tm æ rö A = P çç1 + ÷÷÷ çè mø

3 = 1.014t ln 3 = 4t ln 1.01 ln 3 t = 4 ln 1.01 » 27.6 Because interest is compounded quarterly, round the result up to the nearest quarter, which is 27.75 years or 111 quarters.

Chapter 2 Review 96.

157

r = 7%, m = 4

102. P = $1, r = 0.08

m æ r ö rE = çç 1 + ÷÷÷ - 1 çè mø

A = Pert , A = 3(1) 3 = 1e0.08t ln 3 = 0.08t

æ 0.07 ö÷ = çç 1 + ÷ -1 çè 4 ÷ø

ln 3 =t 0.08 13.7 = t

= 0.0719 = 7.19%

97.

r = 6%, m = 12

It would take about 13.7 years.

æ rö rE = çç1 + ÷÷÷ - 1 çè mø

103. P = $6000, A = $8000, t = 3 A = Pert

12 æ 0.06 ö÷ = çç1 + ÷ -1 çè 12 ÷ø

= 0.0617 = 6.17%

98.

r = 5% compounded continuously rE = er - 1 = e0.05 - 1 = 0.0513 = 5.13%

99.

A = $2000, r = 6%, t = 5, m = 1 -tm æ rö P = A çç1 + ÷÷÷ çè mø -5(1) æ 0.06 ö÷ = 2000 çç1 + ÷ çè 1 ø÷

= 2000(1.06)-5 = $1494.52

100.

A = $10, 000, r = 8%, m = 2, t = 6 -tm æ rö P = A çç1 + ÷÷÷ çè mø

8000 = 6000e3r 4 = e3r 3 ln 4 - ln 3 = 3r ln 4 - ln 3 r = 3 r » 0.0959 or about 9.59% -tm æ rö 104. P = A çç1 + ÷÷÷ çè mø -3(12) æ 0.06 ö÷ P = 25, 000 çç1 + ÷ çè 12 ø÷

= 25, 000(1.005)-36 = $20,891.12

105. (a) n = 1000 - ( p - 50)(10), p ³ 50 = 1000 - 10 p + 500 = 1500 - 10 p (b) R = pn R = p(1500 - 10 p)

-2(6) æ 0.08 ö÷ = 10, 000 çç1 + ÷ çè 2 ÷ø

= 10, 000(1.04)-12 = $6245.97

101. r = 7%, t = 8, m = 2, P = 10, 000

Since n cannot be negative, 1500 - 10 p ³ 0 -10 p ³ -1500 p £ 150.

Therefore, 50 £ p £ 150.

tm æ rö A = P çç1 + ÷÷÷ çè mø 8(2) æ 0.07 ö÷ = 10, 000 çç1 + ÷ çè 2 ÷ø

= 10, 000(1.035)16 = $17,339.86

158

Chapter 2 NONLINEAR FUNCTIONS (d) Since n = 1500 - 10 p, 10 p = 1500 - n n p = 150 - . 10 R = pn æ nö R = çç150 - ÷÷÷ n çè 10 ø

106. C ( x) =

5x + 3 x +1

(a)

(e) Since she can sell at most 1000 tickets, 0 £ n £ 1000. (f ) R = -10 p 2 + 1500 p -b -1500 = = 75 2a 2(-10)

(b) C ( x + 1) - C ( x) 5( x + 1) + 3 5x + 3 ( x + 1) + 1 x +1 5x + 8 5x + 3 = x+2 x +1 (5 x + 8)( x + 1) - (5 x + 3)( x + 2) = ( x + 2)( x + 1) =

The price producing maximum revenue is $75. (g) R = -

1 2 n + 150n 10 -b -150 = = 750 2a 2 -1

5x 2 + 13x + 8 - 5x 2 - 13x - 6 ( x + 2)( x + 1) 2 = ( x + 2)( x + 1) =

( 10 )

The number of tickets producing maximum revenue is 750. (h)

5x +3

R( p) = -10 p 2 + 1500 p

R(75) = -10(75) 2 + 1500(75)

= -56, 250 + 112,500 = 56, 250

The maximum revenue is $56,250.

C ( x) = x +1 x x 5x + 3 = x( x + 1)

(d) A( x + 1) - A( x) 5( x + 1) + 3 5x + 3 ( x + 1)[( x + 1) + 1] x( x + 1) 5x + 8 5x + 3 = ( x + 1)( x + 2) x( x + 1) x(5 x + 8) - (5x + 3)( x + 2) = x( x + 1)( x + 2)

(i)

5x 2 + 8 x - 5x 2 - 13x - 6 x( x + 1)( x + 2) -5 x - 6 = x( x + 1)( x + 2) =

(j) The revenue starts at $50,000 when the price is $50, rises to a maximum of $56,250 when the price is $75, and falls to 0 when the price is $150.

107. C ( x) = x 2 + 4 x + 7 (a)

Chapter 2 Review

159

(b)

(d) f (t ) = 0.0209t 2 + 2.84t + 21.1

C ( x + 1) - C ( x) 2

f (t ) = -0.00119t 3 + 0.124t 2

= ( x + 1) + 4( x + 1) + 7 - ( x + 4 x + 7) 2

= x + 2x + 1 + 4x + 4 + 7 - x - 4x - 7 = 2x + 5

+ 0.624t + 27.3

(e)

C ( x) x2 + 4x + 7 = x x 7 = x+4+ x

(d) A( x + 1) - A( x) æ 7 7ö - çç x + 4 + ÷÷÷ ç x +1 è xø 7 7 = x +1+ 4 + -x-4x +1 x 7 7 =1+ x +1 x 7 x - 7( x + 1) =1+ x( x + 1) 7x - 7x - 7 =1+ x( x + 1) 7 = 1x( x + 1) = ( x + 1) + 4 +

108. (a) y = at

Let y0 = 29.6, the value of y at t = 0. Use the point (58, 251.1) to find a. f (t ) = f0at with f 0 = f (0) = 29.6 251.1 = 29.6a58 a58 =

2511 29.6

109. F ( x) = - 2 x 2 + 14 x + 96 3 3 The maximum fever occurs at the vertex of the parabola. - 14 3 = 7 x = -b = 2a 2 - 43 æ ö2 æ ö y = - 2 çç 7 ÷÷÷ + 14 çç 7 ÷÷÷ + 96 3è 2ø 3 è2ø æ ö = - 2 çç 49 ÷÷÷ + 49 + 96 3è 4 ø 3 49 49 =+ + 96 6 3 = - 49 + 98 + 576 = 625 » 104.2 6 6 6 6 The maximum fever occurs on the third day. It is about 104.2°F.

110. (a) 100% - 87.5% = 12.5% 12.5% = 0.125 = 125 = 1 1000 8

The fraction let in is 1 over the SPF rating.

251.1 a = 58 29.6 » 1.0376

(b)

f (t ) = 29.6(1.0376)t

(b) f (t ) = 32.2(1.0396)t (c) (c) UVB = 1 - 1 SPF

Yes, the answers are close to each other.

(d) 1 - 1 = 87.5% 8 1 - 1 = 75.0% 4 The increase is 12.5%.

160

Chapter 2 NONLINEAR FUNCTIONS and x » 40 for the sleeping infants). So the quadratic functions are both decreasing over this time. Therefore, both respiratory rates are decreasing.

1 = 96.6% 30 1 1= 93.3% 15

(e) 1 -

(c)

The increase is 3.3% or about 3.3%. (f) The increase in percent protection decreases to zero. 111. (a)

(d) When x = 12, the waking respiratory rate is y = 49.23 breaths per minute, and the sleeping respiratory rate is y = 38.55. Therefore, for a 1-year-old infant in the 95th percentile, the waking respiratory rate is approximately 49.23 - 38.55 » 10.7 breaths per minute higher.

(b) y = 2.78t + 34.8 y = 0.0919t 2 + 2.05t + 35.7 y = 0.0635t 3 - 0.670t 2 + 4.34t + 34.6 y = 35.5(1.0628)t

113. This function has a maximum value at x » 187.9. At x » 187.9, y » 345. The largest girth for which this formula gives a reasonable answer is 187.9 cm. The predicted mass of a polar bear with this girth is 345 kg.

(c)

114.

y0 = 26, 000, t = 3, y = 52, 000

(a)

y = y0 2t / k

52, 000 = 26, 000(2)3/ k

The cubic function seems to best capture the behavior of the data. (d) x = 12 corresponds to 2022. y = 2.78(12) + 34.8

= 68.16 » 68.2 y = 0.0919(12)2 + 2.05(12) + 35.7

= 73.5336 » 73.5 y = 0.0635(12)3 - 0.670(12) 2

+ 4.34(12) + 34.6 = 99.928 » 100.0 y = 35.5(1.0628)12

» 73.73045 » 73.7

21 = 23/ k 3 1= k k =3

So, y = 26, 000(2)t /3. (b) 19,540, 000 = 26, 000(2)t /3 19,540 = 2t /3 26 æ 19,540 ÷ö t /3 ln ççç ÷ = ln 2 è 26 ÷ø æ 19,540 ö÷ t ln 2 = ln ççç ÷ è 26 ÷ø 3

112. (a) The first three years of infancy corresponds to 0 months to 36 months, so the domain is [0, 36]. (b) In both cases, the graph of the quadratic function in the exponent opens upward and the x coordinate of the vertex is greater than 36 ( x » 47 for the awake infants

t =

( 26 )

3ln 19,540

ln 2 t » 28.66

It would take about 29 days.

Chapter 2 Review 115.

p(t ) =

161 (b) From the graph, we see that the amount of glucose in the bloodstream decreases from the initial value of 0.08 gram, so it will never increase to 0.1 gram. We can also reach this conclusion by graphing y1 = G(t ) and y2 = 0.1 on the same screen with the window given in (a) and observing that the graphs of y1 and y2 do not intersect.

1.79 ⋅ 1011 (2026.87 - t )0.99

(a) p(2020) » 26.6 billion

This is about 18.8 billion more than, or three times the estimate of 7.795 billion. (b) p(2022) » 37.3 billion p(2025) » 96.3 billion

(c) From the graph, we see that as t increases, the graph of y1 = G(t ) becomes almost horizontal, and G(t ) approaches approximately 0.0769.

116. I ( x) ³ 1 I ( x) = 10e-0.3x

Note that c = 0.1 » 0.0769. a 1.3 The amount of glucose in the bloodstream after a long time approaches 0.0769 grams.

10e-0.3x ³ 1 e-0.3x ³ 0.1 -0.3x ³ ln 0.1 ln 0.1 x £ » 7.7 -0.3 The greatest depth is about 7.7 m.

119. (a) S = 21.35 + 104.6 ln A (b) S = 85.49 A0.3040

117. Graph

S 21.35 104.6 ln A 1200

on a graphing calculator and locate the maximum point. A calculator shows that the x-coordinate of the maximum point is about 0.69, and the y-coordinate is exactly 0.25. Thus, the maximum concentration of 0.25 occurs at about 0.69 minutes. 118. g (t ) =

S 85.49A0.3040 0

(d) S » 742.2 S » 694.7 Neither number is close to the actual number of 421.

æ c cö + çç g0 - ÷÷÷ e-at a çè aø

(a) If g0 = 0.08, c = 0.1, and a = 1.3, the function becomes g (t ) =

2000

0.1 æç 0.1 ö÷ -1.3t + ç 0.08 . ÷e 1.3 çè 1.3 ø÷

Graph this function on a graphing calculator. Use a window with X min = 0, since this represents the time when the drug is first injected. A good choice for the viewing window is [0, 5] by [0.07, 0.11], Xscl = 0.5, Yscl = 0.01. From the graph, we see that the maximum value of g for t ³ 0 occurs at t = 0, the time when the drug is first injected. The maximum amount of glucose in the bloodstream, given by G(0), is 0.08 gram.

120.

y = yoe-kt

(a) 100, 000 = 128, 000e-k (5) 128, 000 = 100, 000e5k 128 = e5k 100 æ ö ln çç 128 ÷÷÷ = 5k è 100 ø 0.05 » k y = 100, 000e-0.05t

(b) 70, 000 = 100, 000e-0.05t 7 = e-0.05t 10 ln 7 = -0.05t 10 7.1 » t It will take about 7.1 years.

162

Chapter 2 NONLINEAR FUNCTIONS ln éê 1 + 8.33( KA ) ùú ë û ln 2 (a) A = 0, K > 0

121. t = (1.26 ´ 109 )

Extended Application: Power Functions 1. a.

ln[1 + 8.33(0)] t = (1.26 ´ 10 ) ln 2 9

= (1.26 ´ 109 )(0) = 0 years ln[1 + 8.33(0.212)] ln 2 ln 2.76596 = (1.26 ´ 109 ) ln 2 = 1,849, 403,169

(b) t = (1.26 ´ 109 )

b. Yes; Y = 5.065 - 0.3289 X

or about 1.85 ´ 109 years (c) As r increases, t increases, but at a slower and slower rate. As r decreases, t decreases at a faster and faster rate 122. (a)

P = kD1

c. It is decreasing; the exponent is negative. As the price increases the demand decreases.

164.8 = k (30.1) k = 164.8 » 5.48 30.1 For n = 1, P = 5.48D. P = kD1.5

d. y = 158.4 x-0.3289

1.5

164.8 = k (30.1) k =

164.8 (30.1)1.5

2. a.

» 1.00

For n = 1.5, P = 1.00D1.5. P = kD 2 164.8 = k (30.1) 2 k = 164.82 » 0.182 (30.1)

b. Yes; Y = 0.7617 X + 1.2874

For n = 2, P = 0.182D 2. (b)

c. y = 3.623x 0.7617

P = 1.00 D1.5 appears to be the best fit.

d. An approximate value for the power is 0.76.

(d) We obtain

Chapter 3

THE DERIVATIVE 3.1 Limits Your Turn 1 f ( x) = x 2 + 2

0.9

0.99

0.999

0.9999

1.0001

1.001

1.01

1.1

f (x)

2.81

2.9801

2.998001

2.99980001

3.00020001

3.002001

3.0201

3.21

The table suggests that, as x get closer and closer to 1 from either side, f (x) gets closer and closer to 3. So, lim ( x 2 + 2) = 3. x 1

Your Turn 2 f ( x) =

( x + 2) ( x - 2) x2 - 4 = x-2 ( x - 2)

y 8

= x + 2, provided x ¹ 2

y 4 −2

2 0 −2

( 3, 3 5)

4 2

( 2, 4) 4

−2 0

( 3, 3 1) 4 x

2 −4

Your Turn 4 2

The graph of y = xx --24 is the graph of y = x + 2,

Find lim

x0

except there is a hole at (2, 4).

Looking at the graph, we see that as x is close to, but not equal to 2, f (x) approaches 4.

6 4

x2 - 4 lim = 4 x 2 x - 2

2 −4

Your Turn 3

ïì 2 x - 1 Find lim f ( x) if f ( x) = ï í ïïî1 x3

2x - 1 . x

if x ¹ 3

−2

0 −2

−4

if x = 3

lim f ( x) = ¥

x  0-

lim f ( x) = -¥

The graph of f is shown in the next column. lim f ( x) = 5

x3

x  0+

Since there is no real number that f (x) approaches as x approaches 0 from either side, nor does f (x) approach either ¥ or -¥, lim 2 xx-1 does not exist. x0

163

164

Chapter 3 THE DERIVATIVE

Your Turn 5

3.1 Warmup Exercises

Let lim f ( x) = 3 and lim g ( x) = 4. x 2

x2

W1. Some trial with the factors of 8 and 15 shows that 8 x 2 + 22 x + 15 = (2 x + 3)(4 x + 5).

ù2

é lim [ f ( x) + g ( x) ]2 = ê lim [ f ( x) + g ( x) ] ú êë x  2 úû x 2

é ù2 = ê lim f ( x) + lim g ( x) ú êë x  2 úû x2

W2. Some trial with the factors of 12 shows that 12 x 2 - 7 x - 12 = (3x - 4)(4 x + 3).

= [ 3 + 4 ]2 = 72 = 49

W3.

Your Turn 6 3 2

lim

x -2

(-2) + 4 x +4 2 = = = x -2 -2 -2 = -1

W4.

3x 2 + x - 14

(3x + 7)( x - 2) 3x + 7 = , ( x + 2)( x - 2) x+2 x -4 provided that x is not equal to 2 or -2. 2

2 x 2 + x - 15

(2 x - 5)( x + 3) 2x - 5 = , ( x + 3)( x - 3) x-3 x -9 provided that x is not equal to 3 or -3. 2

3.1 Exercises

Your Turn 7

True

False. When the limit exists, the functional value may be undefined, it could be something other than the limit, or it could be the limit.

= (-3) - 4

True

= -7

False. If the limit has the indeterminate form (0/0), then there is no conclusion. More information is needed. (See Section 12.7 for more details.)

x +1 x +1

False. Since ¥ is not a real number, the limit in this case does not exist.

( x )2 - 1 x 1 ( x - 1)( x + 1)

True

Since lim f (x) does not equal lim f ( x ),

( x - 4) ( x + 3) x 2 - x - 12 = lim lim x+3 x+3 x -3 x -3

( x ¹ 3)

= lim x - 4 x -3

Your Turn 8 x -1 x -1 = lim ⋅ x 1 x - 1 x 1 x - 1 lim

= lim

x -1 = lim x 1 ( x - 1) ( x + 1) = lim

x 1

x  2+

x  2-

lim f ( x ) does not exist. The answer is c.

x2

1 1 1 = = 1+1 2 x +1

Since lim f ( x) = lim f ( x) = -1, x  2-

x  2+

lim f ( x) = -1. The answer is a.

x 2

Your Turn 9 2 x2 3x 4 + 2 - 2 2 x x x lim = lim 2 5x 7 x ¥ 6 x 2 - 5 x + 7 x ¥ 6 x - 2 + 2 2 x x x 3 4 2+ x - 2 x = lim x ¥ 6 - 5 + 7 x x2

2 x 2 + 3x - 4

2+0-0 2 1 = = 6-0+0 6 3

Since lim f ( x) = lim f ( x) = 6, x  4-

x  4+

lim f ( x) = 6. The answer is b.

x 4

10. Since lim f ( x) = lim f ( x) = -¥, x 1-

x 1+

lim f ( x) = -¥. The answer is b.

x 1

Section 3.1

165

11. (a) By reading the graph, as x gets closer to 3 from the left or right, f (x) gets closer to 3.

(iv) f (-2) does not exist since there is no point on the graph with an x-coordinate of –2.

lim f ( x) = 3

x3

(b) By reading the graph, as x gets closer to 0 from the left or right, f (x) gets closer to 1.

(b)

(i)

By reading the graph, as x gets closer to –1 from the left, f ( x) gets closer to - 12 .

lim f ( x) = 1.

x0

lim

x -1-

12. (a) By reading the graph, as x gets closer to 2 from the left or right, F(x) gets closer to 4.

(ii) By reading the graph, as x gets closer to –1 from the right, f ( x)

lim F ( x) = 4

x 2

gets closer to - 12 .

(b) By reading the graph, as x gets closer to –1 from left or right, F(x) gets closer to 4.

lim

x -1

(iii) Since lim

x -1-

13. (a) By reading the graph, as x gets closer to 0 from the left or right, f (x) gets closer to 0.

lim

x -1+

lim f ( x) = 0

(iv)

x0

(

16. (a)

(i)

By reading the graph, as x gets closer to 1 from the left, f (x) gets closer to 1. lim f ( x) = 1

x 1-

(ii) By reading the graph, as x gets closer to 1 from the right, f (x) gets closer to 1. lim f ( x) = 1

x 1+

x3

(iii) Since lim f ( x ) = 1 and

(b) By reading the graph, as x gets closer to 5 from the left, g(x) gets closer to –2, but as x gets closer to 5 from the right, g(x) gets closer to 1.

x 1-

lim f ( x ) = 1, lim f ( x) = 1.

x 1+

lim g ( x) does not exist.

x5

By reading the graph, as x gets closer to –2 from the left, f (x) gets closer to –1. lim f ( x) = -1 x -2-

(ii) By reading the graph, as x gets closer to –2 from the right, f ( x) gets closer to - 12 .

(iii) Since lim

)

f (-1) = - 12 since -1, - 12 is a point

lim g ( x) = 2

x -2+

f ( x) = - 12 and

of the graph.

14. (a) By reading the graph, as x gets closer to 3 from the left or right, g(x) gets closer to 2.

lim

1 2

x -1

lim f ( x) does not exist.

(i)

f ( x) = -

f ( x) = - 12 , lim f ( x) = - 12 .

x 2

15. (a)

x -1

lim F ( x) = 4

(b) By reading the graph, as x gets closer to 2 from the left, f (x) gets closer to –2, but as x gets closer to 2 from the right, f (x) gets closer to 1.

f ( x) = - 1 2

f ( x) = -

(iv) f (1) = 2 since (1, 2) is part of the graph. (i) By reading the graph, as x gets closer to 2 from the left, f (x) gets closer to 0. lim f ( x) = 0

x  2-

(ii) By reading the graph, as x gets closer to 2 from the right, f ( x) gets closer to 0. lim f ( x) = 0

x  2+

(iii) Since lim f ( x) = 0

1 2

f ( x) = -1 and

(b)

x 1

x  2-

and lim f ( x) = 0, lim f ( x) = 0. lim

x  2+

x -2 x -2 1 f ( x) = - , lim f ( x) does not exist. 2 x -2

x2

(iv) f (2) = 0 since (2, 0) is point of the graph.

166

Chapter 3 THE DERIVATIVE

17. By reading the graph, as x moves further to the right, f ( x) gets closer to 3. Therefore, lim f ( x) = 3.

24.

f ( x) =

2 x3 + 3x 2 - 4 x - 5 ; find lim f ( x). x +1 x -1

x ¥

-1.1

-1.01

-1.001

f ( x) -3.68 -3.969 -3.996

18. By reading the graph, as x moves further to the left, g(x) gets larger and larger. Therefore, lim g ( x ) = ¥ (does not exist).

-0.999 -0.99 -0.9

f ( x) -4.002 -4.02 -4.28

x -¥

19.

As x approaches –1 from the left or the right, f ( x) approaches –4. lim f ( x) = -4

lim F(x) in Exercise 12 exists because

x 2

x -1

lim F ( x) = 4 and lim F ( x) = 4.

x  2-

x  2+

lim f ( x) in Exercise 15 does not exist since

x -2

lim

x -2-

f ( x) = -1, but

25.

f ( x) = - 12 .

lim

x -2+

x -2 ; find lim h( x). x -1 x 1

h( x) =

0.9

0.99

0.999

h( x) 10.51317 100.50126 1000.50013

21. From the table, as x approaches 1 from the left or the right, f (x) approaches 4.

1.001

1.01

1.1

h( x) -999.50012 -99.50124 -9.51191

lim f ( x) = 4

x 1

lim = ¥

x 1-

22.

lim = -¥

f ( x) = 2 x 2 - 4 x + 7; find lim f ( x).

x 1+

x 1

Substitute 0.9 for x in the expression at the right to get f (0.9) = 5.02.

Thus, lim h( x) does not exist. x 1

Continue substituting to complete the table. x

0.9

0.99

0.999

26.

f ( x) =

f ( x) 5.02 5.0002 5.000002 x

1.001

1.01

x -3 ; find lim f ( x). x-3 x3

2.9 2.99 2.999 f ( x) 12.9706 127.0838 1268.237 x

1.1

f ( x) 5.000002 5.0002 5.02

3.001 3.01 3.1 f ( x) -1267.66 -125.506 -12.6795 x

As x approaches 1 from the left or the right, f (x) approaches 5. lim f ( x) = 5

lim f ( x) = ¥

x 1

x  3-

lim f ( x) = -¥

23.

k ( x) =

x  3+

x - 2x - 4 ; find lim k ( x). x-2 x 2

1.9

1.99

Thus, lim f ( x) does not exist. x3

1.999

k ( x) 9.41 9.9401 9.9941 x

2.001

2.01

2.1

27.

lim f ( x) f ( x) 9 1 = x4 = = lim g ( x) 27 3 x  4 g ( x) lim

x 4

k ( x) 10.006 10.0601 10.61

As x approaches 2 from the left or the right, k ( x) approaches 10. lim k ( x) = 10 x 2

28.

lim [( g ( x) ⋅ f ( x)]

x 4

é ù é ù = ê lim g ( x) ú ⋅ ê lim f ( x) ú ëê x  4 ûú ëê x  4 ûú = 27 ⋅ 9 = 243

Section 3.1 29.

167 f ( x ) = lim [ f ( x )1/ 2 ]

lim

x4

37.

x4

é ù1/2 = ê lim f ( x ) ú êë x  4 úû = 91/2 = 3

30.

lim log3 f ( x) = log3 lim f ( x)

x 4

= log3 9 = 2

31.

x+2 x+2 lim 2 = lim x 2 x - x - 6 x  2 ( x - 3)( x + 2) 1 = lim x2 x - 3 1 = 2-3 = -1

38.

lim

x-3

x -3 x 2 + 5 x + 4

-3 - 3 2

(-3) + 5(-3) + 4

lim f ( x)

lim 2 f ( x) = 2 x  4

x 4

39.

= 2

= 512

x2 - 9 ( x - 3)( x + 3) = lim x-3 x3 x - 3 x3 = lim ( x + 3) lim

x3

32.

= 3+3

é ù2 lim [1 + f ( x)] 2 = ê lim (1 + f ( x)) ú x 4 ëê x  4 ûú é ù2 = ê lim 1 + lim f ( x) ú x 4 ëê x  4 ûú

40.

= (1 + 9)2 = 102 = 100

33.

f ( x ) + g ( x) 2 g ( x) x 4 lim [ f ( x) + g ( x)] = x 4 lim 2 g ( x)

x2 - 4 ( x + 2)( x - 2) = lim ( x + 2) x -2 x + 2 x -2 = lim ( x - 2) lim

x -2

= -2 - 2 = -4

lim

41.

lim

5x2 - 7 x + 2

x 1

x -1

x 4

lim f ( x) + lim g ( x)

= x 4

x 4

2 lim g ( x) x4

9 + 27 36 2 = = = 2(27) 54 3

34.

lim [5 g ( x) + 2] 5 g ( x) + 2 = x 4 lim [1 - f ( x)] x  4 1 - f ( x)

42.

lim

x2 - 9

x -3 x 2 + x - 6

x 4

5 lim g ( x) + lim 2 =

x 4

lim 1 - lim f ( x)

x4

5.27 + 2 137 = =1- 9 8

35.

lim (2 x - 4 x + 5) = 2(3) - 4(3) + 5 = 11

x3

43.

(5x - 2)( x - 1) x 1 ( x + 1)( x - 1) 5x - 2 = lim x 1 x + 1 5-2 = 2 3 = 2 = lim

( x - 3)( x + 3) x -3 ( x - 2)( x + 3) x-3 = lim x -3 x - 2 -3 - 3 = -3 - 2 6 -6 = = 5 -5

= lim

x2 - x - 6 ( x - 3)( x + 2) = lim x+2 x+2 x -2 x -2 = lim ( x - 3) lim

x -2

36.

lim ( x - x) =

x 8

8 - 8 = 2 - 8 = -6

= -2 - 3 = -5

168 44.

Chapter 3 THE DERIVATIVE x 2 - 3x - 10 ( x - 5)( x + 2) = lim 5 ( x - 5) x x5 x5 = lim ( x + 2) lim

47.

x5

=5+2= 7

45.

lim

1 - 1 x +3 3

x æ 1 1 öæ 1 ö = lim çç - ÷÷çç ÷÷ ÷ç x ÷ø ç è 3 3 øè x + x0

x0

é 3 x + 3 ùú æç 1 ö÷ = lim ê ç ÷ 3( x + 3) úû èç x ÷ø x  0 êë 3( x + 3) 3- x-3 = lim x  0 3( x + 3)( x) -x = lim x  0 3( x + 3) x -1 = lim x  0 3( x + 3) -1 = 3(0 + 3) 1 =9

46.

lim

-1 + 1 x+2 2

48.

49.

x æ -1 1 öæ 1 ö = lim çç + ÷÷çç ÷÷ ÷ç x ø÷ ç è 2 2 øè x + x0

x0

x -5 x -5 x +5 = lim ⋅ x 25 x 25 x  25 x  25 x +5 x - 25 = lim x  25 ( x - 25)( x + 5) 1 = lim x  25 x + 5 1 = 25 + 5 1 = 10 lim

lim

x - 6 = lim

x  36 x - 36

x -6⋅

x +6 x +6 ( x - 36) = lim x  36 ( x - 36)( x + 6) 1 = lim x  36 x + 6 1 = 36 + 6 = 1 6+6 = 1 12 x  36 x - 36

( x + h) 2 - x 2 h h 0 lim

x 2 + 2hx + h 2 - x 2 h h 0

= lim

é -2 x + 2 ùú æç 1 ö÷ = lim ê + ç ÷ ê 2( x + 2) úû çè x ø÷ x  0 ë 2( x + 2) -2 + x + 2 = lim x  0 2( x + 2)( x ) x = lim x  0 2 x( x + 2) 1 = lim x  0 2( x + 2) 1 1 = = 2(0 + 2) 4

2hx + h 2 h h 0 h(2 x + h) = lim h h 0 = lim (2 x + h) = lim

h 0

= 2x + 0 = 2x

50.

( x + h)3 - x 3 h h 0 lim

( x3 + 3x 2 + 3xh 2 + h3 ) - x3 h h0

= lim

3x 2h + 3xh 2 + h3 h h0

= lim

h(3x 2 + 3xh + h 2 ) h h0

= lim

= lim (3x 2 + 3xh + h 2 ) h0 2

= 3x + 3x(0) + (0)2 = 3x 2

Section 3.1 51.

169

lim ( x 2 + 4 x - 3) = ¥

57.

x ¥

The limit does not exist. 52.

lim ( x3 + 2 x + 8) = -¥

x -¥

3x = lim 7 x x 1 7 x -1 x ¥ x ¥ - x x lim

= lim

58.

3 3 = 7-0 7 8x

54.

+ x 8x + 2 = lim 4xx x -¥ 4 x - 5 x -¥ - 5x x lim

55.

lim

59.

8+0 = 2 4-0

-3x 2 + 2 x

3x 2 + 2 x x2 x2 = lim 2 x -¥ 2 x - 2 x + 1 x2 x2 x2 3 + 2x = lim x -¥ 2 - 2 + 12 x x

56.

lim

x ¥

3x + 2

60.

x2 + 2x - 5 2 2 x2 = lim x 2 x 3x + 2 x ¥ x2 x2 1 + 2x - 52 x = lim x ¥ 3 + 22 x

lim

0-0 0 = =0 3+0 3

2x3 - x - 3

x ¥ 6 x 2 - x - 1

The limit does not exist.

3-0 3 = 2+0+0 2 x2 + 2x - 5

2 x2 - 1 4 x4 lim = lim x 4 4 x ¥ 3x + 2 x ¥ 3x + 2 x4 x4 2 - 1 2 x x4 = lim 2 x ¥ 3 + 4 x

2x2 - 1

2 x3 - x - 3 2 x2 x2 = lim x 2 x ¥ 6 x - x - 1 x2 x2 x2 2 x - 1x - 32 x = ¥ = lim x ¥ 6 - 1 - 12 x x

x -¥ 2 x 2 - 2 x + 1

0+0-0 =0 2-0-0

8 + 2x = lim x -¥ 4 - 5 x =

x ¥ 2 x 4 - 3x3 - 2

x ¥ 7 - 1 x

3x3 + 2 x - 1

3 x3 + 2 x - 1 4 x4 x4 = lim x 4 3 2 3 x x x ¥ - 4 - 24 x4 x x 3 + 2 - 1 x x3 x4 = lim 3 2 x ¥ 2 - 4 x x

The limit does not exist.

53.

lim

1+ 0-0 1 = 3+0 3

lim

x ¥

x 4 - x 3 - 3x 7x2 + 9

x 4 - x 3 - 3x 2 2 x2 = lim x 2 x 7x + 9 x ¥ x2 x2 x 2 - x - 3x = lim x ¥ 7 + 92 x

=¥

The limit does not exist.

61.

2 x2 - 7 x4 x2 x2 lim = lim 2 x ¥ 9 x 2 + 5 x - 6 x ¥ 9 x + 5 x - 6 x2 x2 x2 2

2x2 - 7 x4

= lim

2 - 7x

x ¥ 9 + 5 - 62 x x

The denominator approaches 9, while the numerator becomes a negative number that is larger and larger in magnitude, so 2x2 - 7x4 lim = -¥ (does not exist). x ¥ 9 x 2 + 5 x - 6 Copyright © 2022 Pearson Education, Inc.

170

62.

Chapter 3 THE DERIVATIVE -5 x3 - 4 x 2 + 8

lim

x ¥ 6 x 2 + 3x + 2

-5 x 3 - 4 x 2 + 8 2 x2 x2 = lim x 2 6 3 x x 2 x ¥ + 2 + 2 x2 x x 8 -5 x - 4 + 2 x = lim x ¥ 6 + 3 + 22 x x

The denominator approaches 6, while the numerator becomes a negative number that is larger and larger in magnitude, so -5x 3 - 4 x 2 + 8 lim = -¥ (does not exist). x ¥ 6 x 2 + 3x + 2 63.

lim

x -1-

f ( x ) = 1 and

lim

x -1+

f ( x ) = 1.

Therefore lim f ( x) = 1.

lim

2 x 2 + kx - 9

x  3 x2 - 4x + 3

lim g ( x ) = -1.

x -2+

Therefore lim g ( x) = -1.

numerator contains a factor of x - 3. This occurs when 2(3) 2 + k (3) - 9 = 0, which requires k = -3. The fraction is then 2 x 2 - 3x - 9 (2 x + 3)( x - 3) 2x + 3 = = ( x - 1)( x - 3) ( x - 1)( x - 3) x -1 provided x ¹ 3. The limit of the fraction as 3(3) + 3 9 x  3 is = . 3-1 2

x2 - 9 . x-3

x 2.9 2.99 2.999 3.001 3.01 3.1 f ( x) 5.9 5.99 5.999 6.001 6.01 6.1 x2 - 9 = 6. x3 x - 3

lim f ( x) = lim

x -2

x3

65. (a) lim f ( x) = 2. x3

lim f ( x) does not exist since lim f ( x) = 2

(b)

will exist when the

x3

lim g ( x) = -1 and

x -2-

x 2 - 4 x + 3 = ( x - 1)( x - 3). Thus

69. Find lim f ( x), where f ( x) =

x -1

64.

68. The denominator

x 5

x5

and lim f ( x) = 8. x  5+

66. (a) lim g ( x) does not exist since lim g ( x) = 5 x  0-

x0

x2 - 4 . x+2 x -2 x -2.1 -2.01 -2.001 -1.999 -1.99 -1.9 g ( x) -4.1 -4.01 -4.001 -3.999 -3.99 -3.9

70. Find lim g ( x), where g ( x) =

x2 - 4 = -4. x -2 x + 2

lim g ( x) = lim

x -2

and lim g ( x) = -2. x  0+

71. Find lim f ( x), where f ( x) =

lim g ( x) = 7.

(b)

x 2 - 3x + 2 = ( x - 1)( x - 2). Thus 2

x 0.9 0.99 0.999 1.001 1.01 1.1 f ( x) 1.316 1.482 1.498 1.502 1.517 1.667

67. The denominator 3x + kx - 2

x2 - 1

x 1

x 3

lim

5x2 - 7 x + 2

will exist when the numerator

x  2 x - 3x + 2 contains a factor of x - 2. This occurs when 2

3(2) + k (2) - 2 = 0, which requires k = -5. The fraction is then 3x 2 - 5 x - 2 (3x + 1)( x - 2) 3x + 1 = = ( x - 1)( x - 2) ( x - 1)( x - 2) x - 1 pro

vided x ¹ 2. The limit of the fraction as x  2 3(2) + 1 is = 7. 2 -1

lim f ( x) = lim

x 1

5x 2 - 7 x + 2

x 1

x -1

72. Find lim g ( x), where g ( x) = x3

= 1.5 =

3 . 2

x 2 -9 . x 2 + x -6

x -3.1 -3.01 -3.001 g ( x) 1.196 1.1996 1.19996 x -2.999 -2.99 -2.9 g ( x) 1.20004 1.2004 1.204 lim g ( x) = lim

x -3

x2 - 9

x -3 x 2 + x - 6

= 1.2 =

6 . 5

Section 3.1 73. (a)

171 3x

lim

x -2 ( x + 2)

lim

x -2+ ( x + 2)

x -2- ( x + 2)

x  0+

as the graph gets closer to x = 0, the value of ln x get smaller.

= -¥ 3

and lim

lim ln x = -¥ (does not exist) since,

77. (a)

does not exist since

(b) The graph of y = ln x has a vertical asymptote at x = 0 since lim ln x = -¥.

= ¥. 3

x  0+

(b) Since ( x + 2)3 = 0 when x = -2, x = -2 is the vertical asymptote of the graph of F(x).

78. (a) y = x ln x

From the graph, it appears that

lim x ln x = 0.

x-2

x  0+

F(x) does not exist. 74.

G ( x) =

(a)

x 1

-6

0.5

0.1

0.01

0.001

y 0 -0.347 -0.230 -0.0461 -0.0069

( x - 4)2 lim G( x) = -¥ (does not exist), since by

(b) y = x (ln x)2

x 4

From the graph it appears that

looking at the graph of G(x), we see that as x gets closer to 4 from either the right or the left, g(x) gets smaller.

lim x(ln x)2 = 0.

x  0+

(b) Since ( x - 4)2 = 0 when x = 4, x = 4 is the vertical asymptote of the graph of G(x). (c) The two answers are related. Since x = 4 is a vertical asymptote, we know the lim G( x) does not exist.

x 1

0.5

0.1

0.01

0.001

y 0

0.240

0.53

0.212 0.048

lim x(ln x)n = 0.

(c)

x  0+

x 4

75. (a)

x -¥

x 4 + 4 x3 - 9 x 2 + 7 x - 3 x -1 x 1

further to the left, e x gets closer to 0.

(a)

lim e = 0 since, as the graph goes

(b) The graph of e x has a horizontal asymptote

at y = 0 since 76. (a)

lim e x = 0.

x -¥

81.

lim

x 1.01 1.001 1.0001 0.99 0.999 0.9999 f ( x) 5.0908 5.009 5.0009 4.9108 4.991 4.9991

As x  1- and as x  1+ , we see that f ( x)  5.

y = xe-x From the graph, it appears that

(b) Graph

lim xe-x = 0.

y =

x ¥

y 0.37 0.00045 9.64 ´ 10-21

(b) y = x 2e-x From the graph, it appears that lim x 2e-x = 0.

x ¥

y 0.37 0.0045 4.82 ´ 10-19

(c)

lim x ne-x = 0.

x ¥

x 4 + 4 x3 - 9 x 2 + 7 x - 3 x -1

on a graphing calculator. One suitable choice for the viewing window is [-6, 6] by [ -10, 40] with Xscl = 1, Yscl = 10. Because x - 1 = 0 when x = 1, we know that the function is undefined at this x-value. The graph does not show an asymptote at x = 1. This indicates that the rational expression that defines this function is not written in lowest terms, and that the graph should have an open circle to show a “hole” in the graph at x = 1. The graphing calculator

172

Chapter 3 THE DERIVATIVE doesn’t show the hole, but if we try to find the value of the function at x = 1, we see that it is undefined. (Using the TABLE feature on a TI-84 Plus, we see that for x = 1, the y-value is listed as “ERROR.”) By viewing the function near x = 1, and using the ZOOM feature, we see that as x gets close to 1 from the left or the right, y gets close to 5, suggesting that

Because x + 1 = 0 when x = -1, we know that the function is undefined at this x-value. The graph does not show an asymptote at x = -1. This indicates that the rational expression that defined this function is not written lowest terms, and that the graph should have an open circle to show a “hole” in the graph at x = - 1. The graphing calculator doesn’t show the hole, but if we try to find the value of the function at x = -1, we see that it is undefined. (Using the TABLE feature on a TI-83, we see that for x = -1, the y-value is listed as “ERROR.”)

x 4 + 4 x3 - 9 x 2 + 7 x - 3 = 5. x -1 x 1

x 4 + x - 18

lim

x2 - 4

x 2

(a) x

2.01

2.001

2.0001 1.99

1.999

1.9999

f ( x)

8.29

8.25

8.21

By viewing the function near x = -1 and using the ZOOM feature, we see that as x gets close to -1 from the left or right, y gets close to 0.3333, suggesting that

(b) Graph

y =

x 4 + x - 18 2

x -4

x1/3 + 1 1 = 0.3333 or . 3 x -1 x + 1

lim

One suitable choice for the viewing window is [-5,5] by [0, 20] Because x 2 - 4 = 0 when x = -2 or x = 2, we know that the function is undefined at these two x-values. The graph shows an asymptote at x = -2 . There should be open circle to show a “hole” in the graph at x = 2. The graphing calculator doesn’t show the hole, but if we try to find the value of the function at x = 2, we see that it is undefined. (Using the TABLE feature on a TI-84 Plus, we see that for x = 2, the yvalue is listed as “ERROR.”) By viewing the function near x = 2 and using the ZOOM feature, we verify that the required limit is 8.25. (We may not be able to get this value exactly.)

83.

x1/3 + 1 x -1 x + 1

84.

lim

x3/2 - 8

x  4 x + x1/2 - 6

(a)

x 4.1 4.01 4.001 4.0001 f ( x) 2.4179 2.4018 2.4002 2.4 x 3.9 3.99 3.999 3.9999 f ( x) 2.3819 2.3982 2.3998 2.4

(b) Graph y =

x3/2 - 8 x + x1/2 - 6

This function is undefined at x = 4 because this value would make the denominator equal to 0. However, by viewing the function near x = 4 and using the ZOOM feature, we verify that the required limit is 2.4.

lim

(a)

x1/3 + 1 . x +1

One suitable choice for the viewing window is [-5, 5] by [-2, 2]

lim

82.

y =

(b) Graph

x -1.01 -1.001 -1.0001 f ( x) 0.33223 0.33322 0.33332 x -0.99 -0.999 -0.9999 f ( x) 0.33445 0.33344 0.33334

We see that as x  -1- and as x  -1+ , f ( x)  0.3333 or 13 .

Section 3.1 85.

lim

x ¥

173 (b) As x  -¥,

9x2 + 5 2x

9x2 + 5 3| x |  . 2x 2x

Graph the functions on a graphing calculator. A good choice for the viewing window is [-10, 10] by [-5, 5].

Since x < 0, | x | = -x, so 3 | x| 3(-x) 3 = =- . 2x 2x 2

Thus, 9x2 + 5 3 = - or -1.5. 2x 2 x -¥ lim

(a) The graph appears to have horizontal asymptotes at y = 1.5. We see that as x  ¥, y  1.5, so we determine that lim

x ¥

9x2 + 5 = 1.5. 2x

87.

lim

x -¥

36 x 2 + 2 x + 7 3x

Graph this function on a graphing calculator. A good choice for the viewing window is [-10, 10] by [-5,5].

(b) As x  ¥, 9x2 + 5 3 |x|  . 2x 2x

Since x > 0, | x | = x, so 3 | x| 3x 3 = = . 2x 2x 2

Thus, lim

x ¥

9x2 + 5 3 = or 1.5. 2x 2

(a) The graph appears to have horizontal asymptotes at y = 2. We see that as x  -¥, y  -2, so we determine that lim

x -¥ 2

86.

lim

x -¥

9x + 5 2x

36 x 2 + 2 x + 7 = -2. 3x

(b) As x  -¥,

Graph this function on a graphing calculator. A good choice for the viewing window is [-10,10] by [-5,5].

36 x 2 + 2 x + 7 6|x| .  3x 3x

Since x < 0, | x | = -x, so 6 |x| 6(-x) = = -2. 3x 3x

Thus, (a) The graph appears to have horizontal asymptotes at y = 1.5. We see that as x  -¥, y  -1.5, so we determine that lim

x -¥

9x2 + 5 = -1.5. 2x

36 x 2 + 2 x + 7 = -2. 3x x -¥ lim

174 88.

Chapter 3 THE DERIVATIVE (b) As x  ¥,

36 x 2 + 2 x + 7 3x x ¥ lim

(1 + 5 x1/3 + 2 x5/3 )3 x5

Graph this function on a graphing calculator. A good choice for the viewing window is [-10, 10] by [-5, 5].

Thus, lim

(a) The graph appears to have horizontal asymptotes at y = 2. We see that as x  ¥, y  2, so we determine that

8 x5 x5

(1 + 5x1/3 + 2 x5/3 )3 x5

x ¥

90.



= 8.

(1 + 5 x1/3 + 2 x5/3 )3

lim

x5 Graph this function on a graphing calculator. A good choice for the viewing window is [-60, 60] by [0, 20] with Xscl = 10, Yscl = 10. x -¥

36 x 2 + 2 x + 7 = 2. 3x x ¥ lim

(b) As x  ¥, 36 x 2 + 2 x + 7 6|x| .  3x 3x

Since x > 0, | x | = x, so

(a) The graph appears to have a horizontal asymptote at y = 8. We see that as x  -¥, y  8, so we determine that

6|x| 6x = = 2. 3x 3x

lim

x ¥

(b) As

(1 + 5x1/3 + 2 x5/3 )3

Thus,

(1 + 5x1/3 + 2 x5/3 )3

x5 lim

= 8.

x -¥

36 x 2 + 2 x + 7 = 2. Thus, lim 3x x ¥

89.

(1 + 5x1/3 + 2 x5/3 )3



8 x5 x5

= 8.

(1 + 5x1/3 + 2 x5/3 )3

x -¥

= 8.

Graph this function on a graphing calculator. A good choice for the viewing window is [-20, 20] by [0, 20] with Xscl = 5, Yscl = 5.

92. (a)

lim G(t )

x 12

As t approaches 12 from either direction, the value of G(t) for the corresponding point on the graph approaches 3. Thus, lim G(t ) = 3 which represents x 12

3 million gallons. (b)

lim G(t ) = 1.5

x 16+

lim G(t ) = 2

(a) The graph appears to have a horizontal asymptote at y = 8. We see that as x  ¥, y  8, so we determine that 1/3

lim

x ¥

(1 + 5 x

+ 2x 5

5/3 3

)

x 16-

Since

lim G(t ) ¹ lim G(t ),

x 16+

x 16-

lim G(t ) does not exist.

x 16

= 8.

Section 3.1

175

(c) G(16) is the value of function G(t) when t = 16. This value occurs at the solid dot on the graph G(16) = 2 which represents 2 million gallons. (d) The tipping point occurs at the break in the graph, when t = 16 months. 93.

(a)

lim

x  2017-

63s

63s = lim s s 8 s ¥ s + 8 s ¥ + lim

63 1+ 0 = 63 =

T ( x) = 7.5 cents per dollar

The number of items of work a new employee produces gets closer and closer to 63 as the number of days of training increases.

x  207+

lim T ( x) = does not exist

x  2017

(e) T (2017) = 7.25 cents per dollar

lim

C (t ) = 49 cents

lim

C (t ) = 50 cents

t  2018t  2018+

Q(t ) = 60 - 40e-0.3t

98.

(a) Q(0) = 60 - 40e-0.3(0) = 60 - 40 = 20 units (b) Q(2) = 60 - 40e-0.3(2) » 38 units (c) Q(12) = 60 - 40e-0.3(12) » 59.97 units

lim C (t ) = does not exist

t  2018

(d) C (2018) = 50 cents

(d) 95.

C ( x) = 15,000 + 6 x C ( x) =

15,000 +6= 0+6= 6 x x ¥

x ¥

This means that the average cost approaches $6 as the number of flashdrives produced becomes very large.

t ¥

S (t ) =

99.

90, 000 1 + 4e-0.06t 90, 000

(a) S (1) =

1 + 4e-0.06(1) » 18,880 units

C ( x) = 0.00003964 x + 191.16 C ( x) x 0.00003964 x + 191.16 = x = 0.00003964 + 191.16 x

lim Q(t ) = lim (60 - 40e-0.3t )

t ¥

= 60 units The number of items a new worker should be able to assemble gets closer and closer to 60 units as the number of months increases.

C ( x) 15,000 + 6 x 15,000 = = +6 x x x

lim C ( x) = lim

96.

63 = lim s ¥ 1 + 8 s

lim T ( x) = 7.25 cents per dollar

(d)

(c)

63s s+8

lim T ( x) = 8.25 cents per dollar

(c)

(b)

P( s ) =

x  2010

(b)

94. (a)

97.

(b) S (10) =

C ( x) =

90, 000

1 + 4e-0.06(10) » 28,167 units

90, 000

1 + 4e-0.06(100) » 89,116 units

lim C ( x) = 0.00003964

x ¥

The average cost approaches $0.00003964 per mile as the number of miles becomes very large.

(d)

lim S (t ) = lim

t ¥

90, 000

t ¥ 1 + 4e-0.06t

= 90, 000 units The number of units sold approaches 90,000 as the number of months becomes large.

176

Chapter 3 THE DERIVATIVE

é é ù 1 - (1 + i)-n ùú ú 100. lim êê R êê úú i n ¥ ê ê ûú úû ë ë R = lim éê 1 - (1 + i)-n ùú û i n ¥ ë ù Ré = ê lim 1 - lim (1 + i)-n ú i ëê n ¥ n ¥ ûú R R = [1 - 0 ] = i i

(b)

t ¥

= 155 lim (1 - e-0.0133t ) t ¥

æ ö = 155çç lim 1 - lim e-0.0133t ÷÷÷ è t ¥ ø t ¥ = 155(1) - 155 lim e-0.0133t t ¥

= 155 - 155(0) = 155

Thus,

é ù é æ 1 + g ö÷n ùú ú ê R ê ç lim 1 ÷ ê 101. çç ê úú è 1 + i ÷ø ú ú n ¥ ê i - g ê ë ûû ë é æ 1 + g ö÷n ùú R = lim êê 1 - çç çè 1 + i ÷÷ø úú i - g n ¥ ê ë û é æ 1 + g ÷ön ùú R ê = ÷ ú ê lim 1 - lim ççç i - g ê n ¥ n ¥ è 1 + i ÷ø ú ë û assuming i > g ,

lim D(t ) = 155.

t ¥

Going back in time (t is years before 1990), the depth of the sediment approaches 155 cm. 104.

A(h) =

0.17h

h2 + 2 0.17h lim A(h) = lim 2 x ¥ x ¥ h + 2 = lim

R [1 - 0 ] i-g R = i-g

0.17h h2

2 x ¥ h + 2 h2 h2 0.17 h = lim x ¥ 1 + 22 h

(-0.0685(65))

102. (a) N (65) = 71.8e-8.96e » 64.68 To the nearest whole number, this species of alligator has approximately 65 teeth after 65 days of incubation by this formula. (b) Since lim (-8.96e-0.0685t ) = -8.96 ⋅ 0 = 0, it

t ¥

follows that lim 71.8e-8.96e

(-0.0685t )

= 71.8e0

t ¥

= 71.8 ⋅ 1 = 71.8 So, to the nearest whole number, lim N (t ) » 72. Therefore, by this model a t ¥

newborn alligator of this species will have about 72 teeth. 103. (a)

lim D(t ) = lim 155(1 - e-0.0133t )

t ¥

D(t ) = 155(1 - e-0.0133t )

(

D(20) = 155 1 - e-0.0133(20)

)

= 155(1 - e-0.266 ) » 36.2

The depth of the sediment layer deposited below the bottom of the lake in 1970 was 36.2 cm.

0 = 0 1+ 0 This means that the concentration of the drug in the bloodstream approaches 0 as the number of hours after injection increases. =

105. (a)

æ ö p2 = 1 + çç 0.7 - 1 ÷÷ [1 - 2(0.2)]2 = 0.572 è 2 2ø

æ ö (b) p4 = 1 + çç 0.7 - 1 ÷÷ [1 - 2(0.2)]4 = 0.526 2 è 2ø

æ ö p8 = 1 + çç 0.7 - 1 ÷÷ [1 - 2(0.2)]8 = 0.503 è ø 2 2 é ù æ ö lim Pn = lim ê 1 + çç p0 - 1 ÷÷÷ (1 - 2 p) n ú è ø ê úû 2 2 n ¥ n ¥ ë æ ö = 1 + lim çç p0 - 1 ÷÷÷ (1 - 2 p)n è 2 2ø n ¥ æ ö = 1 + çç p0 - 1 ÷÷÷ lim (1 - 2 p)n è 2 2 ø n ¥ æ ö = 1 + çç p0 - 1 ÷÷÷ ⋅ 0 = 1 è 2 2ø 2

The number in parts (a), (b), and (c) represent the probability that the legislator will vote yes on the second, fourth, and eighth votes. In (d), as the number of roll calls increases, the probability gets close to 0.5, but is never less than 0.5.

Section 3.2

177 W3.

3.2 Continuity

x  3-

lim f ( x) = lim (2 x - 2) = 4

x  3+

Your Turn 1 f ( x) =

lim f ( x) = lim ( x + 1) = 4

x  3-

x  3+

Thus lim f ( x) = 4. x3

5x + 3

The square root function is discontinuous wherever 5x + 3 < 0. There is a discontinuity when 3 5a + 3 < 0, or a < - . 5

W4.

lim f ( x) = lim 2 x - 2 = 8

x  5-

lim f ( x) = lim 4 x + 1 = 21

x  5+

Thus lim f ( x) does not exist. x5

Your Turn 2

W5. Since in an open interval containing 6, f ( x) is defined as 4 x + 1, lim f ( x) = lim (4 x + 1) = 25.

ì 5 x - 4 if x < 0 ï ï ï 2 ï f ( x) = í x if 0 £ x £ 3 ï ï ïï x + 6 if x > 3 ï î y

x6

3.2 Exercises

( 3, 9)

True

False. A function f is continuous at x = c if the three conditions are satisfied: 1. f (c) is defined. 2. lim f ( x) exists, and

( 0, 0) 4

−4

( 0, −4)

−2

xc

3. lim f ( x) = f (c). xc

−8

lim f ( x) = -4

x  0-

True

False. A general logarithmic function, y = log a x, where a > 0 and a ¹ 1, is continuous for all x > 0.

Discontinuous at x = -1

lim f ( x) = 0

x  0+

Because lim f ( x) ¹ lim f ( x), the limit doesn’t x  0-

x  0+

(a)

exist, so f is discontinuous at x = 0.

W1. If x ¹ 2, 2

x - 5x + 6

Thus lim

x 2

( x - 2)(2 x - 7) 2x - 7 . = ( x - 2)( x - 3) x-3

2 x 2 - 11x + 14 2

x - 5x + 6

2x - 7 x2 x - 3 2(2) - 7 = = 3. 2-3

= lim

W2. If x ¹ 4, 3x 2 - 4 x - 32

( x - 4)(3x + 8) 3x + 8 . = = 2 ( x 4)( x 2) x-2 x - 6x + 8 3x 2 - 4 x - 32

3x + 8 x  4 x - 6x + 8 x4 x - 2 3(4) + 8 = = 10. 4-2

Thus lim

= lim

f (-1) does not exist.

f ( x) = 1 2 x -1 (c) lim f ( x) = 1 2 x -1+ (d) lim f ( x) = 1 (since (a) and (b) have 2 x -1 the same answers) (e) f (-1) does not exist.

(b)

3.2 Warmup Exericses 2 x 2 - 11x + 14

x6

lim

Discontinuous at x = -1 (a) (b) (c) (d) (e)

f (-1) = 2 lim f ( x) = 2

x -1-

lim

f ( x) = 4

lim

f ( x) does not exist (since parts (a)

x -1+ x -1

and (b) have different answers). lim f ( x) does not exist. x -1

178 7.

Chapter 3 THE DERIVATIVE Discontinuous at x = 1 (a)

f (1) = 2

(b)

lim f ( x) = -2

(c)

lim f ( x) = -¥ (limit does not exist)

(d)

x0

lim f ( x) = -2 (since parts (a) and (b)

x 2

have the same answer)

x 1-

lim

x 1+

f ( x) = -2

(e) f (0) does not exist and lim f ( x) does not x0

exist. f (2) does not exist.

(d) lim f ( x) = -2 (since (a) and (b) have x 1

the same answers)

11. a = –1; removeable

(e) lim f ( x) ¹ f (1) x 1

12. a = –1; nonremoveable

Discontinuous at x = -2 and x = 3. (a) f (-2) = 1 f (3) = 1 (b) lim f ( x) = -1 lim f ( x) = -1 x -2-

lim

x -2+

x  3-

f ( x) = -1

lim

x  3+

f ( x) = -1

f ( x) = -1 (since parts (a) and (b) x -2 have the same answer) lim

13. a = –5; nonremoveable a = 0; removeable 14. a = 0; nonremoveable a = 2; removeable 15.

lim f ( x) does not exist since lim f ( x) = ¥

(b)

and lim f ( x) = -¥. x  0+

lim

f ( x ) = ¥ (limit does not exist)

lim

f ( x) = 0

x  0-

(c)

x2

16.

f ( x) =

f ( x) = ¥.

-2 x (2 x + 1) (3x + 6)

lim

f ( x) = 0

f (x) is discontinuous at x = - 12 and x = -2

lim

f ( x) does not exist, since the answers

since the denominator equals 0 at these two values. lim f ( x) does not exist since

x -5

x -2

lim

x -2-

x0

since the answers to (a) and (b) are the same.

lim

exist. f (0) does not exist.

(a) f (0) does not exist.

x - 12

f (2) does not exist.

lim f ( x) = -¥ (limit does not exist)

x  0-

lim

f ( x ) = -2

lim

f ( x) = -¥ (limit does not exist)

lim

f ( x) = -2

x  2x  0+ x  2+

lim

x -2+

f ( x) = -¥.

lim f ( x) does not exist since

x -5

10. Discontinuous at x = 0 and x = 2

= + ¥ and

x - 12

(e) f (-5) does not exist and lim f ( x) does not

(c)

x  2+

f ( x) = -¥ (limit does not exist)

to (a) and (b) are different. lim f ( x) = 0,

(b)

f ( x) = -¥ and lim

lim

x  2-

lim

x -5+ x  0+

(d)

lim f ( x) does not exist since

f (-5) does not exist. f (0) does not exist. x -5-

x  0-

x0

x3

Discontinuous at x = -5 and x = 0 (a)

5+x x( x - 2)

f ( x ) is discontinuous at x = 0 and x = 2 since the denominator equals 0 at these two values.

lim f ( x) = -1 (since parts (a) and (b) have the same answer) (e) lim f ( x) ¹ f (-2) lim f ( x ) ¹ f (3) x3

x -2

f ( x) =

17.

f ( x) = -¥ and lim f ( x) = +¥. x  12

x2 - 4 x-2 f ( x) is discontinuous at x = 2 since the denominator equals zero at that value. Since for x ¹ 2 f ( x) =

x2 - 4 ( x + 2)( x - 2) = = x + 2, x-2 x-2 lim f ( x) = 2 + 2 = 4.

x 2

Section 3.2 18.

179

f ( x) =

x 2 - 25 x+5

25. As x approaches 0 from the left or the right,

x goes to -¥. approaches 0 and r ( x) = ln x 1

f (x) is discontinuous at x = -5 since the denominator equals zero at that value.

So lim r ( x) does not exist. As x approaches 1 x0

Since for x ¹ -5

from the left or the right,

x - 25 ( x + 5)( x - 5) = = x - 5, x+5 x+5 f ( x) = -5 - 5 = -10.

lim

x -5

19.

20.

x 1

26. As x approaches -2 from the left or the right, x + 2 approaches 0 and r ( x) = ln x + 2 goes to x -3 x -3

-¥. So lim r ( x) does not exist. As x

Since p(x) is a polynomial function, it is continuous everywhere and thus discontinuous nowhere.

approaches 3 from the left or the right,

x -2

q ( x ) = -3 x + 2 x - 4 x + 1

lim

x -2-

p( x) = -1 and

lim

p( x) does not exist.

r ( x) =

|5 - x | x-5

x -2

lim

x -2+

p( x) = 1,

r(x) is discontinuous at x = 5 since the denominator is zero at that value. Since lim r ( x) = 1 and lim x  5+

lim r ( x) does not exist.

28.

29.

r ( x) = -1,

x  5-

lim r ( x) does not exist.

3x - 2 x+4 f (x) is discontinuous at x = -4 since the denominator is zero at that value. In interval notation: (-¥, -4)  (-4, ¥). f ( x) =

f ( x) =

2x - 5

k ( x) = e x -1

The function is undefined for x < 1, so the function is discontinuous for a < 1. The limit as x approaches any a < 1 does not exist because the function is undefined for x < 1. 24.

f ( x) = 4 x 2 - 3x + 7 Since f (x) is a polynomial function, it is continuous everywhere. In interval notation: (-¥, ¥).

f (x) is continuous when x ³ 5 since the function 2 is not real number when x < 5 . 2 é5 ö÷ In interval notation: ê , ¥ ÷÷. ø êë 2

x5

23.

goes

x3

27.

| x + 2| p ( x) = x+2

since

x+2 x -3

2 . So to ¥ and so does r ( x) = ln xx+ -3

p(x) is discontinuous at x = -2 since the denominator is zero at that value.

22.

goes to ¥ and so

x . So lim r ( x) does not exist. does r ( x) = ln x 1

Since q(x) is a polynomial function, it is continuous everywhere and thus discontinuous nowhere. 21.

x x-1

p( x) = x 2 - 4 x + 11

x x-1

30.

f ( x) = e 1- x f (x) is continuous when x £ 1 since the function is not real number when x > 1 . In interval notation: (-¥, -1].

j ( x) = e1/x

j(x) is discontinuous at x = 0 since the function is undefined there. lim j ( x) does not exist since lim

x  0-

x0

lim

x  0+

j( x) = 0 and

j( x) = ¥.

180

Chapter 3 THE DERIVATIVE

ì 1 if x < 2 ï ï ï 31. f ( x) = ïí x + 3 if 2 £ x £ 4 ï ï ï if x > 4 ï î7 (a)

(d)

f (4) = 0 lim f ( x) = 0 and lim f ( x) = 2 , so

x  4-

x  4+

lim f ( x) does not exist.

x 4

f ( x) is discontinuous at x = 4.

(b)

ìï11 if x < -1 ïï ï 2 33. g ( x) = í x + 2 if -1 £ x £ 3 ïï if x > 3 ïï11 îï (a)

f (2) = 5 lim f ( x) = 1 and lim f ( x) = 5 , so

x  2-

x  2+

lim f ( x) does not exist.

x2

f ( x) is discontinuous at x = 2.

(c)

f (3) = 6 lim f ( x) = 6 and lim f ( x) = 5 , so

x  3-

(b) g ( x) is discontinuous at x = -1.

x  3+

(c)

lim f ( x) = 6

x3

lim g ( x) = (-1)2 + 2 = 3

f ( x) is continuous at x = 3.

(d)

f (4) = 7 lim f ( x) = 7 and lim f ( x) = 7 , so -

x 4

lim f ( x) = 7

x 4

f ( x) is continuous at x = 4.

lim g ( x) = 11

x -1x -1+

ì 0 ï if x < 0 ï ï ï 34. g ( x) = í x 2 - 5x if 0 £ x £ 5 ï ï 5 if x > 5 ï ï î (a)

ìï x - 1 if x < 2 ïï 32. f ( x) = ïí 0 if 1 £ x £ 4 ïï ïïî x - 2 if x > 4 (a)

(b) g ( x) is discontinuous at x = 5. (c)

(b)

x  q+

35.

lim f ( x) = 0

x 1

f ( x) is continuous at x = 1.

(c)

g ( x) = 52 - 5(5)

lim

=0 g ( x) = 5

x  5+

f (1) = 0 lim f ( x) = 0 and lim f ( x) = 0 , so

x 1-

lim

x  5-

ì if ï 4x + 4 h( x) = ï í 2 ï ï î x - 4 x + 4 if (a)

x£0 x>0

f (3) = 0 lim f ( x) = 0 and lim f ( x) = 0 , so

x  3-

x  3+

lim f ( x) = 0

x3

f ( x) is continuous at x = 3.

(b) There are no points of discontinuity.

Section 3.2 36.

181

ì ï x 2 + x - 12 h( x) = ï í ï ï î3 - x (a)

if x £ 1 if x > 1

43.

f ( x) =

x2 + x + 2 3

x - 0.9 x + 4.14 x - 5.4

P( x) Q ( x)

(a) Graph P( x) x2 + x + 2 = 3 Q( x) x - 0.9 x 2 + 4.14 x - 5.4 on a graphing calculator. A good choice for the viewing window is [-3,3] by [-10,10]. Y1 =

(b) h( x) is discontinuous at x = 1. (c)

lim h( x) = 12 + 1 - 12 = -10

x 1-

lim h( x) = 3 - 1 = 2

The graph has a vertical asymptote at x = 1.2, which indicates that f is discontinuous at x = 1.2.

x 1+

37. Find k so that kx 2 = x + k for x = 2.

(b) Graph Y2 = Q( x) = x3 - .09 x 2 + 4.14 x - 5.4

k (2) = 2 + k 4k = 2 + k 3k = 2 2 k = 3

using the same viewing window.

38. Find k so that x3 + k = kx - 5 for x = 3. 33 + k = 3k - 5

We see that this graph has one x-intercept, 1.2. This indicates that 1.2 is the only real solution of the equation Q( x) = 0.

27 + k = 3k - 5 32 = 2k

This result verifies our answer from part (a) because a rational function of the form P( x) f ( x) = Q( x) will be discontinuous wherever Q( x) = 0.

16 = k

39.

2 x 2 - x - 15 (2 x + 5)( x - 3) = = 2x + 5 x-3 x-3

Find k so that 2 x + 5 = kx - 1 for x = 3. 2(3) + 5 = k (3) - 1 6 + 5 = 3k - 1 11 = 3k - 1 12 = 3k 4= k

40.

44.

f ( x) =

x 2 + 3x - 2 3

x - 0.9 x + 4.14 x + 5.4

P ( x) Q ( x)

(a) Graph P( x) x 2 + 3x - 2 = 3 Q( x) x - 0.9 x 2 + 4.14 x + 5.4 on a graphing calculator. A good choice for the viewing window is [-3,3] by [-10,10]. Y1 =

3x 2 + 2 x - 8 (3x - 4)( x + 2) = = 3x - 4 x+2 x+2

Find k so that 3x - 4 = 3x + k for x = -2 . 3(-2) - 4 = 3(-2) + k -6 - 4 = -6 + k -10 = -6 + k -4 = k

182

Chapter 3 THE DERIVATIVE The graph has a vertical asymptote at x » -0.9. It is difficult to read this value accurately from the graph.

47. (a)

As x approaches 6 from the left or the right, the value of P(x) for the corresponding point on the graph approaches 500.

(b) Graph Y2 = Q( x) = x3 - 0.9 x 2 + 4.14 x + 5.4

using the same viewing window.

lim P( x)

x6

Thus, lim P( x) = $500. x6

(b)

P( x) = $1500

lim

x 10-

because, as x approaches 10 from the left, P(x) approaches $1500. (c)

We see that this graph has one x-intercept, » -0.926. This indicates that -0.926 is the only real solution of the equation Q( x) = 0. A rational function of the form f ( x) =

P( x) Q ( x)

will be discontinuous wherever Q( x) = 0, so we see that f is discontinuous at x » -0.926. The result in part (b) is consistent with the result in part (a), but the result in part (b) is more accurate. 45.

approaches 10 from the right, P(x) approaches $1000. (d) Since

lim P( x) ¹ lim P( x),

x 10+

x 10-

lim P( x) does not exist.

x 10

(e) From the graph, the function is discontinuous at x = 10. This may be the result of a change of shifts. (f ) From the graph, the second shift will be as profitable as the first shift when 15 units are produced. 48. In dollars, C ( x) = 4 x + 100 if 0 < x £ 150

x+4

g ( x) =

lim P( x) = $1000 because, as x

x 10+

x2 + 2x - 8 x+4 = ( x - 2)( x + 4) 1 = , x ¹ -4 x-2

C ( x) = 3x + 100 if 150 < x £ 400 C ( x) = 2.5x + 100 if 400 < x.

(a) C (130) = 4(130) + 100 = $620 (b) C (150) = 4(150) + 100 = $700

If g ( x) is defined so that g (-4) = -41-2 = - 16 ,

then the function becomes continuous at -4. It cannot be made continuous at 2. The correct answer is (a).

(d) C (400) = 3(400) + 100 = $1300

46. f is discontinuous at x = -6, -4, and 3 because the limit does not exist there. Also, f (-6) and f (3) are undefined. f is discontinuous at x = 4 because even though lim f ( x) = 1, f (4) is undefined. x 4

f is discontinuous at x = 0 because lim f ( x) ¹ f (0). x0

(e) C (500) = 2.5(500) + 100 = $1350 (f ) C is discontinuous at x = 150 and x = 400 because those represent points of price change. 49. In dollars,

ì1.25x if 0 < x £ 100 ï F ( x) = ï í ï ï î1.00 x if x > 100. (a) F (80) = 1.25(80) = $100 (b) F (150) = 1.00(150) = $150 (c) F (100) = 1.25(100) = $125 (d) F is discontinuous at x = 100.

Section 3.2

183

50. C(t) is a step function. The average cost per day

is A(t ) =

C (t ) for integer t. Thus, A(t) is also t

a step function. 36(4) (a) A(4) = = $36 4 36(5) = $36 (b) A(5) = 5 180 = $30 (c) A(6) = 6 (d) A(7) = (e) (f )

A(8) =

(b) From the graph, we see that lim w(t ) = 147 ¹ 133

t  40-

= lim w(t ),

180 » $25.71 7

t  40+

where w(t) is the weight in pounds t weeks after conception. Therefore, w is discontinuous at t = 40.

180 + 36 = $27 8

lim A(t ) = 36 because as t approaches 5

t  5-

from the left, A(t) is equal to 36. (g )

lim A(t ) = 36 because just to the right of 5,

t  5+

A(t) is equal to 36.

ìï 48 + 3.64t + 0.6363t 2 + 0.00963t 3 , 1 £ t £ 28 53. W (t ) = ïí ïï -1, 004 + 65.8t , 28 < t £ 56 î

(a) W (25) = 48 + 3.64(25) + 0.6363(25)2

(h) A(t) is discontinuous at 1, 2, 3, 4, 7, 8, 9, 10, 11.

+ 0.00963(25)3 » 687.156

51. C(x) is a step function. (a) (b) (c)

A male broiler at 25 days weighs about 687 grams.

lim C ( x ) = $1.40

x  3-

(b) W (t ) is not a continuous function.

lim C ( x) = $1.60

At t = 28

x  3+

lim W (t )

lim C ( x) does not exist.

t  28-

x3

= lim 48 + 3.64t + 0.6363t 2 + 0.00963t 3

(d) C (3) = $1.40 (e) (f ) (g)

t  28-

= 48 + 3.64(28) + 0.6363(28) 2 + 0.00963(28)3

lim C ( x) = $2.60

x  8.5-

» 860.18

lim C ( x) = $2.60

lim W (t ) ¹ lim (-1004 + 65.8t )

and

x  8.5+

t  28-

t  28+

= -1004 + 65.8(28)

lim C ( x) = $2.60

= 838.4

x  8.5

lim W (t ) ¹ lim W (t )

(h) C (8.5) = $2.60

(i) C(x) is discontinuous at 1, 2, 3, …, 11, 12

Thus W(t) is discontinuous.

52. (a) Since t = 0 weeks the woman weighs 120 lbs. and at t = 40 weeks she weighs 147 lbs. graph the line beginning at coordinate (0, 120) and ending at (40, 147), with closed circles at these points. Since immediately after giving birth, she loses 14 lbs. and continues to lose 13 more lbs. over the following 20 weeks, graph the line between the points (40, 133) and (60, 120) with an open circle at (40, 133) and a closed circle at (60, 120).

t  28-

(c)

t  28+

184

Chapter 3 THE DERIVATIVE

Your Turn 4

3.3 Rates of Change

C ( x) = x 2 - 2 x + 12

Your Turn 1

The instantaneous rate of change of cost when x = 4 is

A(t ) = 14.83(1.0159)t

Average rate of change from t = 20 (2020) to t = 30 (2030) is 30

A(30) - A(20) 14.83(1.0159) - 14.83(1.0159) = 30 - 20 10 » 0.347,

lim

h 0 20

or 0.347 million. The U.S. Asian population increased, on average, by 347,000 people per year.

C (h + 4) - C (4) h

[(h + 4)2 - 2(h + 4) + 12] - [42 - 2(4) + 12] h h0

= lim

h 2 + 8h + 16 - 2h - 8 + 12 - 16 + 8 - 12 h h0

= lim

h 2 + 6h h(h + 6) = lim h h h0 h0 = lim h + 6 = 0 + 6 = 6 = lim

h0

Your Turn 2 A(2013) - A(2011) 65.80 - 68.35 = = -1.325 2103 - 2011 2 The average change per year is a decrease of $1.325 billion.

Your Turn 3

When x = 4, the cost increases at a rate of $6 per unit. Your Turn 5 A(t ) = 14.83(1.0159)t , t = 0 corresponds to 2000 14.83(1.0159)20 + h - 14.83(1.0159) 20 h h 0 lim

For t = 2, the instantaneous velocity is s(2 + h )- s(2) feet per second. h h 0

lim

s(2 + h) = 2(2 + h)2 - 5(2 + h) + 40 = 2(4 + 4h + h 2 ) - 10 - 5h + 40 = 8 + 8h + h 2 - 10 - 5h + 40 = h 2 + 3h + 38 s(2) = 2(2) 2 - 5(2) + 40 = 2(4) - 10 + 40 = 38 s(2 + h) - s(2) h 2 + 3h + 38 - 38 lim = lim h h h0 h0 h 2 + 3h h h0 h(h + 3) = lim h h0 = lim h + 3 = 3

Use the TABLE feature on a TI-84 Plus calculator. 14.83(1.0159)20 + h - 14.83(1.0159) 20 h

h 1

0.32326

0.1

0.32097

0.01

0.32075

0.001

0.321

0.0001

0.321

0.00001

0.321

The limit seems to be approaching 0.321 million. The instantaneous rate of change in the U.S. Asian population is about 321,000 people per year in 2020.

= lim

h0

or 3 feet per second.

3.3 Warmup Exercises W1. f ( x ) = 2 x 2 + 3x + 4 f (2 + h) = 2(2 + h)2 + 3(2 + h) + 4 = 2(4 + 4h + h 2 ) + 6 + 3h + 4 = 8 + 8h + 2h 2 + 10 + 3h = 2h 2 + 11h + 18

Section 3.3

185

W2. f ( x ) = 2 x 2 + 3x + 4

f (2 + h) - f (2)

Average rate of change

h 2

2(2 + h) + 3(2 + h) + 4 - (2(2) + 3(2) + 4)

f (1) - f (-2) 1 - (-2) (-4) - (-41) = = -45 = -15 1 - (-2) 3 =

h 2

2h + 8h + 8 + 6 + 3h + 4 - 18 h

2h + 11h = 2h + 11 h

W3. f ( x) =

2 x +1

y = 2 x3 - 4 x 2 + 6 x = f ( x) between x = -1 and x = 4

Average rate of change

2 f (4 + h) = = 2 (4 + h) + 1 5+h 2 x +1 f (4 + h) - f (4) h é ù 2 = 1⋅ê - 2ú h êë (4 + h) + 1 5 úû

f (4) - f (-1) 4 - (-1) 88 - (-12) = 5 100 = 5 = 20 =

W4. f ( x ) =

é (2)(5) - (2)(5 + h) ù = 1⋅ê ú úû (5)(5 + h) h êë

f (4) - f (1) 4 -1 2 1 = 3 = 1 3

3.3 Exercises 1.

True

False. Speed is always positive. Velocity can be positive or negative.

y = x 2 + 2 x = f ( x) between x = 1 and x =3 Average rate of change f (3) - f (1) = = 15 - 3 = 6 3-1 2

y = x = f ( x) between x = 1 and x = 4 Average rate of change

æ ö 2 = 1 ⋅ çç -2h ÷÷ = (5)(5 + h) h èç (5)(5 + h) ÷ø

y = -3x3 + 2 x 2 - 4 x + 1 = f ( x) between x = -2 and x = 1

10.

y = 3x - 2 = f ( x) between x = 1 and x = 2 f (2) - f (1) Average rate of change = 2 -1 = 2 -1 2 -1 1 = =1 1

11.

y = e x = f ( x) between x = -2 and x = 0

Average rate of change

y = -4 x - 6 = f ( x) between x = 2 and x =6 Average rate of change f (6) - f (2) = 6-2 (-150) - (-22) = 6-2 + 22 150 = 4 = -128 = -32 4

f (0) - f (-2) 0 - (-2)

1 - e-2 2 » 0.4323

186

Chapter 3 THE DERIVATIVE y = ln x = f ( x) between x = 2 and x = 4

12.

f (4) - f (2) 4-2 ln 4 - ln 2 = 2 » 0.3466

Average rate of change =

13. lim

h 0

16. s(t ) = 5t 2 - 2t - 7 s(3 + h) - s(3) h h0 lim

[5(3 + h) 2 - 2(3 + h) - 7] - [5(3)2 - 2(3) - 7] h h0

= lim

[45 + 30h + 5h 2 - 6 - 2h - 7] - [45 - 6 - 7] h h0

= lim

32 + 28h + 5h 2 - 32 28h + 5h 2 = lim h h h0 h0 h(28 + 5h) = lim = lim (28 + 5h) = 28 h h0 h 0 = lim

s(6 + h) - s(6) h (6 + h)2 + 5(6 + h) + 2 - [62 + 5(6) + 2] h h0

The instantaneous velocity at t = 3 is 28.

= lim

h 2 + 17h + 68 - 68 h 2 + 17h = lim h h h0 h 0 h(h + 17) = lim = lim (h + 17) = 17 h h0 h 0 = lim

The instantaneous velocity at t = 6 is 17.

17.

s(t ) = t 3 + 2t + 9

lim

s(1 + h) - s(1) h

x 0

= lim

[(1 + h) + 2(1 + h) + 9] - [(1) + 2(1) + 9] h

h0 2

14. s(t ) = t + 5t + 2

= lim

s(1 + h) - s(1) h h0

= lim

[(1 + h)2 + 5(1 + h) + 2] - [(1)2 + 5(1) + 2] h h0

= lim

[1 + 2h + h 2 + 5 + 5h + 2] - [1 + 5 + 2] = lim h h0 8 + 7h + h 2 - 8 7h + h 2 = lim = lim h h h0 h 0 h(7 + h) = lim = lim (7 + h) = 7 h h0 h 0

h + 3h + 5h + 12 - 12 h

h0

= lim

h + 3h + 5h h

h0

= lim

h0

= lim (h + 3h + 5) = 5 h0

s(t ) = t 3 + 2t + 9 s(4 + h) - s(4) lim h h 0 3

(4 + h) + 2(4 + h) + 9 - [4 + 2(4) + 9] h

h0

2 h(h + 3h + 5)

= lim

s(2 + h) - s(2) lim h h0

h0

18.

15. s(t ) = 5t 2 - 2t - 7

= lim

h 3

The instantaneous velocity at t = 1 is 5.

The instantaneous velocity at t = 1 is 7.

h0

[1 + 3h + 3h + h + 2 + 2h + 9] - [1 + 2 + 9]

h0

lim

= lim

[5(2 + h) - 2(2 + h) - 7] - [5(2) - 2(2) - 7] h 2

[20 + 20h + 5h - 4 - 2h - 7] - [20 - 4 - 7] h

9 + 18h + 5h 2 - 9 18h + 5h 2 = lim h h h0 h0 h(18 + 5h) = lim = lim (18 + 5h) = 18 h h0 h 0 = lim

= lim

h0

= lim

h + 12h + 48h + 2h h

h0

= lim

h0

= lim

h0

4 + 3(4 )h + 3(4h ) + h + 8 + 2h + 9 - (4 + 8 + 9)

h(h + 12h + 50) h

h + 12h + 50h h

= lim (h + 12h + 50) = 50 h0

The instantaneous velocity at t = 4 is 50.

The instantaneous velocity at t = 2 is 18.

Section 3.3 19.

187

f ( x) = x 2 + 2 x at x = 0 lim

h 0

22.

f (0 + h) - f (0) h

F ( x) = x 2 + 2 at x = 0 lim

h 0

(0 + h)2 + 2(0 + h) - [02 + 2(0)] h h0

F (0 + h) - F (0) h (0 + h) 2 + 2 - [02 + 2] h h 0

= lim

h 2 + 2h h h0 h(h + 2) = lim h h0 = lim h + 2 = 2

h2 h 0 h = lim h

= lim

h 0

h0

The instantaneous rate of change at x = 0 is 0.

The instantaneous rate of change at x = 0 is 2. 20.

23.

s(t ) = -4t - 6 at t = 2

f ( x) = x x at x = 2 h

s(2 + h) - s(2) lim h h 0

0.01

-4(2 + h) 2 - 6 - [-4(2) 2 - 6] h h0

2.012.01 - 22 0.01 = 6.84

= lim

-4(4 + 4h + h 2 ) - 6 + 16 + 6 h h0

= lim

-16h - 4h h h0 h(-16 - 4h) = lim h h0 = lim (-16 - 4h) = -16 = lim

f (2 + 0.01) - f (2) 0.01

0.001

f (2 + 0.001) - f (2) 0.001 2.0012.001 - 22 0.001 = 6.779 =

0.0001

h0

The instantaneous rate of change at t = 2 is -16.

f (2 + 0.0001) - f (2) 0.00001 2.00012.0001 - 22 0.0001 = 6.773 =

21.

g (t ) = 1 - t 2 at t = -1 lim

h 0

g (-1 + h) - g (-1) h

0.00001

1 - (-1 + h) 2 - [1 - (-1) 2 ] h h 0

2.000012.00001 - 22 0.00001 = 6.7727

= lim

1 - (1 - 2h + h 2 ) - 1 + 1 h h 0

= lim

2h - h h h 0 h(2 - h) = lim h h 0 = lim (2 - h) = 2 = lim

f (2 + 0.00001) - f (2) 0.00001

0.000001

f (2 + 0.000001) - f (2) 0.000001 2.0000012.000001 - 22 0.000001 = 6.7726 =

The instantaneous rate of change at x = 2 is 6.773.

h 0

The instantaneous rate of change at t = -1 is 2.

188 24.

Chapter 3 THE DERIVATIVE f ( x) = x x at x = 3

25.

f ( x) = x ln x at x = 2 h

h 0.01

f (3 + 0.01) - f (3) 0.01

0.01

2.01ln 2.01 - 2ln 2 0.01 = 1.1258

3.013.01 - 33 0.01 = 57.3072

0.001

f (3 + 0.001) - f (3) 0.001

0.001

f (3 + 0.00001) - f (3) 0.00001 3.00001

h 0.0001

3.00001 -3 0.00001 = 56.6632 =

0.000001

3.000001

2.0001ln 2.0001 - 2ln 2 0.0001 = 1.1207 0.00001

3.000001 -3 0.000001 = 56.6626 0.0000001

2.00001ln 2.00001 - 2ln 2 0.00001 = 1.1207

The instantaneous rate of change at x = 2 is 1.121.

3.0000001 -3 0.0000001 = 56.6625 =

f (2 + 0.00001) - f (2) 0.00001 =

f (3 + 0.0000001) - f (3) 0.0000001 3.0000001

f (2 + 0.0001) - f (2) 0.0001 =

f (3 + 0.000001) - f (3) 0.000001 =

f (2 + 0.001) - f (2) 0.001 2.001ln 2.001 - 2ln 2 0.001 = 1.1212

3.0013.001 - 33 0.001 = 56.7265 =

0.00001

f (2 + 0.01) - f (2) 0.01

The instantaneous rate of change at x = 3 is 56.66.

26.

f ( x) = x ln x at x = 3 h 0.01

f (3 + 0.01) - f (3) 0.01 3.01ln3.01 - 3ln3 0.01 = 2.4573 =

0.001

f (3 + 0.001) - f (3) 0.001 3.001ln3.001 - 3ln3 0.001 = 2.4495 =

0.00001

f (3 + 0.00001) - f (3) 0.00001 3.00001ln3.00001 - 3ln3 0.00001 = 2.4486 =

The instantaneous rate of change at x = 3 is 2.449.

Section 3.3

189

28. If the instantaneous rate of change of f (x) with respect to x is positive when x = 1, the function would be increasing.

(c)

P(2 + h) - P(2) h h 0 lim

2(2 + h)2 - 5(2 + h) + 6 - 4 h h0

= lim

29.

P( x) = 2 x 2 - 5 x + 6

8 + 8h + 2h 2 - 10 - 5h + 2 h h0

= lim

(a) P(4) = 18 P(2) = 4

2h 2 + 3h h(2h + 3) = lim h h h0 h0 = lim (2h + 3) = 3, = lim

Average rate of change of profit P(4) - P(2) 4-2 18 - 4 14 = = = 7, 2 2

h0

which is $300 per item. (d)

which is $700 per item.

2(4 + h)2 - 5(4 + h) + 6 - 18 h h0

= lim

(b) P(3) = 9 P(2) = 4

Average rate of change of profit =

P(4 + h) - P(4) h h 0 lim

P(3) - P(2) 9-4 = =5 3-2 1

which is $500 per item.

32 + 16h + 2h 2 - 20 - 5h - 12 h h0

= lim

2h 2 + 11h h h0 h(2h + 11) = lim h h0 = lim 2h + 11 = 11,

= lim

h0

which is $1100 per item. 30.

R = 10 x - 0.002 x 2

(a) Average rate of change =

R(1001) - R(1000) 8005.998 - 8000 = = 5.998 1001 - 1000 1

The average rate of change is $5998. R(1000 + h) - R(1000) h h 0

(b) Marginal revenue = lim

[10(1000 + h) - 0.002(1000 + h)2 ] - [10(1000) - 0.002(1000)2 ] h h 0

= lim

[10,000 + 10h - 0.002(1,000,000 + 2000h + h 2 )] - 8000 h h 0

= lim

10,000 + 10h - 2000 - 4h - 0.002h 2 - 8000 h h 0

= lim

6h - 0.002h 2 h(6 - 0.002h) = lim = lim (6 - 0.002h) = 6 h h h 0 h0 h 0

= lim

The marginal revenue is $6000. This is approximately the revenue generated by the 1000th unit produced. (c)

Additional revenue = R(1001) - R(1000) = [10(1001) - 0.002(1001)2 ] - [10(1000) - 0.002(1000)2 ] = 8005.998 - 8000 The additional revenue is $5998.

(d) The answers to parts (a) and (c) are the same and are approximately equal to the answer to part (b).

190 31.

Chapter 3 THE DERIVATIVE (c)

N ( p) = 80 - 5 p 2 , 1 £ p £ 4

(a) Average rate of change of demand is

Use the TABLE feature on a TI-84 Plus calculator to estimate the limit.

N (3) - N (2) 35 - 60 = 3-2 1 = -25 boxes per dollar.

(b) Instantaneous rate of change when p is 2 is N (2 + h) - N (2) lim h h 0 80 - 5(2 + h)2 - [80 - 5(2)2 ] h h 0

= lim

80 - 20 - 20h - 5h 2 - (80 - 20) = lim h h 0 -5h 2 - 20h = -20 boxes per dollar. h h 0

= lim

(c)

Around the $2 point, a $1 price increase (say, from $1.50 to $2.50) causes a drop in demand of about 20 boxes. Instantaneous rate of change when p is 3 is 80 - 5(3 + h)2 - [80 - 5(3)2 ] h h 0 80 - 45 - 30h - 5h 2 - 80 + 45 h h0

= lim

-30h - 5h 2 h h0 = -30 boxes per dollar. = lim

(d) As the price increases, the demand decreases; this is an expected change. A(t ) = 1000(1.03)t

(a) Average rate of change in the total amount from t = 0 to t = 5:

34.7782

0.1

34.3174

0.01

34.2718

0.001

34.2673

0.0001

34.2668

0.00001

34.2668

A(t ) = 1000e0.03t

(a) Average rate of change in the total amount from t = 0 to t = 5: A(5) - A(0) 1000e0.03(5) - 1000e0.03(0) = 5-0 5 = 32.3668,

which is $32.37 per year. (b) Average rate of change in the total amount from t = 5 to t = 10: A(10) - A(5) 1000e0.03(10) - 1000e0.03(5) = 10 - 5 5 = 37.6049, which is $37.60 per year. (c) Instantaneous rate of change for t = 5: 1000e0.03(5+ h) - 1000e0.03(5) h h0 Use the TABLE feature on a TI-84 Plus calculator to estimate the limit.

which is $31.85 per year. (b) Average rate of change in the total amount from t = 5 to t = 10: 5

A(10) - A(5) 1000(1.03) - 1000(1.03) = 10 - 5 5 = 36.9285,

which is $36.93 per year.

lim

A(5) - A(0) 1000(1.03)5 - 1000(1.03)0 = 5-0 5 = 31.8548,

1000(1.03)5 + h - 1000(1.03)5 h

The limit seems to be approaching 34.27. So, the instantaneous rate of change for t = 5 is about $34.27 per year. 33.

lim

32.

Instantaneous rate of change for t = 5: 1000(1.03)5+ h - 1000(1.03)5 lim h h0

1000e0.03(5 + h ) - 1000e0.03(5) h

35.3831

0.1

34.9074

0.01

34.8603

0.001

34.8556

0.0001

34.8551

0.00001

34.8550

Section 3.3

191

The limit seems to be approaching 34.855. So, the instantaneous rate of change for t = 5 is about $34.86 per year. 34. (a)

37. (a) A(25) - A(20) 14.83(1.015925 - 1.015920 ) = 25 - 20 5 » 0.334 A gives the population in millions so this is an average rate of change of 334,000 people per year.

9.6 - 4.0 = 0.56% per year 2010 - 2000

(b)

(b) Use the TABLE feature on a TI-84 Plus calculator to estimate the limit. 14.83(1.0159

35. Let P(t ) = the price per gallon of gasoline for the year t. (a) P(2010) = 284 (cents) P(2012) = 368 (cents)

Average change in price from 2010 to 2012: P(2012) - P(2010) 368 - 284 = = 42 2012 - 2010 2 On average, the price of gasoline increased about 42 cents per gallon per year. (b) P(2012) = 368 (cents) P(2019) = 269 (cents)

20 + k

- 1.0159 )

0.3233

0.1

0.3210

0.01

0.3210

0.001

0.3210

The limit seems to be approaching 0.321. So, the instantaneous rate of change of the Asian population in 2025 is increasing at about 321,000 people per year. 38. Let P(t ) = world population estimated in billions for year t.

Average change in price from 2012 to 2019: P(2019) - P(2012) 269 - 368 = » -14.1 2019 - 2012 7 On average, the price of gasoline decreased about 14.1 cents per gallon per year. (c) P(2010) = 284 (cents) P(2019) = 269 (cents)

Average change in price from 2010 to 2019: P(2019) - P(2010) 269 - 284 = » -1.7 2019 - 2010 9 On average, the price of gasoline decreased about 1.7 cents per gallon per year. 36. (a) For the period 2015 to 2018 304.7 - 263.3 = 13.8 2018 - 2015 Increases by an average of $13.8 billion per year. (b) For the period 2018 to 2028: 3.1 - 304.7 = -30.16 2028 - 2018 Decreases by an average of $30.16 billion per year.

(a)

P(1990) = 5.3 If replacement-level fertility is reached in 2010, P(2050) = 8.6.

Average rate of change P(2050) - P(1990) = 2050 - 1990 8.6 - 5.3 = = 0.055 60 On average, the population will increase 55 million per year. If replacement-level fertility is reached in 2030, P(2050) = 9.2. Average rate of change P(2050) - P(1990) = 2050 - 1990 9.2 - 5.3 = = 0.065 60 On average, the population will increase 65 million per year. If replacement-level fertility is reached in 2050, P(2050) = 9.8. Average rate of change P(2050) - P(1990) = 2050 - 1990 9.8 - 5.3 = = 0.075 60

192

On average, the population will increase 75 million per year. The projection for replacement-level fertility by 2010 predicts the smallest rate of change in world population. (b) If replacement-level fertility is reached in 2010 P(2090) = 9.3 P(2130) = 9.6

Average rate of change P(2130) - P(2090) = 2130 - 2090 9.6 - 9.3 = 40 = 0.0075 On average, the population will increase 7.5 million per year.

L(t ) = -0.01t 2 + 0.788t - 7.048

40.

(a)

L(28) - L(22) 7.176 - 5.448 = 28 - 22 6 = 0.288 The average rate of growth during weeks 22 through 28 is 0.288 mm per week.

(b) lim

L(t + h) - L(t )

h0

= lim

h L(22 + h) - L(22) h

h0

If replacement-level fertility is reached in 2030, P(2090) = 10.3 P(2130) = 10.6

= lim

Average rate of change

= lim

2 [-0.01(22 + h) + 0.788(22 + h) - 7.048] - 5.448

h0 2

-0.01(h + 44h + 484) + 17.336 + 0.788h - 12.496

h0

P(2130) - P(2090) = 2130 - 2090 - 10.3 10.6 = 40 = 0.0075

= lim

-0.01h + 0.348h

h = lim (-0.01h + 0.348) h0 h0

On average, the population will increase 7.5 million per year. If replacement-level fertility is reached in 2050, P(2090) = 11.35 P(2130) = 11.75 Average rate of change P(2130) - P(2090) = 2130 - 2090 - 11.35 11.75 = 40 = 0.01 On average, the population will increase 10 million per year. From 2090 to 2130 the three projections show almost the same rate of change in world population.

= 0.348

The instantaneous rate of growth at exactly 22 weeks is 0.348 mm per week. (c) L(t) 0.01t 2 0.788t 7.048 9

41. (a) F(t) 10.28 175.9te t/1.3 100

39. (a) 2010 to 2014: 667 - 63 = 151 2014 - 2010 Increase by 151 cases per year. (b) 2014 to 2017: 120 - 667 » -182.3 2017 - 2014 Decrease by about 182.3 cases per year.

(b) The average rate of change during the first hour is F (1) - F (0) » 81.51 1- 0

(c)

kilojoules per hour per hour. Store F(t) in a function menus of a graphing calculator. Store

Y1 (1+ X )-Y1 (1) as Y2 in the X

Section 3.3

193

function menu, where Y1 represents F(t). Substitute small values for X in Y2 perhaps with use of a table feature of the graphing calculator. As X is allowed to get smaller, Y2 approaches 18.81 kilojoules per hour per hour. (d) Through use of a MAX/feature program of a graphing calculator, the maximum point seen in part (a) is estimated to occur at approximately t = 1.3 hours. 42. (a) The average rate of change of M(t) on the interval [105, 115] is M (115) - M (105) 0.8 = = 0.08 115 - 105 10 kilograms per day. M (105 + h) - M (105) h h 0 M (105 + h) = 27.5 + 0.3(105 + h)

(b) Calculate lim

-0.001(105 + h) 2 = 27.5 + 31.5 + 0.3h -(11.025 + 0.21h + 0.001h 2 ) = 47.975 + 0.09h - 0.001h 2 M (105) = 47.975

So, the instantaneous rate of change of M(t) at t = 105 is 2 æ ö ç 47.975 + 0.09h - 0.001h - 47.975 ÷÷ lim çç ÷÷ ø h h 0 è 2ö æ ç 0.09h - 0.001h ÷÷ = lim çç ÷÷ ø h h0 è

= lim (0.09 - 0.001h) h0

(b)

(c)

I (2010) - I (1960) 1043 - 265 = 2010 - 1960 50 = 15.56 The average rate of change is 15,560 immigrants per year. I (2010) - I (1910) 1043 - 1042 = 2010 - 1910 100 = 0.01 The average rate of change is 10 immigrants per year.

(d)

20 -15,540 + 15,560 = = 10 2 2 They are equal. This will not be true for all time periods. (It is true only for time periods of equal length.)

(e)

2018 is 8 years after 2010. 1, 043, 000 + 8(15,560/year) » 1,167, 480 The predicted number of immigrants in 2018 is about 1,167,480 immigrants. The predicted value is about 70,869 more than the actual number of 1,096,611.

44. Let D(t) represent the percent of students (8th, 10th, or 12th graders) who have used marijuana by the year t. (a) 8th graders:

15.6 - 17.3 D(2014) - D(2010) = 2014 - 2010 4 = -0.425 percent per year 15.2 - 15.5 D(2019) - D(2015) = 2019 - 2015 4 = -0.075 percent per year D(2019) - D(2010) 15.2 - 17.3 = 2019 - 2010 9 = -0.233 percent per year

= 0.09 kilograms per day.

(b) 10th graders:

33.7 - 33.4 D(2014) - D(2010) = 2014 - 2010 4 = 0.075 percent per year 5

125

43. Let I(t) represent immigration (in thousands) in year t. I (1960) - I (1910) 265 - 1042 = (a) 1960 - 1910 50 = -15.54 The average rate of change is –15,540 immigrants per year.

34.0 - 31.1 D(2019) - D(2015) = 2019 - 2015 4 = 0.725 percent per year 34.0 - 33.4 D(2019) - D(2010) = 2019 - 2010 9 = 0.067 percent per year

194

Chapter 3 THE DERIVATIVE (c)

12th graders:

44.4 - 43.8 D(2014) - D(2010) = 2014 - 2010 4 = 0.15 percent per year 43.7 - 44.7 D(2019) - D(2015) = 2019 - 2015 4 = -0.25 percent per year

(c)

20 - 14 s(6) - s(4) = = 3 ft /sec 6-4 2

(d)

s(8) - s(6) 30 - 20 = = 5 ft/sec 8-6 2 f ( x0 + h) - f ( x0 - h) 2h f (4 + 2) - f (4 - 2) = (2)(2) f (6) - f (2) = 4 20 10 = 4 10 = = 2.5 ft /sec 4 2+3 (ii) = 2.5 ft /sec 2

(e) (i)

D(2019) - D(2010) 43.7 - 43.8 = 2019 - 2010 9 = -0.011 percent per year 45. (a)

70 - 60 T (2000) - T (1000) = 2000 - 1000 1000 10 = 1000

From 1000 to 3200 ft, the temperature changes about 10o F per 1000 ft; the temperature rises (on the average). (b)

T (8000) - T (5000) 70 - 90 -20 = = 8000 - 5000 3000 3000 - 20 3 = 1000

From 5000 to 9000 ft, the temperature changes about -6.7o F per 1000 ft; the temperature falls (on the average). (d)

(e)

(f)

46. (a) (b)

(i)

90 - 60 30 T (5000) - T (1000) = = 5000 - 1000 4000 4000 7.5 = 1000

From 1000 to 5000 ft, the temperature changes about 7.5o F per 1000 ft; the temperature rises (on the average). (c)

(f )

T (8000) - T (2000) 70 - 70 = =0 8000 - 2000 6000 From 2000 to 8000 ft, the temperature changes about 0o per 1000 ft; the temperature stays constant (on the average).

The temperature is highest at 5000 ft and lowest at 1000 ft. If 7000 ft is changed to 10,000 ft, the lowest temperature would be at 10,000 ft. The temperature at 8500 ft is the same as 1000 ft. s(2) - s(0) 10 - 0 = = 5ft/sec 2-0 2 s(4) - s(2) 14 - 10 = = 2 ft /sec 4-2 2

(ii)

f ( x0 + h) - f ( x0 - h) 2h f (6 + 2) - f (6 - 2) = (2)(2) f (8) - f (4) = 4 30 14 = 4 16 = = 4 ft /sec 4 3+5 = 4 ft /sec 2

47. (a) Average rate of change from 0.5 to 1: f (1) - f (0.5) 55 - 30 = = 50 mph 1 - 0.5 0.5

Average rate of change from 1 to 1.5: f (1.5) - f (1) 80 - 55 = = 50 mph 1.5 - 1 0.5 Estimate of instantaneous velocity is 50 + 50 = 50 mph. 2 (b) Average rate of change from 1.5 to 2: f (2) - f (1.5) 104 - 80 = = 48 mph 2 - 1.5 0.5

Average rate of change from 2 to 2.5 f (2.5) - f (2) 124 - 104 = = 40 mph 2.5 - 2 0.5 Estimate of instantaneous velocity is 48 + 40 = 44 mph. 2

Section 3.3 48.

195

s(t ) = t 2 + 5t + 2

(a) Average velocity s(6) - s(4) 6-4 68 - 38 = 6-4 30 = = 15 ft /sec 2 =

(b) Average velocity s(5) - s(4) 5-4 52 - 38 = 5-4

14 = 14 ft /sec 1

(c)

lim

h 0

s(4 + h) - s(4) h (4 + h)2 + 5(4 + h) + 2 - 38 h h0

= lim

16 + 8h + h 2 + 20 + 5h + 2 - 38 h h0

= lim

h 2 + 13h h h0 h(h + 13) = lim h h0 = lim (h + 13) = lim

h0

= 13 ft /sec

3.4

Definition of the Derivative

Your Turn 1 (a)

f ( x) = x 2 - x, x = -2 and x = 1.

Slope of secant line =

f (1) - f (-2) 0-6 = = -2 1 - (-2) 3

Use the point-slope form and the point (1, f (1)), or (1, 0). y - y1 = m( x - x1) y - 0 = -2( x - 1) y = -2 x + 2

(b)

slope of tangent at (-2, 6) f (-2 + h) - f ( x ) = lim h h0

196

Chapter 3 THE DERIVATIVE [(-2 + h)2 - (-2 + h)] - [(-2)2 - (-2)] h h0 é 4 - 4h + h 2 + 2 - h ù - [ 4 + 2 ] ê úû = lim ë h h0 = lim

-5h + h 2 = lim (-5 + h) h h0 h 0

= lim

= -5 The equation of the tangent line is y - 6 = (-5)( x - (-2)) y = 6 - 5x - 10 y = -5x - 4.

Your Turn 2 f ( x) = x 2 - x f ( x + h) - f ( x) h h0

f ¢( x) = lim

[( x + h) 2 - ( x + h)] - [ x 2 - x] h h0

= lim

x 2 + 2 xh + h 2 - x - h - x 2 + x h h0

= lim

h(2 x - 1 + h) 2 xh - h + h 2 = lim h h h0 h0

= lim

= lim (2 x - 1 + h) = 2 x - 1 + 0 h0

= 2x - 1 f ¢(-2) = 2(-2) - 1 = -5

Section 3.4

197

Your Turn 3 f ( x) = x 3 - 1 f ¢( x) = lim

h 0

f ( x + h) - f ( x) h

[( x + h)3 - 1] - [ x3 - 1] h h 0

= lim

x3 + 3x 2h + 3xh 2 + h3 - 1 - x3 + 1 h h 0

= lim

3x 2h + 3xh 2 + h3 h h 0

= lim

h(3x 2 + 3xh + h 2 ) h h 0

= lim

= lim (3x 2 + 3xh + h 2 ) h 0

= 3x 2 + 0 + 0 = 3x 2 f ¢(-1) = 3(-1)2 = 3

198

Chapter 3 THE DERIVATIVE

Your Turn 4 f ( x) = -

2 x

f ( x + h) - f ( x) h h0

f ¢( x) = lim

2 2 - x + h ) - (- x ) ( = lim

h é 2 2ù 1 = lim ê + ú⋅ x úû h h  0 êë x + h h0

-2 x + 2 x + 2h 1 ⋅ x( x + h) h 2h 1 2 = lim ⋅ = lim h  0 x ( x + h) h h  0 x ( x + h) = lim

h0

= =

2 x( x + 0) 2 x2

Section 3.4

199

Your Turn 5 f ( x) = 2 x f ( x + h) - f ( x) h h0

f ¢( x) = lim

= lim

h0

2 x+h -2 x h

2 x+h -2 x 2 x+h +2 x ⋅ h h0 2 x+h +2 x 4( x + h) - 4 x = lim h0 h 2 x + h + 2 x = lim

(

= lim

h0 h

= lim

h0 h

)

4 x + 4h - 4 x

(2 x + h + 2 x ) 4h

(2 x + h + 2 x )

4 h0 2 x + h + 2 x

= lim =

4 4 = 2 x +2 x 4 x

1 x

200

Chapter 3 THE DERIVATIVE

Your Turn 6 C ( x ) = 10 x - 0.002 x 2 C ( x + h) - C ( x ) h h0

C ¢( x ) = lim

10( x + h) - 0.002( x + h)2 - 10 x + 0.002 x 2 h h0

= lim

10 x + 10h - 0.002 x 2 - 0.004 xh - 0.002h 2 - 10 x + 0.002 x 2 h h0

= lim

10h - 0.004 xh - 0.002h 2 h h0 h(10 - 0.004 x - 0.002h) = lim h h0 = lim

= lim (10 - 0.004 x - 0.002h) h0

= 10 - 0.004 x + 0 = 10 - 0.004 x ¢ C (100) = 10 - 0.004(100) = 10 - 0.4 = 9.60

The rate of change of the cost when x = 100 is $9.60.

Section 3.4

201

Your Turn 7 f ( x) = 2 x at x = 4

From Your Turn 5, we have f ¢( x ) =

1 . x

At x = 4: f ¢(4) =

1 1 = and f (4) = 2 4 = 2(2) = 4 2 4 1

Slope of the tangent line at (4, f (4)), or (4, 4) is 2 . y - y1 = m( x - x1) 1

y - 4 = 2 ( x - 4) 1

y-4= 2x-2 1

y = 2x+2

202

Chapter 3 THE DERIVATIVE

W1.

3. False. A derivative can‘t exist at a point where the function does not exist.

f ( x ) = 3x 2 - 2 x - 5 f ( x + h) - f ( x ) h

4. False. For example, see the graph in Exercise 38. At x = 2, f ( x) is continuous, but f ¢( x) does not exist at x = 2.

3.4

Warmup Exercises

3( x + h)2 - 2( x + h) - 5 - (3x 2 - 2 x - 5) h

3x 2 + 6 xh + 3h 2 - 2 x - 2h - (3x 2 - 2 x - 5) = h =

6 xh + 3h 2 - 2h = 6 x + 3h - 2 h

W2. 3 x-2 f ( x + h) - f ( x ) 1 é 3 3 ù ú = ⋅ê ê h h ëx+h-2 x - 2 úû 1 é (3)( x - 2) - (3)( x + h - 2) ùú = ⋅ê ú h ëê ( x + h - 2)( x - 2) û 1 é 3x - 6 - 3x - 3h + 6 ùú = ⋅ê h êë ( x + h - 2)( x - 2) úû ù -3h 1 é ú = ⋅ê h êë ( x + h - 2)( x - 2) úû 3 =( x + h - 2)( x - 2) f ( x) =

W3. The line 5x + 6 y = 7 has slope -5/6 so any line parallel to this line has equation y = -(5/6) x + b. If the line passes through (4, - 1), then -1 = -(5/6)4 + b b = 1 + 5/3 = 7 /3

5. Using the points (5,3) and (6,5), we have 5-3 2 m= = 6-5 1 = 2. 6. Using the points (2, 2) and (-2, 6), we have 6-2 4 m= = = -1. -2 - 2 -4 7. Using the points (-2, 2) and (2,3), we have m=

3-2 1 = . 2 - (-2) 4

8. Using the points (3, -1) and (-2, 3), we have 3 - (-1) 4 4 m= = =- . -2 - 3 -5 5 9. Using the points (-3, - 3) and (0, - 3), we have m=

-3 - (-3) 0 = = 0. 0-3 -3

10. The line tangent to the curve at (4, 2) is a vertical line. A vertical line has undefined slope. 11. f ( x) = 3x - 7

and the equation of the line is y = -(5/6) x + 7/3.

Step 1 f ( x + h)

W4. The line through (6, 2) and ( 2, 5) has slope equal 2-5 -3 3 to = = - . This line has 6 - (-2) 8 8 equation y = -(3/8) x + b. Since it passes through (6, 2), 2 = -(3/8)(6) + b b = 2 + 9/4 = 17/4 and the equation of the line is y = -(3/8) x + 17/4.

= 3( x + h) - 7 = 3x + 3h - 7 Step 2 f ( x + h) - f ( x) = 3x + 3h - 7 - (3x - 7) = 3x + 3h - 7 - 3x + 7 = 3h

Step 3

f ( x + h) - f ( x ) 3h = =3 h h

Step 4 f ¢( x) = lim

3.4

h h 0 = lim 3 = 3 h 0

Exercises

1. True

f ( x + h) - f ( x )

f ¢(-2) = 3, f ¢(0) = 3 f ¢(3) = 3

2. True

Section 3.4

203

12. f ( x) = -2 x + 5

14. f ( x) = 6 x 2 - 5x - 1

Step 1 f ( x + h) = -2( x + h) + 5 = -2 x - 2h + 5 Step 2 f ( x + h) - f ( x) = -2 x - 2h + 5 - (-2 x + 5) = -2 x - 2h + 5 + 2 x - 5 = -2h

Step 1 f ( x + h) = 6( x + h)2 - 5( x + h) - 1 = 6( x 2 + 2 xh + h 2 ) - 5 x - 5h - 1 = 6 x 2 + 12 xh + 6h 2 - 5x - 5h - 1

Step 2 f ( x + h) - f ( x) = 6 x 2 + 12 xh + 6h 2 - 5x - 5h - 1

f ( x + h) - f ( x) -2h = = -2 h h f ( x + h) - f ( x) Step 4 f ¢( x) = lim h h0

- 6 x 2 + 5x + 1

Step 3

= lim -2 = -2 h0 ¢ f (-2) = -2, f ¢(0) = -2 f ¢(3) = -2

= 6h 2 + 12 xh - 5h = h(6h + 12 x - 5)

Step 3

f ( x + h) - f ( x ) h(6h + 12 x - 5) = h h = 6h + 12 x - 5 f ( x + h) - f ( x ) h h 0 = lim (6h + 12 x - 5)

Step 4 f ¢( x) = lim

13. f ( x) = -4 x + 9 x + 2 Step 1 f ( x + h)

h 0

= -4( x + h) + 9( x + h) + 2 = -4( x 2 + 2 xh + h 2 ) + 9 x + 9h + 2 = -4 x 2 - 8 xh - 4h 2 + 9 x + 9h + 2

Step 2 f ( x + h) - f ( x) = -4 x 2 - 8 xh - 4h 2 + 9 x + 9h + 2 -(-4 x 2 + 9 x + 2)

Step 4

f ¢(-2) = 12(-2) - 5 = -29 f ¢(0) = 12(0) - 5 = -5 f ¢(3) = 12(3) - 5 = 31

15. f ( x) = 12 x f ( x + h) =

= -8 xh - 4h 2 + 9h = h(-8x - 4h + 9)

Step 3

= 12 x - 5

12 x+h 12 - 12 x+h x 12 x - 12( x + h) = x( x + h) = 12 x - 12 x - 12h x ( x + h) h 12 = x( x + h)

f ( x + h) - f ( x) =

f ( x + h) - f ( x) h h(-8x - 4h + 9) = h = -8x - 4h + 9

f ( x + h) - f ( x) h = lim (-8 x - 4h + 9)

f ¢( x) = lim

h 0 h 0

= -8 x + 9 f ¢(-2) = -8(-2) + 9 = 25 f ¢(0) = -8(0) + 9 = 9 f ¢(3) = -8(3) + 9 = -15

f ( x + h) - f ( x ) = -12h h hx( x + h) = -12 x( x + h) = 2-12 x + xh f ( x + h) - f ( x ) h 12 = lim 2 h  0 x + xh = 12 x2

f ¢( x) = lim

h 0

204

Chapter 3 THE DERIVATIVE -12

f ¢(-2) =

(-2)

-12 = -3 4

f ( x + h) - f ( x ) h h 0 1 1 = lim = h 0 x + h + x 2 x

f ¢( x) = lim

f ¢(0) = -12 which is undefined so f ¢(0) does not 2 0

exist.

f ¢(-2) =

f ¢(3) = -12 32 = -12 = - 4 9 3

16. f ( x) = 3 x f ( x + h) =

which is undefined so f ¢(-2)

does not exist. f ¢(0) =

1 2 0

= 10 which is undefined so f ¢(0)

does not exist. 1 f ¢(3) = 2 3

3 x+h

3 - 3 x+h x 3x - 3( x + h) = x( x + h) = -3h x( x + h)

18. f ( x) = -3 x

f ( x + h) - f ( x ) =

Steps 1-3 are combined. f ( x + h) - f ( x ) h =

f ( x + h) - f ( x ) -3h = h hx( x + h) -3 = x ( x + h)

-3 x + h + 3 x h

Rationalize the numerator. =

f ( x + h) - f ( x ) h h0 3 = lim h  0 x( x + h) = -23 x f ¢(-2) = - 3 4 f ¢( x) = lim

f ¢(0) = - 03 which is undefined, so f ¢(0) does not

exist. f ¢(3) = - 32 = - 1 3 3

17. f ( x) = x Steps 1-3 are combined. f ( x + h) - f ( x ) h x+h - x = h x+h - x x+h + =  h x+h + x+h-x = h( x + h + x ) 1 = x+h + x

1 2 -2

-3 x + h + 3 x - 3 x + h - 3 x ⋅ h -3 x + h - 3 x

9( x + h) - 9 x h(-3 x + h - 3 x ) 9 x + 9h - 9 x = h(-3 x + h - 3 x ) 9 3 = = -3 x + h - 3 x - x+h -

f ¢( x) = lim

h 0 -

3 - x -

3 x+h -

x 3 = x -2 x

3 which is undefined so f ¢(-2) -2 - 2 does not exist. f ¢(-2) =

f ¢(0) =

3 3 = which is undefined so 0 -2 0

f ¢(0) does not exist. f ¢(3) = x x

3 3 =-2 3 2 3

Section 3.4

205

19. f ( x) = 2 x3 + 5

21. (a)

Steps 1-3 are combined.

2( x + h)3 + 5 - (2 x3 + 5) h

2( x3 + 3x 2h + 3xh 2 + h3 ) + 5 - 2 x3 - 5 h

f (5) - f (3) 5-3 2 (5) + 2(5) - [(3)2 + 2(3)] = 2 35 15 = 2 = 10

Slope of secant line =

f ( x + h) - f ( x ) h

f ( x) = x 2 + 2 x; x = 3, x = 5

Now use m = 10 and (3, f (3)) = (3,15) in the point-slope form.

2 x + 6 x h + 6 xh + 2h + 5 - 2 x - 5 h

y - 15 = 10( x - 3)

6 x 2h + 6 xh 2 + 2h3 = h

y - 15 = 10 x - 30 y = 10 x - 30 + 15

h(6 x 2 + 6 xh + 2h 2 ) = h

y = 10 x - 15

(b) f ( x) = x 2 + 2 x; x = 3

= 6 x 2 + 6 xh + 2h 2

f ( x + h) - f ( x ) h

f ¢( x) = lim (6 x 2 + 6 xh + 2h 2 ) = 6 x 2 h 0

f ¢(-2) = 6(-2)2 = 24 f ¢(0) = 6(0) 2 = 0 f ¢(3) = 6(3)2 = 54

[( x + h)2 + 2( x + h)] - ( x 2 + 2 x) h

( x 2 + 2hx + h 2 + 2 x + 2h) - ( x 2 + 2 x) h

2hx + h 2 + 2h = 2x + h + 2 h f ¢( x) = lim (2 x + h + 2) = 2 x + 2 =

20. f ( x) = 4 x3 - 3

Steps 1-3 are combined.

h0

f ( x + h) - f ( x ) h

f ¢(3) = 2(3) + 2 = 8 is the slope of the tangent line at x = 3 .

4( x + h)3 - 3 - (4 x3 - 3) h

4( x3 + 3x 2h + 3xh 2 + h3 ) - 3 - 4 x3 + 3 h

4 x 3 + 12 x 2h + 12 xh 2 + 4h3 - 3 - 4 x3 + 3 h 2

12 x h + 12 xh + 4h h

h(12 x 2 + 12 xh + 4h 2 ) h

Use m = 8 and (3,15) in the point-slope form. y - 15 = 8( x - 3) y = 8x - 9 22. (a)

f ( x ) = 6 - x 2 ; x = -1, x = 3

Slope of secant line =

6 - (3)2 - [6 - (-1)2 ] 4 3 5 = 4 = -2 =

= 12 x 2 + 12 xh + 4h 2 f ¢( x) = lim (12 x 2 + 12 xh + 4h 2 ) = 12 x 2 h 0

f ¢(-2) = 12(-2)2 = 48 f ¢(0) = 12(0)2 = 0 f ¢(3) = 12(3)2 = 108

f (3) - f (-1) 3 - (-1)

Now use m = -2 and (-1, f (-1)) = (-1, 5) in the point-slope form. y - 5 = -2[ x - (-1)] y - 5 = -2 x - 2 y = -2 x - 2 + 5 y = -2 x + 3

206

Chapter 3 THE DERIVATIVE (b) f ( x) = 6 - x 2 ; x = -1

(b) f ( x) =

f ( x + h) - f ( x ) h =

5 ;x = 2 x 5

[6 - ( x + h)2 ] - [6 - ( x) 2 ] h 2

f ( x + h) - f ( x) = x+h x h h

[6 - ( x + 2 xh + h )] - [6 - x ] h

6 - x 2 - 2 xh - h 2 - 6 + x 2 h

-2 xh - h 2 h(-2 x - h) = = -2 x - h h h

5 x -5( x + h) ( x + h) x

h 5 x - 5 x - 5h = h( x + h)( x) -5h = h( x + h) x -5 = ( x + h) x

f ¢( x) = lim (-2 x - h) = -2 x h 0

-5 5 =- 2 ( + )( ) x h x h0 x

f ¢( x) = lim

f ¢(-1) = -2(-1) = 2 is the slope of the tangent line at x = -1. Use m = 2 and (-1, 5) in the point-slope form.

f ¢(2) = -25 = - 54 is the slope of the tangent 2

line at x = 2.

y - 5 = 2( x + 1)

Now use m = -

y - 5 = 2x + 2

5 4

( ) in the point-

and 2,

5 2

slope form.

y = 2x + 7

5 5 = - ( x - 2) 2 4 5 5 10 y- =- x+ 2 4 4 5 y =- x+5 4

23. (a)

f ( x) =

5 ; x = 2, x = 5 x

Slope of secant line = =

f (5) - f (2) 5-2 5 5 5 2

3 1 =2

Now use m = -

1 2

5 2

24. (a)

f ( x) = -

3 ; x +1

x = 1, x = 5

Slope of secant line =

and (5, f (5)) = (5, 1) in

f (5) - f (1) 5-1 -

the point-slope form.

1 y - 1 = - [ x - 5] 2 1 5 y -1 = - x + 2 2 1 5 y = - x + +1 2 2 1 7 y =- x+ 2 2

é 3 3 ùú - êê (1+1) ú (5+1) ë û

1 3 - + 2 2 =

1 = 4

Now use m =

1 and (1, f (1)) = 4

(1, - ) in the

point-slope form. æ ö y - çç - 3 ÷÷÷ = 1 ( x - 1) è 2ø 4 y+ 3 = 1x- 1 2 4 4 y = 1x- 1 - 3 4 4 2 1 7 y = x4 4

3 2

Section 3.4 (b)

207

f ( x) =

-3 ; x =1 x +1 -3

-3

- x +1 f ( x + h) - f ( x ) ( x + h) +1 = h h = = =

-3( x +1) + 3( x + h +1) ( x +1)( x + h +1)

h -3 x -3+ 3x + 3h + 3 ( x +1)( x + h +1)

3h h( x + 1)( x + h + 1)

h  0 ( x + 1)( x + h + 1)

f ¢(1) =

3 (1+1)2

16( x + h) - 16 x h  0 h(4 x + h + 4 x ) 16h = lim h  0 h(4 x + h + 4 x ) 4 4 = lim = h 0 ( x + h + x ) 2 x 2 = x

f ¢( x) = lim

(b) f ( x) = 4 x ; x = 9 f ( x + h) - f ( x ) h 4 x+h -4 x 4 x+h +4 x = ⋅ h 4 x+h +4 x 16( x + h) - 16 x = h(4 x + h + 4 x )

3 ( x + 1)2

= 34 is the slope of the tangent

line at x = 1. Use m = 34 and (1, - 32 ) in the

f ¢(9) =

point-slope form.

Use m = 23 and (9, 12) in the point-slope form.

3 3 3 = x2 4 4 3 9 y = x4 4

2 ( x - 9) 3 2 y = x+6 3

y - 12 =

26. (a)

f ( x) = 4 x ; x = 9, x = 16 f (16) - f (9) 16 - 9 4 16 - 4 9 = 7 16 - 12 4 = = 7 7

Slope of secant line =

Now use m = 74 and (9, f (9)) = (9, 12) in the point-slope form. 4 ( x - 9) 7 4 36 y - 12 = x 7 7 4 36 y = x+ 12 7 7 4 48 y = x+ 7 7

y - 12 =

= 32 is the slope of the tangent

line at x = 9 .

æ 3ö 3 y - çç - ÷÷÷ = ( x - 1) çè 2 ø 4

25. (a)

2 9

f ( x) =

x ; x = 25, x = 36

f (36) - f (25) 36 - 25 36 - 25 = 11 6-5 1 = = 11 11

Slope of secant line =

1 and (25, f (25)) = (25, 5) Now use m = 11

in the point-slope form. 1 ( x - 25) 11 1 25 y-5= x11 11 1 25 y = x+5 11 11 1 30 y = x+ 11 11 y-5=

208

Chapter 3 THE DERIVATIVE (b) f ( x) =

f ¢( x) = lim (12 x + 6h - 4) = 12 x - 4

x ; x = 25

f ( x + h) - f ( x) h x+h - x = h x+h - x = ⋅ h

h 0

f ¢(2) = 12(2) - 4 = 24 - 4 = 20 f ¢(16) = 12(16) - 4 = 192 - 4 = 188 f ¢(-3) = 12(-3) - 4 = -36 - 4 = -40

x+h + x+h +

x+h-x h( x + h + x ) h = = h( x + h + x )

x x

29. f ( x) = e x f ( x + h) - f ( x ) ex+h - ex = h h

1 x+h +

ex+h - ex h h 0

f ¢( x) = lim

1 1 = h0 x + h + x 2 x 1 1 1 f ¢(25) = = = 2⋅5 10 2 25 f ¢( x) = lim

1 and (25, 5) in the point-slope form. Use m = 10

1 ( x - 25) 10 1 25 y-5 = x10 10 1 5 y = x+ 10 2 y-5 =

f ¢(2) » 7.3891; f ¢(16) » 8,886,111; f ¢(-3) » 0.0498 30. f ( x) = ln | x | f ( x + h) - f ( x ) h ln | x + h | - ln | x | = h ln | x + h | - ln | x | h h 0 f ¢(2) = 0.5 f ¢(16) = 0.0625 f ¢( x) = lim

f ¢(-3) = -0.3

27. f ( x) = -4 x 2 + 11x 31. f ( x) = -

f ( x + h) - f ( x ) h -4( x + h) 2 + 11( x + h) - (-4 x 2 + 11x) = h -8xh - 4h 2 + 11h = h

2 x

-2 - -2 (x) f ( x + h) - f ( x ) = x+h h h

f ¢( x) = lim (-8 x - 4h + 11) = -8x + 11 h 0

f ¢(2) = -8(2) + 11 = -5 f ¢(16) = -8(16) + 11 = -117 f ¢(-3) = -8(-3) + 11 = 35

28. f ( x) = 6 x 2 - 4 x f ( x + h) - f ( x ) h =

6( x + h)2 - 4( x + h) - (6 x 2 - 4 x) h

12 xh + 6h 2 - 4h = 12 x + 6h - 4 h

-2 x + 2( x + h ) ( x + h) x

h 2h 2 = = h( x + h ) x ( x + h) x 2 2 = 2 f ¢( x) = lim h  0 ( x + h) x x 2 1 f ¢(2) = 2 = 2 2 2 2 1 = f ¢(16) = 2 = . 256 128 16 2 2 = f ¢(-3) = 2 9 (-3)

Section 3.4

209

32. f ( x) =

6 x

3 h 0 - x + h - x 3 3 = =- x - x 2 x 3 f ¢(2) = 2 2 -3 3 =f ¢(16) = 8 2 16 f ¢( x) = lim

6 - 6 f ( x + h) - f ( x ) x = x+h h h -6 6 x - 6( x + h) = = hx( x + h) x( x + h )

-6

f ¢( x ) = lim

-6

x2 -6 -6 3 = =f ¢(2) = 2 4 2 (2) h  0 x( x + h )

f ¢(-3) = -3 , which is not a real number, so 2 -3

f ¢(-3) does not exist.

-6

-6 3 =f ¢(16) = 2 = 256 128 16 -6 -6 2 = =f ¢(-3) = 2 9 3 (-3)

35. At x = 0, the graph of f (x) has a sharp point. Therefore, there is no derivative for x = 0. 36. No derivative exists at x = -6 because the function is not defined at x = -6.

33. f ( x) = x f ( x + h) - f ( x) h =

x+h h

⋅

x+h + x+h +

( x + h) - x h( x + h + x ) h = = h( x + h + x )

37. For x = -3 and x = 0, the tangent to the graph of f (x) is vertical. For x = -1, there is a gap in the graph of f (x). For x = 2, the function f (x) does not exist. For x = 3 and x = 5, the graph of f (x) has sharp points. Therefore, no derivative exists for x = -3, x = -1, x = 0, x = 2, x = 3, and x = 5.

x x

f ¢( x) = lim

h 0

f ¢(2) =

1 x+h +

1 x+h + x

38. For x = -5 and x = 0, the function f (x) is not defined. For x = -3 and x = 2 the graph of f (x) has sharp points. For x = 4, the tangent to the graph is vertical. Therefore, no derivative exists for x = -5, x = -3, x = 0, x = 2, or x = 4.

1 2 x

2 2 1 1 = f ¢(16) = 8 2 16 1 is not a real number, so f ¢(-3) = 2 -3

39. (a)

f ¢(-3) does not exist. 34. f ( x) = -3 x f ( x + h) - f ( x ) -3 x + h + 3 x = h h =

-3 x + h + 3 x -3 x + h - 3 x  h -3 x + h - 3 x

9( x + h) - 9 x h(-3 x + h - 3 x ) 9 3 = = -3 x + h - 3 x - x+h =

The line tangent to g ( x) = 3 x at x = 0 is a vertical line. Since the slope of a vertical line is undefined, g ¢(0) does not exist. 2

40. f ( x) = xx +-21 is not differentiable when x + 2 = 0 or x = -2 because the function is undefined and a vertical asymptote occurs there.

210

Chapter 3 THE DERIVATIVE

41. (a)

f ( x) = 5 is a horizontal line and has slope 0; the derivative is 0.

(b) f ( x) = x has slope 1; the derivative is 1. (c)

f ( x) = -x has slope of -1, the derivative is -1.

46. f ( x) = x x , a = 2 (a)

h 0.01

2.012.01 - 2 2 0.01 = 6.84 =

(d) x = 3 is vertical and has undefined slope; the derivative does not exist. (e) y = mx + b has slope m; the derivative is m.

0.001

42. If the rate of change of f ( x) is zero when x = a, the tangent line at that point must have a slope of zero. Thus, the tangent line is horizontal at that point. 43. (a) The rate of change of f (x) is positive when f (x) is increasing, that is, on (a, 0) and (b, c). (b) The rate of change of f (x) is negative when f (x) is decreasing, that is, on (0, b,). (c) The rate of change is zero when the tangent to the graph is horizontal, that is, at x = 0 and x = b.

f (2 + 0.001) - f (2) 0.001 2.0012.001 - 2 2 0.001 = 6.779 =

0.0001

f (2 + 0.0001) - f (2) 0.0001 2.00012.0001 - 2 2 0.0001 = 6.773 =

0.00001

f (2 + 0.00001) - f (2) 0.00001 2.000012.00001 - 2 2 0 .00001 = 6.7727

44. The zeros of graph (b) correspond to the turning points of graph (a), the points where the derivative is zero. Graph (a) gives the distance, while graph (b) gives the velocity. 45. The zeros of graph (b) correspond to the turning points of graph (a), the points where the derivative is zero. Graph (a) gives the distance, while graph (b) gives the velocity.

f (2 + 0.01) - f (2) 0.01

0.000001

f (2 + 0.000001) - f (2) 0.000001 2.0000012.000001 - 2 2 0.000001 = 6.7726 =

It appears that f ¢(2) » 6.773. (b) Graph the function on a graphing calculator and move the cursor to an x-value near x = 2. A good choice for the initial viewing window is [0, 3] by [0, 10].

Now zoom in on the function several times. Each time you zoom in, the graph will look less like a curve and more like a straight line. When the graph appears to be a straight line, use the TRACE feature to select two points on the graph, and record their coordinates. Use these two points to compute the slope. The result will be very close to the most accurate value found in part (a), which is 6.773.

Section 3.4

211

47. f ( x) = x x , a = 3 (a)

48. f ( x) = x1/x , a = 2 (a)

h 0.01

f (3 + 0.01) - f (3) 0.01 3.01

h 0.01

-3 0.01 = 57.3072

0.001

3.01

3.0013.001 - 33 0.001 = 56.7265 0.00001

3.000013.00001 - 33 0.00001 = 56.6632 f (3 + 0.000001) - f (3) 0.000001 3.0000013.000001 - 33 = 0.000001 = 56.6626 0.0000001

f (2 + 0.001) - f (2) 0.001 0.001

2.0011/2.001 - 21/2 0.001 = 0.1084 f (2 + 0.0001) - f (2) 0.0001

0.0001

2.00011/2.0001 - 21/2 0.0001 = 0.1085 f (2 + 0.00001) - f (2) 0.00001

0.00001

2.0 00011/2.00001 - 21/2 0.00001 = 0.1085 f (2 + 0.000001) - f (2) 0.000001

f (3 + 0.00001) - f (3) 0.00001 =

0.000001

2.011/2.01 - 21/2 0.01 = 0.1071 =

f (3 + 0.001) - f (3) 0.001 =

f (2 + .01) - f (2) 0.01

f (3 + 0.0000001) - f (3) 0.0000001

0.000001

3.00000013.0000001 - 33 0.0000001 = 56.6625

It appears that f ¢(2) = 0.1085.

It appears that f ¢(3) » 56.66. (b) Graph the function on a graphing calculator and move the cursor to an x-value near x = 3. A good choice for the initial viewing window is [0, 4] by [0, 60] with Xscl = 1, Yscl = 10.

Now zoom in on the function several times. Each time you zoom in, the graph will look less like a curve and more like a straight line. Use the TRACE feature to select two points on the graph, and record their coordinates. Use these two points to compute the slope. The result will be close to the most accurate value found in part (a), which is 56.66. Note: In this exercise, the method used in part (a) gives more accurate results than the method used in part (b).

2.0000011/2.000001 - 21/ 2 0.000001 = 0.1085 =

(b) Graph this function on a graphing calculator and move the cursor to an x-value near x = 2. A good choice for the initial viewing window is [0, 5] by [0, 3]. Follow the procedure outlined in the solution for Exercise 42, part (b). The final result will be close to the value found in part (a) of this exercise, which is 0.1085.

212

Chapter 3 THE DERIVATIVE

49. f ( x) = x1/x , a = 3 (a)

f ( x) = e x

h f (3 + 0.01) - f (3) 0.01

0.01

Graph y =

e x + 0.01 - e x . 0.01

3.011/3.01 - 31/3 0.01 = - 0.0160 =

0.001

f (3 + 0.001) - f (3) 0.001 3.0011/3.001 - 31/3 0.001 = - 0.0158 =

0.0001

f (3 + 0.0001) - f (3) 0.0001

f ( x) = x 3

Graph y =

( x + 0.01)3 - x3 . 0.01

3.00011/ 3.0001 - 31/3 0.0001 = - 0.0158 =

It appears that f ¢(3) = -0.0158. (b) Graph the function on a graphing calculator and move the cursor to an x-value near x = 3. A good choice for the initial viewing window is [0, 5] by [0, 3].

Column B Graph y = e x

Follow the procedure outlined in the solution for Exercise 43, part (b). Note that near x = 3, the graph is very close to a horizontal line, so we expect that it slope will be close to 0. The final result will be close to the value found in part (a) of this exercise, which is -0.0158. 50. For Column A, let h = 0.01 f ( x) = ln | x | Graph y =

Graph y = 3x 2

ln | x + 0.01| - ln | x | . 0.01

Section 3.4

213 1 x

Graph y =

f ( x + h) - f ( x ) h h 0 = lim (-8 x + 4h + 11)

f ¢( x) = lim

h 0

= -8 x + 11 f ¢(3) = -8(3) + 11 = -13 f (3 + 0.1) - f (3) 0.1 f (3.1) - f (3) = 0.1

We observe that the graph of

-4(3.1)2 + 11(3.1) - (-4(3)2 + 11(3)) 0.1 -4(9.61) + 11(3.1) + 4(9) - 11(3) = 0.1 -38.44 + 34.1 + 36 - 33 = 0.1 -1.34 = = -13.4 0.1 =

ln | x + 0.01| - ln | x | y = 0.01

is very similar to the graph of y =

1 , x

the graph of y =

e x + 0.01 - e x 0.01

is very similar to the graph of y = e x , and the graph of y =

( x + 0.01) - x 0.01

f (3 + 0.1) - f (3 - 0.1) 2(0.1) f (3.1) - f (2.9) = 0.2 -4(3.1)2 + 11(3.1) - (-4(2.9) 2 + 11(2.9)) 0.2 -4(9.61) + 11(3.1) + 4(8.41) - 11(2.9) = 0.2 -38.44 + 34.1 + 33.64 - 31.9 = 0.2 -2.6 = = -13 0.2 =

is very similar to the graph of y = 3x 2 .

Thus the derivative of In x is 1x , the derivative of e x is e x , and the derivative of x3 is 3x 2.

52. (a) f ( x) = -4 x 2 + 11x f ( x + h) = -4( x + h)2 + 11( x + h) = -4( x 2 + 2 xh + h 2 ) + 11( x + h) = -4 x 2 - 8 xh - 4h 2 + 11x + 11h f ( x + h) - f ( x) = -4 x 2 - 8 xh - 4h 2 + 11x + 11h + 4 x 2 - 11x = -8 xh - 4h 2 + 11h f ( x + h) - f ( x ) -8 xh - 4h 2 + 11h = h h = -8x + 4h + 11

(b) f ¢(3) = -8(3) + 11 = -13 f (3 + 0.01) - f (3) 0.01 f (3.01) - f (3) = 0.01 -4(3.01)2 + 11(3.01) - (-4(3) 2 + 11(3)) 0.01 -4(9.0601) + 11(3.01) + 4(9) - 11(3) = 0.01 -36.2404 + 33.11 + 36 - 33 = 0.01 -0.1304 = = -13.04 0.01 =

214

Chapter 3 THE DERIVATIVE

f (3 + 0.01) - f (3 - 0.01) 2(0.01) f (3.01) - f (2.99) = 0.02

(d) f ¢(3) =

-4(3.01)2 + 11(3.01) - (-4(2.99) 2 + 11(2.99)) 0.02 -4(9.0601) + 11(3.01) + 4(8.9401) - 11(2.99) = 0.02 -36.2404 + 33.11 + 35.7604 - 32.89 = 0.02 -0.26 = = -13 0.02 =

(c)

-2 x -2 f ( x + h) = x+h æ -2 ÷ö -2 - çç f ( x + h) - f ( x) = ÷ x + h çè x ÷ø -2 2 = + x+h x -2 x + 2( x + h) = x( x + h) 2h = x ( x + h) f ( x) =

f ( x + h) - f ( x) = h

2h x ( x + h)

h 2h 1 = ⋅ x ( x + h) h 2 = x ( x + h)

f ( x + h) - f ( x ) h h 0 2 2 = lim = 2 h  0 x( x + h) x 2 2 f ¢(3) = 2 = » 0.222222 9 3

2 2

f (3 + .01) - f (3) f (3.01) - f (3) = 0.01 0.01 - 2 - -2 3 = 3.01 0.01 » 0.221484 f (3 + 0.01) - f (3 - 0.01) f (3.01) - f (2.99) = 2(0.01) 0.02 -2 - -2 2.99 = 3.01 0.02 » 0.222225

(e)

f ( x) =

f ( x + h) - f ( x) h =

x+h h

f (3 + 0.1) - f (3 - 0.1) f (3.1) - f (2.9) = 2(0.1) 0.2

⋅

( x + h) - x h( x + h + x)

h h( x + h +

1 h( x + h +

x+h + x+h +

1 h  0 h( x + h +

f ¢( x) = lim f ¢(3) =

x x

1 2 x

1 » 0.288675 2 3

f (3 + 0.1) - f (3) f (3.1) - f (3) = 0.1 0.1 3.1 - 3 0.1 » 0.286309

f ¢( x) = lim

f (3 + 0.1) - f (3) f (3.1) - f (3) = 0.1 0.1 - 2 - -2 3 = 3.1 0.1 » .215054

2 » 0.222222 9

f (3 + 0.1) - f (3 - 0.1) f (3.1) - f (2.9) = 2(0.1) 0.2 3.1 - 2.9 0.2 » 0.288715

(f ) f ¢(3) =

1 2 3

» 0.288675

f (3 + 0.01) - f (3) f (3.01) - f (3) = 0.01 0.01

-2 - - 2 2.9

= 3.1

0.2 = 0.222469

3.01 - 3 0.01 » 0.288435

Section 3.4

215

f (3 + 0.01) - f (3 - 0.01) f (3.01) - f (2.99) = 2(0.01) 0.02

R(1001) - R(1000) = 20(1001) - 1001 500 2ù é - êê 20(1000) - 1000 úú 500 û ë = 18, 015.998 - 18, 000 = $15.998 or $16.

3.01 - 2.99 0.02 » 0.288676

53. D( p) = -2 p 2 - 4 p + 300

D is demand; p is price. (a) Given that D¢( p) = -4 p - 4, the rate of change of demand with respect to price is -4 p - 4, the derivative of the function D(p). (b) D¢(10) = -4(10) - 4

= -44 The demand is decreasing at the rate of about 44 items for each increase in price of $1. 54. P( x) = 1000 + 32 x - 2 x 2 (a) $8000 is 8 thousands, so x = 8.

(d) The marginal revenue gives a good approximation of the actual revenue from the sale of the 100lst table. 600 x + 20 ( x + 20)(0) - 600(1) (a) C ¢( x) = ( x + 20)2 = - 600 2 ( x + 20)

56. C ( x) =

(b) For x = 50, 600 » -0.12 (50 + 20)2 The average cost is decreasing by $0.12 per unit of yogurt. C ¢(50) = -

P¢(8) = 32 - 4(8) = 32 - 32 = 0

No, the firm should not increase production, since the marginal profit is 0. (b) $6000, x = 6 P¢(6) = 32 - 4(6) = 32 - 24 = 8

Yes, the first should increase production, since the marginal profit is positive, $8000. (c) $12, 000, x = 12 P¢(12) = 32 - 4(12) = 32 - 48 = -16 No, because the marginal profit is negative, -$16,000.

(d) $20, 000, x = 20 P¢(20) = 32 - 4(20) = 32 - 80 = -48 No, because the marginal profit is negative, -$48,000.

55. R( x) = 20 x -

x2 500

18x 106 - x (106 - x)(18) - 18x(-1) f ¢( x ) = (106 - x)2 1908 = (106 - x) 2

57. f ( x) = (a)

(b) For x = 75, 1908 » 1.985 (106 - 75) 2 The cost is increasing by $1985 for every additional percent of pollution removed. f ¢(75) =

(a) R¢( x) = 20 - x 250

1908 » 7.453 (106 - 90) 2 The cost is increasing by $7453 for every additional percent of pollution removed. f ¢(90) =

(b) When x = 1000, R¢(1000) = 20 - 1000 250 = $16 per table. The marginal revenue for the l00lst table is approximately R¢(1000) which is about $16.

(d) For x = 95, 1908 » 15.769 (106 - 95) 2 The cost is increasing by $15,769 for every additional percent of pollution removed. f ¢(95) =

216

Chapter 3 THE DERIVATIVE

58. C ( x) = -0.00375x 2 + 1.5 x + 1000, 0 £ x £ 180

59. (a)

(a) The marginal cost is given by C ¢( x) = -0.0075x + 1.5, 0 £ x £ 180

(b) C ¢(100) = -0.0075(100) + 1.5 = 0.75

This represents the fact that the cost of producing the next (l0lst) taco is approximately $0.75. (c) The exact cost to produce the l0lst taco is given by C (101) - C (100) 2

= [-0.00375(101) + 1.5(101) + 1000] -[-0.00375(100)2 + 1.5(100) + 1000] = (-38.25375 + 151.5 + 1000) -(37.5 + 150 + 1000) = 0.74625

That is, the exact cost of producing the l0lst taco is $0.74625. (d) The exact cost of producing the l0lst taco is $0.00375 less than the approximate cost. They are very close. (e)

C ( x) = ax 2 + bx + c C ¢( x) = 2ax + b

[C ( x + 1) - C ( x)] - C ¢( x) = [a( x + 1)2 + b( x + 1) + c] - [ax 2 + bx + c] - (2ax + b) = ax 2 + 2ax + a + bx + b + c -ax 2 - bx - c - 2ax - b = a

(f )

C ( x) = ax 2 + bx + c C ¢( x) = 2ax + b C ( x + 1) - C ( x) = [a( x + 1) 2 + b( x + 1) + c] - [ax 2 + bx + c] = ax 2 + 2ax + a + bx + b + c - ax 2 - bx - c = 2ax + a + b

f ( x) = 0.0000329 x3 - 0.00450 x 2 + 0.0613x + 2.34 f (10) = 2.54 f (20) = 2.03 f (30) = 1.02

(b) Y1 = 0.0000329 x3 - 0.00450 x 2 + 0.0613x + 2.34

nDeriv(Y1, x, 0) » 0.061 nDeriv(Y1, x,10) » -0.019 nDeriv(Y1, x, 20) » -0.079 nDeriv(Y1, x,30) » -0.120 nDeriv(Y1, x,35) » -0.133

60. (a) From the graph, Vmp is just about at the turning point of the curve. Thus, the slope of the tangent line is approximately zero. The power expenditure is not changing. (b) From the graph, the slope of the tangent line at Vmr is approximately 0.54. The power expended is increasing 0.54 unit per unit increase in speed. (c) The power level first decreases to Vmp, then increases at greater rates. (d) Vmr is the point which produces the smallest slope of a line. 61. The derivative at (2, 4000) can be approximated by the slope of the line through (0, 2000) and (2, 4000). The derivative is approximately 4000 - 2000 2000 = = 1000. 2-0 2

Thus the shellfish population is increasing at a rate of 1000 shellfish per unit time. The derivative at about (10,10,300) can be approximated by the slope of the line through (10,10,300) and (13,12,000). The derivative is approximately 12, 000 - 10,300 1700 = » 570. 13 - 10 3

= 2ax + a + b

The shellfish population is increasing at a rate of about 570 shellfish per unit time. The derivative at about (13, 11,250) can be approximated by the slope of the line through (13, 11,250) and (16, 12,000). The derivative is approximately

æ 1ö Thus, C ( x + 1) - C ( x) = C ¢ çç x + ÷÷÷. çè 2ø

12, 000 - 11, 000 1000 = » 200. 16 - 11 5

æ æ 1ö 1ö C ¢ çç x + ÷÷ = 2a çç x + ÷÷ + b ÷ çè ç è 2ø 2 ÷ø

Section 3.4 62.

217

I (t ) = 27 + 72t - 1.5t 2

(a)

I (t + h) - I (t ) h h0

I ¢(t ) = lim

27 + 72t + 72h - 1.5t 2 - 3th - 1.5h 2 - 27 - 72t + 1.5t 2 h h0

= lim

72h - 3th - 1.5h 2 h h0 = lim 72 - 3t - 1.5h = lim

h0

= 72 - 3t I ¢(5) = 72 - 3(5) = 72 - 15 = 57

The rate of change of the intake of food 5 minutes into a meal is 57 grams per minute. (b)

I ¢(24) = 0 ?

72 - 3(24) = 0 0=0

24 minutes after the meal starts the rate of food consumption is 0. (c) After 24 minutes the rate of food consumption is negative according to the function where a rate of zero is more accurate. A logical range for this function is 0 £ t £ 24.

63. (a) Set M (v)  150 and solve for v. 0.0312443v 2 - 101.39v + 82, 264 = 150 0.0312443v 2 - 101.39v + 82,114 = 0

Solve using the quadratic formula. Let D equal the discriminant. D = b 2 - 4ac = (-101.39) 2 - 4(0.0312443)(82,114) » 17.55 v =

101.39  D 2(0.0312443)

v » 1690 meter per second or v » 1560 meters per second.

Since the functions is defined only for v ³ 1620, the only solution is 1690 meters per second.

218

Chapter 3 THE DERIVATIVE M (1700 + h) - M (1700) h h 0

(b) Calculate lim M (1700 + h)

= 0.0312443(1700 + h) 2 - 101.39(1700 + h) + 82, 264 = 90, 296.027 + 106.23062h + 0.0312443h 2 - 172,363 - 101.39h + 82, 264 = 0.01312443h 2 + 4.84062h + 197.027 M (1700) = 197.027, so the derivative of M (v) at v = 1700 is æ 0.0312443h 2 + 4.84062h + 197.027 - 197.027 ö÷ ç lim çç ÷÷÷ h h  0 çè ÷ø 2 æ ö ç 0.0312443h + 4.84062h ÷÷ = lim çç ÷÷ h h  0 çè ÷ø

= lim ( 0.0312443h + 4.84062 ) h0

= 4.84062 » 4.84 days per meter per second

The increase in velocity for this cheese from 1700 m/s to 1701 m/s indicates that the approximate age of the cheese has increased by 4.84 days.

64. (a) From the graph, T (0.5) » 185. A tangent drawn at this point appears to intersect the T = 1 vertical line at about 320, so the tangent has a slope of about T (1) - T (0.5) 320 - 185 = = 270. 1 - 0.5 0.5 T ¢(0.5) » 270; at 9:00 AM the temperature is increasing at about 270° per hour. (b) A tangent drawn to the curve at T = 3 appears to intersect the T = 2 line at 480 and the T = 4 line at 180, so the tangent has a slope of about T (4) - T (2) 180 - 480 = = -150. 4-2 2 T ¢(3) » -150; at 11:30 AM the temperature is decreasing at about 150° per hour. (c) At T = 4 the graph appears to have a horizontal tangent, so T ¢(4) » 0; the temperature is staying constant at 12:30 PM. (d) At about 11:15 AM.

65. (a) The slope of the graph at 16 looks horizontal. Thus, the derivative for a 16 ounce bat is about 0 mph per oz . The slope of the graph at x = 25 can be estimated using the points (25, 63.4) and (26, 62.8 ). slope =

62.8 - 63.4 = -0.6 26 - 25

Thus the derivative is for a 25-ounce bat is about -0.6 mph per oz. (b) The optimal bat is 16 oz. 66. (a)

At 40 oz the tangent looks horizontal; thus the derivative for a 40-ounce bat is about 0 mph per oz. The slope of the graph at x = 30 can be estimated using the points (30, 79) and (32, 80). 80 - 79 = 0.5 slope = 32 - 30 Thus, the derivative for a 30 ounce bat is about 0.5 mph per oz .

(b) The optimal bat is 40 oz.

Section 3.5

219 (e) The graph of the derivative is

3.5 Graphical Differentiation Your Turn 1

8. (a)

f ¢( x) = 0 nowhere.

(b) f ¢( x) is undefined when x = –2 and x = 2. (c)

f ¢( x) > 0 on the intervals (-¥, -2) and (2, ¥).

Your Turn 2

(d) f ¢( x) < 0 on the interval (–2, 2). (e) The graph of the derivative is

3.5 Warmup Exercises W1. The tangent at (2, 13) goes through the points (0, 9) and (4, 17), so the tangent has slope 17 - 9 = 2. 4-0

f ¢( x) = 0 when x = –1, x = 0, and x = 1.

(b) f ¢( x) is undefined nowhere. (c)

f ¢( x) > 0 on the intervals (-1, 0) and (1, ¥).

W2. The tangent at (3, 4) goes through the point (1, 14), so the tangent has slope 4 - 14 = -5. 3-1

3.5

9. (a)

(d) f ¢( x) < 0 on the intervals (-¥, -1) and (0, 1). (e) The graph of the derivative is

Exercises

1. True 2. True 3. False. If a function is positive at x = a, that means that it is above the x-axis. If a derivative is positive, that means that the slope is positive at that point. For a counterexample to the statement, see Exercise 8, when x = –3. The value of the function is positive, however the value of the derivative at x = –3 is negative. 4. True 7. (a)

10. (a)

f ¢( x) = 0 when x = 0.

(b) f ¢( x) is undefined when x = –2 and x = 2. (c)

f ¢( x) > 0 on the intervals (-2, 0) and (2, ¥).

(d) f ¢( x) < 0 on the intervals (-¥, -2) and (0, 2). (e) The graph of the derivative is

f ¢( x) = 0 when x = 2.

(b) f ¢( x) is undefined when x = 1. (c)

f ¢( x) > 0 on the intervals (-¥, 1) and (2, ¥).

(d) f ¢( x) < 0 on the interval (1, 2).

220

Chapter 3 THE DERIVATIVE

11. Since the x-intercepts of the graph of f ¢ occur whenever the graph of f has a horizontal tangent line, Y1 is the derivative of Y2. Notice that Y1 has 2 x-intercepts; each occurs at an x-value where the tangent line to Y2 is horizontal. Note also that Y1 is positive whenever Y2 is increasing, and that Y1 is negative whenever Y2 is decreasing. 12. Since the x-intercepts of the graph f ¢ occur whenever the graph of f has a horizontal tangent line, Y2 is the derivative of Y1. Notice that Y2 has 2 x-intercepts; each occurs at an x-value where the tangent line to Y1 is horizontal. Note also that Y2 is negative whenever Y1 is decreasing, and Y2 is positive whenever Y1 is increasing. 13. Since the x-intercepts of the graph of f ¢ occur whenever the graph of f has a horizontal tangent line, Y2 is the derivative of Y1. Notice that Y2 has 1 x-intercept which occurs at the x-value where the tangent line to Y1 is horizontal. Also notice that the range on which Y1 is increasing, Y2 is positive and the range on which it is decreasing, Y2 is negative. 14. Since the x-intercepts of the graph f ¢ occur whenever the graph of f has a horizontal tangent line, Y1 is the derivative of Y2. Notice that Y1 has 4 x-intercepts; each occurs at an x-value where the tangent line to Y2 is horizontal. Note also that Y1 is negative whenever Y2 is decreasing and Y1 is positive whenever Y2 is increasing. 15. To graph f ¢, observe the intervals where the slopes of tangent lines are positive and where they are negative to determine where the derivative is positive and where it is negative. Also, whenever f has a horizontal tangent, f ¢ will be 0, so the graph

of f ¢ will have an x-intercept. The x-values of the three turning point on the graph of f become the three x-intercepts of the graph of f. Estimate the magnitude of the slope at several points by drawing tangents to the graph of f.

16. To graph f ¢, observe the intervals where the slopes of lines are positive and where they are negative to determine where the derivative is positive and where it is negative. Also, whenever f has a horizontal tangent, f ¢ will be 0.

Estimate the magnitude of the slope at several points by drawing tangents to the graph of f.

17. On the interval (-¥, -2), the graph of f is a horizontal line, so its slope is 0. Thus, on this interval, the graph of f ¢ is y = 0 on (-¥, -2). On the interval (-2, 0), the graph of f is a straight line, so its slope is constant. To find this slope, use the points (-2, 2) and (0, 0). m=

2-0 2 = = -1 -2 - 0 -2

On the interval (0, 1), the slope is also constant. To find this slope, use the points (0, 0) and (1, 1). m=

1- 0 =1 1- 0

On the interval (1, ¥), the graph is again a horizontal line, so m = 0. The graph of f ¢ will be made up of portions of the y-axis and the lines y = -1 and y = 1 . Because the graph of f has “sharp points” or “corners” at x = -2, x = 0, and x = 1, we know that f ¢(-2), f ¢(0), and f ¢(1) do not exist. We show this on the graph of f ¢ by using open circles at the endpoints of the portions of the graph.

18. On the interval (-¥, - 3), the graph of f is a straight line, so its slope is constant. To find this slope, use the points (-6, - 3) and (-3, 0). m=

0 - (-3) 3 = =1 -3 - (-6) 3

Section 3.5

221

On the interval (3, ¥), the slope of f is also constant. To find this slope, use the points (3, 0) and (6,–3). -3 - 0 -3 m= = = -1 6-3 3 Thus, we have f ¢( x) = 1 on (-¥, -3) and f ¢( x) = -1 on (3, ¥). Because the graph of f has sharp points at x = –3 and x = 3, we know that f ¢(-3) and f ¢(3) do not exist. We show this on

the graph of f ¢( x) by using open circles. We also observe that the slopes of tangent lines are negative on (–3, 0), that the graph has a horizontal tangent at x = 0, and that the slopes of tangent lines are positive on (0, 3). Thus, f ¢ is negative on (–3, 0) 0 at x = 0, and positive on (0, 3). Furthermore, by drawing tangents, we see that on (–3, 3), f ¢ increases from –1 to 1.

20. On the interval (-¥, 0), the graph of f is a horizontal line, so its slope is 0. Thus, the graph of f ¢ is y = 0 (the x-axis) on (-¥, 0).

Since f is discontinuous at x = 0, f ¢(0) does not exist. Thus, the graph of f ¢ has an open circle at x = 0. On the interval (0, ¥), the slopes are positive but decreasing at a slower rate as x gets larger. Therefore, the value of f ¢ will be positive but decreasing on this interval. This value approaches 0, but never becomes 0.

21. We observe that the slopes of tangent lines are positive on the interval (-¥, 0) and negative on the interval (0, ¥), so the value of f ¢ will be positive on (-¥, 0) and negative on (0, ¥). Since f is undefined at x = 0, f ¢(0) does not exist. 19. On the interval (-¥, - 2), the graph of f is a straight line, so its slope is constant. To find this slope, use the points (-4, 2) and (-2, 0). m=

0-2 = -2 = -1 -2 - (-4) 2

On the interval (2, ¥) , the slope of f is also constant. To find this slope, use the points (2, 0) and (3, 2). m = 2-0 = 2 = 2 3-2 1

Notice that the graph of f becomes very flat when | x |  ¥. The value of f approaches 0 and also the slope approaches 0. Thus, y = 0 (the x-axis) is a horizontal asymptote for both the graph of f and the graph of f ¢. As x  0- and x  0+ , the graph of f gets very steep, so | f ¢( x) | ¥. Thus, x = 0 (the y-axis) is a vertical asymptote for both the graph of f and the graph of f ¢.

Thus, we have f ¢( x) = -1 on (-¥, -2) and f ¢( x) = 2 on (2, ¥).

Because f is discontinuous at x = -2 and x = 2, we know that f ¢(-2) and f ¢ (2) do not exist, which we indicate with open circles at (-2, -1) and (2, 2) on the graph of f ¢. On the interval (-2, 2) all tangent lines have positive slopes, so the graph of f ¢ will be above the y-axis. Notice that the slope of f (and thus the y-value of f ¢) decreases on (-2, 0) and increases on (0, 2) with a minimum value on this interval of about 1 at x = 0.

22. The graph of f is a step function. (This is the greatest integer function, f ( x ) = [[ x ]].) The graph is made up of an infinite series of horizontal line segments. Thus, the derivative will be 0 everywhere it is defined. However, since f is discontinuous wherever x is an integer, f ¢( x) does not exist at any integer.

222

Chapter 3 THE DERIVATIVE 26. (a) The curve slants downward up to about age 4.25, where it turns and begins to rise. There is a slight decline in steepness between ages 10 and 18. Correspondingly, the graph of the rate of change lies below the horizontal axis to the left of 4.25 years and above the horizontal axis to the right of that point. Rate of change of BMI

23. The slope of f (x) is undefined at x = -2, -1, 0,1, and 2, and the graph approaches vertical (unbounded slope) as x approaches those values. Accordingly, the graph of f ¢( x) has vertical asymptotes at x = -2, -1, 0,1, and 2. f ( x) has turning points (zero slope) at x = -1.5, - 0.5, 0.5, and 1.5, so the graph of f ¢( x) crosses the x-axis at those values. Elsewhere, the graph of f ¢( x) is negative where f (x) is decreasing and positive where f (x) is increasing.

2.0 1.0 6 10

–1.0

–2.0 –3.0 Age (years)

Elsewhere, the graph of f ¢( x) is negative where f (x) is decreasing and positive where f (x) is increasing. Between x = -1 and x = 1, the slope of f (x) is 0.5; therefore, f ¢( x) = 0.5 between -1 and 1.

25. The graph of G decreases steadily with varying degrees of steepness. The steepness increases (that is, the slopes of the tangent lines becomes more negative) between t = 12 and t = 16. Since G is discontinuous at t = 16, G ¢(16) doesn’t exist. The graph continues to decrease after t = 16, but the slopes of the tangent lines become less negative as the curve gets flatter. So, the derivative values are increasing toward 0.

Rate of change of BMI

24. The slope of f (x) is undefined at x = -1 and 1. The graph approaches vertical (unbounded slope) as x approaches -1 from the left and 1 from the right. Accordingly, the graph of f ¢( x) has vertical asymptotes for x = -1 approached from the left and for x = 1 approached from the right. f ( x) has a turning point (zero slope) near x = 1.7, so the graph of f ¢( x) crosses the x-axis near x = 1.7.

(b) The curve slants downward up to about age 5.75, where it turns and begins to rise. There is a slight decline in steepness between ages 13 and 20. Correspondingly, the graph of the rate of change lies below the horizontal axis to the left of 5.75 years and above the horizontal axis to the right of that point.

1.0 6 10

–1.0

Age (years)

27. The growth rate of the function y = f (t ) is given by the derivative of this function y ¢ = f (t ). We use the graph of f to sketch the graph of f ¢. First, notice as x increases, y increases throughout the domain of f , but at a slower and slower rate. The slope of f is positive but always decreasing, and approaches 0 as t gets large. Thus, y ¢ will always be positive and decreasing. It will approach but never reach 0. To plot point on the graph of f ¢, we need to estimate the slope of f at several points. From the graph of f , we obtain the values given in the following table.

y¢

1000

700

250

Use these points to sketch the graph.

Chapter 3 Review

223

First, we observe that the graph of P has one turning point, at V = Vmp » 8. At this value of V, the graph has a horizontal tangent, so the graph of V ¢ has an x-intercept at this value of V. Since the slopes of tangent lines are negative on the interval (0, Vmp ) and positive when V > Vmp , the value of V ¢ is negative on (0, Vmp ) and

32. (a) Rate of change of transformed scores

28. Let P(V) represent the power corresponding to a given value of the tern’s speed, V. The rate of change of power as a function of time is given by the derivative of this function, P¢(V ). We use the graph of P to sketch the graph of P¢.

0.6 0.5 0.4 0.3 0.2 0.1 0

20 40 60 80 100 Student scores

(b) The marginal increase in test scores is the greatest around a student test score of 50.

Chapter 3 Review Exercises

positive when V > Vmp .

1. True

Use the tangents drawn on the graph and additional tangent as needed to estimate the slope at several points on the graph of V to improve the accuracy of the graph of V ¢.

2. True 3. True 2

4. False; for example, if f ( x) = xx +-24 ,

P'(V) 50

lim f ( x) = -4, but the graph of f ( x) = xx +-24

8 12 16 V

x -2

–50

has a hole at the point (-2, -4).

–100

5. True

Rate of change of growth

29. 11

–1

13 14 15 16 17 Skeletal age (years)

–2 –3

6. False; for example, the rational function 5 is discontinuous at x = -1. f ( x) = x + 1 7. False; the derivative gives the instantaneous rate of change of a function.

–4

About 9 cm; about 2.6 cm less per year

8. True 9. True

30.

10. True

kg/yr.

11. False; the slope of the tangent line gives the instantaneous rate of change.

3 2 1 0

31.

8 12 16 Age (years)

12. False; for example, the function f ( x) = | x | is continuous at x = 0, but f ¢(0) does not exist. The graph of f ( x) = | x | has a “corner” at x = 0. 16. The derivative can be used to find the instantaneous rate of change at a point on a function, and the slope of a tangent line at a point on a function. 17. (a) (b)

lim

= 4

lim

= 4

x -3x -3+

224

Chapter 3 THE DERIVATIVE lim = 4 (since parts (a) and (b) have the

(c)

x-3

24. Let f ( x) =

same answer) (d) f (-3) = 4, since (-3, 4) is a point of the graph.

lim g ( x) = -2

18. (a)

x -1-

x -1+

lim g ( x) does not exist since parts (a) and

(c)

x -1

-2.9 -2.99 -2.999

f ( x)

-8

1002

-98

-998

2 x +5

does not exist.

x 2 - 16 ( x - 4)( x + 4) = lim x-4 x 4 x - 4 x 4 = lim ( x + 4)

lim f ( x) = ¥

x 4

= 4+4

lim f ( x) = -¥

x  4+

lim f ( x) does not exist since limits in (a) and

(c)

x4

(b) do not exist.

x 2 + 3x - 10 ( x + 5)( x - 2) = lim x 2 x-2 x 2 x 2 = lim ( x + 5) = 2 + 5 = 7

26. lim

(d) f (4) does not exist since the graph has no point with an x-value of 4. 20. (a) (b) (c)

x 2

lim h( x) = 1

x  2-

27.

lim h( x) = 1

x  2+

2 x 2 + 3x - 20 (2 x - 5)( x + 4) = lim x+4 x+4 x -4 x -4 = lim (2 x - 5) lim

x -4

lim h( x) = 1

= 2(-4) - 5

x 2

(d) h(2) does not exists since the graph has no point with an x-value of 2. lim g ( x) = ¥ since the y-value gets very large

x -¥

as the x-value gets very small. lim f ( x) = -3 since the line y = -3 is a

= -13 3x 2 - 2 x - 21 x-3 x3 (3x + 7)( x - 3) = lim x-3 x 3 = lim (3x + 7) = 9 + 7 = 16

28. lim

x 3

x ¥

horizontal asymptote for the graph.

x -3 x -3 x +3 = lim ⋅ x x 9 9 x9 x 9 x +3 x-9 = lim x  9 ( x - 9)( x + 3) 1 = lim x 9 x + 3 1 = 9 +3 1 = 6

29. lim 23. lim

102

25. lim

x  4-

(b)

22.

x -3 x + 3

(d) g (-1) = -2, since (-1, -2) is a point on the graph.

21.

-3.1 -3.01 -3.001

Therefore, lim

(b) have different answers.

19. (a)

x f ( x)

As x approaches -3 from the left, f (x) gets infinitely larger. As x approaches -3 from the right, f (x) gets infinitely smaller.

lim g ( x) = 2

(b)

2x + 5 . x+3

x6

2x + 7 2(6) + 7 19 = = x+3 6+3 9

Chapter 3 Review

225

x -4 x 16 x - 16

35. As shown on the graph, f ( x) is discontinuous at x2

30.

lim

and x4.

x -4 x +4 ⋅ x 16 x - 16 x +4 x - 16 = lim x 16 ( x - 16)( x + 4) 1 1 = lim = x 16 x + 4 16 + 4 1 1 = = 4+4 8 = lim

31.

lim

2x2 + 5

x ¥ 5 x 2 - 1

= lim

36. As shown on the graph, f ( x) is discontinuous at x1 and x4. 37. f (x) is discontinuous at x = 0 and x = - 13 since

that is where the denominator of f (x) equals 0. f (0)

( )

and f - 13 do not exist. lim f ( x) does not exist since lim f ( x) = -¥,

2 x2 5 + 2 x x2

but lim f ( x) = ¥. lim f ( x) does not exist

1 2 x x2

since lim

x  0-

x ¥ 5 x 2

= lim

x  0+

x0

x - 13

5 x2

38. f ( x) =

x ¥ 5 - 1

2+0 5-0 2 = 5

lim

x2 + 6x + 8

x ¥ x3 + 2 x + 1

x2 6 x 8 + + x3 x 3 x3

x ¥ x3 x3

2x 1 + x3 x3

0+0+0 =0 1+ 0 + 0

æ3 3 6 ö lim çç + - 2 ÷÷÷ x x -¥ çè 8 x ø

34.

lim

æ 9 ö 10 lim çç 4 + 2 - 6 ÷÷÷ ç è ø x -¥ x x =

lim

x -¥ x 4

lim

and

lim

x -3+

f ( x) = ¥.

lim

x -3-

f ( x) = - ¥

f ( x) = ¥.

lim f ( x) does not exist since lim f ( x) = ¥ and x 1-

x 1

lim f ( x) = -¥.

x 1+

f (-3) and f (1) do not exist since there is no point of the graph that has an x-value of -3 or 1.

39. f ( x) is discontinuous at x = -5 since that is where the denominator of f ( x) equals 0. f (-5) does not exist. lim f ( x) does not exist since

3 3 6 + lim - lim 2 x -¥ 8 x -¥ x x -¥ x 3 = +0-0 8 3 = 8 =

7 - 3x (1 - x)(3 + x)

lim f ( x) does not exist since

1 6 8 + + x x 2 x3 = lim x ¥ 1 + 2 + 1 x2 x3

33.

x - 13

x -3

= lim

= -¥, but lim

The function is discontinuous at x = -3 and x = 1 because those values make the denominator of the fraction equal to zero.

32.

x - 13

x -5

but

lim

x -5+

40. f ( x) =

lim

x -5-

f ( x) = ¥,

f ( x) = -¥ .

x2 - 9 x+3

The function is discontinuous at x = -3 since this value makes the denominator of the fraction equal to zero. x2 - 9 ( x + 3)( x - 3) = lim x+3 x -3 x + 3 x -3 = lim ( x - 3) lim

x -¥ x 2

- lim 6 x -¥

x -3

= 0 + 0 - 6 = -6

= -3 - 3 = -6 f (-3) does not exist since there is no point on the graph with an x-value of –3.

226

Chapter 3 THE DERIVATIVE

41. f ( x) = x 2 + 3x - 4 is continuous everywhere since f is a polynomial function. 2

42. f ( x) = 2 x - 5 x - 3 has no points of discontinuity since it is a polynomial function, which is continuous everywhere. 43. (a)

(b) The graph is discontinuous at x = 1. (c)

lim f ( x) = 0; lim f ( x) = 2

x 1-

x 1+

ì2 if ï ï ï 44. f ( x) = ïí -x 2 + x + 2 if ï ï ï 1 if ï ï î

x<0

(b) Graph y =

x 4 + 2 x3 + 2 x 2 - 10 x + 5 x2 - 1

on a graphing calculator. One suitable choice for the viewing window is [-2, 6] by [-10, 10]. Because x 2 - 1 = 0 when x = -1 or x = 1, this function is discontinuous at these two x-values. The graph shows a vertical asymptote at x = -1 but not at x = 1. The graph should have an open circle to show a “hole” in the graph at x = 1. The graphing calculator doesn’t show the hole, but trying to find the value of the function of x = 1 will show that this value is undefined. By viewing the function near x = 1 and using the ZOOM feature, we see that as x gets close to 1 from the left or the right, y gets close to 2, suggesting that

0£ x£2 x>2

lim

x 4 + 2 x3 + 2 x 2 - 10 x + 5 x2 - 1

x 1

(a) 46. f ( x) =

= 2.

x 4 + 3x3 + 7 x 2 + 11x + 2 x3 + 2 x 2 - 3x - 6

(a) Find values of f (x) when x is close to -2.

x -2.01 -2.001 -2.0001 -1.99 -1.999 -1.9999

(b) The graph is discontinuous at x = 2. (c)

lim f ( x) = -4 + 2 + 2 = 0

x  2-

lim f ( x) = 1

x  2+

45. f ( x) =

x 4 + 2 x3 + 2 x 2 - 10 x + 5

(a) Find the values of f (x) when x is close to 1.

1.1 1.01 1.001 1.0001 0.99 0.999 0.9999

y 2.6005 2.06 2.006 2.0006 1.94 1.994 1.9994

It appears that lim f ( x) = 2. x 1

It appears that lim f ( x) = -13. x -2

x2 - 1

f (x) -12.62 -12.96 -13 -13.41 -13.04 -13

(b) Graph y =

x 4 + 3x3 + 7 x 2 + 11x + 2 x3 + 2 x 2 - 3x - 6

on a graphing calculator. One suitable choice for the viewing window is [-5, 5] by [-10, 10]. By viewing the function near x = -2, we see that as x gets close to -2 from the left on the right, y gets close to -13, suggesting that lim

x -2

x 4 + 3x3 + 7 x 2 + 11x + 2 x3 + 2 x 2 - 3x - 6

= -13.

Chapter 3 Review

227

47. y = 6 x3 + 2 = f ( x); from x = 1 to x = 4

50. y =

f (4) = 6(4)3 + 2 = 386

5+4 9 = 5 -1 4 2+4 f (2) = =6 2 -1 f (5) =

f (1) = 6(1)3 + 2 = 8

Average rate of change: =

x+4 = f ( x) x -1

386 - 8 378 = = 126 4 -1 3

Average rate of change:

y ¢ = 18 x

Instantaneous rate of change at x = 1: y¢ =

f ¢(1) = 18(1) = 18 3

48. y = -2 x - 3x + 8 = f ( x) f (6) = -2(6)3 - 3(6) 2 + 8 = -532

9 -6 4

( x - 1) 2 x -1- x - 4

f ¢(2) =

Average rate of change: f (6) - f (-2) 6 - (-2) -532 - 12 -544 = = = -68 6+2 8

51. (a)

-5 (2 - 1)

-6 = f ( x); from x = 4 to x = 9 3x - 5

-5 = -5 1

f (4) - f (2) 4-2

Now use m = 13 and 2, f (2) = (2,9) in the point-slope form.

3 -6 -6 = =3(9) - 5 22 11 6 -6 f (4) = =3(4) - 5 7 f (9) =

y - 9 = 13( x - 2) y - 9 = 13x - 26 y = 13x - 26 + 9 y = 13x - 17

Average rate of change: -21+ 66 77

45 9 = = = 9-4 5 5(77) 77 (3x - 5)(0) - (-6)(3) 18 y¢ = = 2 (3x - 5) (3x - 5)2

(b) f ( x) = 3x 2 - 5x + 7; x = 2 f ( x + h) - f ( x ) h

[3( x + h)2 - 5( x + h) + 7] - [3x 2 - 5 x + 7] h

3x 2 + 6 xh + 3h 2 - 5 x - 5h + 7 - 3x 2 + 5x - 7 h

Instantaneous rate of change at x = 4: 18 (3 ⋅ 4 - 5)

f ¢(-2) = -6(-2) 2 - 6(-2) = -6(4) + 12 = -12

f ¢(4) =

( x - 1)2

[3(4) 2 - 5(4) + 7] - [3(2) 2 - 5(2) + 7] 2 35 - 9 = 2 = 13

Instantaneous rate of change at x = -2:

-5

f ( x) = 3x 2 - 5x + 7; x = 2, x = 4

y ¢ = -6 x - 6 x

-3 æç 6 ö÷ -ç - ÷ 11 çè 7 ÷ø

Slope of secant line

49. y =

( x - 1)

Instantaneous rate of change at x = 2:

f (-2) = -2(-2)3 - 3(-2)2 + 8 = 12

-15 4

5 =5-2 3 4 ( x - 1)(1) - ( x + 4)(1) =

18 7

18 49

6 xh + 3h 2 - 5h h = 6 x + 3h - 5

228

Chapter 3 THE DERIVATIVE f ¢( x) = lim 6 x + 3h - 5

è2

= 6x - 5 ¢ f (2) = 6(2) - 5

æ ö y - 2 = -4 çç x - 1 ÷÷÷ è 2ø y - 2 = -4 x + 2

Now use m = 7 and (2, f (2)) = (2,9) in the pointslope form.

y = -4 x + 2 + 2 y = -4 x + 4

y - 9 = 7( x - 2) y - 9 = 7 x - 14 y = 7 x - 14 + 9

53. (a)

f ( x) =

y = 7x - 5

12 ; x = 3, x = 7 x -1

Slope of secant line = f ( x) = 1 ; x = 1 , x = 3 x 2

Slope of secant line =

the point-slope form.

52. (a)

æ öö è 2 øø

Now use m = -4 and çççç 1 , f ççç 1 ÷÷÷ ÷÷÷÷ = ççèç 1 , 2 ÷÷ø÷ in

h 0

f (3) - f

2 1 -2 = 3 1 = 2 - 12 18 - 3 32

= -10 = - 2 15 3 æ

æ öö è 2 øø

Now use m = - 23 and çççç 1 , f ççç 1 ÷÷÷ ÷÷÷÷ = èççç 1 , 2 ø÷÷÷ in 2

the point-slope form. æ ö y - 2 = - 2 çç x - 1 ÷÷÷ 3è 2ø

Now use m = -1 and (3, f ( x)) = (3, 6) in the point-slope form. y - 6 = -1( x - 3) y - 6 = -x + 3 y = -x + 3 + 6 y = -x + 9 (b) f ( x) =

12 ; x = 3 x -1 12

f ( x + h) - f ( x ) = x + h -1 x -1 h h 12( x - 1) - 12( x + h - 1) = h( x - 1)( x + h - 1) -12h = h( x - 1)( x + h - 1) 12 =( x - 1)( x + h - 1)

y - 2 = -2 x + 1 3 3 2 1 y =- x+ +2 3 3 y = -2 x + 7 3 3

(b) f ( x) = 1 ; x = 1 x 2 1

12 - 12 7 -1 3-1

4 2 6 = 4 = -1

( 12 )

3- 1

è2

f (7) - f (3) 7-3

f ( x + h) - f ( x ) = x+h x h h x - ( x + h) h = = xh( x + h) xh( x + h) 1 =x( x + h) 1 f ¢( x) = lim h  0 x( x + h) = - 12 x æ ö f ¢çç 1 ÷÷ = - 1 2 è2ø æç 1 ö÷

12 ( x - 1)( x + h - 1) 12

f ¢( x) = lim h 0

=f ¢(3) = -

( x - 1)2 12

(3 - 1)2 = -3 Now use m = -3 and (3, f ( x)) = (3, 6) in the point-slope form. y - 6 = -3( x - 3) y - 6 = -3 x + 9

ççè 2 ÷÷ø

= -4

y = -3 x + 9 + 6 y = -3x + 15

Chapter 3 Review 54. (a)

229

f ( x) = 2 x - 1; x = 5, x = 10

f (10) - f (5) 10 - 5 2 10 - 1 - 2 5 - 1 = 5 2(3) - 2(2) 2 = = 5 5

Slope of secant line =

Now use m =

2 5

and (5, f ( x)) = (5, 4) in the point-slope form.

2 ( x - 5) 5 2 y-4= x-2 5 2 y = x-2+4 5 2 y = x+2 5 y-4=

(b)

f ( x ) = 2 x - 1; x = 5 f ( x + h) - f ( x) h = = = =

2 x + h -1 - 2 x -1 h 2( x + h - 1 -

x - 1)( x + h - 1 +

h( x + h - 1 + 2( x + h - 1 - x + 1) h( x + h - 1 +

x - 1)

x - 1) 2h h( x + h - 1 +

x - 1)

2 x + h -1 +

x -1

2 h 0 x + h - 1 + 1 1 f ¢(5) = = 2 5 -1

f ¢( x) = lim

Now use m =

1 2

x -1

2 = 2 x -1

and (5, f ( x)) = (5, 4) in the point-slope form. 1 ( x - 5) 2 1 5 y-4= x2 2 1 5 y = x- +4 2 2 1 3 y = x+ 2 2 y-4=

1 x -1

230

Chapter 3 THE DERIVATIVE

55. y = 4 x 2 + 3x - 2 = f ( x) f ( x + h) - f ( x) h [4( x + h)2 + 3( x + h) - 2] - [4 x 2 + 3x - 2] = lim h h0 2 2 4( x + 2 xh + h ) + 3x + 3h - 2 - 4 x 2 - 3x + 2 = lim h h0 2 2 x xh h x 4 + 8 + 4 + 3 + 3h - 2 - 4 x 2 - 3x + 2 = lim h h0 2 xh h h 8 + 4 + 3 = lim h h0 h(8 x + 4h + 3) = lim h h0 = lim (8x + 4h + 3)

y ¢ = lim

h0

= 8x + 3

56. y = 5 x 2 - 6 x + 7 = f ( x) f ( x + h) - f ( x) y ¢ = lim h h0 [5( x + h)2 - 6( x + h) + 7] - [5x 2 - 6 x + 7] = lim h h0 5( x 2 + 2 xh + h 2 ) - 6 x - 6h + 7 - 5 x 2 + 6 x = lim h h0 2 2 2 = lim 5x + 10 xh + 5h - 6 x - 6h + 7 - 5 x + 6 x + 7 h h0 2 h(10 x + 5h - 6) = lim 10 xh + 5h - 6h = lim h h h0 h0 = lim (10 x + 5h - 6) = 10 x - 6 h0

57. f ( x) = (ln x) x , x0 = 3 (a) h 0.01 f (3 + 0.01) - f (3) 0.01 (ln 3.01)3.01 - (ln 3)3 = 1.3385 0.01 f (3 + 0.001) - f (3) 0.001 =

0.001

(ln 3.001)3.001 - (ln 3)3 = 1.3323 0.001 f (3 + 0.0001) - f (3) 0.0001 =

0.0001

(ln 3.0001)3.0001 - (ln 3)3 = 1.3317 0.0001 f (3 + 0.00001) - f (3) 0.00001 =

0.00001

(ln 3.00001)3.00001 - (ln 3)3 = 1.3317 0.00001 It appears that f ¢(3) » 1.332. =

Chapter 3 Review

231

(b) Using a graphing calculator will confirm this result. 58. f ( x) = xln x , x0 = 2 (a)

h 0.01

f (2 + 0.01) - f (2) 0.01 =

0.001

f (2 + 0.001) - f (2) 0.001 =

0.0001

2.001ln 2.001 - 2ln 2 = 1.1212 0.001

f (2 + 0.0001) - f (2) 0.0001 =

0.00001

2.01ln 2.01 - 2ln 2 = 1.1258 0.01

2.0001ln 2.0001 - 2ln 2 = 1.1207 0.0001

f (2 + 0.00001) - f (2) 0.00001 =

2.00001ln 2.00001 - 2ln 2 = 1.1207 0.00001

It appears that f ¢(2) = 1.121. (b) Graph the function on a graphing calculator and move the cursor to an x-value near x = 2. A good choice for the viewing window is [0, 10] by [0, 10]. Zoom in on the function until the graph looks like a straight line. Use the TRACE feature to select two points on the graph, and use these points to compute the slope. The result will be close to the most accurate value found in part (a), which is 1.121. 59. On the interval (-¥, 0), the graph of f is a straight line, so its slope is constant. To find this slope, use the points (-2, 2) and (0, 0). m=

0-2 -2 = = -1 0 - (-2) 2

Thus, the value of f ¢ will be -1 on this interval. The graph of f has a sharp point at 0, so f ¢(0) does not exist. To show this, we use an open circle on the graph of f ¢ at (0, -1).

negative from there on. As x  ¥, f ( x)  0 and also f ¢( x) = 0. Use this information to complete the graph of f ¢.

60. On the intervals (-¥, 0) and (0, ¥), the slope of any tangent line will be positive, so the derivative will be positive. Thus, the graph of f ¢ will lie above

the y-axis. The slope of f and thus the value of f ¢ approaches 0 when x  -¥ and x  ¥ and approaches some particular but unknown positive value > 1 when x  0- and x  0+. Because f is discontinuous at x = 0, we know that f ¢(0) does not exist, which we indicate with an open circle at x = 0 on the graph of f ¢.

We also observe that the slope of f is positive but decreasing from x = 0 to about x = 1, and then

232 61.

Chapter 3 THE DERIVATIVE lim

x ¥

cf ( x) - dg ( x) f ( x) - g ( x) lim [cf ( x) - dg ( x)]

(f ) C (100) = $1.50 (g) C (125) = $1.50

= x ¥ lim [ f ( x) - g ( x)]

(h) C (140) = $1.35

x ¥

The marginal cost is given by

lim [cf ( x)] - lim [dg ( x)]

ìï1.50 for 0 < x £ 125 C ( x) = ï í ïïî1.35 for x > 125.

x ¥ = x ¥ lim [ f ( x)] - lim [ g ( x)] x ¥

x ¥

c lim [ f ( x)] - d lim [ g ( x)] =

x ¥

(i)

x ¥

lim [ f ( x)] - lim [ g ( x)]

x ¥

C ¢(100) = 1.50; the l01st pound will cost $1.50.

(j) C ¢(140) = 1.35; the 141st pound will cost $1.35.

x ¥

c⋅c-d⋅d (c + d )(c - d ) = c-d c-d =c+d =

The answer is (e).

64. P( x) = 15x + 25x 2 (a) P(6) = 15(6) + 25(6)2 = 90 + 900 = 990 P(7) = 15(7) + 25(7)2 = 105 + 1225 = 1330

62. R( x) = 5000 + 16 x - 3x 2 (a) R¢( x) = 16 - 6 x

Average rate of change:

(b) Since x is in hundreds of dollars, $1000 corresponds to x = 10.

P(7) - P(6) 1330 - 990 = 7-6 1 = 340 cents or $3.40 =

R¢(10) = 16 - 6(10)

= 16 - 60 = -44 An increase of $100 spent on advertising when advertising expenditures are $1000 will result in the revenue decreasing by $44.

(b) P(6) = 990 P(6.5) = 15(6.5) + 25(6.5) 2 = 97.5 + 1056.25 = 1153.75

ì ï1.50 x for 0 < x £ 125 63. C ( x) = ïí ï ï î1.35 x for x > 125 (a) C (100) = 1.50(100) = $150

Average rate of change: P(6.5) - P(6) 6.5 - 6 1153.75 - 990 = 0.5 = 327.5cents or $3.28 =

(b) C (125) = 1.50(125) = $187.50 (c) C (140) = 1.35(140) = $189 (d) (c)

P(6) = 990 P(6.1) = 15(6.1) + 25(6.1)2 = 91.5 + 930.25 = 1021.75

(e) By reading the graph, C(x) is discontinuous at x = $125.

The average cost per pound is given by C ( x) =

C ( x) . x

Average rate of P(6.1) - P(6) change: = 6.1 - 6 1021.75 - 990 = 0.1 = 317.5 cents or $3.18

ìï1.50 for 0 < x £ 125 C ( x) = ï í ïïî1.35 for x > 125

Chapter 3 Review

233 (f ) For 0 £ x £ 29,300,

(d) P¢( x) = 15 + 50 x P¢(6) = 15 + 50(6)

A( x) =

= 15 + 300 = 315 cents or $3.15

For x > 29,300,

(e) P¢(20) = 15 + 50(20) = 1015¢ or $10.15

T ( x) x 4350 + (0.27)( x - 29,300) = x 0.27 x - 3561 = x 3561 . = 0.27 x

A( x) =

(f) P¢(30) = 15 + 50(30) = 1515¢ or $15.15 (g) The domain of x is [0, ¥) since pounds cannot be measured with negative numbers. (h) Since P¢( x) = 15 + 50 x gives the marginal

profit, and x ³ 0, P¢( x) can never be negative. (i)

P ( x) =

P( x) x

(g)

15x + 25x x = 15 + 25 x =

(j)

T ( x) 0.15x = = 0.15. x x

(h)

lim

x  29,300-

lim

x  29,300

A( x) = 0.15 A( x) = 0.27 -

3561 29,300

= 0.14846

P¢( x) = 25

(i)

(k) The marginal average profit cannot change since P¢( x) is constant. The profit per pound never changes, no matter now many pounds are sold.

lim

x  29,300

A( x) does not exist since parts (g)

and (h) have different answers. (j)

lim A( x) = 0.27 - 0 = 0.27

x ¥

(k) 65. (b) The value of x for which the average cost is smallest is x = 7.5. This can be found by drawing a line from the origin to any point of C(x). At x = 7.5, you will get a line with the smallest slope. (c) The marginal cost equals the average cost at the point where the average cost is smallest. 66. (a)

lim

x  29,300-

T ( x) = (29,300)(0.15)

67.

= $4395

(b) lim

x  29,300+

T ( x) = 4350 + (0.27)(29,300 - 29,300) = $4350

(c)

lim

x  29,300

T ( x) does not exist since parts (a) and

(b) have different answers. (d)

The annual unemployment rate in 2012 was about 8.2%. Our sketch of the rate of change of the unemployment rate indicates that the rate of change in 2012 was approximately -0.5% per year. 68. (a) The slope of the tangent line at x = 2000 is about 0.13; the number of people aged 65 and over with Alzheimer’s Disease is going up at a rate of about 0.13 million per year. (b) The slope of the tangent line at x = 2040 is about 0.34; the number of people aged 65 and

(e) The graph is discontinuous at x = 29,300.

234

Chapter 3 THE DERIVATIVE

(c)

over with Alzheimer’s Disease is going up at a rate of about 0.34 million per year.

and (17:40, 400). The rate the whale is descending at 17:37 is about

A(2040) - A(2000) 11.1 - 4.6 = 2040 - 2000 40 » 0.16 The average rate of change in the number of people 65 and over with Alzheimer’s Disease over this interval is about 0.16 million people per year.

400 - 0 400 = 17:40 - 17:35.5 4.5 » 90 meters per minute.

(ii) The tangent line to the curve at 17:39 can be found by using the endpoints at (17:36.2, 0) and (17:40.8, 400). The rate the whale is descending at 17:39 is about 400 - 0 400 = 17:40.8 - 17:36.2 4.6 » 85 meters per minute.

69. V (t ) = -t 2 + 6t - 4 (a)

(b) The whale appears to have 5 distinct rates at which it is descending.

(b) The x-intercepts of the parabola are 0.8 and 5.2, so a reasonable domain would be [0.8, 5.2], which represents the time period from 0.8 to 5.2 weeks. (c) The number of cases reaches a maximum at the vertex; x =

-b -6 = =3 -2 2a

Interval

Rate (meters per minute)

17:35-17:35.3

f (17:35.3) - f (17:35) 17:35.3 - 17:35 =

17:36-17:37

f (17:37.7) - f (17:37.3) 17:37.7 - 17:37.3 150 - 150 = =0 0.4

17:38.3-17:39.5

f (17:39.5) - f (17:38.3) 17:39.5 - 17:38.3 300 - 200 = » 83 1.2

17:40-17:41

(e) The rate of change in the number of cases at the maximum is

f (14:41) - f (14:40) 17:41 - 17:40 =

V ¢(3) = -2(3) + 6 = 0.

(f) The sign of the rate of change up to the maximum is + because the function is increasing. The sign of the rate of change after the maximum is - because the function is decreasing.

130 - 40 » 90 1

17:37.3-17:37.7

(d) The rate of change function is V ¢(t ) = -2t + 6.

f (17:37) - f (17:36) 17:37 - 17:36 =

V (3) = -32 + 6(3) - 4 = 5

The vertex of the parabola is (3, 5). This represents a maximum at 3 weeks of 500 cases.

10 - 0 » 13 0.3

400 - 333 = 67 1

Making smooth transitions between each interval, we get

70. (a) The rate can be estimated by estimating the slope of the tangent line to the curve at the given time. Answers will vary depending on the points used to determine the slope of the tangent line. (i) The tangent line to the curve at 17:37 can be found by using the endpoints at (17:35.5, 0)

Extended Application

235

Extended Application: A Model for Drugs Administered Intravenously

Rate of change of BMI

71. (a) 2

1. (a) Since the half-life of the drug is 9 hours, the exponential decay constant is

1 6 10

–1

ln 2

–2

k = - 9 » -0.077.

Age (years)

Since the initial amount injected is 500 mg, the model for the amount in the bloodstream (in mg) after t hours is

Rate of change of BMI

(b) 2

A(t ) = 500e-0.077t

1 –1

6 10

(b) The average rate of change over the t-interval from 0 to 2 is

–2 Age (years)

( -0.077(2)

-0.077(0)

500 e -e A(2) - A(0) = 2 2 = -35.68 mg/hr

72.

)

For the interval from t = 9 to t = 11, we get

( -0.077(11)

-0.077(9)

500 e -e A(11) - A(9) = 2 2 = -17.84 mg/hr

For a 10-year old girl, the remaining growth is about 14 cm and the rate of change is about -2.75 cm per year.

)

2. (a) Since the half-life of the drug is 3 hours, the exponential decay constant is ln 2

73. (a) The graph is discontinuous nowhere. (b) The graph is not differentiable where the graph makes a sudden change, namely at x = 50, x = 130, x = 230, and x = 770. (c)

k = - 3 » -0.23.

With an infusion rate of 350 mg/hr, the model is A(t ) =

350 (1 - e-0.23t ) » 1522(1 - e-0.23t ) 0.23

(b) The average rate of change over the t-interval from 0 to 3 is A(3) - A(0) 3 é ù 1522 ê 1 - e-0.23(3) - 1 - e-0.23(0) ú ë û = 3

(

) (

(

1522 e-0.23(0) - e-0.23(3)

)

= 252.9 mg/hr

For the interval from t = 3 to t = 6, we get

( -0.23(3)

-0.23(6)

1522 e -e A(6) - A(3) = 3 3 = 126.8 mg/hr

)

236

Chapter 3 THE DERIVATIVE

3. The exponential decay constant for a half-life of 9 hours is as found in Exercise 1: k = -0.077.

Following Example 4, for a steady level of 240 mg, we want the infusion rate to satisfy 250 =

r , -k r

so, r = 250(-k ) = 240(0.077) = 19.25 mg/hr. With this value of r, -k = 250, so the model is A(t ) = 250e-0.077t + 250(1 - e-0.077t )

t A(t) t A(t) t A(t) t A(t) t A(t)

0 500 5 340.225 10 231.507 15 157.529 20 107.191

0.5 481.116 5.5 327.376 10.5 222.763 15.5 151.579 20.5 103.142

1 462.945 6 315.011 11 214.35 16 145.854 21 99.247

t A(t) t A(t) t A(t)

0 0 4 915.454 8 1280.28

0.5 165.341 4.5 981.345 8.5 1306.539

1 312.72 5 1040.079 9 1329.945

1.5 444.089 5.5 1092.432 9.5 1350.809

2 561.186 6 1139.097 10 1369.406

2.5 665.563 6.5 1180.694

3 758.601 7 1217.771

3.5 841.532 7.5 1250.821

t A(t) t A(t) t A(t)

0 250 4 250 8 250

0.5 250 4.5 250 8.5 250

1 250 5 250 9 250

1.5 250 5.5 250 9.5 250

2 250 6 250 10 250

2.5 250 6.5 250

3 250 7 250

3.5 250 7.5 250

1.5 445.46 6.5 303.114 11.5 206.254 16.5 140.346 21.5 95.498

2 428.636 7 291.666 12 198.464 17 135.045 22 91.891

2.5 412.447 7.5 280.65 12.5 190.968 17.5 129.945 22.5 88.421

3 396.87 8 270.05 13 183.756 18 125.037 23 85.081

3.5 381.881 8.5 259.851 13.5 176.816 18.5 120.314 23.5 81.868

4 367.458 9 250.037 14 170.138 19 115.77 24 78.776

4.5 353.579 9.5 240.593 14.5 163.712 19.5 111.398

The results show that the chosen infusion rate has the desired effect of keeping the drug at a contsant level of 250 mg in the bloodstream.

Chapter 4

CALCULATING THE DERIVATIVE 4.1 Techniques for Finding Derivatives Your Turn 1 f (t ) =

Your Turn 5 Find the marginal revenue of the demand function p = 16 - 1.25q when q = 5. R(q) = qp = q(16 - 1.25q) = 16q - 1.25q 2

1 = t-1/2 t

R(q) = 16 - 2.5q R(5) = 16 - 2.5(5) = 3.5

f ¢(t ) = - 12 t-1/2-1 = - 12 t-3/2

The marginal revenue for 5 units is $3.50 per unit.

1 1 = - 3/2 or 2t 2 t3

4.1 Warmup Exercises W1. A line parallel to 4 x + 2 y = 9 will have slope (-4)/2 = -2 and equation

Your Turn 2

y = -2 x + b.

y = 3 x = 3x1/2

If the line passes through (2, 7),

1 dy = 3 ⋅ x-1/2 2 dx 3 -1/2 = x 2 3 = 2 x

7 = -2(2) + b b = 11

and the equation is y = -2 x + 11. W2. A line parallel to 2 x - 5 y = 3 will have slope 2/5 and equation y = (2/5) x + b. If the line passes through (-3, 4),

Your Turn 3 2

h(t ) = -3t + 2 t +

5 t4

4 = (2/5)(-3) + b b = 4 + 6/5 = 26/5

-7

and the equation is y = (2/5) x + 26/5.

= 3t 2 + 2t1/2 + 5t -4 - 7 h¢(t ) = -6t + t -1/2 - 20t -5 = -6t +

4.1 Exercises

1 20 - 5 t t

Your Turn 4

True

False. The derivative of f ( x) = 1 / x 4 = x-4 is f ¢( x) = -4 x-5 = -4 / x-5.

Find the marginal cost of the cost function C ( x) = 5x3 - 10 x 2 + 75 when x = 100.

C ¢( x) = 15 x 2 - 20 x

False. The derivative of f ( x) = 710 is f ¢( x) = 0 since 710 is a constant.

C ¢(100) = 15(100)2 - 20(100) = 150, 000 - 2000 = 148, 000

The marginal cost when x = 100 is $148, 000.

True

y = 12 x3 - 8 x 2 + 7 x + 5 dy = 12(3x3-1) - 8(2 x 2-1) + 7 x1-1 + 0 dx = 36 x 2 - 16 x + 7

237

238 8.

10.

Chapter 4 CALCULATING THE DERIVATIVE y = 8 x3 - 5 x 2 - x 12 dy = 8(3x3-1) - 5(2 x 2-1) - 1 x1-1 12 dx 1 2 = 24 x - 10 x 12

16.

= -25 x-6 + 12 x-3 - 13x-2 -25 12 13 or + 6 3 x x x2

y = 3x 4 - 6 x3 + x + 5 8 dy = 3(4 x 4-1) - 6(3x3-1) + 1 (2 x 2-1) + 0 8 dx 1 3 2 = 12 x - 18 x + x 4

17.

f ¢(t ) = 7(-1t -1-1) - 5(-3t -3-1) = -7t -2 + 15t -4 or

dy = 5(4 x 4-1) + 9(3x3-1) dx 2-1

1-1

) - 7( x

18. )

f ¢(t ) = 14(-1t

y = 6 x3.5 - 10 x 0.5

5 = 21x 2.5 - 5 x-0.5 or 21x 2.5 - 0.5 x

f ( x ) = -2 x

+ 12 x 1.5-1

f ¢( x) = -2(1.5 x

19.

14.

) + 12(0.5 x

)

9 -1/4 4 9 x or 1/2 + 1/4 2 2x x

15.

2 2

) + 12(-4 t -4-1) + 0 -14 t

48 t5

7 3 - 3 + + 5 x x x 4

= -24 x-5 + 21x-4 - 3x-2 or

20.

3 x

21 3 + 4 - 2 x x

dy = 3(-6 x-7 ) + (-5x-6 ) - 7(-2 x-3 ) dx = -18 x-7 - 5 x-6 + 14 x-3 or

21.

y = 10 x + 5 x - 8 x dy = 10(-3x-3-1) + 5(-4 x-4-1) - 8 x1-1 dx = -30 x-4 - 20 x-5 - 8 or -30 - 205 - 8 x4 x

y =

-24

= 3x-6 + x-5 - 7 x-2

22 x-1/3 22 -50 or 1/2 - 1/3 = -50 x-1/2 3 x 3x -4

+ 3(-1x-1-1) + 0

y = -100 x - 11x 2/3 = -100 x1/2 - 11x 2/3 æ1 ö æ2 ö dy = -100 çç x1/2-1 ÷÷ - 11çç x 2/3-1 ÷÷ ÷ ç ç è2 ø è3 ø÷ dx

-3

y =

dy = 6(-4 x-4-1) - 7(-3x-3-1) dx

0.5-1

y = 8 x + 6 x3/4 = 8x1/2 + 6 x3/4 æ1 ö æ3 ö dy = 8çç x1/2-1 ÷÷ + 6 çç x3/4-4 ÷÷ ÷ ç ç è2 ø è4 ø÷ dx = 4 x-1/2 +

= 6 x-4 - 7 x-3 + 3x-1 + 5

0.5

6 = -3x0.5 + 6 x-0.5 or -3x0.5 + 0.5 x

13.

-1-1

-7

= -14t -2 - 48t -5 or

dy = 6(3.5x3.5-1) - 10(0.5x 0.5-1) dx

12.

14 12 + 4 + t t

f (t ) =

= 14t -1 + 12t -4 +

= 20 x3 + 27 x 2 + 24 x - 7

1.5

7 5 - 3 t t

f (t ) =

= 7t -1 - 5t -3

y = 5 x 4 + 9 x3 + 12 x 2 - 7 x

+ 12(2 x

11.

y = 5 x-5 - 6 x-2 + 13x-1 dy = 5(-5x-5-1) - 6(-2 x-2-1) + 13(-1x-1-1) dx

-18 x

5 x

14 x3

p( x) = -10 x-1/2 + 8 x-3/2 æ 1 ö æ 3 ö p¢( x) = -10 çç - x-3/2 ÷÷ + 8 çç - x-5/2 ÷÷ ÷ø ÷ø çè 2 çè 2 = 5 x-3/2 - 12 x-5/2 or

5 x

3/2

12 x5/2

Section 4.1 22.

239

h( x) = x-1/2 - 14 x-3/2 æ 3 ö 1 h¢( x) = - x-3/2 - 14 çç - x-5/2 ÷÷÷ çè 2 ø 2 =

23.

-x-3/2 -1 21 + 21x-5/2 or 3/2 + 5/2 2 2x x 6

y =

= 6 x-1/4

29.

A quadratic function has degree 2. When the derivative is taken, the power will decrease by 1 and the derivative function will be linear, so the correct choice is (b).

30.

A cubic function has degree 3, so its derivative function will be quadratic. The correct choice is (a).

32.

4 x æ 1ö dy = 6 çç - ÷÷÷ x-5/4 ç è 4ø dx

= 4(3x 2 ) - 6(-2 x-3 ) = 12 x 2 + 12 x-3 = 12 x 2 + 123 x 2 3 12 x ( x ) + 12 = x3 5 = 12 x 3+ 12 x

-3 3 = - x-5/4 or 5/4 2 2x

24.

-2 -2 y = 3 = 1/3 = -2 x-1/3 x x æ 1 -4/3 ö÷ dy = -2 çç - x ÷ çè 3 ø÷ dx =

25.

x3 + 5 = x 2 + 5x-1 x

33.

f ( x) = 2 x 2-1 + 5(-1x-1-1) = 2 x - 5 x-2 or 2 x -

26.

é 2 ùú Dx ê 9 x-1/2 + 3/2 êë x úû

5 x

æ ö æ ö = 9 çç - 1 x-3/ 2 ÷÷÷ + 2 çç - 3 x-5/ 2 ÷÷÷ è 2 ø è 2 ø

3 = - 9 x-3/ 2 - 3x-5/ 2 or -3/9 2 - 5/ 2 x 2 2x

34.

é Dx ê 48 ê x ë

3 ù ú x3 úû

= Dx[8 x-1/4 - 3x-3/2 ]

5 3/2 5 2 x - 2 x-1/2 or x3/2 2 2 x

æ ö æ ö = 8çç - 1 x-5/4 ÷÷÷ - 3çç - 3 x-5/2 ÷÷÷ è 4 ø è 2 ø -5/2

g ( x) = (8 x 2 - 4 x)2

= -2 x-5/4 + 9 x

= 64 x 4 - 64 x3 + 16 x 2 g ¢( x) = 64(4 x 4-1) - 64(3x3-1) + 16(2 x 2-1) = 256 x3 - 192 x 2 + 32 x

28.

h( x) = ( x 2 - 1)3 = x 6 - 3x 4 + 3x 2 - 1 h¢( x) = 6 x6-1 - 3(4 x 4-1) + 3(2 x 2-1) - 0 = 6 x5 - 12 x3 + 6 x

choice (b)

= Dx[9 x-1/2 + 2 x-3/2 ]

x - 4x x - 4x = = x5/2 - 4 x1/2 x x1/2 æ1 ö 5 g ¢( x) = x5/2-1 - 4 çç x1/2-1 ÷÷÷ ç è2 ø 2 g ( x) =

27.

choice (c)

Neither choice (a) nor choice (d) equals d (4 x3 - 6 x-2 ). dx

2 x-4/3 2 or 3 3x 4/3

f ( x) =

d (4 x3 - 6 x-2 ) dx

35.

f ( x) = x - 3 x 6 1 = x 4 - 3x 6 1 ¢ f ( x) = (4 x3 ) - 3 6 = 2 x3 - 3 3 f ¢(-2) = 2 (-2)3 - 3 3 = - 16 - 3 3 25 =3

2 + 9 or -5/4 2 x5/2 x

240 36.

37.

Chapter 4 CALCULATING THE DERIVATIVE 3

40.

f ( x) = x - 7 x 2 9 1 = x3 - 7 x 2 9 f ¢( x) = 1 (3x 2 ) - 7(2 x) 9 1 = x 2 - 14 x 3 1 ¢ f (3) = (3)2 - 14(3) 3 = 3 - 42 = -39

y ¢ = -(-3x-4 ) + (-2 x-3 ) = 3x-4 - 2 x-3 2 - 3 x x 3 2 y ¢(2) = (2) 4 (2)3 3 2 = 16 8 1 =16 =

y = x 4 - 5 x3 + 2; x = 2

1 . The slope of the tangent line at x = 2 is - 16

y ¢ = 4 x3 - 15 x 2 y ¢(2) = 4(2)3 - 15(2) 2 = -28 The slope of tangent line at x = 2 is –28.

41.

Use m = -28 and ( x1, y1) = (2, -22) to obtain the equation. y - (-22) = -28( x - 2) y = -28 x + 34 38.

y = -x-3 + x-2

f ( x) = 9 x 2 - 8 x + 4 f ¢( x) = 18x - 8

Let f ¢( x) = 0 to find the point where the slope of the tangent line is zero. 18x - 8 = 0 18x = 8

y = -3 x 5 - 8 x 3 + 4 x 2

x =

y ¢ = -3(5x 4 ) - 8(3x 2 ) + 4(2 x) = -15x 4 - 24 x 2 + 8x 4

Find the y-coordinate.

y ¢(1) = -15(1) - 24(1) + 8(1) = -15 - 24 + 8 = -31

f ( x) = 9 x 2 - 8 x + 4 æ4ö æ 4 ö2 æ4ö f çç ÷÷ = 9 çç ÷÷ - 8çç ÷÷ + 4 çè 9 ÷ø çè 9 ÷ø çè 9 ÷ø

y(1) = -3(1)5 - 8(1)3 + 4(1)2 = -7

æ 16 ö 32 = 9 çç ÷÷÷ +4 çè 81 ø 9

The slope of the tangent line at x = 1 is -31. Use m = -31 and ( x1, y1) = (1, -7) to obtain the equation. y - (-7) = -31( x - 1) y + 7 = -31x + 31 y = -31x + 24 39.

8 4 = 18 9

y = -2 x1/2 + x3/2 æ ö y ¢ = -2 çç 1 x-1/2 ÷÷÷ + 3 x1/2 è2 ø 2 3 -1/2 = -x + x1/2 2 1 + 3x1/2 = - 1/2 2 x 1/2 3(9) y ¢(9) = - 11/2 + 2 (9) 1 9 =- + 3 2 25 = 6 The slope of the tangent line at x = 9 is 25 . 6

16 32 36 20 + = 9 9 9 9 The slope of the tangent line is zero =

at one point,

42.

( 94 , 209 ).

f ( x) = x3 + 9 x 2 + 19 x - 10 f ¢( x) = 3x 2 + 18x + 19

If the slope of the tangent line is -5, then the f ¢( x) = -5. 3x 2 + 18x + 19 = -5 3x 2 + 18x + 24 = 0 3( x 2 + 6 x + 8) = 0 3( x + 2)( x + 4) = 0

Section 4.1

241 x = -2 or x = -4

x = 8  2 37 6 2(4  37) x = 6 x = 4  37 3 Thus, the tangent line is horizontal

f (-2) = (-2)3 + 9(-2)2 + 19(-2) - 10 = -8 + 36 - 38 - 10 = -20 f (-4) = (-4)3 + 9(-4)2 + 19(-4) - 10 = -64 + 144 - 76 - 10 = -6

Thus, the points where the slope of the tangent line is -5 are (-2, -20) and (-4, -6). 43.

at x = 4 3 37 . 46.

f ( x) = x 3 - 5 x 2 + 6 x + 3 f ¢( x) = 3x 2 - 10 x + 6 If the tangent line is horizontal, then its slope is zero and f ¢( x) = 0.

f ( x) = 2 x3 + 9 x 2 - 60 x + 4 f ¢( x) = 6 x 2 + 18x - 60

If the tangent line is horizontal, then its slope is zero and f ¢( x) = 0.

3x 2 - 10 x + 6 = 0 x = 10  100 - 72 6 x = 10  28 6  10 2 7 x = 6  5 7 x = 3

6 x + 18x - 60 = 0 6( x 2 + 3x - 10) = 0 6( x + 5)( x - 2) = 0 x = -5 or x = 2

Thus, the tangent line is horizontal at x = -5 and x = 2. 44.

The tangent line is horizontal at x = 53 7 .

f ( x) = x + 15 x + 63x - 10 f ¢( x) = 3x 2 + 30 x + 63

47.

If the tangent line is horizontal, then its slope is zero and f ¢( x) = 0.

f ( x) = 6 x 2 + 4 x - 9 f ¢( x) = 12 x + 4

If the slope of the tangent line is -2, f ¢( x) = -2. 12 x + 4 = -2

3x 2 + 30 x + 63 = 0

12 x = -6

3( x 2 + 10 x + 21) = 0

x = -1 2 æ 1 ö÷ 19 f çç - ÷÷ = è 2ø 2

3( x + 3)( x + 7) = 0 x = -3 or x = -7

Therefore, the tangent line is horizontal at x = -3 or x = -7. 45.

(

48.

f ( x) = x 3 - 4 x 2 - 7 x + 8

)

The slope of the tangent line is -2 at - 12 , - 19 . 2 f ( x) = 2 x3 - 9 x 2 - 12 x + 5 f ¢( x) = 6 x 2 - 18x - 12

f ¢( x ) = 3 x 2 - 8 x - 7

If the tangent line is horizontal, then its slope is zero and f ¢( x) = 0.

If the slope of the tangent line is 12, f ¢( x) = 12. 6 x 2 - 18 x - 12 = 12

3x 2 - 8 x - 7 = 0

6 x 2 - 18x - 24 = 0

8  64 + 84 6 8  148 x = 6

6( x 2 - 3x - 4) = 0

x =

6( x - 4)( x + 1) = 0 x = 4 or x = -1

f (4) = -59 and f (-1) = 6 The slope of the tangent line is –12 at (4,–59) and (–1, 6).

242 49.

Chapter 4 CALCULATING THE DERIVATIVE 56.

f ( x) = x3 + 6 x 2 + 21x + 2

Graph the numerical derivative of f ( x) = 1.25 x3 + 0.01x 2 - 2.9 x + 1

f ¢( x) = 3x 2 + 12 x + 21

If the slope of the tangent line is 9, f ¢( x) = 9.

for x ranging from -5 to 5.

3x 2 + 12 x + 21 = 9

(a) When x = 4, the derivative equals 57.18.

3x 2 + 12 x + 12 = 0

(b) The derivative crosses the x-axis at approximately -0.88 and 0.88.

3( x 2 + 4 x + 4) = 0 3( x + 2) 2 = 0 x = -2 f (-2) = -24 The slope of the tangent line is 9 at (–2,–24).

50.

57.

(a)

f ( x) = 3g ( x) - 2h( x) + 3 f ¢( x ) = 3 g ¢( x) - 2h ¢( x ) = 3(12) - 2(-3) = 42

46.

(c)

C (101) = -0.04(101)2 + 80(101) + 75 = 7746.96 7746.96 – 7675 = 71.96

The approximate cost of $72 is close to the actual cost of $71.96. 58.

(a) From the graph, f (1) = 2, because the curve goes through (1, 2). f ¢(1) gives the slope of the tangent line to f at 1. The line goes through (-1, 1) and (1, 2). m = 2 -1 = 1, 1 - (-1) 2

R( x) = 48 x - 0.3x 2

(a)

R¢( x) = 48 - 0.6 x

(b)

R¢(20) = 48 - 0.6(20) = 36 After 20 book bags have been sold, the revenue from the sale of one more bag is approximately $36.

(c)

R(20) = 48(20) - 0.3(20) 2 = 840

f ¢(1) = 1 . 2

f ( x) 1 = ⋅ f ( x) k k Use the rule for the derivative of a constant times a function. ù d é f ( x) ù d é1 ê ú = ê ⋅ f ( x) ú ê ú ê úû dx ë k û dx ë k 1 f ¢( x) k f ¢( x) = k =

C (100) = -0.04(100)2 + 80(100) + 75 = 7675

f ( x) = 1 g ( x) + 1 h( x) 2 4 1 f ¢( x ) = g ¢( x ) + 1 h ¢( x) 2 4 1 1 f ¢(2) = g ¢(2) + h¢(2) 2 4 1 1 = (7) + (14) = 7 2 4

(b)

C ¢( x) = -0.08x + 80

(b) C ¢(100) = -0.08(100) + 80 = 72 After 100 devices have been produced, the cost to produce one more device is approximately $72.

f ¢(5) = 3g ¢(5) - 2h¢(5)

51.

C ( x) = -0.04 x 2 + 80 x + 75

R(21) = 48(21) - 0.3(21) 2 = 875.70 875.70 – 840 = 35.70

The approximate revenue of $36 is close to the actual cost of $35.70. 59.

The price is given by p(q) = (a)

1000 q2

+ 1000

æ 1000 ö R(q) = q çç 2 + 1000 ÷÷÷ ÷ø ççè q =

1000 + 1000q = 1000q-1 + 1000q q

R¢(q) = -1000q-2 + 1000 1000 = - 2 + 1000 q

Section 4.1 (b)

(c)

243 R¢(10) = -

1000 2

+ 1000

10 = 990 After 10 systems have been sold, the revenue from the sale of one more system is approximately $990.

(e)

P¢(500) = 70 - 0.08(500) = 30 After 500 units have been sold, the profit increases by $30 per unit.

(f)

Solve P¢(q) = 0 for q. 70 - 0.08q = 0 q = 875 When 875 units have been produced, the profit will be $0 for producing an extra unit.

C (q) = 0.4q 2 + 160q + 50 C ¢(q) = 0.8q + 160 P ¢( q ) = R ¢( q ) - C ¢( q ) ==-

(d)

60.

1000 q2 1000

P¢(10) = -

+ 1000 - (0.8q + 160)

61. - 0.8q + 840

- 0.8(10) + 840 (10)2 = 822 After 10 systems have been sold, the profit from the sale of one more system is approximately $822.

æ q ö÷ R(q) = q ççç 50 ÷ è 100 ø÷

C ¢( x) = 2

(b)

R ¢( x) = 6 -

(c)

P ( x) = R ( x) - C ( x) æ x 2 ö÷÷ ç = çç 6 x ÷ - 2x 1000 ÷ø÷ çè

2x x = 61000 500

x2 1000 2x x P ¢( x) = 4 = 41000 500 = 4x -

(d)

P ¢( x) = 4 -

x = 0 500

x 500 x = 2000

q2 100 q R¢(q) = 50 50 500 (b) R¢(500) = 50 50 = 40 After 500 units have been sold, the revenue increases by $40 per unit. 1000 (c) R¢(1000) = 50 50 = 30 After 1000 units have been sold, the revenue increases by $30 per unit. = 50q -

(d)

x2 1000

(a)

1000

The demand is given by q = 5000 - 100 p Solve for p. q = 5000 - 100 p 100 p = 5000 - q q p = 50 100 (a)

C ( x ) = 2 x ; R ( x) = 6 x -

C (q) = 3000 - 20q + 0.03q 2 C ¢(q) = -20 + 0.06q

P ¢( q ) = R ¢( q ) - C ¢( q ) q = 50 - - (-20 + 0.06q) 50 = 70 - 0.08q

(e) Using part (d), we can see that marginal profit is 0 when x = 2000 units. (2000)2 1000 = 8000 - 4000 = 4000

P(2000) = 4(2000) -

The profit for 2000 units produced is $4000. 62. (a) y = kn p By the constant rule and the power rule, the marginal product of labor is dy kp = k p ⋅ n p -1 = 1- p . dn n

(

)

(b) Since 1 - p > 0, an increase in n increases the denominator of the marginal product of labor, while the numerator remains constant.

244

Chapter 4 CALCULATING THE DERIVATIVE

63. (a) In 1995 when t = 25:

C (25) = -0.00246(25) + 1.02(25) + 7.44 = 31.4025 » 31.4 cents

M ¢(15) = 0.0966(15)2 + 3.42(15) + 18.2 = 91.235 » $91.2 billion per year

In 2015 when t = 45:

(d) For 2019, t = 19:

C (45) = -0.00246(45) 2 + 1.02(45) + 7.44 = 48.3585 » 48.4 cents

(b)

C (t ) = -0.00246 x 2 + 1.02 x + 7.44 C ¢(t ) = -0.0.00492 x + 1.02

M ¢(19) = 0.0966(19)2 + 3.42(19) + 18.2 = 118.0526 » $118.1 billion per year

65.

N (t ) = 0.013984t 2.2

In 1995 when t = 25: C ¢(25) = -0.0.00492(25) + 1.02 = 0.897 cents Postage is increasing by about 0.897¢. In 2015 when t = 45: C ¢(45) = -0.0.00492(45) + 1.02 » 0.799 cents Postage is increasing by about 0.799¢.

N (t ) = 0.00437t 3.2

66.

(a)

N ¢(5) » 0.4824

(b)

N ¢(10) » 2.216

G( x) = -0.2 x 2 + 450

(a)

G(0) = -0.2(0)2 + 450 = 450

C (t ) = 0.000247 x3 - 0.0203x 2 + 1.35x + 6.06

(b) G(25) = -0.2(25) 2 + 450 = -125 + 450 = 325 ¢ G ( x) = -2(0.2) x = -0.4 x

C ¢(t ) = 0.000741x 2 - 0.0406 x + 1.35

(c)

G ¢(10) = -0.4(10) = -4 After 10 units of insulin are injected, the blood sugar level is decreasing at a rate of 4 points per unit of insulin.

In 1995, when t = 25: C ¢(25) = 0.000741(25) 2 - 0.0406(25) + 1.35 » 0.798 cents per year

(d) G ¢ (25) = -0.4(25) = -10

In 1995, when t = 25: C ¢(45) = 0.000741(45) 2 - 0.0406(45)

After 25 units of insulin are injected, the blood sugar level is decreasing at a rate of 10 points per unit of insulin.

+ 1.35 » 1.024 cents per year

67.

V (t ) = -2159 + 1313t - 60.82t 2

(a) V (3) = -2159 + 1313(3) - 60.82(3) 2

58. M (t ) = 0.0322t 3 + 1.71t 2 + 18.2t + 595

= 1232.62 cm3

M ¢(t ) = 0.0966t 2 + 3.42t + 18.2

(a) For 2005, t = 5:

(b)

M ¢(5) = 0.0966(5)2 + 3.42(5) + 18.2 = 37.715 » $37.7 billion per year

(b) For 2010, t = 10:

= 948.08 cm3 /yr

68.

M ¢(10) = 0.0966(10)2 + 3.42(10) + 18.2 = 62.06 » $62.1 billion per year

V ¢(t ) = 1313 - 121.64t V ¢(3) = 1313 - 121.64(3)

c3 1500 100 c (a) When c = 30, w = 220 g or 0.22 kg. w(c) =

(b)

dw 3c 2 1500 = + 2 dc 100 c

Section 4.1

245 When c = 30, dw = 28 23 g/cm. When dc

73.

the circumference of the brain is 30 cm, it is increasing by 28 23 g with every centimeter

703w

BMI =

6¢8¢¢ = 80 in.

(a)

the circumference increases. BMI = 69.

70.

v = 2.69l dv = (1.86)2.69l1.86-1 » 5.00l 0.86 dl

(b)

BMI =

l ( x) = -2.318 + 0.2356 x - 0.002674 x 2

(c)

802

» 27.5

= 24.9 implies

24.9(80)2 » 227. 703

A 250-lb person needs to lose 23 pounds to get down to 227 lbs. (c) If f (h) =

703(125) h2

= 87,875h-2 , then

f ¢(h) = 87,875(-2h-2-1) = -175, 750h-3 = -

175, 750

l ¢(25) = 0.2356 - 0.005348(25) = 0.1019 cm/week

(d)

t = 0.0588s1.125

f ¢ = -

ht/wt 60 65 70 75

= 0.06615s 0.125 dt » 0.118 sec/m. When s = 100, ds

At 100 meters, the fastest possible time increases by 0.118 seconds for each additional meter.

V = C ( R0 - R) R 2 = CR0 R 2 - C R3

Bm =

dV = 2CR0 R - 3CR 2 = 0 dR CR(2 R0 - 3R) = 0

R =

74.

2 R0 3

Discard R = 0, since a closed windpipe produces no airflow. Velocity is maximized when R = 23 R0.

» -0.64

140 27 23 20 17

160 31 27 23 20

180 35 30 26 22

200 39 33 29 25

(f) Substituting the metric equivalents into the formula 703w we have BMI = h2

R =0

653

(e) Sample Chart

dt = 0.0588(1.125s1.125-1) ds

CR = 0 or 2R0 - 3R = 0

175, 750

For a 125-lb female with a height of 65 in. (5¢ ¢¢), the BMI decreases by 0.64 for each additional inch of height.

(a) When s = 1609, t » 238.1 seconds, or 3 minutes, 58.1 seconds. (b)

703w

w =

(b) l ¢( x) = 0.2356 - (2)0.002674 x = 0.2356 - 0.005348x

72.

802

1.86

(a) The problem states that a fetus this formula concerns is at least 18 weeks old. So, the minimum x value should be 18. Considering the gestation time of a cow in general, a meaningful range for this function is 18 £ x £ 44.

71.

703(250)

æ wm ö÷ 703 çç çè 0.4536 ÷÷ø 2

æ hm ö÷ çç çè 0.0254 ÷÷ø

hm 2

s(t ) = 11t 2 + 4t + 2

(a)

v(t ) = s¢(t ) = 22t + 4

(b)

v(0) = 22(0) + 4 = 4 v(5) = 22(5) + 4 = 114 v(10) = 22(10) + 4 = 224

246 75.

Chapter 4 CALCULATING THE DERIVATIVE s(t ) = 18t 2 - 13t + 8

(a)

79.

(a) v(t ) = s ¢(t ) = -16(2t ) + 64 = -32t + 64 v(2) = -32(2) + 64 = -64 + 64 = 0 v(3) = -32(3) + 64 = -96 + 64 = -32

v(t ) = s ¢(t ) = 18(2t ) - 13 + 0 = 36t - 13

(b)

s(t ) = -16t 2 + 64t

v(0) = 36(0) - 13 = -13

The ball’s velocity is 0 ft/sec after 2 seconds and -32 ft/sec after 3 seconds.

v(5) = 36(5) - 13 = 167 v(10) = 36(10) - 13 = 347

76.

(b) As the ball travels upward, its speed decreases because of the force of gravity until, at maximum height, its speed is 0 ft/sec.

s(t ) = 4t + 8t + t

(a)

v(t ) = s¢(t ) = 4(3t 2 ) + 8(2t ) + 1

In part (a), we found that v(2) = 0.

= 12t + 16t + 1 (b)

It takes 2 seconds for the ball to reach its maximum height.

v(0) = 12(0)2 + 16(0) + 1 = 1

(c)

v(5) = 12(5) + 16(5) + 1

s(2) = -16(2)2 + 64(2) = -16(4) + 128

= 300 + 80 + 1 = 381

= -64 + 128

v(10) = 12(10) 2 + 16(10) + 1

= 64

= 1200 + 160 + 1 = 1361

77.

s(t ) = -3t 3 + 4t 2 - 10t + 5

(a)

It will go 64 ft high. 80.

(a)

d ( x ) = 1.66 - 0.90 x + 0.47 x 2 d (0.5) = 1.66 - 0.90(0.5) + 0.47(0.5)2

v(t ) = s¢(t ) = -3(3t 2 ) + 4(2t ) - 10 + 0

= 1.3275 g/cm3

= -9t 2 + 8t - 10

(b)

v(0) = -9(0)2 + 8(0) - 10 = -10

= -0.90 + 0.94 x ¢ d (0.5) = -0.90 + 0.94(0.5)

v(5) = -9(5) 2 + 8(5) - 10 = -225 + 40 - 10 = -195

= -0.43 g/cm3

v(10) = -9(10) 2 + 8(10) - 10

When the level of the Dead Sea decreases to 50% of the current level, the density of the brine is decreasing at the rate of 0.43 gram per cm3.

= -900 + 80 - 10 = -830

78.

s(t ) = -16t 2 + 144

velocity = s¢(t ) = -32t (a)

81.

s¢(1) = -32 ⋅ 1 = -32 s¢(2) = -32 ⋅ 2 = -64

(a) When the tread length is 0.5 m, m 340 c sec = 340 1 or 340 f = = 2T (2)(0.5 m) sec cycles per second.

The rock’s velocity is -32 ft/sec after 1 second and -64 ft/sec after 2 seconds. (b) The rock will hit the ground when s(t ) = 0. -16t 2 + 144 = 0 t =

d ¢( x ) = -0.90 + 0.47(2 x)

(b)

df c =- 2 dT 2T

At T  0.5 m, 144 =3 16

The rock will hit the ground after 3 seconds. (c) The velocity at impact is the velocity at 3 seconds.

m 340 df c sec = -680 1 ⋅ 1 =- 2 = dT m sec 2T (2)(0.5 m) 2

or -680 cycles/sec/m.

s ¢(3) = -32 ⋅ 3 = -96

Section 4.2

247

82. P( x) = -0.00209 x3 + 0.3387 x 2 - 15.15 x + 208.6

4.2 Derivatives of Products and Quotients Your Turn 1

P¢( x) = 0.00627 x + 0.6774 x - 15.15

y = ( x3 + 7)(4 - x 2 )

(a) (i)

dy = ( x3 + 7)(-2 x) + (4 - x 2 )(3x 2 ) dx

P(45) = -0.00209(45)3 + 0.3387(45)2 - 15.15(45) + 208.6 = 22.266 » 22.3

= -2 x 4 - 14 x + 12 x 2 - 3x 4 = -5 x 4 + 12 x 2 - 14 x

P¢(45) = 0.00627(45)2 + 0.6774(45) - 15.15 = 2.636 » 2.64

Your Turn 2 f ( x) =

(ii) P(75) = -0.00209(75)3 + 0.3387(75) 2 - 15.15(75) + 208.6

f  ( x) =

= 95.8196 » 95.8 =

P¢(75) = 0.00627(75) 2 + 0.6774(75) - 15.15 = 0.3863 » 0.386

83.

3x + 2 5 - 2x (5 - 2 x)(3) - (3x + 2)(-2) (5 - 2 x)2 19 (5 - 2 x)2

y1 = 4.13x + 14.63

Your Turn 3

y2 = -0.033x 2 + 4.647 x + 13.347

é (5x - 3)(2 x + 7) ù ú Dx ê êë úû 3x + 7 é (3x + 7) Dx [(5x - 3)(2 x + 7)] ù ê ú ê - [(5x - 3)(2 x + 7)]D (3x + 7) ú x ë û = (3x + 7)2 é (3x + 7)[(5x - 3)(2) + (2 x + 7)(5)] ù ê ú ê ú 2 êë - (10 x + 29 x - 21)(3) úû = 2 (3x + 7) é (3x + 7)(10 x - 6 + 10 x + 35) ù ê ú ê ú 2 x x (30 87 63) + ê ûú = ë (3x + 7)2

(a) When x = 5, y1 » 35 and y2 » 36. (b)

dy1 = 4.13 dx dy2 = 0.033(2 x) + 4.647 dx = -0.066 x + 4.647 dy

When x = 5, dx1 = 4.13 and dx2 » 4.32. These values are fairly close and represent the rate of change of four years for a dog for one year of a human, for a dog that is actually 5 years old. (c) With the first two points eliminated, the dog age increases in 2-year steps and the human age increases in 8-year steps, for a slope of 4. The equation has the form y = 4 x + b. A value of 16 for b makes the numbers come out right. y = 4 x + b. For a dog of age x = 5 years or more, the equivaltent human age is given by y = 4 x + 16.

= =

60 x 2 + 227 x + 203 - 30 x 2 - 87 x + 63 (3x + 7)2 30 x 2 + 140 x + 266 (3x + 7)2

Your Turn 4 C ( x) =

4 x + 50 x+2

The marginal average cost is d é C ( x) ù d çæ 4 x + 50 ÷ö ÷ ê ú = ç dx êë x úû dx çè x 2 + 2 x ÷÷ø =

( x 2 + 2 x )(4) - (4 x + 50)(2 x + 2)

( x 2 + 2 x)2

248

Chapter 4 CALCULATING THE DERIVATIVE = = =

4 x 2 + 8 x - (8 x 2 + 108 x + 100) 2

( x + 2 x) 2

4 x + 8 x - 8 x 2 - 108 x - 100

= 60 x 2 + 30 x - 4

-4 x 2 - 100 x - 100 ( x 2 + 2 x)2

y = (2 x - 5) 2 = (2 x - 5)(2 x - 5) dy = (2 x - 5)(2) + (2 x - 5)(2) dx = 4 x - 10 + 4 x - 10 = 8x - 20

y = (7 x - 6) 2 = (7 x - 6)(7 x - 6) dy = (7 x - 6)(7) + (7)(7 x - 6) dx = 49 x - 42 + 49 x - 42 = 98x - 84

k (t ) = (t 2 - 1)2 = (t 2 - 1)(t 2 - 1)

f ( x ) = 3x 4 + 4 x 3 - 5 f ¢( x ) = (3)(4) x 4-1 + 4(3) x 3-1 - 0 = 12 x 3 + 12 x 2

W2.

f ( x) =

2 x

+ 6 x = 2 x-3 + 6 x1/ 2 3

f ¢( x ) = (2)(-3) x-3-1 + 6(1/2) x1/ 2-1 = -6 x-4 + 3x-1/ 2 6 3 =- 4 + x x

W3.

y = (5x 2 - 1)(4 x + 3) dy = (5x 2 - 1)(4) + (10 x)(4 x + 3) dx = 20 x 2 - 4 + 40 x 2 + 30 x

( x 2 + 2 x)2

4.2 Warmup Exercises W1.

4 f ( x) = 9 x 2/3 + = 9 x 2/3 + 4 x-1/2 x

k ¢(t ) = (t 2 - 1)(2t ) + (t 2 - 1)(2t ) = 2t 3 - 2t + 2t 3 - 2t

f ¢( x) = (9)(2/3) x(2/3) - 1 + (4)(-1/2) x(-1/2) -1

= 4t 3 - 4t

= 6 x-1/3 - 2 x-3/2

10.

4.2 Exercises

g (t ) = (3t 2 + 2) 2 = (3t 2 + 2)(3t 2 + 2)

True

False. Use the product rule.

g ¢(t ) = (3t 2 + 2)(6t ) + (6t )(3t 2 + 2) = 18t 3 + 12t + 18t 3 + 12t

The derivative of f ( x) = x 2 (3x - 7) is

= 36t 3 + 24t

f ¢( x) = x 2 (3) + (3x - 7)(2 x) = 3x 2 + 6 x 2 - 14 x

11.

= 9 x - 14 x.

False. The derivative of a quotient is the denominator times the derivatie of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

True

y = (3x 2 + 2)(2 x - 1) dy = (3x 2 + 2)(2) + (2 x - 1)(6 x) dx

y = ( x + 1)( x + 2) = ( x + 1)( x1/2 + 2) æ1 ö dy = ( x + 1) çç x-1/2 ÷÷÷ + ( x1/2 + 2)(1) çè 2 ø dx 1 1/2 1 x + x-1/2 + x1/2 + 2 2 2 3 1/2 1 -1/2 = x + x +2 2 2 =

= 6 x 2 + 4 + 12 x 2 - 6 x = 18x 2 - 6 x + 4

3x1/2 1 + 1/2 + 2 2 2x

Section 4.2 12.

249

y = (2 x - 3)( x - 1)

17.

= (2 x - 3)( x1/2 - 1) æ1 ö dy = (2 x - 3) çç x-1/2 ÷÷÷ + 2( x1/2 - 1) çè 2 ø dx = x1/2 -

dy = dx =

3 -1/2 x + 2 x1/2 - 2 2

3x-1/2 -2 2 3 or 3x1/2 - 1/2 - 2 2x

= 3x1/2 -

13.

18.

p¢( y ) = ( y-1 + y-2 )(-6 y-4 + 20 y-5 )

+ (- y 2 - 2 y-3)(2 y-3 - 5 y-4 ) = - 6y

+ 20 y

-6

- 6y

-6

+ 20 y

y = dy = dx

p( y ) = ( y-1 + y-2 )(2 y-3 - 5 y-4 )

-5

y =

-7

- 2 y-5 + 5 y-6 - 4 y-6 + 10 y-7 = -8 y-5 + 15 y-6 + 30 y-7

19. 14.

q( x) = ( x-2 - x-3 )(3x-1 + 4 x-4 ) -2

q¢( x) = ( x

-3

-x

)(-3x

-2

-5

- 16 x

)

q¢( x) = -3x-4 - 16 x-7 + 3x-5 + 16 x-8 - 6 x-4 - 8 x-7 + 9 x-5 + 12 x-8

q¢( x) = -9 x-4 + 12 x-5 - 24 x-7 + 28 x-8

f ( x) = f  ( x) = = =

16.

f ( x) = f ¢( x ) = = =

6x + 1 3x + 10 (3x + 10)(6) - (6 x + 1)(3)

20.

(4 + t )2 -17 (4 + t )2 9 - 7t 1- t (1 - t )(-7) - (9 - 7t )(-1) (1 - t ) 2 -7 + 7t + 9 - 7t (1 - t )2 2 (1 - t ) 2 x2 + x x -1

y =

2x2 + x - 2x - 1 - x2 - x ( x - 1) 2 x2 - 2x - 1 ( x - 1) 2 x2 - 4x x+3

dy ( x + 3)(2 x - 4) - ( x 2 - 4 x)(1) = dx ( x + 3) 2

(3x + 10)2 18x + 60 - 18 x - 3 (3x + 10)2 57

(3x + 10)2

8 x - 11 7x + 3 (7 x + 3)(8) - (8x - 11)(7)

(4 + t )2 -12 - 3t - 5 + 3t

dy ( x - 1)(2 x + 1) - ( x 2 + x)(1) = dx ( x - 1) 2

+ (-2 x-3 + 3x-4 )(3x-1 + 4 x-4 )

15.

y =

5 - 3t 4+t (4 + t )(-3) - (5 - 3t )(1)

21.

2 x 2 + 6 x - 4 x - 12 - x 2 + 4 x ( x + 3) 2 x 2 + 6 x - 12

f (t ) =

(7 x + 3) 56 x + 24 - 56 x + 77 (7 x + 3) 101

f ¢(t ) =

( x + 3)2 4t 2 + 11 t2 + 3 (t 2 + 3)(8t ) - (4t 2 + 11)(2t ) (t 2 + 3)2

8t 3 + 24t - 8t 3 - 22t (t 2 + 3) 2

(7 x + 3) 2 =

2t (t 2 + 3) 2

250 22.

Chapter 4 CALCULATING THE DERIVATIVE y=

-x 2 + 8 x

26.

4x - 5

dy (4 x 2 - 5)(-2 x + 8) - (-x 2 + 8x)(8x) = dx (4 x 2 - 5) 2 = =

23.

(4 x - 5)

-32 x + 10 x - 40

g ( x) =

= =

k ( x) = k ( x) = = =

25.

r ¢(t ) =

-8 x3 + 32 x 2 + 10 x - 40 + 8x3 - 64 x 2

g( x) =

24.

r (t ) =

(4 x - 5)

x2 - 4x + 2

x2 + 3 ( x 2 + 3)(2 x - 4) - ( x 2 - 4 x + 2)(2 x) ( x 2 + 3) 2

p¢(t ) =

(

( x 2 + 3) 2

27.

4 x 2 + 2 x - 12 ( x 2 + 3) 2 x2 + 7 x - 2

28.

x2 - 2 ( x 2 - 2)(2 x + 7) - ( x 2 + 7 x - 2)(2 x) ( x 2 - 2)2

(2t + 3) 2 t1/2 + 32 t -1/2 - 2t1/2 (2t + 3)2

-t1/2 + 31/2 2t

(2t + 3)2 - t + 3

2 t 2

(2t + 3) -2t + 3 or 2 t (2t + 3) 2

5x + 6 5x + 6 = = 5x1/ 2 + 6 x-1/ 2 1/ 2 x x dy 5 -1/ 2 5x - 6 = x - 3x-3/ 2 or dx 2 2x x y =

4x - 3 4x - 3 = = 4 x1/ 2 - 3x-1/ 2 1/2 x x dy 3 -3/ 2 4x + 3 -1/ 2 = 2x + x or dx 2 2x x y =

2 x3 + 7 x 2 - 4 x - 14 - 2 x3 - 14 x 2 + 4 x ( x 2 - 2)2

29.

h( z ) =

-7 x 2 - 14 ( x 2 - 2) 2

t1/2 t -1

h ¢( z ) = = =

(

z 2.2 z

( z 3.2 + 5)2 -z 4.4 + 11z1.2

1 t1/2 - 1 t -1/2 - t1/2 2 2

(t - 1)

- 12 t1/2 - 12 t -1/2 = 2 (t - 1)

(t - 1)

+ 5)(2.2 z1.2 ) - z 2.2 (3.2 z 2.2 )

2.2 z 4.4 + 11z1.2 - 3.2 z 4.4

)

1 2 t 2

3.2

( z 3.2 + 5)2

(t - 1)2

- 2t -

3.2

(t - 1) 12 t -1/2 - t1/2 (1)

= 2

)

(2t + 3) 12 t -1/2 - (t1/2 )(2)

2 x3 - 4 x 2 + 6 x - 12 - 2 x3 + 8x 2 - 4 x

t p(t ) = t -1 =

t t1/2 = 2t + 3 2t + 3

-t - 1 2 t (t - 1) 2

( z 3.2 + 5) 2

Section 4.2

30.

251

g ( y) = g ¢( y ) = =

31.

f ( x) = f ¢( x ) = = = =

32.

g ( x) = g ¢( x ) = = = =

33.

y1.4 + 1 y 2.5 + 2 ( y 2.5 + 2)(1.4 y 0.4 ) - ( y1.4 + 1)(2.5 y1.5 ) ( y 2.5 + 2)2

1.4 y 2.9 + 2.8 y 0.4 - 2.5 y 2.9 - 2.5 y1.5 ( y 2.5 + 2)2

-1.1y 2.9 - 2.5 y1.5 + 2.8 y 0.4 ( y 2.5 + 2) 2 (3x 2 + 1)(2 x - 1) 5x + 4 (5x + 4)[(3x 2 + 1)(2) + (6 x)(2 x - 1)] - (3x 2 + 1)(2 x - 1)(5) (5 x + 4)2 (5x + 4)(18 x 2 - 6 x + 2) - (3x 2 + 1)(10 x - 5) (5 x + 4)2 90 x3 - 30 x 2 + 10 x + 72 x 2 - 24 x + 8 - 30 x3 + 15 x 2 - 10 x + 5 (5x + 4) 2 60 x3 + 57 x 2 - 24 x + 13 (5 x + 4)2 (2 x 2 + 3)(5 x + 2) 6x - 7 (6 x - 7)[(2 x 2 + 3)(5) + (4 x)(5 x + 2)] - (2 x 2 + 3)(5x + 2)(6) (6 x - 7) 2 (6 x - 7)(30 x 2 + 8x + 15) - (2 x 2 + 3)(30 x + 12) (6 x - 7)2 180 x3 + 48 x 2 + 90 x - 210 x 2 - 56 x - 105 - 60 x3 - 24 x 2 - 90 x - 36 (6 x - 7) 2 120 x3 - 186 x 2 - 56 x - 141 (6 x - 7)2

h( x) = f ( x) g ( x) h ¢( x ) = f ( x) g ¢( x ) + g ( x) f ¢( x) h¢(3) = f (3) g ¢(3) + g (3) f ¢(3) = 9(5) + 4(8) = 77

34.

h( x ) = h¢( x) = h¢(3) =

35.

f ( x) g ( x) g ( x ) f ¢( x) - f ( x) g ¢( x) [ g ( x)]2 g (3) f ¢(3) - f (3) g ¢(3) 2

[ g (3)]

4(8) - 9(5) 4

13 16

In the first step, the two terms in the numerator are reversed. The correct work follows. æ 2 x + 5 ö÷ ( x 2 - 1)(2) - (2 x + 5)(2 x) 2 x 2 - 2 - 4 x 2 - 10 x -2 x 2 - 10 x - 2 Dx çç 2 = = ÷= çè x - 1 ø÷ ( x 2 - 1)2 ( x 2 - 1) 2 ( x 2 - 1)2

252 36.

37.

Chapter 4 CALCULATING THE DERIVATIVE

In the first step, the denominator, ( x3 )2 = x 6 , was omitted. The correct work follows. 3 2 2 4 4 2 4 2 2 2 2 æ x 2 - 4 ö÷ ç ÷ = x (2 x) - ( x - 4)(3x ) = 2 x - 3x + 12 x = -x + 12 x = x (-x + 12) = -x + 12 Dx çç ÷ çè x3 ø÷÷ ( x3 ) 2 x6 x6 x2 ( x4 ) x4 f ( x) = (2 x 2 - 4)( x + 2) at (2, 16) m = f ¢( x) = (2 x 2 - 4)(1) + ( x + 2)(4 x) = 6 x 2 + 8 x - 4

At (2, 16), m = 6(2)2 + 8(2) - 4 = 36 Use the point-slope form, y - 16 = 36( x - 2) y = 36 x - 56

38.

f ( x) = ( x + 3)(2 x 2 - 6) at (1, -16) æ1 ö m = f ¢( x) = ( x + 3)(4 x) + (2 x 2 - 6) ççç x-1/ 2 ÷÷÷ è2 ø

At (1,–16), æ1 ö m = ( 1 + 3)(4(1)) + (2(1) 2 - 6) ççç (1)-1/ 2 ÷÷÷ = 14 è2 ø Use the point-slope form, y + 16 = 14( x - 1) y = 14 x - 30

39.

f ( x) =

x at (1, 0.5) 4x - 2 m = f ¢( x) =

æ1 ö (4 x - 2) ççç x-1/ 2 ÷÷÷ è2 ø

x (4)

(4 x - 2) 2

2x - 1 -4 x x = (4 x - 2)2

At (1, 0.5), 2 -1 -4 1 3 1 m= = - = -0.75 2 4 (4(1) - 2) Use the point-slope form, y - 0.5 = -0.75( x - 1) y = -0.75x + 125

40.

f ( x) =

x2 - 3 at (-1, 2) 2x + 1 m = f ¢( x) =

(2 x + 1)(2 x) - ( x 2 - 3)(2) (2 x + 1)

2x2 + 2x + 6 (2 x + 1)2

At (–1, 2), m=

2(-1) 2 + 2(-1) + 6

(2(-1) + 1) 2 Use the point-slope form, y - 2 = 6( x + 1)

y = 6x + 8

Section 4.2 41.

(a)

253 f ( x) =

f ¢( x ) = (b)

3x3 + 6 x2 / 3

( x 2/3 )(9 x 2 ) - (3x3 + 6)( 23 x-1/3 ) ( x 2/3 ) 2

9 x8/3 - 2 x8/3 - 4 x-1/3 x 4/3

7 x8/3 x 4/3

4 x1/3

7 x3 - 4 x5/3

f ( x) = 3x7/3 + 6 x-2/3 æ7 ö æ 2 ö f ¢( x) = 3çç x 4/3 ÷÷ + 6 çç - x-5/3 ÷÷ = 7 x 4/3 - 4 x-5/3 ÷ø ÷ø çè 3 çè 3

f ¢( x) = kg ¢( x) + g ( x) ⋅ 0 = kg ¢( x)

The result is the same as applying the rule for differentiating a constant times a function 43.

f ( x) =

u ( x) v ( x) u ( x + h)

u ( x)

- v ( x) f ( x + h) - f ( x) u ( x + h)v( x) - u( x)v( x + h) v( x + h) = lim = lim f ¢( x) = lim h h hv( x + h)v( x) h0 h 0 h 0 u( x + h)v( x) - u ( x)v( x) + u ( x)v( x) - u( x)v( x + h) = lim hv( x + h)v( x) h0 v( x)[u ( x + h) - u( x)] - u( x)[v( x + h) - v( x)] = lim hv( x + h)v( x) h0 = lim

v( x)

v( x + h)v( x)

h0

44.

u ( x + h ) -u ( x ) v( x + h)- v( x ) - u ( x) h h

v( x) ⋅ u¢( x) - u( x)v¢( x) [v( x)]2

u ( x)

Given f ( x) = v( x) , multiply both sides by v( x) to obtain f ( x)v( x) = u ( x).

Using the product rule, we have f ( x) ⋅ v¢( x) + v( x)  f ¢( x) = u¢( x). u ( x) é u ( x) ù ¢ Substitute v( x) for f ( x) and solve for ê v( x) ú . ë û

u ( x) u ( x) ¢ ⋅ v¢( x) + vx éê v( x) ùú = u ¢( x) ë û v( x) u ( x) ¢ u( x) ⋅ v¢( x) + [v( x)]2 éê v( x) ùú = v( x) ⋅ u ¢( x) ë û u ( x) ¢ [v( x)]2 éê v( x) ùú = v( x) ⋅ u ¢( x) - u ( x) ⋅ v¢( x) ë û é u ( x ) ù ¢ = v ( x ) ⋅ u ¢( x) - u ( x ) ⋅ v ¢( x) ê v( x) ú ë û [v( x)]2

45.

Graph the numerical derivative of f ( x) = ( x 2 - 2)( x 2 - 2) for x ranging from -2 to 2. The derivative crosses the x-axis at 0 and at approximately -1.307 and 1.307.

254 46.

Chapter 4 CALCULATING THE DERIVATIVE Graph the numerical derivative of f ( x) =

x - 2 for x ranging from -5 to 10. The derivative crosses the x-axis at x2 + 4

approximately -0.828 and 4.828. 47. (a) f ( x ) = x 2 + bx and g ( x ) = cx + d f ¢( x ) g ¢( x ) = (2 x + b)(c) = 2cx + bc (b) [ f ( x ) g ( x ) ]¢ = f ( x ) g ¢( x ) + f ( x ) g ¢( x ) = ( x 2 + bx )(c) + (2 x + b)(cx + d ) = 3cx 2 + 2 x(d + bc) + bd

48.

a and g ( x ) = bx 1- x d æ abx ö÷ ÷ [ f ( x) g ( x) ]¢ = ççç dx è 1 - x ø÷ f ( x) =

= =

(1 - x )(ab) - (abx )(-1) (1 - x )2 ab (1 - x )2

æ (1 - x )(0) - (a )(-1) ö÷ ÷÷( b ) f ¢( x ) g ¢( x ) = ççç ÷÷ø çè (1 - x )2 =

49.

ab (1 - x )2

C ( x) = 0.02 x 2 + 300 C ( x) =

C ( x) 0.02 x 2 + 300 = x x

(a)

C (50) = 0.02(50)2 + 300 = 350 The total cost to produce 50 T-shirts is $350.

(b)

C ¢( x) = 2(0.02 x) = 0.04 x C ¢(50) = 0.04(50) = 2 After 50 T-shirts have been produced, the total cost is increasing by $2 per shirt.

(c)

(d)

0.02(50)2 + 300 =7 50 When 50 T-shirts are produced, the average cost is $7 per shirt. C (50) =

C ¢( x) = C ¢(50) =

x(0.04 x) - (0.02 x 2 + 300)(1) x2

0.02 x 2 - 300 x2

0.02(50)2 - 300

= -0.1 (50)2 After 50 T-shirts have been produced, the average cost is decreasing by $0.10 per shirt.

50.

P( x) = 0.52 x 2 - 0.0002 x3 P( x) =

P ( x) 0.52 x 2 - 0.0002 x3 = = 0.52 x - 0.0002 x 2 x x

Section 4.2 (a) (b)

255 P(100) = 0.52(100) 2 - 0.0002(100)3 = 5000 The total profit for 100 watches is $5000. P¢( x) = 1.04 x - 0.0006 x 2 P¢(100) = 1.04(100) - 0.0006(100) 2 = 98 After 100 watches have been sold, the total profit is increasing by $98 per week.

(c)

(d)

P(100) = 0.52(100) - 0.0002(100) 2 = 50 When 100 watches have been sold, the average profit is $50 per week. P ¢ ( x) = 0.52 - 0.0004 x P ¢ (100) = 0.52 - 0.0004(100) = 0.48 After 100 watches have been sold, the average profit is increasing by $0.48 per watch.

51.

3x + 2 x+4 C ( x) 3x + 2 3x + 2 C ( x) = = = 2 x x( x + 4) x + 4x

C ( x) =

(a)

( x 2 + 4 x)(3) - (3x + 2)(2 x + 4)

(b) C ¢( x) = (c)

( x 2 + 4 x) 2

3x 2 + 12 x - (6 x 2 + 16 x + 8) ( x 2 + 4 x) 2

-3 x 2 - 4 x - 8 ( x 2 + 4 x) 2

3(10) + 2

C (10) =

» 0.2286 hundreds (10)2 + 4(10) At 10 units, the average cost is $22.86 per unit. -3(10)2 - 4(10) - 8

C ¢(10) =

» -0.0178 hundreds ((10)2 + 4(10)) 2 At 10 units, the average cost is decreasing by $1.78 per unit. 3(20) + 2

(d) C (20) =

» 0.1292 hundreds (20) 2 + 4(20) At 20 units, the average cost is $12.92 per unit.

C ¢(20) =

-3(20)2 - 4(20) - 8

» -0.0056 hundreds ((20) 2 + 4(20)) 2 At 20 units, the average cost is decreasing by $0.56 per unit.

52.

5x - 6 2x + 3 P( x) 5x - 6 = P ( x) = x 2 x 2 + 3x

P( x) =

(a)

(b)

P ¢( x ) = =

(c)

P (8) =

(2 x 2 + 3x)(5) - (5 x - 6)(4 x + 3) 2

(2 x + 3x)

10 x 2 + 15x - 20 x 2 - 15 x + 24 x + 18 (2 x 2 + 3x)2

-10 x 2 + 24 x + 18 (2 x 2 + 3x) 2 5(8) - 6 2

34 » 0.224 tens 152

2(8) + 3(8) At 8 books, the average profit is $2.24 per book.

256

Chapter 4 CALCULATING THE DERIVATIVE -10(8) 2 + 24(8) + 18

P¢(8) =

» -0.019 tens (2(8) 2 + 3(8))2 At 8 books, the average profit is decreasing by $0.19 per unit.

(d)

5(15) - 6 69 = » 0.139 tens 495 2(15) 2 + 3(15) At 8 books, the average profit is $1.39 per book.

P (15) =

P¢(15) =

-10(15)2 + 24(15) + 18

» -0.08 tens (2(15)2 + 3(15))2 At 15 books, the average profit is decreasing by $0.19 per unit.

53.

M (d ) =

100d 2 3d 2 + 10

(a)

M ¢( d ) =

(b)

M ¢(2) =

(3d 2 + 10)(200d ) - (100d 2 )(6d ) 2

4000 » 8.3 484

(3d + 10) 2000(2) 2

[3(2) + 10]

600d 3 + 2000d - 600d 3 2

(3d + 10)

2000d (3d 2 + 10)2

This means the new employee can assemble about 8.3 additional bicycles per day after 2 days of training. M ¢(5) =

2000(5) 2

[3(5) + 10]

10, 000 » 1.4 7225

This means the new employee can assemble about 1.4 additional bicycles per day after 5 days of training. 54.

The revenue R(q) is given by R(q) = pq = D(q)q. Using the product rule, the derivative of R(q), which is the marginal revenue, is R¢(q) = D(q) ⋅ 1 + qD¢(q) = D(q) + qD¢(q).

55.

C ( x) =

C ( x) x

Let u( x) = C ( x), with u ¢( x) = C ¢( x) Let v( x) = x with v¢( x) = 1. Then, by the quotient rule, C ( x) =

56.

v( x) ⋅ u¢( x) - u ( x) ⋅ v¢( x) 2

[v( x)]

x ⋅ C ¢( x) - C ( x) ⋅ 1 x

xC ¢( x) - C ( x) x2

Let d(t) be the demand as a function of time and p(t) be the price as a function of time. Then R(t ) = d (t ) p(t ) is the revenue as a function of time. Let t = t1 represent the beginning of the year. R¢(t ) = d (t ) p¢(t ) + d ¢(t ) p(t ) R¢(t1) = d (t1) p¢(t1) + d ¢(t1) p(t1) = (500)(0.5) + (30)(15) = 250 + 450 = 700

Revenue is increasing $700 per month.

Section 4.2 57.

257

Let C(t) be the cost as a function of time and q(t) be the quantity as a function of time.

60.

(a) dW ( H - 0.93)(20) - 20H (1) =1+ dH ( H - 0.93)2

C (t )

Then C (t ) = q(t ) is the revenue as a function of time. Let t = t1 represent last month. C ¢(t ) = C ¢ (t1) = =

æ ö÷ 20 20 H W = çç1 + ÷÷ H = H + çè H - 0.93 ø H - 0.93

q(t )C ¢(t ) - C (t )q¢(t )

[ g (t )]2 q(t1)C ¢(t1) - C (t1)q¢(t1)

[ g (t1)]2

(12,500)(1200) - (27, 000)(350) (12,500)2

( H - 0.93)2 ( H - 0.93)2

58.

s ( x) =

(a)

= =

( H - 0.93)

(b)

59.

(a)

f ¢( x ) =

(b)

f ¢( A) =

-18.6 ( H - 0.93) 2

( H - 0.93)2

By the quadratic formula, 1.86  1.862 + 4(17.7351) 2 » -3.38 or 5.24

H =

(m + nx) 2 m

Discarding the negative solution, W is minimized at H = 5.24 m. (c) Crows apply optimal foraging techniques.

Kx A+x

f ¢( x ) =

H 2 - 1.86 H - 17.7351

(m + nx)2 m + nx - nx

x = 50, m = 10, n = 3 m s¢(50) = (m + 50n)2 10 = [10 + 50(3)]2 1 = 2560 » 0.000391 mm per ml

f ( x) =

dW = 0 when H 2 - 1.86 H - 17.7351 = 0 dH

(m + nx)1 - x(n)

(m + nx)

( H - 0.93)2

(b)

x ; m and n constants m + nx

s ¢( x) =

20 H - 18.6 - 20H

H 2 - 1.86 H + 0.8649

= 0.03552

The average cost is increasing at a rate of $0.03552 per gallon per month.

61.

R(w) =

30(w - 4) w - 1.5 30(5 - 4) 5 - 1.5 » 8.57 min

(a)

R(5) =

(b)

R(7) =

30(7 - 4) 7 - 1.5 » 16.36 min

(c)

( A + x) K - Kx(1) ( A + x) AK ( A + x)

R¢( w) =

( A + A)2 AK K = = 2 4A 4A

( w - 1.5)(30) - 30( w - 4)(1) ( w - 1.5)2 30w - 45 - 30w + 120 ( w - 1.5)2 75 ( w - 1.5)2 R¢(5) = R¢(7) =

» 6.12

min 2 kcal

» 2.48 2

min 2 kcal

75 (5 - 1.5)

75 (7 - 1.5)

258

62.

Chapter 4 CALCULATING THE DERIVATIVE

(a)

(b)

63.

-8100

f ¢(1) =

= 2

f ( x) =

= = = =

9202 9202 + a 2

dW (9202 + a 2 )(0) - 9202 (2a) = da (9202 + a 2 )2 =-

x2 2(1 - x)

1, 692,800a (9202 + a 2 ) 2

When a = 640, dW 1, 692,800(640) =» -0.000687 da (9202 + (640)2 )2

-8100

(99 - 90) 92 -8100 = = -100 facts/hr 81 -8100 -8100 f ¢(10) = = 2 [99(10) - 90] (900)2 -8100 = 810, 000 1 =or -0.01 facts/hr 100

f ¢( x) =

64.

(b) W (a) =

90t f (t ) = 99t - 90 (99t - 90)(90) - (90t )(99) f ¢(t ) = (99t - 90)2 -8100 = (99t - 90) 2

(c) According to (b), one more run allowed should reduce the value of W by 0.000687, giving a value of 0.674 - 0.000687 = 0.673.

For a = 641, the formula for W gives 9202 » 0.673 W = 9202 + 6412 in agreement with the derivative -based estimate. W ( a) =

(d)

9201.81 9201.81 + a1.81

2(1 - x)(2 x) - x (-2)

W (640) =

[2(1 - x)]

4x - 4x2 + 2x2 4(1 - x)

9201.81 9201.81 + 6401.81

» 0.659

dW (9201.81 + a1.81)(0) - 9201.81(1.81a 0.81) = da (9201.81 + a1.81)2

4x - 2x2

1.81a 0.81(9201.81)

4(1 - x) x(2 - x)

2(1 - x)2

When a = 640, dW 418,923.3255a 0.81 =» -0.000636 da (9201.81 + 6401.81)2

4(1 - x) 2 x(2 - x)

(a)

f ¢(0.1) =

(b)

f ¢(0.6) =

0.1(2 - 0.1)

» 0.1173

2(1 - 0.1)2 0.6(2 - 0.6) 2(1 - 0.6)2

W =

s 2

s +a

920 2

418,923.3255a 0.81 (9201.81 + a1.81) 2

= 2.625

(a) The 2019 Astros actual winning percentage was 107 » 0.660 107 + 55 2

(9201.81 + a1.81)2

920 + 6402

» 0.674

4.3 The Chain Rule Your Turn 1

Let f ( x) = 2 x - 1 and g ( x) = g (0) =

3⋅0+5 =

f [ g (0) ] = 2 5 - 1 f (0) = 2 ⋅ 0 - 1 = -1 g [ f (0) ] - 3(-1) + 5 =

3x + 5.

Section 4.3

259

Your Turn 2

4.3 Warmup Exercises

Let f ( x) = 2 x - 3 and g ( x) = x 2 + 1. g[ f ( x)] = g (2 x - 3)

W1.

= (2 x - 3)2 + 1

f ( x) = f ¢( x) =

= 4 x 2 - 12 x + 9 + 1 = 4 x 2 - 12 x + 10

Your Turn 3 3

Write h( x) = (2 x - 3) in the form h( x) = f [ g ( x)].

W2. f ( x) =

One possible answer is h( x) = f [ g ( x)]

f ¢( x ) =

where g ( x) = 2 x - 3 and f ( x) = x3.

Your Turn 4 y = (5 x 2 - 6 x)-2 dy = -2(5 x 2 - 6 x)-3 ⋅ (10 x - 6) dx -2(10 x - 6) -20 x + 12 = = 2 3 (5x - 6 x) (5x 2 - 6 x)3

x2 x3 + 1 (2 x)( x3 + 1) - ( x 2 )(3x 2 ) ( x3 + 1)2 2x - x4 ( x3 + 1) 2 6 x x4 + 1

6 x1/ 2 x4 + 1

(3x-1/ 2 )( x 4 + 1) - (6 x1/ 2 )(4 x3 ) ( x 4 + 1)2 3( x 4 + 1) - 6 x(4 x3 ) ( x1/ 2 )( x 4 + 1) 2 3 - 21x 4

W3.

x ( x 4 + 1)2 x 2/3

f ( x) =

x1/3 + 2 æ1 ö 2 -1/3 1/3 x x + 2 - x 2/3 çç x-2/3 ÷÷÷ ç è3 ø 3 f ¢( x ) = 1/3 2 ( x + 2)

(

Your Turn 5 Dx[( x 2 - 7)10 ] = 10( x 2 - 7)9 (2 x) = 20 x( x 2 - 7)9

2 + 4 x-1/3 - 1

Your Turn 6 y = x 2 (5x - 1)3 dy = x 2 ⋅ 3(5x - 1)2 (5) + (5 x - 1)3 ⋅ 2 x dx = 15x 2 (5x - 1)2 + 2 x(5 x - 1)3

3( x1/3 + 2) 2

) (

True

False. Consider f ( x) = x 2 and g ( x) = ln x .

= x(5 x - 1)2 (25 x - 2)

g[ f ( x)] = ln x 2 = 2 ln x.

Thus, f [ g ( x)] ¹ g[ f ( x)]. 3.

2 3 é (4 x - 1)3 ù ú = ( x + 3)[3(4 x - 1) (4)] - (4 x - 1) (1) Dx êê ú ( x + 3) 2 ëê x + 3 ûú 2 3 12( x + 3)(4 x - 1) - (4 x - 1)

( x + 3)

3( x1/3 + 2)2

Then f [ g ( x)] = (ln x)2 and

1 + 4 x-1/3

4.3 Exercises

= x(5 x - 1)2[15 x + 2(5x - 1)]

Your Turn 7

(4 x - 1) [ 12( x + 3) - (4 x - 1) ]

False. Use the chain rule. If f ( x) = (2 x3 + 7 x)5 , then f ¢( x) = 5(2 x3 + 7 x) 4 (6 x + 7).

( x + 3) 2 (4 x - 1) 2 (8 x + 37) ( x + 3)

)

260

Chapter 4 CALCULATING THE DERIVATIVE

In Exercises 5 through 10, f ( x) = 5 x 2 - 2 x and g ( x) = 8x + 3.

é ù g[ f ( x)] = 6 ê x + 7 ú - 1 ëê 8 ûú x 6 = + 42 - 1 8 = 3x + 41 4 3 = x + 164 4 4 x + 3 164 = 4

g (2) = 8(2) + 3 = 19 f [ g (2)] = f [19] = 5(19)2 - 2(19) = 1805 - 38 = 1767

g (-5) = 8(-5) + 3 = -37 f [g(-37)] = f [-37] = 5(-37) 2 - 2(-37)

12.

f ( x) = -8x + 9; g ( x) = x + 4 5 éx ù f [ g ( x)] = -8 ê + 4 ú + 9 êë 5 úû = -8 x - 32 + 9 5 = 8x - 23 5 = -8x - 115 5 +9 +4 8 x g[ f ( x)] = 5 = -8x + 9 + 20 5 5 + x 8 29 = 5

13.

f ( x ) = 1 ; g ( x) = x 2 x f [ g ( x)] = 12 x

= 6845 + 74 = 6919

f (2) = 5(2) 2 - 2(2) = 20 - 4 = 16 g[ f (2)] = g[16] = 8(16) + 3 = 128 + 3 = 131

f (-5) = 5(-5)2 - 2(-5) = 125 + 10 = 135 g[ f (-5)] = f [135] = 8(135) + 3 = 1080 + 3 = 1083

g (k ) = 8k + 3 f [ g (k )] = f [8k + 3]

æ ö2 g[ f ( x)] = çç 1 ÷÷÷ èxø

= 5(8k + 3)2 - 2(8k + 3) = 5(64k 2 + 48k + 9) - 16k - 6

= 12 x

= 320k 2 + 224k + 39

10.

f (5z ) = 5(5 z ) 2 - 2(5z )

14.

= 125z 2 - 10 z g[ f (5z )] = g (125z 2 - 10 z ) = 8(125 z 2 - 10 z ) + 3 = 1000 z 2 - 80 z + 3

11.

x + 7; g ( x) = 6 x - 1 8 6x - 1 f [ g ( x)] = +7 8 6x - 1 56 = + 8 8 6 x + 55 = 8 f ( x) =

15.

f ( x) = 24 ; g ( x) = 2 - x x 2 f [ g ( x)] = (2 - x)4 æ ö g[ f ( x)] = 2 - çç 24 ÷÷ = 2 - 24 çè x ÷ø x f ( x) = f [ g ( x)] = =

x + 2; g ( x) = 8 x 2 - 6 (8 x 2 - 6) + 2 8x2 - 4

g[ f ( x)] = 8( x + 2) 2 - 6 = 8 x + 16 - 6 = 8 x + 10

Section 4.3 16.

261

f ( x) = 9 x 2 - 11x; g ( x) = 2 x + 2

23.

f [ g ( x)] = 9(2 x + 2) 2 - 11(2 x + 2)

y = e5 x - 3

If f ( x) = e x and g ( x) = 5x - 3, then y = f [ g ( x)] = e5 x-3.

= 9[4( x + 2)] - 22 x + 2 = 36( x + 2) - 22 x + 2 = 36 x + 72 - 22 x + 2

24.

g[ f ( x)] = 2 (9 x 2 - 11x) + 2

then y = f [ g ( x)] = e x .

f ( x) = x 2 + 1; g ( x) = e x -1 f [ g ( x)] = (e

If f ( x) = e x and g ( x) = x 2 ,

= 2 9 x 2 - 11x + 2

17.

y = ex

25.

x -1 2

y = ln(4 x + 7)

If f ( x) = ln x and g ( x) = 4 x + 7, then y = f [ g ( x)] = ln(4 x + 7).

) +1

= e2 x -2 + 1 2

g[ f ( x)] = e( x +1)-1 =e

18.

26.

If f ( x) = ln x and g ( x) = 1 + x3, then y = f [ g ( x)] = ln(1 + x3 ).

f ( x) = ln( x 2 + 4); g ( x) = 2 x - 1 f [ g ( x)] = ln((2 x - 1) 2 + 4)

27.

= ln(4 x 2 - 4 x + 1 + 4)

Then (8x 4 - 5x 2 + 1) 4 = f [ g ( x)].

g[ f ( x)] = 2[ln( x 2 + 4)] - 1

Use the alternate form of the chain rule.

= 2 ln( x 2 + 4) - 1

dy = f ¢[ g ( x)] ⋅ g ¢( x) dx

y = (5 - x 2 )3/5

f ¢( x) = 4 x3

If f ( x) = x3/5 and g ( x) = 5 - x 2 , then 2 3/5

y = f [ g ( x)] = (5 - x )

20.

y = (3x - 7)

f ¢[ g ( x)] = 4[ g ( x)]3 = 4(8x 4 - 5x 2 + 1)3

g ¢( x) = 32 x3 - 10 x dy = 4(8x 4 - 5x 2 + 1)3 (32 x3 - 10 x) dx

2/3

If f ( x) = x 2/3 and g ( x) = 3x 2 - 7, then y = f [ g ( x)] = (3x 2 - 7)2/3.

21.

then y = f [g( x)] = - 13 + 7 x .

Let f ( x) = x5 and g ( x) = 2 x3 + 9 x.

f ( x) = 5 x 4 f ¢[ g ( x)] = 5[ g ( x)]4 = 5(2 x3 + 9 x)4 g ¢( x) = 6 x 2 + 9

9 - 4x

If f ( x) =

y = (2 x3 + 9 x)5

dy = f ¢[ g ( x)] ⋅ g ¢( x) dx

g ( x) = 13 + 7 x,

y =

28.

Then (2 x3 + 9 x)5 = f [ g ( x)].

y = - 13 + 7 x

If f ( x) = - x and

22.

y = (8 x 4 - 5 x 2 + 1)4

Let f ( x) = x 4 and g ( x) = 8x 4 - 5x 2 + 1.

= ln(4 x 2 - 4 x + 5)

19.

y = ln(1 + x3 )

dy = 5(2 x3 + 9 x) 4 (6 x 2 + 9) dx

x and

g ( x) = 9 - 4 x,

then y = f [ g ( x)] =

9 - 4 x.

262 29.

Chapter 4 CALCULATING THE DERIVATIVE é1 ù g ¢(t ) = -3 ê (7t 3 - 1)-1/2 ⋅ 21t 2 ú êë 2 úû é 21 ù = -3 ê t 2 (7t 3 - 1)-1/2 ú êë 2 úû -63 2 1 t ⋅ = 3 2 (7t - 1)1/2

k ( x ) = -2(12 x 2 + 5)-6

Use the generalized power rule with u = 12 x 2 + 5, n = -6, and u ¢ = 24 x. k ¢( x ) = -2[-6(12 x 2 + 5)-6-1 ⋅ 24 x] = -2[-144 x(12 x 2 + 5)-7 ] = 288 x(12 x 2 + 5)-7

30.

34.

u = 3x 4 + 2, n = -4 and u ¢ = 12 x3.

f (t ) = 8 4t 2 + 7 = 8(4t 2 + 7)1/2

Use the generalized power rule with

f ¢( x) = -7[-4(3x 4 + 2)-4-1 ⋅ 12 x3 ]

u = 4t 2 + 7, n = 12 , and u¢ = 8t.

= -7[-48x3 (3x 4 + 2)-5 ]

é1 ù f ¢(t ) = 8 ê (4t 2 + 7)-1/2 ⋅ 8 t ú êë 2 úû

= 336 x3 (3x 4 + 2)-5

= 8[4t (4t 2 + 7)-1/8 ]

s(t ) = 45(3t 3 - 8)3/2

= 32t (4t 2 + 7)-1/2

Use the generalized power rule with 3 u = 3t 3 - 8, n = , and u ¢ = 9t 2. 2 é3 ù s¢(t ) = 45 ê (3t 3 - 8)1/ 2 ⋅ 9t 2 ú êë 2 úû é 27 ù = 45 ê t 2 (3t 3 - 8)1/ 2 ú êë 2 úû 1215 2 3 t (3t - 8)1/ 2 = 2

32.

2 7t 3 - 1

f ( x) = -7(3x 4 + 2)-4

Use the generalized power rule with

31.

-63t 2

= =

35.

32t 2

(4t + 7)1/2 32t 4t 2 + 7

m(t ) = -6t (5t 4 - 1) 4 Use the product rule and the power rule.

m¢(t ) = -6t[4(5t 4 - 1)3 ⋅ 20t 3 ] + (5t 4 - 1)4 (-6)

s(t ) = 12(2t 4 + 5)3/ 2

= -480t 4 (5t 4 - 1)3 - 6(5t 4 - 1)4

Use the generalized power rule with

= -6(5t 4 - 1)3[80t 4 + (5t 4 - 1)]

u = 2t 4 + 5, n = 32 , and u¢ = 8t 3.

= -6(5t 4 - 1)3 (85t 4 - 1)

é3 ù s¢(t ) = 12 ê (2t 4 + 5)1/2 ⋅ 8t 3 ú êë 2 úû

36.

r (t ) = 4t (2t 5 + 3) 4

= 12[12t 3 (2t 4 + 5)1/2 ]

Use the product rule and the power rule.

= 144t 3 (2t 4 + 5)1/2

r ¢(t ) = 4t[4(2t 5 + 3)3 ⋅ 10t 4 ] + (2t 5 + 3)4 ⋅ 4 = 160t 5 (2t 5 + 3)3 + 4(2t 5 + 3) 4

33.

= 4(2t 5 + 3)3[40t 5 + (2t 5 + 3)]

g (t ) = -3 7t 3 - 1

= 4(2t 5 + 3)3 (42t 5 + 3)

= -3 (7t 3 - 1)1/ 2

Use generalized power rule with u = 7t 3 - 1, n = 12 , and u ¢ = 21t 2.

Section 4.3 37.

263

y = (3x 4 + 1)4 ( x3 + 4)

42.

Use the product rule and the power rule.

y =

(3x - 4)

= 3x 2 (3x 4 + 1) 4 + 48x3 ( x3 + 4)(3x 4 + 1)3

= -30 x(3x 2 - 4)-6

= 3x 2 (3x 4 + 1)3[3x 4 + 1 + 16 x( x3 + 4)]

-30 x (3x 2 - 4)6

= 3x 2 (3x 4 + 1)3 (3x 4 + 1 + 16 x 4 + 64) = 3x 2 (3x 4 + 1)3 (19 x 4 + 64 x + 1)

43. y = ( x3 + 2)( x 2 - 1)4

dy = ( x3 + 2)[4( x 2 - 1)3 ⋅ 2 x] + ( x 2 - 1)4 (3x 2 ) dx

= 8x( x3 + 2)( x 2 - 1)3 + 3x 2 ( x 2 - 1) 4

= x( x 2 - 1)3[8( x3 + 2) + 3x( x 2 - 1)] = x( x 2 - 1)3 (8x3 + 16 + 3x3 - 3x)

= x( x 2 - 1)3 (11x3 - 3x + 16)

q( y ) = 4 y 2 ( y 2 + 1)5/ 4

Use the product rule and the power rule. 5 q¢( y) = 4 y 2 ⋅ ( y 2 + 1)1/4 (2 y) + 8 y( y 2 + 1)5/4 4 = 10 y 3 ( y 2 + 1)1/4 + 8 y( y 2 + 1)5/4 = 2 y( y 2 + 1)1/4[5 y 2 + 4( y 2 + 1) 4/4 ] = 2 y( y 2 + 1)1/4 (9 y 2 + 4)

40.

r (t ) =

44.

p(t ) = p¢(t ) =

p( z ) = z (6 z + 1) 4/3

Use the product rule and the power rule. 4 p¢( z ) = z ⋅ (6 z + 1)1/3 ⋅ 6 + 1 ⋅ (6 z + 1)4/3 3

= =

= 8z (6 z + 1)1/3 + (6 z + 1)4/3 = (6 z + 1)1/3[8 z + (6 z + 1)1] 1/3

= (6 z + 1)

(14 z + 1) =

41.

y =

-5 (2 x3 + 1) 2

= -5(2 x3 + 1)-2

(5t - 6)4 3t 2 + 4

r ¢(t )

Use the product rule and the power rule.

39.

= (3x 2 - 4)-5

dy = -5(3x 2 - 4)-6 ⋅ 6 x dx

dy = (3x 4 + 1)4 (3x 2 ) + ( x3 + 4)[4(3x 4 + 1)3  12 x3 ] dx

38.

(3t 2 + 4)[4(5t - 6)3 ⋅ 5] - (5t - 6)4 (6t ) (3t 2 + 4) 2 20(3t 2 + 4)(5t - 6)3 - 6t (5t - 6)4 (3t 2 + 4)2 2(5t - 6)3[10(3t 2 + 4) - 3t (5t - 6)] (3t 2 + 4)2 2(5t - 6)3 (30t 2 + 40 - 15t 2 + 18t ) (3t 2 + 4)2 2(5t - 6)3 (15t 2 + 18t + 40) (3t 2 + 4)2 (2t + 3)3 4t 2 - 1 (4t 2 - 1)[3(2t + 3) 2 ⋅ 2] - (2t + 3)3 (8t ) (4t 2 - 1)2 6(4t 2 - 1)(2t + 3)2 - 8t (2t + 3)3 (4t 2 - 1)2 (2t + 3)2[6(4t 2 - 1) - 8t (2t + 3)] (4t 2 - 1)2 (2t + 3)2[24t 2 - 6 - 16t 2 - 24t ] (4t 2 - 1)2 (2t + 3)2[8t 2 - 24t - 6] (4t 2 - 1) 2 2(2t + 3)2 (4t 2 - 12t - 3)

dy = -5[-2(2 x3 + 1)-3 ⋅ 6 x 2 ] dx = -5[-12 x 2 (2 x3 + 1)-3 ] = 60 x 2 (2 x3 + 1)-3 =

60 x 2 (2 x3 + 1)3

(4t 2 - 1) 2

264

Chapter 4 CALCULATING THE DERIVATIVE y =

45.

(b)

3x 2 - x (2 x - 1)

Dx ( f [ g ( x)]) at x = 2 = f ¢[ g (2)] ⋅ g ¢(2) æ ö = f ¢(3) ⋅ çç 3 ÷÷÷ è7ø æ ö = -8 çç 3 ÷÷÷ = - 24 è7ø 7

(2 x - 1)5 (6 x - 1) - (3x 2 - x)[5(2 x - 1) 4 ⋅ 2] dy = dx [(2 x - 1)5 ]2 = = = =

(2 x - 1)5 (6 x - 1) - 10(3x 2 - x)(2 x - 1)4 (2 x - 1)10

50.

(2 x - 1)4[(2 x - 1)(6 x - 1) - 10(3x 2 - x)]

(a)

Dx ( g[ f ( x)]) at x = 1 = g ¢[ f (1)] ⋅ f ¢(1) = g ¢(2) ⋅ (-6) = 3 (-6) = - 18 7 7

(2 x - 1)10 12 x 2 - 2 x - 6 x + 1 - 30 x 2 + 10 x (2 x - 1)6 -18 x 2 + 2 x + 1

(b)

Dx ( g[ f ( x)]) at x = 2

(2 x - 1)6

= g ¢[ f (2)] ⋅ f ¢(2) = g ¢(4) ⋅ (-7) = 5 (-7) = -5 7

46. y = dy dx

x2 + 4x (3x3 + 2) 4 3

48.

51.

(3x + 2) (2 x + 4) - ( x + 4 x)[4(3x + 2) ⋅ 9 x ]

f ¢( x ) =

(3x3 + 2) 4 (2 x + 4) - 36 x 2 ( x 2 + 4 x)(3x3 + 2)3 (3x3 + 2)8

f ¢( x ) =

2(3x3 + 2)3[(3x3 + 2)( x + 2) - 18 x 2 ( x 2 + 4 x)] 3

(3x + 2)

1 2 ( x + 16)-1/2 (2 x) 2 x

f ¢(3) =

2(3x 4 + 6 x3 + 2 x + 4 - 18 x 4 - 72 x3 ) 3

x 2 + 16; x = 3

f ( x) =

f ( x) = ( x 2 + 16)1/2

[(3x3 + 2) 4 ]2

f (3) =

-2(15 x + 66 x - 2 x - 4)

dy = f ¢[ g ( x)] ⋅ g ¢( x). dx

Then, using the power rule, f ¢( x) = nx n-1. dy = n[ g ( x)]n-1 ⋅ g ¢( x). dx

= f ¢[ g (1)] ⋅ g ¢(1) æ2ö = f ¢(2) ⋅ çç ÷÷÷ çè 7 ø æ2ö = -7 çç ÷÷÷ = -2 çè 7 ø

32 + 16 = 5

y - 5 = 3 ( x - 3) 5 y -5 = 3x- 9 5 5 y = 3 x + 16 5 5

Let f ( x) = x . Then y = f ( g ( x)) = [ g ( x)] .

Dx ( f [ g ( x)]) at x = 1

3 5

slope form

(a)

32 + 16

We use m = 53 and the point P(3, 5)in the point-

(3x3 + 2)5

Using the chain rule,

49.

x 2 + 16 3

52.

f ( x) = ( x3 + 7) 2/3; x = 1 f ¢( x) = 2 ( x3 + 7)-1/3 (3x 2 ) 3 2 f ¢( x) = 3 2 x 1/3 ( x + 7) f ¢(1) = 2 = 1 2 f ¢(1) = 82/3 = 4

Section 4.3 53.

265

f ( x) = x( x 2 - 4 x + 5)4 ; x = 2

56.

f ( x) =

f ¢( x) = x ⋅ 4( x 2 - 4 x + 5)3 ⋅ (2 x - 4) + 1 ⋅ ( x 2 - 4 x + 5)4

f ¢( x ) =

= ( x 2 - 4 x + 5)3 =

⋅ [4 x(2 x - 4) + ( x - 4 x + 5)] 2

x 2

( x + 4) 4 ( x 2 + 4) 4 ⋅ 1 - x ⋅ 4( x 2 + 4)3 (2 x) ( x 2 + 4)8 ( x 2 + 4)3[( x 2 + 4) - 8x 2 ] ( x 2 + 4)8

= ( x - 4 x + 5) (9 x - 20 x + 5) =

f ¢(2) = (1) (1) = 1 4

f (2) = 2(1) = 2 We use m = 1 and the point P(2, 2). y - 2 = 1( x - 2)

4 - 7x2 ( x 2 + 4)5

If the tangent line is horizontal, its slope is zero and f ¢( x) = 0. 4 - 7 x2

y-2= x-2

( x 2 + 4)5

y = x

4 - 7 x2 = 0

54. f ( x) = x

x2 =

x - 12; x = 2 1 4 ( x - 12)-1/2 (4 x3 ) + 2 x( x 4 - 12)1/2 2

f ¢( x ) =

1/2

+ 2 x( x 4 - 12)1/2

( x - 12) 64 f ¢(2) = 1/2 + 4(4)1/2 4 f ¢(2) = 32 + 8 = 40

59.

-p + 500; p(c) = 2c - 10 100 The demand in terms of the cost is

D(c) = D [ p(c)] -(2c - 10)2 + 500 100 -4(c - 5)2 = + 500 100 2 = -c + 10c - 25 + 500 25 -c 2 + 10c - 25 + 12,500 = 25 -c 2 + 10c + 12, 475 = . 25 =

x3 - 6 x 2 + 9 x + 1

f ( x) = ( x3 - 6 x 2 + 9 x + 1)1/2 f ¢( x ) =

1 3 ( x - 6 x 2 + 9 x + 1)-1/2 2 ⋅ (3x 2 - 12 x + 9)

f ¢( x ) =

54.

3( x 2 - 4 x + 3)

R( x) = 24( x 2 + x)2/3

(a)

2 x3 - 6 x 2 + 9 x + 1

If the tangent line is horizontal, its slope is zero and f ¢( x) = 0. 3( x 2 - 4 x + 3) 2 x3 - 6 x 2 + 9 x + 1

D( p ) =

We use m = 40 and the point P(2,8). y - 8 = 40( x - 2) y = 40 x - 72 f ( x) =

2 7

The tangent line is horizontal at x =  2 .

f (2) = 4 4 = 8

55.

4 7

x =

f ( x) = x 2 ( x 4 - 12)1/2 f ¢( x ) = x 2 ⋅

(b)

R(100) = 24((100) 2 + 100)2/3 » 11, 213.96 When 100 sets are sold, the total revenue is $11,213.96.

é ù R¢( x) = 24 ê 2 ( x 2 + x)-1/3 ⋅ (2 x + 1) ú ëê 3 ûú = 16( x 2 + x)-1/3 (2 x + 1)

3( x - 4 x + 3) = 0 3( x - 1)( x - 3) = 0 x = 1 or x = 3 The tangent line is horizontal x= 1 and x = 3.

R¢(100) = 16(1002 + 100)-1/3 (2 ⋅ 100 + 1) » 148.78 When 100 sets are sold, the total revenue is increasing by about $148.78.

266

Chapter 4 CALCULATING THE DERIVATIVE 1824 dA = 75æç1 + 8 ö÷ = $111.86. ÷ ççè dr 36,500 ÷ø At 8%, the balance is increasing by about $111.86 per percentage point.

When 100 sets are sold, the total revenue is increasing by about $187.29. R ( x) =

(d)

R( x) 24( x 2 + x) 2 / 3 = x x

= = = =

(e)

(f)

(g)

61.

1824 9 ö÷ dA = 75æç1 + = $117.59. ÷ ç çè dr 36,500 ÷ø At 9%, the balance is increasing by about $117.59 per percentage point.

x[16( x 2 + x)-1/3 (2 x + 1)] - 24( x 2 + x)2/3 x2 16 x(2 x + 1) 2

1/3

( x + x)

- 24( x 2 + x)2/3 ⋅

( x 2 + x)1/3 ( x 2 + x)1/3

32 x 2 + 16 x - 24( x 2 + x)

62.

æ ç çç è

= 150 -

8 x( x - 1) = 2 2 2 2 1/3 x ( x + x) x ( x + x)1/3 8( x - 1)

8x - 8x

24(1002 + 100)2/3 » 112.14 100 When 100 sets are sold, the average revenue is $112.14 per set. 8(100 - 1)

» 0.37 100(1002 + 100)1/3 When 100 sets are sold, the average revenue is increasing by $0.37 per set. 2

1/2

( p + 1)

63.

V =

(1)

= -

30 2

( p + 1)

3/2

60,000 1 + 0.3t + 0.1t 2

The rate of change of the value is

1825 æ r ÷ö A = 1500 çç1 + ÷ çè 36,500 ÷ø dA is the rate of change of A with respect to r. dr

1824 æ 1 ö dA = 1500(1825) æç1 + r ÷ö ÷ ç çççè 36,500 ÷÷÷ø dr 36,500 ÷ø èç 1824 æ r ÷ö = 75çç1 + ÷ çè 36,500 ÷ø

(a) For

-1/2

-30( p + 1)

» 0.23

200(200 + 200) When 200 sets are sold, the average revenue is increasing by $0.23 per set.

30 p 2

= 0 - êê

= 1/3

= 150 -

é ( p 2 + 1)1/2 D (30 ) - (30 )( D ( p 2 + 1)1/2 ) ù p p q p ú ú 2 1/2 2 dp [( p + 1) ] êë úû é ( p 2 + 1)1/ 2 (30) - (30 p) 1 ( p 2 + 1)-1/2 (2 p ) ù (2) ê ú = -ê ú 2 ê ú ( p + 1) ë û é ( p 2 + 1)1/2 (30) - (30 p)( p )( p 2 + 1)-1/2 ) ù ú = - êê ú 2 ( p + 1) êë úû é -30( p 2 + 1)-1/2 ([ p 2 + 1] - p 2 ) ù ú = - êê ú 2 ( p + 1) ëê ûú dq

R (100) =

R¢(200) =

30 p p +1

x( x 2 + x)1/3

8(200 - 1)

ö÷ ÷÷ ÷÷ 2 p + 1 ø÷ p

q = D( p ) = 30 çç 5 -

x 2 ( x 2 + x)1/3

R¢(100) =

r = 9%,

R ¢( x ) =

r = 8%,

(b) For

R¢(200) = 16(2002 + 200)-1/3(2 ⋅ 200 + 1) » 187.29

(c)

r = 6%,

1824 dA = 75æç1 + 6 ö÷ = $101.22. ÷ ç çè dr 36,500 ÷ø At 6%, the balance is increasing by about $101.22

V ¢(t ) =

(1 + 0.3t + 0.1t 2 )(0) - 60, 000(0.3 + 0.2t )

(1 + 0.3t + 0.1t 2 )2 -60, 000(0.3 + 0.2t ) = . (1 + 0.3t + 0.1t 2 )2

(a) 2 years after purchase, the rate of change in the value is -60, 000[0.3 + 0.2(2)] V ¢(2) = [1 + 0.3(2) + 0.1(2)2 ]2 -60, 000(0.3 + 0.4) = (1 + 0.6 + 0.4)2 -42, 000 = 4 = -$10,500 per year.

Section 4.3

267

(b) 4 years after purchase, the rate of change in the value is -60, 000[0.3 + 0.2(4)] V ¢(4) = [1 + 0.3(4) + 0.1(4) 2 ]2 -66, 000 = 14.44 = -$4570.64 per year. 64.

(d) If p = $5000 and

then q =

q 2 = 15, 000 - 1.5 p 2

q - 15, 000 = p -1.5

65.

P( x) = 2 x 2 + 1; x = f (a) = 3a + 2 P[ f (a)] = 2(3a + 2)2 + 1

q 15, 000 + = p -1.5 1.5

= 2(9a 2 + 12a + 4) + 1 = 18a 2 + 24a + 9

-2 q + 10, 000 = p 3

66.

(a)

30, 000q - 2q3 3

A(r ) =  r 2 r (t ) = t 2

æ -2q 2 ö÷ ç R(q) = qp = q çç + 10, 000 ÷÷÷ ÷ø çè 3

A[r (t )] = [t 2 ]2 =  (t 4 ) = t4

This function represents the area of the oil slick as a function of time t after the beginning of the leak. (b)

(b)

Dt A[r (t )] = 4 t 3 Dt A[r (100)] = 4 (100)3

P (q ) = R ( q ) - C (q ) 30, 000q - 2q3 - (2000q + 3500) 3 æ -2q3 ö÷ ç = çç + 10, 000q ÷÷÷ - (2000q + 3500) çè 3 ø÷

= 4, 000, 000

-2 q + 8000q - 3500 3

At 100 minutes the area of the spill is changing at the rate of 4,000,000 ft2/min. 67.

(a)

= 4 t 2

A = 4 t 2 gives the area of the pollution in terms of the time since the pollutants were first emitted.

P(q) is the profit function. dP 2q3 = Dq (8000q - 3500) dq 3

= 8000 - 2q 2 dp dq

or P¢(q) gives the marginal profit.

r (t ) = 2t; A(r ) =  r 2 A[r (t )] =  (2t ) 2

2q 3 = 8000q - 3500 3

(c)

7500.

When the price is $5000, the marginal profit is -$7000.

-2q3 + 30, 000q 3

15, 000 - 7500 =

= 8000 - 15,000 = -7000

q = 15, 000 - 1.5 p

= 8000 - 2(7500)

Solve for p.

-2q3 + 10, 000q 3

15, 000 - 1.5(5000)

P¢( q) = 8000 - 2q 2

q = 15, 000 - 1.5 p

q =

Thus, q 2 = 7500.

C = 2000q + 3500

(a)

15, 000 - 1.5 p ,

(b)

Dt A[r (t )] = 8 t Dt A[r (4)] = 8 (4) = 32 At 12 P.M., the area of pollution is changing at the rate of 32 mi2/hr.

268 68.

Chapter 4 CALCULATING THE DERIVATIVE 1 (b) C ¢(4) = - [2(4) + 1]-3/2 2 1 = - (9)-3/2 2 -1 1 = ⋅ 2 ( 9)3

N (t ) = 2t (5t + 9)1/2 + 12 é1 ù N ¢(t ) = (2t ) ê (5t + 9)-1/2 (5) ú êë 2 úû + 2(5t + 9)1/2 + 0 = 5t (5t + 9)-1/2 + 2(5t + 9)1/2 = (5t + 9)-1/2[5t + 2(5t + 9)]

1 54 » -0.02 After 4 days, the calcium level in the bloodstream is decreasing by 0.2 mg/cm3 per day. =-

= (5t + 9)-1/2 (15t + 18) =

(a)

(b)

15t + 18 (5t + 9)1/2 15(0) + 18

N ¢(0) =

(c)

[5(0) + 9]1/2 = 18 =6 91/2 At 0 hours, the number of bacteria is increasing by 6,000,000 per hour.

( ) (5)

15 75 + 18 æ 7 ö÷ ç ¢ = N ç ÷÷ è5ø [5 7 + 9]1/2

1 128 » -0.008 After 7.5 days, the calcium level in the bloodstream is decreasing by 0.008 mg/cm3 per day. =-

21 + 8 (7 + 9)1/ 2 = 391/2 (16) = 39 = 9.75 4 At 7/5 hours, the number of bacteria is increasing by 9,750,000 per hour. =

(c)

N ¢(8) = =

(d) C is always decreasing because C ¢ = - 12 (2t + 1)-3/2

is always negative for t ³ 0. (The amount of calcium in the bloodstream will continue to decrease over time.)

15(8) + 18 [5(8) + 9]1/2 120 + 18

70.

1/2

(49) 138 = » 19.71 7 At 8 hours, the number of bacteria is increasing by about 19,710,000 per hour.

69.

1 C ¢(7.5) = - [2(7.5) + 1]-3/2 2 1 = - (16)-3/2 2 æ ö 1 çç 1 ÷÷÷ = - çç ÷ 2 çç ( 16)3 ÷÷ è ø

C (t ) = 1 (2t + 1)-1/2 2 æ ö C ¢(t ) = 1 çç - 1 ÷÷÷ (2t + 1)-3/2 (2) è 2 2ø = - 1 (2t + 1)-3/2 2 (a) C ¢(0) = - 1 [2(0) + 1]-3/2 2 1 =2 = -0.5 At 0 days, the calcium level in the bloodstream is decreasing by 0.5 mg/cm3 per day.

(a)

1/2 æ Qö R(Q) = Q ççç C - ÷÷÷ è 3ø -1/2 é æ ö1/2 æ 1 ö÷ ùú æç Qö çç - ÷ ú + ç C - Q ÷÷ (1) R¢(Q) = Q êê 1 ççç C - ÷÷÷ è 3 ø÷ ú èç 3ø 3 ø÷ êë 2 è û

-1/2 1/2 æ æ Qö Qö = - 1 Q ççç C - ÷÷÷ + ççç C - ÷÷÷ è 6 è 3ø 3ø

(

æ ö1/2 çç C - Q ÷÷ + 1/2 çè 3 ÷ø Q

6 C- 3

(b)

)

R¢(Q) = -

(

æ ö1/2 ç C - Q ÷÷ + ç 1/2 çè 3 ÷ø Q

6 C- 3

)

If Q = 87 and C = 59, then

Section 4.4

269 1/2 æ 87 ö÷ 87 R¢(Q) = çç 59 ÷ 1/2 çè 3 ÷ø 6 59 - 87

(

= (30)1/2 -

66.

)

6(30)1/2 87 = 5.48 32.88 = 5.48 - 2.65 = 2.83.

V (r ) =

(a)

r (t ) = 0 when 6 -

y = (( x 2 )2 )2 = x8

(b)

dy = 8x8-1 = 8 x 7 dx

73.

(a)

3 t = 0; 17

y = (( x3 )2 ) 2 dy = 2(( x3 ) 2 ) ⋅ d (( x3 )2 ) dx dx 6 3 = 2 x ⋅ 2( x ) ⋅ d ( x3 ) dx 6 3 = 2 x ⋅ 2 x ⋅ 3x 2

= 12 x11

17(6) = 34 min . 3

(b)

dV dS dr 3 = 4 r 2 , = 8 r , =dr dr dt 17 dV dV dr 12 = ⋅ = -  r2 dt dr dt 17

y = (( x 2 )2 ) 2 dy = 2(( x 2 ) 2 ) ⋅ d (( x 2 )2 ) dx dx 4 2 = 2 x ⋅ 2( x ) ⋅ d ( x 2 ) dx = 2x4 ⋅ 2x2 ⋅ 2x = 8x7

4 3 3  r , S (r ) = 4 r 2 , r (t ) = 6 - t 3 17

t =

(b)

(a)

y = (( x3 )2 ) 2 = x12 dy = 12 x12-1 = 12 x11 dx

2 12 æç 3 ö  ç 6 - t ÷÷÷ 17 çè 17 ø dS dS dr 24 = ⋅ = - r dt dr dt 17 24 æ 3 ÷ö t÷ = -  çç 6 17 çè 17 ÷ø

4.4 Derivatives of Exponential Functions

When t = 17,

(b)

Your Turn 1 (a)

y = 43x dy = (ln 4) ⋅ 43x ⋅ 3 = 3(ln 4)43x dx

ù2 dV 12 é 3 = -  ê6 (17) ú úû dt 17 êë 17

y = e7 x + 5 3 3 dy = e7 x +5 ⋅ 21x 2 = 21x 2e7 x +5 dx

108  mm3 /min 17 ù 24 é 3 dS (17) ú = -  ê6 ê úû 17 ë 17 dt 72 = -  mm 2 /min 17 At t = 17 minutes, the volume is decreasing

Your Turn 2

by 108  mm3 per minute and the surface 17

Your Turn 3

72  mm 2 per minute. area is decreasing by 17

y = ( x 2 + 1)2 e 2 x dy = ( x 2 + 1) ⋅ e 2 x ⋅ 2 + e2 x ⋅ 2( x 2 + 1) ⋅ 2 x dx = 2( x 2 + 1)e2 x ( x 2 + 2 x + 1)

= 2e 2 x ( x 2 + 1)( x + 1)2

f ( x) = f ¢( x ) =

e-x 4x2 + 3

(4 x 2 + 3) ⋅ (-e-x ) - e-x ⋅ (8 x) (4 x 2 + 3)2 -e-x (4 x 2 + 8 x + 3) (4 x 2 + 3) 2

270

Chapter 4 CALCULATING THE DERIVATIVE

Your Turn 4

æ x 2 - 1 ö÷5/ 2 W3. f ( x ) = ççç 2 ÷÷ çè x + 1 ÷÷ø

Q(t ) = 100e-0.421t dQ = 100 ⋅ e-0.421t (-0.421) dt

Let h( x ) = x5/ 2 and g ( x ) =

-0.421t

= -42.1e

h¢( g ( x )) =

dQ = -42.1e-0.421(2) dt = -42.1e-0.842 » -18.1 grams per year

g ¢( x ) =

4.4 Warmup Exericses

W1. Let h( x ) = x1/ 2 and g ( x ) = x 6 + 5. Then f ( x ) = h( g ( x )) and f ¢( x ) = h¢( g ( x )) g ¢( x ). h¢( g ( x )) =

(

2( g ( x) )

2 x6 + 5

(

) (

)

(2 x ) x 2 + 1 - x 2 - 1 (2 x ) 2

( x 2 + 1) 4x 2

( x 2 + 1)

(

10 x x 2 - 1

1/ 2

)

g ¢( x ) = 6 x5

)

3/ 2

)

7/2

( x 2 + 1)

4.4 Exercises 6 x5

f ¢( x ) =

3/ 2 5 æç x 2 - 1 ö÷÷ çç ÷ 2 çè x 2 + 1 ÷÷ø

(

= 1/ 2

ù 3/ 2 ù é é æ 2 ú ê 5 ç x - 1 ÷÷ö ú êê 4x ú f ¢( x ) = ê çç 2 ÷÷ ú ê ú 2 ê 2 çè x + 1 ÷ø ú ê 2 êë úû ê x + 1 úú ë û

x6 + 5

x2 + 1

Then f ( x ) = h( g ( x )) and f ¢( x ) = h¢( g ( x )) g ¢( x ).

After 2 years (t = 2), the rate of change of the quantity present is

f ( x) =

x2 - 1

2 x +5

3x 5

x +5

True

False. The derivative of f ( x) = 5 x is f ¢( x) = (ln x)5 x.

W2.

(

f ( x ) = 4 x 2 + 3x + 2

)

True

False. A logistic function will start growing slowly, then grow more rapidly, and then gradually level off.

f ( x) = 1.2e x

Let h( x ) = x and g ( x ) = 4 x + 3x + 2. Then f ( x ) = h( g ( x )) and f ¢( x ) = h¢( g ( x )) g ¢( x ).

(

h¢( g ( x )) = 10( g ( x ))9 = 10 4 x 2 + 3x + 2

f ¢( x) = 1.2e x

)

g ¢( x ) = 8 x + 3

(

f ¢( x ) = 10 4 x 2 + 3x + 2

f ¢( x) = -7.3e x

) (8x + 3)

f ( x) = -7.3e x

f (t ) = 100t f ¢(t ) = (ln t )100t

f ( x) =  x f ¢( x) = (ln x) x

Section 4.4

271 y = e4 x

9. Let

g ( x) = 4 x,

with

g ¢( x) = 4.

17.

g ¢( x ) = 4 x 2 ö dy æ = 3çç 4 xe2 x ÷÷ è ø dx

= 12 xe 2 x

-2 x

y = e Let

g ( x ) = -2 x,

with

g ¢( x ) = -2,

18.

y = -8e3x

y = 1.2e

= -60 x 2e4 x

19.

y = 4e2 x -4 g ( x) = 2 x 2 - 4 g ¢( x ) = 4 x

dy = 1.2(5e5 x ) = 6e5 x dx

13.

y = -5e4 x

g ¢( x) = 12 x 2 3 dy = (-5)(12 x 2 )e4 x dx

dy = -8(3e3x ) = -24e3x dx

12.

g ( x) = 4 x 3

dy = -2e-2 x . dx

11.

g ( x) = 2 x 2

dy = 4e 4 x dx

10.

y = 3e 2 x

2 dy é ù = 4 ê (4 x)e2 x -4 ú dx ëê ûú

y = -16e 2 x +1

= 16 xe2 x - 4

g ( x) = 2 x + 1 g ¢( x) = 2 dy = -16(2e 2 x +1) = -32e2 x +1 dx

14.

20.

y = -3e3x + 5 g ( x) = 3 x 2 + 5 g ¢( x ) = 6 x

y = -4e-0.3x dy = -4(-0.3e-0.3x ) dx

y¢ = (-3)(6 x)e3x +5 2

= -18 xe3x +5

= 1.2e-0.3x

21. 15.

y = ex g ( x) = x

Use the product rule. dy = xe x + e x ⋅ 1 = e x ( x + 1) dx

g ¢( x) = 2 x 2 dy = 2 xe x dx

y = xe x

22.

y = x 2e-2 x

Use the product rule. 16.

-x 2

y =e

g ( x ) = -x g ¢ ( x ) = -2 x 2 dy = -2 xe-x dx

dy = x 2 (-2e-2 x ) + 2 xe-2 x dx = -2 x 2e-2 x + 2 xe-2 x = 2 x(1 - x)e-2 x

272 23.

Chapter 4 CALCULATING THE DERIVATIVE y = ( x + 3)2 e4 x

y =

28.

Use the product rule.

dy x(e x - (-1)e-x ) - (e x - e-x )(1) = dx x2

dy = ( x + 3) 2 (4)e4 x + e4 x ⋅ 2( x + 3) dx = 4( x + 3) 2 e4 x + 2( x + 3)e4 x

= 2( x + 3)e4 x [ 2( x + 3) + 1 ] =

= 2( x + 3)(2 x + 7)e4 x

24.

y = (3x3 - 4 x)e-5 x dy = (3x3 - 4 x) (-5e-5 x ) + e-5 x (9 x 2 - 4) dx = (-15 x3 + 20 x)e-5 x + (9 x 2 - 4)e-5 x 3

-5 x

y =

30.

ex dy e x (2 x) - x 2e x = dx (e x ) 2

26.

y =

27.

e x ( x - 1) + e-x ( x + 1) x2 10, 000

9 + 4e-0.2t

8000e-0.2t (9 + 4e-0.2t )2 p =

xe x (2 - x) e2 x x(2 - x)

31.

500 12 + 5e-0.5t

1250e-0.5t (12 + 5e-0.5t )2 2

f ( z ) = ( 2 z + e-z ) 2

f ¢( z ) = 2 ( 2 z + e-z ) ( 2 - 2 ze-z ) 2

= 4 ( 2 z + e-z )(1 - ze-z )

ex 2x + 1

dy e x (2 x + 1) - (2)(e x ) = dx (2 x + 1) 2 =

dp (12 + 5e-0.5t ) ⋅ 0 - 500[0 + 5(0.5)e-0.5t ] = dt (12 + 5e-0.5t )2

Use the quotient rule.

xe x + xe-x - e x + e-x

dp (9 + 4e-0.2t ) ⋅ 0 - 10, 000[0 + 4(-0.2)e-0.2t ] = dt (9 + 4e-0.2t ) 2

= (-15 x + 9 x + 20 x - 4)e

25.

p =

29.

Use the product rule.

e x - e-x x

32.

f (t ) = ( et + 5t ) 2

e x (2 x + 1 - 2)

f ¢(t ) = 3( et + 5t ) ⋅ ( et ⋅ 2t + 5 )

(2 x + 1)2

= 3( et + 5t ) ( 2tet + t )

e (2 x - 1) (2 x + 1) 2

33.

y = 73x +1

Let g ( x) = 3x + 1, with g ¢( x) = 3. Then

e x + e-x y = x dy x(e x - e-x ) - (e x + e-x ) = dx x2

dy = (ln 7)(73x +1) ⋅ 3 = 3(ln 7)73x +1 dx

34.

y = 4-5 x + 2

Let g ( x) = -5x + 2, with g ¢( x) = -5. Then dy = (ln 4)(4-5 x + 2 ) ⋅ (-5) = -5(ln 4)4-5 x + 2 dx

Section 4.4 35.

273 2

y = 3 ⋅ 4x +2

40.

Let g ( x) = x 2 + 2, with g ¢( x) = 2 x. Then 2 2 dy = 3(ln 4)4 x + 2 ⋅ 2 x = 6 x(ln 4)4 x + 2 dx

36.

y = -103x -4

dy dx =

Let g ( x) = 3x - 4, with g ¢( x) = 6 x. 2 dy = -(ln10)103x -4 ⋅ 6 x dx 2 æ ö = -6 x çç103x - 4 ÷÷ ln 10 è ø

37.

t , with g ¢(t ) =

1 . Then 2 t

41.

(t + e3t )(2te 2t + t 2 ⋅ 2e 2t ) - t 2e 2t (1 + 3e3t ) (t + e3t ) 2

(t + e3t )(2te 2t + 2t 2e 2t ) - t 2e 2t (1 + 3e3t ) (t + e3t ) 2 (2t 2e 2t + 2t 3e 2t + 2te5t + 2t 2e5t ) - (t 2e 2t + 3t 2e5t ) (t + e3t ) 2 t 2e 2t + 2t 3e 2t + 2te5t - t 2e5t (t + e3t ) 2 (2t 3 + t 2 )e 2t + (2t - t 2 )e5t (t + e3t ) 2

(ln 3)3 t

f ( x) = e x 3 x + 2

Let g ( x) = x 3x + 2.

ds 1 = 2(ln 3)3 t ⋅ dt 2 t =

æ 3 ÷÷ö g ¢( x) = 1 ⋅ 3x + 2 + x ççç ÷ èç 2 3x + 2 ÷ø

3x 2 3x + 2 2(3x + 2) 3x = + 2 3x + 2 2 3x + 2 9x + 4 = 2 3x + 2 æ 9 x + 4 ö÷ ÷ f ¢( x) = e x 3x + 2 ⋅ ççç çè 2 3x + 2 ÷÷ø =

s = 5 ⋅ 2 t -2

Let g (t ) = t - 2, with g ¢(t ) =

(

1 . Then 2 t -2

)

ds 1 = 5(ln 2) 2 t -2 ⋅ dt 2 t-2 =

39.

y =

(5ln 2)2 t -2 2 t-2

42.

tet + 2

e2t + 1 Use the quotient rule and product rule. dy (e2t + 1)(tet + et ⋅ 1) - (tet + 2)(2e2t ) = dt (e2t + 1) 2 = = = =

t 2e2t

t + e3t Use the quotient rule and product rule.

s = 2⋅3 t

Let g (t ) =

38.

y =

(e2t + 1)(tet + et ) - (tet + 2)(2e2t ) (e2t + 1)2 te3t + e3t + tet + et - 2te3t - 4e2t (e2t + 1)2 -te3t + e3t + tet + et - 4e2t (e2t + 1)2

3x + 2 +

f ( x) = e x /( x + 2)

Let g ( x) = g ¢( x ) = =

x3 + 2

( x3 + 2)(2 x) - x 2 (3x 2 ) ( x3 + 2)2 4x - x4 ( x3 + 2)

= 2

x(4 - x3 )

( x3 + 2)2 é x(4 - x3 ) ù 2 3 ú f ¢( x) = e x / ( x + 2) ⋅ êê 3 2ú ( x + 2) ëê ûú 2

x(4 - x3 )e x /( x + 2) ( x3 + 2)2

(1 - t )e3t - 4e2t + (1 + t )et (e2t + 1)2

2 x 4 + 4 x - 3x 4 ( x3 + 2) 2

274 43.

Chapter 4 CALCULATING THE DERIVATIVE æ e3 ö÷ ç ÷÷ Find the slope at çç 3, è 9 ø÷

f ( x) = xe2 x f ¢( x) = x ⋅ 2e2 x + e2 x ⋅ 1 = 2 xe2 x + e2 x

f ¢(3) = 2

Find the slope at (1, e ) f ¢(1) = 2(1)e

2(1)

= 3e

e3 27 Using point slope form, m=

m = 3e Using point slope form,

y - e2 = 3e2 ( x - 1)

e3 e3 = ( x - 3) 9 27

y = 3xe2 - 2e 2

44.

e3 (3 - 2)

y =

f ( x ) = x 2e x

47.

f ¢( x ) = x 2 ⋅ e x + e x ⋅ 2 x

e3 x 27

y = yoe kt dy d é yoekt ùú = yo ke kt = û dx dt ëê

= x 2e x + 2 xe x

Find the slope at (2, 4e2 )

= k ( yoekt )

f ¢(2) = (2) 2 e 2 + 2(2)e 2

= ky

m = 8e Using point slope form,

48.

y - 4e 2 = 8e2 ( x - 2)

e x + 0.0001 - e x 0.0001 on a graphing calculator. A good choice for the viewing window is [-1, 4] by [-1,16] y =

y = 8xe 2 - 12e2

45.

f ( x) = xx e f ¢( x ) = = =

with Xscl = 1, Yscl = 2.

e x ⋅ (3x 2 ) - x3e x (e x )2 x 2e x (3 - x) (e x ) 2 x 2 (3 - x) ex

If we graph y = e x on the same screen, we see that the two graphs coincide. They are close enough to being identical that they are indistinguishable. By the definition of the

Find the slope at (-1, -e) f ¢(-1) =

(-1) 2 (3 - (-1)) e-1

derivative, if f ( x) = e x ,

m = 4e

Using point slope form, y + e = 4e( x + 1) y = 4ex + 3e 46.

f ( x) = f ¢( x ) = =

Graph

f ( x + h) - f ( x) h h0

f ¢( x) = lim

ex+h - ex , h h0

= lim

and h = 0.0001 is very close to 0.

x2 x2 ⋅ e x - e x ⋅ 2x ( x 2 )2

xe x ( x - 2) x4

Comparing the two graphs provides graphical evidence that f ¢( x) = e x .

e x ( x - 2) x3

Section 4.4 49.

275

Graph the function y = e x .

51.

900 - 800 ⋅ 1.1-x

C ( x) =

C ( x) = [900 - 800(1.1-x )]1/2 =

1 [900 - 800(1.1-x )]-1/2 2 ⋅ [-800(ln 1.1)(1.1-x )(-1)]

C ¢( x) =

Sketch the lines tangent to the graph at x = -1, 0, 1, 2.

(a)

(400 ln 1.1)(1.1-x ) 900 - 800(1.1-x )

C¢(0) =

400 ln 1.1 » 3.81 100

The marginal cost is $3.81. (b) C ¢(20) =

(400 ln 1.1)(1.1-20 ) 900 - 800(1.1-20 )

» 0.20

The marginal cost is $.20. (c) As x becomes larger and larger, C ¢( x) approaches zero.

Estimate the slopes of the tangent lines at these points.

52.

S (t ) = 100 - 90e-0.3t S ¢(t ) = -90(-0.3)e-0.3t

At x = -1 the slope is a little steeper than 13 or

= 27e-0.3t

approximately 0.3. At x = 0 the slope is 1.

(a)

S ¢(1) = 27e-0.3(1)

At x = 1 the slope is a little steeper than 52

= 27e-0.3

or 2.5.

» 20 thousand computers per year

At x = 2 the slope is a little steeper than 7 13

(b)

S ¢(5) = 27e-0.3(5)

or 7.3.

= 27e-1.5

Note that e-1 » 0.36787944, e0 = 1,

» 6 thousand computers per year

e1 = e » 2.7182812, and e 2 » 7.3890561. The values are close enough to the slopes of the

tangent lines to convince us that de = e x. dx 50.

(d)

S ¢(t ) = 27e-0.3t ¹ 0, but lim S ¢(t ) = lim 27e-0.3t = 0.

(a) The calculator gives a value for the

t ¥

derivative of 2 x at 0 of 0.6931. This agrees with the value given by the formula, which is ln 2 » 0.693147.

t ¥

Although the rate of change of sales never equals zero, it gets closer and closer to zero as t increases.

(b) The calculator gives a value for the

derivative of 3x at 0 of 1.099. This agrees with the value given by the formula, which is ln 3 » 1.098612. (c) The derivative of a x at x = 0 is equal to 1 when a is equal to 2.718, which is approximately equal to e = 2.71828... .

53.

A(t ) = 5000e0.0575t A¢(t ) = 0.0575(5000)e0.0575t = 287.5e0.0575t

(a)

A¢(1) = 287.5e0.0575(1) » 304.52 After 1 year, the compound amount is increasing by about $304.52 per year.

276

Chapter 4 CALCULATING THE DERIVATIVE (b)

A¢(5) = 287.5e0.0575(5)

56.

» 383.26 After 5 years, the compound amount is increasing by about $383.26 per year.

(c)

54.

A(t ) = 10t 2 2-t A¢(t ) = 10t 2 (ln 2)2-t (-1) + 20t 2-t A¢(t ) = 10t 2-t (-t ln 2 + 2)

A¢(10) = 287.5e0.0575(10)

(a)

» 510.93 After 10 years, the compound amount is increasing by about $510.93 per year.

(b)

A¢(2) = 10(2)(2-2 )(-2 ln 2 + 2) » 3.07 percent per month

A¢(4) = 10(4)(2-4 )(-4 ln 2 + 2) » -1.93 percent per month

A(r ) = 1000e3r /100 A¢(r ) = 0.03(1000)e3r /100

57.

= 30e0.03r

(a)

(b)

(c)

55.

A¢(2) = 30e0.03(2) » 31.86 At 2%, the compound amount is increasing by about $31.86 per percent. A¢(5) = 30e0.03(5) » 34.86 At 5%, the compound amount is increasing by about $34.86 per percent. A¢(10) = 30e0.03(10) » 40.50 At 10%, the compound amount is increasing by about $40.50 per percent.

(a) S c

(a)

G0 = 2, m = 3150, k = 0.0018 G(t ) = =

315 0 - 315(156.5e-0.2961t )(-0.2961) (1 + 156.5e-0.2961t )2 14,596.98975e-0.2961t (1 + 156.5e-0.2961t )2

1995 when t = 5: G(5) =

S(A) c de uA

G¢(5) =

315 1 + 156.5e-0.2961(5)

» 8.6

14,596.98975e-0.2961(5)

» 2.5 (1 + 156.5e-0.2961(5) ) 2 The number of Internet users in 1995 is about 8.6 million, and the growth rate is about 2.5 million users per year.

c–d A

(b) S ( A) = c - de-A S ¢( A) =  de- A

With c = 60,000, d = 50,000, and  = 0.0006, -(2500)(0.0006)

G(15) =

S (2500) = 60,000 - (50,000)e

= 48,843.49 = $48,843.

S ¢(2500) = (0.0006)(50,000)e-(2500)(0.0006) = 6.694

(The units are dollars of sales per dollar of advertising.)

G¢(15) =

-(4000)(0.0006)

S ¢(4000) = (0.0006)(50,000)e

315 1 + 156.5e-0.2961(15)

» 110.8

14,596.98975e-0.2961(15)

» 21.27 (1 + 156.5e-0.2961(15) )2 The number of Internet users in 2005 is about 110.8 million, and the growth rate is about 21.27 million users per year.

(d) 2015 when t = 25: G(25) =

= 55,464.10 = $55, 464. = 2.722

)

1 + 156.5e-0.2961t

(b) G¢(t ) = =

(

315 315 - 1 e-0.00094(315)t 2

G¢(25) =

315 1 + 156.5e-0.2961(25)

» 287.6

14,596.98975e-0.2961(25)

» 7.42 (1 + 156.5e-0.2961(25) )2 The number is Internet users in 2015 is about 287.6 million, and the growth rate is about 7.42 million users per year.

(e) The rate of growth increases for a while and then gradually decreases to 0.

Section 4.4 58.

277

S (t ) = 5000e0.1(e

0.25t

For the cactus wrens, k = 0.0125, G0 = 2.33, and m = 31.4.

)

Let g (t ) = 0.1e0.25t , with

m - G0 31.4 - 2.33 = = 12.4764 » 12.48 G0 2.33

g ¢(t ) = 0.1(e0.25t )(0.25) = 0.025e0.25t S ¢(t ) = 5000e0.1(e

0.25t

)

mt = (31.4)(0.0125) = 0.3925 » 0.393

(0.025e0.25t )

= 125e0.1(e

0.25t

) 0.25t

= 125e0.1(e

0.25t

) + 0.25t

S ¢(8) = 125e0.1(e

0.25.8

The model is G (t ) =

) + 0.25(8)

= 125e0.1(e ) + 2 » 1934

G¢(t ) =

The answer is b. 59.

A(t ) = 14.83(1.0159)

A¢(t ) = 14.83(ln 1.0159)(1.0159)

= 0.234(1.0159)t

A(25) = 14.83(1.0159) 25 » 22.0 The Asian population in the United States will be about 22,000,000 in 2025.

60.

= 50.3072e0.0158t

(a) For 2020, t = 60: A¢(60) = 50.3072e0.0158(60) » 129.8 In 2020, the world population is growing by about 129,800,000 people per year. (b) For 2025, t = 65: A¢(65) = 50.3072e0.0158(65) » 140.5 In 2025, the world population is growing by about 140,500,000 people per year. (a) The general logistic model is mG0 G(t ) = , or, G0 + (m - G0 )e-kmt G(t ) =

m . m - G0 -kmt 1+ e G0

( 2 (1 + 12.5e-0.393t )

-31.4 (12.5)(-0.393)e-0.393t

)

154.25e-0.393t 2

(1 + 12.5e-0.393t )

(d) G(10) = 25.2098 = 25.2 g G¢(10) = 1.9531 = 1.95 g/day (e) The growth rate of cactus wrens increases for a while and then gradually decreases. 62.

A(t ) = 3184e0.0158t

dividing by G0 ,

A¢(25) = 0.234(1.0159)25 » 0.0350 The Asian population in the United States will be increasing at the rate of 3,500,000 people per year at the end of the year 2025.

A¢(t ) = 3184e0.0158t (0.0158)

61.

1 + 12.5e-0.393t

(b) G(1) = 3.3270 = 3.33 g G ¢(1) = 1.1690 = 1.17 g/day

(a) For 2025, t = 25:

(b)

31.4

(a) The general logistic model is m . G(t ) = m - G0 -kmt 1+ e G0 For the whooping cranes, k = 6.01 ´ 10-4 and m = 787, so the model becomes 787 . G(t ) = 787 - G0 -0.0473t 1+ e G0 To find the complete model we solve 787 32 = for X , -(0.0473)(20) 1 + Xe obtaining æ 787 ö X = e0.946 çç - 1÷÷÷ = 60.8. çè 32 ø So our model is 787 . G(t ) = 1 + 60.8e-0.0473t The derivative, which we will need below, is

G ¢(t ) =

2263.3e-0.0473t

(1 + 60.8e-0.0473t )

278

Chapter 4 CALCULATING THE DERIVATIVE goes to 1100 cm3, which corresponds to a

(b) The initial population is G (0)  12.7 or about 13 cranes.

sphere with a radius of 3

(e) By the chain rule, dV = 1100(-4)[1023e-0.2415t + 1]-5 dt

(d) In 1985, t = 47.

⋅ (1023)(e-0.2415t )(-0.02415)

G(47) = 103.78 » 104 cranes G¢(47) = 4.262 » 4.26 cranes per year

= 108, 703.98[1023e-0.02415t + 1]-5 e-0.2415t

(e) In 2025, t = 87.

At t = 240,

G(67) » 395 cranes

(f) It increases for a while but eventually decreases to 0. -0.02415t

V (t ) = 1100[1023e

64.

-4

(a) V (240) = 1100[1023e

(a)

-4

r (3.857) = 3

P(25) = 0.00239e0.0957(25)

+ 1]

» 0.026%

P(50) = 0.00239e0.0957(50) » 0.286%

4 3 3V  r , so r (V ) = 3 3 4

V =

(b)

P(t ) = 0.00239e0.0957t

+ 1]

-0.02415(240)

dV » 0.282. dt

At 240 months old, the tumor is increasing in volume at the instantaneous rate of 0.282 cm3/month.

G ¢(67) » 9.3 cranes per year

» 3.857 cm

» 6.4 cm.

It makes sense that a tumor growing in a person’s body reaches a maximum volume of this size.

G(7) = 17.621 » 18 cranes G ¢(7) = 0.815 = 0.815 cranes per year

63.

3(1100) 4

P(75) = 0.00239e0.0957(75) » 3.130%

3(3.857) » 0.973cm 4

(b)

P¢(t ) = 0.00239e0.0957t (0.0957) = 0.000228723e0.0957t

(c)

P¢(25) = 0.000228723e0.0957(25)

V (t ) = 1100[1023e-0.02415t + 1]-4 = 0.5 [1023e-0.02415t + 1]-4 =

» 0.0025%/year

1 2200

P¢(50) = 0.000228723e0.0957(50)

(1023e-0.02415t + 1)4 = 2200 -0.02415t

1023e

» 0.0274%/year

1/ 4

P¢(75) = 0.000228723e0.0957(75)

+ 1 = 2200

» 0.300%/year

1023e-0.02415t = 22001/ 4 - 1 e-0.02415t =

22001/ 4 - 1 1023

(c) The percentage of people in each of the age groups that die in a given year is increasing as indicated by the answers in parts (a) and (b). A person who is 75 years old has a 3% chance of dying during year and the rate is increasing by almost 0.3%. The formula implies that everyone will be dead by age 112.

æ 22001/ 4 - 1 ö÷ ç ÷÷ -0.02415t = ln çç ÷÷ 1023 çè ø t =

æ 22001/ 4 - 1 ö÷ 1 ç ln çç ÷÷÷ » 214 months -0.02415 çè 1023 ÷ø

The tumor has been growing for almost 18 years. (d) As t goes to infinity, e-0.02415t goes to zero,

65.

s(t ) = 0.881(1.032)t s¢(t ) = 0.881(ln1.032)(1.032)t = 0.02775(1.032)t

Section 4.4

279

(a) For 35 years old, t = 35: s(35) = 0.881(1.032)

(e) » 2.7

2.7 cases per 1000 people s ¢(35) = 0.02775(1.032)35 » 0.084 The number of cases increases for 35-yearolds by about 0.084 cases per 1000 people for each additional year.

Growth is initially rapid, then tapes off.

(b) For 55 years old, t = 55:

(f )

s(55) = 0.881(1.032)55 » 5.0

Day 50 100 150 200 250 300

5.0 cases per 1000 people s ¢(55) = 0.02775(1.032)55 » 0.16 The number of cases increases for 55-yearolds by about 0.16 cases per 1000 people for each additional year.

Weight 991 2122 2734 2974 3059 3088

Rate 24.88 17.73 7.60 2.75 0.94 0.32

67.

9.4 cases per 1000 people s ¢(75) = 0.02775(1.032)75 » 0.29 The number of cases increases for 75-yearolds by about 0.29 cases per 1000 people for each additional year.

66.

W2 (t ) = 498.4(1 - 0.889e-0.00219t )1.25

(a) Both W1 and W2 are strictly increasing functions, so they approach their maximum values as t approaches ¥. lim W1(t ) = lim 509.7(1 - 0.941e-0.00181t )

-0.022(t -56)

M (t ) = 3102e-e

(a)

W1(t ) = 509.7(1 - 0.941e-0.00181t )

t ¥ -e-0.022(200 - 56)

t ¥

= 509.7(1 - 0) = 509.7

M (200) = 3102e

lim W2 (t ) = lim 498.4(1 - 0.889e-0.00219t )1.25

» 2974.15 grams, or about 3 kilograms.

t ¥

-0.022(t -56)

(b) As t gets very large, -e

t ¥

= 498.4(1 - 0)1.25 = 498.4

goes to

-0.022(t -56)

-e

goes to 1, and M(t) zero, e approaches 3102 grams or about 3.1 kilograms.

So, the maximum values of W1 and W2 are 509.7 kg and 498.4 kg respectively. (b) 0.9(509.7) = 509.7(1 - 0.941e-0.00181t ) 0.9 = 1 - 0.941e-0.00181t

0.1 » e-0.00181t 0.941 1239 » t

-0.022(t - 56)

2481.6 = 3102e-e

2481.6 = e-0.022(t -56) 3102 æ 3102 ö÷ = -0.022(t - 56) ln çç ln çè 2481.6 ÷÷ø - ln

0.9(498.4) = 498.4(1 - 0.889e-0.00219t )1.25 0.9 = (1 - 0.889e-0.00219t )1.25

æ 1 3102 ö÷ ln çç ln ÷ + 56 ç 0.022 è 2481.6 ÷ø » 124 days

t =-

(d)

(

)

-0.022(t - 56)

Dt -e-0.022(t -56)

-0.022(t - 56)

( -e-0.022(t -56) )(-0.022)

Dt M (t ) = 3102e-e = 3102e-e

1 - 0.90.8 = e-0.00219t 0.889 1095 » t

Respectively, it will take the average beef cow about 1239 days or 1095 days to reach 90% of its maximum.

-0.022(t - 56)

= 68.244e-e e-0.022(t -56) When t = 200, Dt M (t ) » 2.75 g/day.

280

Chapter 4 CALCULATING THE DERIVATIVE (c)

(b) G ¢(t ) = -(1 + 270e-3.5t )-2 ⋅ 270e-3.5t (-3.5)

W1¢ (t ) = (509.7)(-0.941)(-0.00181)e-0.00181t » 0.868126e-0.00181t

945e-3.5t

(1 + 270e-3.5t ) 2 1 » 0.109 G(1) = 1 + 270e-3.5(1)

W1¢ (750) » 0.868126e-0.00181(750) » 0.22 kg/day W2¢ (t ) = (498.4)(1.25)(1 - 0.889e-0.00219t )0.25

G ¢(1) =

-0.00219t

⋅ (-0.889)(-0.00219)e

» 1.21292e-0.00219t (1 - 0.889e-0.00219t )0.25

(

é 1 + 270e-3.5(1) ù 2 êë úû

» 0.341

The proportion is 0.109 and the rate of growth is 0.341 per century.

W2¢ (750) » 1.12192e-0.00219(750) ⋅ 1 - 0.889e-0.00219(750)

945e-3.5(1)

0.25

)

G(2) =

(c)

1 + 270e-3.5(2) » 0.802

» 0.22 kg/day

Both functions yield a rate of change of about 0.22 kg per day.

G¢(2) =

945e-3.5(2)

[1 + 270e-3.5(2) ]2 » 0.555

(d) Looking at the graph, the growth patterns of the two functions are very similar.

The proportion is 0.802 and the rate of growth is 0.555 per century. (d)

1 + 270e-3.5(3) » 0.993

G¢(3) =

(e) The graphs of the rate of change of the two functions are also very similar.

G(3) =

945e-3.5(2)

[1 + 270e-3.5(2) ]2 » 0.0256

The proportion is 0.993 and the rate of growth is 0.0256 per century. (e) The rate of growth increases for a while and then gradually decreases to 0. 68.

R(c) = 3.19(1.006c )

H ( N ) = 1000(1 - e-kN ), k = 0.1

70.

dR dc = 3.19(ln 1.006)(1.006c ) dt dt

H ¢ = -1000e-0.1N (-0.1) = 100e-0.1N

If c = 180 and dc = 15, then dt dR = 3.19(ln 1.006)(1.006180 )(15) » 0.840 dt

(a)

» 36.8 (b)

69.

(a)

H ¢(10) = 100e-0.1(10) H ¢(100) = 100e-0.1(100)

G0 = 0.00369, m = 1, k = 3.5 G(t ) =

1 æ ö 1 - 1÷÷÷ e-3.5(1)t 1 + çç çè 0.00369 ø 1 -3.5t

1 + 270e

» .00454

(c)

H ¢(1000) = 100e-0.1(1000) »0

(d) 100e-0.1N is always positive since powers of e are never negative. This means that repetition always makes a habit stronger.

Section 4.4 71.

281

G0 = 1.603, m = 6.8, k = 0.0440

(a)

G(t ) = =

(

6.8

)

6.8 - 1 e-0.0440(6.8)t 1.603

6.8 1 + 3.242e-0.2992t

(b) For 2006, t = 4: G(4) =

6.8 1 + 3.242e-0.2992(4)

G(4) =

» 3.435

6.59604e-0.2992(4)

(

1 + 3.242e-0.2992(4)

» 0.509

)

In 2006, about 3.435 million students enrolled in at least one online course and the growth rate is about 0.509 million students per year. (c) For 2013, t = 11: G(11) = G¢(11) =

6.8

» 6.068

1 + 3.242e-0.2992(11) 6.59604e-0.2992(11)

(1 + 3.242e

-0.2992(11)

)

» 0.195

In 2013, about 6.068 million students enrolled in at least one online course and the growth rate is about 0.195 million students per year. (d) For 2020, t = 18: G(18) = G(18) =

6.8 1 + 3.242e-0.2992(18)

» 6.700

6.59604e-0.2992(18)

(

1 + 3.242e-0.2992(18)

)

» 0.029

In 2020, about 6.700 million students enrolled in at least one online course and the growth rate is about 0.029 million students per year. (e) The growth rate increases at first and then decreases toward 0.

72.

A(t ) = 500e-0.31t

(c)

A¢(10) = -155e-0.31(10)

A¢(t ) = 500(-0.31)e-0.31t = -155e-0.31t

(a)

= -155e-3.1 » -6.98 grams per year

A¢(4) = -155e-0.31(4) = -155e-1.24

(d)

» -44.9 grams per year

(b)

A¢(6) = -155e

» -24.1 grams per year

t ¥

As the number of years increases, the rate of change approaches zero.

-0.31(6)

= -155e-1.86

lim A¢(t ) = lim - 155e-0.31t = 0

t ¥

(e)

A¢(t ) will never equal zero, although as t increases, it approaches zero. Powers of e are always positive.

282 73.

Chapter 4 CALCULATING THE DERIVATIVE 75. Using the given value of T = 30, æ 1 é ù 1ö - ÷ 5417.7530 ⋅çç ú çè 273.16 D ÷÷ø 5 êê - 10 ú H ( D ) = 30 + 6.11e ê ú 9ê ú ë û

Q(t ) = CV (1 - e-t /RC )

(a)

Ic =

é æ 1 ö÷ ù dQ ú = CV ê 0 - e-t /RC çç çè RC ÷÷ø ú ê dt ë û

æ 1 ö÷ -t /RC = CV çç e çè RC ø÷÷

H ¢( D) =

(b) When C = 10-5 farads, R = 107 ohms, and V = 10 volts, after 200 seconds

(a) H (278) = 29.239 » 29.24 H ¢(278) = 0.3361

7 -5 10 I c = 7 e-200 /(10 .10 ) 10

(b) H (290) = 35.183 » 35.18 H ¢(290) = 0.6918

» 1.35 ´ 10-7 amps

(a) H (T )

= T - 0.9971e0.02086T éê 1 - e0.445(85-57.2) ùú ë û = T + 2.4385e0.02086T

(b)

76.

H (80) = 80 + 2.4385e0.0286(80)

The heat index when the temperature is 80°F is 92.94°F. H ¢(T ) = 1 + 2.4385e0.2086T (0.02086)

If h = 1000 and

H ¢(80) = 1 + 2.4385e0.2086(80) (0.02086) = 1.2699

(a)

(

s(0) = 693.9 - 34.38 e0.01003(0) + e-0.01003(0)

dh = 800, then dt

dT = -0.0052e-0.000065(1000) (800) dt » -3.90 The temperature is decreasing at 3.90 degrees/hr.

The rate of change of the heat index when the temperature is 80°F is 1.2699°F per degree increase in the temperature. 77.

T (h) = 80e-0.000065h æ dT dh ö = 80e-0.000065h çç -0.000065 ÷÷÷ ç è dt dt ø dh = -0.0052e-0.000065h dt

» 92.94

(c)

æ 1 1ö - ÷÷ 5417.7530 ⋅ççç è 273.16 D ÷ø

⋅e D2 where 18,390.2616 is approximately equal to (5/9)(6.11)(5417.7530)

V = e-t /RC R

74.

18,390.2616

)

= 693.9 - 34.38(2) = 625.14

The height of the Gateway Arch at its center is 625.14ft. (b)

m = s ¢(t ) = -34.38(e0.01003x (0.01003) + e-0.01003x (-0.01003)) = -34.38(0.1003e0.01003x - 0.01003e-0.01003x ) m = s ¢(0) = -34.38(0.01003(1) - 0.01003(1)) = 0

The slope of the line tangent to the curve when x = 0 is 0, which makes sense because the tangent line has to be a horizontal line at the center of the arch (the highest point on the curve) (c)

(

s¢(150) = -34.38 0.01003e0.01003(150) - 0.01003e-0.01003(150)

)

= -1.476

The rate of change of the height of the arch at x = 150 ft is -1.476 ft per foot from the center.

Section 4.5 78.

283

f (t ) = (5.4572 ´ 10-59 ) ⋅ 100.02923t

Your Turn 3

f ¢(t ) = (3.6729 ´ 10-60 ) ⋅ 100.02923t

(a)

y = ln |2 x + 6| dy 1 d = ⋅ (2 x + 6) dx 2 x + 6 dx 2 2 1 = = = 2x + 6 2( x + 3) x+3

where 3.6729 ´ 10-60 is approximately equal to ln(10) ⋅ (5.4572 ´ 10-59 ) ⋅ (0.02923) f (2015) = 4.319 deaths per 100 students f ¢(2015) = 0.2907 deaths per 100 students per year

79.

t (n) = 218 + 31(0.933)n t (70) = 218 + 31(0.933)70 » 218.2 sec

(a)

(b) t ¢(n) = (31 ln 0.933)(0.933) n t ¢(70) = (31 ln 0.933)(0.933)70 » -0.01675

(b) f ( x) = x 2 ln 3x 1 (3) + ln 3x(2 x) f ¢( x ) = x 2 ⋅ 3x = x + 2 x ln 3x (c) s(t ) =

s ¢(t ) =

The record is decreasing by 0.01675 seconds per year at the end of 2020.

ln(t 2 - 1) t +1 (t + 1)

1 t 2 -1

(t + 1)2

= t -1

( ) (2t) - ln(t - 1)(1)

- ln(t 2 - 1) (t + 1)2

2t - (t - 1) ln(t 2 - 1) (t - 1)(t + 1)2

If the estimate is correct, then this is the least amount of time that it will ever take a human to run a mile.

4.5 Warmup Exercises

4.5 Derivatives of Logarithmic Functions

W1.

f ( x) = e x 2 æ d ö 2 f ¢( x ) = çç x 2 ÷÷÷ e x = 2 xe x çè dx ø

W2.

f ( x ) = x 2e4 x æ d ö æ d ö f ¢( x ) = x 2 çç e4 x ÷÷÷ + çç x 2 ÷÷÷ e4 x çè dx ç ø è dx ø

Your Turn 1 f ( x) = log3 x f ¢( x ) =

1 (ln 3) x

(

Your Turn 2 (a)

y = ln(2 x3 - 3) dy 1 d = ⋅ (2 x3 - 3) 3 dx 2 x - 3 dx = =

1 3

2x - 3

= 2 x ( 2 x + 1) e4 x

W3.

f ¢( x ) =

2 x3 - 3

(b) f ( x) = log 4 (5 x + 3x3 )

f ( x) =

(6 x 2 )

6x2

f ¢( x ) =

)

= x 2 4e4 x + ( 2 x ) e4 x

1 ⋅ d (5x + 3x3 ) (ln 4)(5x + 3x3 ) dx

e2 x e3 x + 1 (2eex )(e3x + 1) - (e2 x )(3x3x ) (e3x + 1)2 e2 x (2e3x + 2 - 3e3x ) (e3x + 1) 2 e2 x (2 - e3x )

5 + 9x2 (ln 4)(5x + 3x3 )

(e3x + 1) 2

284

Chapter 4 CALCULATING THE DERIVATIVE

4.5 Exercises

10.

y = ln |-8 x3 + 2 x | g ( x ) = -8 x 3 + 2 x

True

False. The derivative of f ( x) = log x is 1 f ¢( x ) = . (ln10) x

g ¢( x) = -24 x 2 + 2 -24 x 2 + 2 -12 x 2 + 1 dy = = dx -8 x 3 + 2 x -4 x 3 + x

True

g ( x) =

x + 5 = ( x + 5)1/2

y = ln (8 x) dy d = (ln 8x) dx dx d = (ln 8 + ln x) dx d d = (ln 8) + (ln x) dx dx 1 1 =0+ = x x

g ¢( x ) =

1 ( x + 5)-1/2 2

11.

1 ( x + 5)-1/2 dy 1 = 2 = 1/2 dx 2( x + 5) ( x + 5)

12.

g ¢( x ) =

13.

é 2 x(2 x 2 + 5) ù ú = 3 êê 2 2 2 êë x ( x + 5) úúû 3(2 x 2 + 5) = x( x 2 + 5)

-3 dy g ¢( x) 3 = = or dx g ( x) 8 - 3x 3x - 8 y = ln (1 + x3 )

3x 2 dy g ¢( x) = = dx g ( x) 1 + x3

y = ln | 4 x 2 - 9 x | 2

g ( x) = 4 x - 9 x g ¢( x) = 8 x - 9 dy g ¢( x) 8x - 9 = = dx g ( x) 4x2 - 9x

y = ln( x 4 + 5 x 2 )3/2 = 3 ln ( x 4 + 5 x 2 ) 2 dy = 3 Dx[ln ( x 4 + 5 x 2 )] dx 2 g ( x) = x 4 + 5 x 2 g ¢( x) = 4 x3 + 10 x æ 3 ö dy = 3 ççç 4 x4 + 102x ÷÷÷ dx 2 çè x + 5x ÷ø

y = ln (8 - 3x) g ( x) = 8 - 3 x

g ¢( x) = 3x 2

1 (2 x + 1)-1/2 (2) = (2 x + 1)-1/2 2

dy (2 x + 1)-1/2 1 = = 1/2 dx 2 x +1 (2 x + 1)

y = ln (-4 x) dy d = [ln (-4 x)] dx dx d = [ln 4 + ln(-x)] dx d d = ln 4 + ln(-x) dx dx -1 1 =0+ = -x x

g ( x) = 1 + x 3

y = ln 2 x + 1 = ln(2 x + 1)1/2 g ( x) = (2 x + 1)1/2

g ¢ ( x ) = -3

y = ln x + 5

14.

y = ln (5x3 - 2 x)3/2 = 3 ln (5 x3 - 2 x) 2 dy 3 = Dx ln (5 x3 - 2 x) dx 2 g ( x) = 5 x 3 - 2 x g ¢( x) = 15x 2 - 2 3 æ ö dy = 3 ççç 153x - 2 ÷÷÷ 2 è 5 x - 2 x ÷ø dx =

3(15x 2 - 2) 2(5 x3 - 2 x)

Section 4.5 15.

285

y = -5 x ln (3x + 2)

20.

v =

ln u u3

Use the product rule.

Use the quotient rule.

é d ù é d ù dy = -5 x ê ln(3x + 2) ú + ln(3x + 2) ê (-5 x) ú ê ú ê úû dx ë dx û ë dx æ 3 ö÷ = -5 x çç + [ln(3x + 2)](-5) çè 3x + 2 ÷÷ø

u dv = du =

15 x =- 5ln(3x + 2) 3x + 2

16.

y = (3x + 7) ln (2 x - 1)

Use the product rule. æ 2 ö÷ dy = (3x + 7) çç + [ln (2 x - 1)](3) çè 2 x - 1 ÷÷ø dx 2(3x + 7) = + 3 ln (2 x - 1) 2x - 1

17.

18.

21.

(u 3 ) 2

u 2 - 3u 2 ln u u6 u 2 (1 - 3ln u) u6 1 - 3ln u u4

ln x 4x + 7

Use the quotient rule. (4 x + 7) ( 1x ) - (ln x)(4) dy = dx (4 x + 7)2

s = t 2 ln | t | ds 1 = t 2 ⋅ + 2t ln | t | dt t = t + 2t ln | t | = t (1 + 2 ln | t |)

= =

4 x + 7 - 4 ln x x 2

(4 x + 7) 4 x + 7 - 4 x ln x x(4 x + 7) 2

y = x ln 2 - x 2

22.

Use the product rule. æ 1 ö÷ dy = x çç ÷ (-2 x) + ln |2 - x 2 | dx èç 2 - x 2 ÷ø =-

19.

y =

( ) - (ln u)(3u 2 )

3 1 u

y =

2x2 2 - x2

( )

(3x - 1)(-2) 1x - (-2 ln x)(3) dy = dx (3x - 1) 2

+ ln |2 - x |

2 ln ( x + 3) x

Use the quotient rule. x dy = dx

( x +2 3 ) - 2 ln ( x + 3) ⋅ 2x

= =

-2(3x -1) + 6 ln x x 2

(3x - 1) -2(3x - 1) + 6 x ln x x(3x - 1) 2 -2(3x - 1 - 3x ln x) x(3x - 1)2

2 2

(x ) 2

= x +3 =

-2 ln x 3x - 1 Use the quotient rule. y =

- 4 x ln ( x + 3)

23.

2 x 2 - 4 x( x + 3) ln ( x + 3) x 4 ( x + 3) x[2 x - 4( x + 3) ln ( x + 3)] x 4 ( x + 3) 2 x - 4( x + 3) ln ( x + 3)

y =

3x 2 ln x

(ln x)(6 x) - 3x dy = dx (ln x)2 =

6 x ln x - 3x

x3 ( x + 3)

(ln x)2

( )

2 1 x

286 24.

Chapter 4 CALCULATING THE DERIVATIVE y =

x3 - 1 2 ln x

29. 2

(2 ln x)(3x ) - ( x - 1)(2) dy = dx (2 ln x)2 = = =

25.

6 x 2 ln x - 2x ( x3 - 1) (2 ln x) 2

( 1x )

x ⋅ x

= e

30.

6 x3 ln x - 2( x3 - 1) 4 x(ln x)2 3x3 ln x - ( x3 - 1) 2 x(ln x)2

y = (ln | x + 1|)4

y = e x ln x, x > 0 2æ 2 ö dy = e x çç 1 ÷÷÷ + (ln x)(2 x)e x èxø dx

31.

æ 1 ö÷ dy = 4(ln | x + 1|)3 çç çè x + 1 ÷÷ø dx

x y = e ,x>0 ln x Use the quotient rule.

(ln x)e x - e x dy = dx (ln x) 2

32.

p ¢( y ) =

2( x - 3)(ln | x - 3|)1/2 1 = 2( x - 3) ln | x - 3| y = ln |ln x | g ( x) = ln x 1 g ¢( x) = x

= = =

33.

1 x

ln x 1 = x ln x y = (ln 4)(ln |3x |) æ 1 ö dy = (ln 4) çç ÷÷÷ (3) çè 3x ø dx 3ln 4 ln 4 = 3x x (Recall that ln 4 is a constant.) =

xe ln x - e

( 1x ) ⋅ x x

x(ln x) 2

ln y ey e y ⋅ 1y - (ln y)e y (e y ) 2 e y - y(ln y)e y ye 2 y e y (1 - y ln y) ye2 y 1 - y ln y ye y

g ( z ) = (e 2 z + ln z )3 æ ö g ¢( z ) = 3(e 2 z + ln z )2 çç e2 z ⋅ 2 + 1 ÷÷÷ è zø 2z æ ö = 3(e 2 z + ln z )2 ççç 2 ze + 1 ÷÷÷ è ø z

dy g ¢( x) = dx g ( x)

28.

p( y ) =

ln | x - 3| = (ln | x - 3|)1/2

y =

dy 1 d = (ln | x - 3|)1/2 ⋅ (ln | x - 3|) dx 2 dx æ 1 ö÷ 1 ç = 1/2 ççè x - 3 ÷ø÷ 2(ln | x - 3|)

27.

+ 2 xe x ln x

y = e 2 x-1 ln(2 x - 1) dy = (2)e2 x -1 ln(2 x - 1) dx æ ö + (e 2 x -1) çç 1 ÷÷÷ (2) è 2x - 1 ø 2 x -1 = 2e 2 x -1 ln (2 x - 1) + 2e 2x - 1

4(ln | x + 1|)3 = x +1

26.

34.

e-t + ln 2t = (e-t + ln 2t )1/2 æ ö 1 1 s¢(t ) = (e-t + ln 2t )-1/2 çç e-t (-1) + (2) ÷÷÷ ç è ø 2 2t s (t ) =

æ ö = 1 (e-t + ln 2t )-1/2 çç -e-t + 1 ÷÷÷ è 2 tø -t æ ö = 1 (e-t + ln 2t )-1/2 ççç -te + 1 ÷÷÷ è ø 2 t =

1 - te-t 2t e-t + ln 2t

Section 4.5 35.

287

y = log(6 x) g ( x) = 6 x and g ¢( x) = 6. æ ö dy = 1 çç 6 ÷÷÷ dx ln10 è 6 x ø =

36.

y = log3 ( x 2 + 2 x)3/2

41.

g ( x) = ( x 2 + 2 x)3/2 and g( x) =

1 x ln10

= 3( x + 1)( x 2 + 2 x)1/2 dy 1 3( x + 1)( x 2 + 2 x)1/2 = ⋅ dx ln 3 ( x 2 + 2 x)3/2

y = log(4 x - 3) g ( x) = 4 x - 3

g( x) = 4 dy 4 = 1 ⋅ dx ln10 4 x - 3 4 = (ln10)(4 x - 3)

37.

g ¢( x ) =

1 (ln10)( x - 1)

g ( p) = 2 p - 1

y = log |3x |

g ¢( p) = (ln 2)2 p dw 1 (ln 2)2 p = ⋅ p dp ln 8 (2 - 1)

dy = 1 ⋅ 3 = 1 dx ln10 3x x ln10

40.

(ln 2)2 p

y = log5 5x + 2 5x + 2 and g ¢( x) =

1 dy = ⋅ ln 5 dx

5(4 x - 1) (2 ln 2)(2 x 2 - x)

w = log8 (2 p - 1)

43.

g ( x) = 3x and g ¢( x) = 3.

g ( x) =

5 (2 x 2 - x)3/2 ⋅ (4 x - 1). 2

5 (2 x 2 - x)3/2 ⋅ (4 x - 1) dy 1 = ⋅ 2 dx ln 2 (2 x 2 - x)5/2

dy -1 1 = ⋅ dx ln10 1 - x 1 =(ln 10)(1 - x)

39.

(ln 3)( x 2 + 2 x)

g ( x) = (2 x 2 - x)5/2 and

g ( x) = 1 - x and g ¢( x) = -1.

38.

3( x + 1)

y = log 2 (2 x 2 - x)5/2

42.

y = log|1 - x |

3 2 ( x + 2 x)1/2 ⋅ (2 x + 2) 2

5 . 2 5x + 2

z = 10 y log y

44.

5 2 5x + 2

(ln 8)(2 p - 1)

g ( y ) = 10 y and g ¢( y) = (ln10)10 y. dz 1 = 10 y ⋅ + log y ⋅ (ln10)10 y dy (ln10) y

5x + 2

5 2 ln 5(5 x + 2)

10 y + (log y)(ln10)10 y (ln10) y

y = log 7 4 x - 3

45.

g ( x) =

4x - 3

g ¢( x ) =

4 = 2 4x - 3

dy 1 = ⋅ dx ln 7

2 4 x -3

4x - 3 2 = (ln 7)(4 x - 3)

2 4x - 3

f ( x) = e x ln( x + 5)

Use the product rule. f ¢( x ) = e

1 æ ö ç 2 x ÷÷ æ ö ÷÷ + [ln( x + 5)]e x çç 1 ÷÷ çç x + 5 ÷÷ çè 2 x ÷ø÷ çè ø÷

x çç

1 ln( x + 5) ùú e x éê + ú 2 êëê x ( x + 5) x úû

288 46.

Chapter 4 CALCULATING THE DERIVATIVE f ( x) = ln( xe x + 2) g ( x) = xe x + 2 é æ 1 ÷ö ù e x x e x ÷÷ úú + e x (1) = g ¢( x) = x êê e x çç +e x = ( x + 2) çè 2 x ÷ø 2 2 ë û g ¢( x) f ¢( x ) = = g ( x)

47.

f (t ) =

e x ( 2

x + 2)

xe x + 2

e x ( x + 2) 2( xe x + 2)

ln(t 2 + 1) + t ln(t 2 + 1) + 1

Use the quotient rule. u (t ) = ln(t 2 + 1) + t , u ¢(t ) = v(t ) = ln(t 2 + 1) + 1, v¢(t ) =

f ¢t ) =

= = =

48.

[ln(t 2 + 1) + 1]

(

t +1 2t

t2 + 1

2t + 1 t 2 +1 2

) - [ln(t + 1) + t]( ) 2

2t t 2 +1

[ln(t + 1) + 1]2

2 2t ln(t 2 +1) 2t ln(t 2 +1) 2t - 2t 2 + ln(t 2 + 1) + 1 + 22t + ln(t 2 + 1) + 1 - 22t 2 t 2 +1 t +1 t 2 +1 t +1 = t +1 2 2 2 2

[ln(t + 1) + 1]

2t - 2t + (t + 1) ln(t + 1) + t + 1 (t 2 + 1)[ln(t 2 + 1) + 1]2 -t 2 + 2t + 1 + (t 2 + 1) ln(t 2 + 1) (t 2 + 1)[ln(t 2 + 1) + 1]2 2t 3/2

f (t ) =

ln (2t 3/2 + 1) Use the quotient rule. u(t ) = 2t 3/2 , u ¢(t ) = 3t1/2 v(t ) = ln(2t 3/2 + 1), v¢(t ) =

f ¢(t ) =

49.

2t 2

ln(2t

3/2

3t1/2 2t

3/2

é t1/2 ù 6t 2 + 1)(3t ) - 2t 3/2 ê 33/2 (3t1/2 ) ln(2t 3/2 + 1) - 3/2 ú (6t 2 + 3t1/2 ) ln(2t 3/2 + 1) - 6t 2 2t +1 ë 2t +1 û = = 3/2 2 3/2 2 3/2 3/2 2 1/2

[ln(2t

+ 1)]

[ln(2t

f ( x) = ln 5x - 3 = ln(5x - 3)1/ 2 =

+ 1)]

1 ln(5 x - 3) 2

g ( x) = 5 x - 3 and g ¢( x) = 5. 1 5 5 f ¢( x ) = ⋅ = 2 5x - 3 2(5x - 3)

50.

g ( x) = ln 3x 2 - 7 = ln(3x 2 - 7)1/ 2 =

1 ln(3x 2 - 7) 2

h( x) = 3x 2 - 7 and h¢( x) = 6 x. 1 6x 3x g ¢( x ) = ⋅ 2 = 2 2 3x - 7 3x - 7

(2t

+ 1)[ln(2t

+ 1)]

Section 4.5 51.

289

h( x) = ln(2 x3 + 5 x 2 )3/ 2 =

3 ln(2 x3 + 5x 2 ) 2

g ( x) = 2 x3 + 5x 2 and g ¢( x) = 6 x 2 + 10 x. h ¢( x) =

52.

3 6 x 2 + 10 x 3(3x 2 + 5x) ⋅ 3 = 2 2 x + 5x2 2 x3 + 5x 2

f (t ) = ln(3t 2 + 7t - 4)5/3 =

5 ln(3t 2 + 7t - 4) 3

g (t ) = 3t 2 + 7t - 4 and g ¢(t ) = 6t + 7. 5 6t + 7 5(6t + 7) f ¢(t ) = ⋅ 2 = 3 3t + 7t - 4 3(3t 2 + 7t - 4)

53.

f ( x) = ln[ x 2 (e x + 1)] = ln x 2 + ln(e x + 1) = 2 ln x + ln(e x + 1) g ( x) = e x + 1 and g ¢( x) = e x . f ¢( x ) = 2 ⋅

54.

1 ex 2 ex 2e x + 2 + xe x + x = + x = x x e +1 e +1 x(e x + 1)

g ( x) = ln[ x5 ( x 2 + 5)10 ] = ln x5 + ln( x 2 + 5)10 = 5ln x + 10 ln( x 2 + 5) h( x) = x 2 + 5 and h¢( x) = 2 x. g ¢( x ) = 5 ⋅

55.

56.

æé ö ç (t + 4) 2 ùú ÷÷ 2 h(t ) = ln çç êê ú ÷÷ø = ln(t + 4) - ln t = 2 ln(t + 4) - ln t çè ë t û t-4 1 1 2 1 h¢(t ) = 2 ⋅ - = - = t+4 t t+4 t t (t + 4) é x2 - 4 ù ú = ln( x 2 - 4) - ln x3 = 2 ln(t + 4) - 3ln x r ( x) = ln êê ú 3 ë x û 2 g ( x) = x - 4 and g ¢( x) = 2 x. r ¢( x) =

57.

1 2x 5 20 x 25x 2 + 25 + 10 ⋅ 2 = + 2 = x x x +5 x +5 x( x 2 + 5)

2x x2 - 4

-3⋅

-x 2 + 12 1 2x 3 = 2 - = x x x -4 x( x 2 - 4)

y = ln( x 2 + 2) dy 2x = 2 dx x +2 Find the slope at (1, ln 3) 2(1) 2 m= 2 = 3 1 +2 Using the point slope form, 2 y - ln 3 = ( x - 1) 3 2 2 y = x - + ln 3 3 3

290

Chapter 4 CALCULATING THE DERIVATIVE

58.

y = ln( x + 5)2 = 2 ln( x + 5) 2 dy = dx x+5 Find the slope at (–1, ln 16) 2 1 m= = -1 + 5 2 Using the point slope form, 1 y - ln16 = ( x + 1) 2 1 1 y = x + + ln16 2 2

60.

Note that a is a constant. d d ln | ax | = (ln | a | + ln | x | ) dx dx d d = ln | a | + ln x dx dx d = ln | x | dx

Therefore, d d ln | ax | = ln | x |. dx dx

62.

Graph y =

ln | x + 0.0001| - ln | x | 0.0001

on a graphing calculator. A good choice for the viewing window is [-3, 3].

If we graph y = 1x on the same screen, we see that the two graphs coincide. By the definition of the derivative, if f ( x) = ln | x |, f ( x + h) - f ( x ) ln | x + h | - ln | x | = lim , h h h 0 h0

f ¢( x) = lim

and h = 0.0001 is very close to 0. Comparing the two graphs provides graphical evidence that f ¢( x ) =

1 . x

Section 4.5 63.

291

Use the derivative of ln x. d ln[u ( x)v( x)] 1 d [u ( x)v( x)] = ⋅ dx u ( x)v( x) dx d ln u ( x) 1 d [u ( x)] = ⋅ dx u ( x) dx d ln v( x) 1 d [v( x)] = ⋅ dx v( x) dx

Then since ln[u( x)v( x)] = ln u( x) + ln v( x), 1 d[u ( x)v( x)] 1 d[u ( x)] 1 d [v( x)] ⋅ = ⋅ + ⋅ . u ( x)v( x) dx u ( x) dx v( x) dx

Multiply both sides of this equation by u( x)v( x). Then

d [u( x)v( x)] d [u( x)] d [v( x)] = v ( x) + u ( x) . dx dx dx

This is the product rule. 64.

Use the derivative of ln x. u ( x) é u ( x) ù d éê v( x) ùú v( x) d êë v( x) úû ë û = u ( x) ⋅ = ⋅ dx u ( x) dx

u ( x)

d ln v( x)

v( x )

1 d [u( x)] ⋅ u ( x) dx 1 d[v( x)] ⋅ d ln v( x) = v( x) dx

d ln u ( x) =

u ( x)

Then, since ln v( x) = ln u ( x) - ln v( x), é u ( x) ù v( x) d êë v( x) úû d [u( x)] d [v( x)] 1 1 ⋅ = ⋅ ⋅ . u ( x) dx u ( x) dx v( x ) dx u ( x)

Multiply both sides of this equation by v( x) . Then u ( x) d éê v(x) ùú ë û = 1 ⋅ d[u ( x)] - u ( x) ⋅ d [v( x)] dx v( x ) dx dx [v( x)]2

= =

v ( x) 2

[v( x)] v( x ) ⋅

⋅

d[u ( x)] u ( x) d [v( x)] ⋅ 2 dx dx [v( x)]

d [u ( x)] d [v( x)] - u ( x) ⋅ dx dx 2

[v( x)]

This is the quotient rule.

292

Chapter 4 CALCULATING THE DERIVATIVE

65. Graph the function y = ln x. Sketch lines tagent

d ln h( x) = (b) Since dx , h( x )

h ¢( x )

to the graph at x = 12 ,1, 2,3, 4. Estimate the slopes

d ln h( x) dx é v( x)u ¢( x) ù = h( x) ê + (ln u ( x))v¢( x) ú ê u ( x) ú ë û

of the tangent lines at these points.

h¢( x) = h( x)

é v( x)u ¢( x) ù = u ( x) v ( x ) ê + (ln u ( x))v¢( x) ú ê u ( x) ú ë û

68.

u ( x) = x, u ¢( x) = 1 v( x) = x, v¢( x) = 1

slope of tangent

x 1 2

1 2 1 3 1 4

3 4

é x(1) ù h ¢( x) = x x ê + (ln x)(1) ú êë x úû = x x (1 + ln x)

69.

ln x = 1 . Thus we see that d dx x

The change-of-base theorem for logarithms states ln x . Find the derivative of each side. log a x = ln a ln a ⋅ d log a x = dx

d ln x - ln x ⋅ d ln a dx dx 2

(ln a )

ln a ⋅ 1x - ln x ⋅ 0 = 2 (ln a )

67.

(a)

1 x

ln a

1 x ln a

h( x ) = u ( x ) v ( x ) d d ln h( x) = ln[u ( x)v( x) ] dx dx d = [v( x) ln u ( x)] dx d = v( x) ln u ( x) + (ln u ( x))v¢( x) dx u ¢( x) = v( x ) + (ln u( x))v¢( x) u ( x) v( x)u ¢( x) = + (ln u ( x))v¢( x) u ( x)

h( x) = ( x 2 + 1)5 x u ( x) = x 2 + 1, u ¢( x) = 2 x v( x) = 5 x, v¢( x) = 5

The values of the slopes at x are 1x .

66.

h( x ) = x x

70.

é 5x(2 x) ù + ln( x 2 + 1) ⋅ (5) ú h¢( x) = ( x 2 + 1)5 x ê 2 êx +1 ú ë û é 10 x 2 ù = ( x 2 + 1)5 x êê 2 + 5ln( x 2 + 1) úú êë x + 1 úû (a) æ x + h ö÷ ln çç çè x ÷÷ø d (ln x ) ln( x + h) - ln( x ) = lim = lim dx h h h0 h0 1/ h ö ææ æ1 æ x öö xö ÷ ç = lim çç ln çç 1 + ÷÷÷ ÷÷÷ = lim ln çç çç 1 + ÷÷÷ ÷÷÷ h ø ø÷ h ø ÷ø÷ h  0 èç h èç h  0 ççè çè

(b) Substituting h = x / m with m  ¥ for fixed x this last limit becomes 1/ x m/ x ù mù éæ éæ h ö÷ h ÷ö ú ê ú ê ç ç lim ln ê ç 1 + ÷÷ ú = lim ln ê ççè 1 + ÷÷ø ú . xø x ú m ¥ ê çè m ¥ ê úû ë ë û

that x a is continuous, 1/ x 1/ x ù éæ é ö÷m ùú æ h hö ê lim ln ê çç 1 + ÷÷ ú = ln êê lim çç 1 + ÷÷÷ úú mø ú mø ú m ¥ ê èç êë m ¥èç ë û û 1/ x ù éæ æ 1 h öö ú ê = ln ê çç lim çç1 + ÷÷÷ ÷÷÷ ú = ln e1/ x = . ç ÷ ç m øø ú x ê è m ¥ è ë û

Section 4.5 71.

293

p = 100 +

(a)

(d) The manager can use the information from part (c) to decide whether it is profitable to make and sell additional items.

50 ,q >1 ln q

R = pq R = 100q +

50q ln q

73.

C ( x) =

The marginal revenue is dR = 100 + dq = 100 +

(ln q)(50) - 50q (ln q) 50(ln q - 1) (ln q)

C ¢( x ) =

( ) 1 q

for q = 8. 50(ln 8 - 1) (ln 8)2

5 - (ln 2)(5 log 2 10 + 10)

(b) C ¢(20) =

5 - (ln 2)(5 log 2 20 + 10)

74.

(a) The marginal revenue is æ ö÷ 1 60 ÷÷ (2) = R¢( x) = 30 ççç 2x + 1 è ln(2 x + 1) ø÷

C ¢(q) = 100

(b) The profit function is P( x) = R( x) - C ( x)

P (q ) = R (q ) - C ( q) æ 50 ö÷ ÷ - (100q + 100) = q ççç100 + ln q ÷ø÷ èç

= 30 ln(2 x + 1) -

P¢( x) = R ¢( x) - C ¢( x) 60 1 = 2x + 1 2 ¢ P (60) = 60 - 1

2(60) +1

(1)

(d) When 60 items are manufactured, there is not profit from selling additional items.

50 ln q - 50

75.

(ln q)2

50(ln 8 - 1) (ln 8)2

æ N (t ) ö÷ 2.54197 - 0.2167t ÷÷ = 9.8901e-e ln ççç ÷ çè N 0 ø

(a)

When q = 8, the profit from one more unit is

60 - 1 » 0 = 121 2

(ln q)(50) - 50q q dP = dq (ln q)2 (ln q)2 50(ln q - 1)

x 2

50q - 100q - 100 ln q

50q = - 100 ln q

(102 ) ln 2 » -$0.19396

R( x) = 30 ln(2 x + 1) x C ( x) = 2

(a) The marginal cost is given by C ¢(q).

= 100q +

x 2 ln 2

(202 ) ln 2 » -$0.06099

C (q) = 100q + 100

(b)

x2 5 - (ln 2)(5log 2 x + 10)

= $112.48

C ( x) 5log 2 x + 10 = x x 1 1 x ⋅ 5 ⋅ ln 2 ⋅ x - (5log 2 x + 10) ⋅ 1

C ¢(10) =

(a)

(b) The revenue from one more unit is dR dq

100 +

C ( x) = 5log 2 x + 10

-e2.54197 - 0.2167t N (t ) = e9.8901e 1000 -e2.54197 - 0.2167t

N (t ) = 1000e9.8901e = $12.48.

294

Chapter 4 CALCULATING THE DERIVATIVE (b)

A¢(w) 1æ ln w ö÷ = çç 0.8168 - 0.0154 ÷ ç ln10 ÷ø A(w) wè æ -0.0154 ö÷ 1 + ln w çç çè ln10 ÷÷ø w

N ¢(20) » 1,307, 416 bacteria/hour

Twenty hours into the experiment, the number of bacteria is increasing at a rate of 1,307,416 per hour

0.8168 0.0308 ln w ln10 w w ö 1æ 0.0308 = çç 0.8168 ln w ÷÷÷ ø ln10 w çè ö 1æ 0.0308 ln w ÷÷÷ A¢(w) = çç 0.8168 ç ø ln10 wè

æ N (t ) ö÷ ÷ S (t ) = ln ççç çè N 0 ÷÷ø

(c)

(

⋅ 4.688w0.8168 - 0.0154 log w

(d)

A¢(4000) » 0.4571 » 0.46 g/cm 2 When the infant has a mass of 4000 g, the BSA is increasing by 0.46 square centimeter for each gram of mass.

20,000,000

(c)

A(w) 5 4.688w0.816820.0154 log w

6000

The two graphs have the same general shape, but N(t) is scaled much larger. (e) lim S (t ) = lim 9.8901e-e

t ¥

2000

2.54197 - 0.2167t

10,000

77.

log y = 1.54 - 0.008x - 0.658log x

(a)

y( x) = 10(1.54-0.008 x-0.658 log x)

= 101.54 (10-0.008 x )(10-0.658 log x )

N (t ) = N 0e S (t )

= 101.54 (100.008 )-x (10log x )-0.658

lim S (t )

lim N (t ) = N 0et  ¥

t ¥

» 34.7(1.0186)-x x-0.658

= 1000e9.8901

(b) (i)

» 19, 734, 033 bacteria

y(20) = 34.7(1.0186)-20 20-0.658 » 3.343 imagoes per mated female per day

A (w) = 4.688w0.8168 - 0.0154 log w

(ii)

(a) A(4000) = 4.688(4000)

y(40) = 34.7(1.0186)-40 40-0.658 » 1.466 imagoes per mated female per day

0.8168 - 0.0154 log 4000

» 2590 cm 2

(b)

t ¥

= 9.8901 æ N (t ) ö÷ ÷ S (t ) = ln ççç çè N ÷÷ø

76.

)

A(w) = w0.8166 - 0.0154 log w 4.688

æ ln w ö÷ ÷ ln A( w) - ln 4.688 = (ln w) ççç 0.8168 - 0.0154 ln10 ø÷÷ è

(i) dy (20) = -0.17160... » -0.172 dx imagoes per mated female per day per flies per bottle (ii) dy (40) = -0.051120... » -0.0511 dx imagoes per mated female per day per flies per bottle

Section 4.5 78.

295 (c) For 2020 t = 55: P(55) = 30.60 - 5.79 ln(55) » 7.398 5.79 P¢(55) = » -0.0144 55 In 2020, the percent of persons 65 years and over with family income below the poverty level was about 7.398% and the rate of change was decreasing by 0.0144% per year.

F ( x) = 0.774 + 0.727 log( x)

(a)

(b)

F (25, 000) = 0.774 + 0.727 log(25, 000) = 3.9713... » 4 kJ/day F ¢( x) = 0.727

1 x ln10

0.727 -1 x ln10 0.727 F ¢(25, 000) = 25, 000-1 ln10 » 0.000012629...

(d) The rate of change is approaching 0.

81.

» 1.3 ´ 10-5

M =

(a)

When a fawn is 25 kg in size the rate of change of the energy expenditure of the fawn is about 1.3 ´ 10-5 kJ/day per gram. (c)

2 E log 3 0.007

2 E log 3 0.007 E 13.35 = log 0.007 E 13.35 10 = 0.007 8.9 =

E = 0.007(1013.35 ) » 1.567 ´ 1011 kWh

(b) 10, 000, 000 ´ 247 kWh/month = 2, 470, 000, 000 kWh/month

79.

80.

(a)

1.567 ´ 1011 kWh » 63.4 months 2, 470, 000, 000 kWh/month

P(t ) = (t + 100) ln(t + 2) æ ö P¢(t ) = (t + 100) ç 1 ÷÷÷ + ln(t + 2) çè t + 2 ø æ ö P¢(2) = (102) çç 1 ÷÷ + ln(4) » 26.9 ants /day è4ø æ ö P ¢(8) = (108) çç 1 ÷÷÷ + ln(10) » 13.1 ants /day è 10 ø

(c)

When E = 70, 000, 2 dM = (3 ln 10)70, 000 dE

P(t ) = 30.60 - 5.79 ln t æ ö P¢(t ) = -5.79 çç 1 ÷÷÷ = - 5.79 èt ø t For 1970, t = 5: P(5) = 30.60 - 5.79 ln(5) » 21.28 P¢(5) = - 5.79 » -1.158 5 In 1970, the percent of persons 65 years and over with family income below the poverty level was about 21.28% and the rate of change was about -1.158% per year.

(b) For 1995, t = 30: P(30) = 30.60 - 5.79 ln(30) » 10.91 P¢(30) = - 5.79 » -0.193 30 In 1995, the percent of persons 65 years and over with family income below the poverty level was about 10.91% and the rate of change was decreasing by 0.193% per year.

2 2 log E - log 0.007 3 3 æ ö÷ dM 2 1 2 ÷= = ççç dE 3 è (ln 10) E ÷ø÷ (3 ln 10) E M =

» 4.14 ´ 10-6

(d)

dM dE

varies inversely with E, so as E

increases, dM decreases and approaches dE zero. 82.

f ( x) =

29,000(2.322 - log x) x

(

)

é ù 1 x ê 29,000 - (ln10) - 29,000(2.322 - log x)(1) x úúû êë f ¢( x ) = x2 - 1 - (2.322 - log x) = 29,000 ⋅ ln10 x2 1 + (ln10)(2.322 - log x) = -29,000 ⋅ x 2 ln10

296

Chapter 4 CALCULATING THE DERIVATIVE (a)

11.

29, 000(2.322 - log 30) 30 » 817 1 + (ln10)(2.322 - log 30) f ¢(30) = -29, 000 (30)2 ln10 » -41.2 For a street 30 ft wide, the maximum traffic flow is 817 vehicles/hr, and the rate of change is -41.2 vehicles/hr per foot. f (30) =

(b) 29, 000(2.322 - log 40) 40 » 522 1 + (ln10)(2.322 - log 40) f ¢(40) = -29, 000  (40) 2 ln10 » -20.9 For a street 40 ft wide, the maximum traffic flow is 522 vehicles/hr, and the rate of change is -20.9 vehicles/hr per foot.

= 15x 2 - 14 x - 9

12.

False; the derivative of  3 is 0 because  3 is a constant.

True

False; the derivative of a product u( x) ⋅ v( x) is

y = 7 x3 - 4 x 2 - 5 x + 2 dy = 7(3x 2 ) - 4(2 x) - 5 + 0 dx = 21x 2 - 8x - 5

13.

f (40) =

Chapter 4 Review Exercises

y = 5 x3 - 7 x 2 - 9 x + 5 dy = 5(3x 2 ) - 7(2 x) - 9 + 0 dx

y = 9 x8/3 æ8 ö dy = 9 çç x5/3 ÷÷÷ ç è3 ø dx = 24 x5/3

14.

y = -4 x-3 dy = -4(-3x-4 ) dx 12 = 12 x-4 or 4 x

15.

f ( x) = 3x-4 + 6 x = 3x-4 + 6 x1/2

(

f ¢( x) = 3(-4 x-5 ) + 6 12 x-1/2

12 3 = -12 x-5 + 3x-1/2 or - 5 + 1/2 x x

u( x) ⋅ v¢( x) + v( x) ⋅ u ¢( x).

16.

f ( x) = 19 x-1 - 8 x = 19 x-1 - 8 x1/2

True

False; the chain rule is used to take the derivative of a composition of functions.

False; the derivative ce x is ce x for any constant c. x

(

f ¢( x) = 19(-x-2 ) - 8 12 x-1/2

17.

k ( x) =

False; the derivative of 10 is (ln10)10 .

True

1 False; the derivative of log x is (ln10) , whereas x

1 the derivative of ln x is . x

)

19 4 = -19 x-2 - 4 x-1/2 or - 2 - 1/2 x x

10.

)

k ¢( x) =

3x 4x + 7 (4 x + 7)(3) - (3x)(4) (4 x + 7)2 12 x + 21 - 12 x (4 x + 7)2 21 (4 x + 7) 2

Chapter 4 Review 18.

19.

297

r ( x) = -8 x 2x + 1 (2 x + 1)(-8) - (-8 x)(2) r ¢( x) = (2 x + 1)2 = -16 x - 8 +2 16 x (2 x + 1) 8 = (2 x + 1)2

25.

y = 3x(2 x + 1)3 dy = 3x(3)(2 x + 1)2 (2) + (2 x + 1)3 (3) dx = (18 x)(2 x + 1)2 + 3(2 x + 1)3 = 3(2 x + 1) 2[6 x + (2 x + 1)] = 3(2 x + 1) 2 (8 x + 1)

26.

y = x - x +1 x -1 ( x - 1)(2 x - 1) - ( x 2 - x + 1)(1) dy = dx ( x - 1)2

y = 4 x 2 (3x - 2)5 dy = (4 x 2 )[5(3x - 2) 4 (3)] + (3x - 2)5 (8x) dx = 60 x 2 (3x - 2)4 + 8x(3x - 2)5 = 4 x(3x - 2)4[15 x + 2](3x - 2)]

2 2 = 2 x - 3x + 1 - 2x + x - 1 ( x - 1)

= 4 x(3x - 2)4 (15x + 6 x - 4) = 4 x(3x - 2)4 (21x - 4)

= x - 22x ( x - 1)

20.

y = 2 x - 5x x+2

27. 2

r (t ) = 2

( x + 2)(6 x - 10 x) - (2 x - 5 x )(1) dy = dx ( x + 2) 2 3

x - 2 x + 5x = 6 x + 12 x - 10 x - 20 ( x + 2)2 3

r ¢(t ) = 2

20 x = 4x + 7x ( x + 2)2

21.

f ( x) = (3x 2 - 2) 4

f ¢( x) = 4(3x 2 - 2)3[3(2 x)] = 24 x(3x 2 - 2)3

22.

k ( x) = (5 x3 - 1)6 = 90 x 2 (5x3 - 1)5

s(t ) =

y = 2t 7 - 5 = (2t 7 - 5)1/2 dy = 1 (2t 7 - 5)1/2[2(7t 6 )] 2 dt

s¢(t ) =

= 7t 6 (2t 7 - 5)-1/2 or

24.

(3t + 1)3 (3t + 1)3 (10t - 7) - (5t 2 - 7t )(3)(3t + 1)2 (3) [(3t + 1)3 ]2 (3t + 1)3 (10t - 7) - 9(5t 2 - 7t )(3t + 1) (3t + 1)6 (3t + 1)(10t - 7) - 9(5t 2 - 7t ) (3t + 1) 4 30t 2 - 11t - 7 - 45t 2 + 63t (3t + 1) 4 -15t 2 + 52t - 7 (3t + 1) 4

28.

k ¢( x) = 6(5 x3 - 1)5[5(3x 2 )]

23.

5t 2 - 7t

7t 6 (2t - 5)1/2 7

y = -3 8t 4 - 1 = -3(8t 4 - 1)1/2 dy = -3 12 (8t 4 - 1)-1/2[8(4t 3 )] dx

( )

= -48t 3 (8t 4 - 1)-1/2 or

-48t 3 (8t 4 - 1)1/2

= = =

t 3 - 2t (4t - 3)4 (4t - 3)4 (3t 2 - 2) - (t 3 - 2t )(4)(4t - 3)3 (4) [(4t - 3)4 ]2 (4t - 3)4 (3t 2 - 2) - 16(t 3 - 2t )(4t - 3)3 (4t - 3)8 (4t - 3)3[(4t - 3)(3t 2 - 2t )16(t 3 - 2t )] (4t - 3)8 (4t - 3)3 (12t 3 - 9t 2 - 8t + 6 - 16t 3 + 32t ) (4t - 3)8 -4t 3 - 9t 2 + 24t + 6 (4t - 3)5

298 29.

Chapter 4 CALCULATING THE DERIVATIVE p(t ) = t 2 (t 2 + 1)5/2 5 p¢(t ) = t 2 ⋅ (t 2 + 1)3/2 ⋅ 2t + 2t (t 2 + 1)5/2 2

y = -7 x 2e-3x

36.

Use the product rule. dy = (-7 x 2 )(-3e-3x ) + e-3x (-14 x) dx

= 5t 3 (t 2 + 1)3/2 + 2t (t 2 + 1)5/2 = t (t 2 + 1)3/2[5t 2 + 2(t 2 + 1)1]

= 21x 2e-3x - 14 xe-3x

= t (t 2 + 1)3/2 (7t 2 + 2)

or 7 xe-3x (3x - 2) y = ln (2 + x 2 )

37.

30.

g ( x) = 2 x + x 2 g ¢( x ) = 2 x

g (t ) = t 3 (t 4 + 5)7/2 7 g ¢(t ) = t 3 ⋅ (t 4 + 5)5/2 (4t 3 ) + 3t 2 ⋅ (t 4 + 5)7/2 2 6 4

= 14t (t + 5) 2 4

= t (t + 5)

5/2

2 4

5/2

= t (t + 5)

31.

32.

33.

5/2

2 4

+ 3t (t + 5)

[14t 4 + 3(t 4 + 5)] (17t + 15)

dy 5 = dx 5x + 3

y = -6e2 x dy = -6(2e2 x ) = -12e2 x dx

y =

39.

ln |3x | x-3

( x - 3) ( 31x ) (3) - (ln |3x |)(1) dy = dx ( x - 3)2

y = 8e0.5 x dy = 8(0.5e0.5 x ) = 4e0.5 x dx

x - 3 - x ln |3x | x( x - 3)2

-2 x 3

y =e

g ¢ ( x ) = -6 x

40.

y =

y ¢ = -6 x 2e-2 x y = -4e x

g ( x) = x g ¢( x) = 2 x 2 dy = (2 x)(-4e x ) dx

= -8xe x

ln |2 x - 1| x+3

( x + 3) ( 2 x2-1 ) - (ln |2 x - 1) (1) 2 x - 1 dy = ⋅ dx 2x - 1 ( x + 3)2

35.

y = ln(5 x + 3) g ( x) = 5 x + 3 g ¢( x ) = 5

38.

g ( x ) = -2 x 3

34.

2x dy = dx 2 + x2

7/2

41.

2( x + 3) - (2 x - 1) ln 2 x - 1 (2 x - 1)( x + 3)2 y =

xe x ln(x 2 - 1)

ln(x 2 - 1)[xe x +e x ] - xe x dy = dx [ln(x 2 - 1)]2

y = 5x ⋅ e2 x

(

1 x 2 -1

) (2x)

2 x

Use the product rule. dy = 5x(2e2 x ) + e2 x(5) dx

e x ( x + 1) ln( x 2 - 1) - 2 x2 e

x -1 ⋅ x

[ln(x - 1)]

x -1

e ( x + 1)( x - 1) ln( x - 1) - 2 x 2e x x

= 10 xe2 x + 5e2 x = 5e2 x (2 x + 1)

-1

(x 2 - 1)[ln(x 2 - 1)]2

Chapter 4 Review

299

42.

49. 2

( x + 1)e ln x æ ln x[( x 2 + 1)(2e2 x ) + (e2 x )(2 x)] ö÷ çç ÷÷ çç 2 2 x 1 ÷÷ ç ( 1) x e + ( x )÷ø x çè dy  = dx x (ln x)2 y =

= =

43.

Use the product rule. æ e x + xe x ö÷ ç ÷÷ + [ln( xe x + 1)](2e2 x ) f ¢( x) = e2 x çç çè xe x + 1 ø÷÷ =

e [2 x(ln x)( x + 1 + x) - ( x + 1)]

50.

x(ln x)2

f ( x) =

v( x) = ln( x + 1), v¢( x) =

q = (e2 p +1 - 2) 4

f ¢( x ) =

= 8e2 p +1(e 2 p +1 - 2)3 2

51.

(a)

dy = 3 ⋅ (ln10)10-x (-2 x) dx = -6 x(ln10)  10-x

47.

48.

(b)

1/2

g ( z ) = log 2 ( z 3 + z + 1) 1 3z 2 + 1 ⋅ 3 ln 2 z + z + 1

2 x ( x + 1)[ln( x + 1)]2 Dx ( f [ g ( x)]) at x = 2

ez (ln10)(1 + e z )

Dx ( f [ g ( x)]) at x = 3

(ln 2)( z 3 + z + 1)

1 ez  h ¢( z ) = ln10 1 + e z

3 2

æ4ö = -6 çç ÷÷÷ çè 11 ø

3z 2 + 1

h( z ) = log(1 + e z )

e x [( x + 1) ln( x + 1) - 1]

= f ¢[ g (3)]g ¢(3) æ4ö = f ¢(2) çç ÷÷÷ çè 11 ø

[ln( x + 1)]2

5(ln 2)2 x

g ¢( z ) =

æ x ö æ ö÷ 1 [ln( x + 1)]çç e ÷÷÷ - e x çç ÷ è2 x ø è 2 x ( x +1) ø÷

æ 3ö = -5çç ÷÷÷ çè 10 ø

dy 1 = 10  (ln 2)  2 x  x-1/2 dx 2 =

1 2 x ( x + 1)

= f ¢[ g (2)]g ¢(2) æ 3ö = f ¢(1) çç ÷÷÷ çè 10 ø

y = 10 ⋅ 2

e x 2 x

u ( x ) = e x , u ¢( x ) =

dq = 4(e2 p +1 - 2)3[2e 2 p +1] dp

46.

e x ln( x + 1)

Use the quotient rule.

s = (t 2 + et )2

y = 3 ⋅ 10-x

+ 2e 2 x ln( xe x + 1)

xe x + 1

x(ln x)2

Use the chain rule.

45.

(1 + x)e3x

x ln x[2e2 x ( x 2 + 1) + 2 xe2 x ] - ( x 2 + 1)e2 x

s ¢ = 2(t 2 + et )(2t + et )

44.

f ( x) = e 2 x ln( xe x + 1)

52.

(a)

24 11

Dx ( g[ f ( x)]) at x = 2 = g ¢[ f (2)] f ¢(2) = g ¢(4)(-6) 6 (-6) 13 36 =13 =

300

Chapter 4 CALCULATING THE DERIVATIVE (b)

æ ö y - çç - 3 ÷÷÷ = - 3 [ x - (-1)] è 2ø 4

Dx ( g[ f ( x)]) at x = 3 = g ¢[ f (3)] f ¢(3) = g ¢(2)(-7)

y + 3 = - 3 ( x + 1) 2 4 6 3 y+ =- x- 3 4 4 4 y = -3 x - 9 4 4

3 (-7) 10 21 =10 =

53.

y = x 2 - 6 x; tangent at x = 2

56.

dy = 2x - 6 dx

y =

x 2

x -1

;x = 2

dy ( x 2 - 1)  1 - x(2 x) = dx ( x 2 - 1)2

Slope = y ¢(2) = 2(2) - 6 = -2

Use (2, -8) and point-slope form. y - (-8) = -2( x - 2) y + 8 = -2 x + 4

-x 2 - 1 ( x 2 - 1)2 dy

The value of dx when x = 2 is the slope.

y + 2 x = -4

y = -2 x - 4

-(22 ) - 1 2

(2 - 1)

-5 5 =9 9

When x = 2, 54.

y = 8- x ;x =1 y = 8- x

y =

(

dy = -2 x dx

y - 2 = - 5 ( x - 2) 3 9 6 y - = - 5 x + 10 9 9 9 5 16 y =- x+ 9 9

Use (1, 7) and m = -2 in the point-slope form. y - 7 = -2( x - 1) y - 7 = -2 x + 2 2x + y = 9 y = -2 x + 9

57.

6 x - 2; tangent at x = 3

= 3(6 x - 2)-1/ 2 slope = y ¢(3) = 3(6 ⋅ 3 - 2)-1/2

dy = 3(-1)( x - 1)-2 (1) dx

= 3(16)-1/2

-2

= -3( x - 1)

Slope = y ¢(-1) = -3(-1 - 1)-2 = -

)

6 x - 2 = (6 x - 2)1/ 2

dy 1 = (6 x - 2)-1/ 2 (6) dx 2

3 = 3( x - 1)-1 x -1

y =

3 ; tangent at x = -1 y = x1

(

)

Use m = - 95 with P 2, 23 .

slope = y ¢(1) = -2(1) = -2

55.

2 2 = . 4 -1 3

Use -1, - 32 and point-slope form.

3 4

3 1/2

3 4

16 Use (3, 4) and point-slope form. y - 4 = 3 ( x - 3) 4 16 = 3x- 9 y4 4 4 3 y = x+ 7 4 4

Chapter 4 Review 58.

301 61.

y = - 8 x + 1; x = 3

dy 1 = dx x

y = -(8x + 1)1/2 1 dy = - (8 x + 1)-1/2 (8) 2 dx 4 dy =dx (8 x + 1)1/2

The value of dx when x = 1 is the slope m = 11 = 1. When x = 1, y = ln 1 = 0. Use m = 1 with P(1, 0).

The value of dx when x = 3 is the slope. m=-

4 1/ 2

(24 + 1)

4 5

When x = 3,

y - 0 = 1( x - 1) y = x -1 y = x ln x; x = e

62.

y = - 24 + 1 = -5.

dy 1 = x     ln x dx x = 1 + ln x

Use m = - 54 with P(3, -5). 4 y + 5 = - ( x - 3) 5 25 4 12 y+ =- x+ 5 5 5 4 13 y =- x5 5

59.

y = ln x; x = 1

The value of dx when x = e is the slope. m = 1 + ln e = 1 + 1 = 2

When x = e, y = e ln e = e  1 = e. Use m = 2 with P(e, e). y - e = 2( x - e) y = 2x - e

y = ex; x = 0 dy = ex dx

63. dy

The value of dx when x = 0 is the slope m = e0 = 1.

The slope of the graph of y = x + k is 1. First, we find the point on the graph of f ( x) = 2 x - 1 at which the slope is also 1. f ( x) = (2 x - 1)1/2

When x = 0, y = e0 = 1. Use m = 1 with P(0, 1).

1 (2 x - 1)-1/2 (2) 2 1 f ¢( x ) = 2x - 1 f ¢( x ) =

y - 1 = 1( x - 0) y = x +1

60.

y = xe x ; x = 1

The slope is 1 when

dy = xe x + 1 ⋅ e x dx

1 =1 2x - 1

= e x ( x + 1)

1= 2x = 2

m = e1(1 + 1) = 2e

When x = 1, y = 1e1 = e. Use m = 2e with P(1, e). y - e = 2e( x - 1) y = 2ex - e

2x - 1

1 = 2x - 1

dy The value of dx when x = 1 is the slope.

x = 1,

and f (1) = 1.

Therefore, at P(1,1) on the graph of f ( x) = 2 x - 1, the slope is 1. An equation of the tangent line is

302

Chapter 4 CALCULATING THE DERIVATIVE 64.

y - 1 = 1( x - 1)

(a) Use the chain rule.

y -1 = x -1

Let g ( x) = ln x. Then g ¢( x) = 1x .

y = x + 0.

Let y = g [ f ( x)].

Any tangent line intersects the curve in exactly one point.

Then dx = g ¢ [ f ( x)] ⋅ f ¢( x).

From this we see that if k = 0, there is one point of intersection.

1 d ln f ( x) f ¢( x) . = ⋅ f ¢( x ) = dx f ( x) f ( x)

The graph of f is below the line y = x + 0.

f ¢( x ) g ¢( x ) fˆ = , gˆ = f f ( x) g ( x) ( fg )¢( x)  fg = ( fg )( x) f ( x) ⋅ g ¢( x) + g ( x) ⋅ f ¢( x) = f ( x) g ( x ) f ( x) ⋅ g ¢( x) g ( x) ⋅ f ¢( x) = + f ( x) ⋅ g ( x) f ( x) ⋅ g ( x) g ¢( x ) f ¢( x ) = + g ( x) f ( x) = gˆ + fˆ or fˆ + gˆ

(b)

Therefore, if k > 0, the graph of y = x + k will not intersect the graph. Consider the point Q ( 12 ,0 ) on the graph. We find an equation of the line through Q with slope 1. æ 1ö y - 0 = 1çç x - ÷÷÷ çè 2ø y = x-

1 2

The line with a slope of 1 through Q ( 12 ,0 ) will intersect the graph in two points. One is Q and the other is some point on the graph to the right of Q. The graph of y = x + 0 intersects the graph in

65.

one point, while the graph of y = x - 12

Let T = tuition per person before the increase and S the number of students before the increase. Then the new tuition is 1.03T and the new numbers of students is 1.02S, so the total amount of tuition collected is (1.03T)(1.02S) = 1.0506TS, which is an increase of 5.06%.

intersects it in two points. If we use a value of k in y = x + k with - 12 < k < 0, we will have a line with a y-intercept between - 12 and a 0 and a slope of 1 which will intersect the graph in two points. If k, the y-intercept, is less than - 12 , the graph of y = x + k will be below point Q and will intersect the graph of f in exactly one point. To summarize, the graph of y = x + k will intersect the graph of f ( x) =

2 x - 1 in

(1) no points if k > 0; (2) exactly one point if k = 0 or if k < - 12 ; (3)

exactly two points if - 12 £ k < 0

Using the result  fg = fˆ + gˆ , the total amount of tuition collected goes up by approximately 2% + 3% = 5%.

67. Let f ( x) = a(n - x)-n and g ( x) = bx n for n ¹ 0.

(

)

( )

f ¢( x) g ¢( x) = a(n - x)-n nbx n-1 + (an)(n - x)-n-1 bx n -1 ù úû ë abn 2 x n-1 -1 - é n ù = anbx n (n - x) n ê ú= ëê x - n ûú (n - x) n+1

= anbx

n-1

(n - x)-n éê 1 + x ( n - x )

[ f ( x) g ( x) ]¢ = éê ( a(n - x)-n )( bx n ) ùú ë

n ù¢ é æ æ x ö÷n -1 æ x ö÷¢ x ö÷ ú ç = êê ab çç = abn ÷ ÷ ÷ çç çç ÷ ú è n - x ÷ø èç n - x ÷ø êë èç n - x ø úû æ x ö÷n -1 æç (1)(n - x) - ( x)(-1) ö÷ ÷÷ çç = abn çç ÷ çè èç n - x ø÷ ÷÷ø (n - x ) 2

abn 2 x n -1 ( n - x ) n +1

Chapter 4 Review

303

68.

70. Let f ( x) = ae

k 2 x /(k -1)

and g ( x) = be

for k ¹ 1.

C ( x) 3x + 2 (3x + 2)1/2 = = x x x -1/2 é ù 1 x ê 2 (3x + 2) (3) ú - (3x + 2)1/2 (1) û C ¢( x ) = ë x2

2 ak e k x / (k -1) - kx b(k - 1) ak = e kx / (k -1) b(k - 1)

3 x(3x + 2)-1/2 - (3x + 2)1/2

= 2 =

é k 2 x / (k -1) ù ¢ é f ( x) ù ¢ ú ê ú = êê ae ú kx ê g ( x) ú ê ú be ë û ë û aæ 2 a kx / (k -1) ¢ ö¢ = çç e k x / (k -1) - kx ÷÷ = e è ø b b

(

69.

x +1

C ( x) =

C ( x) = x

71.

x +1 x

C ¢( x ) =

= = = =

x[ 12 ( x + 1)-1/2 ] - ( x + 1)1/2 (1) x2

1 x( x + 1)-1/2 - ( x + 1)1/2 2 2

-1/2

x( x + 1)

3x(3x + 2)-1/2 - 2(3x + 2)1/2 2x2 (3x + 2)-1/2[3x - 2(3x + 2)]

C ( x) = ( x 2 + 3)3 C ( x) =

( x + 1)1/2 = x C ¢( x ) =

( x 2 + 3)3 C ( x) = x x x[3( x 2 + 3)2 (2 x)] - ( x 2 + 3)3 (1) x2 6 x 2 ( x 2 + 3) 2 - ( x 2 + 3)3 x2 ( x 2 + 3)2[6 x 2 - ( x 2 + 3)] x2 ( x 2 + 3)2 (5 x 2 - 3) x2

1/2

- 2( x + 1)

2x2 ( x + 1)-1/2[ x - 2( x + 1)] 2x2

2x2 3x - 6 x - 4 -3 x - 4 = = 2 1/2 2 2 x (3x + 2) 2 x (3x + 2)1/2

)

ak e kx / (k -1) b(k - 1)

C ( x) =

3x + 2

C ( x) =

2 æ 2 f ¢( x) ÷÷ö çæ 1 ÷ö ç ak = çç e kx / (k -1) ÷ç ÷ ÷ g ¢( x) ÷ø çè bkekx ÷ø çè k - 1

C ( x) =

72.

C ( x) = (4 x + 3)4 C ( x) =

( x + 1)-1/2 (-x - 2) 2x2 -x - 2

C ¢( x ) =

2 x 2 ( x + 1)1/2

= = =

(4 x + 3)4 C ( x) = x x x[4(4 x + 3)3 (4)] - (4 x + 3) 4 (1) x2 16 x(4 x + 3)3 - (4 x + 3)4 x2 (4 x + 3)3[16 x - (4 x + 3)] x2 (4 x + 3)3 (12 x - 3)

304 73.

Chapter 4 CALCULATING THE DERIVATIVE C ( x) = 10 - e-x C ( x) =

(a)

C ( x) x

74.

75.

(b)

e-x ( x + 1) - 10

= 30 x-1/2 + 12 =

30 + 12 x

dS 30 30 (9) = + 12 = + 12 = 22 dx 3 9 Sales will increase by $22 million when $1000 more is spent on research.

[2(12) + 1]2 312 = = 0.4992 625

In dollars, this is 312 (100) » $49.92, 625 which is the approximate increase in profit from selling the thirteenth unit. (c)

P¢(20) =

dS 30 30 (25) = + 12 = + 12 = 18 dx 5 25 Sales will increase by $18 million when $1000 more is spent on research.

P ¢( x ) =

x2 2x + 1 (2 x + 1)(2 x) - x 2 (2) (2 x + 1)2

2(20)2 + 2(20)

[2(20) + 1]2 840 = » 0.4997 1681

(d) As the number sold increases, the marginal profit increases. (e) The average profit is defined by x2

P( x) x = 2 x +1 = P ( x) = x x 2x + 1

The marginal average profit is given by d (2 x + 1)(1) - ( x)(2) ( P ( x)) = dx (2 x + 1)2

dS 30 30 (16) = + 12 = + 12 = 19.5 dx 4 16 Sales will increase by $19.5 million when $1000 more is spent on research.

P( x) =

2(12)2 + 2(12)

which is the approximate increase in profit from selling the twenty-first unit.

(d) As more money is spent on research, the increase in sales is decreasing. 76.

[2(4) + 1]

40 » 0.4938 81

840 (100) » $49.97, In dollars, this is 1681

= 1000 + 60 x1/2 + 12 x æ1 ö dS = 60 çç x-1/2 ÷÷÷ + 12 çè 2 ø dx

(c)

P¢(12) =

S ( x) = 1000 + 60 x + 12 x

(b)

is the approximate increase in profit from selling the fifth unit.

x(e-x ) - (10 - e-x ) ⋅ 1

C ( x) = ln( x + 5) ln( x + 5) C ( x) = x  - ln( x + 5)  1 x  x+ 5 C ¢( x ) = x2 x - ( x + 5) ln( x + 5) = x 2 ( x + 5)

(a)

2(4) 2 + 2(4)

In dollars, this is 40 (100) » $49.38, which 81

-x C ( x) = 10 - e x

C ¢( x ) =

P¢(4) =

= =

2x + 1 - 2x (2 x + 1) 2 1 (2 x + 1)2

The marginal average profit when 4 units are sold is P¢(4) =

1 [2(4) + 1]2

1 » 0.0123 81

The average profit is going up at a rate of

( )

1 » $1.23 per unit when 4 units are 100 81

sold.

2 2 = 4 x + 2 x -2 2 x (2 x + 1) 2

= 2 x + 22x (2 x + 1)

Chapter 4 Review 77.

T ( x) = T ¢( x ) = = =

(a)

305

1000 + 60 x 4x + 5 (4 x + 5)(60) - (1000 + 60 x)(4)

80.

-3700 (4 x + 5)2

2 r ù (100 + r ) éê ln ( 1 + 100 ) ûú ë ln 2 T ¢(5) = » -2.77 105 (ln 1.05)2

-3700

-3700 = » -2.201 2 1681 [4(9) + 5]

The doubling time decreases by approximately 2.77 years for every 1% increase in the interest rate when the interest rate is 5%.

-3700

[4(19) + 5]2 -3700 = » -0.564 6561 Costs will decrease $564 for the next $100 spent on training.

81.

P(t ) = -0.003584t 4 - 0.01051t 3 + 0.7432t 2 - 6.508t + 171.7 P¢(t ) = -0.014336t 3 - 0.03153t 2 + 1.4864t - 6.508

(a)

P¢(0) = -6.508 » -6.51 In 2010, the volume of mail is decreasing by about 6,510,000,000 pieces per year.

(b)

P¢(8) = -0.014336(8)3 - 0.03153(8)2

will always be negative. 48

æ r ÷ö A(r ) = 1000 çç1 + ÷ çè 400 ÷ø

+ 1.4864(8) - 6.8508 = -3.974752 » -3.97 In 2018, the volume of mail is decreasing by about 3,970,000,000 pieces per year.

47 æ r ö÷ 1 A¢(r ) = 1000  48çç1 +  ÷ çè 400 ÷ø 400 47 æ r ö÷ = 120 çç1 + ÷ çè 400 ø÷

82.

47 æ 5 ö÷ » 215.15 A¢(5) = 120 çç1 + ÷ çè 400 ÷ø The balance increases by approximately $215.15 for every 1% increase in the interest rate when the rate is 5%.

79.

A(r ) = 1000e12r /100 A¢(r ) = 1000e12r /100 

- ln 2

T ¢( r ) =

Costs will decrease $2201 for the next $100 spent on training.

78.

r ln ( 1 + 100 )

-2 1 é æ r ö÷ ùú 100 ç ê ¢ T ( r ) = (ln 2)(-1) ln ç 1 + ⋅ ÷ r ê çè 100 ÷ø úû 1 + 100 ë

(4 x + 5) 2

(b) T ¢(19) =

ln 2

-1 é æ r ö÷ ùú T (r ) = ln 2 ê ln çç 1 + ÷ ê çè 100 ÷ø úû ë

(4 x + 5)2 240 x + 300 - 4000 - 240 x

T ¢(9) =

T (r ) =

12 100

(a)

y = 0.0001365t 3 - 0.02294t 2 + 0.5698t + 52.31

(b)

y ¢ = 0.0004095t 2 - 0.04588t + 0.5698 For 1965, t = 65:

y ¢ = 0.0004095(65) 2 - 0.04588(65) + 0.5698 » -0.68 In 1965, the percent of men aged 65 and older in the workforce was decreasing by about 0.68% per year.

For 2015, t = 115: ¢ y = 0.0004095(115)2 - 0.04588(115) + 0.5698

= 120e12r /100 A¢(5) = 120e0.6 » 218.65

The balance increases by approximately $218.65 for every 1% increase in the interest rate when the rate is 5%.

» 0.71 In 2015, the percent of men aged 65 and older in the workforce was increasing by about 0.71% per year.

306 83.

Chapter 4 CALCULATING THE DERIVATIVE (a) Using the regression feature on a graphing calculator, a cubic function that models the data is

86.

L(t ) = 71.5(1 - e-0.1t ) and W ( L) = 0.01289  L2.9

y = 0.000001135t 3 + 0.003147t 2

(a)

L(5) = 71.5(1 - e-0.5 ) » 28.1 The approximate length of a 5-year-old monkey-face is 28.1 cm.

(b)

L¢(t ) = 71.5(0.1e-0.1t )

- 0.2019t + 4.099

(b)

y ¢ = 0.000003405t 2 + 0.006294t - 0.2019 For 1955, t = 55:

L¢(5) = 71.5(0.1e-0.5 )

y ¢ = 0.000003405(55)2 + 0.006294(55) - 0.2019

» 4.34 The length is growing by about 4.34 cm/year.

» 0.15 In 1955, the number of dollars required to equal $1 in 1913 is increasing by about $0.15 per year.

(c)

For 2015, t = 115: ¢ y = 0.000003405(115) 2 + 0.006294(115) - 0.2019

(d)

» 0.57

W ¢[ L(5)] » 0.037381(28.1)1.9 » 21.2 The rate of change of the weight with respect to length is 21.2 grams/cm.

P(t ) = ae0.05t P¢(t ) = ae0.05t (0.05) P¢(t ) = P(t )(0.05)

(e)

If P(t ) = 1,000,000, then P¢(t ) = 1, 000, 000(0.05) = 50, 000. The population is growing at a rate of 50,000 per year.

85.

G (t ) =

m G0

G0 + (m - G0 )e-kmt

87.

, where

W ¢( L) = 0.01289(2.9) L1.9 = 0.037381L1.9

In 2015, the number of dollars required to equal $1 in 1913 is increasing by about $0.57 per year. 84.

W [ L(5)] » 0.01289(28.1) 2.9 » 205 The approximate weight is 205 grams.

dW dW dL =  dt dL dt » (21.2) (4.34) » 92.0 The weight is growing at about 92.0 grams/year. -0.020(t - 66)

M (t ) = 3583e-e

-0.020(250 - 66)

m = 30, 000, Go = 2000, and k = 5.10-6.

M (250) = 3583e-e » 3493.76 grams,

(a)

or about 3.5 kilograms

G(t ) = =

(a)

(30, 000)(2000) -5.10-6

2000 + (30, 000 - 2000)e 30, 000

(b) As t  ¥, -e-0.020(t -66)  0, (30, 000)t

1 + 14e-0.15t

-0.020(t - 66)

e-e  1, and M (t )  3583 grams or about 3.6 kilograms.

(b) G(t ) = 30, 000(1 + 14e-0.15t )-1 G(6) = 30, 000(1 + 14e-0.90 )-1 » 4483 G ¢(t ) = -30, 000(1 + 14e-0.15t )-2 (-2.1e-0.15t ) = G ¢(6) =

63, 000e-0.15t (1 + 14e-0.15t )2

-0.020(t - 66)

1791.5 = 3583e-e æ 1791.5 ö÷ ln çç = -e-0.020(t -66) çè 3583 ÷÷ø æ 3583 ö÷ ln çç ln = -0.020(t - 66) çè 1791.5 ÷ø÷

63, 000e-0.90

» 572 (1 + 14e-0.90 )2 The population is 4483, and the rate of growth is 572.

æ 3583 ö÷ 1 ln çç ln ÷ + 66 0.020 çè 1791.5 ÷ø » 84 days

t =-

Chapter 4 Review

307 (b) For 2015, t = 15: C ¢(15) = 54,531.1864 » 54,531 megawatts/year

(d) Dt M (t ) -0.020(t - 66)

= 3583e-e

-e-0.020(t - 56)

= 3583e

(

Dt -e-0.020(t -66)

)

( -e )(-0.020) ( e-0.020(t -66) ) -0.020(t -66)

» 108,982 megawatts/year

-0.020(t - 66)

= 71.66e-e

When t = 250, Dt M (t ) » 1.76 g/day.

90.

(e)

8 20 + 2 t +1 t +1 (a) The average velocity from t = 1 to t = 3 is given by f (3) - f (1) average velocity = 3-1 f (t ) =

Growth is initially rapid, then tapers off.

88.

Weight 904 2159 2974 3346 3494 3550

Belmar’s average velocity between 1 sec and 3 sec is –5 ft/sec.

Rate 24.90 21.87 11.08 4.59 1.76 0.66

(b) f (t ) = 8(t + 1)-1 + 20(t 2 + 1)-1 f ¢(t ) = -8(t + 1)-2 ⋅ 1 - 20(t 2 + 1)-2 ⋅ 2t 8 =- 2 40t 2 (t + 1) 2 (t + 1) f ¢(3) = - 8 - 120 16 100 = -0.5 - 1.2 = -1.7 Belmar’s instantaneous velocity at 3 sec is –1.7 ft/sec.

y(t ) = 10, 000(1.149)t y ¢(t ) = 10, 000(ln1.149)(1.149)t

(a) For 2 hours, t = 2: y ¢(2) = 10, 000(ln1.149)(1.149) 2 » 1834 After 2 hours, the number of bacteria is increasing by 1834 bacteria per hour. (b) For 4 hours, t = 4: y ¢(4) = 10, 000(ln1.149)(1.149) 4 » 2421 After 4 hours, the number of bacteria is increasing by 2421 bacteria per hour. (c) For 10 hours, t = 10: y ¢(10) = 10, 000(ln1.149)(1.149)10 » 5570 After 10 hours, the number of bacteria is increasing by 5570 bacteria per hour.

89.

91.

p(t ) = 1.757(1.0248)t p¢(t ) = (1.757)(ln1.0248)(1.0248)t = 0.043042(1.0248)t p¢(90) = 0.390313 » 0.390 The production of corn is increasing at a rate of 0.390 billion bushels per year in 2020.

92.

(a) N (t ) = N 0e-0.217t , where t = 1 and N 0 = 210 N (1) = 210e-0.217(1) » 169 The number of words predicted to be in use in 1950 is 169, and the actual number in use was 167.

C (t ) = 49,330.3(1.14853)t C ¢(t ) = 49,330.3(ln1.14853)(1.14853)t (a) For 2010, t = 10: C ¢(10) = 27, 285.58 » 27, 286 megawatts/year

= 4 - 14 2 = -5

(f )

Day 50 100 150 200 250 300

( 84 + 1020 ) - ( 82 + 202 )

(b)

N (1.1) = 210e-0.217(1.1) » 165 In 2050 the will be about 165 words still being used.

308

Chapter 4 CALCULATING THE DERIVATIVE (c)

N (t ) = 210e-0.217t

N ¢(t ) = 210e-0.217t ⋅ (-0.217)

V =

-0.217t

= -45.57e

In the year 2050 the number of words in use will be decreasing by 36 words per millennium. 3.

f ( x) = k ( x - 49)6 + .8 f ¢( x) = k  6( x - 49)5 = (3.8 ´ 10-9 )(6)( x - 49)5 = (2.28 ´ 10-8 )( x - 49)5

(a)

f ¢(20) = (2.28 ´ 10-8 )(20 - 49)5 » -0.4677

fatalities per 1000 licensed drivers per 100 million miles per year.

fatalities per 1000 licensed drivers per 100 million miles per year. At the age of 60, each extra year results in an increase of 0.003672 fatalities per 1000 licensed drivers per 100 million miles

Extended Application: Electric Potential and Electric Field 1.

E =-

dV dz æ z 2 ö÷÷ ç V = k1 çç R - z + ÷ çè 2 R ÷ø÷ æ ö 1 dV  2 z ÷÷÷ = k1 çç 0 - 1 + ç è ø 2R dz æz ö÷ = k1 çç - 1 ÷÷ çè R ø E =-

æz ö dV = -k1 çç - 1 ÷÷÷ ç èR ø dz æ ö z = k1 çç 1 - ÷÷÷ çè Rø

f ¢(60) = (2.28 ´ 10-8 )(60 - 49)5 » 0.003672

k2 = k 2 z -1 z

E =-

At the age of 20, each extra year results in a decrease of 0.4677 fatalities per 1000 licensed drivers per 100 million miles. (b)

dV dz

dV k = k2 (-1) z-2 = - 22 dz z dV k2 E == 2 dz z

N ¢(1.1) = -45.57e-0.217(1.1) » -36

93.

E =-

If z is very close to the disk, then 2zR approaches

(

zero, and V = k1 R - z + 2zR

)

becomes V = k1( R - z ). V = k1( R - z ) dV = k1(0 - 1) = -k1 dz = -(-k1) = k1, Therefore, E = - dV dz

where k1 is a constant.

dV dz

V = k1[( z 2 + R 2 )1/2 - z ] dV d = k1  [( z 2 + R 2 )1/2 - z ] dz dx é1 ù = k1 ê ( z 2 + R 2 )-1/2 (2 z + 0) - 1ú êë 2 úû é ù z = k1 êê - 1úú êë z 2 + R 2 úû é ù dV z = -k1 êê - 1úú E =dz êë z 2 + R 2 úû

Chapter 5 GRAPHS AND THE DERIVATIVE 5.1

Increasing and Decreasing Functions

Your Turn 1 Scanning across the x-values from left to right we see that the function is increasing on (-1, 2) and (4, ¥) and decreasing on ( -¥, -1) and (2, 4). Your Turn 2 f ( x) = -x3 - 2 x 2 + 15x + 10 f ¢( x) = -3x 2 - 4 x + 15

Set f ¢( x) equal to 0 and solve for x. -3x 2 - 4 x + 15 = 0 3x 2 + 4 x - 15 = 0

Your Turn 3 f ( x) = (2 x + 4) 2/5 2 f ¢( x) = (2 x + 4)-3/5 (2) 5 4 = 5(2 x + 4)3/5

Find the critical numbers. f ¢( x) is never 0 so we find

(3x - 5)( x + 3) = 0 5 x = or x = -3 3

any values where f ¢( x) fails to be defined by setting the denominator equal to 0.

The only critical numbers are -3 and 5/3. These points determine three intervals: (-¥, -3), (-3, 5/3), and (5/3, ¥). Determine the sign of f ¢ in each interval by

5(2 x + 4)3/5 = 0 (2 x + 4)3/5 = 0 2 x + 4 = 05/3 = 0

picking a test point and evaluating f ¢( x). f ¢(-4) = -(3(-4) - 5)((-4) + 3) = -(-17)(-1) < 0 f ¢(0) = -(3(0) -5)((0) + 3) = -(-5)(3) > 0 f ¢(2) = -(3(2) - 5)((2) + 3) = -(1)(5) < 0

The arrows in each interval in the figure below indicate where f is increasing or decreasing.

2 x = -4 x = -2

Since f ¢(-2) does not exist but f (-2) is defined, x = -2 is a critical number. Choosing -3 and -1 as test points, we find that f ¢(-3) < 0 and f ¢(-1) > 0. Thus the function f

is decreasing on (-¥, -2) and increasing on (-2, ¥). Using the fact that f (-2) = 0 and f (0) = 42/5 » 1.74 we can sketch the following graph. Note that the graph is not smooth at x = -2, where the derivative is not defined.

The function f ( x) = -x3 - 2 x 2 + 15 x + 10 is increasing on (-3, 5/3) and decreasing on (-¥, -3) and (5/3, ¥). » 24.8, the ( 53 ) = 670 27 . graph goes through (-3, -26) and ( 53 , 670 27 )

Since f (-3) = -26 and f

309

310

Chapter 5 GRAPHS AND THE DERIVATIVE

Your Turn 4 f ( x) = f ¢( x ) = =

W2. -2 x ; the domain of f excludes -2 x+2 ( x + 2)(-2) - (-2 x)(1)

12 x 3 - 41x 2 - 15x = 0 ( x )(3x + 1)(4 x - 15) = 0 x = 0 or 3x + 1 = 0 or 4 x - 15 = 0 1 15 x = 0 or x = - or x = 3 4 The roots are 0, -1/3, 15/4.

( x + 2) 2 -2 x - 4 + 2 x

( x + 2)2 4

W3.

( x + 2) 2

f ( x ) = 12 x 3 - 41x 2 - 15x

f ( x ) = e x -9 - 1 2

e x -9 - 1 = 0

f ¢( x) is never 0 and fails to exist at x = -2; however, f is undefined at x = -2 so there are no critical numbers. Apply the first derivative test for numbers on either side of x = -2. f ¢(-3) = f ¢(-1) = -

4 (-3 + 2) 2 4 (-1 + 2)

e x - 9 = 1 = e0 x2 - 9 = 0 x = 3 or -3 The roots are -3, 3.

<0 2

W4.

f ( x ) = x 7 / 3 - 4 x1/ 3 x 7 / 3 - 4 x1/ 3 = 0

Since f ¢( x) is negative wherever it is defined, the function is never increasing and is decreasing on (-¥, -2) and (-2, ¥).

x1/ 3 ( x 2 - 4) = 0

Next find any asymptotes. Since x = -2 makes the denominator of f equal to 0, the line x = -2 is a vertical asymptote.

The roots are 0, -2, 2.

x1/ 3( x + 2)( x - 2) = 0 x = 0 or x = -2 or x = 2

W5.

-2 x -2 lim = lim = -2 x ¥ x + 2 x ¥ 1 + 2 x

f ( x) =

x2 + 4

x3 + 5 Use the quotient rule.

x 3 + 5 )( 2 x ) - ( x 2 + 4 )( 3x 2 ) ( f ¢( x ) = 2 ( x3 + 5 )

The line y = -2 is a horizontal asymptote. The x-intercept is (0, 0). The graph is shown below.

2 x 4 + 10 x - 3x 4 - 12 x 2 2

( x3 + 5 )

-x 4 - 12 x 2 + 10 x

5.1

Warmup Exercises

W1.

f ( x ) = 8 x 2 + 2 x - 15

( x3 + 5 ) (-x ) ( x 3 + 12 x - 10 ) = 2 ( x3 + 5 ) W6.

(

3 - x2 = 3 - x2

f ( x) =

1/ 2

)

8 x 2 + 2 x - 15 = 0 (4 x - 5)(2 x + 3) = 0 4 x - 5 = 0 or 2 x + 3 = 0

Use the chain rule with f ( x ) = h( g ( x ))

5 3 or x = x = 4 2 The roots are 5/4, -3/2.

f ¢( x ) =

where h( x ) = x1/ 2 and g ( x ) = 3 - x 2 .

-1/ 2 1 3 - x2 ( -2 x ) 2 -x

(

3 - x2

)

Section 5.1 W7.

311

f ( x ) = x 2 e5 x

Use the product and chain rules. 3ö æ æ 3ö f ¢( x ) = x 2 çç 15x 2e5 x ÷÷÷ + ( 2 x )çç e5x ÷÷÷ è ø è ø

( ) = (15x 4 + 2 x ) e5x = x (15x 3 + 2 ) e5x

W8.

(

f ( x ) = 2ln x 3/ 2 + 5

(b) decreasing on (-¥, 3). 9.

By reading the graph, g is (a) increasing on (3, ¥) and

By reading the graph, h is (a) increasing on (-¥, -4) and (-2, ¥) and

(b) decreasing on (-4, -2).

)

10.

Use the formula for the derivative of the natural logarithm and the chain rule.

By reading the graph, h is (a) increasing on (1, 5) and (b) decreasing on (-¥, 1) and (5, ¥).

3/ 2

Let h( x ) = 2ln x and g ( x ) = x + 5. ¢ Then f ( x ) = h( g ( x )) and f ( x ) = h¢( g ( x )) g ¢( x ).

11.

By reading the graph, f is (a) increasing on (-7, -4) and (-2, ¥) and

2 2 h¢( x ) = so h¢( g ( x )) = 3/ 2 x x +5 3 g ¢( x ) = x1/ 2 2 æ 2 ÷ö÷æç 3 x1/ 2 ö÷÷ h¢( g ( x )) g ¢( x ) = çç 3/ 2 ÷ç çè x ø÷ + 5 ø÷èç 2 =

3x1/ 2 x

3/ 2

(b) decreasing on (-¥, -7) and (-4, -2). 12.

(a) increasing on (-3, 0) and (3, ¥) and (b) decreasing on (-¥, -3) and (0, 3). 13.

5.1

Exercises

False. If f ¢( x) < 0 for every x on an interval, then the slope of f is negative on that interval.

True

False. The derivative in regions separated by critical numbers can alternate between positive and negative, but not necessarily.

False. If f ¢( x) > 0 for each x in an interval, then f is increasing on that interval.

By reading the graph, f is

(a) Since the graph of f ¢( x ) is positive for x < -1 and x > 3, the intervals where f ( x) is increasing are (-¥, -1) and (3, ¥). (b) Since the graph of f ¢( x ) is negative for -1 < x < 3, the interval where f ( x) is decreasing is (-1, 3).

14.

(a) Since the graph of the f ¢( x ) is positive for 3 < x < 5, the intervals where f(x) is increasing is (3, 5). (b) Since the graph of f ¢( x ) is negative for x < 3 and x > 5, the intervals where f ( x) is decreasing are (-¥, 3) and (5, ¥).

(a) increasing on (1, ¥) and 15.

(b) decreasing on (-¥, 1). 6.

By reading the graph, f is

the intervals where f ( x) is increasing are (-¥, -8), (-6, -2.5), and (-1.5, ¥).

(a) increasing on (-¥, 4) and (b) decreasing on (4, ¥). 7.

(a) Since the graph of f ¢( x) is positive for x < -8, -6 < x < -2.5 and x > -1.5,

By reading the graph, g is (a) increasing on (-¥, -2) and

(b) Since the graph of f ¢( x ) is negative for -8 < x < -6 and -2.5 < x < -1.5, the intervals where f ( x) is decreasing are (-8, -6) and (-2.5, -1.5).

(b) decreasing on (-2, ¥).

312 16.

Chapter 5 GRAPHS AND THE DERIVATIVE Test a point in each interval.

(a) Since the graph of f ¢( x ) is positive for x < -3, -3 < x < 3, and x > 3, the

When x = -1, y ¢ = -0.3 - 0.6(-1) = 0.3 > 0.

intervals where f ( x) is increasing are (-¥, -3), (-3, 3), and (3, ¥).

When x = 0, y ¢ = -0.3 -0.6(0) = -0.3 < 0.

(b) Since the graph of f ¢( x ) is never negative,

( ) (c) The function is decreasing on ( - 12 , ¥ ).

there are no intervals where f ( x) is decreasing. 17.

(b) The function is increasing on -¥, - 12 .

y = 2.3 + 3.4 x - 1.2 x 2

(a)

y¢ = 3.4 - 2.4 x

19.

y¢ is zero when 3.4 - 2.4 x = 0 x = 3.4 = 17 2.4 12

f ( x) =

(a)

Test a point in each interval. f ¢(-4) = 16 > 0 f ¢(0) = -24 < 0 f ¢(5) = 16 > 0

Test a point in each interval. When x = 0, y¢ = 3.4 - 2.4(0) = 3.4 > 0.

(b) f is increasing on (-¥, -3) and (4, ¥). (c) f is decreasing on (-3, 4).

When x = 2, y¢ = 3.4 - 2.4(2) = -1.4 < 0.

)

(b) The function is increasing on -¥, 17 . 12 (c) The function is decreasing on 18.

f ¢( x) = 2 x 2 - 2 x - 24 = 2( x 2 - x - 12) = 2( x + 3)( x - 4) f ¢ x  is zero when x = -3 or x = 4, so the critical numbers are -3 and 4.

and there are no values of x where y¢ does not exist, so the only critical number is x = 17 . 12

(

2 3 x - x 2 - 24 x - 4 3

( 1712 , ¥ ).

20.

f ( x) =

(a)

2 3 x - x2 - 4x + 2 3

f ¢( x ) = 2 x 2 - 2 x - 4 = 2( x 2 - x - 2) = 2( x + 1)( x - 2)

y = 1.1 - 0.3x - 0.3x 2

(a)

f ¢( x) is zero when x = -1 or x = 2, so the critical numbers are -1 and 2.

y ¢ = -0.3 - 0.6 x y¢ is zero when -0.3 - 0.6 x = 0

x = - 0.3 = - 1 0.6 2

and there are no values of x where y¢ does not exist, so the only critical number is x = - 12 .

Test a point in each interval. f ¢(-2) = 8 > 0 f ¢(0) = -4 < 0 f ¢(3) = 8 > 0

(b) f is increasing on (-¥, -1) and (2, ¥). (c) f is decreasing on (-1, 2).

Section 5.1 21.

313 f ¢( x) is zero when x = 0, x = -2, or x = -1, so the critical numbers are 0, -2, and -1.

f ( x) = 4 x3 - 15x 2 - 72 x + 5

(a)

f ¢( x) = 12 x 2 - 30 x - 72 = 6(2 x 2 - 5 x - 12) = 6(2 x + 3)( x - 4) f ¢( x) is zero when x = - 32 or x = 4, so

the critical numbers are - 32 and 4.

Test a point in each interval. f ¢(-3) = -12(-1)(-2) = -24 < 0 f ¢(-1.5) = -6(.5)(-.5) = 1.5 > 0 f ¢(-.5) = -2(1.5)(.5) = -1.5 < 0 f ¢(1) = 4(3)(2) = 24 > 0

f ¢(-2) = 36 > 0 f ¢(0) = -72 < 0 f ¢(5) = 78 > 0

(

(b) f is increasing on (-2, -1) and (0, ¥). (c) f is decreasing on (-¥, -2) and (-1, 0).

)

(b) f is increasing on -¥, - 32 and (4, ¥).

(

24.

)

f ( x) = 3x 4 + 8 x3 - 18 x 2 + 5

(a)

f ¢( x) = 12 x3 + 24 x 2 - 36 x = 12 x( x 2 + 2 x - 3)

22.

f ( x) = 4 x3 - 9 x 2 - 30 x + 6

(a)

= 12 x( x + 3)( x - 1) f ¢( x) is zero when x = 0, x = -3 or x = 1, so the critical numbers are 0, -3, and 1.

f ¢( x) = 12 x 2 - 18 x - 30 = 6(2 x 2 - 3x - 5) = 6(2 x - 5)( x + 1) f ¢( x) is zero when x = 52 or x = -1, so

the critical numbers are 52 and -1. Test a point each interval. f ¢(-4) = -240 < 0 f ¢(-1) = 48 > 0

( )

f ¢ 1 = - 21 < 0 2 2 f ¢(2) = 120 > 0

Test a point in each interval. f (-2) = 54 > 0 f (0) = -30 < 0 f (3) = 24 > 0

(b) f is increasing on (-3, 0) and (1, ¥).

(

)

(b) f is increasing on (-¥, -1) and 52 , ¥ .

(

)

f ( x) = x 4 + 4 x 3 + 4 x 2 + 1

(a)

f ¢( x) = 4 x3 + 12 x 2 + 8 x 2

= 4 x( x + 3x + 2) = 4 x( x + 2)( x + 1)

y = -3 x + 6

(a)

y ¢ = -3 < 0

There are no critical numbers since y¢ is never 0 and always exists. (b) Since y¢ is always negative, the function is increasing on no interval.

314

Chapter 5 GRAPHS AND THE DERIVATIVE (c)

26.

y¢ is always negative, so the function is decreasing everywhere, or on the interval (-¥, ¥).

y = 6x - 9

(a)

7 <0 16 f ¢(5) = -7 < 0 f ¢(0) = -

y¢ = 6 > 0

There are no critical numbers since y¢ can never be 0 and always exists. (b) Since y¢ is always positive, the function is increasing everywhere, or on the interval (-¥, ¥). (c)

27.

(b)

f ¢( x) is always negative, so f  x  is increasing on no interval.

(c)

f ¢( x) is always negative, so f  x  is decreasing everywhere that is defined. Since f  x  is not defined at x = 4, these intervals

y¢ is never negative, so the function is decreasing on no interval.

x+2 x +1 ( x + 1)(1) - ( x + 2)(1) (a) f ¢( x) = ( x + 1)2 -1 = ( x + 1) 2 The derivative is never 0, but it fails to exist at x = -1. Since –1 is not in the domain of f , however, –1 is not a critical number. f ( x) =

are (-¥, 4) and (4, ¥). 29.

y =

x2 + 1

= ( x 2 + 1)1/2

(a)

y¢ =

1 2 ( x + 1)-1/2 (2 x) 2

= x( x 2 + 1)-1/2 =

x 2

x +1

y¢ = 0 when x = 0.

Since y does not fail to exist for any x, and since y¢ = 0 when x = 0, 0 is the only critical number.

f ¢(-2) = -1 < 0 f ¢(0) = -1 < 0

(b) f is increasing on no interval. (c) f is decreasing everywhere that it is defined, on (-¥, -1) and on (-1, ¥).

1 >0 2 -1 y ¢(-1) = <0 2 y ¢(1) =

28.

f ( x) =

(a)

x+3 x-4

f ¢( x ) = =

(1)( x - 4) - (1)( x + 3) ( x - 4)2 x-4- x-3 ( x - 4)2

-7 ( x - 4) 2

f ¢ is never 0, but it fails to exist when x = 4. Since 4 is not in the domain of f , 4 is not a critical number. Thus, there are no critical numbers. However, the line x = 4 is an asymptote of the graph, so the function might change direction from one side of the asymptote to the other.

30.

(b)

y is increasing on (0, ¥).

(c)

y is decreasing on (-¥, 0).

y = - 9 + x 2 = (9 + x 2 )1/2

(a) Use the chain rule. 1 y ¢ = - (9 + x 2 )-1/2 (2 x) 2 x =9 + x2

Section 5.1

315 Critical numbers occur when y ¢ = 0 or when y ¢ fails to exist.

f ¢(-2) = f ¢(0) =

y ¢ = 0 when x =0

4 <0 5

4 >0 5

(b) f is increasing on (-1, ¥).

y ¢ exists for all values of x.

Thus, the only critical number is 0. This one value determines two intervals.

33.

g ( x) = x 3 + 3 x 2 + 3 x + 1

(a)

g ¢( x ) = 3 x 2 + 6 x + 3 = 3( x 2 + 2 x + 1) = 3( x + 1) 2 g ¢( x) is zero when x = -1, so the critical number is –1.

f ¢(-4) = 0.8 > 0 f ¢(4) = -0.8 < 0

(b) f is increasing on (-¥, 0). (c) f is decreasing on (0, ¥). 31.

g ¢(-2) = 3 > 0

f ( x) = x 2/3

(a)

g ¢(0) = 3 > 0

2 -1/3 2 = 1/3 x 3 3x f ¢( x) is never zero, but fails to exist when x = 0, so 0 is the only critical number. f ¢( x ) =

(b) g is increasing on (-¥, -1) and (-1, ¥). (c) g is decreasing nowhere. 34.

g ( x) = -x3 + 6 x 2 - 12 x + 8

(a)

g ¢( x) = -3x 2 + 12 x - 12 = -3( x 2 - 4 x + 4)

f ¢(-1) = f ¢(1) =

2 <0 3

= -3( x - 2)2 g ¢( x) is zero when x = 2, so the critical number is 2.

2 >0 3

(b) f is increasing on (0, ¥). (c) f is decreasing on (-¥, 0). 32.

f ( x) = ( x + 1)4/5

(a)

f ¢( x ) =

g ¢(0) = -12 < 0 g ¢(3) = -3 < 0

4 4 ( x + 1)-1/5 = 5 5( x + 1)1/5

(b) g is increasing nowhere. (c) g is decreasing on (-¥, 2) and (2, ¥).

f ¢( x) is never zero, but fails to exist when x = -1, so the critical number is -1.

35.

h( x) = 5 - x1/3

(a)

1 1 h¢( x) = - x-2/3 = - 2/3 3 3x

316

Chapter 5 GRAPHS AND THE DERIVATIVE f ¢(4) = -3 < 0

h¢( x) is never zero, but fails to exist when x = 0, so 0 is the only critical number.

f (8) =

1 >0 5

(b) The function is increasing on (7, ¥). (c) The function is decreasing on (3, 7). 1 <0 3 1 h¢(1) = - < 0 3

h¢(-1) = -

38.

f ( x) = ln

5x2 + 4 x2 + 1

= ln(5 x 2 + 4) - ln(x 2 + 1)

(b) h is increasing nowhere.

(a)

f ¢( x ) =

36.

h( x) = ( x - 1)3/5

(a)

h¢( x) =

3 3 ( x - 1)-2/5 = 5 5( x - 1) 2/5

h¢( x) is never zero, but fails to exist when x = 1, so 1 is the only critical number.

10 x

5x + 4

2x 2

x +1

10 x( x + 1) 2

(5 x + 4)( x + 1)

2 x(5x 2 + 4) (5 x 2 + 4)( x 2 + 1)

10 x3 + 10 x - 10 x3 - 8 x (5 x 2 + 4)( x 2 + 1) 2x 2

(5 x + 4)( x + 1)

f ¢( x) is zero when x = 0 and there are no values of x where f ¢( x) does not exist, so the critical number is 0.

3 >0 5 3 h¢(2) = > 0 5 h¢(0) =

Test a point in each interval.

(b) h is increasing on (-¥, ¥).

f ¢(-1) =

f ¢(1) =

y = x - 4 ln (3x - 9)

(a)

12 4 y¢ = 1 = 13x - 9 x-3 x-7 = x-3 y¢ is zero when x = 7. The derivative does not exist at x = 3, but note that the domain of f is (3, ¥).

2(-1) 2

(5(-1) + 4) ((-1) + 1) 2(1) 2

(5(1) + 4) ((1) + 1)

1 <0 9

1 >0 9

(b) The function is increasing on (0, ¥). (c) The function is decreasing on (-¥, 0). 39.

f ( x) = xe-3x

(a)

f ¢( x) = e-3x + x(-3e-3x ) = (1 - 3x) e-3x

Thus, the only critical number is 7.

1 - 3x e3 x

f ¢( x) is zero when x = 13 and three are no

values of x where f ¢( x) does not exist, so Choose a value in the intervals (3, 7) and (7, ¥).

the critical number is 13 .

Section 5.1

317 f ¢( x) is zero when x = 0 or x = ln22

and there are no values of x where f ¢( x) does not exist. The critical numbers are 0 and ln22 . Test a point in each interval. 1 - 3(0)

f ¢(0) =

e3(0) 1 - 3(1)

f ¢(1) =

3(1)

=1> 0 2 =- 3 <0 e

Test a point in each interval.

(

)

(b) The function is increasing on -¥, 13 .

(

)

y = xe x

(a)

[ x(2 x - 3) + 1]

(

(2 x - 1)( x - 1)

y¢ is zero when x = 12 or x = 1, so the

critical numbers are 12 and 1.

42.

)

2 ,¥ . ln 2

f ( x) = x2-x

(a)

2 2 ö é æ öù æ f ¢ x  = x ê ln 2 çç 2-x (-2 x) ÷÷ ú + çç 2-x ÷÷ è ø úû è ø êë

= 2-x =

Test a point in each interval. f ¢0 = 1(-1)(-1) = 1 > 0 f ¢(0.6) = e

(0.2)(-0.4) =

e1.44

(

1 - 2 x 2 ln 2 2x

x2 =

)

(1, ¥).

( 12 , 1).

1 2 ln 2

x =

1 2 ln 2

and there are no values of x where f ¢( x) does not exist. The critical numbers are  1 . 2 ln 2

f ( x) = x 2 2-x

(a)

( -2x2 ln 2 + 1)

1 - 2 x 2 ln 2 = 0

(b) The function is increasing on -¥, 12 and

41.

f ¢( x) = 0 when

-0.08

3 f ¢ 2 = e-2 (3)(1) = 2 > 0 e

)

= e x -3x (2 x 2 - 3x + 1)

-1.44

(b) The function is increasing on 0, ln22 .

(

x 2 -3 x

= -2(2 + ln 2) < 0

2 - ln 2 >0 2 2 (3)(2 - (3) ln 2) 3(2 - 3ln 2) = <0 f ¢(3) = 3 8 2

y ¢ = xe x -3x (2 x - 3) + e x -3x (1) x 2 -3 x

2-1 (1)(2 - (1) ln2)

f ¢(1) =

2 - 3x

(-1)(2 - (-1) ln 2)

f ¢(-1) =

f ¢( x) = x 2[ln2(2-x )(-1)] + (2-x )2 x = 2-x (-x 2 ln2 + 2 x) x(2 - x ln2) = 2x

318

Chapter 5 GRAPHS AND THE DERIVATIVE Test a point in each interval. f ¢(-1) = f ¢(0) = f ¢(1) =

1 - 2(-1)2 ln 2 2-1 1 - 2(0)2 ln 2 20 1 - 2(1)2 ln 2 1

= 2(1 - 2 ln 2) < 0

=1> 0 =

Test a number in each interval. y¢(-1) = -1 < 0 æ 1ö 2 y¢ çç - ÷÷÷ = > 0 çè 8 ø 3

1 - 2 ln 2 <0 2

(b) The function is increasing on æ ö 1 1 ççç - 2 ln 2 , 2 ln 2 ÷÷÷. è ø

y¢ (1) =

(

(0, ¥) and the function is defined at x = 0. Thus, y is increasing on

( - 14 , ¥ ).

y = x 2/3 - x5/3 2 5 2 - 5x (a) y¢ = x-1/3 - x 2/3 = 3 3 3x1/3 y¢ = 0 when x = 52 . The derivative does

not exist at x = 0. So the critical numbers are 0 and 52 .

(

45.

f ( x) = ax 2 + bx + c, a > 0 f ¢( x) = 2ax + b

Let f ¢( x) = 0 to find the critical number.

Test a point in each interval. 7 <0 -3

b . Choose a value in the interval -¥, 2a

(

(5)

y ¢ (1 ) =

1 -b -b - 1 -b . = < 2a 2a 2a 2a æ -b - 1 ö÷ æ -b - 1 ö÷ f ¢ çç = 2a çç ÷ ÷+b ÷ èç 2a ø èç 2a ø÷

-3 = -1 < 0 3

(

)

(b) y is increasing on 0, 52 .

= -1 < 0

y = x

y¢ =

)

Since a > 0,

æ1ö 1 51/3 = >0 y¢ çç ÷÷÷ = 1/3 çè 5 ø 2 3 1

(a)

)

y is decreasing on -¥, - 14 .

(c)

2ax + b = 0 2ax = -b -b x = 2a

y¢(-1) =

44.

)

(b) The derivative is positive on - 14 , 0 and

5 >0 3

( 52 , ¥ ).

(

Since a > 0,

4/3

-b 1 -b + 1 -b + = > . 2a 2a 2a 2a æ -b + 1 ö÷ f ¢ çç =1> 0 çè 2a ø÷÷

1 -2/3 4 1 + 4x x + x1/3 = 3 3 3x 2/3

y¢ = 0 when x = - 14 . The derivative does

)

b,¥ . Choose a value in the interval 2a

not exist at x =  So the critical numbers

f ( x) is increasing on

are - 14 and 0.

b . on -¥, 2a

(

)

( -2ab , ¥ ) and decreasing

Section 5.1

319

This tells us that the curve opens upward and b is the x-coordinate of the vertex. x = 2a

æ -b ö÷ æ -b ÷ö2 æ -b ÷ö ç = f çç a ÷ ÷ + b ççç ÷+c çç èç 2a ÷ø è 2a ÷ø è 2a ÷ø

æ -b ö÷ æ -b ö÷2 æ -b ö÷ f çç = a çç + b çç ÷ ÷+c çè 2a ÷ø çè 2a ø÷÷ èç 2a ø÷

b2 +c 2a

b2 = +c 2a 4a 2

b2 2b 2 4bc + 4a 4a 4a

b2 2b 2 4ac = + 4a 4a 4a

4ac - b 2 4a

ab 2

The vertex is 46.

ab 2

(

b , 4ac -b The vertex is 2a 4a

4ac - b 2 4a

( , -b 2a

4ac -b 2 4a

) or ( - , b 2a

4ac -b 2 4a

47.

) or ( - , b2 2a

4ac -b 2 4a

f ( x) = e x f ¢( x ) = e x > 0

f ( x) = ax 2 + bx + c, a < 0

f ( x) = e x is increasing on (-¥, ¥).

f ¢( x) = 2ax + b

f ( x) = e x is decreasing nowhere.

Let f ¢( x) = 0 to find the critical number.

Since f ¢( x) is always positive, it never equals zero. Therefore, the tangent line is horizontal nowhere.

2ax + b = 0 2ax = -b x =

-b 2a

48.

(

b Choose a value in the interval -¥, - 2a

f ( x) = ln x f ¢( x) =

Since a < 0,

1 x

f ¢( x) is undefined at x = 0. f ¢( x) never

equals zero. Note that f ( x) has a domain of (0, ¥). Pick a value in the interval (0, ¥).

-b -1 -b + 1 -b . = < 2a 2a 2a 2a æ -b + 1 ö÷ æ -b + 1 ÷ö f ¢ çç = 2a çç +b çè 2a ÷÷ø çè 2a ÷÷ø

f ¢ 2 =

=1<0

Choose a value in the interval

f ( x) is increasing on (0, ¥).

( -2ab , ¥ ).

f ( x) is never decreasing.

Since a < 0, -b -1 -b - 1 -b . = < 2a 2a 2a 2a æ -b - 1 ÷ö æ -b - 1 ÷ö f ¢ çç = 2a çç +b ÷ ÷ çè 2a ÷÷ø èç 2a ø

Since f ( x) never equals zero, the tangent line is horizontal nowhere. 49.

(a) The critical number are -3 and 4. The average is -3 + 4 = 0.5. 2

= -1 > 0

(

b f is increasing on -¥, 2a

1 >0 2

) and decreasing on

( -2ab , ¥ ).

(b) Use a graphing calculator to find the zeros of the function f ( x) = 23 x3 - x 2 - 24 x - 4.

This tells us that the curve opens downward and b is the x-coordinate of the vertex. x= 2a

The roots are -5.2005, -0.1680, and 6.8685. The average is

320

Chapter 5 GRAPHS AND THE DERIVATIVE -5.2005 + (-0.1680) + 6.8685 = 0.5. 3

Since r is a mortgage rate (in percent), it is always positive. Thus, H ¢(r ) is always negative.

(a) H is increasing on nowhere.

(d) The critical numbers are - 32 and 4.

(b) H is decreasing on (0, ¥).

The average is - 32 + 4 2

54.

= 1.25.

C ( x) = x3 - 2 x 2 + 8x + 50 C ¢( x ) = 3 x 2 - 4 x + 8

(e) Use a graphing calculator to find the zeros

Since C ( x) is a polynomial function, the only critical numbers are found by solving C ¢( x) = 0.

of the function f ( x) = 4 x3 -15 x 2 -72 x +5. The roots are -2.8112, 0.0685, and 6.4927. The average is

3x 2 - 4 x + 8 = 0

-2.8112 + 0.0685 + 6.4927 = 1.25. 3

4  16 - 96 6 4  -80 = 6

x =

(f) The answers are the same. 51.

f ( x) = e0.001x - ln x f ¢( x) = 0.001e0.001x -

Since -80 is not a real number, there are no critical points. The function either always increases or always decreases.

1 x

Note that f ( x) is only defined for x > 0. Use a graphing calculator to plot f ¢( x) for x > 0. (a) (b)

Test any point, say x = 0. C¢(0) = 8 > 0

f ¢( x) > 0 on about (567, ¥), so f ( x) is

increasing on about (567, ¥).

(a)

C ( x) is decreasing nowhere.

f ¢( x) < 0 on about (0, 567), so f ( x) is

(b)

C ( x) is increasing on (-¥, ¥) .

decreasing on about (0, 567). 55. 52.

R( x) = 0.848 x 2 - 0.0002 x3

f ( x) = ln(x 2 + 1) - x 0.3 f ¢ x  =

2x 2

x +1

P ( x) = R ( x) = C ( x )

- 0.3x-0.7

= (0.848x 2 - 0.0002 x3 )

Note that f ( x) is only defined for x > 0. Use a graphing calculator to plot f ¢( x) for x > 0. (a)

f ¢( x) > 0 on about (0, 558), so f ( x) is

increasing on about (0, 558). (b)

f ¢( x) < 0 on (558, ¥), so f ( x) is

decreasing on (558, ¥). 53.

H (r ) =

300 1 + 0.03r

= 300(1 + 0.03r 2 )-1

-18r (1 + 0.03r 2 ) 2

- (0.32 x 2 - 0.00004 x3 ) = 0.528 x 2 - 0.00016 x3 P¢( x) = 1.056 x - 0.00048 x 2 1.056 x - 0.00048 x 2 = 0 x(1.056 - 0.00048 x) = 0 x = 0 or x = 2200

H ¢(r ) = 300[-1(1 + 0.03r 2 )-2 (0.06r )] =

C ( x) = 0.32 x 2 - 0.00004 x3

Choose x = 1000 and x = 3000 as test points.

P¢1000 = 1.056(1000) - 0.00048(1000)2 = 576 P¢3000 = 1.056(3000) - 0.00048(3000)2 = -1152

The function is increasing on (0, 2200).

Section 5.1 56.

321

P( x) = -( x - 4)e x - 4(0 < x £ 3.9)

59.

A( x) = 0.003631x3 - 0.03746 x 2 + 0.1012 x + 0.009

P¢ x  = -( x - 4)e x + e x (-1) = -e x[( x - 4) + 1]

A¢( x) = 0.010893x 2 - 0.07492 x + 0.1012

= -e x ( x - 3)

Solve for A¢( x) = 0.

P¢( x) = 0 when x = 3.

x » 1.85 or x » 5.03

Choose x = 1 and x = 4 as test points. A¢(1) = 0.010893(1)2 - 0.07492(1) + 0.1012 = 0.037173

P¢(1) = -e1(-2) = 2e > 0 P¢(3.5) = -e3.5 (0.5) =

-0.5 e3/5

A¢(4) = 0.010893(4) 2 - 0.07492(4) + 0.1012 = - 0.024192

(a) The function is increasing on (0, 1.85).

(a) The profit is increasing on (0, 3), which represents production between 0 and 300 phones. (b) The profit is decreasing on (3, 3.9), which represents production between 300 and 390 phones. 57.

(b) The function is decreasing on (1.85, 5). 60.

K ( t) = K ¢t  =

A(t ) = 0.0000329t 3 - 0.00450t 2 + 0.0613t + 2.34

A¢(t ) = 0.0000987t 2 - 0.009t + 0.0613

Set A¢(t ) equal to 0 and solve for t.

4t 2

3t + 27 (3t 2 + 27)4 - 4t (6t ) (3t 2 + 27)2 108 - 12t 2 (3t 2 + 27) 2

K ¢t is zero when 108 - 12t 2 = 0

0.0000987t 2 - 0.009t + 0.0613 = 0 t =

0.009 

12(9 - t 2 ) = 0

(0.009)2 - 4(0.0000987)(0.0613) (2)(0.0000987)

(3 - t )(3 + t ) = 0 t = 3 or t = -3.

t » 83.8 or t = 7.4

However, in this problem t represents time, which cannot be negative, so 3 is the only critical number.

Since 0 £ t £ 50, t » 7.4. Check the sign of A¢ on either side of this critical number. A¢(7) » 0.00314 > 0 A¢(8) » -0.00438 < 0

K ¢(0) =

(a) The projected year-end assets function is increasing on the interval (0, 7.4), or from 2000 to about the middle of 2007. (b) The projected year-end assets function is decreasing on the interval (7.4, 50), or from about the middle of 2007 to 2050. 58.

Following the graph from left to right we find that the function is increasing on (2000, 2003) and (2007, 2010); the function is decreasing on (2003, 2007) and (2010, 2019).

K ¢(4) =

(a)

108

>0 27 2 -84 (48 + 27)2

K (t ) is increasing on (0, 3).

(Note: t must be at least 0.) (c)

K (t ) is decreasing on (3, ¥).

322 61.

Chapter 5 GRAPHS AND THE DERIVATIVE K (t ) = K ¢(t ) = = =

63.

(a)

t +1

F (t ) = -10.28 + 175.9te-t /1.3 F ¢(t ) = (175.9)(e-t /1.3 ) æ 1 ö + (175.9.9t ) çç - e-t /1.3 ÷÷÷ çè 1.3 ø

5(t + 1) - 2t (5t ) (t 2 + 1) 2 5t 2 + 5 - 10t 2 2

(t + 1)

æ t ö÷ = (175.9)(e-t /1.3 ) çç1 ÷ çè 1.3 ÷ø

5 - 5t 2

» 175.9e-t /1.3 (1 - 0.769t )

(t 2 + 1) 2

(b)

K ¢(t ) = 0 when

5 - 5t 2 (t 2 + 1) 2

F ¢(t ) is equal to 0 at t = 1.3. Therefore, 1.3 is a critical number. Since the domain is (0, ¥), test values in the intervals from (0, 1.3) and (1.3, ¥).

F ¢ » 18.83 > 0 and F ¢(2) » -20.32 < 0

5 - 5t 2 = 0

F ¢t  is increasing on (0, 1.3) and

5t 2 = 5 t = 1.

decreasing on (1.3, ¥).

Since t is the time after a drug is administered, the function applies only for [0, ¥), so we discard t = -1. Then 1 divides the domain into two intervals.

W1(t ) = 619(1 - 0.905e-0.002t )1.2386

64.

W ¢1(t ) = 1.388e-0.002t (1 - 0.905e-0.002t ) 0.2386 > 0

Since W ¢1(t ) > 0 and t is a positive number, this function is increasing on (0, ¥). N0 - N 2 DN Use the chain rule and quotient rule.

(a) R =

65.

K ¢(0.5) = 2.4 > 0 K ¢ 2 = -0.6 < 0

(a) K is increasing on (0, 1).

dR dN

(b) K is decreasing on (1, ¥).

D¢ p  = 0.000006 p 2 - 0.0016 p + 0.1141

æ ö çç (2 DN )(-1) - ( N 0 - N )(2 D) ÷÷ ÷÷ ç 2 N 0 - N èç (2 DN ) ø÷ 2 2 DN æ -N 1 ÷ö çç 0 = ÷÷÷ ç 2 N 0 - N çè (2 DN ) ÷ø 2 2 DN

By the quadratic formula, there are no real number solutions for p when D¢( p) = 0.

62.

D( p) = 0.000002 p3 - 0.0008 p 2 + 0.1141 p + 16.683

Choose a value in the interval (55, 130). =

D¢(60) = 0.0397 > 0

The function is increasing on the entire interval (55, 130), so it is decreasing nowhere.

-N 0 N0 - N 2N ( 2 DN )( 2 D ) 2 DN -N 0 2 N 3/ 2 2 D( N 0 - N )

(b) As the number of plants increases, the maximum plant radius decreases. (c) As N approaches 0, dR/dN approaches -¥. As N approaches N 0 , dR/dN approaches -¥.

Section 5.1 66.

323

P(t ) =

10 ln (0.19t + 1) 0.19t + 1

67.

1 æ (0.19t + 1) ⋅ ⋅ 0.19 - 0.19 ln (0.19t + 1) ö÷÷ ç 0.19t +1 ÷÷ P¢t = 10 çç çç (0.19t + 1)2 ÷÷ø è

æ 0.19 - 0.19 ln (0.19t + 1) ö÷ ÷÷ = 10 ççç (0.19t + 1)2 èç ø÷÷ æ 1 - ln (0.19t + 1) ö÷ ÷÷ = 1.9 ççç 2 ÷ø÷ èç (0.19t + 1)

1 -x 2 /2 e 2 1 -x 2 /2 (-x ) f ¢( x) = e 2 -x -x 2 /2 e = 2 f ( x) =

f ¢( x) = 0 when x = 0.

Choose a value from each of the intervals (-¥, 0) and (0, ¥).

Note that P(t ) is defined when 0.19t + 1 > 0,

1 -1/2 e >0 2 -1 -1/2 <0 f ¢(1) = e 2 f ¢(-1) =

-1 » -5.26, so the which implies that t > 0.19 100 » -5.26}. domain of P is {t t > -0.19

Find critical numbers. Set numerator equal to 0. 1 - ln (0.19t + 1) = 0 ln (0.19t + 1) = 1 0.19t + 1 = e e -1 t = » 9.04 0.19

The function is increasing on (-¥, 0) and decreasing on (0, ¥). 68.

So, P¢(t ) = 0 when t = 100 (e - 1) » 9.04. Set 19

(b) The total inventories of both countries were decreasing on the following approximate interval: (1987, 2019).

denominator equal to 0. (0.19t + 1) 2 = 0 0.19t + 1 = 0 -1 t = » -5.26 0.19

69.

(a) (2500, 5750) (b) (5750, 6000) (c) (2800, 4800)

100 » -5.26 is not in the domain of P, Since -19 100 (e - 1) is the only critical number. 19

(a) The total inventories of both countries were increasing on the following approximate intervals: (1945, 1965) and (1970, 1973).

(d) (2500, 2800) and (4800, 6000) 70.

(a) As x increases, f ( x) decreases, so f is decreasing for these x-values. Therefore, f ¢ x  is negative. (b)

mpg , or miles per gallon per pound. lb

Test a number in each interval. æ 1 - ln(1) ö÷ P¢(0) = 1.9 çç ÷÷ = 1.9 > 0 çè 12 ø æ 1 - ln(2.9) ö÷ P¢(10) = 1.9 çç ÷<0 ÷ø çè 2.92

since ln (2.9) > ln (e) = 1. (e - 1) ) ( -19100 , 100 19 and decreases on ( 100 (e - 1), ¥ ) , so the 19

Therefore, P increases on

324

Chapter 5 GRAPHS AND THE DERIVATIVE

5.2 Relative Extrema Your Turn 1

There is an open interval containing x1 for which f ( x) ³ f ( x1). There is an open interval containing x2 for which f ( x) £ f ( x2 ). There is an open interval containing x3 for which f ( x) ³ f ( x3 ). There are relative minima of f ( x1) at x = x1 and f ( x3 ) at x = x3 and a relative maximum of f ( x2 ) at x = x2. Your Turn 2 f ( x) = -x3 - 2 x 2 + 15x + 10 f ¢( x) = -3x 2 - 4 x + 15

Set f ¢( x ) equal to 0 and solve for x to find the critical numbers.

By the first derivative test, we know that the function has a relative minimum of f (-3) = -26 at x = -3 and a relative maximum of f (5/3) = 670/27 » 24.8 at x = 5/3. Your Turn 3 f ( x) = x 2/3 - x5/3 2 5 f ¢( x) = x-1/3 - x 2/3 3 3 1 æç 2 - 5x ö÷ = ç 1/3 ÷ ÷ø 3 çè x x = 0 is a critical number because f ( x ) is defined at

-3x - 4 x + 15 = 0

x = 0 but f ¢( x ) is not. To find any other critical

-(3x 2 + 4 x - 15) = 0 -(3x - 5)( x + 3) = 0

numbers, assume x ¹ 0 and set f ¢( x ) = 0 and solve for x.

3x - 5 = 0 or x + 3 = 0 x = 5/3 or x = -3

1 æç 2 - 5x ö÷ ÷= 0 ç 3 çè x1/3 ø÷ 2 - 5x = 0 2 x = 5

The critical numbers are -3 and 5/3. They divide the domain of f into the three intervals (-¥, -3), (-3, 5/3), and (5/3, ¥). Pick test points in each interval to determine the sign of f ¢ on that interval. For example, choose x = -4, x = 0, and x = 2. f ¢(-4) = -17 f ¢(0) = 15

Use the critical numbers to divide the line into three intervals, (-¥, 0), (0, 2/5), and (2/5, ¥). Choose test points in each interval; for example, choose -1, - 1/5, and 1. f ¢(-1) = -2.33 f ¢(1/5) = 0.57

f ¢(2) = -5

Thus the derivative is negative on (-¥, - 3), positive on (-3, 5/3), and negative on (5/3, ¥). The graph below shows this information and indicates where f is increasing or decreasing.

f ¢(1) = -1

Thus the derivative is negative on (-¥, -1), positive on (-1, 2/5), and negative on (2/5, ¥). The graph below shows this information and indicates where f is increasing or decreasing.

Section 5.2

325

The first derivative test identifies f (2/5) as a relative maximum, and there is a relative minimum at x = 0. Thus the function has a relative maximum of 2/3 æ2ö 3æ 2 ö 2 f çç ÷÷ = çç ÷÷ » 0.3257 at x = ÷ ç èç 5 ÷ø è ø 5 5 5

The function has a relative maximum of 4e-2 » 0.5413 at x = -2 and a relative minimum of f (0) = 0 at x = 0. Your Turn 5 C (q) = 100 + 10q p = D(q) = 50 - 2q P (q ) = R (q ) - C (q )

and a relative minimum of f (0) = 0 at x = 0.

= qD(q) - C (q) = q(50 - 2q) - (100 + 10q)

Your Turn 4 f ( x ) = x 2e x

= 50q - 2q 2 - 100 + 10q

Using the product rule: 2 x

f ¢( x ) = x e + 2 xe

= -2q 2 + 40q - 100

P¢(q) = -4q + 40 = -4(q - 10)

= e x ( x2 + 2x)

P ¢ is 0 only when q = 10, and this is the only critical number. P¢(5) = 20 and P(20) = -40 so P is increasing as q approaches 10 from the left and decreasing to the right of q = 10. The value of P at q = 10 is P(10) = -2(100) + 40(1) - 100 = 100.

Find the critical values: e x ( x 2 + 2 x) = 0 x2 + 2x = 0 x( x + 2) = 0 x + 2 = 0 or x = 0 x = -2 or x = 0

There are two critical numbers, dividing the line into the intervals (-¥, - 2), (-2, 0) and (0, ¥). Choose a test point in each interval, for example -3, -1, and 1. f ¢(-3) = 0.149 f ¢(-1) = -0.368

Thus the maximum weekly profit is $100 when the demand is 10 items per week. The company should charge D(10) = 50 - 2(10) = 30, or $30 per item.

5.2

Warmup Exercises

W1.

f ( x ) = 2 x 3 - 3x 2 - 36 x + 4 f ¢( x ) = 6 x 2 - 6 x - 36 = 6( x 2 - x - 6) = 6( x - 3)( x + 2)

f ¢(1) = 8.155

Thus the derivative is positive on (-¥, - 2), negative on (-2, 0), and positive on (0, ¥). The graph below shows this information and indicates where f is increasing or decreasing.

f ¢( x ) = 0 when ( x - 3)( x + 2) = 0, that is, when x = 3 and x = -2. These points divide the real line into three intervals: (-¥, -2), (-2, 3), (3, ¥)

326

Chapter 5 GRAPHS AND THE DERIVATIVE 8.

As shown on the graph, the relative minimum of -4 occurs when x = 3.

As shown on the graph, the relative maximum of 3 occurs when x = -4 and the relative minimum of 1 occurs when x = -2.

10.

As shown on the graph, the relative minimum of -6 occurs when x = 1 and the relative maximum of 2 occurs when x = 5.

11.

that is, when x = -3/2, x = 0, and x = 4. These points divide the real line into four intervals: (-¥, -3/2), (-3/2, 0), (0, 4), (4, ¥)

As shown on the graph, the relative maximum of 3 occurs when x = -4; the relative minimum of -2 occurs when x = -7 and x = -2 .

12.

As shown on the graph, the relative maximum of 4 occurs when x = 0; the relative minimum of 0 occurs when x = -3 and x = 3.

Evaluate f ¢ at a test point in each interval, say x = -2, x = -1, x = 1, and x = 5.

13.

Since the graph of the function is zero at x = -1 and x = 3, the critical numbers are -1 and 3.

f ¢(-4) = 84 > 0 f ¢(0) = -36 < 0 f ¢(4) = 36 > 0 Thus f is increasing on (-¥, -2) and (3, ¥) and is decreasing on (-2, 3).

W2.

f ( x) = 3x 4 - 10 x3 - 36 x 2 - 72 f ¢( x) = 12 x3 - 30 x 2 - 72 x = 6 x(2 x 2 - 5x - 12) = 6 x(2 x + 3)( x - 4)

f ¢( x) = 0 when 6 x(2 x + 3)( x - 4) = 0,

f ¢(-2) = -72 < 0 f ¢(-1) = 30 > 0

Since the graph of the derivative is positive on (-¥, -1) and negative on (-1, 3), there is a relative maximum at -1. Since the graph of the function is negative on (-1, 3) and positive on (3, ¥), there is a relative minimum at 3.

f ¢(1) = -90 < 0 f ¢(5) = 390 > 0

Thus f is increasing on (-3/2, 0) and (4, ¥) and is decreasing on (-¥, - 3/2) and (0, 4).

5.2

Exercises

False. An absolute maximum is the greatest possible value of a function.

True

False. A function has relative at most relative extrema at the critical numbers.

False. A function can have only one absolute maximum and one absolute minimum.

As shown on the graph, the relative minimum of -4 occurs when x = 1.

As shown on the graph, the relative maximum of 1 occurs when x = 4.

As shown on the graph, the relative maximum of 3 occurs when x = -2.

14.

Since the graph of the derivative is zero at x = 3 and x = 5, the critical numbers are 3 and 5. Since the graph of the derivative is negative on (-¥, 3) and positive on (3, 5), there is a relative minimum at 3. Since the graph of the derivative is positive on (3, 5) and negative on (5, ¥), there is a relative maximum at 5.

15.

Since the graph of the derivative is zero at x = -8, x = -6, x = -2.5 and x = -1.5, the critical numbers are -8, -6, -2.5, and -1.5. Since the graph of the derivative is positive on (-¥, -8) and negative on (-8, -6), there is a relative maximum at -8. Since the graph of the derivative is negative on (-8, -6) and positive on (-6, -2.5), there is a relative minimum at -6. Since the graph of the derivative is positive on (-6, -2.5) and negative on (-2.5, -1.5), there is a relative maximum at -2.5. Since the graph of the derivative is negative on (-2.5, -1.5) and positive on (-1.5, ¥), there is a relative minimum at -1.5.

Section 5.2 16.

327

Since the graph of the derivative is zero at x = -3 and x = 3, the critical numbers are -3 and 3.

f ¢( x) is zero when x = -1 or x = -3.

Since the graph of the derivative is positive on (-¥, -3), (-3, 3), and (3, ¥), there are no relative extrema. 17.

f ¢(-4) = 9 > 0 f ¢(-2) = -3 < 0 f ¢(0) = 9 > 0

f ( x) = x - 10 x + 33 f ¢( x) = 2 x - 10 f ¢( x) is zero when x = 5.

Thus, f is increasing on (-¥, -3), decreasing on (-3, -1), and increasing on (-1, ¥). f has a relative maximum at -3 and a relative minimum at -1. f (-3) = -8 f (-1) = -12

f ¢(0) = -10 < 0 f ¢(6) = 2 > 0

Relative maximum of –8 at –3 relative minimum of –12 at –1

f is decreasing on (-¥, 5) and increasing on (5, ¥). Thus, a relative minimum occurs at x = 5.

f ( x) = x3 + 3x 2 - 24 x + 2 f ¢( x) = 3x 2 + 6 x - 24 = 3( x 2 + 2 x - 8) = 3( x + 4)( x - 2)

f (5) = 8

Relative minimum of 8 at 5. 18.

20.

f ¢( x) is zero when x = -4 and x = 2.

f ( x) = x + 8 x + 5 f ¢( x ) = 2 x + 8 f ¢( x) is zero when x = -4. f ¢(-5) = 21 > 0 f ¢(0) = -24 < 0 f ¢(3) = 21 > 0

f ¢(-5) = -2 < 0 f ¢(0) = 8 > 0

f is decreasing on (-¥, -4) and increasing on (-4, ¥). Thus, a relative minimum occurs at x = -4. f (-4) = -11

Relative minimum of -11 at -4 19.

f ( x) = x 3 + 6 x 2 + 9 x - 8 f ¢( x) = 3x 2 + 12 x + 9 = 3( x 2 + 4 x + 3) = 3( x + 3)( x + 1)

f is increasing on (-¥, -4) and decreasing on (-4, 2). Thus, a relative maximum occurs at x = -4. f ( x) is decreasing on (-4, 2) and increasing on (2, ¥). Thus, a relative minimum occurs at x = 2. f (-4) = (-4)3 + 3(-4)2 - 24(-4) + 2 = 82 f (2) = (2)3 + 3(2) 2 - 24(2) + 2 = -26

Relative maximum of 82 at -4; relative minimum of -26 at 2.

328 21.

Chapter 5 GRAPHS AND THE DERIVATIVE f is decreasing on (-¥, -5), increasing on

g ( x) = x 3 + 3x 2 + 3 x + 4 g ¢( x ) = 3 x 2 + 6 x + 3

( -5, - 14 ), and decreasing on ( - 14 , ¥ ). f has a

= 3( x 2 + 2 x + 1)

relative minimum at -5 and a relative maximum at - 14 .

= 3( x + 1) 2

f ¢( x) is zero when x = –1.

377 6 æ 1 ö÷ 827 f çç - ÷÷ = çè 4 ø 96

f (-5) = -

Relative maximum of 827 at - 14 ; relative 96

g ¢(-2) = 3 > 0 g ¢(0) = 3 > 0

minimum of - 377 at -5. 6

g is increasing on (-¥, -1) and (-1, ¥) and decreasing nowhere. Thus, there are no relative extrema. 22.

24.

2 1 f ( x) = - x 3 - x 2 + 3 x - 4 3 2 f ¢ ( x ) = -2 x 2 - x + 3 = -(2 x 2 + x - 3)

g ( x) = -x3 + 6 x 2 - 12 x - 3

= -(2 x + 3)( x - 1)

g ¢( x) = -3x 2 + 12 x - 12

f ¢( x) is zero when x = - 32 or x = 1.

= -3( x - 4 x + 4) = -3( x - 2) 2

g ¢( x) is zero when x = 2. f ¢(-2) = -3 < 0 f ¢(0) = 3 > 0 f ¢(2) = -7 < 0

g ¢(0) = -12 < 0 g ¢(3) = -3 < 0

(

f is decreasing on -¥, - 32

g is decreasing on (-¥, 2) and (2, ¥) and increasing nowhere. Thus, there are no relative extrema. 23.

4 21 2 f ( x) = - x 3 x - 5x + 8 3 2

) and increasing on

( - 32 , 1). Thus, a relative minimum occurs at x = - 32 . f ( x) is increasing on ( - 32 , 1) and decreasing on (1, ¥). Thus, a relative maximum occurs at x = 1. æ 3ö 59 f çç - ÷÷÷ = çè 2 ø 8

f ¢( x) = -4 x 2 - 21x - 5 = (-4 x - 1)( x + 5)

f (1) = -

f ¢( x) is zero when x = -5, or x = - 14 .

13 6

Relative maximum of - 13 at 1; relative 6 minimum of - 59 at - 32 . 8 f ¢(-6) = -23 < 0 f ¢(-4) = 15 > 0 f ¢(0) = -5 < 0

Section 5.2 25.

329

f ( x) = x 4 - 18x 2 - 4

f ( x) = 3 - (8 + 3x)2/3

27.

2 f ¢( x) = - (8 + 3x)-1/3 (3) 3 2 =(8 + 3x)1/3

f ¢( x) = 4 x3 - 36 x = 4 x( x 2 - 9) = 4 x( x + 3)( x - 3) f ¢( x) is zero when x = 0 or x = -3 or x = 3.

Critical number: 8 + 3x = 0 x =-

8 3

f ¢(-4) = 4(-4)3 - 36(-4) = -112 < 0 f ¢(-1) = -4 + 36 = 32 > 0 f ¢(1) = 4 - 36 = -32 < 0 f ¢(4) = 4(4)3 - 36(4) = 112 > 0

f ¢(-3) = 2 > 0

f is decreasing on (-¥, -3) and (0, 3); f is increasing on (-3, 0) and (3, ¥).

f ¢(0) = -1 < 0

f (-3) = -85

f is increasing on -¥, - 83 and decreasing on

(

( - 83 , ¥ ).

f (0) = -4 f (3) = -85

æ 8ö f çç - ÷÷÷ = 3 çè 3 ø

Relative maximum of –4 at 0; relative minimum of –85 at 3 and –3. 26.

)

Relative maximum of 3 at - 83 .

f ( x) = x - 8 x - 9 f ¢( x) = 4 x3 - 16 x

28.

= 4 x( x 2 - 4) = 4 x( x + 2)( x - 2) f ¢( x) is zero when x = 0 or x = -2 or x = 2.

(5 - 9 x)2/3 +1 7 æ 2 ö (5 - 9 x)-1/3 (-9) f ¢( x) = çç ÷÷÷ çè 3 ø 7 f ( x) =

6 7(5 - 9 x)1/3

Critical number: 5 - 9x = 0 x =

f ¢(-3) = -60 < 0 f ¢(-1) = 12 > 0

5 9

f ¢(1) = -12 < 0 f (3) = 60 > 0

f is increasing on (-2, 0) and (2, ¥); f is decreasing on (-¥, -2) and (0, 2).

f ¢(0) » -0.5013 < 0

f (-2) = -7

f ¢(1) » 0.5400 > 0

f (0) = 9

(

)

f is decreasing on -¥, 95 and increasing on

f (2) = -7

Relative maximum of 9 at 0; relative minimum of –7 at –2 and 2.

(

)

5,¥ . 9

330

Chapter 5 GRAPHS AND THE DERIVATIVE æ5ö f çç ÷÷÷ = 1 çè 9 ø

30.

f ¢( x) = 5x 2/3 - 10 x-1/3

Relative minimum of 1 at 95 . 29.

f ( x) = 2 x + 3x

f ( x) = 3x5/3 - 15x 2/3 = 5x-1/3 ( x - 2) =

2/3

5( x - 2)

x1/3 Find the critical numbers:

f ¢( x) = 2 + 2 x-1/3 2 = 2+ 3 x

5( x - 2) = 0

x1/3 = 0

x = 2

x =0

Find the critical numbers. f ¢( x) = 0 when 2 2+ 3 = 0 x 2 = -2 3x

f ¢(-1) = 15 > 0 f ¢(1) = -5 < 0 f ¢(3) = 3.47 > 0

x = (-1)3 x = -1.

f is increasing on (-¥, 0) and (2, ¥).

f ¢( x) does not exist when

f is decreasing on (0, 2).

3 x = 0, that is, when x = 0.

f ( x) = 3x 2/3 ( x - 5) f (0) = 3 ⋅ 0(0 - 5) = 0 f (2) = 3 ⋅ 22/3 (2 - 5) = -9 ⋅ 22/3 » -14.287

f ¢(-2) = 2 + 3

Relative maximum of 0 at 0; relative minimum

2 -2 2

of -9 ⋅ 22/3 » -14.287 at 2

» 0.41 > 0

æ 1ö f ¢ çç - ÷÷÷ = 2 + çè 2 ø 3-1

31.

23 2 » -0.52 < 0 -1 2 f ¢(1) = 2 + 3 = 4 > 0 1 = 2+

h( x) = 5 - x1/3 1 1 h¢( x) = - x-2/3 = - 2/3 3 3x

Find the critical numbers. h¢( x) is never zero, but fails to exist when x = 0, so 0 is the only critical number.

f is increasing on (-¥, -1) and (0, ¥). f is decreasing on (-1, 0). f (-1) = 2(-1) + 3(-1)2/3 = 1 f (0) = 0

1 <0 3 1 h¢(1) = - < 0 3

h¢(-1) = -

Relative maximum of 1 at -1; relative minimum of 0 at 0.

h is decreasing on (-¥, ¥) and increasing nowhere. Thus, there are no relative extrema.

Section 5.2 32.

331

(

h( x) = ( x - 1)3/5 3 3 h¢( x) = ( x - 1)-2/5 = 5 5( x - 1) 2/5

f ( x) is decreasing on -¥, 24

) and increasing

( , ¥ ). 3

4 2

Find the critical numbers. æ 3 4 ö÷ æ 3 4 ö÷2 æ 3 4 ö÷-1 ÷÷ = çç ÷÷ + çç ÷ f ççç çç 2 ÷÷÷ çè 2 ÷÷ø ççè 2 ÷÷ø è ø ù æ 3 4 ö÷-1 éê æ 3 4 ö÷3 ú ÷÷ ê çç ÷÷ + 1 ú = ççç ç ÷ ÷ ú çè 2 ÷ø ê çè 2 ÷ø ëê ûú

h¢( x) is never zero, but fails to exist when x = 1, so 1 is the only critical number.

ö 2 æ4 2 æ3ö 32 = 3 çç + 1 ÷÷ = 3 çç ÷÷ ⋅ 3 ø÷ 4 èç 8 4 çè 2 ÷ø 2

3 >0 5 3 h¢(2) = > 0 5 h¢(0) =

h is increasing on (-¥, ¥) and decreasing nowhere. Thus, there are no relative extrema. 33.

f ( x) = x -

1 x x

at x = 0 Since f ( x) also fails to exist at x = 0, there are no critical numbers and no relative extrema. f ( x) = x 2 +

1 x 1

35.

f ( x) =

x2 - 2x + 1 x-3 ( x - 3)(2 x - 2) - ( x 2 - 2 x + 1)(1) ( x - 3)2 x2 - 6x + 5 ( x - 3) 2

Find the critical numbers: x2 - 6x + 5 = 0 ( x - 5)( x - 1) = 0 x = 5 or x = 1

Note that f ( x) and f ¢( x) do not exist at x = 3, so the only critical numbers are 1 and 5.

f ¢( x) = 2 x - 2 x =

Relative minimum of 3 22 » 1.890 at 24 .

f ¢( x ) =

f ¢( x) = 1 + 12 is never zero, but fails to exist

34.

33 2 » 1.890 2

2 x3 - 1

x2 Find the critical number: 2 x3 - 1 = 0 1

5 >0 9 f ¢(2) = -3 < 0 5 f ¢(6) = > 0 9

f ¢(0) =

4 x = 3 = » 0.79 2 2

Note that both f ( x) and f ¢( x) do not exist at x = 0, so 0 is not a critical number.

f ( x) is increasing on (-¥, 1) and (5, ¥). f ( x) is decreasing on (1, 5).

f ¢(-1) = -3 < 0 f ¢(1) = 1 > 0

f (1) = 0 f (5) = 8

332

36.

Chapter 5 GRAPHS AND THE DERIVATIVE

f ( x) = f ¢( x ) = = =

f is increasing on (-¥, -2) and (0, ¥).

x2 - 6x + 9 x+2

f is decreasing on (-2, 0). 2

(2 x - 6)( x + 2) - (1)( x - 6 x + 9)

f (0) = 0 ⋅ e0 - 3 = -3

( x + 2)2

f (-2) = (-2) 2 e-2 - 3 4 = 2 -3 e » -2.46

x 2 + 4 x - 21 ( x + 2) 2 ( x + 7)( x - 3) ( x + 2) 2

Relative minimum of -3 at 0; relative maximum of -2.46 at -2.

Find the critical numbers: ( x + 7)( x - 3) = 0 x = -7 or x = 3

38.

Note that f ( x) and f ¢( x) do not exist at x = -2, so the only critical numbers are -7 and 3.

f ( x) = 3xe x + 2 f ¢( x) = 3xe x + 3e x = 3e x ( x + 1)

Find the critical numbers: 3e x = 0 e =0

11 >0 36 f ¢(-3) = -24 < 0 21 f ¢(0) = <0 4 11 f ¢(4) = >0 36 f is increasing on (-¥, -7) and (3, ¥). f ¢(-8) =

= -3e-2 < 0 f ¢(0) = 3e0 (0 + 1) =3>0

f is increasing on (-1, ¥) and decreasing on (-¥, -1). f (-1) = 3(-1) e-1 + 2

f ¢( x) = x 2e x + 2 xe x

-3 + 2 » 0.9 e Relative minimum of 0.90 at -1 =

= xe ( x + 2) f ¢( x) is zero at x = 0 and x = -2.

39.

f ( x) = 2 x + ln x f ¢( x ) = 2 +

f ¢(-3) = 3e

x = -1

f ¢(-2) = 3e-2 (-2 + 1)

f ( x ) = x 2e x - 3

-3

e x is always positive, so the only critical number is -1.

f is decreasing on (-7, -2) and (-2, 3). f (-7) = -20 f (3) = 0 Relative maximum of -20 at -7; relative minimum of 0 at 3. 37.

or x + 1 = 0

>0 e3 -1 f ¢(-1) = -e-1 = <0 e f ¢(1) = 3e1 > 0

1 2x + 1 = x x

f ¢( x) is zero at x = - 12 . The domain of f ( x)

is (0, ¥). Therefore f ¢( x) is never zero in the domain of f ( x). f ¢(1) = 3 > 0. Since f ( x) is always increasing, f has no relative extrema.

Section 5.2

40.

333

f ( x) = f ¢( x ) = =

Since f is defined for x = 0, 0 is not a critical

x2 ln x

number. x = ln1 2 » 1.44 is the only critical

( ) = 2x ln x - x

(ln x) 2 x - x 2 1x 2

(ln x) x(2 ln x - 1)

(ln x)

number.

(ln x)2

Find the critical numbers: x = 0 or 2 ln x - 1 = 0 or ln x = 0 2 ln x = 1 x =1 1 ln x = 2

f ¢(1) » -0.6137 < 0 f ¢(2) » 0.3863 > 0

(

)

f is decreasing on 0, ln12 and increasing on

x = e1/2

(

» 1.65

)

1 ,¥ . ln 2 1/ln2

Since the domain of y = ln x is (0, ¥), 0 is not a critical number. Since ln 1 = 0, we see that 1 is also not in the domain of f . Thus is the only critical number.

( ln21 ) = 2ln21

e » 1.65

(

= ln 2 eln 2

1/ ln 2

)

= e ln2

Relative minimum of e ln 2 at ln12 .

f ¢(1.5) = f ¢(2) =

1.5(2ln1.5 - 1) (ln 1.5)2 2(2ln2 - 1) (ln2)2

42. <0

f ( x) is increasing on (1.65, ¥) and decreasing

on (1, 1.65). f (1.65) =

(1.65)2 » 5.44 ln (1.65)

f ( x) = x + 8-x f ¢( x) = 1 + (ln 8)8-x (-1) = 1-

8x Find the critical numbers: ln 8 1- x = 0 8 ln 8 1= x 8 8 x = ln 8

Relative minimum of 5.44 at 1.65 41.

f ( x) = f ¢( x ) = =

ln 8

x ln 8 = ln (ln 8)

2 x

ln (ln 8) » 0.35 ln 8

x =

( x) ln 2(2 x ) - 2 x (1) x2 2 x ( x ln 2 - 1)

x2 Find the critical numbers: x ln2 - 1 = 0 1 x = ln2

f ¢(0) » -1.0794 < 0

x2 = 0 x =0

f ¢(1) » 0.7401 > 0

(

f is decreasing on -¥,

( ln(lnln 88) , ¥ ).

ln(ln 8) ln 8

) and increasing

334

Chapter 5 GRAPHS AND THE DERIVATIVE Note that

Graph f ¢ on a graphing calculator. A suitable choice for the viewing window is [-4, 4] by [-50, 50], Yscl = 10.

8-ln(ln8) / ln8 = e(ln8)(-ln(ln8) / ln8) = e-ln(ln8) =

1 . ln 8

Use the calculator to estimate the x-intercepts of this graph. These numbers are the solutions of the equation f ¢( x) = 0 and thus the critical numbers for f . Rounded to three decimal places, these x-values are 0.085 and 2.161.

æ ln(ln 8) ö÷ ln(ln 8) = + 8-ln(ln 8)/ln 8 f çç çè ln 8 ø÷÷ ln 8 ln (ln 8) 1 + ln 8 ln 8 ln (ln 8) + 1 = ln 8 =

Relative minimum of 43.

Examine the graph of f ¢ near x = 0.085 and x = 2.161. Observe that f ¢( x) > 0 to the left of x = 0.085 and f ¢( x) < 0 to the right of x = 0.085. Also observe that f ¢( x) < 0 to the left of x = 2.161 and f ¢( x) > 0 to the right of x = 2.161. The first derivative test allows us to conclude that f has a relative maximum at x = 0.085 and a relative minimum at x = 2.161. f (0.085) » 6.211 f (2.161) » -57.607

ln (ln 8) +1 ln (ln 8) at ln 8 . ln 8

y = -2 x 2 + 12 x - 5 y¢ = -4 x + 12 = -4( x - 3)

The vertex occurs when y¢ = 0 or when x-3= 0

Relative maximum of 6.211 at 0.085; relative minimum of -57.607 at 2.161.

x =0

When x = 3,

46.

y = -2(3) 2 + 12(3) - 5 = 13

The vertex is (3, 13). 44.

[-100, 100], Yscl = 20.

y ¢ = 2ax + b

Use the calculator to estimate the x-intercepts of this graph. These numbers are the solutions of the equation f ¢( x) = 0 and thus the critical numbers for f. Rounded to three decimal places, these x-values are 0.183 and -2.703.

The vertex occurs when y¢ = 0. 2ax + b = 0 x =

-b 2a

æ -b ö÷2 æ -b ö÷ ab 2 b2 a çç + b çç +c = +c ÷ ÷ çè 2a ø÷ èç 2a ø÷ 2a 4a 2

45.

( , -b 2a

f ¢( x) = -5 x 4 - 4 x3 + 6 x 2 - 50 x + 9

Graph f ¢ on a graphing calculator. A suitable choice for the viewing window is [-4, 4] by

y = ax 2 + bx + c

The vertex is at

f ( x) = -x5 - x 4 + 2 x3 -25 x 2 + 9 x + 12

b2 2b 2 4ac + 4a 4a 4a

4ac - b 2 . 4a

4ac -b 2 4a

f ( x) = x5 - x 4 + 4 x3 - 30 x 2 + 5x + 6 f ¢( x) = 5 x 4 - 4 x3 + 12 x 2 - 60 x + 5

Examine the graph of f ¢ near x = -2.703 and x = 0.183. Observe that f ¢( x) < 0 to the left of x = -2.703 and f ¢( x) > 0 to the right of x = 2.703. Also observe that f ¢( x) > 0 to the left of x = 0.183 and f ¢( x) < 0 to the right of x = 0.183. The first derivative test allows us to conclude that f has a relative minimum at x = -2.703 and a relative maximum at x = 0.183. f (0.183) » 12.821 f (-2.703) » -143.572 Relative maximum of 12.821 at 0.183; relative minimum of -143.572 at -2.703

Section 5.2 47.

335

f ( x ) = 2 | x + 1| + 4 | x - 5| - 20 Graph this function in the window [–10, 10] by [-15, 30], Yscl = 5.

49.

f ¢ = 0 when x = -1 and when x = 3. f ¢ is positive just to the left of -1 and negative just to the right of -1, so f has a relative

maximum at x = -1. f ¢ is negative just to the left of 3 and positive just to the right of 3, so f has a relative minimum at x = 3. 50.

The graph shows that f has no relative maxima, but there is a relative minimum at x = 5. (Note that the graph has a sharp point at (5,–8), indicating that f ¢(-5) does not exist.) 48.

f ¢ = 0 at x = 1, x = 2, and x = 4. The sign

of f ¢ in each of the intervals into which these points divide the real line is indicated in the following sign graph.

(a) When graphing g ( x) in the standard window, no graph seems to appear. (b)

æ 1000 ö÷6 çç 2 çè x ÷÷ø x12 1

g ( x) =

Thus f has relative minima at x = 1 and x = 4 and a relative maximum at x = 2.

æ 1000 ö÷5 æç -1000 ö÷ g ¢( x) = 13 - 12 çç ÷ ÷ ç èç x ø÷ èç x 2 ÷ø x -12

= = =

-12

+ x13

-12

+ x13

1.2 ´ 1019 x7

51. 19

(1.2 ´ 10 ) x

C (q) = 80 + 18q; p = 70 - 2q P(q) = R(q) - C (q) = pq - C (q) = (70 - 2q) q - (80 + 18q)

x13

= -2q 2 + 52q - 80

-12 + (1.2 ´ 1019 ) x 6 x13

(a) Since the graph of P is a parabola that opens downward, we know that its vertex is a maximum point. To find the q-value of this point, we find the critical number.

Find the critical numbers: -12 + (1.2 ´ 1019 ) x 6 = 0 or x13 = 0 (1.2 ´ 1019 ) x 6 = 12 12 x6 = 1.2 ´ 1019 10 x 6 = 19 = 10-18 10

x = 0

P¢(q) = -4q + 52 P¢(q) = 0 when -4q + 52 = 0 4q = 52 q = 13

x = 10-18 = 0.001

g ¢(-1) = g ¢(-0.0001) = g ¢(0.0001) = g ¢(1) =

The number of units that produce maximum profit is 13.

x =0

-12 + (1.2 ´ 1019 )(-1)6 (-1)13

-12 + (1.2 ´ 1019 )(-0.0001)6 (-0.0001)13 -12 + (1.2 ´ 1019 )(0.0001)6 (0.0001)13 19 6

-12 + (1.2 ´ 10 )1 113

(b) If q = 13, p = 70 - 2(13) = 44 The price that produces maximum profit is $44. (c)

P(13) = -2(13) 2 + 52(13) - 80 = 258

The maximum profit is $258.

g (0.001) = -10-36 ; g (-0.001) = -1036

336 52.

Chapter 5 GRAPHS AND THE DERIVATIVE (b) If q = 100,

C (q) = 25q + 5000; p = 90 - 0.02q

p = 40e-0.01(100)

P(q) = R(q) - C (q) = pq - C (q) = (90 - 0.02q)q - (25q + 5000)

= 40e-1

= -0.02q 2 + 65q - 5000

» 14.72

(a) Since the graph of P is a parabola that opens downward, we know that its vertex is a maximum point. To find the q-value of this point, we find the critical number.

The price per unit that produces maximum profit is $14.72. (c)

P¢ q  = q + 65

= 2000e-1 - 100

P¢ q  =  when -0.04q + 65 = 0 0.04q = 65 q = 1625

» 635.76

The maximum profit is $635.76. 54.

C (q) = 21.047q + 3; p = 50 - 5 ln(q + 10) P(q) = R(q) - C (q) = pq - C (q)

The number of units that produces maximum profit is 1625.

= [50 - 5 ln(q + 10)] q - (21.047q + 3) = 28.953q - 3 - 5q ln(q + 10)

(b) If q = 1625,

(a)

p = 90 - 0.02(1625)

The price per unit that produces maximum profit is $57.50. P(13) = -0.02(1625) 2 + 65(1625) - 5000 = 47,812.5

(b) If q = 120, p = 50 - 5ln (120 + 10) » 25.66 The price per unit that produces maximum profit is $25.66.

The maximum profit is $47,812.50. 53.

C (q) = 100 + 20qe-0.01q ; p = 40e-0.01q P(q) = R(q) - C (q) = pq - C (q) = (40e-0.01q )q - (100 + 20qe-0.01q )

(c)

= 20qe-0.01q - 100

(a)

P(120) = 28.953(120) - 3 - 5(120) ln(102 + 10) » 550.84 The maximum profit is $550.84.

P¢(q) = 20e-0.01q + 20qe-0.01q (-0.01) = (20 - 0.2q)e-0.01q

P¢(q) = 28.953 - 5 ln (q + 10) - q +10

Use a graphing calculator to find the critical numbers for q > 0. The critical number is 120. For q < 120, P¢(q) > 0; for q > 120, P¢(q) < 0. Therefore, the number of units that produces maximum profit is 120.

= 57.5

(c)

P(100) = 20(100)e-0.01(100) - 100

55.

P(t ) = -0.006785t 3 + 0.1858t 2

Solve P¢(q) = 0.

- 0.6549t + 20.62 P¢(t ) = -0.020355t 2 + 0.3716t - 0.6549

(20 - 0.2q)e-0.01q = 0 20 - 0.2q = 0 q = 100

Since e-0.01q > 0 for all values of q, the sign of P¢ q  is the same as the sign of 20 - 0.2q. For q < 100, P¢(q) > 0; for q > 100, P¢(q) < 0. Therefore, the number of units that produces maximum profit is 100.

Use a graphing calculator to find the critical numbers for 0 £ t £ 24. The critical numbers are t » 1.976 and t » 16.280. These divide the domain of P into three intervals, and we pick a test point in each: Pick t = 1, t = 10, and t = 20. P ¢(1) » -0.304 P¢(10) » 1.0256 P¢(20) » -1.3649

Section 5.2

337

P is decreasing on (0, 1.976) and (16.280, 24) and increasing on (1.976, 16.280). There will be relative maxima at the left endpoint (t = 0) and at t = 16.280; there will be relative minima at

57.

p = D (q) = 200e-0.1q R(q) = pq = 200qe-0.1q R¢(q) = 200qe-0.1q (-0.1) + 200e-0.1q

t = 1.976 and at the right endpoint (t = 24).

= 20e-0.1q (10 - q)

P(0) = 20.62

R¢ q  =  when q = 1 the only critical number. Use the first derivative test to verify that q = 10 gives the maximum revenue.

P(1.976) » 20 P(16.280) » 29.93

R¢(9) = 20e-0.9 > 0

P(24) » 18.13

R¢(11) = -20e-1.1 < 0

We convert the time values to hours and minutes by multiplying the decimal parts by 60: (0.976)(60) » 59 and (0.280)(60) » 17.

The maximum revenue results when q = 10 p = D (10) = 200 » 73.58, or when e

telephones are sold at $73.58.

Recalling that the function Pgives the power in thousands of megawatts, we have the final result: 58.

Relative maximum of 20,620 megawatts at midnight (t = 0) ; relative minimum of 20,000 megawatts at 1:59 A.M.; relative maximum of 29,930 megawatts at 4:17 P.M.; relative minimum of 18,130 megawatts at midnight (t = 24). 56.

= 500q 2e-0.0016q

P ¢( x) = =

+ 500q2 e-0.0016q (-0.0016 ⋅ 2 q) = (1000q - 1.6q3 )e-0.0016q

Since e-.0016q > 0 for all values of q,

-x3 + 3x 2 + 72 x + 1

R¢( q) = 0

-3( x - 6)( x + 4)

(1000q - 1.6q3)e-0.0016q = 0

-x + 3x + 72 x + 1

1.6q3 = 1000q

P¢ x  =  when x = 6 or x = -4, but

q 2 = 625

4 is not in the domain of [0, 10]. P¢( x) fails to exist for x » -0.14 and x » 10.123, neither of which are in the domain of [0, 10]. Thus, the critical number is 6. Use the first derivative test to verify that x = 6 gives a maximum profit. P¢(5) = 27 > 0 311 P¢(7) = - 11 < 0 103 The maximum profit results when 6 units are sold. (b)

P(6) » 5.784 The maximum profit is about $5784.

-3x 2 + 6 x + 72

R¢(q) = 500e-0.0016q (2q)

P( x) = ln(-x3 + 3x 2 + 72 x + 1), for x in [0, 10].

(a)

p = D(q) = 500qe-0.0016q 2ö æ R(q) = pq = çç 500qe-0.0016q ÷÷ q è ø

q = 25

If q < 25, R¢(q) > 0; if q > 25, R ¢(q) < 0. So p has a maximum value at q = 25. 2

p = D(25) = 500(25)e-0.0016(25) » 4600

Maximum revenue occurs when the price is about $4600 and 25 computer systems are sold. 59.

C ( x) = 0.002 x3 = 9 x + 6912 C ( x) = 0.002 x 2 + 9 + 6912 x x C ¢ ( x) = 0.004 x - 6912 x2 C ( x) =

C ¢ ( x) = 0 when

338

Chapter 5 GRAPHS AND THE DERIVATIVE 0.004 x -

6912 x2

bt b-1 - ct b = 0

bt b-1 = ct b

0.004 x = 6912

b =t c

x3 = 1, 728, 000 x = 120

Let t = bc in M (t ).

A product level of 120 units will produce the minimum average cost per unit. 60.

æ b ö÷

æbö æ b öb -cççç ÷÷ M çç ÷÷÷ = a çç ÷÷÷ e è c ø èç c ø èç c ø

Relative minimum of 4.0% in 2000; relative maximum of 6% in 2003; relative minimum of 4.6% in 2006–2007; relative maximum of 9.6 in 2010; relative minimum of 3.7% in 2019.

æ b öb = a çç ÷÷÷ e-b çè c ø

The maximum daily consumption is 61.

a(b/c)b e-b kg and it occurs at b /c weeks.

a(t ) = 0.008t 3 - 0.288t 2 + 2.304t + 7 a¢(t ) = 0.024t 2 - 0.576t + 2.304

Set a¢ = 0 and solve for t.

M (t ) = 369(0.93)t (t )0.36

0.024t 2 - 0.576t + 2.304 = 0

M ¢(t ) = (369)(0.93)t ln(0.93)(t 0.36 )

0.024(t 2 - 24t + 96) = 0

+ 369(0.93)t (0.36)(t )-0.64 = (369t 0.36 )(0.93t ln 0.93)

t 2 - 24t + 96 = 0 t » 5.07 or t » 18.93

t = 5.07 = 5 hours + 0.07 ⋅ 60 minutes corresponds to 5:04 P.M.

(a)

M (t ) = 6.281t

Verify that t » 4.96 gives a maximum. M ¢(4) > 0 M ¢(5) < 0

0.242 -0.025t

M ¢(t ) = (1.520002t

-0.758

)(e-0.025t )

M (4.96) = 369(0.93) 4.96 (4.96)0.36 » 485.22

+ (6.281t 0.242 )(-0.025e-0.025t )

The female moose reaches a maximum weight of about 458.22 kilograms at about 4.96 years.

= e-0.025t (1.520002t -0.758 - 0.157025t 0.242 )

64.

M ¢(t ) = 0 when 1.520002t -0.758 - 0.157025t 0.242 = 0 t = 9.68

Let t = 9.68 in M (t ). M (9.68) = 6.281(9.68)0.242 e-0.025(9.68) » 8.54 kg

The maximum daily consumption is 8.54 kg and it occurs at 9.68 weeks. (b)

M (t ) = at be-ct

t 0.64

M ¢(t ) = 0 when t » 4.96.

t = 18.93 = 18 hours + 0.93 ⋅ 60 minutes corresponds to 6:56 A.M.

62.

132.84(0.93)t

F (t ) = -10.28 + 175.9te-t /1.3 æ ö F ¢(t ) = (175.9t ) çç - 1 ÷÷÷ e-t /1.3 è 1.3 ø + (175.9)(e-t /1.3 ) æ ö = e-t /1.3 çç - 175.9t + 175.9 ÷÷÷ è ø 1.3 F ¢(t ) = 0 when - 175.9t + 175.9 = 0 1.3 t = 1.3 At 1.3 hours, the termal effect of the food is maximized.

M ¢(t ) = (bat b-1)(e-ct ) + (at b )(-ce-ct ) = ae-ct (bt b-1 - ct b ) M ¢(t ) = 0 when

Section 5.3 65.

339 (b) The cork remains in the air as long as s(t ) > 0. Use the quadratic formula to solve s(t ) = 0.

D( x) = -x 4 + 8x3 + 80 x 2 D¢( x) = -4 x3 + 24 x 2 + 160 x = -4 x( x + 4)( x - 10)

-16t 2 + 40t + 3 = 0

D¢( x) = 0 when x = 0, x = -4, or x = 10. Disregard the nonpositive values.

t =

Verify that x = 10 gives a maximum.

D¢(11) = -660 < 0

The speaker should aim for a degree of discrepancy of 10. 20t R(t ) = 2 t + 100 R¢(t ) = =

Only the positive solution is relevant, so the cork stays in the air for about 2.57 seconds.

5.3

20(t 2 + 100) - 20t (2t ) (t 2 + 100) 2

Higher Derivatives, Concavity, and the Second Derivative Test

Your Turn 1

2000 - 20t 2

f ( x) = 5 x 4 - 4 x 3 + 3 x

(t 2 + 100) 2

f ¢( x) = 20 x3 - 12 x 2 + 3

R¢(t ) = 0 when

f ¢¢( x) = 60 x 2 - 24 x

2000 - 20t 2 = 0

f ¢¢(1) = 60(12 ) - 24(1) = 36

-20t 2 = -2000 t = 10.

Disregard the negative value.

Your Turn 2

Use the first derivative test to verify that t = 10

(a) Use the chain rule. f ( x) = ( x3 + 1)2

gives a maximum rating. R¢(9) = 0.0116 > 0 R¢(11) = -0.0086 < 0

f ¢( x) = (2)( x3 + 1)(3x 2 ) = 6 x 2 ( x3 + 1)

The film should be 10 minutes long. 67.

402 - 4(-16)(3) 2(-16)

52 7 4 » -0.073, 2.573

D¢(9) = 468 > 0

66.

-40 

s(t ) = -16t 2 + 40t + 3 s¢(t ) = -32t + 40

= 6 x5 + 6 x 2 f ¢¢( x) = 30 x 4 + 12 x

(b) Use the product rule.

(a) when s¢(t ) = 0,

f ( x ) = xe x

-32t + 40 = 0 32t = 40

f ¢( x ) = xe x + (1)e x

40 5 = 32 4 Verify that t = 5/4 gives a maximum.

= xe x + e x

t =

s¢(1) = 8 s¢(2) = -24

Now find the height when t = 5/4. æ5ö æ 5 ö2 æ5ö s çç ÷÷ = -16 çç ÷÷ + 40 çç ÷÷ + 3 çè 4 ÷ø çè 4 ÷ø èç 4 ø÷

Now note that we have already found the derivative of the first term of f ¢( x ), which is the original function f ( x). Thus f ¢¢( x ) = ( xe x + e x ) + e x = 2e x + xe x

= 28

340

Chapter 5 GRAPHS AND THE DERIVATIVE

( x) (

1 x

) - (ln x)(1) x2

= =





a(t)





net result







x4 -x - 2 x + 2 x ln x

The car speeds up (in the backward direction) for 0 < t < 1; it slows down (still in the backward direction) for 1 < t < 4; it speeds up (in the forward direction) for t > 4. (All times are in seconds.)

x4 -3 + 2 ln x

Your Turn 4

( x ) ( - 1x ) - (1 - ln x )(2 x) 2

h¢¢( x) =



1 - ln x x

v(t)

f ( x) = x5 - 30 x3 f ¢( x) = 5x 4 - 90 x 2

Your Turn 3

f ¢¢( x) = 20 x3 - 180 x

s(t ) = t 3 - 3t 2 - 24t + 10

Factor f ¢¢( x ) and create a number line.

v(t ) = s ¢(t ) = 3t - 6t - 24 a(t ) = v¢(t ) = s ¢¢(t ) = 6t - 6

f ¢¢( x ) = 20 x 3 - 180 x = 20( x 3 - 9 x ) = 20( x )( x - 3)( x + 3)

First find when v changes sign. v(t ) = 3t 2 - 6t - 24 = 0 3(t 2 - 2t - 8) = 0 (t - 4)(t + 2) = 0 t = 4 or t = -2

The zeros of f ( x) are -3, 0, and 3, and they divide the domain of f into four intervals:

(-¥, -3), (-3, 0), (0, 3), (3, ¥)

Only the positive value of t is relevant. Check the velocity at times before and after 4, say at t = 2 and t = 6.

Choose a test point in each interval and find the sign of f ¢¢( x ) at the test points. For example, choose -4, -1, 1, and 4.

v(2) = -24 < 0 v(6) = 48 > 0

f ¢¢(-4) = -560 f ¢¢(-1) = 160

Thus the car backs up for the first four seconds and then goes forward. Now find where a changes sign.

f ¢¢(1) = -160 f ¢¢(4) = 560

Here is the corresponding number line.

a(t ) = 6t - 6 = 0 6t - 6 = 0 t =1

Check the acceleration at times before and after 1, say at t = 1/2 and t = 2. a(1/2) = -3 < 0 a(2) = 6 > 0

Now we construct the following graph showing the signs of v and a.

The figure shows that f is concave downward on (-¥, -3) and (0, 3) and concave upward on

Section 5.3

341 f (-3) = 567 f (0) = 0

W2. f ( x ) = x 4 - 8 x 3 - 32 x 2 + 10 f ¢( x ) = 4 x 3 - 24 x 2 - 64 x

f (3) = -567

Thus the inflection points are (-3, 567), (0, 0) and (3, -567). Your Turn 5

= 4 x( x 2 - 6 x - 16) = 4 x( x - 8)( x + 2) The critical numbers are -2, 0 and 8, and they divide the domain into four intervals: (-¥, -2), (-2, 0), (0, 8), (8, ¥)

f ( x ) = -2 x 3 + 3x 2 + 72 x

Evaluate f ¢ at a test point in each interval.

f ¢( x ) = -6 x 2 + 6 x + 72

f ¢(-3) = -132 f ¢(-1) = 36

Solve the equation f ¢( x) = 0.

f ¢(1) = -84 f ¢(10) = 960

-6 x 2 + 6 x + 72 = 0 x 2 - x - 12 = 0

At x = -2 and x = 8, f ¢ changes from negative to positive, so f has a relative minimum at these

( x + 3)( x - 4) = 0 x = -3 or x = 4

two points. At x = 0, f ¢ changes from positive to negative, so f has a relative maximum at this point. The values of the relative minima are f (-2) = -38 and f (8) = -2038; the value of the relative maximum is f (0) = 10.

Now use the second derivative test. f ¢¢( x ) = -12 x + 6 f ¢¢(-3) = 42 f ¢¢(4) = -42

Since f ¢¢(-3) > 0, x = -3 leads to a relative minimum; since f ¢¢(4) < 0, x = 4 leads to a relative maximum. Thus we have a relative minimum of f (-3) = -135 at x = -3 and a relative maximum of f (4) = 208 at x = 4.

5.3 Warmup Exercises W1.

f ( x ) = x 3 - 3x 2 - 72 x + 20 f ¢( x ) = 3x 2 - 6 x - 72 = 3( x 2 - 2 x - 24) = 3( x - 6)( x + 4) The critical numbers are 6 and -4, and they divide the domain into three intervals: (-¥, -4), (-4, 6) and (6, ¥)

5.3 Exercises 1.

False. If a function is concave upward on an interval, the shape is like a U, so it can be increasing or decreasing on the interval.

True

False. If a function has an inflection point at x = c, then f (c) = 0 or does not exist.

False. If f ¢(c) = 0 and f (c) > 0 , then by the second derivative test, f (c) is a relative minimum.

f ( x) = 5 x 3 - 7 x 2 + 4 x + 3 f ¢( x) = 15 x 2 - 14 x + 4 f ( x) = 30 x - 14 f (0) = 30(0) - 14 = -14 f (2) = 30(2) - 14 = 46

Evaluate f ¢ at a test point in each interval. f ¢(-5) = 33 f ¢(0) = -72 f ¢(8) = 72

At x = -4, f ¢ changes from positive to negative, so f has a relative maximum at x = -4; the relative maximum value is f (-4) = 196. At x = 6, f ¢ changes from negative to positive, so f as a relative minimum at x = 6; the relative minimum value is f (6) = -304.

f ( x) = 4 x 3 + 5 x 2 + 6 x - 7 f ¢( x) = 12 x 2 + 10 x + 6 f ( x) = 24 x + 10 f (0) = 24(0) + 10 = 10 f (2) = 24(2) + 10 = 58

342 9.

Chapter 5 GRAPHS AND THE DERIVATIVE f ( x) = 4 x 4 - 3 x 3 - 2 x2 + 6

14.

f ( x) =

f ¢( x) = 16 x3 - 9 x 2 - 4 x f ( x) = 48 x 2 - 18 x - 4

-x

1 - x2

f ¢( x ) =

f (0) = 48(0)2 - 18(0) - 4 = -4

10.

f (2) = 48(2) - 18(2) - 4 = 152

x2 f ( x) = - x 4 + 7 x 3 2

f ¢( x) = -4 x3 + 21x 2 - x

(1 - x 2 )2 -1 + x 2 - 2 x 2

(1 - x 2 )2 -1 - x 2

(1 - x 2 )2

f ( x ) é (1 - x 2 )2 (-2 x ) ù ê ú ê ú 2 2 êë - (-1 - x )(2)(1 - x )(-2 x ) úû = (1 - x 2 )4

f ( x) = -12 x + 42 x - 1 f (0) = -12(0) 2 + 42(0) - 1 = -1 f (2) = -12(2)2 + 42(2) - 2 = 35

11.

(1 - x 2 )(-1) - (-x)(-2 x)

f ( x) = 3 x 2 - 4 x + 8 f ¢( x ) = 6 x - 4

f ( x) = 6

(1 - x 2 )[-2 x(1 - x 2 ) + 4 x(-1 - x 2 )] (1 - x 2 )4 -2 x( x 2 + 3) (1 - x 2 )3

f (0) = 6

12.

f (2) = 6

f (0) =

f ( x) = 8 x 2 + 6 x + 5

f (2) =

f ( x) = 16 f (0) = 16 f (2) = 16 f ( x) = f ¢( x ) = =

f ( x ) = = =

x2 1+ x

15.

(1 + x)(2 x) - x 2 (1)

2x + x2 (1 + x) 2 (1 + x)2 (2 + 2 x) - (2 x + x 2 )(2)(1 + x) (1 + x)4 2

(1 + x)(2 + 2 x) - (2 x + x )(2) (1 + x)3

(1 + x)3 f (0) = 2 2 f (2) = 27

x 2 + 4 = ( x 2 + 4)1/2

f ( x) = f ¢( x ) =

(1 + x)2

(1 - 02 )3 -2(2)(4 + 3)

(1 - 22 )3 -28 = -27 28 = 27

f ¢( x) = 16 x + 6

13.

-2(0)(02 + 3)

1 2 ( x + 4)-1/2 ⋅ 2 x 2 x ( x 2 + 4)1/2

( x 2 + 4)1/2 (1) - x éê 12 ( x 2 + 4)-1/2 ùú 2 x ë û f  ( x) = x2 + 4 ( x 2 + 4)1/2 = = =

x2 ( x 2 + 4)1/2

x2 + 4 ( x 2 + 4) - x 2 ( x 2 + 4)3/2 4 ( x 2 + 4)3/2

Section 5.3

343

f (0) =

4 2

18.

3/2

(0 + 4) 4 4 1 = 3/2 = = 8 2 4 4 f (2) = 2 (2 + 4)3/2 4 4 1 = 3/2 = = 16 2 4 2 8

f ¢( x) = -2 x-2/3 f ( x) = 4 x-5/3 = 45/3 3 3x f (0) does not exist. f (2) =

19. 16.

f ( x) =

2x2 + 9

2(2 x 2 + 9)1/2 -

= f (0) =

= 80e-4 - 10e-4 = 70e-4 » 1.282

20.

f ( x) = 0.5e x

2 2

f ¢( x) = (0.5)(2 x)e x = xe x

3/2

f ( x) = xe x (2 x) + e x (1) = 2 x 2e x + e x 2

or e x (2 x 2 + 1)

(2 x 2 + 9)3/2 18

f (0) = e0 [2(0) 2 + 1] = 1(1) = 1

3/2

f (2) = e 2 [2(2)2 + 1] = e 4 (9) = 9e 4 » 491.4

21.

f ( x) = f ¢( x) =

ln x 4x

( )

4 x 1x - (ln x)(4)

(4 x) 2 4 - 4 ln x 1 - ln x = = 2 16 x 4x2

f ( x) = 32 x3/4 f ¢( x) = 24 x-1/4 f  ( x ) = -6 x

f ¢¢(2) = 20(22 )e-(2 ) - 10e-(2 )

(2 x + 9) 18

-5/4

f ¢¢(0) = 20(02 )e-0 - 10e = 0 - 10 = -10

4 x2 (2 x + 9)1/2

[2(0) + 9] 18 18 2 = 3/2 = = 27 3 9 18 f (2) = [2(2) 2 + 9]3/2 18 18 = 3/2 = 17 17 17

17.

= 20 x 2e-x - 10e-x

2(2 x 2 + 9) - 4 x 2

f ( x) = -10 xe-x (-2 x) + e-x (-10)

2x2 + 9 2

f ( x) = 5e-x

(2 x 2 + 9)1/2 (2) - 2 x éê 12 (2 x 2 + 9)1/2 ùú 4 x ë û f ( x ) = 2 2x + 9

4 21/3 = 3 3(25/3 )

f ¢( x) = 5e-x (-2 x) = -10 xe-x

= (2 x 2 + 9)1/2 1 f ¢( x) = (2 x 2 + 9)-1/2 ⋅ 4 x 2 2x = 2 (2 x + 9)1/2

f ( x) = -6 x1/3

= - 5/4 x

f (0) does not exist. 6 f (2) = - 5/4 2 3 = - 1/4 2

f  ( x) = = = = =

( )

4 x 2 - 1x - (1 - ln x)8x 16 x 4 -4 x - 8x + 8x ln x 16 x 4 -12 x + 8x ln x 16 x 4 4 x(-3 + 2 ln x) 16 x 4 -3 + 2 ln x

4 x3

344

Chapter 5 GRAPHS AND THE DERIVATIVE f (0) does not exist because ln 0 is undefined. f (2) =

22.

-3 + 2 ln 2 4(2)

f ( x) = ln x +

27.

-3 + 2 ln 2 » 0.050 32

f ¢( x) = f ( x) =

1 x

1 1 - 2 x x 2 -1 f  ( x) = 2 + 3 x x 2 2-x -x = 3 + 3 = x x x3

23.

( x + 2) -3(2)( x + 2)

f (4) ( x) =

0 =0 8

f ( x) = 7 x 4 + 6 x 3 + 5 x 2 + 4 x + 3

(-6)(-3)( x + 2)2 ( x + 2)6

28.

f ( x) = f ¢( x ) =

f ( x) = 84 x + 36 x + 10 f ¢¢¢ ( x) = 168x + 36

f ¢( x) = -8x3 + 21x 2 + 8x + 1

25.

29. 2

f ( x) = 5 x - 3 x + 2 x + 7 x + 4 f ¢( x) = 25 x 4 - 12 x3 + 6 x 2 + 14 x 3

= -x-2

f ¢( x) = f ( x ) =

f ¢( x) = 10 x + 12 x - 15 x + 9 f ( x) = 40 x + 36 x - 30 x f ¢¢¢ ( x) = 120 x2 + 72 x - 30

f (4) ( x) =

-12(3)( x - 2) 2 ( x - 2)

-6 ( x - 2) 2 12

( x - 2)3

= -36( x - 2)-4

-36

( x - 2) 4

f (4) ( x) = 600 x - 72

6 x4

( x - 2) -6(-2)( x - 2)

f ¢¢¢( x) =

f ( x) = 2 x 5 + 3 x 4 - 5 x 3 + 9 x - 2

f ¢¢¢ ( x) = 300 x2 - 72 x + 12

3x x-2 ( x - 2)(3) - 3x(1)

f ( x) =

f ( x) = 100 x - 36 x + 12 x + 14

26.

f (4) ( x) = 24 x-5 or

f ( x) = -24 x 2 + 42 x + 8 f ¢ ( x) = -48x + 42

( x + 2)5

f ¢¢¢ ( x) = -6 x-4 or -

f ( x ) = -2 x 4 + 7 x 3 + 4 x 2 + x

-72

x +1 x (1)( x) - 1( x + 1)

f ( x) = 2 x-3 or

f (4) ( x) = -48

( x + 2)4

( x + 2)8

f (4) ( x) = 168

-18(4)( x + 2)3

( x + 2)3

= -72( x + 2)-5 or

f ¢( x) = 28x3 + 18x 2 + 10 x + 4

24.

( x + 2)2 -6

( x + 2)4

= 18( x + 2)-4 or

f (0) does not exist since the denominator is 0.

2-2

( x + 2) - ( x - 1)

f ¢¢¢( x) =

f  ( x) =

f (2) =

x -1 x+2

f ( x) =

( x - 2)4 -36(-4)( x - 2)3

f (4) ( x) = 240 x + 72

( x - 2) 144 ( x - 2)5

= 144( x - 2)-5

Section 5.3

345

f ( x) =

30.

f ¢( x ) =

x 2x + 1 (1)(2 x + 1) - (2)( x)

34.

(

f ¢( x ) = e(ln a ) x

(2 x + 1) 2 1 = = (2 x + 1)-2 2 (2 x + 1)

)¢ = (ln a)( e(ln a) x )

= a x (ln a ) Each differentiation introduces a factor of ln a, so f ¢¢( x ) = a x (ln a )2

f ( x) = -2(2 x + 1)-3 (2)

f ¢¢¢( x ) = a x (ln a )3 

= -4(2 x + 1)-3 f ¢¢¢( x) = 12(2 x + 1)-4 (2) = 24(2 x + 1)-4 or

f ( x) = a x

f (n) ( x ) = a x (ln a )n

24 (2 x + 1)4

f (4) ( x) = -96(2 x + 1)-5(2) = -192(2 x + 1)-5 -192 or (2 x + 1)5

35.

Concave upward on (2, ¥) Concave downward on (-¥, 2) Inflection point at (2, 3)

36.

31. (a) Any term with power less than n will be 0 after being differentiated n times, so only the

Concave upward on (-¥, 3) Concave downward on (3, ¥) Inflection point at (3, 7)

term x n survives, and its derivative is n !. Thus f (n) ( x ) = n !.

(b) For k > n each term has derivative 0 and

37.

Concave upward on (-¥, -1) and (8, ¥) Concave downward on (-1, 8)

f (k ) ( x ) = 0

Inflection points at (-1, 7) and (8, 6)

f ( x) = ln x

32.

f ¢ ( x) =

(a)

38.

1 = x-1 x -2

f  ( x) = - x

f ¢¢¢( x) = 2 x-3 =

f ( n) ( x) =

(b)

f ( x) = e x

33.

f ¢( x ) = e

f ( x ) = e x f ¢¢¢( x) = e x f

( n)

( x) = e

Concave downward on (-¥, -2) and (6, ¥)

-1

Inflection point at (-2, -4) and (6, -1)

x2 2

x3 -6 f (4) ( x) = -6 x-4 = 4 x 24 (5) -5 f ( x) = 24 x = 5 x

39.

Concave upward on (2, ¥) Concave downward on (-¥, 2) No points inflection

40.

(-1) n-1(n - 1)! x

Concave upward on (-2, 6)

Concave upward on (-¥, 0) Concave downward on (0, ¥)

No inflection points 41.

f ( x) = x 2 + 10 x - 9 f ¢( x) = 2 x + 10 f ( x) = 2 > 0 for all x.

Always concave upward No inflection points

346 42.

Chapter 5 GRAPHS AND THE DERIVATIVE f ( x) = -6 x - 24 < 0 when

f ( x) = 8 - 6 x - x 2 f ¢ ( x ) = -6 - 2 x f ( x) = -2 < 0 for all x.

-6( x + 4) < 0 x+4>0 x > -4.

Always concave downward No inflection points 43.

Concave downward on (-4, ¥) f ( x) = -6 x - 24 = 0 when

f ( x) = - 2 x3 + 9 x 2 + 168 x - 3

-6( x + 4) = 0

f ¢( x) = -6 x + 18x + 168 f ( x) = -12 x + 18 f ( x) = -12 x + 18 > 0 when - 6(2 x - 3) > 0 2x - 3 < 0 3 x< . 2

(

Concave upward on -¥, 32

x = -4. f (-4) = 54

Inflection point at (-4, 54) 45.

f ¢( x) = -6( x + 2)5

)

f ( x) = -30( x + 2)4 f ¢¢ x   -30( x + 4)4 < 0 for all values of x

f ¢¢( x) = -12 x + 18 < 0 when -6(2 x - 3) < 0

Concave downward on (-¥, ¥)

2x - 3 > 0 3 x> . 2

Concave downward on

f ( x) ¹ 0 for any value for x. There are no changes of concavity. There are no inflection points.

( 32 , ¥ )

f ¢¢( x) = -12 x + 18 = 0 when -6(2 x + 3) = 0 2x + 3 = 0

46.

f ( x) = 12(3 - x)2

3 . 2 æ3ö 525 f çç ÷÷÷ = çè 2 ø 2

44.

f ¢¢ x   12(3 - x)2 > 0 for all values of x

Concave upward on (-¥, ¥) f ( x) ¹ 0 for any value for x. There are no changes of concavity. There are no inflection points.

( 32 , 5252 )

f ( x) = -x3 - 12 x 2 - 45 x + 2 f ¢( x) = -3x 2 - 24 x - 45 f ( x) = -6 x - 24 f ( x) = -6 x - 24 > 0 when -6( x + 4) > 0 x+4<0 x < -4.

f ( x) = (3 - x) 4 f ¢( x) = -4(3 - x)3

x =

Inflection point at

f ( x) = -( x + 2)6

47.

3 x-5 -3

f ( x) = f ¢( x) = f  ( x) =

( x - 5) 2 -3(-2)( x - 5) ( x - 5)

f ¢¢ x  

Concave upward on (-¥, -4)

  x - 

6 ( x - 5)3

> 0 when

( x - 5)3 > 0 x-5> 0 x > 5.

Section 5.3

347 f ( x ) =

6 ( x - 5)

f ( x) = 6 x + 20 < 0 when

< 0 when

2(3x + 10) < 0 3x < -10 10 x<- . 3

( x - 5)3 < 0 x-5< 0 x < 5.

(

Concave downward on -¥, - 10 3

Concave downward on (-¥, 5)

)

æ 10 ö ö2 10 æ 10 f çç - ÷÷ = - çç - + 5 ÷÷ ÷ø çè 3 ÷ø 3 çè 3

f ( x) ¹ 0 for any value for x; it does not exist when x = 5. There is a change of concavity there, but no inflection point since f (5) does not exist.

-10 æç -10 + 15 ö÷ ÷÷ ç ø 3 çè 3 10 25 250 =- ⋅ =3 9 27 =

48.

f ( x) =

-2 x +1

= -2( x + 1)-1

(

, - 250 Inflection point at - 10 3 27

f ¢( x) = 2( x + 1)-2 f ¢¢( x) = -4( x + 1)-3 = f ¢¢( x) =

-4 ( x + 1)3

50.

> 0 when ( x + 1)3 x +1< 0 x < -1.

f ( x) = -2( x - 3) - 4 x + 6

Concave upward on (-¥, -1)

= -6 x + 12 f ( x) = -6 x + 12 > 0 when

= -( x - 3)2 - 2 x 2 + 6 x = -2 x + 6 - 4 x + 6

-4

-6(x - 2) > 0

< 0 when ( x + 1)3 x +1> 0

x-2>0 x < 2.

x > -1.

Concave upward on (-¥, 2)

Concave downward on (-1, ¥)

f ( x) = -6 x + 12 < 0 when

f ( x) ¹ 0 for any value for x; it does not exist when x = -1. There is a change of concavity there, but no inflection point since f (-1) does not exist.

49.

f ( x) = -x( x - 3)2 f ¢( x) = -1( x - 3) 2 + 2( x - 3)(-x)

-4

f  ( x) =

)

- 6( x - 2) < 0 x-2<0 x > 2.

Concave downward on (2, ¥)

f ( x) = x ( x + 5) 2

f ( x) = -6 x + 12 = 0 when x = 2.

f ¢( x) = x(2)( x + 5) + ( x + 5) 2

f (2) = -2

= ( x + 5)(2 x + x + 5)

Inflection point at (2, -2)

= ( x + 5)(3x + 5) f ( x) = ( x + 5)(3) + (3x + 5) = 3x + 15 + 3x + 5 = 6 x + 20 f ( x) = 6 x + 20 > 0 when 2(3x + 10) > 0

(

f ( x) = 18x - 18e-x f ¢( x) = 18 - 18e-x (-1) = 18 + 18e-x f ( x) = 18e-x (-1) = -18e-x

3x > -10 10 x>- . 3

Concave upward on - 10 ,¥ 3

51.

f ( x) = -18e-x < 0 for all x

)

f ( x) is never concave upward and always concave downward. There are no points of

348 52.

Chapter 5 GRAPHS AND THE DERIVATIVE f ( x) = 2e-x

f (-1) =

f ¢( x) = 2e-x (-2 x) = -4 xe-x -x 2

-x 2

(-2 x) + e

-x 2

f ( x) = -4 xe = -4e

( 7 ) = - 70 < 0 9 (2)

40 - 8 æ1ö f  çç ÷÷÷ = çè 8 ø 9 1

(-4)

(-2 x + 1)

f (8) =

f ( x) = 0 when -2 x 2 + 1 = 0 1 = x2 2 2  = x 2

f (0) = (0)8/3 - 4(0)5/3 = 0 f (1) = (1)8/3 - 4(1)5/3 = -3

Inflection points at (0, 0) and (1, -3)

Check the sign of f ( x) in each of the intervals 2 2

and x = - 22 using

test points.

-4 4 (-1) = > 0 e e 2

f (0) = -4e-0 [-2(0)2 + 1] = -4(1) = -4 < 0

f (1) = -4e

f ( x) = 0 when x = -8

[-2(1) + 1]

f ( x) fails to exist when x = 0

-4 4 = (-1) = > 0 e e

(

) and ( , ¥ ) ; concave downward on ( - , ). Concave upward on -¥, - 22 2 2

2 2

æ 2 ö÷÷ 2 e-1/2 = f çç çç 2 ÷÷ = 2 e è ø

(

Inflection points at - 22 , 2 53.

8/3

f ( x) = x7/3 + 56 x 4/3 7 4/3 224 1/3 x + x 3 3 28 1/3 224 -2/3 f  ( x) = x + x 9 9 28( x + 8) = 9 x 2/3

f (-1) = -4e-(-1) [-2(-1)2 + 1]

-12

54.

f ¢( x) =

40(7) 140 = >0 9(2) 9

Concave upward on (-¥, 0) and (1, ¥); concave downward on (0, 1)

1 = 2x2

determined by x =

40(-2) 80 = >0 9(-1) 9

) and ( , ) 2 2

2 e

5/3

f ( x) = x - 4 x 8 20 2/3 f ¢( x) = x5/3 x 3 3 40 2/3 40 -1/3 40( x - 1) f  ( x) = x x = 9 9 9 x1/3

Note that both f ( x) and f ¢( x) exist at x = 0. Check the sign of f ( x) in three intervals determined by x = -8 and x = 0 using test points. 28(-19) 532 =<0 9(9) 81 28(7) 196 f (-1) = = >0 9(1) 9 28(9) f (1) = = 28 > 0 9(1)

f (-27) =

Concave upward on (-8, ¥); concave downward on (-¥, -8) f (-8) = (-8)7/3 + 56(-8)4/3 = -128 + 896 = 768 Inflection point at (-8, 786)

f ( x) = 0 when x = 1 f ( x) fails to exist when x = 0

Note that both f ( x) and f ¢( x) exist at x = 0. Check the sign of f ( x) in the three intervals determined by x = 0 and x = 1 using test points.

Section 5.3 55.

349

f ( x) = ln ( x 2 + 1)

= f  ( x) =

x2 + 1

( x + 1)(2) - (2 x)(2 x)

8 ( x + 1)2

( x 2 + 1) 2 -2 x 2 + 2

( x 2 + 1) 2

-3 < x < 1

> 0 when

Note that f ( x) is not defined at x = -1. Concave downward on (-3, -1) and (-1, 1) f (-3) = (-3)2 + 8 ln | - 3 + 1| = 9 + 8 ln 2

x2 < 1

f (1) = (1)2 + 8 ln |1 + 1| = 1 + 8 ln 2

-1 < x < 1

Inflection points at (-3, 9 + 8 ln 2) and (1, 1 + 8 ln 2)

Concave upward on (-1, 1) -2 x 2 + 2 ( x 2 + 1) 2

< 0 when

57.

-2 x 2 + 2 < 0

f ( x ) = x 2 log | x | æ 1 ÷ö f ¢( x ) = 2 x log | x | + x 2 ççç ÷÷ è x ln10 ÷ø

- 2 x < -2 x2 > 1

x ln10 æ 1 ö÷ 1 ÷÷ + f ( x ) = 2 log | x | + 2 x ççç è x ln10 ø÷ ln10 = 2 x log | x | +

x > 1 or x < -1

Concave downward on (-¥, -1) and (1, ¥) f (1) = ln [(1)2 + 1] = ln 2 2

f (-1) = ln [(-1) + 1] = ln 2

Inflection points at (-1, ln 2) and (1, ln 2)

= 2 log | x | +

f ( x) = x 2 + 8ln | x + 1|

f  ( x) = 2 f  ( x) = 2 2-

( x + 1)2

8 ( x + 1)2

3 ln 10

3 2 ln 10 ln | x | 3 >ln10 2 ln10

log | x | > > 0 when

ln | x | > -

3 >0 ln 10

2 log | x | > -

8 x +1 8

( x + 1)2 8

3 ln10

f ( x ) > 0 when 2 log | x | +

f ¢( x) = 2 x +

( x + 1)2

-2 < x + 1 < 2

-2 x 2 > -2

56.

( x + 1) 2 < 4

( x 2 + 1) 2 -2 x 2 + 2

< 0 when

-2 x 2 + 2 > 0

f  ( x) =

( x + 1)2

f ¢( x) = f  ( x) =

f  ( x) = 2 -

3 2

| x | > e-3/ 2 8

x > e-3/2 or x < -e-3/2

( x + 1) 2

Concave upward on (-¥, -e-3/2 )

( x + 1) 2 > 4 x + 1 > 2 or x + 1 < -2

and (e-3/2 , ¥)

x > 1 or x < -3

350

Chapter 5 GRAPHS AND THE DERIVATIVE f ( x ) < 0 when

f ( x) = (2 ln 5)5-x [2 ln 5( x 2 ) - 1] < 0 when

2 log | x | + 3 < 0 ln 10

2 ln 5( x 2 ) - 1 < 0 2 ln 5( x 2 ) < 1

2 log | x | < - 3 ln 10

x2 <

log | x | < - 3 2 ln 10 ln | x | <- 3 ln10 2 ln10

| x | < e-3/2

Note that f ( x) is not defined at x = 0. Concave downward on (-e-3/2 , 0)

æ Inflection points at çç çè

and (0, e-3/2 ). f (-e-3/2 ) = (-e-3/2 )2 log |- e-3/2 | 3e 2 ln10

59.

( )

3e-3 2 ln10

-3

e Inflection points at -e-3/2 , - 23ln 10

(

-3

e and e-3/2 , - 23In 10

f ( x) = 5-x

58.

60.

Since the graph of f ¢( x) is increasing on (0, 5), the function is concave upward on (0, 5). Since the graph of f ¢( x) is decreasing on (-¥, 0) and (5, ¥), the function is concave downward on (-¥, 0) and (5, ¥). The inflection points are at 0 and 5.

61.

Since the graph of f ¢( x) is increasing on (-7, 3) and (12, ¥), the function is concave upward on (-7, 3) and (12, ¥). Since the graph of f ¢( x) is decreasing on (-¥, -7) and (3, 12), the function is concave downward on (-¥, -7) and (3, 12). The inflection points are at -7, 3, and 12.

62.

Since the graph of f ¢( x) is increasing on

f ¢( x) = (ln 5)5-x ⋅ (-2 x) 2 2

f ( x) = -2 ln 5 ⋅ 5-x + (-2 ln 5) x ⋅ (ln 5)(5-x )(-2 x) 2

= (2 ln 5)5-x [2 ln 5(x 2 ) - 1] 2

f ( x) = (2 ln 5)5-x [2 ln 5(x 2 ) - 1] > 0 when

2 ln 5( x 2 ) - 1 > 0 2 ln 5( x 2 ) > 1 1 2 ln 5 1 1 x> or x < 2 ln 5 2 ln 5

x2 >

(

Concave upward on -¥ -

(

1 ,¥ 2 ln 5

)

1 2 ln 5

Since the graph of f ¢( x) is increasing on (-¥, 0) and (4, ¥), the function is concave upward on decreasing on (0, 4), the function is concave downward on (0, 4). The inflection points are at 0 and 4.

)

ö 1 , e-1/2 ÷ and ÷÷ 2 ln 5 ø

(-¥, 0) and (4, ¥). Since the graph of f ¢( x) is

= -(2 x ln 5)5-x

1 ÷ ÷ 2 ln 5 ÷ø

æ 1 ö çç , e-1/2 ÷÷÷ çè 2 ln 5 ø

-3

f (e-3/2 ) = (e-3/2 )2 log |e-3/2 = e-3 log e-3/2 = -

1 , 2 ln 5

1 2 ln 5

æ ö -(-1/ 2 ln 5)2 = 5-1/(2 ln 5) = e-1/ 2 f çç - 1 ÷÷÷ = 5 çè 2 ln 5 ø æ ö -(1/ 2 ln 5) 2 = 5-1/(2 ln 5) = e-1/ 2 f çç 1 ÷÷÷ = 5 çè 2 ln 5 ø

-e-3/2 < x < e-3/2

= e-3 log e-3/2 = -

1 < x< 2 ln 5

æ Concave downward on çç çè

ln | x | < - 3 2

1 2 ln 5

) and

(-¥, -5) and (1, 9), the function is concave on

upward (-¥, -5) and (1, 9). Since the graph of f ¢( x) is decreasing on (-5, 1) and (9, ¥), the function is concave downward on (-5, 1) and (9, ¥). The inflection points are at -5, 1, and 9.

Section 5.3 63.

351

Choose f ( x) = x k , where 1 < k < 2.

66.

The slope of the tangent line to f ( x) = ln x as x  ¥ approaches 0. The slope of the tangent line to the graph of f ( x) = ln x as x  0 approaches ¥. For example, the slope of the tangent line at x = 0.00001 is 100,000.

67.

f ( x) = -x 2 - 10 x - 25

If k = 43 , then f ¢( x ) =

4 1/3 x 3

4 -2/3 4 = x 9 9 x 2/3

f  ( x) =

Critical number: 0 Since f ¢( x) is negative when x < 0 and positive when x > 0, f ( x) = x 4/3 has a relative minimum at x = 0.

f ¢( x) = -2 x - 10

= -2( x + 5) = 0

Critical number: -5

If k = 53 , then f ( x) =

5 2/3 x 3

f ( x) =

f ( x) = -2 < 0 for all x.

10 -1/3 10 = 1/3 x 9 9x

f ( x) is never 0, and does not exist when x = 0; so, the only candidate for an inflection point is at x = 0.

The curve is concave downward, which means a relative maximum occurs at x = -5. 68.

f ( x) = x 2 - 12 x + 36 f ¢( x) = 2 x - 12

Since f ( x) is negative when x < 0 and positive

f ¢( x) = 0 when

when x > 0, f ( x) = x5/3 has an inflection point at x = 0.

2 x - 12 = 0 x = 6.

64.

(a)

Critical number: 6 f ( x) = 2 > 0 for all x.

The curve is concave upward, which means a relative minimum occurs at x = 6. 69.

f ¢( x ) = 9 x 2 - 6 x

(b) Both f ( x) and g ( x) are concave down on (-¥, 0) and concave up on (0, ¥). Thus, they both have a point of inflection at (0, 0). (c)

f ( x) = 3 x 3 - 3 x 2 + 1

= 3x(3x - 2) = 0

Critical numbers: 0 and 23

f ( x) = x 7/3

g ( x) = x5/3

f ( x) = 18 x - 6

7 4/3 x 3 28 1/3 f  ( x) = x 9 f (0) = 0

5 2/3 x 3 10 -1/3 g ( x) = x 9 g (0) is undefined

f (0) = -6 < 0, which means that a relative maximum occurs at x = 0.

f ¢( x) =

g ¢( x ) =

( )

f  23 = 6 > 0, which means that a relative

minimum occurs at x = 23 .

(d) No. 65.

(a) The slope of the tangent line to f ( x) = e as x  -¥ is close to 0 since the tangent line is almost horizontal, and a horizontal line has a slope of 0.

70.

f ( x) = 2 x 3 - 4 x 2 + 2 f ¢( x) = 6 x 2 - 8 x f ¢( x) = 0 when

(b) The slope of the tangent line to f ( x) = e as x  0 is close to 1 since the first derivative represents the slope of the tangent line, f ¢( x) = e x , and e0 = 1.

6 x 2 - 8x = 0 2 x(3x - 4) = 0 4 x = 0 or x = . 3

352

Chapter 5 GRAPHS AND THE DERIVATIVE Critical numbers: 0, 43

f ¢(-1) = 3(-1) 2 = 3 > 0

f ( x) = 12 x - 8

This indicates that f is increasing on (-¥, 0).

f (0) = -8 < 0, which means that a relative maximum occurs at x = 0.

f ¢(1) = 3(1) 2 = 3 > 0

( )

This indicates that f is increasing on (0, ¥). Neither a relative maximum nor relative minimum occurs at x = 0.

f  43 = 8 > 0, which means that a relative

minimum occurs at x = 43 . 71.

f ( x) = ( x + 3) 4

73.

f ¢( x) = 4( x + 3)3 = 0

Critical number: x = -3 f ( x) = 12( x + 3)

f ( x) = x7/3 + x 4/3 7 4 f ¢( x) = x 4/3 + x1/3 3 3 f ¢( x) = 0 when

7 4/3 4 x + x1/3 = 0 3 3

f (-3) = 12(-3 + 3) 2 = 0

x1/3 (7 x + 4) = 0 3

The second derivative test fails. Use the first derivative test.

4 x = 0 or x = - . 7

Critical numbers: - 74 , 0 f ¢(-4) = 4(-4 + 3) 2 = 4(-1)3 = -4 < 0

This indicates that f is decreasing on (-¥, -3). f ¢(0) = 4(0 + 3)3 3

= 4(3) = 108 > 0

This indicates that f is increasing on (-3, ¥). A relative minimum occurs at -3. 72.

f  ( x) =

28 1/3 4 x + x-2/3 9 9

-1/3 -2/3 æ 4ö 28 æç 4 ö÷ 4 æç 4 ö÷ f  çç - ÷÷ = + » -1.9363 ÷ ÷ ç ç èç 7 ÷ø 9 èç 7 ø÷ 9 èç 7 ø÷

Relative maximum occurs at - 74 . f (0) does not exist, so the second derivative test fails. Use the first derivative test.

f ( x) = x 3 f ¢( x) = 3x 2 f ¢( x) = 0 when

3x 2 = 0 x = 0. Critical number: 0 f ( x ) = 6 x f (0) = 0 The second derivative test fails. Use the first derivative test.

4/3 1/3 æ 1ö 7æ 1ö 4æ 1 ö f ¢ çç - ÷÷ = çç - ÷÷ + çç - ÷÷ » -0.1323 çè 2 ÷ø 3 èç 2 ø÷ 3 èç 2 ø÷

(

)

This indicates that f is decreasing on - 74 , 0 . f ¢(1) =

7 4/3 4 11 (1) + (1)1/3 = 3 3 3

This indicates that f is increasing on (0, ¥). Relative minimum occurs at 0.

Section 5.3 74.

353

f ( x) = x8/3 + x5/3 8 5 f ¢( x) = x5/3 + x 2/3 3 3 f ¢( x) = 0 when 8 5/3 5 x + x 2/3 = 0 3 3 x 2/3 (8x + 5) = 0 3

Graph of f  :

5 x = 0 or x = - . 8

Critical numbers: - 85 , 0 f  ( x) =

40 2/3 10 -1/3 x + x 9 9 -1/3

2/3

æ 5ö 40 æç 5 ö÷ f  çç - ÷÷ = ç- ÷ èç 8 ø÷ 9 èç 8 ø÷

10 çæ 5 ÷ö ç- ÷ 9 çè 8 ÷ø

» 1.9493

Relative minimum occurs at - 85 . f (0) does not exist, so the second derivative test fails. Use the first derivative test.

5/3 2/3 æ 1ö 8æ 1 ö 5æ 1 ö f ¢ çç - ÷÷÷ = çç - ÷÷÷ + çç - ÷÷÷ » 0.2100 èç 2 ø 3 èç 2 ø 3 èç 2 ø

(

)

This indicates that f is increasing on - 85 , 0 . f ¢(1) =

8 5/3 5 13 (1) + (1)2/3 = 3 3 3

This indicates that f is increasing on (0, ¥). Neither a relative minimum or maximum occurs at 0. 75.

f ¢( x) = x 3 - 6 x 2 + 7 x + 4 f ( x) = 3x 2 - 12 x + 7

Graph f ¢ and f  in the window [-5, 5] by [-5, 15], Xscl = 0.5. Graph of f ¢:

(a) f has relative extrema where f ¢( x) = 0. Use the graph to approximate the x-intercepts of the graph of f ¢. These numbers are the solutions of the equation f ¢( x) = 0. We find that the critical numbers of f are about -0.4, 2.4, and 4.0. By either looking at the graph of f ¢ and applying the first derivative test or by looking at the graph of f  and applying the second derivative test, we see that f has relative minima at about -0.4 and 4.0 and a relative maximum at about 2.4. (b) Examine the graph of f ¢ to determine the intervals where the graph lies above and below the x-axis. We see that f ¢( x) > 0 on about (-0.4, 2.4) and (4.0, ¥), indicating that f is increasing on the same intervals. We also see that f ¢( x) < 0 on about (-¥, -0.4) and (2.4, 4.0), indicating that f is decreasing on the same intervals. (c) Examine the graph of f . We see that this graph has two x-intercepts, so there are two x-values where f ( x) = 0. These x-values are about 0.7 and 3.3. Because the sign of f  changes at these two values, we see that the x-values of the inflection points of the graph of f are about 0.7 and 3.3. (d) We observe from the graph of f  that f ( x) > 0 on about (-¥, 0.7) and (3.3, ¥), so f is concave upward on the same intervals. Likewise, we observe that f ( x) < 0 on about (0.7, 3.3), so f is concave downward on the same interval.

354 76.

Chapter 5 GRAPHS AND THE DERIVATIVE (c) Examine the graph of f  We see that this graph has three x-intercepts, so there are three values where f ( x) = 0. These x-values are 0, about 0.36, and about 0.84. Because the sign of f  and thus the concavity of f changes at these three values, we see that the x-values of the inflection points of the graph of f are 0, about 0.36, and about 0.84.

f ¢( x) = 10 x 2 ( x - 1)(5x - 3) = 10 x 2 (5x 2 - 8 x + 3) = 50 x 4 - 80 x3 + 30 x 2 f ( x) = 200 x3 - 240 x 2 + 60 x = 20 x(10 x 2 - 12 x + 3)

Graph f ¢ in window [-1, 1.5] by [-2, 2], Xscl = 0.1.

(d) We observe from the graph of f  that f ( x) > 0 on (0, 0.36) and (0.84, ¥), so f is concave upward on the same intervals. Likewise, f ( x) < 0 on (-¥, 0) and (0.36, 0.84), so f is concave downward on the same intervals.

This window does not give a good view of the graph of f , so we graph f  in the window [-1, 1.5] by [-20, 20], Xscl = 0.1. Yscl = 5.

77.

f ¢( x ) = f ( x) = = =

(a) The critical numbers of f are the x-intercepts of the graph of f ¢ (Note that there are no values where f ¢( x) does not exist.) From the graph or by examining the factored expression for f ¢, we see that the critical numbers of f are 0, 0.6, and 1.

By either looking at the graph of f ¢ and applying the first derivative test or by looking at the graph of f  and applying the second derivative test, we see that f has a relative minimum at 1 and a relative maximum at 0.6.

1 - x2 ( x 2 + 1)2 ( x2 + 1)2 (-2 x) - (1 - x 2 )(2)( x2 + 1)2 x ( x2 + 1)4 -2 x( x2 + 1) 2[( x2 + 1) + 2(1 - x2 )] ( x 2 + 1)4 -2 x(3 - x 2 ) ( x 2 + 1)3

Graph f ¢ and f  in the window [-3, 3] by [-1.5, 1, 5], Xscl = 0.2. Graph of f ¢:

Graph of f :

(At x = 0, the second derivative test fails since f (0) = 0, and the first derivative does not change sign, so there is no relative extremum at 0.) (b) Examine the graph of f ¢ to determine the intervals where the graph lies above and below the x-axis. We see that f ¢( x) ³ 0 on (-¥, 0.6), f ¢( x) < 0 on (0.6, 1), and f ¢( x) > 0 on (1, ¥). Therefore, f is increasing on (-¥, 0.6) and (1, ¥) and decreasing on (0.6, 1).

(a) The critical numbers of f are the x-intercepts of the graph of f ¢ (Note that there are no values where f ¢ does not exist.) We see from the graph that these x-values are -1 and 1.

Section 5.3

355 By either looking at the graph of f ¢ and applying the first derivative test or by looking at the graph of f  and applying the second derivative test, we see that f has a relative minimum at -1 and a relative maximum at 1.

Graph of f  :

(b) Examine the graph of f ¢ to determine the intervals where the graph lies above and below the x-axis. We see that f ¢( x) > 0 on (-1, 1), indicating that f is increasing on the same interval. We also see that f ¢( x) < 0 on (-¥, -1) and (1, ¥), indicating that f is decreasing on the same intervals.

(a) The critical number of f is the x-intercept of the graph of f ¢. Using the graph, we find a critical number of f is about 0.5671. By looking at the graph of f  and applying the second derivative test, we see f has a minimum at 0.5671.

(c) Examine the graph of f . We see that the graph has three x-intercepts, so there are three values where f ( x) = 0. These xvalues are about -1.7, 0, and about 1.7.

(b) Examine the graph of f ¢ to determine the intervals where the graph lies above and below the x-axis. We see that f ¢( x) > 0 on about (0.5671, ¥), indicating that f is increasing on about (0.5671, ¥). We also see that f ¢( x) < 0 on about (0, 0.5671), indicating that f is decreasing on about (0, 0.5671).

Because the sign of f  and thus the concavity of f changes at these three values, we see that the x-values of the inflection points of the graph of f are about -1.7, 0, and about 1.7.

(c) Examine the graph of f . We see that the graph has one x-intercept, so there is one x-value where f ( x) = 0. This value is about 0.2315. Because the sign of f  changes at this value, we see that x-value of the inflection point of the graph of f is about 0.2315.

(d) We observe from the graph of f  that f ( x) > 0 on about (-1.7, 0) and (1.7, ¥), so f is concave upward on the same intervals. Likewise, we observe that f ( x) < 0 on about (-¥, -1.7) and (0, 1.7), so f is concave downward on the same intervals. 78.

(d) We observe from the graph f  that f  > 0 on about (0.2315, ¥), so f is concave upward on about (0.2315, ¥). Likewise, we observe from the graph f  that f  < 0 on about (0, 0.2315), so f is concave downward on about (0, 0.2315).

f ¢( x) = x 2 + x ln x æ1ö f ( x) = 2 x + ( x) çç ÷÷÷ + (1)(ln x) çè x ø = 2 x + 1 + ln x

Graph f ¢ and f  in the window [0, 1] by [-2, 3], Xscl = 0.1, Yscl = 1. Graph of f ¢:

79.

There are many examples. The easiest is f ( x) = x . This graph is increasing and concave downward. f ¢( x ) =

1 -1/2 1 x = 2 2 x

f ¢(0) does not exist, while f ¢( x) > 0 for all x > 0. (Note that the domain of f is [0, ¥).)

As x increases, the value of f ¢( x) decreases, but remains positive. It approaches zero, but never becomes zero or negative.

356 80.

Chapter 5 GRAPHS AND THE DERIVATIVE R( x) = 10, 000 - x3 + 42 x 2 + 800 x, 0 £ x £ 20

82.

R( x) = -0.3x3 + x 2 + 11.4 x, 0 £ x £ 6 R¢( x) = -0.9 x 2 + 2 x + 11.4

R¢( x) = -3x 2 + 84 x + 800

R( x) = -1.8 x + 2

R( x) = -6 x + 84 A point of diminishing returns occurs at a point of inflection, or where R( x) = 0.

A point of diminishing returns occurs at a point of inflection, or where R( x) = 0. -1.8x + 2 = 0

-6 x + 84 = 0

2 = 1.8x

x = 14 Test R x  to determine whether concavity changes at x = 14.

1.11 » x

R(12) = 12 > 0

Test R( x) to determine whether concavity changes at x = 1.11.

R(16) = -12 < 0

R(2) = -1.8(2) + 2 = -3.6 + 2 = -1.8 < 0

R( x) is concave upward on (0, 14) and concave

R(1) = -1.8(1) + 2 = 0.2 > 0

downward on (14, 20).

R( x) is concave upward on (0, 1.11) and

R(14) = 10, 000 - (14)3 + 42(14) 2 + 800(14)

concave downward on (1.11, 6).

= 26, 688

R(1.11) = -0.3(1.11)3 + (1.11)2 + 11.4(1.11)

The point of diminishing returns is (14, 26, 688).

= 13.476 » 13.5

81.

4 R( x) = (-x3 + 66 x 2 + 1050 x - 400) 27 0 £ x £ 25 4 R¢( x) = (-3x 2 + 132 x + 1050) 27 4 R ( x) = (-6 x + 132) 27 A point of diminishing returns occurs at a point of inflection, or where R( x) = 0.

The point of diminishing returns is (1.11, 13.5). 83.

R( x) = -0.6 x3 + 3.7 x 2 + 5x, 0 £ x £ 6 R¢( x) = -1.8 x 2 + 7.4 x + 5 R( x) = -3.6 x + 7.4

A point of diminishing returns occurs at a point of inflection or where R( x) = 0. -3.6 x + 7.4 = 0 -3.6 x = -7.4

4 (-6 x + 132) = 0 27 -6 x + 132 = 0

x =

6 x = 132 x = 22

Test R( x) to determine whether concavity changes at x = 22. 4 16 >0 (-6 ⋅ 20 + 132) = 27 9 4 16 R(24) = (-6 ⋅ 24 + 132) = <0 27 9 R(20) =

R( x) is concave upward on (0, 22) and concave

downward on (22, 25). 4 [-(22)3 + 66(22) 2 + 1060(22) - 400] 27 » 6517.9

R(22) =

The point of diminishing returns is (22, 6517.9).

-7.4 » 2.0556 » 2.06 -3.6 x

Test R( x) to determine whether concavity changes at x = 2.06. R(2) = -3.6(2) + 7.4 = -7.2 + 7.4 = 0.2 > 0 R(3) = -3.6(3) + 7.4 = -10.8 + 7.4 = -3.4 < 0 R( x) is concave upward on (0, 2.06) and concave downward on (2.06, 6). R(2.0556) = -0.6(2.0556)3 + 3.7(2.0556) 2 + 5(2.0556) » 20.7

The point of diminishing returns is (2.06, 20.7).

Section 5.3 84.

357 -U (M ) U ¢(M )

I (M ) =

For U (M ) =

86.

+ 0.0613t + 2.34

M = M

1/ 2

A¢(t ) = 0.0000987t 2 - 0.009t + 0.0613 A¢¢(t ) = 0.0001974 x - 0.009

1 -1/2 M 2 1 U (M ) = - M -3/2. 4 U ¢(M ) =

Assets will decrease most rapidly when A¢ has a minimum. A has a single inflection point when A¢¢(t ) = 0, which occurs when

Then I (M ) =

A(t ) = 0.0000329t 3 - 0.00450t 2

(

- - 14 M

-3/2

)

0.0001974t - 0.009 = 0 0.0001974t = 0.009

1 M -1/2 2

t =

1 -1 M 2 1 = . 2M =

This t-value corresponds to a minimum of A¢ and the minimum rate of decrease is A¢(45.6) » -0.144. Since t = 0 represents the year 2000, the maximum rate of decrease of Social Security assets will occur in around the middle of 2045.

For U (M ) = M 2/3, 2 -1/3 M 3 2 U (M ) = - M -4/3. 9 U ¢( M ) =

87. (a)

I (M ) =

(

R( x) = Cx 1 - e-kx

)

2 M -1/3 3

(kx - 1)e-kx ³ -1

1 -1 M 3 1 . = 3M

kx - 1 ³ -ekx

U (M ) =

1 - kx £ ekx The left and right sides are equal at x = 0 an 1 - kx is a decreasing function

and ekx is an increasing function for positive k, so R¢ will be nonnegative on [0, 1] and R will be increasing on [0, 1]

M indicates a greater risk aversion

1 for 0 < M < 1. because 21M > 3M

85.

)

R¢( x) = C éê x ⋅ (kx)e-kx + 1 ⋅ (1 - e-kx ) ùú ë û -kx ù é = C ê 1 + (kx - 1)e ú ë û The derivative will be non-negative when

Then - - 92 M -4/3

0.009 » 45.6. 0.0001974

Let D(q) represent the demand function. The revenue function, R(q), is R(q) = qD(q). The marginal revenue is given by R¢(q) = qD¢(q) + D(q)(1) = qD¢(q) + D(q). R(q) = qD(q) + D¢(q)(1) + D¢(q) = qD(q) + 2 D¢(q)

gives the rate of decline of marginal revenue. D¢(q) gives the rate of decline of price. If marginal revenue declines more quickly than price,

(b)

R¢( x) = C êé 1 + (kx - 1)e-kx ùú ë û kx + (kx - 1)(-ke-kx ) ùú R¢¢( x) = C éê ke ë û -kx é ù = C ê ke ( 2 - kx ) ú ë û R¢¢ will be positive when 2  kx or x  2 . k So R will be concave up on [0, 2 / k ) if k  2 and on [0, 1] if k < 2.

358 88.

Chapter 5 GRAPHS AND THE DERIVATIVE f (c) = f ¢(c) = = f ¢¢(c) =

a-c b-c (b - c)(-1) - (a - c)(-1)

96 = - 3 < 0 implies that K (2) = 512 16

K (t ) is maximized at t = 2. Thus, the concentration is a maximum after 2 hours.

(b - c)2 a-b

(b)

(b - c)2 2(a - b)

92.

K (t ) =

so f is increasing and concave upward. (b) If b  a  c, then both numerators are negative and both denominators remain positive, so f is decreasing and concave downward.

90.

(a)

f 0 represents initial population (t = 0).

(b)

(a, f (a)) is the point where the graph changes concavity or the inflection point.

(c)

f M is the maximum carrying capacity.

(a)

t +4

3(t 2 + 4) - (2t )(3t )

K ¢(t ) =

(3t 2 + 27)(4) - 4t (6t ) (3t 2 + 27)2 -12t 2 + 108 (3t 2 + 27)2

(3t 2 + 27)2 (-24t )

(t 2 + 4) 2 (t 2 + 4)2

3t + 27

For this application, the domain of K is [0, ¥), so the only critical number is 3. To determine whether a relative maximum or minimum occurs at t = 3, we find K (3) and use the second derivative test. To find K (t ), apply the quotient rule to find the derivative of K ¢(t ). The numerator of K (t ) will be

-3t 2 + 12

-12(t 2 - 9) = 0 t = -3 or t = 3

K ¢(t ) =

(2) + 4

- (-12t 2 + 108)(2)(3t 2 + 27)(6t )

= 0

= (3t 2 + 27)2 (-24t ) -(12t )(-12t 2 + 108)(3t 2 + 27).

- 3t 2 + 12 = 0 t2 = 4

The denominator of K  is (3t 2 + 27)4 , so

t = 2 or t = -2 K (3) =

For this application, the domain of K is [0, ¥), so the only critical number is 2. K (t ) = = =

(t 2 + 4)4 2

-6t (t + 4) - 4t (-3t + 12) 6t - 72t (t 2 + 4)3

(54)4 72 542

< 0.

This indicates that the concentration is a maximum after 3 hours.

(t 2 + 4)3

(54) 2 (-72) - (36)(0)(54)

(t + 4) (-6t ) - (-3t + 12)(2)(t + 4)(2t )

3 4

Set K ¢(t ) = 0.

The chemicals are accumulating, but at a slower rate. Therefore, c(t ) is increasing and concave

K (t ) =

downward. Thus, c¢(t ) > 0 and c¢¢(t ) < 0. 91.

3(2)

The maximum concentration is 34 %.

( b - c )3

(a) If a  b  c, then both numerator and denominator of f and f ¢ are positive,

89.

K (2) =

(b)

K (3) =

4(3) 3(3)2 + 27

2 9

The maximum concentration is 92 %.

Section 5.3

93.

359

31.4

G(t ) =

1 + 12.5e-0.393t The solution will be easier to follow if we replace the given constants with letters and derive a general solution. Use the quotient rule and the chain rule. a G(t ) = 1 + be-ct Use the power rule and the chain rule. G¢(t ) =

G ¢¢(t ) =

(

) = abc ⋅ e-ct 2 2 (1 + be-ct ) (1 + be-ct ) 2ö æ ( -abc2 ⋅ e-ct )çççè(1 + be-ct ) ÷÷÷÷ø - ( abc ⋅ e-ct )( 2(1 + be-ct )( -bc ⋅ e-ct ))

a(-1) -bc ⋅ e-ct

2 2ù é ê 1 + be-ct ú êë úû

(

)

-abc 2 ⋅ e-ct )( 1 + be-ct ) + ( -abc 2 ⋅ e-ct )( -2be-ct ) ( = 3 (1 + be-ct ) abc 2 ⋅ e-ct )( be-ct - 1) ( = 3 (1 + be-ct ) G ¢¢(t ) will be 0 when be-ct = 1 b = ect ln(b) c For the cactus wren function, b = 12.5 and c = 0.393, so ln(12.5) = 6.427. G(6.427) = 15.7, so the inflection t = 0.393 point is (6.427, 15.7). t =

94. G(t ) =

787

1 + 60.8e-0.0473t Using the solution derived for Exercise 85, the inflection point will ln(60.8) occur when t = = 86.8. G(86.8) = 393, so the 0.0473 inflection point is (86.8, 393).

360 95.

Chapter 5 GRAPHS AND THE DERIVATIVE -kt

)

-kt

)

= ec(1-e

-kt

)

= ckec-ce

-kt

N (t ) = ckec-ce

-kt

(c - ce-kt - kt )¢

[-ce-kt (-kt )¢ - k ]

= ckec -ce

-kt

[-ce-kt (-kt )¢ - k ]

(cke-kt - k )

= ckec-ce

-kt

(cke-kt - k )

L(t ) = Be-ce

-kt

L¢(t ) = Be-ce

-kt

(-ce-kt )¢

= Be-ce

-kt

[-ce-kt (-kt )¢]

= Bcke-ce

-kt

L(t ) = Bcke-ce

-kt

(-ce-kt - kt )¢

= Bcke-ce

-kt

= Bcke-ce

-kt

96.

= Bck 2e-ce

-kt

N (t ) = ec (1-e N ¢(t ) = ec(1-e

= ck 2ec -ce

(ce-kt - 1)

[c(1 - e-kt )]¢ (-ce-kt )(-kt )¢

-kt

(ce-kt - 1)

L(t ) = 0 when ce-kt - 1 = 0

N (t ) = 0 when ce-kt - 1 = 0

ce-kt - 1 = 0

ce-kt - 1 = 0 c =1 ekt

ekt

ekt = c kt = ln c ln c t = k

e kt = c kt = ln c ln c k Letting c = 7.267963 and k = 0.670840 t =

ln 7.267963 » 2.96 years 0.670840 Verify that there is a point of inflection at

t =

ln c

-kt

ln 27.3 » 301 0.011 Verify that there is a point of inflection at

t =

t = k » 2.96. For L(t ) = Bck 2e-ce

Letting c = 27.3 and k = 0.011

(ce-kt - 1),

we only need to test the factor ce-kt - 1 on the intervals determined by t » 2.96 since the other factors are always positive. L(1) has the same sign as

7.267963e-0.670840(1) - 1 » 2.72 > 0. L(3) has the same sign as 7.267963e-0.670840(3) - 1 » -0.029 < 0.

( ) and concave down on ( , ¥ ) , so there is a ln c

Therefore L, is concave up on 0, k » 2.96 ln c k

ln c point of inflection at t = k » 2.96 years.

This signifies the time when the rate of growth begins to slow down since L changes from concave up to concave down at this inflection point.

ln 27.3 . 0.011 -kt

For N (t ) = ck 2ec -ce

- kt

(ce-kt - 1), we

only need to test the factor ce-kt - 1 on the intervals determined by t =

ln 27.3 » 301 since 0.011

the other factors are always positive. N (200) has the same sign as

27.3e-0.011(200) - 1 = 27.3e-2.2 - 1 » 2.02 > 0. N(400) has the same sign as 27.3e-0.011(400) - 1 = 27.3e-4.4 - 1 » -0.66 < 0.

(

ln 27.3

Therefore, N is concave up on 0, 0.011 is concave down on

(

) and N

)

ln 27.3 , ¥ , so there is a 0.011

point of inflection at t =

ln 27.3 » 301 days. 0.011

This signifies the time when the rate of growth begins to slow down since L changes from concave up to concave down at this inflection point.

Section 5.3 97.

361 100. Since the rate of violent crimes is decreasing but at a slower rate than in previous years, we know that f ¢(t ) < 0 but f (t ) > 0. Note that since f ¢(t ) < 0, f is decreasing, and since f (t ) > 0, the graph of f is concave upward.

v¢( x) = -35.98 + 12.09 x - 0.4450 x 2 v¢( x) = 12.09 - 0.89 x v( x) = -0.89

Since -0.89 < 0, the function is always concave down.

101.

-8.96e -0.0685t

N (t ) = 71.8e

98.

N ¢(t ) = 71.8(-8.96)(-0.0685)e-0.0685t -0.0685t ö æ ÷ ´ çç e-8.96e è ø÷

= (44.067968e-0.0685t )(e-8.96e

-0.0685t

)

-0.0685t

N (t ) = (44.067968)(0.61376e

- 0.0685)e

s(t ) = -16t 2 v(t ) = s¢(t ) = -32t

(a)

v(3) = -32(3) = -96 ft/s ec

(b)

v(5) = -32(5) = -160 ft/sec

(c)

v(8) = -32(8) = -256 ft/sec

(d)

a(t ) = v¢(t ) = s(t )

(-8.96e -0.0685t -0.0685t)

= -32 ft/sec2

N (t ) = 0 when 0.61376e-0.0685t - 0.0685 = 0

102.

0.61376e-0.0685t - 0.0685 = 0 e-0.0685t » 0.11162

s(t ) = 1.5t 2 + 4t

(a)

- 0.0685t » ln 0.11162 t » 32.01

s(10) = 1.5(102 ) + 4(10) = 150 + 40 = 190

The car will move 190 ft in 10 seconds.

N (t ) has an inflection point at (32.01, N (32.01)) or (32.01, 26.41).

(b)

v(t ) = s¢(t ) = 3t + 4

s¢(5) = 3 ⋅ 5 + 4 = 19 s¢(10) = 3 ⋅ 10 + 4 = 34

99. P( R) =

1 1 + 2 DR

P ¢( R ) =

The velocity at 5 sec is 19 ft/sec, and the

velocity at 10 sec is 34 ft/sec.

-4 DR

( 1 + 2 DR ) 2

) - (-4 DR)(2) ( 1 + 2 DR ) (4 DR) ( 1 + 2 DR ) (-4 D ) ( 1 + 2 DR ) - (-4 DR )(2)(4 DR ) = ( 1 + 2 DR ) (4 D ) ( 8 DR - 1 - 2 DR ) (4 D ) ( 6 DR - 1 ) = = ( 1 + 2 DR ) ( 1 + 2 DR )

P ¢¢( R ) =

(

(-4 D ) 1 + 2 DR

(d)

The acceleration at 5 sec is 3 ft/sec2 , and the

acceleration at 10 sec is also 3 ft/sec2.

a(t ) = v¢(t ) = s(t ) = 3

The inflection point will occur when 1 6 DR2 = 1 or R = . 6 D 1 = 0.022 implies 6 D 1 = 109.61 » 110. D = (0.022)2 ⋅ 6

(e) As t increases, the velocity increases, but the acceleration is constant.

103.

s(t ) = 256t - 16t 2 v(t ) = s¢(t ) = 256 - 32t a(t ) = v¢(t ) = s(t ) = -32

To find when the maximum height occurs, set s¢(t ) = 0. 256 - 32t = 0

t =8 Find the maximum height. s(8) = 256(8) - 16(82 ) = 1024

362

Chapter 5 GRAPHS AND THE DERIVATIVE The maximum height of the ball is 1024 ft. The ball hits the ground when s = 0.

The value where the second derivative is 0 divides the line into two intervals, (-¥, 1) and (1, ¥). Evaluate

256t - 16t 2 = 0

f ¢¢( x) at a point in each interval.

16t (16 - t ) = 0

f ¢¢(0) = 6 f ¢¢(2) = -6

t =0

(initial moment)

t = 16 (final moment)

The ball hits the ground 16 seconds after being shot. 104. The car was moving most rapidly when t » 6, because acceleration was positive on (0, 6) and negative after t = 6, so velocity was a maximum at t = 6.

5.4

This shows that f is concave upward on (-¥, 1) and concave downward on (1, ¥). The graph has an inflection point at (1, f (1)), or (1, 1). This information is summarized in the following table. Interval Sign of

f¢

Sign of

f ¢¢

f increasing or decreasing Concavity of f Shape of graph

Curve Sketching

(∞, 1) 

(1, 1) +

(1, 3) +

(3, ∞) 



Decreasing

Increasing

Increasing Decreasing

Upward

Downward Downward

Your Turn 1 f ( x) = -x3 + 3x 2 + 9 x - 10

Now sketch the graph using this information.

f ¢( x) = -3x + 6 x + 9 f ¢¢( x) = -6 x + 6

Use the first derivative to find intervals where the function is increasing or decreasing. -3 x 2 + 6 x + 9 = 0 -3( x 2 - 2 x - 3) = 0 ( x + 1)( x - 3) = 0 x = -1 or x = 3

These critical numbers divide the line into three intervals, (-¥, -1), (-1, 3) and (3, ¥). Evaluate the derivative at a test point in each region. f ¢(-2) = -15

Your Turn 2 f ( x) = 4 x +

1 4x2 + 1 = x x

Because x = 0 makes the denominator 0 but not the numerator, the line x = 0 (that is, the y-axis) is a vertical asymptote. Neither lim f ( x) nor lim f ( x)

f ¢(0) = 9 f ¢(4) = -4

This shows that f is decreasing on (-¥, -1), increasing on (-1, 3), and decreasing on (3, ¥). By the first derivative test, f has a relative minimum of f (-1) = -15 at x = -1 and a relative maximum of f (3) = 17 at x = 3. Use the second derivative to find the intervals where the function is concave upward or downward. -6 x + 6 = 0 6x = x x =1

x -¥

x ¥

exists, so there is no horizontal asymptote. But since the term 1/x gets very small as | x| gets large, the graph of f approaches the line y = 4 x as | x | becomes larger and larger, so this line is an oblique asymptote. f (-x) = - f ( x), so the graph of the left side can be found by rotating the right side around the origin by 180°.

Find any critical numbers. 1 f ¢( x) = 4 - 2 x 1 4- 2 = 0 x

Section 5.4

363 so the line y = 4 is a horizontal asymptote. Note that f (-x) = f ( x), so the graph of f is symmetrical around the y-axis,

1 4 1 1 x = or 2 2

x2 =

Find any critical numbers.

The critical numbers are -1/2 and 1/2, which together with the location of the vertical asymptote divide the line into four regions, (-¥, -1/2), (-1/2, 0), (0, 1/2), and (1/2, ¥). Evaluate the derivative at a test point in each region. f ¢(-1) = 3 æ 1ö f ¢ çç- ÷÷÷ = -12 çè 4 ø æ1ö f ¢ çç ÷÷÷ = -12 çè 4 ø f ¢(1) = 3

This shows that f has a relative maximum of f (-1/2) = -4 at x = -1/2 and a relative minimum of f (1/2) = 4 at 0 x = 1/2. Now find the intervals where the graph is concave upward or concave downward. 2 f ¢¢( x) = 3 x The second derivative is never 0, but concavity might change where the second derivative is undefined, at x = 0. In fact, f ¢¢( x) < 0 for x < 0, f ¢¢( x) > 0 for x > 0,

so the graph is concave downward to the left of the origin and concave upward to the right of the origin. This provides enough information to sketch the graph.

f ¢( x ) = = =

( x 2 + 4)(8x) - (4 x 2 )(2 x) ( x 2 + 4) 2 8 x3 + 32 x - 8x3 ( x 2 + 4) 2 32 x ( x 2 + 4) 2

The derivative is 0 only when x = 0, and the denominator of the derivative is never 0. Therefore the critical number divides the line into just two regions, (-¥, 0) and (0, ¥). Evaluate the derivative at a test point in each region. f ¢(-1) = -1.28 f ¢(1) = 1.28

Thus f is decreasing as x approaches 0 from the left, and increasing to the right of x = 0, so f has a relative minimum of f (0) = 0 at x = 0. Now find the intervals where the graph is concave upward or concave downward. 32 x f ¢( x) = 2 ( x + 4) 2 f ¢¢( x) = = =

( x 2 + 4) 2 (32) - (32 x)(2)( x 2 + 4)(2 x) ( x 2 + 4)4

( x 2 + 4)[( x 2 + 4)(32) - 128 x 2 ] ( x 2 + 4)4 128 - 96 x 2 ( x 2 + 4)3

The denominator is never 0, but the numerator is 0 when 128 - 96 x 2 = 0 96 x 2 = 128 3x 2 = 4 x2 =

Your Turn 3 f ( x) =

x=

x +4

The denominator of f is never 0, so f has no vertical asymptotes. However, lim

4x2

x -¥ x 2 + 4

4 3

= 4 and lim

4x2

x ¥ x 2 + 4

= 4,

4 » 1.155 3

The zeros of the second derivative divide the line into three regions, æ ö æ ö æ ö çç-¥, - 4 ÷÷ , çç- 4 , 4 ÷÷ and çç 4 , ¥÷÷. ÷ ÷ ÷÷ çç çç 3 3 ÷ø ççè 3 3 ÷ø è è ø

364

Chapter 5 GRAPHS AND THE DERIVATIVE Thus f is increasing on the interval (-¥, -1) and decreasing on the interval (-1, ¥), and has a relative maximum of f (-1) » 2.718 at x = -1.

Evaluate f ¢¢( x) at a point in each interval. f ¢¢(-2) = -0.5 f ¢¢(0) = 2 f ¢¢(2) = -0.5

Now find the intervals where the graph is concave upward or concave downward.

According to this information, the graph is

(

4 3

concave downward on the interval -¥, -

(

concave upward on the interval -

4, 3

and concave downward on the interval

4 3

)

( , ¥). 4 3

Thus there will be two inflection points. Find the corresponding values of f. æ 4 ÷ö ÷ f çççç 3 ÷÷ = 1 è ø æ 4 ö÷ f ççç ÷÷ = 1 èç 3 ÷ø

(

The inflection points are at -

f ¢( x) = -(1 + x)e-x f ¢¢( x) = (1 + x)e-x + (-1)e-x = xe-x

The second derivative is 0 only at x = 0. Evaluate f ( x) at points on either side of x = 0. f ¢¢(-1) » -2.718 f ¢¢(1) » 0.368

The graph is concave downward to the left of x = 0 and concave upward to the right of x = 0. There is an inflection point at (0, f (0)), or (0, 2). This provides enough information to sketch the graph.

) ( , 1).

4 , 1 and 3

4 3

This provides enough information to sketch the graph.

5.4 Warmup Exercises W1.

f ( x) = x 4 - 2 x3 - 12 x 2 + 4 x + 13 f ¢( x) = 4 x3 - 6 x 2 - 24 x + 4

Your Turn 4

f ¢¢( x) = 12 x 2 - 12 x - 24 -x

f ( x) = ( x + 2)e

Since lim ( x + 2)e-x = 0, the line y = 0 (the x-axis) x ¥

is a horizontal asymptote for the graph. Neither f (-x) = - f ( x) nor f (-x) = f ( x) is true, so the graph has no symmetry. Find any critical numbers. f ¢( x) = ( x + 2)(-e-x ) + (1)(e-x )

= 12( x 2 - x - 2) = 12( x - 2)( x + 1)

f ¢¢( x) will be 0 at x = 2 and x = -1. The factorization shows that each of these values is a single root at which f ¢¢ changes sign, so each root locates an inflection point. f (-1) = 0 and f (2) = -27, so the inflection points for f are (-1, 0) and (2, -27).

= -(1 + x)e-x

The derivative is 0 at only one point, where 1 + x = 0 or x = -1. This critical number divides the line into two regions, (-¥, -1) and (-1, ¥). Evaluate the derivative at a test point in each region. f ¢(-2) » 7.389 f ¢(0) = -1

Section 5.4 W2.

365 The calculator shows no y-value when x = 0 because 0 is not in the domain of this function. However, we see from the graph that

f ( x) = x3 - 21x 2 - 72 x + 72 x ln x f ¢( x) = 6 x 2 - 42 x - 71 æ ö 1 + 72 çç x ⋅ + ln x ÷÷÷ çè ø x = 6 x 2 - 42 x + 72 ln x 72 f ¢¢( x) = 12 x - 42 + x Set f ¢¢( x) equal to 0, multiply through by x (which will not be 0 since 0 is not in the domain of f ) , and factor. 72 6 x - 42 + =0 x

lim x ln | x| = 0

x  0-

and lim x ln | x| = 0.

x  0+

Thus, lim x ln | x| = 0.

x0

f ( x) = -2 x3 - 9 x 2 + 108x - 10

Domain is (-¥, ¥).

6 x 2 - 42 x + 72 = 0

f (-x) = -2(-x)3 - 9(-x) 2 + 108(-x) - 10

6( x 2 - 7 x + 12) = 0 6( x - 3)( x - 4) = 0 Thus f ¢¢ is 0 at x = 3 and x = 4.

= 2 x3 - 9 x 2 - 108x - 10 No symmetry f ¢( x) = -6 x 2 - 18x + 108

Note that f ¢¢(2) = 6, f ¢¢(7/2) = -3/7,

= -6( x 2 + 3x - 18) = -6( x + 6)( x - 3)

and f ¢¢(5) = 12/5, so the concavity of the graph of f changes at each of the roots. f (3) = -140.6997 » 140.7 and f (4) = -160.7472 » -160.7, so the inflection points for f are (3, -140.7) and (4, -160.7).

f ¢( x) = 0 when x = -6 or x = 3.

Critical numbers: -6 and 3 Critical points: (-6, -550) and (3, 179) f ( x) = -12 x - 18 f (-6) = 54 > 0 f (3) = -54 < 0

5.4 Exercises (a)

Relative maximum at 3, relative minimum at -6

False. Consider f ( x) = x 4.

False. Consider f ( x) = x + 1.

Decreasing on (-¥, -6) and (3, ¥)

False. The second derivative test is used to test concavity.

f ( x) = -12 x - 18 = 0 -6(2 x + 3) = 0

False. Consider f ( x) = x 4 at x = 0.

Graph y = x ln | x | on a graphing calculator. A suitable choice for the viewing window is [-1, 1] by [-1, 1], Xscl = 0.1, Yscl = 0.1.

Increasing on (-6, 3)

3 2 Point of inflection at (-1.5, -185.5) x =-

(b)

Concave upward on (-¥, -1.5) Concave downward on (-1.5, ¥) (c)

y-intercept: y = -2(0)3 - 9(0)2 + 108(0) - 10 = -10

366 8.

Chapter 5 GRAPHS AND THE DERIVATIVE f ( x) = x 3 -

15 2 x - 18 x - 1 2

f ¢( x) = -9 x 2 + 12 x - 4

= -(3x - 2) 2

Domain is (-¥, ¥).

(3x - 2)2 = 0

15 (-x 2 ) - 18(-x) - 1 2 15 2 = -x 3 x + 18x - 1 2 No symmetry f (-x) = (-x)3 -

f ¢( x )

x =

Critical number: 23 3 2 f 23 = -3 23 + 6 23 - 4 23 - 1 = - 17 9

( )

( ) ( ) Critical point: ( 23 , -17 9 )

= 3x 2 - 15 x - 18 = 3( x 2 - 5x - 6)

f ¢(1) = -9(1)2 + 12(1) - 4 = -1 < 0

Critical numbers: 6 and -1 Critical points: (6, -163) and (-1, 8.5)

(a)

f ( x) = 6 x - 15

( 23 , -179 )

f ¢¢( x) = -18x + 12 = -6(3x - 2) 3x - 2 = 0 2 x = 3

f (-1) = -21 < 0

Relative maximum at x = -1, relative minimum at x = 6 Increasing on (-¥, -1) and (6, ¥) Decreasing on (-1, 6)

(b)

f ( x) = 6 x - 15 = 0 5 x = 2

Point of inflection at

(c)

y-intercept:

Point of inflection at

( 23 , -179 )

f (0) = -18(0) + 12 = 12 > 0 f (1) = -18(1) + 12 = -6 < 0

( ) Concave upward on ( 23 , ¥ ) Point of inflection at ( 23 , -17 9 )

Concave upward on -¥, 23

( 52 , -77.25 ) Concave upward on ( 52 , ¥ ) Concave downward on ( -¥, 52 )

(b)

(c)

10.

f ( x ) = -3 x + 6 x - 4 x - 1

y-intercept: y = -3(0)3 + 6(0) 2 - 4(0) - 1 = -1

15 y = (0) - (0)2 - 18(0) - 1 = -1 2 3

No relative extremum at Decreasing on (-¥, ¥)

f (6) = 21 > 0

( )

f ¢(0) = -9(0)2 + 12(0) - 4 = -4 < 0

= 3( x - 6)( x + 1) = 0

(a)

2 3

f ( x) = x3 - 6 x 2 + 12 x - 11

Domain is (-¥, ¥).

f (-x) = (-x)3 - 6(-x)2 + 12(-x) - 11

f (-x) = -3(-x)3 + 6(-x) 2 - 4(-x) - 1

= -x3 - 6 x 2 - 12 x - 11 No symmetry

= 3x3 + 6 x 2 + 4 x - 1

Section 5.4

367

f ¢( x) = 3x 2 - 12 x + 12 = 3( x 2 - 4 x + 4) = 3( x - 2) 2

12( x 2 - 4) = 0 x = 2 (b) Points of inflection at (-2, 0) and (2, 0)

Critical number: 2 Critical point: (2, -3) f ( x) = 6 x - 12 f (0) = 6(0) - 12 = -12 < 0 f (3) = 6(3) - 12 = 6 > 0 (a) (b)

(c)

12 x 2 - 48 = 0

Concave upward on (-¥, -2) and (2, ¥) Concave downward on (-2, 2) (c)

No relative extrema Increasing on (-¥, ¥)

Let u = x 2.

Point of inflection at (2, -3) Concave upward on (2, ¥); concave downward on (-¥, 2)

u 2 - 24u + 80 = 0 (u - 4) (u - 20) = 0 u = 4

y-intercept:

x = 2 or x = 2 5

y = 03 - 6(0)2 + 12(0) - 11 = -11

11.

f ( x) = x 4 - 24 x 2 + 80

or u = 20

y-intercept: y = (0)4 - 24(0)2 + 80 = 80

12.

Domain is (-¥, ¥).

f ( x ) = -x 4 + 6 x 2

Domain is (-¥, ¥).

f (-x) = (-x)4 - 24(-x) 2 + 80 = x 4 - 24 x 2 + 80 = f ( x)

f (-x) = -(-x) 4 + 6(-x) 2 = -x 4 + 6 x 2 = f ( x)

The graph is symmetric about the y-axis.

f ¢( x) = 4 x3 - 48 x

f ¢( x) = -4 x3 + 12 x

4 x3 - 48x = 0

-4 x3 + 12 x = 0

4 x( x 2 - 12) = 0

-4 x( x 2 - 3) = 0

4 x( x - 2 3)( x + 2 3) = 0

-4 x( x - 3)( x + 3) = 0

Critical numbers: -2 3, 0, and 2 3

Critical numbers: - 3, 0, and 3

Critical points: (-2 3, -64), (0, 80), and

Critical points: (- 3, 9), (0, 0) and ( 3, 9)

(2 3, -64)

f ( x) = -12 x 2 + 12

f ( x) = 12 x 2 - 48

f (- 3) = -12( 3)2 + 12 = -24 < 0

f (-2 3) = 12(-2 3)2 - 48 = 96 > 0

f (0) = -12(0)2 + 12 = 12 > 0

f (0) = 12(0)2 - 48 = -48 < 0

f ( 3) = -12( 3)2 + 12 = -24 < 0

f (2 3) = 12(2 3) 2 - 48 = 96 > 0

(a)

x-intercepts: 0 = x 4 - 24 x 2 + 80

Relative maximum at 0, relative minima at -2 3 and 2 3 Increasing on (-2 3, 0) and (2 3, ¥)

(a)

Relative minimum at 0, relative maxima at - 3 and 3 Increasing on (¥, - 3) and (0, 3) Decreasing on (- 3, 0) and ( 3, ¥)

368

Chapter 5 GRAPHS AND THE DERIVATIVE -12 x 2 + 12 = 0

-12( x 2 - 1) = 0 x = 1 (b) Points of inflection at (-1, 5) and (1, 5)

(c)

12 x 2 - 24 x = 0 12 x( x - 2) = 0 x = 0 or x = 2

(b)

Points of inflection at (0, 0) and (2, -16)

Concave upward on (-1, 1)

Concave upward on (-¥, 0) and (2, ¥)

Concave downward on (-¥, -1) and (1, ¥)

Concave downward on (0, 2)

-x4 + 6x2 = 0

x-intercepts:

(c)

x-intercepts: x 4 - 4 x3 = 0 x3 ( x - 4) = 0 x = 0 or x = 4

- x 2 ( x 2 - 6) = 0 x = 0 or x =  6

y-intercepts: y = (0) 4 - 4(0)3 = 0

y-intercept: y = -(0) 4 + 6(0) 2 = 0

13.

f ( x) = x 4 - 4 x 3

14.

Domain is (-¥, ¥).

f (-x) = (-x)5 - 15(-x)3

f (-x) = (-x) 4 - 4(-x)3

= -x5 + 15 x3 = - f ( x)

= x 4 + 4 x3 ¹ f ( x) or - f (x)

The graph is symmetric about the origin.

The graph is not symmetric about the y-axis or the origin. 3

f ¢( x) = 4 x - 12 x

f ¢( x) = 5x 4 - 45 x 2 = 0

5 x 2 ( x 2 - 9) = 0

4 x3 - 12 x 2 = 0

5 x 2 ( x + 3)( x - 3) = 0

4 x ( x - 3) = 0

Critical numbers: 0, -3, and 3

Critical numbers: 0 and 3

Critical points: (0, 0), (-3, 162), and (3, -162)

Critical points: (0, 0) and (3, -27)

f ( x) = 20 x3 - 90 x = 10 x( x 2 - 9) f (0) = 0 f (-3) = -270 < 0 f (3) = 270 > 0

f ( x) = 12 x - 24 x f (0) = 12(0)2 - 24(0) = 0 f (3) = 12(3)2 - 24(3) = 36 > 0

Second derivative test fails for 0. Use first derivative test. f ¢ =  - 12(-1)2 = -16 < 0 f ¢ = 4(1)3 - 12(1) 2 = -8 < 0

(a)

f ( x) = x5 - 15x3

Neither a relative minimum nor maximum at 0 Relative minimum at 3 Increasing on (3, ¥) Decreasing on (-¥, 3)

(a)

Relative maximum at -3 Relative minimum at 3 No relative extremum at 0 Increasing on (-¥, -3) and (3, ¥) Decreasing on (-3, 3) f  ( x) = 20 x3 - 90 x = 0 10 x(2 x 2 - 9) = 0

3 2

Section 5.4 (b)

369

(

)

and

(- 5, 0), (0, 5), and ( 5, ¥).

( , -100.23) 3 2

(

)

Concave upward on - 3 , 0 and 2

(

Concave downward on -¥, -

3 2

f ¢(-3) = 2 - 10(-3)-2 =

( , ¥) ) and 3 2

( 0, ) 3 2

(c)

Test a point in the intervals (-¥, - 5),

Points of inflection at (0, 0), - 3 , 100.23 ,

f ¢(-1) = 2 - 10(-1)-2 = -8 < 0 f ¢(1) = 2 - 10(1)-2 = -8 < 0 f ¢(3) = 2 - 10(3)-2 =

(a) 5

x-intercepts: 0 = x - 15 x

8 >0 9

Relative maximum at - 5 Relative minimum at

0 = x3 ( x 2 - 15)

8 >0 9

Increasing on (-¥, - 5) and ( 5, ¥)

x = 0, x =  15

Decreasing on (- 5, 0) and (0, 5)

y-intercepts: y = 05 - 15(0)3 = 0

(Recall that f ( x) does not exist at x = 0.) f ( x) = 20 x-3 = f  ( x) =

(b)

20 x3

is never equal to zero. x3 There are no inflection points.

Test a point in the intervals (-¥, 0) and (0, ¥).

15.

f (-1) =

= -20 < 0 (-1)3 20 f (1) = 3 = 20 > 0 (1)

10 f ( x) = 2 x + x = 2 x + 10 x-1

Since f ( x) does not exist when x = 0, the domain is (-¥, 0) È (0, ¥). -1

f (-x) = 2(-x) + 10(-x)

Concave upward on (0, ¥) Concave downward on (-¥, 0) (c)

= -(2 x + 10 x-1)

f ( x) is never zero, so there are no x-intercepts.

The graph is symmetric about the origin.

f ( x) does not exist for x = 0, so there is no y-intercept. Vertical asymptote at x = 0

f ¢( x) = 2 - 10 x-2

y = 2 x is an oblique asymptote.

= - f ( x)

10 2- 2 = 0 x 2( x 2 - 5) x2

f (-x) = 2(-x) + 10(-x)-1 = -(2 x + 10 x-1) = - f ( x)

x = 5

Critical numbers: - 5 and

Critical points: (- 5, -4 5) and ( 5, 4 5)

370 16.

Chapter 5 GRAPHS AND THE DERIVATIVE f ( x) = 16 x +

f ( x) does not exist for x = 0, so there is no x-intercept. Vertical asymptote at x = 0

1 x

= 16 x + x-2

y = 16 x is an oblique asymptote.

Since f ( x) does not exist when x = 0, the domain is (-¥, ) È (0, ¥). f (-x) = 16(-x) + (-x)-2 = -16 x + x-2

The graph is not symmetric about the y-axis or the origin. 2 f ¢( x) = 16 - 2 x-3 = 16 - 3 x 2 16 - 3 = 0 x 2(8 x3 - 1) x3

17.

Since f ( x) does not exist when x = -2, the domain is (-¥, - 2) È (-2, ¥).

1 x = 2

( 12 , 12 ) (

The graph is not symmetric about the y-axis or the origin.

)

Test a point in the intervals (-¥, 0), 0, 12 , and

f ¢(-x) =

( 12 , ¥ ).

f ¢(-1) = 16 - 2(-1)-3 = 18 > 0

æ1ö æ 1 ö-3 f ¢ çç ÷÷ = 16 - 2 çç ÷÷ = -38 < 0 èç 3 ø÷ èç 3 ø÷ f ¢(1) = 16 - 2(1)-3 = 14 > 0

(a)

( x + 2) 2 -6 ( x + 2)2

f ¢( x) < 0 and is never zero. f ¢( x) fails to exist for x = -2.

(a)

No critical numbers; no relative extrema Decreasing on (-¥, -2) and (-2, ¥)

Increasing on (-¥, 0) and ( 12 , ¥)

f  ( x) =

Decreasing on (0, 12 )

f ( x) fails to exist for x = -2.

f  ( x) =

(c)

( x + 2)(-1) - (-x + 4)(1)

Relative minimum at 12

f ( x) = 6 x-4 =

(b)

-(-x) + 4 x+4 = (-x) + 2 -x + 2

f (-x) =

Critical number: 12 Critical point:

-x + 4 x+2

f ( x) =

(b)

6 x

12 ( x + 2)3

No points of inflection Test a point in the intervals (-¥, -2) and (-2, ¥).

> 0 and is never zero. x4 There are no inflection points.

f (-3) = -12 < 0

Concave upward on (-¥, 0) and (0, ¥)

Concave upward on (-2, ¥)

x-intercept: 16 x + x 3

-2

16 x + 1 x2

f (-1) = 12 > 0

= 0

Concave downward on (-¥, -2)

= 0

(c)

1 x =- 3 2 2

x-intercept:

-x + 4 =0 x+2 x= 4

y-intercept: y =

-0 + 4 = 2 0+2

Section 5.4

18.

371

Vertical asymptote at x = -2

Vertical asymptote at x = 2

Horizontal asymptote at y = -1

Horizontal asymptote at y = 3

f ( x) =

3x x-2

19.

Since f ( x) does not exist when x = 2, the domain is (-¥, 2) È (2, ¥). f (-x) =

( x - 2)2 -6 ( x - 2) 2

f ¢(-4) =

[(-4 + 3)(-4 + 1)]

(

4 >0 9

)

)(

(

f (3) = 12 > 0

Concave upward on (2, ¥) Concave downward on (-¥, 2)

3(0) =0 0-2

)

-2 - 32 + 2 æ 3ö 16 f ¢ çç - ÷÷÷ = =<0 2 çè 2 ø 9 é -3 + 3 -3 +1 ù êë 2 úû 2 4 -2(0 + 2) f ¢(0) = =- <0 2 9 [(0 + 3)(0 + 1)]

f (1) = -12 < 0

y-intercept: y =

(

Test a point in the intervals (-¥, 2) and (2, ¥).

(c)

-2(-4 + 2)

-2 - 52 + 2 æ 5ö 16 f ¢ çç - ÷÷÷ = = >0 2 çè 2 ø 9 é -5 + 3 -5 + 1 ù êë 2 úû 2

No points of inflection

3x x-intercept: =0 x-2 x =0

1 x2 - 4x + 3

Test a point in the intervals (-¥, -3), (-3, -2), (-2, -1), and (-1, ¥).

f ( x) fails to exist for x = 2.

(b)

Critical number: -2

Decreasing on (-¥, 2) and (2, ¥) ( x - 2)

1 (-x) 2 + 4(-x) + 3

The graph is not symmetric about the y-axis or the origin. 0 - (2 x + 4) -2( x + 2) f ¢( x ) = 2 = 2 ( x + 4 x + 3) [( x + 3)( x + 1)]2

No critical numbers; no relative extrema

f  ( x) =

x + 4x + 3 1 = ( x + 3)( x + 1)

f (-x) =

( x - 2)(3) - (3x)(1)

f ¢( x) < 0 is never zero. f ¢( x) fails to exist for x = 2.

(a)

1 2

Since f ( x) does not exist when x = -3 and x = -1, the domain is (-¥, -3) È (-3, -1) È (-1, ¥).

-3 x 3(-x) = -x - 2 (-x) - 2

The graph is not symmetric about the y-axis or the origin. f ¢( x ) =

f ( x) =

f (-2) =

(a)

)(

)

1 = -1 (-2 + 3)(-2 + 1)

Relative maximum at (-2, -1) Increasing on (-¥, -3) and (-3, -2) Decreasing on (-2, -1) and (-1, ¥)

372

Chapter 5 GRAPHS AND THE DERIVATIVE

f ( x)= = = = =

(b)

( x 2 + 4 x + 3)2 (-2) - (-2 x - 4)(2)( x 2 + 4 x + 3)(2 x + 4) -2( x 2 + 4 x + 3)[( x 2 + 4 x + 3) + (-2 x - 4)(2 x + 4)] -2( x 2 + 4 x + 3 - 4 x 2 - 16 x - 16) ( x 2 + 4 x + 3)3

f ¢( x ) =

( x 2 + 4 x + 3)3

2(3x 2 + 12 x + 13) 3

[( x + 3)( x + 1)]

Since 3x + 12 x + 13 = 0 has no real solutions, there are no x-values where f ( x) = 0. f ( x) does not exist where x = -3 and x = -1. Since f ( x) does not exist at these x-values, there are no points of inflection.

f (0) =

2[3(-4) 2 + 12(-4) + 13] 3

[(-4 + 3)(-4 + 1)]

2[3(-2) 2 + 12(-2) + 13] [(-2 + 3)(-2 + 1)]3 2[3(0) 2 + 12(0) + 13] 3

[(0 + 3)(0 + 1)]

0 - (-8)(2 x - 6) ( x 2 - 6 x - 7) 2 16( x - 3)

  



     -8 1 f (3) = = (3 + 1)(3 - 7) 2

(a)

80 <0 81

(

Relative minimum at 3, 12

80 >0 81

)

Increasing on (3, 7) and (7, ¥) Decreasing on (-¥, -1) and (-1, 3) f  ( x) = =

Vertical asymptotes where f ( x) is undefined at x = -3 and x = -1.

Horizontal asymptote at y = 0

= =

(b)

f ¢ =

= -2 < 0 26 >0 27

16(-2 - 3)

[(-2 + 1)(-2 - 7)] 16(0 - 3) 48 f ¢(0) = =<0 2 49 [(0 + 1)(0 - 7)] 16(6 - 3) 48 f ¢(6) = = >0 49 [(6 + 1)(6 - 7)]2

26 >0 27

f ( x) is never zero, so there are no x-intercepts. 1 1 y-intercept: y = = (0 + 3)(0 + 1) 3

-8

[( x + 1)( x - 7)]2

f ¢(-2) =

Concave downward on (-3, -1)

x2 - 6x - 7

x + 6x - 7

Test a point in the intervals (-¥, -1), (-1, 3), (3, 7), and (7, ¥).

Concave upward on (-¥, -3) and (-1, ¥)

f ( x) =

8 2

Critical number: 3

f (-2) =

20.

(-x) - 6(-x) - 7

-2(-3x 2 - 12 x - 13)

f (-4) =

The graph is not symmetric about the y-axis or the origin.

( x 2 + 4 x + 3)4

Test a point in the intervals (-¥, -3), (-3, -1), and (-1, ¥).

(c)

f (-x) =

( x 2 + 4 x + 3)4

-8 ( x + 1)( x - 7)

Since f ( x) does not exist when x = -1 and x = 7, the domain is (-¥, -1) È (-1, 7) È (7, ¥).

( x 2 - 6 x - 7)2 (16) - 16( x - 3)(2)( x 2 - 6 x - 7)(2 x - 6) ( x 2 - 6 x - 7)4 2

16( x - 6 x - 7)[( x - 6 x - 7) - 4( x - 3)( x - 3)] ( x 2 - 6 x - 7) 4 16( x 2 - 6 x - 7 - 4 x 2 + 24 x - 36) ( x 2 - 6 x - 7)3 16(-3x 2 + 18 x - 43) ( x 2 - 6 x - 7)3 -16(3x 2 - 18 x + 43) [( x + 1)( x - 7)]3

Since 3x 2 - 18 x + 43 = 0 has no real solutions, there are no x-values where f ( x) = 0. f ( x) does not exist where x = -1 and x = 7. Since f ( x) does not exist at these x-values, there are no points of inflection.

Test a point in the intervals (-¥, -1), (-1, 7), and (7, ¥).

Section 5.4 f (-2) = f (0) = f (8) =

(c)

373 -16[3(-2) 2 - 18(-2) + 43) [(-2 + 1)(-2 - 7)]3 -16[3(0) 2 - 18(0) + 43) [(0 + 1)(0 - 7)]3 -16[3(8)2 - 18(8) + 43) [(8 + 1)(8 - 7)]3

1456 <0 729

(a)

Relative maximum at 1 Relative minimum at -1 Increasing on (-1, 1)

688 >0 343

Decreasing on (-¥, -1) and (1, ¥)

1456 <0 729

f  ( x) =

2 x3 - 6 x

( x 2 + 1)3

Concave upward on (-1, 7)

2 x3 - 6 x = 0

Concave downward on (-¥, -1) and (7, ¥)

2 x( x 2 - 3) = 0

f ( x) is never zero, so there are no x-intercepts.

x = 0, x =  3

y-intercept: y =

-8 8 = (0 + 1)(0 - 7) 7

(b)

Inflection points at (0, 0),

( 3, ) and 3 4

(- 3, - ) 3 4

Vertical asymptotes where f ( x) is undefined at x = -1 and x = 7.

Concave upward on (- 3, 0) and ( 3, ¥)

Horizontal asymptote at y = 0

Concave downward on (-¥, - 3) and (0, 3) (c)

x-intercept: 0 =

x 2

x +1 0= x 0 y-intercept: y = 2 =0 0 +1

Horizontal asymptote at y = 0 21.

f ( x) =

x 2

x +1

Domain is (-¥, ¥) f (-x) =

-x (-x) 2 + 1

x x2 + 1

= - f ( x)

The graph is symmetric about the origin. ( x 2 + 1)(1) - x(2 x)

f ¢( x ) =

( x 2 + 1) 2

22.

1 - x2

f ( x) =

1 2

x +4

( x 2 + 1) 2

Domain is (-¥, ¥).

1 - x2 = 0

f (-x) =

Critical numbers: 1 and -1

( )

(

Critical points: 1, 12 and -1, - 12 f ( x) = =

( x 2 + 1) 2 (-2 x) - (1 - x 2 )(2)( x 2 + 1)(2 x) ( x 2 + 1) 4 3

-2 x - 2 x - 4 x + 4 x 2

( x + 1)

f  =  f  =

)

1 2

(-x) + 4

2x - 6x 2

( x + 1)

1 2

x +4

= f ( x)

The graph is symmetric about the y-axis. -2 x f ¢( x ) = 2 ( x + 4) 2 Critical number: 0

(

Critical point: 0, 14

 <0 

 >0 

)

374

Chapter 5 GRAPHS AND THE DERIVATIVE f  ( x) = = = = =

( x 2 + 4)2 (-2) - (-2 x)(2)( x 2 + 4)(2 x) ( x 2 + 4)4 ( x 2 + 4)4 ( x 2 + 4)3 -2(-3x 2 + 4) 2(3x 2 - 4)

[(0) + 4]

(

Relative maximum at 0, 14

f (0) =

1 2

(0) + 4 f ( x) = 0 when

1 <0 8

(

1 2

x -9

= f ( x)

)

Test a point in the intervals (-¥, -3), (-3, 0), (0, 3), and (3, ¥). 8 -2(-4) f ¢(-4) = = >0 2 2 49 [(-4) - 9]

1 4

1 -2(-4) = >0 32 [(-1) 2 - 9]2 1 -2(1) f ¢(1) = =<0 2 2 32 [(1) - 9] 8 -2(4) f ¢(4) = =<0 2 2 49 [(4) - 9]

f ¢(-1) =

2 3

) and ( ) Concave upward on ( -¥, - ) and ( , ¥ ) Concave downward on ( - , )

Inflection points at

(-x) - 9

Critical point: 0, - 19

)

4 3

x =

Critical number: 0

2(3x 2 - 4) = 0 x2 =

The graph is symmetric about the y-axis. -2 x f ¢( x) = 2 ( x - 9)2

( x 2 + 4)3

Decreasing on (0, ¥)

(

3 - 2 , 16 3

2 , 3 3 16

2 3

(c)

x -9 1 = ( x + 3)( x - 3)

f (-x) =

( x 2 + 4)3

2[3(0)2 - 4]

1 2

Since f ( x) does not exist when x = -3 and x = 3, the domain is (-¥, -3) È (-3, 3) È (3, ¥).

-2( x 2 + 4 - 4 x 2 )

Increasing on (-¥, 0)

(b)

f ( x) =

-2( x 2 + 4)[( x 2 + 4) + (-2 x)(2 x)]

f (0) =

(a)

23.

(a)

)

Increasing on (-¥, -3) and (-3, 0) Decreasing on (0, 3) and (3, ¥)

2 3

(

Relative maximum at 0, - 19

f  ( x) = =

f ( x) is never zero, so there are no x-intercepts. 1

1 y-intercept: y = 2 = 4 0 +4

Horizontal asymptote at y = 0 = =

(b)

( x 2 - 9) 2 (-2) - (-2 x)(2)( x 2 - 9)(2 x) ( x 2 - 9) 4 -2( x 2 - 9)[( x 2 - 9) + (-2 x)(2 x)] ( x 2 + 4) 4 -2( x 2 - 9 - 4 x 2 ) ( x 2 - 9)3 -2(-3x 2 - 9) ( x 2 - 9)3 6( x 2 + 3) [( x + 3)( x - 3)]3

Since x 2 + 3 = 0 has no solutions, there are no x-values where f ( x) = 0. f ( x) does not exist where x = -3 and x = 3. Since f ( x) does not exist at these x-values, there are no points of inflection.

Section 5.4

375

Test a point in the intervals (-¥, - 3), (-3, 3), and (3, ¥). f (-4) = f (0) = f (4) =

f  ( x) =

6[(-4) 2 + 3]

114 = >0 3 343 [(-4 + 3)(-4 - 3)] 6[(0) 2 + 3]

2 <0 81

114 = >0 3 343 [(4 + 3)(4 - 3)]

[(0 + 3)(0 - 3)]

6[(4)2 + 3]

1 2

0 -9

4 x( x 2 - 4)[( x 2 - 4) - (2 x 2 + 8)] ( x 2 - 4)4 4 x(-x 2 - 12) ( x 2 - 4)3 -4 x( x 2 + 12) ( x 2 - 4)3

-4 x( x 2 + 12) = 0

f ( x) is never zero, so there are no x-intercepts.

y-intercept: y =

( x 2 - 4) 4

f  x  =  when

Concave upward on (-¥, -3) and (3, ¥) Concave downward on (-3, 3) (c)

( x 2 - 4)2 (4 x) - (2 x 2 + 8)(2)( x 2 - 4)(2 x)

x =0 Test a point in the intervals (-¥, -2), (-2, 0), (0, 2), and (2, ¥).

1 9

Vertical asymptotes where f ( x) is undefined at x = -3 and x = 3.

f (-3) =

Horizontal asymptote at y = 0 f (-1) = f (1) = f (3) =

(b) 24.

f ( x) =

(c)

-2(-x) (-x) 2 -4

2x x 2 -4

= - f ( x)

The graph is symmetric about the origin.

[(-3) - 4]

-4(-1)[(-1)2 + 12] 2

[(-1) - 4]

-4(1)[(1)2 + 12] 2

[(1) - 4]

-4(3)[(3)2 + 12] 2

[(3) - 4]

252 >0 125

52 <0 27

52 >0 27

252 <0 125

Inflection at (0, 0)

( x 2 - 4)(-2) - (-2 x)(2 x)

x-intercept:

-2 x

=0 x2 - 4 -2 x = 0 x =0

y-intercept: y =

-2(0) 02 - 4

Vertical asymptotes where f ( x) is undefined at x = -2 and x = 2. Horizontal asymptote at y = 0

( x 2 - 4)2 2x2 + 8 ( x 2 - 4)2

f ¢( x) > 0 and is never zero.

(a)

Concave downward on (-2, 0) and (2, ¥)

Since f ( x) does not exist when x = -2 and x = 2, the domain is (-¥, -2) È (-2, 2) È (2, ¥).

f ¢( x ) =

Concave upward on (-¥, -2) and (0, 2)

-2 x

x2 - 4 -2 x = ( x + 2)( x - 2)

f (-x) =

-4(-3)[(-3) 2 + 12]

No critical points, no relative extrema Increasing on (-¥, -2), and (-2, 2) and (2, ¥)

376

Chapter 5 GRAPHS AND THE DERIVATIVE f ( x) = x ln | x |

25.

The domain of this function is (-¥, 0) È (0, ¥). f (-x) = -x ln -x = -x ln | x | = - f ( x)

The graph is symmetric about the origin. 1 f ¢( x) = x ⋅ + ln | x | x = 1 + ln | x |

26.

Domain is (-¥, 0) È (0, ¥)

f ¢( x) = 0 when

f (-x ) = -x - ln |- x | = -x - ln | x |

0 = 1 + ln | x | -1 = ln | x | -1

The graph has no symmetry. 1 x -1 f ¢( x) = 1 - = x x Critical number: x = 1

= | x|

x =

1 » 0.37. e

Critical numbers:  1e » 0.37.

(Recall that f ( x) does not exist at x = 0.)

f ¢(-1) = 1 + ln | -1| = 1 > 0 f ¢(-0.1) = 1 + ln | -0.1| » -1.3 < 0 f ¢(0.1) = 1 + ln |0.1| » -1.3 < 0 f ¢(1) = 1 + ln |1| = 1 > 0 æ1ö 1 1 1 f çç ÷÷÷ = ln =çè e ø e e e

f ¢(-1) =

æ 1ö 1 1 1 f çç - ÷÷÷ = - ln - = çè e ø e e e

(a)

(

-1 - 1 = 2>0 -1 1 -1 æ1ö f ¢ çç ÷÷ = 2 1 = -1 < 0 çè 2 ÷ø 2

2 -1 1 = >0 2 2 f (1) = 1 - ln |1| = 1

f ¢(2) =

(a)

)

Relative minimum at (1, 1) Increasing on (-¥, 0) and (1, ¥)

Relative maximum of - 1e , 1e ; relative

(

)

minimum of 1e , - 1e .

Decreasing on (0, 1)

( ) ( ) decreasing on ( - 1e , 0 ) and ( 0, 1e ).

f  ( x) =

Increasing on -¥, - 1e and 1e , ¥ and

(b)

f ( x) = x - ln | x |

1 f ( x ) = x 1 = -1 < 0 f (-1) = -1 1 f (1) = = 1 > 0 1 Concave downward on (-¥, 0);

(b)

x(1) - ( x - 1)(1)

x No inflection point.

(c)

There is no y-intercept x-intercept: x - ln | x | = 0 x = ln | x | x » -0.567 Vertical asymptote at x = 0

x-intercept: 0 = x ln | x | or

1 x2

Concave upward on (-¥, 0) and (0, ¥)

There is no y-intercept. x =0

f ( x) is positive everywhere it is defined,

Concave upward on (0, ¥). (c)

ln | x | = 0 | x | = e0 = 1 x = 1

Section 5.4 27.

377

f ( x) =

ln x x

(b)

) » (4.48, 0.33)

Note that the domain of this function is (0, ¥).

Concave downward on (0, e1.5 ); concave upward

ln(-x) does not exist when x ³ 0, -x no symmetry.

on (e1.5 , ¥)

f (-x) =

( ) f ¢( x ) = =

f (e1.5 ) =

x-intercept: f ( x) = 0 when ln x = 0 x = e0 = 1

Vertical asymptote at x = 0 Horizontal asymptote at y = 0

e1 = x ln e 1 f (e) = = e e

(

Critical points: e, 1e 1 - ln1 2

1 1 - ln 3 32

)

1 =1> 0 1

= -0.01 < 0

(

)

28.

There is a relative maximum at e, 1e . The function is increasing on (0, e) and decreasing on (e, ¥). f  ( x) = = = = =

( - 1x ) - (1 - ln x)2 x

x4 -x - 2 x(1 - ln x)

f ¢( x ) =

x3 -3 + 2 ln x

1.5

x =e

f (5) =

x2 ln(-x)2 (-x) 2 ln x 2

= f ( x) x2 The graph is symmetric about the y-axis.

x4 -(1 + 2 - 2 ln x)

13 -3 + 2 ln 5

ln x 2

Domain is (-¥, 0) È (0, ¥)

x4 -x[1 + 2(1 - ln x)]

-3 + 2 ln1

f ( x) =

f (-x) =

x f ( x) = 0 when -3 + 2 ln x = 0 2 ln x = 3 3 ln x = = 1.5 2

f (1) =

e 3

1.5 = 1.5 e

» 0.33 2e1.5 Since x ¹ 0, there is no y-intercept.

1 - ln x = 0 1 = ln x

f ¢(3) =

1.5

x2 1 - ln x

x2 Critical numbers:

f ¢(1) =

ln e1.5

(c)

x 1x - ln x(1)

(a)

(

Inflection point at e1.5 , 1.5 1.5

» 4.48.

x 2 ⋅ 12 ⋅ 2 x - (ln x 2 )2 x x

2 x(1 - ln x 2 ) x4

2(1 - ln x 2 )

f ¢( x) = 0 when 1 - ln x 2 = 0

ln x 2 = 1 x2 = e x =  e.

Critical numbers: - e , e

(

) ( e , 1e )

Critical points: - e , 1e ,

= -3 < 0 » 0.0018 > 0

378

Chapter 5 GRAPHS AND THE DERIVATIVE

(

f  ( x) =

x 2[-4 - 6(1 - ln x 2 )] x6 -10 + 6 ln x 2

f ( x)( e ) =

(a)

= =

)

x3 -2 ⋅ 12 ⋅ 2 x - 2(1 - ln x 2 )3x 2

x4 -10 + 6 ln e e

29. -4

Domain is (-¥, ¥).

(

f (-x) = -xe x

)

There are relative maxima at  e , 1e .

The graph has no symmetry.

Increasing on (-¥, - e ) È (0, e )

f ¢( x) = -xe-x + e-x

Decreasing on (- e , 0) È ( e , ¥)

= e-x (1 - x) f ¢( x) = 0 when e-x (1 - x) = 0

f ( x) = 0 when -10 + 6 ln x 2 = 0 6 ln x

= 10

ln x

x =1

Critical numbers: 1

5 3

( )

Critical points: 1, 1e

x 2 = e5/3 x = e f (3) = f (1) = f (-1) = f (-3) =

(b)

-10 + 6 ln 9 34 -10 + 6 ln12 14

= e

f ¢(2) = e-2 (1 - 2) =

(a) = -10 < 0

(-1)4 -10 + 6 ln(-3)

(

f ¢(0) = e-0 (1 - 0) = 1 > 0

5/6

= 0.0393 > 0

-10 + 6 ln(-1)2

(-3)

5/3

Inflection points at e5/6 ,

f  = 0 when -e-x (2 - x) = 0

)

x = 2. f (0) = -e-0 (2 - 0) = -2 < 0 1 f (3) = -e-3 (2 - 3) = 3 > 0 e

(b)

There is a vertical asymptote at x = 0.

(

Inflection point at 2, 22 e

)

Concave downward on (-¥, 2), concave upward on (2, ¥)

There is no y-intercept. There are x-intercepts when f ( x) = 0:

x = 1.

( )

Relative maximum at 1, 1e

= -e-x (2 - x)

= 0.0393 > 0

Concave downward on (-e5/6 , 0) and (0, e5/6 )

x2 = 1

= -e-x (1 + 1 - x)

( e5/6 , ¥ )

ln x 2 = 0

f ( x) = e-x (-1) + (1 - x) (-e-x )

= -10 < 0

5 3e5/3

-1

Increasing on (-¥, 1); decreasing on (1, ¥)

Concave upward on (-¥, - e5/6 ) and

(c)

f ( x) = xe-x

(c)

x-intercept: 0 = xe-x x =0 y-intercept: y = 0 ⋅ e-0 = 0

Horizontal asymptote at y = 0

Section 5.4

379

(b) 30.

2 -x

f ( x) = x e

f (0) = e-0[02 - 4(0) + 2] = 2 > 0 1 f (1) = e-1(12 - 4(1) + 2) = - < 0 e 2 f (4) = e-4[42 - 4(4) + 2] = 4 > 0 e Inflection points at (0.6, 0.19) and (3.4, 0.38)

Concave upward on (-¥, 0.6) and (3.4, ¥); concave downward on (0.6, 3.4)

Domain is (-¥, ¥) f (-x) = (-x) 2 e x

(c)

= x 2e x The graph has no symmetry.

y-intercept: y = 02 e-0 = 0 x-intercept: 0

f ¢( x) = x 2e-x (-1) + e-x (2 x) = xe-x (-x + 2)

Critical numbers x = 0 or -x + 2 = 0 x = 2. f ¢(-1) = (-1)e-(-1)[-(-1) + 2] = -3e < 0 1 >0 e -3 f ¢(3) = 3e-3 (-3 + 2) = 3 < 0 e The function is decreasing on (-¥, 0) and f ¢(1) = (1)e-1(-1 + 2) =

(a)

31.

f ( x) = ( x - 1)e-x

Domain is (-¥, ¥) f (-x) = (-x - 1)e x

The graph has no symmetry.

(2, ¥) and increasing on (0, 2).

f ¢( x) = -( x - 1)e-x + e-x (1)

f (0) = 02 e-0 = 0

= e-x[-( x - 1) + 1]

4 f (2) = 22 e-2 = 2 e

= e-x (2 - x) f ¢( x) = 0 when e-x (2 - x) = 0 x = 2.

Relative minimum at (0, 0); relative maximum at

( 2, ) 4 e2

Critical number: 2

(

f ( x) = xe-x (-1) + (-x + 2)(-xe-x + e-x ) -x

2 -x

= -xe

+x e

2 -x

- 4 xe

-x

= x e =e

-x

- xe

-x

- 2 xe

Critical point: 2, 12 -x

+ 2e

)

f ( x) = -e-x + (2 - x)(-e-x )

-x

+ 2e

= -e-x[1 + (2 - x)]

( x - 4 x + 2)

= -e-x (3 - x)

f ( x) = 0 when

f (2) = -e-2 (3 - 2) =

x2 - 4x + 2 = 0 4  16 - 4(1)(2) 2 4 8 = = 2  2. 2

x =

x = 2+

2 » 3.4 or 2 - 2 » 0.6

(a)

(

-1 e2

Relative maximum at 2, 12 e

)

f ¢(0) = e-0 (2 - 0) = 2 > 0 -1 f ¢(3) = e-3 (2 - 3) = 3 < 0 e

Increasing on (-¥, 2); decreasing on (2, ¥). f ( x) = 0 when -e-x (3 - x) = 0 x = 3.

380

Chapter 5 GRAPHS AND THE DERIVATIVE f (0) = -e-0 (3 - 0) = -3 < 0 1 f (4) = -e-4 (3 - 4) = 4 > 0 e

(b)

f ¢¢( x ) = e x + e-x =

( ) e

-3

f (3) = (3 - 1)e

(b)

e2 x + 1 ex

Since f ¢¢( x ) ¹ 0 (e2 x + 1 is always positive), there is no inflection point. f (0) = e0 + e-0 = 2 > 0

The entire graph is concave upward on (-¥, ¥).

y-intercept: y = (0 - 1)e-0 = (-1)(1) = -1 x-intercept:

Relative minimum at (0, 2) Decreasing on (-¥, 0); increasing on (0, ¥)

Inflection point at 3, 23

Concave downward on (-¥, 3); concave upward on (3, ¥) (c)

(a)

(c)

y-intercept: y = e0 + e-0 = 1 + 1 = 2

The y-intercept is 2. There is no x-intercept.

0 = ( x - 1)e-x x -1 = 0 x =1

Horizontal asymptote at y = 0

32.

33.

-x

f ( x) = e + e

Domain is (-¥, ¥).

Domain is (-¥, ¥) -x

f (-x) = e

f (-x) = x 2/3 + x5/3

+ e = f ( x)

The graph has no symmetry.

The graph has symmetry about the y-axis. f ¢( x) = e x - e-x

Critical numbers: -x

e -e

f ¢( x ) = =

2 -1/3 5 x - x 2/3 3 3 2 - 5x

3x1/3 f ¢( x) = 0 when 2 - 5x = 0

1 ex - x = 0 e e2 x - 1

f ( x) = x 2/3 - x5/3

Critical number: x =

æ 2 ö æ 2 ö2/3 æ 2 ö5/3 f çç ÷÷ = çç ÷÷ - çç ÷÷ èç 5 ÷ø èç 5 ø÷ èç 5 ø÷

e2 x - 1 = 0 e2 x = 1 2 x = ln 1 2x = 0 x =0 The only critical number is 0.

3 ⋅ 22/3

» 0.326 55/3 Critical point: (0.4, 0.326)

The only critical point is (0, 2). f ¢(-1) = e-1 - e-(-1) =

2 5

1 - e » -2.35 < 0 e

f ¢(1) = e1 - e-1 » 2.35 > 0

Section 5.4

381

= = =

(

( )

3x1/3 (-5) - (2 - 5 x)(3) 13 x-2/3

f  ( x) =

(3x1/3 ) 2

Inflection point at - 15 , (c)

-15x1/3 - (2 - 5 x) x-2/3

6 55/3

) » (-0.2, 0.410)

y-intercept: y = 02/3 - 05/3 = 0 x-intercept: 0 = x 2/3 - x5/3

9 x 2/3 -15x - (2 - 5x)

= x 2/3 (1 - x)

4/3

9x -10 x - 2

x = 0 or x = 1

9 x 4/3

( ) (5)

-10 52 - 2 æ2ö f  çç ÷÷÷ = 4/3 çè 5 ø 9 2 » -2.262 < 0

(a)

(

Relative maximum at

2 , 3⋅ 22/3 5 55/3

) » (0.4, 0.326)

34.

f ( x) = x1/3 + x 4/3

Domain is (-¥, ¥).

f ¢( x) does not exist when x = 0

The graph has no symmetry.

Since f (0) is undefined, use the first derivative test.

f ¢( x ) =

f ¢(-1) =

2 - 5(-1) 1/3

3(-1)

7 <0 -3

3x 2/3 f ¢( x) = 0 when 1 + 4 x = 0

(1) (8)

2-5 8 æ1ö 11 f ¢ çç ÷÷÷ = = >0 1/3 çè 8 ø 12 3 1 f ¢(1) =

x =-

= -1 < 0 3 ⋅ 11/3 Relative minimum at (0, 0)

æ 1 ö æ 1 ö1/3 æ 1 ö4/3 f çç - ÷÷ = çç - ÷÷ + çç - ÷÷ çè 4 ÷ø çè 4 ÷ø çè 4 ÷ø

)

f increases on 0, 52 . f decreases on (-¥, 0) and

3 = - 4/3 » -0.472 4

( 52 , ¥ ).

(

x =-

1 5

f  ( x) =

f ( x) undefined when 9 x 4/3 = 0 x =0 f (-1) = æ 1ö f  çç - ÷÷÷ = çè 8 ø f (1) =

(b)

9(-1) 4/3

( ) 4/3 9 ( - 18 )

-10 - 18

-10(1) - 2 9(1)

4/3

8 = >0 9

-2

4 <0 3

( )

3x 2/3 (4) - (1 + 4 x)3 23 x-1/3 (3x 12 x

2/3

2/3 2

)

- 2(1 + 4 x) x-1/3

9 x 4/3 12 x - 2(1 + 4 x)

9 x5/3 4x - 2

9 x5/3

( 1) ( 4)

4 <0 3

( ) Concave upward on ( - 15 , ¥ )

)

Critical points: - 14 , -0.472 , (0, 0)

f ( x) = 0 when -10 x - 2 = 0

-10(-1) - 2

1 4

Critical number: x = - 14 and x = 0

2-5

(

1 -2/3 4 x + x1/3 3 3 1 + 4x

4 -4 - 2 æ 1ö 45/3 f  çç - ÷÷÷ = = 5/3 çè 4 ø 3 9 -1

Concave upward on -¥, - 15

» 3.360 > 0

(a)

(

3 Relative minimum at - 14 , - 4/3

» (-0.25, -0.472)

)

382

Chapter 5 GRAPHS AND THE DERIVATIVE

f ¢( x) =

1 + 4x 3x 2/3

undefined at x = 0.

35.

Test sign of f ¢( x) on intervals defined by

For Exercise 15, the default window shows only a small portion of the graph. For Exercise 19, the default window does not allow the graph to properly display the vertical asymptotes.

x = - 14 , x = 0. -3 = -1 < 0 3 1- 1 æ 1ö f ¢ çç - ÷÷÷ = 3 2 > 0 çè 8 ø f ¢(-1) =

36.

(

)

f increases on - 14 , ¥ , f decreases on

(

-¥, - 14

For Exercise 20, the default window does not allow the graph to properly display the vertical asymptotes.

)

No extreme point at (0, f (0)) = (0, 0)

37.

For Exercises 21, 23, 27, 29, and 31, the y-coordinate of the relative minimum, relative maximum, or inflection points is so small, it may be hard to distinguish.

38.

For Exercises 22 and 28, the y-coordinate of the relative minimum, relative maximum, or inflection point is so small, it may be hard to distinguish. For Exercise 24, the default window does not allow the graph to properly display the vertical asymptotes.

f ( x) = 0 when 4 x - 2 = 0 1 2

x =

f ( x) is undefined when 9 x5/3 = 0

x =0 f (-1) =

4(-1) - 2 9(-1)5/3

2 >0 3

(1) (8)

4 8 -2 æ1ö 16 =<0 f  çç ÷÷÷ = 5/3 çè 8 ø 3 1 9 f (1) =

(b)

4(1) - 2 9(1)5/3

2 >0 9

Inflection points at (0, 0) and

For Exercises 39–43 other graphs are possible. 39.

(a) indicates a smooth, continuous curve except where there is a vertical asymptote. (b) indicates that the function decreases on both sides of the asymptote, so there are no relative extrema.

( , ) 3 1 2 24/3

» (0.5, 1.191)

( 12 , ¥ )

Concave upward on (-¥, 0) and

(

Concave downward on 0, 12 (c)

For Exercise 10, the y-intercept is outside the vertical window of -10 £ y £ 10. For Exercises 8, 14, and 16, the relative maxima or minima are outside the vertical window of -10 £ y £ 10.

5 >0 3

f ¢(1) =

For Exercises 7, 11, and 13, the relative maxima or minima are outside the vertical window of -10 £ y £ 10.

(d) and (e) indicate that concavity does not change left of the asymptote, but that the right portion of the graph changes concavity at x = 2 and x = 4.

)

y-intercept: y = 01/3 + 04/3 = 0 1/3

1/3

(1 + x)

x-intercept: 0 = x = x

There are inflection points at 2 and 4.

4/3

x = 0 or x = -1

40.

(a) indicates that the curve may not contain breaks. (b) and (c) indicate relative minima at -6 and 3 and a relative maximum at 1.

Chapter 5 Review

383

(d) and (e), when combined with (b) and (c) show that concavity does not change between relative extrema.

(e) and (f ) indicate (combined with (c) and (d)) a relative maximum at (1, 5), and (combined with (b)) a relative minimum at 4.

(f ) gives the y-intercept.

(g) is consistent with (b). (h) indicates the curve is concave upward on (2, 3). (i) 41.

(a) indicates that there can be no asymptotes, sharp “corners”, holes, or jumps. The graph must be one smooth curve.

indicates the curve is concave downward on (-¥, 2), (3, 4) and (4, ¥).

(b) and (c) indicate relative maxima at -3 and 4 and a relative minimum at 1. (d) and (e) are consistent with (g). (f ) indicates turning points at the critical numbers -3 and 4.

Chapter 5 Review Exercises

42.

(a) indicates that the curve may not contain breaks. (b) and (c) indicate relative maxima at -2 and 3 and a relative minimum at 0.

True

False: The function is increasing on this interval.

False: The function could have neither a minimum nor a maximum; consider f ( x) = x3 at c = 0.

True

False: Consider f ( x) = x3 / 2 at c = 0.

True

False: The function is concave upward.

Thus, a sharp corner must exist at 0.

False: Consider f ( x) = x 4 at c = 0.

indicates that concavity changes just once, at (5, 1).

False: Consider f ( x) = x 4 , which has a relative minimum at c = 0.

10.

False: Polynomials are rational functions, and nonconstant polynomials have neither vertical nor horizontal asymptotes.

11.

True

12.

False: Consider f ( x) = x 2 on (-1, 1) with c = 0.

17.

f ( x) = x 2 + 9 x + 8

(d) shows that concavity does not change at 0. (d) and (e) are consistent with (i). (f ) shows critical values. (g) and (h) indicate that the function is not differentiable at 0, and is differentiable everywhere else. (i)

f(x)

3 –2

43.

(a) indicates that the curve may not contain breaks. (b) indicates that there is a sharp “corner” at 4. (c) gives a point at (1, 5). (d) shows critical numbers.

f ¢( x ) = 2 x + 9

384

Chapter 5 GRAPHS AND THE DERIVATIVE f ¢( x) = 0 when x = - 92 and f ¢ exists

20.

everywhere.

f ( x) = 4 x3 + 8 x 2 - 16 x + 11 f ¢ ( x) = 12 x 2 + 16 x - 16

Critical number: - 92

= 4(3x 2 + 4 x - 4)

(

Test an x-value in the intervals -¥, - 92

) and

= 4(3x - 2)( x + 2) f ¢ ( x) = 0 when x = 23 or x = -2 and

( - 92 , -¥ ).

f ¢ exists everywhere.

f ¢(-5) = -1 < 0

Critical numbers: -2 and 23

f ¢(-4) = 1 > 0

(

)

Test an x-value in the intervals (-¥, -2),

f is increasing on - 92 , ¥ and decreasing on

( -¥, - 92 ).

( -2, 23 ), and ( 23 , ¥ ). f ¢ (-3) = 44 > 0

f ( x) = -2 x + 7 x + 14

f ¢ (0) = -16 < 0

f ¢ ( x ) = -4 x + 7

f ¢ (1) = 12 > 0

18.

f ¢( x) = 0 when x = 74 and f ¢ exists

f is increasing on (-¥, -2) and

everywhere.

decreasing on -2, 23 .

(

Critical number: 74

(

)

Test an x-value in the intervals -¥, 74 and

21.

( 74 , ¥ ).

f ¢( x ) =

f ¢(0) = 7 > 0

(

)

f is increasing on -¥, 74 and decreasing on

16 9 - 3x 16(-1)(-3) (9 - 3x)

(9 - 3x)2

f is increasing on (-¥, 3) and (3, ¥) and never decreasing.

( 74 , ¥ ). 3

)

f ¢( x) > 0 for all x ( x ¹ 3), and f is not defined for x = 3.

f ¢(2) = -1 < 0

22.

f ( x) = -x + 2 x + 15 x + 16

19.

f ( x) =

( 23 , ¥ ) and

f ¢( x) = -3x 2 + 4 x + 15

f ( x) = f ¢( x ) =

= -(3x 2 - 4 x - 15)

15 2x + 7 15(-1) (2) (2 x + 7)

-30 (2 x + 7) 2

(

)

f ¢( x) < 0 for all x x ¹ - 72 , and f is not

= -(3x + 5)( x - 3) f ¢( x) = 0 when x = - 53 or x = 3 and f ¢

defined for x = - 72 .

exists everywhere.

f is never increasing, and decreasing on -¥, - 72

Critical numbers: - 53

(

and 3

Test an x-value in the intervals

(

-¥, - 53

23.

2x 2

x -1 f is not defined for x = -1 and x = 1.

f ¢(0) = 15 > 0 f ¢(4) = -17 < 0

(

)

f ¢( x) = 0 when x = 0.

f is increasing on - 53 , 3 and decreasing on

(

f ( x) = ln | x 2 - 1| f ¢( x ) =

f ¢(-2) = -5 < 0

)

and - 72 , ¥ .

( - 53 , 3 ), and (3, ¥).

-¥, - 53

(

)

) and (3, ¥).

Test an x-value in the intervals (-¥, -1), (-1, 0), (0, 1), and (1, ¥).

Chapter 5 Review

385 f (2) = -7

f ¢(-2) = - 4 < 0 3 æ 1 ö÷ 4 f ¢ çç - ÷÷ = > 0 è 2ø 3 æ ö f ¢ çç 1 ÷÷÷ = - 4 < 0 è2ø 3 f ¢(2) = 4 > 0 3

Relative minimum of -7 at 2 28.

f ¢ ( x ) = -6 x + 2 f ¢( x) = 0 when x = 13 .

Critical number: 1 3

f is increasing on (-1, 0) and (1, ¥) and decreasing on (-¥, -1) and (0, 1). 24.

Since f ( x) = -6 < 0 for all x, f

f ( x) = 8 xe

f ¢( x) = 8(e-4 x ) + 8x(-4e-4 x )

at 13 Relative maximum of - 14 3

= 8e-4 x - 32 xe-4 x = 8e-4 x (1 - 4 x)

29.

f ¢( x) = 0 when x = 14 .

(

)

intervals -¥, 14 and

( 14 , ¥ ).

f ¢(1) = -24e-4 < 0 f is increasing on

( -¥, 14 ) and decreasing on ( 14 , ¥ ). f ( x ) = -x 2 + 4 x - 8

f ¢ ( x ) = -2 x + 4 = 0 Critical number: x = 2 f ( x) = -2 < 0 for all x, so f (2) is a relative maximum. f (2) = -4 Relative maximum of –4 at 2

26.

6( x 2 + x - 6) = 0 ( x + 3)( x - 2) = 0 Critical numbers: –3 and 2 f ( x) = 12 x + 6 f (-3) = -30 < 0, so a maximum occurs at x = -3. f (2) = 30 > 0, so a minimum occurs at x = 2. f (-3) = 101 f (2) = -24 Relative maximum of 101 at -3

Relative minimum of -24 at 2 30.

f ( x) = 2 x3 + 3x 2 - 12 x + 5 f ¢( x) = 6 x 2 + 6 x - 12

f ( x) = x 2 -6 x + 4

= 6( x 2 + x - 2) = 6( x + 2)( x - 1)

f ¢( x ) = 2 x - 6 f ¢( x) = 0 when x = 3.

f ¢( x) = 0 when x = -2 or x = 1.

Critical number: 3 f ( x) = 2 > 0 for all x, so f (3) is a relative minimum.

Critical numbers: -2, 1 f ( x) = 12 x + 6 f (-2) = 18 < 0, so a maximum occurs at x = -2.

f (3) = -5

Relative minimum of -5 at 3 27.

f ( x) = 2 x3 + 3x 2 - 36 x + 20 f ¢ ( x) = 6 x 2 + 6 x - 36 = 0

f ¢(0) = 8 > 0

25.

( 13 ) is a

relative maximum. æ 1 ÷ö f çç ÷÷ = - 14 è3ø 3

-4 x

Test an x-value in the

f ( x ) = -3 x 2 + 2 x - 5

f (1) = 18 > 0, so a minimum occurs at x = 1.

f ( x) = 2 x - 8 x + 1

f (-2) = 25 f (1) = -2

f ¢( x) = 4 x - 8 = 0

Critical number: x = 2 f ( x) = 4 > 0 for all x, so f (2) is a relative minimum.

Relative maximum of 25 at –2 Relative minimum of –2 at 1

386 31.

Chapter 5 GRAPHS AND THE DERIVATIVE f ( x) = f ¢( x ) = = = =

1 - 2 ln 3x = 0 1 = 2 ln 3x 1 = ln 3x 2 e1/2 = 3x

xe x x -1 ( x - 1)( xe x + e x ) - xe x (1) ( x - 1)2 x 2e x - xe x - xe x - e x - xe x

e1/2 = x 3 e » 0.55 3 f ¢(1) = 1 - 2 ln 3 » -0.6 < 0 2 æ 1 ÷ö 1 2 ln1 = 1 = 27 > 0 f ¢ çç ÷÷ = 3 1 è3ø 2 2 ⋅ 27 1 2 3

( x - 1)2 x 2e x - xe x - e x ( x - 1)2 e x ( x 2 - x - 1) ( x - 1)2

( )

f ¢( x) is undefined at x = 1, but 1 is not in the domain of f ( x).

æ ö f ççç e ÷÷÷ » 0.83 è 3 ø

f ¢( x) = 0 when x 2 - x - 1 = 0 1  1 - 4(1)(-1) 2 1 5 = 2

Relative maximum at

( , 0.83) or (0.55, 0.83) e 3

x =

33.

f ¢( x) = 12 x3 - 10 x - 11 f ( x) = 36 x 2 - 10

1+ 5 1- 5 » 1.618 or = -0.618 2 2 Critical numbers are -0.618 and 1.618. f ¢(1.4) = f ¢(2) = f ¢(-1) = f ¢(0) =

1.4

(1.4 - 1.4 - 1) (1.4 - 1) 2

e (2 - 2 - 1) (2 - 1)

f (1) = 36(1)2 - 10 = 26 f (-3) = 36(-3) 2 - 10 = 314

» -11.15 < 0

34. = e2 » 7.39 > 0

e-1[(-1) 2 - (-1) - 1] (-1 - 1) 2

f ( x) = 54 x + 2 x-3 = 54 x + 23 x 2 f (1) = 54(1) + 3 = 56 (1) f (-3) = 54(-3) + 2 3 (-3) = -162 - 2 = - 4376 27 27

» 0.09 > 0

e0 (02 - 0 - 1)

and a relative minimum at (1.618, 13.203). y = y¢ = = =

ln(3x) 2x2 2 x 2 ⋅ 1x - (ln 3x) ⋅ 4 x 4x4 2 x(1 - 2 ln 3x) 4x4 1 - 2 ln 3x 2 x3

y ¢ = 0 when

f ( x) = 9 x3 + 1 = 9 x3 + x-1 x f ¢( x) = 27 x 2 - x-2

= -1 < 0 (0 - 1) 2 There is a relative maximum at (-0.618, 0.206)

32.

f ( x) = 3x 4 - 5 x 2 - 11x

35.

f ( x) = 4 x + 2 3x - 6 (3x - 6)(4) - (4 x + 2)(3) f ¢( x ) = (3x - 6)2 -30 = 12 x - 24 - 122x - 6 = (3x - 6) (3x - 6)2 = -30(3x - 6)-2 f ( x) = -30(-2)(3x - 6)-3 (3) = 180(3x - 6)-3 or

180 (3x - 6)3

Chapter 5 Review

387

f (1) = 180[3(1) - 6]-3 = - 20 3 -3

f (-3) = 180[3(-3) - 6]

f (1) =

=- 4 75

f (-3) =

f ( x) = 1 - 2 x 4x + 5 (4 x + 5)(-2) - (1 - 2 x)(4) f ¢( x ) = (4 x + 5)2 = -8x - 10 - 42 + 8x (4 x + 5) 14 = = -14(4 x + 5)-2 2 (4 x + 5)

36.

39.

Critical points:

æ 1ö f  çç - ÷÷÷ = 5 > 0 çè 2 ø

(a)

f (t ) = (t 2 + 1)-1/2 (1) éæ ù ö + t ê çç - 1 ÷÷÷ (t 2 + 1)-3/2 (2t ) ú êë è 2 ø úû

Relative maximum at 13 Relative minimum at - 12

( ) Decreasing on ( -¥, - 12 ) and ( 13 , ¥ ) Increasing on - 12 , 13

= (t 2 + 1)-1/2 - t 2 (t 2 + 1)-3/2 1 t2 t2 + 1 - t2 = (t 2 + 1)1/2 (t 2 + 1)3/2 (t 2 + 1)3/2 1 = (t 2 + 1)-3/2 or 2 (t + 1)3/2 =

1 1 » 0.354 = 3/2 (1 + 1)3/2 2 1 f (-3) = = 13/2 » 0.032 3/2 (9 + 1) 10 f (1) =

f ( x) = -12 x - 1 = 0 x =-

(b)

= -(5 - t )

1 12

( ) 1 Concave upward on ( -¥, -12 ) 1 ,¥ Concave downward on ( - 12 ) 1 , -3.09 Point of inflection at - 12

y-intercept: y = -2(0)3 -

1 f ¢(t ) = - (5 - t 2 )-1/2 (-2t ) = t (5 - t 2 )-1/2 2 é 1 ù f (t ) = (1)(5 - t 2 )-1/2 + t ê - (5 - t 2 )-3/2 (-2t ) ú êë 2 úû = (5 - t 2 )-1/2 + t[t (5 - t 2 )-3/2 ] = (5 - t 2 )-3/2 (5 - t 2 + t 2 ) =

( 13 , -2.80 ) and ( - 12 , -3.375 )

f ( x) = -12 x - 1 æ1ö f  çç ÷÷÷ = -5 < 0 çè 3 ø

t 2 + 1 = (t 2 + 1)1/2

f (t ) = - 5 - t

1 2 x + x-3 2 Domain is (-¥, ¥) f ( x ) = -2 x 3 -

Critical numbers: 13 and - 12

f ¢(t ) = 1 (t 2 + 1)-1/2 (2t ) = t (t 2 + 1)-1/2 2

38.

(5 - 9)3/2

(3x - 1)(2 x + 1) = 0

112 = - 16 49 [4(-3) + 5]3

5 8

f ¢ ( x ) = -6 x 2 - x + 1 = 0

112 = 112(4 x + 5) or (4 x + 5)3 112 f (1) = = 112 3 729 [4(1) + 5]

f (t ) =

The graph has no symmetry.

-3

37.

(5 - 1) 5

3/2

This value does not exist since (-4)3/2 does not exist. (In fact, f is undefined at t = -4.)

f ( x) = -14(-2)(4 x + 5)-3 (4)

f (-3) =

5 (5 - t 2 )3/2

1 2 (0) + (0) - 3 = -3 2

388 40.

Chapter 5 GRAPHS AND THE DERIVATIVE 4 f ( x) = - x3 + x 2 + 30 x - 7 3 Domain is (-¥, ¥)

f ¢( x) = 4 x 3 - 4 x 2 - 8 x = 0 4 x( x 2 - x - 2) = 0 4 x( x - 2)( x + 1) = 0

The graph has no symmetry.

Critical numbers: 0, 2, and -1

f ¢( x) = -4 x 2 + 2 x + 30

Critical points: (0, 1),

= -2(2 x + 5)( x - 3) = 0

f ( x) = 12 x 2 - 8 x - 8

Critical numbers: - 52 and 3

(

= 4(3x 2 - 2 x - 2)

)

f (-1) = 12 > 0

Critical points: - 52 , -54.91 and (3, 56) f  ( x) = - 8 x + 2 æ 5ö f  çç - ÷÷÷ = 22 > 0 çè 2 ø

f (0) = -8 < 0 f (2) = 24 > 0

(a)

f (3) = -22 < 0

(a)

Decreasing on (-¥, -1) and (0, 2) f ( x) = 4(3x 2 - 2 x - 2) = 0

( ) Decreasing on ( -¥, - 52 ) and (3, ¥) Increasing on

- 52 , 3

41.

x = =

f  ( x ) = -8 x + 2 = 0 1 x = 4

(c)

(b)

( 14 , 0.54 ) Concave upward on ( -¥, 14 ) Concave downward on ( 14 , ¥ )

Domain is (-¥, ¥) The graph has no symmetry.

4 - (-24) 6

1 7 3

1- 7 , 0.11 3

(

1 7 , -5.12 3

) and

) (

) and ( , )

Concave upward on -¥, 1-3 7

y-intercept: 4 y = - (0)3 + (0)3 + 30(0)2 - 7 3 = -7

4 3 x - 4x2 + 1 3

2

Points of inflection at

(

Point of inflection at

f ( x) = x 4 -

Relative maximum at 0 Relative minima at -1 and 2 Increasing on (-1, 0) and (2, ¥)

Relative maximum at 3 Relative minimum at - 52

(b)

( 2, - 293 ) and ( -1, - 23 )

= -2(2 x 2 - x - 15)

Concave downward on (c)

1- 7 1+ 7 3 3

y-intercept: y = (0) 4 -

42.

(

4 3 (0) - 4(0) 2 + 1 = 1 3

2 9 f ( x) = - x 3 + x 2 + 5 x + 1 3 2

Domain is (-¥, ¥) The graph has no symmetry.

1+ 7 ,¥ 3

)

Chapter 5 Review

389

f ¢( x ) = -2 x 2 + 9 x + 5 = 0

(a)

-(2 x + 1)( x - 5) = 0

number because -  is not in the domain of f .

Critical numbers: - 12 and 5

(

Thus, there are no critical numbers, so f ( x) has no relative extrema.

)

Critical points: - 12 , -0.29 and (5, 55.17)

(

f  ( x) =

(b)

(

) Concave downward on ( - 12 , ¥ )

( ) Decreasing on ( -¥, - 12 ) and (5, ¥) (c)

f  ( x ) = -4 x + 9 = 0

(c)

No inflection points Concave upward on -¥, - 12

Increasing on - 12 , 5

x-intercept:

9 4

x -1 =0 2x + 1 x =1

y-intercept: y =

( ) Concave upward on ( -¥, 94 ) Concave downward on ( 94 , ¥ ) Point of inflection at

-12

f (-1) = 12 > 0

Relative maximum at 5 Relative minimum at - 12

(b)

(  , ¥ )

(2 x + 1)3 f (0) = -12 < 0

f (5) = -11 < 0

x =

)

Increasing on -¥,  and

f  ( x ) = -4 x + 9 æ 1 ÷ö f  çç - ÷÷ = 11 > 0 çè 2 ø

(a)

( )

f ¢ - 12 does not exist, but - 12 is not a critical

9 , 27.44 4

0 -1 = -1 2(0) + 1

Vertical asymptote at x = - 12 Horizontal asymptote at y = 12

y-intercept: 2 9 y = - (0)3 + (0)2 + 5(0) + 1 = 1 3 2

44.

f ( x) =

2x - 5 x+3

Domain is (-¥, -3) È (-3, ¥) 43.

f ( x) =

x-1 2x + 1

(

f ¢( x ) =

) (

Domain is -¥, - 12 È - 12 , ¥

)

The graph has no symmetry. f ¢( x ) = =

(2 x + 1) 3 (2 x + 1)2

f ¢ is never zero.

( x + 3)2 11 ( x + 3) 2

f ¢ is never zero.

(2 x + 1)(1) - ( x - 1)(2) 2

2( x + 3) - (2 x - 5)

(a)

f ( x) has no extrema. f  ( x) =

-22

( x + 3)3 f (-4) = 22 > 0 f (-2) = -22 < 0

390 (b)

Chapter 5 GRAPHS AND THE DERIVATIVE

(

Concave upward on (-¥, -3)

1 Concave upward on -¥, -12

Concave downward on (-3, ¥) (c)

(

1 ,¥ Concave downward on - 12

2x - 5 x-intercept: =0 x+3 5 x = 2

y-intercept:

)

(c)

)

y-intercept: y = -4(0)3 - (0)2 + 4(0) + 5 = 5

2 (0) - 5 5 =0+3 3

Vertical asymptote at x = -3 Horizontal asymptote at y = 2

46.

f ( x) = x 3 +

5 2 x - 2x - 3 2

Domain is (-¥, ¥) The graph has no symmetry. f ¢ ( x) = 3x 2 + 5x - 2 = (3x - 1) ( x + 2) = 0

Critical numbers: 13 and -2 45.

Critical points:

Domain is (-¥, ¥)

f ( x ) = 6 x + 5 æ1ö f  çç ÷÷÷ = 7 > 0 çè 3 ø

The graph has no symmetry. f ¢( x) = -12 x 2 - 2 x + 4

f (-2) = -7 < 0

= -2(6 x 2 + x - 2) = 0 (3x + 2)(2 x - 1) = 0

(a)

(

)

Critical points: - 23 , 3.07 and

Relative maximum at -2 Relative minimum at 13

Critical numbers: - 23 and 12

(

1 , 6.25 2

Increasing on (-¥, -2) and

)

(

Decreasing on -2, 13

f ( x) = -24 x - 2 = -2(12 x + 1)

x =-

æ1ö f  çç ÷÷÷ = -14 < 0 çè 2 ø

(b)

Relative maximum at 12 Relative minimum at - 23

( ) Decreasing on ( -¥, - 23 ) and ( 12 , ¥ ) Increasing on - 23 , 12

(b)

(

)

(c)

5 6

( ) Concave upward on ( - 56 , ¥ ) Concave downward on ( -¥, - 56 ) Point of inflection at - 56 , -0.18

y-intercept: -3

f ( x) = -2(12 x + 1) = 0 x =-

( 13 , ¥ )

f  ( x) = 6 x + 5 = 0

æ 2ö f  çç - ÷÷÷ = 14 > 0 çè 3 ø

(a)

( 13 , -3.35 ) and (-2, 3)

f ( x ) = -4 x 3 - x 2 + 4 x + 5

1 12

1 , 4.66 Point of inflection at - 12

Chapter 5 Review 47.

391

( ) Decreasing on ( 92 , ¥ )

f ( x) = x 4 + 2 x 2

Increasing on -¥, 92

Domain is (-¥, ¥) f (-x) = (-x)4 - 2(-x)2

f ( x) = 12 x(3 - x) = 0 x = 0 or x = 3

= x 4 + 2 x 2 = f ( x)

The graph is symmetric about the y-axis.

(b)

f ¢( x ) = 4 x + 4 x

Points of inflection at (0, 0) and (3, 81) Concave upward on (0, 3)

= 4 x( x 2 + 1) = 0

Concave downward on (-¥, 0) and (3, ¥)

Critical number: 0

(c)

Critical point: (0, 0) 2

x-intercept: 6 x3 - x 4 = 0 x3 (6 - x) = 0

f ( x) = 12 x + 4 = 4(3x + 1)

x = 0, x = 6

f (0) = 4 > 0

(a)

The x-intercepts are 0 and 6. y-intercept: 0

Relative minimum at 0 Increasing on (0, ¥) Decreasing on (-¥, 0) f ( x) = 4(3x 2 + 1) ¹ 0 for any x

(b)

No points of inflection f (-1) = 16 > 0 f (1) = 16 > 0 49.

Concave upward on (-¥, ¥) (c)

x-intercept: 0; y-intercept: 0

f ( x) =

x2 + 4 x

Domain is (-¥, 0) È (0, ¥) f (-x) =

(-x)2 + 4 -x

x2 + 4 = - f ( x) -x The graph is symmetric about the origin. =

48.

f ¢( x ) =

f ( x) = 6 x 3 - x 4

Domain is (-¥, ¥)

The graph has no symmetry.

(

f  ( x) =

)

(a)

x3 f (-2) = -1 < 0

f ( x) = 36 x - 12 x 2 = 12 x(3 - x) f (0) = 0 æ9ö f  çç ÷÷÷ = -81 < 0 çè 2 ø

x2 - 4

Critical points: (-2, -4) and (2, 4)

Critical number: 0 and 92 Critical points: (0, 0) and

Critical numbers: -2 and 2

f ¢( x) = 18x 2 - 4 x3 = 2 x 2 (9 - 2 x) = 0

9 , 136.7 2

x(2 x) - ( x 2 + 4)

f (2) = 1 > 0

(a)

Relative maximum at -2 Relative minimum at 2 Increasing on (-¥, -2) and (2, ¥)

Relative maximum at 92

Decreasing on (-2, 0) and (0, 2)

No relative extrema at 0

f ¢¢( x) = 83 > 0 for all x. x

392 (b)

Chapter 5 GRAPHS AND THE DERIVATIVE No inflection points Concave upward on (0, ¥) Concave downward on (-¥, 0)

(c)

No x- or y-intercepts Vertical asymptote at x = 0 Oblique asymptote at y = x 51.

f ( x) =

2x 3- x

Domain is (-¥, 3) È (3, ¥) The graph has no symmetry. f ¢( x ) =

50.

f ( x) = x +

8 x

Domain is (-¥, 0) È (0, ¥) f (-x) = -x + 8 -x æ ö = -çç x + 8 ÷÷÷ = - f ( x) è xø

(a)

The graph is symmetric about the origin. f ¢( x) = 1 - 82 x

(b)

Relative maximum at -2 2 Increasing on (-¥, -2 2) and (2 2, ¥)

(3 - x) 2

12 (3 - x)3

f ( x) is never zero. f (3) does not exist, but since 3 is not in the domain of f , there is no inflection point at x = 3. 12 >0 27 f (4) = -12 < 0 f (0) =

Concave upward on (-¥, 3) Concave downward on (3, ¥) (c)

Relative minimum at 2 2

Increasing on (-¥, 3) and (3, ¥)

Critical points: (2 2, 4 2), (-2 2, -4 2) f ( x) = 163 x f (-2 2) = - 2 < 0 2 f (2 2) = 2 > 0 2 (a)

(3 - x)2

f ¢( x) is never zero. f ¢(3) does not exist, but since 3 is not in the domain of f, it is not a critical number. No critical numbers, so no relative extrema 2 f ¢(0) = > 0 3 f ¢(4) = 6 > 0

f  ( x) =

2 8 =0 = x 2 x Critical numbers: x = 2 2

(3 - x)(2) - (2 x)(-1)

x-intercept: 0; y-intercept: 0 Vertical asymptote at x = 3 Horizontal asymptote at y = -2

Decreasing on (-2 2, 0) and (0, 2 2) f ( x) = 163 > 0 for all x. x

(b)

No inflection points Concave upward on (0, ¥) Concave downward on (-¥, 0)

(c)

Vertical asymptote at x = 0 Oblique asymptote at y = x

Chapter 5 Review 52.

393

f ( x ) = -4 x 1 + 2x

Domain is

(

f ¢(-1) = e 2(-1)[2(-1) + 1] = -e-2 < 0

-¥, - 12

)È(

- 12 , ¥

f ¢(0) = e 2(0)[2(0) + 1] = 1 > 0

)

(a)

The graph has no symmetry. f ¢( x ) =

(

Relative minimum at - 12 , - 21e

-4(1 + 2 x) - 2(-4 x)

( -¥, - 12 ) and increasing on ( - 12 , ¥ )

f ( x) = 2e2 x (2 x + 1) + e2 x (2) = 4e2 x ( x + 1) f ( x) = 0 when x = -1.

f ¢( x) is never zero.

f  ( x) =

(b)

(

-¥, - 12

) and (

- 12 , ¥

f  (0) = 4e2(0)[(0) + 1] = 4 > 0

)

(b)

Inflection point at (-1, -e-2 ) Concave upward on (-1, ¥)

Concave downward on (-¥, -1)

(1 + 2 x)3

f ( x) is never zero; no points of inflection. f (0) = 16 > 0 f (-1) = -16 < 0

(

Concave upward on - 12 , ¥

(c)

(

x-intercept: xe2 x = 0 x=0 y-intercept: y = (0)e2(0) = 0

)

Concave downward on -¥, - 12 (c)

f (-2) = 4e 2(-2)[(-2) + 1] = -4e-4 < 0

No critical values; no relative extrema f ¢(0) = -4 < 0 f ¢(-1) = -4 < 0 Decreasing on

)

Decreasing on

(1 + 2 x)2 = -4 - 8 x +2 8x (1 + 2 x) -4 = (1 + 2 x)2

(a)

No relative maximum

Since

)

lim xe 2 x = 0, there is a horizontal

x -¥

asymptote at y = 0.

x- intercept: 0; y-intercept: 0 Vertical asymptote at x = - 12 Horizontal asymptote at y = -2

54.

f ( x ) = x 2e 2 x

Domain is (-¥, ¥). 53.

f ( x) = xe

f (-x) = (-x)2 e2(-x) = x 2e-2 x

Domain is (-¥, ¥).

The graph is not symmetric about the y-axis or origin.

-2 x

f (-x) = -xe

The graph has no symmetry.

f ¢( x) = (2 x)(e2 x ) + ( x 2 )(2e2 x )

f ¢( x) = (1) (e 2 x ) + ( x) (2e2 x )

= 2( x 2 + x)e2 x

= e 2 x (2 x + 1)

= 2 x( x + 1)e 2 x

f ¢( x) = 0 when x = - 12 .

Critical number: - 12

(

Critical point: - 12 , - 21e

f ¢( x) = 0 when x = -1 and x = 0.

Critical numbers: –1 and 0 Critical points: (-1, e-2 ) and (0, 0)

)

f ¢(-2) = 2[(-2)2 + (-2)]e2(-2) = 4e-4 > 0

394

Chapter 5 GRAPHS AND THE DERIVATIVE éæ æ ö ö2 æ öù f ¢ çç - 1 ÷÷÷ = 2 êê çç - 1 ÷÷÷ + çç - 1 ÷÷÷ úú e2(-1/2) è 2ø è 2 øû ëè 2 ø

55.

Domain is (-¥, ¥). f (-x) = ln[(-x)2 + 4] = ln( x 2 + 4) = f ( x) The graph is symmetric about the y-axis. f ¢( x ) = 2 2 x x +4 f ¢( x) = 0 when x = 0. Critical number: 0 Critical point: (0, ln 4) 2(-1) f ¢(-1) = = -2 < 0 2 5 (-1) + 4

= - 1 e-1 < 0 2 f ¢(1) = 2[(1) 2 + (1)]e 2(1) = 4e 2 > 0

(a)

f ( x) = ln( x 2 + 4)

Relative maximum at (-1, e-2 ) Relative minimum at (0, 0) Increasing on (-¥, -1) and (0, ¥) Decreasing on (–1, 0) f ( x) = 2(2 x + 1)e 2 x + 2( x 2 + x)e2 x (2) f ( x) = 0 when

(a)

2x + 4x + 1 = 0 x =

-4  16 - 4(2)(1) 2(2)

= -4  8 4 = -1  2 2

( ( -1 +

-1 -

(

2 e

)

(

2 e-2 + 2

-2 - 2

)

2 , 23 2

) ) » (-0.293, 0.04775)

(-¥, -1 - ) and (-1 + 2 2

2 ,¥ 2

( -1 (c)

2 , -1 + 22 2

(c)

y-intercept: y = (0) 2 e 2(0) = 0 lim x 2e 2 x = 0, horizontal asymptote

x -¥

f (3) =

-2[(-3)2 - 4] 2

[(-3) + 4]

-2[(0)2 - 4] [(0) 2 + 4]2 -2[(3)2 - 4] [(3) 2 + 4]2

= - 10 < 0 169

= 1 >0 2 = - 10 < 0 169

Inflection points at (-2, ln 8) and (2, ln 8) Concave downward on (-¥, -2) and (2, ¥)

x=0

at y = 0.

( x 2 + 4)2

Concave upward on (-2, 2)

x-intercept: x 2e2 x = 0

Since

f (0) =

(b)

)

-2 ( x 2 - 4)

f (-3) =

)

Concave downward on

( x 2 + 4) 2

x -4=0 x = 2

» (-1.707, 0.09588)

Concave upward on

( x 2 + 4) (2) - (2 x) (2 x)

f (0)=2[2(0) 2 + 4(0) + 1]e2(0) = 2 > 0 2 , 32 + 2

= 2 >0 5

f ¢¢ ( x) = 0 when

f (-1)=2[2(-1) 2 + 4(-1) + 1]e2(-1) = -2e-2 < 0

Inflection points at

(1)2 + 4

No relative maximum Relative minimum at (0, ln 4) Increasing on (0, ¥) Decreasing on (-¥, 0) f  ( x) =

f (-2)=2[2(-2) 2 + 4(-2) + 1]e2(-2) = 2e-4 > 0

(b)

2(1)

f ¢(1) =

= 2(2 x 2 + 4 x + 1)e2 x

Since f ( x) never equals zero, there are no x-intercepts. y-intercept: y = ln[(0)2 + 4] = ln 4

No horizontal or vertical asymptotes.

Chapter 5 Review 56.

395

f ( x) = x 2 ln x

Domain is (0, ¥). The graph is not symmetric about the y-axis or origin since f ( x) is not defined for x £ 0. f ¢( x) = 2 x(ln x) x 2 ⋅

1 x

= 2 x ln x + x

57.

= x(2 x ln x + 1) f ¢( x =  when 2 ln x + 1 = 0 ln x = -

f (-x) = 4(-x)1/3 + (-x)4/3 = -4 x1/3 + x 4/3 The graph is not symmetric about the y-axis or origin. 4 4 f ¢( x) = x-2/3 + x1/3 3 3 f ¢( x) = 0 when

1 2

x = e-1/2

Critical number: e-1/2

(

Critical point: e-1/2 , - 21e

)

æ1ö ö 1æ 1 f ¢ çç ÷÷÷ = çç 2 ln + 1÷÷÷ » -0.1931 < 0 èç 2 ø ø 2 èç 2 f ¢(1) = 1(2 ln1 + 1) = 1 > 0

(a)

No relative maximum

(

Relative minimum at e-1/2 , - 21e

)

Increasing on (e-1/2 , ¥) -1/2

Decreasing on (0, e

)

f ¢(1) = 4 (1)-2/3 + 4 (1)1/3 = 8 > 0 3 3 3 (a) No relative maximum Relative minimum at (–1,–3)

= 2 ln x + 3 f ( x) = 0 when 2 ln x + 3 = 0 ln x = - 3 2 -3/2 x =e

Increasing on (-1, ¥) Decreasing on (-¥, -1)

Concave downward on (0, e-3/2 )

8 4 f ( x) = - x-5/3 + x-2/3 9 9 f ( x) = 0 when 8 4 - x-5/3 + x-2/3 = 0 9 9 4 -5/3 x (-2 + x) = 0 9 x = 2

x-intercept: x 2 ln x = 0 ln x = 0 x =1 Since f ( x) is not defined at x = 0, there is no y-intercept. No horizontal or vertical asymptotes

f ( x) is not defined when x = 0 8 4 4 f (-1) = - (-1)-5/3 + (-1)-2/3 = > 0 9 9 3 8 -5/3 4 -2/3 4 f (1) = - (1) + (1) =- <0 9 9 9 8 -5/3 4 -2/3 1 f (8) = - (8) + (8) = >0 9 9 12

-2

f (e

-2

) = 2 ln e

+ 3 = -1 < 0

f (1) = 2 ln1 + 3 = 3 > 0

(

Inflection point at e-3/2 , -

3 2e3

)

Concave upward on (e-3/2 , ¥)

(c)

4 x-2/3 + 4 x1/3 = 0 3 3 4 x-2/3 (1 + x) = 0 3 x = -1 f ¢( x) is not defined when x = 0 Critical numbers: –1 and 0 Critical points: (–1,–3) and (0, 0) f ¢(-8) = 4 (-8)-2/3 + 4 (-8)1/3 = - 7 < 0 3 3 3 æ ö æ ö-2/3 æ ö1/3 + 4 çç - 1 ÷÷÷ = 14 > 0 f ¢ çç - 1 ÷÷÷ = 4 çç - 1 ÷÷÷ è 8ø 3è 8ø 3è 8ø 3

æ2ö f ( x) = (1) (2 ln x + 1) + x çç ÷÷÷ çè x ø

(b)

f ( x) = 4 x1/3 + x 4/3 Domain is (-¥, ¥).

396 (b)

Chapter 5 GRAPHS AND THE DERIVATIVE Decreasing on (–2, 0) 10 10 -1/3 f ( x) = - x-4/3 + x 9 9

Inflection points at (0, 0) and (2, 6 ⋅ 21/3 ) Concave upward on (-¥, 0) and (2, ¥) Concave downward on (0, 2)

(c)

f ( x) = 0 when

x-intercept: 4 x1/3 + x 4/3 = 0

x1/3 (4 + x) = 0 x = 0 or x = -4

y-intercept: y = 4(0)1/3 + (0) 4/3 = 0

No horizontal or vertical asymptotes

10 -4/3 10 -1/3 + =0 x x 9 9 10 -4/3 x (-1 + x) = 0 9 x =1

f ( x) is not defined when x = 0 f (-1) = - 10 (-1)-4/3 + 10 (-1)-1/3 9 9 20 =<0 9 æ ö æ ö-4/3 10 æ 1 ö÷-1/3 + çç ÷÷ f  çç 1 ÷÷÷ = - 10 çç 1 ÷÷÷ è8ø 9 è8ø 9 è8ø

58.

f ( x) = 5 x

2/3

= - 140 < 0 9 10 f (8) = - (8)-4/3 + 10 (8)-1/3 9 9 = 35 > 0 72

5/3

Domain is (-¥, ¥). f (-x) = 5(-x)2/3 + (-x)5/3 = 5 x 2/3 - x5/3

The graph is not symmetric about the y-axis or origin. 10 -1/3 5 f ¢( x ) = x + x 2/3 3 3 f ¢( x) = 0 when 10 x-1/3 + 5 x 2/3 = 0 3 3 5 x-1/3 (2 + x) = 0 3 x = -2

(b)

Concave upward on (1, ¥) Concave downward on (-¥, 0) and (0, 1) Inflection point at (1, 6)

(c)

x-intercept: 5x 2/3 + x5/3 = 0 x 2/3 (5 + x) = 0 x = 0 or x = -5

y-intercept: y = 5(0) 2/3 + (0)5/3 = 0

No horizontal or vertical asymptotes.

f ¢( x) is not defined when x = 0

Critical numbers: –2 and 0 Critical points: (0, 0) and (-2, 3 ⋅ 22/3 ) f ¢(-8) = 10 (-8)-1/3 + 5 (-8) 2/3 3 3 =5>0 f ¢(-1) = 10 (-1)-1/3 + 5 (-1)2/3 3 3 5 =- <0 3 10 f ¢(1) = (1)-1/3 + 5 (1)2/3 3 3 =5>0

(a)

59.

Relative maximum at (-2, 3 ⋅ 22/3 ) Other graphs are possible.

Relative minimum at (0, 0) Increasing on (-¥, -2) and (0, ¥)

Chapter 5 Review 60.

397

Other graphs are possible.

64.

p(t ) = 1.884t 3 - 82.56t 2 + 1168t - 5013 for 10 £ t £ 19, t in months and p in cents p¢(t ) = 5.652t 2 - 165.12 x + 1168

Find the critical numbers. p¢(t ) = 5.652t 2 - 165.12 x + 1168 61.

62.

63.

Using a graphing calculator, we find that the two zeros of p¢ in the interval 10 £ t £ 19 are t » 12.02 and t » 17.20.

(a)-(b) If the price of the stock is falling faster and faster, P(t ) would be decreasing, so P¢(t ) would be negative. P(t ) would be concave downward, so P(t ) would also be negative.

Evaluate p¢(t ) on either side of these critical numbers. p¢(11) = 35.572 p¢(15) = -37.1

(a)-(b) When a stock reaches its highest price of the day, P(t ) is at a maximum. Maxima and minima occur when P¢(t ) = 0. Since this is a maximum, the graph would be concave down. Therefore, P(t ) < 0.

p¢(18) = 27.088 (a) The price of gasoline increased on the intervals (10, 12.02) and (17.20, 19); that is, increasing from 2010 to 2012 and from early 2017 to 2019

(a) Profit = Income - Cost P(q) = qp - C (q)

(b) The price of gasoline decreased on the interval (12.02, 17.20); that is, decreasing from 2012 to early 2017.

= q(-q 2 - 3q + 299) - (-10q 2 + 250q) = -q3 - 3q 2 + 299q

(c) Evaluate p(t ) at 10, 12.02, 17.20, and 19. p(10) = 295 p(12.02) » 370 p(17.20) » 239 p(19) » 297 The price had a relative maximum of about $3.70 in 2012.

+ 10q 2 - 250q = -q3 + 7q 2 + 49q

(b)

P¢(q) = -3q 2 + 14q + 49 = (-3q - 7)(q - 7) = -(3q + 7)(q - 7) q = 73 (nonsensical) or q = 7

f (t ) is increasing and concave downward.

P(q) = -6q + 14

f ¢(t ) is positive and decreasing.

P(7) = -6 ⋅ 7 + 14 = -28 < 0

f ¢¢(t ) is negative.

(indicates a maximum) 7 brushes would produce the maximum profit.

(c)

65.

67.

(a) Since the second derivative has many sign changes, the graph continually changes from concave upward to concave downward. Since there is a nonlinear decline, the graph must be one that declines, levels off, declines, levels off, etc. Therefore, the first derivative has many critical numbers where the first derivative is zero. (b) The curve is always decreasing except at frequent points of inflection.

68.

(a)

p = -7 2 - 3(7) + 299 = -49 - 21 + 299 = 229 $229 is the price that produces the maximum profit.

(d)

P(7) = -73 + 7(7 2 ) + 49(7) = 343 The maximum profit is $343.

(e)

P(q) = 0 when -6q + 14 = 0 q = 7 3 P(2) = -6(2) + 14 = 2 > 0 P(3) = -6 ⋅ 3 + 14 = -4 < 0

The point of diminishing returns is q = 73

Y (M ) = Y0 M b Y ¢(M ) = bY0 M b-1 Y (M ) = b(b - 1)Y0 M b- 2

(between 2 and 3 brushes).

398

Chapter 5 GRAPHS AND THE DERIVATIVE Critical numbers v » 0.12 and v » 0.72.

When b > 0, Y ¢   so metabolic rate and life span are increasing function of mass.

f ¢(0.1) = -0.03 < 0 f ¢(0.5) = 0.25 > 0 f ¢(1) = -0.75 < 0

When b < 0, Y ¢ < 0, so heartbeat is a decreasing function of mass.

The function is decreasing on (0, 0.12) and (0.72, 1) and increasing on (0.12, 0.72).

When 0 < b < 1, b(b - 1) < 0, so metabolic rate and life span have graphs that are concave downward.

f (0.12) » -0.01 f (0.72) » 0.09

When b < 0, b(b - 1) > 0, so heartbeat has a graph that is concave upward. (b)

69.

Relative minimum at (0.12, -0.01), relative maximum at (0.72, 0.09).

dY = bY M b-1 = b Y M b 0 0 dM M = b Y M

f (v) = -6v + 2.5 f (v) = 0 when

e0.33v ¹ 0

-6v + 2.5 = 0 v » 0.42 f (0.1) = 1.9 > 0 f (0.5) = -0.5 < 0

l1(v ) has no critical points.

Concave upward on (0, 0.42)

l1¢¢(v) = 0.008712e0.33v

Concave downward on (0.42, 1)

Sketch the curve for l1(v) = 0.08e0.33v l1¢(v) = 0.0264e0.33v

e0.33v ¹ 0 l1(v ) has no inflection points.

Sketch the curve for l2 = -0.87v 2 + 28.17v -211.41 l2¢ (v) = -1.74v + 28.17

71.

-1.74v + 28.17 = 0

y = 34.7(1.0186)-x ( x-0.658 )

In chapter 4, the function was originally defined as

v » 16.19

log y = 1.54 - 0.008 x - 0.658 log x

Critical point: (16.19, 16.62)

so, 0 < x < ¥, and 0 < y < ¥.

l2¢¢(v) = -1.74

The function will have a vertical asymptote at x = 0 and a horizontal asymptote at y = 0.

l2 (v) has no inflection points. l2 (v) has a relative maximum at (16.19, 16.62).

é ù dy = -34.7(1.0186)-x x-0.658 ê ln(1.0186) + 0.658 ú dx x ûú ëê dy

For every value in the domain, dx < 0, so y has no critical points and is decreasing on (0, ¥).

70.

Let a = 0.25. f (v) = v (0.25 - v) (v - 1) f ¢ (v) = -v3 + 1.25v 2 - 0.25v

72.

= -3v 2 + 2.5v - 0.25

By the quadratic formula, f ¢(v) = 0 when v = 0.12 and v » 0.72.

(a) Set the two formulas equals to each other. 1486S 2 - 4106S + 4514 = 1486S - 825 1486S 2 - 5592S + 5339 = 0

Chapter 5 Review

399

Take the derivative. 2972S - 5592 = 0

t =-

ln[- ln(0.3982 ⋅ 10-291)] ln 0.4252 By properties of logarithms,

Solve for S. S » 1.88

For males with 1.88 square meters of surface area, the red cell volume increases approximately 1486 ml for each additional square meter of surface area. (b) Set the formulas equal to each other. 995e0.6085S = 1578S

- ln(0.3982 ⋅ 10-291) = -[ln(0.3982) + ln(10-291)] = -[ln(0.3982) - 291 ln(10)] = - ln(0.3982) + 291 ln(10)

So, ln[- ln(0.3982) + 291 ln(10)] ln(0.4252) » 7.6108 At about 7.6108 years the rate of learning to pass the test begins to slow down.

995e0.6085S - 1578S = 0

t =-

Take the derivative. 605.4575e0.6085S - 1578 = 0 e0.6085S » 2.6063

74.

0.6085S » ln 2.6063 S » 1.57 square meters

(c) For males with 1.57 square meters of surface area, the red cell volume increases approximately 1578 ml for each additional square meter of surface area. (d) When f and g are closest together, their absolute difference is minimized. d [ f ( x ) - g ( x )] = 0 dx x = x 0

f ¢( x0 ) - g ¢( x0 ) = 0 f ¢( x0 ) = g ¢( x0 )

y(t ) = Ac

(a) The U.S. total inventory was at a relative maximum in 1965, 1973, 1976, 1983, 1986, and 1988. (b) The U.S. total inventory was at its largest relative maximum from 1965 to 1967. During this period, the Soviet total inventory was concave upward. This means that the total inventory was increasing at an increasingly rapid rate.

By plugging the exact value of S into the two formulas given for PV, we get about 2593 ml (Hurley) and 2484 for Pearson et al.

73.

ln(- ln A) ln c

75.

(a)

s(t ) = 512t - 16t 2 v(t ) = s¢(t ) = 512 - 32t a(t ) = v¢(t ) = s(t ) = -32

(b) The maximum height is attained when v(t ) = 0. 512 - 32t = 0 t = 16 v(0) = 512 > 0 v(20) = 512 - 640 = -128 < 0

The height reaches a maximum when t = 16.

d y¢(t ) = (ln A) A ⋅ ct dt ct

s(16) = 512 ⋅ 16 - 16(162 ) = 4096

t ct

= (ln A) (ln c)c A

y(t ) = (ln A) (ln c) ⋅ [(ln c)ct Ac

The maximum height is 4096 ft.

+ ct (ln A) (ln c)ct Ac ] t

= (ln A) (ln c) 2 ct Ac [1 + (ln A)ct ] y(t ) = 0 when 1 + ln( A)ct = 0 1 ct = ln A æ 1 ö÷ t ln c = ln çç çè ln A ÷÷ø

512 t - 16t 2 = 0 16t (32 - t ) = 0 t = 0 or t = 32 v(32) = 512 - 32(32) = -512

The projectile hits the ground after 32 seconds with a velocity of -512 ft/sec.

400 76.

Chapter 5 GRAPHS AND THE DERIVATIVE (a)

G( x ) =

3 -c( x -50)

1+ e

-c( x -70)

1+ e

1 -c( x -90)

1+ e

Each term of the derivative looks like

Extended Application: A Drug Concentration Model for Orally Administered Medications 1.

-c( x - b)

kce

(1 + e

-c( x - b)

Let C (t ) stand for the steady-state concentration in mcg/mL at time t measured in hours, with 0 £ t £ 12.

)

C (t ) = 1.99(500)e-0.14t - 0.162(500)e-2.08t

for positive k and c, so both numerator and denominator are positive and the sum of three positive terms is positive. We could simply note that each denominator in the formula for G( x) is a decreasing function of x.

C ¢(t ) = -139.3e-0.14t + 1685e-2.08t

Using a graphing calculator we find that C ¢(t ) = 0 when t » 1.28. Testing values of C¢ on either side of this time, we find C ¢(0.5) = 465.62 C ¢(2) = -78.99

(b) Here is the graph of G( x) with the constant c set to 0.7.

Since C is increasing before t = 1.28 and decreasing after, t = 1.28 represents a local maximum. The value of C at this time is C (1.28) » 775.2 so the maximum steady-state concentration is 775.2 mcg/mL. [0, 100] by [0, 5]

The minimum concentration will occur at one of the endpoints, so we evaluate C (0) and C (12).

The derivative is close to 0 on the intervals [0, 44], [56, 65], [75, 84], and [97, 100].

C (0) = 185 C (12) » 185

(c) From the graph in (b), the inflection points appear to be at the x-values 50, 60, 70, 80 and 90. The corresponding points on the graph are (50, 1.5), (60, 3.0), (70, 3.5), (80, 4.0), and (90, 4.5).

Thus the minimum steady state concentration of 185 mcg/mL occurs immediately after a dose and immediately before the next dose.

(d) The first term of the derivative can be written as follows: 3ce-c( x -50) 2

(1 + e-c( x -50) )

If the dose is increased to 1500 mg, the factor of 500 in the formula for C changes to 1500. This does not affect the zero of the derivative, but it multiplies all the values of C by a factor of 3. Thus the maximum concentration is now (775.2)(3) = 2325.6 at t = 1.28, and the minimum concentration is (185)(3) = 555 at t = 0 and t = 12.

400 » 0.52 and 80 » 0.43. So if we multiply 775.2 185

3c 2

(1 + e-c( x-50) ) ( ec( x-50) ) 3c

(1 + e

-c( x -50)

) ( ec( x-50) / 2 )

(

ec( x -50) / 2 + e-c( x -50) / 2

)

In whichever direction x moves from 50, one term in the denominator quickly becomes very large, so the fraction only contributes significantly to the derivative when x is close to 50. The other two fractions behave similarly at x = 70 and x = 90.

the original 500 mg dose by a fraction between these two values, we will keep the concentration within the required range. For example, if we give a dose of (0.5)(500) = 250 mg, the maximum concentration will be (0.5)(775.2) = 387.6 mcg/mL and the minimum concentration will be (0.5)(185) = 92.5 mcg/mL. 500(80 / 185) » 216 and 500(400 / 775.2) » 258 so the dose should be between 216 mg and 258 mg.

Chapter 6

APPLICATIONS OF THE DERIVATIVE 6.1 Absolute Extrema

Setting f ¢( x) equal to 0, we have x = 0, x = -4, or x = 1.

Your Turn 1 2/3

5/3

To find the absolute extrema of f ( x) = 3x - 3x on [0, 8], first find the critical numbers in the interval (0,8).

f ¢( x) = 2 x-1/3 - 5x 2/3 2 = 1/3 - 5 x 2/3 x 2 5x = 1/3 - 1/3 x x 1 = 1/3 (2 - 5 x) x

W1.

f ( x) = 4 x3 - 7 x 2 - 40 x + 5 f ¢( x) = 12 x 2 - 14 x - 40 = 2(6 x 2 - 7 x - 20)

f ¢ is defined everywhere, so the only critical

numbers will occur where f ¢( x) = 0. 2(6 x 2 - 7 x - 20) = 0 2(3x + 4)(2 x - 5) = 0

0.4 0.977 8

148

Warmup Exercisess

critical value 0.4 and at the endpoints 0 and 8. 0

-4

6.1

x = 52 = 0.4. Now evaluate the function f at the f ( x)

f ( x)

The absolute maximum is 148, at x = -4. As noted above, there is no absolute minimum.

f ¢( x) is 0 when 2 - 5 x = 0, that is, when

3x + 4 = 0 or 2 x - 5 = 0

-84

x =-

The absolute maximum of approximately 0.977 occurs when x = 2/5; the absolute minimum of -84 occurs when x = 8.

4 5 The critical numbers are - and . 3 2

Your Turn 2

W2.

The domain of the function

f ( x) = 15x7 / 2 - 7 x9 / 2 105 5 / 2 63 7 / 2 x x 2 2 æ 21 ö = çç x5 / 2 ÷÷÷ ( 3x - 5 ) çè 2 ø

f ( x) = - x 4 - 4 x3 + 8x 2 + 20

f ¢( x) =

is the open interval (-¥, ¥). As x approaches +¥ and -¥ the dominant term is -x 4 , and this approaches -¥. Thus the function has no absolute minimum. To find the absolute maximum, we evaluate f at the critical points.

f ¢ is defined everywhere, so the only critical

numbers will occur where f ¢( x) = 0, which happens at x = 0 and at x = 5/3.

f ¢( x) = -4 x3 - 12 x 2 + 16 x = -4 x( x 2 + 3x - 4) = -4 x( x + 4)( x - 1)

4 5 or x = 3 2

6.1

Exercises

1. True 2. True

401

402

Chapter 6 APPLICATIONS OF THE DERIVATIVE

3. False. It is possible for a continuous function to not have an absolute maximum or an absolute minimum.

16. f ( x) = x3 - 3x 2 - 24 x + 5; [-3,6]

Find critical numbers: f ¢( x) = 3x 2 - 6 x - 24 = 0

4. True

3( x 2 - 2 x - 8) = 0

5. As shown on the graph, the absolute maximum occurs at x3; there is no absolute minimum. (There is no functional value that is less than all others.)

3( x + 2)( x - 4) = 0 x = -2 or x = 4

6. As shown on the graph, the absolute minimum occurs at x1; there is no absolute maximum. (There is no functional value that is greater than all others.)

f ( x)

-3

-2

Absolute maximum

-75

Absolute minimum

-31

7. As shown on the graph, there are no absolute extrema. 17. f ( x) = 8. As shown on the graph, there are no absolute extrema.

Find critical numbers: f ¢( x ) = x 2 + 3 x - 4 = 0

9. As shown on the graph, the absolute minimum occurs at x1; there is no absolute maximum. 10. As shown on the graph, the absolute maximum occurs at x1; there is no absolute minimum. 11. As shown on the graph, the absolute maximum occurs at x1; the absolute minimum occurs at x2. 12. As shown on the graph, the absolute maximum occurs at x2 ; the absolute minimum occurs at x1.

( x + 4)( x - 1) = 0 x = -4 or

f ( x)

x -3

17.5 7 1 - » -1.17 6 5 2 » 1.67 3

f ¢( x) = 3x -12 x + 9 = 0

1 3 1 x - x 2 - 6 x + 3; [-4, 2] 3 2

( x + 2)( x - 3) = 0

x2 - 4x + 3 = 0

f ( x)

-8

-4

-8

Absolute minimum

Absolute maximum

Absolute minimum

f ¢( x ) = x 2 - x - 6 = 0

Absolute maximum

Find critical numbers:

( x - 3)( x - 1) = 0 x = 1 or x = 3

x =1

But –4 is not in [–3, 2]

18. f ( x) = 15. f ( x) = x3 - 6 x 2 + 9 x - 8; [0,5]

1 3 3 x + x 2 - 4 x + 1; [-3, 2] 3 2

x = -2 or x = 3

But 3 is not in [–4, 2] x

f ( x)

7 » -2.3 3 31 -2 » 10.33 3 25 2 » -8.33 3 -4 -

Absolute maximum Absolute minimum

Section 6.1 19.

403

f ( x) = x 4 - 18x 2 + 1; [-4, 4]

23.

f ( x) =

f ¢( x) = 4 x3 - 36 x = 0 4 x( x 2 - 9) = 0

f ¢( x ) =

4 x( x + 3)( x - 3) = 0

x2 + 1

; [1, 5]

-x 2 + 2 x + 1 ( x 2 + 1)2

f ¢( x) = 0 when

x = 0 or x = -3 or x = 3

x -1

-x 2 + 2 x + 1 = 0

f ( x)

x = 1  2,

-4 - 31 -3 - 80 0

Absolute minimum

Absolute maximum

3 - 80

Absolute minimum

but 1 - 2 is not in [1, 5].

4 - 31

20.

f ( x)

2 » 0.15 13 2 -1 » 0.21 Absolute maximum 2

f ( x) = x - 32 x - 7; [-5, 6] f ¢( x) = 4 x3 - 64 x = 0

4 x( x - 16) = 0 4 x( x - 4)( x + 4) = 0

24.

f ( x) =

x = 0 or x = 4 or x = -4 f ¢( x ) =

0 3

22.

x 2

x +2

; [0, 4]

( x 2 + 2)1 - x(2 x)

f ( x)

-5 -4

-182 -263

-7

-263

Absolute minimum

-x 2 + 2 = 0

137

Absolute maximum

x2 = 2

Absolute minimum

f ( x) 1 3 -1 3

( x 2 + 2) 2 -x 2 + 2 ( x 2 + 2)2

2 or x = - 2, but - 2 is not in [0, 4].

x =

21. f ( x) = 1 - x ; [0, 3] 3+ x -4 f ¢( x ) = (3 + x) 2 No critical numbers x

Absolute minimum

f ¢( x) is defined for all x. x

f ( x)

Absolute maximum 4

Absolute minimum

f ( x) = 8 + x ; [4, 6] 8-x (8 - x)(1) - (8 + x)(-1) f ¢( x) = (8 - x) 2 16 = (8 - x) 2 f ¢( x) is never zero. Although f ¢( x) fails to exist

if x = 8, 8 is not in the given interval. x f ( x) 4 3 Absolute minimum 6 7 Absolute maximum

25.

Absolute minimum

2 » 0.35 Absolute maximum 4 2 9 » 0.22

f ( x) = ( x 2 - 4)1/3; [-2, 3] 1 f ¢( x) = ( x 2 - 4)-2/3 (2 x) 3 2x = 2 3( x - 4) 2/3 f ¢( x) = 0 when 2 x = 0 x =0 f ¢( x) is undefined at x = -2 and x = 2, but f ( x) is defined there, so -2 and 2 are also critical numbers:

404

Chapter 6 APPLICATIONS OF THE DERIVATIVE x -2 0 2 3

26.

f ( x) 0 1/3

(-4)

28.

» -1.587 0

51/3 » 1.710 2

f ¢( x) = 1 + 2 x-1/3 2 =1+ 3 x

Absolute minimum Absolute maximum

2/3

f ( x) = ( x - 16) ; [-5, 8] 2 4x f ¢( x) = ( x 2 - 16)-1/3 (2 x) = 2 3 3( x - 16)1/3

-4

0 4

Absolute minimum

(-16)2/3 » 6.350 Absolute minimum

8 482/3 » 13.208

Absolute maximum

= = =

29.

f ( x) = x 2 - 8 ln x; [1, 4] 8 f ¢( x) = 2 x x

10 -1/3 10 2/3 + x x 3 3

8 =0 x 8 2x = x

2x2 = 8

10 x 2/3 + 3 3x1/3 10 x + 10 10

x2 = 4 x = -2 or

3x1/3 10( x + 1)

x = 2

but x = -2 is not in the given interval.

33 x

f ¢( x) = 0 when 10( x + 1) = 0 x +1 = 0 x = -1. f ¢( x) is undefined at x = 0, but f ( x) is defined at x = 0, so 0 is also a critical number. x f ( x) -2 1.587 -1 3 0 0 1 7

x = -2 x = -8

Absolute maximum Absolute minimum Absolute maximum

f ¢( x) = 0 when 2x -

f ( x) = 5 x 2/3 + 2 x5/3; [-2, 1] f ¢( x ) =

x f ( x) -10 3.925 -8 4 0 0 1 4

f ( x) 92/3 » 4.327

f ¢( x) is undefined at x = 0, but f ( x) is defined at x = 0, so 0 is also a critical number.

f ¢( x) is undefined at x = -4 and x = 4; but f ( x) is defined there, so -4 and 4 are also critical numbers. -5

x +2

x =0

f ¢( x) = 0 when 3 x + 2 = 0

f ¢( x) = 0 when 4 x = 0

27.

f ( x) = x + 3x 2/3; [-10, 1]

Although f ¢( x) fails to exist at x = 0, 0 is not in the specified domain for f ( x), so 0 is not a critical number. x 1

f ( x) 1

2 -1.545 Absolute minimum 4

4.910 Absolute maximum

Absolute minimum Absolute maximum

Section 6.1 30.

405

f ( x) = f ¢( x ) = = =

ln x x

f ( x) = x 2e-0.5 x ;[2, 5] æ 1 ö f ¢( x) = x 2 çç - e-0.5 x ÷÷÷ + 2 xe-0.5 x èç 2 ø

32.

; [1, 4]

x 2 ⋅ 1x - ln x ⋅ 2 x

æ 1 ö = e-0.5 x çç 2 x - x 2 ÷÷÷ çè 2 ø

x4 x - 2 x ln x

f ¢( x) = 0 when æ 1 ö e-0.5 x çç 2 x - x 2 ÷÷÷ = 0 çè 2 ø x = 0 or x = 4

x4 x(1 - 2 ln x) x4

1 - 2 ln x x3

but x = 0 is not in the given interval.

f ¢( x) = 0 when 1 - 2 ln x = 0 - 2 ln x = -1 1 ln x = 2

x f ( x) 2 1.472 Absolute minimum 4 2.165 Absolute maximum 5 2.052

1/2

x =e

Although f ¢( x) fails to exist at x = 0, 0 is not in the specified domain for f ( x), so 0 is not a critical number.

31.

x 1

f ( x) 0

e1/2 4

0.1839 Absolute maximum 0.0866

33.

f ¢( x) = 1 - 3e-3x

34.

-1 19.09

; [-1, 1]

f ( x) =

x3 + 2 x + 5 x 4 + 3x3 + 10

; [-3, 0]

The indicated domain tells us the x-values to use for the viewing window, but we must experiment to find a suitable range for the y-values. In order to show the absolute extrema on [-3, 0] , we find that a suitable window is [-3, 0] by [-9, 1] . From the graph, we see that on [-3, 0], f has an absolute maximum of 0.5 at 0 and an absolute minimum of 8.10 at about -2.35.

f ( x) Absolute maximum

ln 3 0.6995 Absolute minimum 3 3 3.000

x 4 - x3 + x 2 + 7

From the graph, we see that on [-1, 1], f has an absolute maximum of 1.356 at about 0.6085 and an absolute minimum of 0.5 at -1.

f ( x) = x + e-3x ; [-1, 3]

-3e-3x = -1 1 e-3x = 3 1 -3x = ln 3 ln 3 x = 3

-5 x 4 + 2 x 3 + 3 x 2 + 9

The indicated domain tells us the x-values to use for the viewing window, but we must experiment to find a suitable range for the y-values. In order to show the absolute extrema on [-1, 1], we find that a suitable window is [-1, 1] by [0, 1.5] with Xscl = 0.1, Yscl = 0.1.

Absolute minimum

f ¢( x) = 0 when 1 - 3e-3x = 0

f ( x) =

35.

8 f ( x) = 2 x + 2 + 1, x  0 x 16 f ¢( x ) = 2 - 3 x = =

2 x3 - 16 x3 2( x - 2)( x 2 + 2 x + 4)

406

Chapter 6 APPLICATIONS OF THE DERIVATIVE Since the specified domain is (0, ¥), a critical number is x = 2. x

f ( x)

f ( x) = 12 - x -

39.

9 , x>0 x

9 f ¢ ( x ) = -1 + 2 x = =

x2 (3 + x)(3 - x)

f ¢( x) = 0 when x = -3 or x = 3, and f ¢( x) does not exist when x = 0. However, the specified domain for f is (0, ¥). Since -3 and 0 are not in the domain of f, the only critical number is 3. x

f ( x)

= =

x -1 2

x + 2x + 6 ( x 2 + 2 x + 6)(1) - ( x - 1)(2 x + 2) ( x 2 + 2 x + 6)2 x2 + 2x + 6 - 2x2 + 2 ( x 2 + 2 x + 6)2 -x 2 + 2 x + 8 ( x 2 + 2 x + 6)2 -( x 2 - 2 x - 8) ( x 2 + 2 x + 6)2 -( x - 4)( x + 2) ( x 2 + 2 x + 6)2

Critical numbers are 4 and -2. x -2

f ( x) = -3x 4 + 8 x3 + 18x 2 + 2 f ¢( x) = -12 x3 + 24 x 2 + 36 x = -12 x( x 2 - 2 x - 3)

40.

f ( x) =

= -12 x( x - 3)( x + 1) f ¢( x ) =

Critical numbers are 0, 3, and -1. x

f ( x)

-1

f ( x)

1 2 4 0.1 There is an absolute maximum of 0.1 at x = 4 and an absolute minimum of 0.5 at x = -2. This can be verified by looking at the graph of f.

There is an absolute maximum of 6 at x = 3. There is no absolute minimum, as can be seen by looking at the graph of f. 37.

f ( x) = f ¢( x ) =

9 - x2

f ( x) 1

2 1 There is no absolute maximum, as can be seen by looking at the graph of f. There is an absolute minimum of 1 at x = 0 and x = 2.

There is an absolute minimum of 7 at x = 2; there is no absolute maximum, as can be seen by looking at the graph of f. 36.

x 0

137

There is an absolute maximum of 137 at x = 3; there is no absolute minimum, as can be seen by looking at the graph of f.

x 2

x +1 ( x 2 + 1) - x(2 x) ( x 2 + 1)2 x2 + 1 - 2x2 ( x 2 + 1) 2 1 - x2 ( x 2 + 1)2 (1 + x)(1 - x) ( x 2 + 1) 2

The critical numbers are –1 and 1. 38.

f ( x) = x 4 - 4 x 3 + 4 x 2 + 1 3

f ( x)

-1 -0.5

f ¢( x) = 4 x - 12 x + 8x = 4 x( x 2 - 3x + 2) = 4 x( x - 2)( x - 1)

The critical numbers are 0, 1, and 2.

1 0.5 There is an absolute maximum of 0.5 at x = 1 and an absolute minimum of -0.5 at x = -1. This can be verified by looking at the graph of f.

Section 6.1 41.

407

f ( x) = f ¢( x ) = = = =

ln x

f ( x)

-1

-5

x x

⋅ 1x - 3x 2 ln x 6 x

-1

0.5

-0.88988

x - 3x ln x x6

Absolute minimum of -5 at x = -1; absolute maximum of 0 at x = 0

x (1 - 3ln x) x6 1 - 3ln x

(b) On [0.5, 2] x

0.5 -0.88988

f ¢( x) = 0 when x = e1/3 , and f ¢( x) does not 1/3

exist when x £ 0. The only critical number is e . x

f ( x)

e1/3

1 -1 e » 0.1226 3

-1

-0.7622

Absolute maximum of about -0.76 at x = 2; absolute minimum of -1 at x = 1.

There is an absolute maximum of 0.1226 at x = e1/3 . There is no absolute minimum, as can be seen by looking at the graph of f. 42.

f ( x)

44. Let P( x) be the perimeter of the rectangle with

vertices (0, 0), ( x, 0), ( x, f ( x)), and (0, f ( x)) for x >  when f ( x) = e-2 x .

f ( x) = x ln x 1 + 1 ⋅ ln x x = 1 + ln x

f ¢( x ) = x ⋅

f ¢( x) = 0 when x = e-1 , and f ¢( x) does not

exist when x £ 0. The only critical number is e-1. x

f ( x)

e-1 -e-1 » -0.3679

There is an absolute minimum of -0.3679 at x = e-1. There is no absolute maximum, as can be seen by looking at the graph of f. 43.

The length of the rectangle is x and the width is given by e-2 x . Therefore, an equation for the perimeter is P( x) = x + e-2 x + x + e-2 x = 2( x + e-2 x ).

f ( x) = 2 x - 3x 2/3 2 23 x - 2 f ¢( x) = 2 - 2 x-1/3 = 2 - 3 = 3 x x f ¢( x) = 0 when 2 3 x - 2 = 0

P¢( x) = 2 - 4e-2 x P¢( x) = 0 when 2 - 4e-2 x = 0 - 4e-2 x = -2

2 x = 2 3

e-2 x =

x =1

1 2

e2 x = 2

x =1

f ¢( x) is undefined at x = 0, but f ( x) is defined at x = 0. So the critical numbers are 0 and 1. (a) On [-1, 0.5]

2 x = ln 2 x =

ln 2 2

408

Chapter 6 APPLICATIONS OF THE DERIVATIVE x

P( x)

48.

ln 2 1 + ln 2 » 1.693 2

There is an absolute minimum of 1.693 at ln 2

x = 2 . There is no absolute maximum, as can be seen by looking at the graph of P. Therefore, the correct statement is a.

P( x) = -0.02 x3 + 600 x - 20, 000, [50,150] P¢( x) = -0.06 x 2 + 600 = -0.06( x 2 - 10, 000) = -0.06( x + 100)( x - 100)

There are critical numbers at x = -100 and x = 100. Only x = 100 is in the domain of P( x).

45. (a) Looking at the graph we see that there are relative maxima of 5546 in 2010 and 4185 in 2016. There are relative minima of 3879 in 2014 and 2975 in 2018. (b) The absolute maximum is 5546 (in 2010) and the absolute minimum is 2975 (in 2018). 46. (a) Looking at the graph we see that there are relative maxima of 74 in 2010, 74 in 2012, 71 in 2014, and 65 in 2016. There are relative minima of 60 in 2011, 47 in 2013, 53 in 2015, and 56 in 2017 and 2018. (b) The absolute maximum is 74 (in 2010) and the absolute minimum is 47 (in 2013). 47. P( x) = -x3 + 9 x 2 + 120 x - 400, x ³ 5

P ( x)

7500

100

20, 000

150

2500

The maximum profit is $20,000, which occurs when 100 units per week are made. 49. C ( x) = x3 + 37 x + 250 (a) 1 £ x £ 10 C ( x) x3 + 37 x + 250 = x x 250 = x 2 + 37 + x

C ( x) =

C ¢( x ) = 2 x -

P¢( x) = -3x 2 + 18x + 120 =

= -3( x - 6 x - 40)

250 x2

2 x3 - 250

= -3( x - 10)( x + 4) = 0 x = 10 or x = -4

= 0 when

2 x3 = 250 x3 = 125

-4 is not relevant since x ³ 5, so the only critical number is 10.

x = 5.

The graph of P¢( x) is a parabola that opens

Test for relative minimum.

downward, so P¢( x) > 0 on the interval [5, 10)

C ¢(4) = -7.625 < 0

and P¢( x) < 0 on the interval (10, ¥). Thus, P( x) is a maximum at x = 10.

C ¢(6) » 5.0556 > 0

Since x is measured in hundreds thousands, 10 hundred thousand or 1,000,000 tires must be sold to maximize profit. Also,

C (1) = 1 + 37 + 250 = 288

P(10) = -(10)3 + 9(10)2 + 120(10) - 400 = 700. The maximum profit is $700 thousand or $700,000.

C (5) = 112 C (10) = 100 + 37 + 25 = 162

The minimum for 1 £ x £ 10 is 112. (b) 10 £ x £ 20

There are no critical values in this interval. Check the endpoints. C (10) = 162 C (20) = 400 + 37 + 12.5 = 449.5

The minimum for10 £ x £ 20 is 162.

Section 6.1

409

50. C ( x) = 81x 2 + 17 x + 324

54. The value x = 280 maximizes

is the point where the line from the origin to the curve is tangent to the curve. A production level of 280 units results in the maximum profit per item produced.

(a) 1 £ x £ 10 C ( x) =

81x + 17 x + 324 C ( x) = x x 324 = 81x + 17 + x 324 C ¢( x ) = 81 - 2 = 0 when x 324 81 = 2 x x = 2.

56. P(i) = ( p - r ) AS ait k b X 1-c - W0 X - W1 Xi - C (a) Substituting the given values into the function ( p - r ) AS a k b X 1-c = 150 W0 X + C = 10 W1 X = 300 t = 0.5 results in the simplified function P(i) = 150i 0.5 - 10 - 300i

–2 is not meaningful is this application and is not in the given domain. Test 2 for minimum. C ¢(1) = -243 < 0

= 150 i - 300i - 10

C ¢(3) = 45 > 0

(b) Find the derivative of P(i).

C ( x) is minimum when x = 2. 324 C (2) = 81(2) + 17 + = 341 2 The minimum on the interval 1 £ x £ 10 is 341.

P(i) = 150 i - 300i - 10 1 P¢(i) = (150i-1/2 ) - 300 2 75 = - 300 i Find maximum when P¢(i) = 0, 75 - 300 = 0 i 75 = 300 i

(b) 10 £ x £ 20

There are no critical numbers in this interval. Test the endpoints. C (10) = 859.4 C (20) = 1620 + 17 + 16.2

0.25 = i

= 1653.2 The minimum on the interval 10 £ x £ 20 is 859.4.

51. The value x = 11.5 minimizes

f ( x) because this is x

the point where the line from the origin to the curve is tangent to the curve. A production level of 20 units results in the minimum cost per unit. 53. The value x = 100 maximizes

1 16 The administrative intensity that maximizes profit is 1/16 (1 manager per 16 workers). i = 0.0625 =

f ( x) because this x

is the point where the line from the origin to the curve is tangent to the curve. A production level of 11.5 units results in the minimum cost per unit. 52. The value x = 20 minimizes

f ( x) because this x

is the point where the line from the origin to the curve is tangent to the curve. A production level of 100 units results in the maximum profit per item produced.

57.

f ( x) = f ¢( x ) = = = =

x 2 + 36 , 1 £ x £ 12 2x 2 x(2 x) - ( x 2 + 36)(2) (2 x)2 4 x 2 - 2 x 2 - 72 4x2 2 x 2 - 72

2( x 2 - 36)

4x2 4x2 ( x + 6)( x - 6) 2x2

f ¢( x) = 0 when x = 6 and when x = -6. Only 6 is in the interval 1 £ x £ 12.

410

Chapter 6 APPLICATIONS OF THE DERIVATIVE Test for relative maximum or minimum. (11)(-1) <0 f ¢(5) = 50 (13)(1) >0 f ¢(7) = 98 The minimum occurs at x = 6, or at 6 months. Since f (6) = 6, f (1) = 18.5, and f (12) = 7.5, the minimum percent is 6%.

61.

When M ¢( x) = 0, 2x +2=0 45 2x 2= 45 45 = x. x M ( x) 30 20 45 25 145 65 » 16.1 9 The absolute maximum miles per gallon is 25 at 45 mph and the absolute minimum miles per gallon is about 16.1 at 65 mph. -

58. S (T ) = -T 3 + 3T 2 + 360T + 5000; 6 £ T £ 20 S ¢(T ) = -3T 2 + 6T + 360 = -3(T 2 - 2T - 120) S ¢(T ) = -3(T - 12)(T + 10) = 0 T = 12 or T = -10 (not in the interval) T 6 12 10

f (T ) 7052 8024 7900

12° is the temperature that produces the maximum number of salmon.

1 2 x + 2 x - 20, 30 £ x £ 65 45 1 2x M ¢( x) = - (2 x) + 2 = +2 45 45 M ( x) = -

62.

M ( x) = -0.015x 2 + 1.31x - 7.3, 30 £ x £ 60 M ¢( x) = -0.03x + 1.31 = 0 x » 43.7

59. Since we are only interested in the length during weeks 22 through 28, the domain of the function for this problem is [22, 28]. We now look for any critical numbers in this interval. We find

x M ( x) 30 18.5 43.7 21.3 60 17.3

L¢(t ) = 0.788 - 0.02t = 39.4. There is a critical number at t = 0.788 0.02

which is not in the interval. Thus, the maximum value will occur at one of the endpoints.

The absolute maximum of 21.3 mpg occurs at 43.7 mph. The absolute minimum of 17.3 mpg occurs at 60 mph.

t L(t ) 22 5.4 28 7.2

æ x ö÷2 æ 12 - x ö÷2 63. Total area A( x) =  çç ÷ + ççç ÷ èç 2 ÷ø è 4 ÷ø

The maximum length is about 7.2 millimeters.

60. The function is defined on the interval [15, 46]. We look first for critical numbers in the interval. We find R¢(T ) = -0.00021T 2 + 0.0802T - 1.6572

Using our graphing calculator, we find one critical number in the interval at about 21.92 T R(T ) 15 81.01 21.92 79.29 46 98.89 The relative humidity is minimized at about 21.92C.

x2 (12 - x)2 + 4 16

12 - x x =0 2 8 4 x -  (12 - x) =0 8 12 » 5.28 x = 4+

A¢( x) =

x 0 5.28 12

Area 9 5.04 11.46

The total area is minimized when the piece used to form the circle is 412+ feet, or about 5.28 feet long.

Section 6.2

411 (b) - ln p + ln (1 - p) = 0 ln (1 - p) = ln p 1- p = p 1 = 2p 1 = p 2

64. Total area = A( x) æ x ö÷2 æ 12 - x ö÷2 =  çç + çç ÷ çè 2 ÷÷ø èç 4 ÷ø =

x2 (12 - x)2 + 4 16 12 - x x =0 2 8 4 x -  (12 - x) =0 8 12 » 5.28 x = 4+

I ¢(0.25) = 1.0986 I ¢(0.75) = -1.099

A¢( x) =

x 0 5.28 12

Area 9 5.04 11.46

There is a relative maximum of 0.693 at p = 12 .

6.2

Applications of Extrema

Your Turn 1

Assign a variable to the quantity to be minimized:

The total area is maximized when all 12 feet of wire are used to form the circle. 65. For the solution to Exercise 63, the piece of length x used to form the circle is 412+ feet. The circle

M = x2 y

Use the given condition to express M in terms of one variable, say x : x + 3 y = 30 30 - x y = 3 æ 30 - x ö÷ 1 M = x 2 çç = 10 x 2 - x3 çè 3 ÷÷ø 3

can be inscribed inside the square if the side of the square equals the diameter of the circle (that is, twice the radius). side of the square = 2 (radius) æ x ö÷ 12 - x = 2 çç çè 2 ÷÷ø 4 x 12 - x =  4 4 x = 12 -  x x(4 +  ) = 12 12 x = 4+

Find the domain of M: Since x and y must be nonnegative, we have x ³ 0 and 303- x ³ 0 which requires x £ 30. Thus the domain of M is [0, 30]. Find the critical numbers:

Therefore, the circle formed by piece of length x = 412+ can be inscribed inside the square. 66. (a)

I ( p) = - p ln p - (1 - p) ln (1 - p) æ1ö I ¢( p) = - p çç ÷÷÷ + (ln p)(-1) çè p ø÷

dM = 20 x - x 2 = x(20 - x) dx dM dx

is 0 when x = 0 (already found as an endpoint)

and when x = 20. Evaluate M at the critical numbers and endpoints: x

é ù -1 - ê (1 - p) + [ln (1 - p)](-1) ú ê ú 1- p ë û = -1 - ln p + 1 + ln (1 - p) = - ln p + ln (1 - p)

4000 3

The maximum of M occurs at x = 20. The corresponding value of y is 30-3 20 or 10 . 3

412

Chapter 6 APPLICATIONS OF THE DERIVATIVE This derivative has two positive roots, x = 4 and x = 4/3. We have already identified one of these as an endpoint of the domain and the other lies within the domain. Evaluating V at 0, 4/3 and 4 gives the following table.

Your Turn 2

Repeat Example 2 using 270m for 800m. T ( x) = =

300 - x + 160

2702 + x 2 80

300 x (2702 + x 2 )1/ 2 + 160 160 80

As before, the domain of T is [0, 300]. The derivative of T is T ¢( x ) = -

1 1 æç 1 ö÷ 2 2 -1/2 (2 x) + ç ÷ (270 + x ) 160 80 çè 2 ÷ø

Now find the critical numbers by setting the derivative of T equal to 0. -

1 1 æç 1 ö÷ 2 2 -1/2 (2 x) = 0 + ç ÷ (270 + x ) 160 80 çè 2 ÷ø 1 x = 2 2 160 80 270 + x 2x =

2702 + x 2

4 x 2 = 2702 + x 2 2

3x = 270

270 3 x » 155.88 x =

Calculate the total time at this critical number and at the endpoints of the domain. x

4/3

1024 27

The maximum volume is 1024/27 m3 and occurs when x = 4/3 m. Note that we can solve this problem efficiently using a scaling argument. Since the proportions of the largestvolume box should not depend on the linear scale we adopt (that is, on the units in which we measure length), we can just note that our piece of metal is 8/12 or the 2/3 size of the one in Example 3. Our minimizing length x will be 2/3 of the value we found in Example 3, or (2/3)(2) = 4/3 m. Your Turn 4

Follow the procedure of Example 4 with a volume of 500 cm3 instead of 1000 cm3. V =  r 2h = 500

so h=

500

r2

and

0 5.25 155.88

S = 2 r 2 + 2 r

4.80

300 5.24

The minimum travel time occurs for x = 155.88. The professor runs 300 – 155.88 meters or about 144 meters along the trail and then head into the woods.

= 2 r 2 +

V ( x) = x(8 - 2 x) 2 = 4(16 x - 8x 2 + x3 ).

For nonnegative side lengths we require x ³ 0 and 8 - 2 x ³ 0 or x £ 4; the domain of V is thus [0, 4].

Set the derivative of V equal to 0 and solve for x. V ¢( x) = 4(16 - 16 x + 3x 2 ) = 4(3x - 4)( x - 4)

r2

1000 . r

Excluding a radius of 0 gives a domain for S of (0, ¥). Now find the critical numbers of S.

Your Turn 3

We follow the procedure of Example 3, but now our volume function is

500

S ¢ = 4 r 4 r = r3 =

1000 r2

1000 r2 250



æ 250 ö÷1/3 r = çç » 4.301 çè  ÷÷ø

We can verify that S ¢ is negative to the left of 4.3 and positive to the right (for example, S ¢(4) » -12.2 and S ¢(5) » 22.8). Thus the function S is decreasing as we

Section 6.2

413

move toward 4.3 from left and increasing as we move past 4.3 to the right, so there is a relative minimum at 4.3. Since there is only one critical number, the critical point theorem tells us that it corresponds to an absolute minimum of the area function. Then h »

500

f ¢( x ) = 0 when x 2 + x - 3 = 0. The roots of this equation are given by -1  13 2 -1  13 so the critical numbers include . 2 the quadratic formula as x =

» 8.604. The minimum surface

 (4.301)2

area is obtained with a radius of 4.3 cm and a height of 8.6 cm.

Since f ¢ is defined everywhere, these will be the only critical numbers.

As with the box in Your Turn 3, we could obtain this answer directly from Example 4 using a scaling argument. Changing the volume by a factor of 1/2 (from 1000 to 500) scales all linear measures by a factor of (1/2)1/3, so we can find the new r and h by dividing the values found in Example 4 by (1/2)1/3 : r = h=

6.2 W1.

5.419 21/3 10.84 21/3

» 8.604

3. False. For a function on an open interval, check every critical number to determine absolute extrema. 2

f ( x ) = 5x - 18 x - 28 x + 12 f ¢( x ) = 20 x 3 - 54 x 2 - 56 x

(

= (2 x ) 10 x 2 - 27 x - 28

)

= (2 x )(5x + 4)(2 x - 7)

4. False. It is possible that a function on an open interval can have both an absolute maximum and an absolute minimum. 5.

x + y = 180, P = xy

(a) y = 180 - x

f ¢( x ) = 0 when

(b) P = xy = x(180 - x)

2 x = 0 or 5x + 4 = 0 or 2 x - 7 = 0.

(c) Since y = 180 - x and x and y are nonnegative numbers, x ³ 0 and 180 - x ³ 0 or x £ 180. The domain of P is [0, 180].

Thus the critical numbers are 4 7 0, - , . 5 2

W2.

1. True 2. False. If a function on a closed interval has only one critical number at x = c, then don’t forget to evaluate the function at the end points in addition to the critical number to determine which yields the absolute maximum or minimum.

» 4.301

Warmup Exercises 4

6.2 Exercises

f ( x) = f ¢( x ) = = =

(d) P¢( x) = 180 - 2 x

2x + 1

180 - 2 x = 0 2(90 - x) = 0 x = 90

x2 + 3 (2)( x 2 + 3) - (2 x + 1)(2 x ) ( x 2 + 3)2 2x 2 + 6 - 4x2 - 2x ( x 2 + 3)2 (-2)( x 2 + x - 3) ( x 2 + 3)2

(e)

x 0 90 180

P 0 8100 0

(f) From the chart, the maximum value of P is 8100; this occurs when x = 90 and y = 90.

414 6.

Chapter 6 APPLICATIONS OF THE DERIVATIVE x + y = 140

(f) The maximum value of x 2 y occurs when x = 60 and y = 30. The maximum value is 108,000.

Minimize x 2 + y 2. (a) y = 140 - x 8.

(b) Let 2

P = x + y = x + (140 - x)

= x 2 + 19,600 - 280x + x 2 = 2 x 2 - 280 x + 19, 600.

(a) y = 105 - x (b) Let P = xy 2 = x(105 - x)2 = x(11, 025 - 210 x + x 2 ) = 11, 025x - 210 x 2 + x3.

(d) P¢ = 11, 025 - 420 x + 3x 2

x P 0 19, 600 70 9800 140 19, 600

3x 2 - 420 x + 11, 025 = 0

(f) The minimum value of x 2 + y 2 occurs when x = 70 and y = 140 - x = 140 -70 = 70. The minimum value is 9800. 7.

Maximize xy 2.

4 x - 280 = 0 4 x = 280 x = 70

(e)

x + y = 105

x + y = 90

3( x 2 - 140 x + 3675) = 0 3( x - 35) ( x - 105) = 0 x = 35 or x = 105

(e)

x P 0 0 35 171,500 105 0

(f) The maximum value of xy 2 occurs when x = 35 and y = 70. The maximum value is 171,500.

Minimize x y. (a) y = 90 - x (b) Let P = x 2 y = x 2 (90 - x) 2

= 90 x - x .

1 3 x + 2 x 2 - 3x + 35 2 The average cost function is

9. C ( x) =

A( x) = C ( x) =

1 x3 + 2 x 2 - 3x + 35

= 2

x 1 2 35 = x + 2x - 3 + 2 x 1 2 x + 2 x - 3 + 35x-1. or 2

180 x - 3x 2 = 0 3x(60 - x) = 0 x = 0 or x = 60

(e)

x P 0 0 60 108, 000 90 0

C ( x) x

Then A¢( x) = x + 2 - 35x-2 or x + 2 -

35 x2

Section 6.2

415

Graph y = A¢( x) on a graphing calculator. A suitable choice for the viewing window is [0,10] by [-10,10]. (Negative values of x are not meaningful in this application.) Using the calculator, we see that the graph has an x-intercept at x » 2.722. Thus, 2.722 is a critical number. Now graph y = A( x) and use this graph to confirm that a minimum occurs at x » 2.722. Thus, the average cost is smallest at x » 2.722. At this value of x, A » 19.007. 10. C ( x) = 10 + 20 x1/2 + 16 x3/2 C ( x) x

10 + 20 x1/2 + 16 x3/2 x 10 -1/2 = + 20 x + 16 x1/2 x =

or 10 x-1 + 20 x-1/2 + 16 x1/2.

Then A¢( x) = -10 x-2 - 10 x-3/2 + 8x-1/2.

Graph y = A¢( x) on a graphing calculator. A suitable choice for the viewing window is [0, 10] by [-10,10]. (Negative values of x are not meaningful in this application.) We see that this graph has one x-intercept. Using the calculator, we find that this x-value is about 2.110, which shows that 2.110 is the only critical number of A. Now graph y = A( x) and use this graph to confirm that a minimum occurs at x » 2.110. Thus, the average cost is smallest at x » 2.110. At this value of x, A » 41.749. 11. p( x) = 160 -

The maximum revenue is 64,000,000 cents or $640, 000. x 8 (a) Revenue from x thousand compact discs: R( x) = 1000 xp æ xö = 1000 x çç12 - ÷÷÷ çè 8ø

12. p( x) = 12 -

= 12,000 x - 125 x 2

The average cost function is A( x) = C ( x) =

(b) R¢( x) = 12, 000 - 250 x 12,000 - 250 x = 0 12,000 = 250 x 48 = x

The maximum revenue occurs when 48 thousand compact discs are sold. (c) R(48) = 12,000(48) - 125(48)2 = 288,000 The maximum revenue is $288,000. 13. Let x = the width and y = the length. (a) The perimeter is P = 2x + y = 1400,

so y = 1400 - 2 x.

(b) Area = xy = x(1400 - 2 x) A( x) = 1400 x - 2 x 2

x 10

1400 - 4 x = 0 1400 = 4 x

(a) Revenue from sale of x thousand candy bars: R( x) = 1000 xp æ x ö = 1000 x çç160 - ÷÷÷ çè 10 ø = 160,000 x - 100 x 2

350 = x A¢¢ = -4, which implies that x = 350 m leads to the maximum area.

(d) If x = 350, y = 1400 - 2(350) = 700.

(b) R¢( x) = 160, 000 - 200 x

The maximum area is (350)(700)

160, 000 - 200 x = 0

= 245, 000 m 2.

800 = x

416

Chapter 6 APPLICATIONS OF THE DERIVATIVE 1800 - 4 x = 0

14. Let x = length of field y = width of field.

1800 = 4 x 450 = x

Perimeter:

A¢¢ = -4, which implies that x = 450 is the location of a maximum.

P = 2 x + 2 y = 300 x + y = 150

If x = 450, y = 1800 - 2(450) = 900.

y = 150 - x

The maximum area is

Area:

(450)(900) = 405, 000 m 2.

A = xy = x(150 - x)

16.

= 150 x - x 2

1200  (3/2)x

Thus, A( x) = 150 x - x 2 A¢( x) = 150 - 2 x.

A¢( x) = 0 when

1200  (3/2)x

150 - 2 x = 0 x = 75. A¢¢( x) = -2, so A¢¢(75) = -2 < 0, which confirms that a maximum value occurs at x = 75. If x = 75.

There are three fence pieces of length x, which leaves 2400 - 3x for the two remaining sides. Each of these remaining sides thus has length 1200 - (3/2) x. The area enclosed is A( x) = x[1200 - (3/2) x] = 1200 x -

y = 150 - x = 150 - 75 = 75.

A maximum area occurs when the length is 75 m and the width is 75 m. 15. Let x = the width of the rectangle y = the total length of the rectangle.

3x 2 , 2

measured in square meters. Both x and 1200 - (3/2) x must be nonnegative, so the domain of A is [0,800]. Now set the derivative equal to zero to solve for any critical numbers: A¢( x) = 1200 - 3x 1200 - 3x = 0 1200 = 3x x = 400

Evaluate A at the endpoints of the domain and at the single critical value x = 400. x 0

An equation for the fencing is

400 240, 000

3600 = 4 x + 2 y 2 y = 3600 - 4 x y = 1800 - 2 x. Area = xy = x(1800 - 2 x) A( x) = 1800 x - 2 x 2

A 0

800

The maximum area enclosed by the pen is 240,000 m 2. To achieve this, the farmer will use three fence sections of length 400 m and two of length 1200 - (3/2) (400) = 600 m.

A¢ = 1800 - 4 x

Section 6.2

417

17. Let x = length at $1.50 per meter y = width at $3 per meter.

128, 000 x2

xy = 25,600 25,600 y = x

Minimize cost:

1.5 -

153,600 x2

153, 600

x = 160 20, 000 y = = 125 160

320 ft at $2.50 per foot will cost $800. 250 ft at $3.20 per foot will cost $800. The entire cost will be $1600.

= 144, 000 + 700 x - 10 x 2

153,600 x2

1.5 x = 153,600 x 2 = 102, 400 x = 320 25,600 = 80 320

320 m at $1.50 per meter will cost $480. 160 m at $3 per meter will cost $480. The total cost will be $960. 18. Let x = the length at $2.50 per foot y = the width at $3.20 per foot. xy = 20,000 20,000 y = x 40, 000 x 40, 000 Cost = C ( x) = 2 x(2.5) + (3.2) x 128, 000 = 5x + x

Perimeter = 2 x + 2 y = 2 x +

Minimize cost: C ¢( x ) = 5 -

x 2 = 25, 600

(a) Revenue = R( x) = (90 + x)(1600 - 10 x)

y =

19. Let x = the number of refunds. Then 90 + x = the number of passengers.

1.5 =

128, 000

5x 2 = 128, 000

51, 200 Perimeter = x + 2 y = x + x 51, 200 Cost = C ( x) = x(1.5) + (3) x 153, 600 = 1.5x + x

C ¢( x) = 1.5 -

= 0

128, 000 x2

Assume that the number of refunds is nonnegative and that the number of refunds is limited to 160 so that the revenue will be nonnegative. Thus the domain of R is [0,160]. Now set the derivative of R equal to 0 and solve. R¢ = 700 - 20 x = 0 x = 35

Check the value of R at this critical number and at the endpoints of the domain: x

0 144, 000 35 156, 250 160

Thus the maximum revenue is obtained with 35 refunds, which happens when there are 125 passengers. (b) The maximum revenue is $156,250. 20. Let x = the number of seats.

Profit is 8 dollars per seat for 0 £ x £ 50. Profit (in dollars) is 8 - 0.1( x - 50) per seat for x > 50. We except that the total number of seats which makes the total profit a maximum will be greater than 50 because after 50 the profit is still increasing, though at a slower rate. (Thus we know the function is concave down and its one extremum will be a maximum.)

418

Chapter 6 APPLICATIONS OF THE DERIVATIVE

(a) The total profit for x seats is P( x) = [8 - 0.1( x - 50)]x = (8 - 0.1x + 5) x = (13 - 0.1x) x = (13x - 0.1x 2 )

P¢( x ) = 13 - 0.2 x 13 - 0.2 x = 0 x = 65

65 seats will produce the maximum profit. (b) P(65) = 13(65) - 0.1(652 ) = 845 - 422.5 = 422.5 The maximum profit is $422.50. 21. Let x = the number of days to wait. 12, 000 = 120 = the number of 100-lb groups 100 collected already.

R¢¢ = -1 < 0, so when 250 + 35 = 285 tables are ordered, revenue is maximum.

Thus, the maximum revenue is R(35) = 40,000 + 35(35) - 0.50(35)2 = 40,612.5

The maximum revenue is $40,612.50. Minimum revenue is found by letting R = 0. (160 - 0.50 x)(250 + x) = 0 160 - 0.50 x = 0 or 250 + x = 0 x = 320 or x = -250 (impossible)

So when 250 + 320 = 570 tables are ordered, revenue is 0, that is, each table is free. I would fire the assistant. 23. Let x = a side of the base h = the height of the box.

Then 7.5 - 0.15 x = the price per 100 lb; 4 x = the number of 100-lb groups collected per day; 120 + 4 x = total number of 100-lb groups collected. Revenue = R( x) = (7.5 - 0.15 x)(120 + 4 x) = 900 + 12 x - 0.6 x 2

An equation for the volume of the box is V = x 2h,

R¢( x) = 12 - 1.2 x = 0 x = 10 R¢¢( x) = -1.2 < 0 so R( x) is maximized at x = 10.

The scouts should wait 10 days at which time their income will be maximized at R(10) = 900 + 12(10) - 0.6(10)2 = $960. x = the number of additional tables. Then 160 - 0.50 x = the cost per table and 250 + x = the number of tables ordered.

22. Let

so 32 = x 2h h=

32 x2

The box is open at the top so the area of the surface material m ( x) in square inches is the area of the base plus the area of the four sides. m( x) = x 2 + 4 xh æ 32 ö = x 2 + 4 x çç 2 ÷÷÷ èç x ø = x2 + m¢( x) = 2 x -

R = (160 - 0.50 x)(250 + x) = 40, 000 + 35x - 0.50 x 2 R¢ = 35 - x = 0 x = 35

128 x 128 x2

Section 6.2

419 2 x 3 - 128 x2

x = the length of the side of the cutout square.

25. Let

= 0

Then 3 - 2 x = the width of the box and 8 - 2 x = the length of the box.

2 x 3 - 128 = 0 2( x 3 - 64) = 0 x = 4

V ( x) = x(3 - 2 x)(8 - 2 x) = 4 x3 - 22 x 2 + 24 x

m¢( x) = 2 + 256 > 0 since x > 0. 3 x

So, x = 4 minimizes the surface material.

The domain of V is ( 0, 32 ) .

If x = 4.

Maximize the volume.

32 = = 2. 16 x2 The dimensions that will minimize the surface material are 4 in. by 4 in. by 2 in. h=

V ¢( x) = 12 x 2 - 44 x + 24 12 x 2 - 44 x + 24 = 0 4(3x 2 - 11x + 6) = 0 4(3x - 2)( x - 3) = 0

24. Let x = the width. Then 2 x = the length and h = the height.

x =

3 is not in the domain of V.

An equation for volume is

V ¢¢( x) = 24 x - 44 æ2ö V ¢¢ çç ÷÷÷ = -28 < 0 çè 3 ø

V = (2 x)( x)h = 2 x 2h 36 = 2 x 2h.

So, h = 182 . x

The surface area S ( x) is the sum of the areas of the base and the four sides. S ( x) = (2 x)( x) + 2 xh + 2(2 x)h = 2 x 2 + 6 xh æ 18 ö = 2 x 2 + 6 x çç 2 ÷÷÷ èç x ø

the total area of all four walls is 4 x(12 - 2 x) = -8x 2 + 48x. Since the box has maximum volume when x = 2, the area of the base is

108 x

4 x - 108 x2

4(2)2 - 48(2) + 144 = 64 square inches

and the total area of all four walls is -8(2) 2 + 48(2) = 64 square inches. So, both are 64 square inches.

= 0

4( x 3 - 27) = 0 x = 3 S ¢¢( x) = 4 + = 4+

The box will have maximum volume when x = 23 ft or 8 in.

(12 - 2 x)(12 - 2 x) = 4 x 2 - 48x + 144 and

108 = 2x + x

This implies that V is maximized when x = 23 .

26. (a) From Example 3, the area of the base is

S ¢( x ) = 4 x -

2 or x = 3 3

(b) From Exercise 25, the are of the base is (3 - 2 x)(8 - 2 x) = 4 x 2 - 22 x + 24 and

108(2) x3 216 x3

the total area of all four walls is 2 x(3 - 2 x)

> 0 since x > 0

So x = 3 minimizes the surface material. If x = 3,

+ 2 x(8 - 2 x) = -8 x 2 + 22 x. Since the box

has maximum volume when x = 23 , the area 2

( ) - 22( 23 ) + 24 = 1009

of the base is 4 23

square feet and the total area of all four walls 18

18 h= 2 = = 2. 9 x

( ) + 22( 23 ) = 1009 square feet.

is -8 23

The dimensions are 3 ft by 6 ft by 2 ft.

So, both are 100 square feet. 9

420

Chapter 6 APPLICATIONS OF THE DERIVATIVE (c) Based on the results from parts (a) and (b), it appears that the area of the base and the total area of the walls for the box with maximum volume are equal. (This conjecture is true.)

Also, x + 2 = the width of a page and

y + 3 = the length of a page.

The area of a page is A = ( x + 2)( y + 3)

27. Let

the length of a side of the top and bottom.

Then

the area of the top and bottom

and

(3)(2x2)

the cost for the top and bottom

= xy + 2 y + 3x + 6 æ 36 ö = 36 + 2 çç ÷÷÷ + 3x + 6 çè x ø = 42 +

72 + 3x. x

72 A¢ = - 2 + 3 = 0 x

Let

depth of box.

x 2 = 24

Then

the area of one side,

x =

4xy

the total area of the sides.

(1.50)(4xy)

The cost of the sides

and

= 2 6

(We discard x = -2 6 since must have x > 0.) A¢¢ = 216 > 0 when x = 2 6, which implies 3 x

The total cost is C ( x) = (3)(2 x 2 ) + (1.50)(4 xy) = 6 x 2 + 6 xy.

The volume is

that A is minimized when x = 2 6. y =

V = 16,000 = x 2 y. y =

96, 000 x2

36 36 18 18 6 = = = =3 6 x 6 2 6 6

The width of a page is

16,000 2

x+2= 2 6 +2

æ 16, 000 ö÷ 96, 000 C ( x) = 6 x 2 + 6 x çç ÷ = 6x2 + x èç x 2 ÷ø C ¢( x) = 12 x -

» 6.9 in.

The length of a page is

y+3= 3 6 +3 » 10.3 in.

x3 = 8000 x = 20 > 0 at x = 20, which C ¢¢ x      3 x

implies that C ( x) is minimized when x = 20. y =

16,000 (20)2

= 40

29. (a) S = 2 r 2 + 2 rh, V =  r 2h 2V S = 2 r 2 + r Treat V as a constant. 2V S ¢ = 4 r - 2 r

So the dimensions of the box are x by x by y, or 20 cm by 20 cm by 40 cm. 96,000 C (20) = 6(20) 2 + = 7200 20 The minimum total cost is $7200. 28. Let x = the width of printed material and y = the length of printed material.

4 r -

2V r2

4 r 3 - 2V r2

=0 =0

4 r 3 - 2V = 0 2 r 3 - V = 0

2 r 3 = V 2 r 3 =  r 2h 2r = h

Section 6.2

421

30. V =  r 2h = 16 16 h= r2

The total cost is the sum of the cost of the top and bottom and the cost of the sides. C = 2(2)( r 2 ) + 1(2 rh) æ 16 ö = 4( r 2 ) + 1(2 r ) çç 2 ÷÷÷ çè  r ø 32 r

= 4 r 2 +

Minimize cost. C ¢ = 8 r 8 r -

32 r

32 r2

32. As in Exercise 13, the dimensions of the lot are x meters perpendicular to the river and 1400 - 2 x meters along the river. The number sold will be half the river frontage, or 700 - x; since this number must be an integer, x must be an integer. The revenue will be the number sold times the lot area in square meters times the price per square meter, or r( x ) = (700 - x )( x )(1400 - 2 x )(0.50)

8 r 3 = 32

 r3 = 4 r = 3



» 1.08 h=

 (1.08) 2

Since r¢¢( x ) = -180 x + 760 æ 94 ö and r¢¢ çç ÷÷÷ = -1120 < 0, çè 9 ø 94/9 represents a local maximum for r. Since x must be a multiple of 0.5, we evaluate r at x = 10 and x = 10.5. r(10) = 28,800 r(10.5) = 28,906.25 Check the value of r at the endpoints of the domain of r, which are also possible locations for an absolute maximum. r(0) = 2000 and r(16.5) = 1711.25 Thus the maximum revenue of $28,906.25 is obtained when the carpets have length and width equal to 2 + 10.5 = 12.5 feet.

» 4.34

= x(700 - x )2 .

The radius should be 1.08 ft and the height should be 4.34 ft. If these rounded values for the height and radius are used, the cost is $2(2)( r 2 ) + $1(2 rh) = 4 (1.08) 2 + 2 (1.08)(4.34) = $44.11.

31. Let x be the number of feet over the minimum 2foot length and width. The revenue r( x ) is then the number sold times the area in square feet times the price per square foot: r( x ) = (100 - 6 x )(2 + x )2 (5) = -30 x 3 + 380 x 2 + 1880 x + 2000 r ¢( x ) = -90 x 2 + 760 x + 1880 = (-10)( x + 2)(9 x - 94) The domain of both r and r ¢ is [0, 16.5] since 2-by2 is the minimum size and the number sold must be nonnegative. The only positive root of r ¢( x ) is 94/9 » 10.444.

r ¢( x ) = (1)(700 - x )2 + ( x )(-2)(700 - x ) = (700 - x )(700 - 3x ) ¢¢ r ( x ) = 6 x - 2800 r¢ has two positive roots, 700 and 700/3 » 233.333. Evaluate r¢¢ at each of these roots: r¢¢(700) = 1400 r ¢¢(700/3) = -1400 < 0 Thus r has a local maximum at x = 700/3. We should also check the endpoints of the domain of r, which are x = 0 and x = 700, but clearly the revenue is 0 at both of these points. Thus x = 700/3 delivers the maximum revenue. Since x must be an integer, evaluate r( x ) at x = 233 and x = 234. r(233) = 50,814,737 r(234) = 50,814,504 Thus the maximum revenue of $50,814,737 occurs when the lot dimensions are 233 meters by 1400 - 2(233) = 934 meters. The number sold will be 700 - 233 = 467.

422

Chapter 6 APPLICATIONS OF THE DERIVATIVE

33. From Example 4, we know that the surface area of the can is given by S = 2 r 2 +

34. In Exercise 33, we found that the cost of the aluminum to make the can is æ 2000 ö÷ 60 2 0.03çç 2 r 2 + . ÷ = 0.06 r + çè r ø÷ r

2000 . r

2 , so the cost of the Aluminum costs 3c/cm / aluminum to make the can is

æ 2000 ö÷ 60 2 0.03çç 2 r 2 + . ÷ = 0.06 r + çè r ÷ø r

The perimeter (or circumference) of the circular charge to seal top is 2 r. Since there is a 2c/cm / the top and bottom, the sealing cost is

The cost for the vertical seam is 0.01h. From Example 4, we see that h and r are related by the equation h=

= 0.06 r 2 + 60r -1 + 0.08 r.

y = 0.12 x -

1000

 (5.206)

10 -2 r .



C ¢( x) = 0.12 r - 60r -2 or 0.12 r -

60 r

20 -3 r



 r3

Graph y = 0.12 r -

60 r

 r3

on a graphing calculator. Since r must be positive, our window should not include negative values of x. A suitable choice for the viewing window is [0, 10] by [-10, 10]. From the graph, we find that C ¢ x  =  when x » 5.454. Thus, the cost is minimized when the radius is about 5.454 cm. We can find the corresponding height by using the equation.

r2

from Example 4. If r = 5.206. h=

60 10 + 2 r r

Then

+ 0.08 2

on a graphing calculator. Since r must be positive in this application, our window should not include negative values of x. A suitable choice for the viewing window is [0,10] by [-10, 10]. From the graph, we find that C ¢ x  =  when x » 5.206. Thus, the cost is minimized when the radius is about 5.206 cm. We can find the corresponding height by using the equation

or 0.06 r 2 + 60r -1 +

60 x

r2

C (r ) = 0.06 r 2 +

-2

Graph

Thus, the total cost is given by the function

Then C ¢(r ) = 0.12 r - 60r + 0.08 60 = 0.12 r - 2 + 0.08 . r

æ 1000 ö 0.01h = 0.01çç 2 ÷÷÷ çè  r ø

Thus, the total cost is given by the function 60 + 0.08 r r

r2

so the sealing cost is

0.02(2)(2 r ) = 0.08 r.

C (r ) = 0.06 r 2 +

1000

» 11.75. 2

To minimize cost, the can should have radius 5.206 cm and height 11.75 cm.

h= h=

1000

r2 1000

 (5.454) 2

» 10.70.

To minimize cost, the can should have radius 5.454 cm and height 10.70 cm.

Section 6.2

423

35. In Exercise 33 and 34, we found that the cost of the

aluminum to make the can is 0.06 r

+ 60 , the r

cost to seal the top and bottom is 0.08 r , and the cost to seal the vertical seam is

10 . Thus, the

 r2

total cost is now given by the function 60 10 C (r ) = 0.06 r 2 + + 0.08 r + r  r2 2

or 0.06 r + 60r

-1

10 -2 r . + 0.08 r +



If r = 0, there is no box, so we discard this value. V ¢¢   - 12 r < 0 for r = 10, which implies that r = 10 gives maximum volume. When r = 10, h = 30 - 2(10) = 10. The volume is maximum when the radius and height are both 10 cm. 37. Distance on shore: 9 - x miles

Cost on shore: $400 per mile Distance underwater:

Then C ¢(r ) = 0.12 r - 60r -2 + 0.80 or 0.12 r -

60 r2

+ 0.08 -

20 -3 r



 r3

Graph y = 0.12 r -

60 r

+ 0.08 -

 r3

on a graphing calculator. A suitable choice for the viewing window is [0, 10] by [–10, 10]. From the graph, we find that C ¢( x) = 0 when x » 5.242. Thus, the cost is minimized when the radius is about 5.242 cm. To find the corresponding height, use the equation 1000 h= r2 from Example 4. If r = 5.242, 1000 » 11.58. h=  (5.242) 2 To minimize cost, the can should have radius 5.242 cm and height 11.58 cm. 36. 120 centimeters of ribbon are available; it will cover 4 heights and 8 radii. 4h + 8r = 120 h + 2r = 30 h = 30 - 2r V =  r 2h V =  r 2 (30 - 2r ) = 30 r 2 - 2 r 3

Maximize volume. V ¢ = 60 r - 6 r 2 60 r - 6 r 2 = 0 6 r (10 - r ) = 0 r = 0 or r = 10

x 2 + 36

Cost underwater: $500 per mile Find the distance from A, that is, (9 - x), to minimize cost, C ( x). æ ö C ( x) = (9 - x)(400) + çç x 2 + 36 ÷÷÷ (500) è ø = 3600 - 400 x + 500( x 2 + 36)1 2 æ1ö C ¢( x) = -400 + 500 çç ÷÷÷ ( x 2 + 36)-1 2 (2 x) çè 2 ø = -400 +

500 x x 2 + 36

If C ¢( x) = 0, 500 x

= 400 x 2 + 36 5x = x 2 + 36 4 25 2 x = x 2 + 36 16 9 2 x = 36 16 64 x = = 8. 3 (Discard the negative solution.) Then the distance should be 9- x = 9-8 = 1 mile from point A.

38. Distance on shore: 7 - x miles

Cost on shore: $400 per mile Distance underwater:

x 2 + 36

Cost underwater: $500 per mile Find the distance from A, that is, 7 - x. to minimize cost, C ( x).

424

Chapter 6 APPLICATIONS OF THE DERIVATIVE C ( x) = (7 - x)(400) + ( x 2 + 36)(500) = 2800 - 400 x + 500( x 2 + 36)1/2 æ1ö C ¢( x) = -400 + 500 çç ÷÷÷ ( x 2 + 36)1/2 (2 x) çè 2 ø = -400 +

500 x

p(t ) = 10te-t /8 , [0, 40] æ 1ö p¢(t ) = 10te-t /8 çç - ÷÷÷ + e-t /8 (10) çè 8 ø æ t ö = 10e-t /8 çç - + 1÷÷÷ çè 8 ø

Critical numbers:

x 2 + 36

p¢(t ) = 0 when

If C ¢( x) = 0, 500 x

40. (a)

= 400

x + 36 5 x = x 2 + 36 4 25 x 2 = x 2 + 36 16 9 x 2 = 36 16 x 2 = 36 ⋅ 16 9 6 4 = 8.  x = 3 (Discard the negative solution.) x = 8 is impossible since Point A is only 7 miles from point C. Check the endpoints. x C ( x) 0 5800 7 4610

The cost is minimized when x = 7.

t +1 = 0 8 t = 8.

p(t )

8 29.43 40

2.6952

The percent of the population infected reaches a maximum in 8 days. (b) P(8) = 29.43% æ t æ t öö 41. N (t ) = 20 ççç - ln çç ÷÷÷ ÷÷÷ + 30; çè 12 ø ø÷ è 12 1 £ t £ 15

é 1 12 æç 1 ö÷ ùú N ¢(t ) = 20 ê ç ÷ ê 12 t èç 12 ø÷ úû ë æ1 1ö = 20 çç - ÷÷÷ çè 12 tø

7 – x = 7 – 7 = 0, so the company should angle the cable at Point A.

20(t - 12) 12t

N ¢(t ) = 0 when t - 12 = 0 t = 12.

20t 3 - t 4 , [0, 20] 1000 3 2 1 3 (a) p¢(t ) = t t 50 250 1 2é 1 ù = t ê3 - t ú ê 50 ë 5 úû

39. p(t ) =

N ¢¢t  does not exist at t = 0, but 0 is not in the domain of N. Thus, 12 is the only critical number. To find the absolute extrema on [1, 15], evaluate N at the critical number and at the endpoints. t N (t ) 1 81.365 12 50 15 50.537

Critical numbers: 1 2 1 t = 0 or 3 - t = 0 50 5 t = 0 or t = 15 t p(t ) 0 0 15 16.875 20 0

Use this table to answer the questions in (a)-(d).

The number of people infected reaches a maximum in 15 days.

(a) The number of bacteria will be a minimum at t = 12, which represents 12 days.

(b) P(15) = 16.875%

Section 6.2

425

(b) The minimum number of bacteria is given by N (12) = 50, which represents 50 bacteria per mL.

S 2 + 4S - 46 = 0 -4  16 + 184 2 = 5.071.

S =

(c) The number of bacteria will be a maximum at t = 1, which represents 1 day. (d) The maximum number of bacteria is give by N(1) = 81.365, which represents 81.365 bacteria per mL. 42.

(Discard the negative solution.) The number of creatures needed to sustain the population is S0 = 5.071 thousand. H ¢¢ =

H (S ) = f (S ) - S f (S ) = 12S 0.25

(S + 2)2 (-2S - 4) - (S 2 - 4S + 46)(2S + 4) (S + 2)4

so H is a maximum at S0 = 5.071.

H (S ) = 12S 0.25 - S

25(5.071) - 5.071 7.071 » 12.86

H ¢(S ) = 3S -0.75 - 1

H ( S0 ) =

H ¢(S ) = 0 when

The maximum sustainable harvest is 12.86 thousand.

3S -0.75 - 1 = 0 1 S -0.75 = 3 1 1 = 3 S 0.75

44. (a)

S = 34/3 S = 4.327.

Note that f (S ) = Ser (1- S /P) æ rö f ¢(S ) = Ser (1- S /P) çç - ÷÷÷ + er (1- S /P). çè P ø

The number of creatures needed to sustain the population is S0 = 4.327 thousand. < 0 when S = 4.327, so H (S ) H ¢¢(S ) = -2.25 1.75 S

is maximized. H (4.327) = 12(4.327)0.25 - 4.327 » 12.98

The maximum sustainable harvest is 12.98 thousand. H (S ) = f (S ) - S 25S f (S ) = S +2  S +  - S H ¢ S  = -1 (S + 2)2 = = =

H (S ) = f (S ) - S = Ser (1- S /P) - S æ rö H ¢(S ) = Se r (1- S /P) çç - ÷÷÷ + er (1- S /P) - 1 çè P ø

S 3/4 = 3

43.

25S + 50 - 25S - (S + 2)2

æ rö H ¢(S ) = Se r (1- S /P) çç - ÷÷÷ + er (1- S /P) - 1 = 0 çè P ø æ rö Ser (1- S /P) çç - ÷÷÷ + er (1- S /P) = 1 çè P ø f ¢ S  = 

(b)

f (S ) = Ser (1- S /P) æ rö f ¢(S ) = çç - ÷÷÷ Ser (1- S /P) + er (1- S /P) çè P ø

Set f ¢(S0 ) = 1 é -rS0 ù er (1- S0 /P) ê + 1ú = 1 êë P úû er (1- S0 /P) = -rS

(S + 2) 2 50 - (S 2 + 4S + 4) (S + 2)2

< 0,

1 0

Using H (S ) from part (a), we get

-S 2 - 4S + 46 (S + 2) 2

H ¢ S  =  when

426

Chapter 6 APPLICATIONS OF THE DERIVATIVE H ( S0 ) = S0e r (1- S0 /P) - S0

(

= S0 e r (1- S0 /P ) - 1

47. Let x = distance from P to A.

)

æ ö÷ ç 1 ÷÷. 1 = S0 ççç ÷÷÷ çè 1 - rS0 ø P

45. r = 0.1, P = 100 f (S ) = Ser (1- S /P) 1 f ¢(S ) = ⋅ Se0.1(1- S /100) + e0.1(1- S /100) 1000 f ¢(S0 ) = -0.001S0e0.1(1- S0 /100) + e0.1(1- S0 /100)

Energy used over land: 1 unit per mile Energy used over water: 43 units per mile Distance over land: (2 - x) mi Distance over water: 1 + x 2 mi

Graph Y1 = -0.001xe0.1(1- x /100) + e0.1(1- x /100)

Find the location of P to minimize energy used. 4 1 + x 2 , where 0 £ x £ 2. 3 4æ1ö E ¢( x) = -1 + çç ÷÷÷ (1 + x 2 )-1/2 (2 x) 3 çè 2 ø

E ( x) = 1(2 - x) +

and Y2 = 1

on the same screen. A suitable choice for the viewing window is [0, 60] by [0.5, 1.5] with Xscl = 10 and Yscl = 0.5. By zooming or using the “intersect” option, we find the graphs intersect when x » 49.37. The maximum sustainable harvest is 49.37. 46. r = 0.4, P = 500 f (S ) = Ser (1- S /P) 0.4 0.4(1- S /500) f ¢(S ) = Se + e0.4(1- S /500) 500

If E ¢( x) = 0, 4 x(1 + x 2 )-1/2 = 1 3 4x =1 3(1 + x 2 )1/2 4 x = (1 + x 2 )1/2 3 16 2 x = 1 + x2 9 7 2 x =1 9 9 x2 = 7 3 3 7 . x = = 7 7

f ¢(S0 ) = -0.0008S0e0.4(1- S0 /500) + e0.4(1- S0 /500)

Graph Y1 = -0.0008 x e0.4(1- x /500) + e0.4(1- x /500)

and Y2 = 1

on the same screen. A suitable choice for the viewing window is [0, 300] by [0.5, 1.5] with Xscl = 50, Yscl = 0.5. By zooming or using the “intersect” option, we find that the graphs intersect when x » 237.10. The maximum sustainable harvest is 237.10.

0 1.134 2

E ( x) 3.3333 2.8819 2.9814

The absolute minimum occurs at x » 1.134. Point P is 3 77 » 1.134 mi from Point A.

Section 6.2

427

48. Let x = distance from P to A.

49. (a)

f (S ) = aSe-bS

f (S ) = Ser (1- S /P) = Se r -rS /P = Se r e-rS /P = e r Se-(r /P) S

Comparing the two terms, replace a with e r and b with r /P.

(b) Shepherd:

Energy used over land: 1 unit per mile Energy used over water: 10 units per mile 9 Distance over land: (2 - x) mi Distance over water:

f (S ) = f ¢( S ) =

1 + x 2 mi

aS 1 + (S /b)c [1 + (S /b)c ](a) - (aS )[c(S /b)c-1(1/b)] [1 + (S /b)c ]2 a + a(S /b)c - (acS /b)(S /b)c-1 [1 + (S /b)c ]2

Find the location of P to minimize energy used. 10 1 + x 2 , where 0 £ x £ 2. 9 10 æç 1 ö÷ 2 -1/2 (2 x) E ¢ ( x ) = -1 + ç ÷ (1 + x ) 9 çè 2 ÷ø

E ( x) = 1(2 - x) +

a + a(S /b)c - ac(S /b)c [1 + (S /b)c ]2

a[1 + (1 - c)(S /b)c ] [1 + (S /b)c ]2

If E ¢( x) = 0, 10 x(1 + x 2 )-1/2 = 1 9 10 x =1 9(1 + x 2 )1/2 10 x = (1 + x 2 )1/2 9 100 2 x = 1 + x2 81 19 2 x =1 81 81 x2 = 19 9 x = 19 9 19 = 19 » 2.06.

E ( x)

0 3 19 » 3.1111 2

2.4845

f (S ) = aSe-bS f ¢(S ) = ae-bS + aSe-bS (-b) = ae-bS (1 - bS )

Berverton-Holt: f (S ) = f ¢( S ) = = =

aS 1 + (S /b)

[1 + (S /b) ](a) - aS (1/b) [1 + (S /b)]2 a + a(S /b) - a(S /b) [1 + (S /b)]2 a [1 + (S /b)]2

This value cannot give the absolute maximum since the total distance from A to L is just 2 miles. Test the endpoints of the domain. x

Ricker:

f ¢(0) =

a[1 + (1 - c)(0/b)c ] [1 + (0/b)c ]2

= a

Ricker: f ¢(0) = ae-b(0)[1 - b(0)] = a

Beverton-Holt: f ¢(0) =

a [1 + (0/b)]2

= a

428

Chapter 6 APPLICATIONS OF THE DERIVATIVE The constant a represents the slope of the graph of f (S ) at S = 0.

æ abS ö¢ f ¢( S ) = çç - rS ÷÷÷ çè b + S ø

(d) First find the critical numbers by solving f ¢(S ) = 0.

Shepherd:

f ¢( S ) = 0 a[1 + (1 - c)(S /b)c ] = 0 (1 - c)(S /b) = -1

ab2

(c - 1)(S /b)c = 1

(b + S ) 2

æ 1 ÷ö1/3.24 » 193.914 S = 248.72 çç çè 2.24 ÷÷ø

Using the Shepherd model, next year’s population is maximized when this year’s population is about 194,000 tons. This can be verified by examing the graph of f (S ). (e) First find the critical numbers by solving f ¢(S ) = 0.

Ricker: f ¢( S ) = 0 (1 - bS ) = 0 1 - bS = 0 1 b

Substitute b = 0.0039 and solve for S. 1 0.0039 S » 256.410 S =

Using the Ricker model, next year’s population is maximized when this year’s population is about 256,000 tons. This can be verified by examining the graph of f (S ).

( a / r - 1)

-2ab3

< 0 everywhere (b + S )3 on the domain of f , assuming a and b are positive, there is a local maximum of f at

1/3.24

- rS S 1+ b Find the critical numbers:

= r

Since f ¢¢(S ) =

æ 1 ö÷ S = çç çè 2.24 ÷ø÷ 248.72

-r

S = -b + b a / r = b

æ S ö÷3.24 1 çç = çè 248.72 ø÷÷ 2.24

50. f (S ) =

(b + S ) 2

(b + S ) 2 =

(3.24 - 1)( S /248.72)3.24 = 1

S =

ab2

-r

ab2 r b + S = b a / r (we assume b > 0)

Substitute b = 248.72 and c = 3.24 and solve for S.

(b + S ) 2

f ¢(S ) = 0 when

-bS

(ab)(b + S ) - (abS )(1)

S = b

51.

( a / r - 1 ).

The goal is to minimize surface area for a fixed volume, that is, to minimize 2 r 2 + 2 rh when  r 2h = 67. h = 672 so we can write the surface as a r function of r alone: æ ö S (r ) = 2 r 2 + 2 r çç 672 ÷÷ çè  r ÷ø = 2 r 2 + 134 r Find the critical numbers. S ¢(r ) = 4 r - 134 r2 S ¢(r ) = 0 implies 4 r = 134 r2 r 3 = 134 4 æ ö1/3 r = çç 134 ÷÷÷ » 2.2 è 4 ø

æ 67 ö The formula S (r ) = 2 r 2 + 2 r çç 2 ÷÷÷ èç  r ø

shows that letting r go to 0 and increasing r without limit both increase the surface area without limit, so the critical value found above represents a local minimum for the surface area. r = 2.2 implies Copyright © 2022 Pearson Education, Inc.

Section 6.2

429

67 » 4.4  (2.2)2 Thus the radius and height of the cell of minimum surface area are 2.2  m and 4.4  m. h=

52. (a) Solve the given equation for effective power for T, time.

The rate on the river is 5 mph, the rate on land is 2 mph. Using t = dr ,

kE = aSv3 + I T kE aSv3 + I

8- x = the time on the river, 5

Since distance is velocity, v, times time, T, we have D(v) = v

(b) D¢(v) = = =

kE aSv3 + I kEv 3

aSv + I

9 - x2 = the time on land. 2

The total time is

8- x 9 + x2 + 5 2 8 1 1 = - x + (9 + x 2 )1/2. 5 5 2 1 1 T ¢ = - + ⋅ 2 x(9 + x 2 )-1/2 5 4

T ( x) =

(aSv3 + I )kE - kEv(3aSv 2 ) (aSv3 + I ) 2 kE (aSv3 + I - 3aSv3 ) (aSv3 + I ) 2

kE ( I - 2aSv )

1 x + = 0 5 2(9 + x 2 )1/2

(aSv3 + I )2

Find the critical numbers by solving D¢(v) = 0 for v.

1 x = 5 2(9 + x 2 )1/2 2(9 + x 2 )1/2 = 5x

I - 2aSv3 = 0

4(9 + x 2 ) = 25x 2

2aSv3 = I

36 + 4 x 2 = 25x 2

v3 =

36 = 21x 2

I 2aS

6 = x 21

1/3

æ I ö÷ v = çç çè 2aS ÷÷ø

53. Let Then

6 21 2 21 = = x 21 7

8 - x = the distance the hunter will travel on the river. 9+x

= the distance he will travel on land.

T ( x)

3.1

2 21 7

2.98

4.27

Since the maximum time is 2.98 hr, the hunter should travel 8 - 2 721 = 56-72 21 or about 6.7 miles along the river.

430

Chapter 6 APPLICATIONS OF THE DERIVATIVE

54. Let 8 - x = the distance the hunter will travel on the river.

T ( x) is minimized when x = 8.

The distance along the river is given by 8 - x, so the hunter should travel 8 - 8 = 0 miles along the river. He should complete the entire trip on land. 55. Let x = width. Then x = height and 108 - 4 x = length.

(since length plus girth = 108 ) V ( x) = l ⋅ w ⋅ h = (108 - 4 x) x ⋅ x

Then

192 + x 2 = the distance he will travel on land.

Since the rate on the river is 5 mph, the rate on land is 2 mph, and t = dr ,

V ¢( x) = 216 x - 12 x 2

Set V ¢( x) = 0, and solve for x.

8- x = the time on the river 5

216 x - 12 x 2 = 0

361 + x 2 = the time on the land. 2

x = 0 or x = 18

The total time is 8-x 361 + x 2 + T ( x) = 5 2 8 1 1 = - x + (361 + x 2 )-1/2. 5 5 2 1 1 T ¢( x) = - + ⋅ 2 x(361 + x 2 )-1/2 5 4 -

= 108x 2 - 4 x3

x 1 + =0 5 2(361 + x 2 )1/2 1 x = 5 2(361 + x 2 )1/2 2(361 + x 2 )1/2 = 5 x 4(361 + x 2 ) = 25x 2 1444 + 4 x 2 = 25x 2 1444 = 21x 2 38 = x 21 8.29 = x

8.29 is not possible, since the cabin is only 8 miles west. Check the endpoints. x T ( x) 0 11.1 8 10.3

12 x(18 - x) = 0

0 is not in the domain, so the only critical number is 18. Width = 18 Height = 18 Length = 108 - 4(18) = 36

The dimensions of the box with maximum volume are 36 inches by 18 inches by 18 inches. 56. (a) The area of the semicircle with radius r is 1 Ac =  r 2 , where 0 £ r £ 10. 2 If the width of the rectangle is 20 m, then the total length of the bases of triangles is 20 – 2r. So the base of one triangle is 10 – r. If the area of one triangle is 1 1 At = bh = (10 - r )10, 2 2 then the area of both triangles is twice that, A = 10(10 - r ). Thus, the area of the curtain is 1 A(r ) =  r 2 + 10(10 - r ), 2 for 0 £ r £ 10 (b) Find the radius that will minimize the area. 1 A(r ) =  r 2 + 10(10 - r ) 2 A¢(r ) =  r - 10

Section 6.2

431

Set A¢(r ) = 0 and solve for r.  r - 10 = 0  r = 10 10 r =



A radius of 10 /  m will minimize the covered area. (c) Find the arc length of the semicircle. We use the formula for half a circle’s circumference and substitute the value for r, 10 /  m . 10 C = r =  = 10



The arc length of the semicircle is 10 m. The height of the stage is 10m. They are equal. (d) The area of 2 semicircles with radius r is Ac =  r 2 , where 0 £ r £ 5. If the width of the rectangle is 20 m, then the total length of the bases of triangles is 20 – 4r. So the base of one triangle is 10 – 2r. If the area of one triangle is 1 1 At = bh = (10 - 2r )10, 2 2 then the area of both triangles is twice that, A = 10(10 - 2r ). Thus, the area of the curtain is A(r ) =  r 2 + 10(10 - 2r ), for 0 £ r £ 5

Find the radius that will minimize the area. A(r ) =  r 2 + 10(10 - 2r ) A¢(r ) = 2 r - 20

(e) The area of 3 semicircles with radius r is 3 10 Ac =  r 2 , where 0 £ r £ . 2 3 If the width of the rectangle is 20 m, then the total length of the bases of triangles is 20 – 6r. So the base of one triangle is 10 – 3r. If the area of one triangle is 1 1 At = bh = (10 - 3r )10, 2 2 then the area of both triangles is twice that, A = 10(10 - 3r ). Thus, the area of the curtain is 3 A(r ) =  r 2 + 10(10 - 3r ), 2 10 for 0 £ r £ 3

Find the radius that will minimize the area. 3 A(r ) =  r 2 + 10(10 - 3r ) 2 A¢(r ) = 3 r - 30 Set A¢(r ) = 0 and solve for r. 3 r - 30 = 0 3 r = 30 10 r =



A radius of 10 /  m will minimize the covered area. Find the arc length of each semicircle. We use the formula for half a circle’s circumference and substitute the value for r, 10 /  m . 10 C = r =  = 10



Set A¢(r ) = 0 and solve for r. 2 r - 20 = 0 2 r = 20 10 r =

The arc length of each semicircle is 10 m. Thus, the length of each semicircle, 10 m, should equal the stage height, 10 m.



The arc length of each semicircle is 10 m.

432

6.3

Chapter 6 APPLICATIONS OF THE DERIVATIVE

Further Business Applications: Economic Lot Size; Economic Order Quantity; Elasticity of Demand

Your Turn 1

Use Equation (3) with k = 3, M = 18, 000 and f = 750.

2(750)(18,000) 3

(900)(10,000)

Your Turn 2

Use Equation (3) with k = 2, M = 320 and f = 30.

80 p e-0.4 p

q = 3600 - 3 p 2 dq = -6 p dp p dq E =q dp =

2(30)(320) 2

= 9600 » 97.98

Since q is very close to 98 we expect a q of 98 to minimize costs, but we will check both 97 and 98 fM q

Your Turn 5

To minimize production costs there should be 3000 cans per batch, requiring 18, 000/3, 000 = 6 batches per year.

using T (q) =

dq = -80e-0.4 p dp p dq E =q dp

For p = 100 we have E = 0.4(100) = 40, which corresponds to elastic demand.

= 3000

q =

q = 200e-0.4 p

200e-0.4 p = 0.4 p

2fM k

q =

Your Turn 4

6 p2 3600 - 3 p 2 2 p2 1200 - p 2

The demand has unit elasticity when E = 1. 2 p2 1200 - p 2

+ 2 . T (97) » 195.969 and T (98)

2 p 2 = 1200 - p 2

» 195.959, so the company should order 98 units in each batch. The time between orders will be about 12

3 p 2 = 1200

( ) = 3.675 months. 98 320

Your Turn 3 q = 24,000 - 3 p

p 2 = 400 p = 20

Testing value for p smaller and larger than 20 (say 15 and 25) we find

dq = -6 p dp E =-

E (15) < 1 E (25) > 1

p dq 6 p2 = q dp 24,000 - 3 p 2

When p = 50, E =

6(50 )

Thus demand is inelastic when p < 20 and elastic when p > 20. Revenue is maximized at the price corresponding to unit elasticity, which is p = $20. At this price the revenue is pq = (20)[3600 - 3(202 )] = (20)(2400) = 48,000

24,000 - 3(502 ) » 0.909,

which corresponds to inelastic demand. or $48,000.

Section 6.3

433

6.3

Warmup Exercises

W1.

f ( x) =

k = kx-1 x

f ¢( x) = k (-1) x-2 = -

W2.

f ( x) =

a x2

k x2

2a x3

Exercises

1. True 2. True

the demand curve on the interval 0 £ p £ mn .

3. True 4. False. If the demand has unit elasticity at price p, then the total revenue is maximized at the price. Increasing the price will decrease the total revenue. 5. When q <

when q >

2fM , T ¢(q) < - k2 + k2 = 0; and k 2fM , T ¢(q) > - k2 + k2 = 0. Since k

the function T (q) is decreasing before q =

2fM k

and increasing after q =

2fM , there must be a k

relative minimum at q =

2fM . By the critical k

point theorem, there is an absolute minimum there. 7. The economic order quantity formula assumes that M, the total units needed per year, is known. Thus, c is the correct answer. 9. The demand function q( p) is positive and dq

increasing, so dp is positive. Since p0 and q0 are p

12. (a)

q = Cp-k dq = -Ckp-k -1 dp - p dq ⋅ E = q dp -p (-Ckp-k -1) = -k Cp =

kp-k p-k

= k

13. (a) Use equation (3) with k = 1, M = 100,000, and f = 500. q =

2fM = k

2(500)(100,000) 1

= 100,000,000 = 10,000 10,000 lamps should be made in each batch to minimize production costs.

(b) From (a), M = 100,000, and q = 10,000. The number of batches per year is = 100,000 = 10. 10,000

M q

also positive, the elasticity E = - q 0 ⋅ dp is 0

negative. 10. E = -

m n

dq = -n dp p dq E =- ⋅ q dp p (-n) E =m - np pn E = =1 m - np pn = m - np 2np = m m p = 2n m Thus, E = 1 when p = 2n , or at the midpoint of

= ax-2

f ¢( x) = a(-2) x-3 = -

6.3

11. q = m - np for 0 £ p £

14. (a) Use equation (3) with k = 9, M = 13,950, and f = 31. q =

p dq ⋅ q dp

dq Since p ¹ 0, E = 0 when dp = 0. The

derivative is zero, which implies that the demand function has a horizontal tangent line at the value of p where E = 0.

2fM k 2(31)(13,950) 9

= 96,100 = 310 310 cases should be made in each batch to minimize production costs.

434

Chapter 6 APPLICATIONS OF THE DERIVATIVE (b) From (a), M = 13,950 and q = 310. The number of batches per year is 13,950 M = = 45 q 310

15. (a) Here k = 0.50, M = 100,000, and f = 60. We have 2fM 2(60)(100,000) q = = k 0.50 = 24,000,000 » 4898.98 T (4898) = 2449.489792 and T (4899) = 2449.489743, so ordering 4899 copies per order minimizes the annual costs.

(b) From (a), M = 100,000 and q = 4899. The number of orders per year is 100, 000 M = » 20.4 times. q 4899 16. (a) Here k = 1, M = 900, and f = 5. We have q = =

Since lim T (q) = ¥, q0

q =

fM k

is the only critical value in (0, ¥),

q =

fM k

is the number of unit that should be

ordered or manufactured to minimize total costs. fM k

18. Use q =

2(5)(900) 1

= 9000 » 94.9 T (94) » 94.872 and T (95) » 94.868, so ordering 95 bottles per order minimizes the annual costs.

(b) From (a), M = 900 and q = 95. The number of orders per year is M = 900 » 9.5 times. q 95

M = 5000, and f = 1000. fM = k

q =

fM q

with T (q) =

+ kq (assume g = 0 since the

fM kq + gM + 1 + k2q; (0, ¥) 2 q k -f M T ¢( q ) = + 1 + k2 2 2 q T (q ) =

Set the derivative equal to 0. -f M q

k1 + k2 = 0 2

k1 + 2k2 fM = 2 2 q q 2 (k1 + 2k2 ) = fM 2 2f M q2 = k1 + 2k2

fM + gM + kq; (0, ¥) q -f M T ¢( q ) = +k q2 T ( q) =

q k = fM fM q2 = k fM q = k

(1000)(5000) » 912.9 6

subsequent cost per book is so low that it can be ignored), T (912) » 10,954.456 and T (913) » 10,954.451. So, 913 books should be printed in each print run.

17. Using maximum inventory size,

from Exercise 15 with k = 6,

19. Assuming an annual cost, k1, for storing a single unit, plus an annual cost per unit, k2, that must be paid for each unit up to the maximum number of units stored, we have

2f M k

Set the derivative equal to 0. -f M +k =0 q2 fM k = 2 q

lim T (q) = ¥, and

q ¥

q =

Since lim T (q) = ¥, q0

2f M k1 + 2k2

lim T (q) = ¥, and

q ¥

q =

2f M k1 + 2k2

is the only critical value in (0, ¥),

q =

2f M k1 + 2k2

is the number of units that should

be ordered or manufactured to minimize the total cost in this case.

Section 6.3

435 2f M k1 + 2k2

20. Use q =

from Exercise 19 with k1 = 1,

k2 = 2, M = 30, 000, and f = 750. Also, note that g = 8.

22. q = 25,000 - 50 p (a)

2fM k1 + 2k2

q =

p (-50) 25,000 - 50 p p = 500 - p =-

2(750)(30,000) 1 + 2(2)

9,000,000 = 3000

The number of production runs each year to minimize her total costs is

(b)

30,000 M = = 10. q 3000 21. q = 50 (a)

R = pq dR = q(1 - E ) dp

When R is maximum, q(1 - E ) = 0. Since q = 0 means no revenue, set 1 - E = 0.

p 4

E =1

dq 1 =dp 4 p dq E =- ⋅ q dp =-

dq = -50 dp p dq E =- ⋅ q dp

From part (a), p =1 500 - p p = 500 - p p = 250. q = 25,000 - 50 p

æ ö ç - 1 ÷÷ ÷ 4ø

pç 50 - 4 çè

p æ 1ö = - 200- p çç - ÷÷÷ çè 4 ø

= 25,000 - 50(250) = 12,500

p = 200 - p

(b)

Total revenue is maximized if q = 12,500.

R = pq dR = q(1 - E ) dp

23. (a)

When R is maximum, q(1 - E ) = 0. Since q = 0 means no revenue, set 1 - E = 0, so E = 1. From (a), p =1 200 - p p = 200 - p p = 100. p q = 50 4 100 = 50 4 = 25

q = 37,500 - 5 p 2 dq = -10 p dp - p dq ⋅ E = q dp -p = (-10 p) 37,500 - 5 p 2 ==

(b)

Total revenue is maximized if q = 25.

10 p 2 37,500 - 5 p 2 2 p2

7500 - p 2

R = pq dR = q(1 - E ) dp

When R is maximum, q(1 - E ) = 0. Since q = 0 means no revenue, set 1 - E = 0. E =1

436

Chapter 6 APPLICATIONS OF THE DERIVATIVE From (a),

q = 48, 000 - 10 p 2

2 p2 7500 - p

= 48, 000 - 10(40)2 = 48, 000 - 10(1600) = 48, 000 - 16, 000 = 32, 000

=1 2

2 p 2 = 7500 - p 2 3 p 2 = 7500 p 2 = 2500 p = 50.

25. p = 400e-0.2q dq

Since p must be positive, p = 50. q = 37,500 - 5 p

In order to find the derivative dp , we first need to solve for q in the equation p = 400e-0.2q . (a)

= 37,500 - 5(50)2 = 37,500 - 5(2500) = 37,500 - 12,500 = 25,000.

p = e-0.2q 400 æ p ÷ö = ln (e-0.2q ) = -0.2q ln çç çè 400 ÷÷ø p

q =

24. q = 48,000 - 10 p 2 (a)

dq = -20 p dp p dq E =- ⋅ q dp -p = (-20 p) 48,000 - 10 p 2 ==

(b)

20 p

æ p ö÷ = -5 ln çç çè 400 ÷÷ø -0.2

Now dq 1 1 -5 = -5 p ⋅ = , and dp 400 p 400

E =-

(b)

48, 000 - 10 p 2 2 p2

4800 - p

ln 400

p dq p -5 5 ⋅ =- ⋅ = . q dp q p q

R = pq dR = q(1 - E ) dp

When R is maximum, q(1 - E ) = 0. Since q = 0 means no revenue, set 1 - E = 0.

E =1

R = pq dR = q(1 - E ) dp

From part (a), 5 =1 q 5= q

When R is maximum, q(1 - E ) = 0. Since q = 0 means no revenue, set 1 - E = 0. E =1

26. q = 10 - ln p

From part (a), 2 p2 4800 - p

(a)

=1 2

2 p 2 = 4800 - p 2 3 p 2 = 4800 2

p = 1600 p = 40.

1 dq =dp p p dq E =- ⋅ q dp =

- p æç 1 ö÷ ç- ÷ 10 - ln p çè p ÷÷ø

1 10 - ln p

Since p must be positive, p = 40.

Section 6.3 (b)

437

R = pq dR = q(1 - E ) dp

28.

When R is maximum, q(1 - E ) = 0. Since q = 0 means no revenue, set 1 - E = 0. E =1

From part (a), 1 =1 10 - ln p 1 = 10 - ln p ln p = 9

q = 300 - 2 p dq = -2 dp p dq E =- ⋅ q dp p(-2) E =300 - 2 p 2p = 300 - 2 p

(a) When p = $100, 200 300 - 200 = 2.

E =

p =e q = 10 - ln p

Since E > 1, demand is elastic. This indicates that a percentage increase in price will result in a greater percentage decrease in demand.

= 10 - ln e9 = 10 - 9 =1

Note that E = 1q , thus we would expect E to

(b) When p = $50, 100 300 - 100 1 = < 1. 2

E =

be maximum when q = 1. 27.

q = 400 - 0.2 p 2 dq = 0 - 0.4 p dp p dq E =- ⋅ q dp P (-0.4 p) E =400 - 0.2 p 2 =

Since E < 1, supply is inelastic. This indicates that a percentage change in price will result in a smaller percentage change in demand. 29. D( p) = q = -25 p 2 + 63, 075 (a)

0.4 p 2 400 - 0.2 p 2

(a) If p = $20, E =

dq = -50 p dp p dq E =- ⋅ q dp E =-

(0.4)(20) 2

400 - 0.2(20) 2 = 0.5.

Since E < 1, demand is inelastic. This indicates that total revenue increases as price increases.

-25 p + 63, 075 50 p 2

-25 p 2 + 63, 075

(-50 p)

2 p2 - p 2 + 2523

(b) When p = $25, E =

(b) If p = $40, E =

p 2

(0.4)(40) 2

2 p2 - p 2 + 2523 2(25)2

-(25) 2 + 2523 » 0.66

400 - 0.2(40) 2 = 8.

Since E > 1, demand is elastic. This indicates that total revenue decreases as price increases.

Since E < 1, supply is inelastic. This indicates that an increase in price will result in an increase in total revenue. Yes.

438

Chapter 6 APPLICATIONS OF THE DERIVATIVE Since E < 1, supply is inelastic. This indicates that an increase in price will result in an increase in total revenue. Yes.

2 p2 - p 2 + 2523

2(35) 2

-(35)2 + 2523 » 1.89

E =

2 p2

400 - p » 2.57

Since E > 1, supply is elastic. This indicates that an increase in price will result in a decrease in total revenue. No.

Since q = 0 means no revenue, set 1 – E = 0, so E = 1. From (a), 2 p2 =1 400 - p 2

2 p 2 = - p 2 + 2523

2 p 2 = 400 - p 2

3 p = 2523

3 p 2 = 400

p 2 = 841 p = 29

400 3 p » 11.55

p2 =

Since p must be positive, p = 29. q = -25 p 2 + 63, 075

Since p must be positive, p = 11.55.

q = 400 - p 2

= -25(29) + 63, 075 = 42, 050

= 400 - (11.55) 2 » 266.5975 hundred or 26,659.75

Total revenue is maximized if q = 42,050. Calculate the revenue, R. R = pq = 29(42, 050) = 1, 219, 450

Total revenue is maximized if q = 26,659.75. Calculate the revenue, R. R = pq = 11.55(26659.75) = 307,920

The revenue is maximized when p = $29. The maximum revenue is $1,219,450.

31.

400 - p » 0.13

2, 431,129 p = 0.06

400 - p 2

= 2

p dq q dp

2 p2

q = 2, 431,129 p-0.06 dq = (-0.06)[2, 431,129 p-1.06 ] dp E =-

(b) When p = $5, E =

The revenue is maximized when p = $11.55. The maximum revenue is $307,920.

dq = -2 p dp p dq E =- ⋅ q dp p E =(-2 p) 400 - p 2 =

400 - (15)2

(d) When R is maximum, q(1 - E ) = 0.

Since q = 0 means no revenue, set 1 – E = 0, so E = 1. From (a), 2 p2 =1 - p 2 + 2523

(a)

2(15) 2

Since E > 1, supply is elastic. This indicates that an increase in price will result in a decrease in total revenue. No.

(d) When R is maximum, q(1 - E ) = 0.

30. D( p) = q = 400 - p

= 2

2(5)2

-0.06

(-0.06)[2, 431,129 p-1.06 ]

At any price (including $100/barrel) the elasticity is 0.06 and the demand is inelastic.

400 - (5)2

Section 6.3 32.

33.

439

q = Ap-0.13 dq = (-0.13)[ Ap-1.13 ] dp p dq E =q dp p = - -0.13 (-0.13)[ Ap-1.13 ] Ap = 0.13 At any price the elasticity is 0.13 and the demand for rice is inelastic.

When p = $166.10, E =

3.6542 » 0.071. 55.2 - 3.6542

(b) Since E < 1, the demand for airfare is inelastic at this price. (c)

R = pq dR = q(1 - E ) dp

When R is maximum, q(1 - E ) = 0. Since q = 0 means no revenue, set 1 - E = 0.

q = 3,751, 000 p-2.826 dq = (-2.826)[3,751,000 p-3.826 ] dp p dq E =q dp p =(-2.826) 3,751, 000 p-2.826 [3,751, 000 p-3.826 ] = 2.826

E =1

From (a), 0.022 p =1 55.2 - 0.022 p 0.022 p = 55.2 - 0.022 p 0.044 p = 55.2 p » 1255

At any price the elasticity is 2.826 and the demand is elastic. 34. q = 342.5 p-0.5314 dq = 342.5(-0.5314) p-0.5354-1 dp -182 = 1.5314 p p dq E =- ⋅ q dp -p -182 = ⋅ 342.5 p-0.5314 p1.5314 182 = 342.5 » 0.5314

Since E < 1, the demand is inelastic.

Total revenue is maximized if p » $1255. 36. (a) p = 0.604q 2 - 20.16q + 263.067 dq = 1.208q - 20.16 dp p dq p E =- ⋅ = dp q dp -q

( dq )

0.604q 2 - 20.16q + 263.067 -q(1.208q - 20.16)

(b) Since q = 11, 114.391 E = » 1.51 -11(-6.872) (c) Since E > 1, demand is elastic. (d) As q approaches 16.6887, the denominator of E approaches zero, so E approaches infinity.

35. (a) q = 55.2 - 0.022 p dq = -0.022 dp p dq E =- ⋅ q dp -p = ⋅ (-0.022) 55.2 - 0.022 p 0.022 p = 55.2 - 0.022 p

440

Chapter 6 APPLICATIONS OF THE DERIVATIVE

6.4

37.

Implicit Differentiation

Your Turn 1 5 x3 + 2 y 3 = 7 d d (5x3 + 2 y 3 ) = (7) dx dx dy 15 x 2 + 6 y 2 =0 dx dy 6 y2 = -15 x 2 dx dy 5x 2 =- 2 dx 2y

In the figure, we label P0 as P. The slope of the tangent line is OB - q0 dq BR OB - OR == = -RP - p0 dp RP

or dq = OB - q0 dp p dq OB - 0 ⋅ = -1 q0 dp q0 - p0

OB -1 OR

Because triangles AOB and PRB are similar,

Your Turn 2 xe y + x 2 = ln y

d d ( xe y + x 2 ) = ln y dx dx d d 2 d xe y + x = ln y dx dx dx dy 1 dy e y + xe y + 2x = dx y dx æ y ö çç xe - 1 ÷÷ dy = -2 x - e y çè y ø÷÷ dx ( xye y - 1)

p dq AB - 0 ⋅ = -1 q0 dp AP AB - AP AP PB = PA

dy ye y - 2 xy -2 xy - ye y = = dx 1 - xye y xye y - 1

p dq But E = - q 0 ⋅ dp so the ratio PB equals the PA 0

elasticity E. 38. (a) q( p) = mp + b The p and q intercepts are (-b / m, 0) and (0, b). (b) The midpoint of the line joining these intercepts is (-b /(2m), b /2). (c) The elasticity is p dq p = - ⋅ m. q dp q At the point found in (b) the elasticity is -b /(2m) ⋅ m = 1. b /2

dy = -2 xy - ye y dx

Your Turn 3 y 4 - x4 - y 2 + x2 = 0 d 4 d ( y - x4 - y 2 + x2 ) = (0) dx dx dy dy - 4 x3 - 2 y + 2x = 0 4 y3 dx dx dy = 4 x3 - 2 x (4 y 3 - 2 y) dx

At the point (1, 1), 4 y3 - 2 y ¹ 0 so we can divide both sides of the equation above by this factor. dy 4 x3 - 2 x = dx 4 y3 - 2 y

At (1, 1), dy 4 x3 - 2 x 4-2 = = =1 3 4 dx -2 4y - 2y

Section 6.4

441

so the slope of the tangent line is m = 1. Use the pointslope form of the equation of a line. y - y1 = m( x - x1)

y(-2) - (-2 x) d æç dy ö÷ dx ççè ÷÷ø = dx dx y2 d2y

y - 1 = 1( x - 1) y = x

dx 2

The equation of the tangent line at the point (1, 1) is y = x.

d2y dx

Your Turn 4 p =

100,000 q 2 + 100q

-1

-2 y + 2 x y2

æ ö çè y ø

2 y + 2 x ççç -2 x ÷÷÷÷ y

d d p = [100, 000 (q 2 + 100q)-1] dp dp 2q + 100 dq 1 = -100, 000 2 (q + 100q) 2 dp

æ ö çè y ø

dy -2 x = dx y

y y

⋅

2 y - 4x2 y3 -(4 x 2 - 2 y) y3

Since 4x 2 - 2 y = 5

5 =- 3 y

dq (q 2 + 100q) 2 =dp 100, 000 (2q + 100)

Substitute q = 200 in this expression for the derivative. 2

Since

2 y + 2 x ççç -2 x ÷÷÷÷

= 100,000(q + 100q)

dy dx

dq (q + 100q) =dp 100, 000 (2q + 100) [2002 + 100(200)]2 =(100, 000)[2(200) + 100] = -72

When the price is 200, the rate of change of demand with respect to price is –72 units per unit change in price. 4x2 + 2 y 2 = 5 Take the first derivative of both sides.

6.4

Warmup Exercises

W1.

f ( x) = x ln( x 2 + 1) Use the product rule and write the factor ln( x 2 + 1) as h( g ( x)) with h( x) = ln x and g ( x) = x 2 + 1 and apply the chain rule. 2x f ¢( x) = (1) ln( x 2 + 1) + ( x) 2 x +1 = ln( x 2 + 1) +

2x2 x2 + 1

W2. Use the quotient and chain rules, writing 3

e x as h( g ( x )) with h( x ) = e x and

d d (4 x 2 + 2 y 2 ) = (5) dx dx dy 8x + 4 y =0 dx dy -8 x -2 x = = dx 4y y

g ( x ) = x 3.

f ( x) =

Take the derivative of both sides; use the quotient rule.

x2 æ 2 x 3 ö÷ 2 æ 3ö çç 3x e ÷÷ x - çç e x ÷÷÷( 2 x ) è ø è ø

( ) 2 ( x2 ) e x ( 3x 4 - 2 x ) =

f ¢( x ) =

(

e x 3x 3 - 2

)

442

6.4

Chapter 6 APPLICATIONS OF THE DERIVATIVE

Exercises

7. 8 x 2 - 10 xy + 3 y 2 = 26

1. False. The equation xy 2 + 3 y - ln y = e xy is written implicitly in terms of y. 2. True 3. False. If r is a function of t, and 10 = r 2 + t 2 , then d d 2 (10) = (r + t 2 ) dt dt d d 2 d (10) = (r ) + (t 2 ) dt dt dt dr 0 = 2r + 2t dt 4. True 5. 6 x 2 + 5 y 2 = 36 d d (6 x 2 + 5 y 2 ) = (36) dx dx d d d (6 x 2 ) + (5 y 2 ) = (36) dx dx dx dy 12 x + 5  2 y =0 dx dy 10 y = -12 x dx 6x dy =5y dx

6. 7 x 2 - 4 y 2 = 24 d d (7 x 2 - 4 y 2 ) = (24) dx dx d d d (7 x 2 ) (4 y 2 ) = (24) dx dx dx dy 14 x - 8 y =0 dx dy 8y = 14 x dx 7x dy = 4y dx

d d (8 x 2 - 10 xy + 3 y 2 ) = (26) dx dx d d 16 x (10 xy ) + (3 y 2 ) = 0 dx dx dy d dy 16 x - 10 x - y (10 x) + 6 y =0 dx dx dx dy dy 16 x - 10 x - 10 y + 6 y =0 dx dx dy (-10 x + 6 y) = -16 x + 10 y dx dy -16 x + 10 y = dx -10 x + 6 y 8x - 5 y dy = 5x - 3 y dx

8. 7 x 2 = 5 y 2 + 4 xy + 1 d d (7 x 2 ) = (5 y 2 + 4 xy + 1) dx dx d 14 x = (5 y 2 + 4 xy + 1) dx d d d 14 x = 10 y y + 4 x ( y) + y (4 x) dx dx dx dy dy 14 x = 10 y + 4 x (1) + 4 y dx dx dy 14 x - 4 y = (10 y + 4 x) dx 14 x - 4 y dy = 10 y + 4 x dx 7x - 2 y dy = 2x + 5 y dx

9. 5x3 = 3 y 2 + 4 y d d (5x3 ) = (3 y 2 + 4 y) dx dx d d 15 x 2 = (3 y 2 ) + (4 y) dx dx dy dy 15 x 2 = 6 y +4 dx dx 15 x 2 dy = 6y + 4 dx

Section 6.4

443

10. 3x3 - 8 y 2 = 10 y

13. 2 x + 4 y = 5 y d d (2 x1/2 + 4 y1/2 ) = (5 y) dx dx dy dy x-1/2 + 2 y-1/2 =5 dx dx dy -1 2 (2 y - 5) = -x-1/2 dx

d d (3x3 - 8 y 2 ) = (10 y) dx dx d d dy (3x3 ) (8 y 2 ) = 10 dx dx dx dy dy = 10 9 x 2 - 16 y dx dx 9 x 2 = (16 y + 10)

dy x-1/2 çæ x1/2 y1/2 ö÷÷ çç = ÷ dx 5 - 2 y-1/2 çè x1/2 y1/2 ÷÷ø

dy dx

9x2 dy = 16 y + 10 dx

11. 3x 2 =

y1/2 = 1/2 1/2 x (5 y - 2)

2- y 2+ y

d d æç 2 - y ö÷ ÷÷ (3x 2 ) = ç dx dx çèç 2 + y ø÷ 6x =

6x =

(2 + y)

( )

dy dy - (2 - y) (2 + y) dx dx

(2 + y)2

d d (4 x1/2 - 8 y1/2 ) = 6 ( y 3/2 ) dx dx 3 dy -1/2 -1/2 dy - 4y = 6 ⋅ y1/2 2x 2 dx dx 2 x-1/2 = (9 y1/2 + 4 y-1/2 )

dy -4 dx

2 x-1/2

(2 + y) 2 dy 6 x(2 + y)2 = -4 dx

1/2

+ 4y

-1/2

2 y

3x(2 + y) 2 dy = 2 dx

x (9 y + 4)

dy dx

15. x 4 y 3 + 4 x3/2 = 6 y 3/2 + 5

5+ x 12. 2 y = 5- x 2

d d æç 5 + x ö÷ (2 y 2 ) = ÷ ç dx dx çè 5 - x ÷ø 4y

(1)(5 - x) - (-1)(5 + x) dy = dx (5 - x) 2

5-x+5+ x dy = dx (5 - x)2 10 (5 - x)2 10

dy = dx 4 y(5 - x) 2 =

x (5 y - 2)

14. 4 x - 8 y = 6 y 3/2

(2 + y) d (2 - y) - (2 - y) d (2 + y)

d 4 3 d ( x y + 4 x3/2 ) = (6 y 3/2 + 5) dx dx d 4 3 d d d (x y ) + (4 x3/2 ) = (6 y 3/2 ) + (5) dx dx dx dx dy dy 4 x3 y 3 + x 4 ⋅ 3 y 2 + 6 x1/2 = 9 y1/2 +0 dx dx dy dy - 3x 4 y 2 dx dx dy 4 x3 y 3 + 6 x1/2 = (9 y1/2 - 3x 4 y 2 ) dx 4 x3 y 3 + 6 x1/2 = 9 y1/2

4 x3 y 3 + 6 x1/2 1/2

4 2

- 3x y

2 y(5 - x) 2

dy dx

444

Chapter 6 APPLICATIONS OF THE DERIVATIVE ( xy) 4/3 + x1/3 = y 6 + 1

16.

18. x 2e y + y = x3

d d 6 [( xy)4/3 + x1/3 ] = ( y + 1) dx dx d 4/3 4/3 d 1/3 d 6 d (x y ) + (x ) = (y ) + (1) dx dx dx dx x 4/3 ⋅

4 1/3 dy 4 1 + x1/3 y 4/3 + x-2/3 y 3 3 3 dx dy = 6 y5 +0 dx

4 1/3 4/3 1 4 dy dy + x-2/3 = 6 y5 - x 4/3 y1/3 x y 3 3 3 dx dx dy dy - 4 x 4/3 y1/3 4 x1/3 y 4/3 + x-2/3 = 18 y 5 dx dx dy 4 x1/3 y 4/3 + x-2/3 = (18 y 5 - 4 x 4/3 y1/3 ) ⋅ dx 4 x1/3 y 4/3 + x-2/3 5

18 y - 4 x x 2/3 x

⋅ 2/3

4/3 1/3

4 x1 3 y 4/3 + x-2/3 5

18 y - 4 x 4 xy

18x

4/3

2/3 5

4/3 1/3

+1 2 1/3

y - 4x y

dy dx

d 2 y d 3 ( x e + y) = (x ) dx dx d 2 y d (x e ) + ( y ) = 3x 2 dx dx dy dy 2 xe y + x 2e y + = 3x 2 dx dx dy dy + = 3x 2 - 2 xe y x 2e y dx dx dy ( x 2e y + 1) = 3x 2 - 2 xe y dx 3x 2 - 2 xe y dy = dx x 2e y + 1

19. x + ln y = x 2 y 3 d d 2 3 ( x + ln y) = (x y ) dx dx 1 dy dy 1+ = 2 xy 3 + 3x 2 y 2 y dx dx 1 dy dy - 3x 2 y 2 = 2 xy3 - 1 y dx dx æ1 ö çç - 3x 2 y 2 ÷÷ dy = 2 xy3 - 1 ÷÷ dx çè y ø

17. e x y = 5 x + 4 y + 2 d x2 y d (e ) = (5 x + 4 y + 2) dx dx 2 d d d d e x y ( x 2 y) = (5 x) + (4 y) + (2) dx dx dx dx 2 æ dy ö dy e x y çç 2 xy + x 2 ÷÷÷ = 5 + 4 +0 çè dx ø dx 2 2 dy dy 2 xye x y + x 2e x y = 5+4 dx dx 2 dy 2 dy x 2e x y -4 = 5 - 2 xye x y dx dx 2 dy 2 x2 y (x e - 4) = 5 - 2 xye x y dx 2

5 - 2 xye x y dy = 2 dx x 2e x y - 4

2 xy 3 - 1 dy = 1 - 3x 2 y 2 dx y

y(2 xy 3 - 1) 1 - 3x 2 y 3

20. y ln x + 2 = x3/2 y 5/2 d d 3/2 5/2 ( y ln x + 2) = (x y ) dx dx 3 5 dy y dy ln x + + 0 = x1/2 y 5/2 + x 3/2 y 3/2 2 2 dx x dx 5 3/2 3/2 dy 3 1/2 5/2 dy y ln x - x y = x y 2 2 dx dx x 5 3/2 3/2 ö÷ 3 1/2 5/2 dy æç y çç ln x - x y ÷÷ = x y ø 2 2 dx è x 3 x1/2 y 5/2 - y dy x = 2 dx ln x - 5 x 3/2 y 3/2 2

3x 3/2 y 5/2 - 2 y x(2 ln x - 5x 3/2 y 3/2 )

Section 6.4

445

21. x 2 + y 2 = 25; tangent at (-3, 4) d 2 d (x + y2) = (25) dx dx dy =0 2x + 2 y dx dy = -2 x 2y dx dy x =dx y

23. x 2 y 2 = 1; tangent at (-1, 1) d 2 2 d (x y ) = (1) dx dx d d x2 ( y 2 ) + y 2 ( x2 ) = 0 dx dx dy + y 2 (2 x) = 0 x 2 (2 y) dx dy 2x2 y = -2 xy 2 dx dy -2 xy 2 y = =2 dx x 2x y

3 x -3 == 4 4 y y - y1 = m( x - x1)

m=-

3 y - 4 = [ x - (-3)] 4 4 y - 16 = 3x + 9

y 1 ==1 x -1 y - 1 = 1[ x - (-1)] m=-

y = x +1+1

4 y = 3x + 25

y = x+2

3 25 y = x+ 4 4

24. x 2 y 3 = 8; tangent at (-1, 2)

22. x 2 + y 2 = 100; tangent at (8, - 6) d 2 d (x + y2) = (100) dx dx dy =0 2x + 2 y dx dy x =dx y m=-

x 8 4 == y -6 3

d 2 3 d (x y ) = (8) dx dx dy 2 xy 3 + 3x 2 y 2 =0 dx 2 xy 3 dy =- 2 2 dx 3x y 2(-1)(2)3 = 3x 2 y 2 3(-1)2 (2)2 16 4 = = 12 3 4 y - 2 = ( x + 1) 3 3 y - 6 = 4x + 4

m =-

y - y1 = m( x - x1) 4 ( x - 8) 3 3 y + 18 = 4 x - 32 3 y = 4 x - 50 4 50 y = x3 3 y+6=

2 xy 3

3 y = 4 x + 10 4 10 x+ 3 3 1 y - 2 = - ( x - 4) 4 1 y =- x+3 4 y =

446 25.

Chapter 6 APPLICATIONS OF THE DERIVATIVE 2 y2 -

x = 4; tangent at (16, 2)

26. y +

d d (2 y 2 - x ) = (4) dx dx 1 dy 4y - x-1/2 = 0 2 dx 1 dy 4y = 1/2 dx 2x 1 dy = dx 8 yx1/2 1

x = 3; tangent at (4, 2) y d æç x ö÷÷ d çç y + (3) ÷= dx çè y ÷ø dx dy d æç x ö÷÷ ç + ÷=0 dx dx çèç y ÷ø

( )

x dx

-1/2

- yx dy = 2 dx

= 8 yx1/2 8(2)(16)1/2 1 1 = = 8(2)(4) 64

( y2 -

-y

( y2 - x ) -2 = 2(2)(4 - 2) 1 =4

x 7 + 64 4

1/2

e x + y = xe5 y - y 2e5 x /2 ; tangent at (2,1)

d x2 + y 2 d (e )= ( xe5 y - y 2e5 x /2 ) dx dx é dy 2 2 d 2 d d æ 5x ö ù ex + y ⋅ ( x + y 2 ) = e5 y + x (e5 y ) - ê 2 y e5 x /2 + y 2e5x /2 çç ÷÷÷ ú ê dx dx dx èç 2 ø úû ë dx 2 2æ dy ö dy dy 5 e x + y çç 2 x + 2 y ÷÷÷ = e5 y + x  5e5 y - 2 ye5 x /2 - y 2e5x /2 çè ø dx dx dx 2 2 2 2 2 5 æ ö dy ç 2 ye x + y - 5xe5 y + 2 ye5 x /2 ÷÷ = -2 xe x + y + e5 y - y 2e5 x /2 çè ø dx 2 2

-2 xe x + y + e5 y - 52 y 2e5x /2 dy = 2 2 dx 2 ye x + y - 5xe5 y + 2 ye5 x /2

-4e5 + e5 - 52 e5 5

2e - 10e + 2e

- 11 e5 2 5

-6e

11 12

11 ( x - 2) 12 11 5 y = x12 6

y -1 =

x dx

1 dy dy = - yx-1/2 + x 2 dx dx 1 -1/2 dy x) = - yx 2 dx dy -y = 1/2 2 dx 2x ( y - x )

64 y = x + 112

27.

1 y - 2 = - ( x - 4) 4 1 y =- x+3 4 1 y-2= ( x - 16) 64 64 y - 128 = x - 16 y =

y 12 x-1/ 2 dy + dx y2

Section 6.4

447 3

28. 2 xe xy = e x + ye x ; tangent at (1, 1)

(

)

2 d d x3 (2xe xy ) = e + ye x dx dx 3 d 2 dy x2 d 2 xy xy d 2e + 2x ⋅ e ( xy) = e x ( x3) + (x ) ⋅ 2 + ye x ⋅ dx dx dx dx 3 2 æ dy ö dy 2 2e xy + 2xe xy çç y + x ÷÷÷ = e x ⋅ 3x2 + e x + ye x ⋅ 2x çè dx ø dx 2 dy xy x xy 2 ( 2x e - e ) = -2e - 2xyexy + 2xyex2 + 3x2e x3 dx 2

dy -2e xy - 2xye xy + 2xye x + 3x2e x = 2 dx 2x2e xy - e x

m = -2e - 2e + 2e + 3e = e = 1 2e - e e y - 1 = 1( x - 1) y = x

29. ln( x + y ) = x3 y 2 + ln( x 2 + 2) - 4; tangent at (1, 2) d d 3 2 [ln( x + y)] = [ x y + ln( x 2 + 2) - 4] dx dx d dy d 2 d 1 1 ( x + y ) = 3x 2 y 2 + x3  2 y + 2 ( x + 2) (4)   x + y dx dx dx dx x +2 æ 1 ö dy 2x 1 çç - 2 x3 y ÷÷÷ = 3x 2 y 2 + 2 ÷ø dx çè x + y x+ y x +2 3x 2 y 2 + 22 x - x +1 y dy x +2 = 1 - 2 x3 y dx x+ y m=

3  1  4 + 231 - 13 1 - 2 1 2 3

= -311 = 3

37 11

y - 2 = - 37 ( x - 1) 11 37 y = - x + 59 11 11

30. ln( x 2 + y 2 ) = ln(5 x) +

y - 2; tangent at (1, 2) x d [ln( x 2 + y 2 )] = d êé ln(5x) + y - 2 ùú dx dx ëê x ûú dy

x -y 1 ⋅ d ( x 2 + y 2 ) = 1 ⋅ d (5x) + dx 2 - d (2) 2 2 dx dx 5x dx x + y x æ ö 1 ç 2 x + 2 y dy ÷÷ = 1 + 1 ⋅ dy - y 2 2 ççè dx ø÷ x x dx x + y x2 æ 2y ö÷ dy y çç - 1 ÷÷ = 1 - 2 - 2 2x 2 çç 2 2 ÷ ÷ x dx x x x + y èx + y ø 1 - y - 2x x2 x2 + y 2 2y - 1x x2 + y 2

x dy = dx

1 - 2 - 52 4 -1 5

-7

= -51 = 7 5

y - 2 = 7( x - 1) y = 7x - 5

448

Chapter 6 APPLICATIONS OF THE DERIVATIVE

31. y 3 + xy - y = 8 x 4 ; x = 1

33.

First, find the y-value of the point.

y 3 + xy 2 + 1 = x + 2 y 2 ; x = 2

Find the y-value of the point.

y 3 + (1) y - y = 8(1)4

y3 + 2 y 2 + 1 = 2 + 2 y 2

y3 = 8

y3 + 1 = 2

y = 2

y3 = 1 y =1

The point is (1, 2). dy

Find dx . 3y2

dy dy dy +x + y= 32 x3 dx dx dx dy = 32 x3 - y (3 y 2 + x - 1) dx dy 32 x3 - y = dx 3y2 + x - 1

At (1, 2),

The point is (2, 1). dy Find . dx dy dy dy + x2 y + y2 = 1 + 4 y dx dx dx dy dy dy + 2 xy - 4y = 1 - y2 3y2 dx dx dx dy = 1 - y2 (3 y 2 + 2 xy - 4 y) dx 3y2

dy 1 - y2 = dx 3 y 2 + 2 xy - 4 y

dy 32(1)3 - 2 30 5 = = = . 2 dx 12 2 3(2) + 1 - 1

At (2, 1),

5 ( x - 1) 2 5 5 y-2 = x2 2 5 1 y = x2 2 y-2 =

dy 1 - 12 = = 0. 2 dx 3(1) + 2(2)(1) - 4(1) y - 0 = 0( x - 2) y =1

32. y 3 + 2 x 2 y - 8 y = x3 + 19, x = 2 3y2

dy dy dy + 2x2 + 4 xy - 8 = 3x 2 dx dx dx dy = 3x 2 - 4 xy (3 y 2 + 2 x 2 - 8) dx dy 3x 2 - 4 xy = dx 3 y 2 + 2x2 - 8

Find y when x = 2.

Find the y-value. y 4 (1 - 1) + y = 2 y = 2

The point is (1, 2). dy

Find dx . dy dy +x + y = 0 dx dx dy dy dy - y 4 + 4 y3 - 4 xy3 +x + y = 0 dx dx dx dy = y4 - y (4 y3 - 4 xy3 + x) dx y 4 (-1) + 4 y 3 (1 - x)

y + 8 y - 8 y = 8 + 19 y 3 = 27 y =3 dy 12 - 24 -12 4 = = =dx 27 + 8 - 8 27 9 4 y - 3 = - ( x - 2) 9 4 8 y-3 = - x+ 9 9 4 35 y =- x+ 9 9

34. y 4 (1 - x) + xy = 2, x = 1

dy y4 - y = dx 4 y3 - 4 xy3 + x

dy at (1, 2) is dx 24 - 2 3

4 ⋅ 2 - 1(1)2 + 1

16 - 2 = 14. 32 - 32 + 1

Section 6.4

449 y - 2 = 14( x - 1)

Find

y - 2 = 14 x - 14 y = 14 x - 12

dy . dx

y 1 2 dy 1 dy (2 x) + ( x - 64) + x 2/3 y-2/3 18 18 dx 3 dx 2 -1/3 1/3 y =0 + x 3

35. 2 y3 ( x - 3) + x y = 3; x = 3

Find the y-value of the point.

æ 2 -2 xy x 2/3 y-2/3 ÷÷ö dy 2 x-1/3 y1/3 çç x - 64 + = ÷ ç 18 ÷÷ dx 3 18 3 çè ø

2 y 3 (3 - 3) + 3 y = 3 3 y =3

-1/3 1/3

-2 xy - 2x 3 y -2 xy - 12 x-1/3 y1/3 dy 18 = 2 2/3 -2/3 = dx x - 64 + x y x 2 - 64 + 6 x 2/3 y-2/3

y =1 y =1

dy at (8, 27) is dx

The point is (3, 1) dy

Find dx . dy 2 y 3 (1) + 6 y 2 ( x - 3) dx æ 1 ö÷ -1/2 dy + x çç ÷÷ y + y =0 çè 2 ø dx dy x dy + = -2 y 3 6 y 2 ( x - 3) dx 2 y dx é ù ê 6 y 2 ( x - 3) + x ú dy = -2 y3 ê 2 y úúû dx êë

-2(8)(27) - 12(8)-1/3(27)-1/3 64 - 64 + 6(8)

-432 - 18 24 9

-675 4

675 ( x - 8) 4 675 y - 27 = x + 1350 4 675 y =x + 1377 4 y - 27 = -

12 y 5/2 ( x - 3) + x

y - 1 = -2( x - 3)

-4 y 7/2 - 2 y

dy -4(1) - 2 -6 = = = -2. dx 12(1)(3 - 3) + 3 3

(27)

2 y

At (3, 1),

-2/3

æ 9 ö = (-450) çç ÷÷÷ çè 24 ø

-2 y 3 - y dy = dx 6 y 2 ( x - 3) + x =

2/3

37.

x 2/3 + y 2/3 = 2; (1, 1)

Find

dy . dx

2 -1/3 2 dy x + y-1/3 =0 3 3 dx 2 -1/3 dy 2 y = - x-1/3 3 dx 3 2 -1/3

y - 1 = -2 x + 6

- x dy = 23 dx y-1/3

y = -2 x + 7

36.

y1/3 = - 1/3 x

y 2 ( x - 64) + x 2/3 y1/3 = 12, x = 8 18

Find the y-value of the point.

At (1, 1) dy 11/3 = - 1/3 = -1 dx 1 y - 1 = -1( x - 1) y - 1 = -x + 1 y = -x + 2

y (64 - 64) + 82/3 y1/3 = 12 18 4 y1/3 = 12 y1/3 = 3 y = 27

The point is (8, 27).

450

Chapter 6 APPLICATIONS OF THE DERIVATIVE

38. 3( x 2 + y 2 )2 = 25( x 2 - y 2 ); (2, 1)

Find

dy . dx d 2 d ( x + y 2 ) = 25 ( x 2 - y 2 ) dx dx æ æ dy ö dy ö 6( x 2 + y 2 ) çç 2 x + 2 y ÷÷÷ = 25 çç 2 x - 2 y ÷÷÷ èç èç dx ø dx ø 6( x 2 + y 2 )

dy dy dy + 12 xy 2 + 12 y 3 = 50 x - 50 y dx dx dx dy 2 dy 3 dy 3 12 x y + 12 y + 50 y = -12 x - 12 xy 2 + 50 x dx dx dx dy (12 x 2 y + 12 y3 + 50 y) = -12 x3 - 12 xy 2 + 50 x dx

12 x3 + 12 x 2 y

dy -12 x3 - 12 xy 2 + 50 x = dx 12 x 2 y + 12 y 3 + 50 y

At (2,1), dy -12(2)3 - 12(2)(1)2 + 50(2) = dx 12(2)2 + 12(1)3 + 50(1) -20 110 2 =11 2 y - 1 = - ( x - 2) 11 2 4 y -1 = - x + 11 11 2 15 y =- x+ 11 11 =

39. y 2 ( x 2 + y 2 ) = 20 x 2 ; (1, 2)

Find

dy . dx æ dy dy ö + y 2 çç 2 x + 2 y ÷÷÷ = 40 x ç è dx dx ø dy dy dy + 2 y3 + 2 xy 2 + 2 y 3 = 40 x 2x2 y dx dx dx dy dy + 4 y3 = -2 xy 2 + 40 x 2x2 y dx dx æ dy ö (2 x 2 y + 4 y 3 ) çç ÷÷÷ = -2 xy 2 + 40 x çè dx ø 2 y( x 2 + y 2 )

-2 xy 2 + 40 x dy = dx 2x 2 y + 4 y3

Section 6.4

451

At (1, 2), dy -2(1)(2) 2 + 40(1) = dx 2(1)2 (2) + 4(2)3 32 8 = 36 9 8 y - 2 = ( x - 1) 9 8 8 y-2= x9 9 8 10 y = x+ 9 9 =

40. 2( x 2 + y 2 ) 2 = 25 xy 2 ; (2, 1)

Find

dy . dx d 2 d ( x + y 2 ) = 25 ( xy 2 ) dx dx æ æ dy ö dy ö 4( x 2 + y 2 ) çç 2 x + 2 y ÷÷ = 25 çç y 2 + 2 xy ÷÷ ÷ èç èç dx ø dx ÷ø 4( x 2 + y 2 )

dy dy dy + 8xy 2 + 8 y 3 = 25 y 2 + 50 xy dx dx dx dy dy dy + 8 y3 - 50 xy = -8x3 - 8xy 2 + 25 y 2 8x 2 y dx dx dx æ dy ö (8x 2 y + 8 y3 - 50 xy) çç ÷÷÷ = -8x3 - 8xy 2 + 25 y 2 çè dx ø

8x3 + 8 x 2 y

-8 x3 - 8 xy 2 + 25 y 2 dy = dx 8x 2 y + 8 y 3 - 50 xy

At (2,1), dy -8(2)3 - 8(2)(1)2 + 25(1) 2 = dx 8(2)2 (1) + 8(1)3 - 50(2)(1) -55 11 = -60 12 11 ( x - 2) y -1 = 12 11 11 y -1 = x12 6 11 5 y = x12 6 =

452

Chapter 6 APPLICATIONS OF THE DERIVATIVE

41. x 2 + y 2 = 100

43. 10 x 2 + 3 y 2 = 8

(a) Lines are tangent at points where x = 6. By substituting x = 6 in the equation, we find that the points are (6, 8) and (6, - 8).

d d (10 x 2 + 3 y 2 ) = (8) dx dx dy 20 x + 6 y =0 dx dy 20 x 10 x ==dx 6y 3y

d 2 d (x + y2) = (100) dx dx dy 2x + 2 y =0 dx dy 2y = -2 x dx x dy = y

d æç dy ö÷ d æç 10 x ö÷ ÷ ççè ÷÷ø = çdx dx dx çè 3 y ÷÷ø d2y dx 2

9 y2

(

30 y - 30 x -

x 6 3 =- =y 8 4 x 6 3 m2 = - = = y -8 4 m1 = -

(

3 y - 8 = - ( x - 6) 4 3 25 y =- x+ 4 2

==-

Second tangent:

x +

44.

1/ 2

10 x 3y

)

10 x 3y

) ⋅ 3y

9 y2 90 y 2 + 300 x 2 27 y 3 30(8) 27 y 80

dy dx

30 y - 30 x -

First tangent:

3 y - (-8) = ( x - 6) 4 3 18 y+8 = x4 4 3 25 y = x4 2

3 y(10) - 10 x(3)

30(10 x 2 + 3 y 2 ) 27 y 3

10x 2 + 3 y 2 = 8

9 y3

y = 2 1/ 2

+ y

= 2

d 1/ 2 d (x + y1/ 2 ) = (2) dx dx 1 1 dy + =0 2 x 2 y dx

(b)

y dy y1/ 2 == - 1/ 2 dx x x

42.

y 2 = x3 + ax + b d 2 d 3 (y ) = ( x + ax + b) dx dx dy 2y = 3x 2 + a dx 3x 2 + a dy = 2y dx

Section 6.4

453

d æç dy ö÷ d æç y1/ 2 ö÷÷ ÷ ççè ÷ø÷ = çç dx dx dx çè x1/ 2 ÷ø

46.

1 dy - y x 2 y dx 2 x d2y =x dx 2 æ y ÷÷ö y x ç çç ÷÷ ç 2 yè xø 2 x =x

x4 + y 4 = 7 d 4 d (x + y4) = (7) dx dx dy 4 x3 + 4 y 3 =0 dx

y 1 + 2 2 x ⋅ 2 x = x 2 x x +

2 x3/ 2 1

45.

3/ 2

d æç dy ö÷ d æç x3 ö÷÷ çè ÷ø÷ = çç - ÷ ç dx dx dx çè y 3 ÷÷ø d2y

dx 2

2x x 2

4 x3 dy x3 =- 3 =- 3 dx 4y y

x +

y = 2

= x-3/ 2 =-

4 x + y = -1

d d (4 x3 + y 3 ) = (-1) dx dx dy 12 x 2 + 3 y 2 =0 dx

dy 12 x 2 4x2 =- 2 =- 2 dx 3y y d æç dy ö÷ d æç 4 x 2 ÷ö÷ ÷ çèç ÷ø÷ = çç dx dx dx çè y 2 ÷ø÷

47. (a) 2

( ) 4x

8xy 2 - 8x 2 y - 2 y

===

8xy 3 + 32 x3 y5 8x(-1)

8x y5

æ x3 ö 3x 2 y 3 - 3x3 y 2 ççç - 3 ÷÷÷ è y ø y6 æ x3 ö 3x 2 y 3 - 3x3 çç - ÷÷÷ è yø y

2 4

3x y + 3x 6 y7 3x 2 (7) y

y y

3x 2 ( x 4 + y 4 ) y7

x4 + y4 = 7

u + 2v + 1 = 5

2u1/2 du =dv (2v + 1)1/2

4x y

⋅

21x 2

( )⋅ y

8xy 2 - 8x 2 =-

dy dx

du du ( u + 2v + 1) = (5) dv dv 1 -1/2 du 1 + (2v + 1)-1/2 (2) = 0 u 2 2 dv 1 -1/2 du 1 =u 2 dv (2v + 1)1/2

dy y (8x) - 4 x (2 y) d2y dx = 2 4

y 3 (3x 2 ) - x3 (3 y 2 )

8 x(4 x3 + y 3 ) y5

4x3 + y 3 = -1

(b)

u + 2v + 1 = 5

dv dv ( u + 2v + 1) = (5) du du 1 -1/2 1 dv + (2v + 1)-1/2 (2) =0 u 2 2 du 1 dv (2v + 1)-1/2 = - u-1/2 2 du (2v + 1)1/2 dv =du 2u1/2

454

Chapter 6 APPLICATIONS OF THE DERIVATIVE 50. x = w( x)e w( x) Differentiate both sides with respect to x using the product and chain rules. dw w dw w e + w⋅ e 1= dx dx dw w e + we w 1= dx dw 1 = w dx e (1 + w)

48. (a) eu - v - v = 1 d æ u2 -v d ö çç e - v ÷÷÷ = (1) è ø dv dv d æ u 2 - v ö÷ d d çe ÷÷ø - dv v = dv (1) dv çè 2 d eu - v (u 2 - v ) - 1 = 0 dv 2 æ du ö eu - v çç 2u - 1 ÷÷÷ = 1 çè dv ø 2u

(

du 1 -1 = 2 dv eu - v ö÷ du 1 æç 1 çç 2 = + 1 ÷÷ dv 2u çè eu - v ø÷÷ 2

du 1 + eu - v = 2 dv 2ueu - v

(b) e

u2 - v

)

When w = 1, 1 1 dw . = 1 = 2e dx e (1 + 1) 51. C 2 = x 2 + 100 x + 50 (a) 2C

1 dC = 2 x + (100) x-1/2 2 dx 2 x + 50 x-1/2 dC = 2C dx

-v =1

d æ u2 - v d ö çe - v ÷÷÷ = (1) ø du çè du d æ u 2 - v ÷ö d d çe ÷÷ø - du v = du (1) du çè 2 d 2 dv = 0 (u - v ) eu - v du du 2 æ dv ö÷ dv = 0 eu - v çç 2u ÷çè du ÷ø du

dC x + 25 x-1/2 x1/2 = ⋅ 1/2 dx C x dC x3/2 + 25 = dx Cx1/2

When x = 5, the approximate increase in cost in dollars of an additional unit is (5)3/2 + 25 2

1/2

(5 + 100 5 + 50) dv dv = -2ueu - v du du 2 dv æ 2 ö - çç eu - v + 1 ÷÷÷ = -2ueu - v è ø du 2

1/2

(5)

(b) 900( x - 5) 2 + 25R 2 = 22, 500 R 2 = 900 - 36( x - 5) 2

u2 -v

2ue dv = 2 du eu - v + 1

49. x 2 + y 2 + 1 = 0 d 2 d (x + y2) = (-1) dx dx dy 2x + 2 y =0 dx dy -2 x x = =dx 2y y

36.18 (17.28) 5

» 0.94.

eu - v

dR = -72( x - 5) dx dR -36( x - 5) 180 - 36 x = = dx R R

When x = 5, the approximate change in revenue for a unit increase in sales is 180 - 36(5) 0 = = 0. R R

If x and y are real numbers, x 2 and y 2 are nonnegative; 1 plus a nonnegative number cannot equal zero, so there is no function y = f ( x) that satisfies x 2 + y 2 + 1 = 0. Copyright © 2022 Pearson Education, Inc.

Section 6.4

455

(a) 4 p + 2q

dq =0 dp 4 p = -2q

ln q = C - 0.678ln p 1 dq 0.678 =q dp p dp q = -0.678 dq p

54. (a)

52. 2 p 2 + q 2 = 1600

dq dp

p dq q dp pæ qö = - ççç -0.678 ÷÷÷ qè p ø÷

2p dq = q dp

E =-

This is the rate of change of demand with respect to price. (b) 4 p

= 0.678

dp + 2q = 0 dq dp q =dq 2p

E is less than 1, so the demand is inelastic. (b) Solving for q first: ln q = C - 0.678ln p

This is the rate of change of price with respect to demand. 53. (a)

ln q = D - 0.44 ln p 1 dq 0.44 =q dp p dp q = -0.44 dq p

eln q = eC -0.678ln p q = eC p-0.678 dq = eC (-0.678) p-1.678 dp E =-

æ ö÷ p = ççç - C -0.678 ÷÷ (eC (-0.678) p-1.678 ) ÷ø÷ èç e p

p dq q dp pæ qö = - ççç -0.44 ÷÷÷ qè p ÷ø

E =-

= 0.678

This is the same answer as found in part (a).

= 0.44

55. First note that

E is less than 1, so the demand is inelastic. (b) Solving for q first: ln q = D - 0.44 ln p e

ln q

p dq q dp

log R( w) = 1.83 - 0.43log(w)

then

R(w) = 101.83-0.43log( w) = 101.8310-0.43log( w)

D -0.44 ln p

= 101.83[10log( w) ]-0.43

D -0.44

q =e p dq = e D (-0.678) p-1.44 dp E =-

p dq q dp

æ ö÷ p = ççç - D -0.44 ÷÷ (e D (-0.44) p-1.44 ) çè e p ÷÷ø

= 101.83 w-0.43

(a)

d d [log R(w)] = [1.83 - 0.43log(w)] dw dw 1 1 dR 1 1 = 0 - 0.43 ln10 R(w) dw ln10 w dR R(w) = -0.43 dw w

= 0.44

= -0.43

This is the same answer as found in part (a).

101.83 w-0.43 w

» -29.0716w-1.43

456

Chapter 6 APPLICATIONS OF THE DERIVATIVE (b)

R(w) = 101.83 w-0.43 d d [ R(w)] = [101.83 w-0.43 ] dw dw dR = 101.83 (-0.43)w-1.43 dw

b+a =

-(b - a) = - 1

3 3

2a =

» -29.0716w-1.43 xy a = k d d ( xy a ) = (k ) dx dx

56.

a =

58.

d a ( y ) + y a (1) = 0 dx æ dy ö x çç ay a -1 ÷÷÷ + y a = 0 çè dx ø x

axy

a -1 dy

= -y

1 3

2 3 3

1 3 3

f ( x) g ( N ( x)) - m - s( x) = 0 Differentiate both sides with respect to x. f ¢( x ) g ( N ( x )) + f ( x ) g ¢( N ( x )) N ¢( x ) - s¢( x ) = 0

Solve for N ¢( x). f ( x ) g ¢( N ( x )) N ¢( x ) = s¢( x ) - f ¢( x ) g ( N ( x )) s¢( x ) - f ¢( x ) g ( N ( x )) N ¢( x ) = f ( x ) g ¢( N ( x )) dN s¢( x ) - f ¢( x ) g ( N ) = dx f ( x ) g ¢( N )

dy ya =dx axy a -1 dy y =dx ax

59.

57. b - a = (b + a)3 d d (b - a) = [(b + a)3 ] db db da d = 3(b + a)2 1(b + a) db db æ da da ÷ö = 3(b + a)2 çç1 + 1÷ çè db db ÷ø da da = 3(b + a)2 + 3(b + a) 2 db db da 2 da 2 - 3(b + a) = 3(b + a) - 1 db db da = 3(b + a)2 - 1 [-1 - 3(b + a) 2] db 1-

da 3(b + a) 2 - 1 = db -1 - 3(b + a) 2 da =0 db

Differentiate f ( R) twice, treating t as a function of R. f ¢¢( R) C = (1 + hcR)3 é d 2t ù 2 3 2 2 ê ú + + 2 R R hC R h C ê 2 ú dR ê ú ´ ê ú dt ê ú + + 2 1 2 hcR thC ( ) ê ú dR ë û C = (1 + hcR)3 é d 2t ù 2 ê ú ê 2 R ( 1 + hcR ) ú dR ú ´ê ê ú dt ê + 2 ( 1 + hcR ) - 2thC úú ê dR ë û For the graph of f ( R) to be concave down, f ( R) must be negative, which will be true when the expression in brackets above is negative. This is what the exercise claims.

3(b + a )2 - 1 = 0 1 b+a = 3

æ 1 ö÷3 1 Since b - a = (b + a)3 = çç ÷ = . çè 3 ÷÷ø 3 3

(

)

Section 6.5

457

60. 2s 2 + 4s

st - 4 = 3t

æ ds 1 ds ö + (st )-1/2 çç s + t ÷÷÷ = 3 ç è dt 2 dt ø ds

s + t dt ds + =3 dt 2 st

8s( st ) ds + s + t ds dt dt 2 st (8s st + t ) ds +s dt 2 st

61.

æ -1 ö÷ = 6çç çè -2 ø÷÷ = -3

Your Turn 2

s 3 - 4st + 2t 3 - 5t = 0 ö ds æç ds - ç 4t + 4s ÷÷÷ + 6t 2 - 5 = 0 3s 2 ø dt çè dt 3s 2

æ 3(12 ) + 2(-2) ÷ö ç = 6çç ÷÷ çè 2(1) + 2(-2) ÷ø÷

ds = 6 st - s dt ds -s + 6 st = dt 8s st + t

(8s st + t )

dy dx çæ 3x 2 + 2 y ÷÷ö ç= ÷ dt dt ççè 2 x + 2 y ÷ø÷

ds ds - 4t - 4s + 6t 2 - 5 = 0 dt dt

ds 2 (3s - 4t ) = 4s - 6t 2 + 5 dt ds 4s - 6t 2 + 5 = dt 3s 2 - 4t

Ladder y

25 ft

x 2 + y 2 = 252 , where both x and y are functions of t.

We are interested in what happens when x = 7 ft. At this time, y = 252 - 7 2 = 24, and since the bottom of the ladder is slipping away from the building at 3 ft/min, dx /dt = 3.

(

)

d 2 d x + y2 = (252 ) dt dt dx dy + 2y =0 2x dt dt

Now substitute the known values.

6.5

2(7)(3) + 2(24)

Related Rates

Your Turn 1 x3 + 2 xy + y 2 = 1, where both x and y are functions of t. Given x = 1, y = -2, and dx /dy = 6, find dy /dt.

(

)

d 3 d x + 2 xy + y 2 = (1) dt dt dx dy dx dy 3x 2 + 2x + 2y + 2y = 0 dt dt dt dt æ dy dy dx dx ö 52 x + 2y = -çç 3x 2 + 2 y ÷÷÷ çè dt dt dt dt ø dy dx æçç 3x 2 + 2 y ö÷÷ = ÷ çdt dt ççè 2 x + 2 y ÷÷ø

Now substitute the given values.

dy =0 dt dy -2(7)(3) = dt 2(24) 7 =8

The latter is sliding down the side of the building at 7/8 ft/min. Your Turn 3

In Example 4, differentiating the formula for the volume of a cone gives the following result: dV 1 é dh dr ù =  ê r2 + (h)(2r ) ú dt 3 ëê dt dt ûú

For this problem, dV dr = -10, = -0.4, r = 4, and h = 20. dt dt

458

Chapter 6 APPLICATIONS OF THE DERIVATIVE

Substitute these values above and solve for dh /dt.

W2.

dV dr ù 1 é dh =  ê r2 + (h)(2r ) ú ê dt dt úû 3 ë dt ù dh 1 é -10 =  ê (42 ) + (20)(2)(4)(-0.4) ú úû 3 êë dt dh + 64 = 16 dt  30 - + 64 dh =  16 dt dh » 3.4 dt

x 2 + y 2 = 3xy3 Differentiate both sides with respect to x. æ dy dy ö = (3x ) çç 3 y 2 ÷÷÷ + 3 y 3 2x + 2 y çè dx dx ø

(

dy 3y3 - 2x = dx 2 y - 9 xy 2

6.5

Exercises

1. True

The length increases at a rate of 3.4 cm per hour. Your Turn 4

The revenue equation is æ q 2 ö÷÷ q3 ç R = qp = q çç 2000 . ÷÷ = 2000q çè 100 ø÷ 100

Differentiate with respect to t.

2. False. In an applied extrema problems, you are trying to maximize or minimize something. In related rates problems, you are trying to find how fast something is changing. 3. False. Suppose x and y are functions of t. If y = 3x, then we need more information to find dy/dt. 4. True 5.

dR dq 3q dq = 2000 dt dt 100 dt æ 3q 2 ö÷÷ dq ç = çç 2000 ÷ 100 ÷ø÷ dt èç

y 2 - 8x3 = -55; 2y

Substitute the know values for q and dq /dt , which are the same as in Example 5: q = 200, dq /dt = 50. æ dR 3q 2 ÷÷ö dq ç = çç 2000 ÷ dt 100 ÷÷ø dt çè æ 3(2002 ) ö÷÷ ç = çç 2000 ÷ (50) çè 100 ÷÷ø

24 y 2

Revenue is increasing at the rate of $40,000 per day.

6.5

Warmup Exercises

W1.

x3 y + y 4 x = 5 Differentiate both sides with respect to x. ù é dy ù é æ dy ö ê x3 + 3x 2 y ú + ê x çç 4 y 3 ÷÷÷ + y 4 ú = 0 ú êë dx úû êë èç dx ø û dy 3 x + 4 xy 3 = - 3x 2 y + y 4 dx

)

3x 2 y + y 4 dy =- 3 dx x + 4 xy 3

(

)

dx = -4, x = 2, y = 3 dt

dy dx - 24 x 2 =0 dt dt dy dx y = 12 x 2 dt dt dy 3 = 48(-4) dt dy = -64 dt

6. 8 y 3 + x 2 = 1;

= (800)(50) = 40, 000

(

)

dy 2 y - 9 xy 2 = 3 y 3 - 2 x dx

dx = 2, x = 3, y = -1 dt

dy dx + 2x = 0 dt dt -2 x dx x dx dy dt = - dt = dt 24 y 2 12 y 2 (3)(2) 1 ==2 2 12(-1) dx = -6, dt x = 3, y = -2

2 xy - 5 x + 3 y 3 = -51;

dy dx dx dx + 2y -5 + 9 y2 =0 dt dt dt dt dy dx (2 x + 9 y 2 ) + (2 y - 5) =0 dt dt dy dx (2 x + 9 y 2 ) = (5 - 2 y ) dt dt

Section 6.5

459

dy dx 5 - 2y = ⋅ 2 dt dt 2x + 9 y 5 - 2(-2)

⋅ (-6) 2(3) + 9(-2)2 -54 9 = ⋅ (-6) = 42 42 9 =7

dx = 3, dt x = -3, y = 4

4 x3 - 6 xy 2 + 3 y 2 = 228;

dx æç 2 dx dy ö dy - ç6y + 12 xy ÷÷÷ + 6 y =0 çè dt dt dt ø dt dx dx dy dy 12 x 2 - 6 y2 - 12 xy + 6y =0 dt dt dt dt dy dx (6 y - 12 xy) = (6 y 2 - 12 x 2 ) dt dt

12 x 2

dy y 2 - 2 x 2 dx = ⋅ dt y - 2 xy dt 42 - 2 ⋅ (-3) 2 ⋅3 4 - 2 ⋅ (-3) ⋅ 4 2 -6 =⋅3= 28 28 3 =14

x2 + y dx = 9; = 2, x = 4, y = 2 x-y dt

(

)

(

+ dt - ( x 2 + y) dx - dt ( x - y) 2 x dx dt dt ( x - y)

)=0

+ ( x - y) dt - ( x 2 + y) dx + ( x 2 + y) dt 2 x( x - y) dx dt dt

=0 ( x - y)2 dx dy + [( x - y) + ( x 2 + y)] =0 [2 x( x - y) - ( x 2 + y)] dt dt [( x 2 + y ) - 2 x( x - y)] dx dy dt = dt ( x - y) + ( x 2 + y) (-x 2 + y + 2 xy) dx dy dt = dt x + x2 = =

[-(4)2 + 2 + 2(4)(2)](2) 4 + 42 4 1 = 20 5

460

Chapter 6 APPLICATIONS OF THE DERIVATIVE

10.

y - 4x

(ln x + xe y )

44 dx ; = 5, 31 dt

x + 2y x = -3, y = -2

31y 3 - 124 x 2 = 44 x3 + 88 y dy dx dx dy - 248 x = 132 x 2 + 88 dt dt dt dt dy dx 2 2 (93 y - 88) = (132 x + 248x) dt dt

93 y 2

132 x 2 + 248x dx dy = ⋅ dt dt 93 y 2 - 88 132(-3) 2 + 248(-3)

⋅5 93(-2)2 - 88 444 2220 555 = ⋅5 = = 284 284 71

11. xe y = 3 + ln x; y dx

dx = 6, x = 2, y = 0 dt

= =

[1 - (2)e0 ](6) 2 2 e0 3 -6 =4 2

dx = 12 dt

dC R dR = ⋅ 225,000 dx dx 25,000 dR 15 = ⋅ 225,000 dx 1 dR 15 = ⋅ 9 dx dR 135 = dx

Revenue is changing at a rate of $135 per month. 15. R = 50 x - 0.4 x 2 ; C = 5 x + 15; dx x = 40; = 10 dt

dx = 5, x = 1, y = 0 dt

d d d ( y ln x) + ( xe y ) = (1) dt dx dt dy y dx dx dy ln x + + ey + x ⋅ ey =0 dt x dt dt dt ö dx dy æç y (ln x + xe y ) + ç + e y ÷÷÷ =0 çè x ø dt dt

= -5

R2 dC + 12,000; = 15 450,000 dx R = 25,000

(a) 12. y ln x + xe y = 1;

(1) ln1 + 12 e0

14. C =

xe y x 2e y

[0 + (1)e0 ](5)

The cost is changing at a rate of $384 per month.

( 1x - e y ) dxdt (1 - xe y ) dx dt

x ln x + x 2e y

dC dx = 0.2(2 x) = 0.2(160)(12) = 384 dt dt

1 dx e + xe = 0+ dt dt x dt æ1 ö dx y dy xe = çç - e y ÷÷÷ çè x ø dt dt

( y + xe y ) dx dt

13. C = 0.2 x 2 + 10,000; x = 80,

y dy

dy = dt

)

+ e y dx x dt dy =dt ln x + xe y

31( y 3 - 4 x 2 ) = 44( x3 + 2 y)

æy ö dx dy = -çç + e y ÷÷÷ ç èx ø dt dt

dR dx dx = 50 - 0.8 x dt dt dt = 50(10) - 0.8(40)(10) = 500 - 320 = 180 Revenue is increasing at a rate of $180 per day.

(b)

dC dx =5 = 5(10) = 50 dt dt Cost is increasing at a rate of $50 per day.

Section 6.5

461 æ 1 ö R = çç - q + 70 ÷÷÷ q çè 10 ø

16. R = 50 x - 0.4 x 2 , C = 5 x + 15; dx x = 80, = 12 dt (a)

1 2 q + 70q 10 dR dq 1 dq =- q + 70 dt dt 5 dt 1 = - (20)(25) + 70(25) 5 = -100 + 1750 =-

dR dx dx = 50 - 0.8x dt dt dt = 50(12) - 0.8(80)(12) = 600 - 768 = -168 Revenue is decreasing at a rate of $168 per day.

dC dx (b) = (5) = (5)(12) = 60 dt dt Cost is increasing at a rate of $60 per day.

(c)

= $1650

Revenue is increasing at a rate of $1650 per day. 19. V = k ( R 2 - r 2 ); k = 555.6, R = 0.02 mm, dR = 0.003 mm per minute; r is constant. dt

P = R-C dP dR dC = dt dt dt = -168 - 60

V = k (R2 - r2 ) V = 555.6( R 2 - r 2 ) æ ö dV dR = 555.6 çç 2 R - 0 ÷÷÷ ç è ø dt dt

= -228 Profit is decreasing at a rate of $228 per day.

17. pq = 8000; p = 3.50,

= 555.6(2)(0.02)(0.003) = 0.067 mm/min

dp = 0.15 dt pq = 8000

20. y = nx m

dq dp +q =0 dt dt

Note that n is a constant. ln y = ln(nx m )

-q dt dq = dt p =

(

)

(0.15) - 8000 3.50 3.50

» -98

ln y = ln n + ln x m ln y = ln n + m ln x

Now take the derivative of both sides with respect to t. 1 dy 1 dx =0+m y dt x dt 1 dy 1 dx = m y dt x dt

Demand is decreasing at a rate of approximately 98 units per unit time. 18. R = pq;

dq = 25 dt

Find the relationship between p and q by finding the equation of the line through (0, 70), and (100, 60). 70 - 60 10 1 = =6 - 100 10 -100 1 p - 70 = - (q - 0) 10 1 p - 70 = - q 10 1 p = - q + 70 10 m =

21. b = 0.22m0.87 db dm = 0.22(0.87)m-0.13 dt dt -0.13 dm = 0.1914m dt dm m0.13 db = 0.1914 dt dt 250.13 (0.25) 0.1914 » 1.9849

462

Chapter 6 APPLICATIONS OF THE DERIVATIVE

22. E = 429m-0.35

26. W (t ) =

dE dm = 429(-0.35)m-1.35 dt dt dm = -150.15m-1.35 dt = -150.15(10)-1.35 (0.001) » -0.0067

The rate of change of the energy expenditure is about -0.0067 cal/g/hr 2.

-0.02t 2 + t t +1

dW (-0.04t + 1)(t + 1) - (1)(-0.02t 2 + t ) = dt (t + 1)2

If t = 5, dW (-0.2 + 1)(6) - (-0.5 + 5) = dt 62 4.8 - 4.5 = 36 = 0.008.

23. r = 140.2m0.75 (a)

(b)

27. Let x = The distance of the base of the ladder from the base of the building y = The distance up the side of the building to the top of the ladder

dr dm = 140.2(0.75)m-0.25 dt dt -0.25 dm = 105.15m dt dr = 105.15(250)-0.25 (2) dt » 52.89

The rate of change of the average daily metabolic rate is about 52.89 kcal/day 2. 24. E = 26.5w-0.34

dE dw = 26.5(-0.34)w-1.34 dt dt -1.34 dw = -9.01w dt -1.34

= -9.01(5) » -0.0521

1 (T - 60)2 + 100 10 dC 1 dT = (T - 60) dt 5 dt C =

If T = 76 and

Since y = 17 2 - x 2 , when x = 8, y = 15. By the Pythagorean theorem,

(0.05)

The rate of change of the energy expenditure is about -0.0521 kcal/kg/km/day. 25.

Find dt when x = 8 ft and dx = 9 ft/min. dt

dT = 8, dt

x 2 + y 2 = 17 2. d 2 d (x + y2) = (17 2 ) dx dt dx dy 2x + 2y =0 dt dt dy dx 2y = -2 x dt dt dy -2 x dx x dx = ⋅ =- ⋅ 2 y dt dt y dt 8 (9) 15 24 =5 =-

dC 1 1 = (76 - 60)(8) = (16)(8) dt 5 5 = 25.6.

The crime rate is rising at the rate of 25.6 crimes/month.

The ladder is sliding down the building at the rate of 24 ft/min. 5

Section 6.5

463

28. (a) Let x = the distance one car travels west; y = the distance the other car travels north s = the distance between the two cars

The distance between the cars after the second car has traveled 1 hour is about 72.11 mi. 72.11

ds = (40)(40) + (60)(30) dt 3400 ds = » 47.15 dt 72.11

The distance between the two cars is changing at a rate of about 47.15 mph. 29. Let r = the raius of the circle formed by the ripple.

when r = 4 ft and dr = 2 ft/min. Find dA dt dt

s2 = x2 + y 2

A =  r2

ds dx dy = 2x + 2y dt dt dt ds dx dy s = x + y dt dt dt

dA dr = 2 r dt dt = 2 (4)(2) = 16

Use d = rt to find x and y.

The area is changing at the rate of 16 ft 2 /min.

x = (40)(2) = 80 mi y = (30)(2) = 60 mi s = =

x + y

30. V = 43  r 3 , r = 4 in, and dr = - 14 in/hr dt

dV dr = 4 r 2 dt dt æ 1ö = 4 (4)2 çç - ÷÷÷ = -16 in 3 /hr çè 4 ø

(80)2 + (60)2

= 100

The distance between the cars after 2 hours is 100 mi. dx = 40 mph and dy = 30 mph dt dt

(100)

ds = (80)(40) + (60)(30) dt 5000 ds = = 50 dt 100

The distance between the two cars is changing at the rate of 50 mph. (b) From part (a), we have s

31.

V = x3, x = 3 cm, and dV = 2 cm3 /min dt dV dx = 3x 2 dt dt dx 1 dV = dt 3x 2 dt 1 2 = (2) = cm/min 2 27 3⋅3

32. Let r = the radius of the base of the conical pile.

ds dx dy = x + y dt dt dt

Use d = rt to find x and y. When the second car has traveled 1 hour, the first car has traveled 2 hours. x = 40(1) = 40 mi y = 30(2) = 60 mi s = =

x2 + y2 (40) 2 + (60) 2

= 72.11

464

Chapter 6 APPLICATIONS OF THE DERIVATIVE V =

 3

(b) Find how fast the tip of the shadow moving. dx dy 10 25 + = +5= dt dt 3 3

r 2h



r 2 (2r ) 3 2 3 = r 3

V =

The tip of the shadow is moving at the rate of 25/3 ft per second. 34. Let x = one-half the width of the triangular cross section; h = the height of the water; V = the volume of the water. dV = 4 cu ft per min dt

dV 3 ⋅ 2 r 2 dr = ⋅ dt 3 dt dV 2 = 2 (6) (0.75) dt = 54

The volume is changing at the rate of 54 in 3 /min.

33. Let y = the length of the man’s shadow; x = the distane of the man from the lamp post; h = the height of the lamp post = 15ft. dx = 5 ft/sec dt

Find dh when h = 4. dt æ Area of ö÷ ç V = çç triangular ÷÷÷ ⋅ (length) ççè cross section ÷ø Area of triangular cross section 1 (base)(height) 2 1 = (2 x)(h) = xh 2 =

(a) Find

dy when x = 25 ft. dt

By similar triangles, 6 6 = , 2x h h so x = . 2

= 6y , by similar triangles. Now x15 +y 15 6 = x+ y y 15 y = 6 x + 6 y

V = ( xh) (16) æh ö = çç h ÷÷÷ 16 çè 2 ø

9 y = 6x y =

2 x 3

= 8h 2

d d æç 2 ÷ö ( y) = ç x÷ dt dt çè 3 ø÷ 2 dx dy = 3 dt dt Recall dx/dt = 5. dy 2 10 = (5) = dt 3 3

The length of the shadow is increasing at the rate of 10/3 ft per second.

dV dh = 16h dt dt dh 1 dV = dt 16h dt dh 1 (4) = dt 16(4) dh 1 = dt 16

The height of the water is increasing at a rate of 1 ft/min. 16

Section 6.5

465

35. Let x = the distance from the docks s = the length of the rope.

dr dx = 2x +0 dt dt dr dx r = x dt dt x dx dr = dt dt r dr 100 3(50) = = 25 3 dt 200 » 43.3

She must let out the string at a rate of 25 3 » 43.3 ft/min.

ds = 1 ft/sec dt

37. Let x = the distance from home to first base; y = the distance from first base to the runner s = the distance from home to the runner

s 2 = x 2 + (8) 2 ds dx = 2x +0 dt dt ds dx s = x dt dt

If x = 8, s =

(8)2 + (8)2 = 128 = 8 2.

Then, 8 2(1) = 8 dx = dt

dx dt

x2 + y 2 = s 2

2 » 1.41

The boat is approaching the dock at ft/sec.

2 » 1.41

36. Let x = the horizontal length; r = the rope length.

d 2 ( x + y2 ) = d ( s2 ) dt dt dy ds = 2s 0 + 2y dt dt

Substitute x = 90 and y = 90 – 18 = 72, to solve for s. s 2 = x2 + y 2 s = =

x2 + y 2 (90)2 + (72) 2

» 115.3

Substitute x = 90, y = 90 – 18 = 72, and dy/dt = 20, and solve for ds/dt. 2y

dx = 50 ft/min dt

2(72)(20) = 2(115.3)

By the Pythagorean theorem,

x =

30, 000 = 100 3

ds dt

ds » 12.5 dt

x 2 + 1002 = 2002 r = x + 1002

dy ds = 2s dt dt

The runner’s distance from home plate is increasing at the rate of about 12.5 ft per second.

466

6.6

Chapter 6 APPLICATIONS OF THE DERIVATIVE

Differentials: Linear Approximation

W2. Use the chain rule, writing f ( x ) = ln(e2 x + 1) as h( g ( x )) with

Your Turn 1

h( x ) = ln x and g ( x ) = e2 x + 1.

y = 300 x-2/3 , x = 8, dx = 0.05

f ¢( x ) = f ¢( g ( x )) g ¢( x )

æ 2ö dy = ççç - ÷÷÷ (300) x-5/3 = -200 x-5/3 è 3ø dx

dy = -200 x-5/3dx

= (-200)(8-5/3 )(0.05) æ öæ ö = (-200) çç 1 ÷÷÷çç 1 ÷÷÷ è 32 øè 20 ø 5 =16

6.6

)

2e 2 x e

Exercises

1. True 2. False. Differentials can be used to find an approximate value of 65.

Your Turn 2

Use the approximation formula for f ( x) = developed in Example 2: 1 f ( x +  x) » x + dx 2 x

3. False. When using differentials to estimate the value of f ( x), the error does not change as x increases.

4. True

For this problem, x = 100 and dx = -1.

» 100 +

y = 2 x3 - 5x; x = -2, x = 0.1 dy = (6 x 2 - 5)dx

f (99) = f (100 - 1) 1 (-1) 2 100

y » (6 x 2 - 5)x » [6(-2) 2 - 5](0.1) » 1.9

= 10 - 1 20 = 9.95

y = 4 x3 - 3x; x = 3, x = 0.2 dy = (12 x3 - 3)dx

y » (12 x 2 - 3)x » [12(3)2 - 3](0.2) » 21

Your Turn 3

Use the approximation derived in Example 5:

dV = 4 r dr

y » (3x 2 - 4 x )x

dV = 4 (1.25) 2 (0.025)

= [3(12 ) - 4(1)](-0.1)

» 0.491

= 0.1

» 0.5 3

The maximum error in the volume is about 0.5 mm .

y » (6 x 2 + 2 x - 4)x

W1. Use the chain rule, writing f ( x ) = 1/ 2

as h( g ( x )) with h( x ) = x g ( x ) = x 4 + 2. f ¢( x) = f ¢( g ( x)) g ¢( x)

( )

2 x3

y = 2 x 3 + x 2 - 4 x; x = 2, x = -0.2 dy = (6 x 2 + 2 x - 4)dx

Warmup Exercises

= 1 g ( x)-1/ 2 4 x3 2

y = x 3 - 2 x 2 + 3, x = 1, x = -0.1 dy = (3x 2 - 4 x )dx

For this problem, r = 1.25 and dr = r = 0.025.

6.6

(

1 2e 2 x g ( x)

and

= [6(2)2 + 2(2) - 4](-0.2)

x +2

= (24 + 4 - 4)(-0.2) = -4.8

y =

3x + 2, x = 4, x = 0.15

æ1 ö dy = 3çç (3x + 2)-1/ 2 ÷÷÷ dx çè 2 ø

y »

3 3 (0.15) » 0.060 x » 2(3.74) 2 3x + 2

x4 + 2

Section 6.6 10.

y =

467 4 x - 1; x = 5,  x = 0.08

14.

1 dy = (4 x - 1)-1/2 (4) dx 2

We know 25 = 5, f ( x) = dx = -2.

= 2(4 x - 1)-1/2 dx = 2[4(5) - 1]-1/2 (0.08) = 2(19)-1/2 (0.08) 2(0.08) = = 0.037 (19)-1/2

23 » f ( x) + dy = 5 - 0.2 = 4.8

By calculator,

15.

0.99

We know 1 = 1, so f ( x) = dx = -0.01.

y = 6 x - 3 ; x = 3, x = -0.04 2x + 1 6(2 x + 1) - 2(6 x - 3) dy = dx (2 x + 1)2 12 dx = (2 x + 1) 2 12 y » x (2 x + 1) 2 12 = (-0.04) [2(3) + 1]2 = -0.48 = -0.010 49

0.99 » f ( x) + dy = 1 - 0.005 = 0.995

By calculator,

0.99 » 0.9950.

The difference is |0.995 - 0.9950| = 0. 16.

17.02

We know 16 = 4, f ( x) = dx = 1.02.

x , x = 144,

dy 1 = x-1/2 dx 2 1 dy = dx 2 x 1 1 dy = (1) = 24 2 144

x , x = 16, and

dy 1 = x-1/2 dx 2 1 1 dy = dx = (1.02) 2 x 2 16 1 = (1.02) = 0.1275 8

145

We know 144 = 12, so f ( x) = dx = 1.

x , x = 1,

dy 1 = x-1/2 dx 2 1 dy = dx 2 x 1 dy = (-0.01) = -0.005 2 1

7 x ( x + 1)2 7 = (-0.03) (2 + 1)2 = -0.023

13.

23 » 4.7958.

The difference is |4.8 - 4.7958| = 0.0042.

y = 2 x - 5 ; x = 2, x = -0.03 x +1 ( x + 1)(2) - (2 x - 5)(1) dy = dx ( x + 1)2 7 = dx ( x + 1)2

y »

12.

x , x = 25, and

dy 1 = x-1 2 dx 2 1 1 1 dy = dx = (-2) = - = -0.2. 5 2 x 2 25

y » 2(4 x - 1)-1/2  x

11.

17.02 » f ( x) + dy = 4 + 0.1275 = 4.1275

By calculator, 17.02 » 4.1255. The difference is | 4.1275 - 4.1255 | = 0.0020.

145 » f ( x) + dy = 12 + 1 24 » 12.0417 By calculator, 145 » 12.0416. The difference is |12.0417 – 12.0416| = 0.0001.

17. e0.01

We know e0 = 1, so f ( x) = e x , x = 0, dx = 0.01.

468

Chapter 6 APPLICATIONS OF THE DERIVATIVE 21. Let D = the demand in thousands of pounds; x = the price in dollars.

dy = ex dx

dy = e x dx

D(q) = -3q3 - 2q 2 + 1500

dy = e0 (0.01) = 0.01

(a)

0.01

dD = (-9q 2 - 4q)dq

» f ( x) + dy = 1 + 0.01 = 1.01

D » (-9q 2 - 4q) q

By calculator, e0.01 » 1.0101.

» [-9(4) - 4(2)](0.10)

The difference is |1.01 - 1.0101| = 0.0001.

» -4.4 thousand pounds

-0.002

18. e

We know e0 = 1, f ( x) = e x , x = 0, and dx = -0.002. dy = ex dx e

(b)

q = 6, q = 0.15

D » [-9(36) - 4(6)](0.15) » -52.2 thousand pounds

22. A( x) = 0.04 x3 + 0.1x 2 + 0.5x + 6

dy = e x dx = e0 (-0.002) = -0.002 -0.002

q = 2, q = 0.10

(a)

x = 3, x = 1 dA = (0.12 x 2 + 0.2 x + 0.5)dx

» f ( x) + dy = 1 - 0.002 = 0.998

By calculator, e-0.002 » 0.9980.

= (0.12 x 2 + 0.2 x + 0.5) x

The difference is |0.9980 - 0.998| = 0.

= [(0.12)(3) 2 + (0.2)(3) + (0.5)](1) = 2.18 The change in average cost is about $2.18.

19. ln 1.05

We know ln 1 = 0, so f ( x) = ln x, x = 1, dx = 0.05. dy 1 = dx x 1 dy = dx x 1 dy = (0.05) = 0.05 1 ln 1.05 » f ( x) + dy = 0 + 0.05 = 0.05

By calculator, ln 1.05 » 0.0488. The difference is |0.05 - 0.0488| = 0.0012. 20. ln 0.98

We know ln1 = 0, f ( x) = ln x, x = 1, and dx = -0.02. dy 1 = dx x 1 1 dy = dx = (-0.02) = -0.02 x 1 ln 0.98 » f ( x) + dy = 0 - 0.02 = -0.02

By calculator, ln 0.98 » -0.0202. The difference is | -0.02 - (-0.0202)| = 0.0002.

(b)

x = 5, x = 1 dA = [(0.12)(5) 2 + (0.2)(5) + (0.5)](1) = 4.5 The change in average cost is about $4.50.

23. R( x) = 12,000 ln(0.01x + 1) x = 100, x = 1 12,000 (0.01)dx dR = 0.01x + 1 120 R » x 0.01x + 1 120 » (1) 0.01(100) + 1 » $60 24. P = R - C = 12,000 ln(0.01x + 1) - (150 + 75x) x = 100, x = 1 12, 000 (0.01)dx - 75dx 0.01x + 1 12, 000 P » (0.01) x - 75 x 0.01x + 1 12, 000 » (0.01)(1) - 75(1) 0.01(100) + 1 » 60 - 75 dP =

» -15 The change in profit is a loss of about of $15.

Section 6.6

469

25. If a cube is given a coating 0.1 in. thick, each edge increases in length by twice that amount, or 0.2 in. because there is a face at both ends of the edge.

28.

C =

9 + t2 5(9 + t 2 ) - 2t (5t )

dC =

V = x , x = 4, x = 0.2 dV = 3x 2dx

(9 + t 2 ) 2 45 + 5t 2 - 10t 2

(9 + t 2 )2

V » 3x x = 3(42 )(0.2)

45 - 5t 2

(9 + t 2 )2

= 9.6

(9 + t 2 )2

V = the volume of x beach balls. Then

dC »

hollow. 4 3 r x 3 r = 6 in., x = 5000,  r = 0.03 in.

V =

45 - 5(1) 2

(0.5)

(b) t = 2, t = 0.25 dC »

4 dV =  (3r 2 x + r 3 )  r 3 4 =  (3 ⋅ 36 ⋅ 5000 + 216)(0.03) 3 = 21,608.64

45 - 5(2)2

(9 + 4) 2 = 0.037

29. P(t ) =

21, 608 in.3 of material would be needed.

27. (a) A( x) = y = 0.003631x - 0.03746 x + 0.1012 x + 0.009

(9 + 1)2 40 = (0.5) 100 = 0.2

dV » the volume of material in dr beach balls since they are

(a) t = 1, t = 0.5

Let x = the number of beach balls;

26.

45 - 5t 2

For 1000 cubes 9.6(1000) = 9600 in.3 of coating should be ordered.

dP =

25t 8 + t2 (8 + t 2 )(25) - 25t (2t ) (8 + t 2 ) 2

(8 + t 2 )(25) - 25t (2t ) (8 + t 2 )2

Let x = 1, dx = 0.2.

dP =

 y » (0.010893x 2 - 0.07492 x + 0.1012)x » (0.010893 ⋅ 12 - 0.07492 ⋅ 1 + 0.1012) ⋅ 0.2

[(8 + 4)(25) - (25)(2)(4)](0.5)

(b) t = 3, t = 0.25

» 0.007435

dP =

The alcohol concentration increases by about 0.74 percent.

[(8 + 9)(25) - 25(3)(6)]0.25

(8 + 9)2 » -0.022 million

(b)

The alcohol concentration decreases by about 0.51 percent.

t

(8 + 4)2 = 0.347 million

dy = (0.010893x - 0.07492 x + 0.1012)dx

» -0.005105

(a) t = 2,  t = 0.5

dy = 0.010893x 2 - 0.07492 x + 0.1012 dx

y » (0.010893 ⋅ 32 - 0.07492 ⋅ 3 + 0.1012) ⋅ 0.2

(0.25)

30.

A =  r 2 , r = 1.7 mm, r = -0.1 mm dA = 2 r dr  A » 2 r r

= 2 (1.7)(-0.1) = -0.34 mm 2

470

Chapter 6 APPLICATIONS OF THE DERIVATIVE

31. r changes from 14 mm to 16 mm, so Dr = 2.

35. W (t ) = -3.5 + 197.5e-e

4 3 r 3 4 dV = (3) r 2 dr 3 V =

(a) -e

dW = 197.5e

-e

= 4 (14)2 (2)

32.

W (90) - W (80) » 50.736 - 41.202 = 9.534

33. r increases from 20 mm to 22 mm, so Dr = 2. A =  r2 dA = 2 rdr

36.

V » 4 r 2 r

= 2 (20)(2)

= 4 (4) 2 (0.2)

= 80 mm 2

= 12.8 cm3

1.181 p 94.359 - p

37. r = 3 cm, r = -0.2 cm

(a) Since values for p must be non-negative and the denominator can’t be zero, a sensible domain would be from 0 to about 94. (94.359 - p)(1.181) - 1.181 p(-1) (94.359 - p)2 111.437979 - 1.181 p + 1.181 p

(94.359 - p) 111.437979 dp = (94.359 - p)2

V =

4 3 r 3

dV = 4 r 2dr

V » 4 r 2 r = 4 (9)(-0.2) = -7.2 cm3

38. V = x3, V = 27, x = 3, V = 0.1

We are given p = 60 and dp = 65 - 60 = 5. (94.359 - 60) 2

or about 9.5 kg. 4 V =  r 3, r = 4 cm, r = 0.2 cm 3 dV = 4 r 2dr

 A » 2 r r

111.437979

(b) The actual weight gain is calculated as

= 0.48 mi 2

dA »

-0.01394(t -108.4)

(-0.01394)dt

The pig will gain about 9.3 kg.

= 2 (1.2)(0.2)

-0.01394( t - 108.4)

dW » 9.258

 A » 2 r  r

(b) dA =

-0.01394(t -108.4)

(-1)e

We are given t = 80 and dt = 90 - 80 = 10.

A =  r 2 , r = 1.2 mi,  r = 0.2 mi dA = 2 r dr

34. A( p) =

-0.01394( t - 108.4)

= 2.75315e

V » 4 r 2  r

= 1568 mm

-0.01394(t -108.4)

(5) » 0.472

It will take about 0.47 years. The actual value is

dV = 3x 2dx

V » 3x 2 x V x » 2 3x 0.1 » 3.32 » 0.0037 mm

A(65) - A(60) » 2.615 - 2.062 = 0.553

or about 0.55 years.

Section 6.6 39. V =

471 1 2  r h; h = 13, dh = 0.2 3

V = = dV = =

V » »

43.

2 1 æç h ö÷  ç ÷÷ h 3 çè 15 ø

 775

 225



225

= 4 (5.81)2 (0.003)

⋅ 3h 2dh

= 0.405 » 1.273 in.3 The relative error of the volume is

h 2dh h 2h (132 )(0.2)

» 0.472 cm3

40.

4 3  r ; r = 5.81, r = 0.003 3 4 dV =  (3r 2 )dr 3 4 V »  (3r 2 )r 3 V =

A = x 2 ; x = 3.45, x = 0.002 dA = 2 x dx A » 2 x x = 2(3.45)(0.002)

dV 4 r 2dr = 4 3 V r 3 3dr = r 3(0.003) = 5.81 » 0.001549

44. V = x3; x = 5, dV = 0.3 dV = 3x 2dx

= 0.0138 in.2

V » 3 x 2  x

41. A = x ; x = 4, dA = 0.01

x »

dA = 2 x dx

A » 2 x x 0.01 A x » » » 0.00125 cm 2x 2(4)

3(52 ) » 0.004 ft

1 2 r h 3 2 dV =  rh dr 3 2 V »  rh r 3 2 =  (1.09)(7.284)(0.007) 3 V =

A =  r2 dA = 2 r dr

 A » 2 r r = 2 (4.87)(0.040)

The relative error of the area is dA 2 rdr = A  r2 2dr = r 2(0.040) = 4.87 » 0.01643

3x 2 0.3

45. h = 7.284 in., r = 1.09  0.007 in.

42. r = 4.87 in., r = 0.040

= 1.224 in.2

V

= 0.116 in.3

472

Chapter 6 APPLICATIONS OF THE DERIVATIVE

Chapter 6 Review Exercises

13.

f ( x) = x3 + 2 x 2 - 15x + 3; [-4, 2] f ¢( x) = 3x 2 + 4 x - 15 = 0 when

1. False: The absolute maximum might occur at the endpoint of the interval of interest.

(3x - 5)(x + 3) = 0 x =

2. True

5 3

or x = -3.

f (-4) = 31

3. False: It could have either. For example

f (-3) = 39 æ5ö 319 f çç ÷÷÷ = çè 3 ø 27

f ( x) = 1/(1 - x 2 ) has an absolute minimum of

1 on (-1, 1).

f (2) = -11

4. True

Absolute maximum of 39 at -3; absolute at 53 . minimum of - 319 27

5. True 6. True

14.

f ( x) = -2 x3 - 2 x 2 + 2 x - 1; [-3, 1] f ¢( x ) = - 6 x 2 - 4 x + 2 f ¢( x) = 0 when

7. True

3x 2 + 2 x - 1 = 0

8. True

(3x - 1)( x + 1) = 0

9. True

1 or x = -1. 3

x =

10. True

f (-3) = 29

11.

f (-1) = -3 æ1ö 17 f çç ÷÷÷ = çè 3 ø 27

f ( x) = -x + 6 x + 1; [-1, 6] f ¢( x) = -3x 2 + 12 x = 0 when x = 0, 4. f (-1) = 8

f (1) = -3

f (0) = 1

Absolute maximum of 29 at -3; absolute minimum of -3 at -1 and 1.

f (4) = 33 f (6) = 1

Absolute maximum of 33 at 4; absolute minimum of 1 at 0 and 6.

17. (a)

f ( x) =

2ln x

f ¢( x ) =

12.

; [1, 4]

f ( x) = 4 x3 - 9 x 2 - 3; [-1, 2]

( 2x ) - (2 ln x)(2 x)

x4 2 x - 4 x ln x

f ¢( x) = 12 x 2 - 18 x = 0 when x = 0, 32 .

f (-1) = -16

f (0) = -3 æ3ö f çç ÷÷÷ = -9.75 çè 2 ø

f ¢( x) = 0 when

x4 2 - 4 ln x x3

2 - 4ln x = 0

f (2) = -7

Absolute maximum of -3 at 0; absolute minimum of -16 at -1.

2 = 4ln x 0.5 = ln x e0.5 = x x » 1.6487.

Chapter 6 Review

473 x e

f ( x)

0.5

0.36788

0.17329

21. x 2 - 4 y 2 = 3x3 y 4

Maximum is 0.37; minimum is 0. (b) [2, 5]

d 2 d (x - 4 y2 ) = (3x3 y 4 ) dx dx dy dy 2x - 8 y = 9 x 2 y 4 + 3x3 ⋅ 4 y 3 dx dx dy (-8 y - 3x3 ⋅ 4 y3 ) = 9x2 y 4 - 2x dx

Note that the critical number of f is not in the domain, so we only test the endpoints. x

f ( x)

2 0.34657

22. x 2 y 3 + 4 xy = 2

5 0.12876

d 2 3 d ( x y + 4 xy) = (2) dx dx æ dy ö dy =0 2 xy 3 + 3 y 2 çç ÷÷÷ x 2 + 4 y + 4 x çè dx ø dx dy = -2 xy 3 - 4 y (3x 2 y 2 + 4 x) dx

Maximum is 0.35, minimum is 0.13. 18. f ( x) =

e2 x x2

First find the critical numbers. f ¢( x ) = =

dy 2x - 9x2 y 4 = dx 8 y + 12 x3 y 3

-2 xy 2 - 4 y dy = dx 3x 2 y 2 + 4 x

2e2 x x 2 - e2 x (2 x) ( x 2 )2

23. 2 y - 1 = 9 x 2/3 + y

2e2 x ( x 2 - x) x4 2

f ¢( x) = 0 when x - x = ( x)( x - 1) = 0, that is, at x = 1. (Both the derivative and the original function are undefined at x = 0.) (a) The function f is continuous on the interval

[1/2, 2 ] and the critical number lies in this interval, so we evaluate the function at three points: x

d d [2( y - 1)1/2 ] = (9 x 2/3 + y) dx dx dy dy 1 = 6 x-1/3 + 2 ⋅ ⋅ ( y - 1)-1/2 dx dx 2 dy = 6 x-1/3 [( y - 1)-1/2 - 1] dx 1-

y -1 y -1

⋅

6 y -1 dy = 1/3 dx x (1 - y - 1)

f ( x)

1/2 10.873 1

24. 9 x + 4 y 3 = 2 y

7.389

2 13.650

The absolute minimum is 7.39 and the absolute maximum is 13.65. (b) The function f is continuous on the interval [1, 3]. The critical number coincides with an endpoint so we need to look at only two function values: x

f ( x)

7.389

dy 6 = 1/3 dx x

d d 1/2 (9 x + 4 y 3 ) = 2  (y ) dx dx 9 -1/2 dy 1 dy + 12 y 2 = 2  y-1/2  x 2 dx 2 dx æ ö 9 çç 1 - 12 y 2 ÷÷ dy = ÷ çç 1/2 ÷ø÷ dx 2 x1/2 èy 9

= 2 x1/2

3 44.825

The absolute minimum is 7.39 and the absolute maximum is 44.83.

1 - 12 y 5/2 dy dx y1/2

9 y1/2 dy = 1/2 dx 2 x (1 - 12 y 5/2 ) =

9 y 2 x (1 - 12 y5/2 )

474 25.

Chapter 6 APPLICATIONS OF THE DERIVATIVE 6 + 5x 1 = 2 - 3y 5x

27. ln( xy + 1) = 2 xy3 + 4

5x(6 + 5x) = 2 - 3 y 30 x + 25x 2 = 2 - 3 y d d (30 x + 25 x 2 ) = (2 - 3 y) dx dx dy 30 + 50 x = -3 dx 30 + 50 x dy = 3 dx

26.

d d [ln( xy + 1)] = (2 xy3 + 4) dx dx d dy d 1  ( xy + 1) = 2 y 3 + 2 x  3 y 2 + (4) xy + 1 dx dx dx ö dy d dy 1 æç (1) ÷÷ = 2 y 3 + 6 xy 2 + çy + x xy + 1 èç dx dx ÷ø dx y x dy dy  + = 2 y 3 + 6 xy 2 xy + 1 xy + 1 dx dx æ x ö dy y çç - 6 xy 2 ÷÷÷ = 2 y3 ÷ø dx ççè xy + 1 xy + 1 y

2 y3 dy xy + 1 = x dx - 6 xy 2

x + 2y = y1/2 x - 3y

xy + 1

x + 2 y = y1/2 ( x - 3 y) =

d d 1/2 ( x + 2 y) = [ y ( x - 3 y)] dx dx æ dy dy ö 1+ 2 = y1/2 çç1 - 3 ÷÷÷ çè dx dx ø

dy 1 ( x - 3 y) y-1/2 dx 2 dy dy 1 dy 1+ 2 = y1/2 - 3 y1/2 + xy-1/2 dx dx 2 dx 3 1/2 dy - y 2 dx +

1 -1/2 3 dy xy + y1/2 ) = y1/2 - 1 2 2 dx æ ö 9 1 2 y1/2 çç 2 + y1/2 - xy-1/2 ÷÷÷ çè ø dy 2 2 = y1/2 - 1 1/2 dx 2y

(2 + 3 y1/2 -

æ 4 y1/2 + 9 y - x ö÷ dy çç ÷÷ = y1/2 - 1 çç ÷ çè 2 y1/2 ø÷ dx

2 y3 ( xy + 1) - y x - 6 xy 2 ( xy + 1) 2 xy 4 + 2 y 3 - y x - 6 x 2 y 3 - 6 xy 2

28. ln( x + y) = 1 + x 2 + y3 d d [ln( x + y)] = (1 + x 2 + y3 ) dx dx d dy 1  ( x + y) = 2 x + 3 y 2  x + y dx dx dy ö÷ 1 æç 2 dy ÷÷ = 2 x + 3 y  çç1 + x + yè dx ø dx æ 1 ö 1 2 ÷÷ dy çç = 2x y 3 ÷ çè x + y ÷ø dx x+ y

2 y - 2 y1/2 dy = dx 4 y1/2 + 9 y - x

2 x - x +1 y dy = 1 - 3y2 dx x+y

= = =

2 x( x + y ) - 1 1 - 3 y 2 ( x + y) 2 x 2 + 2 xy - 1 1 - 3xy 2 - 3 y 3 1 - 2 x 2 - 2 xy 3xy 2 + 3 y 3 - 1

Chapter 6 Review 29.

475 To find the slope m of the tangent line, substitute -3 for x and 1 for y.

2 y - 4 xy = -22, tangent line at (3, 2). d dx

( 2 y - 4xy ) = dxd (-22)

1 dy æç dy ö - ç 4 y + 4 x ÷÷÷ = 0 (2)(2 y)-1/2 ç 2 dx è dx ø dy = 4y ((2 y)-1/ 2 - 4 x) dx 4y dy = 1 dx - 4x

6(12 ) - 2(-3)(1) 1 = 12

The equation of the tangent line is y - y1 = m( x - x1)

To find the slope m of the tangent line, substitute 3 for x and 2 for y.

1 ( x - (-3)) 12 3 1 y -1= x 12 12 1 5 y = x+ . 12 4 y -1 =

4y 1 - 4x 2 2y

4(2)

6 y 2 - 2 xy

1 - 4(3) 2(2)

We can also write this as 1x - 12 y = -15.

8 12 2

= 1

33. y = 8x3 - 7 x 2 , dx = 4, x = 2 dt

16 16 = =1 - 24 23

dy d = (8x3 - 7 x 2 ) dt dt dx dx = 24 x 2 - 14 x dt dt

The equation of the tangent line is y - y1 = m( x - x1)

= 24(2) 2 (4) - 14(2)(4)

16 ( x - 3) 23 48 16 y-2=- x 23 23 16 94 y =- x+ . 23 23 y-2 =-

= 272

34.

We can also write this equation as 16 x + 23 y = 94.

30. 8 y 3 - 4 xy 2 = 20, tangent line at (-3, 1).

(

9 - 4 x dx ; = -1, x = -3 3 + 2 x dt dy (-4)(3 + 2 x) - (2)(9 - 4 x) dx = dt dt (3 + 2 x)2 y =

-30

(3 + 2 x) -30

2 dt 2

[3 + 2(-3)]

(-1) =

30 10 = 9 3

)

d d 8 y 3 - 4 xy 2 = (20) dx dx ö dy æç dy 24 y 2 - ç 8 xy + 4 y 2 ÷÷÷ = 0 ç è ø dx dx dy (24 y 2 - 8xy) = 4 y2 dx 4 y2 dy = dx 24 y 2 - 8xy =

y2 2

6 y - 2 xy

35.

1+ 1-

x dx , = -4, x = 4 x dt dy d éê 1 + x ùú = dt dt êêë 1 - x úúû é ù 1 -1/2 dx ê 1- x 2 x ú dt ê ú ê -1/2 dx ú 1 ê - 1 1 + x (- 2 ) x ú dt úû ê = ë 2 1- x y =

)(

(

) (

)

(

)

476

Chapter 6 APPLICATIONS OF THE DERIVATIVE

= =

é (1 - 2) 1 (-4) ù ( 2⋅2 ) ê ú ê ú ê - (1 + 2) -1 (-4) ú ( 2⋅2 ) úû êë

38. y = e

ö÷ dy d æç 1 ÷÷ çç 2 = dt dt çè e x + 1 ÷÷ø

(1 - 2)2

1- 3 = -2 1

36.

x + 5y = 2; x - 2y dx = 1, x = 2, y = 0 dt

= =

x 2 + 5 y = 2( x - 2 y )

x2 + 5 y = 2 x - 4 y 9 y = -x 2 + 2 x

1 (-x 2 + 2 x) 9 1 2 y = - x2 + x 9 9 æ 2 dy 2 ö dx = çç - x + ÷÷÷ çè 9 dt 9 ø dt éæ 2 ö 2ù = ê çç - ÷÷÷ (2) + ú (1) ê çè 9 ø 9 úû ë 4 2 2 =- + =9 9 9 y =

37. y = xe3x ;

dx = 3, x = 1 + 1 dt ;

41.

= (1 + 3 ⋅ 1)e3(1) (-2) = -8e3

d x2 (e + 1) dt

é x2 d 2 ù d ê e ⋅ ( x ) + (1) ú ê dt dt úû + 1)2 ë

-1 (e x

é x2 dx ù ê e ⋅ 2x ú dt úû + 1)2 êë

-1 2

(e x -1

(e + 1) 2 -6e

[e ⋅ 2 ⋅ 1 ⋅ 3]

(e + 1) 2

3x - 7 ; x = 2, x = 0.003 2x + 1 (3)(2 x + 1) - (2)(3x - 7) dy = dx (2 x + 1)2

y »

dy d = ( xe3x ) dt dt dx 3x d = ⋅ e + x ⋅ (e3x ) dt dt dx 3x dx 3x = ⋅ e + xe ⋅ 3 dt dt dx = (1 + 3x)e3x dt

(e x + 1)2

⋅

y =

dy =

dx = -2, x = 1 dt

-1 2

17 (2 x + 1)2 17 (2 x + 1)2 17

dx x

(2[2] + 1)2 = 0.00204

(0.003)

42. y = 8 - x 2 + x3, x = -1,  x = 0.02 dy = (-2 x + 3x 2 )dx y » (-2 x + 3x 2 ) x = [-2(-1) + 3(-1)2 ](0.02) = 0.1

Chapter 6 Review

477

43. -12 x + x3 + y + y 2 = 4

45. (a)

dy d (-12 x + x3 + y + y 2 ) = (4) dx dx dy dy -12 + 3x 2 + + 2y =0 dx dx dy = 12 - 3x 2 (1 + 2 y) dx dy 12 - 3x 2 = dx 1 + 2y

(a) If

P( x) = -x3 + 10 x 2 - 12 x P¢( x) = -3x 2 + 20 x - 12 = 0 3x 2 - 20 x + 12 = 0 (3x - 2)( x - 6) = 0 3x - 2 = 0 2 x = 3

x-6= 0

x =6

P¢¢( x) = -6 x + 20 æ2ö P¢¢ çç ÷÷÷ = 16, çè 3 ø

dy = 0, dx

which implies that x = 23 is the location of

12 - 3x 2 = 0

the minimum.

12 = 3x 2 2 = x.

P¢¢(6) = -16,

x = 2:

which implies that x = 6 is the location of the maximum. Thus, 600 boxes will produce a maximum profit.

-24 + 8 + y + y 2 = 4 y + y 2 = 20

(b) Maximum profit = P(6)

y + y - 20 = 0 ( y + 5)( y - 4) = 0

= -(6)3 + 10(6)2 - 12(6) = 72

The maximum profit is $720.

y = -5 or y = 4 (2, - 5) and (2, 4) are critical points. x = -2:

46. Let x = the length and width of a side of the base; h = the height.

The volume is 32 m3; the base is square and there is no top. Find the height, length, and width for minimum surface area.

24 - 8 + y + y 2 = 4 y + y 2 = -12 y 2 + y + 12 = 0 y =

Volume = x 2h 2

-1  1 - 48 2

This leads to imaginary roots. x = -2 does not produce critical points.

(b)

x y1 y2 1.9 -4.99 3.99 2 4 -5 2.1 -4.99 3.99

x 2h = 32 32 h= 2 x Surface area = x 2 + 4 xh æ 32 ö A = x 2 + 4 x çç 2 ÷÷ çè x ÷ø = x 2 + 128 x-1 A¢ = 2 x - 128x-2

The point (2, - 5) is a relative minimum.

If A¢ = 0,

The point (2, 4) is a relative maximum. (c) There is no absolute maximum or minimum for x or y.

2 x3 - 128 x2

x3 = 64 x = 4.

478

Chapter 6 APPLICATIONS OF THE DERIVATIVE A¢¢( x) = 2 + 2(128) x-3 A¢¢(4) = 6 > 0

16 r -

The minimum is at x = 4, where h=

C ¢(r ) = 16 r -

= 2. 2

r » 1.684 C ¢¢(r ) = 16 + 480 > 0, so r = 1.684 3

Surface area of cylinder open at one end = 2 rh +  r 2. 2

V =  r h = 27 27 27 = 2 h= 2 r r æ 27 ö÷ A = 2 r çç 2 ÷÷ +  r 2 çè r ø = 54 r -1 +  r 2 A¢ = -54 r -2 + 2 r

minimizes cost. h=

 (1.684) 2

= 4.490

2f M = k

2(12)20,000 0.15

3, 200,000 » 1789

Ordering 1789 rolls each time minimizes annual cost.

r = 27 r = 3.

If r = 3, A¢¢ = 108 r -3 + 2 > 0,

so the value at r = 3 is a minimum.

50. M = 180,000 cases sold per year k = $12, cost to store 1 case for 1 yr f = $20, fixed cost for order x = the number of cases per order

For the minimum cost, the radius of the bottom should be 3 inches. 2

48. V =  r h = 40, so h =

 r2

2f M k

2(20)(180,000) 12

600,000

= 100 60 » 775

The store should order 775 cases each time.

æ 80 ö Cost = C (r ) = 4(2 r ) + 3çç ÷÷÷ çè r ø 2

240 r

x =

A = 2 r 2 + 2 rh æ 40 ö = 2 r 2 + 2 r çç 2 ÷÷ çè  r ø÷ 80 = 2 r 2 + r

= 8 r 2 +

r

49. Here k = 0.15, M = 20,000, and f = 12. We have

54

The radius should be 1.684 in. and the height should be 4.490 in.

q =

If A¢ = 0, 2 r =



The dimensions are 2 m by 4 m by 4 m. 47. Volume of cylinder =  r 2h

16 r 3 = 240 15 r3 =

32 4

240

51. Use equation (3) from Section 6.3 with k = 1,

M = 128, 000, and f = 10. q = =

2f M = k

2(10)(128,000) 1

2,560,000 = 1600

Chapter 6 Review

479

The number of lots that should be produced annually is

55. A =  r 2 ; dr = 4 ft/min, r = 7 ft dt dA dr = 2 r dt dt dA = 2 (7)(4) dt dA = 56 dt

M 128,000 = = 80. q 1600 52. Use equation (3) from Section 6.3 with k = 2, M = 240,000, and f = 15. q = =

2f M = k

The rate of change of the area is 56 ft 2 /min.

2(15)(240,000) 2

56.

3,600,000 » 1897.4

T (1897) » 3794.7333 and T (1898) » 3794.7334. Since T (1897) < T (1898), then the number of batches per year should be

dx = rx( N - x) dt = rxN - rx 2 d 2x dt

dx dx - 2rx dt dt dx = r ( N - 2 x) dt = r[rx( N - x)]( N - 2 x) = rN

M 240,000 = » 127. q 1897

53.

ln q = D - 0.447 ln p 1 dq 0.47 =q dp p dp q = -0.47 dq p p dq E =q dp pæ qö = - çç -0.47 ÷÷÷ ç qè p ÷ø

= r 2 x( N - x)( N - 2 x) d 2 x = 0 when x = 0, x = N , or x = N . 2 dt 2

(

( N2 , N ), ddt x < 0; therefore, the curve is 2

concave downward.

E is less than l, so the demand is inelastic. q =

concave upward.

= 0.47

54.

) dt

On 0, N2 , d 2x > 0; therefore, the curve is

Hence x = N2 is a point of inflection. 57. (a)

7.5

p dq A = -k k +1 dp p

p dq E =-  q dp æ ö æ A ö÷ çç p ÷÷ ç = ç - A ÷÷÷ (-k ) ççç k +1 ÷÷ ç ÷ èç p ø÷÷ ççè p k ÷ø

(b) We use a graphing calculator to graph M ¢(t ) = -0.4321173 + 0.1129024t - 0.0061518t 2 + 0.0001260t 3 - 0.0000008925t 4

k +1 öæ æ ö ç p ÷÷÷çç -k A ÷÷÷ = çç ç 1 + k ÷ ÷÷ø A ÷øèç p èç

on [5, 51] by [0, 7.5]. We find the maximum value of M ¢(t ) on this graph at about 15.41, or on about the 15th day.

= k

The demand is elastic when k > 1 and inelastic when k < 1.

480

Chapter 6 APPLICATIONS OF THE DERIVATIVE

58. (a)

60.

dV = 0.9 ft 3 /min dt find dr when r = 1.7 ft dt 4 3 r 3 dV dr 4 =  (3)r 2 dt dt 3 2 dr = 4 r dt dr 0.9 = 4 (1.7 2 ) dt dr 0.9 = dt 4 (1.7 2 ) V =

(b) To find where the maximum and minimum numbers occur, use a graphing calculator to locate any extreme points on the graph. One critical number is formed at about 87.78. t P(t ) 0 237.09 87.78 43.56 95 48.66

The maximum number of polygons is about 237 at birth. The minimum number is about 44. 59. Let x = the distance from the base of the ladder to the building; y = the height on the building at the top of the ladder.

dr 0.9 = » 0.0248 dt 11.56

The radius is changing at the rate of 0.9 » 0.0248 ft/min. 11.56 61. Let x = one-half of the width of the triangular cross section; h = the height of the water; V = the volume of the water. dV dt

= 3.5 ft 3 /min.

when h = Find dV dt

æ Area of ö÷ ç V = çç triangular ÷÷÷ (length) ççè side ÷ø

dy = -2 dt 502 = x 2 + y 2 dx dy + 2y 0 = 2x dt dt dx y dy =dt x dt

When x = 30, y =

1 . 3

Area of triangular cross section

2500 - (30)2 = 40.

1 (base)(altitude) 2 1 = (2 x)(h) = xh 2 =

So dx -40 80 8 = (-2) = = dt 30 30 3

The base of the ladder is slipping away from the building at a rate of 83 ft/min.

By similar triangles, 2hx = 12 , so x = h.

Chapter 6 Review

481 V = ( xh)(4)

Set

= h ⋅4 = 4h 2

488 + 212h - 12h 2 = 0 12h 2 - 212h - 488 = 0

dV dh = 8h dt dt 1 dV dh ⋅ = 8h dt dt 1 dh (3.5) = dt 8 1

3h 2 - 53h - 122 = 0 53  2809 + 1464 6 » -2.06 or h = 19.73 h can’t be negative, so h » 19.73. h=

(3)

dh 21 = = 1.3125 dt 16

The depth of water is changing at the rate of 1.3125 ft/min. 62.

V =

dV = 0. dh

4 3  r , r = 4in., r = 0.02in. 3

dV = 4 r 2dr

Thus, l » 122 - 4h » 43.1.

The length that produces the maximum volume is about 43.1 inches. 65. We need to minimize y. Note that x > 0. dy x 2 = dx 8 x

V » 4 r 2 r

= 4 (4) 2 (0.02) = 1.28 or about 4.021

Set the derivative equal to 0. x 2 - =0 x 8 x 2 = 8 x

The volume of the coating is 1.28 in.3 or about 4.021 in.3.

63.

x 2 = 16 x = 4

A = s 2 ; s = 9.2, s = 0.04 ds = 2s ds A » 2s s = 2(9.2)(0.04)

Since lim y = ¥, lim y = ¥, and x = 4 is x0

= 0.736in.2

64.

x ¥

the only critical value in (0, ¥), x = 4 produces a minimum value.

V = l ⋅ w⋅h w= 4+h l + g = 130; g = 2(w + h) l + 2(w + h) = 130 l + 2w + 2h = 130 l = 130 - 2w - 2h = 130 - 2(4 + h) - 2h = 122 - 4h

42 1 - 2 ln 4 + + 2 ln 6 16 4 = 1.25 + 2(ln 6 - ln 4) = 1.25 + 2 ln1.5

y =

The y coordinate of the Southern most point of the second boat’s path is1.25 + 2 ln 1.5. 66. Let x = width of play area; y = length of play area.

V = l ⋅ w⋅h = (122 - 4h)(4 + h)h = 488h + 106h 2 - 4h3 dV = 488 + 212h - 12h 2 dh

482

Chapter 6 APPLICATIONS OF THE DERIVATIVE An equation describing the amount of fencing is 900 = 2 x + y or y = 900 - 2 x.

Then A = xy so

40 - 30 + 5

302 + 402 » 18.67 seconds. 3

68. Distance on shore: 25 - x feet

Speed on shore: 5 feet per second

A( x) = x(900 - 2 x) = 900 x - 2 x 2.

x 2 + 402 feet

Distance in water:

If A¢( x) = 900 - 4 x = 0, x = 225.

Speed in water: 3 feet per second

Then y = 900 - 2(225) = 450.

The total travel time t is t = t1 + t2 = v1 + v 2 .

A¢¢( x) = -4 < 0, so the area is maximized if the dimensions are 225 m by 450 m.

67. Distance on shore: 40 - x feet

x 2 + 402 feet

Speed in water: 3 feet for second The total travel time t is d d t = t1 + t2 = 1 + 2 . v1 v2 40 - x t ( x) = + 5

25 - x + 5

x 2

3 x + 1600

x 2 + 402 3

25 2 x = x 2 + 1600 9 1600 ⋅ 9 x2 = 16 40 ⋅ 3 x = = 30 4

(Discard the negative solution.) To minimize the time, he should walk 40 - x = 40 - 30 = 10 ft along the shore before paddling toward the desired destination. The minimum travel time is

1 5

25 2 x = x 2 + 1600 9 16 2 x = 1600 9 1600 ⋅ 9 x2 = 16 40 ⋅ 3 x = = 30 4

1 5

5x = 3 x 2 + 1600 5x = x 2 + 1600 3

Minimize the travel time t ( x). If t ¢( x) = 0: 3 x 2 + 1600

Minimize the travel time t ( x). If t ¢( x) = 0:

x 2 + 402 3

x x 2 + 1600 + 5 3 1 1 1 2 t ¢( x) = - + ⋅ ( x + 1600)-1/2 (2 x) 5 3 2 x 1 =- + 2 5 3 x + 1600

x x 2 + 1600 = 8- + 5 3 1 1 1 2 t ¢( x) = - + ⋅ ( x + 1600)-1/2 (2 x) 5 3 2 1 x =- + 2 5 3 x + 1600

= 5-

Speed on shore: 5 feet per second Distance in water:

t ( x) =

(Discard the negative solution.) x = 30 is impossible since the closest point on the shore to the desired destination is only 25 ft from where he is standing. Check the end points. x 0 25

t ( x) 18.33 15.72

The time is minimized when x = 25. 25 - x = 25 - 25 = 0 ft, so the mathematician should start paddling where he is standing. The travel time is t (25) = 15.72 sec.

Extended Application

483

Extended Application: A Total Cost Model for a Training Program 1.

æ m - 1 ÷ö C1 + DtC2 + DC3 çç ÷ ç è 2 ÷ø m DC3 C Z ¢(m) = - 12 + 0 + 2 m DC3 C = - 12 + 2 m Z (m) =

Z ¢(m) = 0 when

N = mD = 3 ⋅ 3 = 9 There should be 9 trainees per batch.

7. The graphs below show total cost per batch of trainees in dollars for four scenarios in which one of the values used Exercise 3 is allowed to vary. Demand D for trainees per month varies from 1 to 10, with other values as in Exercise 3:

Total cost

DC3 C = 12 2 m 2C1 2 m = DC3

6. Since Z (3) < Z (4), the optimal time interval is 3 months.

2C1 . DC3

1 10

5000 0

D = 3, t = 12, C1 = 15,000, C3 = 900 m =

2(15,000) = 3(900)

100 » 3.33 9

2 10

é m(m-1) ù DC3 ê 2 ú C1 + mDtC2 ë û Z ( m) = + m m æ m - 1 ö÷ C = 1 + DtC2 + DC3 çç çè 2 ÷÷ø m C1 = 15,000; D = 3, t = 12, C2 = 100; C3 = 900

Total cost

= 4 and m

1 10

0 0

4 4 1 10 2 10 Fixed cost

3 10

Marginal cost per trainee per month C2 varies from $50 to $250, with other values as in Exercise 3:

Z (m ) = Z (4) =

5 10 Trainee demand

æ 4 - 1 ö÷ 15,000 + 3(12)(100) + 3(900) çç ÷ ç è 2 ÷ø 4

= 3750 + 3600 + 4050 = $11,400 Z (m- ) = Z (3) 15,000 (900)(3 - 1) = + 3600 + 3 3 2 = 5000 + 3600 + 2700 = $11,300

Total cost

Fixed cost C1 of training a batch varies from $5000 to $20,000, with other values as in Exercise 3:

4. 3 < 3.33 < 4

1.5 10

2 10

1 10

100 150 200 Marginal cost

250

484

Chapter 6 APPLICATIONS OF THE DERIVATIVE Salary for jobless trainee C3 varies from $100 to $1800, with other values as in Exercise 3: 4

Total cost

2 10

1 10

500 1000 1500 Jobless salary

2000

Chapter 7

INTEGRATION 7.1

Antiderivatives

Your Turn 4

Your Turn 1

Find an antiderivative for the function f ( x) = 8x 7 .

x3 - 2 dx = x

æ 3 2 ÷÷ö ç x ÷ ççç x ÷÷ø è x

1/2

5/ 2

-1/2

general antiderivative is x8 + C.

ò x dx - ò x dx = ò x dx - 2ò x dx

Your Turn 2

Since the derivative of x n is nx n-1, the derivative of x8

is 8x7 . Thus x8 is an antiderivative of 8x 7 . The

Find

æ2 ö 2 7/2 x - 2 çç x1/2 ÷÷÷ + C ç è ø 7 1 2 = x7/2 - 4 x1/2 + C 7

ò t dt. 4

Use the power rule with n = -4. 1

òt

dt = 4

Your Turn 5 æ3 ö çç + e-3x ÷÷ dx = ÷ø çè x

t -4dt

t -4 + 1 +C -4 + 1

t -3 +C -3 1 =- 3 +C 3t

1 = 3ln | x | - e-3x + C 3 Your Turn 6

Suppose an object is thrown down from the top of the 2717-ft tall Burj Khalifa with an initial velocity of -20 ft/sec. Find when it hits the ground and how fast it is traveling when it hits the ground.

Your Turn 3

ò (6x + 8x - 9) dx. 2

In Example 11 (b) we derived the formulas

Use the sum or difference rule and the constant multiple rule.

6 x 2 dx +

Now use the power rule on each term. 6

v (t ) = -32t - 20

ò ò ò = 6 x 2 dx + 8 x dx - 9 dx ò ò ò

(6 x 2 + 8 x - 9) dx =

-3 x

Find

ò x dx + ò e dx 1 =3 ò x dx + ò e dx

ò x2 dx + 8ò x dx - 9ò dx æ x 3 ö÷ æ x 2 ö÷ ç ç = 6çç ÷÷÷ + 8çç ÷÷÷ - 9 x + C çè 3 ÷ø èç 2 ø÷

8 x dx -

9 dx

s (t ) = -16t 2 - 20t + 1100

for the velocity v and distance above the ground s for an object thrown down from the Willis Tower. The only change required for the new problem is to change 1100 to the height of the Burj Khalifa, 2717, so we have s(t ) = -16t 2 - 20t + 2717.

To find when the object hits the ground we solve s(t ) = -16t 2 - 20t + 2717 = 0.

= 2 x3 + 4 x2 - 9 x + C

485

486

Chapter 7 INTEGRATION

Use the quadratic formula.

202 - 4(-16)(2717) 2(-16) t » -13.62 or t » 12.42 t =

20 

Only the positive root is relevant. To find the speed on impact, substitute t = 12.42 into the formula for v.

7.1

Exercises

False. If F ( x) and G( x) are both antiderivatives of f ( x), then there is a constant C such that F ( x ) - G ( x ) = C.

False.

True

If F ( x) and G( x) are both antiderivatives of f ( x), then there is a constant C such that

v(12.42) = -32(12.42) - 20 » -417 The object hits the ground after 12.42 sec, traveling downward at 417 ft/sec. Your Turn 7 f ¢( x) = 3x1/2 + 4

x3dx =

x4 +C 4

a x dx =

ax + C , for a > 0, a ¹ 1 ln a

F ( x ) - G ( x ) = C.

ò f ¢( x) dx = ò (3x1/2 + 4) dx = 3 x1/2 dx + 4 dx ò ò

The two functions can differ only by a constant.

f ( x) =

ò 6 dk = 6ò 1 dk =6

æ2 ö = 3çç x 3/2 ÷÷÷ + 4 x + C çè 3 ø

1 = 6 ⋅ k 0+1 + C 1 = 6k + C

= 2 x 3/2 + 4 x + C

Since the graph of f is to go through the point (1, -2), f (1) = -2.

ò k 0dk

10.

ò 9 dy = 9ò 1 dy = 9ò y dy 0

9 0 +1 y +C 1 = 9y + C =

2(1)3/2 + 4(1) + C = -2 2 + 4 + C = -2 C = -8

Thus,

11. f ( x) = 2 x3/2 + 4 x - 8.

ò (2z + 3) dz = 2 z dz + 3 z dz ò ò 0

1 1+1 1 z + 3⋅ z 0 +1 + C 1+1 0 +1

7.1

Warmup Exercises

= 2⋅

W1.

f ( x) = 5x 4 + 6 x = 5 x 4 + 6 x1/ 2 æ1 ö f ¢( x) = (5) 4 x3 + (6) çç x-1/ 2 ÷÷÷ çè 2 ø

= z 2 + 3z + C

( )

W2.

= 20 x3 +

3 x

5/ 2

f ( x ) = 7e

- 8x

)

ò (3x - 5) dx = 3 x dx - 5 x dx ò ò 0

= 3⋅

æ5 ö f ¢( x) = (7) 3e3x - (8) çç x(5 / 2) -1 ÷÷÷ çè 2 ø

(

12.

1 2 1 x -5⋅ x +C 2 1

3x 2 - 5x + C 2

= 21e3x - 20 x3 / 2

Section 7.1 13.

487

ò (6t - 8t + 7) dt = 6 t dt - 8 t dt + 7t t dt ò ò ò 2

18.

t1/4 +1 (t1/4 +  1/4 ) dt = 1 +  1/4t + C + 1 4 =

6t 8t + 7t + C 3 2

= 2t - 4t + 7t + C

14.

ò (5x - 6x + 3) dx = 5 x dx - 6 x dx + 3 x dx ò ò ò 2

19.

5 x3 6x2 + 3x + C 3 2

5 x3 - 3x 2 + 3x + C 3

20. 15.

ò (4z + 3z + 2z - 6) dz = 4 z dz + 3 z dz + 2 z dz ò ò ò -6 z dz ò 3

4z 4 3z 3 2z 2 + + - 6z + C 4 3 2

21.

= z 4 + z3 + z 2 - 6z + C

16.

ò (16 y + 9 y - 6 y + 3) dy = 16 y dy + 9 y dy ò ò - 6 y dy + 3 dy ò ò 3

22.

= 4 y 4 + 3 y3 - 3 y 2 + 3 y + C

17.

ò (5 z + 2) dz = 5ò z dz + 2 ò dz 1/2

5 z 3/2 3 2

2z + C

æ2ö = 5 çç ÷÷÷ z 3/2 + çè 3 ø =

10 z 3/2 + 3

2z + C

5x 4 40 x 2 +C 4 2

5x 4 - 20 x 2 + C 4

x7 4x4 3x3 + + +C 7 4 3

x7 + x 4 + x3 + C 7

ò (4 v - 3v3/2 ) dv = 4 v1/2dv - 3 v 3/2dv ò ò =

16 y 4 9 y3 6 y2 + + 3y + C 4 3 2

+  1/4t + C

4t 5

ò x2 (x4 + 4x + 3) dx = ò (x6 + 4x3 + 3x2 ) dx

+  1/4t + C

5 4 5/4

ò 5x(x2 - 8) dx = ò (5x3 - 40x) dx

t 5/4

4v 3/2 3 2 3/2

8v 3

3v 5/2 5 2 5/2

6v 5

+C +C

ò (15x x + 2 x ) dx = 15 x( x1/2 ) dx + 2 x1/2 dx ò ò = 15 x3/2 dx + 2 x1/2 dx ò ò =

15x5/2 5 2

2 x3/2 3 2

æ2ö æ2ö = 15 çç ÷÷ x5/2 + 2 çç ÷÷ x3/2 + C çè 5 ÷ø èç 3 ÷ø = 6 x5/2 +

4 x3/2 +C 3

488 23.

Chapter 7 INTEGRATION

ò (10u = 10 =

24.

3/2

27.

3/2

10u 5/2 5 2

5/2

14u 7/2 7 2

æ3 çç çç 3 èy

÷ dy y ÷÷ø

3 -3

-1/2

-3

-1/2

dy dy

æ2ö æ2ö = 10 çç ÷÷ u 5/2 - 14 çç ÷÷ u 7/2 + C çè 5 ÷ø çè 7 ø÷

æ -1/2 ö÷ æ y-2 ö÷ ç ÷÷ + C ÷÷ -  ççç y =  3 çç çç 1 ÷÷ çè -2 ÷÷ø è 2 ø

= 4u 5/2 - 4u 7/2 + C

ò (56t =

5/2

+ 18t 7/2 ) dt

ò t dt + 18ò t dt 5/2

56t

7/2

7 2 7/2

= 16t

7/2

18t

+ 4t

9/2

9 2 9/2

28.

3 2 y2

ò 7 z-2dz = 7 z-2dz ò 7 z -1 +C -1

ò (-9t

30.

4 x-2 +C -2

= -2 x-2 + C x2

òt

ò (10x

-3.5

= 10 =

-3

-2

3 2 3/2

2u 3

u -1 +C -1

1 +C u

-2.5

dt - 2

ò t dt -1

òt

= 6t -1.5 - 2 ln | t | + C + 4 x-1) dx

òx

-3.5

dx + 4

ò x dx -1

10 x-2.5 + 4 ln | x | + C -2.5

= -4 x-2.5 + 4 ln | x | + C

u 3/2

- 2t -1) dt

-9t -1.5 -2 -1.5

ò 4x dx = 4 x 3dx ò =

-2.5

= -9

7 +C z

æ 4 ö÷ çç ÷ dx = èç x3 ÷ø

-2

1/2

29.

1 ö

æ 7 ö ççç 2 ÷÷÷ dz = èz ø

-2 y +C

ò çççè u + u ÷ø÷÷ du = ò u du + ò u du

æ -2 + 1 ÷ö çz = 7 çç ÷÷ + C çè -2 + 1 ÷÷ø

26.

 ö÷÷

ò  y dy - ò  y = ò y dy -  ò y

ò u du - 14ò u du

= 56

25.

- 14u 5/2 ) du

31.

1 -2

ò 3x dx = ò 3x dx 1 = x dx 3ò 2

-2

1 æç x-1 ö÷÷ ç ÷+C 3 ççè -1 ÷÷ø

1 = - x-1 + C 3 1 =+C 3x

Section 7.1 32.

489 2

2 -4

ò 3x dx = ò 3 x dx 2 = x dx 3ò

37.

-4

2 = - x-3 + C 9 2 =- 3 +C 9x

38.

3e-0.2 x dx = 3

e-0.2 x dx

3(e ) +C -0.2

= -15e

-4e0.2v dv = -4

= (-4)

1 1 æç t 3 ö÷ ln | t | + çç ÷÷÷ + C 4 2 çè 3 ø÷

1 t3 ln | t | + +C 4 6

2 y1/2 dy 6y -1/2

-0.4 x

39.

0.1

40.

æ9

-0.4 x ö

-0.4 x

ò y dy

41.

2 y1/2 y2 +C 3 4

e2u + 2u 2 + C 2

ò (v - e ) dv = ò v dv - ò e dv 2

v3 e3v +C 3 3

v3 - e3v +C 3

ò (x + 1) dx = ò (x + 2x + 1) dx 2

æ 1 ö÷ -0.4 x = 9 ln | x | - 3çç +C e çè 0.4 ÷ø÷ = 9 ln | x | +

1 2

e2u 4u 2 + +C 2 2

4e-0.4 x + e0.1x + C -0.4

ò çççè x - 3e ÷÷÷ø dx 9 = ò x dx - 3ò e

(e2u + 4u ) du =

= -3 ln | x | - 10e-0.4 x + e0.1x + C

36.

1 0.2v e +C 0.2

1 æçç y1/2 ÷÷ö y2 +C ÷ ç 3 ççè 12 ÷÷ø 4

0.1 ö

-0.4 x

ò t dt + 2 ò t dt

3y2 dy 6y

dy -

e0.2v dv

ò çççè - x + 4e + e ÷÷÷ø dx dx = -3 ò x + 4ò e dx + e ò dx = -3 ln | x | +

æ t 2 ÷÷ö çç 1 çç + ÷÷ dt 2 ÷ø è 4t

æ 2 y1/2 - 3 y 2 ö÷ çç ÷÷ dy çç ÷÷ 6 y è ø

= -20e0.2v + C

35.

1 4

ò 1 = y 3ò

-0.2 x

æ 1 ö÷ -0.2 x = 3çç e +C çè -0.2 ÷÷ø

34.

2 æç x-3 ÷÷ö çç ÷+C 3 çè -3 ÷÷ø

33.

æ 1 + 2t 3 ÷ö çç ÷÷ dt = çç ÷ è 4t ÷ø

15e-0.4 x +C 2

x3 2x2 + +x+C 3 2

x3 + x2 + x + C 3

490 42.

43.

Chapter 7 INTEGRATION

ò (2 y - 1) dy = ò (4 y - 4 y + 1) dy

f (1) =

4 y3 4 y2 + y+C 3 2

f (1) =

4 y3 - 2 y2 + y + C 3

x +1

æ x 1 ö÷÷ çç çç 3 + 3 ÷÷ dx xø è x

dx =

(1/2-1/3)

-1/3

1/6

-1/3

x 7/6

44.

1- 2 z 3

7 6

dz =

3 ö æ çç 1 + 2 z ÷÷ dz ÷ çç 3 3 z ø÷÷ è z

46.

ò ò

32 x dx =

ò (z

-1/3

z 2/3 2 3 2/3

f ( x) =

48.

3x5/3 . 5

Find f ( x) such that f ¢( x) = 6 x 2 - 4 x + 3, and (0,1) is on the curve.

6 x7/6 3x 2/3 + +C 7 2

10 x dx =

2 3

45.

x 2/3

3(1)5/3 3 +C = 5 5 3 3 +C = 5 5 C = 0.

Thus,

+x ) dx ò (x = ò x dx + ò x dx =

3 . 5

f ( x) =

ò (6x - 4x + 3) dx

6 x3 4x2 + 3x + C 3 2

= 2 x3 - 2 x 2 + 3x + C

Since (0,1) is on the curve, then f (0) = 1. f (0) = 2(0)3 - 2(0)2 + 3(0) + C = 1

- 2) dz

C =1

Thus, - 2z + C

3z 2

f ( x) = 2 x3 - 2 x 2 + 3x + 1.

- 2z + C

49.

10 x +C ln 10 32 x +C 2(ln 3)

C ¢( x) = 4 x - 5; fixed cost is $8. C ( x) =

ò (4x - 5) dx

4x2 - 5x + k 2

= 2 x 2 - 5x + k C (0) = 2(0) 2 - 5(0) + k = k

47.

( )

Since C (0) = 8, k = 8.

Find f ( x) such that f ¢( x) = x 2/3, and 1, 53 is

Thus,

on the curve.

ò x dx = 2/3

f ( x) =

Since

x5/3 5 3 5/3

3x 5

+C +C

( ) is on the curve, 1, 53

C ( x) = 2 x 2 - 5 x + 8.

50.

C ¢( x) = 0.2 x 2 + 5x; fixed cost is $10. C ( x) =

ò (0.2x + 5x) dx

0.2 x3 5x 2 + +k 3 2

C (0) =

0.2(0)3 5(0) 2 + +k = k 3 2

Section 7.1

491

Since C (0) = 10, k = 10.

96 = k 5 114 = k. 5

58 - 16 -

Thus, C ( x) =

51.

0.2 x3 5x 2 + + 10. 3 2

Thus,

C ¢( x) = 0.03e0.01x ; fixed cost $8.

ò 0.03e = 0.03 e ò

C ( x) =

0.01x 0.01x

C ( x) =

54.

C ¢( x) = x + 12 ; 2 units cost $5.50, so x

C (2) = 5.50. æ ö çç x + 1 ÷÷ dx C ( x) = 2 ÷ø çè x

æ 1 0.01x ÷ö = 0.03çç e ÷÷ + k çè 0.01 ø

ò = ò (x + x ) dx

= 3e0.01x + k

C (0) = 3e0.01(0) + k = 3(1) + k

= 3+k

Since C (0) = 8, 3 + k = 8, and k = 5. C ( x) =

Thus, C ( x) = 3e0.01x + 5.

52.

x1/2 dx =

x3/2 3 2

+k =

2 3/2 x +k 3

Since C (2) = 5.50, 5.50 - 1.5 = k 4 = k.

Thus,

Since C (16) = 45,

C ( x) =

128 + k = 45 3 7 k = . 3 Thus,

53.

55.

2 3/2 7 x + . 3 3

ò (x 5/3

3x 5

2/3

x2 1 - + 4. 2 x

C ¢( x ) = 5 x -

1 ; 10 units cost $94.20, so x

C (10) = 94.20. C ( x) =

C ¢( x) = x 2/3 + 2; 8 units cost $58. C ( x) =

1 x2 - +k 2 x

(2) 2 1 - +k 2 2 1 = 2- +k 2

2 2 128 C (16) = (16)3/2 + k = (64) + k = +k 3 3 3

C ( x) =

x2 x-1 + +k 2 -1

C (2) =

C ¢( x) = x1/2 , 16 units cost $45. C ( x) =

3x5/3 114 + 2x + . 5 5

+ 2)dx

+ 2x + k

3(8)5/3 + 2(8) + k 5 3(32) = + 16 + k 5

C (8) =

2 æ ö çç 5x - 1 ÷÷ dx = 5x - ln | x | +k çè x ÷ø 2

5(10)2 - ln (10) + k 2 = 250 - 2.30 + k.

C (10) =

Since C (10) = 94.20, 94.20 = 247.70 + k -153.50 = k.

Thus, C ( x) =

5x2 - ln | x | - 153.50. 2

492 56.

Chapter 7 INTEGRATION C ¢( x) = 1.2 x (ln1.2); 2 units cost $9.44

50 x - 3x5/3 = xp

(Hint: Recall that a x = e x ln a .)

50 - 3x 2/3 = p,

C ( x) =

which gives the demand function.

1.2 x (ln 1.2) dx

59.

ò 1.2xdx = ln 1.2 e x ln 1.2dx ò = ln 1.2

R =

x ln 1.2

If x = 0, R = 0 (no items sold means no revenue), and

C (2) = e

2 ln 1.2

+ k = 1.44 + k

0 = 500(0) - 0.1(0)3/2 + C

Since C (2) = 9.44,

0 = C.

144 + k = 9.44

Thus, R = 500 x - 0.1x3/2 gives the revenue function. Now, recall that R = xp, where p is the demand function. Then

k = 8.

Thus, C ( x ) = e x ln 1.2 + 8 = 1.2 x + 8.

500 x - 0.1x3/2 = xp 500 - 0.1 x = p, the demand function.

R¢( x) = 175 - 0.02 x - 0.03x 2 R =

ò (175 - 0.02x - 0.03x2 ) dx

= 175 x - 0.01x 2 - 0.01x3 + C.

If x = 0, then R = 0 (no items sold means no revenue), and 0 = 175(0) - 0.01(0)2 - 0.01(0)3 + C 0 = C.

60.

R¢( x) = 600 - 5e0.0002 x R =

ò (600 - 25, 000e

0.0002 x

) dx

= 600 x - 25, 000e0.0002 x + C.

If x = 0, then R = 0 (no items sold means no revenue), and

Thus, R = 175x - 0.01x 2 - 0.01x3 gives the revenue function. Now, recall that R = xp, where p is the demand function. Then

0 = 600(0) - 25, 000e0.0002(0) + C

175x - 0.01x 2 - 0.01x 3 = xp

Thus, R = 600 x - 25, 000e0.0002 x + 25, 000

175 - 0.01x - 0.01x 2 = p, the demand function. 58.

ò (500 - 0.015 x ) dx

= 500 x - 0.1x3/2 + C.

æ 1 ö = ln 1.2 ççç e x ln 1.2 ÷÷÷ + k ÷ø çè ln 1.2

57.

R¢( x) = 500 - 0.15 x

R¢( x) = 50 - 5 x 2/3 R =

ò (50 - 5x2/3) dx

0 = -25, 000 + C 0 = 25, 000.

= 600 x - 25, 000(1 - e0.0002 x )

gives the revenue function. Now, recall that R = xp, where p is the demand function. Then 600 x + 25,000(1 - e0.0002 x ) = xp

= 50 x - 3x5/3 + C

25, 000 (1 - e0.0002 x ) = p, x Which gives the demand function.

600 +

If x = 0, then R = 0 (no items sold means no revenue), and 0 = 50(0) - 3(0)5/3 + C 0=C

Thus, R = 50 x - 3x5/3 gives the revenue function. Now, recall that R = xp, where p is demand function. Then

Section 7.1

493

61. (a) p¢(t ) = 0.01056t + 1.94 p(t ) =

P(2) =

(b)

ò p(t) dt

æ1 ö = (0.01056) çç t 2 ÷÷÷ + 1.94t + C çè 2 ø = 0.00528t 2 + 1.94t + C In 2010, when t = 10, there were 29,059 patent applications, or 29.059 thousand. p(10) = 29.059

The profit from selling 200 lbs of Brie cheese is $240. 64.

f ¢(t ) = 0.01e-0.01t

(a)

f (t ) =

(b) In 2025, t = 25.

Since f (0) = 0, 0 = -1 + k k = 1.

p(25) = 0.00528(25)2 + 1.94(25) + 9.131 = 60.931 thousand or approximately 60,931 design patent applications

f (t ) = -e-0.01t + 1

x + 12 ; profit is -1 when 0

f (10) = -e-0.01(10) + 1

hamburgers are sold. æ

= -e-0.1 + 1 = -0.905 + 1 = 0.095 0.095 unit is excreted in 10 min.

1ö

2 x 3/2 x + +k 3 2

P(0) =

2(0)3/2 0 + +k 3 2

65.

Since P(0) = -1, k = -1. P( x) =

63.

(a)

x 2 3/2 x + -1 3 2

= P(0) =

Since x represents a positive quantity, the absolute value sign can be dropped. g ( x) dx = a ln x - bx + C x

ò (50x3 + 30x2 ) dx

25 x + 10 x3 + k 2 4

25(0) + 10(0)3 + k 2

66.

(a)

c(t ) = (c0 - C )e-kAt /V + M æ -kA ö÷ -kAt /V c¢(t ) = (c0 - C ) çç e çè V ÷÷ø

Since P(0) = -40, -40 = k.

Thus, P( x) =

25 x 4 + 10 x3 - 40. 2

a - bx

g ( x)

ò x dx = ò x dx æa ö = ò ççèç x - b ø÷÷÷ dx dx = a ò x - bò dx = a ln | x | - bx + C

P¢( x) = 50 x3 + 30 x 2 ; profit is -40 when no cheese is sold.

P( x) =

0.01e-0.01t +k 0.01

f (0) = -e-0.01(0) + k = -e0 + k = -1 + k

Thus, p(t ) = 0.00528t 2 + 1.94t + 9.131

ò çççè x + 2 ÷÷÷ø dx

(b)

C = 9.131

P( x ) =

-0.01t

= -e-0.01t + k

19.928 + C = 29.059

P ¢( x) =

ò 0.01e

0.00528(10) 2 + 1.94(10) + C = 29.059

62.

25(2)4 + 10(2)3 - 40 = 240 2

-kA (c0 - C )e-kAt /V V

(b) According to (1) and (2), kA é c¢(t ) = C - (c0 - C )e-kAt /V + C ùú û V êë kA (C - c0 )e-kAt /V . = V

494

Chapter 7 INTEGRATION This is also what we get for c¢(t ) by differentiating Equation (2).

67.

k k P0e-mt + V0 - P0 m m kP0 -mt kP0 = + V0 . e m m

V (t ) =

N ¢(t ) = Aekt

(a)

A kt e +C k

N (t ) =

69.

(a)

A = 50, N (t ) = 300 when t = 0. 50 0 e + C = 300 k N ¢(5) = 250 N (0) =

N ¢(5) = 50e5k = 250 e5 k = 5 5k = ln 5

ò (0.02937t - 0.202t + 8.95) dt

0.02937 3 0.202 2 t t + 8.95t + C 3 2

B(t ) = 0.00979t 3 - 0.101t 2 + 8.95t + 840

ln 5 k = . 5

(b) To project the number of bachelor’s degrees conferred in 2025 we set t equal to 55 and evaluate B(55).

N (0) = ln 5 + C = 300 5

B(55) = 0.00979(55)3 - 0.101(55)2 +8.95(55) + 840 » 2656

250 + C = 300 ln 5 250 » 144.67 ln 5

The formula predicts that 2656 thousand or about 2,656,000 bachelor’s degrees will be conferred in 2025.

50 N (t ) = ln 5 e(ln 5/5)t + 144.67 5

= 155.3337e0.321888t + 144.67 » 155.3e0.3219t + 144.7

(b)

B(t )

= 0.00979t 3 - 0.101t 2 + 8.95t + C In 1970, when t = 0, 840,000 or 840 thousand degrees were conferred, so B(0) = 840 and thus C = 840 and the formula for B is

Therefore,

C = 300 -

B¢(t ) = 0.02937t 2 - 0.202t + 8.95

70.

D¢(t ) = 56.486e0.03537t

(a) D(t ) =

N (12) = 155.3337e0.321888(12) + 144.67

» 7537

There are 7537 cells present after 12 days. 68.

V ¢(t ) = -kP(t ) P(t ) = P0e-mt V ¢(t ) = -kP0e-mt k P0e-mt + C m k V (0) = P0e0 + C m k V0 - P0 = C m V (t ) =

Therefore,

ò 56.486e0.03537t dt 56.486 0.03537t e +C 0.03537

= 1597e0.03537t + C In 2001, when t = 1, D = 1695, so 1597e0.03537(1) + C = 1695 1654.5 + C = 1695 C = 40.5 Thus D(t ) = 1597e0.03537t + 40.5

(b) To project the number of dentistry degrees conferred in 2025 we set t equal to 25 and evaluate D(25).

D(25) = 1597e0.03537(25) + 40.5 » 3907 The formula predicts that about 3907 dentistry degrees will be conferred in 2025.

Section 7.1 71.

495

a (t ) = 5t 2 + 4 v (t ) =

-16t 2 + 6400 = 0 -16t 2 = -6400

(5t 2 + 4) dt

t 2 = 400 t = 20

5t 3 = + 4t + C 3 v (0) =

Discard -20 since time must be positive.

5(0)3 + 4(0) + C 3

The object hits the ground in 20 sec.

Since v(0) = 6, C = 6.

74.

5t 3 v(t ) = + 4t + 6 3

72.

v(t ) =

ò (18t + 8) dt

= 9t 2 + 8t + C1

v(t ) = 9t 2 - 3 t

v(1) = 9(1)2 + 8(1) + C1 = 17 + C1

ò = ò (9t 2 - 3 t ) dt

v(t ) = 9t 2 + 8t - 2

= 3t 3 - 2t 3/2 + C

s(t ) =

s =

v(t )dt

Since v(1) = 15, C1 = -2.

s = 3t 3 - 2t 3/2 + C 3

8 = 3(1) - 2(1)

3/2

= 5 + C2

Since s(1) = 19, C2 = 14. Thus,

Thus,

s(t ) = 3t 3 + 4t 2 - 2t + 14.

s(t ) = 3t 3 - 2t 3/2 + 7. a(t ) = -32

75.

ò -32 dt = -32t + C1

15 t = 3e-t 2 æ 15 ö v(t ) = t + 3e-t ÷÷÷ dt ççç è 2 ø æ 15 1/2 ö çç t + 3e-t ÷÷ dt = ÷ø çè 2 æ ö æ 1 -t ö÷ 15 çç t 3/2 ÷÷ e ÷÷ + C1 = ÷ çç 3 ÷ + 3ççç è -1 ø 2 çè 2 ÷ø

a(t ) =

ò ò

v(0) = -32(0) + C1

Since v(0) = 0, C1 = 0. v(t ) = -32t s(t ) =

s(1) = 3(1)3 + 4(1) 2 - 2(1) + C2

8 =1+C 7 = C.

v(t ) =

ò (9t + 8t - 2)dt

= 3t 3 + 4t 2 - 2t + C2

Since s(1) = 8,

73.

a(t ) = 18t + 8

ò -32t dt

= 5t 3/2 - 3e-t + C1

-32t 2 = + C2 2

v(0) = 5(0)3/2 - 3e-0 + C1 = -3 + C1

= -16t 2 + C2

Since v(0) = -3, C1 = 0.

At t = 0, the plane is at 6400 ft. That is, s(0) = 6400. s(0) = -16(0) 2 + C2 6400 = 0 + C2 C2 = 6400 s(t ) = -16t 2 + 6400

When the object hits the ground, s(t ) = 0.

496

Chapter 7 INTEGRATION 0 = -32(5) + k

v(t ) = 5t 3/2 - 3e-t s(t ) =

160 = k

(5t 3/2 - 3e-t ) dt

and

æ 5/2 ö÷ æ 1 ö çt = 5 çç 5 ÷÷÷ - 3çç - e-t ÷÷÷ + C2 çç èç 1 ø è 2 ÷ø = 2t 5/2 + 3e-t + C2 s(0) = 2(0)5/2 + 3e-0 + C2 = 3 + C2

v(t) = -32t + 160.

Now integrate v(t) to find h(t). h(t ) =

Thus,

s(t ) = 2t 5/2 + 3e-t + 1.

h(t ) = -16t 2 + 160t + 12.

The acceleration of gravity is a constant with

Therefore, from the equation given in Exercise 72, the initial velocity v0 is 160 ft/sec and the initial height of the rocket h0 is 12 ft.

value -32 ft/sec2 ; that is, a(t ) = -32.

First find v(t ) by integrating a(t): v(t ) =

78 .

(a) First find v(t) by integrating a(t):

ò (-32)dt = -32t + k.

When t = 0, v(t ) = v0:

v(t) =

v0 = -32(0) + k

v0 = k

and v(t) = -32t + v0. v(t ) = -32t + v0.

Now integrate v(t) to find s(t).

Now integrate v(t) to find h(t).

s(t ) =

ò (-32t + v0 )dt = -16t + v0t + C 2

Since h(t ) = h0 when t = 0, we can substitute these values into the equation for h(t ) to get C = h0 and

h(t ) = -16t 2 + υ0t + h0. Recall that g is a constant with value

h(t ) =

v(t) =

ò (-32)dt = -32t + k.

Since s(t ) = 0 when t = 0, we can substitute these values into the equation for s(t) to get C = 0 and s(t ) = -16t 2 + v0t.

(b) When t = 14, s(t ) = 0, so 0 = -16(14)2 + v0 (14) v0 = 224

1 2 gt + v0t + h0. 2

First find v(t ) by integrating a(t):

ò (-32t + v )dt

= -16t 2 + v0t + C

-32 ft/sec2 , so 12 g has value -16 ft/sec2 , and

77.

ò (-32)dt = -32t + k.

When t = 0, v(t) = v0:

and

h(t ) =

Since h(t ) = 412 when t = 5, we can substitute these values into the equation for h(t ) to get C = 12 and

Since s(0) = 4, C2 = 1.

76.

ò (-32t + 160)dt = -16t + 160t + C

The velocity was 224 feet per second at time t = 0. (c)

v0t = 224(14) = 3136

The distance the rocket would travel horizontally would be 3136 feet.

When t = 5, v(t) = 0:

Section 7.2

7.2

497 Your Turn 3

Substitution

Your Turn 1

Find

ò 8x(4x + 8) dx.

Let

u = 4 x 2 + 8.

x +1

ò (4x2 + 8x)3 dx.

Let u = 4 x 2 + 8 x. Then du = (8 x + 8) dx = 8( x + 1) dx

Then du = 8 x dx.

Multiply the integral by 88 to introduce the factor of

Now substitute.

8 needed for du, and then substitute.

ò 8x(4x2 + 8)6 dx = ò (4x2 + 8)6 (8x dx) = ò u6du =

x +1

1 7 u +C 7

Now replace u with 4 x 2 + 8.

ö 1 æç 1 -2 ç- u + C ÷÷÷ ø 8 çè 2 1 =+C 16u 2 =

ò 8x(4x2 + 8)6 dx = 7 u7 + C =

1 (4 x 2 + 8)7 + C 7

Now replace u with 4 x 2 + 8 x. x +1

Let

ò x3 3x4 + 10 dx.

u = 3x 4 + 10.

to introduce the factor of Multiply the integral by 12 12 12 needed for du, and then substitute. 1

ò x3 3x4 + 10 dx = 12 ò 12x3 3x4 + 10 dx 1 = 3x 4 + 10 (12 x3 dx) ò 12 1 = u1/2du 12 ò ö 1 æç 2 3/2 + C ÷÷÷ çç u è ø 12 3 1 3/2 = +C u 18 =

Find

x+3

Let u = x 2 + 6 x. Then du = 2 x + 6 dx = 2( x + 3) dx Multiply the integral by 22 to introduce the factor of 2 needed for du, and then substitute. x+3

1 3/2 3x + 10 dx = u +C 18 1 (3x 4 + 10)3/2 + C = 18

2( x + 3)

ò x + 6x dx = 2 ò x + 6x dx 1 1 = du ò u 2 2

1 (ln | u | + C ) 2 1 = ln | u | + C 2 =

we can also call C.) Now replace u with 3x 4 + 10.

òx

16(4 x + 8 x) 2

ò x + 6x dx.

(Note that (1/12) C is just a different constant, which

1 2

Your Turn 4

Then du = 12 x3 dx.

ò (4x2 + 8x)3 dx = - 16u 2 + C

Your Turn 2

Find

8( x + 1)

ò (4x2 + 8x)3 dx = 8 ò (4x2 + 8x)3 dx 1 1 du = 8 ò u3 1 = u-3du 8ò

Now replace u with x 2 + 6 x. x+3

ò x + 6x dx = 2 ln | u | +C 2

1 2 | x + 6x | + C 2

498

Chapter 7 INTEGRATION

Your Turn 5

7.2 Warmup Exercises

Find

Let

u = x 4.

x3e x dx.

W1.

ò (10x + 6 x ) dx = ò (10x + 6x ) dx 4

1/ 2

æ 1 3/ 2 ö÷ æ1 ö = 10 çç x5 ÷÷ + 6 ççç x ÷÷ + C çè 5 ÷ø ÷ø èç 3/2

Then du = 4 x dx.

= 2 x5 + 4 x3/ 2 + C

Multiply the integral by 44 to introduce the factor of 4 needed for du, and then substitute. 1

W2.

ò x3ex dx = 4 ò 4x3e x dx 1 = e x (4 x3 dx) 4ò 1 = eu du ò 4 4

æ5

1 u e +C 4

1 u

-3 ö

2 = 5ln x - 2 + C x

e6 x dx =

1 6x e6 x e +C = +C 6 6

7.2 Exercises 1

ò x3e dx = 4 e + C = 4 e + C x4

æ5

æ 1 ö = 5ln x - 4 çç - x-2 ÷÷÷ + C çè 2 ø

W3.

Now replace u with x 4.

4 ö

ò ççèç x + x3 ÷÷÷ø dx =ò ççèç x + 4x ÷÷÷ø dx

Your Turn 6

True

False. To use substitution to determine

ò 4x3(x4 + 12)5 dx, let u = x + 12. 4

Find

Let

u = 3 + x.

x 3 + x dx. 4.

Now substitute.

ò x 3 + x dx = ò (u - 3) u du = ò u u du - 3ò u du = ò u du - 3ò u du 3/ 2

False. To use substitution to determine

ò 2x x2 - 5 dx, let u = x2 - 5.

Then du = dx and x = u - 3. 5.

æ2 ö 2 5/2 u - 3çç u 3/2 ÷÷÷ + C ç è ø 5 3 2 = u 5/2 - 2u 3/2 + C 5

2 5/2 - 2u 3/2 + C u 5 2 = (3 + x)5/2 - 2(3 + x)3/2 + C 5

ò 1 - x dx Let u = 1 - x; then du = -dx.

ò 2x3 + 1 dx Let u = 2 x3 + 1; then du = 6 x 2dx.

Now replace u with 3 + x.

ò (3x2 - 5)4 2x dx Let u = 3x 2 - 5; then du = 6 x dx.

1/2

ò 4 x e x dx 3

Let u = x 4 ; then du = 4 x3 dx.

x 3 + x dx =

ò 4(2x + 3) dx = 2ò 2(2x + 3) dx 4

Let u = 2 x + 3, so that du = 2 dx.

Section 7.2

499

ò u du 4

ò w dw

2 ⋅ u5 +C 5

w-3 +C -3

2(2 x + 3)5 +C 5

= 2

10.

( x 2 + 2 x - 4)-3 +C 3 1 =- 2 +C 3( x + 2 x - 4)3

ò (-4t + 1) dt = - 4 ò -4(-4t + 1) dt 3

Let u = -4t + 1, so that du = -4dt. =-

1 4

14.

u 3 du

11.

-u 4 +C 16

-(-4t + 1) 4 +C 16

2 dm

6 x 2dx

ò (2x + 7) 3

1 u4 =- ⋅ +C 4 4 =

ò u- du

u-2 -(2m + 1)-2 +C = +C 2 -2 3(3u - 5)-1/2 du

15.

1 2 1/2

u-1/2 +C - 12

ò (2x + 2)(x + 2x - 4) dx 2

Let w = x 2 + 2 x - 4, so that dw = ( 2 x + 2 ) dx.

1/2

1 8

1 u 3/2 ⋅ 3 +C 8 2

1 æç 2 ö÷ 3/2 ⋅ ç ÷u +C 8 çè 3 ÷ø

(4 z 2 - 5)3/2 +C 12

2x + 2

ò (x + 2x - 4) dx 2

= 2(3u - 5)1/2 + C

13.

-3/2

ò z 4z - 5dz = ò z(4z - 5) dz 1 = 8 z (4 z - 5) dz 8ò

= 2w

òu

1/2

Let u = 4 z 2 - 5, so that du = 8z dz.

w-1/2 dw

w1/2

Let w = 3u - 5, so that dw = 3 du. =

= -2u-1/2 + C -2 = 1/2 + C u -2 = +C 3 (2 x + 7)1/2

-3/2

Let u = 2 x3 + 7, so that du = 6 x 2dx.

ò (2m + 1) = ò 2(2m + 1)- dm

3 du = 3u - 5

3/2

ò 6x (2x + 7)

Let u = 2m + 1, so that du = 2 dm.

12.

-4

16.

ò u du 1/2

ò r 5r + 2 dr = ò r(5r + 2) dr 1 = 10r (5r + 2) dr 10 ò 2

1/2

Let u = 5r 2 + 2, so that du = 10r dr.

1/2

500

Chapter 7 INTEGRATION =

1 10

1 u 3/2 ⋅ 3 +C 10 2

= =

17.

3x e

3/2

u +C 15

x3 - 3 x

eu du =

eu +C 3

1 3

e x -3 x +C 3

2 2 x3

2 ⋅ 3x e

21.

e1/z

1/z

e1/z

1 eu du 2 1 u = e +C 2 =

18.

ò z dz = -ò e du

2 x3

= -eu + C = -e1/z + C

22.

e dy = 2 y =

Let u = -r 2 , so that du = -2r dr. 1

19.

-r

-e -e +C = 2 2

23.

1 du 2 u 1 = ln | u | + C 2

ò e du u

x3 - 3 x

= 2

+C =e

ò (x - 1)e

Let t 2 + 2 = u, so that 2t dt = du.

eu e 2t - t = +C = +C 2 2

20.

ò t + 2 dt

2(1 - t )e2t - t dt

1 2

1 -1/2 y1/2 y e dy 2

Let u = 2t - t 2 , so that du = (2 - 2t )dt. =

1/2

1 2

ò 2 y dy

ò e du = e + C

= ey

(1 - t )e2t - t dt =

1/2

Let u = y1/2 , so that du = 12 y-1/2dy.

ò re- dr = - 2 ò -2re- dr 1 =e du 2ò r2

ò re dr -r

-1

ò z dz = -ò e ⋅ z dz Let u = 1z , so that du = -21 dx.

Let u = 2 x3 , so that du = 6 x 2dx. =

(5r 2 + 2)3/2 +C 15

x3 - 3 x

1/2

1 dx = 2

2 2 x3

ò (x - 1)e dx 1 3( x - 1)e = 3ò

ò u du

24.

ln (t 2 + 2) +C 2

-4 x

ò x + 3 dx 2

Let u = x 2 + 3, so that du = 2 x dx.

Let u = x3 - 3x, so that du = (3x 2 - 3) dx = 3( x 2 - 1) dx.

-4 x

2 x dx

ò x + 3 dx = -2ò x + 3 = -2ò u = -2 ln | u | +C = -2 ln ( x 2 + 3) + C

Section 7.2 25.

501 x3 + 2 x

ò x + 4x + 7 4

1 6 1 = 6 =

u = x 4 + 4 x 2 + 7.

Let

Then du = (4 x3 + 8 x)dx = 4( x3 + 2 x)dx. x3 + 2 x

ò x + 4x + 7 4

1 (4 x 3 + 2 x) dx 4 x4 + 4 x2 + 7 1 1 1 = du = ln | u | +C 4 4 u 1 4 2 = ln ( x + 4 x + 7 ) + C 4

ò ò

dx =

Since x 4 + 4 x 2 + 7  0 for all x, we can write 1 this answer as ln( x 4 + 4 x 2 + 7) + C. 4 26.

29.

1 u1/3 ⋅ 1 +C 6 3

u1/3 +C 2

(2 y 3 + 3 y 2 + 1)1/3 +C 2

ò p( p + 1)5dp Let u = p + 1, so that du = dp; also, p = u - 1.

t2 + 2

ò t + 6t + 3 dt

ò (u - 1)u du = ò (u - u )du

Let u = t 3 + 6t + 3, so that du = (3t 2 + 6)dt.

27.

t2 + 2

1 (3t 2 + 6)dt dt = 3 t 3 + 6t + 3 t 3 + 6t + 3 du 1 = u 3 1 = ln | u | + C 3 1 = ln | t 3 + 6t + 3| + C 3

ò ò

30.

2x + 1

ò (2x + 1)(x + x) dx -3

28.

-3

+C =

ò 4r(8 - r) dr 12

-1 2

2( x + x)2

Let u = 2 y 3 + 3 y 2 + 1, so that du = (6 y 2 + 6 y) dy.

-2/3

3/2

æ 3/2 ö u 5/2 ÷ ç 8u = -4 çç 3 - 5 ÷÷÷ + C çç ÷ è 2 2 ø

2/3

1/2

ò (2 y + 3 y + 1) dy = ò ( y + y) (2 y + 3y + 1) 2

( p + 1)7 ( p + 1)6 +C 7 6

1/2

-2

y2 + y

ò -r(8 - r) dr = -4 (8 - u )u du ò = -4 (8 u - u )du ò

ò u du = -2 + C -1

u7 u6 +C 7 6

= -4

Let u = x 2 + x, so that du = (2 x + 1) dx. =

Let u = 8 - r , so that du = -dr; also, r = 8 - u.

ò 4r 8 - r dr =

ò (x + x) dx 2

ò 6( y 2 + y)(2 y3 + 3y2 + 1)-2/3 dy ò u-2/3du

31.

8(8 - r )5/2 64(8 - r )3/2 +C 5 3 u

ò u - 1 du = ò u(u - 1)

-1/ 2

502

Chapter 7 INTEGRATION

ò (w + 1)w dw = ò (w + w ) dw -1/2

-1/2

= =

3/2

3 2

2(u - 1) 3

1/2

1 2 3/2

32.

ò (x + 5) =

1 2

ò (x - 6x) 2(x - 3) dx

1 2

æ ö 1 ç u 3/2 ÷ u1/2du = çç 3 ÷÷÷ + C 2 ççè 2 ÷ø

u 3/2 ( x 2 - 6 x)3/2 +C = +C 3 3

dx 6

Let u = 1 + 3 ln x, so that du = 3x dx.

ò 2x(x + 5) dx = 2ò x(x + 5) dx -6

-6

1 3 1 = 3 =

du = dx; also, u - 5 = x.

(u - 5)u-6du = 2

(u-5 - 5u-6 )du

æ u-4 ö÷ æ u-5 ö÷ ç ç = 2 çç ÷÷÷ - 10 çç ÷÷ + C çè -4 ÷ø çè -5 ÷÷ø

u -4 + 2u-5 + C 2 -1 2 = + +C 4 2( x + 5) ( x + 5)5 =-

33.

36.

ò ò u 2du

1 u3 ⋅ +C 3 3

(1 + 3 ln x)3 +C 9

2 + In x dx x

2 + ln x dx = x

34.

( x 2 + 12 x)3/2 +C 3

37.

ò u1/2 du

e2 x

ò e + 5 dx 2x

Let u = e2 x + 5, so that du = 2e2 x dx.

æ 2 ö çç x - 6 x ÷÷÷ ( x - 3) dx è ø

1 du 2 u 1 = ln | u | + C 2 1 = ln | e2 x + 5| + C 2 1 = ln (e2 x + 5) + C 2 =

ò (x - 6x) (x - 3) dx 2

ò u du

(2 x + 12) dx = du 2( x + 6) dx = du. 1æ 2ö u1 2du = çç ÷÷÷ u 3/2 + C 2 çè 3 ø

1 2

1 dx. x

u 3/2 +C 3/2 2 = u 3/2 + C 3 2 = (2 + ln x)3/2 + C 3

Let x 2 + 12 x = u, so that

3(1 + 3 ln x)2 dx x

Let u = 2 + ln x, so that du =

ò ççè x2 + 12 x ÷÷÷ø( x + 6) dx = ò ( x2 + 12 x)1/2( x + 6) dx

1/2

3(1 + 3 ln x)2 dx x

Let u = x + 5, so that

= 2

+ 2(u - 1)1/2 + C

35. 2x

1/2

Let u = x 2 - 6 x, so that du = (2 x - 6) dx = 2( x - 3) dx.

Section 7.2 38.

503 1

1 6 1 = 6

ò x(ln x) dx

Let u = ln x, so that 1 dx. x

du = 1

42.

log x dx = (ln 10) (ln 10) x æ u 2 ö÷ ç = (ln 10) çç ÷÷÷ + C çè 2 ÷ø

2 5 2 = 5 =

ò u du

(ln 10) (log x)2 = +C 2

40.

(a)

ò x8

2 10u 2 ⋅ 105 x + 2 ⋅ +C = +C 5 ln10 5ln10

Let u = x 2 + 27, 000, so that du = 2 x dx.

5[log 2 (5x + 1)]

ln 2 æç u 3 ÷÷ö ç ÷+C = 5 ççè 3 ÷ø÷

41.

ò ò 10u du

ò 4x(x2 + 27, 000)-2/3 dx = 2 2 x( x 2 + 27, 000)-2/3 dx ò

R = 2

5 ⋅ 105 x + 2 dx 2 x

R¢( x) = 4 x( x 2 + 27, 000)-2/3

ò (ln 2) (5x + 1) dx ln 2 u du = 5 ò =

5 dx. 2 x

R ( x) =

Let u = log 2 (5 x + 1), so that 5 du = dx. (ln 2) (5 x + 1) ln 2 5

= 45.

[log 2 (5 x + 1)]2 dx 5x + 1

105 x + 2 dx x

Let u = 5 x + 2, so that du =

1 Let u = log x, so that du = (ln 10) dx. x

log x dx x

log x dx = (ln 10) x

83x +1 = +C 6 ln 8

= ln| u | + C = ln |ln x | + C

39.

3 x 2 +1

1 æç 8u ÷÷ö ç ÷+C 6 ççè ln 8 ÷÷ø

ò x(ln x) dx = ò u du

ò 6x ⋅ 8 ò 8 du

ò u-2/3du

= 2 ⋅ 3u1/3 + C = 6( x 2 + 27, 000)1/3 + C R(125) = 6(1252 + 27, 000)1/3 + C

(ln 2)[log 2 (5x + 1)]3 +C 15

3 x 2 +1

Since R(125) = 29.591, 6(1252 + 27, 000)1/3 + C = 29.591

C = -180

Let u = 3x 2 + 1, so that du = 6 x dx.

Thus, R( x) = 6( x 2 + 27, 000)1/3 - 180.

(b) R( x) = 6( x 2 + 27,000)

1/3

- 180 ³ 40

6( x + 27,000)1/3 ³ 220 ( x2 + 27,000)

1/3

³ 36.6667

x + 27,000 ³ 49,296.43 x 2 ³ 22,296.43 x ³ 149.4

504

Chapter 7 INTEGRATION For revenue of at least $40,000, then 150 gaming devices must be sold.

46.

ò C¢( x) dx 60 x = ò 5x2 + e dx du = 6 ò u = 6 ln |u | + C

C ( x) =

D¢(t ) = 90(t + 6) t 2 + 12t

(a)

= 90(t + 6)(t 2 + 12t )1/2

ò 90(t + 6)(t + 12t) dt = 45 (2t + 12)(t + 12t ) dt ò 2

D(t ) =

1/2

= 6 ln (5x 2 + e ) + C

1/2

Since C (0) = 10, C = 4. Therefore, C ( x) = 6 ln |5 x 2 + e | + 4

Let u = t 2 + 12t , so that du = (2t + 12) dt.

= 6 ln(5 x 2 + e) + 4.

ò u du 1/2

D = 45

= 45 ⋅

(b) C (5) = 6 ln(5 ⋅ 52 + e) + 4 » 33.099

2u 3/2 +C 3

Since this represents $33,099 dollars which is greater than $20,000, a new source of investment income should be sought.

= 30u 3/2 + C D(t ) = 30(t 2 + 12t )3/2 + C

48.

(a)

P¢(t ) = te-t

Let -t 2 = u, so that -2t dt = du, or

Since D(4) = 16, 260,

t dt = - 12 du.

30(42 + 12 ⋅ 4)3/2 + C = 16, 260 15,360 + C = 16, 260 C = 900

P(t ) =

D(t ) = 30(t + 12t )

3/2

+ 900.

t 2 + 12t = 119.317

P(3) = -

e-9 2 = 0.01006 » 0.01.

-12  122 - 4(1)(-119.317) 2(1) -12  24.93 x » 2(1) x » 6.465 or x » -18.464

C = 0.01 + 2

-e-t P(t ) = + 0.01 2

7 years must pass. (b)

5x 2 + e

e-9 +C 2

e-9 + C = 0.01 2

x =

60 x

eu +C 2

Since 10,000 = 0.01 million and P(3) = 0.01,

t 2 + 12t - 119.317 = 0

C ¢( x) =

eu du = -

30(t 2 + 12t )3/2 = 39,100 (t 2 + 12t )3/2 = 1303.333

1 2

e-t =+C 2

(b) D(t ) = 30(t 2 + 12t )3/2 + 900 = 40, 000

47.

ò te-t dt

Thus, 2

æ -t 2 ÷ö ç -e + 0.01÷÷÷ lim (t ) = lim ççç ÷÷ t ¥ t ¥ ç 2 è ø æ ö÷ 1 = lim ççç - 2 + 0.01÷÷ = 0.01 t ¥ çè 2et ø÷÷

(a) Let u = 5 x + e, so that du = 10 x dx.

Section 7.2

505

f ¢(t ) = 4.0674 ⋅ 10-4 t (t - 1970)0.4

49.

(a) Let u = t - 1970. To get the t outside the parentheses in terms of u, solve u = t - 1970 for t to get t = u + 1970. Then dt = du and we can substitute as follows. f (t ) =

ò f ¢(t )dt = ò 4.0674 ⋅ 10-4 t(t - 1970)0.3dt = ò 4.0674 ⋅ 10-4(u + 1970)(u)0.3du = 4.0674 ⋅ 10-4 (u + 1970)(u )0.3 du ò = 4.0674 ⋅ 10-4 (u1.3 + 1970u 0.3 )du ò æ u 2.3 1970u1.3 ö÷÷ ç = 4.0674 ⋅ 10-4 çç + ÷+C çè 2.3 1.3 ÷ø÷ é (t - 1970)2.3 1970(t - 1970)1.3 ùú = 4.0674 ⋅ 10-4 êê + ú+C 1.3 2.3 êë úû

Since f (1970) = 61.298, C = 61.298. é (t - 1970)2.3 1970(t - 1970)1.3 ùú + Therefore, f (t ) = 4.0674 ⋅ 10-4 êê ú + 61.298. 2.3 1.3 ëê ûú

(b)

é (2020 - 1970)2.3 1970(2020 - 1970)1.3 ùú f (2020) = 4.0674 ⋅ 10-4 êê + ú + 61.298 » 162.0 2.3 1.3 ëê ûú

In the year 2020, there will be about 162,000 local transit vehicles.

50.

f ¢(t ) = 0.001483t (t - 1980)0.75

(a) Let u = t - 1980. To get the t outside the parentheses in terms of u, solve u = t - 1980 for t to get t = u + 1980. Then dt = du and we can substitute as follows.

ò f ¢(t)dt = ò 0.001483t(t - 1980) = ò 0.001483(u + 1980)(u)0.75 du = 0.001483 (u + 1980)(u )0.75 du ò = 0.001483 (u1.75 + 1980u 0.75 )du ò

f (t ) =

0.75

æ u 2.75 1980u1.75 ö÷÷ ç = 0.001483çç + ÷+C çè 2.75 1.75 ÷ø÷ é (t - 1980)2.75 1980(t - 1980)1.75 ù ú = 0.001483 êê + ú+C 2.75 1.75 êë úû

506

Chapter 7 INTEGRATION Since f (1980) = 262.951, C = 262.951. Therefore, é (t - 1980)2.75 1980(t - 1980)1.75 ù ú + 262.951. f (t ) = 0.001483 êê + ú 2.75 1.75 êë úû é (2020 - 1980)2.75 1980(2020 - 1980)1.75 ùú f (2020) = 0.001483 êê + ú + 262.951 » 1344.188 2.75 1.75 ëê ûú

(b)

In the year 2020, there will be about 1,344,000,000 outpatient visits. 2

51. (a) P¢(t ) = 500te-t / 5 P (t ) =

P¢(t ) dt =

500te-t / 5 dt

Let u = -t /5; then du = (-2/5)t dt.

500te-t / 5 dt

ò e ( (-2/5)t dt ) = -1250 e du ò -t 2 / 5

= (-1250)

= -1250eu + C 2

= -1250e-t / 5 + C Since P(2000) is 0, 2

2000 = -1250e-(0) / 5 + C C = 3250 2

P(t ) = 3250 - 1250e-t / 5

(b)

P(3) = 3250 - 1250e-(3) / 5 = 3043 to the nearest integer

52. (a)

N ¢(t ) = N (t ) =

100t 2

t +2 100t

ò t + 2 dt 2

Let u = t 2 + 2; then du = 2t dt. 100t

ò t + 2 dt = 50ò t + 2 ( 2t dt ) 1 = 50 ò u du 2

= 50ln u + C

(

)

= 50ln t 2 + 2 + C

Section 7.3

507

If 37 people were infected at t = 0, then

(

)

37 = 50ln (0)2 + 2 + C C = 37 - 50ln 2 » 2.343

(

)

N (t ) = 50ln t 2 + 2 + 2.343

(b)

(

)

N (21) = 50ln (21)2 + 2 + 2.343 = 307 to the nearest integer

7.3

Area and the Definite Integral

Your Turn 1 5

Approximate

ò 4x dx using four rectangles. 1

Find the area of the shaded region: y f(x) = 4x

Build a table giving the heights of the rectangles, which are the values of f ( x) = 4 x at the midpoint of each interval.

i xi 1 1.5

f ( xi ) 6.0

2 2.5 3 3.5 4 4.5

10.0 14.0 18.0

For each interval, x = 1. The sum of the areas of the rectangles is 4

å f (x ) x = f (1.5)x + f (2.5)x + f (3.5)x + f (4.5) x i

i =1

= 1(6) + 1(10) + 1(14) + 1(18) = 48. Thus our approximation to the integral is 48. In this case the approximation is exact.

508

Chapter 7 INTEGRATION

Your Turn 2 A driver has the following velocities at various times:

Time (hr)

0 0.5 1

Velocity (mph) 0 50

= f ( x1)D x + f ( x2 )D x + f ( x3 )D x + f ( x4 )D x = f (0) (2) + f (2) (2) + f (4) (2) + f (6) (2) = [2(0) + 5](2) + [2(2) + 5](2)

1.5 2

56 40

+ [2(4) + 5](2) + [2(6) + 5](2)

= 10 + 9(2) + 13(2) + 17(2)

Approximate the total distance traveled during the 2-hour period.

= 88

(b)

Using left endpoints:

distance = 0(0.5) + 50(0.5) + 56(0.5) + 40(0.5) = 73 miles Using right endpoints:

distance = 50(0.5) + 56(0.5) + 40(0.5) + 48(0.5) = 97 miles Averaging these two estimates: distance =

73 + 97 = 85 miles 2

The sum of these rectangles approximates 8

ò0 (2 x + 5) dx.

7.3

Exercises

False. As x = (b - a) / 2 decreases, the sum n

å f ( xi )Dx always gives a better approximation

f ( x) = 1x and x1 = 12 , x2 = 1, x3 = 32 ,

x4 = 2, and Dx = 12

i =1

to the area under f ( x) from a to b. In other words, the smaller the rectangles, the better the approximation. 2.

å f (x )D x = f (x1)D x + f ( x2 )D x i

i =1

+ f ( x3 )D x + f ( x4 )D x

False. It is usually the case that the midpoint rule gives a more accurate answer than either the left or right sum.

True

(a)

æ1ö 1 f ( x1) = f çç ÷÷÷ = 1 = 2 çè 2 ø 2

1 f ( x2 ) = f (1) = = 1 1 æ 3 ö÷ 1 2 f ( x3 ) = f çç ÷÷ = 3 = çè 2 ø 3

( x 2 + 3) Dx, å ò0 (x + 3)dx = lim ¥ =1 2

1 f ( x4 ) = f (2) = 2

0 = 4 and x is any value where Dx = 4i n n

of x in the ith interval. 7.

Thus,

f ( x) = 2 x + 5, x1 = 0, x2 = 2, x3 = 4, x4 = 6, and x = 2 4

(a)

å f (x )Dx i

i =1

å f (x )Dx i

i =1

æ1ö æ1ö æ2ö æ1ö æ1ö æ1ö = (2) çç ÷÷÷ + (1) çç ÷÷÷ + çç ÷÷÷ çç ÷÷÷ + çç ÷÷÷ çç ÷÷÷ èç 2 ø èç 2 ø èç 3 ø èç 2 ø èç 2 ø èç 2 ø 1 1 1 + + 2 3 4 12 + 6 + 4 + 3 25 = = . 12 12 =1+

Section 7.3

509

(b)

f ( x) = 3x + 2 from x = 1 to x = 3

10.

For n = 4 rectangles: x =

3 -1 = 0.5 4

(a) Using the left endpoints:

The sum approximates the integral 9.

5/2

ò1/2 1 dx. x

f ( x) = 2 x + 5 from x = 2 to x = 4

= 5(0.5) + 6.5(0.5) + 8(0.5) + 9.5(0.5) = 14.5

(b) Using the right endpoints:

4-2 = 0.5 4

i 1 2 3 4

f (xi)

1 2 3 4

2 2.5 3 3.5

9 10 11 12

(c)

f ( xi )x

Average =

14.5 + 17.5 32 = = 16 2 2

(d) Using the midpoints:

= 21

(b) Using the right endpoints:

i 1 2 3 4

xi 2.5 3 3.5 4

f (xi) 10 11 12 13

(d) Using the midpoints:

xi 2.25 2.75 3.25 3.75

f (xi)

1.25 1.75 2.25 2.75

5.75 7.25 8.75 10.25

å f ( x ) x i

21 + 23 44 Average = = = 22 2 2

i 1 2 3 4

i 1 2 3 4 4

= 5.75(0.5) + 7.25(0.5) + 8.75(0.5) + 10.25(0.5)

A = 10(0.5) + 11(0.5) + 12(0.5) + 13(0.5) = 23

= 16

11.

f ( x) = -x 2 + 4 from x = -2 to x = 2

For n = 4 rectangles: f (xi) 9.5 10.5 11.5 12.5

x =

2 - (-2) =1 4

(a) Using the left endpoints:

f (xi) 6.5 8 9.5 11

= 17.5

xi 1.5 2 2.5 3

A = 6.5(0.5) + 8(0.5) + 9.5(0.5) + 11(0.5)

= 9(0.5) + 10(0.5) + 11(0.5) + 12(0.5)

(c)

5 6.5 8 9.5

i =1

f (xi)

1 1.5 2 2.5

å f ( x )x

(a) Using the left endpoints:

For n = 4 rectangles: x =

i 1 2 3 4

f ( xi )x = 9.5(0.5) + 10.5(0.5)

+ 11.5(0.5) + 12.5(0.5) = 22

f (xi)

-2

-(-2)2 + 4 = 0

-1

-(-1)2 + 4 = 3

-(0)2 + 4 = 4

-(1)2 + 4 = 3

510

Chapter 7 INTEGRATION (b) Using the right endpoints:

f ( xi )x

i 1 2 3 4

i =1

= (0) (1) + (3) (1) + (4) (1) + (3) (1) = 10

(b) Using the right endpoints:

i 1 2 3 4

xi -1 0 1 2

(c)

(d) Using the midpoints:

i 1 2

(d) Using the midpoints:

f (xi)

3 2 1 2 1 2 3 2

7 4 15 4 15 4 7 4

2 3 4

3 4

13.

9 4 25 4 49 4 81 4

å f (x )x i

f ( x) = e x + 1 from x = -2 to x = 2

For n = 4 rectangles:

i =1

2 - (-2) =1 4 (a) Using the left endpoints:

x =

f ( x) = x 2 from x = 1 to 5

For n = 4 rectangles:

5 -1 x = =1 4

(a) Using the left endpoints: i xi f (xi) 1 1 1 2 2 4 3 3 9 4 4 16

f (xi)

-2

-1

e +1 = 2

e1 + 1

f ( xi )x =

i =1 -2

f ( xi )(1) =

i =1 -1

= (e + 1) + (e » 8.2215 » 8.22

3 2 5 2 7 2 9 2

= 9 (1) + 25 (1) + 49 (1) + 81 (1) 4 4 4 4 = 41

å f (x )x

f (xi)

i =1

7 15 15 7 = (1) + (1) + (1) + (1) 4 4 4 4 = 11

12.

30 + 54 84 = = 42 2 2

Average =

10 + 10 Average = = 10 2

f (xi) 4 9 16 25

A = 4(1) + 9(1) + 16(1) + 25(1) = 54

f (xi) 3 4 3 0

Area = 1(3) + 1(4) + 1(3) + 1(0) = 10 (c)

xi 2 3 4 5

i =1

= 1(1) + 4(1) + 9(1) + 16(1) = 30

+ 1) + 2 + e + 1

f (xi)

1 2 3 4

-1 0 1 2

e-1 + 1 2 e +1

i =1 1

(b) Using the right endpoints:

f ( xi )x

å f (x )

e2 + 1

Section 7.3

511

Area = 1(e-1 + 1) + 1(2) + 1(e + 1) + 1(e2 + 1)

(c)

» 15.4752 » 15.48 (c)

(d) Using the midpoints:

8.2215 + 15.4752 2 = 11.84835 » 11.85

Average =

1 2 3 4

-3/2

-1/2

3 4

e1/2 + 1

15.

+ 1) (1) + (e-1/2 + 1) (1)

1/2

+ (e + 1) (1) + (e » 10.9601 » 10.96

14.

3/2

f ( x) = 2 from x = 1 to x = 9 x

For n = 4 rectangles: x = 9 - 1 = 2 4

+ 1) (1)

(a) Using the left endpoints:

4-0 =1 4

(a) Using the left endpoints:

For n = 4 rectangles:

i 1 2 3 4

xi 0 1 2 3

f (xi) 0 1.718 6.389 19.086

f (xi) 2 = 2 1 2 3 2 = 0.4 5 2 7

å f ( x ) x i

i =1

32.115

å f ( x ) x

f ( x) = e x - 1 from x = 0 to x = 4

x =

11.182

= 0.649(1) + 3.482(1) + 11.182(1) + 32.115(1) » 47.43

f ( xi ) x

= (e

3.482

i =1

e3/2 + 1

i =1 -3/2

0.649

æ ö = (2)(2) + 2 (2) + (0.4)(2) + çç 2 ÷÷÷ (2) è7ø 3

å f (x )x i

i =1

= 0(1) + 1.718(1) + 6.389(1) + 19.086(1) » 27.19

» 6.7048 » 6.70

(b) Using the right endpoints:

f (xi) 1.718 6.389 19.086 53.598

i 1 2 3 4

xi 1 2 3 4

f (xi)

1 2 3 2 5 2 7 2

f (xi)

3 2 1 2 1 2 3 2

(d) Using the midpoints:

27.19 + 80.79 = 53.99 2

Average =

A = 1.718(1) + 6.389(1) + 19.086(1) + 53.598(1) » 80.79

f (xi) 2 3 2 5 2 7 2 9

512

Chapter 7 INTEGRATION æ2ö æ2ö æ2ö æ2ö Area = 2 çç ÷÷ + 2 çç ÷÷ + 2 çç ÷÷ + 2 çç ÷÷ çè 3 ÷ø çè 5 ÷ø çè 7 ÷ø çè 9 ÷ø 4 4 4 4 + + + » 3.1492 » 3.15 3 5 7 9

(c)

1.283 + 0.95 2 2.233 = = 1.117 2

Average =

(c)

(d) Using the midpoints:

6.7 + 3.15 Average = = 4.93 2

i 1 2 3 4

(d) Using the midpoints:

i 1

xi 2

f (xi) 1 1 2 1 3 1 4

= 0.8(0.5) + 0.5714(0.5) + 0.4444(0.5) + 0.3636(0.5) = 1.090

å f ( x ) x

17.

(a)

Width =

æ5ö æ7ö + 1 ⋅ f çç ÷÷ + 1 ⋅ f çç ÷÷ çè 2 ÷ø èç 2 ø÷

1 from x = 1 to x = 3 x

For n = 4 rectangles: x =

4-0 x = 1; f ( x) = 4 2

æ1ö æ3ö Area = 1 ⋅ f çç ÷÷÷ + 1 ⋅ f çç ÷÷÷ èç 2 ø èç 2 ø

1 1 1 (2) + (2) + (2) 2 3 4 » 4.1667 » 4.17

= 1(2) +

f ( x) =

å f (x )x 1

i =1

16.

f (xi) 0.8 0.5714 0.4444 0.3636

xi 1.25 1.75 2.25 2.75

1 3 5 7 16 + + + = = 4 4 4 4 4 4

(b)

3 -1 = 0.5 4

(a) Using the left endpoints:

i 1 2 3 4

xi 1 1.5 2 2.5

f (xi) 1 0.6667 0.5 0.4

å f ( x ) x

= 1(0.5) + 0.6667(0.5) + 0.5(0.5) + 0.4(0.5) = 1.283

xi 1.5 2 2.5 3

f (xi) 0.6667 0.5 0.4 0.3333

18.

A = 0.6667(0.5) + 0.5(0.5) + 0.4(0.5) + 0.3333(0.5) = 0.95

1 = (4)(2) = 4 2

(b) Using the right endpoints:

i 1 2 3 4

ò f (x)dx = ò 2 dx = 2 (base)(height)

(a)

Width =

5-0 = 1; f ( x) = 5 - x. 5

æ1ö æ 3ö æ5ö Area = f çç ÷÷ (1) + f çç ÷÷ (1) + f çç ÷÷ (1) çè 2 ÷ø çè 2 ø÷ èç 2 ø÷ æ7ö æ9ö + f çç ÷÷÷ (1) + f çç ÷÷÷ (1) çè 2 ø èç 2 ø = 4.5 + 3.5 + 2.5 + 1.5 + 0.5 = 12.5

Section 7.3

513

(b)

(5 - x) dx

ò f (x) dx is the area of a triangle with

Graph y = 5 - x.

base 4 and height 2. The triangle has area 1 ⋅ 4 ⋅ 2 = 4. 2

Therefore,

ò f ( x) dx = 4 +  . 0

21.

-4

16 - x 2 dx

Graph y = 16 - x 2 . 5

ò (5 - x) dx is the area of a triangle with base 0

= 5 - 0 = 5 and altitude = 5. 1 (altitude)(base) 2 1 = (5) (5) = 12.5 2

Area =

19.

(a) Area of triangle is 12 ·base · height.

ò- 16 - x dx is the area of the portion of the 4

The base is 4; the height is 2. 4

circle in the second quadrant, which is one-fourth of a circle. The circle has radius 4.

ò f (x) dx = 2 ⋅ 4 ⋅ 2 = 4 0

Area =

(b) The larger triangle has an area of 1 ⋅ 3 ⋅ 3 = 9 . The smaller triangle has an 2 2 22.

area of 12 ⋅ 1 ⋅ 1 = 12 . The sum is

(a)

1 2 1  r =  (4)2 = 4 4 4

ò- 9 - x dx 2

9 + 1 = 10 = 5. 2 2 2

20.

Graph y =

9 - x2 .

ò f (x) dx is the area of a rectangle with 0

width x = 2 and length y = 4. The rectangle has area 2 ⋅ 4 = 8.

ò f (x) dx 2

is the area of one-fourth of a circle that has radius 4. The area is 1  r 2 = 1  (4) 2 = 4 . 4 4

Therefore,

ò f ( x) dx = 8 + 4 . 0

(b)

ò f (x) dx is the area of one-fourth of a

-3

9 - x 2 dx is the area of a semicircle with

radius 3 centered at the origin. Area =

circle that has radius 2. The area is 1  r 2 = 1  (2) 2 =  . 4 4

1 2 1 9  r =  (3) 2 =  2 2 2

514 23.

Chapter 7 INTEGRATION

25.

(1 + 2 x) dx

-0 = 0.1, and (a) With n = 10,  x = 110

x1 = 0 + 0.1 = 0.1, use the command

Graph y = 1 + 2 x.

seq (X 2 , X, 0.1, 1, 0.1) →L1. The resulting screen is:

(

)

(b) Since å in= 1 f ( xi )x = x åin= 1 f ( xi ) , use

the command 0.1sum (L1) to approximate

(1 + 2 x) dx is the area of the trapezoid with

ò0 x dx. The resulting screen is:

B = 11, b = 5, and h = 3. The formula for the area is 1 ( B + b)h, 2

so we have

ò x dx » 0.385

1 A = (11 + 5)(3) = 24. 2

-0 = 0.01 and (c) With n = 100,  x = 1100

24.

ò (5 + x) dx

x1 = 0 + 0.01 = 0.01, use the command

seq (X 2 , X, 0.01, 1, 0.01)→ L1. The resulting screen is:

Graph y = 5 + x. y

f ( x) = 5 + x

10 8

( 3, 8)

( 1, 6) 4

Use the command 0.01sum(L1) to approximate

ò0 x dx. The resulting screen is:

1

5 x

ò (5 + x) dx is the area of a trapezoid with bases 1

of length 6 and 8 and height of length 2. 1 (height) (base1 + base2 ) 2 1 = (2)(6 + 8) = 14 2

ò x dx » 0.33835

Area =

(d) With n = 500,  x =

1- 0 = 0.002, and 500

x1 = 0 + 0.002 = 0.002, use the command

seq( X 2 , X, 0.002, 1, 0.002)→ L1. The resulting screen is:

Section 7.3

515

Use the command 0.002sum(L1) to approximate 1

Use the command 0.01sum(L1) to approximate 1

ò0 x dx. The resulting screen is:

ò x dx » 0.255025

x 2dx » 0.334334

-0 = 0.002, and (d) With n = 500,  x = 1500

(e) As n gets larger the approximation for 1

ò0 x dx seems to be approaching 0.333333

x1 = 0 + 0.002 = 0.002, use the command

seq ( X ^ 3, X, 0.002, 1, 0.002)→L1. The resulting screen is:

or 13 . We estimate ò0 x 2dx = 13 . 26.

-0 = 0.1, and (a) With n = 10, Dx = 110

x1 = 0 + 0.1 = 0.1, use the command seq( X^3, X, 0.1, 0.1, 0.1) → L1. The resulting screen is:

Use the command 0.002sum(L1) to approximate 1

ò0 x dx. The resulting screen is: n

æ n

i =1

çè i = 1

ø÷

(b) Since å f ( xi )Dx = Dxççç å f ( xi ) ÷÷÷÷, use the

command 0.1sum(L1) to approximate 1

ò0 x dx. The resulting screen is:

ò x dx » 0.251001 3

(e) As n gets larger the approximation for 1

ò0 x dx seems to be approaching 0.25 or 14 . 1

We estimate ò0 x3dx = 14 .

ò x dx » 0.3025 3

For Exercises 28–34, reading on the graphs and answers may vary.

-0 = 0.01, and (c) With n = 100, D x = 1100

x1 = 0 + 0.01 = 0.01, use the command seq( X^3, X, 0.01, 0.1, 0.01) → L1. The resulting screen is:

28.

Left endpoints: Read values of the function from the graph for every 2 hours from midnight to 10 P.M. These values give the heights of 12 rectangles. The width of each rectangle is Dx = 2. We estimate the area under the curve as

516

Chapter 7 INTEGRATION 6

f ( xi )Dx

= 142(3) + 341(3) + 923(3) +1595(3) + 2096(3) + 2736(3)

= 3.0(2) + 3.2(2) + 3.5(2) + 4.2(2) + 5.2(2) + 6.2(2) + 8.0(2) + 11.0(2) + 11.8(2) + 10.0(2) + 6.0(2) + 4.4(2) = 153.0

= 23, 499 trillion BTUs

Average: 15,501 + 23, 499 = 19,500 trillion BTUs 2 We estimate the total wind energy consumption over the 18-year period from 2001 to 2019 as 19,500 trillion BTUs.

Right endpoints: Read values of the function from the graph for every 2 hours from 2 A.M. to midnight. Now we estimate the area under the curve as 12

å f ( x )Dx i

30.

i =1

= 3.2(2) + 3.5(2) + 4.2(2) + 5.2(2) + 6.2(2) + 8.0(2) + 11.0(2) + 11.8(2) + 10.0(2) + 6.0(2) + 4.4(2) + 3.8(2) = 154.6 153.0 + 154.6 307.6 = 2 2 = 153.8 million kilowatt-hours

The area under the curve represents the total electricity usage. We estimate this usage as about 153.8 million kilowatt hours.

D (qi )Dx = Dx

i =0

å D(qi ) i =0

= (10)(260 + 170 + 100 + 50 + 20) = 6000 or a total revenue of $6000

Left endpoints:

Right endpoints:

Read values of the function on the graph every three years from 2001 to 2016, that is, at x0 = 2001, x1 = 2004, x2 = 2007, x3 = 2010, x4 = 2013, and x5 = 2016. These values give us the heights of six rectangles. The width of each rectangle is Dx = 3. We estimate the area under the curve as follows:

Compute values of the demand function every 10 liters from 10 to 50, that is, at qi = 10 ⋅ i for i = 1, 2, 3, 4, 5. We estimate the area under the demand curve as follows: 5

A = x

The demand function is 1 2 D( q ) = q - 10q + 260 10 Left endpoints: Compute values of the demand function every 10 liters from 0 to 40, that is, at qi = 10 ⋅ i for i = 0, 1, 2, 3, 4. These values give us the heights of five rectangles. The width of each rectangle is Dx = 10. We estimate the area under the demand curve as follows:

Average:

29.

i =1

å f ( x ) x

å D (q ) i

i =1

å f (xi ) x

= (10)(170 + 100 + 50 + 20 + 10)

i =0

= 3500 or a total revenue of $3500

= 70(3) + 142(3) + 341(3) + 923(3)

6000 + 3500 = 4750 or $4750 2 We estimate the total revenue as $4750.

+1595(3) + 2096(3)

Average:

= 15,501 trillion BTUs

Right endpoints: Read values of the function on the graph every three years from 2004 to 2019, that is, discard the point x0 = 2001 and include the point x6 = 2019. We estimate the area under the curve as follows:

31.

The demand function is 2 2 4 D( q) = q - q + 15 125 5 Left endpoints: Compute values of the demand function every 5 liters from 0 to 20, that is, at qi = 5 ⋅ i for i = 0, 1, 2, 3, 4. These values give us the heights of five rectangles. The width of

Section 7.3

517

each rectangle is Dx = 5. We estimate the area under the demand curve as follows: A=

i =0

å f (x )Dx i

i =1

å D(qi )Dx = Dxå D(qi )

= 2.4(1) + 2.9(1) + 3.1(1) + 3.2(1) + 3.3(1) + 3.3(1) + 3.3(1) + 3.3(1) + 3.3(1) + 3.3(1)

= (5)(15 + 11.4 + 8.6 + 6.6 + 5.4) = 235 or a total revenue of $235

+ 1.1(1) + 0.7(1) + 0.6(1) + 0.5(1) + 0.4(1) + 0.3(1) + 0.3(1) + 0.2(1) + 0.2(1) + 0.2(1)

Right endpoints:

» 35.9.

Compute values of the demand function every 10 liters from 5 to 25, that is, at qi = 5 ⋅ i for i = 1, 2, 3, 4, 5. We estimate the area under the demand curve as follows:

Average:

35.7 + 35.9 = 35.8 liters 2 The area under the curve represents the total volume of oxygen inhaled. We estimate this volume as about 35.8 liters.

A = x

å D (q ) i

33.

i =1

= (5)(11.4 + 8.6 + 6.6 + 5.4 + 5) = 185 or a total revenue of $185

235 + 185 = 210 or $210 2 We estimate the total as $210.

Average:

32.

Left endpoints: Read values of the function from the graph for every minute from 0 minutes to 19 minutes. These values give the height of 20 rectangles. The width of the each rectangle is Dx = 1. We estimate the area under the curve as

Read values of the function on the graph every five days from 3/13 to 4/7, These values give us the heights of six rectangles. The width of each rectangle is Dx = 5. We estimate the area under the curve as follows: 5

å f ( x ) x i

i =0

= 600(5) + 2800(5) + 10, 000(5) +19,800(5) + 30, 400(5) + 30,800(5) = 472, 000 cases

Right endpoints: Read values of the function on the graph every five days from 3/18 to 4/12. We estimate the area under the curve as follows:

å f (x )Dx i

i =1

= 0(1) + 2.4(1) + 2.9(1) + 3.1(1) + 3.2(1) + 3.3(1) + 3.3(1) + 3.3(1) + 3.3(1) + 3.3(1) + 3.3(1) + 1.1(1) + 0.7(1) + 0.6(1) + 0.5(1) + 0.4(1) + 0.3(1) + 0.3(1) + 0.2(1) + 0.2(1) » 35.7.

å f ( x ) x i

i =1

= 2800(5) + 10, 000(5) + 19,800(5) +30, 400(5) + 30,800(5) + 28,900(5) = 613,500 cases

Right endpoints: Read values of the function from the graph for every minute from 1 minute to 20 minutes. Now we estimate the area under the curve as

34.

Left endpoints:

Average: 472, 000 + 613,500 = 542, 750 cases 2 We estimate the total cases over the 30-day period from 3/13 to 4/12 as 542,750 cases.

Left endpoints: Read values of the function on the graph every two years from 2009 to 2015. These values give us the heights of four rectangles. The width of each rectangle is  x = 2. We estimate the area under the graph as follows: 3

å f (x )x = 2805(2) + 2628(2) + 2853(2) + 3168(2) = 22,908 i

i=0

518

Chapter 7 INTEGRATION Right endpoints: Read values of the function on the graph every two years from 2011 to 2017. We estimate the area under the graph as follows. 4

å f (x )x = 2628(2) + 2853(2) + 3168(2) + 3582(2) = 24, 462 i

i =1

22,908 + 24, 462 = 23, 685 accidents 2 The area under the graph represents the number of fatal automobile accidents in California from 2009 to 2017. We estimate this number as 23,685 fatal accidents.

Average:

35.

Read the value of the function at x = 1.5 sec, 4.5 sec, 7.5 sec, and 10 sec. These are the midpoints of rectangles with widths x equal to 3, 3, 3 and 2. We estimate the function values as 22, 76, 106 and 122, all in mph, which we will convert to feet per second after taking the sum, which will give us an answer in feet. 4

å f ( x )x » 22(3) + 76(3) + 106(3) + 122(2) = 856 i

i =1

856 (5280) » 1300 3600

The Lamborghini traveled about 1300 ft. 36.

Read the value for the speed at x = 2 sec, 6 sec, 10 sec and 13.75 sec. These are the midpoints of rectangles with widths x equal to 4, 4, 4, and 3.5. We estimate the function values as 30, 77, 102 and 124, all in mph, which we will convert to feet per second after taking the sum, which will give us an answer in feet. 4

å f ( xi )x » 30(4) + 77(4) + 102(4) + 124(3.5) = 1270 i =1

1270 (5280) » 1900 3600

The Alfa Romeo traveled about 1900 ft. In the last 3.5 seconds, the car traveled at an average speed of about 122 mph, covering about 626 feet, so the circle is at around 1224 feet, roughly a quarter of a mile. 37.

This problem is best solved using a calculator that can handle lists and summations. Extend the given table with a speed of 0 mph at 0 seconds and then make a list of the time differences: d1 = 1.7 - 0 = 1.7 d 2 = 2.3 - 1.7 = 0.6 d3 = 3.1 - 2.3 = 0.8

 d14 = 27.0 - 21.7 = 5.3 These are the widths of the 14 time intervals. There are 15 speeds, including the first speed, 0 mph and the final speed, 160 mph. List these speeds: s1 = 0 s2 = 30 s3 = 40

 s15 = 160

For left-endpoint rectangles, we pair d1 through d14 with s1 through s14 . Noting that 1 second is 1/3600 hr and 1 mile is 5280 feet, the left endpoint estimate in feet is

Section 7.3 5280 3600

519 14

å d ⋅ s = 4160.933 » 4161 ft k

k =1

For the right endpoint estimate we pair we pair d1 through d14 with s2 through s15. The right endpoint estimate is 5280 3600

åd ⋅ s k

k +1 = 4606.800 » 4607 ft

k =1

The average of these two estimates is 4160.933 + 4606.800 = 4383.867 2 » 4384 ft, so our best estimate of the distance traveled by the Mercedes Benz C63 AMG Edition 507 is 4384 ft. 38. This problem is best solved using a calculator that can handle lists and summations. Extend the given table with a speed of 0 mph at 0 seconds and then make a list of the time differences: d1 = 1.7 - 0 = 1.7 d 2 = 2.3 - 1.7 = 0.6 d3 = 3.3 - 2.3 = 1.0  d14 = 27.8 - 23.3 = 4.5 These are the widths of the 14 time intervals. There are 15 speeds, including the first speed, 0 mph and the final speed, 160 mph. List these speeds: s1 = 0

s2 = 30 s3 = 40  s15 = 160

For left-endpoint rectangles, we pair d1 through d14 with s1 through s14 . Noting that 1 second is 1/3600 hr and 1 mile is 5280 feet, the left endpoint estimate in feet is 5280 3600

å dk ⋅ sk = 4279.733 » 4280 ft k =1

For the right endpoint estimate we pair we pair d1 through d14 with s2 through s15. The right endpoint estimate is 5280 3600

å dk ⋅ sk +1 = 4737.333 » 4737 ft k =1

The average of these two estimates is 4279.733 + 4737.333 = 4508.533 2 » 4509 ft, so our best estimate of the distance traveled by the BMW M3 is 4509 ft.

520 39.

Chaper 7 INTEGRATION (a) Read values of the function on the plain glass graph every 2 hr from 6 to 6.These are at midpoints of the widths  x = 2 and represent the heights of the rectangles.

å f ( x )x = 80(2) + 110(2) + 80(2) + 26(2) + 21(2) + 18(2) + 8(2) = 684 i

The total heat gain was about 680 BTUs per square foot. (b) Read values on the triple-glazed graph every 2 hr from 6 to 6.

å f ( x )x = 35(2) + 54(2) + 38(2) + 12(2) + 10(2) + 8(2) + 3(2) = 320 i

The total heat gain was about 320 BTUs per square foot. 40.

(a) Read the value for a plain glass window facing south for every 2 hr from 6 to 6. These are the heights, at the midpoints, of rectangles with width  x = 2.

å f ( x )x = 7(2) + 17(2) + 39(2) + 53(2) + 39(2) + 17(2) + 8(2) = 360 i

The heat gain is about 360 BTUs per square foot. (b) Read the value for the triple-glazed facing south graph for every 2 hr from 6 to 6. These are the heights, at the midpoints, of rectangles with width  x = 2.

å f ( x )x = 3(2) + 8(2) + 19(2) + 24(2) + 19(2) + 8(2) + 4(2) = 170 i

The heat gain is about 170 BTUs per square foot. 41.

(a) Then area of a trapezoid is A=

1 1 h(b1 + b2 ) = (6)(1 + 2) = 9. 2 2

Car A has traveled 9 ft. (b) Car A is furthest ahead of car B at 2 sec. Notice that from t = 0 to t = 2, v(t ) is larger for car A than for car B. For t > 2, v(t ) is larger for car B than for car A. (c) As seen in part (a), car A drove 9 ft in 2 sec. The distance of car B can be calculated as follows: 2-0 1 = = width 4 2 Distance = 1 ⋅ v (0.25) + 1 v (0.75) + 1 v (1.25) + 1 v (1.75) 2

1 1 1 1 = (0.2) + (1) + (2.6) + (5) 2 2 2 2 = 4.4 9 - 4.4 = 4.6

The furthest car A can get ahead of car B is about 4.6 ft. (d) At t = 3, car A travels 12 (6)(2 + 3) = 15 ft and car B travels approximately 13 ft.

At t = 3.5, car A travels 12 (6)(2.5 + 3.5) = 18 ft and car B travels approximately 18.25 ft. Therefore, car B catches up with car A between 3 and 3.5 sec. (e) Cars A and B are going the same speed where their graphs intersect. From reading the values, the graphs intersect at 0 seconds and approximately 1.9 seconds. (f)

Section 7.4 42.

521

Using the left endpoints:

7.4 The Fundamental Theorem of Calculus

Distance = v0 (1) + v1(1) + v2 (1) + v3 (1)

Your Turn 1

= 0 + 8 + 13 + 17 = 38 ft Using the right endpoints:

Find

43.

Using the left endpoints: Distance = v0 (1) + v1(1) + v2 (1) = 10 + 6.5 + 6 = 22.5 ft Using the right endpoints:

ò 3x dx. 2

Distance = v1(1) + v2 (1) + v3 (1) + v4 (1) = 8 + 13 + 17 + 18 = 56 ft

ò 3x dx = x + C. 2

The indefinite integral is

By the Fundamental Theorem,

ò1

3x 2dx = x3

= 33 - 13

Distance = v1(1) + v2 (1) + v3 (1)

= 27 - 1

= 6.5 + 6 + 5.5 = 18 ft

= 26

44.

(a) Using the left endpoints: Your Turn 2

Distance =

å f ( x ) x i

i =1

= 0(2.88) + 22.8(1.76) + 27.0(1.65) + 27.3(1.63) + 27.6(1.66) = 0 + 40.128 + 44.55 + 44.499 + 45.816

Since we multiplied the units of seconds by miles per hour, we need to divide by 3600 (the number of seconds in an hour) to get a distance in miles. 174.993 » 0.0486 3600 The estimate of the distance is 0.0486 miles.

ò 3 (2x3 - 3x + 4) dx.

ò 3 (2 x3 - 3x + 4) dx 5

ò3

x dx + 4

ò 3 dx

5 2 4 5 3 5 x - x2 + 4 x 3 3 3 4 2 1 3 = (54 - 34 ) - (52 - 32 ) + 4(5 - 3) 2 2 = 256

Find

å f (x ) x

i =1

= 22.8(2.88) + 27.0(1.76) + 27.3(1.65) + 27.6(1.63) + 27.3(1.66) = 65.664 + 47.52 + 45.045 + 44.988 + 45.318 = 248.535

Divide by 3600 (the number of seconds in an hour) to get a distance in miles. 248.535 » 0.0690 3600

3æ

yö

3æ

0.5 y ö÷

ò1 ççèç y + e0.5 ÷ø÷÷÷ dy. ò ççèç y + e 1

÷÷ dy ø÷

æ 1 0.5 y ö÷ = ççç 2 ln | y | + e ÷÷ è ø 0.5 = ( 2 ln |3| + 2e

0.5(3)

100 » 0.0622 1609 Bolt actually ran 0.0622 miles. The answer to part b is closer.

3 1

) - ( 2 ln |1| + 2e0.5(1) )

= 2 ln 3 + 2e1.5 - 2e0.5

The estimate of the distance is 0.0690 miles. (c)

x 3 dx - 3

Your Turn 3

(b) Using the right endpoints: i

= 2

= 174.993

Distance =

Find

522

Chapter 7 INTEGRATION

Your Turn 4 4

ò 2x 16 - x dx .

Evaluate

( x 2 - 9) dx +

ò (x - 9) dx 2

æ1 ö æ1 ö = çç x3 - 9 x ÷÷ + çç x3 - 9 x ÷÷ ÷ø ÷ø çè 3 çè 3 0 3

Using Method 1: u = 16 - x 2.

Let

|9 - 27 | + (27 - 54 - (9 - 27) = 18 + 18 + 18 = 54.

Then du =-2 x. If x = 4, then u = 16 - 42 = 0. If x = 0, then u = 16 - 02 = 16.

7.4 Warmup Exercises

Now substitute.

W1.

ò 6x x + 4 dx 2

Let u = x 2 + 4; then du = 2 x dx.

2 x 16 - x 2 dx

ò 6x x + 4 dx = 3 u du ò 2

u du

1/ 2

0 16

2 3/ 2 2 = (64 - 0) u 3 3 0

128 3

æ2ö = 3çç ÷÷÷ u 3/ 2 + C çè 3 ø

(

= 2 x2 + 4

Your Turn 5

W2.

Find the area between the graph of the function f ( x) = x 2 - 9 and the x-axis from x = 0 to x = 6. Here is a graph of the function and the area to be found.

f ( x)

ò x + 2 dx 2

Let u = x 2 + 2; then du = 2 x dx. 4x dx 2 x +2 1 = 2 du u = 2ln u + C

3/ 2

)

x 9

= 2ln x 2 + 2 + C

(

)

= 2ln x 2 + 2 + C since x 2 + 2 is 10

always positive

W3. 6

15x 2e x dx

Let u = x 3; then du = 3x 2 dx.

ò 15x e dx = 5 e du ò 2 x3

Since the curve is below the x-axis on the interval (0, 3), the definite integral will count this area as negative. The total positive area is thus

= 5eu + C 3

= 5e x + C

Section 7.4

523

7.4 Exercises 1. 2.

ò (5x - 4x + 2) dx

True

False. A definite integral is a real number and an indefinite integral is a family of antiderivatives of a function.

False. If f ( x) ³ 0, then ò f ( x)dx gives the

ò dp = -3 ⋅ p

-2

4 -2

= -3[4 - (-2)] = -18

-4

2 dx =

ò dx = 2 ⋅ x

2[1 - (-4)]

= 5 2 » 7.071 2

ò-

(5t - 3)dt = 5

ò-

ò- dt

t dt - 3

1 2

5 2 2 t - 3t -1 2 -1

5 = [22 - (-1) 2 ] - 3[2 - (-1)] 2 5 = (4 - 1) - 3(2 + 1) 2 15 = -9 2 15 18 3 = =2 2 2

ò-

(4 z + 3) dz = 4

= 2z

ò-

- 2x2 0

-2 3

ò-

-2

+ 3z

ò-

x 2dx - 3

-2

= 2[22 - (-2) 2 ] + 3[2 - (-2)] = 2(4 - 4) + 3(4) = 12

2 0

ò- dx

x dx + 5

1 3 3 = - x3 - x2 + 5x -2 3 2 -2 -2 1 3 = - [33 - (-2)3 ] - [32 - (-2)2 ] + 5[3 - (-2)] 3 2 1 3 = - (27 + 8) - (9 - 4) + 5(5) 3 2 35 15 35 =+ 25 = 3 2 6

11.

ò 3 4u + 1 du 0

Let 4u + 1 = x, so that 4 du = dx. When u = 0, x = 4(0) + 1 = 1. When u = 2, x = 4(2) + 1 = 9. 2

ò 3 4u + 1 du 0

dz 2

ò dx 0

+ 2x

2 3

ò-

z dz + 3

-4

x dx + 2

ò (-x - 3x + 5) dx

10. =-

5 x3 3

True (-3) dp = -3

x 2dx - 4

5 3 (2 - 03 ) - 2(22 - 02 ) + 2(2 - 0) 3 5 40 - 24 + 12 = (8) - 2(4) + 2(2) = 3 3 28 = 3

area between the graph of f ( x) and the x-axis, from x = a to x = b. 4.

3 4 3 4

4u + 1 (4 du)

0 9

ò x dx 1/ 2

3 x 3/ 2 = ⋅ 4 3/2 1 3 2 3/ 2 ⋅ (9 - 13/ 2 ) 4 3 1 26 = (27 - 1) = = 13 2 2 =

524 12.

Chapter 7 INTEGRATION

2r - 2 dr =

= 2 y 5/2

(2r - 2)1/2 dr

5/2

- 1) + 2(43/2 - 1)

If r = 9, u = 2 ⋅ 9 - 2 = 16. If r = 3, u = 2 ⋅ 3 - 2 = 4.

= 2(32 - 1) + 2(8 - 1)

1/2

= 62 + 14 = 76

1 dr = 2 =

ò (2r - 2) 2dr 1/2

3 16

1 2

16.

u1/2du

ò (4 r - 3r r ) dr 4

4 3/2

1 u = ⋅ 3 2 2

2(t1/2 - t ) dt = 2

= 2⋅

r 3/ 2 = 4 3

t1/2dt - 2

0 4

t 3/2

t2 2

ò -(3x

3/2

1/2

) dx

= -3

x 3/2dx -

= -3

ò t dt 4 0

17.

x1/2dx

x 3/2 3 2

Let u = 2 x -7, so that du = 2 dx. When x = 6, u = 2 ⋅ 6 - 7 = 5. When x = 4, u = 2 ⋅ 4 - 7 = 1.

æ 5/2 ö÷ çy = 5 çç 5 ÷÷÷ çç è 2 ÷ø÷

ò u du -2

= -u-1

1/2

æ 3/2 ÷ö çy + 3 çç 3 ÷÷÷ çç è 2 ÷÷ø

1 5 1

( ) = -( - 54 ) =-

ò y dy

dx =

y 3/2dy + 3

u -1 = -1

1 4

ò (2x - 7)

(5 y y + 3 y ) dy

= 5

4 9

6 2 = - (32) - (8) 5 3 192 16 656 ==5 3 15 4

4 9

ò4 (2x - 7)2 dx

x 5/2 4 5 2

r 5/ 2 -3 5

8 3/ 2 6 r - r 5/ 2 3 5 4 4

8 6 (27 - 8) - (243 - 32) 3 5 8 6 = ⋅ 19 - (211) 3 5 760 3798 3038 = =15 15 15

4 = (43/2 - 03/2 ) - (42 - 02 ) 3 32 16 = - 16 = 3 3

ò4 r3/2dr

-2⋅

3 2

ò4

r1/2dr - 3

= 4

16 1 ⋅ u 3/2 4 3 1 3/2 = (16 - 43/2 ) 3 1 56 = (64 - 8) = 3 3 4

15.

= 2(4

ò (2r - 2)

14.

+ 2 y 3/2

Let u = 2r - 2, so that du = 2 dr.

13.

4 5

1 -1 5

Section 7.4 18.

525

-3

ò (2 p + 1) 1

dp 2

ò (2 p + 1) dp

= -3

3æ 1 1 ö 5æ 1 1ö = - çç - ÷÷ + çç - ÷÷ ÷ ç ç 2è 9 4 ø 3 è 27 8 ÷ø

-2

Let u = 2 p + 1, so that du = 2 dp.

If p = 4, u = 2 ⋅ 4 + 1 = 9. If p = 1, u = 2 ⋅ 1 + 1 = 3. 4

3 =2

21.

ò3 u du -2

= 2

19.

3 2u 3 3 3 1 - =18 6 3

-2 e + 3 ln | y| -0.1 -3 -3

= 20e

0.3

-2

+ 3ln | y|

-3

- 20e

-3

0.2

+ 3ln 2 - 3ln 3

22.

ò1 n dn - ò1 n dn -2

n-1 = 6⋅ -1

n-2 -2

-3

-6 6 1 1 = - + 5 1 50 2 108 = 25

-3

-4

-1 -2

3 e 0.3

-1

ò 2 0.3e0.3 dt t

-1 0.3t

-2

ö÷ 1 çç e 4u ÷÷ du çç (u + 1) 2 ÷÷ø 1 è

e4u = 4

ò x dx - 5ò x dx

x-2 = 3⋅ -2

x-3 -5 -3

e4u du -

1 3 dt + 0.3 -2 t

2æ

-4

-1

» 3.306

23.

ò (3x - 5x ) dx =3

= 2 ln 2 + 10e-0.3 - 10e-0.6

-6 æç -6 ö÷ éê 1 1 ùú = -ç + ÷ çè 1 ÷ø ê 2(25) 5 2(1) úû ë

tö

-2

ò 2 çççè t + 3 e0.3 ÷÷÷ø dt = -2 ln |t |

-6 1 + 2 n 1 2n 1

-3

-1æ

= -2

20.

ò 3 y dy

-0.1 y - 2

= -2 0e-0.1y

ò1 (6n 2 - n 3) dn

ò3

-2

e-0.1y dy +

» 1.353

-2

= 2⋅

-2 æ

ö çç 2e-0.1y + 3 ÷÷ dy ç y ø÷÷ -3 è

3 u -1 =- ⋅ 2 -1 =

1 3 5 5 + + 6 8 81 24

5 81

ò1 (2 p + 1) 2 dp

-3

3 1 5 1 ⋅ 2 + ⋅ 3 2 x 2 3 x 2

ò (u + 1) du

-1 u +1 1

1 1 e e4 + 4 4 2 +1 1+1

1 e8 e4 4 4 6 » 731.4

= 2

526 24.

Chapter 7 INTEGRATION

( p3 - e4 p )dp =

0.5

p3dp -

0.5 4

p = 4 =

e4 p dp

0.5

1 12

110

u 3 du =

110

1 u4 ⋅ 12 4

110

146, 410, 000 16 48 48 146, 409,984 = 48 9,150, 624 = 3 » 3, 050, 208 =

0.5

æ e4 e2 ö÷÷ 1 1 ç - çç ÷ 4 64 çè 4 4 ÷ø÷

15 e4 e2 + 64 4 4 » -11.57

27. 25.

ò- y(2 y - 3) dy 2

-1

1 u6 ⋅ 4 6

u5du =

z 7/6

7 6

-3

-1 -3

28. 2

Let u = 4m + 2, so that

Also, when m = 3, u = 4(33 ) + 2 = 110,

and when m = 0, u = 4(03 ) + 2 = 2.

3 - y1/3 y 2/3

ò (3 y

-2/3

- y-1/3 ) dy

1 du = 12m dm and du = m2 dm. 12 2

2 3

-1/3

1 64

6(64)7/6 6(1)7/6 7 7

ò z

z 2/3

-2 1

-3(642/3 - 12/3) 6(128) 6 = - - 3(16 - 1) 7 7 768 - 6 - 315 447 = = » 63.86 7 7

1 1 (-3)6 (-1)6 24 24 729 1 = 24 24 728 91 = = 24 3 =

ò m (4m + 2) dm

z1/6dz - 2

64 6 z 7/6 64 -3z 2/3 1 7 1

1 6 u = 24 -1

26.

When y = 0, u = 2(0)2 - 3 = -3 .

ö çç z -1/3 ÷÷ 2 z ÷÷ dz ç 1/2 ÷ø çè z

When y = -1, u = 2(-1)2 - 3 = -1.

-3

64 æ 1/2

Let u = 2 y 3 - 3, so that 1 du = 4 y dy and du = y dy. 4

1 4

z -2

1 4 u 48 2

3 y-2/3dy -

3 y1/3 1 3

ò y

-1/3

1 8

= 9 y1/3 1

y 2/3 2 3

3 y 2/3 2

1 8

3 9 9 = 9(2 - 1) - (4 - 1) = 9 - = 2 2 2

Section 7.4 29.

527

When x = 8, u = 84/3 + 9 = 25.

ln x dx x

3 4

Let u = ln x, so that du =

ò9

1 dx. x

(ln 2)2 -0 2

(ln 2)2 2 » 0.2402

30.

32.

du =

when x = 1, u = 1 + ln 1 = 1.

1 dx. x When x = 3, u = ln 3, and du =

ò1

ò0

ln 3

u 3/2 3 2

= 3 ln (1 + ln 2) - 3 ln 1

u1/2du

= 3 ln (1 + ln 2) » 1.580

ln 3

33.

2 3/2 u 3 0

ò x

1/3

e 2t

ò0 (3 + e2t )2 dt Let u = 3 + e 2t , so that du = 2e2t dt.

2 2 = (ln 3)3/2 - (0)3/2 3 3 2 = (ln 3)3/2 3 » 0.7677

31.

When x = 1, u = 3 + e2⋅1 = 3 + e2. When x = 0, u = 3 + e2⋅0 = 4. 1

e2t

1 dt = 2t 2 2 0 (3 + e )

3 + e2

ò4

1 u -1 = ⋅ 2 -1

x 4/3 + 9 dx

u-2du

3 + e2

3 4 1/3 x dx and du = x1/3dx. 4 3

When x = 0, u = 04/3 + 9 = 9.

» 0.07687

3 + e2

-1 = 2u 4

1 1 = 8 2(3 + e2 )

Let u = x 4/3 + 9, so that du =

3 du u 1+ ln 2

ln 3

1+ ln 2

= 3 ln |u| 1

when x = 1, u = ln 1 = 0. u du =

1 dx. x

When x = 2, u = 1 + ln 2, and

Let u = ln x, so that

ln 3

ò1 x(1 + ln x) dx Let u = 1 + ln x, so that

ln x dx x

1 3/2 u 9 2 1 1 = (25)3/2 - (9)3/2 2 2 125 27 = = 49 2 2

ln 2

u2 u du = 2 =

ò9 u1/2du 2

When x = 2, u = ln 2.

3 u 3/2 = ⋅ 3 4

When x = 1, u = ln 1 = 0.

ln 2

3 4

udu =

528 34.

Chapter 7 INTEGRATION e2 z

ò0 1 + e2z

36.

f ( x) = 4 x - 32; [5, 10]

Let u = 1 + e 2 z , so that

du = 2e2 z dz and

1 du = e 2 z dz. 2

When z = 1, u = 1 + e2 , and when z = 0, u = 1 + e0 = 2. 1 2

1+ e2

1 1 du = 2 u

1+ e2

The graph crosses the x-axis at 0 = 4 x - 32 4 x = 32 x = 8.

2 1+ e2

1 u1/2 ⋅ 1 2 2

u-1/ 2du

1+ e

= u1/2

2 2

This location is in the interval. The area of the region is

= 1 + e2 - 2

» 1.482

35.

(4 x - 32) dx +

ò (4x - 32) dx 8

= (2 x 2 - 32 x)

f ( x) = 2 x - 14; [6, 10]

8 5

+ (2 x 2 - 32 x)

10 8

= | (128 - 256) - (50 - 160) | + (200 - 320) - (128 - 256) = | -18| + (-120) + 128 = 26.

37.

The graph crosses the x-axis at 0 = 2 x - 14 2 x = 14 x = 7. This location is in the interval. The area of the region is

(2 x - 14) dx +

(2 x - 14) dx

f ( x) = 2 - 2 x 2 ; [0, 5]

Find the points where the graph crosses the x-axis by solving 2 - 2 x 2 = 0.

= |( x 2 - 14 x )| 2

7 6

+ ( x 2 - 14 x)

2 - 2x2 = 0

2x2 = 2

x2 = 1 x = 1.

= |(7 - 98) - (6 - 84)| + (102 - 140) - (72 - 98) = |- 1| + (-40) - (-49) = 10.

Section 7.4

529

The only solution in the interval [0, 5] is 1. The total area is 1

ò0

39.

f ( x) = x3; [-1, 3]

ò2 (2 - 2x2 ) dx

(2 - 2 x 2 ) dx + 1

æ æ 2 x3 ö÷÷ 2 x3 ö÷÷ ç ç = çç 2 x ÷÷ + çç 2 x ÷ çè 3 ÷ø 3 ø÷÷ çè 0 1 = 2-

2 2(53 ) 2 + 10 -2+ 3 3 3

The solution

4 -224 + 3 3 228 = 3 = 76. =

x3 = 0 x=0 indicates that the graph crosses the x-axis at 0 in the given interval [-1, 3]. The total area is

38.

f ( x) = 9 - x ; [0, 6]

ò1 -

x3dx +

x4 = 4

0 -1

ò0 x3dx x4 + 4

æ 4 æ 1ö ÷ö ç3 = çç 0 - ÷÷ + çç - 0 ÷÷÷ èç 4 ÷ø èç 4 ø÷ 1 81 82 + = 4 4 4 41 = ⋅ 2 =

The graph crosses the x-axis at 0 = 9 - x2 x2 = 9 x = 3. In the interval, the graph crosses at x = 3. The area of the region is

40.

f ( x) = x3 - 2 x; [-2, 4]

ò (9 - x ) dx

(9 - x 2 ) dx +

æ æ x3 ÷÷ö x3 ÷÷ö ç ç = çç 9 x ÷÷ + çç 9 x ÷ 3 ÷ø 3 ÷÷ø çè çè 0 3

= (27 - 9)+ | (54 - 72) - (27 - 9) | = 18 + | - 36 |

The graph crosses the x-axis at 0 = x3 - 2 x

= 18 + 36

= x( x 2 - 2)

= 54. x = 0, x =

2, and x = - 2.

These location are all in the interval. The area of the region is

530

Chapter 7 INTEGRATION

- 2

-2

( x3 - 2 x) dx +

- 2

( x3 - 2 x) dx +

( x3 - 2 x) dx 4

42.

f ( x) = 1 - e-x ; [-1, 2]

ò (x - 2x) dx 3

æ x4 ö÷ ç = çç - x 2 ÷÷÷ çè 4 ø÷

- 2

æ 4 ÷ö çx + çç - x 2 ÷÷÷ çè 4 ÷ø

-2

- 2

æ x4 ö÷ 2 æ x4 ö÷ ç ç + çç - x 2 ÷÷÷ + çç - x 2 ÷÷÷ ÷ø ÷ø çè 4 çè 4 0

The graph crosses the x-axis at 2

0 = 1 - e-x

= | (1 - 2) - (4 - 4) | + | 0 - (1 - 2) | + | (1 - 2) - 0 + (64 - 16) - 1 - 2| = | - 1| + |1| + | -1| + | 49| = 1 + 1 + 1 + 49 = 52.

e-x = 1 -x ln e = ln 1 -x = 0 x = 0. The area of the region is 0

41.

ò1

f ( x) = e - 1; [-1, 2]

(1 - e-x ) dx +

= ( x + e-x )

0 -1

ò0 (1 - e x ) dx -

+ ( x + e-x ) 1

= e - 2 + 1 + e-2 = e - 1 + e-2 » 1.854.

Solve e x - 1 = 0.

43.

e =1 x ln e = ln 1 x =0

f ( x) =

1 1 - ; [1, e2 ] x e

The graph crosses the x-axis at 0 in the given interval [-1, 2]. The total area is

-1

(e x - 1) dx +

ò (e - 1) dx x

= (e - x)

-1 -1

= | (1 - 0) - (e -1

+ (e - x) 2

2 0

+ 1) | + (e - 2) - (1 - 0)

= |1 - e - 1| + e2 - 2 - 1 1 = + e2 - 3 e » 4.757.

The graph crosses the x-axis at 1 1 0= x e 1 1 = x e x = e.

-2

= (1) - (-1 + e ) + (2 + e = 2 - e + 2 + e-2 - 1

) - (e 0 )

Section 7.4

531

This location is in the interval. The area of the region is eæ

ö çç 1 - 1 ÷÷ dx + ç e ÷ø 1 èx

òe

y = 4 - x 2 ; [0, 3]

45.

From the graph, we see that the total area is

e2 æ

ö çç 1 - 1 ÷÷ dx çè x e ÷ø

ò0

e æ x xö = ln x + çç ln x - ÷÷÷ çè e 1 eø e

ò2 (4 - x2 ) dx

æ x3 ÷÷ö ç + çç 4 x ÷ 3 ÷÷ø çè

éæ ù 8ö = ê çç 8 - ÷÷÷ - 0 ú ê çè ú 3ø ë û

1 + |2 - e | e 1 = e-2+ . e =

f ( x) = 1 -

(4 - x 2 ) dx +

æ x3 ö÷÷ ç = çç 4 x ÷ 3 ÷ø÷ èç

æ 1ö = 0 - çç - ÷÷÷+ |(2 - e) - 0| çè e ø

44.

é æ 8 öù + ê (12 - 9) - çç 8 - ÷÷÷ ú çè ê 3 ø úû ë 16 16 = + 33 3 16 7 = + 3 3 23 = 3

1 ; [e-1, e] x

f ( x) = x 2 - 2 x; [-1, 2]

46.

From the graph, we see that the total area is 0

ò1 -

The graph crosses the x-axis at 1 0 = 1x 1 =1 x x = 1. This location is in the interval. The area of the region is 1 æ

1ö çç 1 - ÷÷÷ dx + -1 ç xø e è

eæ

1ö

ò1 ççèç1 - x ÷÷ø÷ dx

= ( x - ln x ) -1 + ( x - ln x ) -1

= |(1 - 0) - [e

- (-1)]| + (e - 1) - (1 - 0)

æ1 ö = 1 - çç + 1÷÷÷ + (e - 1) - 1 çè e ø = e-2+

1 . e

ò0 (x - 2x) dx

( x 2 - 2 x) dx +

æ x3 ö÷ ç = çç - x 2 ÷÷÷ çè 3 ÷ø

-1

æ x3 ö÷ ç + çç - x 2 ÷÷÷ ÷ø èç 3

æ 1 ö 8 = - çç - - 1÷÷÷ + -4 çè 3 ø 3 =

4 4 4 4 8 + = + = . 3 3 3 3 3

y = e x - e; [0, 2]

47.

From the graph, we see that total area is 1

ò0

(e x - e) dx +

= (e x - xe) 1

1 0

ò1 (e x - e) dx

+ (e x - xe) 0

2 1 2

= |(e - e) - (e + 0)| + (e - 2e) - (e1 - e) = |-1| + e 2 - 2e = 1 + e 2 - 2e » 2.952.

532 48.

Chapter 7 INTEGRATION ln x é 1 ù ; ê , eú x êë e úû

y =

52.

From the graph, we can see that the total area is 1 æ

ö çç ln x ÷÷ dx + ÷ ç 1/ e è x ø

(ln x ) = 2

1 1/ e

ö çç ln x ÷÷ dx ÷ ç 1 è x ø

(ln x ) + 2

òa

k f ( x) dx = k

ò k f (x) dx = kF (x)

49.

= kF (b) - kF (a)

= k[ F (b) - F (a)] = k

ò f (x) dx a

ò f (x) dx = ò f (x) dx + ò f (x) dx

53.

ò f (x) dx

Prove:

ò f (x) dx.

f ( x) dx +

50.

The equation for Exercise 49 is c

Let F(x) be an antiderivative of f ( x).

ò f (x) dx = ò f (x) dx + ò f (x) dx.

f ( x) dx +

ò f (x) dx c

= F ( x) + F ( x)

(a) If b is replaced by d, we get c

ò f (x) dx = ò f (x) dx + ò f (x) dx.

= [ F (c) - F (a)] + [ F (b) - F (c)]

This is a correct statement.

= F (b) - F (a)

= F (c) - F (a) + F (b) - F (c)

(b) If b is replaced by g, we get c

òa

f ( x) dx =

f ( x) dx +

òg

ò f (x) dx a

f ( x) dx.

This is a correct statement.

54.

Prove:

ò0

f ( x) dx =

ò0 +

f ( x) dx +

ò2

f ( x) dx

ò5 f (x) dx + ò8 f (x) dx

ò f (x) dx.

f ( x) dx = -

51.

b a

1 1 + -0 2 2 1 1 = + =1 2 2 b

òa f (x) dx.

Then k F ( x) is an antiderivative of k f ( x).

= 0-

Assume F ( x) is an antiderivative of f ( x).

eæ

Prove:

Assume F ( x) is an antiderivative of f ( x). b

ò f (x) dx = [F (x)] a

1  (3 ) ⋅ 2(1 + 3) + 2 4

 (32 )

1 - (3)(8) 4 2 9 9 = 4 +  -  - 12 = -8 4 4

b a

= F (b) - F (a) = (-1)[ F (a) - F (b)] a

ò f (x) dx

= -1

b a

ò f (x) dx

Section 7.4 55.

533

-1

f ( x) dx 0

ò-1

(2 x + 3) + dx

= ( x 2 + 3x )

f ( x) =

4æ

ö çç - x - 3 ÷÷ dx ÷ø ç 0 è 4

æ x2 ö÷ ç + çç - 3x ÷÷÷ -1 ÷ø çè 8 0

et dt and f (1.01) =

1.01

et dt.

Use the fnInt command in the Math menu of your calculator to find

ò e dx and x2

= 2 - 14

= -12

56.

f (1) =

= - (1 - 3) + (-2 - 12)

et dt.

1.01

e x dx. The resulting screens are:

ò0 e dx = 1.46265; ò0 e dx = 16.45263 x

(a) Since e x is symmetric about the y-axis, 1

ò-1

e x dx = 2

ò0 e x dx 2

= 2(1.46265) = 2.92530.

(b)

e x dx =

e x dx -

ò e dx x2

= 16.45263 - 1.46265 = 14.98998

57.

(a)

f (1) » 1.46265 f (1.01) » 1.49011

g (t ) = t 4 and c = 1, use substitution.

f ( x) =

f (1 + h) - f (1) to approximate h f ¢(1) with h = 0.01

ò g(t)dt

Use

ò t dt 4

t 5

f (1 + h) - f (1) f (1.01) - f (1) = 0.01 h 1.49011 - 1.46265 » 0.01 = 2.746

x 1

(1)5 = x 5 5 5

So f ¢(1) » 2.746, and g (1) = e1 = e » 2.718.

5 = x -1 5 5

(b)

f ¢( x) = d ( f ( x)) dx æ 5 ö = d ççç x - 1 ÷÷÷ dx è 5 5ø

58.

ò x(x + 3) dx 2

-5

Let u = x2 + 3, so that du = 2 x dx.

æ ö = 1 ⋅ d ( x5 ) - d çç 1 ÷÷÷ 5 dx dx è 5 ø = 1 ⋅ 5x4 - 0 = x4 5 4

(a)

When x = 5, u = 52 + 3 = 28. When x = -5, u = (-5)2 + 3 = 28. 4

Since g (t ) = t , then g ( x) = x and we

-5

1 2 = 0

x( x 2 + 3)7dx =

ò u du 28

534 59.

Chapter 7 INTEGRATION P¢(t ) = (3t + 3)(t 2 + 2t + 2)1/3

(a)

60.

ò 3(t + 1)(t + 2t + 2) dt 2

1/3

H ¢( x) = 20 - 2 x is the rate of change of the number of hours it takes a worker to produce the xth item.

(a) The total number of hours required to produce the first 5 item is

Let u = t + 2t + 2, so that

du = (2t + 2) dt and 12 du = (t + 1) dt.

(20 - 2 x) dx = (20 x - x 2 )

It would take 75 hr to produce 5 items.

When t = 3, u = 32 + 2 ⋅ 3 + 2 = 17.

(b) The total number of hours required to produce the first 10 items is

3 u 4/3 17 u1/3du = ⋅ 4 2 2 9 4/3 17 u 8 2 9 9 4/3 = (17) - (2) 4/3 8 8 » 46.341

(20 - 2 x)dx = (20 x - x 2 )

(b)

Total profits for the first 3 yr were 9000 (17 4/3 - 24/3 ) » $46, 341. 8

It would take 100 hr to produce the first 10 items. 61.

P¢(t ) = 140t 5/2 4

ò 140t dt = 140 ⋅ 5/2

t 7/2 7 2

= 40t 7/2

Let u = t 2 + 2t + 2, so that du = (2t + 2) dt = 2(t + 1) dt and 3 du = 3(t + 1) dt. 2

9 u1/3du = u 4/3 8 17 17

9 9 (26) 4/3 - (17) 4/3 » 37.477 8 8 Profit in the fourth year was 9000 (264/3 - 17 4/3 ) » $37, 477. 8 (c) lim P¢(t ) =

x ¥

= lim (3t + 3)(t 2 + 2t + 2)1/3

62.

The tanker is leaking oil at a rate in barrels per hour of 80 ln (t + 1) L¢(t ) = . t +1 (a)

80 ln (t + 1) dt t +1

1 dt. Let u = ln (t + 1), so that du = t + 1

When t = 24, u = ln 25. When t = 0, u = ln 1 = 0. 80

The annual profit is slowly increasing without bound.

Since 5120 is above the total level of acceptable pollution (4850), the factory cannot operate for 4 years without killing all the fish in the lake.

x ¥

=¥

= 5120

When t = 3, u = 32 + 2 ⋅ 3 + 2 = 17.

= 100.

3 2

= (200 - 100) - (0)

3(t + 1)(t 2 + 2t + 2)1/3 dt

When t = 4, u = 42 + 2 ⋅ 4 + 2 = 26.

= 100 - 25 = 75.

When t = 0, u = 0 + 2 ⋅ 0 + 2 = 2.

3 2

ln 25

ò0

u2 u du = 80 2 = 4 0u 2

ln 25

0 ln 25 0

= 40(ln 25) 2 - 40(0) 2 » 414

Section 7.4 (b)

535 64.

80 ln (t + 1) dt t +1 24

3.5

ò0

Let u = ln (t + 1), so that the limits of integration with respect to u are ln 25 and ln 49. 80

ln 49

òln 25

Total growth after 3.5 days is

u du = 40 u 2

R¢( x)dx =

ln 25

= 750e0.2 x

= 40(ln 49)2 - 40(ln 25)2

= 750e

» 191

Growth rate is 0.6 +

4 (t +1)3

65.

R¢(t ) =

5 + t +1

12 æ

3.5

0 3.5 0

- 750e0

2 t +1

ò1 ççççè t + 1 + t + 1 ÷÷÷÷ø dt

ft/yr.

= éê 5 ln(t + 1) + 4 t + 1 ùú ë û 1

= (5 ln 13 + 4 13) - (5 ln 2 + 4 2)

ò1 êêë 0.6 + (t + 1)3 úúû dt

» 18.12.

é ù 4 = ê 0.6t + ú 2 êë -2(t + 1) úû

(b) Total reaction from t = 12 to t = 24 is 24 æ

5 + ççç ç 12 è t + 1

é ù 2 = ê 0.6(2) ú êë (2 + 1)2 úû é ù 2 - ê 0.6(1) ú 2 êë (1 + 1) úû

2 ö÷ ÷÷ dt t + 1 ÷ø 24

= éê 5 ln(t + 1) + 4 t + 1 ùú ë û 12

= (5 ln 25 + 4 25) - (5 ln 13 + 4 13) » 8.847.

= 44 - 1 45 10 » 0.8778ft.

66.

F (T ) =

(b) Total growth in the third year is =

ù 4 ê 0.6 + ú dt ê ú (t + 1)3 ûú 2 ëê

0.7

(a) Total reaction from t = 1 to t = 12 is

(a) Total growth in the second year is 2

0.2 x

» 760.3.

The number of barrels of oil leaking per day is decreasing to 0. 63.

ò0 150e

e0.2 x = 150 ⋅ 0.2

ln 49

About 191 barrels will leak on the second day. 80 ln (t + 1) (c) lim L¢(t ) = lim = 0 t +1 t ¥ t ¥

3.5

3é

é ù 4 ú = êê 0.6t + 2ú 2( + 1) t ëê ûú 2 é ù 2 ú = êê 0.6(3) ú (3 + 1)2 ûú ëê é ù 2 ú - êê 0.6(2) 2ú + (2 1) ëê ûú 67 44 = 40 45 » 0.6972 ft.

ò0 f (x) dx T

ò0 kb xdx T

ò0 ke(ln b)xdx

= k

ò0 e(ln b)xdx

T k ⋅ e(ln b) x dx 0 ln b k é (ln b)T = - 1ùú êe û ln b ë k = [bT - 1] ln b

536 67.

Chapter 7 INTEGRATION (b)

(c)

ò0

(b) Q(0.4) =

n( x) dx

2 = 0.04k mm/min

5x + 1 dx

70.

Let u = 5 x + 1. Then du = 5 dx.

1 5

ò (0.1762 x - 3.986x + 22.68) dx 9

æ 0.1762 3 3.986 2 ö = çç x x + 22.68 x ÷÷÷ çè 3 ø 3 2

ò26 u1/2du 1 u 3/2 = ⋅ 3 5 2

= 85.5036 - 51.6888 = 33.8148

The total increase in the length of a ram’s horn during the period is about 33.8 cm.

71.

26 51

2 3/2 u 15 26

E (t ) = 753 t -0.1321

(a) Since t is the age of the beagle in years, to convert the formula to days, let T = 365t , T . or t = 365

2 (513/2 - 263/2 ) 15 » 30.89 million

æ T ö÷-0.1321 E (T ) = 753 çç çè 365 ÷÷ø » 1642T -0.1321 Now, replace T with t.

w¢(t ) = (3t + 2)1/3 w(t ) =

When x = 5, u = 26; when x = 10, u = 51.

68.

ò0 (3t + 2)

1/3

1 (3t + 2)4/3 dt = ⋅ 4 3 3 =

(3t + 2) 4

4/3

E (t ) = 1642t -0.1321

(b) The beagle’s age in days after one year is 365 days and after 3 years she is 1095 days old.

3 0

1095

ò365 1642t-0.1321dt

1 = (114/3 - 24/3 ) 4 » 5.486 mg

69.

1095

= 164 2

v = k (R2 - r 2 )

(a)

 k (0.4) 4

Q( R ) = =

» 1892 (1, 0950.8679 - 3650.8679 ) » 505,155

ò 2 vr dr 0 R

ò0

= 2 k

The beagle’s total energy requirements are

2 k ( R 2 - r 2 ) r dr

about 505,000 kJ/W0.67 , where W represents weight.

ò0 (R2r - r 2 ) dr

æ = 2 k ççç R r - r ÷÷÷ è 2 4 ø 2 2

4ö

4ö æ 4 = 2 k ççç R - R ÷÷÷ è 2 4 ø

æ 4ö = 2 k ççç R ÷÷÷ è 4 ø =  kR 2

1 t 0.8679 0.8679 365

72.

100

0.85e0.0133x dx

Let u = 0.0133x, so that du = 0.0133dx, or 1 dx = 0.0133 du

When x = 100, u = 1.33. When x = 0, u = 0.

Section 7.4

537

1.33

1.33 æ 1 ö÷ 0.85 u du = e 0.85eu çç ÷ çè 0.0133 ø÷ 0.0133 0

(d)

20 = 30e0.04T - 30. 50 = 30e0.04T 5 = e0.04T 3 5 ln = 0.04T ln e 3

ò (38.5 + 5.66x - 1.10x ) dx 2

= (38.5x + 2.83x 2 - 0.367 x3 )

T =

The oil will last about 12.8 years.

The integral represents the population aged 0 to 90, which is about 308 million.

(b)

(e)

(38.5 + 5.66 x - 1.10 x ) dx

ò 1.2e

74.

4 = e0.02T 3 4 ln = 0.02T ln e 3 ln 43 T = » 14.4 0.02 The oil will last about 14.4 years.

(0.00462 x - 0.160 x + 1.3x + 2.85) dx 7.5 5

c¢(t ) = ke rt

76.

c¢(t ) = 1.2 e0.04t

ò 1.2e

0.04t

(c)

ò 1.2e 0

0.04t

ò 1.2e

0.04t

dt.

1.2 0.04t e 0.04 0

= 30 (e0.04T - e0 )

C ¢(t ) = 1.2e0.04t C (T ) =

(b) The amount of oil that the company will sell in the next ten years is given by the integral 10

- 60

80 = 60e0.02T

(a)

20 = 60e0.02T - 60.

» 14.5 About 14.5 million, or 14,500,000 families have incomes between $50,000 and $75,000.

75.

0.02T

Solve

5.5

= (0.001155 x 4 - 0.0533x3 + 0.65 x 2 + 2.85 x)

dt = 60e0.02t = 60e

7.5

» 57 The number of baby boomers is about 57 million. 7.5

0.02t

5.5

= (38.5 x + 2.83x 2 - 0.367 x3 )

ln 53

0.04 » 12.8

» 308

7.5

Solve

f ( x) = 38.5 + 5.66 x - 1.10 x 2 9

= 30e0.04T - 30

0.85 1.33 - e0 ) (e 0.0133 » 177.736 The total mass of the column is about 178 g.

(a)

1.2e0.04t dt = 30e0.04t

73.

dx =

0.04t

1.2e 0.04

= 30e0.04t

= 30 (e0.04T - 1)

In 5 yr, 0 10

C (5) = 30(e0.04(5) - 1)

= 30e0.4 - 30 » 14.75

= 30(e0.2 - 1) » 6.64 billion barrels.

538

Chapter 7 INTEGRATION The corresponding y values are 0 and 1, respectively, so the curves intersect at (0, 0) and (1, 1), as shown in

7.5 The Area Between Two Curves

the graph below. Here f ( x) = x1/4 and g ( x) = x 2.

Your Turn 1

Find the area bounded by f ( x) = 4 - x 2 , g ( x) = x + 2, x = -2, and x = 1. A sketch such as the one below shows that the two graphs intersect at the points (-2, 0) and (1, 3).

g ( x)  x

2 1 4

f ( x)  x

y 4 f ( x)  4  x

g ( x)  x  2

Over the interval [0, 1], f ( x) ³ g ( x), so the area is 4

given by the integral

ò [ f (x) - g (x)] dx. 0

Over the interval [-2, 1], f ( x) ³ g ( x), so the area will be given by 1

ò -2

ò-

[ f ( x) - g ( x)] dx =

ò (x

1/ 4

- x 2 ) dx

æ4 1 ö = çç x5/2 - x3 ÷÷÷ çè 5 3 ø0

[ f ( x) - g ( x) ] dx.

[ f ( x) - g ( x) ] dx =

æ4 1ö 7 = çç - ÷÷÷ - ( 0 - 0 ) = çè 5 3ø 15

ò -2[4 - x2 - (x + 2)] dx 1

ò -2 (2 - x - x2 ) dx 1

æ 1 1 ö = çç 2 x - x 2 - x3 ÷÷÷ çè 2 3 ø -2

Your Turn 3

Find the area enclosed by y = x 2 - 3x and y = 2 x on [0, 6]. First find where the two graphs intersect.

æ 1 1ö æ 8ö = çç 2 - - ÷÷ - çç -4 - 2 + ÷÷ ÷ çè ç 2 3ø è 3 ÷ø

x 2 - 3x = 2 x

9 = 2

x( x - 5) = 0

x2 - 5x = 0 x = 0 or

Your Turn 2

x = 5.

Find the area between the curves y = x1/4 and

The intersection points are (0, 0) and (5, 10), so we will need to use two integrals. On (0, 5), 2 x is the larger

y = x 2. First find where these two curves intersect by setting the two righthand sides equal.

function and on (5, 6), x 2 - 3x is the larger function, as illustrated in the following graph.

x1/4 = x 2

x = x

y x

 3x y  2x

x8 - x = 0 x( x 7 - 1) = 0

x = 0 or x 7 - 1 = 0 x = 0 or x = 1. 0

Section 7.5

Area =

539 5

[2 x - ( x 2 - 3x)] dx +

0 5

[( x 2 - 3x) - 2 x] dx

ò (5x - x ) dx + ò (x - 5x) dx

7.5 Warmup Exercises 8

ò1 x1/ 3 + x 4 / 3 dx

W1.

æ3 ö 3 = çç x 4 / 3 + x 7 / 3 ÷÷÷ çè 4 ø1 7

æ5 æ1 1 ö 5 ö = çç x 2 - x3 ÷÷ + çç x3 - x 2 ÷÷ ÷ çè 2 ç 3 ø 0 è3 2 ÷ø 5

æ3 ö æ3 ö 3 3 = çç 84 / 3 + 87 / 3 ÷÷÷ - çç 14 / 3 + 17 / 3 ÷÷÷ çè 4 ç ø è4 ø 7 7 3 3 3 3 = 16 + 128 - 4 7 4 7 1839 = » 65.68 28

æ 125 125 ö æ 216 180 125 125 ÷ö = çç - 0 ÷÷÷ + çç + ÷ èç 2 ø èç 3 3 2 3 2 ÷ø 71 3

Your Turn 4

Find the consumers’ surplus and the producers’ surplus for oat bran when the price in dollars per ton is D(q) = 600 - e q /3 when the demand is q tons, and the price in dollars per ton is S (q) = eq /3 - 100 when the demand is q tons.

W2.

e2 x dx =

First find the equilibrium quantity. e q /3 - 100 = 600 - eq /3 2e q /3 = 700

q /3

q » 17.57380

The equilibrium price is S (17.57380) = e17.57380 / 3 - 100

(

)

1 6 e - e0 2

e6 - 1 » 201.2 2

7.5 Exercises 1.

False. To find the area between f and g on an interval, both functions do not have to be positive on the interval.

True

x = -2, x = 1, y = 2 x 2 + 5, y = 0

= 350

q = ln 350 3 q = 3 ln 350

1 2x e 2 0

» 250.00

The consumers’ surplus is given by the following integral:

17.57380

(600 - e q /3 - 250) dq

(

= 350q - 3e q /3

17.57380

= (350(17.57380) - 3e17.57380/3) - (0 - 3) » 5103.83

The consumers’ surplus is $5103.83. As in Example 5, the producers’ surplus has the same value, $5103.83.

æ 2 x3 ö÷ ç + 5 x ÷÷÷ [(2 x + 5) - 0] = çç çè 3 ÷ø -2 -2

æ2 ö æ 16 ö = çç + 5 ÷÷÷ - çç - - 10 ÷÷÷ èç 3 ø çè 3 ø = 21

540 6.

Chapter 7 INTEGRATION

x = 1, x = 2, y = 3x3 + 2, y = 0

-1

ò-

[0 - ( x3 + 1) dx] +

ò- [(x + 1) - 0] dx 3

æ -x 4 ö÷ ç = çç - x ÷÷÷ çè 4 ø÷

-1

-3

æ x4 ö÷ ç + çç + x ÷÷÷ èç 4 ø÷

-1

æ 1 ö æ 81 ö æ1 ö æ1 ö = çç - + 1÷÷ - çç - + 3 ÷÷ + çç + 1÷÷ - çç - 1÷÷ ÷ø çè 4 ÷ø èç 4 ÷ø èç 4 çè 4 ø÷ = 20

x = -3, x = 0, y = 1 - x 2 , y = 0

æ 3x 4 ö÷ ç [(3x + 2) - 0] dx = çç + 2 x ÷÷÷ çè 4 1 ø÷ 1

æ3 ö = (12 + 4) - çç + 2 ÷÷÷ çè 4 ø =

53 4

x = -3, x = 1, y = x3 + 1, y = 0

To find the points of intersection of the graphs in [3, 0], substitute for y. 1 - x2 = 0 x2 = 1 x = -1 or x = 1

The region is composed of two separate regions because y = 1 - x 2 intersects y = 0 at x = -1. Let f ( x) = 1 - x 2 , g ( x) = 0. In the interval [-3, -1], g ( x) ³ f ( x). To find the points of intersection of the graphs, substitute for y. 3

x +1= 0

In the interval [-1, 0], f ( x) ³ g ( x).

-1

ò [(1 - x ) - 0]dx

-1 æ x3 ö÷÷ çç = ç -x + ÷ 3 ÷÷ø çè -3

æ x3 ö÷÷ ç + çç x ÷ 3 ÷÷ø èç -1

-3

x 3 = -1 x = -1

The region is composed of two separate regions because y = x3 + 1 intersects y = 0 at x = -1. 3

Let f ( x) = x + 1, g ( x) = 0.

[0 - (1 - x 2 )]dx +

-1

æ æ 1ö 1ö = çç1 - ÷÷ - (3 - 9) + 0 - çç -1 + ÷÷ ÷ çè ç è 3ø 3 ÷ø =

22 3

In the interval [-3, -1], g ( x) ³ f ( x). In the interval [-1, 1], f ( x) ³ g ( x).

Section 7.5

541 5 x = 3x + 10

x = -2, x = 1, y = 2 x, y = x 2 - 3

2 x = 10

x =5

If x = 5, y = 5(5) = 25. The region is composed of two separate regions because y = 5 x and y = 3x + 10 intersect at x = 5 (that is, (5, 25). Let f ( x) = 3x + 10, g ( x) = 5x. In the interval [0, 5], f ( x) ³ g ( x) .

Find the points of intersection of the graphs of

In the interval [5, 6], g ( x ) ³ f ( x ).

y = 2 x and y = x 2 - 3 by substituting for y.

ò (3x + 10 - 5x) dx

2x = x2 - 3

0 = x2 - 2x - 3

0 = ( x - 3) ( x + 1) =

In the interval [-2, -1], ( x 2 - 3) ³ 2 x.

ò-

ò- [(2 x) - ( x - 3)] dx

ò-

( x 2 - 3 - 2 x ) dx +

æ x3 ö÷ ç = çç - 3x - x 2 ÷÷÷ èç 3 ø÷

ò-

-2

æ ö÷ 1 x3 ç + çç x 2 + 3x ÷÷÷ 3 èç ø÷

-1

æ 8 ö 1 1 + 3 - 1 - çç - + 6 - 4 ÷÷÷ + 1 - + 3 ç è 3 ø 3 3 æ ö 1 - çç1 + - 3 ÷÷÷ çè ø 3

10.

= (-x + 10 x )

(2 x - x 2 + 3) dx

ò (-2x + 10) dx + ò (2x - 10)

5 0

6 5

æ 2x2 ö÷ ç + çç - 10 x ÷÷÷ èç 2 ø÷ 2

+ ( x - 10 x)

6 5

= -25 + 50 + (36 - 60) - (25 - 50)

-1

æ -2 x 2 ö÷ ç = çç + 10 x ÷÷÷ èç 2 ø÷

-1

In the interval [-1, 1], 2 x ³ ( x 2 - 3). [( x 2 - 3) - (2 x )] dx +

ò [5x - (3x + 10)] dx 5

The only intersection in [-2, 1] is at x = -1.

-1

= 26

11.

y = x 2 - 30 y = 10 - 3x

5 23 +6 = 3 3 x = 0, x = 6, y = 5 x, y = 3x + 10

Find the points of intersection. x 2 - 30 = 10 - 3x x 2 + 3x - 40 = 0 ( x + 8)( x - 5) = 0 x = -8 or x = 5

To find the intersection of y = 5 x and y = 3x + 10, substitute for y.

542

Chapter 7 INTEGRATION 4

ò [(x - 6) - (x - 18)] dx

Let f ( x) = 10 - 3x and g ( x) = x 2 - 30.

-3

The area between the curves is given by 5

ò-8 = =

[ f ( x) - g ( x)] dx 5

ò-8

[(10 - 3x) - ( x 2 - 30)] dx

ò-8 (-x2 - 3x + 40) dx

ò (x - 6 - x + 18) dx

ò (-x + x + 12) dx

-3 4

-3

æ -64 ö æ ö 9 = çç + 8 + 48 ÷÷ - çç 9 + - 36 ÷÷ ÷ ÷ èç 3 ø çè ø 2

æ -64 ö æ 9ö = çç + 56 ÷÷ - çç -27 + ÷÷ ÷ø èç çè 3 2 ÷ø

-53 3(5) 2 + 40(5) 3 2 é - (-8)3 ù 3(-8) 2 - êê + 40(-8) úú 3 2 ëê ûú

75 512 192 -125 + 200 + + 320 3 2 3 2 » 366.1667.

12.

æ 3 x2 ÷ö ç -x = çç + 12 x ÷÷÷ + çè 3 2 ø÷

æ -x 3 ö÷ 3x3 ç = çç + 40 x ÷÷÷ 2 ÷ø çè 3 -8 =

9 -64 + 83 3 2 343 = 6 » 57.167 =

13.

y = x2, y = 2x

y = x 2 - 18, y = x - 6

Find the points of intersection. x2 = 2x x2 - 2x = 0 x ( x - 2) = 0 x = 0 or x = 2

Find the intersection points. x 2 - 18 = x - 6

Let f ( x) = 2 x and g ( x) = x 2.

x 2 - x - 12 = 0 ( x - 4)( x + 3) = 0

The area between the curves is given by

The curves intersect at x = -3 and x = 4.

ò0

[ f ( x) - g ( x)] dx =

ò0 (2x - x2 ) dx 2

æ 2x2 x3 ö÷÷ ç = çç ÷ 3 ÷÷ø çè 2 0 = 4-

8 4 = . 2 3

Section 7.5 14.

543

y = x 2 , y = x3

2æ

1 1ö ççç - ÷÷÷ dx + è x 2ø 1

6æ

1ö

ò çççè 2 - x ÷÷÷ø dx 2

æ æx ö xö = çç ln x - ÷÷ + çç - ln x ÷÷ ÷ ÷ø çè ç è2 2ø 1 2 æ 1ö = (ln 2 - 1) - çç 0 - ÷÷÷ + (3 - ln 6) - (1 - ln2) çè 2ø = 2 ln 2 - ln 6 +

x = 0, x = 4, y =

16.

Find the intersection points.

3 » 1.095 2 1 x -1 ,y = x +1 2

x 2 = x3 x 2 - x3 = 0 x 2 (1 - x) = 0

The curves intersect at x = 0 and x = 1. In the interval [0, 1], x 2 > x3. 1

æ x3 x 4 ö÷÷ ç ( x - x ) dx = çç ÷ çè 3 4 ÷ø÷ 0 0

15.

Find the intersection points. 1 x -1 = 2 x +1

1 1 1 - = 3 4 12

x2 - 1 = 2

1 1 x = 1, x = 6, y = , y = x 2

x2 - 3 = 0

In the interval [0, 4], the only intersection point is at x = 3.

To find the points of intersection of the graphs, substitute for y. 1 1 = x 2 x = 2

3æ

1 - x - 1 ö÷ dx + ÷ ççç è x +1 2 ÷ø

4æ

Let f ( x) = 1x , g ( x) = 12 . In the interval [1, 2], f ( x) ³ g ( x). In the interval [2, 6], g ( x) ³ f ( x).

2 æ ö = ççç ln | x + 1| - x + x ÷÷÷ è 4 2ø

3 0

æ 2 ö + ççç x - x - ln | x + 1| ÷÷÷ è 4 ø 2

The region is composed of two separate regions because y = 1x intersects y = 12 at x = 2.

x -1

ò çèçç 2 - x + 1 ÷÷ø÷ dx 4 3

= ln ( 3 + 1) - 3 + 3 4 2 ì é ù ïü ï + ïí (4 - 2 - ln 5 )- ê 3 - 3 - ln ( 3 + 1) ú ïý ê úû ïï ï 2 ë4 ï î þ 3 3 = ln ( 3 + 1) - + +2 4 2 - ln 5 - 3 + 3 + ln ( 3 + 1) 4 2 = ln ( 3 + 1) + ln ( 3 + 1) - ln 5 + 1 + 3 2 ( 3 + 1)2 1 = ln + + 3 » 2.633 5 2

544 17.

Chapter 7 INTEGRATION x = -1, x = 1, y = e x , y = 3 - e x

18.

The total area between the curves from x = -1 to x = 2 is

To find the point of intersection, set x

e = 3 - e and solve for x. ex = 3 - ex

-1

2e = 3 ex =

x = -1, x = 2, y = e-x , y = e x

(e-x - e x ) dx +

ò (e - e ) dx 0 -1

+ (e x + e-x ) -1

= [(-1 - 1) - (-e - e

3 2 3 x ln e = ln 2 3 x = ln 2 ln e x = ln

-2

+ [(e + e 2

-2

= e +e

19.

-x

= (-e-x - e x ) +

3 2

2 0

)]

) - (1 + 1)]

+ e + e-1 - 4 » 6.611.

x = -1, x = 2, y = 2e2 x , y = e2 x + 1

The area of the region between the curves from x = -1 to x = 1 is

ln 3/2

-1

[(3 - e x ) - e x ] dx

[e x - (3 - e x )] dx

ln 3/2

-1

(3 - 2e x ) dx +

= (3x - 2e x )

(2e x - 3) dx

ln 3/2

ln 3/2 -1

+ (2e x - 3x)

1 ln 3/2

éæ ù ö 3 = ê çç 3 ln - 2eln 3/2÷÷÷ - [3(-1) - 2e-1] ú ê çè ú ø 2 ë û é 1 æ ln 3/2 ö÷ ù 3 + ê 2e - 3(1) - çç 2e - 3 ln ÷÷ ú çè ê 2 ø úû ë éæ ö æ 3 2 öù = ê çç3 ln - 3 ÷÷ - çç -3 - ÷÷ ú ÷ ç ê èç ø è 2 e ÷ø úû ë é æ 3 öù + ê 2e - 3 - çç3 - 3 l n ÷÷÷ ú ê èç 2 ø úû ë 3 2 = 6 ln + + 2e - 6 » 2.605. 2 e

To find the points of intersection of the graphs, substitute for y. 2e 2 x = e2 x + 1 e2 x = 1 2x = 0 x =0

The region is composed of two separate regions because y = 2e2 x intersects y = e2 x + 1 at x = 0. Let f ( x) = 2e2 x , g ( x) = e2 x + 1. In the interval [-1, 0], g ( x) ³ f ( x).

Section 7.5

545

In the interval [0, 2], f ( x) ³ g ( x).

ò2

ò-1 (e2x + 1 - 2e2x )dx ò0 [2e2x - (e2x + 1)] dx

é x( x - 2) ù +ê - ln | x - 1| ú 8 ëê ûú 3

ö÷ æ 1 ö æç e-2 = çç - + 0 ÷÷ - çç - 1÷÷÷ ÷ø ç çè 2 2 è ø÷ æ e4 ö÷ æ 1 ö ç + çç - 2 ÷÷÷ - çç - 0 ÷÷ ÷ø ç çè 2 ÷ø è 2

20.

æ x -1

ò3 ççè 4 - x - 1 ÷÷÷ø dx 3

æ e2 x ö÷ 0 æ e2 x ö÷ çç ç = ç+ x ÷÷÷ + çç - x ÷÷÷ çè 2 ÷ø çè 2 ø÷ 0 -1

é x( x - 2) ù = ê ln | x - 1| ú 8 ëê ûú 2

æ 1 ö çç - x - 1 ÷÷÷ dx + è x -1 4 ø

æ ö æ ö = çç ln 2 - 3 ÷÷÷ - 0 + (1 - ln 3) - çç 3 - ln 2 ÷÷÷ è è8 ø 8ø = 2 ln 2 - ln 3 + 1 » 0.5377 4 y = x3 - x 2 + x + 1, y = 2 x 2 - x + 1

21.

Find the points of intersection.

e-2 + e4 - 2 » 25.37 2

x3 - x 2 + x + 1 = 2 x 2 - x + 1 x 3 - 3x 2 + 2 x = 0 x( x 2 - 3x + 2) = 0

x -1 1 x = 2, x = 4, y = ,y = x -1 4

x( x - 2)( x - 1) = 0

The points of intersection are at x = 0, x = 1, and x = 2 .

To find the points of intersection of the graphs in [2, 4], substitute for y.

Area between the curves is 1

x -1 1 = 4 x -1 ( x - 1)( x - 1) = 4 x2 - 2x + 1 = 4 x2 - 2x - 3 = 0 x = -1or x = 3

The region is composed of two separate regions 1 intersects y = 1 at because y = x4 x -1 x = 3. 1 , g ( x) = 1 . Let f ( x) = x 4 x -1

In the interval [2, 3], g ( x) ³ f ( x). In the interval [3, 4], f ( x) ³ g ( x).

ò0 [(x3 - x2 + x + 1) - (2x2 - x + 1)] dx 2 + ò0 [(2x2 - x + 1) - ( x3 - x2 + x + 1)] dx 1 2 = ( x3 - 3x 2 + 2 x) dx + ò0 ò1 (-x3 + 3x2 - 2x)] dx æ 4 ö = ççç x - x3 + x 2 ÷÷÷ ÷ø èç 4

1 0

æ 4 ö + ççç -x + x3 - x 2 ÷÷÷ ÷ø èç 4

éæ ù ö = ê çç 1 - 1 + 1÷÷÷ - (0) ú ø êë è 4 úû é æ öù + ê (-4 + 8 - 4) - çç - 1 + 1 - 1÷÷÷ ú è 4 ø úû êë = 1 + 1 4 4 1 = . 2

2 1

546

Chapter 7 INTEGRATION The area between the curves is

y = 2 x3 + x 2 + x + 5,

22.

y = x3 + x 2 + 2 x + 5

ò [( x + ln (x + 10)) - (x + ln (x + 10))] dx 3

ò (x - x ) dx 3

æ x4 x5 ö÷÷ ç = çç ÷ 5 ÷÷ø çè 4 0

æ1 1ö 1 . = çç - ÷÷÷ - (0) = çè 4 5ø 20

To find the points of intersection, substitute for y. 3

2x + x + x + 5 = x + x + 2x + 5

y = x5 - 2 ln ( x + 5),

24.

y = x3 - 2 ln ( x + 5)

x -x =0

To find the points of intersection, substitute for y.

x( x 2 - 1) = 0

x5 - 2 ln ( x + 5) = x3 - 2 ln ( x + 5)

The points of intersection are at x = 0, x = -1, and x = 1.

x5 - x3 = 0 x3 ( x 2 - 1) = 0

The area of the region between the curves is

-1

The points of intersection are at x = 0 and x = 1 and x = -1.

[(2 x3 + x 2 + x + 5) - ( x3 + x 2 + 2 x + 5)] dx 1

In the interval [-1, 0],

ò [(x + x + 2x + 5) - (2x + x + x + 5)] dx 3

x5 - 2 ln ( x + 5) > x3 - 2 ln ( x + 5).

ò (x - x) dx +ò (-x + x) dx 3

-1

In the interval [0, 1] ,

2ö 0

æx x ÷÷ ç = çç ÷ 2 ø÷÷ èç 4 4

2ö

æ x ÷÷ ç x + çç ÷ 2 ÷÷ø çè 4 -1 0 4

é éæ 1 ù æ1 1 öù 1ö = ê 0 - çç - ÷÷ ú + ê çç - + ÷÷ - 0 ú ÷ ÷ çè 4 ç ê ú ê ú 2 øû 2ø ë ëè 4 û 1 1 1 = + = . 4 4 2

x5 - 2 ln ( x + 5) < x3 - 2 ln ( x + 5).

The area between the curves is 0

ò [( x - 2 ln ( x + 5)) - (x - 2 ln (x + 5))] dx 5

-1

ò [(x - 2 ln (x + 5)) - (x - 2 ln (x + 5))] dx 3

23.

y = x 4 + ln ( x + 10), =

y = x + ln ( x + 10)

ò-

( x5 - x3 ) dx +

Find the points of intersection.

ò (x - x ) dx 3

4ö 0

x 4 + ln (x + 10) = x3 + ln (x + 10) x 4 - x3 = 0 3

x ( x - 1) = 0 x = 0 or x = 1

The points of intersection are at x = 0 and x = 1.

æ x6 æ x4 x ÷÷ x6 ö÷÷ ç ç = çç + çç ÷÷ ÷ 4 ÷ø 6 ø÷÷ çè 6 çè 4 -1 0 é éæ 1 ù æ1 1 öù 1ö = ê 0 - çç - ÷÷ ú + ê çç - ÷÷ - 0 ú ÷ ÷ çè 6 ç ê ú ê ú 4 øû 6ø ë ëè 4 û 1 1 1 = + = . 12 12 6

Section 7.5

547 The area between the curves is

y = x 4/3, y = 2 x1/3

25.

ò0

Find the points of intersection. x 4/3 = 2 x1/3 x

4/3

1/3

- 2x

( x - x x ) dx =

ò0 (x1/2 - x3/2 ) dx 1

æ x3/2 x5/2 ö÷÷ ç = çç ÷ 5/2 ÷÷ø çè 3/2 0

= 0

1/3

x ( x - 2) = 0 x = 0 or x = 2

æ2 ö 2 = çç x3/2 - x5/2 ÷÷÷ çè 3 ø 0 5

The points of intersection are at x = 0 and x = 2.

é2 2 ù = ê (1) - (1) ú - 0 êë 3 5 úû 4 = . 15

27.

x = 0, x = 3, y = 2e3x , y = e3x + e6

The area between the curves is 2

ò (2x - x ) dx = 2 1/3

4/3

x 4/3 4 3

x 7/3 7 3

0 2

3 4/3 3 x - x7/3 2 7 0

é3 ù 3 = ê (2)4/3 - (2)7/3 ú - 0 êë 2 úû 7 3(24/3 ) 3(27/3 ) 2 7 » 1.62.

26.

y =

To find the points of intersection of the graphs, substitute for y. 2e3x = e3x + e6

x, y = x x

e3 x = e 6

To find the points of intersection, substitute for y.

3x = 6

x = x x x x -

x = 2

x =0

The region is composed of two separate regions

x ( x - 1) = 0

The points of intersection are at x = 0 and x = 1. In [0, 1], x > x x .

because y = 2e3x intersects y = e3x + e6 at x = 2. Let f ( x) = 2e3x , g ( x) = e3x + e6. In the interval [0, 2], g ( x) ³ f ( x). In the interval [2, 3], f ( x) ³ g ( x).

548

Chapter 7 INTEGRATION

(e3x + e6 - 2e3x ) dx +

[2e3x - (e3x + e6 )] dx

29.

æ e3x ö÷ æ e3x ö÷ ç ç = çç + e6 x ÷÷÷ + çç - e6 x ÷÷÷ ÷ø ÷ø çè 3 çè 3 0 2

Graph y1 = e x and y2 = -x 2 - 2 x on your graphing calculator. Use the intersect command to find the two intersection points. The resulting screens are:

æ 6 ö÷ æ 1 ö ç e = çç + 2e6 ÷÷÷ - çç - + 0 ÷÷ ç ø÷ ÷ø è 3 çè 3 æ e9 ö÷ æ e6 ö÷ ç ç + çç - 3e6 ÷÷÷ - çç - 2e6 ÷÷÷ èç 3 ø÷ èç 3 ø÷ e9 + e 6 + 1 3 » 2836

28.

These screens show that e x = -x 2 - 2 x when x » -1.9241 and x » -0.4164.

x = 0, x = 3, y = e x , y = e4- x

In the interval [–1.9241,–0.4164], e x < -x 2 - 2 x. The area between the curves is given by -0.4164

ò [(-x - 2x) - e ] dx. 2

-1.9241

Use the fnInt command to approximate this definite integral. The resulting screen is:

To find the points of intersection of the graphs, substitute for y. e x = e4- x x = 4-x x = 2

The last screen shows that the area is approximately 0.6650.

The region is composed of two separate regions x

because y = e intersects y = e

4- x

at x = 2.

Let f ( x) = e x , g ( x) = e4- x .

30.

In the interval [0, 2], g ( x) ³ f ( x). In the interval [2,3], f ( x) ³ g ( x).

ò0

(e4- x - e x ) dx +

= (-e4- x - e x ) 2

2 0

Graph y1 = ln x and y2 = x3 - 5 x 2 + 6 x - 1 on your graphing calculator. Use the intersect command to find the two intersection points. The resulting screens are:

ò2 (e x - e4- x ) dx + (e x + e4- x )

3 2

= (-e - e ) - (-e - 1) + (e3 + e) - (e2 + e2 ) = e4 + e3 - 4e2 + e + 1 » 48.85

Section 7.5

549

These screens show that ln x = x3 - 5 x 2 + 6 x - 1 when x » 1.4027 and x » 3.4482.

(b) Since 150 - t 2 > t 2 + 11 t , in the interval 4 [0, 8], the net total saving in the first year are

In the interval [1.4027, 3.4482],

1é

The area between the curves is given by 3.4482

ò [ln x - (x - 5x + 6x - 1)] dx. 3

11 ö ù

æ 2

ò êêë (150 - t ) - ççè t + 4 t ÷÷÷ø úúû dt æ ö 11 = ò ççè -2t - 4 t + 150 ÷÷ø÷ dt

ln x > x3 - 5 x 2 + 6 x -1.

3 2 æ ö = ççç -2t - 11t + 150t ÷÷÷ è 3 ø 8

1.4027

1 0

= - 2 - 11 + 150 » $148. 3 8

Use the fnInt command to approximate this definite integral. The resulting screen is:

æ 2

11 ö ù

ò ëêê (150 - t ) - çèç t + 4 t ÷÷ø÷ ûúú dt 2

The last screen shows that the area is approximately 3.3829. 31.

2t 2 +

11 t 4

-2(8 ) 11(8 ) + 150(8) 3 8 = -1024 - 704 + 1200 » $771. 3 8

32.

11 t - 150 = 0 4

(a) It is profitable to use the machine until S ¢(t ) = C ¢(t ). 150 - t 2 = t 2 +

3 2 æ ö = ççç -2t - 11t + 150t ÷÷÷ ÷ø çè 3 8

(a)

S ¢(t ) = -t 2 + 4t + 8, C ¢(t ) =

3 2 t 25

S ¢(t ) = C ¢(t )

8t 2 + 11t - 600 = 0

-t 2 + 4t + 8 =

-11  121 - 4(8)(-600) 16 -11  139 = 16 t = 8 or t = -9.375

t =

It will be profitable to use this machine for 8 years. Reject the negative solution.

3 2 t 25

-25t 2 + 100t + 200 = 3t 2

0 = 28t 2 - 100t - 200 0 = 7t 2 - 25t - 50 0 = (7t + 10)(t - 5) t =-

10 or t = 5 7

Since time would not be negative, 5 is the only solution. It will pay to use the device for 5 yr. (b) The total savings over 5 yr is given by 5

ò (-t + 4t + 8) dt 2

æ -t 3 ö÷ ç = çç + 2t 2 + 8t ÷÷÷ çè 3 ø÷ =

-125 + 90 = 48.33. 3

550

Chapter 7 INTEGRATION The total cost over 5 yr is given by

0.7et /2 > 104

t3 3 2 t dt = 25 0 25 5

0.3et /2 > 104 - 0.4et /2

= 5.

et /2 > 104 0.7 æ ö ln et /2 > ln çç 104 ÷÷÷ è 0.7 ø æ ö t > 2 ln çç 104 ÷÷÷ è 0.7 ø

Net savings = $48.33 million - $5 million = $43.33 million

33.

(a)

E ¢( x) = e0.1x and I ¢( x) = 98.8 - e0.1x

t > 10

To find the point of intersection, where profit will be maximized, set the functions equal to each other and solve for x. e

0.1x

= 98.8 - e

0.1x

= 98.8

It will no longer be profitable to use process after 10 yr. (b) The total net saving is

0.1x

ò [(104 - 0.4e ) - 0.3e ] dt t /2

e0.1x = 49.4

0.1x

= (104t - 1.4et /2 ) = [(104t - 1.4e

æ e0.1x ö÷÷ ç = çç 98.8 x ÷ 0.1 ø÷÷ çè 0

(

= 98.8 x - 10e0.1x

35. 39

3.9

] - (0 - 10)

ò e 0

e0.1x dx = 0.1 = 10e = 10e

34.

(a)

ò [ p - S (q)] dq, 0

where p0 is the equilibrium price and q0 is equilibrium supply. p0 = S (16) = (16)5/2 + 2(16)3/2 + 50

Therefore, the producers’ surplus is 16

0 39 0

- 10

ò0 [1202 - (q5/2 + 2q3/2 + 50)] dq 16 = ò0 (1152 - q5/2 - 2q3/2 ) dq 16

= $484.02.

(d)

) - (0 - 1.4)]

= 1202

0.1x

3.9

S (q) = q5/2 + 2q3/2 + 50; q = 16 is the equilibrium quantity.

Producers surplus =

= $3369.18.

0.1x

» 834. The net total saving will be $834,000.

t /2

= 1041.4 - 1.4e

) dx

= [98.8(39) - 10e

t /2

(b) The total income for 39 days is

ò (98.8 - e

ò (104 - 0.7e ) dt

æ 0.7et /2 ö÷÷ ç = çç104t ÷ çè 1/2 ø÷÷ 0

ln 49.4 » 39 0.1

The optimum number of days for the job to last is 39.

0.1x = ln 49.4 x =

t /2

Profit = Income - Expense = 3369.18 - 484.02 = $2885.16 R¢(t ) = 104 - 0.4et /2 ; C ¢(t ) = 0.3et /2

It will no longer be profitable when C ¢(t ) > R¢(t ). Find t when C ¢(t ) > R¢(t ).

æ ö = çç1152q - 2 q 7/2 - 4 q5/2 ÷÷÷ è ø 0 7 5

= 1152(16) - 2 (16)7/2 - 4 (16)5/2 7 5 32, 768 4096 = 18, 432 7 5 = 12,931.66. The producers’ surplus is $12,931.66.

Section 7.5 36.

551

S (q) = 100 + 3q3/2 + q5/2 ; equilibrium quantity is q = 9. Producers’ surplus =

ò [424 - (100 + 3q

3/ 2

ò (324 - 3q

3/2

-6

200 200 + -6 30 3 = $54

+ q 5/2 )] dq

- q5/2 )] dq

38.

D (q ) =

32, 000 (2q + 8)3

; q = 6 is the equilibrium

æ ö 6 2 = çç 324q - q5/2 - q 7/2 ÷÷÷ çè ø 0 5 7

quantity.

éæ ù ö 6 2 = ê çç 324(9) - (9)5/2 - (9)7/2 ÷÷÷ - 0 ú ê çè ú ø 5 7 ë û 1458 4374 = 2916 5 7 = 1999.54

Consumers’ surplus =

D (q ) =

200 (3q + 1)

quantity.

ò | D(q) - p | dq 0

ù ê ú dq 2 ê ú 2 0 ëê (3q + 1) ûú

200

6é

ò (2q + 8)

du = 2 dq

Let u = 3q + 1, so that

ò (2q + 8) 0

1 2

ò (3q + 1) 0

= =

1 3

200 3

du -

ò 4 dq. 0

ò 2 dq 0

ò 4 dq 0

du -

8000

ò 4 dq 0

u-3du -

u-2 -2

ò 4 dq 0 6

-4q 8

- 24 u2 8 8000 8000 =+ - 24 400 64 = $81

ò 2 dq

u-2du -

dq -

1 du = dq. 2

2 dq

32, 000

= 16, 000 ⋅

dq 2

200

= 16, 000

1 du = 3 dq and du = dq. 3 200

dq 3

and

32, 000

ò (3q + 1) dq - ò 2 dq.

32, 000

200

203

Therefore, the consumers’ surplus is

32, 000

p0 = D (6) =

p0 = D(3) = 2

Let u = 2q + 8, so that

3é

ò | D(q) - p | dq 0

; q = 3 is the equilibrium 2

Consumers’ surplus =

ù ê 32, 000 - 4 ú dq ê ú 3 0 êë (2q + 8) úû

The producers’ surplus is $1999.54. 37.

200 3u 1

p0 = S (9) = 424 9

-2q 1

ò [ p - S (q)] dq 0

200 u-1 = ⋅ -1 3

= 4

552 39.

Chapter 7 INTEGRATION 15

S (q) = q 2 + 10q D (q) = 900 - 20q - q

ò [375 - (q + 10q)] dq

(a) The graphs of the supply and demand functions are parabolas with vertices at (-5, -25) and (-10, 1900), respectively.

ò (375 - q - 10q) dq 2

æ ö 1 = çç 375q - q3 - 5q 2 ÷÷÷ çè ø 0 3

é ù 1 = ê 375(15) - (15)3 - 5(15) 2 ú - 0 êë úû 3 = 3375

The producer’s surplus is $3375. (b) The graphs intersect at the point where the y-coordinates are equal. q 2 + 10q = 900 - 20q - q 2 2q 2 + 30q - 900 = 0 q 2 + 15q - 450 = 0 (q + 30)(q - 15) = 0 q = -30 or q = 15

40.

S (q) = (q + 1)2 1000 D (q ) = q +1

(a) The graph of the supply function is a parabola with vertex at (–1, 0) The graph of the demand function is the graph of a rational function with vertical asymptote of x = -1 and horizontal asymptote of y = 0.

Disregard the negative solution. The supply and demand functions are in equilibrium when q = 15. S (15) = 152 + 10 (15) = 375

The point is (15,375).

(b) Find the equilibrium point by setting the two functions equal. (q + 1)2 =

ò [D (q) - p )] dq

(q + 1)3 = 1000

q3 + 3q 2 + 3q + 1 = 1000

p0 = D (15) = 375 15

ò [(900 - 20q - q ) - 375] dq 2

ò (525 - 20q - q ) dq 0

æ 1 ö = çç 525q - 10q 2 - q3 ÷÷÷ çè 3 ø 0

The consumer’s surplus is $4500.

ò [ p - S (q)] dq 0

p0 = S (15) = 375

(q - 9)(q 2 + 12q + 111) = 0

S (9) = (9 + 1)2 = 100.

é ù 1 = ê 525(15) - 10(15)2 - (15)3 ú - 0 = 4500 êë úû 3

q3 + 3q 2 + 3q - 999 = 0

Since q 2 + 12q + 111 has no real roots, q = 9 is the only root. At the equilibrium point where the supply and demand are both 9 items, the price is

(d) Find the producers’ surplus.

1000 q +1

The equilibrium point is (9, 100). (c) The consumers’ surplus is given by 9æ ö çç 1000 - 100 ÷÷ dq ÷÷ ø 0 çè q + 1

= (1000 ln | q + 1| - 100q) 0 = 1000 ln (9 + 1) - 100(9) - 0 » 1402.59 Here the consumers’ surplus is $1402.59.

Section 7.5

553

(d) The producers’ surplus is given by

(d) For the equilibrium price, the total consumers’ and producers’ surplus is

ò0 [100 - (q + 1)2 ] dq 9 = ò0 (99 - q2 - 2q) dq

4500 + 3375 = $7875

For the government price, the total consumers’ and producers’ surplus is 5616 + 1872 = $7488.

æ ö = çç 99q - 1 q3 - q 2 ÷÷÷ è ø0 3 1 3 = 99(9) - (9) - (9)2 - 0 3 = 567 Here the producers’ surplus is $567.

41.

(a)

The difference is 7875 - 7488 = $387. 42.

(a)

S (q) = q 2 + 10q; S (q) = 264 is the price the government set.

0 = (q - 12)(q + 22) q = 12 or q = -22

» 290.0 The area under the curve approximately $290 thousand. The total income, from 2000 to 2018, of a family with an income in the lowest fifth of the U.S. population was about $290,000.

Only 12 is a meaningful solution here. Thus, 12 units of oil will be produced. (b) The consumers’ surplus is given by

ò (900 - 20q - q - 264) dq = ò (636 - 20q - q ) dq 2

æ ö = çç 636q - 10q 2 - 1 q3 ÷÷÷ è 3 ø 0

= 636(12) - 10(12) - 1 (12)3 - 0 3 = 5616 2

Here the consumer’ surplus is $5616. In this case, the consumers’ surplus is 5616 – 4500 = $1116 larger. (c) The producers’ surplus is given by

ò [264 - (q + 10q)] dq 2

ò (264 - q - 10q) dq 2

æ ö 1 = çç 264q - q3 - 5q 2 ÷÷÷ çè ø 0 3 = 264(12) -

f (t ) dt

æ 0.00354 4 0.0762 3 ö÷ çç t t ÷÷ çç ÷÷ 4 3 = çç ÷÷ 0.608 2 çç ÷÷ + + 13.71 t t ÷ø çè 2 0

0 = q 2 + 10q - 264

264 = q 2 + 10q

f (t ) = 0.00354t 3 - 0.0762t 2 + 0.0608t + 13.71 The area under the graph of f is

1 (12)3 - 5(12)2 - 0 3

= 1872

Here the producers’ surplus is $1872. In this case, the producers’ surplus is 3375 – 1872 = $1503 smaller.

(b)

g (t ) = 0.0465t 3 - 0.888t 2 + 11.3t + 278 The area under the graph of g is

g (t ) dt

0 18

æ 0.04654 4 0.888 3 ÷ö çç t t ÷÷ çç ÷÷ 4 3 = çç ÷÷ 11.3 2 çç ÷÷ + + 278 t t çè ÷ø 2 0

» 6329.7 » 6330 The area under the curve approximately $6330 thousand. The total income, from 2000 to 2018, of a family with an income in the top 5% of the U.S. population was about $6,330,000.

(c) A graph of f and g on the same coordinate system shows that g lies above f on the entire interval 0 £ t £ 18, so the area between the curves is 6330 - 290 » 6040, or $6040 thousand.

The difference in the total income of the top 5% and the bottom fifth was approximately $6,040,000.

554 43.

Chapter 7 INTEGRATION (a) The pollution level in the lake is changing at the rate f (t ) - g (t ) at any time t. We find the amount of pollution by integrating. 12

ò [ f (t) - g (t)] dt 0

ò [10(1 - e- ) - 0.4t] dt

A suitable window for the graph is [0, 50] by [0, 110].

0.5t

æ 1 -0.5t 1 ö = çç10t - 10 ⋅ - 0.4 ⋅ t 2 ÷÷÷ e çè -0.5 2 ø 0 12

= (20e-0.5t + 10t - 0.2t 2 ) 0

= [20e-0.5(12) + 10(12) - 0.2(12)2 ] - [20e-0.5(0) + 10(0) - 0.2(0)2 ]

Use the calculator’s features to approximate where the graph intersects the x-axis. These are at 0 and about 47.91. Therefore, the pollution will be removed from the lake after about 47.91 hours.

= (20e-6 + 91.2) - (20) = 20e-6 + 71.2 » 71.25

After 12 hours, there are about 71.25 gallons. (b) The graphs of the functions intersect at about 25.00. So the rate that pollution enters the lake equals the rate the pollution is removed at about 25 hours. (c)

ò [ f (t) - g(t)] dt 2

+ 10t - 0.2t )

25 0

= [20e-0.5(25) + 10(25) - 0.2(25) 2 )] - 20 = 20e-12.5 + 105 » 105

After 25 hours, there are about 105 gallons. (d) For t > 25, g (t ) > f (t ), and pollution is being removed at the rate g (t ) - f (t ). So, we want to solve for c, where

[ f (t ) - g (t )] dt = 0.

Alternatively, we could solve for c in c

ò [g(t) - f (t)] dt = 105. 25

One way to do this with a graphing calculator is to graph the function y =

= (20e

(a) The pollution level in the lake is changing at the rate f (t ) - g (t ) at any time t. We find the amount of pollution by integrating.

ò [ f (t) - g(t)] dt

-0.5t

44.

ò [15(1 - e

-0.05t

) - 0.3t ] dt

æ 1 1 ö = çç15t - 15 e-0.05t - 0.3 t 2 ÷÷÷ çè -0.05 2 ø 0 = (300e-0.05t + 15t - 0.15t 2 ) -0.05(12)

= [300e

12 0

+ 15(12) - 0.15(12) 2 ]

- [300e-0.05(0) + 15(0) - 0.15(0) 2 ] = (300e-0.6 + 158.4) - (300) = 300e-0.6 - 141.6 » 23.04

After 12 hours, there are about 23.04 gallons. (b) The graphs of the functions intersect at about 44.63. So the rate that pollution enters the lake equals the rate the pollution is removed at about 44.63 hours.

ò [ f (t) - g(t)] dt 0

Section 7.5 (c)

555

44.63

45.

[ f (t ) - g (t )] dt

0 -0.05t

= (300e

+ 15t - 0.15t )

I ( x) = 0.9 x 2 + 0.1x

I (0.1) = 0.9(0.1) 2 + 0.1(0.1)

(a)

44.63

= 0.019

The lower 10% of income producers earn 1.9% of total income of the population.

= [300e-0.05(44.63) + 15(44.63) - 0.15(44.63)2 ] - 300

(b)

= (300e-2.2315 + 370.674465) - 300 = 300e-2.2315 + 70.674465 » 102.88

After 44.63 hours, there are about 102.88 gallons.

I (0.4) = 0.9(0.4)2 + 0.1(0.4) = 0.184 The lower 40% of income producers earn 18.4% of total income of the population.

(d) For t > 44.63, g (t ) > f (t ), and pollution is being removed at the rate g (t ) - f (t ). So we want to solve for c, where

ò [x - (0.9x + 0.1x)] dx = 2 ò (0.9x - 0.9x ) dx 1

ò [ f (t) - g(t)] dt = 0

2 3ö æ = 2 ççç 0.9 x - 0.9 x ÷÷÷ è 2 3 ø

(Alternatively, we could solve for c in

[ g (t ) - f (t )] dt = 102.88.)

46.

y =

1 0

æ ö = 2 çç 0.9 - 0.9 ÷÷÷ = 0.30. è 2 3 ø

44.63

One way to do this with a graphing calculator is to graph the function

ò [ f (t) - g(t)] dt 0

and determine the values of x for which y = 0.

331 2 91 x x 240 240 331 91 (0.6)2 (0.6) (a) I (0.6) = 240 240 = 0.269 The lower 60% of the Hispanic income producers earn about 26.9% of total income.

I ( x) =

(b) The Gini index of inequality is 1é æ öù ê x - çç 331 x 2 - 91 x ÷÷÷ ú dx 2 è 240 240 ø ûú 0 ëê

The first window shows how the function can be defined.

= 2

æ 331

331

2ö

ò ççè 240 x - 240 x ÷÷÷ø dx 0

æ ö = 2 çç 1 ⋅ 331 x 2 - 1 ⋅ 331 x3 ÷÷÷ è 2 240 3 240 ø 0 æ ö = 2 çç 331 - 331 ÷÷÷ = 0.46. è 480 720 ø

A suitable window for the graph is [0.75] by [0, 110]. 47.

y =

x, y =

Use the calculator’s features to approximate where the graph intersects the x-axis. These are at 0 and about 73.47. Therefore, the pollution will be removed from the lake after about 73.47 hours. Copyright © 2022 Pearson Education, Inc.

x 2

556

Chapter 7 INTEGRATION To find the points of intersection, substitute for y.

Use Simpson’s rule with n = 4 to approximate

x 2

x =

x - x =0 2 x-2 x =0 x ( x - 2) = 0 x = 0 or x = 4 Area =

4æ

Your Turn 2

xö

ò çççè x - 2 ÷÷÷ø dx 4æ

i 0 1 2 3 4

xö

ò çççè x - 2 ÷÷÷ø dx 1/2

æ2 1 ö = çç x3/2 - x 2 ÷÷÷ çè 3 4 ø 0 =

16 4 -4= 3 3

Your Turn 1

x 2 + 3 dx

1 [ 2 + 4(2.29129) + 2(2.64757) + 4(3.04138) + 3.46410 ] 6 » 5.3477.

x 2 + 3 dx.

7.6 Exercises 2

Here f ( x) = x + 3, a = 1, b = 3, and n = 4. The subintervals have length (3 - 1)/4 = 1/2. The following table summarizes the information required. i 0 1 2 3 4

xi 1 3/2 2 5/2 3

f ( xi ) f (1) = 2 f (3/2) » 2.29129 f (2) » 2.64575 f (5/2) » 3.04138 f (3) » 3.46410

False. Simpson’s rule approximates the definite integral with portions of parabolas, rather than line segments of the trapezoidal rule.

True

False. Although the trapezoidal rule is usually less accurate than Simpson’s rule, there are cases for which the trapezoidal rule gives exceptional accuracy.

False. Simpson’s rule does not require an antiderivative of the integrand.

The trapezoidal rule gives

f ( xi ) f (1) = 2 f (3/2) » 2.29129 f (2) » 2.64575 f (5/2) » 3.04138 f (3) » 3.46410

Use the trapezoidal rule with n = 4 to approximate

xi 1 3/2 2 5/2 3

For Simpson’s rule, the factor in front is (b - a)/3n = (3 - 1)/12 = 1/6. Simpson’s rule thus gives

7.6 Numerical Integration

x 2 + 3 dx.

Here f ( x) = x 2 + 3, a = 1, b = 3, and n = 4. The subintervals have length (3 - 1)/4 = 1/2. The following table summarizes the information required; it is the same as the table used in Your Turn 1.

ò (3x + 2) dx 2

n = 4, b = 2, a = 0, f ( x) = 3x 2 + 2

x 2 + 3 dx

3 -1é1 ê (2) + 2.29129 + 2.64757 » 2 ëê 2 ù 1 + 3.04138 + (3.46410) ú » 5.3552. úû 2

0 1 2 3 4

xi 0 1 2 1 3 2 2

f ( xi ) 2

2.75 5 8.75 14

Section 7.6

557

(a) Trapezoidal rule: 2

(b)Simpson’s rule: 2

ò (2 x + 1) dx

(3x 2 + 2) dx

ù 2 - 0é1 1 ê (2) + 2.75 + 5 + 8.75 + (14) ú úû 4 êë 2 2 = 0.5 (24.5)

2-0 [1 + 4(1.5) + 2(3) + 4(5.5) + 9] 3(4) 2 = (44) » 7.333 12

= 12.25

(b) Simpson’s rule: 2

æ 2 x3 ö÷ ç (2 x + 1) dx = çç + x ÷÷÷ çè 3 0 ø÷ 0

ò (3x + 2) dx

2-0 [2 + 4(2.75) + 2(5) + 4(8.75) + 14] 3(4) 2 = (72) 12 = 12

æ 16 ö = çç + 2 ÷÷÷ - 0 çè 3 ø =

ò (3x + 2) dx = (x + 2x) 2

3 dx x 5 -1

n = 4, b = 3, a = -1, f ( x) =

= (8 + 4) - 0 = 12

0 1 2 3 4

ò (2x + 1) dx 2

n = 4, b = 2, a = 0, f ( x) = 2 x 2 + 1 i

xi 0 1 2 1 3 2 2

0 1 2 3 4

f ( xi ) 1

f ( xi ) 0.5 0.6 0.75 1 1.5

(a) Trapezoidal rule: 3

-1

ù 3 - (-1) é 1 1 ê (0.5) + 0.6 + 0.75 + 1 + (1.5) ú êë 2 úû 4 2 = 1(3.35) = 3.35 »

1.5 3 5.5

(b) Simpson’s rule:

ò 5 - x dx -1

3 - (-1) [0.5 + 4(0.6) + 2(0.75) + 4(1) + 1.5] 3(4) 1 æ 99 ö = çç ÷÷÷ 3 çè 10 ø »

(2 x 2 + 1) dx

ù 2-0é1 1 ê (1) + 1.5 + 3 + 5.5 + (9) ú úû 4 êë 2 2 = 0.5(15) = 7.5

xi -1 0 1 2 3

3 5-x

ò 5 - x dx

(a) Trapezoidal rule: 2

22 » 7.333 3

33 » 3.3 10

558

Chapter 7 INTEGRATION (c) Exact value:

3 dx = -3 ln | 5 - x | -1 5 - x

ò (2x + 1) dx

-1

n = 4, b = 2, a = -1, f ( x) = 2 x3 + 1

-1

= -3(ln | 2| -ln | 6 |) = 3 ln 3 » 3.296

xi –1 1 4 1 2

ò 2x + 1 dx 1

6 n = 4, b = 5, a = 1, f ( x) = 2 x+ 1

xi 1

(a) Trapezoidal rule: 2

ò- (2x + 1) dx »

2 - (-1) é 1 31 5 ê (-1) + + êë 2 4 32 4 +

» 11.34

» 3.997

(b) Simpson’s rule: 5

6 dx 2 x +1 1 æ6ö æ6ö æ 2 ö æ 6 öù 5 - 1 êé 2 + 4 çç ÷÷ + 2 çç ÷÷÷ + 4 çç ÷÷ + çç ÷÷ ú » ÷ ç ç ç ê è5ø è7ø è 3 ÷ø çè 11 ÷ø úû 3(4) ë 1æ 24 12 8 6ö = çç 2 + + + + ÷÷÷ ç 3è 5 7 3 11 ø

(b) Simpson’s rule: 2

ò (2 x + 1) dx 3

-1

ù æ 31 ö æ5ö æ 157 ÷ö 2 - (-1) éê -1 + 4 çç ÷÷ + 2 çç ÷÷ + 4 çç ÷÷ + 17 ú ÷ ÷ ç ç ç ê ú è ø è ø è ø 3(4) ë 32 4 32 û

1 (42) = 10.5 4

5 1

= 3(ln |11| - ln |3|) 11 = 3 ln » 3.898 3

æ x4 ö÷ ç = çç + x ÷÷÷ çè 2 ÷ø -1 æ1 ö = (8 + 2) - çç - 1÷÷÷ çè 2 ø

21 2 = 10.5 =

ò 2x + 1 dx = 3 ln | 2x + 1| 1

ò- (2x + 1) dx

» 3.909

ù 157 1 + (17) ú úû 32 2

= 0.75(15.125)

6 dx 2 x +1 1 5 - 1 éê 1 6 6 2 1 æ 6 öù » (2) + + + + çç ÷÷÷ ú ê 4 ë2 5 7 3 2 çè 11 ø úû æ 6 6 2 3ö = 1çç1 + + + + ÷÷÷ çè 5 7 3 11 ø

2 3 6 11

(a) Trapezoidal rule:

5 4 2

f ( xi ) 2 6 5 6 7

f ( x) –1 31 32 5 4 157 32 17

Section 7.6 10.

559

(2 x3 + 1) dx

11.

n = 4, b = 3, a = 0, f ( x) = 2 x3 + 1 i

xi 0 3 4 3 2

0 1 2

n = 4, b = 5, a = 1, f ( x) = i

f ( x) 1 59 32 31 4 761 32 55

9 4 3

ò x dx xi 1 2 3 4 5

0 1 2 3 4

5 -1é1 ê (1) + 0.25 + 0.1111 4 êë 2 ù 1 + 0.0625 + (0.04) ú úû 2

ò0 (2x3 + 1) dx ù 3-0é1 59 31 761 1 ê (1) + + + + (55) ú úû 4 êë 2 32 4 32 2

» 0.9436

(b) Simpson’s rule:

3 æç 491 ö÷ ÷ ç 4 çè 8 ÷ø » 46.03

5-1 [1 + 4(0.25) + 2(0.1111) 12 + 4(0.0625) + 0.04)] » 0.8374 »

ò0 (2x3 + 1) dx æ 59 ö æ 31 ö 3 - 0 éê 1 + 4 çç ÷÷ + 2 çç ÷÷ ÷ ç ê è ø èç 4 ø÷ 3(4) ë 32

ò1

ù æ 761 ö÷ ú + 4 çç + 55 ÷ ú èç 32 ø÷ û

x-2dx = -x -1 =-

1 (174) 4 = 43.5 =

æ x4 ö÷ ç (2 x3 + 1) dx = çç + x ÷÷÷ çè 2 0 ø÷ 0

ò1 x2 dx

(b) Simpson’s rule:

f ( xi ) 1 0.25 0.1111 0.0625 0.04

ò x dx

(a) Trapezoidal rule:

æ 81 ö = çç + 3 ÷÷÷ - 0 çè 2 ø

87 2 = 43.5 =

12.

5 1

1 +1 5

4 = 0.8 5

ò2 x3 dx n = 4, b = 4, a = 2, f ( x) = i

0 1 2 3 4

xi 2 2.5 3 3.5 5

f ( xi ) 0.125 0.064 0.03703 0.02332 0.015625

1 x3

560

Chapter 7 INTEGRATION (a) Trapezoidal rule: 4

4 - 2 é1 ê (0.125) + 0.064 + 0.03703 4 ëê 2 ù 1 + 0.02332 + (0.015625) ú úû 2

ò2 x

(b) Simpson’s rule:

» 3

ò0 4xe-x dx + 4(3e-9/16 ) + 4e-1]

1 (4e-1/16 + 4e-1/4 + 12e-9/16 + 4e-1) 12 » 1.265

(b) Simpson’s rule: 4

4-2

ò x » 3(4) [0.125 + 4(0.064) + 2(0.03703) 2

1- 0 [0 + 4(e-1/16 ) + 2(2e-1/4 ) 3(4)

1 (0.19466) » 0.0973 2

+ 4(0.02332) + 0.015625]

1 (0.056397) 6 » 0.0940 »

4 xe-x dx = - 2e-x

ò x ò = 3

x-3dx =

= (-2e-1) - (-2) = 2 - 2e-1 » 1.264 4

x-2 -1 = -2 2 2x2 2

-1 1 3 + = = 0.09375 32 8 32

13.

14.

ò x 2x + 1 dx

ò 4xe

n = 4, b = 4, a = 0, f ( x) = x 2 x 2 + 1

-x 2

n = 4, b = 1, a = 0, f ( x) = 4 xe-x i

xi 0 1 4 1 2 3 4 1

0 1 2 3 4

f ( xi ) 0

f ( xi ) 0 3 6

3 19

4 33

ò x 2x + 1 dx

2e-1/ 4

3e-9 /16

4 - 0é1 ê (0) + 4 êë 2

3 + 6 + 3 19 +

ù 1 (4 33) ú úû 2

= 1( 3 + 6 + 3 19 + 2 33) » 32.30

4e-1

(b) Simpson’s rule:

ò0 4xe-x dx 2

1- 0é1 ê (0) + e-1/16 + 2e-1/4 4 êë 2 ù 1 + 3e-9/16 + (4e-1) ú úû 2

1 -1/16 + 2e-1/4 + 3e-9/16 + 2e-1) (e 4 » 1.236

0 1 2

xi 0 1 2

(a) Trapezoidal rule:

e-1/16

(a) Trapezoidal rule:

1 0

ò x 2x + 1 dx 2

4-0 [0 + 4( 3) + 2(6) 3(4)

+ 4(3 19) + 4 33] 1 = (4 3 + 12 + 12 19 + 4 33) 3 » 31.40

Section 7.6

561

Exact value:

(c)

(2 x 2 + 1)3/2 x 2 x + 1 dx = 6 0

y =

15.

1 2 1  r =  (2)2 2 2 » 6.283

Area of semicircle =

Simpson’s rule is more accurate.

333/2 - 1 » 31.43 6

16.

4 x 2 + 9 y 2 = 36 36 - 4 x 2 9 1 y = 36 - 4 x 2 3

y2 =

4 - x2

An equation of the semiellipse is y =

n = 8, b = 2, a = -2, f ( x) = i

0 1 2 3 4 5 6 7 8

xi -2.0 -1.5 -1.0 -0.5 0 0.5 1.0 1.5 2.0

4 - x2

0 1.32289 1.73205 1.93649 2 1.93649 1.73205 1.32289 0

n = 12, b = -3, a = 3 i

(a) Trapezoidal rule:

-2

4 - x 2 dx

2 - (-2) 8 é1 1 ù ⋅ ê (0) + 1.32289 + 1.73205 + ⋅ ⋅ ⋅ + (0) ú êë 2 2 úû » 5.991 »

Simpson’s rule:

(b)

-2

1 36 - 4 x 2 . 3

0 1 2 3 4 5 6 7 8 9 10 11 12

xi -3 -2.5 -2 -1.5 -1 -0.5

0 0.5 1 1.5 2 2.5 3

0 1.1055 1.4907 1.7321 1.8856 1.9720 2 1.9720 1.8856 1.7321 1.4907 1.1055 0

(a) Trapezoidal rule:

4 - x dx

2 - (-2) 3(8) ⋅ [0 + 4(1.32289) + 2(1.73205) + 4(1.93649) + 2(2)

+ 4(1.93649) + 2(1.73205) + 4(1.32289) + 0] » 6.167

A 6 é1 ê (0) + 1.1055 + 1.4907 + 1.7321 = 2 12 êë 2 + 1.8856 + 1.972 + 2 + 1.972 + 1.8856 ù + 1.7321 + 1.4907 + 1.1055 + 2(0) ú ú û » 9.186

562

Chapter 7 INTEGRATION (b) Simpson’s rule: A 6 [0 + 4(1.1055) + 2(1.4907) = 2 3(12) + 4(1.7321) + 2(1.8856) + 4(1.972)

19.

æ1ö x 4dx = çç ÷÷÷ x5 çè 5 ø 0 0

(a)

1 5 = 0.2 =

+ 2(2) + 4(1.972) + 2(1.8856) + 4(1.7321) + 2(1.4907) + 4(1.1055) + 0] 1

(c) The trapezoidal rule gives the area of the region as 9.1859. Simpson’s rule gives the area of the region as 9.3304. The actual area is 3 » 9.4248. Simpson’s rule is a better approximation. 17.

1- 0é1

T >

n = 8, b = 1, a = 0, f ( x) = x 4 1

ò x dx » 4

1- 0é1 1 1 81 ê (0) + + + 8 ëê 2 4096 256 4096

1 625 81 2401 1 ù + + + + (1) ú 16 4096 256 4096 2 úû 1 æ 6724 ö÷ = çç ÷ 8 çè 4096 ÷ø +

òa f (x) dx.

» 0.205200 n = 16, b = 1, a = 0, f ( x) = x 4

òa f (x) dx

By looking at the graph of y = x and dividing the area between 0 and 3 into an even number of trapezoids, you can see that each trapezoid has an area greater than the actual area [case (b)]. f ( x) =

1- 0é1

81 1 625 + + 65, 536 256 65, 536 81 2401 1 + + + 4096 65, 536 16 6561 625 14, 641 + + + 65, 536 4096 65, 536 81 28, 561 2401 + + + 256 65, 536 4096 50, 625 1 ù + + (1) ú 65, 536 2 úû +

ò f (x) dx a

(c) You can’t say which is larger because some trapezoids are greater than the given area and some are less than the given area [case (c)].

By looking at the graph of y = x and dividing area between 0 and 9 into an even number of trapezoids, you can see that each trapezoid has an area less than the actual area [case (a)].

ò x dx » 16 êêë 2 (0) + 65, 536 + 4096

x ; [0, 9]

T <

1 æç 226 ö÷ ÷ ç 4 çè 256 ÷ø » 0.220703

(b)

f ( x) = x 2 ; [0, 3] T >

The correct choice is (b). (a)

ò x dx » 4 êêë 2 (0) + 256 + 16 + 256 + 2 (1) úúû

Since f ( x) > 0 and f ¢¢( x) > 0 for all x between a and b, we know the graph of f ( x) on the interval from a to b is concave upward. Thus, the trapezoid that approximates the area will have an area greater than the actual area Thus,

18.

n = 4, b = 1, a = 0, f ( x) = x 4

(b)

1 (55.982) 6 » 9.330

1 æç 211, 080 ö÷ ÷ ç 16 çè 65, 536 ÷ø÷

» 0.201302

Section 7.6

563 (d) As n changes from 4 to 8, for example, the error changes from 0.020703 to 0.005200. 0.020703a = 0.005200 1 a » 4 Similar results would be obtained using other values for n.

n = 32, b = 1, a = 0, f ( x) = x 4 1

ò x dx 4

é 1 1 » 1 - 0 ê 1 (0) + + ê 32 ë 2 1, 048,576 65,536 81 1 625 + + + 1, 048,576 4096 1, 048,576 81 2401 + + + 1 + 6561 65,536 1, 048,576 256 1, 048,576 14, 641 28,561 + 625 + + 81 + 65,536 1, 048,576 4096 1, 048,576 50, 625 83,521 + 2401 + + 1 + 65,536 1, 048,576 16 1, 048,576 130,321 194, 481 + 6561 + + 625 + 65,536 1, 048,576 4096 1, 048,576 14, 641 279,841 390, 625 + + + 81 + 65,536 1, 048,576 256 1, 048,576 28,561 531, 441 707, 281 + + + 2401 + 65,536 1, 048,576 4096 1, 048,576 ù 50, 625 923,521 + + + 1 (1) ú 65,536 1, 048,576 2 úû æ 6, 721,808 ÷ö ÷ » 0.200325 » 1 çç 32 çè 1, 048,576 ÷÷ø

To find error for each value of n, subtract as indicated. n = 4: (0.220703 - 0.2) = 0.020703 n = 8: (0.205200 - 0.2) = 0.005200 n = 16: (0.201302 - 0.2) = 0.001302 n = 32: (0.200325 - 0.2) = 0.000325

The error is multiplied by 14 . 20.

1 51 x 5 0 1 = 5 = 0.2

(a)

ò0

(b)

n = 4, b = 1, a = 0, f ( x) = x 4

x 4dx =

x 4dx »

ù æ 81 ö÷ ú 1 + 4 çç + ÷ ú èç 256 ø÷ û æ ö 1 ç 77 ÷ = ç ÷ 12 çè 32 ÷ø » 0.2005208 n = 8, b = 1, a = 0, f ( x) = x 4

x 4 dx »

ù æ 18 ö÷ æ 2401 ÷ö + 2 çç + 4 çç + 1ú çè 256 ÷÷ø çè 4096 ÷÷ø ú û

41(0.020703) = 4(0.020703)

1 æç 4916 ö÷ ÷ ç 24 çè 1024 ø÷ » 0.2000326

= 0.082812

81(0.005200) = 8(0.005200) = 0.0416

æ 1 ö÷ æ 1 ö÷ 1 - 0 éê 0 + 4 çç + 2 çç ÷ ÷ ÷ èç 4096 ø èç 256 ÷ø 3(8) êë æ 81 ö÷ æ 1 ö æ 625 ö÷ + 4 çç + 2 çç ÷÷ + 4 çç çè 4096 ÷÷ø çè 16 ÷ø çè 4096 ÷÷ø

Since these are not the same, try p = 2.

æ 1 ÷ö æ 1 ö 1 - 0 êé 0 + 4 çç ÷ + 2 ççç ÷÷÷ èç 256 ÷ø è 16 ø 3(4) êë

n = 16, b = 1, a = 0, f ( x) = x 4

p = 2: 42 (0.020703) = 16(0.020703) = 0.331248 82 (0.005200) = 64(0.005200) = 0.3328 162 (0.001302) = 256(0.001302) = 0.333312 2

564

Chapter 7 INTEGRATION

To find error for each value of n, subtract as indicated. n = 4: (0.2005208 - 0.2) = 0.0005208 n = 8: (0.2000326 - 0.2) = 0.0000326 n = 16: (0.2000020 - 0.2) = 0.0000020 n = 32: (0.2000001 - 0.2) = 0.0000001

x 4dx

æ 1 ÷ö æ 1 ö÷ 1 - 0 êé ÷ + 2 çç 0 + 4 çç ÷ ÷ ê èç 4096 ø÷ 3(16) êë èç 65,536 ø÷ æ 81 ö÷ æ 625 ÷ö æ 1 ÷ö ÷ + 2 çç ÷ + 4 çç + 4 çç ÷ ÷ ÷ çè 65,536 ÷ø÷ èç 256 ø èç 65,536 ø÷

æ 2401 ÷ö æ 81 ö÷ æ 1ö ÷ + 2 çç ÷÷ + 2 çç + 4 çç çè 65,536 ÷ø÷ çè 4096 ÷÷ø çè 16 ÷ø

81(0.0000326) = 8(0.0000326) = 0.0002608

æ 14, 641 ÷ö æ 6561 ö÷ æ 625 ö÷ ÷ + 2 çç ÷ + 4 çç + 4 çç çè 65,536 ÷÷ø çè 4096 ÷÷ø çè 65,536 ÷÷ø

Try p = 2:

æ 28,561 ö÷ æ 81 ÷ö æ 2401 ÷ö ÷ + 2 çç + 2 çç + 4 çç çè 256 ÷÷ø çè 4096 ÷÷ø çè 65,536 ÷÷ø

82 (0.0000326) = 64(0.0000326) = 0.0020864

42 (0.0005208) = 16(0.0005208) = 0.0083328

Try p = 3:

æ 50, 625 öù + 4 çç + 1÷÷÷ úú ÷ø ú çè 65,536 û

43 (0.0005208) = 64(0.0005208) = 0.0333312 83 (0.0000326) = 512(0.0000326) = 0.0166912

1 æç 157, 288 ö÷ ÷ » 0.2000020 = ç 48 çè 16,384 ÷÷ø n = 32, b = 1, a = 0, f ( x) = x 4 1

Try p = 4: 4 (0.0005208) = 256 (0.0005208) = 0.1333248 4

ò x dx

84 (0.0000326) = 4096 (0.0000326) = 0.1335296 164 (0.0000020) = 65536 (0.0000020) = 0.131072

æ 1 - 0 éê 1 ÷÷ö + 2 æçç 1 ÷÷ö 0 + 4 çç » ê ç çè 65, 536 ÷÷ø 3(32) ë è 1, 048, 576 ÷÷ø æ ö÷ æ 625 ö÷ æ 1 ö÷ 81 ÷ + 2 çç ÷ + 4 çç + 4 çç çè 4096 ÷ø÷ çè 1, 048, 576 ÷ø÷ èç 1, 048, 576 ÷÷ø æ 625 ÷ö æ 14, 641 ÷ö æ 81 ö÷ ÷÷ + 4 çç ÷ + 2 çç + 2 çç ÷ èç 4096 ÷ø èç 65, 536 ø÷ èç 1, 048, 576 ø÷÷

324 (0.0000001) = 1048576 (0.0000001) = 0.1048576

These are the closest values we can get; thus, p = 4. 21.

Midpoint rule: n = 4, b = 5, a = 1, f ( x) = i

æ 28, 561 ö÷ æ 2401 ö÷ æ 50, 625 ö÷ ÷÷ + 2 çç ÷÷ + 4 çç ÷ + 4 çç èç 1, 048, 576 ø÷ èç 65, 536 ø÷ èç 1, 048, 576 ø÷÷

æ 83, 521 ö÷ æ 6561 ö÷ æ 1 ö ÷ + 2 çç ÷ + 2 çç ÷÷÷ + 4 çç çè 1, 048, 576 ÷÷ø çè 65, 536 ÷÷ø çè 16 ø

æ 130, 321 ö÷ æ 194, 481 ö÷ æ 625 ö÷ ÷÷ + 2 çç ÷ + 4 çç + 4 çç ÷ ÷ çè 1, 048, 576 ÷ø÷ èç 4096 ø èç 1, 048, 576 ø÷

æ 14, 641 ÷ö æ 279, 841 ÷ö æ 81 ÷ö ÷ + 4 çç ÷ + 2 çç + 2 çç çè 65, 536 ÷ø÷ çè 1, 048, 576 ÷ø÷ çè 256 ø÷÷

æ 531, 441 ö÷ æ 390, 625 ÷ö æ 28, 561 ö÷ ÷ + 4 çç ÷ ÷ + 2 çç + 4 çç çè 65, 536 ÷÷ø çè 1, 048, 576 ÷÷ø çè 1, 048, 576 ÷÷ø

ò x 1

1 æç 50, 033,168 ÷ö ÷ » 0.2000001 ç 96 çè 262,144 ÷÷ø

f ( xi )

3 2 5 2 7 2

4 9 4 25 4 49

9 2

4 81

,x = 1

dx » 2

æ 707, 281 ÷ö æ 50, 625 ÷ö æ 2401 ö÷ ÷ + 2 çç ÷ + 2 çç + 4 çç çè 4096 ø÷÷ çè 1, 048, 576 ø÷÷ çè 65, 536 ÷ø÷ ù æ 923, 521 ÷ö ÷÷ + 1úú + 4 çç çè 1, 048, 576 ÷ø û

å f (xi ) Dx i =1

4 4 4 4 (1) + (1) + (1) = (1) + 9 25 49 81 » 0.7355

Simpson’s rule: m = 8, b = 5, a = 1, f ( x) =

1 x2

Section 7.6

565

xi 1 3 2

0 1 2

5 2

7 2

9 2

f ( xi ) 1 4 9 1 4 4 25 1 9 4 49 1 16 4 81 1 25

ò x dx 3

å f ( xi )  x i =1

æ ö æ ö æ ö æ ö = 64 çç 1 ÷÷÷ + 64 çç 1 ÷÷÷ + 64 çç 1 ÷÷÷ + 64 çç 1 ÷÷÷ 729 è 2 ø 1331 è 2 ø 2197 è 2 ø 3375 è 2 ø » 0.09198

Simpson’s rule: n = 8, b = 4, a = 2, f ( x) =

To verify the formula evaluate 2M3+T . 2M + T 2(0.7355) + 0.9436 » 3 3 » 0.8048

Midpoint rule: n = 4, b = 4, a = 2, f ( x) = x = 1 2 i xi

f ( xi )

9 4 11 4 13 4 15 4 15 4

64 729 64 1331 1 4 64 2197 64 3375

1 2 3 4 4

1 x3

f ( xi )

9 4 5 2

1 8 64 729 8 125

11 4

64 1331

1 27

13 4

64 2197

7 2

8 343

15 4

64 3375

1 64

From #7 part a, T » 0.9436, when n = 4.

22.

1 dx 2 1 x é æ ö æ ö æ ö » 5 - 1 ê 1 + 4 çç 4 ÷÷÷ + 2 çç 1 ÷÷÷ + 4 çç 4 ÷÷÷ è4ø è9ø è 25 ø 3(8) ëê æ 1 ö÷ æ 4 ö÷ æ 1 ÷ö + 2 çç ÷÷ + 4 çç ÷÷ + 2 çç ÷÷ è9ø è 49 ø è 16 ø ù æ 4 ö÷ + 4 çç ÷÷ + 1 ú è 81 ø 25 ûú 1 » (4.82906) 6 » 0.8048

ò x dx 2

é æ ö æ ö æ ö » 4 - 2 ê 1 + 4 çç 64 ÷÷÷ + 2 çç 8 ÷÷÷ + 4 çç 64 ÷÷÷ è 729 ø è 125 ø è 1331 ø 3(8) êë 8 æ ö æ ö æ ö + 2 çç 1 ÷÷÷ + 4 çç 64 ÷÷÷ + 2 çç 8 ÷÷÷ è 27 ø è 2197 ø è 343 ø ù æ ö + 4 çç 64 ÷÷÷ + 1 ú è 3375 ø 64 úû » 1 (1.125223) » 0.09377 12

From #8 part a, T » 0.0973, when n = 4. To verify the formula evaluate

2M + T . 3

2M + T » 2(0.09198) + 0.0973 3 3 » 0.09377

566 23.

Chapter 7 INTEGRATION (a)

Using the trapezoidal rule, the total wind energy consumption is æ1 ö çç ⋅ 70 + 142 + 341 + 923 ÷÷ ÷÷ ç 2019 - 2001 ç 2 ÷÷ çç 6 çç +1595 + 2096 + 1 ⋅ 2736 ÷÷÷ çè ø÷ 2 = 19,500 trillion BTUs.

27.

y = e-t +

1 t +1

The total reaction is 9æ

ò çççè e

-t 2

1 ö÷ ÷ dt. t + 1 ø÷ 2

(b)

1 n = 8, b = 9, a = 1, f (t ) = e-t + t + 1

Using Simpson’s rule, the total wind energy consumption is æ 70 + 4 ⋅ 142 + 2 ⋅ 341 ö÷ 2019 - 2001 ççç ÷÷ ç + 4 ⋅ 923 + 2 ⋅ 1595 ÷÷÷ ç 3⋅6 çç ÷ è + 4 ⋅ 2096 + 2736 ÷ø = 19,322 trillion BTUs.

24.

(a)

7 -1é1 ê (9) + 9.2 + 9.5 + 9.4 A= 6 ëê 2 ù 1 + 9.8 + 10.1 + (10.5) ú úû 2 = 57.75 The total cost for the given period is about $57,750.

(b)

f ( xi )

0.8679

1 2

2 3

0.3516 0.2501

0.2000

0.1667

0.1429

6 7 8

7 8 9

0.1250 0.1111 0.1000

(a) Trapezoidal rule: 9æ

ò çççè e

7 -1 A = [9.0 + 4(9.2) + 2(9.5) 3(6) + 4(9.4) + 2(9.8) + 4(10.1) + 10.5]

1 (172.9) 3 = 57.63

0.01x

9æ

ò çççè e »

Using a calculator program for Simpson’s rule with n = 20, we obtain 3979.242 as the value of this integral. This indicates that the total revenue between the twelfth and thirty-sixth months is about 3979.

3.922t 0.242e-0.00357t dt

Using a calculator program for Simpson’s rule with n = 20, we obtain 1400.88 as the value of this integral. This indicates that the total amount of milk consumed by a calf from 7 to 182 days is about 1401 kg.

1 ö÷ ÷ dt t + 1 ÷ø

9 -1 [0.8679 + 4(0.3516) + 2(0.2501) 3(8) + 4(0.2000) + 2(0.1667) + 4(0.1429) + 2(0.1250) + 4(0.1111) + 0.1000]

1 (5.2739) 3 » 1.758

We need to evaluate 182

-t 2

+ 32) dx.

26.

9 -1é 1 ê (0.8679) + 0.3516 + 0.2501 8 êë 2 ù 1 + ⋅ ⋅ ⋅ + (0.1000) ú úû 2

(b) Simpson’s rule:

We need to evaluate

ò (105e

1 ö÷ ÷ dt t + 1 ÷ø

» 1.831

The total cost for the given period is about $57,630.

25.

-t 2

28.

y =

2 2 + e-t /2 t +2

The total growth is 2 ö çç 2 + e-t /2 ÷÷ dt . ÷ø ç 1 èt + 2

7æ

Section 7.6

567 2

n = 12, b = 7, a = 1, f (t ) = t +2 2 + e-t /2 i

f ( xi )

0 1 2 3 4 5 6 7 8 9 10 11 12

1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7

1.2732 0.8961 0.6353 0.4884 0.4111 0.3658 0.3337 0.3077 0.2857 0.2667 0.2500 0.2353 0.2222

29.

7æ

-t 2 /2 ö÷

ò çççè t + 2 + e 1

÷÷ dt ø

7 -1é 1 ê (1.2732) + 0.8961 + 0.6353 12 êë 2 + 0.4884 + 0.4111 + 0.3658 + 0.3337 + 0.3077 + 0.2857 + 0.2667 + 0.2500 ù 1 + 0.2353 + (0.2222) ú úû 2

1 (5.2234) » 2.612 2

(b) Simpson’s rule 7æ

-t 2 /2 ö÷

ò çççè t + 2 + e 1

÷÷ø dt

7 -1 [1.2732 + 4(0.8961) + 2(0.6353) + 4(0.4884) 3(12) + 2(0.4111) + 4(0.3658) + 2(0.3337) + 4(0.3077) + 2(0.2857) + 4(0.2667) + 2(0.2500) + 4(0.2353) + 0.2222]

1 (15.5670) 6 » 2.595 =

f ( xi )

0 1 2 3 4 5 6 7 8 9

1 2 4 6 8 10 12 14 16 18

0 5 3 2 1.5 1.2 1 0.5 0.3 0.2

0.2

Area under curve for Formulation A é = 20 - 0 ê 1 (0) + 5 + 3 + 2 + 1.5 + 1.2 10 êë 2 ù + 1 + 0.5 + 0.3 + 0.2 + 1 (0.2) ú úû 2 = 2(14.8) » 30 mcg(h)/ml

Trapezoidal rule:

(a)

Note that heights may differ depending on the readings of the graph. Thus, answers may vary. n = 10, b = 20, a = 0

This represents the total amount of drug available to the patient for each ml of blood. 30.

n = 10, b = 20, a = 0 y i xi

0 1 2 3 4 5 6 7 8 9 10

0 2 4 6 8 10 12 14 16 18 20

0 2.0 2.9 3.0 2.5 2.0 1.75 1.0 0.75 0.50 0.25

é A = 20 - 0 ê 1 (0) + 2 + 2.9 + 3 + 2.5 + 2 10 êë 2 ù + 1.75 + 1.0 + 0.75 + 0.5 + 1 (0.25) ú úû 2 = 33.05 (This answer may vary depending upon readings from the graph.)

The area under the curve, about 33 mcg(h)/ml, represents the total amount of drug available to the patient for each ml of blood.

568 31.

Chapter 7 INTEGRATION As in Exercise 29, readings on the graph may vary, so answers may vary. The area both under the curve for Formulation A and above the minimum effective concentration line is on the interval éê 12 , 6 ùú . ë û

Let AB = area under Formulation B curve between t = 2 and t = 10. AB =

Area under curve for Formulation A on éê 12 , 1 ùú , ë û with n = 1 =

1 - 12 é 1 ù ê (2 + 6) ú úû 1 êë 2

1 (4) = 2 2

AB = 20.4

Let AME = area under minimum effective concentration curve between t = 2 and t = 10. AME = (10 - 2)(2) = 16

So the area between AB and AME between t = 2 and t = 10 is 20.4 - 16 = 4.4.

Area under curve for Formulation A on [1, 6 ], with n = 5

This area, about 4.4 mcg(h)/ml, represents the total effective amount of the drug available to the patient for each ml of blood.

ù 6 -1é 1 1 ê ( 6) + 5 + 4 + 3 + 2.4 + (2) ú = 5 ëê 2 2 ûú = 18.4

Notice that between t = 0 and t = 12, the graph for Formulation B is below the line.

Area under minimum effective concentration line é 1 , 6ù êë 2 úû

Thus, no area exists under the curve for Formulation B and above the minimum effective concentration line in the intervals (0, 2) and (10, 12).

= 5.5(2) = 11.0

Area under the curve for Formulation A and above minimum effective concentration line = 2 + 18.4 - 110 » 9 mcg(h)/ml This represents the total effective amount of drug available to the patient for each ml of blood. 32.

The area both under the curve for Formulation B and above the minimum effective concentration line is on the interval (2, 10). n = 8, b = 10, a = 2 i

0 1 2 3 4 5 6 7 8

2 3 4 5 6 7 8 9 10

2.0 2.4 2.9 2.8 3.0 2.6 2.5 2.2 2.0

10 - 2 é 1 ê (2) + 2.4 + 2.9 + 2.8 + 3 8 êë 2 ù 1 + 2.6 + 2.5 + 2.2 + (2) ú 2 úû

33.

y = b0 wb1 e-b2 w

(a) if t = 7w then w =

t . 7

æ t öb 1 y = b0 çç ÷÷÷ e-b2t /7 çè 7 ø

(b) Replacing the constants with the given values, we have æ t ö0.233 -0.027t /7 y = 5.955 çç ÷÷÷ e dt çè 7 ø

In 25 weeks, there are 175 days. 175

ò 0

æ t ö0.233 -0.027t /7 e dt 5.955 çç ÷÷÷ çè 7 ø

n = 10, b = 175, a = 0, æ t ö0.233 -0.027t /7 f (t ) = 5.955 çç ÷÷÷ e çè 7 ø

Section 7.6

569 i

f (ti )

0 1 2 3 4 5 6 7 8 9 10

0 1.75 35 52.5 70 87.5 105 122.5 140 157.5 175

0 6.89 7.57 7.78 7.77 7.65 7.46 7.23 6.97 6.70 6.42

Trapezoidal rule: æ t ö0.233 -0.027t / 7 e dt 5.955 çç ÷÷÷ çè 7 ø 0 175 - 0 é 1 ê ( 0) + 6.89 + 7.57 + 7.78 + 7.77 » 10 êë 2

175

+ 7.65 + 7.46 + 7.23 + 6.97 + 6.70 +

ù 1 (6.42) ú úû 2

The total milk consumed is about 1212 kg.

f (ti )

0 1 2 3 4 5 6 7 8 9 10

0 17.5 35 52.5 70 87.5 105 122.5 140 157.5 175

0 8.74 8.80 8.50 8.07 7.60 7.11 6.63 6.16 5.71 5.28

0.143

ætö 8.409 çç ÷÷÷ çè 7 ø

e-0.037t / 7 dt

175 - 0 é 1 ê ( 0) + 8.74 + 8.80 + 8.50 êë 2 10 + 8.07 + 7.60 + 7.11 + 6.63 ù 1 + 6.16 + 5.71 + (5.28) ú úû 2

= 17.5 (69.96) = 1224.30

Simpson’s rule:

175

= 1211.525

Trapezoidal rule:

= 17.5(69.23)

175

The total milk consumed is about 1224 kg. Simpson’s rule:

æ t ö0.233 -0.027t / 7 e dt 5.955 çç ÷÷÷ çè 7 ø

175 - 0 [0 + 4(6.89) + 2(7.57) + 4(7.78) 3 (10)

175

+ 2(7.77) + 4(7.65) + 2(7.46) + 4(7.23)

+ 2(6.97) + 4(6.70) + 6.42]

The total milk consumed is about 1231 kg.

æ t ö0.143 -0.037t 7 8.409 çç ÷÷÷ dt e çè 7 ø

175 - 0 [0 + 4(8.74) + 2(8.80) + 4(8.50)] 3(10) + 2(8.07) + 4(7.60) + 2(7.11) + 4(6.63) + 2(6.16) + 4(5.71) + 5.28]

35 (214.28) 6 = 1249.97 =

æ t ö0.143 -0.037t /7 y = 8.409 çç ÷÷÷ e . çè 7 ø

The total milk consumed is about 1250 kg.

In 25 weeks, there are 175 days.

175

æ t ö0.143 -0.037t /7 e dt 8.409 çç ÷÷÷ çè 7 ø

n = 10, b = 175, a = 0, æ t ö0.143 -0.037t /7 f (t ) = 8.409 çç ÷÷÷ e çè 7 ø

34.

Using Simpson’s rule, the total number of cases for this time period is æ 600 + 4 ⋅ 2800 + 2 ⋅ 10, 000 ö÷ ÷÷ 30 ççç çç + 4 ⋅ 19,800 + 2 ⋅ 30, 400 ÷÷÷ 3 ⋅ 6ç ÷ èç + 4 ⋅ 30,800 + 28,900 ø÷ » 539,800 cases.

570 35.

Chapter 7 INTEGRATION 4.

True

False: The derivative gives the instantaneous rate of change.

= 71.5

7 -1 [4 + 4(7) + 2(11) + 4(9) 3(6) + 2 (15) + 4 (16) + 23] = 69.0

False: If the function is positive over the interval of integration the definite integral gives the exact area.

True

10.

False: The definite integral may be positive, negative, or zero.

11.

True

12.

False: Sometimes true, but not in general.

13.

False: The trapezoidal rule allows any number of intervals.

14.

True

19.

7 -1 1 6 2

(a)

é ù 1 ê (4) + 7 + 11 + 9 + 15 + 16 + (23) ú êë úû 2

(b)

36.

(a) A=

7 -1 é1 ê (12) + 16 + 18 + 21 + 24 6 ëê 2 ù 1 + 27 + (32) ú úû 2

A = 1 (128) = 128 A =

(b)

7 -1 [12 + 4(16) + 2(18) + 4(21) 3(6) + 2(24) + 4(27) + 32]

A = 128

37.

Use a calculator program for Simpson’s rule with n = 20 to evaluate each of the integrals in this exercise. 1æ ö 2 çç 1 e-x /2 ÷÷ dx » 0.6827 (a) ÷÷ø ç -1 è 2 The probability that a normal random variable is within 1 standard deviation of the mean is about 0.6827.

2æ

3æ

-x 2 /2 ö÷

ò ççèç 2 e

÷ dx » 0.9973 ÷÷ø -3 The probability that a normal random variable is within 3 standard deviations of the mean is about 0.9973.

21.

ò (x - 3x + 2) dx

(5x - 1) dx =

3x 2 x3 + 2x + C 3 2

22.

23.

ò 3 t dt = 3ò t dt

(6 - x 2 ) dx = 6 x -

Chapter 7 Review Exercises 24.

True

False: The statement is false for n = –1.

False: For example, if f ( x ) = 1 the first

x3 +C 3

1/2

3t 3/2

= 2t

5x 2 -x+C 2

20.

The probability that a normal random variable is within 2 standard deviation of the mean is about 0.9545. (c)

= x 2 + 3x + C

ò-2 ççèç 2 e-x /2 ÷÷ø÷÷ dx » 0.9545

(b)

2x2 + 3x + C 2

(2 x + 3) dx =

3 2 3/2

+C +C

1 1/2

ò 2 dt = ò 2 t dt 1 t 3/2 = 2 3 +C

expression is equal to x 2 /2 + C and the second =

2 3/2

Chapter 7 Review

25.

573

x3/2

3 2 3/2

2x = 3

ò (2x

26.

33.

( x1/2 + 3x-2/3 ) dx

4/3

7 3 7/3

6x 7

3x1/3 1 3

3 dw 2 w 3 = ln | w | + C 2

+ 9x

= 1/3

x1/2

3 ln | u 2 - 1| +C 2

1 2

34.

+ 2 x1/2 + C

-4

-4 y -2

1 1 dw = ln | w | + C 2 2 w 1 = ln | 2 - u 2 | + C 2

35.

ò y dy = ò 5 y dy

28.

-4

ò -3e dx = 2x

29.

-3e 2

ò 5e dx = -5e + C

31.

ò xe dx = 6 ò 6xe dx

1 3 1 = 3 =

3x 2

1 eu du 6 1 = eu + C 6

ò (x + 5) 3

36.

òu ò u du 4

-4

1 æç u-3 ö÷÷ çç ÷+C 3 çè -3 ÷÷ø

-( x3 + 5)-3 +C 9

-x

3x 2

ò (x - 5x) (2x - 5) dx 2

Let u = x 2 - 5 x, so that du = (2 x - 5) dx.

ò (x - 5x) (2x - 5) dx = ò u du 2

e3 x = +C 6

3x 2 dx

1 3

= 4

du = 3x 2dx.

Let u = 3x 2 , so that du = 6 x dx.

32.

ò (x + 5)

30.

-x

x 2 dx

Let u = x3 + 5, so that

5 y -3 = +C -3 5 =- 3 +C 3y 2x

= 2 y -2 + C 5

2 xe x dx = e x + C

Let w = 2 - u 2 , so that dw = -2u du.

-3

-2

-2u du

-u

ò 2 - u du = - 2 ò 2 - u

ò y dy = ò -4 y dy

27.

2u du

Let w = u 2 - 1, so that dw = 2u du.

+ x-1/2 ) dx

2 x 7/3

æ1ö

ò u 2 - 1 du = 3 çççè 2 ÷÷÷ø ò u 2 - 1

u5 +C 5

( x 2 - 5 x) 5 +C 5

574

Chapter 7 INTEGRATION x3

òe

37.

dx =

x3e-3x

1 12

40.

ò -12x e

3 -3 x 4

5 ln z + 3 dz z

Let u = 5 ln z + 3 so that

du =

Let u = -3x , so that du = -12 x dx. 1 12 1 =12 =-

ò ò e +C u

-3 x 4

-e 12

òe

38.

3x 2 + 4

1 2u 3 / 2 ⋅ +C 5 3

2(5 ln z + 3)3 / 2 +C 15

1 x dx = 6 1 = 6

f ( x) = 3x + 1, x1 = -1, x2 = 0, x3 = 1,

41.

x4 = 2, x5 = 3

du = 6 x dx.

òe

5 5 ln z + 3

ò z ò u1/ 2 du

x dx

Let u = 3x 2 + 4 so that 3x 2 +4

(5 ln z + 3) 1 dz = z 5 1 = 5

eu du

5 dz. z

f ( x1) = -2, f ( x2 ) = 1, f ( x3 ) = 4,

ò (6x) (e ) dx ò e du 3x 2

f ( x4 ) = 7, f ( x5 ) = 10 5

å f ( xi )

1 e3x + 4 = eu + C = +C 6 6

i =1

= f (1) + f (2) + f (3) + f (4) + f (5) = -2 + 1 + 4 + 7 + 10 = 20

39.

(3 ln z + 2) dz z

42.

Let u = 3 ln z + 2 so that

(a)

ò f ( x) dx = 0, since the area above the 0

x-axis from 0 to 2 is identical to the area below the x-axis from 2 to 4.

3 du = dz. x (3 ln z + 2)4 1 dz = z 3 1 = 3

3(3 ln z + 2) 4 dz z

ò ò u 4 du

(b)

ò f ( x) dx can be computed by 0

1 u5 ⋅ +C 3 5

(3 ln z + 2)5 +C 15

calculating the area of the rectangle and triangle that make up the region shown in graph. Area of rectangle = (length) (width) = (3)(1) = 3 1 Area of triangle = (base) (height) 2 1 3 = (1) (3) = 2 2 4

ò f (x) dx = 3 + 2 = 2 = 4.5 0

Chapter 7 Review

575

f ( x) = 2 x + 3, from x = 0 to x = 4

43. x =

4-0 =1 4 i

1 2 3 4

46.

ò f (x) dx = F ( x) a

xi 0 1 2 3

f ( xi ) 3 5 7 9

b = F (b) - F (a), a

where f is continuous on [a, b] and F is any antiderivative of f.

47.

The Fundamental Theorem of Calculus states that

æ 3x3 ö÷ ç (3x + 5) dx = çç + 5x ÷÷÷ çè 3 1 ø÷ 1 2

å f (xi ) D x

= (23 + 10) - (1 + 5)

i =1

= 18 - 6

= 3(1) + 5(1) + 7(1) + 9(1)

= 12

= 24

44.

48.

ò (2x + 3) dx

ò (2x + x) dx 2

6 æ 2 x3 x 2 ö÷÷ çç =ç + ÷ 2 ÷÷ø çè 3 1

Graph y = 2 x + 3.

é 2(6)3 (6)2 ùú éê 2(1)3 (1)2 ùú = êê + -ê + ú 2 ú ê 3 2 úú êë 3 û ë û 2 1 = 144 + 18 - 3 2 2 1 965 = 162 - - = » 160.83 3 2 6 4

ò (2x + 3) dx is the area of a trapezoid with

49.

5 æ x-2 ö÷÷ ç (3x-1 + x-3 ) dx = çç 3ln | x | + ÷÷ -2 ÷ø ççè 1 1 5

B = 11, b = 3, h = 4. The formula for the area is

1 ( B + b)h. 2 1 A = (11 + 3)(4) = 28, 2 so

æ 1 ö÷ æç 1ö = çç 3ln 5 ÷ - ç 3ln1 - ÷÷÷ èç 50 ÷ø èç 2ø = 3ln 5 +

50.

ò (2x + 3) dx = 28.

3 æ x-1 ö÷÷ ç (2 x-1 + x-2 ) dx = çç 2 ln x + ÷ -1 ÷ø÷ çè 1 1

æ 1ö = çç 2 ln 3 - ÷÷÷ - (2 ln1 - 1) çè 3ø

45.

(a) Since s(t) represents the odometer reading, the distance traveled between t = 0 and t = T will be s(T ) - s(0). (b)

12 » 5.308 25

ò v (t) dt = s (T ) - s (0) is equivalent to 0

the Fundamental Theorem of Calculus with a = 0, and b = T because s(t) is an antiderivative of v(t ).

= 2 ln 3 +

51.

2 » 2.864 3

ò x 5x + 4 dx 2

Let u = 5 x 2 + 4, so that du = 10 x dx and

1 du = x dx. 10

When x = 0, u = 5(02 ) + 4 = 4.

576

Chapter 7 INTEGRATION

upper quarter of a circle centered at the origin with a radius of 1.

When x = 1, u = 5(12 ) + 4 = 9. =

1 10

ò4

u du =

1 10

ò4 u1/2du

Area of circle =  r 2 =  (12 ) =  1

52.

53.

ò0 x (3x + 1) 2

ò 3e

-2 w

254/3 14/3 12 12

254/3 - 1 » 6.008 12

ò0 4x 4 - x2 dx Then du = 2 x dx. When x = 0, u = 0. When x = 2

-3e 2

3(1 - e-4 ) » 1.473 2

3 2

ò 0.4e

0.4 w

ò dw

Let u = 4 x 2. Then du = 8 x dx. When x = 0, u = 0, and when x = 12 , u = 1.

Thus,

ò0

1 1 1 - u 2 du. 8 0 Note that this integral represents the area of right

x 1 - 16 x 4 dx =

4 x 4 - x 2 dx = 2 .

25 - (ln x) 2 dx x

Let u = ln x. Then du = 1x dx. When x = e5, u = ln (e5) = 5. When x = 1, u = ln(1) = 0. Thus,

1/2

4 - u 2 du

x 1 - 16 x 4 dx

57.

25(e2 - e0.4 ) = » 36.86 4 1/ 2

The integral on the right gives the area of a semicircle of radius 2 (the bottom half of a circle of radius 2 centered at the point (0, 2). This area is 1  (2)2 = 2 . Thus 2

5 5 5 = ⋅ ⋅ e 2w/5 2 2 1 25 2 (e - e0.4 ) = 4

55.

4 x 4 - x 2 dx = 2

2, u = 2.

Substitute:

5 0.4w 5 5 e dw = ⋅ 2 2 2



(3x3 + 1) 4/3 dx = 12 0

-4

1 

ò0 1 - u 2 du = 8 ⋅ 4 = 32

Let u = x 2.

-3e 2

dw =

56.

-2 w

54.

1 8

1/3



ò0 1 - u du = 4

1 u 3/2 9 1 3/2 9 = ⋅ = u 10 3/2 4 15 4 1 1 (9)3/2 (4)3/2 = 15 15 27 8 = 15 15 19 = 15

5 25 - (ln x)2 dx = 25 - u 2 du. x 0 Note that this integral represents the area of a right upper quarter of a circle centered at the origin with a radius of 5.

Area of circle =  r 2 =  (5)2 = 25

25 - u 2 du =

25 4

Chapter 7 Review 7

58.

577 2

2 x 36 - ( x - 1) dx

1 2

Let u = x - 1. Then du = 2 x dx. When x =

f ( x) = xe x ; [0, 2]

61.

Area =

x = e 2

2 x 36 - ( x 2 - 1)2 dx =

36 - u 2 du.

36 - u 2 du =

59.

f ( x) =

Area =

36 = 9 4

f ( x) = 1 + e-x ; [0, 4]

62.

ò0

(1 + e-x ) dx = ( x - e-x )

4 x - 3 dx

4 0

= (4 - e-4 ) - (0 - e0 ) = 5 - e-4 » 4.982

4 x - 3; [1, 3]

4 = e -1 2 » 26.80

Area of circle =  r =  (6) = 36

= e -1 2 2

Note that this integral represents the area of a right upper quarter of a circle centered at the origin with a radius of 6. 6

Thus,

ò0 xe dx 2

7, u = ( 7 )2 - 1 = 6.

When x = 1, u = ( 1 )2 - 1 = 0. 7

63.

f ( x ) = 5 - x 2 , g ( x) = x 2 - 3

ò (4x - 3) dx 1/2

3 2 1 ⋅ ⋅ (4 x - 3)3/2 3 4 1 1 1 = (9)3/2 - (1)3/ 2 6 6 1 = (26) 6 13 = 3

60.

f ( x) = (3x + 2) ; [-2, 0]

Area =

Points of intersection: 5 - x2 = x2 - 3 2x2 - 8 = 0 2( x 2 - 4) = 0 x = 2 Since f ( x) ³ g ( x) in [–2, 2], the area between the graphs is

ò- (3x + 2) dx 6

(3x + 2)7 = 21

-2

27 (-4)7 = 21 21 5504 = 7

-2

[ f ( x) - g ( x)] dx = =

ò [(5 - x ) - ( x - 3) ] dx -2 2

ò (-2x + 8) dx 2

-2

3 æ ö = ççç -2 x + 8 x ÷÷÷ ÷ø çè 3

2 -2

= - 2 (8) + 16 + 2 (-8) - 8(-2) 3 3 32 64 = + 32 = . 3 3

578

Chapter 7 INTEGRATION Points of intersection: x2 - 4x = x + 6

f ( x) = x 2 - 4 x; g ( x) = x - 6

64.

x2 - 5x - 6 = 0 ( x + 1)( x - 6) = 0 x = -1 or x = 6 Thus, the area is -1

ò [x - 4x - (x + 6)] dx + ò [x + 6 - (x - 4x)] dx 2

-2

-1

Find the points of intersection.

2 æ 3 ö = ççç x - 5x - 6 x ÷÷÷ è 3 ø 2

x2 - 4x = x - 6 x2 - 5x + 6 = 0 ( x - 3)( x - 2) = 0 x = 2 or x = 3 Since g ( x) ³ f ( x) in the interval [2, 3], the area between the graphs is

-1

3 2 æ ö + ççç - x + 5 x + 6 x ÷÷÷ è 3 ø 2

-2

æ ö æ ö = çç 19 + 2 ÷÷÷ + çç 128 + 19 ÷÷÷ = 149 è 6 3ø è 3 6ø 3 f (t ) = 5 - t 2 , g (t ) = t 2 - 3, t = 0, t = 4

66.

ò [g( x) - f (x)] dx 2

= =

ò [(x - 6) - (x - 4x)] dx 2 3

ò (-x + 5x - 6) dx 2

æ 3 ö÷ 5x2 ç -x = çç + - 6 x ÷÷÷ çè 3 2 ø÷

Find the points of intersection.

-27 -8 5(9) = + - 6(3) 3 2 3 5(4) + 6(2) 2 19 25 1 =+ -6 = . 3 2 6

65.

5 - t = t2 - 3 8 = 2t 2 4 = t2 2 = t The curves intersect at t = 2 and t = -2.

f ( x) = x 2 - 4 x, g ( x) = x + 6, x = -2, x = 4

Thus, area is 2

ò [(5 - t ) - (t - 3)] dt 2

ò [(t - 3) - (5 - t )] dt

(-2t 2 + 8) dt +

3 æ ö = ççç -2t + 8t ÷÷÷ èç 3 ø÷

ò (2t - 8) dt 2

2 0

æ 3 ö + ççç 2t - 8t ÷÷÷ èç 3 ø÷

4 2

æ ö æ ö = -16 + 16 + çç 128 - 32 ÷÷÷ - çç 16 - 16 ÷÷÷ è 3 ø è 3 ø 3 = 32 + 128 - 32 - 16 + 16 3 3 3 = 32.

-1

Chapter 7 Review

579 Then

ln x dx x 1

67.

ò2

Trapezoidal Rule: n = 4, b = 3, a = 1, f ( x) = lnxx i

f ( xi )

0 1 2 3 4

1 1.5 2 2.5 3

0 0.27031 0.34657 0.36652 0.3662

ln x 3-1é1 ê (0) + 0.27031 + 0.34657 dx » x 4 êë 2 1 ù 1 + 0.36652 + (0.3662) ú 2 ûú

x dx = x -1

ö çç 1 + 1 ÷÷ du = ÷ ç uø 1 è

= u

+ ln u

ò0 e x e x + 4 dx

69.

Trapezoidal Rule: n = 4, b = 1, a = 0, f ( x) = e x e x + 4

f ( xi )

0 1 2 3 4

0 0.25 0.5 0.75 1

2.236 2.952 3.919 5.236 7.046

x dx x -1

Trapezoidal Rule:

xi 2

1- 0 é1 ê (2.236) + 2.952 4 êë 2 + 3.919 + 5.236 +

x n = 4, b = 10, a = 2, f ( x) = x 1

ù 1 (7.046) ú úû 2

» 4.187 Exact value:

f ( xi ) 2 4 3 6 5 8 7 10 9

e x e x + 4 dx =

ò e (e + 4) dx x

1/2

2 x (e + 4)3/2 3 0

2 2 (e + 4)3/2 - (5)3/2 3 3 » 4.155

x dx x 1 2 10 - 2 êé 1 4 6 8 1 æ 10 ö ù (2) + + + + çç ÷÷÷ ú » ê 4 ë2 3 5 7 2 çè 9 ø úû » 10.46 Exact Value:

ò e e + 4 dx

» 0.6035 2

ò1 u du

= 8 + ln 9 » 10.20.

1 1 1 (ln x )2 = (ln 3)2 - (ln 1)2 2 2 2 1 10

du +

= (9 - 1) + (ln 9 - ln1)

ln x dx 1 x

68.

ò1

u +1 du u

9æ

= 0.5833 Exact Value:

ò1

70.

xe-x dx

Trapezoidal rule: n = 4, b = 2, a = 0, f ( x) = xe-x

Let u = x - 1 , so that du = dx and x = u + 1.

580

Chapter 7 INTEGRATION i 0 1 2 3 4

xi 0 0.5 1 1.5 2

f ( xi ) 0 0.3894 0.3679 0.1581 0.0366

Simpson’s Rule: i

f ( xi )

2 - 0 é1 ê (0) + 0.3894 + 0.3679 » 4 êë 2

2 4 3 6 5 8 7 10 9

ò xe

-x 2

+ 0.1581 +

ù 1 (0.0366) ú úû 2

= 0.4688

xe-x dx = -

1 2

ò e- (-2x dx) x2

2 1 = - e-x 2 0

1 1 = - e-4 + e0 2 2 1 = (1 - e-4 ) 2 » 0.4908

71.

x dx 1 x 2 æ4ö æ6ö æ 8 ö æ 10 ö ù 10 - 2 éê » 2 + 4 çç ÷÷ + 2 çç ÷÷ + 4 çç ÷÷ + çç ÷÷ ú ÷ ÷ ç ç çè 7 ø÷ çè 9 ø÷ ú ê è3ø è5ø 3(4) ë û » 10.28 This answer is close to the answer of 10.20 obtained from the exact integral in Exercise 68.

Exact value: 2

x dx

ò x -1

72.

ò e e + 4 dx

73.

Simpson’s rule: n = 4, b = 1, a = 0, f ( x) = e x e x + 4

ln x dx x

Simpson’s rule: n = 4, b = 3, a = 1, f ( x) = lnxx

f ( xi )

0 1 2 3 4

1 1.5 2 2.5 3

0 0.27031 0.34657 0.36652 0.3662

ln x dx 1 x 3-1 » [0 + 4(0.27031) + 2(0.34657) 3(4)

f ( xi )

0 1 2 3 4

0 0.25 0.5 0.75 1

2.236 2.952 3.919 5.236 7.046

ò e e + 4 dx x

1- 0 [2.236 + 4(2.952) + 2(3.919) 3(4) + 4(5.236) + 7.046

» 4.156 This answer is close to the answer of 4.155 obtained from the exact integral in Exercise 69.

+ 4(0.36652) + 0.3662 ] » 0.6011 This answer is close to the value of 0.6035 obtained from the exact integral in Exercise 67.

Chapter 7 Review 2

74.

581

xe-x dx

(c)

5æ

æ 5 - 1 ö÷ ÷÷ [0 + 4(0.5) + 2(0.41421) = ççç è 3(4) ø÷

Simpson’s rule: n = 4, b = 2, a = 0, f ( x) = xe-x i

+ 4 (0.23205) + 0] æ1ö = çç ÷÷÷ (3.75662) çè 3 ø

f ( xi )

0 0

0.5 0.3894

2 1

0.1581

4 2

0.0366

= 1.252

0.3679

3 1.5 2

76.

(a)

5é

æ x - 1 öù

ò1 êêë x - 1 - çççè 2 ÷ø÷÷ úúû dx

1ö

ò ççè x - 1 - 2 + 2 ÷÷÷ø dx 1

2 æ ö 5 3/ 2 = ççç 2 ( x - 1) - x + x ÷÷÷ è3 4 2ø 1 æ ö æ ö = çç 16 - 25 + 5 ÷÷÷ - çç 0 - 1 + 1 ÷÷÷ è ø è 3 ø 4 2 4 2 16 4 = -6+2= 2 3

(b)

x -1 - x + 1 2 2 xi f ( xi )

0 1 2 3 4

1 2 3 4 5

0 0.5 0.41421 0.23205 0

1ö

ò ççè x - 1 - 2 + 2 ÷÷÷ø dx 1

= ( 9 - ln 5 ) - (1 - 0) = 8 - ln 5 » 6.391

(b) Trapezoidal rule: n = 4, b = 4, a = 0, f ( x) = x + 2 - 1 2 x +1 i 0 1 2 3 4

æ öé = çç 5 - 1 ÷÷÷ ê 1 (0) + 0.5 + 0.41421 è 4 ø êë 2 ù + 0.23205 + 1 (0) ú 2 úû = 1.146

xi f ( xi ) 0 0 1 1 2 1.6667 3 2.25 4 2.8 4æ

x+2

ò çççè 2 - x + 1 ÷÷÷ø dx 0

é ù » 4 - 0 ê 1 (0) + 1 + 1.6667 + 2.25 + 1 (2.8) ú 4 ëê 2 2 ûú » 6.317

n = 4, b = 5, a = 1,

f ( x) =

» 0.4937 The answer is close to the exact value obtained in Exercise 70, which is approximately 0.4908.

æ ( x + 2)2 ÷ö = ççç - ln x + 1 ÷÷ çè ÷÷ø 4 0

2-0 [0 + 4(0.3894) + 2(0.3679) 3(4) + 4(0.1581) + 0.0366]

(a)

x+2

75.

4æ

ò çççè 2 - x + 1 ÷÷÷ø dx

xe-x dx

1ö

ò çççè x - 1 - 2 + 2 ÷÷ø÷ dx

0 1 2 3 4

xi f ( xi ) 0 0 1 1 2 1.6667 3 2.25 4 2.8 4æ

x+2

ò çççè 2 - x + 1 ÷÷ø÷ dx 0

» 4 - 0 [ 0 + 4(1) + 2(1.6667) + 4(2.25) + 2.8 ] 4(3) » 6.378

582

Chapter 7 INTEGRATION 2

77.

ò- [x(x - 1) (x + 1) (x - 2) (x + 2)] dx 2

ò 3(2x - 1) dx 3 = 2(2 x - 1) dx 2ò 1/2

C ( x) =

(a) Trapezoidal Rule:

1/2

n = 4, b = -2, a = 2, f ( x) = [ x( x - 1) ( x + 1) ( x - 2) ( x + 2)]2 i xi f ( xi ) -2 -1 0 1 2

0 1 2 3 4

C ¢( x) = 3 2 x - 1; 13 units cost $270.

79.

Let u = 2 x - 1, so that du = 2dx.

0 0 0 0 0

ò [x( x - 1) (x + 1) (x - 2) (x + 2)] dx ù 2 - (-2) é 1 1 ê (0) + 0 + 0 + 0 + (0) ú » êë 2 úû 4 2 = 0

3 2

3 æç u 3/2 ö÷÷ ç ÷+C 2 ççè 3/2 ÷÷ø

ò u du 1/2

= (2 x - 1)3/2 + C

-2

C (13) = [2(13) - 1]3/2 + C

Since C (13) = 270, 270 = 253/2 + C 270 = 125 + C

C = 145. Thus,

(b) Simpson’s Rule: n = 4, b = 2, a = 2, f ( x) = [ x( x - 1) ( x + 1) ( x - 2) ( x + 2)]2 i

f ( xi )

2

1 0 1 2

0 0 0 0

2 3 4

-2

[ x( x - 1) ( x + 1) ( x - 2) ( x + 2)]2dx

2 - (-2) [0 + 4(0) + 2(0) + 4(0) + 0] 3(4) = 0 »

78.

f (2 x) dx =

1 2

1 = 2

C ¢( x) = 2 x8+1 ; fixed cost is $18.

80.

C ( x) =

If x = 0, C ( x) = 18. Thus 18 = ln |1| + k k = 18. Thus, C ( x) = 4 ln ( 2 x + 1 ) + 18 since x  0.

The total amount of debt accumulated from 2009 to 2019 is 2(1290 + 1080 + 480 + 580 + 780) = $8420 billion.

81.

0 2

ò f (x) dx 0

ò 2x + 1 dx

= 4 ln | 2 x + 1| +k

f ( x) dx

4 1 + f ( x) dx 2 2 1 1 = (3) + (5) 2 2 = 4

The answer is c.

C ( x) = (2 x - 1)3/2 + 145.

Total amount

82. =

ò 100,000e t dt 0.03

100,000e0.03t 0.03 0 10,000,000 0.03T (e = - 1) 3 =

Chapter 7 Review

583 Equilibrium supply

Set this expression equal to 4,000,000.

= (10)2 + 5(10) + 100 = 250

10,000,000 0.03T - 1) = 4,000,000 (e 3

Equilibrium demand

e0.03T - 1 = 1.2

= 350 - (10)2 = 250

0.03T = ln 2.2 T » 26.3 It will take him 26.3 years to use up the supply.

83.

S ¢( x ) = 3 2 x + 1 + 3

S ( x) =

ò (3 2x + 1 + 3) dx

f (t ) =

0 10

ò (-q - 5q + 150) dq 2

æ -q3 ö÷ 10 5q 2 ç = çç + 150q ÷÷ ÷ø÷ 2 èç 3 0

4 0

= -1000 - 500 + 1500 3 2 $2750 = » $916.67 3

= (27 + 12) - (1 + 0) = 38 Total sales = $38,000.

(a)

84.

ò éêë 250 - (q + 5q + 100) ùúû dq

= =

= [(2 x + 1)3 / 2 + 3x]

(a) Producers’ surplus 10

ò (0.1344t - 0.9852) dt

(b) Consumers’ surplus

0.1344 2 = t - 0.9852t + C 2

= 0.0672t 2 - 0.9852t + C

ò [(350 - q ) - 250] dq ò (100 - q ) dq 10

= 1000 - 1000 = $2000 » $666.67 3 3

86.

S ¢( x) = 225 - x 2 , C ¢( x) = x 2 + 25x + 150 S ¢( x) = C ¢( x)

The productivity at the end of 2019 was approximately 107.6.

æ q3 ö÷÷ = ççç100q ÷ çè 3 ÷÷ø 0

f (t ) = 0.0672t - 0.9852t + 102.1

(b) f (19) = 0.0672(19) - 0.9852(19) + 102.1 = 107.6404 » 107.6

0 10 0

Since t is the number of years since 2000, and the productivity in 2012 is 100, we know that f (12) = 100, so f (12) = 0.0672(12) 2 - 0.9852(12) + C 100 = 9.6768 - 11.8224 + C C = 102.1456 » 102.1 Thus,

225 - x 2 = x 2 + 25x + 150 2 x 2 + 25x - 75 = 0 (2 x - 5) ( x + 15) = 0 5 = 2.5 2 The company should use the machinery for 2.5 years. x =

85.

S (q) = q 2 + 5q + 100 D(q) = 350 - q 2 S (q) = D(q) at the equilibrium point. 2

q + 5q + 100 = 350 - q

2.5

2q + 5q - 250 = 0

[(225 - x 2 ) - ( x 2 + 25 x + 150)]dx

2.5

(-2 x3 - 25x + 75) dx

(-2q + 25) (q - 10) = 0

3 2 æ ö = ççç -2 x - 25x + 75 x ÷÷÷ çè 3 2 ø÷

25 q =or q = 10 2

Since the number of units produced would not be negative, the equilibrium point occurs when q = 10.

2.5 0

-2(2.5 ) 25(2.5 ) + 75(2.5) 3 2 » 98.95833 » 99, 000 The net savings are about $99,000.

584

Chapter 7 INTEGRATION (a) Total amount »

87.

1 (2.002) + 2.068 2 + 2.386 + 2.735

The total number of additional spiders in the first ten months is

+ 3.207 + 3.445 + 3.235 + 3.413 + 4.012 +

1 (4.465) 2

= 27.7345

The estimate is 27.7345 billion barrels. (b) The left endpoint sum is 26.503 and the right endpoint sum is 28.966. Their average is 26.503 + 28.966 = 27.7345. 2 The estimate is 27.7345 billion barrels. (d) The line of best fit has the equation y = 0.2585t – 0.6511

(0.2585t - 0.6511) dt » 27.87

The integral yields an estimate of 27.87 billion barrels. 88.

ò (100 - t 0.4t + 1) dt, 2

where t is the time in months.

100dt -

19 = e

ò t 0.4t + 1 dt.

Let u = 0.4t + 1 , so that 1 du = t dt. du = 0.8t dt and 0.8 When t = 10, u = 41. When t = 0, u = 1.

100 dt -

0 10

41 1 u1/2du 0.8 1

= 100t 0

3T = ln19

0 T

The total number of infected people over the first

If t = 4, u = 17, so 4

100t

ò t + 1 dt 0

= 50

æ 1 ö = 15 çç 20t - e3t ÷÷÷ çè 3 ø 0

1 du = 50 ln | u | u 1

= 50 ln 17 - 50 ln |1| = 50 ln 17 » 141.66.

æ 1 1ö = 15 çç 20T - e3T + ÷÷÷ çè 3 3ø æ 20 19 1ö = 15 çç ln19 + ÷÷÷ çè 3 3 3ø » 204. The correct choice is (c).

100t dt , where t is time in 2 0 t +1

du = 2t dt and 50 du = 100t dt.

» 15(13.63)

Let u = t 2 + 1, so that

ò (20 - e ) dt

= 15

5 3/2 u 6 1

months.

T =

ò [19 - S (t)] dt = 15ò [19 - (e - 1)] dt

= 1000 -

The total number of additional spiders in the first 10 months is about 782.

four months is

» 782

90.

5 u 3/2 - ⋅ 3 4 2

1 ln19 3 Therefore, the inventory carrying cost from 0 to T week is

0 2

The inventory must be replenished when 19 - S (T ) = 1.

19 - (e3T - 1) = 1

f (t ) = 100 - t 0.4t 2 + 1

89.

Approximately 142 people are infected. (a) The total area is the area of the triangle on [0, 12] with height 0.024 plus the area of the rectangle on [12, 17.6] with height 0.024. 1 A = (12 - 0) (0.024) + (17.6 - 12) (0.024) 2 = 0.144 + 0.1344 = 0.2784

91.

Chapter 7 Review

585

-0 = 0.002 and y-intercept 0. (b) On [0, 12] we define the function f(x) with slope 0.024 12-0

f ( x) = 0.002 x

On [12, 17.6], define g (x) as the constant value. g ( x) = 0.024. The area is the sum of the integrals of these two functions. A=

0.002 xdx +

17.6

0.024dx

= 0.001x 2

17.6

+ 0.024 x 12

= 0.001(122 - 02 ) + 0.024 (17.6 - 12) = 0.144 + 0.1344 = 0.2784

92.

Since answers are found by estimating values on the graph, exact answers may vary slightly; how-ever when rounded to the nearest hundred, all answers should be the same. Sample solution: (a) Left endpoints: Read the values of the function from the graph, using the open circles for the functional values. The values of f (x) are listed in the table. x 0 f (x) 30

2 50

5 15 30 60 105 85

45 60 70 55

The values give the heights of 6 rectangles. The width of each rectangle is found by subtracting subsequent values of x. We estimate the area under the curve as 6

f ( xi )  xi = 30(2) + 50(3) + 60(10) å i =1 + 105(15) + 85(15) + 70(15) = 4710.

Right endpoints: We estimate the area under the curve as 6

å f ( x )  x = 50(2) + 60(3) + 105(10) i

i =1

+ 85(15) + 70(15) + 55(15) = 4480.

Average: 4710 + 4480 = 4595 » 4600 pM. 2 (b) Read the values of the function from the graph, using the closed circles for the functional values. The values of x and g(x) are listed in the table.

g (x)

The values give the heights of 6 rectangles. The width of each rectangle is found by subtracting subsequent values of x. We estimate the area under the curve as 6

å g( x )  x = 20(2) + 42(3) + 42(10) + 70(15) + 52(15) + 40(15) = 3016. i

i =1

586

Chapter 7 INTEGRATION Right endpoints: We estimate the area under the curve as 6

å g( x )  x = 42(2) + 42(3) + 70(10) + 52(15) + 40(15) + 20(15) = 2590. i

i =1

Average: 3016 + 2590 = 2803 » 2800 pM. 2 (c)

4600 - 2800 » 0.6428 2800

The area under the curve is about 64% more for the fasting sheep. 93.

(a)

321

1.87t1.49e-0.189(ln t ) dt

Trapezoidal rule: n = 8, b = 321, a = 1, f (t ) = 1.87t1.49e-0.189(ln t ) i

0 1 2 3 4 5 6 7 8

xi 1 41 81 121 161 161 201 281 321

f (ti ) 1.87 34.9086 33.9149 30.7147 27.5809 24.8344 22.4794 20.4622 18.7255

é1 ù ê (1.87) + 34.9086 + 33.9149 + 30.7147 + 27.5809 ú 321 1 2 ê ú Total amount » ê ú 1 8 ê + 24.8344 + 22.4794 + 20.4622 + (18.7255) ú 2 ëê ûú » 8208

The total milk production from t = 1 to t = 321 is approximately 8208 kg. (b) Simpson’s rule: n = 8, b = 321, a = 1, f (t ) = 1.87t1.49e-0.189(ln t ) i 0 1 2 3 4 5 6 7 8

f (ti )

1.87

34.9086

33.9149

121

30.7147

161

27.5809

201

24.8344

241

22.4794

281

20.4622

321

18.7255

Chapter 7 Review

587

321 - 1 éê 1.87 + 4 (34.9086) + 2 (33.9149) + 4 (30.7147) + 2 (27.5809) ùú ú 8(3) êë + 4(24.8344) + 2(22.4794) + 4(20.4622) + 18.7255 û » 8430 The total milk production from t = 1 to t = 321 is approximately 8430 kg. Total amount »

94. (a)

Total amount

321

1.87t1.49e-0.189(ln t ) dt » 8558, or 8558 kg

1 (259,889) + 252,876 2 + 241,886 + 233, 447 + 224, 078 + 233,941 + 256,104 + 292, 672 +

1 (289, 279) 2

= 2, 009,588 The estimate of the total amount of property damage is $2,009,588.

(b) The left endpoint sum is 1,994,893 and the right endpoint sum is 2,024,283. Their average is 1,994,893 + 2, 024, 283 = 2, 009,588 2 The estimate of the total amount of property damage is $2,009,588. (d) The line of best fit has the equation y = 4431t + 196,190. 17

ò 9 (4431t + 196,190) dt » 2, 030, 000

The integral yields an estimated total property damage of $2,030,000. 95.

v (t ) = t 2 - 2t t

s (t ) =

ò0 (t - 2t) dt

s (t ) =

t3 - t 2 + s0 3

If t = 3, s = 8. 8 = 9 - 9 + s0 8 = s0

Thus, s(t ) =

t3 - t 2 + 8. 3

588

Chapter 7 INTEGRATION For t = 50,

Extended Application: Estimating Depletion Dates for Minerals 1.

k (t ) =

3, 000, 000 ¸ 16,900 » 177

This gives a growth rate of 1% for 2020.

The reserve would last about 177 yr. 3, 000, 000 =

(

16, 900 0.02T1 e -1 0.02

(b) Use the form of the function k (t ) = t +a b ,

where a and b are both constants. Since a k (0) = 0.03, k (t ) = , where t+b a a 0.03 = = . Or a = 0.03b. 0+b b

)

3, 000, 000 (0.02) = e0.02T1 - 1 16, 900 3.55 + 1 = e0.02T1

Also, since k (50) = 0.02, a 0.02 = . = Or a = 0.02(50 + b). 50 + b

4.55 + 1 = e0.02T1 ln 4.55 = 0.02T1 ln 4.55 0.02 T1 » 75.8 The reserves would last about 75.8 yr. T1 =

15, 000, 000 =

(

63, 000 0.06T 1 e -1 0.06

Solve: 0.03b = 0.02(50 + b) 0.03b = 1 + 0.02b 0.01b = 1 b = 100 Find a using substitution. a = 0.03b a = 0.03(100) a =3

)

15, 000, 000(0.06) + 1 = e0.06T 1 63, 000 15.286 » e0.06T 1

The function that satisfies these conditions is 3 k (t ) = . t + 100

ln 15.286 » 0.06T1 ln 15.286 0.06 = 45.4 The depletion time for bauxite is about 45.4 yr. T1 »

6. 2, 000, 000 =

(

2200 0.04T1 e -1 0.04

)

2, 000, 000(0.04) + 1 = e0.04T1 2200 37.36 = e0.04T1 ln 37.36 = 0.04T1

The depletion time for bituminous coal is about 90.5 yr. k (t ) =

1 t + 50

(a) For t = 0, k (0) =

(a) The function from 5(b) is k (t ) = t +3100 .

The instantaneous growth rate at time t is 3t

then 16,900e k (t )⋅t = 16,900e t +100 . According to this model, world petroleum consumption from 1970 to time T is represented by the integral

T1 » 90.5

1 = 0.01. 50 + 50

1 = 0.02. 0 + 50

3t + t 16,900e 100 dt.

(b) Trying various values of T and using a numerical integrator, we find that according to this model the current reserve of 3,000,000 million barrels will be exhausted in about 77 years from 1970, that is, in about 2047.

This gives a growth rate of 2% for 1970.

Chapter 8

FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION 8.1

Your Turn 3

Integration by Parts

Your Turn 1

ò (3x + 4) e dx

ò xe

Choose 3x 2 + 4 as the part to be differentiated and

-2x

dx -2 x

u = x and dv = e

Let

Then du = dx and v =

e2x as the part to be integrated. dx.

e-2 x dx

3x 2 + 4

e2 x

1 = - e-2 x . 2

ò u dv = uv - ò v du ò xe

-2 x

1 2x e 2

6x -

æ 1 -2 x ÷ö çç - e ÷÷ dx çè 2 ø

1 dx = - xe-2 x 2 1 1 = - xe-2 x + e-2 x dx 2 2 1 1æ 1ö = - xe-2 x + çç - ÷÷÷ e-2 x + C 2 2 çè 2 ø

1 1 = - xe-2 x - e-2 x + C 2 4 1 -2 x = - e (2 x + 1) + C 4

1 2x e 4

1 2x e 8

ò (3x + 4)e dx 2

æ1 ö æ1 ö æ1 ö = (3x 2 + 4) çç e 2 x ÷÷÷ - (6 x) çç e2 x ÷÷÷ + çç e2 x ÷÷÷ + C èç 2 ø èç 4 ø èç 8 ø 1 2x e [4(3x 2 + 4) - 2(6 x) + 6] + C 8 1 = e2 x (12 x 2 + 16 - 12 x + 6) + C 8 1 = e 2 x (6 x 2 - 6 x + 11) + C 4 =

Your Turn 2

ò ln 2x dx u = ln 2 x and dv = dx. 1 1 Then du = (2) = and v = 2x x Let

ò dx = x.

ò u dv = uv - ò v du

ò x ln x dx 2

ò ln 2x dx = (ln 2x)(x) - ò x ⋅ x dx = x ln 2 x - dx ò = x ln 2 x - x + C

Your Turn 4

Let

u = ln x and dv = x 2dx.

Then du =

1 dx and v = x

x 2dx =

x3 . 3

ò u dv = uv - ò v du

589

590

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION

x 2 ln x dx = =

x3 ln x 3

x3 1 ln x 3 3

æ x3 1 ö÷ çç ⋅ ÷÷÷ dx çç è 3 x ø÷

W3.

æ 5 ö çç 3x - 2 ÷÷ dx ÷ 3 èç x ø =

x 2dx

æ1 ö æ 1 ö = 3çç x 6 ÷÷÷ - 2 çç - x-2 ÷÷÷ + C èç 6 ø èç 2 ø

x3 1 æç x3 ÷ö = ln x - çç ÷÷÷ + C 3 3 çè 3 ÷ø =

x ç 1ö ç ln x - ÷÷÷ + C ç 3è 3ø e

W4.

æ 2x

5 ö

ò çççè e + e ÷÷÷ø dx = ò ( e + 5e ) dx x

e ç 1 ö 1æ 1ö = ç1 - ÷÷÷ - ççç 0 - ÷÷÷ 3 èç 3 ø 3è 3ø 2 1 = e3 + 9 9 2e3 + 1 9

Your Turn 5

(

Use formula 10 from the table of integrals with a = 2.

1 2x e + 5 -e-x + C 2

e2 x 5 - x +C 2 e

W5. Let u = x 5; then du = 5x 4 . 5 5 1 x 4e x dx = 5x 4e x dx 5 1 1 eu du = eu + C = 5 5

ò 5

ex +C 5

ò x 4 + x dx = ò x 2 + x dx 2

)

ò x 4 + x2 dx 1

-x

3æ

x6 1 + 2 +C 2 x

3æ

x3 æç 1ö x ln x dx = ç ln x - ÷÷÷ ç 3è 3ø 1 e

ò ( 3x5 - 2x-3 ) dx

1 2 + 4 + x2 = - ln 2 x

W6.

ò0 ( 6x2 - 5 + e x ) dx

8.1 Warmup

(

)

é ù = ê 54 - 15 + e3 - ( 0 - 0 + 1) ú ë û

( )

W1. ln(3x ) + ln x 2 = ln 3 + ln x + 2ln x = ln 3 + 3ln x

( ) = ln 3 + ln x + 2ln x

ln(3x ) + ln x

= ln 3 + 3ln x

W2.

æ 5 ö çç 3x - 2 ÷÷ dx 3 ÷ø çè x =

ò ( 3x5 - 2x-3 ) dx

= 38 + e3 » 58.09

8.1 Exercises 1.

True

False. Integration by parts can be used in a definite integral.

True

æ1 ö æ 1 ö = 3çç x 6 ÷÷÷ - 2 çç - x-2 ÷÷÷ + C çè 6 ø çè 2 ø =

(

= 2 x 3 - 5x + e x

x6 1 + 2 +C 2 x

Section 8.1

591

ò xe xdx

Let

dv = e x dx and u = x.

Then v =

ò (6x + 3) e

-2 x

Let dv = e-2 x dx and u = 6 x + 3

ò e dx and du = dx. x

v =e

Then v =

v =

Use the formula

ò u dv = uv - ò v du. x

e-2 x +C -2

ò e dx and du = dx.

v = ex

Use the formula

u dv = uv -

v du.

( x + 6)e x dx = ( x + 6)e x -

ò x ln dx Let

dv = x dx and u = ln x.

Then

v = x2

and du = 1x dx.

x ln dx =

x2 ln x 2

e x dx

= xe x + 6e x - e x + C

= ( x + 5)e x + C

10.

(4 x - 12) e-8 x dx

òe

-8 x

Then v =

(4 x - 12)e

dx and du = 4 dx.

x4 4

and du =

x3 ln x dx =

x4 ⋅ ln x 4

x 4 ln x 1 4 4

ò x dx

x 4 ln x x4 +C 4 16

æ e-8 x ö÷ ç ÷÷ = (4 x - 12) çç çè -8 ÷ø÷

æ e-8 x ö÷ çç ÷÷ ⋅ 4 dx çç ÷ è -8 ø÷

4 x -8 x 12 -8 x æç 4 e-8 x ö÷÷ e e + - çç - ⋅ ÷÷ + C 8 8 èç 8 -8 ø÷

x 3 1 -8 x e = - e-8 x + e-8 x +C 2 2 16 æ x 23 ö÷ -8 x = çç - + +C ÷e çè 2 16 ø÷

x 2 ln x x2 +C 2 4

ò x ln x dx Then v =

e-8 x v = -8 -8 x

ò 2 dx

Let dv = x3dx and u = ln x.

Let dv = e-8 x dx and u = 4 x - 12

6e-2 x dx -2

Then v =

1 3e-2 x = - (6 x + 3)e-2 x + +C 2 -2 1 3 = - (6 x + 3) e-2 x - e-2 x + C 2 2

Let dv = e x dx and u = x + 6.

dx and du = 6 dx.

(6 x + 3)e-2 x -2

-2 x

ò (x + 6)e dx

òe

ò (6x + 3) e

ò xe dx = xe - ò e dx = xe - e + C x

11.

2x + 1

ò (2x + 1) e-x dx 0

1 dx. x x 4 æç 1 ö÷ ç ÷ dx 4 çè x ÷ø 3

592

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION dv = e-x dx

Let

Then v =

and u = 2 x + 1.

e-x dx and du = 2 dx.

= (9 ln 27 - 9) - (ln 3 - 1)

v = -e-x

= 9 ln 33 - 9 - ln 3 + 1

2x + 1

= 27 ln 3 - ln 3 - 8

= 26 ln 3 - 8 » 20.56

= - (2 x + 1)e-x + -x

= - (2 x + 1)e 1

ln 3x dx = ( x ln 3x - x)

2x + 1

ò0 e x

ò 2e dx -x

-x

- 2e

ò ln 5x dx 1

Let dv = dx and u = ln 5 x.

= [- (2 x + 1)e-x - 2e-x ] -1

= [- (3)e

-1

= -5e

14.

-1

- 2e

1 dx. x

Then v = x and du =

æ1

] - (-1 - 2)

» 1.161

ln 5 x dx = ( x ln 5 x - x)

12.

3- x 3e

dx =

1 3

= 2 ln 10 - 2 - ln 5 + 1

ò (3 - x) e dx -x

= ln

Let dv = e-x dx and u = 3 - x. Then v = -e-x 1 3

and du = -dx.

15.

-x

(3 - x) e-x dx =

ò (x - 2) e dx. x

dv = e x dx and u = x - 2.

Let

The area is

102 - 1 = ln 20 - 1 » 1.996 5

ò (3 - x) e dx ù 1é = ê -(3 - x) e-x e-x dx ú 3 ëê ûú 1 = [-(3 - x) e-x + e-x dx] 3 1 = ( x - 2) e-x 3

ò ln 5x dx = x ln 5x - ò x çççè x dx ÷÷÷ø = x ln 5x - x

Then v = e x

ò (x - 2) e dx = (x - 2)e - ò e dx x

ln 3x dx

16.

ò ( x + 1) ln x dx ò

x ln x dx +

ò ln x dx 1

For the first integral,

1 Then v = x and du = dx. x

1 x2 dx and u = . x 2 Then v = ln x and du = x dx. let dv =

ò ln 3x dx = x ln 3x - ò dx

= e 4 + e2 » 61.99

Let dv = dx and u = ln 3x.

= (2e 4 - e 4 ) - (0 - e2 )

13.

ò1 (x - 2) e dx = [(x - 2)e - e ] 2

1 ( x - 2)e-x 3 0 e-3 + 2 » 0.6833 3

and du = dx.

= x ln 3x - x

Section 8.1

593

ò ( x + 1) ln x dx

19.

x2 = ln x 2

e2 x2 = 2 4 =

Let u = x 2 and dυ = ( x + 4)1/2. Use column integration.

x dx 2 1

ò x x + 4 dx D

e 1

1 e2 e2 + 2 4 4

ò ln x dx = ( x ln x - x) 1

e 1

ò x e dx 2 2x

Let u = x 2 and dv = e2 x dx. Use column integration.

20.

( 23 )( 25 ) ( x + 4)5/2 ( 23 )( 25 )( 27 )( x + 4)7/2

2 2 8 x ( x + 4)3/2 x( x + 4)5/2 3 15 16 + ( x + 4)7/2 + C 105

ò (2x - 1) ln (3x) dx Let dv = (2 x - 1) dx

e2 x 2

Then v = x 2 - x and

e2 x 4

ò (2x - 1) ln(3x)dx

ò 2 x3 + 1

1 6

= ( x 2 - x) ln 3x -

21.

æ1

x2 +x+C 2

Let dv = (8 x + 10) dx and u = ln (5 x).

6 x 2dx

ò 2 x3 + 1

Then v = 4 x 2 + 10 x and du =

Then du = 6 x 2 dx. 1

1 dx. x

ò (8x + 10) ln (5x) dx

u = 2 x3 + 1.

x 2dx

du =

x 2e 2 x xe2 x e2 x + +C 2 2 4

and u = ln 3x.

ò (x - x)çççè x dx ÷÷÷ø = ( x - x) ln 3x ò (x - 1) dx = ( x 2 - x) ln 3x -

e2 x 8

2x æ e2 x ö÷ æ 2x ö ç ÷÷ - 2 x çç e ÷÷÷ + 2e + C x 2e2 x dx = x 2 çç ç ÷ 8 çè 2 ÷÷ø èç 4 ø÷

Let

2 ( x + 4)3/2 3

e2 x

x 2dx

18.

æ2ö æ 2 öæ 2 ö = x 2 ( x + 4)3/2 çç ÷÷ - 2 x( x + 4)5/2 çç ÷÷ çç ÷÷ çè 3 ÷ø çè 3 ÷øèç 5 ÷ø æ 2 öæ 2 öæ 2 ö + 2( x + 4)7/2 çç ÷÷ çç ÷÷ çç ÷÷ + C ÷ ç 7 ø÷ çè 3 ÷øèç 5 øè

1 e2 e2 e2 + 5 + +1 = 2 4 4 4 » 3.097.

( x + 4)1/2

Thus the original integral is equal to

ò x x + 4 dx

17.

Using the result of Example 2, e

ò 2 x + 1 = 6 ln |2 x + 1| + C 3

1 dx. x

594

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION

ò (8x + 10) ln(5x) dx

= (4 x + 10 x) ln(5 x) -

ö e2 x æç 1 2 ç - x + x ÷÷÷ ç ø1 2 è2

e4 æç 3 ö÷ e2 æç 1 ö÷ ç- ÷ ç ÷ 2 èç 2 ø÷ 2 èç 2 ø÷

= (4 x 2 + 10 x) ln(5 x) - 2 x 2 - 10 x + C

24.

22.

ò 2x + 1 3

u = 2 x3 + 1.

Let

and

du = 4 x3 dx. 1

Then du = 6 x 2 dx.

ò x3 e x dx = 4 ò 4x3 e x dx 1 = eu du 4ò 4

x 2dx

ò0 2x + 1 3

1 u e +C 4 1 4 = ex + C 4 =

25.

3+ x

Then v =

-2 0

1ö 1 æç 3 çç ln |2 x + 1| ÷÷÷÷ 0ø 6è

ò x (3x + x )

2(3 + x 2 )1 2 2 2 1/2

v = (3 + x )

x3dx

e 2

e2 x 4 e2 x 8

ò (1 - x2 ) e2xdx (1 - x 2 ) e2 x (-2 x) e2 x (-2) e2 x + 2 4 8

xe2 x e2 x (1 - x 2 ) e2 x + 2 2 4 2x æ ö ö e ç e2 x çæ 1 1 2 2 = çç1 - x + x - ÷÷÷ = çç - x + x ÷÷÷ ø 2 è 2ø 2 è2

ò 3+ x

and

du = 2 x dx.

= x 2 (3 + x 2 )1/2 -

ò 2x(3 + x ) dx

= x 2 (3 + x 2 )1/2 -

2 (3 + x 2 )3/2 3

x3dx

3 + x2

2 -1/2

I e

ò0 2 x3 + 1

dv = x(3 + x 2 )-1/2 dx and u = x 2.

Use column integration.

-2 x

x3dx

Let

1- x

Let u = 1 - x 2 and dv = e2 x dx.

6 x 2dx

1 6

1 (ln 3) 6 » 0.1831

ò1 (1 - x ) e dx

23.

e2 (3e2 + 1) » -42.80 4

x 2 dx

x3 ex 4 dx

Let u = x 4

(1 - x 2 ) e2 x dx =

æ1ö (4 x 2 + 10 x) çç ÷÷÷ dx çè x ø

ò = (4 x + 10 x) ln(5 x) ò (4x + 10) dx 2

2 1/2

é ù 2 = ê x 2 (3 + x 2 )1/2 - (3 + x 2 )3/2 ú 3 ëê ûú 2 2 = 41/2 - (43/2 ) - 0 + (33/2 ) 3 3 2 2 3/2 = 2 - (8) + (3 ) 3 3 10 =- +2 3 3 » 0.1308

1 0

Section 8.1 26.

595

æ ö çç 1 11 + 121 - x 2 ÷÷÷ = 3çç ln ÷÷ + C x çç 11 ÷ø÷ è

x 3 x 2 + 2 dx

ò x(x + 2) dx 2

1/2

1 2

ò 2x(x + 2) dx 2

1/3

Let u = x 2 + 2. Then du = 2 x dx.

30.

If x = 5, u = 27. If x = 0, u = 2. 1 = 2

ò u du 2

31.

4/3

243 3(2 ) 8 8

-6

ò x(4x + 6) dx 2

-6

ò x(4x + 6) dx 2

ò x + 16 dx

= -6

Use formula 5 from the table of integrals with a = 4. 16

= 16ln x +

x 2 + 16 + C

32.

ò x - 25 dx

ò x + 15 dx 2

Use formula 15 from table of integrals with a = 15.

Use formula 8 from the table of integrals with a = 5. 1

ò x2 - 25 dx = 10ò x2 - 52 dx é 1 x - 5 ùú = 10 ê ⋅ ln +C ê 2(5) x + 5 úû ë x-5 = ln +C x+5 3

ò x(4x + 6) dx

é ù 1 1 x ú+C = -6 ê + 2 ln ê 6(4 x + 6) 4 x + 6 úû 6 ë -1 1 x = - ln +C (4 x + 6) 6 4x + 6

ò x2 + 16 dx = 16ò x2 + 42 dx

29.

2 x ln +C 15 3x - 5

Use formula 14 from the table of integrals with a = 4 and b = 6.

2 æç 1 x ÷÷ö + C ç - ln ÷ 3 èç 5 3x - 5 ÷ø

243 33 2 = » 29.43 8 4

28.

Use formula 13 from table with a = 3, b = -5.

3 3 = (27 4/3 ) - (2) 4/3 8 8 =

ò 3x(3x - 5) dx = 3 ò x(3x - 5) dx

1/3

1 æç u 4/3 ö÷÷çæ 3 ö÷ = çç ÷ç ÷ 2 çè 1 ÷÷øèç 4 ÷ø

27.

3 11 + 121 - x 2 +C ln x 11

ò x 121 - x2 dx = 3ò x 112 - x2

ò x + 15 dx = ò x + ( 15) dx 2

x 2 x + ( 15)2 2

( 15) 2 ln x + x 2 + ( 15)2 + C 2 x 2 15 ln x + x 2 + 15 + C = x + 15 + 2 2 +

596 35.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION First find the indefinite integral using integration by parts.

ò x ⋅ ln | x | dx n

é1 ù 1 1 ê ⋅ x n +1 ln | x | x n +1 ú dx êë x n + 1 úû n +1 1 1 n x n +1 ln | x | x dx = n +1 n +1 1 1 x n +1 ln | x | x n +1 + C = n +1 (n + 1)2

ò u dv = uv - ò v du Now substitute the given values. 1

u dv = uv

ò v du 0

= [u(1) v (1) - u (0) v (0)] - 4

é ln | x | ù 1 ú+C = x n +1 êê 2ú êë n + 1 (n + 1) úû

= (3)(-4) - (2)(1) - 4 = -18

36.

First find the indefinite integral using integration by parts.

ò x e dx, a ¹ 0 n ax

40.

ò u dv = uv - ò v du

Let dv = eax dx and u = x n ; then v = 1a eax

Now substitute the given values.

ò1

u dv = uv

ò1

ò ò

v du

x neax dx =

= [u (20) v (20) - u (1) v (1)] - (-1) = (15)(6) - (5)(-2) + 1 = 101

37.

ò r ds = rs - ò s dr ò0

r ds = rs 0 -

41. 2

ò u dv = uv - ò v du 3

Let u = ln | x | and dv = x n dx.

ln | x |

n a

ò x e dx + C n -1 ax

ò x x + 1 dx D

x + 1 dx. I

x +1 æ 2 ö÷ çç ÷ ( x + 1)3/2 çè 3 ÷ø æ 4ö 5/2 ççç ÷÷÷ ( x + 1) è 15 ø

ò x x + 1 dx (b) Let u = x + 1; then u - 1 = x and du = dx.

Use column integration.

æ 1 ax ö çç e ⋅ nx n-1 ÷÷ dx ÷ çè a ø

æ2ö æ 4ö = çç ÷÷÷ x( x + 1)3/2 - çç ÷÷÷ ( x + 1)5/2 + C èç 3 ø èç 15 ø

x n ⋅ ln | x | dx, n ¹ -1

1 x

x e a

15 = u(3)v(3) - u(1)v(1) - 20 -35 = u(1)v(1)

n ax

ò1 u dv = uv 1 - ò1 v du

39.

x neax a

(a) Let u = x and dv = Use column integration.

ò 0 s dr

10 = r(2) s(2) - r(0)s(0) - 5 15 = r( s)s(2)

38.

du = nx n -1dx.

and

1 x n +1 n +1

Section 8.1

597

ò x x + 1 dx = ò (u - 1)u1/2du = ò (u3/2 - u1/2 ) du 2 2 = u 5/2 - u 3/2 + C 5 3 2 2 = ( x + 1)5/2 - ( x + 1)3/2 + C 5 3

equivalent. 42.

The integration constant is missing.

43.

R =

ò [D(q) - p ] dq 0

e4 -2

[10 - ln(q + 2) - 6] dq

e4 -2

[4 - ln(q + 2)] dq

e4 -2 = [ 4q - (q + 2) ln(q + 2) ] + 0

= [ 5q - (q + 2) ln(q + 2) ]

e4 -2 0

= éêë 5(e4 - 2) - (e 4 - 2 + 2) ln(e4 - 2 + 2) ùúû

( x + 1) ln ( x + 1) dx

- [ 0 - 2 ln 2 ] 4

Let u = ln ( x + 1) and dv = ( x + 1) dx.

= 5e - 10 - e 4 ln e4 + 2 ln 2

1 dx and v = 1 ( x + 1) 2. Then du = x + 1 2

= 5e 4 - 10 - 4e4 + 2 ln 2 = e4 + 2 ln 2 - 10

ò (x + 1) ln (x + 1) dx 1 = ( x + 1) 2 ln ( x + 1) 2 é1 1 ù ê ( x + 1) 2 ⋅ ú dx êë 2 x + 1 úû 1 1 ( x + 1) dx = ( x + 1) 2 ln ( x + 1) 2 2 1 1 = ( x + 1) 2 ln ( x + 1) - ( x + 1) 2 + C 2 4

D(q) = (q + 20)e-0.1q , q0 = 10 p0 = D(q0 )

45.

= (q0 + 20)e-0.1q0

= (10 + 20)e-0.1(10)

ò ( x + 1) ln (x + 1) dx 0

é1 ù 1 = ê ( x + 1)2 ln ( x + 1) - ( x + 1)2 ú êë 2 úû 4

44.

D(q) = 10 - ln(q + 2), q0 = e4 - 2 p0 = D(q0 ) = 10 - ln(q0 + 2) = 10 - ln(e 4 - 2 + 2)

= 30e-1 = 30/e

Find the consumers’ surplus. q0

ò0 [D(q) - p0 ] dq 10

ò0 éêë (q + 20)e-0.1q - 30e-1 ùúû dq

Use column integration. 10

169 ln 13 - 42 » $174.74 2

= 10 - ln e 4 = 10 - 4

q+2 dq q+2

e4 -2 = [ 4q - (q + 2) ln(q + 2) + q ] 0

e4 - 2

é ù æ -0.1q ö÷ e-0.1q ÷= êê (q + 20) ççç -e - 30 q úú ÷ è 0.1 ø 0.01 e û0 ë é ù æ -0.1(10) ö÷ e-0.1(10) 30 (10) ú ÷= êê (10 + 20) ççç -e ú ÷ø è 0.1 0.01 e ë û é ù æ -0.1(0) ÷ö e-0.1(0) ÷- êê (0 + 20) ççç -e - 30 (0) úú è 0.1 ÷ø 0.01 e ë û æ ö = 30 çç -10 ÷÷÷ - 100 - 300 - [ 20(-10) - 100 - 0 ] è e ø e e = - 700 + 300 e

Find the consumers’ surplus.

598

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION First, evaluate A.

S (q) = qe0.2q + 10, q0 = 5 p0 = S (q0 )

46.

ò [ 20 - 4 ln 8 - 5q ]dq

= q0e0.2q0 + 10

= 5e + 10

Find the producers’ surplus.

ò0 [ p0 - S (q)] dq ò0

ò0

= 80 - 16 ln 8 - 40 = 40 - 16 ln 8

é 5e + 10 - (qe0.2q + 10) ù dq êë úû

Next integrate B by parts. 4

é 5e - qe0.2q ù dq êë úû

ò q ln(12 - q) dq 0

Use column integration. é 0.2q ù qe0.2q ú = êê 5eq + e 0.2 0.04 úû ë

Let 5

Then v =

ò q ln(12 - q) dq 0

S (q) = 5q - q ln(12 - q) + 6, q0 = 4

é q2 ù = êê ln(12 - q) úú ë 2 û

Find the producers’ surplus. q0

ò0 [ p0 - S (q)] dq

é q2 ù = êê ln(12 - q) úú ë 2 û

ò0 [ 26 - 4 ln 8 - {5q - q ln(12 - q) + 6} ]dq 4

ò0 [ 20 - 4 ln 8 - 5q + q ln(12 - q) ]dq ò0

q2 dq 0 q - 12

q2 = q + 12 + 144 q - 12 q - 12

= 26 - 4 ln 8

[ 20 - 4 ln 8 - 5q ] dq +

-1 2 0

q2 is improper, we use 12 - q long division to rewrite the expression as

= 5(4) - 4 ln(12 - 4) + 6

Since the fraction

= 5q0 - q0 ln(12 - q0 ) + 6

-1 dt. 12 - q 1 du = dt. q - 12

and du =

p0 = S (q0 )

q2 2

= éêë 5e(5) - 5(5)e0.2(5) + 25e0.2(5) ùúû - éë 0 - 0 + 25e0.2(0) ùû = 25e - 25

dv = q dq and u = ln(12 - q).

= éëê 5eq - 5qe0.2q + 25e0.2q ùûú

47.

é 5(4)2 ùú [ = êê 20(4) - 4(4) ln 8 - 0 - 0 - 0] 2 úû ë

é 5q 2 ùú = êê 20q - 4q ln 8 2 úû ë

= 5e0.2(5) + 10

ò0 q ln(12 - q) dq

A B We look at integrals A and B separately.

-1 2 0

144 ù

ò êëê q + 12 + q - 12 úûú dq 0

é q2 ù = êê ln(12 - q) - 1 q 2 - 6q - 72 ln q - 12 úú 4 ë 2 û

4 0

= 8ln 8 - 4 - 24 - 72 ln 8 - [0 - 0 - 0 - 72 ln12] = -64 ln 8 - 28 + 72 ln12

Thus, A + B q0

ò0 [ p0 - S (q)] dq =

ò0

[ 20 - 4 ln 8 - 5q ] dq +

ò0 q ln(12 - q) dq

= [40 - 16 ln 8] + [-64 ln 8 - 28 + 72 ln12] = 12 - 80 ln 8 + 72 ln 2

Section 8.1 48.

599

r (t ) =

ò1

2t 2e-t dt

dv = e-t dt

Let

Then v = -e-t

50. and u = 2t 2.

Then du =

Use column integration. +

e-t

-e-t

e-t

2t e dt

51.

-t 2

= -2e (t + 2t + 2) 6

ò1 2t 2e dt = -2e (t 2 + 2t + 2) 1 -t

ò0 (-10.28 + 175.9te-t /1.3)dt = -10.28t + 175.9 te-t /1.3dt ò Evaluate this integral using integration by parts.

= -100e-6 + 10e-1 » 3.431

Let

The total reaction to the drug from t = 1 to t = 6 is about 3.431.

dv = e3t dt

Then v =

e3t 3

ò (-1.3e-t /1.3)dt

= (t )(-1.3e-t /1.3 ) -

= -1.3te-t /1.3 - 1.69e-t /1.3 + C

Substitute this expression in the earlier expression.

and u = 27t. and

du = 27dt.

-10.28t + 175.9(-1.3te-t /1.3 - 1.69e-t /1.3 ) -t /1.3

= -10.28t - 228.67te

e ⋅ 27dt 3

= (-61.68 - 1669.291e » 219.07

ò0 27 te3t dt = (9te3t - 3e3t ) 0

= 15e6 + 3 » 6054

52.

(a)

) - (-297.271)

ò ke (1 - t)dt -kt

Let u = 1 - t

and

dv = e-kt dt.

Then du = -dt

and

1 v = - e-kt k

- 297.271e

The total thermic energy is about 219 kJ.

= (18e6 - 3e6 ) - (0 - 3)

-t /1.3

-6/1.3

= 9t e3t - 3 e3t 2

v = -1.3e-t /1.3.

e 27 te dt = 27t ⋅ 3 3t

and dv = e-t /1.3dt.

ò te-t /1.3dt

ò0 27t e3 dt. Let

u =t

Then du = dt and

The total accumulated growth of the microbe population during the first 2 days is given by 2

2 3/2 4 t ln t - t 3/2 + C 3 9

16 76 ln 4 3 9 » 23.71 sq cm

= -2t 2e-t - 4te-t - 4e-t

1/2

t ln t dt = 18ln 9 -

= 2t 2 (-e-t ) - 4t (e-t ) + 4(-e-t )

49.

æ 2 3/2 1 ö

3/2

2 -t

-t

v = 23 t 3/2.

and

ò t ln t dt = 3 t ln t - ò çççè 3 t ⋅ t ÷÷÷ø dt 2 2 = t ln t ò 3 t dt 3

-e-t

1 dt t

t dt = t1/2 dt.

dv =

and

2 3/2

D 2t2

ò4 t ln t dt

Let u = ln t

du = 4t dt.

and

6 0

600

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION 1

ò0

æ6 t 1 ö÷ -e-kt çç - ÷ çè 5 5 5k ÷ø 1

ke-kt (1 - t ) dt

é æ 1 ö = k ê (1 - t ) çç - e-kt ÷÷÷ ê èç k ø ë é 1 1 = k ê - e-kt (1 - t ) êë k k = -e-kt (1 - t ) +

ù æ 1 -kt ö ççç - e ÷÷÷ (-dt ) úú è k ø û 0

ù e-kt dt ú úû 0

1 -kt e k 0

é æ 1 öù é æ 1 öù = ê -e-k çç - ÷÷ ú - ê -e0 çç1 - ÷÷ ú ÷ çè ê èç k ø úû êë k ÷ø úû ë 1 1 = 1 - + e-k k k

1 : 12 1 -t /12 (1 - t )dt

ò 12 e

= 12e-1/12 - 11 » 0.0405 1 k = : 24 1 -t /24 (1 - t ) dt

ò 24 e

1 1 -1/24 + e 1/24 1/24

k =

1 : 24

1 -t /24 6 - t dt 5 1 é 1 1 ùú -1/24 = e-6(1/24) + ê 1 e ê 5(1/24) 5(1/24) úû ë =

24 -1/4 19 -1/24 e e » 0.0933 5 5 1 : 48

1 -t /48 6 - t dt 5 1 é 1 1 ùú -1/48 = e-6(1/ 48) + ê 1 e ê 5(1/48) 5(1/48) úû ë

ò 48 e

= 24e-1/24 - 23 » 0.0205

12 -1/2 7 e - e-1/12 » 0.1676 5 5

k =

ò 24 e

1 1 -1/12 = 1+ e 1/12 1/12

= 1-

1 -t /12 6 - t dt 5 1 é 1 1 ùú -1/12 e-6(1/12) + ê 1 e = ê 5(1/12) 5(1/12) úû ë

ò 12 e

1 : 12

k =

æ 1ö = -e-kt çç1 - t - ÷÷÷ çè kø0

k =

é æ6 6 1 ö÷ ùú = ê -e-6k çç - ÷ ê èç 5 5 5k ø÷ úû ë é æ6 1 1 ö÷ ùú - ê -e-k çç - ÷ ê èç 5 5 5k ÷ø úû ë æ 1 -6 k 1 ö÷ -k = + çç1 e ÷e ç è 5k 5k ÷ø

1 : 48

48 -1/8 43 -1/48 e e 5 5 » 0.0493

1 -t /48 (1 - t )dt

ò 48 e 0

1 1 -1/48 + e 1/48 1/48

= 1-

= 48e-1/48 - 47 » 0.0103 6

(b)

-kt 6 - t

ò ke 1

The integral, easily found by comparing it to the integral in part (a), is

Section 8.2

601 1 4 -1

8.2 Volume and Average Value Your Turn 1

ö÷ 1 æç x 2 2 x ) dx = çç + x3/2 ÷÷÷ 2 çè 2 3 ÷ø

ò1 (x +

by y = x 2 + 1, y = 0, x = -1, and x = 1. The region and the solid are shown below.

8.2 Warmup Exercises 2

f ( x)  x  1

W1.

ò0 ( x + 5 ) dx 2

x 1 0

f ( x)  x  1 W2. 2 x

1 3 ( x + 5) 3 0

73 53 343 - 125 = 3 3 3 218 = 3

2

ò0 ( 6 - x2 ) dx =

ò0 ( 36 - 12x2 + x4 ) dx 3

V =

æ 1 ö = çç 36 x - 4 x 3 + x5 ÷÷÷ çè 5 ø0

ò -1 (x2 + 1)2 dx

=

æ 243 ö÷ = çç 108 - 108 + ÷ - (0) çè 5 ÷ø

ò -1(x4 + 2x2 + 1) dx

243 5

æ x5 ö÷ 2 x3 ç =  çç + + x ÷÷÷ çè 5 3 ø÷ -1

W3.

éæ 1 ö æ 1 öù 2 2 =  ê çç + + 1÷÷ - çç - - - 1÷÷ ú ÷ ÷ ê èç 5 ø çè 5 ø úû 3 3 ë æ 56 ö 56 =  çç ÷÷÷ = çè 15 ø 15

ò-2 x + 3 dx 1

2 ( x + 3)3/ 2 3 -2

2 ⋅ 43/ 2 2 ⋅ 13/ 2 3 3 16 2 14 = - = 3 3 3

Your Turn 2

Find the average value of the function f ( x) = x + on the interval [1, 4].

1æ 16 1 2ö = çç 8 + - - ÷÷÷ 3 çè 3 2 3ø 1 æ 73 ö 73 = çç ÷÷÷ = ç 3è 6 ø 18

Find the volume of the solid of revolution formed by rotating about the x-axis the region bounded

x  1

Use the formula for average value with a = 1 and b = 4.

602 W4.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION

ò2

7 7 dx = 7 ( x + 2)-1/ 2 dx x+2 2

(

f ( x) = x, y = 0, x = 0, x = 3

)2 = 14 ( 91/ 2 - 41/ 2 ) = 7 2( x + 2)1/ 2

= 14

W5.

W6.

ò0 ( e x ) dx = ò0 e2xdx e

1 0

(

1 2 = e - e0 2

V =

) 6.

e2 - 1 » 3.1945 2

 x3

ò0 x dx = 2 0 = 2

 (27) 3

- 0 = 9

f ( x) = 3x, y = 0, x = 0, x = 2

Graph f ( x) = 3x. Then show the solid of revolution formed by rotating about the x-axis the region bounded by f ( x), x = 0, and x = 2.

ò1 x ln x dx Let v = ln x and du = x dx. Then dv =

x2 1 dx and u = . x 2

ò x ln x dx = ò v du = uv ò u dv =

x 2 ln x 2

ò 2 dx

V =

ò1

2æ

1ö x ç x ln x dx = çç ln x - ÷÷÷ 2 è 2ø

(3x )2 dx = 

x 2 ln x x2 = +C 2 4 3

9 x 3 3

ò 9 x dx 2

= 3 (8 - 0) = 24 0

f ( x) = 2 x + 1, y = 0, x = 0, x = 4

æ 9 öæ 1 ö æ 1 öæ 1ö = çç ÷÷÷çç ln 3 - ÷÷÷ - çç ÷÷÷çç ln1 - ÷÷÷ çè 2 øèç ç ç 2 ø è 2 øè 2ø =

9ln 3 - 2 » 2.9438 2

8.2 Exercises 1.

True

False. The average value of a function on an interval is the integral of the function over the interval, divided by the length of the interval.

False. The average value of a function on an interval can be calculated, provided the indicated definite integral exists.

V =

ò (2x + 1) dx 2

Let u = 2 x + 1. Then du = 2 dx. If x = 4, u = 9. If x = 0, u = 1.

Section 8.2

603 V = =

4 1  2(2 x + 1)2 dx 2 0

V =

 æçç u 3 ö÷÷

÷ 2 ççè 3 ÷ø÷ 1

= 3

ò (

1x + 2 3

343 386 = 27 = 27 27

10.

f ( x) =

1 x + 4, y = 0, x = 0, x = 5 2

V =

= 2



ö2 çç 1 x + 4 ÷÷ dx ç ø÷ 0 è2

[(6)3 - 0] 3 = 72 =

)

æ1 ö =  çç x + 2 ÷÷÷ çè 3 ø

ò (x - 4) dx

 ( x - 4)

f ( x) =

ö 1 æç 1 çç x + 2 ÷÷÷ dx ø 1 3è 3

f ( x) = x - 4, y = 0, x = 4, x = 10

 æç 729

V =

= 3

1ö - ÷÷÷ çç 2è 3 3ø 728 = 6 364 = 3 =

ö çç 1 x + 2 ÷÷ dx ÷ø ç 1 è3

1 u 2du  2 1

3æ

5æ

ö 1 æç 1 çç x + 4 ÷÷÷ dx ø 0 2è 2

( 1 x + 4) = 2 2

1 x + 2, y = 0, x = 1, x = 3 3

é ù ö3 2 ê çæ 5 3ú ÷ + 4 (4) ÷ êç ú ÷ø 3 ê çè 2 úû ë 3 é ù 2 ê æç 13 ö÷ ú = ê ççè ø÷÷ - 64 ú 3 ê 2 úû ë 2 æç 2197 512 ö÷ 1685 = ÷= çç 3 è 8 8 ÷ø 12 =

604 11.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION f ( x) =

x , y = 0, x = 1, x = 4

V =

( x )2 dx = 

 x2 2

f ( x) =

2 x + 1, y = 0, x = 1, x = 4

ò x dx

ò1 ( 2x + 1)2 dx

V =

= 1



ò1 (2x + 1) dx

15 2

æ 2x2 ö÷ ç =  çç + x ÷÷÷ ÷ø çè 2 1

x + 5, y = 0, x = 0, x = 3

=  [(16 + 4) - 2]

= 8 -

12.

13.

= 18 

14.

V = =

f ( x) =

4 x + 2, y = 0, x = 0, x = 2

ò ( x + 5) dx 2

1 3

ò (x + 5) dx

ò0 ( 4x + 2)2 dx

V =

æ x2 ö÷ ç =  çç + 5 x ÷÷÷ çè 2 ÷ø 1

=

éæ 9 ö æ1 öù = ê çç + 15 ÷÷÷ - çç + 5 ÷÷÷ ú ç ê çè 2 ø è2 ø úû ë æ 39 11 ÷ö =  çç - ÷÷ çè 2 2ø = 14

ò0 (4x + 2)dx

=  (2 x 2 + 2 x)

2 0

=  (8 + 4) = 12

15.

f ( x) = e x ; y = 0, x = 0, x = 2 V =

ò0

e 2 x dx = =

 e2 x 2

e 

 2

(e 4 - 1) 2 » 84.19

Section 8.2 16.

605

f ( x) = 2e x , y = 0, x = -2, x = 1 V =

ò2 -

(2e x ) 2 dx = 

21.

f ( x) = 1 - x 2 , y = 0

ò 2 4x2 dx x

4 2 x (e ) = 2 (e2 - e-4 ) 2 -2

» 46.31

17.

2 , y = 0, x = 1, x = 3 x

f ( x) =

ö2 çç 2 ÷÷ dx V = ÷ ç 1 è x ø÷

Since f ( x) = 1 - x 2 intersects y = 0 where

x = 1,

=

3æ

1 - x2 = 0

ò1 x dx

a = -1 and b = 1.

3 = 4 ln | x | 1

V =

= 4 (ln 3 - ln1) = 4 ln 3 » 13.81

18.

=

ò-1 x + 2 dx

= 2 -

= 4 (ln 4 - ln 1) = 4 ln 4 » 17.42

22.

4 2 + 3 5

16 15

f ( x) = 2 - x 2 , y = 0

f ( x) = x 2 , y = 0, x = 1, x = 5

V =

x 4 dx =

20.

æ æ 2 1ö 2 1ö =  çç1 - + ÷÷ -  çç -1 + - ÷÷ ÷ çè ç è 3 5ø 3 5 ø÷

= 4 ln | x + 2| -1

19.

ò- (1 - 2x + x ) dx 1

2æ

=

1 1

2 2

æ 2 x3 x5 ö÷÷ ç =  çç x + ÷ çè 3 5 ÷÷ø -1

ö÷2 2 çç ÷÷ dx ç -1èç x + 2 ø÷

V =

ò- (1 - x ) dx 1

2 , y = 0, x = 1, x = 2 x+2

f ( x) =

 x5 5

= 625 1

 3124 = 5 5

x2 , y = 0, x = 0, x = 4 2

V =

=

2 4æ 2 ö

Since f ( x) = 2 - x 2 intersects y = 0 where

çç x ÷÷ ç ÷÷ dx 0 èç 2 ø÷

 20

 æç x5 ö÷ x dx = çç ÷÷÷ 4 4 çè 5 ø÷

4 4

(45 ) =

2 - x2 = 0

x =  2, a = - 2 and b =

256 5

606

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION V = =

ò- 2

25.

(2 - x 2 )2 dx

V =

ò- 2 (4 - 4x + x ) dx 2

=

æ 4 x3 x5 ö÷÷ ç =  çç 4 x + ÷ 3 5 ø÷÷ çè - 2

æ 2r 3 ö÷ ç ÷÷ = 2r 3 - çç çè 3 ø÷÷ =

26.

ò-1( 1 - x2 )2 dx

r =

=

ò a (a - x ) dx -a

x3 ö÷÷ = 2 ça x ÷ 3 ÷÷ø a çè

ò-1(1 - x2 ) dx

ù a3 ÷ö÷ ú  b ê çç 3 a ÷÷ æçç 3 a a + ÷ ÷ êç ú ç a 2 êë çè

3 ÷÷ø

æ æ 1ö 1ö =  çç1 - ÷÷ -  çç -1 + ÷÷ ÷ çè ç è 3ø 3 ø÷

 b2 æçç

2a 3 ÷÷ö ÷ 3 ÷÷ø

 b2 æçç 4a 3 ö÷÷

2 4  = 3 3

36 - x 2

27.

2 éæ

3ö

-a

æ x3 ÷÷ö ç =  çç x ÷ 3 ÷÷ø çè -1

36 = 6

V =

a é

 b 2 æçç 2

= 2 -

f ( x) =

4 r 3 3

ù b 2 ê a - x 2 ú dx úû -a êë a

V = =

24.

ò- (r2 - x2 ) dx

æ æ r 3 ö÷ r 3 ö÷÷ ç ç =  çç r 3 - ÷÷÷ -  çç -r 3 + ÷ çè 3 ø÷ 3 ø÷÷ èç

r = 1 =1

=

r r

64 2 = 15

V =

ò- ( r 2 - x2 )2 dx

æ x 3 ö÷÷ ç =  çç r 2 x ÷ 3 ÷÷ø çè -r

æ 32 ö 32 2 + 2 ÷÷ =  çç ÷ çè 15 ø 15

f ( x) = 1 - x 2

éæ ö 8 4 2 + 2 ÷÷÷ =  ê çç 4 2 ê èç ø 3 5 ë æ öù 8 4 22 ÷÷÷ ú -çç -4 2 + èç ø úû 3 5

23.

r 2 - x2

f ( x) =

2 çç

a è

÷= ç a 2 çè 3 ÷÷ø

3 ÷÷ø ú û

4ab2 3

f ( x) = r , x = 0, x = h

Graph f ( x) = r; then show the solid of revolution formed by rotating about the x-axis the region bounded by f ( x), x = 0, x = h.

ò-6 ( 36 - x2 )2 dx 6

ò-6 (36 - x2 ) dx

æ x3 ö÷÷ ç =  çç 36 x ÷ çè 3 ÷ø÷

2a 3 -

çè

-6

éæ 216 ÷ö æç 216 ö÷ ùú =  ê çç 216 ÷÷ - çç -216 + ÷ ê çè 3 ø è 3 ø÷ úû ë = 288

Section 8.2

607 h 2

31.

 r dx =  r x

f ( x) =

Average value

=  r 2h - 0

=  r 2h =

28.

f ( x) = 2 - 3x 2 ; [1, 3]

1 3 -1

1 (2 x - x3 ) 2 1

1 [(6 - 27) - (2 - 1)] 2 = -11

1 5

ò (x + 1) dx 1/2

1 2 ⋅ ( x + 1)3/2 5 3 3

32.

f ( x) = e0.1x ; [0, 10]

Average value = f ( x) = x 2 - 4; [0, 5]

1 5-0

33.

0.1x

1 ⋅ 10(e0.1x ) 10 0

f ( x) = e x /7 ; [0, 7]

Average value

ù ö 1 éê æç 125 - 20 ÷÷÷ - 0 ú ç ú ø 5 êë èç 3 û

13 » 4.333 3

1 7-0

ò e dx x /7

1 ⋅ 7e x /7 7 0

= e x /7

7 0

= e1 - e0

e - 1 » 1.718

Average value 1 13 - 1

ò e

f ( x) = (2 x - 1)1/3; [1, 13]

= e1 - 1 = e - 1 » 1.718

ò (x - 4)dx

ö÷ 1 æç x3 = çç - 4 x ÷÷÷ 5 çè 3 ÷ø

30.

1 10 - 0

Average value =

x + 1 dx

29.

2 3/2 (9 - 43/2 ) 15 2 38 = » 2.533 (27 - 8) = 15 15

ò (2 - 3x ) dx 3

1 8-3

Average value =

x + 1; [3, 8]

ò (2x - 1) dx

1 æç 1 ö÷ = ç ÷ 12 çè 2 ÷ø

1/2

34.

f ( x) = x ln x; [1, e]

ò 2(2x - 1) dx 1/2

1 2 = ⋅ (2 x - 1)3/2 24 3 1

Average value =

1 e -1

ò x ln x dx 1

Let u = ln x and dv = x dx. Use column integration. D

1 (253/2 - 1) 36 1 1 = (125) 36 36 124 31 = = » 3.444 36 9 =

ln x

1 x

1 x2 2

608

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION

ò x ln x dx

36.

æ1 1 ö 1 2 x ln x - çç ⋅ x 2 ÷÷÷ dx çè x 2 ø 2 1 1 = x 2 ln x x dx 2 2 1 1 = x 2 ln x - x 2 2 4

ò ò

1 e -1

f ( x) =

V =

4 + x2

ö2 çç 1 ÷÷ dx =  2÷ -2 çè 4 + x ÷ø 2æ

ò x ln x dx

37.

1 æç 1 2 1 2ö çç x ln x - x ÷÷÷ 4 ø1 e - 1è 2

1 æç e 2 1 ÷ö e2 çç + ÷÷÷ 4 4 ÷ø e - 1 çè 2

1 æç e 2 + 1 ö÷÷ ç ÷ e - 1 ççè 4 ÷÷ø

38.

e +1 » 1.221 4(e - 1)

Average value =

1 2-0

1 5-0

Use column integration. e2 x

1 e2 x 2 1 e2 x 4 1 e2 x 8

39.

2 2x

ò (25t - 5t + 18) dt 2

125 125 + 18 » $38.83 2 3

ò (37 + 6e

-0.03t

)dt

5e - 1 » 34.00 8

Use the formula for average value with a = 0 and b = 6. 1 6-0

ù æ1ö æ1ö æ1ö 1é = ê ( x 2 ) çç ÷÷ e2 x - (2 x) çç ÷÷ e2 x + 2 çç ÷÷ e2 x ú ú èç 2 ø÷ èç 4 ø÷ èç 8 ø÷ 2 êë û

-2 x 2

-2

ö 1 æ 625 625 = çç + 90 ÷÷÷ ø 5 çè 2 3

ò x e dx

ò e

ö÷ 1 æç 25t 2 5t 3 = çç + 18t ÷÷÷ 5 çè 2 3 ø÷

1 æç 4 1 4 1ö 4 çç 2e - e + e - ÷÷÷ 2è 4 4ø

ò [t(25 - 5t) + 18] dt

1 5

2 2x

Let u = x 2 and dυ = e2 x dx.

-1

(e-x )2dx = 

Use the formula for average value with a = 0 and b = 5. The average price is

ò x e dx

Using an integration feature on a graphing calculator to evaluate the integral, we obtain 3.758249634 » 3.758.

f ( x) = x 2e2 x ; [0, 2]

2 -2

-2

f ( x) = e-x , y = 0, x = -1, x = 1

V =

ò (4 + x ) dx

Using an integration feature on a graphing calculator to evaluate the integral, we obtain 0.5047746784 » 0.5048.

1 2-0

35.

, y = 0, x = -2, x = 2

2 0

ö 1 æç 6 e-0.03t ÷÷÷ çç 37t + ø 0 -0.03 6è

1 (37t - 200e-0.03t ) 6 0

1 [(222 - 200e-0.18 ) - (0 - 200)] 6 1 = (422 - 200e-0.18 ) 6 » 42.49 =

The average price is $42.49.

Section 8.2 40.

609

At the end of any given business day, CFFC has 400 - 80t cases of perfume on hand, where t = 1 represents Monday, t = 2 represents Tuesday, and so on. The average daily number of cases in inventory is 1 5-0

ò0

(400 - 80t )dt =

1 æç 80t 2 ö÷÷ çç 400t ÷ 5 çè 2 ø÷÷

æ -0.0000448t 5 + 0.00515t 4 ö÷ çç ÷÷ çç ÷ çè -0.1897t 3 + 2.81t 2 + 23.7t ÷ø

1 (400t - 40t 2 ) 5 0

43.

ò p(t) dt = ò 1.757(1.0248) dt t

æ 1.757 ö çç ⋅ (1.0248)t ÷÷÷ ÷ø çè ln(1.0248)

ò0 (600 - 20 30t ) dt

(b) The average corn production from 2000 to 2010 is

æ 1.757 ö çç ⋅ (1.0248)t ÷÷÷ ÷ø çè ln(1.0248)

1 = (18, 000 - 12, 000) 30 = 200 cases

42.

44.

(a)

- 0.1897t 3 + 2.81t 2 + 23.7t + C

(a) The average public debt as a percent of GDP, from 1990 to 2000 is æ -0.0000448t 5 + 0.00515t 4 ö÷ çç ÷÷ çç ÷ 3 2 èç -0.1897t + 2.81t + 23.7t ÷ø

ò0 2 rk (R2 - r 2 ) dr = 2 k

= -0.0000448t 5 + 0.00515t 4

» 38.6%

» 11.06 billion bushels.

æ -0.000224t 4 + 0.0206t 3 ö÷ çç ÷÷ dt çç ÷ çè -0.569t 2 + 5.62t + 23.7 ÷ø

20 - 10

80 - 70 » 11.0609

ò y(t) dt =

» 2.27 billion bushels.

1 æç 40 30 3/2 ö÷÷ çç 600t t ÷ ÷ø 30 çè 3

20 - 0 2.2672 »

1 æç 2 3/2 ö = çç 600t - 20 30 ⋅ t ÷÷÷ è ø0 30 3 =

1.757 ⋅ (1.0248)t + C ln(1.0248)

(a) The average corn production from 1930 to 1950 is

Use the formula for average value with a = 0 and b = 6. The average price is 1 30 - 0

» 54.3%

1 = (2000 - 1000) 5 = 200 cases

41.

38 - 20

0 5

(b) The average public debt as a percent of GDP, from 2000 to 2028 is

(b)

ò0 r(R2 - r 2 ) dr

ò0 2 rk (R2 - r2 ) dr = 2 k

ò0 (rR2 - r3)dr R

é r2R2 r 4 ùú = 2 k êê 4 úú êë 2 û 0 4 é R4 ù R = 2 k êê - 0 úú 2 4 ëê ûú 4 æ R 4 ö÷ ç ÷÷ =  kR = 2 k çç 2 çè 4 ÷ø÷

610 45.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION ii. Robin: a = 0.95, b = 0.75

R(t ) = te-0.1t

“During the nth hour” corresponds to the interval (n - 1, n).

V =

The average intensity during nth hour is 1 n - (n - 1)

ò te

-0.1t

n -1

dt =

-0.1t

Let u = t and dv = e

dt.

The volume is about 2.238 cubic cm.

ò te

-0.1t

n -1

iii. Turtledove: a = 1.55, b = 1.15 V =

iv. Hummingbird: a = 0.5, b = 0.5

e-0.1t

-10 e-0.1t

V =

100 e-0.1t

4 (0.5)(0.5)2 » 0.5236 3

The volume is about 0.5236 cubic cm.

òn-1te

-0.1t

v. Raven: a = 2.5, b = 1.65

= (-10te-0.1t - 100e-0.1t )

= -10e

(b) Using the formula -0.1

(12) + 10e

(11)

= 110e-0.1 - 120e-0.2 » 1.284

V =

Average intensity = -10e-1.2 (12 + 10) + 10e-1.1(11 + 10) = 210e-1.1 - 220e-1.2 » 3.640

V =

Average intensity = -10e-2.4 (24 + 10) + 10e-2.3 (23 + 10) = 330e

-2.4

- 340e

» 2.241

From problem 26, the volume of an ellipsoid with major axis 2a and minor axis 2b is 4 V = ab 2 . 3

wö 4 æç l öæ ç ÷÷÷ççç ÷÷÷  3 èç 2 øè 2ø 4 æç 1.585w - 0.487 ö÷æç w2 ö÷÷ ÷÷çç ÷ ç øçè 4 ÷ø÷ 3 èç 2

 (1.585w3 - 0.487 w2 ) 6

i. Canada goose: w = 5.8 V =

 [1.585(5.8)3 - 0.487(5.8) 2 ] 6

» 153.3

8.6 (a) i. Canada goose: a = = 4.3, 2 5.8 b= = 2.9 2 V =

4 2 ab  3

a is half the length, l, and b half the width, w, of a bird egg. Substituting these expressions and writing l in terms of w, we have

(b) Twelfth hour, n = 12

-2.3

4 (2.5)(1.65)2 » 28.51 3

The volume is about 28.51 cubic cm.

Average intensity -0.2

V =

n -1

(a) Second hour, n = 2

46.

4 (1.55)(1.55)2 » 8.586 3

The volume is about 8.586 cubic cm.

0 n

4 (0.95)(0.75)2 » 2.238 3

4 (4.3)(2.9) 2 » 151.51 3

The volume is about 153.3 cubic cm. ii. Robin: w = 1.5 V =

 [1.585(1.5)3 - 0.487(1.5)2 ] 6

» 2.227

The volume is about 2.227 cubic cm.

Section 8.2

611 iii. Turtledove: w = 2.3

(a) The average number of items produced daily after 5 days is

 [1.585(2.3)3 - 0.487(2.3) 2 ]

V =

45 [(t + 1) ln(t + 1) - t ] 5-0 0

» 8.749

= 9[(6 ln 6 - 5) - (ln1 - 0)]

The volume is about 8.749 cubic cm.

= 9(6 ln 6 - 5) » 51.76.

iv. Hummingbird: w = 1.0

(b) The average number of items produced daily after 9 days is

 [1.585(1.0)3 - 0.487(1.0)2 ]

V =

45 [(t + 1) ln (t + 1) - t ] 9-0 0

» 0.5749

The volume is about 0.5749 cubic cm.

= 5(10 ln10 - 9) » 70.13.

v. Raven: w = 3.3 V =

 [1.585(3.3)3 - 0.487(3.3)2 ] 6

45 [(t + 1) ln(t + 1) - t ] 30 - 0 0

» 27.051

The volume is about 27.05 cubic cm.

47.

For each part below, use Average value

ò 45ln(t + 1)dt

45 b-a

ò ln(t + 1)dt.

48.

1 = b-a

(a) W (0) = -3.75(0) 2 + 30(0) + 40 W (0) = 40 words/minute

(b)

u = ln(t + 1) 1 Then du = dt t +1 Let

and

dv = dt.

and

v = t.

ò ln (t + 1) dt ò = t ln (t + 1) ò

= t ln (t + 1) -

t dt t +1 æ ö çç1 - 1 ÷÷ dt ÷ çè t + 1ø

If 0 £ t < 4, W ¢(t ) > 0. If 4 < t £ 5, W ¢(t ) < 0.

W (4) = -3.75(4) 2 + 30(4) + 40 W (4) = 100

A maximum speed of 100 words per minute occurs when t = 4 minutes. (c) The average value of W over [0, 5] is given by 5 1 (-3.75t 2 + 30t + 40) dt 5-0 0 5 1 = (-1.25t 3 + 15t 2 + 40t ) 5 0 1 = [(-156.25 + 375 + 200) - 0] 5 = 83.75.

Therefore Average value b

ò 45ln(t + 1) dt a

W ¢(t ) = -7.50t + 30 -7.50t + 30 = 0 t = 4

Therefore, a maximum occurs when t = 4.

= t ln (t + 1) - t + ln (t + 1) + C = (t + 1)ln (t + 1) - t + C

1 b-a

W (t ) = -3.75t 2 + 30t + 40

a b

Evaluate the integral using integration by parts.

3 (31ln 31 - 30) » 114.7. 2

45 [(t + 1) ln(t + 1) - t ] . b-a a

The average value is 83.75 words per minute.

612 49.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION From Exercise 26, the volume of an ellipsoid with horizontal axis of length 2a and vertical axis of length 2b is V =

ò0

50, 000te-0.035t dt = 50, 000

ò 0 te-0.035 dt t

Use integration by parts.

4ab 2 . 3

u = t and dv = e-0.035t dt. 1 e-0.035t Then du = dt and v = 0.035 Let

For the Earth, a = 6,356,752.3142 and b = 6,378,137. 4(6,356, 752.3142) (6,378,137) 2 V = 3 » 1.083 ´ 10

P =

The volume of Earth is about 1.083 ´ 1021 cubic meters (m3 ).

= -28.57143e-0.035t . 5

ò 0 te-0.035 dt t

= -28.57143te-0.035t -

28.57143 -0.035t +C e 0.035

= (-28.57143t - 816.32657) e-0.035t + C P = 50, 000

8.3 Continuous Money Flow

ò 0 te-0.035 dt t

= 50, 000(-28.57143t - 816.32657)e-0.035t

Your Turn 1

0 -0.035(5)

= 50, 000[(-28.57143(5) - 816.32657)e - (0 - 816.32657)] = 50, 000(-805.19351 + 816.32657) » 556, 653

f (t ) = 810ekt

Use f (1) = 797.94 to find k. f (1) = 810ek (1)

The present value of the income is $556,653.

797.94 = 810ek 797.94 ek = » 0.9851 810 k » ln 0.9851 » -0.015

Your Turn 3

Find the accumulated amount of money flow for an income given by f (t ) = 50,000t over the next 5 years if the interest rate is 3.5%.

f (t ) = 810e-0.015t Total income =

ò0

810e-0.015t dt

A = e0.035(5)

810 -0.015t e 0.015 0

= -54,000e-0.015t -0.03

= -54,000(e » 1595.94

ò 0 50,000te-0.035 dt t

The integral was computed in Your Turn 2. Using this value, the accumulated value is e0.175 (556,653) » 663,111.

2 0

The accumulated amount of money flow is $663,111.

- 1)

Your Turn 4

The total income is $1595.94

Find the present value at the end of 8 years of the continuous flow of money given by

Your Turn 2

Find the present value of an income given by f (t ) = 50,000t over the next 5 years if the interest rate is 3.5%. f (t ) = 50,000t , 0 £ t £ 5

f (t ) = 200t 2 + 100t + 50

at 5% compounded continuously. P =

ò 0 (200t 2 + 100t + 50) e-0.05 dt

Section 8.3

613

Using integration by parts as in Example 5, you can verify that

ò (200t 2 + 100t + 50) e-0.05 dt t

-0.05t

= -1000e

= -1000(e » 28,156.02

(4t + 162t + 3241) + C. 8 0

(4793) - 3241)

The present value at the end of 8 years of this flow of money is $28,156.02.

8.3 Warmup Exercises W1.

æ 1

ò 300e0.01t dt = 300çççè 0.01 e0.01t ø÷÷÷ + C = 30,000e0.01t + C

W2.

ò0

100e0.04t dt 5

100 0.04t e 0.04 0

(

= 2500 e0.2 - 1

)

» 553.51

W3.

400te-0.04t dt = 400

ò te-0.04 dt æ 5 öæ ö = çç ÷÷÷ççç 400 te-0.04 dt ÷÷÷ çè 2 øè ò ø t

1000

Thus P = -1000e-0.05t (4t 2 + 162t + 3241) -0.4

W4. Using the result in W3,

te-0.04t dt

æ 5 öé ù = çç ÷÷÷ ê -10,000e-0.04t ( t + 25 ) + C ú çè 2 øë û

(

)

= -25,000e-0.04t ( t + 25 ) + C ¢ 5

ò0 1000te-0.04 dt t

(

= -25,000 éê (5 + 25)e-0.2 - ( 0 + 25 ) e-0 ùú ë û » 10,951.94

8.3 Exercises 1.

True

False. The total money flow is found by taking the integral of the function giving the rate of money flow over the time interval.

False. The present value of money flow for T years is the integral from 0 to T of the function giving the rate of continuous money flow.

True

f (t ) = 1000

(a)

ò0 1000e-0.08 dt t

1000 -0.08t e -0.08 0

= -12,500(e-0.8 - e0 ) = -12,500(e-0.8 - 1) » 6883.387949

ò te dt = 400 v du ò æ ö = 400 ççç uv ò u dv ÷÷ø÷ è

(We will use this value for P in part (b). Store it in your calculator without rounding.) The present value is $6883.39.

é æ öù = 400 ê -25te-0.04t - ççç -25 e-0.04t dt ÷÷÷ ú êë è ø úû æ 1 -0.04t ö÷ e = -10,000 çç te-0.04t + ÷÷ø + C çè 0.04

)

P = =

-0.04t

(

)

é ù = ê -25,000e-0.04t ( t + 25 ) ú ë û 0

Let v = t and du = e-0.04t . Then dv = dt and 1 -0.04t u =e = -25e-0.04t . 0.04 400

)

= -10,000e-0.04t ( t + 25 ) + C

(b)

A = e0.08(10)

ò 1000e

-0.08t

= e0.8 P » 15,319.26161

The accumulated value is $15,319.26.

614 6.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION f (t ) = 300

(a)

P =

9. 10

ò (300)e

-0.08t

f (t ) = 400e0.03t

(a)

P =

300 -0.08t e -0.08 0 -0.8

= -3750(e

-e )

ò0

e-0.05t dt =

400 -0.05t e -0.05 0

» 3147.754722

The present value is $3147.75. (b)

The present value is $2065.02. 10

= -8000(e-0.5 - e0 )

(We will use this value for P in part (b). Store it in your calculator without rounding.)

A = e0.08(10)

ò0 400e0.03te-0.08tdt

= 400

= -3750(e-0.8 - 1) » 2065.016385

(b)

ò 300e

-0.08t

A = e0.08(10)

ò0 400e0.03t e-0.08t dt

= e0.8 P » 7005.456967 The accumulated value is $7005.46.

= e0.8 P » 4595.778482 The accumulated value is $4595.78.

10.

f (t ) = 800e0.05t

(a)

f (t ) = 500

(a)

P =

ò 500e

-0.08t

= -6250(e-0.8 - e0 ) » 3441.693974

The present value is $6911.51.

The present value is $3441.69. A = e0.08(10)

ò 500e

-0.08t

(b)

A = e0.08(10)

11.

P =

2000e

(a)

= 5000

2000 -0.08t = e -0.08 0

-0.01t -0.08t

ò e

-0.09t

= -25,000(e-0.8 - e0 ) » $13,766.7759

The present value is $13,766.78. 10

ò 5000e

A = e0.08(10)

P =

(b)

f (t ) = 5000e-0.01t

f (t ) = 2000

(a)

0.05t -0.08t

The accumulated value is $15,381.86.

The accumulated value is $7659.63.

-0.08t

ò 800e

= e0.8 P » 15,381.85754

= e0.8P » 7659.630803

ò0 e-0.03t dt

800 -0.03t 10 e -0.03 0 800 -0.03 =(e - e0 ) 0.03 » 6911.514115

500 -0.08t e -0.08 0

(b)

ò0 800e0.05t e-0.08t dt

= 800

P =

ò 2000e

-0.08t

5000 -0.09t e -0.09 0

5000 -0.9 (e - e0 ) 0.09 » 32,968.35224

The present value is $32,968.35.

= e0.8P » $30,638.52321

Section 8.3 (b)

615 A = e0.08(10)

Therefore

5000e-0.01t e-0.08t dt

P = 25[-(12.5t + 156.25)e-0.08t ]

= e0.8 P

= [-25(12.5t + 156.25)e-0.08t ]

» 73,372.41725

-0.8

The accumulated value is $73,372.42.

= (-7031.25e

10 0 10 0

) - (-3906.25e0 )

» 746.9057211.

12.

-0.02t

f (t ) = 1000e

(a)

The present value is $746.91.

ò 1000e

P =

-0.02t -0.08t

(b)

= 1000

ò e

-0.1t

14.

The present value is $6321.21. 10

ò 1000e

f (t ) = 50t

(a)

-0.02t -0.08t

P =

f (t ) = 25t 10

ò 25te

-0.08t

ò te

u =t

-0.08t

te-0.08t dt

u =t

and dv = e-0.08t dt.

Then du = dt

1 e-0.08t and v = -0.08

= -12.5e-0.08t . -0.08t

and dv = e-0.08t dt. 1 e-0.08t and v = -0.08

= -12.5e-0.08t .

Find the antiderivative using integration by parts.

ò te

Let

Then du = dt

= 25

Find the antiderivative using integration by parts. Let

P =

-0.08t

The accumulated value is $14,068.10.

(a)

ò 50te

= 50

» 14,068.10175

13.

The accumulated value is $1662.27.

» 6321.205588

0.8

-0.08t

» 1662.269252

1000 -0.1t 10 e -0.1 0

A = e0.08(10)

ò 25te

= e0.8 P

= -10,000(e-1 - e0 )

(b)

A = e0.08(10)

ò (-12.5e )dt + 12.5 e dt ò

= t (-12.5e-0.08t ) = -12.5te-0.08t

-0.08t

= -12.5te-0.08t +

12.5 -0.08t +C e -0.08

= -(12.5t + 156.25)e-0.08t + C

Therefore:

ò (-12.5e )dt + 12.5 e dt ò

= t (-12.5e-0.08t ) -0.08t

= -12.5te

-0.08t

= -12.5te

-0.08t

P = 50[-(12.5t + 156.25)e-0.08t ]

-0.08t

= [-50(12.5t + 156.25)e-0.08t ]

12.5 -0.08t + +C e -0.08

-0.8

= (-14, 062.5e

The present value is $1493.81.

0 10 0

) - (-7812.5e0 )

» 1493.811442.

= -(12.5t + 156.25)e-0.08t + C

616

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION (b)

A = e0.08(10) =e

0.8

ò0

16.

50te-0.08t dt

f (t ) = 0.05t + 500

(a)

ò (0.05t + 500)e

P =

-0.08t

» 3324.538504

The accumulated value is $3324.54.

ò 500e

0.05te-0.08t dt +

15.

f (t ) = 0.01t + 100

(a)

P = =

ò (0.01t + 100)e

-0.08t

ò 0.01te

-0.08t

ò te

-0.08t

ò 100e

-0.08t

ò te

= 0.01

-0.08t

ò e

òe

-0.08t

From Exercise 13, we know that -0.08t

dt = -(12.5t + 156.25)e-0.08t + C

= (-6264.0625e

(b)

A = e0.08(10)

ò (0.05t + 500)e

-0.08t

(0.01t + 100)e

f (t ) = 1000t - 100t 2

(a)

P =

ò (1000t - 100t )e 2

-0.08t

= 1000

-0.08t

» 7662.955342 The accumulated value is $7662.96.

17.

0.8

= e0.8 P

The present value is $688.64. A=e

) - (-6257.8125e0 )

» 688.6375571

(b)

The present value is $3443.19.

= (-1252.8125e-0.8 ) - (-1251.5625e0 )

0.08(10)

» 3443.187786

= [-(0.125t + 1251.5625)e-0.08t ]

dt = -12.5e-0.08t + C

-0.8

P = {0.01[-(12.5t + 156.25)e-0.08t ] )}

= [-(0.625t + 6257.8125)e-0.08t ]

Substitute the given expressions and simplify.

+100 (-12.5e

+500(-12.5e-0.08t )}

e-0.08t dt = -12.5e-0.08t + C

-0.08t

P = { 0.05[- (12.5t + 156.25) e-0.08t ]

From Exercise 5, we know that

ò e

Substitute the given expression and simplify.

ò te

From Exercise 6, we know that

+ 100

te-0.08t dt + 500

= -(12.5t + 156.25)e-0.08t + C

From Exercise 14, we know that

-0.08t

= 0.05

0 10

» 1532.591068 The accumulated value is $1532.59.

ò te

-0.08t

- 100

ò te 0

2 -0.08t

Section 8.3

617 From Exercise 13, we know that

ò te

-0.08t

18.

dt = -(12.5t + 156.25)e-0.08t + C

Evaluate the antiderivative

f (t ) = 2000t - 150t 2

P =

(a)

ò (2000t - 150t )e

-0.08t

òt e

2 -0.08t

= 2000

ò te

-0.08t

using column integration. (Note that 1 = -12.5.) -0.08

- 150

ò te

2 -0.08t

-0.08t

- 12.5 e-0.08t

156.25 e-0.08t

From Exercise 14, we know that

ò te

-0.08t

Evaluate the antiderivative ò t 2e-0.08t dt using column integration. (Note that 1 = -12.5.) -0.08

-0.08t

- 1953.125 e

Thus,

dt = -(12.5t + 156.25)e-0.08t + C

-0.08t

-12.5e-0.08t

- (2t )(156.25e-0.08t )

156.25e-0.08t

+ (2)(-1953.125e-0.08t ) + C

t 2e-0.08t dt

0 -0.08t

= (t )(-12.5e

)

= -(12.5t 2 + 312.5t -0.08t

+ 3906.25)e

-1953.125e-0.08t

Thus, + C.

ò te

Therefore:

2 -0.08t

= (t 2 ) (-12.5e-0.08t ) - (2t )(156.25e-0.08t )

P = {1000[-(12.5t + 156.25)e-0.08t ]

+ (2) (-1953.125e-0.08t ) + C

- 100[-(12.5t 2 + 312.5t + 3906.25)e-0.08t ]}

= -(12.5t 2 + 312.5t + 3906.25)e-0.08t + C.

10 0

Therefore:

Collect like terms and simplify.

P = {2000[-(12.5t + 156.25)e-0.08t ]

P = [(1250t 2 + 18, 750t + 234,375)e-0.08t ]

10 0

= (546,875e-0.8 ) - (234,375e0 )

-150[-(12.5t 2 + 312.5t + 3906.25)e-0.08t ]}

P = [(1875t 2 + 21, 875t + 273, 437.5)e-0.08t ]

The principal value is $11,351.78.

-0.8

(b)

A=e

Collect like terms and simplify.

» 11,351.77725

0.08(10)

ò (1000t - 100t )e 2

-0.08t

= e0.8 P » 25, 263.84488

= (679, 687.5e » 31, 965.7803

) - (273, 437.5e )

The principal value is $31,965.78. (b) A = e0.08(10)

The accumulated value is $25,263.84.

ò (2000t - 150t ) e 2

-0.08t

= e0.8 P » 71,141.1524

The accumulated value is $71,141.15.

618 19.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION A = e0.04(3)

(b) Final amount

20, 000e-0.04t dt

= e0.08(8)

æ 20, 000 -0.04t ö÷ = e0.12 çç e ÷÷ çè -0.04 ø0

22.

(a) Present Value

8000e-0.02t dt

ò0 1000e0.05 e 0.035 dt

8000 -0.02t e -0.02 0

t -

= 1000

ò0 e0.015 dt t

= -400, 000(e-0.12 - 1) » $45, 231.83

1000 0.015t e 0.015 0

1000 0.06 - 1) (e 0.015 » 4122.436

(b) Present value =

t -

$54, 075.81 - $28,513.76 = $25,562.05

(a) Present value 6

» $54,075.81

» $63, 748.43

ò0 5000e 0.01 e 0.08 dt

» e0.64 (28,513.76)

æ 20, 000 -0.12 20, 000 ö÷ e = e0.12 çç + ÷ çè -0.04 0.04 ø÷

20.

ò0 8000e

-0.05t

» $4122.44

8000 -0.05t e -0.05 0

(b) Final amount

= -160, 000(e-0.3 - 1)

= e0.035(4)

» $41, 469.08

ò0 8000e 0.08 dt -

-0.48

= -100,000(e

$4741.93 - $4122.44 = $619.49

23.

- 1)

» $38,121.66

ò0 5000e 8

P = =

(a) Present value =

t -

» $4741.93

8000 -0.08t = e -0.08 0

21.

ò0 1000e0.05 e 0.035 dt

= e0.14 (4122.436)

-0.01t -0.08t

ò0 (1500 - 60t 2 )e 0.05 dt 5

ò0

ò0 60t e

1500e-0.05t dt -

= 1500

ò0

e-0.05t dt - 60

2 -0.05t

ò0 t 2e 0.05 dt -

Find the second integral by column integration.

ò0 5000e 0.09 dt -

æ 5000 -0.09t ÷ö = çç e ÷÷ çè -0.09 ø0 5000e-0.72 5000 + -0.09 0.09 = $28,513.76 =

Section 8.4

619

8.4 Improper Integrals

e-0.05t

Your Turn 1

(a)

-0.05t

e -0.05

ò8

dx = lim b ¥ x1/3

e-0.05t 0.0025

æ3 ö 3 = lim çç b 2/3 - (4) ÷÷÷ ç ø 2 b ¥ è 2

As b  ¥, b 2/3  ¥, so the limit above does not exist and the integral diverges.

e-0.05t -0.000125

t 2e-0.05t 2te-0.05t 2e-0.05 + +C -0.05 -0.000125 0.0025 æ t2 ö÷ 2t 2 ç = -e-0.05t çç + + ÷÷ + C çç 0.05 0.0025 0.000125 ÷÷ø è

t 2e-0.05t dt =

(b)

ò0

e-0.05t dt - 60

æ 2 ö÷ 2t 2 ÷÷ + + çç 0.0025 0.000125 ø÷÷ çè 0.05 0

1 4/3

dx b

Your Turn 2

5e-2 x dx = lim

b ¥

ò 5e- dx 2x

= $4560.94

A = e0.05(3) = e0.15

æ 5 ö = lim çç - e-2 x ÷÷÷ ø0 b ¥ çè 2

ò0 (1000 - t 2 )e 0.05 dt -

ò0

e-0.05t 1000e0.15

) -0.05

é 5 æ 5 öù = lim ê - e-2b - çç - e-2(0) ÷÷÷ ú ç ê è 2 ø úû b ¥ ë 2

1000e-0.05t dt - e0.15

ò0 t 2e 0.05 dt -

Using the integral we found in Exercise 23, we can write this as follows:

(

ò x

1500 -0.25 (e = - 1) -0.05 é ù æ 25 ö 10 2 2 ú + 60 ê e-0.25 çç + + ÷÷ ê ÷ çè 0.05 0.0025 0.000125 ø 0.000125 ûú ë » 6636.977 - 2075.037

24.

é æ 3 öù = lim ê -3b-1/3 - çç - 1/3 ÷÷ ú ê ç è 8 ÷ø úû b ¥ ë 1 As b  ¥, b-1/3 = 1/ 3  0, so the b integral is convergent and its value is 3 = 3. 1/3 2

1500 -0.05t e -0.05 0 + 60 e

( b ¥

-0.05t çç t

= lim -3x-1/3

1 4/3

b ¥

ò 0 t 2e 0.05 dt -

= lim

Now add the first integral to this result. 1500

ò 8 x1/3 dx

æ3 ö = lim çç x 2/3 ÷÷÷ ç è ø8 b ¥ 2

As b  ¥, e-2b = 1b  0, so the integral is e

convergent and its value is 52 e0 = 52 .

8.4 Warmup Exercises 0

æ ç

æ 2 ö÷ 2t 2 ÷÷ + + çç ÷ 0.0025 0.000125 ÷ø÷ èç 0.05

ç ç t - (e0.15 ) çç -e-0.05t ç çç çèç

3 ö÷

÷÷÷ ÷÷ ÷÷ 0 ÷ø

W1. lim

x ¥

3 - x2 x4

» 3236.685 - 9.348 » $3227.34

x ¥ x

x ¥ x 4

= lim

- lim 2

where both limits exist = 0+0 = 0

620 W2.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION ¥

4 x 2 + 3x - 7

lim

x ¥ 2 x 5 + 3x - 5

x2 x 1 +3⋅ 5 -7⋅ 5 5 x x x = lim x ¥ 1 x5 x 2⋅ 5 + 3⋅ 5 -5⋅ 5 x x x æ 1 1 1 ö lim çç 4 ⋅ 3 + 3 ⋅ 4 - 7 ⋅ 5 ÷÷÷ ç x ¥ è x x x ø = æ ö 1 1 lim çç 2 + 3 ⋅ 4 - 5 ⋅ 5 ÷÷÷ x ¥ èç x x ø

W3.

= lim

é æ ö÷ ù 1 1 çç ÷ú = lim êê çç 2 ÷÷÷ ú b ¥ ê 2(b + 1) 2 2(3 1) + è ø úû ë

As b  ¥,

lim e-4 x = 0

lim e-0.2 x = ¥

ò4

1 2(3+1)2

> M by taking x < -5 ln M .

1 . = 32

ò4 2x-1/2dx b

b ¥

= lim (4 b - 4 4) b ¥

= lim 4 b - 8 b ¥

8.4 Exercises 1.

False. It is possible for an integral from 0 to ¥ to have a finite value.

False. An integral from 0 to ¥ is an improper integral, by definition.

True

False. If an integral from -¥ to ¥ is written as the sum of two integrals, one from -¥ to c and the other from c to ¥ , the integral is convergent as long as both of the two integrals in the sum exist.

 0. The integral is

= lim 4 x1/2

For any given M greater than 1 we can make

(b + 1)2

2 dx = lim b ¥ x

x -¥

convergent and its value is

x ¥

-0.2x

æ 1 ö÷ 1 ÷÷ = lim ççç 2 b ¥ èç 2 ( x + 1) ø÷÷ 0

0 =0 2

For any given  less than 1 we can make - ln  . 0 < e-4 x <  by taking x > 4 W4.

dx b ¥ ò3 ( x + 1)3

4⋅

ò3 ( x + 1)3 dx

As b  ¥, 4 b  ¥. The integral diverges.

b -3 -3x-1/2dx dx = lim b ¥ 100 x 100

æ -3x1/2 ö÷ ç = lim çç 1 ÷÷÷ ÷ø b ¥ ççè 2 100 = lim (-6 b - 6 100) b ¥

= lim (-6 b - 60) b ¥

As b  ¥, (-6 b - 60)  -¥, so the integral is divergent.

x-2dx ò3 x2 dx = blim ¥ ò3 = lim -x-1 b ¥

b 3

æ 1 1ö = lim çç - + ÷÷÷ ç 3ø b ¥ è b æ 1 ö÷ 1 = lim çç - ÷÷ + lim b ¥ çè b ø b ¥ 3

-1

-¥ x

dx = 3

As b  ¥, - b1  0. The integral is convergent.

-1

-¥

lim

2 x-3dx

a -¥

-1

2 x-3dx

a -1

æ 2 x-2 ö÷ ç ÷÷ = lim çç a -¥ çè -2 ÷÷ø a =

æ 1 ö lim çç -1 + 2 ÷÷ ç a -¥ è a ÷ø

Section 8.4

621

As a  -¥, 12  0. The integral is a

12.

convergent.

ò ò

dx = lim 0.999

b ¥

-4

= lim (1000 x0.001)

dx =

lim

a -¥

3x-4dx

b ¥

æ 1 ö = lim çç - 3 ÷÷ ç a -¥ è x ø÷

The integral is divergent. 13.

-10

-¥

x-2dx = =

-¥ x

dx =

1 1 +0= 64 64

dx x1.0001

-1.0001

14.

= lim

b ¥

-1.0001

-1

-¥

1 0.0001 b0.0001

-10

lim (-x-1)

a -¥

1 1 +0= 10 10 1 . 10

( x - 2)-3 dx lim

a -¥

-1

( x - 2)-3 dx

é 1 æ ö÷ ù 1 ÷÷ ú lim êê - - ççç çè 2(a - 2)2 ÷÷ø úú a -¥ ê 18 ë û

As a  -¥,

 0.

1 2(a - 2) 2

 0. The integral is

1 . convergent and its value is - 18

The integral is convergent. 1

x-2dx

-1

æ 1 1 ö÷÷ = lim ççç + ÷ 0.0001 ÷ø÷ b ¥ çè (0.0001)b0.0001

æ 1 ö÷ 1 = lim ççç ÷÷÷ 2 a -¥ çè 2 ( x - 2) ÷ø a

æ x-0.0001 ö÷ ç ÷÷ = lim çç b ¥ çè -0.0001 ÷ø÷ 1

As b  ¥, -

a -¥

-10

The integral is convergent and its value is

lim

æ 1 1ö = lim çç + ÷÷÷ ç aø a -¥ è 10

- 1000)

As b  ¥, (1000b0.001 - 1000)  ¥.

As a  ¥, 13  0. The integral is convergent. -4

0.001

b ¥

-4 a

= lim (1000b

æ 1 1 ö lim çç + 3 ÷÷÷ a -¥ çè 64 a ø

-4

-0.999

dx = -1 + 0 = -1

-¥ x

ò x

æ x0.001 ö÷ ç ÷÷ = lim çç b ¥ çè 0.001 ÷ø÷ 1

æ 3x-3 ö÷ ç ÷÷ = lim çç a -¥ çè -3 ÷÷ø a

11.

-1

-¥ x

10.

1 dx = 0 + = 10,000 1.0001 0.0001 x

15.

-1

-¥

x-8/3dx =

lim

a -¥

-1

x-8/3dx

-1

æ 3 ö = lim çç - x-5/3 ÷÷÷ ç è øa 5 a -¥ =

æ3 3 ö lim çç + 5/3 ÷÷ ç a -¥ è 5 5a ÷ø

3 3 +0 = 5 5

3 . 5

622 16.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION -27

ò-¥

x-5/3dx

lim

a -¥

-27

19.

-¥

1000e x dx =

x-5/3dx

-27

æ -2/3 ö÷ çx = lim çç ÷÷÷ ç a -¥ çè - 2 ÷ø 3 a

-27

æ 1 3 ö lim çç - + 2/3 ÷÷÷ a -¥ èç 6 a ø

-¥

20.

-¥

5e60 x dx =

17.

1 1 dx = - + 0 = 6 6

8e-8 x dx = lim

b ¥

æ 8e-8 x ÷ö ç = lim çç ÷÷ ç b ¥ è -8 ÷÷ø 0

21.

-1

-¥

b ¥

æ 1 ö = lim çç - 8b + 1÷÷÷ ø b ¥ èç e

lim

a -¥

ò 5e

60 x

æ 5 ö lim çç e60 x ÷÷÷ ç øa a -¥ è 60

æ 1 1 60a ö÷ e ÷÷ lim çç ç ø 12 a -¥ è 12

b ¥

-50b

-1 e50b

-¥

ln | x| dx =

+ e0 )

ln | x| dx

v = x.

lim ( x ln | x| -x)

a -¥

-1 a

lim (- ln1 + 1 - a ln |a| +a)

a -¥

lim (1 + a - a ln |a|)

a -¥

= lim (-e-50b + 1)

The integral is divergent, since as a  -¥.

 0. The integral is

(a - a ln |a|) = -a (-1 + ln |a|)  ¥.

22.

ò 50e

-1

convergent. ¥

b ¥

As b  ¥,

a -¥

-1

= x ln | x| - x + C

= lim (-e-50 x ) b ¥

lim

ò ln |x| dx = x ln |x| - ò x dx

50e-50 x dx

= lim (-e

ln | x| dx =

1 1 -0 = 12 12

Then du = 1x dx and

The integral is convergent, and its value is 1. b

5e60 x dx =

u = ln | x| and dv = dx.

Let

= 0 +1=1

50e-50 x dx = lim

-¥

= lim (-e-8b + 1)

lim (1000 - 1000ea )

a -¥

converges.

8e-8 x dx b

1 e60a  0. The integral As a  -¥, 12

18.

a -¥

-5/3

lim (1000e x )

convergent. -27

-¥

3  0. The integral is a 2/3

As a  -¥,

ò 1000e dx = 1000 - 0 = 1000

é 3 ù 3 = lim ê - (-27)-2/3 + (a)-2/3 ú ê úû 2 a -¥ ë 2 =

ò 1000e dx

a -¥

As a  ¥, -1000ea  0. The integral is convergent.

æ 3 ö lim çç - x-2/3 ÷÷÷ ç ø a a -¥ è 2

lim

-50 x

dx = 0 + 1 = 1

ò1

Let

ln | x| dx = lim

b ¥

ò1 ln |x| dx

u = ln| x and dv = dx.

Then du = 1x dx and v = x.

Section 8.4

623 x

ò ln |x| dx = xln|x| -ò x dx = x ln|x| - x + C

25.

-1

2x - 1

-¥ x

-x

ò1 ln |x| dx

= lim ( x ln | x | - x) b ¥

= lim [(b ln b - b) - (-1)]

b ¥

= lim [b ln b - b + 1] b ¥

23.

26.

ò0 ( x + 1)2 b

ò0 ( x + 1)2

b ¥

= lim -( x + 1)-1 b ¥

a -¥

4x + 6

ò1 x2 + 3x

Use substitution

4x + 6

ò1 x2 + 3x dx

4x + 6

ò1 x2 + 3x dx

b ¥

= lim 2 b ¥

2x + 3

ò1 x2 + 3x dx

= lim (2 ln | x 2 + 3x|) b ¥

= lim (2 ln |b + 3b| -2 ln 4) b ¥

= 0 +1 = 1 2

As b  ¥, ln |b 2 + 3b|  ¥. The integral diverges.

ò0 (4x + 1)3

27.

ò0 (4x + 1)3 ö÷ 4(4 x + 1)-3 dx ÷÷÷ 0 ø÷

ò2 x ln x dx = lim

b ¥

é 1 ù = lim ê - (4 x + 1)-2 ú úû b ¥ êë 8 0 é 1 ù 1 = lim ê - (4b + 1)-2 + (1)-2 ú úû 8 b ¥ êë 8

As b  ¥, ln (ln b)  ¥. The integral is divergent.

As b  ¥, - 18 (4b + 1)-2  0. The integral converges. 1

ò2 x ln x dx Use substitution

b ¥

bù é = lim ê ln (ln x) 2 ú û b ¥ ë = lim [ln (ln b) - ln (ln 2)]

é 1 (4 x + 1)-2 ù ú = lim êê ⋅ ú -2 b ¥ ê 4 ë ûú 0

ò0 (x + 1)

æ1 = lim ççç b ¥ çè 4

b ¥

lim

b ¥

dx = lim

Note that if u = x 2 + 3x then du = (2 x + 3)dx. Also, 4 x + 6 = 2(2 x + 3).

convergent.

= lim

a 2

lim (ln 2 - ln |a - a|)

1  0. The integral is As b  ¥, - b + 1

-1

a -¥

æ -1 ö = lim çç + 1 ÷÷÷ ç ø b ¥ è b + 1

24.

2x - 1

ò0 x2 - x dx Use substitution

lim ln | x 2 - x|

= lim

a -¥

As a  -¥, ln |a 2 - a|  ¥. The integral is divergent.

As b  ¥, (b ln b - b + 1)  ¥. The integral is divergent. ¥

-1

lim

624

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION ¥

28.

ò xe

ò2 x(ln x)2 dx = lim

b ¥

ò2 x(ln x)2 dx Use substitution

lim

a -¥

òa xe0.2xdx

lim [5( x - 5)e0.2 x ]

a -¥ 0

= -25e -

1 dx = 0 + 2 ln 2 x(ln x)

1 ln 2 b

b ¥

-¥

ò xe dx 4x

31.

Let

dv = e

Then

v = 14 e4 x dx and du = dx.

é 1 1 ù ú = lim ê (4b - 1)e4b + 16 úû b ¥ êë 16

-¥

dx =

lim

a -¥

òa xe

0.2 x

u = x

Then

du = dx and

1 e0.2 x = 5e0.2 x . v = 0.2

x3e-x dx

ò x e dx Use substitution

b -¥

3 -x 4

é 1 4 0 ù ú = lim êê - e-x ú b -¥ ê 4 b ú ë û é 1 ù 1 ú = lim êê - + 4 ú 4 b -¥ ê 4eb úû ë

As b  -¥,

1 4eb

 0. The integral is

convergent. 0

x3e-x dx = -

1 1 +0=4 4 b

x3e-x dx = lim

b ¥

ò xe

3 -x 4

dx Use substitution

é 1 4 = lim êê - e-x b ¥ ê 4 ë

b ù

ú ú û

0 ú

é 1 1 ùú = lim êê + 4 4 ûúú b ¥ ê 4eb ë

and dv = e0.2 x dx.

Let

x3e-x dx +

is divergent.

0.2 x

-¥

-¥ ¥

1 (4b - 1)e4b  ¥. The integral As b  ¥, 16

x 3e-x dx = lim

é 1 ù = lim ê (4 x - 1)e4 x ú ê úû b ¥ ë 16 0 é 1 ù 1 (-1)(1) ú = lim ê (4b - 1)e4b ê úû 16 b ¥ ë 16

xe0.2 x dx = - 25 - 0 = -25.

ò0 xe4x dx

-¥

lim [5(a - 5)e0.2a ]

a -¥

x3e-x dx

We evaluate each of two improper integrals on the right.

x 1 4x e +C = e4 x 4 16 1 = (4 x - 1)e4 x + C 16

-¥

dx and u = x.

ò xe dx = 4 e - ò 4 e dx

30.

fraction so that 5(a - 5) e0.2a  0. The integral converges.

xe4 x dx = lim

As a  -¥, e0.2a is in the denominator of a

0.2 x

= 5 ( x - 5) e0.2 x + C

convergent. ¥

òe

= 5xe0.2 x - 25e0.2 x + C

As b  ¥, - ln1 b  0. The integral is

29.

dx = 5xe0.2 x - 5

bö æ ç 1 ÷÷ = lim çç ÷÷ b ¥ ççè ln x 2 ÷÷ø æ -1 1 ö÷ = lim çç + ÷ ln 2 ÷ø b ¥ çè ln b

0.2 x

As b  -¥, convergent.

1 4eb

 0. The integral is

Section 8.4

625 ¥

Since each of the improper integrals converges, the original improper integral converges. ¥ 4 1 1 x3e-x dx = - + = 0 4 4 -¥

32.

-¥

e-| x | dx =

-¥

e-| x| dx +

e-| x | dx

We evaluate each of the two improper integrals on the right.

-¥

e-| x|dx =

-¥

e x dx =

lim

b -¥

34.

2x + 4

-¥ x

As b  -¥, eb  0. The integral is convergent.

-¥ ¥

ò e dx = ò e dx = lim ò e dx -| x |

-| x |

x ¥

é = lim ê -e-x b ¥ ëê

b ù 0 úú

-x

2x + 4

-¥ x ¥

-b

2x + 4 x + 4x + 5 0

lim

a -¥

2x + 4

ò x + 4x + 5 dx

b ¥

2x + 4

ò x + 4x + 5 dx 0

Evaluate the first of the two proper integrals in the

As b  ¥, e-b  0. The integral is convergent.

last equation first. If we let u = x 2 + 4 x + 5, then du = (2 x + 4) dx so that the integral is of

lim [-e

+ 4x + 5 2

+ lim

b -¥

ò e dx = -0 + 1 = 1 -| x |

the form ò du = ln | u | + C. u

Since each of the improper integrals converges, the original improper integral converges.

-¥

33.

+ 1]

+ 4x + 5

= b

ò +

e-| x |dx = 1 - 0 = 1 ¥

2x + 4

-¥ x

+ 4x + 5

This is an improper integral at each limit of integration. Therefore, rewrite the doubly improper integral as two separate improper integrals. Then rewrite each of the improper integrals as limits of proper integrals.

e x dx

é 0ù lim ê e x ú = lim (1 - eb ) b úû b -¥ êë b -¥

As b  -¥, ln (b 2 + 1)  ¥. The integral is divergent. Since one of the two improper integrals on the right diverges, the original improper integral diverges.

1 1 = 4 4

x3e-x dx = 0 +

-¥ x

e-| x|dx = 1 + 1 = 2

-¥ x

lim

b -¥

0ö æ lim çç ln | x 2 + 4 x + 5 | ÷÷÷ ç a ÷ø a -¥ è

dx +

x x2 + 1

As a  -¥, ln(a 2 + 4a + 5)  ¥. The integral diverges. Since one of the two integrals diverges, the original improper integral diverges.

dx 0

ò x + 1 dx Use substitution b

0 ù é1 ú = lim êê ln( x 2 + 1) ú b -¥ ê 2 b ú ë û

a -¥

We evaluate the first improper integrals on the right. 0

2x + 4

ò x + 4 x + 5 dx

= ln 5 - lim ln | a 2 + 4a + 5|

-¥ x

x -¥

lim

é ù 1 lim ê 0 - ln(b 2 + 1) ú úû 2 b -¥ êë

626 35.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION

38.

Ce-kt dt , k > 0

f ( x) = e-x for (-¥, e]

ò Ce dt = lim ò Ce dx -kt

b ¥

-¥

-kx

æ Ce-kb Ce-ka ö÷÷ ç = lim çç ÷ -k ÷ø b ¥ è -k

ò t dt, k > 1 C

b ¥

-k

æ Ct -k +1 ö÷ ç ÷÷ = lim çç b ¥ çè -k + 1 ø÷ a

= -k +1 ö

æ Cb Ca ÷ ç = lim çç ÷÷ ç -k + 1 ÷ø b ¥ è -k + 1 -k +1

f ( x) =

Ca-k +1 -k + 1 C

+ e-a )

for (-¥, 0]

lim

a -¥

ò (x - 1) a

lim - ( x - 1)-1

a -¥

Use substitution

0 a

(

= 1+ 0 = 1

Therefore, the area is 1. 40.

f ( x) = 0

dx ln | x - 1| x -1 a

0 = lim ln | x - 1| a a -¥

But

lim (-e

a -¥

convergent.

1 for (-¥, 0] x -1

-e

1  0. The integral is As a  -¥, a 1

1 dx = lim x 1 a -¥ -¥

a -¥

æ 1 1 ö÷ = lim çç + ÷ ç a - 1 ÷ø a -¥ è -1

(k - 1)a k -1 The integral is convergent.

37.

-x

-8

ò t dt = lim ò Ct dx k

ò e dx

lim (-e-x )

( x - 1) 2

ò (x - 1)

a -¥

f ( x) = 0

(e-e + e-a )  ¥. The integral is divergent, so the area cannot be found.

39.

lim

As a  -¥, e-a  ¥ and

Ce-ka = k The integral is convergent.

36.

e-x =

æ Ce-kx ö÷ ç ÷÷ = lim çç b ¥ è -k ø÷

)

lim (ln |- 1| - ln | a - 1|)

a -¥

lim (ln | a - 1|) = ¥.

a -¥

3 ( x - 1)3

for (-¥, 0]

ò (x - 1) dx 3

-8

lim

a -¥

ò (x - 1) dx a

æ 3( x - 1)-2 ö÷ ç ÷ = lim çç ÷÷÷ -2 a -¥ çè ø a 0

The integral is divergent, so the area cannot be found.

æ ö÷ 3 ÷ = lim ççç ÷ a -¥ çè 2( x - 1) 2 ÷ø÷ a =

æ ö 3 çç - 3 + ÷÷ ÷ ç a -¥ èç 2 2(a - 1)2 ÷ø÷ lim

3 2

Section 8.4 41.

-¥

627 2

xe-x dx

43.

dx xp Case 1a p < 1: 1

Let u = -x 2 , so that du = -2 x dx.

æ 1 0 ö÷ 2 = lim ççç -2 xe-x dx ÷÷÷ ÷ø a -¥ çè 2 a æ 1 b ö÷ 2 + lim ççç -2 xe-x dx ÷÷÷ b ¥ çè 2 0 ø÷

lim

a -¥

xdx

x b

dx = p

é ù 1 -1 ú lim êê + ú 2 2 a -¥ ê 2(1 + 0 ) 2(1 + a ) ûú ë ù é 1 -1 ú + lim êê + 2 ú b ¥ ê 2(1 + b 2 ) 2(1 0 ) + úû ë 1 1 =- +0+0+ = 0 2 2

ò1 x dx a

dx a ¥ ò1 x a = lim ( ln | x | 1 ) a ¥ = lim

xdx

2 2

æ ö÷ æ ö÷ -1 -1 ÷ + lim çç ÷ = lim ççç ÷ ÷ ç a -¥ çè 2 (1 + x 2 ) ÷ø÷ b ¥ èç 2(1 + x 2 ) ø÷÷ a =

ò (1 + x )

p = 1:

ò1 x

(1 + x 2 )2

+ lim b ¥ (1 + x 2 )2

a1- p  ¥. The integral diverges.

ò x- dx

Since p < 1, 1 - p is positive and, as a  ¥,

dx + 2 2 -¥ (1 + x )

é ù 1 (a- p +1 - 1) ú = lim ê ê ú a ¥ ë (- p + 1) û é ù 1 1 ú a1- p = lim ê ê (- p + 1) úû a ¥ ë (- p + 1)

Case 1b

é - p +1 a ù ê x ú ú = lim ê a ¥ ê (- p + 1) ú êë 1 úû

æ 1 1 ö÷÷ = lim ççç - + 2 ÷ ÷ a -¥ çè 2 2e-a ÷ø æ 1 1 ö÷÷ + lim ççç + ÷ 2 2 ÷÷ø b ¥ èç 2eb 1 1 =- + =0 2 2

ò x- dx a ¥

2 2 -¥ (1 + x )

= lim

2 ö æ 1 + lim çç - e-x ÷÷÷ ç è ø 0 2 b ¥

42.

2 ö æ 1 = lim çç - e-x ÷÷÷ ç è ø a 2 a -¥

= lim ( ln | a | - ln 1)

a ¥

= lim ln | a | a ¥

As a  ¥, ln | a |  ¥. The integral diverges. ¥

Therefore, ò1

ò1 x

diverges when p £ 1.

p > 1:

Case 2 ¥

1 xp

dx = lim

ò x- pdx

x ¥ 1

æ - p +1 a ö÷ çç x ÷÷ = lim çç ÷ a ¥ çç - p + 1 ÷÷÷ 1ø è é a- p +1 ù 1 ú = lim êê (- p + 1) úú a ¥ ê (- p + 1) ë û

628

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION Since p > 1, - p + 1 < 0; thus as a  ¥,

(d) Since the values of the integrals in part c appear to be approaching some fixed finite number, the integral

- p +1

a  0. (- p +1)

ò1

Hence,

dx 1 + x4 appears to be convergent.

é a- p +1 ù 1 1 ú = 0lim êê (- p + 1) úú (- p + 1) a ¥ ê (- p + 1) ë û -1 = -p + 1 1 . = p -1

(e) For large x, we may consider 1 + x 2 » x 2

and 1 + x 4 » x 4. Thus, 1

The integral converges. 45.

(a) Use the fnInt feature on a graphing utility to obtain

ò1 ò1

1 + x2 50

100

ò1 ò1 ò1

1 1 1 + x2 1 1+ x

1000

1 1 + x2

10,000

1 1 + x2

ò1

1 1 + x2

1+ x

ò ò

100

1 1 + x4

1000

10,000

1 + x4 1 1+ x

x 1

= =

1 and x 1 x2

ò1

1+ x 2

dx diverges as well. In Exercise 1, ¥

we saw that ò2

1 dx converges. Thus, we x2 ¥

dx » 4.417;

might guess that ò1

1 dx converges as 1+ x 4

well. dx » 6.720;

46. dx » 9.022.

dx » 0.8770;

(a) Use the fnInt feature on a graphing utility to obtain

ò ò

0 10

e-x dx » 0.7468;

ò e

-x 2

dx » 0.8862;

e-x dx » 0.8862;

e-x dx » 0.8862.

(b) Since the approximate values of the last three integrals in part a are all 0.8862, it appears ¥

that the integral ò0 e- x dx converges with an approximate value of 0.8862. 2

diverges. Thus, we might guess that

dx » 3.724;

dx appears to be divergent.

1+ x

1+ x 1

¥ In Example 1(a) we showed that ò1 1x dx

dx » 2.808;

(b) Since the values of the integrals in part a do not appear to be approaching some fixed finite number but get bigger, the integral ¥

1 1 + x4

dx » 0.9070;

between the graph of y = e-x and the x-axis on the interval [1, ¥] is less than the area between y = e-x and the x-axis. Since

dx » 0.9170;

e-x dx = lim

dx » 0.9260; dx » 0.9269.

b ¥

ò e dx -x

= lim [-e-x b ¥

b 1

é 1 1ù = lim ê - b + ú e úû b ¥ êë e 1 1 = 0+ = , e e

Section 8.4

629 the area between y = e-x and the x-axis on the interval [1, ¥) is finite. It follows

1  0. The integral converges. As b  ¥, 0.06 b e

ò0 225, 000e-0.06t dt

that the area between y = e-x and the x-axis, being smaller, must also be finite, so ¥

= -3, 750,000(0 - 1) = 3, 750,000

the integral ò1 e- x dx converges. Since the 2

area between y = e-x and the x-axis on the interval [0, 1] is finite as well, the ¥

The capital value is $3,750,000.

integral ò0 e- x dx converges.

49.

ò 1, 000, 000e

-0.05t

47.

(a) Use the fnInt feature on a graphing utility to obtain

0 50

ò ò

0 1000

-.00001x

As b  ¥, e-0.05b =

e-.00001x dx » 99.9500;

ò e

¥ 0

 0. The integral

ò0 1, 000,000e-0.05t dt = -20,000,000(0 - 1) = 20,000,000

e-0.00001x dx

The capital value is $20,000,000.

-0.00001x

50.

= lim

b ¥

ò e

-0.00001x

(a)

ò0 7200e-0.05t dt

= lim

b ¥

é e-0.00001x b ù ú = lim êê ú b ¥ ê -0.00001 0 ú ë û é ù 1 1 ú = lim ê + b 0.00001 0.00001 úû b ¥ êë 0.00001e = 0 + 100, 000 = 100, 000 ¥

ò 225,000e

-0.06t

ò0 7200e-0.05t dt b

æ 7200 -0.05t ö÷ e = lim çç ÷÷ ø 0 b ¥ çè -0.05 é ù = -144, 000 ê lim (e-0.05b ) - e0 ú êë b ¥ úû

As b  ¥, e-0.05b =

1 e0.05b

 0.

The integral converges. dt

ò0 7200e-0.05t dt = -144,000(0 - 1)

b ¥

1 e0.05b

converges. e-.00001x dx » 995.0166.

appears to be divergent.

= lim

é ù = -20,000,000 ê lim (e-0.05b ) - e0 ú êë b ¥ úû

dx » 49.9875;

finite number, the integral ò

48.

-0.05t

(b) Since the values of the integrals in part a do not appear to be approaching some fixed

ò 1,000,000e

æ 1,000,000 -0.05t ö÷ e = lim çç ÷÷ø b ¥ çè -0.05 0

(c)

b ¥

e-.00001x dx » 9.9995;

ò e 0 100

= lim

ò 225, 000e

-0.06t

= 144,000

The capital value is $144,000.

æ 225, 000 -0.06t ö÷ b e = lim çç ÷÷ø b ¥ çè -0.06 0 é ù = -3, 750, 000 ê lim (e-0.06b ) - e0 ú ëê b ¥ ûú

630

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION ¥

ò (6000 + 200t)e

-0.05t

ò0 7200e-0.10t dt

(b)

= lim

b ¥

ò0

= -20e-0.05t (6000 + 200t ) +

7200e-0.10t dt

æ 7200 -0.10t ö÷ e = lim çç ÷ ç ø÷ 0 b ¥ è -0.10

é ù = -72,000 ê lim (e-0.10b ) - e0 ú êë b ¥ úû

As b  ¥, e

ò 7200e

-0.10t

-0.05t

= lim

b ¥

ò (6000 + 200t)e

-0.05t

= lim [-20e-0.05t (6000 + 200t ) - 80, 000e-0.05t )] b ¥

dt = -72,000(0 - 1)

= 72,000

b ¥

-0.05b

= lim [e b ¥

b 0

(-120,000 - 4000b - 80,000)

+ 200, 000] = 0 + 200,000 = $200,000

ò 1200e

0.03t -0.07t

= lim

b ¥

1200e-0.04t dt

53.

ò 4200e

-0.075t

æ 1200 -0.04t ö÷ = lim çç e ÷÷ ø 0 b ¥ çè -0.04

= lim

é ù = -30,000 ê lim (e-0.04b ) - e0 ú ëê b ¥ ûú

4200e-0.075b = lim -0.075 b ¥

b ¥

ò 4200e

-0.075t

b 0

æ 4200e-0.075b 4200 ö÷÷ = lim ççç + ÷ -0.075 0.075 ÷ø÷ b ¥çè

converges.

1  0. The integral As b  ¥, e-0.04b = 0.04 b

= 0 + 560,000

1200e0.03t e-0.07t dt = -30,000(0 - 1)

= $560,000

= 30,000

The capital value is $30,000. 52.

54.

¥ B

R(t ) = 6000 + 200t , r = 0.05

The capital value is given by ¥

ò (6000 + 200t) e

-0.05t

= lim

b ¥

( y - B)

k 2 Be-ky

dy 1 - e-kB a æ ö÷ çç lim k 2 B ye-ky dy ÷÷÷ ç æ öç a ¥ 1 ÷÷ B ÷ç ÷ç ççç ÷ ÷÷÷ a è 1 - e-kB ø ççç 2 2 -ky çç - lim k B e dy ÷÷÷ çè a ¥ ø÷ B

ò (6000 + 200t) e

dt.

Let u = 6000 + 200t and dv = e-0.05t dt.

ky + 1 -ky e +C 2

ò ye dy = - k -ky

-0.05t

b 0

= lim [e-0.05t ((-20)(6000 + 200t ) - 80,000)]

The capital value is $72,000.

51.

-0.05t

0.10b  0.

The integral converges. ¥

ò (6000 + 200t)e

ò 4000e

= -20e-0.05t (6000 + 200t ) - 80,000e-0.05t + C

-0.10b

using integration by parts, and 1 e-ky dy = - e-ky + C k . Thus the expected sales lost are

Section 8.4

631

é 2 æ kB + 1 -kB ak + 1 -ka ö÷ ùú ê k B çç e e ÷÷ 2 ê èç k ø úú k2 ê lim ê a ¥ ê 1 -ka ö÷ úú 2 2 æç 1 -kB k B e e ÷÷ ú ç ê çè k ø û k ë . -kB 1- e Since k is positive, the two terms in the

numerator containing e  ka have limit 0 as a  , and the other two terms are constant, so the limit is é 2 æ kB + 1 -kB ö ù ÷÷ - k 2 B 2 æç 1 e-kB ÷ö÷ ú ê k B çç e ÷ø ú ççè k ÷ø çè k 2 ê ë û 1 - e-kB e-kB éê B(kB + 1) - kB 2 ùú ë û = 1 - e-kB =

2 x e-xdx ò b ¥ bù é = lim ê (-2 x 2e-x - 4 xe-x - 4e-x ) ú 0 úû b ¥ êë

= lim [(-2b 2e-b - 4be-b - 4e-b ) - (0 - 4 ⋅ 1)] b ¥

= -(-4) Use hint to evaluate limits = 4.

ò e-rt (at + b)K dt 0

= K lim

c ¥

Evaluate

ò (at + b)e-rt dt 0

ò (at + b)e-rt dt using integration by 0

parts. Let u = at + b and dv = e-rt dt.

-0.042t

Then du = a dt and v = - 1r e-rt .

= lim

b ¥

ò (at + b)e rt dt

663e-0.042t dt

æ 663 -0.042t ö÷ e = lim çç ÷ ø÷ 30 b ¥ çè -0.042

é æ 1 ö = ê (at + b) çç - e-rt ÷÷ ÷ø çè r ê ë

é ù = -15, 786 ê lim (e-0.042b ) - e-0.042(30) ú ëê b ¥ ûú 1  0. The integral As b  ¥, e-0.042b = 0.042 b e

converges.

ò 663e

-0.042t

dt = -15, 786(0 - e-0.042(30) )

= 4478 Annual production can be estimated to be 4478 million pounds.

é at + b -rt a = êe + êë r r

r ( x) =

2 x e dx ò 2x e dx = blim ¥ ò 2 -x

-x

Let u = 2 x and dυ = e

dx.

Use column integration to obtain

2 -x

ò e rt dt úúû -

c 0

é at + b -rt ù a = ê- 2 e-rt ú e êë úû r r 0 æ ac + b -rc ö æ b a a ö = çç - 2 e-rc ÷÷÷ - çç - e0 - 2 e0 ÷÷÷ e èç ø èç r ø r r r

Therefore, c ¥

r ¢( x) = 2 x 2e-x

ù æ 1 -rt ÷ö çç - e ÷ a dt ú ÷ø çè r ú û

K lim

56.

P =

57.

. 1 - e-kB

ò 663e

Be-kB

55.

lim

ò (at + b)e rt dt -

æ ac + b -rc a b a ö = K lim çç e - 2 e-rc + + 2 ÷÷÷ ç r r c ¥ è r r ø æ b a ö = K çç 0 - 0 + + 2 ÷÷÷ çè r r ø =

K (a + br ) r2

632 58.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION S = N

a(1 - e-kt ) -bt e dt k

Na lim k c ¥

Na = lim k c ¥

60.

ò0

e-0.06t b ¥ -0.06

(1 - e-kt )(e-bt )dt

0 c

ò (e

-bt

-(b + k )t

-e

)dt

é 1 ù 1 Na lim ê - e-bt + e-(b + k )t ú úû k c ¥ êë b b+k 0

éæ 1 ö 1 Na lim ê çç - e-bc + e-(b + k )c ÷÷÷ ç ê è ø k c ¥ ë b b+k

æ 1 öù 1 - çç - e0 + e0 ÷÷÷ ú èç b b + k ø úû 1 1 ö÷ Na æç = ÷ çç 0 + 0 + k è b b + k ÷ø Na (b + k ) - b ⋅ = k b(b + k ) Nak = kb(b + k ) Na = b(b + k )

61.

ò0 50e-0.04t dt b

e b ¥ ò0

= 50 lim

-0.04t

e-0.04t = 50 lim b ¥ -0.04

-0.03725t

æ 1 ö 50 lim çç 0.04b - 1÷÷÷ = ç -0.04 b ¥ è e ø As b  ¥,

1 e

 0.

0.04b

ò0 50e-0.04tdt = - -0.04

= 1250

= lim

b ¥

ò 2632e

50 = lim (e-0.06b - e0 ) -0.06 b ¥ 50 50 =(0 - 1) = 0.06 0.06 » 833.3

59.

= 50 lim

e-0.06t dt ò b ¥ 0

50e-0.06t dt = 50 lim

ò 2632e

-0.03725t

62.

(a) (1) =

ò0 e1-1e-xdx

æ 2632 ö e-0.03725t ÷÷÷ = lim çç ø 20 b ¥ çè -0.03725

= lim

é ù = -70, 658 ê lim (e-0.03725b ) - e-0.03725(20) ú ëê b ¥ ûú

aö æ = lim çç -e-x ÷÷÷ = lim 1 - e-a ç 0 ÷ø a ¥ è a ¥

1  0. The As b  ¥, e-0.03725b = 0.03725 b

a ¥

(b)  (2) =

2632e-0.03725t dt

= -70, 658(0 - e-0.03725(20) ) = 33,544 Annual deaths can be estimated to be 33,544.

ò0 e-xdx (

integral converges. ¥

ò0

= lim

a ¥

x 2 -1e-x dx =

ò0 xe-xdx

Let u = x and dv = e-x dx. Then du = dx and v = -e-x .

)

Chapter 8 Review

633

(2) = lim

a ¥

ò0

10.

xe-x dx

æ a = lim ççç -xe-x + 0 a ¥ çè

lim

a ¥

ò0 e dx

(c)

 (3) =

of these integrals diverges so ò

3 -1 -x

14.

(3) = lim

a ¥

ò x(8 - x)3/2 dx 2 2 = - x(8 - x)5/2 + (8 - x)5/2 dx 5 5 2 2æ 2ö = - x(8 - x)5/2 - çç ÷÷÷ (8 - x)7/2 + C 5 5 çè 7 ø

ò0 x2e-xdx

æ a = lim ççç -x 2e-x + 0 a ¥ çè

÷ö 2 xe-x dx ÷÷÷ ÷ø 0

xe-x dx a ¥ ò0

= 0 + 2 lim

u = x and dv = (8 - x)3/2.

Then du = dx and v = - 52 (8 - x)5/2.

Let u = x 2 and dv = e-x dx. Then du = 2 x dx and v = -e

ò x(8 - x)3/2 dx Let

2 -x

ò0 x2e-xdx -x

xe-2x dx

diverges.

a ¥

-¥

ò0 x e dx = ò0 x e dx

= lim

-x

= (1) = 1 ¥

ò xe-2 xdx + blim ò xe-2 xdx. The first ¥ c

a -¥ a

ö÷ e-x dx ÷÷÷ ÷ø 0

= (0 - 0) + lim

False: We must write the integral as

15.

= 2(2) = 2

ò x - 2 dx = ò 3x(x - 2)-1/2 dx Let

(d) The integration by parts argument remains the

2x 4 (8 - x)5/2 (8 - x)7/2 + C 5 35

u = 3x and dv = ( x - 2)-1/2 dx.

Then du = 3 dx and v = 2( x - 2)1/2.

same, but now u = x3 and du = 3x 2dx. We find that  (4) = 3 (3) = 6.

ò x - 2 dx = 6 x( x - 2)1/2 - 6

Chapter 8 Review Exercises

= 6 x( x - 2)1/2 -

False: This integral is best evaluated by substitution.

True

False: Using the substitution u = x 2 , 2

dv = xe-x this integral requires only one integration by parts.

True

False: The integrand should be just 2 x 2 + 3.

False: The integrand should be  ( x 2 + 1).

True

ò (x - 2)1/2 dx

6( x - 2)3/2 3 2

= 6 x( x - 2)1/2 - 4( x - 2)3/2 + C

16.

ò xe xdx Let

u = x and dv = e x dx.

Then du = dx and v = e x .

ò xe xdx = xe x - ò e xdx

= xe x - e x + C

634 17.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION

ò (3x + 6)e-3xdx

20.

Use substitution.

u = 3x + 6 and dv = e-3x dx.

Let

Then du = 3dx

and

v = -13 e-3x .

Let u = 25 - 9 x 2. Then du = -18x dx.

æ 1

ò 25 - 9x2 dx = - 18 ò 25 - 9x2 dx 1 du =18 ò u

ò (3x + 6)e-3xdx æ 1 ö = (3x + 6) çç - e-3x ÷÷ ÷ø çè 3

18.

ò 25 - 9x2 dx

ò çççè - 3 e-3x ÷÷÷ø 3 dx

= -( x + 2)e-3x +

ò e-3xdx

= -( x + 2)e-3x -

1 -3 x e +C 3

1 ln | u | +C 18 1 = - ln | 25 - 9 x 2 | + C 18 =-

ò ln | 4x + 5| dx

21.

ò 16 + 8x2 dx

First, use substitution.

Use substitution.

Let w = 4 x + 5. Then dw = 4 dx. 1 ln | 4 x + 5| dx = ln | w | dw 4

Let u = 16 + 8x 2. Then du = 16 x dx.

Then du = w1 dw

and

1 1 du 16 u 1 = u-1/2du 16 1 = (2)u1/2 + C 16 1 = (16 + 8x 2 )1/2 + C 8 1 = 16 + 8 x 2 + C 8

v = w. 1

= w ln | w | - w + C

Finally, resubstitute 4 x + 5 for w.

ò ln | 4x + 5| dx 1 = [(4 x + 5) ln | 4 x + 5| - (4 x + 5)] + C 4 1 = (4 x + 5)(ln | 4 x + 5| - 1) + C 4

ò ò

ò ln | w | dw = w ln | w | -ò w w dw = w ln | w | - dw ò 1 4

16 x

ò 16 + 8x2 dx = 16 ò 16 + 8x2 dx

Now use integration parts. Let u = ln w and dv = dw.

19.

-18 x

22.

ò1 x3 ln x dx Let u = ln x and dv = x3 dx . Use column integration. D

( x - 1) ln | x | dx

Let

ln x 1 x

u = ln | x | and dv = ( x - 1) dx.

Then du = 1x dx

I x3

+ -

1 x4 4

v = x2 - x.

and

ò (x - 1) ln x dx æ x2 æx ö ÷ö ç = çç - x ÷÷÷ ln | x | - çç - 1÷÷ dx ÷ø ç çè 2 è2 ÷ø æ x2 ö÷ x2 ç = çç - x ÷÷÷ ln | x | +x+C 4 ÷ø çè 2

ò x ln x dx 3

æ1 1 4 1ö x ln x - çç x 4 ⋅ ÷÷÷ dx çè 4 xø 4 1 1 x3 dx = x 4 ln x 4 4 1 1 4 x +C = x 4 ln x 4 16 =

ò ò

Chapter 8 Review e

635 D

ò x ln x dx 1

e2 x

e2 x 2

e2 x 4 e2 x 8

e4 e4 1 + 4 16 16

e æ1 1 4 ÷ö = çç x 4 ln x x ÷÷ çè 4 16 ø 1 æ e4 ö÷ e4 1 ç = çç ÷÷÷ (1) -0+ 16 16 çè 4 ø÷

3e4 + 1 = 16 » 10.30

x 2 e2 x dx = x 2 =

e2 x e2 x e2 x - 2x +2 2 4 8

x 2e 2 x xe2 x e2 x + 2 2 4 1

23.

æ3 x 2e 2 x xe2 x e2 x ö÷÷ ç A = çç e2 x + + ÷ 2 2 4 ÷÷ø çè 2 0

ò x e x dx 2

3 2 e2 e2 e2 æç 3 1ö e + + - ç + ÷÷÷ ç è 2 2 2 4 2 4ø æ 6 + 2 - 2 + 1 ÷ö 2 æ 7 ÷ö = çç ÷÷ e - ççç ÷÷ çè ø è4ø 4

e x /2

2e x /2

4e x /2

Let u = x 2 and dv = e x /2 dx. Use column integration. D x

x /2

x 2e x /2 dx = (2 x 2e x /2 - 8xe x /2 + 16e x /2 )

= 2e1/2 - 8e1/2 + 16e1/2 - 16 = 10e1/2 - 16 » 0.4872

24.

A= =

ò (3 + x )e dx 0 1

ò 3e xdx +ò x e xdx 2

3 2

ò 3e xdx = 2 e x + C Use column integration.

ò x (x - 1) dx 3

1/3

u = x 2 and dv = x( x 2 - 1)1/3 dx.

Let

Then du = 2 x dx and v = 83 ( x 2 - 1)4/3.

ò x (x - 1) dx 3

1/3

3x 2 2 3 ( x - 1) 4/3 8 4

ù 3x 2 2 3é1 3 ( x - 1) 4/3 - ê ⋅ ( x 2 - 1)7/3 ú ê úû 8 4ë2 7

3x 2 2 9 2 ( x - 1) 4/3 ( x - 1)7/3 + C 8 56

ò x(x - 1) dx 2

4/3

For the second integral,  ò x 2e2 x dx, let u = x 2 and dv = e2 x dx.

2 2

25.

7 2 (e - 1) » 11.18 4

é 3x 2 ù 9 2 A = êê ( x 2 - 1) 4/3 ( x - 1)7/3 úú 56 êë 8 úû 0 3 9 = (144) (128) 8 56 144 234 = 54 = » 33.43 7 7

636 26.

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION 2 =

vdu = ( uv )

1 0

29.

ò udv

f ( x) =

x - 4; y = 0; x = 13

ò udv = ( u(1)v(1) - u(0)v(0) ) - 2 0

= ( (5)(-1) - (-3)(4) ) - 2 = 5

27.

-11 =

vdu = ( uv )

2 0

Since f ( x) = x - 4 intersects y = 0 at x = 4, the integral has lower bound a = 4.

udv

udv = ( u (2)v(2) - u (0)v(0) ) + 11

V =

= ( (3)(-2) - (-6)(-4) ) + 11

=

= -19

ò ( x - 4) dx 2

4 13

ò (x - 4)dx 4

28.

æ x2 ö÷ ç =  çç - 4 x ÷÷÷ çè 2 ø÷ 4

f ( x) = 3x - 1, y = 0, x = 2

é æ 169 ù ö =  ê çç - 52 ÷÷÷ - (8 - 16) ú ê çè 2 ú ø ë û æ 65 ö 81  » 127.2 =  çç + 8 ÷÷÷ = çè 2 ø 2

30.

f ( x) = e-x , y = 0, x = -2, x = 1 V = =

Since f ( x) = 3x - 1 intersects y = 0

=

ò (3x - 1) dx 1/3 2

31.

ò (9x - 6x + 1)dx

=  (3x3 - 3x 2 + x)

f ( x) =

V =

2 1/3

é æ1 1 1 öù =  ê (24 - 12 + 2) - çç - + ÷÷÷ ú ê èç 9 3 3 ø úû ë æ 1ö 125 =  çç14 - ÷÷÷ = » 43.63 çè 9ø 9

-2

e

1/3

ò-

e-2 x dx =

-2 » 85.55

at x = 13 , the integral has a lower bound a = 13 . V =

=

e

-2

 (e - e-2 ) 4

1 , y = 0, x = 2, x = 4 x -1 ö2 çç 1 ÷÷ dx ÷ 2 çè x - 1 ø÷

4æ

ò x -1 2

=  (ln | x - 1|) 2 =  ln 3 » 3.451

 e-2 x

Chapter 8 Review 32.

637

f ( x) = 4 - x 2 , y = 0, x = -1, x = 1 V = =

V =

ò- (4 - x ) dx 2 2

1 1

æ 8x3 x 5 ÷÷ö ç =  çç 16 x + ÷ çè 3 5 ÷÷ø -1 æ 8 1 8 1ö =  çç 16 - + + 16 - + ÷÷÷ çè 3 5 3 5ø æ 16 2ö =  çç 32 + ÷÷÷ çè 3 5ø

33.

406 » 85.03 15

x2 , y = 0, x = 4 f ( x) = 4 2

36.

the integral has a lower bound, a = 0. V =

2 4æ 2 ö

4 4

çx ÷

ò çççè 4 ÷÷÷ø÷ dx =  ò 16 0

 ççæ x5 ÷÷ö

÷ 16 ççè 5 ÷ø÷

f ( x) =

x +1

1 b-a

òa

f ( x ) dx = =

= 0

 æç 1024 ö÷

1 8-0 1 8

x + 1 dx

ò ( x + 1) dx 1/2

÷ ç 16 çè 5 ÷ø

64 = » 40.21 5

34.

2 h æç 7r 3 ö÷÷ çç ÷ 3r çè 8 ÷ø÷

7 r 2h 12

Since f ( x) = x4 intersects y = 0 at x = 0,

é ù ö÷3 2 h ê æç r 3ú (0 ) r r =+ + ÷ ç ê ú ÷ø 3r ê çè 2 úû ë 3 é ù 2 h ê çæ r ÷ö 3ú =ê çç ÷÷ - r ú 3r ê è 2 ø úû ë ö÷ 2 h æç r 3 çç =- r 3 ÷÷÷ 3r çè 8 ø÷

ö÷3 2 h æç r x r =+ ÷÷ ç ø 3r çè 2h

ò- (16 - 8x + x ) dx 2

ö2 çç - r x + r ÷÷ dx ÷ø ç 0 è 2h hæ

1 æç 2 ö÷ 3/2 ç ÷ ( x + 1) 8 çè 3 ÷ø 0

1 1 (9)3/2 (1) 12 12 27 1 26 = = 12 12 12 13 = 6 =

The frustum may be shown as follows.

37.

Average value = =

1 2-0 7 2

ò 7 x (x + 1) dx 2

ò x (x + 1) dx 2

Let u = x3 + 1. Then du = 3x 2dx. Use the two points given to find f ( x) = -

r x + r. 2h

ò x (x + 1) dx = 3 ò 3x ( x + 1) dx 1 = u du 3ò 2

1 1 7 ⋅ u +C 3 7 1 3 = ( x + 1)7 + C 21 =

638

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION 7 2

7 1 3 ⋅ ( x + 1)7 2 21 0

x 2 ( x 3 + 1)6 dx =

41.

6e-x dx = lim

b ¥

1 7 1 (9 - 17 ) = (4,782,969 - 1) 6 6 2,391, 484 = 3

x-1dx = lim

b ¥

æ -6 6ö = lim çç b + ÷÷ ç e ÷ø b ¥ è e

As b  ¥, eb  ¥, so -b6  0. The integral e

converges.

42.

lim ò-¥ x- dx = a -¥ òa x- dx 2

æ x-1 ÷ö ç ÷÷ = lim çç a -¥ çè -1 ÷÷ø a

1 a -¥ 2

1 (ln | x 2 + 3|) a -¥ 2 a

-5

æ1ö 1 + lim çç ÷÷÷ 5 a -¥ çè a ø

ò-¥ ò

dx (3x + 1)

1 1 +0= 5 5

= lim

b ¥

43.

lim

(ln 3 - ln | a 2 + 3|)

ln(5 x)dx = lim

b ¥

ò ln(5x)dx 4

u = ln(5x) and dv = dx. and

v = x. 1

ò ln(5x)dx = x ln(5x) - ò x ⋅ x dx = x ln(5x) ò dx = x ln(5x) - x + C

é -1 ùú = lim ê b ¥ êë 3(3x + 1) úû 0 æ ö÷ 1 1 ÷÷ + = lim ççç 3 b ¥ è 3(3b + 1) ø÷ 1 As b  -¥, 3(3-b +  0. The integral 1)

converges. 1

ò (3x + 1) = 0 + 3 = 3 0

a -¥ 2

Then du = 1x dx

(3x + 1)-2 dx

é 1 (3x + 1)-1 ù ú = lim êê ⋅ ú -1 b ¥ ê 3 úû 0 ë

lim

2 x dx

òa x + 3

The integral is divergent.

Let

40.

lim

As a  -¥, 12 (ln 3 - ln | a 2 + 3|)  ¥.

As a  -¥, 1a  0. The integral converges. x-2dx =

æ 1ö = lim çç - ÷÷÷ a -¥ çè x ø a

-5

ò-¥ x + 3 dx

-5

As b  ¥, ln | b |  ¥. The integral diverges. -5

ò 6e xdx = 0 + e = e » 2.207

b ¥

39.

+ 6e-1)

= lim (-6e

= lim ln b - ln 10

-5

b 1

-b

ò x- dx

= lim ln | x |

b ¥

b ¥ 10

ò 6e-xdx

= lim - 6e-x

38.

lim

b ¥

ò ln(5x)dx 4

= lim [ x ln(5 x) - x] b ¥

b 4

= lim [b ln(5b) - b] - (4 ln 20 - 4) b ¥

As b  ¥, b ln(5b) - b  ¥. The integral diverges.

Chapter 8 Review 44.

A= =

639

ò-¥ (x - 2) 1

òa

lim

a -¥

f ( x) = 5000; 8 yr; 9%

48.

P =

5( x - 2)-2 dx

æ 5 ö÷ = 5 + lim çç ÷ a -¥ çè a - 2 ÷ø

P =

ò 3e-xdx

f (t ) = 150e0.04t ; 5 yr; 6%

50.

-x

= lim (-3e b ¥

P =

) 0

æ -3 ö = lim çç b + 3 ÷÷÷ ø b ¥ èç e

ò 150e 0 5

0.04t -0.06t

ò 150e- t dt 0.02

As b  ¥, -b3  0. e

150 -0.02t e -0.02 0

= -7500(e-0.1 - e0 )

A= 0+3= 3

= -7500(e-0.1 - 1) » 713.7193647 The present value is $713.72.

R¢ = x( x - 50)1/2

ò x(x - 50) dx 1/2

Let

» $174,701.45

ò 3e dx

-0.10t

» 250,000(-0.3012 + 1)

-x

R =

ò 25,000e

47.

æ e-0.10t ö÷ ç ÷÷ = 25,000 çç ÷ èç -0.10 ÷ø 0

f ( x) = 3e-x for [0, ¥)

b ¥

A= 5+0 = 5

= lim

f (t ) = 25,000; 12 yr; 10%

49.

5  0. The integral converges. As a  -¥, a 2

5000 -0.09 x dx = e -0.09

-5000 -0.72 5000 e = + » $28,513.76 0.09 0.09

ò 5000e

-0.09 x

1 é 5( x - 2)-1 ù ú = lim éê -5 ùú = lim êê ú -1 a -¥ ê a -¥ êë x - 2 úû a úû ë a

45.

1/2

u = x and dv = ( x - 50)

f (t ) = 15t; 18 mo; 8%

51. .

P =

Then du = dx and v = 23 ( x - 50)3/2.

1.5

ò 15te- t dt 0.08

ò x(x - 50) dx 1/2

= 15

1.5

ò te- t dt 0.08

2 2 = x( x - 50)3/2 ( x - 50)3/2dx 3 3 2 2 2 = x( x - 50)3/2 - ⋅ ( x - 50)5/2 3 3 5

é2 ù 4 R = ê x( x - 50)3/2 - ( x - 50)5/2 ú êë 3 úû 15 50 2 4 = (75)(253/2 ) - (255/2 ) 3 15 2500 = 6250 3 16, 250 = » $5416.67 3

Find the antiderivative using integration by parts. Let

u =t

Then du = dt

and dv = e-0.08t dt. and v =

1 e-0.08t -0.08

= -12.5e-0.08t

ò te

-0.08t

dt = -12.5te-0.08t -

ò (-12.5e

-0.08t

)dt

= -12.5te-0.08t - 156.25e-0.08t + C

640

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION P = 15

1.5

= 15(-12.5te-0.08t - 156.25e-0.08t )

and dv = e-0.04t dt.

u =t

Let

te-0.08t dt

Then du = dt

1.5

= -25e-0.04t

= 15[(-18.75e-0.12 - 156.25e-0.12 ) - (0 - 156.25)]

ò te

-0.04t

= 15(-175e-0.12 + 156.25)

dt = -25te-0.04t -

ò (-25e

-0.04t

)dt

= -25te-0.04t - 625e-0.04t + C

» 15.58385362 20e0.24

The present value is $15.58.

1 e-0.04t -0.04

and v =

ò te

-0.04t

52.

f (t ) = 1000; 5 yr; 6% per yr A = e0.06(5)

= 20e0.24 (-25te-0.04t - 625e-0.04t )

ò 1000e

-0.06t

æ 1000 -0.06t ÷ö e = e0.3 çç ÷÷ çè -0.06 ø0 é 1000 -0.3 ù (e = e0.3 ê - 1) ú êë -0.06 úû » 5830.980126

The accumulated value is $5830.98.

= 20e

0.24

-0.24

[(-150te

- 625e

6 0

) - (0 - 625)]

= 20(625e0.24 - 775) » 390.614379

The accumulated value is $390.61. 55.

f (t ) = 1000 + 200t; 10 yr; 9% per yr

e(0.09)(10)

ò (1000 + 200t)e

-0.09t

53.

f (t ) = 500e-0.04t ; 8 yr; 10% per yr A = e0.1(8)

ò 500e

-0.04t

⋅ e-0.1t dt

= e0.8

ò 500e

-0.14t

é ù 200 -0.09t ú 0.09t + ( 0.09 t 1) e ê -0.09 e ú (0.09)2 ëê ûú 0

0.9 ê 1000

é 1000 ù 200 -0.9 ú e = e0.9 êê + (e-0.9 - 1) + ( 1.9 1) ú (0.09)2 êë -0.09 úû » $30, 035.17

æ 500 -0.14t ö÷ e = e0.8 çç ÷÷ çè -0.14 ø0

56.

é 500 -1.12 ù (e = e0.8 ê - 1) ú êë -0.14 úû » 5354.971041

The accumulated value is $5354.97. 54.

f (t ) = 20t; 6 yr; 4% per yr A = e0.04(6)

ò 20te

-0.04t

= 20e0.24

ò te

-0.04t

f (t ) = 1000e0.05t

Use P =

ò f (t)e dt with -rt

r = 0.11 T = 7. P =

ò 1000e

0.05t

⋅ e-0.11t dt

= 1000

f (t ) = Cekt where C = 1000, k = 0.05

ò e

-0.06t

Find the antiderivative using integration by parts.

æ 1 ö e-0.06t ÷÷÷ = 1000çç çè -0.06 ø0 =

1000 -0.42 (e - 1) » $5715.89 -0.06

Chapter 8 Review e0.105(10)

57.

641

Average value

-0.105t ö æ ÷÷ 1.05 çç 10,000e

çç è

÷÷ ÷ø 0

-0.105

» $176,919.15 b

-kt

ò 50,000e

-0.09t

æ 50,000e-0.09t ÷ö ç = lim çç ÷÷ -0.09 b ¥ çè ÷ø÷ 0 é 50,000 -0.09(b) 50,000 ù = lim ê + e (1) ú 0.09 b ¥ ëê -0.09 ûú

61.

é -50,000 50,000 ù ú = lim ê + 0.09 b ê 0.09 úû b ¥ ë 0.09e

1 ( -0.89t 3 + 16.05t 2 + 926t ) 17 1

ò te dt

Let

u = t and dv = e-t dt.

 0.

0.5

(0.0137t 2 - 2.63t + 143)dt

ò0 te dt = 0.5(-te - e ) 0 -t

The total reaction over the first 5 hr is 0.4798.

Approximately 24% of men aged 65 and older in the labor force between 1950 and 2018.

-t

= 0.5(-5e-5 - e-5 + e0 ) » 0.4798

118

éæ 3 öù ê çç 0.0137(118) - 1.315(118)2 + 143(118) ÷÷ ú ê ÷÷ø ú 1 ê çè 3 ú = ê ö÷ úú 68 ê æç 0.0137(50)3 ê -ç - 1.315(50) 2 + 143(50) ÷÷÷ ú êë çè ø úû 3 » 24

e-t -1 5

-t

ö÷ 1 æç 0.0137t 3 2.63 2 ç t + 143t ÷÷÷ ç ø 68 è 3 2

ò e-t dt

= -te-t +

Average value

te-t + -1

te-t dt =

f (t ) = 0.0137t 2 - 2.63t + 143; [50, 118] 118

-t

As b  ¥, e0.09b  ¥, so -50,000 0.09e0.09b

1 118 - 50

0.5te-t dt = 0.5

Then du = dt and v = e-1 .

» $555,555.56

1 18

59.

é ù 1 ê ( -0.89(18)3 + 16.05(18)2 + 926(18) ) ú ê ú 17 ê -( -0.89 + 16.05 + 926 ) úû ë » 926 The average amount of coal consumed is about 926 thousand short tons.

ò Re dt =

18 1 (-2.67t 2 + 32.1t + 926)dt 18 - 1 1

ö÷ 1 æç -2.67t 3 32.1 2 ÷÷ ç 926 t t = + + ø÷ 17 çè 3 2

10,000e1.05 -1.05 (e = - 1) -0.105 » -272,157.25(-0.65006)

58.

f (t ) = -2.67t 2 + 32.1t + 926; [1, 18]

60.

10,000e-0.105t dt

62.

f (t ) = 125e-0.025t

The total amount of oil is found by evaluating the improper integral. ¥

ò0 125e

-0.025t

= lim

b ¥

ò0 125e-0.025t dt b

æ 125 -0.025t ö÷ e = lim çç ÷÷ ø0 b ¥ çè -0.025 = lim (-5000)(e-0.025b - e0 ) b ¥

æ ö 1 = -5000 çç lim 0.025 - 1÷÷÷ b è b ¥ e ø

642

Chapter 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION 1 As b  ¥, 0.025  0. The integral converges. b e

ò 125e

-0.025t

65.

(a)

dt = -5000(0 - 1) = 5000

T =

1 10 - 0

ò0 (160 - 0.05x2 ) dx

1 æç 0.05 x3 ö÷÷ çç160 x = ÷ 10 çè 3 ÷ø÷

The total amount of oil that will enter the bay is 5000 gallons.

ù 1 é 0.05 ê 160(10) (10)3 ú úû 10 êë 3 1 » (1583.3) » 158.3 10

= ¥

63.

ò40 237e-0.0277t dt b

237e-0.0277t dt ò b ¥ 40

= lim

(b) T = b

æ 237 ö e-0.0277t ÷÷÷ = lim çç ç ø 40 b ¥ è -0.0277

integral converges. ¥

ò40 277e-0.0277tdt

64.

ò40 55,545e

-0.05056t

= lim

b¥

ò40 55,545e

-0.05056t

dt b

é ù = -1, 098,596 ê lim (e-0.05056b ) - e-0.05056(40) ú êë b¥ úû

(c)

T =

1 40 - 0

ò0 (160 - 0.05x2 )dx

1 (5, 333.33) 40 » 133.3 »

integral converges. 0.05056t

é 1 ê æç (0.05)(40)3 ö÷÷ ç = (160)(40) ÷÷ êç 40 ê çè 3 ø÷ ë

1  0. The As b  ¥, e-0.05056b = 0.05056 b

ò 55,545e-

1 æç 0.05x3 ö÷÷ çç160 x = ÷ 40 çè 3 ÷÷ø

æ 55,545 -0.05056t ö÷ e = lim çç ÷ ø÷ 40 b¥ çè -0.05056

ò10 (160 - 0.05x2 )dx

é 1 ê æç 0.05(40)3 ö÷÷ = êçç160(40) ÷÷÷ 30 ê çè 3 ø ë ù æ 0.05(10)3 ö÷÷ ú ç - çç160(10) ÷÷ ú çè 3 ø÷ úû 1 » (5333.33 - 1583.33) 30 = 125

1  0. The As b  ¥, e-0.0277b = 0.0277 b

1 æç 0.05 x3 ö÷÷ ç160 x = ÷ ç 30 çè 3 ÷ø÷

é ù = -8556 ê lim ( e-0.0277b ) - e-0.0277(40) ú ëê b ¥ ûú

= -8556(0 - e-0.0277(40) ) = 2825 Annual emissions can be estimated to be 2825 million tons.

1 40 - 10

= -1, 098,596(0 - e-0.05056(40) ) = 145,385 Annual emissions can be estimated to be 145,000 million tons, or 145 billion tons.

Extended Application

643

Extended Application: Estimating Learning Curves in Manufacturing with Integrals 1.

Thus, 1 - 15 1 dx + 2 1 x is a slight underestimate of the area of the rectangles, which is 1 + 12 + 13 + 14 .

» 271,000 ⋅ 280-0.152 » $115,000

No; using the Change of Base formula for logarithms, we have ln r log r . = log 2 r = ln 2 log 2

C (280) » C (1) ⋅ 280-0.152

If C ( x) = axb is a solution to the function equation C (2n) = r ⋅ C (n), then

(b)

1- 5 1 dx + = ln x 2 1 x

= ln 5 + 1+

ln r ln 2

ln r

Thus, choose b = ln 2 . The only condition on a is that it be nonzero. Thus, choose a = 1. (a) The sum 1 + 12 + 13 + 14 is the area of the

region comprising the four rectangles in the 5 1

picture. The integral ò1

dx is the

area of the part of the region lying below the graph of y = 1x . The graph at the right in Figure 16 shows the parts of the region above the graph of y = 1x stacked vertically inside a rectangle of height 1. The lowest point in the shaded region is at a height 15 ; thus, the height of the shaded region is 1 - 15 . Since the width of the shaded region is 1 at its widest points, the shaded regions lies in a rectangle of area 1 - 15 . By observation, we see that slightly more than half of that rectangle is shaded, 1- 15 2

2 » 2.009 5

1 1 1 25 + + = » 2.083 2 3 4 12

2b = r, a ¹ 0, n ¹ 0 b ln 2 = ln r

2 5

2.009 - 2.083 » -0.037 » -3.7%. 2.009

a ⋅ 2b ⋅ n b = a ⋅ r ⋅ n b

The percent error is approximately

a(2n)b = r ⋅ an b

b =

is a slight underestimate of the area

of the shaded region.

Chapter 9

MULTIVARIABLE CALCULUS 9.1 Functions of Several Variables

9.1 Warmup Exercises

Your Turn 1

W1. 3 y

f ( x, y) = 4 x 2 + 2 xy +

f (2, 3) = 4(2)2 + 2(2)(3) +

3 3

= 16 + 12 + 1 = 29

W2.

Your Turn 2 f ( x, y, z ) = 4 xz - 3x 2 y + 2 z 2 f (1, 2, 3) = 4(1)(3) - 3(1)2 (2) + 2(3) 2 = 12 - 6 + 18 = 24

Your Turn 3 W3.

Your Turn 4 Use the Cobb-Douglas production function z = x1/4 y 3/4. 27 = x1/4 y 3/4 27 = y 3/4 x1/4

W5. f (3a) = 2(3a) 2 + 3(3a) - 6

( )

= 2 9a 2 + 9a - 6 = 18a 2 + 9a - 6

4 æ 27 ö÷4 çç ÷ = y 3/4 çè x1/4 ÷ø

(

y3 = 1/3

( ) y3

)

W6.

(27)4 x

f ( x + h) - f ( x) h é 2( x + h)2 + 3( x + h) - 6 ù - 2 x 2 + 3x - 6 ê úû = ë h

(

æ (27)4 ö÷1/3 ç ÷÷ = çç çè x ÷÷ø

y =

(27)4/3 1/3

W4. f (-2) = 2(-2)2 + 3(-2) - 6 = 8-6-6 = -4

)

81 x1/3

645

646

Chapter 9 MULTIVARIABLE CALCULUS

( 2x2 + 4xh + 2h2 + 3x + 3h - 6 ) - ( 2 x 2 + 3x - 6 ) =

(a)

f ( x, y) =

4 xh + 2h 2 + 3h = 4 x + 2h + 3 h

(b)

False. The expression z = f ( x, y) is a function of two variables if a unique value of z is obtained from each ordered pair of real numbers (x, y).

True

False. Substitute 4 for x 2 + y 2 into the ellipsoid equation:

100 1 = 10 9(100) + 5(1) f (100, 1) = log100

(c)

f (1000, 0) = =

(d)

4+ z =1 4

æ1 ö f çç , 5 ÷÷÷ = çè 10 ø

z 2 = -12 There is no real value for z that makes this equation true. Thus, the statement is false.

9(1000) + 5(0) log 1000 9000 3

( )

1 + 5(5) 9 10 1 log 10

25.9 -1 = - 25.9 =

f ( x, y) = 2 x - 3 y + 5

(a)

f (2, -1) = 2(2) - 3(-1) + 5 = 12

(b)

f (-4, 1) = 2(-4) - 3(1) + 5 = -6

(c)

f (-2, -3) = 2(-2) - 3(-3) + 5 = 10

(d)

f (0, 8) = 2(0) - 3(8) + 5 = -19

g ( x, y ) = x 2 - 2 xy + y3

f ( x, y) = e x + ln( x + y)

(a)

f (1, 0) = e1 + ln(1 + 0) = e

(b)

f (2, -1) = e 2 + ln(2 + (-1)) = e 2

(c)

f (0, e) = e0 + ln(0 + e) = 1 + 1 = 2

(d)

f (0, e 2 ) = e0 + ln(0 + e2 ) = 1 + 2 = 3

10. f ( x, y) = xe x + y

(a)

g (-2, 4) = (-2) 2 - 2(-2)(4) + 43 = 84

(a)

f (1, 0) = (1)e1+ 0 = e

(b)

g (-1, -2) = (-1) 2 - 2(-1)(-2) + (-2)3

(b)

f (2, -2) = (2)e 2 + (-2) = 2

(c)

f (3, 2) = (3)e3 + 2 = 3e5

(d)

f (-1, 4) = (-1)e-1+ 4 = -e3

= -11 2

(c)

g (-2,3) = (-2) - 2(-2)(3) + 3 = 43

(d)

g (5, 1) = (5)2 - 2(5)(1) + (1)3 = 16

11. 7.

905 2

= 10 10

(x2 + y2 ) + z = 1 4

9(10) + 5(2) log10

f (10, 2) = =

9.1 Exercises 1.

9x + 5 y log x

h( x, y) =

x2 + 2 y2

(a)

h(5, 3) =

25 + 2(9) =

(b)

h(2, 4) =

4 + 32 = 6

(c)

h(-1, -3) = 1 + 18 = 19

(d)

h(-3, -1) =

x+ y+z =9 If x = 0 and y = 0, z = 9. If x = 0 and z = 0, y = 9. If y = 0 and z = 0, x = 9.

9 + 2 = 11

Section 9.1 12.

647

x + y + z = 15

15.

x+y=4 If x = 0, y = 4. If y = 0, x = 4. There is no z-intercept.

16.

y+z =5

To find x-intercept, let y = 0, z = 0. x + 0 + 0 = 15 x = 15

To find y-intercept, let x = 0, z = 0. y = 15

To find z-intercept, let x = 0, y = 0. z = 15

Sketch the portion of the plane in the first octant that contains these intercepts.

x-intercept: y = 0, z = 0 0=5 Impossible, so no x-intercept y-intercept: x = 0, z = 0 y =5

13.

z-intercept: x = 0, y = 0 z =5

2 x + 3 y + 4 z = 12

Sketch the portion of the plane in the first octant that contains these intercepts and is parallel to the x-axis.

If x = 0 and y = 0, z = 3. If x = 0 and z = 0, y = 4. If y = 0 and z = 0, x = 6.

17.

x =5

The point (5, 0, 0) is on the graph. 14.

There are no y- or z-intercepts. The plane is parallel to the yz-plane.

4 x + 2 y + 3z = 24

x-intercept: y = 0, z = 0 4 x = 24 x =6 y-intercept: x = 0, z = 0 2 y = 24 y = 12 18.

z-intercept: x = 0, y = 0 3z = 24 z =8 Sketch the portion of the plane in the first octant that contains these intercepts.

z = 4 No x-intercept, no y-intercept Sketch the portion of the plane in the first octant that passes through (0, 0, 4) parallel to the xy-plane.

648 19.

Chapter 9 MULTIVARIABLE CALCULUS 3x + 2 y + z = 24

For z = 0, 3x + 2 y = 24. Graph the line 3x + 2 y = 24 in the xy-plane.

22.

2y -

x2 = z 3

If z = 0, we have y = 16 x 2. The level curve is

For z = 2 , 3x + 2 y = 22. Graph the line 3x + 2 y = 22 in the plane z = 2.

a parabola in the xy-plane with vertex at the origin.

For z = 4, 3x + 2 y = 20. Graph the line 3x + 2 y = 20 in the plane z = 4.

curve is a parabola in the plane z = 2 with vertex at the point (0,1, 2).

If z = 2, we have y = 16 x 2 + 1. The level

If z = 4, we have y = 16 x 2 + 2. The level curve is a parabola in the plane z = 4 with vertex at the point (0, 2, 4). Sketch portions of these curves in the first octant.

20.

3x + y + 2 z = 8

For z = 0, 3x + y = 8. Graph the line 3x + y = 8 in the xy-plane. For z = 2, 3x + y = 4. Graph the line 3x + y = 4 in the plane z = 2.

24.

x-intercept: y = 0, z = 0 (a ¹ 0)

For z = 4, 3x + y = 0. Graph the line 3x + y = 0 in the plane z = 4.

ax = d x =

d a

y-intercept: x = 0, z = 0 (b ¹ 0) by = d y =

d b

z-intercept: x = 0, y = 0 (c ¹ 0) 21.

y 2 - x = -z

cz = d

For z = 0, x = y 2. Graph x = y 2 in the xy-plane.

z =

For z = 2, x = y 2 + 2. Graph x = y 2 + 2 in the plane z = 2. For z = 4, x = y 2 + 4. Graph x = y 2 + 4 in the plane z = 4.

d c

If d ¹ 0, then for the plane to have a portion in the first octant, one of da ,

d , or d c b

must be

positive, so at least one of a, b, or c must have the same sign as d. If d = 0, then the trace in the xy-plane is the line x = - ba y, the trace in the yz-plane is y = - bc z, and the trace in the xz-plane is

x = - ac z. - ac , - bc , or - ba must be positive, so a, b, and c cannot all have the same sign.

Section 9.1 27.

649

z = x2 + y 2

30.

z = y 2 - x2

If z = 0,

The xz-trace is 2

z = x + 0= x . The yz-trace is

x2 = y 2 x =  y.

z = 0 + y 2 = y 2.

xy-trace: two intersecting lines. If x = 0,

Both are parabolas with vertices at the origin that open upward. The xy-trace is

z = y 2.

yz-trace: parabola, opening upward If y = 0,

0 = x 2 + y 2.

This is a point, the origin. The equation is represented by a paraboloid, as shown in (c). 28.

z = -x 2. xz-trace: parabola, opening downward Hyperbolic paraboloid

z 2 - y2 - x2 = 1

If z = 0, -( y 2 + x 2 ) = 1. This is impossible so there is no xy-trace. If x = 0, z 2 - y 2 = 1.

Both (a) and (e) are hyperbolic paraboloids, but only (a) has traces described by this function.

31.

xz-trace:

yz-trace: hyperbola If y = 0,

x2 z2 + = 1, an ellipse 16 4

z 2 - x 2 = 1. xz-trace: hyperbola The equation is represented by a hyperboloid of two sheets, as shown in (f ) .

29.

x2 y2 z2 + + =1 16 25 4

yz-trace: y2 z2 + = 1, an ellipse 25 4

xy-trace:

x2 - y 2 = z

x2 y2 + = 1, an ellipse 16 25

The xz-trace is

The graph is an ellipsoid, as shown in (b).

x 2 = z,

which is a parabola with vertex at the origin that opens upward. The yz-trace is

32.

z = 5( x 2 + y 2 )-1/2 =

5 2

x + y2

Note that z > 0 for all values of x and y xz-trace: y = 0

- y 2 = z,

which is a parabola with vertex at the origin that opens downward. The xy-trace is x2 - y2 = 0 x2 = y 2 x = y or x = - y,

z =

5 x

This gives one branch of the hyperbola xz = 5, where z > 0 and x > 0, and one branch of the hyperbola xz = -5, where z > 0 and x < 0.

which are two lines that intersect at the origin. The equation is represented by a hyperbolic paraboloid, as shown in (e).

650

Chapter 9 MULTIVARIABLE CALCULUS (b)

f ( x, y + h) - f ( x, y) h [4 x 2 - 2( y + h)2 ] - [4 x 2 - 2 y 2 ] h

4 x 2 - 2 y 2 - 4 yh - 2h2 - 4 x 2 + 2 y 2 h h(-4 y - 2h) = h = -4 y - 2h =

yz-trace: x = 0 z =

5 y2

5 | y|

(c)

f ( x + h, y) - f ( x, y) h = lim (8 x + 4h)

lim

h 0

h0

This gives one branch of the hyperbola yz = 5, where z > 0 and y > 0, and one branch of the hyperbola yz = -5, where z > 0 and y < 0.

= 8 x + 4(0) = 8x

(d)

f ( x, y + h) - f ( x, y) h = lim (-4 y - 2h)

lim

h 0

h0

= -4 y - 2(0) = -4 y

34.

f ( x, y) = 5x3 + 3 y 2

(a)

Level curves on planes z = k , where k > 0, are 5

k =

x + y2 25 k2

radii k5 . The graph of z = 5( x 2 + y 2 )-1/2 is (d). f ( x, y) = 4 x - 2 y

(a)

[5( x + h)3 + 3 y 2 ] - (5x3 + 3 y 2 ) h

5x3 + 15 x 2h + 15 xh 2 + 5h3 + 3 y 2 - 5 x3 - 3 y 2 h

h(15x 2 + 15 xh + 5h 2 ) h

These are circles with centers (0, 0, k ) and

33.

x2 + y 2 =

f ( x + h, y) - f ( x, y) h

= 15 x 2 + 15 xh + 5h 2

(b)

f ( x, y + h) - f ( x, y) h

f ( x + h, y) - f ( x, y) h

[4( x + h)2 - 2 y 2 ] - [4 x 2 - 2 y 2 ] = h

4 x 2 + 8xh + 4h 2 - 2 y 2 - 4 x 2 + 2 y 2 h h (8 x + 4h) = = 8 x + 4h h

[5x3 + 3( y + h)2 ] - [5x3 + 3 y 2 ] h

5x3 + 3 y 2 + 6 yh + 3h2 - 5x3 - 3 y 2 h h(6 y + 3h) = h = 6 y + 3h

(c)

lim

h 0

f ( x + h, y) - f ( x, y) h

= lim (15x 2 + 15 xh + 5h 2 ) h0 2

= 15 x + 15 x(0) + 5(0) 2 = 15 x 2

Section 9.1 (d)

651 The slope of the tangent line in the direction

f ( x, y + h) - f ( x, y) h h 0 = lim (6 y + 3h) lim

of y at (1, 1) is 3e 2.

h0

36.

= 6 y + 3(0) = 6 y

35.

f ( x, y) = xye x + y

The y = 1 column gives

f1( x) = 7.045 + 1.98 x.

f (1 + h, 1) - f (1, 1) h h 0

(a) lim

= lim

A linear regression on the y = 0 column gives f 0 ( x) = 4.019 + 1.989 x.

The y = 2 column gives

1+ 2h + h 2 +1

(1 + h)(1)e

1+1

- (1)(1)e

h0

(1 + h) e 2 + 2h + h - e2 = lim h h0 2

(1 + h) e 2h + h - 1 h h 0

= e2 lim

f 2 ( x) = 9.978 + 2.013x.

The y = 3 column gives f3 ( x) = 12.989 + 2.019 x.

Use the nearest integer values. f 0 ( x) = f ( x, 0) = 4 + 2 x = a + bx + c(0) = a + bx f1( x) = f ( x, 1) = 7 + 2 x = a + bx + c(1)

The graphing calculator indicates that 2

= a + bx + c f 2 ( x) = f ( x, 2) = 10 + 2 x

(1+ h)e2 h + h -1 lim = 3, thus h h 0

= a + bx + c(2) = a + bx + 2c

f (1+ h, 1)- f (1,1) lim = 3e 2. h h 0

f3 ( x) = f ( x, 3) = 13 + 2 x = a + bx + c(3)

The slope of the tangent line in the direction

= a + bx + 3c

of x at (1, 1) is 3e2. (b)

Thus, b = 2 and we have

f (1, 1 + h) - f (1, 1) h h 0 lim

4= a 7= a+c

(1)(1 + h)e1+1+ 2h + h - (1)(1)e1+1 h h 0

= lim

10 = a + 2c 13 = a + 3c

(1 + h)e2 + 2h + h - e2 = lim h h 0 2

(1 + h)e2h + h - 1 h h 0

= e2 lim

so a = 4 and c = 3. f ( x, y) = 4 + 2 x + 3 y

So, this limit reduces to the exact same limit as in part a. Therefore, since 2

(1 + h) e 2h + h - 1 lim = 3, h h 0

then f (1, 1 + h) - f (1, 1) = 3e 2. h h 0 lim

652 37.

Chapter 9 MULTIVARIABLE CALCULUS é3 ù -5 2 P( x, y) = 100 ê x-2/5 + y-2/5 ú 5 ëê 5 ûú

(a)

38.

z = 1.01x3/4 y1/4

500 = 1.01x3/4 y1/4

P(32, 1)

y1/4 =

é3 ù -5 2 = 100 ê (32)-2/5 + (1)-2/5 ú êë 5 úû 5

y =

é 3 æ 1 ö 2 ù -5 = 100 ê çç ÷÷÷ + (1) ú ê 5 çè 4 ø 5 ú ë û

500 1.01x3/4 (500)4 (1.01x3 / 4 ) 4

æ 500 ö÷4 1 = çç çè 1.01 ÷÷ø x3

æ 11 ö-5 = 100 çç ÷÷÷ çè 20 ø

6 ⋅ 1010 x3

æ 20 ö = 100 çç ÷÷÷ çè 11 ø » 1986.95

The production is approximately 1987 cameras. (b)

P(1, 32)

é3 ù -5 2 = 100 ê (1)-2/5 + (32)-2/5 ú êë 5 úû 5

39.

z = x 0.6 y 0.4 where z = 500 500 = x3/5 y 2/5

-5 é3 2 æç 1 ö÷ ùú ê = 100 (1) + ç ÷÷ ê5 5 èç 4 ø úû ë

500 x

æ 7 ö- 5 = 100 çç ÷÷÷ çè 10 ø

3/5

= y 2/5

æ 500 ÷ö5/2 çç ÷ = ( y 2/5 )5/2 çè x3/5 ÷ø

æ 10 ö 5 = 100 çç ÷÷÷ » 595 çè 7 ø

y =

The production is approximately 595 cameras.

y »

(500)5/2 x3/2 5,590,170 x3/2

P(32, 243)

é3 ù -5 2 = 100 ê (32)-2/5 + (243)-2/5 ú 5 ëê 5 ûú é 3 æ 1 ö 2 æ 1 ö ù -5 = 100 ê çç ÷÷÷ + çç ÷÷÷ ú ê 5 èç 4 ø 5 èç 9 ø ú ë û

40.

æ 7 ö-5 = 100 çç ÷÷÷ çè 36 ø

z = x 0.7 y 0.3

If x is doubled, z is multiplied by 20.75. If y is doubled, z is multiplied by 20.25. If both are doubled, z is multiplied by

æ 36 ö5 = 100 çç ÷÷÷ çè 7 ø

20.75 ⋅ 20.25 = 21.0 = 2. Thus, z is doubled.

» 359, 767.81

The production is approximately 359,768 cameras.

41.

The cost function, C, is the sum of the products of the unit costs times the quantities x, y, and z. Therefore, C ( x, y, z ) = 250 x + 150 y + 75z.

Section 9.1 42.

653

M = f (25, 0.05, 0.33) = =

(1 + 0.05)25 (1 - 0.33) + 0.33

46.

F =

v2 gl

(a)

2.56 =

[1 + (1 - 0.33)(0.05)]25 25

(1.05) (0.67) + 0.33

[1 + (0.67)(0.05)]25 » 1.1403

v2 9.81 ⋅ 1.2 v = 5.5 m/sec

2.56 =

The multiplier is 1.14. Since M > 1, the IRA account grows faster. 43.

M = f (40, 0.06, 0.28) = =

(b)

(1 + 0.06)40 (1 - 0.28) + 0.28 [1 + (1 - 0.28)(0.06)]40

47.

(1.06)40 (0.72) + 0.28

[1 + (0.72)(0.06)]40 » 1.416

(b)

= 48 - 2.43(5) - 1.81(15) - 1.22(0)

P(0, 0, 0) = 48 - 2.43(0) - 1.81(0) - 1.22(0) = 48

H =

48% of fish will be intolerant to pollution.

0.67

15.2(29) (38 - 16) 10.23 ln 29 - 10.74

A = 0.024265(178)0.3964 (72)0.5378

P(W , 0, 0) = 48 - 2.43W - 1.81(0) - 1.22(0) = 48 - 2.43W . 48 - 2.43W = 0

A = 0.024265(140)0.3964 (65)0.5378 » 1.62 m

(c)

(c) Any combination of values of W, R, and A that result in P = 0 is a scenario that will drive the percentage of fish intolerant to pollution to zero.

If R = 0 and A = 0:

» 1.89 m 2

(b)

8.7% of fish will be intolerant to pollution.

15.2(21)0.67 (36 - 4) 10.23ln 21 - 10.74 » 183 W

H =

A = 0.024265h0.3964m0.5378 (a)

= 8.7

(b) The maximum percentage will occur when the variable factors are a minimum, or when W = 0, R = 0, and A = 0.

» 135 W

45.

P(5, 15, 0)

15.2m0.67 (T - A) 10.23ln m - 10.74

H (m, t , A) =

(a)

v2 9.81 ⋅ 4 v = 1 m/sec

2.56 =

P (W , R, A) = 48 - 2.43W - 1.81R - 1.22 A

(a)

The multiplier is 1.416. Since M > 1, the IRA account grows faster.

44.

v2 9.81 ⋅ 0.09 v = 1.5 m/sec

48 2.43 » 19.75

W =

A = 0.024265(160)0.3964 (70)0.5378 » 1.78 m 2

(d) Answers will vary.

So W = 19.75, R = 0, A = 0 is one scenario. If W = 10 and R = 10: P(10, 10, A) = 48 - 2.43(10) - 1.81(10) - 1.22 A = 5.6 - 1.22 A 5.6 - 1.22 A = 0 5.6 1.22 » 4.59

654

Chapter 9 MULTIVARIABLE CALCULUS So W = 10, R = 10, A = 4.59 is another scenario.

(c)

(d) Since the coefficient of W is greater than the coefficients of R and A, a change in W will affect the value of P more than an equal change in R or A. Thus, the percentage of wetland (W) has the greatest influence on P. (d) 48.

I ( p, a, m, n, e) = (25.54 + 0.04 p - 7.92a + 2.62m + 4.46n + 0.15e)2

(a) Replace the variables with the given values. I (80, 23,34,16,50) = [25.54 + 0.04(80) - 7.92(23)

51.

+ 2.62(34) + 4.46(16) + 0.15(50)]2 = (14.52)

ln(T ) = 5.49 - 3.00 ln( F ) + 0.18ln(C ) eln(T ) = e5.49-3.00 ln( F ) + 0.18ln(C )

(a)

T = e5.49e-3.00 ln( F )e0.18ln(C )

= 210.8304

49.

The incidence of a dengue fever outbreak is about 211.

(b) Since the coefficient of a, the average temperature, is negative, it has a negative influence on the incidence of dengue fever. If all other variables are held constant, an increase in a results in a decrease in I, and vice-versa.

T »

A( 266, 107,484, 31,697, 24,870 ) = 53.02 + 0.383(266) + 0.0015(107,484) + 0.0028(31,697) - 0.0003(24,870)

52.

The estimated number of accidents is 397. 50.

)

eln( F ) 242.257C 0.18 F3

242.257(40)0.18 (2)3

» 58.82

T is about 58.8%. In other words, a tethered sow spends nearly 59% of the time doing repetitive behavior when she is fed 2 kg of food per day and neighboring sows spend 40% of the time doing repetitive behavior.

+ 0.0028U - 0.0003C

» 397

0.18

(b) Replace F with 2 and C with 40 in the preceding formula.

A( L, T , U , C ) = 53.02 + 0.383L + 0.0015T

(a)

e5.49eln(C

L( E , P) = 23E 0.6 P-0.267 (a) For the black rat: E = (23)(7.35)0.6 (150)-0.267 » 19.975 » 20.0 yr (b) For humans:

R = 3.968 - 0.0383T - 0.0224 H

(a) T = 10°, H = 30% R = 3.968 - 0.0383(10) - 0.0224(30) = 2.913 » 2.91

E = (23)(14,100)0.6 (68,700)-0.267 » 362.90 » 363 yr

T = 20°, H = 30% R = 3.968 - 0.0383(20) - 0.0224(30) = 2.53

53.

The girth is 2H + 2W . Thus, f ( L, W , H ) = L + 2 H + 2W .

T = 10°, H = 55% R = 3.968 - 0.0383(10) - 0.0224(55)

54.

Let the area be given by g ( L, W , H ).

= 2.353 » 2.35 (b) It decreases.

Then, g ( L, W , H ) = 2LW + 2WH + 2 LH ft 2.

Section 9.2 55.

655

f ( H , D) =

H 2 + D 2 with D = 3.75 in

Your Turn 3 2

(a)

¶ æ x 2 + y 3 ö÷ æ x 2 + y 3 ö÷ ¶ çe ( xy ) ÷ø + ççè e ø÷ ¶x ¶x èç 2 3ö æ æ 2 3ö = xy çç 2 xe x + y ÷÷ + çç e x + y ÷÷ y è ø è ø

f x ( x, y ) = xy ⋅

= (2 x 2 y + y )e x + y

(2.8125)2 + (3.75) 2

The length of the ellipse is approximately 4.69 inches, and its width is 3.75 inches. H 2 = D 5 2 H = D 5 2 H = (3.75) 5 H = 1.5 f (1.5, 3.75) =

» 4.69

(b)

f ( x, y ) = xye x + y Use the product rule.

H 3 = D 4 3 H = D 4 3 H = (3.75) 4 H = 2.8125 f (2.8125, 3.75)

f x (2, 1) = [2(2)2 (1) + 1]e2 +1 = 9e 5

¶ æ x 2 + y 3 ö÷ æ x 2 + y 3 ö÷ ¶ çe ( xy ) + çe ø÷÷ çè ø÷÷ ¶y ¶y çè 2 3ö æ æ 2 3ö = xy çç 3 y 2e x + y ÷÷÷ + çç e x + y ÷÷÷ x è ø è ø

f y ( x, y ) = xy ⋅

= (3xy 3 + x )e x + y 2

3 3

f y (2, 1) = [3(2)(1)3 + 2]e 2 +1 = 8e5

Your Turn 4 f ( x, y) = x 2e7 y + x 4 y5

(1.5)2 + (3.75)2

f x ( x, y) = 2 xe7 y + 4 x3 y 5

» 4.04

f y ( x, y) = 7 x 2e7 y + 5 x 4 y 4

The length of the ellipse is approximately 4.04 inches, and its width is 3.75 inches.

¶ f x ( x, y) ¶x ¶ = (2 xe7 y + 4 x3 y5 ) ¶x

f xx ( x, y ) =

= 2e7 y + 12 x 2 y 5

9.2 Partial Derivatives

¶ f y ( x, y) ¶y ¶ = (7 x 2e7 y + 5 x 4 y 4 ) ¶y

f yy ( x, y ) =

Your Turn 1 f ( x, y) = 2 x 2 y 3 + 6 x5 y 4 f x ( x, y) = 4 xy 3 + 30 x 4 y 4 2 2

= 49 x 2e7 y + 20 x 4 y 3

5 3

f y ( x, y) = 6 x y + 24 x y

¶ f x ( x, y) ¶y ¶ = (2 xe7 y + 4 x3 y 5 ) ¶y

f xy ( x, y) =

Your Turn 2 2

f ( x, y) = e3x y ¶ (3x 2 y) ¶x

f x ( x, y) = e3x y ⋅ 2

= 6 xye3x y 2

f y ( x, y) = e3x y ⋅

= 14 xe7 y + 20 x3 y 4 ¶ f y ( x, y) ¶x ¶ = (7 x 2e7 y + 5 x 4 y 4 ) ¶x

f yx ( x, y) =

¶ (3x 2 y) ¶y

= 14 xe7 y + 20 x3 y 4

= 3 x 2e 3 x y

656

Chapter 9 MULTIVARIABLE CALCULUS

9.2 Warmup Exercises W1.

W7.

f ( x) = 2 x3 + 7 x - 12

f ( x) = ln 2 x3 + 5 f ¢( x ) =

( )

f ¢( x) = 2 3x 2 + 7(1) - 0

= 6x2 + 7

f ( x) =

x +

(

5 x2

)

(

)

(

2 x3 + 5

1 2 x

10 x3

(

)

f ( x) = xe x Use the product rule. æ d ö÷ x æ d ö f ¢( x) = çç x ÷ e + ( x) çç e x ÷÷ èç dx ø÷ èç dx ø÷

9.2 Exercises 1.

True

Use the quotient rule.

True

12 x3 )( x 2 - 5 ) - ( 3x 4 )( 2 x ) ( ¢ f ( x) = 2 ( x2 - 5 )

True

z = f ( x, y) = 6 x 2 - 4 xy + 9 y 2

= e x + xe x

f ( x) =

x -5

6 x5 - 60 x3

(a)

¶z = 12 x - 4 y ¶x

(b)

¶z = -4 x + 18 y ¶y

(c)

¶f (2,3) = 12(2) - 4(3) = 12 ¶x

(d)

f y (1, -2) = -4(1) + 18(-2)

( x2 - 5 )

f ( x) =

2 x + 8 Use the chain rule. ö÷ æ d æ 1 öæ ö 1 ÷ ç (2 x + 8) ÷÷ f ¢( x) = çç ÷÷ ççç ÷ø çè 2 ÷ø èç 2 x + 8 ø÷÷ ççè dx =

)

( )

+ 12e2 x

( )

W5.

6x2

Use the product rule and the chain rule. æ d ö æ d ö f ¢( x) = ççç ( x + 1) ÷÷÷ (ln x 2 ) + ( x + 1) ççç ln x 2 ÷÷÷ è dx ø è dx ø æ 1 d 2 ö÷ x ÷ = ln x 2 + ( x + 1) çç 2 çè x dx ÷ø ( x + 1)(2 x) 2( x + 1) = ln x 2 + = ln x 2 + 2 x x

+ 6e2 x = x1/ 2 + 5x-2 + 6e2 x

f ¢( x ) = (1/2) x-1/ 2 + 5 -2 x-3 + 6 2e2 x

W4.

æ d ö çç 2 x3 + 5 ÷÷÷ ç ø 2 x + 5 è dx 1

W8. f ( x) = ( x + 1) ln x 2

W2.

W3.

Use the chain rule.

1 2x + 8

= -40

æ 2 ö3 W6. f ( x) = 9 çç e x + 5 x ÷÷ Use the chain rule twice. è ø é æ 2 ö2 æ d æ 2 öö ù f ¢( x) = 9 êê (3) çç e x + 5 x ÷÷ çç çç e x + 5x ÷÷ ÷÷ úú è ø çè dx è ø ÷ø ú ëê û ù æ 2 ö2 é 2 æ d 2 ÷ö x ÷÷ + 5 ú = 27 çç e x + 5 x ÷÷ ê e x çç ú è ø êë èç dx ø û 2

2 æ 2 ö æ ö = 27 çç e x + 5 x ÷÷ çç 2 xe x + 5 ÷÷ è ø è ø

z = g ( x, y) = 8x + 6 x 2 y + 2 y 2

(a)

¶g = 8 + 12 xy ¶x

(b)

¶g = 6x2 + 4 y ¶y

(c)

(d)

¶z ¶g (-3, 0) = (-3, 0) = 6(-3)2 + 4(0) ¶y ¶y = 54 g x (2, 1) =

¶g (2, 1) = 8 + 12(2)(1) = 32 ¶x

Section 9.2 7.

657

f ( x, y) = -4 xy + 6 y3 + 5 f x ( x, y) = -4 y

12.

f x ( x, y) = 12e3x + 2 y

f y ( x, y) = -4 x + 18 y 2

f y ( x, y) = 8e3x + 2 y

f x (2, -1) = -4(-1) = 4 f y (-4, 3) = -4(-4) + 18(3)

f x (2, -1) = 12e3(2) + 2(-1) = 12e 4

f y (-4, 3) = 8e3(-4) + 2(3) = 8e-6

= 16 + 18(9) = 178

13.

f ( x, y) = -6e4 x -3 y

f ( x, y) = 9 x 2 y 2 - 4 y 2

f x ( x, y) = -24e4 x-3 y

f x ( x, y) = 18 xy 2

f y ( x, y) = 18e 4 x -3 y

f y ( x, y) = 18 x 2 y - 8 y

f x (2, -1) = -24e 4(2)-3(-1) = -24e11

f x (2, -1) = 18(2)(-1)2 = 36

f y (-4, 3) = 18e 4(-4)-3(3) = 18e-25

f y (-4, 3) = 18(-4)2 (3) - 8(3) = 840

14. 9.

f ( x, y) = 4e3x + 2 y

f ( x, y) = 8e7 x- y

f ( x, y) = 5x 2 y 3

f x ( x, y) = 56e7 x - y

f x ( x, y) = 10 xy 3

f y ( x, y) = -8e7 x - y

f y ( x, y) = 15 x 2 y 2

f x (2, -1) = 56e7(2)-(-1) = 56e15

f x (2, -1) = 10(2)(-1)3 = -20

f y (-4, 3) = -8e7(-4)-3 = -8e-31

f y (-4, 3) = 15(-4)2 (3) 2 = 2160

15. 10.

f ( x, y ) =

f ( x, y) = -3x 4 y 3 + 10 f x ( x, y) = -12 x3 y 3

f x ( x, y) =

f y ( x, y) = -9 x 4 y 2 f x (2, -1) = -12(2)3 (-1)3 = 96

f y (-4, 3) = -9(-4) 4 (3) 2 = -20, 736 =

11.

f ( x, y) = e

x+ y

f x ( x, y) = e x + y f y ( x, y) = e

x+ y

f x (2, -1) = e2-1 = e1 = e f y (-4, 3) = e-4 + 3 = e-1 =

f y ( x, y) =

1 e

= = f x (2, -1) =

x 2 + y3 x3 - y 2 2 x( x 3 - y 2 ) - 3 x 2 ( x 2 + y 3 ) ( x3 - y 2 ) 2 2 x 4 - 2 xy 2 - 3x 4 - 3x 2 y3 ( x3 - y 2 ) 2 -x 4 - 2 xy 2 - 3x 2 y 3 ( x3 - y 2 ) 2 3 y 2 ( x3 - y 2 ) - (-2 y)( x 2 + y 3 ) ( x3 - y 2 ) 2 3x3 y 2 - 3 y 4 + 2 x 2 y + 2 y 4 ( x3 - y 2 ) 2 3x3 y 2 - y 4 + 2 x 2 y ( x3 - y 2 ) 2 -24 - 2(2)(-1) 2 - 3(22 )(-1)3 [23 - (-1)2 ]2

=f y (-4, 3) =

3(-4)3 (3) 2 - 34 + 2(-4) 2 (3)

8 49 [(-4)3 - 32 ]2 1713 5329

658 16.

Chapter 9 MULTIVARIABLE CALCULUS f ( x, y) = f x ( x, y) = = f y ( x, y) = =

3x 2 y 3 x2 + y 2

2(2)3 - (-2)(-1)2 15 = 7 2(3) f y (-4, 3) = 2(-4)2 - (3)2 6 =23

( x 2 + y 2 ) ⋅ 6 xy 3 - 3x 2 y3 ⋅ 2 x ( x 2 + y 2 )2 6 xy5 ( x 2 + y 2 )2 ( x 2 + y 2 ) ⋅ 9 x 2 y 2 - 3x 2 y 3 ⋅ 2 y ( x 2 + y 2 )2 9 x 4 y 2 + 3x 2 y 4

19.

( x2 + y 2 )2

f y ( x, y) = x3e x y

2 2

9(-4) (3) + 3(-4) (3)

f y (-4, 3) = -64e 48

20.

f ( x, y ) = ln |1 + 5x y |

f y ( x, y) = f x (2, -1) =

1 1 + 5x3 y 2 1 1 + 5x3 y 2

⋅ 15 x 2 y 2 = ⋅ 10 x y =

1 + 5(2)3 (-1)

= 2

= f y ( x, y) =

f y (-4,3) = 3e-4 +3(3)[3(3) + 2]

1 + 5 x3 y 2

60 41

f ( x, y) = ln |4x 4 - 2 x 2 y 2 | 1 f x ( x, y) = ⋅ (16 x3 - 4 xy 2 ) 4 4x - 2x2 y 2 16 x - 4 xy

f x (2, -1) = (-1)2 e2 +3(-1) = e-1

10 x3 y

1920 f y (-4, 3) = = 3 2 2879 1 + 5(-4) (3)

= ye x +3 y (3 y + 2)

1 + 5 x3 y 2

10(-4)3 (3)

f y ( x, y) = y 2 ⋅ 3e x +3 y + e x +3 y ⋅ 2 y

15 x 2 y 2

15(2) 2 (-1)2

f ( x, y) = y 2e x + 3 y f x ( x, y) = y 2e x +3 y

3 2

f x ( x, y) =

18.

f x (2, -1) = e-4 (1 - 8) = -7e-4

[(-4) + (3) ] 24, 624 = 625

17.

= 33e5

21.

f ( x, y) =

f x ( x, y) = f y ( x, y) = f x (2, -1) =

4x4 - 2x2 y 2 =

8x 2 - 2 y 2 2 x3 - xy 2 1 4x4 - 2x2 y 2

==-

= e x y (1 + 2 x 2 y)

2 2

[(2) + (-1) ] 12 =25

f y (-4, 3) =

f ( x, y) = xe x y f x ( x, y) = e x y ⋅ 1 + x(2 xy)(e x y )

6(2)(-1)5

f x (2, -1) =

8(2)2 - 2(-1)2

f x (2, -1) =

⋅ (-4 x y)

f y (-4,3) =

4x2 y

4x4 - 2x2 y 2 2y 2x2 - y 2

x 4 + 3xy + y 4 + 10

4 x3 + 3 y 2 x 4 + 3xy + y 4 + 10 3x + 4 y 3 2 x 4 + 3xy + y 4 + 10 4(2)3 + 3(-1) 2 24 + 3(2)(-1) + (-1) 4 + 10 29 2 21 3(-4) + 4(3)3 2 (-4)4 + 3(-4)(3) + 34 + 10 48 311

Section 9.2 22.

659

f ( x, y) = (7 x 2 + 18 xy 2 + y 3 )1/3

24.

14 x + 18 y 2

f x ( x, y) =

f x ( x, y) = 7e x + 2 y (e x + y 2 + 2)

3(7 x 2 + 18 xy 2 + y 3 )2/3 36 xy + 3 y

f y ( x, y) =

+ 2 xe x (7e x + 2 y + 4)

f y ( x, y) = 14e x + 2 y (e x + y 2 + 2)

3(7 x + 18 xy 2 + y 3 )2/3

f x (2, -1) = =

f ( x, y) = (7e x + 2 y + 4)(e x + y 2 + 2)

14(2) + 18(-1)2

+ 2 y (7e x + 2 y + 4)

3(7(2)2 + 18(2)(-1)2 + (-1)3 )2/3 46

f x (2, -1) = 7e2 + 2(-1) (e2 + (-1) 2 + 2)

3(63)

+ 2(2)e 2 (7e2 + 2(-1) + 4)

2/3

= 7e0 (e4 + 3) + 4e4 (7e0 + 4)

36(-4)(3) + 3(3)2

f y (-4, 3) =

= 51e4 + 21

3(7(-4)2 + 18(-4)(3) 2 + (3)3 )2/3 135 =(509) 2/3

f y (-4, 3) = 14e-4 + 2(3) (e(-4) + 32 + 2) + 2(3)(7e-4 + 2(3) + 4) = 14e2 (e16 + 11) + 6(7e2 + 4)

23. f ( x, y) = f x ( x, y) = = f y ( x, y) = = f x (2, -1) = = =

f y (-4, 3) = = =

3x 2 y

25.

e xy + 2

f ( x, y) = 4 x 2 y 2 - 16 x 2 + 4 y

6 xy(e xy + 2) - ye xy (3x 2 y)

f x ( x, y) = 8xy 2 - 32 x

(e xy + 2)2

f y ( x, y) = 8x 2 y + 4

6 xy(e xy + 2) - 3x 2 y 2e xy

f xx ( x, y) = 8 y 2 - 32

(e 2

3x (e

+ 2)

f yy ( x, y) = 8x 2

+ 2) - xe (3x y)

f xy ( x, y) = f yx ( x, y) = 16 xy

(e xy + 2) 2 3x 2 (e xy + 2) - 3x3 ye xy

26.

(e xy + 2) 2

g x ( x, y) = 20 x3 y 2 - 9

6(2)(-1)(e2(-1) + 2) - 3(2)2 (-1)2 e2(-1)

g y ( x, y) = 10 x 4 y + 36 y 2

(e 2(-1) + 2)2

g xx ( x, y) = 60 x 2 y 2

-12e-2 - 24 - 12e-2 (e-2 + 2) 2

g yy ( x, y) = 10 x 4 + 72 y

-24(e-2 + 1)

g xy ( x, y) = g yx ( x, y) = 40 x3 y

(e-2 + 2)2

(

)

3(-4)2 e(-4)(3) + 2 - 3(-4)3 (3)e(-4)(3) (e(-4)(3) + 2) 2 48e-12 + 96 + 576e-12 (e-12 + 2)2 -12

624e

-12

+ 96

+ 2)2

g ( x, y) = 5 x 4 y 2 + 12 y3 - 9 x

27.

R( x, y) = 4 x 2 - 5xy 3 + 12 y 2 x 2 Rx ( x, y) = 8 x - 5 y 3 + 24 y 2 x R y ( x, y) = -15xy 2 + 24 yx 2 Rxx ( x, y) = 8 + 24 y 2 R yy ( x, y) = -30 xy + 24 x 2 Rxy ( x, y) = -15 y 2 + 48xy = R yx ( x, y)

660 28.

Chapter 9 MULTIVARIABLE CALCULUS k yy ( x, y) = 21x(-2)(2 x + 3 y)-3 (3)

h( x, y) = 30 y + 5 x 2 y + 12 xy 2 hx ( x, y) = 10 xy + 12 y 2

(2 x + 3 y)3 k xy ( x, y) = k yx ( x, y)

hxx ( x, y) = 10 y hxy ( x, y) = 10 x + 24 y

= -21y(-2)(2 x + 3 y)-3 (3)

hy ( x, y) = 30 + 5 x 2 + 24 xy hyy ( x, y) = 24 x hyx ( x, y) = 10 x + 24 y

29.

r ( x, y) = rx ( x, y) =

6y x+ y ( x + y)(0) - 6 y(1)

31.

( x + y) 2

z xy = z yx = 9e x

12 y ( x + y )3

32. (1)

z y = -6 xe y z xx = 0

= -6 y(-2)( x + y)-3 (1) + ( x + y)-2 (-6) 12 y - 6( x + y) = ( x + y )3 6 y - 6x = ( x + y )3

k ( x, y) =

k x ( x, y) =

z = -6 xe y z x = -6e y

12 x

( x + y )3 rxy ( x, y) = ryx ( x, y)

30.

z = 9 ye x

z yy = 0

rxx ( x, y) = -6 y(-2)( x + y)-3 (-1)

-7 x 2x + 3y

(2 x + 3 y)(-7) - (-7 x)(2) (2 x + 3 y) 2

= -21y(2 x + 3 y)-2 (2 x + 3 y)(0) - (-7 x)(3) k y ( x, y) = (2 x + 3 y) 2 = 21x(2 x + 3 y)-2

z yy = -6 xe y z xy = z yx = -6e y

33.

r = ln | x + y | 1 rx = x+ y 1 ry = x+ y -1 rxx = ( x + y)2 -1 ryy = ( x + y)2 -1 rxy = ryx = ( x + y) 2

k xx ( x, y) = -21y(-2)(2 x + 3 y)-3 (2) =

(2 x + 3 y)3

z xx = 9 ye x

= 6 x( x + y )

ryy ( x, y) = 6 x(-2)( x + y)

(2 x + 3 y)3 63 y - 42 x

z y = 9e x

-2

-3

+ (2 x + 3 y)-2 (-21) 126 y + (-21)(2 x + 3 y)

z x = 9 ye x

= -6 y( x + y)-2 ( x + y)(6) - 6 y(1) ry ( x, y) = ( x + y) 2

126 x

84 y (2 x + 3 y)3

Section 9.2 34.

661

k = ln |5x - 7 y |

z xx ( x, y ) = -3x-2 ( y + 1) =

x2 y ⋅ 1 - ( y + 1) ⋅ 1 1 z yy ( x, y ) = + 2 y y

5 k x ( x, y) = = 5(5x - 7 y)-1 5x - 7 y -7 k y ( x, y) = = -7(5x - 7 y)-1 5x - 7 y

k xx ( x, y) = -5(5 x - 7 y)-2 ⋅ 5 -2

= -25(5 x - 7 y)

-49

37.

35.

35 (5x - 7 y) 2

We must solve the system 12 x + 6 y + 36 = 0 12 y + 6 x = 0.

35 (5x - 7 y) 2

Multiply both sides of the first equation by -2 and add.

z = x ln | xy | z x = ln | xy | + 1

-24 x - 12 y - 72 = 0 =0 6 x + 12 y

x y 1 z xx = x

-18x

zy =

z yy = -xy-2 = z xy = z yx =

36.

1 y

f ( x, y) = 6 x 2 + 6 y 2 + 6 xy + 36 x - 5

k yx ( x, y) = 7(5x - 7 y)-2 ⋅ (5) = 35(5 x - 7 y)-2 or

First, f x = 12 x + 6 y + 36 and f y = 12 y + 6 x.

k xy ( x, y) = -5(5x - 7)-2 ⋅ (-7) = 35(5 x - 7 y)-2 or

3 x 1 3 z yx ( x, y ) = 3 ⋅ = x x

(5 x - 7 y)2

(5x - 7 y) 2

z xy ( x, y ) = 3x-1 =

-25

k yy ( x, y) = 7(5x - 7 y)-2 ⋅ (-7) = -49(5x - 7 y)-2 or

-3( y + 1)

Substitute into either equation to get y = 2. The solution is x = -4, y = 2.

-x y2

38.

1 y

z = ( y + 1) ln | x3 y | = ( y + 1)(3 ln | x | + ln | y |) æ 1ö z x ( x, y) = ( y + 1) ⋅ çç 3 ⋅ ÷÷÷ çè xø =

f ( x, y) = 50 + 4 x - 5 y + x 2 + y 2 + xy f x ( x, y) = 4 + 2 x + y, f y ( x, y) = -5 + 2 y + x

Solve the system 4 + 2x + y = 0 -5 + x + 2 y = 0.

Multiply the second equation by -2 and add.

3( y + 1) x

4 + 2x + y = 0 10 - 2 x - 4 y = 0 - 3y = 0 14 14 y = 3

= 3x-1( y + 1) æ1ö z y ( x, y ) = ( y + 1) ⋅ çç ÷÷÷ + (3 ln x + ln y ) ⋅ 1 çè y ÷ø =

- 72 = 0 x = -4

y +1 + 3 ln x + ln | y | y

Substitute into the second equation to get x = - 13 . The solution is x = - 13 , y = 14 . 3 3 3

662 39.

Chapter 9 MULTIVARIABLE CALCULUS x = 0 or

f ( x, y) = 9 xy - x3 - y 3 - 6

First, f x = 9 y - 3x 2 and f y = 9 x - 3 y 2.

Since y = - 92 x,

We must solve the system

if x = 0, y = 0;

x = 4

9 y - 3x 2 = 0

if x = 4, y = - 92 (4) = -18.

9 x - 3 y 2 = 0.

The solutions are x = 0, y = 0, and x = 4, y = -18.

From the first equation, y = 13 x 2. Substitute into the second equation to get

41.

f x ( x, y, z ) = 4 x3

æ 1 ö2 9 x - 3 çç x 2 ÷÷÷ = 0 çè 3 ø

f y ( x, y, z ) = 2 z 2

æ1 ö 9 x - 3 çç x 4 ÷÷÷ = 0 çè 9 ø 9x -

f ( x, y, z ) = x 4 + 2 yz 2 + z 4

f z ( x, y, z ) = 4 yz + 4 z 3 f yz ( x, y, z ) = 4 z

1 4 x = 0. 3

42.

Multiply by 3 to get

f ( x, y, z ) = 6 x3 - x 2 y 2 + y5 f x ( x, y, z ) = 18x 2 - 2 xy 2

27 x - x 4 = 0.

f y ( x, y, z ) = -2 x 2 y + 5 y 4

Now factor.

f z ( x, y, z ) = 0

x(27 - x3 ) = 0

f yz ( x, y, z ) = 0

Set each factor equal to 0. x = 0 or 27 - x3 = 0

43.

x =3 2

Substitute into y = x3 . y = 0 or y = 3

The solutions are x = 0, y = 0 and x = 3, y = 3. 40.

f ( x, y) = 2200 + 27 x3 + 72 xy + 8 y 2 f x ( x, y) = 81x 2 + 72 y, f y ( x, y) = 72 x + 16 y

Solve the system

44.

6x - 5 y 4z + 5 6 f x ( x, y, z ) = 4z + 5 -5 f y ( x, y, z ) = 4z + 5 -4 (6 x - 5 y) f z ( x, y, z ) = (4 z + 5) 2 20 f yz ( x, y, z ) = (4 z + 5)2 f ( x, y, z ) =

2 x 2 + xy yz - 2 1 = (2 x 2 + xy) yz - 2

f ( x, y, z ) =

81x 2 + 72 y = 0 (1) 72 x + 16 y = 0. (2)

Divide equation (1) by 9 and equation (2) by 8. 9 x 2 + 8 y = 0 (3) 9 x + 2 y = 0 (4)

From equation (4), y = - 92 x. Substitute into equation (3). 9 x 2 - 36 x = 0 9 x( x - 4) = 0

= (2 x 2 + xy) ( yz - 2)-1

Section 9.2

663 1 (4 x + y) yz - 2 4x + y = yz - 2

f z ( x, y , z ) =

f x ( x, y, z ) =

f y ( x, y, z ) = =

( yz - 2) ⋅ x - (2 x 2 + xy) ⋅ z ( yz - 2)

f yz ( x, y, z ) =

-2 x - 2 x 2 z

( yz - 2) 2

f z ( x, y, z ) = -(2 x 2 + xy)( yz - 2)-2 ⋅ y =-

47.

(2x 2 y + xy 2 )

( yz - 2)2 f yz ( x, y, z ) = =

( yz - 2) 4 4 x 2 + 4 xy + 2 x 2 yz

8 xy + 5 yz - x 3 5y 8 xy + 5 yz - x 3

(8 xy + 5 yz - x 3 ) ⋅ 5 - (8 x + 5z ) ⋅ 5 y (8 xy + 5 yz - x 3)2 -5x 3 (8 xy + 5 yz - x 3 )2

f (1 + h, 2) - f (1, 2) h h 0 We will use a small value for h. Let h = 0.00001. f (1.00001, 2) - f (1, 2) f x (1, 2) » 0.00001 f x (1, 2) = lim

1.00001+ 2/2

(1.00001 + 22 ) »

( yz - 2)3

f x ( x, y, z ) = f y ( x, y, z ) = f z ( x, y, z ) = f yz ( x, y, z ) = =

46.

-2 » 2.00001 0.00001 » 6.773

f ( x, y, z ) = ln | x 2 - 5 xz 2 + y 4 | 2 x - 5z 2 x 2 - 5xz 2 + y 4 4y

f y ( x, y , z ) =

x - 5xz 2 + y 4 -10 xz x 2 - 5xz 2 + y 4

h 0

1+ 2.00001/2

4 y (10 zx) 2

( x - 5 xz 2 + y 4 )2

8xy + 5 yz - x3 8 xy + 5 yz - x 8 x + 5z

8 xy + 5 yz - x 3

⋅ (8 x + 5z )

1+ 2/2

)

0.00001

- 22 » 2.000005 0.00001 » 3.386

( x 2 - 5 xz 2 + y 4 )2

8 y - 3x 2

(

- 1 + 22

2.000005

40 xy 3 z

f y (1, 2) = lim

(1 + 2.00001 ) 2 »

1+ 2/2

)

f (1, 2 + h) - f (1, 2) h Again, let h = 0.00001. f (1, 200001) - f (1, 2) f y (1, 2) » 0.00001

(b)

f ( x, y, z ) = ln |8xy + 5 yz - x3 | 1 f x ( x, y, z ) = ⋅ (8 y - 3x 2 ) 8xy + 5 yz - x3 =

(

- 1 + 22

0.00001

2.00001

45.

⋅ 5y

x + y /2 æ yö f ( x, y) = çç x + ÷÷÷ çè 2ø

(a)

( yz - 2) 2 (-2 x 2 ) - (-2 x - 2 x 2 z ) ⋅ 2( yz - 2) ⋅ y

f ( x, y) = ( x + y 2 )2 x + y

48.

f (2 + h, 1) - f (2, 1) h h 0 We will use a small value for h. Let h = 0.00001. f x (2, 1)

(a)

f x (2, 1) = lim

f (2.00001, 1) - f (2, 1) 0.00001

(2.00001 + 12 ) 2(2.00001) +1 - (2 + 12 ) 2(2) +1 0.00001

(3.00001)5.00002 - 35 0.00001 » 938.9 »

664

Chapter 9 MULTIVARIABLE CALCULUS (b)

f (2, 1 + h) - f (2, 1) h h0

f y (2, 1) = lim

51.

P( x, y) = 250 x 2 + y 2 = 250( x 2 + y 2 )1/2

Again, let h = 0.00001. f y (2, 1) »

f (2, 1.00001) - f (2, 1) 0.00001

[2 + (1.00001)2 ]2(2) +1.00001 - (2 + 12 ) 2(2) +1 0.00001 2 5.00001

[2 + (1.00001) ] 0.00001 » 1077 »

49.

¶P (6,8) = ¶x

-3

M ( x, y) = 45 x 2 + 40 y 2 - 20 xy + 50

(a)

æ1ö ¶P = 250 çç ÷÷÷ ( x 2 + y 2 )-1/2 (2 x) ç è2ø ¶x 250 x = x2 + y2

(a)

(b)

M y ( x, y) = 80 y - 20 x M y (4, 2) = 80(2) - 20(4) = 80

(b)

M x (3, 6) = 90(3) - 20(6) = 150

(c)

¶M (2, 5) = 90(2) - 20(5) = 80 ¶x

(d)

¶M (6, 7) = 80(7) - 20(6) = 440 ¶y

(e)

M x ( x, y) = 90 x - 20 y M xx ( x, y) = 90

(f)

1500 = 150 10

250(8) (6)2 + (8)2

2000 = 200 10

52.

æ1 ö-4 3 f ( x, y) = çç x-1/4 + y-1/4 ÷÷÷ çè 4 ø 4

é1 ù -4 3 f (16,81) = ê (16)-1/4 + (81)-1/4 ú 4 ëê 4 ûú

(a)

M y ( x, y) = 80 y - 20 x

æ 1 1 3 1 ö-4 = çç ⋅ ⋅ ⋅ ÷÷÷ çè 4 2 4 3 ø

M yy ( x, y) = 80 (g)

(6)2 + (8)2

æ1ö ¶P = 250 çç ÷÷÷ ( x 2 + y 2 )-1/2 (2 y) çè 2 ø ¶x 250 y = x2 + y2 ¶P (6,8) = ¶y

M x ( x, y) = 90 x - 20 y

250(6)

æ 3 ö-4 = çç ÷÷÷ » 50.56790123 çè 8 ø

M x ( x, y) = 90 x - 20 y M xy ( x, y) = -20

50.57 hundred units are produced. 50.

R( x, y) = 5x + 9 y - 4 xy

(a)

(b)

R(9, 5) = 5(9)2 + 9(5)2 - 4(9)(5) = 450 Rx ( x, y) = 10 x - 4 y

ù æ1 ö-5 é 1 æ 1 ö 3 f x ( x, y ) = -4 çç x-1/4 + y-1/4 ÷÷ ê çç - ÷÷ x-5/4 ú ÷ ÷ ú èç 4 ø êë 4 èç 4 ø 4 û =

Rx (9, 5) = 10(9) - 4(5) = 70

ö-5 1 -5/4 æç 1 -1/4 3 + y-1/4 ÷÷÷ x çç x è4 ø 4 4

R would increase by $70. (b)

f x (16,81) =

R y ( x, y) = 18 y - 4 x R y (9, 5) = 18(5) - 4(9) = 54 R would increase by $54.

é1 ù -5 1 3 (16)-5/4 ê (16)-1/4 + (81)-1/4 ú êë 4 úû 4 4 -5

1 æç 1 ÷ö çæ 3 ÷ö 256 çç ÷÷ çç ÷÷ = è ø è ø 4 32 8 243 » 1.053497942

Section 9.2

665 f x (16,81) = 1.053 hundred units and is the rate at which production is changing when labor changes by one unit (from 16 to 17) and capital remains constant.

55.

æ ö2/3 f x (64,125) = çç 125 ÷÷÷ = 25 = 1.5625, è 64 ø 16 which is the approximate change in production (in thousands of units) for a 1 unit change in labor.

-5 3 -5/4 æç 1 -1/4 3 -1/4 ÷ö + y x y çç ÷÷ø è4 4 4

f y (16,81) =

é1 ù 3 3 (81)-5/4 ê (16)-1/4 + (81)-1/4 ú ê úû 4 4 ë4

-5

f y ( x, y) = 2 x1/3 y-1/3 = 2

-5

3 æç 1 öæ 3ö 8192 ÷÷÷ççç ÷÷÷ = çç 4 è 243 øè 8 ø 19, 683 » 0.4161967180

which is the approximate change in production (in thousands of units) for a 1 unit change in capital.

the rate at which production is changing when capital changes by one unit (from 81 to 82) and labor remains constant.

(b) Increasing to 65 units of labor would result in an increase of approximately 25 (1000) » 1563 batteries. 16

(c) An increase of approximately 16 (1000) = 3200 batteries 5 would be the effect of increasing capital to 126 while holding labor at 64. Increasing capital is the better option.

z = x 0.7 y 0.3, where x is labor, y is capital.

Marginal productivity of labor is ¶z = 0.7 x-0.3 y 0.3. ¶x

Marginal productivity of capital is q( f , M , k ) =

56.

¶z = 0.3x 0.7 y-0.7 . ¶y

54.

2 fM k 2(32)(40, 000) 4 = 800

(a)

q(32, 40, 000, 4) =

(b)

q( f , M , k ) =

z = Cx 0.65 y 0.35

The marginal productivity of labor is ¶z = C (0.65 x-0.35 y 0.35 + x 0.65 ⋅ 0) ¶x = 0.65Cx-0.35 y 0.35.

q f ( f , M , k) =

The marginal productivity of capital is ¶z = C ( x 0.65 (0.35 y-0.65 ) + y 0.35 ⋅ 0) ¶y = 0.35x 0.65 y-0.65.

æ x ö÷1/3 çç ÷ = 2 çè y ÷÷ø y1/3

x1/3

æ 64 ö÷1/3 8 f y (64,125) = 2 çç = = 1.6, çè 125 ÷÷ø 5

f y (16,81) = 0.4162 hundred units and is

53.

æ y ö÷2/3 çç ÷ = çè x ÷ø x 2/3 y 2/3

f x ( x, y) = x-2/3 y 2/3 =

(a)

ù æ1 ö-5 é 3 æ 1 ö 3 f y ( x, y) = -4 çç x-1/4 + y-1/4 ÷÷ ê çç - ÷÷ y-5/4 ú ÷ ÷ çè 4 ç ê ú ø ë 4è 4ø 4 û =

f ( x, y) = 3x1/3 y 2/3, where x is labor, y is capital.

2M ⋅ f 1/ 2 k 1 2

2M -1/ 2 f = k

q f (32, 40, 000, 4) =

M 2 fk

40, 000 = 12.5 2(32)(4)

When the quantity produced is held constant at 40,000, and the storage cost is held constant at $4, the number of gallons produced when the setup cost is $32 goes up by approximately 12.5 gallons for each $1 increase in the setup cost.

666

Chapter 9 MULTIVARIABLE CALCULUS (c)

q( f , M , k ) =

2 f M ⋅ k -1/ 2

(

1 q f ( f , M , k) = 2 f M k -3/ 2 2 f M =2k 3

q f (32, 40, 000, 4) = -

f y ( x, y ) =

2(4)3

(

)

( ( )

= =

(

59.

)

( (1 + x )( x + y ) - ( y + xy ) )( e-x ) ( x + y )2 x(1 + x )e-x ( x + y )2

f (w, v) = 25.92w0.68 +

3.62w0.75 v

(a) f (300,10) = 25.92(300)0.68 +

))

(

3.62(300)0.75 10

» 1279.46 The value is about 1279 kcal/hr.

)

(b)

f w (w, v) = 25.92(0.68)w-.32 3.62(0.75)w-0.25 v 17.6256 2.715 = + 0.25 0.32 w w v 17.6256 2.715 f w (300,10) = + 0.32 (300) (300)0.25 (10) » 2.906 The value is about 2.906 kcal/hr/g. This means the instantaneous rate of change of energy usage for a 300 kg animal traveling at 10 kilometers per hour to walk or run 1 kilometer is about 2.906 kcal/hr/g. +

( ( -xy )( x + y) - ( y + xy ) )( e-x ) ( x + y )2

)( )

y -x 2 - x - xy - 1 e-x ( x + y)

)

Since x and y are positive, f y ( x, y) > 0.

)

(

)

(

y(1 + x)e-x x+ y

(

(32)(40, 000)

é ¶ ù ê ú y(1 + x )e-x ( x + y ) ê ¶x ú ê ú ê ú -x ¶ y (1 x ) e ( x y ) + + ê ú êë úû x ¶ f x ( x, y ) = 2 ( x + y) é ù -x + ( y(1 + x ) ) -e-x ( x + y ) ú ê ( y) e ê ú ê ú -x ê ú (1) - y(1 + x )e êë úû = 2 ( x + y) =

)

( x + y )2 é ù -x ( x + y) ê (1 + x )e ú ê ú ê ú -x ê (1) ú - y(1 + x )e êë úû = 2 ( x + y)

= -100 When the quantity produced is held constant at 40,000, and the fixed setup cost is held constant at $32, the number of gallons produced when the storage cost is $4 goes down by approximately 100 gallons for each $1 increase in the storage cost.

57. (c) f ( x, y) =

é ¶ ù ê ú y(1 + x )e-x ( x + y ) ê ¶y ú ê ú ê ú ¶ ê - y(1 + x )e-x ( x + y) ú êë úû ¶y

Since x and y are positive, f x ( x, y) < 0.

60.

H (m, T , A) =

(a)

15.2m0.67 (T - A) 10.23ln m - 10.74

HT (m, T , A) =

15.2m0.67 10.23ln m - 10.74

15.2(24)0.67 10.23ln 24 - 10.74 » 5.87 The approximate change in the rate of heat loss is 5.87 W. HT (24, 37, 8) =

Section 9.2 (b)

667 H A (m, T , A) =

Cb (160, 200, 125) = (160 - 125)-1

-15.2m0.67 10.23ln m - 10.74

1 35 » 0.0286

-15.2(26)0.67 H A (26, 37, 10) = 10.23ln 26 - 10.74 » -5.97 The approximate change in the rate of heat loss is -5.97 W.

61.

(d) Cv (a, b, v) = -b(a - v)-2 ⋅ (-1) =

A = 0.024265h0.3964m0.5378

b ( a - v) 2

Cv (160, 200,125) =

(a) Am = (0.024265)(0.5378)h0.3964m(0.5378-1) =

= 0.013050h0.3964m-0.4622

(e) Changing a by 1 unit produces the greatest decrease in the liters of blood pumped, while changing v by 1 unit produces the same amount of increase in the liters of blood pumped.

0.013050(180)0.3964 (72)-0.4622 » 0.0142

or about 0.0142 m2. Ah = (0.024265)(0.3964)h(0.3964-1) m0.5378 = 0.0096186h-0.6036m0.5378

63.

f (n, c) =

(a)

When the height h increases from 160 to 161 while the mass m remains at 70, the approximate change in body surface area is 0.0096186(160)-0.6036 (70)0.5378 » 0.00442

(b)

or about 0.00442 m .

C (a, b, v) =

(a)

b (a - v ) 2

Ca (160, 200,125) = =-

200 (160 - 125)2 200

Cb (a, b, v) = (a - v)-1

64.

¶f 1 1 1 = (2n) - (0) + 0 = n, ¶n 8 5 4

1 3 (3) = lb 4 4 represents an additional weight loss by adding the fourth workout. f n (3,1100) =

B(w, h) =

(a) (b)

352 » -0.1633

(c)

C (160, 200,125) =

(b) Ca (a, b, v) = -b (a - v)-2 ⋅ 1

1 1 1937 (4) - (1200) + 8 5 8 1937 = 2 - 240 + = 4.125 8

f (4,1200) =

which represents the rate of change of weight loss per unit change in number of workouts.

b = b(a - v)-1 a-v 200 160 - 125 200 = 35 » 5.714

1 2 1 1937 n - c+ 8 5 8

The client could expect to lose 4.125 lb.

62.

(160 - 125)2 200

352 » 0.1633

When the mass m increases from 72 to 73 while the height h remains at 180 cm, the approximate change in body surface area is

(b)

200

703w h2

B(315, 78) =

703(315) (78) 2

» 36.4

¶B 703 = 2 ¶w h ¶B -2(703)w 1406w = =- 3 3 ¶h h h B is positive, as w increases, so Since ¶¶w

does B. Since ¶¶Bh is negative, B decreases as h increases.

668

Chapter 9 MULTIVARIABLE CALCULUS (c) If wm and hm represent a person’s weight and height, in kilograms and meters, respectively, then wm = 0.4536w, where w is weight in pounds, and hm = 0.0254h, where h is height in inches. To transform the formula B = 7032w to handle metric inputs,

(b)

¶ L( E , P ) ¶E = 23(0.6)(14,100)-0.4 (68,700)-0.267

» 0.015 so for a 1-gram increase in brain mass we expect a 0.015-year increase in life span. Computing the change directly gives L(14,101, 68,700) - L(14,100, 68,700) » 0.015 years.

m make the substitutions w = 0.4536 and

m h = 0.0254 .

Bm =

Since 65.

ABSI =

m 703 0.4536

(

hm 0.0254

)

¶ L( E , P) = 23(0.6) E -0.4 P-0.267 ¶E When E = 14,100 and P = 68, 700,

67.

703(0.0254)2 wm ⋅ 2 . 0.4536 hm

w 703(0.0254) 2 » 1, use Bm = 2m . 0.4536 hm

R( x, t ) = x 2 (a - x)t 2e-t = (ax 2 - x3 )t 2e-t

(a)

¶R = (2ax - 3x 2 )t 2e-t ¶x

(b)

¶R = x 2 (a - x) ⋅ [t 2 ⋅ (-e-t ) + e-t ⋅ 2t ] ¶t = x 2 (a - x)(-t 2 + 2t )e-t

w b 2 / 3h1/ 2

(c)

(a) For w = 0.864, b = 23.1, and h = 1.85, 0.864 ABSI = » 0.0783. 2/3 (23.1) (1.85)1/ 2 ¶ 1 ABSI = 2 / 3 1/ 2 ¶w b h For the person in part (a) this is 1 = 0.0906 per m. (23.1)2 / 3(1.85)1/ 2

(b)

¶ 1 w ABSI = (c) ¶h 2 b 2 / 3h3 / 2 For the person in part (a) this is æ 1 ö÷ 0.864 çç - ÷ = -0.0212 per m. çè 2 ø÷ (23.1)2 / 3(1.85)3/ 2

66. L( E , P) = 23E 0.6 P-0.267

¶2R ¶x

= (2a - 6 x)t 2e-t

(d)

¶2R = (2ax - 3x 2 )(-t 2 + 2t )e-t ¶x¶t

(e)

¶R ¶x

gives the rate of change of the reaction

per unit of change in the amount of drug administered. ¶R ¶t

gives the rate of change of the reaction

for a 1-hour change in the time after the drug is administered. 68.

l (1 - 2-t /5 ) 33 33 (1 - 2-10/5 ) p(33, 10) = 1 + 33

p(l , t ) = 1 +

(a)

¶ L( E , P) = 23(-0.267) E 0.6 P-1.267 ¶P When E = 7.35 and P = 150,

= 1 + (1)(1 - 2-2 )

(a)

=1+

¶ L( E , P ) ¶P

= 1.75 The pressure at 33 feet for a 10-minute dive is 1.75 atmospheres.

= 23(-0.267)(7.35)0.6 (150)-1.267 » -0.036 so for a 1-gram increase in body mass we expect a 0.036-year decrease in life span. Computing the change directly gives L(7.35, 151) - L(7.35, 150) » -0.035 years.

3 4

(b)

1 (1 - 2-t /5 ) 33 æ 1 ö l (-(ln 2)) çç - 2-t /5 ÷÷÷ pt (l , t ) = çè 5 ø 33 l (ln 2) (2-t /5 ) = 165

pl (l , t ) =

Section 9.2

669 1 1 æç 3 ö÷ (1 - 2-10/5 ) = ç ÷ 33 33 çè 4 ø÷ » 0.0227 atm/ft

pl (33,10) =

Increasing the depth of the dive by 1 foot (to 34 feet), while keeping the dive length at 10 minutes, increases the pressure by approximately 0.0227 atmospheres.

The wind speed is approximately 15 mph. (c)

33 (ln 2)(2-10/5 ) 165 1 ln 2 = 20 » 0.0347 atm/min

pt (33, 10) =

Increasing the dive time by 1 minute (to 11 minutes), while keeping the depth of the dive to 33 feet increases the pressure by approximately 0.0347 atmospheres. (c) Solve p(66, t ) = 2.15 66 1+ (1 - 2-t /5 ) = 2.15 33

(91.4 - 10) » -1.114

When the temperature is held fixed at 10 F, the wind chill decreases approximately 1.1 degrees when the wind velocity increases by 1 mph.

2(1 - 2-t /5 ) = 1.15 1 - 2-t /5 = 0.575

WT (20,10) =

2-t /5 = 0.425

1 [10.45 + 6.69 20 - 0.447(20)] » 1.429 22

t - ln 2 = ln 0.425 5 5 ln 0.425 t = ln 2

When the wind velocity is held fixed at 20 mph, the wind chill increases approximately 1.429  F when the temperature increases from 10 F to 11 F.

t » 6.172 The maximum dive length is 6.172 minutes.

69.

(d) A sample table is

W (V ,T )

T V 30 20 10 0

(10.45 + 6.69 V - 0.447V )(91.4 - T ) = 91.4 22

(a) W (20,10) = 91.4 -

ö 1 æç 6.69 - 0.447 ÷÷÷ (91.4 - T ) çç 22 è 2 V ø÷ 1 WT = - (10.45 + 6.69 V 22 - 0.447V )(-1) 1 = (10.45 + 6.69 V 22 - 0.447V ) ö 1 æç 6.69 - 0.447 ÷÷÷ WV (20, 10) = çç ÷ 22 è 2 20 ø WV = -

5 27 16 6 5

10 16 3 9 21

15 9 5 18 32

20 4 11 25 39

(10.45 + 6.69 20 - 0.447(20))(91.4 - 10) 22

70.

» -24.9

The wind chill is -24.9F when the wind speed is 20 mph and the temperature is 10F.

(a)

f (90, 30) » 90

(b)

f (90, 75) » 109

(c)

f (80, 75) » 86

(d)

(b) Solve (10.45 + 6.69 V - 0.447V )(91.4 - 5) 22 for V.

-25 = 91.4 -

f (95, 30) - f (90, 30) 5 95 - 90 » =1 5

fT (90, 30) »

670

Chapter 9 MULTIVARIABLE CALCULUS (e)

(f )

(g)

71.

f (90, 35) - f (90, 30) 5 92 - 90 » = 0.4 5

p = f (4,5,300)

(b)

f H (90, 30) »

= 0.05(300) + 6[(4)(5)]1/2 = 15 + 6(20)1/2 » 41.83281573

f (95, 75) - f (90, 75) fT (90, 75) » 5 130 - 109 » = 4.2 5

The student’s probability of passing is 41.83%. æ1ö f n (s, n, a) = 6 çç ÷÷÷ (sn)-1/2 (s) çè 2 ø

(c)

f (90, 80) - f (90, 75) 5 114 - 109 » =1 5

= 3s(sn)-1/2

f H (90, 75) »

f n (4,5, 480) = 3(4)[(4)(5)]-1/2 = 12(20)-1/2 » 2.683281573

The rate of change in lung capacity with respect to age can be found by comparing the change in two lung capacity measurements to the difference in the respective ages when the height is held constant. So for a woman 58 inches tall, at age 20 the measured lung capacity is 1900 ml, and at age 25 the measured lung capacity is 1850 ml. So the rate of change in lung capacity with respect to age is

f n (4,5, 480) » 2.683% is the rate at which the probability of success is changing when the number of semesters of mathematics passed in high school change by one unit (from 5 to 6) and the other quantities remain constant. f a (s, n, a) = 0.05 f a (4,5, 480) = 0.05% is the rate at which the probability of success is changing when the student’s SAT score changes by one unit (from 480 to 481) and the other quantities remain constant.

1900 - 1850 50 = 20 - 25 -5 = -10 ml per year.

The rate of change in lung capacity with respect to height can be found by comparing the change in two lung capacity measurements to the difference in the respective heights when the age is held constant. So for a 20-year old woman the measured lung capacity for a woman 58 inches tall is 1900 ml and the measured lung capacity for a woman 60 inches tall is 2100 ml. So the rate of change in lung capacity with respect to height is

73.

F =

(a)

Fm =

= mgR 2r -2 gR 2

is the approximate rate of

change in gravitational force per unit change in mass while distance is held constant. Fr =

1900 - 2100 -200 = 58 - 60 -2 = 100 ml per in.

The two rates of change remain constant throughout the table.

mgR 2

-2mgR 2 r3

is the approximate rate of

change in gravitational force per unit change in distance while mass is held constant. (b)

Fm =

gR 2 r2

, where all quantities are

positive. Therefore, Fm > 0. 72.

1/2

p = f (s, n, a) = 0.05a + 6(sn)

(a)

Fr =

p = f (6, 4, 460)

-2mgR 2 r3

, where m, g , R 2 , and r 3

are positive.

= 0.05(460) + 6[(6) (4)]1/2 1/2

= 23 + 6(24)

» 52.39387691

The student’s probability of passing is 52.39%.

Therefore, Fr < 0. These results are reasonable since gravitational force increases when mass increases (m is in the numerator) and gravitational force decreases when distance increases (r is in the denominator).

Section 9.3 74.

671 x+ y

w( x, y) =

(a)

(b)

xy 1+ 2 c

w(50, 000, 150, 000) =

50,000 + 150,000 (50,000) (150,000) (186,282)2

» 164, 456

(1 + ) - ( x + y) (b) w = (1 + ) xy c2

y c2

xy c2

xy xy y2 - 2 - 2 2 c c c 2 xy 1+ 2 c

xy c2

It the width of the target area is increased by 1 foot, while keeping the distance fixed at 3 feet, the movement time decreases by approximately 764.5 msec.

150,000 ( 186,282 )

2 æ (50,000) (150,000) ö÷ ÷÷ ççèç1 + 2 (186,282) ø

» 0.2377

The instantaneous rate of change is 0.238 m/sec per m/sec. (c)

w(c, c) =

265 æç 1 ÷ö ç ÷ ln 2 çè w ÷ø 265 » 127.4 msec/ft Ts (3, 0.5) = 3ln 2 If the distance the object is being moved increases from 3 feet to 4 feet, while keeping w fixed at 0.5 foot, the time to move the object increases by approximately 127.4 msec. 265 0.5ln 2 » -764.5 msec/ft

Wx (50,000, 150,000) 1-

ln 2 265 = 105 + [ln(2s) - ln(w)] ln 2 265 æç 1 ö÷ Ts (s, w) = ç ÷ ln 2 çè s ÷ø

Tw (3, 0.5) = -

( ) 1-( ) = (1 + ) y c

c+c (c)(c) 1+ 2 c

9.3 Maxima and Minima Your Turn 1 f ( x, y) = 4 x3 + 3xy + 4 y 3 f x ( x, y) = 12 x 2 + 3 y, f y ( x, y) = 3x + 12 y 2

2c =c 2

12 x 2 + 3 y = 0 3x + 12 y 2 = 0

The speed is the speed of light, c. 75.

æ 2s ö T = (s, w) = 105 + 265log 2 çç ÷÷÷ çè w ø

(a)

( )

ln 2ws

Tw (s, w) = -

The rocket is traveling 164,500 m/sec relative to the stationary observer.

T (s, w) = 105 + 265

é 2(3) ù ú T (3, 0.5) = 105 + 265log 2 ê êë 0.5 úû = 105 + 265log 2 12

Solve this system by substitution. From the first equation, y = -4 x 2 .

Substituting for y in the second equation gives

(

3x + 12 -4 x 2

» 1055

) = 0.

x + 64 x 4 = 0 x(1 + 64 x3 ) = 0 x = 0 or 1 + 64 x3 = 0 x = 0 or

x =-

1 4

672

Chapter 9 MULTIVARIABLE CALCULUS f ¢( x ) = 0

If x = 0, y = -4(0)2 = 0.

3( x 2 - 2 x - 8) = 0

æ 1 ö2 1 1 If x = - , y = -4 çç - ÷÷÷ = - . ç è 4ø 4 4

( x - 4)( x + 2) = 0 The critical values are 4 and –2. The second derivative is positive at 4 so there is a relative minimum at x = 4. The second derivative is negative at –2, so there is a relative maximum at x = –2. f (4) = -75, f (-2) = 33. Thus there is a relative minimum at (4,–75) and a relative maximum at (–2, 33).

So the critical points are æ 1 1ö (0, 0) and çç - , - ÷÷÷. çè 4 4 ø

Your Turn 2

Use the information from Your Turn 1: f ( x, y ) = 4 x 3 + 3xy + 4 y 3

W2.

f ( x) = x 4 - 8 x 2 + 1

f x ( x, y ) = 12 x 2 + 3 y

f ¢( x) = 4 x3 - 16 x

f y ( x, y ) = 3x + 12 y 2

= 4 x( x 2 - 4)

The critical points are (0, 0) and (-1/4, -1/4).

f ¢¢( x) = 12 x 2 - 16

Now compute the second partial derivatives.

Find the critical points: f ¢( x ) = 0

f xx ( x, y ) = 24 x, f yy ( x, y) = 24 y, f xy ( x, y) = 3

4 x( x 2 - 4) = 0 4 x( x - 2)( x + 2) = 0 The critical numbers are –2, 0, and 2. The second derivative is negative at 0, so there is a relative maximum for x = 0. The second derivative is positive at –2 and 0, so there are relative minima at these values. f (0) = 1, f (2) = f (-2) = -15. Thus there is a relative maximum at (0, 1) and relative minima at (–2,–15) and (2,–15).

D(a, b) = f xx (a, b) ⋅ f yy (a, b) - [ f xy (a, b)]2 = (24a)(24b) - 9 = 576ab - 9

At the critical point (0, 0), D(0, 0) = 576(0)(0) - 9 = -9 < 0, so this critical point is a saddle point.

At the critical point (-1/4, -1/4), æ 1 1ö æ 1 öæ 1 ö D çç - , - ÷÷÷ = 576 çç - ÷÷÷çç - ÷÷÷ - 9 èç 4 4 ø èç 4 øèç 4 ø = 36 - 9 = 24

and æ 1 1ö æ 1ö 1 f xx çç - , - ÷÷÷ = 24 çç - ÷÷÷ = - . èç 4 4 ø èç 4 ø 6

Since D > 0 and f xx < 0, there is a relative maximum at (-1/4, -1/4).

9.3 Warmup Exercises W1.

f ( x) = x3 - 3x 2 - 24 x + 5

9.3 Exercises 1.

False. There could be a saddle point at (a, b).

True

False. If D = 0 at (a, b), then the test gives no information.

True

f ( x, y) = xy + y - 2 x f x ( x, y) = y - 2, f y ( x, y) = x + 1

If f x ( x, y) = 0, y = 2. If f y ( x, y) = 0, x = -1. Therefore, (–1, 2) is the critical point. f xx ( x, y) = 0 f yy ( x, y) = 0

f ¢( x) = 3x 2 - 6 x - 24 = 3( x 2 - 2 x - 8)

f ¢¢( x) = 6 x - 6 Find the critical points:

f xy ( x, y) = 1

For (–1, 2),

Section 9.3 6.

673

f ( x, y) = 3xy + 6 y - 5x

2x + y - 6 = 0

f x ( x, y) = 3 y - 5

x + 2y

f y ( x, y) = 3x + 6

2x + y - 6 = 0

If f x ( x, y) = 0, y = 53 .

-2 x - 4 y = 0 - 3y - 6 = 0

If f y ( x, y) = 0, x = -2.

(

y = -2 x + 2(-2) = 0

)

Therefore, -2, 53 is a critical point.

x = 4

f xx ( x, y) = 0

f xx ( x, y) = 2, f yy ( x, y) = 2, f xy ( x, y) = 1

f yy ( x, y) = 0

D = 2 ⋅ 2 - 12 = 3 > 0 and f xx ( x, y) > 0

f xy ( x, y) = 3

Relative minimum at (4, -2)

D = f xx (a, b) ⋅ f yy (a, b) - [ f xy (a, b)]2 = 0 ⋅ 0 - 32 = -9 < 0

(

f y ( x, y) = -x + 2 y + 2

f ( x, y) = 3x - 4 xy + 2 y + 6 x - 10

Solve the system f x ( x, y) = 0, f y ( x, y) = 0.

f x ( x, y) = 6 x - 4 y + 6

2x -

f y ( x, y) = -4 x + 4 y

2x -

6x - 4 y + 6 = 0 2x

y+2=0

-2 x + 4 y + 4 = 0 3y + 6 = 0

+6=0

y = -2

x = -3 -4(-3) + 4 y = 0

-x + 2(-2) + 2 = 0 x = -2

y = -3

(-2, -2) is the critical point.

Therefore, (-3, -3) is a critical point.

f xx ( x, y ) = 2

f xx ( x, y) = 6

f yy ( x, y ) = 2

f yy ( x, y) = 4

f xy ( x, y ) = -1

f xy ( x, y) = -4

For (-2, -2),

D = 6 ⋅ 4 - (-4)2 = 8 > 0

D = (2)(2) - (-1)2 = 3 > 0.

Since f xx ( x, y) = 6 > 0, there is a relative minimum at (-3, -3). 8.

y+2=0

- x + 2y + 2 = 0

Solve the system f x ( x, y ) = 0, f y ( x, y ) = 0. -4 x + 4 y

f ( x, y) = x 2 - xy + y 2 + 2 x + 2 y + 6 f x ( x, y) = 2 x - y + 2,

)

There is a saddle point at -2, 53 . 2

f ( x, y) = x 2 + xy + y 2 - 6 x - 3 f x ( x, y) = 2 x + y - 6, f y ( x, y) = x + 2 y

Since f xx ( x, y) > 0, a relative minimum is at (-2, -2). 10.

f ( x, y) = 2 x 2 + 3xy + 2 y 2 - 5x + 5 y f x ( x, y) = 4 x + 3 y - 5 f y ( x, y) = 3x + 4 y + 5

Solve the system f x ( x, y) = 0, f y ( x, y) = 0.

674

Chapter 9 MULTIVARIABLE CALCULUS 4x + 3 y - 5 = 0 3x + 4 y + 5 = 0

12.

-12 x - 9 y + 15 = 0 12 x + 16 y + 20 = 0 7 y + 35 = 0 y = -5

f ( x, y ) = 5xy - 7 x 2 - y 2 + 3x - 6 y - 4 f x ( x, y ) = 5 y - 14 x + 3 f y ( x, y ) = 5 x - 2 y - 6

5 y - 14 x + 3 = 0 -2 y + 5 x - 6 = 0

4 x + 3(-5) - 5 = 0 4 x = 20

10 y - 28 x + 6 = 0 -10 y + 25 x - 30 = 0

x =5

- 3x - 24 = 0 x = -8

Therefore, (5, -5) is a critical point. f xx ( x, y) = 4

-2 y + 5(-8) - 6 = 0 -2 y = 46 y = -23

f yy ( x, y) = 4 f xy ( x, y) = 3 D = 4 ⋅ 4 - 32 = 7 > 0

f xx ( x, y) = -14, f yy ( x, y) = -2,

Since f xx ( x, y) = 4 > 0, there is a relative minimum at (5, -5). 11.

f xy ( x, y) = 5 D = (-14)(-2) - 52 = 3 > 0 and f xx ( x, y) < 0

f ( x, y) = x 2 + 3xy + 3 y 2 - 6 x + 3 y f x ( x, y) = 2 x + 3 y - 6,

Relative maximum at (-8, -23)

f y ( x, y) = 3x + 6 y + 3

Solve the system f x ( x, y) = 0, f y ( x, y) = 0.

13.

f ( x, y ) = 4 xy - 10 x 2 - 4 y 2 + 8x + 8 y + 9 f x ( x, y ) = 4 y - 20 x + 8

2x + 3y - 6 = 0 3x + 6 y + 3 = 0 -4 x - 6 y + 12 = 0 3x + 6 y + 3 = 0 - x + 15 = 0 x = 15

f y ( x, y ) = 4 x - 8 y + 8 4 y - 20 x + 8 = 0 4x - 8 y + 8 = 0 4 y - 20 x + 8 = 0

3(15) + 6 y + 3 = 0 6 y = -48 y = -8

-4 y + 2 x + 4 = 0 - 18 x + 12 = 0 x =

(15, -8) is the critical point. f xx ( x, y) = 2

2 3

æ2ö 4 y - 20 çç ÷÷÷ + 8 = 0 çè 3 ø

f yy ( x, y) = 6 f xy ( x, y) = 3

The critical point is

For (15, -8),

( 23 , 43 ).

D = 2 ⋅ 6 - 9 = 3 > 0.

f xx ( x, y) = -20

Since f xx ( x, y) > 0, a relative minimum is at (15, -8).

f yy ( x, y) = -8

f xy ( x, y) = 4

Section 9.3 For

675

( 23 , 43 ),

16.

D = (-20)(-8) - 16 = 144 > 0.

f ( x, y ) = x 2 + xy + y 2 - 3x - 5 f x ( x, y ) = 2 x + y - 3 f y ( x, y ) = x + 2 y

Since f xx ( x, y) < 0, a relative maximum is at 14.

(

2,4 3 3

2x + y - 3 = 0 x + 2y =0

2x + y - 3 = 0 -2 x - 4 y =0 - 3y - 3 = 0 y = -1

f ( x, y) = 4 y + 2 xy + 6 x + 4 y - 8 f x ( x, y) = 2 y + 6 f y ( x, y) = 8 y + 2 x + 4

x + 2(-1) = 0 x = 2

If f x ( x, y) = 0, y = -3. Substitute -3 for y in f y = ( x, y) = 0 and solve for x.

f xx ( x, y) = 2, f yy ( x, y) = 2, f xy ( x, y) = 1

8(-3) + 2 x + 4 = 0 2 x = 20

D = 2 ⋅ 2 - 12 = 3 > 0 and f xx ( x, y ) > 0

x = 10

Relative minimum at (2, -1)

Therefore, (10, -3) is a critical point. f xx ( x, y) = 0

17.

f yy ( x, y) = 8

f x ( x, y) = 6 x - 18 y

f xy ( x, y) = 2 2

D = 0 ⋅ 8 - 2 = -4 < 0

There is a saddle point at (10, - 3). 15.

f ( x, y) = 3x 2 + 2 y 3 - 18xy + 42

f ( x, y ) = x 2 + xy - 2 x - 2 y + 2

f y ( x, y) = 6 y 2 - 18x

If f x ( x, y) = 0, 6 x - 18 y = 0, or x = 3 y. Substitute 3y for x in f y ( x, y) = 0 and solve for y. 6 y 2 - 18(3 y) = 0

f x ( x, y ) = 2 x + y - 2

6 y ( y - 9) = 0

f y ( x, y ) = x - 2

2x + y - 2 = 0 x

Then

-2=0 x = 2

y =9

x = 0 or

x = 27.

Therefore, (0, 0) and (27, 9) are critical points.

2(2) + y - 2 = 0

f xx ( x, y) = 6

y = -2

f yy ( x, y) = 12 y

The critical point is (2, -2).

f xy ( x, y) = -18

f xx ( x, y) = 2

For (0, 0),

f yy ( x, y) = 0

D = 6 ⋅ 12(0) -(-18)2 = -324 < 0.

f xy ( x, y) = 1

For (2, -2), D = 2 ⋅ 0 - 12 = -1 < 0.

A saddle point is at (2, -2).

y = 0 or

There is a saddle point at (0, 0). For (27, 9), D = 6 ⋅ 12(9) - (-18)2 = 324 > 0.

Since f xx ( x, y) = 6 > 0, there is a relative minimum at (27, 9).

676 18.

Chapter 9 MULTIVARIABLE CALCULUS f ( x, y) = 7 x3 + 3 y 2 - 126 xy - 63

f xx ( x, y) = 2

f x ( x, y) = 21x 2 - 126 y

f yy ( x, y) = 24 y f xy ( x, y) = -6

f y ( x, y) = 6 y - 126 x

For (0, 0),

If f y ( x, y) = 0, 6 y - 126 x = 0, or y = 21x. Substitute 21x for y in f x ( x, y) = 0 and solve for x.

D = 2 ⋅ 24(0) - (-6) 2 = -36 < 0.

21x - 126(21x) = 0 21x( x - 126) = 0 x = 0 or y = 0 or

Then

A saddle point is at (0, 0).

(

)

For 92 , 32 ,

x = 126 y = 2646.

æ3ö D = 2 ⋅ 24 çç ÷÷÷ - (-6) 2 çè 2 ø

Therefore, (0, 0) and (126, 2646) are critical points.

= 36 > 0.

f xx ( x, y) = 42 x

Since f xx ( x, y) > 0, a relative minimum is at

( 92 , 32 ).

f yy ( x, y) = 6 f xy ( x, y) = -126

For (0, 0),

20.

D = 42(0) ⋅ 6 - (-126)2 = -15,876 < 0.

There is a saddle point at (0, 0). For (126, 2646), D = 42(126) ⋅ 2646 - (-126)2 = 13,986, 756 > 0.

f ( x, y ) = 3x 2 + 7 y 3 - 42 xy + 5 f x ( x, y ) = 6 x - 42 y f y ( x, y ) = 21 y 2 - 42 x 6 x - 42 x = 0

(1)

21y 2 - 42 x = 0

(2)

Divide equation (1) by 6 and equation (2) by 21.

Since f xx (126, 2646) = 42(126) = 5292 > 0, there is a relative minimum at (126, 2646). 19.

f ( x, y) = x 2 + 4 y 3 - 6 xy - 1

y2 - 2x = 0

(4)

x = 7y

Substituting in equation (4), we have

Solve f x ( x, y) = 0 for x.

y 2 - 2(7 y) = 0

2x + 6 = 0 x = 3y

y 2 - 14 y = 0 y = 0 or y = 14 x = 0 or x = 98.

Substitute for x in 12 y 2 - 6 x = 0. 12 y 2 - 6(3 y) = 0 6 y (2 y - 3) = 0 3 y = 0 or y = 2 9 x = 0 or x = . 2

The critical points are (0, 0) and

(3)

Solve equation (3) for x.

f x ( x, y) = 2 x - 6 y, f y ( x, y) = 12 y 2 - 6 x

Then

x - 7y = 0

( 92 , 32 ).

f xx ( x, y) = 6, f yy ( x, y) = 42 y, f xy ( x, y) = -42

At (0, 0), D = 6 ⋅ 0 - (-42) 2 = -1764 < 0.

Saddle point at (0, 0) At (98, 14), D = 6 ⋅ 588 - (-42)2 = 1764 > 0 and f xx ( x, y) > 0.

Section 9.3

677 Solve the first equation for y, substitute into the second, and solve for x.

f ( x, y) = e x( y +1)

21.

f x ( x, y) = ( y + 1)e x( y +1) f y ( x, y) = xe

y = x2

x( y +1)

-x - x4 = 0

f x ( x, y) = 0 ( y + 1)e

x( y +1)

x(1 + x3 ) = 0

y +1 = 0

Then

y = -1.

y = 1.

For (0, 0), D = 0 ⋅ 0 - (-3) 2 = -9 < 0.

f yy ( x, y) = x 2e x( y +1)

A saddle point is at (0, 0). ⋅x+e

x( y +1)

For (-1, 1),

⋅1

= ( xy + x + 1)e x( y +1)

D = -6(-6) - (-3)2 = 27 > 0.

f xx ( x, y) = 6(-1) = -6 < 0.

For (0, -1),

f (-1, 1) = -3(-1)(1) + (-1)3 - 13 +

f xx (0, -1) = (0)2 e0 = 0 f yy (0, -1) = (0)2 e0 = 0

f xy (0, -1) = (0 + 0 + 1)e0 = 1

The equation matches graph (a).

There is a saddle point at (0, -1). 26.

f ( x, y) = y 2 + 2e x f x ( x, y) = 2e x f y ( x, y) = 2 y

If f y ( x, y) = 0, y = 0. But f x ( x, y) = 2e x = 0 has no solution. There are no extrema and no saddle points. z = -3xy + x3 - y3 +

3 1 1 y - y3 - x2 y + 2 2 16 ¶z ¶z 3 3 = -2 xy, = - y 2 - x2 ¶x ¶y 2 2 - 2 xy = 0 x = 0 or y = 0 3 3 2 - y - x2 = 0 2 2 If x = 0, z =

1 8

3 3 - y2 = 0 2 2 y = 1.

f x ( x, y) = -3 y + 3x 2 , f y ( x, y) = -3x - 3 y 2

Solve the system f x = 0, f y = 0. -3 y + 3 x 2 = 0 -3 x - 3 y 2 = 0

9 8

A relative maximum of 98 is at (-1, 1).

D = 0 ⋅ 0 - 12 = -1 < 0

25.

f xy ( x, y) = -3

f xx ( x, y) = ( y + 1) 2 e x( y +1)

22.

y =0

f yy ( x, y) = -6 y

Therefore, (0, -1) is a critical point.

f xy ( x, y) = ( y + 1)e

x = -1

f xx ( x, y) = 6 x

x = 0.

x( y +1)

The critical points are (0, 0) and (-1, 1).

f y ( x, y) = 0 x( y +1)

x =0

If y = 0, 3 - x2 = 0 2 x =

-y + x2 = 0 -x - y 2 = 0

3 6 = . 2 2

1 8

678

Chapter 9 MULTIVARIABLE CALCULUS ¶2z

¶2z

¶x

¶y

= -2 y , 2

= -3 y , 2

D = (-2 y)(-3 y) - (-2 x) = 6 y - 4 x At (0,1), D = 6 > 0 and

¶2z 2

For (0, -1),

¶2 z = -2 x ¶y¶x

D = 2[12(-1)2 - 4] - 0 = 16 > 0.

f xx ( x, y) = 2 > 0 f (0, -1) = (-1)4 - 2(-1) 2 + 02 -

= -2 < 0.

¶x 3 1 1 17 z = - + = 2 2 16 16

1 = -2 16

A relative minimum of -

17 at (0, 1) Relative maximum of 16

D = 2(12 ⋅ 12 - 4) - 0 = 16 > 0

¶ z

= 2 > 0. ¶x 2 3 1 1 15 z =- + + =2 2 16 16

f xx ( x, y ) = 2 > 0 f (0,1) = 14 - 2 ⋅ 12 + 02 -

15 at (0, -1) Relative minimum of - 16

(

)

At  26 , 0 , D = -6 < 0. Saddle points at

z = y 4 - 2 y 2 + x2 -

28.

z = -2 x3 - 3 y 4 + 6 xy 2 +

1 16

¶z ¶z = -6 x 2 + 6 y 2 , = -12 y 3 + 12 xy ¶x ¶y

17 16

f x ( x, y) = 2 x, f y ( x, y) = 4 y 3 - 4 y

Solve the system f x = 0, f y = 0. 2x = 0

(1)

4y - 4y = 0

(2)

33 16

The equation matches graph (b).

6 2

The correct graph is (d). 27.

17 16

33 is at (0, 1). A relative minimum of - 16

( ,0 ) and ( - ,0 ) 6 2

33 is at (0, -1). 16

For (0, 1),

At (0, -1), D = 6 > 0 and

17 16

-6 x 2 + 6 y 2 = 0 -12 y + 12 xy = 0

(2)

Divide equation (1) by 6 and equation (2) by 12.

4 y( y - 1) = 0 4 y( y + 1)( y - 1) = 0

(1)

-x 2 + y 2 = 0

(3)

- y 3 + xy = 0

(4)

Equation (1) gives x = 0 and equation (2) gives y = 0, y = -1, or y = 1.

Solve equation (3) for y. Substitute y =  x in equation (4). If y = x,

The critical points are (0, 0), (0, -1) and (0, 1).

-x 3 + x 2 = 0 -x 2 (-x + 1) = 0 x = 0 or x = 1.

f xx ( x, y) = 2, f yy ( x, y) = 12 y 2 - 4,

If x = 0, y = 0.

f xy ( x, y) = 0

If x = 1, y = 1.

For (0, 0), D = 2(12 ⋅ 02 - 4) - 0 = -8 < 0.

If y = -x, x3 - x 2 = 0 x 2 ( x - 1) = 0 x = 0 or x = 1.

A saddle point is at (0, 0).

Section 9.3

679

If x = 0, y = 0

For (0, 0), D = 4(-4) - 0 = -16 < 0.

If x = 1, y = -1. ¶2 z

¶2 z

¶x

= -12 x, 2

¶y

= -36 y 2 + 12 x,

For (0, -1), D = 4(8) - 0 = 32 > 0,

¶ z = 12 y ¶y ¶x

and f xx ( x, y) = 4 > 0.

D = (-12 x)(-36 y 2 + 12 x) - (12 y 2 )

f (0, -1) = -

At (0, 0), D = 0, which gives no information. For (0, 1),

At (1, 1),

D = 4(8) - 0 = 32 > 0,

D = -12(-36 + 12) - 144 2

= 144 > 0 and ¶ 2z = -12 < 0. ¶x

z = -2 - 3 + 6

and f xx ( x, y) = 4 > 0.

1 17 = 16 16

f (0, 1) = 17 at (1, 1). 16

so there is a relative maximum of

= 144 > 0 and ¶ 2z = -12 < 0,

For (-1, 0),

and f xx ( x, y) = -8 < 0.

¶x

17 so there is a relative maximum of at (1, -1). 16 The correct graph is (c). 4

f x ( x, y) = -4 x + 4 x, f y ( x, y) = 4 y - 4 y

Solve f x ( x, y) = 0, f y ( x, y) = 0. -4 x + 4 x = 0 3

4y - 4y = 0

(1) (2)

-4 x( x - 1) = 0

For (-1, -1),

D = -8(8) - 0 = -64 < 0.

For (1, 0), D = -8(-4) = 32 > 0,

and f xx ( x, y) = -8 < 0.

(1)

1 f (1, 0) = 1 16

-4 x( x + 1)( x - 1) = 0 4 y( y 2 - 1) = 0

(2)

4 y( y + 1)( y - 1) = 0

Equation (2) gives y = 0, -1 , or 1. Critical points are (0, 0), (0, 1), (0, 1), (1, 0), (1, 1), (1, 1), (1, 0), (1, 1), (1, 1). f xx ( x, y) = -12 x 2 + 4, f xy ( x, y) = 0

For (1, -1), D = -8(8) - 0 = -64 < 0.

Equation (1) gives x = 0, -1, or 1.

f yy ( x, y) = 12 y 2 - 4

17 16

For (-1, 1), 3

f (-1, 0) =

D = -8(8) - 0 = -64 < 0.

1 z = -x + y + 2 x - 2 y + 16 4

15 16

D = -8(-4) - 0 = 32 > 0,

At (1, -1), D = -12(-36 + 12) - 144

29.

15 16

For (1, 1), D = -8(8) - 0 = -64 < 0.

Saddle points are at (0, 0), (-1, -1), (-1, 1), (1, -1), and (1, 1). Relative maximum of 17 is at (-1, 0) and (1, 0). 16 15 is at (0, -1) and (0, 1). Relative minimum of - 16

The equation matches graph (e).

680 30.

Chapter 9 MULTIVARIABLE CALCULUS z = - y 4 + 4 xy - 2 x 2 +

Therefore, the test gives no information. Examine a graph of the function drawn by using level curves.

1 16

¶z ¶z = 4 y - 4 x, = -4 y 3 + 4 x ¶x ¶y 4 y - 4x = 0

(1)

( 2)

-4 y + 4 x = 0

If f ( x, y) = 1, then x 4 + y 4 = 0. The level curve is the point (0, 0, 1). If f ( x, y) = 0, then x 4 + y 4 = 1. The level curve is the circle with center (0, 0, 0) and radius 1.

Divide equation (1) by 4 and equation (2) by -4.

If f ( x, y) = -15, then x 4 + y 4 = 16. The level curve is the curve with center (0, 0, -15) and radius 2. The xz-trace is

y - x = 0 (3) - y 3 + x = 0 (4)

Solve equation (3) for y. y = x

z = 1 - x 4.

Substituting, we have

This curve has a maximum at (0, 0, 1) and opens downward. The yz-trace is

-x 3 + x = 0 x(-x 2 + 1) = 0 x =0

z = 1 - y 4.

x = 1 or x = -1.

This curve also has a maximum at (0, 0, 1) and opens downward.

If x = 0, y = 0. If x = 1, y = 1. If x = -1, y = -1. ¶2z

¶2 z

¶x

¶y

= -4, 2

= -12 y 2 , 2

¶2 z = 4 ¶ y¶ x

D = -4(-12 y 2 ) - 42 = 48 y 2 - 16

At (0, 0), D = -16 < 0. Saddle point at (0, 0) At (1, 1) and (-1, -1),

If f ( x, y) > 1, then x 4 + y 4 < 0, which is impossible, so the function does not exist. Thus, the function has a relative maximum of 1 at (0, 0).

2 D = 48 - 16 = 32 > 0 and ¶ 2z = -4 < 0. ¶x

17 Relative maximum of at (1, 1) and at 16 (-1, -1). The correct graph is (f ) .

31.

32.

f ( x, y ) = x 3 + ( x - y ) 2 f x ( x, y ) = 3x 2 + 2( x - y )

f ( x, y) = 1 - x 4 - y 4 3

f x ( x, y) = -4 x , f y ( x, y) = -4 y

f y ( x, y ) = -2( x - y ) 3x 2 + 2 x - 2 y = 0 -2 x + 2 y = 0

The system f x ( x, y) = -4 x3 = 0, f y ( x, y) = -4 y3 = 0

3x 2 = 0 x =0

gives the critical point (0, 0). f xx ( x, y) = -12 x 2 f yy ( x, y) = -12 y

0 + 2y = 0 y =0

f xy ( x, y) = 0

f xx ( x, y) = 6 x + 2, f yy ( x, y) = 2, f xy ( x, y) = -2

For (0, 0), D = 0 ⋅ 0 - 02 = 0.

D = (6 x + 2)(2) - (-2)2 = 12 x

Section 9.3

681 For (5, 25),

At (0, 0), D = 0, which gives no information. Examine the graph of z = x3 + ( x - y) 2: in the

f xx (5, 25) = -4(25) + 24(5) + 40 = 60 f yy (5, 25) = 2

yz-plane, the trace is z = y , which has a minimum at (0, 0, 0); in the xz-plane, the trace is 3

f xy (5, 25) = -4(5) = -20

z = x + x , which has a minimum at (0, 0, 0).

But in the plane y = x, the trace is z = x3, which has neither a maximum nor a minimum at (0, 0, 0). So the function has no relative extrema. Notice the orientation of axes in the following figure: This is a back view.

D = (60) ⋅ (2) - (-20) 2 = -280 < 0

There is a saddle point at (5, 25). The correct answer choice is (e). 35.

f ( x, y) = x 2 ( y + 1)2 + k ( x + 1)2 y 2

(a)

f x ( x, y) = 2 x + 2ky 2 ( x + 1) f y ( x, y) = 2 x 2 ( y + 1) + 2k ( x + 1) 2 y f x (0, 0) = 2(0) + 2k (0) 2 (0 + 1) =0 f y (0, 0) = 2(0) 2 (0 + 1) + 2k (0 + 1)2 (0) =0 Thus, (0, 0) is a critical point for all values of k.

34.

f ( x, y) = y 2 - 2 x 2 y + 4 x3 + 20 x 2

(b)

f x ( x, y) = -4 xy + 12 x 2 + 40 x f y ( x, y) = 2 y - 2 x

f xx ( x, y) = 2 + 2ky 2 f yy ( x, y) = 2 x 2 + 2k ( x + 1)2

f xy ( x, y) = 4ky( x + 1)

f xx ( x, y) = -4 y + 24 x + 40

f xx (0, 0) = 2 + 2k (0)2 = 2

f yy ( x, y) = 2

f yy (0, 0) = 2(0)2 = 2k (0 + 1) 2 = 2k

f xy ( x, y) = -4 x

f xy (0, 0) = 4k (0)(0 + 1) = 0

For (-2, 4),

D = 2 ⋅ 2k - 02 = 4k

f xx (-2, 4) = -4(4) + 24(-2) + 40 = -24 f yy (-2, 4) = 2

(0, 0) is a relative minimum when 4k > 0, hence when k > 0. When k = 0, D = 0 so the test for relative extrema gives no information. But if k = 0, f ( xy ) =

f xy (-2, 4) = -4(-2) = 8 D = (-24) ⋅ (2) - 82 = -112 < 0

There is a saddle point at (-2, 4).

x 2 ( y + 1)2 , which is always greater than or

For (0, 0),

equal to f (0,0) = 0. So (0, 0) is a relative minimum for k ³ 0.

f xx (0, 0) = -4(0) + 24(0) + 40 = 40 f yy (0, 0) = 2

36.

f xy (0, 0) = -4(0) = 0 D = (40) ⋅ (2) - 02 = 80 > 0

Since f xx (0, 0) = 40 > 0, there is a relative minimum at (0, 0).

å (mx + b - y) S (m, b) = å 2(mx + b - y) ( x) = 2å (mx + bx - xy) S (m, b) = å 2(mx + b - y)(1) = 2å (mx + b - y) 2

S (m, b) =

682

Chapter 9 MULTIVARIABLE CALCULUS 3x - 2 - 2 y = 0

If Sm (m, b) = 0, then

-2 x - 2 + 2 y = 0 x-4 =0 x = 4

å (mx + bx - xy) = 0 å (mx ) + å (bx) = å (xy) ( å x ) m + å (x)b = å ( xy). (1) 2

-2(4) - 2 + 2 y = 0 2 y = 10

If Sb (m, b) = 0, then

y =5

å (mx + b - y) = 0 å (mx) + å (b) - å y = 0 ( å x ) m + nb = å y. (2) 2

Let Lxx ( x, y) = 3, L yy ( x, y) = 2, Lxy ( x, y) = -2. D = 3(2) - (-2)2 = 2 > 0 and Lxx ( x, y) > 0.

Equations (1) and (2) give us the following system of equations.

Relative minimum at (4, 5) is L(4,5) =

( å x )b + ( å x ) m = å xy nb + ( å x ) m = å y 2

38.

3 2 (4) + (5)2 - 2(4) - 2(5) 2 - 2(4)(5) + 68

= 59.

So $59 is a minimum cost, when x = 4 and y = 5.

P( x, y) = 1500 + 36 x - 1.5 x 2 + 120 y - 2 y 2 Px ( x, y) = 36 - 3x Py ( x, y) = 120 - 4 y

40.

C ( x, y) = 2 x 2 + 2 y 2 - 3xy

Therefore, (12, 30) is a critical point.

+ 4 x - 94 y + 4200 C x ( x, y) = 4 x - 3 y + 4

Pxx ( x, y) = -3

C y ( x, y) = 4 y - 3x - 94

Pyy ( x, y) = -4

Solve the system C x ( x, y) = 0, C y ( x, y) = 0.

If Px = 0, x = 12. If Py = 0, y = 30.

Pxy ( x, y) = 0

4x - 3y + 4 = 0

D = (-3) ⋅ (-4) - 0 = 12 > 0

- 3x + 4 y - 94 = 0

Since Pxx = -3 < 0, there is a relative maximum at (12, 30). P(12,30) = 1500 + 36(12) - 1.5(12)2 + 120(30) - 2(30)2

12 x - 9 y + 12 = 0 -12 x + 16 y - 376 = 0 7 y - 364 = 0 y = 52

= 3516 (hundred dollars)

4 x - 3(52) + 4 = 0 4 x = 152 x = 38

The maximum profit is $351,600 when the cost of a unit of labor is 12 and the cost of a unit of goods is 30. 39.

3 2 x + y 2 - 2 x - 2 y - 2 xy + 68, 2 where x is the number of skilled hours and y is the number of semiskilled hours. L( x, y) =

Lx ( x, y ) = 3x - 2 - 2 y, L y ( x, y ) = 2 y - 2 - 2 x

Therefore, (38, 52) is a critical point. C xx = 4 C yy = 4 C xy = -3 D = (4)(4) - (-3)2 = 7 > 0

Since C xx = 4 > 0, there is a relative minimum at (38, 52).

Section 9.3

683 Px ( x, y) = 0

C (38,52) = 2(38) 2 + 2(52) 2 - 3(38)(52) + 4(38) - 94(52) + 4200 = 1832

38 units of electrical tape and 52 units of packing tape should be produced to yield a minimum cost of $1832. 41.

36 y - 3x 2 = 0 36 y = 3x 2 1 2 y = x 12 Px ( x, y) = 0 36 x - 24 y 2 = 0 36 x = 24 y 2 2 x = y2 3 Use substitution to solve the system of equations

R( x, y) = 15 + 169 x + 182 y - 5x 2 - 7 y 2 - 7 xy Rx ( x, y) = 169 - 10 x - 7 y R y ( x, y) = 182 - 14 y - 7 x

1 2 x 12 2 x = y 2. 3

y =

Solve the system Rx = 0, R y = 0. -10 x - 7 y + 169 = 0 -7 x - 14 y + 182 = 0

2 1 æç 2 2 ö÷ y ÷ ç 12 çè 3 ÷ø 1 æç 4 ö÷ 4 y = ç ÷y 12 çè 9 ø÷

y =

20 x + 14 y - 338 = 0 -7 x - 14 y + 182 = 0 13x - 156 = 0 x = 12

y =

-10(12) - 7 y + 169 = 0 -7 y = -49 y =7

1 4 y -y =0 27 æ 1 3 ö çç y - 1÷÷÷ y = 0 çè 27 ø

(12, 7) is a critical point. Rxx = -10 R yy = -14 Rxy = -7 2

D = (-10)(-14) - (-7) = 91 > 0

Since Rxx = -10 < 0, there is a relative maximum at (12, 7). R(12, 7) = 15 + 169(12) + 182(7) - 5(12) 2 - 7(7)2 - 7(12)(7) = 1666 (hundred dollars)

12 spas and 7 solar heaters should be sold to produce a maximum revenue of $166,600.

1 4 y 27

1 3 y -1 = 0 27 1 3 y =1 27

y =0

y3 = 27 or

y =0

y =3

y =0

2 2 (3) = 6. 3 2 If y = 0, x = (0)2 = 0. 3 The critical points are (6, 3) and (0, 0). If y = 3, x =

Pxx ( x, y) = -6 x Pyy ( x, y) = -48 y

42.

P( x, y) = 36 xy - x - 8 y

Pxy ( x, y) = 36 Pxx (6,3) = -36

Px ( x, y) = 36 y - 3x 2

Pyy (6,3) = -144

Py ( x, y) = 36 x - 24 y 2

Pxy (6,3) = 36 D = (-36)(-144) - (36)2 = 3888

684

Chapter 9 MULTIVARIABLE CALCULUS Here D > 0 and Pxx < 0, so there is a relative maximum at (6, 3).

If y = 0, x = 2(0)3 = 0. If y = 1, x = 2(1)3 = 2.

Pxx (0, 0) = 0

The critical points are (0, 0) and (2, 1).

Pyy (0, 0) = 0

Txx ( x, y) = 12 x 2

Pxy (0, 0) = 36

Tyy ( x, y) = 192 y 2

D = 0 ⋅ 0 - 362 = -1296

Txy ( x, y) = -32 Txx (0, 0) = 0

Since D < 0, there is a saddle point at (0, 0).

Tyy (0, 0) = 0

P(6,3) = 36(6)(3) - (6)3 - 8(3)3 = 648 - 216 - 216 = 216

Txy (0, 0) = -32 D = 0 ⋅ 0 - (-32)2 = -1024

Since D < 0, there is a saddle point at (0, 0).

So 6 tons of steel and 3 tons of aluminum produce a maximum profit of $216,000. 43.

Txx (2,1) = 48 Tyy (2,1) = 192

T ( x, y) = x 4 + 16 y 4 - 32 xy + 40

Txy (2,1) = -32

Tx ( x, y) = 4 x3 - 32 y

D = 48 ⋅ 192 - (-32)2 = 8192

Ty ( x, y) = 64 y 3 - 32 x Tx ( x, y) = 0

Since D > 0 and Txx > 0, there is a relative minimum at (2, 1).

4 x3 - 32 y = 0

T (2,1) = 24 + 16(1) 4 - 32(2)(1) + 40

4 x = 32 y 1 3 x = y 8

= 16 + 16 - 64 + 40 =8

Spend $2000 on quality control and $1000 on consulting, for a minimum time of 8 hours.

Ty ( x, y) = 0 64 y 3 - 32 x = 0

44.

64 y 3 = 32 x

P( , r , s) =  (3r 2 (1 - r ) + r 3 ) + (1 -  ) (3s 2 (1 - s) + s3 )

2y = x

Use substitution to solve the system of equations 1 3 x = y 8 3

2y = x 1 (2 y 3 )3 8 1 y = (8) y 9 8 y =

y = y

y -y =0 y( y8 - 1) = 0 y = 0 or

y8 - 1 = 0

y = 0 or

y8 = 1

y = 0 or

y =1

(a) P(0.9,0.5,0.6) = 0.9 [3(0.5) 2 (1 - 0.5) + (0.5)3 ] + (1 - 0.9)[3 (0.6)2 (1 - 0.6) + (0.6)3 ] = 0.5148

P(0.1,0.8,0.4) = 0.1[3(0.8)2 (1 - 0.8) + (0.8)3 ] + (1 - 0.1)[3 (0.4)2 (1 - 0.4) + (0.4)3 ] = 0.4064

The jury is less likely to make the correct decision in the second situation. (b) If r = s = 1 then P( ,1,1) = 1, so the jury always makes a correct decision. These values do not depend on  , but in a real-life situation  is likely to influence r and s.

Section 9.3

685

a = er = e3.98375 = 53.718 and

P ( , r, s) = 3r 2 (1 - r ) + r 3 2

b = e s = e0.272021 = 1.31261.

- (3s (1 - s) + s )

Thus an exponential fit to the original data

= 3r 2 - 2r 3 - (3s 2 - 2s3 )

has the form y = a ⋅ bt with a = 53.718 and b = 1.31261:

Pr ( , r, s) =  (6r (1 - r ) - 3r 2 + 3r 2 ) = 6 r (1 - r ) 2

y = 53.718(1.31261)t

Ps ( , r, s) = (1 -  )(6s(1 - s) - 3s + 3s )

(b) Same as (a).

= 6s(1 -  )(1 - s)

P ( , r, s) = 0 when r = s.

Since Pr ( , r , s) = 6 r (1 - r ), and Ps ( , r , s) = 6(1 -  )s (1 - s), then P , Pr , and Ps are simultaneously 0 at the points ( ,1,1) and ( , 0, 0). So ( ,1,1) and ( ,0,0) are critical points. P( , 0, 0) = 0 while P( ,1,1) = 1

46.

E (t , T ) = 436.16 - 10.57t - 5.46T - 0.02t 2 + 0.02T 2 + 0.08tT

(a)

The value of E before cooking is 436.16 kJ/ mol. (b)

- 5.46(180) - 0.02(10)2 + 0.02(180)2 + 0.08(10)(180) = 137.66

Using the procedure suggested, take the natural logarithm of the number-of-transistors data. Time in years since 1970, t 4 9 15 19 23 29 36 42 46 49

Natural log of number of transistors in millions w

E (10,180)

= 436.16 - 10.57(10)

Since P( , r, s) represents a probability, 0 £ P( , r , s) £ 1. Thus, P( ,1,1) = 1 is a maximum value of the function. 45.

E (0, 0) = 436.16.

After cooking for 10 minutes at 180 C , the total change in color is 137.66 kJ/mol. (c)

Et = -10.57 - 0.04t + 0.08T ET = -5.46 + 0.04T + 0.08t Solve the system Et = 0, ET = 0.

ln(300) ln(879) ln(2644) ln(6822) ln(10,544) ln(74,219) ln(2,234,568) ln(5,689,529) ln(15,798,474) ln(36,341,912)

» » » » » » » » » »

5.704 6.779 7.880 8.828 9.263 11.215 14.620 15.556 16.575 17.408

(a) A linear fit of the data in the table above gives w = r + st w = 3.98375 + 0.272021t

where w is the natural logarithm of the number of transistors in millions and t is the number of years since 1970. To convert this back to an exponential fit for the original data, compute

-0.04t + 0.08T - 10.57 = 0 0.08t + 0.04T - 5.46 = 0 - 0.04t + 0.08T - 10.57 = 0 - 0.16t - 0.08T + 10.92 = 0 - 0.20t + 0.35 = 0 t = 1.75 -0.04(1.75) + 0.08T - 10.57 = 0 0.08T - 10.64 = 0 T = 133

(1.75, 133) is a critical point. Ett = -0.04 ETT = 0.04 EtT = 0.08 D = (-0.04)(0.04) - (0.08)2 = -0.008 < 0 (1.75, 133 C ) is a saddle point.

686

Chapter 9 MULTIVARIABLE CALCULUS

9.4 Lagrange Multipliers Your Turn 1

Find the minimum value of f ( x, y) = x 2 + 2 x + 9 y 2 + 3 y + 6 xy subject to the constraint 2 x + 3 y = 12.

Then 2(11) + 3 y = 12, so y = -10/3. f (11, -10/3) = 13, which is larger than our candidate. We conclude that the minimum value of f subject to the given constraint is 12. Your Turn 2

Step 1 Rewrite the constraint as g ( x, y) = 2 x + 3 y - 12. Step 2 Form the Lagrange function. F ( x, y,  ) = f ( x, y) -  ⋅ g ( x, y) = x2 + 2x + 9 y 2 + 3 y + 6 xy -  (2 x + 3 y - 12) = x 2 + 9 y 2 + 6 xy - 2 x - 3 y + 2 x + 3 y + 12

Step 3 Find the first partial derivatives of F . Fx ( x, y,  ) = 2 x + 6 y - 2 + 2

As in Example 3, let x, y, and z be the dimensions of the box. If the front and top are missing, the surface area is 2 xy + xz + yz so the constraint is 2 xy + xz + yz = 6. As in Example 3, the volume is xyz. The problem is thus to maximize f ( x, y, z ) = xyz subject to 2 xy + xz + yz = 6. Step 1 Rewrite the constraint as g ( x, y, z ) = 2 xy + xz + yz - 6.

Step 2 Form the Lagrange function. F ( x, y, z,  ) = f ( x, y, z ) -  ⋅ g ( x, y, z )

Fy ( x, y,  ) = 6 x + 18 y - 3 + 3

= xyz -  ( 2 xy + xz + yz - 6 )

F ( x, y,  ) = -2 x - 3 y + 12

= xyz - 2 xy - xz - yz + 6

Step 3 Find the first partial derivatives of F .

Step 4 Form the system of equations Fx ( x, y,  ) = 0, Fy ( x, y,  ) = 0, F ( x, y,  ) = 0.

Fx ( x, y, z,  ) = yz - 2 y - z 

2 x + 6 y - 2 + 2 = 0 6 x + 18 y - 3 + 3 = 0

Fy ( x, y, z,  ) = xz - 2 x - z 

-2 x - 3 y

F ( x, y, z,  ) = -2 xy - xz - yz + 6

+ 12 = 0

Step 5 Solve the system of equations from Step 4.

Fz ( x, y, z,  ) = xy - x - y

Step 4 Form the system of equations

First solve the first two equations for  and then set these results equal to each other (remove the common factors of 2 from the first equation and 3 from the second equation). x + 3y + 1 =  2x + 6 y + 1 = 

Fx ( x, y, z,  ) = 0, Fy ( x, y, z,  ) = 0, Fz ( x, y, z,  ) = 0, F ( x, y, z,  ) = 0. yz - 2 y - z  = 0

x + 3y = 2x + 6 y x + 3y = 0

xz - 2 x - z  = 0 xy - x - y = 0

x = -3 y

-2 xy - xz - yz + 6 = 0

Now substitute for x in the last equation. -2(-3 y) - 3 y + 12 = 0 6 y - 3 y = -12 3 y = -12 y = -4

Step 5 Solve the system of equations from Step 4. First solve each of the first three equations for . This gives three expressions for . yz xz xy  = ,  = ,  = x+ y 2y + z 2x + z

Since x = -3 y we have x = 12 and y = -4. So a candidate for a minimum value of the function f will be f (12, -4) = 12.

Set the second and third expressions equal. xz xy = 2x + z x+ y

To see if this is really a minimum we can evaluate f at a few nearby points that satisfy the constraint 2 x + 3 y = 12. For example, let x = 11.

x 2 z + xyz = 2 x 2 y + xyz

x 2 z = x 2 (2 y)

Section 9.4

687

Since x is not zero, this shows that z = 2 y.

Exactly the same calculation using the first and third expressions will show that z = 2 x. Thus x = y. Now use the fourth equation:

4y -  = 0 4x -  = 0 x + y - 16 = 0

Equations (1) and (2) give  = 4y and  = 4 x. Thus, 4 y = 4x y = x.

-2 xy - xz - yz + 6 = 0 -2 x 2 - x(2 x) - x(2 x) + 6 = 0 6x2 = 6 x = 1 (since x is positive)

(1) (2) (3)

Substituting into equation (3), x + ( x) - 16 = 0 x = 8. y = 8. So

Now we know that x = 1, y = 1, and z = 2. The box should be 2 ft. wide and 1 ft. high and long.

Maximum is f (8,8) = 4(8) (8) = 256.

9.4 Warmup Exercises 3 2

f ( x, y) = 4 x y - 6 x - 3 y

W1.

f x ( x, y) = 12 x 2 y 2 - 24 x3 3

f y ( x, y) = 8x y - 15 y

W2.

Maximize f ( x, y) = 2 xy + 4, subject to x + y = 20. 1.

g ( x, y) = x + y - 20

F ( x, y,  ) = 2 xy + 4 -  ( x + y - 20)

Fx ( x, y,  ) = 2 x -  Fy ( x, y,  ) = 2 x - 

f ( x, y, z ) = x 2 y 3 z 5 - 4 x 2 y 5 + 3xz 6 - 2 y8 z 7

F ( x, y,  ) = -( x + y - 20)

f x ( x, y, z ) = 2 xy 3 z 5 - 8 xy 5 + 3z 6 f y ( x, y, z ) = 3x 2 y 2 z 5 - 20 x 2 y 4 - 16 y 7 z 7

2 y -  = 0 (1) 2 x -  = 0 ( 2) x + y - 20 = 0 (3)

Equations (1) and (2) give  = 2y and  = 2 x. Thus, 2 y = 2x y = x

f z ( x, y, z ) = 15 x 2 y 2 z 4 + 18xz 5 - 14 y8 z 6

9.4 Exercises 1.

True

False. The second derivative test from the previous section does not apply to the solutions found by Lagrange multipliers.

True

Maximize f ( x, y) = 4 xy,

subject to x + y = 16. 1.

g ( x, y) = x + y - 16

2. 3.

Substitution into equation (3), x + ( x) - 20 = 0 x = 10. y = 10. So Maximum is f (10,10) = 2(10) (10) + 4 = 204. Maximize f ( x, y) = xy 2 , subject to x + 2 y = 15. 1.

g ( x, y) = x + 2 y - 15

F ( x, y,  ) = 4 xy -  ( x + y - 16).

F ( x, y,  ) = xy 2 -  ( x + 2 y - 15)

Fx ( x, y,  ) = 4 y - 

Fx ( x, y,  ) = y 2 - 

Fy ( x, y,  ) = 4 x -  F ( x, y,  ) = - ( x + y - 16)

Fy ( x, y,  ) = 2 xy - 2 F ( x, y,  ) = -( x + 2 y - 15)

688

Chapter 9 MULTIVARIABLE CALCULUS 4.

y2 -  = 0 2 xy - 2 = 0 + x 2 y - 15 = 0

Substituting x = 0 into equation (3),

(1) (2) (3)

3(0) - y - 9 = 0 y = -9.

Equations (1) and (2) give  = y 2 and  = xy. Thus,

Substituting x = - 3 into equation (3), æ 2y ö 3çç - ÷÷÷ - y - 9 = 0 çè 3 ø

y 2 = xy y ( y - x) = 0 y = 0 or y = x

y = -3.

f (0, -9) = 8(0)2 (-9) = 0, and f (2, -3) = 8(2)2 (-3) = -96.

Since f (0, -9) > f (2, -3), f (0, -9) = 0 is a maximum.

Thus,

f (15, 0) = 15(0) 2 = 0, and

Minimize f ( x, y ) = x 2 + 2 y 2 - xy, subject to x + y = 8.

f (5,5) = 5(5)2 = 125.

Since f (5,5) > f (15, 0), f (5,5) = 125 is a maximum.

g ( x, y ) = x + y - 8

F ( x, y,  ) = x 2 + 2 y 2 - xy -  ( x + y - 8)

Maximize f ( x, y) = 8x 2 y, subject to 3x - y = 9.

g ( x, y) = 3x - y - 9

F ( x, y,  ) = 8 x 2 y -  (3x - y - 9)

Fx ( x, y,  ) = 16 xy - 3

F ( x, y,  ) = -( x + y - 8)

Fy ( x, y,  ) = 8x 2 + 

(1)

8x2 +  = 0 3x - y - 9 = 0

(2) (3)

2x - y -  = 0 4y - x -  = 0 x + y-8 = 0

F ( x, y,  ) = -(3x - y - 9) 16 xy - 3 = 0

Fx ( x, y,  ) = 2 x - y -  Fy ( x, y,  ) = 4 y - x - 

Subtracting the second equation from the first equation to eliminate  gives the new system of equations x+ y =8 3x - 5 y = 0.

Equations (1) and (2) give  =

16 xy 3

and

Solve this system.

 = -8x . Thus, 16 xy = -8 x 2 3 16 xy =0 8x2 + 3 æ 2 y ÷ö 8x çç x + ÷=0 çè 3 ÷ø x = 0 or x = -

2(-3) = 2. 3

Thus,

Substituting y = x into equation (3) x + 2( x) - 15 = 0 x = 5. = x = 5. y So

x =-

Substituting y = 0 into equation (3), x + 2(0) - 15 = 0 x = 15.

2y . 3

5 x + 5 y = 40 3x - 5 y = 0 8 x = 40 x =5

But x + y = 8, so y = 3. Thus, f (5,3) = 25 + 18 - 15 = 28 is a minimum.

Section 9.4 10.

689 5.

Minimize f ( x, y) = 3x 2 + 4 y 2 - xy - 2, subject to 2 x + y = 21. 1.

g ( x, y) = 2 x + y - 21

F ( x, y,  )

Adding the first two equations to eliminate  gives 2 x - 20 y = 0 x = 10 y.

Substituting x = 10 y in the third equation gives 10 y - y = 18

= 3x 2 + 4 y 2 - xy

y = 2

- 2 -  (2 x + y - 21)

x = 20.

Fx ( x, y,  ) = 6 x - y - 2

Thus,

Fy ( x, y,  ) = 8 y - x - 

f (20, 2) = 202 - 10 (2)2 = 400 - 40 = 360. f (20, 2) = 360 is a maximum.

F ( x, y,  ) = -(2 x + y - 21)

6 x - y - 2 = 0 8y - x -  = 0

12.

2 x + y - 21 = 0

 =

Maximize f ( x, y) = 12 xy - x 2 - 3 y 2 , subject to x + y = 16.

6x - y and  = 8y - x 2 6x - y = 8y - x 2 8 y = x 17

g ( x, y) = x + y - 16

F ( x, y,  ) = 12 xy - x 2 - 3 y 2 -  ( x + y - 16)

Fy ( x, y,  ) = 12 x - 6 y - 

Substituting into 2 x + y - 21 = 0, we have 2x +

F ( x, y,  ) = -( x + y - 16)

8 x - 21 = 0, 17

Minimum is æ

ö2

 = 12 y - 2 x and  = 12 x - 6 y 12 y - 2 x = 12 x - 6 y

( 172 , 4 ) = 3ççèç 172 ÷ø÷÷ + 4(4)2 - 172 (4) - 2

y =

979 = = 244.75. 4

11.

12 y - 2 x -  = 0 12 x - 6 y -  = 0 x + y - 16 = 0

so x = 17 and y = 4. 2

Fx ( x, y,  ) = 12 y - 2 x - 

7 x 9

Substituting into x + y - 16 = 0, we have 7 x - 16 = 0, 9 so x = 9 and y = 7. x+

Maximize f ( x, y) = x 2 - 10 y 2 , subject to x - y = 18. 1.

g ( x, y) = x - y - 18

Maximum is

F ( x, y,  )

f (9, 7) = 12(9)(7) - (9) 2 - 3(7) 2 = 528.

= x 2 - 10 y 2 -  ( x - y - 18)

Fx ( x, y,  ) = 2 x - 

13.

Maximize f ( x, y, z ) = xyz 2 ,

Fy ( x, y,  ) = -20 y - 

subject to x + y + z = 6.

F ( x, y,  ) = -( x - y - 18) 2x -  = 0

g ( x, y, z ) = x + y + z - 6

F ( x, y,  )

-20 +  = 0 x - y - 18 = 0

= xyz 2 -  ( x + y + z - 6)

690

Chapter 9 MULTIVARIABLE CALCULUS 14.

Fx ( x, y, z,  ) = yz 2 - 

Fy ( x, y, z,  ) = xz 2 - 

Fz ( x, y, z,  ) = 2 zxy - 

g ( x, y, z ) = xyz - 32

F ( x, y, z,  ) = -( x + y + z - 6)

F ( x, y, z,  ) = xy + 2 xz + 2 yz -  ( xyz - 32)

Fx ( x, y, z,  ) = y + 2 z -  yz

Setting Fx, Fy , Fz and F equal to zero yields yz 2 -  = 0

(1)

xz 2 -  = 0

(2)

2 xyz -  = 0

(3)

Fy ( x, y, z,  ) = x + 2 z -  xz Fz ( x, y, z,  ) = 2 x + 2 y -  xy F ( x, y, z,  ) = -( xyz - 32)

x + y + z - 6 = 0. (4)

Maximize f ( x, y, z ) = xy + 2 xz + 2 yz, subject to xyz = 32.

y + 2 z -  yz = 0 x + 2 z -  xz = 0 2 x + 2 y -  xy = 0 xyz - 32 = 0

 =

 = yz 2 ,  = xz 2 , and  = 2xyz yz 2 = xz 2 z 2 ( y - x) = 0 x = y

z =0

y + 2z x + 2z , = , and yz xz 2x + 2 y  = xy

yz = 2 xyz 2 xyz - yz 2 = 0

y + 2z x + 2x = yz xz

yz (2 x - z ) = 0 y = 0 or z = 0 or z = 2 x

xyz + 2 xz 2 = xyz + 2 yz 2

In a similar way, the third equation

2 yz 2 - 2 xz 2 = 0

xz 2 = 2 xyz

2 z 2 ( y - x) = 0

implies that x = 0 or z = 0 or z = 2 y. By the nature of the function to be maximized, f ( x, y, z ) = xyz 2 , a nonzero maximum can come only from those points with nonzero coordinates. Therefore, assume y = x and z = 2 y = 2 x.

If y = x and z = 2 x are substituted into equation (4), then

z = 0 or

and y + 2z 2x + 2 y = yz xy xy 2 + 2 xyz = 2 xyz + 2 y 2 z 2 y 2 z - xy 2 = 0

y 2 (2 z - x) = 0

x + x + 2x - 6 = 0 x =

y = 0

3 . 2

z =

x 2

impossible. Substituting y = x and z = 2x into xyz - 32 = 0, we have

æ3 3 ö 3 3 f çç , ,3 ÷÷÷ = ⋅ ⋅ 9 çè 2 2 ø 2 2

So, f

Since xyz = 32, z = 0 and y = 0 are

Thus, y = 32 and z = 3, and

y = x

81 > 0. 4

æxö x( x) çç ÷÷÷ = 32, so x = 4, y = 4, and z = 2. çè 2 ø

Maximum is

( 32 , 32 ,3 ) = 814 = 20.25 is a maximum.

f (4, 4, 2) = 4(4) + 2(4)(2) + 2(4)(2)

= 48.

Section 9.4 15.

691

The problem can be restated as

Fx ( x, y,  ) = 10 xy - 

Maximize f ( x, y) = 3xy 2,

Fy ( x, y,  ) = 5 x 2 - 

subject to x + y = 24, x > 0, y > 0.

F ( x, y,  ) = -( x + y - 48)

g ( x, y ) = x + y - 24

10 xy -  = 0

F ( x, y,  ) = 3xy 2 -  ( x + y - 24)

Fx ( x, y,  ) = 3 y 2 - 

3y2 -  = 0

(1)

6 xy -  = 0

(2)

x + y - 24 = 0

(3)

( 2)

x + y - 48 = 0

(3)

 = 5x 2. Thus,

F ( x, y,  ) = -( x + y - 24)

5x -  = 0

Equations (1) and (2) give  = 10xy and

Fy ( x, y,  ) = 6 xy - 

5 x 2 = 10 xy 5x 2 - 10 xy = 0 5 x( x - 2 y ) = 0 x = 0 or x = 2 y.

Equations (1) and (2) give  = 3y 2 and  = 6 xy. Thus,

Substituting x = 0 into equation (3), (0) + y - 48 = 0 y = 48.

3 y = 6 xy 2

Substituting x = 2 y into equation (3),

3 y - 6 xy = 0 3 y ( y - 2 x) = 0 y = 0

(1)

(2 y) + y - 48 = 0

y = 2 x.

3 y - 48 = 0 y = 16.

Substituting y = 0 into equation (3),

x = 2 y = 32.

x + (0) - 24 = 0

Thus,

x = 24.

f (0, 48) = 5(0) 2 (48) + 10 = 10, and

Substituting y = 2 x into equation (3), x + (2 x) - 24 = 0

f (32,16) = 5(32) 2 (16) + 10 = 81,930.

3x - 24 = 0

Since f (32,16 ) > f (0, 48), x = 32 and

x = 8.

y = 16 will maximize f ( x, y ) = 5x 2 y + 10.

y = 2 x = 16.

So Thus,

17.

Let x, y, and z be three number such that

f (24, 0) = 3(24)(0) = 0, and 2

x + y + z = 90 f ( x, y, z ) = xyz.

f (8,16) = 3(8) (16) = 6144.

and

Since f (8,16 ) > f (24, 0), x = 8 and y = 16

g ( x, y, z ) = x + y + z - 90

F ( x, y, z )

will maximize f ( x, y) = 3xy .

= xyz -  ( x + y + z - 90)

16.

The problem can be restated as Maximize f ( x, y ) = 5 x 2 y + 10, subject to x + y = 48, x > 0, y > 0.

Fx ( x, y, z,  ) = yz -  Fy ( x, y, z,  ) = xz -  F ( x, y, z,  ) = xy -  F ( x, y, z,  ) = -( x + y + z - 90)

g ( x, y) = x + y - 48

F ( x, y,  ) = 5 x 2 y + 10 -  ( x + y - 48)

692

Chapter 9 MULTIVARIABLE CALCULUS 4.

yz -  = 0

(1)

xz -  = 0

( 2)

xy -  = 0

(3)

x + y + z - 90 = 0

(4)

Thus,

x = y = z = 80. The three numbers are 80, 80, and 80.

19.

 = yz,  = xz, and  = xy yz = xz yz - xz = 0 ( y - x) z = 0 y - x = 0 or z = 0

Find the maximum and minimum of f ( x, y) = x3 + 2 xy + 4 y 2 subject to x + 2 y = 12.

g ( x, y) = x + 2 y - 12

F ( x, y ) = x 3 + 2 xy + 4 y 2 -  ( x + 2 y - 12)

xz - xy = 0

x ( z - y) = 0

Fy ( x, y,  ) = 2 x + 8 y - 2

x = 0 or z - y = 0

Since x = 0 or z = 0 would not maximize f ( x, y, z ) = xyz, then y - x = 0 and z - y = 0 imply that y = x = z. Substituting into equation (4) gives x + x + x - 90 = 0 x = 30. x = y = z = 30 will maximize f ( x, y, z ) = xyz. The numbers are 30, 30, and 30.

Fx ( x, y,  ) = 3x 2 + 2 y -  F ( x, y,  ) = -x - 2 y + 12

3x 2 + 2 y -  = 0 2 x + 8 y - 2 = 0 -x - 2 y + 12 = 0

Solve the second equation for  and substitute into the first equation. 2 x + 8 y - 2 = 0  = x + 4y The first equation is now

18.

Let x, y, and z be three positive numbers such that x + y + z = 240. Maximize f ( x, y, z ) = xyz, subject to x + y + z = 240. 1.

g ( x, y) = x + y + z - 240

F ( x, y, z,  ) = xyz -  ( x + y + z - 240)

Fx ( x, y, z,  ) = yz -  Fy ( x, y, z,  ) = xz - 

or 3x 2 - x - 2 y = 0

Solve the last equation for –2y and substitute into this new equation. -x - 2 y + 12 = 0 -2 y = x - 12 3x 2 - x + ( x - 12) = 0

Fz ( x, y, z,  ) = xy -  F ( x, y, z,  ) = -( x + y + z - 240)

3x 2 + 2 y - ( x + 4 y ) = 0

yz -  = 0 xz -  = 0

3x 2 - 12 = 0

Now solve for x. 3x 2 - 12 = 0 x2 = 4

xy -  = 0 x + y + z - 240 = 0

x = 2 or x = -2

 = yz,  = xz,  = xy

Find the corresponding y.

yz = xz

When x = 2, -2 - 2 y + 12 = 0 so y = 5. When x = -2, -(-2) - 2 y + 12 = 0 so y = 7. Thus our candidates for the locations of maxima or minima of f subject to the given constraint are (2, 5) and (-2, 7).

z = 0 (impossible) or x = y xz = xy x = 0 (impossible) or z = y Thus, x = y = z. x + x + x - 240 = 0 x = 80

Section 9.4

693 f (2, 5) = 128 and f (-2, 7) = 160, so

25.

Minimize f ( x, y ) = x 2 + 2 x + 9 y 2 + 4 y + 8 xy subject to x + y = 1.

probably the maximum is 160 at (-2, 7) and the minimum is 128 at (2, 5). Try some nearby points that satisfy the constraints to check.

(a) 1. g ( x, y) = x + y = 1

When x = -2.2, y = (12 + 2.2)/2 = 7.1; f (-2.2, 7.1) = 159.752.

+ 8 xy -  ( x + y - 1)

When x = 2.2, y = (12 - 2.2)/2 = 4.9; f (2.2, 4.9) = 128.248.

Fx ( x, y,  ) = 2 x + 8 y -  + 2 Fy ( x, y,  ) = 18 y + 8 x -  + 4

This confirms our answer: There is a maximum value of 160 at (-2, 7) and a minimum value of 128 at (2, 5). 22.

F ( x, y) = x 2 + 2 x + 9 y 2 + 4 y

F ( x, y,  ) = -x - y + 1 2x + 8 y -  + 2 = 0

18 y + 8 x -  + 4 = 0

Consider the constraint and solve of x in terms of y. x + 2 y = 15 x = 15 - 2 y Then

-x - y + 1 = 0

Solve the system in Step 4. Solve the first two equations for  and eliminate  . 2 x + 8 y + 2 = 18 y + 8 x + 4 6 x + 10 y + 2 = 0

f ( x, y) = xy 2 = (15 - 2 y) y 2

3x + 5 y + 1 = 0

= -2 y 3 + 15 y 2 3

Now combine this equation in x and y with the last equation to form a system.

So, f ( x, y) = -2 y + 15 y = f ( y ). Notice that f is unbounded; more specifically, lim f ( y) = -¥

3x + 5 y + 1 = 0 -x - y + 1 = 0

y ¥

and

lim

y -¥

f ( y) = ¥.

The solution of this system is (3, -2). f (3, -2) = -5. Test nearby points that satisfy the constraint to see if -5 is indeed a minimum.

Therefore f, subject to the given constraint, has neither and absolute maximum nor an absolute minimum. 23.

When x = 3.2, y = 1 - 3.2 = -2.2.

Consider the constraint and solve for y in terms of x. 3x - y = 9 y = 3x - 9 Then

When x = 2.8, y = 1 - 2.8 = -1.8. f (3.2, -2.2) = f (2.8, -1.8) = -4.92, so subject to the given constraint f has a minimum of -5 at (3, -2).

f ( x, y) = 8 x 2 y

(d)

= 8x (3x - 9)

f xy ( x, y) = 8.

= 24 x3 - 72 x 2

So, f ( x, y) = 24 x3 - 72 x 2 = f ( x). Notice that f is unbounded; more specifically, lim f ( x) = ¥ y ¥

and

lim

y -¥

f ( x) = -¥.

Therefore f , subject to the given constraint, has neither an absolute maximum nor an absolute minimum.

f xx ( x, y) = 2, f yy ( x, y) = 18, D(3, -2) = 18 ⋅ 2 - 64 = -28, so (3,  2) is a saddle point of the function f .

27.

Maximize f ( x, y) = xy 2 subject to x + 2 y = 60. 1.

g ( x, y) = x + 2 y - 60

F ( x, y) = xy 2 -  ( x + 2 y - 60)

694

Chapter 9 MULTIVARIABLE CALCULUS 3.

If either x or y equals 0, the utility will have a minimum value of 0. So we can assume

Fx ( x, y,  ) = y 2 -  Fy ( x, y,  ) = 2 xy - 2

xy ¹ 0 and divide by xy 2 to find that y = 3x. Substitute for y in the last equation.

F ( x, y,  ) = -x - 2 y + 60

y2 -  = 0 2 xy - 2 = 0 -x - 2 y + 60 = 0 Solve the first two equations for  and eliminate . y2 =  xy = 

-2 x - 3x + 80 = 0 x = 16, y = 3x = 48. f (16, 48) = 28,311,552.

The maximum utility is 28,311,552, obtained by purchasing 16 units of x and 48 units of y. 29.

y 2 = xy

Either y = 0 or x = y. If y = 0, then x = 60; if x = y, then -3x = 60 and x = y = 20. So our candidates for a maximum of f are (60, 0) and (20, 20). But f (60, 0) = 0 so the maximum must be f (20, 20) = 8000. We can confirm this by solving the constraint for x. x = 60 - 2 y

Maximize f ( x, y) = x 4 y 2 subject to 2 x + 4 y = 60. 1.

g ( x, y) = 2 x + 4 y - 60

F ( x, y) = x 4 y 2 -  (2 x + 4 y - 60)

Fx ( x, y,  ) = 4 x3 y 2 - 2 Fy ( x, y,  ) = 2 x 4 y - 4 F ( x, y,  ) = -2 x - 4 y + 60

2 x 4 y - 4 = 0 -2 x - 4 y + 60 = 0

f ( x, y) = xy 2 = (60 - 2 y) y 2 = 60 y 2 - 2 y3.

This is now a function of one variable. The second derivative with respect to y is 120 - 12y which is negative when y = 20, so the value of 8000 found above is the maximum utility, obtained by purchasing 20 units of x and 20 units of y. 28.

Maximize f ( x, y) = x 2 y3 subject to 2x + y = 80. 1.

g ( x, y) = 2 x + y - 80

F ( x, y) = x 2 y 3 -  (2 x + y - 80)

Fx ( x, y,  ) = 2 xy3 - 2

4 x3 y 2 - 2 = 0

Solve the first two equations for 2 and eliminate 2. 4 x3 y 2 = 2 x 4 y = 2 4 x3 y 2 = x 4 y

If either x or y equals 0, the utility will have a minimum value of 0. So we can assume xy ¹ 0 and divide by x3 y to find that 4 y = x. Substitute for x in the last equation.

Fy ( x, y,  ) = 3x 2 y 2 - 

-2 x - x + 60 = 0 x = 20, y = 20/4 = 5 f (20,5) = 4, 000, 000.

F ( x, y,  ) = -2 x - y + 80

2 xy 3 - 2 = 0

The maximum utility is 4,000,000, obtained by purchasing 20 units of x and 5 units of y.

3x 2 y 2 -  = 0 -2 x - y + 80 = 0

Solve the first two equations for  and eliminate . xy3 =  3x 2 y 2 =  xy3 = 3x 2 y 2

Section 9.4 30.

695

Maximize f ( x, y) = x3 y 4 subject to 3x + 3 y = 42 or x + y = 14. 1.

g ( x, y) = x + y - 14

F ( x, y) = x3 y 4 -  ( x + y - 14)

Fx ( x, y,  ) = 3x 2 y 4 - 

y - 10 = 0 x - 10 = 0 10 x + 10 y - 1200 = 0

10 = y and 10 = x y = x

Substituting into the third equation gives

3 3

Fy ( x, y,  ) = 4 x y - 

10 x + 10 x - 1200 = 0 20 x - 1200 = 0 x = 60 y = 60.

F ( x, y,  ) = -x - y + 14

3x 2 y 4 -  = 0 4 x3 y 3 -  = 0 -x - y + 14 = 0

These dimensions, 60 feet by 60 feet, will maximize the area.

Solve the first two equations for  and eliminate . 3x 2 y 4 =  4 x3 y 3 =  3x 2 y 4 = 4 x 3 y 3

If either x or y equals 0, the utility will have a minimum value of 0. So we can assume 2 3

xy ¹ 0 and divide by x y to find that 3 y = 4 x or y = (4/3) x. Substitute for y in the last equation. 4 x + 14 = 0 3 7 x = 42 æ4ö x = 6, y = çç ÷÷÷ 6 = 8 çè 6 ø

-x -

f (6,8) = 884, 736.

The maximum utility is 884,736, obtained by purchasing 6 units of x and 8 units of y. 31.

32.

Let x be the length of the fence opposite the building and y be the length of each end. The area is then xy and the total cost is 25 x + 15(2 y). Restate the problem as follows. Maximize xy, subject to 25 x + 30 y = 2400. 1.

g ( x, y) = 25x + 30 y - 2400

F ( x, y,  ) = xy -  (25 x + 30 y - 2400)

Fx ( x, y,  ) = y - 25 Fy ( x, y,  ) = x - 30 F ( x, y,  ) = -(25 x + 30 y - 2400)

y - 25 = 0 x - 30 = 0 25 x + 30 y - 2400 = 0

Equations (1) and (2) give  = 25 and

x . Thus,  = 30

y x = 25 30 5 y = x. 6

Let x be the width and y be the length of a field such that the cost in dollars to enclose the field is 6 x + 6 y + 4 x + 4 y = 1200 10 x + 10 y = 1200.

(1) (2) (3)

Substituting y = 56 x into equation (3) gives æ5 ö 25 x + 30 çç x ÷÷÷ - 2400 = 0 çè 6 ø

The area is f ( x, y) = xy.

g ( x, y) = 10 x + 10 y - 1200

2. 3.

F ( x, y) = xy -  (10 x + 10 y - 1200) Fx ( x, y,  ) = y - 10 Fy ( x, y,  ) = x - 10 F ( x, y,  ) = -(10 x + 10 y - 1200)

50 x - 2400 = 0 x = 48. 5 y = x = 40. 6

The dimensions are 48 ft (opposite the building) by 40 ft (the ends).

696 33.

Chapter 9 MULTIVARIABLE CALCULUS Maximize C ( x, y) = 2 x 2 + 6 y 2 + 4 xy + 10, subject to x + y = 10.

35.

Maximize f ( x, y) = 3x1/3 y 2/3, subject to 80 x + 150 y = 40, 000.

g ( x, y) = x + y - 10

g ( x, y) = 80 x + 150 y - 40, 000

F ( x, y)

= 3x1/3 y 2/3 -  (80 x + 150 y - 40,000)

= 2 x + 6 y + 4 xy + 10 -  ( x + y - 10)

3,4. Fx ( x, y,  ) = x-2/3 y 2/3 - 80 = 0

Fx ( x, y,  ) = 4 x + 4 y -  Fy ( x, y,  ) = 12 y + 4 x - 

Fy ( x, y,  ) = 2 x1/3 y-1/3 - 150 = 0

F ( x, y,  ) = -( x + y - 10)

F ( x, y,  ) = -(80 x + 150 y - 40, 000) =0

4x + 4 y -  = 0 12 y + 4 x -  = 0 x + y - 10 = 0

x-2/3 y 2/3 2 x1/3 y-1/3 = 80 150 15 y = x 16

 = 4 x + 4 y and  = 12 y + 4 x. 4 x + 4 y = 12 y + 4 x 8y = 0 y =0

Substitute into the third equation. 80

( 1516y ) + 150 y - 40, 000 = 0 y = 178(rounded) 15(178) x = 16 » 167

Since x + y = 10, x = 10. 10 large kits and no small kits will maximize the cost. 34.

Use about 167 units of labor and 178 units of capital to maximize production.

Maximize P( x, y) = -x 2 - y 2 + 4 x + 8 y, subject to x + y = 6. 1.

g ( x, y) = x + y - 6

F ( x, y,  )

36.

subject to 100 x + 180 y = 25, 200.

= -x 2 - y 2 + 4 x + 8 y -  ( x + y - 6)

Maximize f ( x, y) = 12 x3/4 y1/4 ,

g ( x, y) = 100 x + 180 y - 25, 200

F ( x, y,  )

Fx ( x, y,  ) = -2 x + 4 -  Fy ( x, y,  ) = -2 y + 8 - 

F ( x, y,  ) = -( x + y - 6) -2 x + 4 -  = 0 -2 y + 8 -  = 0 x+ y-6= 0

 = -(2 x - 4) and  = -(2 y - 8) 2x - 4 = 2 y - 8 y = x+2

= 12 x3/4 y1/4 -  (100 x + 180 y - 25, 200) 3 Fx ( x, y,  ) = (12 x-1/4 y1/4 ) - 100 4

3x3/4

- 180 y 3/4 F ( x, y,  ) = -(100 x + 180 y - 25, 200) =

Substituting into x + y - 6 = 0, we have x + ( x + 2) - 6 = 0 so x = 2 and y = 4.

9 y1/4

- 100 x1/4 1 Fy ( x, y,  ) = (12 x3/4 y-3/4 ) - 180 4 =

Section 9.4 4.

697 9 y1/4 x1/4

38.

- 100 = 0

3x3/4

- 180 = 0 y 3/4 100 x + 180 y - 25, 200 = 0

 =

1/4

100 x

and  =

1/ 4

g ( x, y) = x + 2 y - 600

F ( x, y,  ) = xy -  ( x + 2 y - 600)

3/4

180 y 3/4

If x and y are the dimensions of the field, we must maximize f ( x, y) = xy subject to x + 2 y = 600.

60 y 3/4 x

F ( x, y,  ) = -( x + 2 y - 600)

3/4

100 x 60 y 3/4 100 x = 540 y 27 y x = 5

Substitute into 100 x + 180 y - 25, 200 = 0. æ 27 y ö÷ + 180 y = 25, 200 100 çç çè 5 ÷÷ø 540 y + 180 y = 25, 200 720 y = 25, 200 y = 35 27(35) x = = 189 5 Production will be maximized with 189 units of labor and 35 units of capital.

Fx ( x, y,  ) = y -  Fy ( x, y,  ) = x - 2

3/4

x- = 0 x - 2 = 0 x + 2 y - 600 = 0

 = y and  =

x 2

y =

x 2

Substituting into x + 2 y - 600 = 0, we have

( )

x + 2 2x - 600 = 0, so x = 300 and y = 150.

The largest area is (300)(150) = 45, 000 m 2. 39.

Let x be the radius r of the circular base and y the height h of the can, such that the volume is

 x 2 y = 250 . 37.

Let x and y be the dimensions of the field such that 2 x + 2 y = 500, and the area is f ( x, y) = xy. 1.

g ( x, y) = 2 x + 2 y - 500

F ( x, y ) = xy -  (2 x + 2 y - 500)

Fx ( x, y,  ) = y - 2 Fy ( x, y,  ) = x - 2 F ( x, y,  ) = -(2 x + 2 y - 500)

y - 2 = 0 x - 2 = 0 2 x + 2 y - 500 = 0

2 = y and 2 = x, so x = y. 2 x + 2 x - 500 = 0 4 x - 500 = 0 x = 125 Thus, y = 125.

Dimensions of 125 m by 125 m will maximize the area.

The surface area is f ( x, y) = 2 xy + 2 x 2.

g ( x, y) =  x 2 y - 250

F ( x, y) = 2 xy + 2 x 2 -  ( x 2 y - 250 )

Fx ( x, y,  ) = 2 y + 4 x -  (2 xy) Fy ( x, y,  ) = 2 x -  ( x 2 ) F ( x, y,  ) = -( x 2 y - 250 )

698

Chapter 9 MULTIVARIABLE CALCULUS 4.

2 y + 4 x -  (2 xy) = 0

Fx ( x, y,  ) = 2 y + 4 x - 2 xy

2 x -  x 2 = 0

Fy ( x, y,  ) = 2 x -  x 2

 x 2 y - 250 = 0

F ( x, y,  ) = -( x 2 y - 25)

Simplifying these equations gives

y + 2 x - 1 xy = 0

2 y + 4 x - 2 xy = 0 2 x -  x 2 = 0

2 x - 1 x = 0

 x 2 y - 25 = 0

x 2 y - 250 = 0.

From the second equation,

 =

x(2 -  x) = 0 x = 0 or  =

2x + y 2 = xy x

2 . x

2 x 2 + xy = 2 xy

If x = 0, the volume will be 0, which is not possible.

2 x 2 - xy = 0

Substituting x = 2 into the first equation gives

x = 0 or

æ2ö æ2ö y + 2 çç ÷÷÷ -  çç ÷÷÷ y = 0 èç  ø èç  ø y+



Substituting y = 2 x into  x 2 y - 25 = 0, we have  x 2 (2 x) - 25 = 0, so x = 3 225 25 » 3.169 inches. » 1.585 inches and y = 2 3 2π

= y

 =

The can with minimum surface area will have a radius of approximately 1.585 inches and a height of approximately 3.169 inches.

4 . y

Since  = 2x , y = 2 x. Substituting into third equation gives 2

x (2 x) - 250 = 0

y = 2x

x = 0 is impossible.

- 2y = 0



2 2x + y and  = x xy

41.

Let x, y, and z be the dimensions of the box such that the surface area is xy + 2 yz + 2 xz = 500

2 x3 - 250 = 0

and the volume is

x =5 y = 10.

f ( x, y, z ) = xyz.

Since g (1, 250) = 0 and f (1, 250) = 502 > f (5,10) = 150 ,

a can with radius of 5 inches and height of 10 inches will have a minimum surface area. 40.

Let x be the radius of the can and y be the height.

g ( x, y, z ) - 500 = 0

Minimize surface area f ( x, y) = 2 xy + 2 x 2 ,

F ( x, y, z )

= xyz -  ( xy + 2 yz + 2 xz - 500)

subject to the constraint that  x 2 y = 25. 1.

g ( x, y) =  x 2 y - 25

F ( x, y,  )

3,4.

= 2 xy + 2 x 2 -  ( x 2 y - 25)

Fx ( x, y, z,  ) = yz -  ( y + 2 z ) = 0

(1)

Fy ( x, y, z,  ) = xz -  ( x + 2 z ) = 0

(2)

Fz ( x, y, z,  ) = xy -  (2 y + 2 x) = 0

(3)

F ( x, y, z,  ) = -( xy + 2 xz + 2 yz - 500)

(4)

Section 9.4

699

Multiplying equation (1) by x, equation (2) by y, and equation (3) by z gives

4 x + 4 y - 2 xy = 0 4x -  x2 = 0

xyz -  x( y + 2 z ) = 0

x 2 y - 185 = 0

xyz -  y ( x + 2 z ) = 0 xyz -  z (2 y + 2 z ) = 0.

 =

Subtracting the first equation from the second equation gives

 x( y + 2 z ) -  y ( x + 2 z ) = 0 2 xz - 2 yz = 0  z( x - y) = 0,

2 x 2 + 2 xy = 4 xy 2 x 2 - 2 xy = 0 2 x( x - y ) = 0

x = y.

x = 0 or

Subtracting the third equation from the second equation gives

Substituting y = x into x 2 y - 185 = 0, we have

y . 2 Substituting into the fourth equation gives

y = x = 3 185 » 5.698. The dimensions are 5.698 inches by 5.698 inches by 5.698 inches.

z =

æ yö æyö y 2 + 2 y çç ÷÷ + 2 y çç ÷÷ - 500 = 0 çè 2 ÷ø çè 2 ÷ø 3 y 2 = 500

43.

Let x, y, and z be the dimensions of the box. The surface area is 2 xy + 2 xz + 2 yz. We must minimize f ( x, y, z ) = 2 xy + 2 xz + 2 yz subject to xyz = 125.

500 3 » 12.9099

g ( x, y, z ) = xyz - 125

x » 12.9099

F ( x, y, z )

y =

12.9099 2 » 6.4549.

= 2 xy + 2 xz + 2 yz -  ( xyz - 125)

z »

surface area f ( x, y) = 2 x 2 + 4 xy, subject to

Fz ( x, y, z,  ) = 2 x + 2 y -  xy F ( x, y, z,  ) = -( xyz - 125)

x 2 y = 185.

g ( x, y ) = x 2 y - 185

F ( x, y,  )

= 2 x 2 + 4 xy -  ( x 2 y - 185)

Fx ( x, y,  ) = 4 x + 4 y - 2 xy

Fx ( x, y, z,  ) = 2 y + 2 z -  yz Fy ( x, y, z,  ) = 2 x + 2 z -  xz

The dimensions are 12.91 m by 12.91 m by 6.455 m. If the box is x by x by y, we must minimize

2 y + 2 z -  yz = 0

(1)

2 x + 2 z -  xz = 0

( 2)

2 x + 2 y -  xy = 0

(3)

xyz - 125 = 0

(4)

Equations (1) and (2) give 2x + 2z 2 y + 2z = . =  and xz yz Thus, 2 y + 2z 2x + 2z = yz xz

Fy ( x, y,  ) = 4 x -  x 2 F ( x, y,  ) = -( x 2 y - 185)

y = x

x = 0 is impossible.

 z(2 y + 2 x) -  y( x + 2 z) = 0 2 xz -  xy = 0  x(2 z - y) = 0,

42.

2x + 2 y 4 and  = xy x 2x + 2 y 4 = xy x

2 xyz + 2 xz 2 = 2 xyz + 2 yz 2 2 xz 2 - 2 yz 2 = 0

700

Chapter 9 MULTIVARIABLE CALCULUS 2z 2 = 0

x-y =0

z = 0 (impossible) or

x = y.

Equations (2) and (3) give 2x + 2z 2x + 2 y =  and = . xz xy 2x + 2z 2x + 2 y = xz xy Thus,

xyz = 32 y + 2z x + 2z = yz xz xyz + 2 xz 2 = xyz + 2 yz 2

2 z 2 ( x - y) = 0 z 2 = 0 or x - y = 0 z = 0 (impossible) or x = y

2 x 2 y + 2 xyz = 2 x 2 z + 2 xyz 2x y - 2x z = 0

x + 2z 2x + 2 y = xz xy

2 x 2 ( y - z) = 0

x 2 y + 2 xyz = 2 x 2 z + 2 xyz

2x2 = 0

y-z =0

x = 0 (impossible) or

y = z.

Therefore, x = y = z. Substituting into equation (4) gives x3 - 125 = 0 x3 = 125

x 2 ( y - 2 z) = 0 x 2 = 0 or

y - 2z = 0

x = 0 (impossible) or

Since x = y and y = 2 z and since xyz = 32, we have (2 z )(2 z ) z = 32

x = 5.

z3 = 8

Thus,

y = 5 and z = 5. The dimensions that will minimize the surface area are 5 m by 5 m by 5 m.

44.

Let the dimensions of the bottom be x by y, and let the height be z. We must minimize f ( x, y, z ) = xy + 2 xz + 2 yz subject to xyz = 32. 1.

g ( x, y, z ) = xyz - 32

F ( x, y, z,  )

= xy + 2 xz + 2 yz -  ( xyz - 32)

Fx ( x, y, z,  ) = y + 2 z -  yz Fy ( x, y, z,  ) = x + 2 z -  xz Fz ( x, y, z,  ) = 2 x + 2 y -  xy F ( x, y, z,  ) = -( xyz - 32)

y + 2 z -  yz = 0 x + 2 z -  xz = 0

2 x + 2 y -  xy = 0

xyz - 32 = 0 y + 2z  = yz x + 2z  = xz 2x + 2 y  = xy

y = 2z

z = 2.

If z = 2, y = 4 and x = 4. The dimensions are 4 feet by 4 feet for the base and 2 feet for the height. 45. (a) The total area of the two sides and bottom made out of free material is 2 xz  xy. This material does not add to the cost, so we would like to make at as large as possible, which corresponds to the constraint 2 xz  xy  8. The cost of the two reinforced

ends will be (1000)(2 yz ). The number of trips will be 400 divided by the volume of the box, which is xyz. Thus the shipping 400 and the total cost will xyz 4000  2000 yz. be f ( x, y , z )  xyz

cost will be 10 

(b) The solver gives x  4 yd, y  1 yd, z 

for a minimum cost of $3000.

1 yd, 2

Section 9.4 46.

701 P(r , s, t ) = rs(1 - t ) + (1 - r )st

(a)

+ r (1 - s)t + rst g ( r , s, t ) = r + s + t -  F (r , s, t ,  ) = rs(1 - t ) + (1 - r )st

+ r (1 - s)t + rst -  (r + s + t -  )

(b)

Fr (r , s, t ,  ) = s(1 - t ) - st + (1 - s)t + st - 

= s + t - 2st -  Fs (r , s, t ,  ) = r (1 - t ) + (1 - r )t - rt + rt -  = r + t - 2rt -  Ft (r , s, t ,  ) = -rs + (1 - r )s + r (1 - s) + rs -  = r + s - 2rs -  F (r , s, t ,  ) = -(r + s + t -  )

s + t - 2st -  = 0 r + t - 2rt -  = 0 r + s - 2rs -  = 0 r + s + t - = 0 - = 2st - s - t

- = 2rt - r - t - = 2rs - r - s r+s+t = 2st - s - t = 2rt - r - t s(2t - 1) = r (2t - 1) (s - r )(2t - 1) = 0 1 2 2rt - r - t = 2rs - r - s

s = r

t =

t (2r - 1) = s(2r - 1) (t - s)(2r - 1) = 0 1 2 2st - s - t = 2rs - r - s t (2s - 1) = r (2s - 1)

t = s

or r =

(t - r )(2s - 1) = 0 t = r

s =

1 2

Since r, s and t are probabilities, 0 £ r , s, t £ 1. also, r + s + t =  = 0.75. If t = 12 , then either t = s = 12 or r = 12 , both of which are impossible (the third value would have to be -0.25 to get a sum

of 0.75). Thus, r = s = t = 0.25. (c) Now we have r + s + t =  = 3. If t = 12 , then either t = s = 12 or r = 12 , both of which are

702

Chapter 9 MULTIVARIABLE CALCULUS

9.5 Total Differentials and Approximations

5.032 + 11.992

Your Turn 1 f ( x, y) = 3x 2 y 4 + 6 x 2 - 7 y 2 2

x - 7y

f y ( x, y) = 12 x 2 y 3 -

(a)

42 y 2

x - 7y

Your Turn 3

ö÷ ÷÷ ÷ dx 2 2 ÷÷ x - 7y ø 6x

æ ç + ççç12 x 2 y 3 ççè

As calculated in Example 3, dV = 2 dr + dh . Thus V r h

ö÷ 42 y ÷ ÷÷ dy 2 2 ÷÷ x - 7y ø

Evaluate dz for x = 4, y = 1, dx = 0.02, dy = -0.03. æ ç dz = ççç 6(4)(1)4 + çèç

0.15 0.12 13 13 » 13.0023

= 13 +

dz = f x ( x, y) dx + f y ( x, y) dy æ ç = ççç 6 xy 4 + çèç

(b)

æ ö÷ ç 12 ÷÷ (-0.01) + çç çç 2 2 ÷÷÷ è 5 + 12 ø

f x ( x, y) = 6 xy 4 +

the maximum percent error in the volume if the errors in the radius and length are at most 4% and 2% is dV = 2(0.04) + (0.02) = 0.10 or 10%. V

9.5 Warmup Exercises W1.

÷÷ö ÷÷ (0.02) ÷ (4) 2 - 7(1) 2 ÷ø

æ ç + ççç12(4) 2 (1)3 ççè

6(4)

(

= (6(2) + 3)(0.1) = 2.7.

W2.

x 2 + y 2 . As we

found in Example 2,

æ 2 x ö÷ ÷ x dy = çç 2 çè x + 1 ÷ø÷

æ ö÷ æ ö÷ ç ç x y ÷÷ ÷÷ dz = ççç dx + ççç dy ÷ ÷ çèç x 2 + y 2 ÷÷÷ø çèç x 2 + y 2 ÷÷÷ø

Then f ( x + dx, y + dy ) » f ( x, y ) + dz

y = ln( x 2 + 1) dy = y ¢( x) x æ 2 x ÷ö ÷ x = çç 2 çè x + 1 ÷÷ø When x = 2 and x = -0.02,

Your Turn 2

To approximate 5.032 + 11.992 we let x = 5, y = 12, dx = 0.03, dy = -0.01.

)

dy = 6 x 2 + 3 x

æ æ 24 ö÷ 42 ö÷ = çç 24 + (0.02) + çç192 ÷ ÷ (-0.03) ÷ èç èç 3 ø 3 ø÷ = -4.7

)

When x = 2 and x = 0.1,

ö÷ ÷÷ ÷ (-0.03) 2 2 ÷÷ (4) - 7(1) ø

= (32)(0.02) - 178(0.03)

y = 2 x3 + 3x dy = y ¢( x) x = 6 x 2 + 3 x

42(1)

As in Example 2, we let f ( x, y) =

æ ö÷ ç 5 ÷÷ (0.03) 52 + 122 + çç çç 2 2 ÷÷÷ è 5 + 12 ø

æ 2(2) ö÷ ÷÷ (-0.02) = ççç 2 çè (2) + 1 ÷÷ø = -0.016.

W3.

x with x = 25 and x = -1. 1 y » y ¢(5)x = (-1) = -0.1 2 25

Use y =

Thus

24 »

25 - 0.1 = 4.9.

W4. Use y = e x with x = 0 and x = 0.01. y » y ¢(0)x = e0 (0.01) = 0.01

Thus e0.01 » e0 + 0.01 = 1.01.

Section 9.5

703

9.5 Exercises

z = ln( x 2 + y 2 )

True

x = 2, y = 3, dx = 0.02, dy = -0.03

False. The differential for a function z = f ( x, y) is used to approximate a function by its tangent plane.

dz =

True

z = f ( x, y) = 2 x 2 + 4 xy + y 2

2 x dx x2 + y 2

Substitute the given information. dz =

2(3)(-0.03) 22 + 32

f x ( x, y) = 4 x + 4 y

5x 2 + y 2 z +1 x = -2, y = 1, z = 1

f y ( x, y) = 4 x + 2 y

dx = 0.02, dy = -0.03, dz = 0.02

dz = (4 x + 4 y)dx + (4 x + 2 y)dy = [4(5) + 4(-1)(0.03)

f x ( x, y) =

+ [4(5) + 2(-1)](-0.02) = 0.48 - 0.36 = 0.12

z = f ( x, y) = 5 x3 + 2 xy 2 - 4 y

f y ( x, y) =

x = 1, dx = 0.01, y = 3, dy = 0.02

f x ( x, y) = 15 x 2 + 2 y 2

f y ( x, y) = 4 xy - 4 2

f z ( x, y) =

dz = (15 x + 2 y ) dx + (4 xy - 4) dy 2

= [15(1) + 2(3) ](0.01) + [4(1)(3) - 4](0.02) = 0.33 + 0.16 = 0.49

2(2)(0.02)

2 +3 = -0.00769

x = 5, dx = 0.03, y = -1, dy = -0.02

2 y dy + 2 x + y2

z =

y 2 + 3x y2 - x

dw =

, x = 4, y = -4,

( y 2 - x)  3 - ( y 2 + 3x)  (-1) ( y 2 - x) 2 +

( y 2 - x) 2 2

( y - x)

4(-4)2

dx -

( z + 1)(2 y) - (5 x 2 + y 2 )(0) ( z + 1)2 2y z +1 ( z + 1)(0) - (5 x 2 + y 2 )(1) ( z + 1) 2 -5 x 2 - y 2 ( z + 1) 2

10 x 2y -5 x 2 - y 2 dx + dy + dz z +1 z +1 ( z + 1)2

8xy 2

( y - x) 2

-20 2 (0.02) + (-0.03) 2 2 [-5(4) - 1](0.02) + (2) 2 21 = -0.2 - 0.03 (0.02) 4 = -0.335

8(4)(-4) (0.01) (0.03) 2 2 [(-4) - 4] [(-4)2 - 4]2 » 0.0311

10 x z +1

dw =

( y 2 - x)  2 y - ( y 2 + 3 x)  2 y 4 y2

( z + 1)2

Substitute the given values.

dx = 0.01, dy = 0.03 dz =

( z + 1)10 x - (5x 2 + y 2 )(0)

10.

æxö w = x ln ( yz ) - y ln çç ÷÷÷ çè z ø = x(ln y + ln z ) - y(ln x - ln z ), x = 2, y = 1, z = 4, dx = 0.03, dy = 0.02,

704

Chapter 9 MULTIVARIABLE CALCULUS dz = -0.01 æ yö dw = çç ln y + ln z - ÷÷÷ dx çè xø

12.

Let z = f ( x, y) = Then

dz = f x ( x, y)dx + f y ( x, y) dy

æx ö + çç - ln x + ln z ÷÷÷ dy ÷ø çè y

1 2 ( x + y 2 )-1/2 (2 x) dx 2 1 + ( x 2 + y 2 )-1/2 (2 y) dy 2 x dx + y dy = . x2 + y 2 =

æx yö + çç + ÷÷÷ dz çè z zø æ 1ö = çç ln 1 + ln 4 - ÷÷÷ (0.03) çè 2ø æ2 ö + çç - ln 2 + ln 4 ÷÷÷ (0.02) çè 1 ø

To approximate 4.962 + 12.062 , we let x = 5, dx = -0.04, y = 12, and dy = 0.06.

æ2 1ö + çç + ÷÷÷ (-0.01) çè 4 4ø » 0.0730

11.

Let z = f ( x, y) =

dz = 2

x + y .

dz = f x ( x, y)dx + f y ( x, y)dy

f (4.96,12.06) = f (5,12) + Dz

1 2 ( x + y 2 )-1/2 (2 x)dx 2 1 + ( x 2 + y 2 )-1/2 (2 y)dy 2 xdx + ydy = . x2 + y2 =

» f (5,12) + dz =

Thus,

» f (8, 6) + dz =

82 + 62 + 0.222

8.052 + 5.97 2 » 10.022.

Using a calculator,

4.962 + 12.062

The absolute value of the difference of the two results is |13.04 - 13.0401| = 0.0001. 13.

Let z = f ( x, y) = ( x 2 + y 2 )1/3. Then dz = f x ( x, y )dx + f y ( x, y )dy 1 2 ( x + y 2 )-2 / 3(2 x)dx 3 1 + ( x 2 + y 2 )-2/3(2 y )dy 3 2x 2y = dx + dy 2 2 2/3 2 3( x + y ) 3( x + y 2 )2/3

dz =

= 10.022

Thus,

4.962 + 12.062 » 13.04.

Using a calculator, » 13.0401.

8(0.05) + 6(-0.03)

f (8.05,5.97) = f (8, 6) + D z

52 + 122 + 0.04

= 13.04

To approximate 8.052 + 5.97 2 , we let x = 8, dx = 0.05, y = 6 and dy = -0.03.

82 + 62 4 3 = (0.05) + (-0.03) 5 5 = 0.04 - 0.018 = 0.022

5(-0.04) + 12(0.06)

52 + 122 5 12 = (-0.04) + (0.06) 13 13 = 0.04

Then

dz =

x2 + y 2 .

8.052 + 5.97 2 » 10.0221.

The absolute value of the difference of the two results is |10.022 - 10.0221| = 0.0001.

To approximate (1.922 + 2.12 )1/3 , we let x = 2, dx = -0.08, y = 2, and dy = 0.1.

Section 9.5

705 dz =

2(2) 2

2 2/3

(-0.08)

15.

3[(2) + (2) ] 2(2) (0.1) + 2 3[(2) + (2) 2 ]2/3 4 4 = (-0.08) + (0.1) 12 12 = 0.006

Let z = f ( x, y) = xe y . Then dz = f x ( x, y)dx + f y ( x, y)dy = e y dx + xe y dy.

To approximate 1.03e0.04, we let x = 1, dx = 0.03, y = 0, and dy = 0.04.

f (1.92, 2.1) = f (2, 2) + D z

dz = e0 (0.03) + 1 ⋅ e0 (0.04)

» f (2, 2) + dz

= 0.07

= 2 + 0.006

f (1.03, 0.04) = f (1, 0) + D z

f (1.92, 2.1) » 2.0067

» f (1, 0) + dz

Using a calculator, (1.922 + 2.12 )1/3 » 2.0080.

= 1 ⋅ e0 + 0.07

The absolute value of the difference of the two results is |2.0067 - 2.0080| = 0.0013. 14.

= 1.07

Thus, 1.03e0.04 » 1.07.

Let z = f ( x, y) = ( x 2 - y 2 )1/3.

Using a calculator, 1.03e0.04 » 1.0720.

Then

The absolute value of the difference of the two results is |1.07 - 1.0720| = 0.0020.

dz = f x ( x, y)dx + f y ( x, y) dy. 1 2 ( x - y 2 )-2/3 (2 x)dx 3 1 + ( x 2 - y 2 )-2/3 (-2 y)dy 3 2x 2y dz = dx dy 2 2 2/3 2 3( x - y ) 3( x - y 2 ) 2/3 dz =

16.

Then dz = f x ( x, y) dx + f y ( x, y) dy = e y dx + xe y dy.

To approximate 0.98e-0.04, we let x = 1, dx = -0.02, y = 0, and dy = -0.04.

To approximate (2.932 - 0.942 )1/3, we let x = 3, dx = -0.07, y = 1, and dy = -0.06. dz =

Let z = f ( x, y) = xe y .

dz = e0 (-0.02) + 1 ⋅ e0 (-0.04)

2(3)

(-0.07) 3[(3) 2 - (1) 2 ]2/3 2(1) (-0.06) 2 3[(3) - (1) 2 ]2/3 1 1 dz = (-0.07) - (-0.06) 2 6 = -0.025

= -0.06 f (0.98, -0.04) = f (1, 0) + Dz » f (1, 0) + dz = 1 ⋅ e0 - 0.06 = 0.94

Thus, 0.98e-0.04 » 0.94.

f (2.93, 0.94) = f (3,1) + D z

Using a calculator, 0.98e-0.04 » 0.9416. The absolute value of the difference of the two results is | 0.94 - 0.9416| = 0.0016.

» f (3,1) + dz = 2 + (-0.025) f (2.93, 0.94) » 1.975

Using a calculator, (2.932 - 0.942 )1/3 » 1.9748. The absolute value of the difference of the two result is |1.975 - 1.9748| = 0.0002.

17.

Let z = f ( x, y) = x ln y. Then dz = f x ( x, y) dx + f y ( x, y) dy = ln y dx +

x dy y

706

Chapter 9 MULTIVARIABLE CALCULUS To approximate 0.99 ln 0.98, we let x = 1, dx = -0.01, y = 1, and dy = -0.02.

20.

Let r be the radius inside the tumbler and h be the height inside.

1 dz = ln (1) ⋅ (-0.01) + (-0.02) 1 = -0.02 f (0.99, 0.98) = f (1,1) + Dz » f (1,1) + dz = 1 ⋅ ln (1) - 0.02 » -0.02

Thus, 0.99 ln 0.98 » -0.02.

V =  r 2h, r = 1.5, h = 9, dr = dh = 0.2

Using a calculator, 0.99 ln 0.98 » -0.0200.

dV = 2 rhdr +  r 2dh

The absolute value of the difference of the two results is | - 0.02 - (-0.0200)| = 0.

= 2 (1.5)(9)(0.2) +  (1.5) 2 (0.2) » 18.4

18.

Let z = f ( x, y) = x ln y.

Approximately 18.4 cm3 of material is needed.

Then dz = f x ( x, y)dx + f y ( x, y) dy = ln y dx +

21.

V = LWH

x dy. y

with L = 10, W = 9, and H = 18. Since 0.1 inch is applied to each side and each dimension has a side at each end,

To approximate 2.03 ln 1.02, we let x = 2, dx = 0.03, y = 1, and dy = 0.02. dz = ln(1) ⋅ (0.03) +

The volume of the box is

dL = dW = dH = 2(0.1) = 0.2

2 (0.02) 1

dV = WH dL + LH dW + LW dH .

= 0.04

Substitute.

f (2.03,1.02) = f (2,1) + D z

dV = (9)(18)(0.2) + (10)(18)(0.2) + (10)(9)(0.2)

» f (2,1) + dz = 2 ⋅ ln(1) + 0.04

= 86.4

= 0.04

Approximately 86.4 in3 are needed.

Thus, 2.03ln1.02 » 0.04. Using a calculator, 2.03 ln 1.02 » 0.0402. The absolute value of the difference of the two results is |0.04 - 0.0402| = 0.0002. 19.

22.

M ( x, y) = 45x 2 + 40 y 2 - 20 xy + 50 x = 8, dx = 8.25 - 8 = 0.25, y = 14, dy = 13.75 - 14 = -0.25 M x ( x, y) = 90 x - 20 y

The volume of the can is

M y ( x, y) = 80 y - 20 x

V =  r h,

dM = (90 x - 20 y )dx + (80 y - 20 x)dy

With r = 2.5cm, h = 14 cm, dr = 0.08, dh = 0.16.

= [90(8) - 20(14)](0.25) + [80(14) - 20(8)](-0.25)

dV = 2 rhdr +  r 2dh = 2 (2.5)(14)(0.08) +  (2.5)2 (0.16) » 20.73

= 110 - 240 = -130

The change in cost will be a decrease of $130.

Section 9.5 23.

707 26.

z = x 0.65 y 0.35 x = 50, y = 29,

Assume that blood vessels are cylindrical. V =  r 2h, r = 0.8, h = 7.9, dr = dh = 0.15

dx = 52 - 50 = 2 dy = 27 - 29 = -2

f x ( x, y) = y

0.35

-0.35

(0.65)( x

dV = 2 rhdr +  r 2dh = 2 (0.8)(7.9)(0.15)

)

æ y ö0.35 = 0.65 çç ÷÷÷ çè x ø f y ( x, y) = ( x

0.65

)(0.35)( y

+  (0.8)2 (0.15) » 6.258 cm3 -0.65

)

æ x ö0.65 = 0.35 ççç ÷÷÷ è y ø÷

27.

æ x ö0.65 æ y ö0.35 dz = 0.65 çç ÷÷ dx + 0.35 ççç ÷÷÷ dy çè x ÷ø è y ÷ø

Substitute. æ 29 ö0.35 æ 50 ö0.65 dz = 0.65 çç ÷÷ (2) + 0.35 çç ÷÷ (-2) çè 50 ÷ø çè 29 ÷ø

dC = -b(a - v)-2 da 1 + db + b(a - v)-2 dv a-v -b 1 b = da + db + dv 2 a v ( a - v) ( a - v) 2

= 0.07694 unit

24.

z = x 0.8 y 0.2 , x = 20, y = 18, dx = 21 - 20 = 1, dy = 16 - 18 = -2

dz = 0.8 x-0.2 y 0.2dx + 0.2 x 0.8 y-0.8dy æ y 0.2 ö÷ æ x 0.8 ö÷ ç ç = 0.8 çç 0.2 ÷÷÷ dx + 0.2 çç 0.8 ÷÷÷ dy çè x ÷ø èç y ø÷ æ x ö0.2 æ x ö0.8 = 0.8 ççç ÷÷÷ dx + 0.2 ççç ÷÷÷ dy è y ø÷ è y ø÷ 0.2

æ 18 ö = 0.8 çç ÷÷ çè 20 ÷ø

0.8

æ 20 ö (1) + 0.2 çç ÷÷ èç 18 ÷ø

(-2)

» 0.348

The change in production is 0.348 unit. 25.

The maximum possible error is 6.26 cm3. b = b(a - v)-1 C = a-v a = 160, b = 200, v = 125 da = 145 - 160 = -15 db = 190 - 200 = -10 dυ = 130 - 125 = 5

-200 2

(-15) +

(160 - 125) 200 + (5) (160 - 125)2 » 2.98 liters

28.

1 (-10) 160 - 125

Using the partial derivatives H T (m, T , A) and H A(m, T , A) found in Exercise 56 of Section 2, we have 15.2(250.67 ) 10.23ln 25 - 10.74 = 5.92

H T (25, 36, 12) =

and H A(25, 36, 12) = -5.92.

The volume of the bone is V =  r 2h,

with h = 7, r = 1.4, dr = 0.09, dh = 2(0.09) = 0.18 dV = 2 rh dr +  r 2dh = 2 (1.4)(7)(0.09) +  (1.4)2 (0.18) = 6.65

6.65 cm3 of preservative are used.

H m (m, T , A) é (15.2)(0.67)m-0.33(10.23ln m - 10.74 ù ê ú ê ú æ ö ê ú 0.67 ç 10.23 ÷ 15.2 m ÷ ê ú ççè m ÷ø êë úû = 2 (10.23ln m - 10.74) 24(78.1156 - 53.7525 H M (25, 36, 12) = 492.3561 = 1.19

708

Chapter 9 MULTIVARIABLE CALCULUS 42 (5.12 + 5.1 ⋅ 2.9 + 2.9 2 ) 3 » 2164.37

The estimated change is (1)(1.19) + (0.5)(5.92) + (-2)(-5.92) = 15.99 The approximate change in H is 16.0 W. 29.

L( E , P ) = 23E 0.6 P-0.267 As we found in Exercise 62 of Section 9.2, ¶ L( E , P ) = 23(0.6)E -0.4 P-0.267 ¶E ¶ L( E , P ) = 23(-0.267)E 0.6 P-1.267 ¶P The total differential is thus dL = 23(0.6)E -0.4 P-0.267dE

V (42,5.1, 2.9) =

The volume found by using the original formula is about 2164.37 cm3. 31.

P( A, B, D) =

30.

V =

h 2 r 1 + r1r2 + r 22 3

(

P(150,1, 20) =

40 2 (5 + 5 ⋅ 3 + 32 ) » 2052.51 3

(a) V =

The volume is about 2052.5 cm3.

1 3.68-0.016(150)-0.77(1)-0.12(20)

1+ e 1

1 + e-1.89

» 0.8688

The probability is about 87%. (b) Since bird pecking is not present, B = 0. P(150, 0, 20) = =

)

1+ e

(a) Since bird pecking is present, B = 1.

0.6 -1.267

+ 23(-0.267)E P dP To approximate the change called for we take E = 14,100, P = 68,700, dE = 200, and dP = -300. For these inputs the value of dL is then 3.5117 » 3.51 years. For comparison, L(14,300, 68, 400) - L(14,100, 68,700) = 3.5077 » 3.51 years.

1 3.68-0.016 A-0.77 B -0.12 D

1 3.68-0.016(150)-0.77(0)-0.12(20)

1+ e 1

1 + e-1.12

» 0.7540

The probability is about 75%. (c) Let B = 0. To simplify the notation, let X = 3.68 - 0.016 A - 0.12D. Then

(b)

 3

(

1 + e3.68-0.016 A-0.12 D 1 . = 1 + eX

h (2r1 + r 2)dr1 r 21+r1r 2 + r 22 dh +

)

h (r1 + 2r 2)dr 2 + 3

Some other values that we will need are

dh = 42 - 40 = 2 dr1 = 5.1 - 5 = 0.1 dr2 = 2.9 - 3 = -0.1 dV (40,5,3) =

 3

P( A, 0, D) =

dV = Vh dh + Vr1d r1 + Vr d r

(52 + 5 ⋅ 3 + 32 )(2)

40 (2 ⋅ 5 + 3)(0.1) 3 40 + (5 + 2 ⋅ 3)(-0.1) 3 » 111.00 V (42,5.1, 2.9) » V (40,5,3) + dV » 2052.51 + 111.00 » 2163.51 +

dA = 160 - 150 = 10 dD = 25 - 20 = 4 X (150, 20) = 3.68 - 0.016(150) - 0.12(20) = -1.12 ¶X XA = = -0.016 ¶A ¶X XD = = -0.12. ¶D PA ( A, 0, D) = PD ( A, 0, D) =

X Ae X (1 + e X ) X De X (1 + e X )

= 2 = 2

0.016e X (1 + e X )2 0.12e X (1 + e X )2

Section 9.5

709

dP = PA ( A, 0, D) dA + PD ( A, 0, D) dD =

0.016e X

0.12e X dA + dD (1 + e X )2 (1 + e X ) 2

Substituting the given and calculated values, dP =

t ( x, y, p, C ) =

(a)

x 2 + ( y - p) 2 331.45 + 0.6C

t (5, - 2, 20, 20) =

52 (-2 - 20)2 331.45 + 0.6(20)

509 » 0.06569 343.45

t (5, - 2,10, 20) =

52 + (-2 - 10)2 331.45 + 0.6(20)

0.016e-1.12

0.12e-1.12 (10) (5) + (1 + e-1.12 ) 2 (1 + e-1.12 )2

= (0.016.10 + 0.12 ⋅ 5)

e-1.12

(1 + e-1.12 )2 » 0.76 ⋅ 0.1855 » 0.14.

Therefore, P(160, 0, 25) = P(150, 0, 20) + DP » P(150, 0, 20) + dP = 0.75 + 0.14 = 0.89.

The probability is about 89%. Using a calculator, P(160, 0, 25) » 0.8676, or about 87%. 32.

33.

169 » 0.0379 343.45 In a close race, this difference could certainly affect the outcome. (b) Since the starter remains stationary, dx = dy = 0, so t x ( x, y, p, C ) and t y ( x, y, p, C ) do not need to be computed. =

t p ( x, y, p, C )

C (t , g ) = 0.6(0.96)(210t /1500)-1

gt é 1 - (0.96)(210t /1500)-1 ù + ê úû 126t - 900 ë

(a)

-2( y - p) 2(331.45 + 0.6C ) x 2 + ( y - p)2 p-y (331.45 + 0.6C ) x 2 + ( y - p)2

C (180,8) = 0.6(0.96)(210(180)/1500)-1 (8)(180) é 1 - (0.96)(210(180)/1500) -1 ù + úû 126(180) - 900 êë » 0.2649

(b) Ct (t , g ) æ 210 ö÷ (0.96)(210t /1500)-1 = 0.6(ln 0.96) çç çè 1500 ÷÷ø +

g (126t - 900) - 126( gt ) (126t - 900)2

´ [1 - (0.96)(210t /1500)-1] æ 210 ö÷ gt (ln 0.96) çç (0.96)(210t /1500)-1 çè 1500 ø÷÷ 126t - 900 C g (t , g ) t é 1 - (0.96)(210t /1500)-1 ù úû 126t - 900 êë C (180 - 10,8 + 1) » C (180,8) + Ct (180,8) ⋅ (-10) =

tC ( x, y, p, C ) = -

0.6 x 2 + ( y - p)2 (331.45 + 0.6C )2

dx = 0, dy = 0, dp = 0.5, dC = -5 dt = t x (5, -2, 20, 20) ⋅ 0 + t y (5, -2, 20, 20) ⋅ 0 + t p (5, -2, 20, 20) ⋅ 0.5 + tC (5, -2, 20, 20) ⋅ (-5) =

20 - (-2) (331.45 + 0.6(20)) 52 + (-2 - 20)2 -

0.6 52 + (-2 - 20)2

(331.45 + 0.6(20))2 » 0.001993 sec

⋅ 0.5

⋅ (-5)

This is the approximate change in the time when the swimmer stands 0.5 m farther from the starter in the y direction and the temperature decreases by 5C.

+ C g (180,8) ⋅ (1) » 0.2649 + (-0.00115)(-10) + 0.00519(1) » 0.2816 C (170,9) » 0.2817 The approximation is very good.

710 34.

Chapter 9 MULTIVARIABLE CALCULUS 1 2  r h, r = 3.2, h = 9.3, 3 dr = dh = 0.1

37.

V =

Let z = f ( L, W , H ) = LWH Then dz = f L ( L, W , H ) dL + fW ( L, W , H ) dW + f H ( L, W , H ) dH

2 1  rhdr +  r 2dh 3 3 2 1 =  (3.2)(9.3)(0.1) +  (3.2)2 (0.1) 3 3 » 7.305 The maximum possible error is 7.305 cm3. dV =

= WHdL + LHdW + LWdH .

A maximum 1% error in each measurement means that the maximum values of dL, dW , and dH are given by dL = 0.01L, dW = 0.01W , and dH = 0.01H . Therefore, dz = WH (0.01L) + LH (0.01W ) + LW (0.01H )

35.

The area is A = 12 bh with b = 15.8 cm,

= 0.01LWH + 0.01LWH + 0.01LWH = 0.03LWH .

h = 37.5 cm, db = 1.1 cm, and dh = 0.8 cm.

Thus, an estimate of the maximum error in calculating the volume is 3%.

1 1 b dh + h db 2 2 1 1 = (15.8)(0.8) + (37.5)(1.1) 2 2 = 26.945

dA =

38.

dV 2 rh  r2 dr dh = + V  r 2h  r 2h 2 1 = dr + dh r h

The maximum possible error is 26.945 cm . 36.

Let z = f (r , h) =

1 2  r h. 3

We need to use the same units for both dimensions; in inches, r = 0.5 and h = (20)(12) = 240.

Then dz = f r ( r, h)dr + f h ( r, h) dh 2 1  rh dr +  r 2dh. 3 3 Since there is a maximum error of a % in measuring the radius, the maximum value of dr is a r. Similarly, the maximum value of dh is 100 dz =

When r changes by a, the relative volume 2 a = 4a. When h changes by a, the change is 0.5 1 a. So the radius relative volume change is 240

has a greater effect by a factor of 4a = 960. (1/240)a

b h. Therefore, the maximum value of dz is 100

æ a ö÷ 1 2 æ b ö÷ 2 r +  r çç h÷  rh çç çè 100 ÷÷ø 3 èç 100 ø÷ 3 b çæ 1 2 ÷ö 2a æç 1 2 ö÷ = çç  r h ÷÷ + ç  r h ÷÷ è ø ø 100 3 100 çè 3

dz =

æ 2a + b öæ ÷ç 1  r 2h ÷÷ö. = çç ÷ø çè 100 ÷÷øèçç 3

The volume of a cylinder is V =  r 2h.

35.

The volume of a cone is V =

Since the volume of a cylinder is  r h, the maximum percent error in calculating the volume of the cylinder is the same.

r 2h.

2 rh  r2 dV = 3 2 dr + 3 2 dh  r h  r h V 3

The maximum percent error in calculating the volume is (2a + b)%.



2 1 = dr + dh r h

When r = 1 and h = 4, a 1% change in radius changes the volume by 2%, and a 1% change in height changes the volume by 14 %. So the change produced by changing the radius is 8 times the change produced by changing the height.

Section 9.6

711

9.6 Double Integrals

Your Turn 3

Your Turn 1

Let the region R be defined by 0 £ x £ 5 and 1 £ y £ 6.

ò (6x y + 4xy + 8x + 10 y + 3) dy 2 2

= (2 x 2 y 3 + 2 xy 2 + 8 x3 y + 2 y 5 + 3 y) 2

y =3

òò x + y + 3 dx dy R

y =1

= (2 x (3) + 2 x(3) + 8 x (3) + 2(3) + 3(3)) 2

ò ò

- (2 x (1) + 2 x(1) + 8x (1) + 2(1) + 3(1)) = 16 x3 + 52 x 2 + 16 x + 490

1 dx dy x+ y+3

x=5

ò1 ( 2 x + y + 3 ) x=0 dy

= 52 x 2 + 16 x + 16 x3 + 484 + 6

= 2

ò ( y + 8 - y + 3 ) dy 1

Your Turn 2 2é

ò ò ê ê 0 êë

ù (6 x 2 y 2 + 4 xy + 8x3 + 10 y 4 + 3) dy úú dx 1 úû 3

Use the result from Your Turn 1 to evaluate the inner integral. 2

4 [(143/2 - 93/2 ) - (93/2 - 43/2 )] 3 4 = (14 14 - 46) 3

ò (16x + 52x + 16x + 490) dx 3

2 = 2 [( y + 8)3/2 - ( y + 3)3/2 ] 3 1

56 14 - 184 3

æ ö 52 3 = çç 4 x 4 + x + 8 x 2 + 490 x ÷÷÷ çè ø0 3 = 4(16) +

52 (8) + 8(4) + 490(2) 3

3644 = 3

ò ò ê ê 1 êë = =

ò (16 y + 8 y + 32 + 20 y + 6) dy 2

ò (20 y + 16 y + 8 y + 38) dy 4

ò 0 ò 0 (4 - x3 - y3) dx dy =

æ ö 16 3 = çç 4 y 5 + y + 4 y 2 + 38 y ÷÷÷ çè ø1 3 æ è

( 972 + 144 + 36 + 114 ) - ççç 4 +

ò 0 ò 0 (4 - x3 - y3) dx dy .

ù (6 x 2 y 2 + 4 xy + 8 x3 + 10 y 4 + 3) dx úú dy úû 0 3é x=2 ù ê (2 x3 y 2 + 2 x 2 y + 2 x 4 + 10 xy 4 + 3x) ú dy ê x = 0 úû 1 ë

surface z = 4 - x3 - y 3 over this region is 3

The function 4 - x3 - y 3 is positive over the region 0 £ x £ 1 and 0 £ y £ 1, so the volume under the

òò (4 - x - y )dx dy

Integrating in the other order: 3é

Your Turn 4

1æ

ö x4 çç 3 ÷÷ ç 4 x - 4 - xy ÷÷÷ 0 çè ø 1æ

ò 0 çççè 4 - y3 ÷÷÷ø dy 1

ö 16 + 4 + 38 ÷÷÷ ø 3

3644 3

æ 15 y 4 ö÷÷ ç = çç y ÷ 4 ÷÷ø çè 4 0 =

15 1 7 - = 4 4 2

x =1

dy x =0

712

Chapter 9 MULTIVARIABLE CALCULUS

Your Turn 5

Find

W2.

òò ( x3 + 4 y ) dy dx over the region bounded by R

ò x + 6 dx = 3 ( x + 6)3/ 2 + C 3

y = 4 x and y = x3 for 0 £ x £ 2.

ò-2

The region is shown in the figure below.

x + 6 dx =

10 5

( 2, 8) y x

W3.

ò x dx = 3 ln x + C 4

ò2 x dx = ( 3 ln x ) 2

Note that throughout the region, 4 x ³ x .

= 3 ( ln 4 - ln 2 )

òò (x3 + 4 y)dy dx

= 3 ln 2 = ln 8 » 2.0794

= = =

ò 0 ò x (x3 + 4 y) dy dx

W4.

y=4x

ò 0 (x3 y + 2 y2 ) y x dx =

ò 0 [(4x + 32x ) - (x + 2x )] dx 4

e6 x dx =

e6 - e 0 6

e6 - 1 » 67.0715 6

ò 0 (4x4 + 32x2 - 3x6 ) dx 2

W5. Use the substitution u = x3, du = 3x 2dx. 3 1 x 2e x dx = eu du 3 1 1 3 = eu + C = e x + C 3 3

128 256 384 + 5 3 7 5888 = 105

ò1

9.6 Warmup Exercises W1.

ò0 ( 3x - 6x )

(

1 x3 1 8 e = e - e1 3 3 1

e8 - e » 992.7466 3

x 2e x dx =

ò ( 3x3 - 6x2 ) dx = 4 x4 - 2x3 + C 3

e6 x +C 6

e6 x e dx = 6 0 0

æ4 32 3 3 ö = çç x5 + x - x7 ÷÷÷ çè 5 3 7 ø0

)

(

2 (3 + 6)3/ 2 - (-2 + 6)3/ 2 3 2 38 = ( 27 - 8 ) = 3 3 =

y y  4x

2 ( x + 6)3/ 2 3 -2

æ3 ö dx = çç x 4 - 2 x3 ÷÷÷ çè 4 ø0

æ3 ö = çç (16) - 2(8) ÷÷÷ - ( 0 - 0 ) çè 4 ø = -4

)

Section 9.6

713

W6. Use the substitution u = x 2 + 9, du = 2 x dx. 1 x x 2 + 9 dx = u du 2 1 2 = ⋅ ⋅ u 3/ 2 + C 2 3 3/ 2 1 2 = x +9 +C 3

(

)

(

)

253/ 2 - 93/ 2 3 125 - 27 98 = = 3 3 =

ò4 x x2 + 3 ydy 5

ò4 x(x2 + 3 y)1/2 dy

2x 2 [( x + 3 y)3/2 9 4

2x 2 [( x + 15)3/2 - ( x 2 + 12)3/2 ] 9

3/ 2 1 2 x x + 9 dx = x +9 3 0 0 2

ò3 x x2 + 3 y dx Let u = x 2 + 3 y. Then du = 2 x dx When x = 6, u = 36 + 3 y. When x = 3, u = 9 + 3 y.

9.6 Exercises 1.

False. The double integral of f ( x, y) over the region a £ x £ b and c £ y £ d can be b

written as

ò ò a

f ( x, y) dy dx or

True

False. For this course, the variable limits always go on the inner integral sign. Double integrals over more complicated regions are discussed in more advanced books

1 [(36 + 3 y)3/2 - (9 + 3 y)3/2 ] 3

9 3 + 5y dx = (3 + 5 y) x-1/2dx x 4 4

= (3 + 5 y)2 x1/2

æ x5 y ö÷ ç ( x 4 y + y)dx = çç - xy ÷÷÷ ÷ø çè 5 0 0

10.

ò2

3 + 5y dy x 7

æx ö = (4 x - 2 x) - çç - x ÷÷÷ çè 4 ø 11 x = 4

æ 3y 5 y 2 ö÷÷ ç = çç + ÷ çè x 2 x ø÷÷ 2

æ xy 4 ö÷ ç ( xy - x)dy = çç + xy ÷÷÷ çè 4 ÷ø 1 1 2

= (3 + 5 y)2 éê 9 - 4 ùú ë û = 6 + 10 y

False. Fubini’s Theorem cannot be used to interchange the order of integration when the limits contain variables.

= (625 y + 5 y) - 0 = 630 y

1 2 3/2 ⋅ (u ) 2 3 9 +3 y

ò9+3 y u1/2du 36 + 3 y

ò ò f (x, y) dx dy.

1 2

36 + y

11.

1 éê æç 245 ÷ö çæ 20 ÷ö ùú - ç6 + 21 + ÷ ÷ ç ÷ 2 ø çè 2 ÷ø úû x êë èç

255 2 x

ò2

e 2 x + 3 y dx = =

1 2 x +3 y e 2 2 1 12 + 3 y - e4 +3 y ) (e 2

714

Chapter 9 MULTIVARIABLE CALCULUS 1

12.

-1

e2 x +3 y dy = =

16.

ò ò ( xy - x) dy dx 0

From Exercise 6.

1 2 x +3 (e - e 2 x -3 ) 3

ò (xy - x)dy = 4 x. 3

ò ye

13.

1 2 x +3 y e 3 -1

4x+ y2

Therefore,

ò0 ò1

Let u = 4 x + y 2 ; then du = 2 y dy. If y = 0 then u = 4 x.

[( xy 3 - x)dy] dx =

eu ⋅

4 x +9

1 1 du = eu 2 2 4x =

14.

ò ye

4x + y2

1 4 x +9 - e4 x ) (e 2

Let u = 4 x + y 2 ; then du = 4 dx.

4+ y

2 2 ö 1 æ = y çç e20 + y - e4 + y ÷÷ è ø 4

15.

ò ò (x y + y) dx dy 1

From Exercise 5 5

ò0 (x y + y) dx = 630 y. 4

Therefore, 2é

ù ( x y + y) dx úú dy 0 ûú

ò ò ê ê 1 ëê

ò 630 y dy

ù x x 2 + 3 ydx úú dy 3 úû From Exercise 8,

= 315 y 2

= 315(4 - 1) = 945.

ò ò ê ê 0 êë

ò x x + 3 y dx 2

ò 3 [(36 + 3 y)

3/2

- (9 + 3 y)3/2 ] dy

Let u = 36 + 3 y. Then du = 3 dy. When y = 0, u = 36. When y = 1, u = 39. Let z = 9 + 3 y. Then dz = 3 dy. When y = 0, z = 9. When y = 1, z = 12. 1 éê 9 êêë

u 3/2du -

ù z 3/2dz úú úû

1 2 ⋅ [(39)5/2 - (36)5/2 9 5 - (12)5/2 + (9)5/2 ]

99 . 8

ò ò

20 + y 2

1 1 ye ⋅ du = yeu 4 4 4 + y2

1 [(36 + 3 y)3/2 - (9 + 3 y)3 ]. 3 1é 6 ù 2 ê ú dy x x + y dx 3 ê ú 0 ëê 3 ûú

If x = 5 then u = 20 + y .

1é

11 2 x 8 0

17.

If x = 1 then u = 4 + y 2.

20 + y 2

ò0 4 x dx 3

If y = 3 then u = 4 x + 9. 4 x +9

2 [(39)5/2 - (12)5/2 - 65 + 35 ] 45 2 (395/2 - 125/2 - 7533) = 45 =

Section 9.6

715

(See Exercise 7.)

18. 3é

ù x x 2 + 3 y dy úú dx 4 ûú 5

ò ò ê ê 0 ëê =

21.

dy dx = xy 1

ò ò 1

2x 2 [( x + 15)3/2 - ( x 2 + 12)3/2 ] dx 0 9

2 [( x 2 + 15)]5/2 - ( x 2 + 12)5/2 45 0

2 (245/2 - 215/2 - 155/2 + 125/2 ) 45

19.

2é

3 + 5y ùú dx ú dy x 4 úû

ò ò ê ê 1 êë

22.

dx dy = y 2

ò ò 1

ò ò

= =

ò (6 + 10 y)dy + 5y

ln 3 dx x 1

5é

1 ùú dx ú dy 2 y úû

5æ

ò ò ê ê 1 êë

ö çç 1 x ÷÷ dy ÷ ç 1 è y ø÷ 2

5é

ò êêë y (4 - 2) úúû dy 5

ò y dy 1

= 6y

ö çç 1 ln | y | ÷÷ dx ÷ø ç 1 èx 1

= (ln 3) 2

3 + 5y dx = 6 + 10 y. x 4 2é 9 3 + 5 y ùú ê dx ú dy ê x 1 ëê 4 ûú

3æ

= (ln 3)(ln 3 - 0)

ù 1 dy úú dx xy úû

ò ò ê ê 1 êë

= (ln 3) ln | x | 1

From Exercise 9,

3é

= 2 ln | y | 1

= 2 ln 5

= 6(2 - 1) + 5(4 - 1) = 6 + 15 = 21

20.

(See Exercise 10.) 25 é

3 + 5 y ùú ê dy ú dx ê x 16 êë 2 ûú

ò2 ò =

ò ò =

x yö çç + ÷÷÷ dx dy 3 ÷ø 3 çè y

23.

255 dx 16 2 x

255 -1/2 x dx 2 16

= 255x1/2 = 255

25 16

= 255(5 - 4)

5æ

yx ö÷÷ çç x + ÷ dy ç 3 ø÷÷ 2 çè 2 y

4 æ 2

æ öù ê 25 + 5 y - çç 9 + 3 y ÷÷ ú dy ê 2y ç 3 3 ÷ø÷ úûú è 2y 2 ëê

4é

16 2 y ö÷ ÷ dy + çç 3 ø÷÷ 2 çè 2 y

4æ

æ y 2 ÷÷ö ç = çç 8 ln | y | + ÷ çè 3 ÷÷ø 2 = 8 (ln 4 - ln 2) + 4 12 + 2 3 = 8 ln 2 + 4 = 8 ln

16 4 3 3

716 24.

Chapter 9 MULTIVARIABLE CALCULUS 4

2æ

yö

ò3 ò1 çççè 5 + x ÷÷ø÷ dx dy = = =

4æ

3x 2

27.

ò3 çèçç 5 + y ln x ÷ø÷÷ 1 dy

òò x + y dy dx R

æ 12 3 ö÷ ç çè 5 + y ln2 - 5 ÷÷ø dy 3

æ9

òò x + y dy dx; 1 £ x £ 3, 0 £ y £ 1

ò ò (x + y) dy dx

1/2

3 é

ù 2 ê ( x + y)3/2 ú dx êë 3 úû 0

1 3

ò 3 [(x + 1)

3/2

- x3/2 ] dx

= 36 + 8 ln2 - 27 - 9 ln2 5 5 2 = 9 + 7 ln2 5 2

25.

ò ççè 5 + y ln2 ÷÷÷ø dy

æ 9y ö÷ y2 ln2 ÷÷÷ = ççç + è 5 ø 2

3 2 2é ⋅ ê ( x + 1)5/2 - x5/2 ùú û1 3 5ë 4 5/2 = (4 - 35/2 - 25/2 + 15/2 ) 15 4 = (32 - 35/2 - 25/2 + 1) 15 4 = (33 - 35/2 - 25/2 ) 15

òò (3x + 4 y)dx dy; 2

0 £ x £ 3,1 £ y £ 4 (3x + 4 y)dx dy òò R 2

x x + 2 y dx dy òò R 2

( x3 + 4 xy) dy 0

(27 + 12 y)dy

1 2

= (27 y + 6 y )

= (108 + 96) - (27 + 6) = 171

26.

òò (x + 4 y )dy dx; 2

ò ò x (x + 2 y) dx dy 0 3

òò ( x + 4 y ) dy dx 3

ò ( x y + y ) dx 2

ò (3x + 81) dx 2

= ( x 3 + 81x )

ò 9 [(8 + 2 y)

1/2

ò ò ( x + 4 y ) dy dx 1

2 2 3 ( x + 2 y)3/2 dy 0 0 9

3/2

- (2 y)3/2 ] dy

1 £ x £ 2, 0 £ y £ 3

3 2 [(8 + 2 y )5/2 - (2 y)5/2 ] = 0 45 2 5/2 5/2 5/2 (14 - 6 - 8 ) = 45

0 £ x £ 2, 0 £ y £ 3

òò x x + 2 y dx dy; R

ò ò (3x + 4 y)dx dy 1

28.

2 1

= (8 + 162) - (1 + 81) = 88

Section 9.6 29.

717 3

dy dx; 2 £ x £ 4, 1 £ y £ 6 òò ( x + y) 2 R

31.

ò2 ò1 ò2

= -3

+ 2

ye( x y ) dx dy òò R

dy dx òò ( x + y)2 R = -3

ye( x y )dx dy; 2 £ x £ 3, 0 £ y £ 2 òò R + 2

( x + y)-2 dy dx

( x + y)-1 dx

4 æ

ö çç 1 - 1 ÷÷ dx ç x + 1 ÷ø 2 èx + 6

= -3

= -3( ln | x + 6| - ln | x + 1| ) æ x + 6 ö÷ ÷ = -3 ççç ln x + 1 ø÷÷ è

32.

( ye3+ y - ye2 + y ) dy

ò0

ye y dy - e2

ò0

ye y dy

2 2 2 e e (e y ) (e y ) 2 2 0 0

x3 + 2 y

2 x3 + 2 y

dx dy; 1 £ x £ 2,1 £ y £ 3

dx dy

ò ò xe 1

dx dy òò 2 x y 2 + 5 R

òò x2e

ò1 ò0 y(2x + 5 y2 )-1/2 dx dy

dx dy

1 x3 + 2 y e dy 1 3 1

ò 3(e

8+ 2 y

- e1+ 2 y ) dy 3

ò1 [ y(4 + 5 y2 )1/2 - y(5 y2 )1/2 ] dy

é 1 2 ù 1 2 = ê ⋅ (4 + 5 y 2 )3/2 ⋅ (5 y 2 )3/2 ú 10 3 ëê 10 3 ûú

2 x3 + 2 y

1 y ⋅ ⋅ 2(2 x + 5 y 2 )1/2 dy 2 1

3 2

òò x e

0 £ x £ 2,1 £ y £ 3

dx dy; òò 2 x + y 2 5 R

ò0

ye x + y

e3 4 e2 4 (e - e0 ) (e - e0 ) 2 2 1 = (e7 - e6 - e3 + e2 ) 2

3 4 or 3 ln = -3ln 4 3

ò0

+ 2

2 = -3ln 8

ò0 ò2 ye x y dx dy

= e3

æ 8ö = -3 çç ln 2 - ln ÷÷÷ çè 3ø

30.

1 8+ 2 y (e - e1+ 2 y ) 6 1

1 14 (e - e7 - e10 + e3 ) 6

3 1

1 = [(493/2 - 453/2 ) - (93/2 - 53/2 )] 15 1 = (343 - 135 5 - 27 + 5 5) 15 1 = (316 - 130 5) 15

718 33.

Chapter 9 MULTIVARIABLE CALCULUS z = 8 x + 4 y + 10; -1 £ x £ 1, 0 £ y £ 3

V =

ò ò -1 1

(24 x + 18 + 30 - 0) dx

ò (24 x + 48) dx

= (12 x 2 + 48 x)

37.

-1

ò (3xy + 5 y + 20 y) 0

1 -2

[(3x + 5 + 20) - (-6 x + 20 - 40)] dx

(9 x + 45) dx

æ9 ö = çç x 2 + 45x ÷÷÷ çè 2 ø0

ò ò

ò ò ê ê 0 ëê

1+ y y

ù u1/2du úú dy ûú 1+ y

1 æç 2 3/2 ö÷ çç u ÷÷ øy 0 2è 3

ò 3 [(1 + y)

3/2

- y 3/2 ] dy

1 2 ⋅ [(1 + y)5/2 - y 5/2 ] 3 5 0

x 2 y dx

1é

2 5/2 (2 - 1 - 1) 15 2 5/2 = (2 - 2) 15

ò ò x dy dx 0

ò ò x x + ydx dy

z = x 2 ;0 £ x £ 2, 0 £ y £ 5 2

81 351 + 135 - 0 = 2 2

V =

When x = 1, u = 1 + y.

35.

ò0 18 dx = 18x 0

z = x x 2 + y ; 0 £ x £ 1, 0 £ y £ 1

When x = 0, u = y.

0 3

ò0 3 (27) dx

Let u = x + y. Then du = 2 x dx.

-2

ò ò (3x + 10 y + 20) dy dx 3

2 3/2 y dx = 0 3 0

V =

= 72

z = 3x + 10 y + 20; 0 £ x £ 3, -2 £ y £ 1 3

ò0 ò0 y1/2dy dx

= (12 + 48) - (12 - 48) = 96

V =

-1

34.

y ;0 £ x £ 4, 0 £ y £ 9

(8 xy + 2 y 2 + 10 y ) dx

-1 1

z = V =

(8 x + 4 y + 10) dy dx

-1 1

36.

5x 2dx

38.

z = yx x 2 + y 2 ;0 £ x £ 4,0 £ y £ 1 V =

ò ò yx( x + y ) dx dy 0

5 3 x 3 0

40 3

2 1/2

y 2 ( x + y 2 )3/2 dy 3 0 0

ò 3 [(16 + y ) 0

2 3/2

- ( y 2 )3/2 ] dy

Section 9.6

719 1

1 [ y(16 + y 2 )3/2 - y 4 ] dy 3 0

40.

V =

1é1 y 5 ùú = êê (16 + y 2 )5/2 3ê5 5 úú ë û0 1 = (175/2 - 1 - 165/2 ) 15 1 = (175/2 - 1 - 1024) 15 1 = (175/2 - 1025) 15

V =

ò ò (x + y ) dy dx 2

2 é

ò ò ê ê 1 êë 2 é ê ê 1 ëê

1 4

ò ò ò ò

ù 1 x( x 2 + y 2 )-2 (2 y)dy úú dx 2 ûú 4

1 + 2

òe 1

ò (e

2 1 1 ⋅ ln x 2 + 1 2 2 1

1+ y

- e y ) dy 1 0

= e -e-e +1 = e2 - 2e + 1

òò xe dx dy; 0 £ x £ 2; 0 £ y £ 1 xy

òò xe dx dy xy

ò ò xe dy dx xy

x xy e 0 x

dx 0

ò (e - e ) dx x

2 0

= e - 2 - e0 + 0 = e2 - 3

1 1 ⋅ ln x 2 + 16 2 2 1

dy 0

1 2 ( x + 1)-1(2 x) dx 2

dx dy

x+ y

= (e x - x)

1 2 ( x + 16)-1(2 x) dx 2

x+ y

= (e1+ y - e y )

ù 1 1 ê - x( x 2 + 16)-1 + x( x 2 + 1)-1 ú dx êë 2 úû 2 2

òòe

2 é

ù xy( x 2 + y 2 )-2 dy úú dx ûú

ù 1 ê - x( x 2 + y 2 )-1 ú êë 2 úû

1 =2

41.

2 2

2 é

; 1 £ x £ 2, 1 £ y £ 4 ( x2 + y 2 )2

z =

39.

z = e x + y ; 0 £ x £ 1, 0 £ y £ 1

1 1 ⋅ ln 20 + ln 17 4 4 1 1 + ln 5 - ln 2 4 4 1 = (-ln 20 + ln 17 + ln 5 - ln 2) 4 1 (17)(5) = ln 4 (20)(2) 1 17 = ln 4 8 =-

720 42.

Chapter 9 MULTIVARIABLE CALCULUS 2 x3e x y dx dy;0 £ x £ 1, 0 £ y £ 1 òò R 2

òò

44.

ò0 ò0 (x2 + y) dx dy ö÷ çç x ÷ çç 3 + xy ÷÷ 0 è ø÷

2 x3e x y dx dy

ò ò 2x e dy dx 0

= =

3 x2 y

1æ

2 ö çç 2 x3 ⋅ 1 e x y ÷÷ dx 2 ÷ø ç x 0 è 0

ò ò

dy 0

ò0 (9 y3 + 3y2 ) dy

2æ 3

æ9 ö = çç y 4 + y 3 ÷÷÷ çè 4 ø0

= 36 + 8 - 0 = 44

(2 xe x - 2 xe0 ) dx

0 2

= (e x - x 2 )

45.

= (e1 - 1) - (e0 - 0)

ò ò 2

( x 2 + y 2 )dy dx

y 3 ö÷÷ ç 2 ÷ dx ççç x y + 3 ÷÷ø 2 è 2

4 æ

8 ö÷÷ x6 çç 4 2 + x x 2 ÷ dx ç 3 3 ÷÷ø 2 çè

4 é

ê 2( xy) ê 3x 0 êë

ò 2 3

2 = 3

1024 16,384 2 8 + - (64) - (4) 5 21 3 3 æ 32 128 16 16 ö÷ - çç + - ÷÷ çè 5 21 3 3ø 1024 32 16,384 - 128 = + 5 5 21 128 32 æç -32 ÷ö -ç çè 3 ÷÷ø 3 3

ú dx ú úû 0

2 x3 x dx = ⋅ 3 3 0

= 0

2 (64) 9

128 = 9

992 16, 256 128 + 5 21 3 20,832 81, 280 4480 = + 105 105 105 97, 632 = 105

3/2 ù

3 éæ ù ê ç x 2 ö÷ ú ÷ 4 êç ú ÷ 0 è ø ê - ú dx ê x x úú 0 ê ê ú ë û

4æ

ò0 ò0 (xy)1/2 dy dx

æ x5 x7 2 8 ÷ö ç = çç + - x3 - x ÷÷÷ 21 3 3 ÷ø çè 5 2

ò0 ò0 xy dy dx =

= e-2

43.

46.

ò1 ò0 x + y dy dx = =

ò1 ò0 (x + y)1/2 dy dx x

2 ( x + y)3/2 dx 1 3 0

ò1 3 [(2x)3/2 - x3/2 ] dx

ù 2é1 2 ê (2 x)5/2 - x5/2 ú úû 3 êë 5 5 1

2 5/2 (8 - 2(4)5/2 - 25/2 + 2) 15

2 5/2 (8 - 64 - 25/2 + 2) 15 2 5/2 = (8 - 62 - 25/2 ) 15 =

Section 9.6 47.

721 4y

1 dx dy 2y x

ò ò 2

= =

ò (ln |4 y | -ln |2 y |) dy

ò2 ln 2 y dy 6

ò2

50.

= (ln 2)(6 - 2) = 4 ln 2

Note: We can write 4 ln 2 as ln 2 , or ln 16.

48.

ò ò 1

1 dy dx y

ò0 (x ln e - x ln 1) dx ò0

x 2 dx =

x3 3

= 0

64 3

ò0 e

4x x+ y

dx 2x

ò0 (e5x - e3x ) dx 1

ò (ln x - ln x) dx 2

æ1 ö 1 = çç e5 x - e3x ÷÷÷ çè 5 ø0 3

ò ln y dx 1

ò0 (x ln | y |) 1 dx

ò0 ò2x e x y dy dx

ln 2 dy

= (ln 2) y 2

ò0 ò1 y dy dx

ò (ln | x |) dy 2 6

49.

ò (2 ln x - ln x) dx

1 5 1 3 1 1 e - e - + 5 3 5 3

2 e5 e3 + 5 3 15

ò ln x dx

51.

= ( x ln x -

ò dx)

òò (5x + 8 y) dy dx; 1 £ x £ 3, R

0 £ y £ x -1

òò (5x + 8 y) dy dx

Integration by parts u = ln x, dv = dx

du = 1x dx, v = x

ò ò 1

= ( x ln x - x) 1

= 4 ln 4 - 4 + 1 = 4 ln 4 - 3

x -1

ò (5xy + 4 y )

x -1

Note: We can write ln y instead of ln | y | since x is in [1, 4] and y is in [x, x2], so y > 0.

(5x + 8 y) dy dx

dx 0

ò [5x(x - 1) + 4( x - 1) - 0] dx 2

ò (9x - 13x + 4) dx 2

æ ö 13 2 = çç 3x3 x + 4 x ÷÷÷ çè ø1 2

æ ö æ ö 117 13 = çç 81 + 12 ÷÷ - çç 3 + 4 ÷÷ ÷ ÷ø çè ç ø è 2 2 = 34

722 52.

Chapter 9 MULTIVARIABLE CALCULUS

òò (2x + 6 y) dy dx;

54.

2 £ x £ 4, 2 £ y £ 3x

dy dx

òò x R

(2 x + 6 y) dy dx òò R = = = =

dy dx

; 1 £ x £ 2, 0 £ y £ x - 1 òò x R

ò2 ò2 (2x + 6 y) dy dx

ò2 [2xy + 3 y ] 2 dx 2

ò2 [2x(3x) + 3(3x)2 - (4x + 12)] dx ò2 [33x2 - 4x - 12] dx

= (11x3 - 2 x 2 - 12 x)

y 1 x 0

ò1

0 £ y £ 2 - 2x

1ù

ò1 êêë1 - x úúû dx

= 1 - ln 2

òò e

x /y 2

òò e

x /y 2

dx dy; 1 £ y £ 2, 0 £ x £ y 2

dx dy

(4 - 4 x 2 ) dy dx

ò ò

2- 2 x

4(1 - x ) dy dx

ò [4(1 - x ) y]

ò ò

ò [y e

2 x /y

]

dy 0

ò (y e

2 y 2 /y 2

- y 2e0 ) dy

4(1 - x 2 )(2)(1 - x) dx

e x /y dx dy

2(1- x)

x -1 dx x

2é

= 2 - ln 2 - 1

òò (4 - 4x ) dy dx; 0 £ x £ 1,

x -1

dy dx x

55.

òò

ò1 ò0

x -1

= ( x - ln x) 1

= (704 - 32 - 48) - (88 - 8 - 24) = 568

53.

ò (ey - y ) dy 2

(1 - x - x 2 + x3 ) dx

= (e - 1)

y3 3

2 1

æ x2 x3 x 4 ö÷÷ ç = 8 çç x + ÷ çè 2 3 4 ø÷÷ 0

æ8 1ö = (e - 1) çç - ÷÷÷ çè 3 3ø

æ 1 1 1ö = 8 çç1 - - + ÷÷÷ çè 2 3 4ø

7(e - 1) 3

æ1 1 ö = 8 çç - ÷÷÷ çè 2 12 ø = 8⋅

5 10 = 12 3

Section 9.6 56.

òò

723 ( x 2 - y ) dy dx; -1 £ x £ 1,

2 4 ù ê x 3 (4 x ) - x 3 ( x ) ú dx ê 2 2 úú 0 êë û 7ö 2æ x ÷÷ çç 5 çç 2 x - 2 ÷÷÷ dx 0 è ø

-x2 £ y £ x2

òò

( x - y ) dy dx

æ1 1 8 ö÷ x ÷ = çç x 6 çè 3 16 ÷ø 0

-1

ò ò

-x

( x 2 - y ) dy dx

1 6 1 ⋅2 ⋅ 28 3 16 64 = - 16 3 16 = . 3 =

2ù ê x2 y - y ú dx ê 2 úú 2 -1 ëê û -x 4 4ù 1 é ê x 4 - x + x 4 + x ú dx ê 2 2 úú -1 êë û

1 é

2 x5 2 x dx = 5 -1

58.

òò x y dy dx; R bounded by y = x, 2 2

-1

y = 2 x, and x = 1

2 2 4 + = 5 5 5

57.

2é

x3 y dy dx; R bounded by y = x , y = 2 x òò R 2

ò ò x y dy dx 0

The points of intersection can be determined by solving the following system for x.

2 2

1 2 3

x y 3 0

y = x y = 2x x = 2x x ( x - 2) = 0 x = 0 or x = 2

7 x5 dx 0 3 1

7 x6 18

7 = 18

Therefore,

òò x y dx dy 3

ò ò 0

2x x2

x y dy dx =

2ö çç 3 y ÷÷ x ÷ dx ç 2 ÷÷ø 2 0 çè

2æ

dx x

x5 ö÷÷ çç 8 x çç 3 - 3 ÷÷ dx ÷ø 0 è

1æ

724 59.

Chapter 9 MULTIVARIABLE CALCULUS For the first integral, use integration by parts with

1 dy dx ; R bounded by y = x, y = , x y

òò R

u = x and dv = e 2 x dx. Then du = dx and 1 v = e2 x . 2

x = 2.

1 2

The graphs of y = x and y =

1 intersect at (1, x

1). 2

= =

ò xe dx 2x

2 æ 1 çç 1 2 x 1 xe ç 2 çèç 2 2 0

dy dx = 1/x y

ö 1 1æ1 2e 4 - çç e 2 x ÷÷÷ ç ø0 4 4è 2

1 4 1 e - (e4 - 1) 2 8

1 2

ò ln y x dx 1

1æ1 ö x dx = - çç x 2 ÷÷÷ 2 çè 2 ø 0

ò 2 ln x dx

(

)

1 4 1 3 7 e - e4 - 1 - 1 = e4 2 8 8 8

= 2[(2 ln 2 - 2) - (ln 1 - 1)] = 4 ln 2 - 2

61.

e y x dy dx; R bounded by y = x , òò R 2

2 /

ò ò e 0

and

2æ

x 2 y /x ö÷ ÷÷ ø

2æ

x 2x

ò ççèç 2 e

dx y =0

ò1 ò0

ò1 (1 - 0) dx 2

= x1

= 2 -1 = 1

xö

ò ççèç 2 e - 2 ÷ø÷÷ dx 0

1 2

xe 2 x dx -

ln x

1 dy dx ln x ln x ù 2é ê 1 ú y ê ú dx x ln 1 êê 0 úûú ë

y = x2

dy dx

ò0 òe ln x dx dy

2 y /x

ò0 òe ln x dx dy

dy dx

ln 2

The region lies between the curve y = x the x-axis between x = 0 and x = 2.

ln 2

3e4 - 7 8

Changing the order of integration,

y = 0, x = 2.

2 y /x

Combine.

òò e

= -1

ö çç ln x - ln 1 ÷÷ dx ç x ÷ø 1 è

= 2( x ln x - x) 1

( )

2æ

60.

ö÷ e2 x dx ÷÷÷ ÷ø÷ 0 2

Evaluate the second integral.

ò ò 1

1 2 x dx 2 0

Section 9.6 62.

725 1

ò0 ò y/2

e x dx dy

Change the order of integration. 2

ò0 ò y/2

e x dx dy =

= =

ò0 ò0 ex dy dx 1

ò0

ex y

dx 0

ò0 (2xe x - 0) dx

= ex

2 1

0 1

= e - e0 = e -1

65.

f ( x, y) = 6 xy + 2 x; 2 £ x £ 5, 1 £ y £ 3

The area of region R is A = (5 - 2) (3 - 1) = 6.

The average value of f over R is 1 A

f ( x, y ) dy dx òò R 5

1 6

ò2

1 6

1 14 x 2 6 2

7 (25 - 4) = 49. 3

ò2 ò1 (6xy + 2x) dy dx 5

(3xy 2 + 2 xy) dx 1

ò2 [(27x + 6x) - (3x + 2x)] dx 5

ò2 28x dx 5

66.

f ( x, y) = x 2 + y 2 ; 0 £ x £ 2, 0 £ y £ 3

The area of region R is A = (2 - 0) (3 - 0) = 6.

The average value of f ( x, y) = x 2 + y 2

726

Chapter 9 MULTIVARIABLE CALCULUS 1 A

òò ( x + y ) dy dx 2

1 = 6

1 6

1 ( x + y ) dy dx = 6 0

ò ò 0

y 3 ö÷÷ çç 2 + x y ÷ dx ç 3 ÷ø÷ 0 èç

ò (3x + 9) dx = 6 ( x + 9 x) 2

2æ

2 0

1 1 13 = (8 + 18 - 0)02 = ⋅ 26 = . 6 6 3

f ( x, y) = e-5 y + 3x ; 0 £ x £ 2, 0 £ y £ 2

67.

The area of region R is (2 - 0) (2 - 0) = 4.

The average value of f over R is 2

1 4

ò ò 0

e-5 y + 3x dy dx =

1 4

1 - e-5 y + 3x 5 0

dx = 0

1 4

ò - 5 [e - - e ] dx 3x 10

68.

1 é 1 3x -10 1 3x ù 1 e6 + e-10 - e-4 - 1 ê e . - e ú = - [e-4 - e6 - e-10 + 1] = úû 20 êë 3 3 60 60 0

f ( x, y) = e2 x + y ; 1 £ x £ 2, 2 £ y £ 3

The area of region R is A = (2 - 1) (3 - 2) = 1.

The average value of f over R is 1 A

òò e

2x+ y

dy dx

òò

e2 x + y dy dx =

ò (e

2x+ y

dy dx

(e 2 x + y ) dx =

ò ò e 1

2x+3

- e2 x + 2 ) dx

1 2x+3 1 (e - e 2 x + 2 ) = (e 4 + 3 - e 2 + 3 - e 4 + 2 + e 4 ) 2 2 1

e 7 - e 6 - e5 + e 4 . 2

Section 9.6 69.

727

The plane that intersects the axes has the equation z = 6 - 2 x - 2 y.

f ( x, y) dA òò R

V =

-x + 3

ò0 ò0

(6 - 2 x - 2 y) dy dx -x + 3

ò0 (6 y - 2xy - y ) 0

ò0 [-6x + 18 - 2x(-x + 3) - (3 - x)2 ] dx

ò (-6x + 18 + 2x - 6x - 9 + 6x - x ) dx 2

æ x3 ö÷ ç ( x - 6 x + 9) dx = çç - 3x 2 + 9 x ÷÷÷ çè 3 ÷ø 0 0

= (9 - 27 + 27) - 0 = 9

The volume is 9 in3. 70.

1 2 x + 2 x + y 2 + 5 y + 100 9 The area of region R is C ( x, y) =

A (80 - 40) (70 - 30) = 1600.

The average cost is 1 A

òò C (x, y) dy dx R

80 æ

1 1600

ò30 ò40

1 1600

70 æ

ò30

ö 1 2 2 ççç x + 2 x + y + 5 y + 100 ÷÷÷ dy dx è9 ø

ö çç 1 x3 + x 2 + xy 2 + 5xy + 100 x ÷÷ ÷ø çè 27

dy 40

70 é æ

ò30

ö æ öù ê çç 320 + 4 + 1 y 2 + 1 y + 5 ÷÷ - çç 40 + 1 + 1 y 2 + 1 y + 5 ÷÷ ú dy ÷ ÷ ç ê çè 27 ø è 27 20 4 40 8 2 ø úû ë

728

Chapter 9 MULTIVARIABLE CALCULUS =

70 æ

857 ö

ò30 çççè 40 y2 + 8 y + 54 ÷÷÷ø dy 70

æ 1 3 1 2 857 ö÷ = çç y + y + y÷ çè 120 16 54 ø÷ 30

æ 8575 1225 29,995 ö÷ æç 225 4285 ö÷ = çç + + + ÷÷ - çç 225 + ÷ çè 3 4 27 ø è 4 9 ø÷ 94,990 27 » 3518.

The average cost is about $3518. 71.

P ( x, y) = 500 x0.2 y 0.8 , 10 £ x £ 50, 20 £ y £ 40 A = 40 ⋅ 20 = 800

Average production: 1 800

5 8

ò10 ò20 500x0.2 y0.8 dy dx =

ò10

x 0.2 y1.8 1.8

dx = 20

25 (401.8 - 201.8 ) x1.2 ⋅ 72 1.2

25 72

ò10 x0.2(401.8 - 201.8) dx

= 10

125 (401.8 - 201.8 ) (501.2 - 101.2 ) 432

» 14,753 units

72.

P( x, y) = -( x - 100)2 - ( y - 50)2 + 2000

Area = (150 - 100)(80 - 40) = (50)(40) = 2000

The average weekly profit is 1 2000

[-( x - 100) 2 - ( y - 50)2 + 2000] dy dx òò R =

1 2000

150

ò100 ò40 [-( x - 100)2 - ( y - 50)2 + 2000] dydx

3 ù 80 ê -( x - 100)2 y - ( y - 50) + 2000 y ú dx ê ú 3 100 ëê ûú 40 3 3 150 é ù 1 ê -( x - 100)2 (80 - 40) - (80 - 50) + (40 - 50) + 2000(80 - 40) ú dx = ú 2000 100 êê 3 3 ë ûú 3 3 150 é ù 1 ê -40( x - 100)2 - 30 + (-10) + 2000(40) ú dx = ú 2000 100 êê 3 3 ë ûú

1 2000

150 é

ò ò

Section 9.6

729 150 é ù 1 28,000 ê -40( x - 100)2 + 80, 000 ú dx úû 2000 100 êë 3 é -40( x - 100)3 ù 150 1 28,000 = ⋅ êê x + 80,000 x úú 2000 ê 3 3 úû 100 ë

ù 1 é 40 40 28, 000 ê - (150 - 100)3 + (100 - 100)3 (150 - 100) + 80,000(150 - 100) ú úû 2000 êë 3 3 3 ù 1 é 40 40 28,000 ê - (50)3 + (50) + 80,000(50) ú = ⋅0ê úû 2000 ë 3 3 3

1 æç -5,000,000 - 1, 400,000 + 12, 000, 000 ö÷ ÷ ç ø÷ 2000 çè 3 1 æç 5,600,000 ÷ö = ÷÷ = $933.33. ç ø 2000 çè 3 =

73.

R = q1 p1 + q2 p2 where q1 = 300 - 2 p1, q2 = 500 - 1.2 p2 , 25 £ p1 £ 50, and 50 £ p2 £ 75. A = 25 ⋅ 25 = 625 R = (300 - 2 p1) p1 + (500 - 1.2 p2 ) p2 R = 300 p1 - 2 p12 + 500 p2 - 1.2 p22

Average Revenue: 1 625

1 625

ò ò (300 p - 2 p + 500 p - 1.2 p )dp dp

1 625

2 1

2 2

ò (300 p p - 2 p p + 250 p - 0.4 p ) 25

2 1

2 2

3 2

75 50

dp1

- 150 p12 + 1, 406, 250 - 168,750) ùú 1 ê ú dp1 25 ê - (15, 000 p - 100 p12 + 625, 000 - 50,000) ú 1 ëê ûú

50 éê 22,500 p

ò (662,500 + 7500 p - 50 p )dp 25

2 1

50 1 æç 50 p13 ö÷÷ 2 ç 662,500 p1 + 3750 p1 = ÷ 3 ÷ø÷ 25 625 ççè

1 æç 6, 250,000 781, 250 ö÷ - 16,562,500 - 2, 343,750 + ÷÷ çç 33,125, 000 + 9,375,000 ø 625 è 3 3 » $34,833

730

Chapter 9 MULTIVARIABLE CALCULUS T ( x, y) = x 4 + 16 y 4 - 32 xy + 40

74.

1 8

ò êêëê

1 32

òò ( x + 16 y - 32 xy + 40)dy dx 4

1 8

1 = 8 =

1 8

ò ò ( x + 16 y - 32 xy + 40) dy dx 0

4é

ò êêêë x (2 - 0) + 4

16(2 - 0) 5

ù 32 x(2 - 0)2 + 40(2 - 0) úú dx 2 úû 4æ ö 1 512 4 = - 64 x + 80 ÷÷÷ dx çç 2 x + ç ø 8 0 è 5 -

P( x, y) = 36 xy - x3 - 8 y3

Areas = (8 - 0)(4 - 0) = 32 The average profit is 1 32

Chapter 9 Review Exercises True

True

False: f ( x + h, y) = 3( x + h) 2

1 é 2(4 - 0)5 512 (4 - 0) = êê + 8ê 5 5 ë ù 64(4 - 0)2 + 80(4 - 0) úú 2 úû ö 1 æ 2048 2048 = çç + - 512 + 320 ÷÷÷ ç ø 8è 5 5 = 78.4 hours

75.

+ 2( x + h) y + y 2

False: (a, b) could be a saddle point.

False: No; near a saddle point the function takes on values both larger and smaller than its value at the saddle point.

True

False: We need to test values of the function at nearby points that satisfy the constraints to tell if the point found represents a maximum or minimum.

10.

False: When dx and dy are interchanged, the limits on the first integral must be exchanged with the limits on the second integral.

11.

True

òò (36xy - x - 8 y ) dy dx 3

ö÷ 8 1 æçç 288 x 2 4x4 512 x ÷÷÷ ç ÷ø 0 32 çè 2 4

ö÷ 64 x 2 1 æç 2 x5 512 = çç + + 80 ÷÷÷ 2 8 çè 5 5 ÷ø

ö÷ 16 y 5 32 xy 2 çç 4 x y + + 40 y ÷÷÷ dx ç 5 2 ÷ø 0 çè

ò (288x - 4x - 512)dx

ù 1 éê 288(8 - 0)2 4(8 - 0)4 ú 512(8 0) ú 32 êê 2 4 úû ë 1 = (9216 - 4096 - 4096) 32 = $32,000

4æ

36 x(4 - 0)2 8(4 - 0)4 ùú - x 3(4 - 0) ú dx 2 4 úû 0

1 32

Area = (4 - 0)(2 - 0) = 8 The average time is

8é

1 32

1 = 32

ò ò

0 0 8æ

(36 xy - x3 - 8 y3 ) dy dx 2

8y çç 36 xy - x3 y çç 2 4 0 èç

4ö4

÷÷ ÷÷ dx ÷ø 0

Chapter 9 Review 12.

17.

731 z-intercept: x = 0, y = 0

False: The two integrals are over different regions, and neither region is a simple region of the sort that we deal with in this chapter.

6z = 6 z =1

f ( x, y) = -4 x 2 + 6 xy - 3 f (-1, 2) = -4(-1) 2 + 6(-1)(2) - 3 = -19 f (6, -3) = -4(6) 2 + 6(6)(-3) - 3 = -4(36) + (-108) - 3 = -255

18.

19.

20.

21.

f ( x, y) = 2 x 2 y 2 - 7 x + 4 y f (-1, 2) = 2(1)(4) - 7(-1) + 4(2) = 23 f (6, -3) = 2(36)(9) - 7(6) + 4(-3) = 594

23.

The plane 5 x + 2 y = 10 intersects the x- and yaxes at (2, 0, 0) and (0, 5, 0). Note that there is no z-intercept since x = y = 0 is not a solution of the equation of the plane.

24.

4 x + 3z = 12

x - 2y x + 5y (-1) - 2(2) -5 5 f (-1, 2) = = =(-1) + 5(2) 9 9 (6) - 2(-3) 12 4 f (6, -3) = = =(6) + 5(-3) -9 3 f ( x, y) =

f ( x, y) =

x2 + y2 x-y

No y-intercept x-intercept: y = 0, z = 0

f (-1, 2) =

1+ 4 5 =-1 - 2 3

f (6, -3) =

36 + 9 = 6+3

4 x = 12 x =3

45 5 = 9 3

z-intercept: x = 0, y = 0 3z = 12 z = 4

The plane x + y + z = 4 intersects the axes at (4, 0, 0), (0, 4, 0), and (0, 0, 4).

25. 22.

x + 2 y + 6z = 6

x =3

The plane is parallel to the yz-plane. It intersects the x-axis at (3, 0, 0).

x-intercept: y = 0, z = 0 x =6

y-intercept: x = 0, z = 0 2y = 6 y =3

732 26.

Chapter 9 MULTIVARIABLE CALCULUS y = 4

29.

No x-intercept, no z-intercept The graph is a plane parallel to the xz-plane.

f ( x, y ) = 6 x 2 y 3 - 4 y f x ( x, y ) = 12 xy 3 f y ( x, y ) = 18 x 2 y 2 - 4

30.

f ( x, y) = 5 x 4 y 3 - 6 x5 y f x ( x, y) = 20 x3 y3 - 30 x 4 y f y ( x, y) = 15x 4 y 2 - 6 x5

27.

z = f ( x, y) = 3x3 + 4 x 2 y - 2 y 2

(a) (b)

31.

¶z = 9 x 2 + 8xy ¶x

f x ( x, y) =

¶z = 4x2 - 4 y ¶y æ ¶z ö÷ çç ÷ (-1, 4) = 4(-1) 2 - 4(4) = -12 çè ¶y ø÷÷

4x2 + y2 1 (4 x 2 + y 2 )-1/2 (8 x) 2 4x

(4 x 2 + y 2 )-1/2 1 f y ( x, y) = (4 x 2 + y 2 )-1/2 (2 y) 2 y = 2 (4 x + y 2 )1/2

f xy ( x, y) = 8x

(c)

f ( x, y) =

f xy ( x, y)(2, -1) = 8(2) = 16

28.

z = f ( x, y) =

32.

x + y2 x - y2 2

(a)

4 xy 2 2

(x - y ) 2

(b)

f x ( x, y) =

( x - y ) ⋅ 2 y - ( x + y )(-2 y) ¶z = ¶y ( x - y 2 )2 =

f y ( x, y) =

¶z (x - y ) ⋅ 1 - (x + y ) ⋅ 1 = ¶x ( x - y 2 )2 =

-2 y

2 2 2

2x + 5 y 2 3x 2 + y 2 (3x 2 + y 2 ) ⋅ 2 - (2 x + 5 y 2 ) ⋅ 6 x (3x 2 + y 2 )2 2 y 2 - 6 x 2 - 30 xy 2 (3x 2 + y 2 ) 2 (3x 2 + y 2 ) ⋅ 10 y - (2 x + 5 y 2 ) ⋅ 2 y (3x 2 + y 2 ) 2 30 x 2 y - 4 xy (3x 2 + y 2 ) 2

(x - y )

= -2 y 2 ( x - y 2 )-2

33.

f xx ( x, y) = 4 y 2 ( x - y 2 )-3 =

4 y2 2 3

(x - y )

f xx (-1, 0) =

0 =0 1

f ( x, y) = x3e3 y f x ( x, y) = 3x 2e3 y

æ ¶z ö 1 -8 =ççç ÷÷÷ (0, 2) = 2 è ¶x ø 2 (-4) (c)

f ( x, y) =

f y ( x, y) = 3x3e3 y

34.

f ( x, y) = ( y - 2)2 e x + 2 y f x ( x, y) = ( y - 2)2 e x + 2 y f y ( x, y) = e x + 2 y ⋅ 2( y - 2) + ( y - 2)2 ⋅ 2e x + 2 y = 2( y - 2)[1 + ( y - 2)]e x + 2 y = 2( y - 2)( y - 1)e x + 2 y

Chapter 9 Review 35.

36.

40.

f ( x, y) = ln |2 x 2 + y 2 | 1 f x ( x, y) = ⋅ 4x 2 2x + y 2 4x = 2 2x + y 2 1 f y ( x, y) = ⋅ 2y 2 2x + y 2 2y = 2 2x + y 2

f y ( x, y) = =

2 3

2-x y

= f xy ( x, y) =

41.

( x - 1)2 -3 - y ( x - 1)2

= (-3 - y)( x - 1)-2

2(3 + y) ( x - 1)3 -1 ( x - 1)2

f ( x, y) = 4 x 2e2 y f x ( x, y) = 8 xe2 y f xx ( x, y) = 8e2 y f xy ( x, y) = 16 xe2 y

⋅ (-3x 2 y 2 )

42.

-3 x 2 y 2 2 - x2 y3

f ( x, y) = ye x

f x ( x, y) = 2 xye x

2 2

f ( x, y) = 5 x3 y - 6 xy 2

f xx ( x, y) = 2 xy ⋅ 2 xe x + e x ⋅ 2 y

f x ( x, y) = 15 x 2 y - 6 y 2

= 2 ye x (2 x 2 + 1)

f xy ( x, y) = 2 xe x

f xy ( x, y) = 15 x 2 - 12 y

43.

f ( x, y) = -3x 2 y3 + x3 y f x ( x, y) = -6 xy3 + 3x 2 y f xx ( x, y) = -6 y3 + 6 xy f xy ( x, y) = -18 xy 2 + 3x 2

39.

( x - 1) ⋅ 3 - (3x + y) ⋅ 1

f xx ( x, y) = -2(-3 - y)( x - 1)-3

f xx ( x, y) = 30 xy

38.

3x + y x -1

-2 xy 3 2 - x2 y3 1

f ( x, y) = f x ( x, y) =

f ( x, y) = ln |2 - x 2 y 3 | 1 ⋅ (-2 xy3 ) f x ( x, y) = 2 - x2 y3 =

37.

733

f xx ( x, y ) = =

f xy ( x, y ) = =

f ( x, y) = ln |2 - x 2 y | 1 f x ( x, y) = ⋅ (-2 xy) 2 - x2 y 2 xy = 2 x y-2

f xx ( x, y) =

3x f ( x, y ) = 2x - y (2 x - y ) ⋅ 3 - 3x ⋅ 2 f x ( x, y ) = (2 x - y )2 -3 y = (2 x - y )2

= =

(2 x - y ) ⋅ 0 - (-3 y ) ⋅ 2(2 x - y ) ⋅ 2 (2 x - y )4

12 y (2 x - y )3 é (2 x - y )2 ⋅ (-3) ù ê ú ê ú - (-3 y ) ⋅ 2(2 x - y ) ⋅ (-1) ûú ëê (2 x - y )4

-6 x - 3 y (2 x - y )3

( x 2 y - 2)2 y - 2 xy(2 xy) ( x 2 y - 2)2 2 y [( x 2 y - 2) - 2 x 2 y] ( x 2 y - 2)2 2 y(-x 2 y - 2) ( x 2 y - 2)2 -2 x 2 y 2 - 4 y (2 - x 2 y)2

734

Chapter 9 MULTIVARIABLE CALCULUS f xy ( x, y) =

z xx ( x, y) = 2, z yy ( x, y) = 2, z xy ( z, y) = 0

( x 2 y - 2)2

D = 2(2) - (0) 2 = 4 > 0 and z xx ( x, y ) > 0.

2 x[( x 2 y - 2) - x 2 y]

( x 2 y - 2)2 2 x(-2)

(

Relative minimum at - 92 , 4 47.

( x 2 y - 2)2 -4 x

44.

2 x( x 2 y - 2) - x 2 (2 xy)

f ( x, y ) = x 2 + 3xy - 7 x + 5 y 2 - 16 y f x ( x, y) = 2 x + 3 y - 7 f y ( x, y) = 3x + 10 y - 16

(2 - x 2 y)2

Solve the system f x ( x, y) = 0, f y ( x, y) = 0.

f ( x, y) = ln |1 + 3xy | f x ( x, y) = =

1 1 + 3xy 3y

2x + 3 y - 7 = 0 3x + 10 y - 16 = 0

⋅ 3y2

-6 x - 9 y + 21 = 0

1 + 3xy

6 x + 20 y - 32 = 0 11y - 11 = 0

y =1

= 3 y 2 (1 + 3xy 2 )-1 f xx ( x, y) = 3 y 2 ⋅ (-3 y 2 )(1 + 3xy 2 )-2 = f xy ( x, y) = =

45.

-9 y

2 x + 3(1) - 7 = 0 2x = 4

x = 2

(1 + 3xy 2 ) 2 (1 + 3xy 2 ) ⋅ 6 y - 3 y 2 (6 xy)

Therefore, (2, 1) is a critical point.

2 2

(1 + 3xy )

f xx ( x, y) = 2

f yy ( x, y) = 10

(1 + 3xy 2 ) 2

f xy ( x, y) = 3 D = 2 ⋅ 10 - 32 = 11 > 0

z = 2 x 2 - 3 y 2 + 12 y

Since f xx = 2 > 0, there is a relative minimum at (2,1).

z x ( x, y) = 4 x z y ( x, y) = -6 y + 12

If z x ( x, y) = 0, x = 0. If z y ( x, y) = 0, y = 2. Therefore, (0, 2) is a critical point.

48.

z = x3 - 8 y 2 + 6 xy + 4 z x ( x, y) = 3x 2 + 6 y, z y ( x, y) = -16 y + 6 x

z xx ( x, y) = 4

3x 2 + 6 y = 0

z yy ( x, y) = -6

x2 + 2 y = 0

z xy ( x, y) = 0

y =-

D = 4(-6) - 02 = -24 < 0

z = x2 + y 2 + 9x - 8 y + 1 z x ( x, y) = 2 x + 9, z y ( x, y) = 2 y - 8

-8 y + 3 x = 0

Substituting, we have

2x + 9 = 0 x =-

x2 2

-16 y + 6 x = 0

There is a saddle point at (0, 2). 46.

)

9 2

æ x 2 ö÷ ç -8çç - ÷÷÷ + 3x = 0 çè 2 ø÷ 4 x 2 + 3x = 0

2y - 8 = 0

x(4 x + 3) = 0

y = 4

Chapter 9 Review

735 4 x + 4(-2) - 3 = 0

3 4 9 y = 0 or y = - . 32 x = 0 or x = -

4 x = 11 x =

z xx ( x, y ) = 6 x, z yy ( x, y) = -16, z xy ( x, y) = 6

Therefore,

D = 6 x(-16) - (6) = -96 x - 36

f yy ( x, y) = 8 f xy ( x, y) = 4

)

9 , D = 36 > 0 and z ( x, y ) At - 34 , - 32 xx

D = 4 ⋅ 8 - 42 = 16 > 0

= - 92 < 0.

Relative maximum at 49.

z =

(

9 - 34 , - 32

( 114 , -2 ) is a critical point.

f xx ( x, y) = 4

At (0,0), D = -36 < 0. Saddle point at (0,0).

(

Since f xx = 4 > 0, there is a relative minimum

)

1 2 1 x + y 2 + 2 xy - 5 x - 7 y + 10 2 2

at 51.

( 114 , -2 ).

z = x3 + y 2 + 2 xy - 4 x - 3 y - 2 z x ( x, y) = 3x 2 + 2 y - 4

z x ( x, y) = x + 2 y - 5

z y ( x, y) = 2 y + 2 x - 3

z y ( x, y ) = y + 2 x - 7

Setting z x = z y = 0 and solving yields

Setting z x ( x, y) = z y ( x, y) = 0 yields

x + 2y = 5

3x 2 + 2 y - 4 = 0 (1)

2x + y = 7

2 y + 2 x - 3 = 0. (2)

Solving for 2y in equation (2) gives 2 y = -2 x + 3.

-2 x - 4 y = -10 2x + y = 7 - 3y = - 3

Substitute into equation (1). 3x 2 + (-2 x) + 3 - 4 = 0

y = 1, x = 3.

3x 2 - 2 x - 1 = 0

z xx ( x, y ) = 1, z yy ( x, y ) = 1, z xy ( x, y) = 2

(3x + 1)( x - 1) = 0

For (3, 1), D = 1 ⋅ 1 - 4 = -3 < 0.

1 or x = 1 3 11 1 y = or y = 6 2 x =-

Therefore, z has a saddle point at (3, 1). 50.

11 4

f ( x, y) = 2 x 2 + 4 xy + 4 y 2 - 3x + 5 y - 15 f x ( x, y) = 4 x + 4 y - 3

z xx ( x, y) = 6 x, z yy ( x, y) = 2, z xy ( x, y) = 2

f y ( x, y) = 4 x + 8 y + 5

, For - 13 , 11 6

(

æ ö D = 6 çç - 1 ÷÷÷ (2) - 4 è 3ø

Solve the system f x ( x, y) = 0, f y ( x, y) = 0. 4x + 4 y - 3 = 0 4x + 8 y + 5 = 0

-4 x - 4 y + 3 = 0 4x + 8 y + 5 = 0 4y + 8 = 0 y = -2

)

= -4 - 4 = -8 < 0,

(

)

so z has a saddle point at - 13 , 11 . 6 D = 6(1)(2) - 4 = 8 > 0.

(

)

z xx 1, 12 = 6 > 0, so z has a relative minimum

(

)

at 1, 12 .

736

Chapter 9 MULTIVARIABLE CALCULUS f ( x, y ) = 7 x 2 + y 2 - 3x + 6 y - 5xy

52.

55.

f x ( x, y ) = 14 x - 3 - 5 y f y ( x, y ) = 2 y + 6 - 5 x 14 x - 5 y - 3 = 0 -5 x + 2 y + 6 = 0

f ( x, y) = x 2 + y 2 ; x = y - 6.

g ( x, y) = x - y + 6

F ( x, y,  ) = x 2 + y 2 -  ( x - y + 6)

Fx ( x, y,  ) = 2 x -  Fy ( x, y,  ) = 2 y + 

28x - 10 y - 6 = 0

F ( x, y,  ) = -( x - y + 6)

-25 x + 10 y + 30 = 0 3x + 24 = 0

2 x -  = 0 (1) 2 y +  = 0 ( 2) x - y + 6 = 0 (3)

Equations (1) and (2) give  = 2 x, and  = -2 y. Thus, 2 x = -2 y x = - y.

x = -8 -5(-8) + 2 y + 6 = 0 2 y = -46 y = -23 f xx ( x, y) = 14, f yy ( x, y) = 2, f xy ( x, y) = -5 D = 14(2) - (-5)2 = 3 > 0 and f xx ( x, y) > 0.

Substituting into equation (3),

Relative minimum at (–8,–23) 54.

(- y) - y + 6 = 0 y = 3.

f ( x, y) = x y; x + y = 4

x = -3.

g ( x) = x + y - 4

F ( x, y,  ) = x 2 y -  ( x + y - 4)

Fx ( x, y,  ) = 2 xy - 

f (-3,3) = (-3)2 + (3)2 = 18.

And

Fxx ( x, y,  ) = 2 Fyy ( x, y,  ) = 2

Fy ( x, y,  ) = x 2 - 

Fxy ( x, y,  ) = 0

F ( x, y,  ) = -( x + y - 4)

D = 2 ⋅ 2 - 02 = 4 > 0

2 xy -  = 0 (1)

Since Fxx > 0, there is a relative minimum of 18 at (–3, 3).

x 2 -  = 0 (2) x + y - 4 = 0 (3)

56. Let x and y be the numbers such that x + y = 80

 = 2 xy

and f ( x, y) = x 2 y.

 = x2 2 xy = x 2 2 xy - x 2 = 0 x(2 y - x) = 0 x = 0 or 2 y = x

(

and a maximum of

(

8, 4 3 3

F ( x, y,  ) = x 2 y -  ( x + y - 80)

Fx ( x, y,  ) = 2 xy -  F ( x, y,  ) = -( x + y - 80)

2 xy -  = 0 (1) x 2 -  = 0 (2)

)

The critical points are (0, 4) and 83 , 43 . f (0, 4) = 0 æ ö f çç 8 , 4 ÷÷÷ = 64 ⋅ 4 è3 3ø 9 3 256 = 27 Therefore, f has a minimum of 0 at (0,4) 256 at 27

g ( x) = x + y - 80

Fy ( x, y,  ) = x 2 - 

Substituting into equation (3) gives y = 4 or y = 4 . 3 If y = 43 , x = 83 .

x + y - 80 = 0 (3)

 = 2 xy  = x2 2 xy = x 2

2 xy - x 2 = 0 x(2 y - x) = 0 x = 0 or x = 2 y

Chapter 9 Review

737

Substituting into equation (3) gives y = 80 or 2 y + y - 80 = 0

58.

maximizing x 2 y with x + y = 80. If x < 0, then y = 80 - x > 80. The values of y and

3 y = 80

and

y = 80 3 160 x = . 3

x 2 y will increase as the value of | x | increases, so there is no maximum. A similar situation

occurs by xy 2 when x + y = 50, by taking y < 0 and considering what happens to the

f (0,80) = 0 ⋅ 802 = 0 æ ö (160)2 (80) f çç 160 , 80 ÷÷÷ = è 3 3 ø 9 3 =

values of x and xy 2 as | y | increases.

2, 048, 000 > f (0,80) 27

f has a maximum at

59.

( 1603 , 803 ).

Therefore, if x = 160 3

f y ( x, y) = -14 y + 4 x dz = (12 x + 4 y) dx + (-14 y + 4 x) dy = [12(3) + 4(-1)](0.03) + [-14(-1) + 4(3)](0.01) = 0.96 + 0.26 = 1.22

Maximize f ( x, y) = xy 2 , subject to x + y = 75. 1.

g ( x, y) = x + y - 75

F ( x, y,  ) = xy 2 -  ( x + y - 75)

Fx ( x, y,  ) = y 2 -  Fy ( x, y,  ) = 2 xy -  F ( x, y,  ) = -( x + y - 75) y 2 -  = 0 (1)

60.

x + 5y x - 2y x = 1, y = -2, dx = -0.04, dy = 0.02 z = ( x, y) =

f x ( x, y) = =

2 xy -  = 0 (2) x + y - 75 = 0 (3)

z = f ( x, y) = 6 x 2 - 7 y 2 + 4 xy x = 3, y = -1, dx = 0.03, dy = 0.01 f x ( x, y) = 12 x + 4 y

and y = 80 , then x 2 y 3

is maximized. 57.

No, a maximum does not exist without the requirement that x and y are positive. Consider

Equations (1) and (2) give  = y 2 and  = 2 xy. Thus,

f y ( x, y) = =

y 2 = 2 xy y ( y - 2 x) = 0

dz =

y = 0 or y = 2 x

Substituting y = 0 into equation (3), x + (0) - 75 = 0 x = 75. Substituting y = 2x into equation (3), x + (2 x) - 75 = 0 x = 25. So y = 2 x = 50.

( x - 2 y)(1) - ( x + 5 y)(1) ( x - 2 y) 2 -7 y ( x - 2 y) 2 ( x - 2 y)(5) - ( x + 5 y)(-2) ( x - 2 y) 2 7x ( x - 2 y) 2 -7 y 2

( x - 2 y) -7(-2)

dx +

7x ( x - 2 y)2

(-0.04) [1 - 2(-2)]2 7(1) + (0.02) [1 - 2(-2)]2 = -0.0224 + 0.0056 = -0.0168

Thus, f (75, 0) = 75(0) 2 = 0, and f (25,50) = 25(50)2 = 62,500.

Since f (25,50) > f (75, 0), x = 25 and y = 50 will maximize f ( x, y) = xy 2.

738 61.

Chapter 9 MULTIVARIABLE CALCULUS f (4.06, 0.04) = f (4, 0) + z

x2 + y 2 .

Let z = f ( x, y) =

» f (4, 0) + dz

Then

= 4e0 + 0.095 f (4.06, 0.04) » 2.095

dz = f x ( x, y) dx + f y ( x, y) dy. 1 2 ( x + y 2 )-1 2 (2 x) dx 2 1 + ( x 2 + y 2 )(2 y) dx 2 x y = dx + dy 2 2 2 x + y x + y2

dz =

Using a calculator, 4.06e0.04 » 2.0972. The absolute value of the difference of the two results is |2.095 - 2.0972| = 0.0022. 63.

= (4 y - 3)(2 x )

To approximate 5.1 + 12.05 , we let x = 5, dx = 0.1, y = 12, and dy = 0.05. 5 2

(0.1) +

5 + 12 5 + 122 5 12 = (0.1) + (0.05) » 0.0846. 13 13

(0.05)

64.

ò e

65.

» f (5,12) + dz

Let z = f ( x, y) =

1 3x +5 y 5 e 1 3 1 = (e15+5 y - e3+5 y ) 3

dx =

6x 2

4x + 2 y 2

When x = 0, u = 2 y 2. When x = 5, u = 100 + 2 y 2.

5.12 + 12.052 » 13.0848 .

The absolute value of the difference of the two results is |13.0846 - 13.0848| = 0.0002. 62.

Let u = 4 x 2 + 2 y 2 ; then du = 8x dx.

= 52 + 122 + 0.0846 f (5.1,12.05) » 13.0846

Using a calculator,

3x +5 y

Therefore, f (5.1,12.05) = f (5,12) + z

= (4 y - 3)(2 ⋅ 2 - 2 ⋅ 1) = 8y - 6

Then, dz =

4y - 3 dx x

3 = 4

xe y .

100 + 2 y 2

2 y2

u-1/2du 2

3 1/2 100 + 2 y (2u ) 4 2 y2 3 = ⋅ 2[(100 + 2 y 2 )1/2 - (2 y 2 )1/2 ] 4 3 = [(100 + 2 y 2 )1/2 - (2 y 2 )1/2 ] 2 =

Then dz = f x ( x, y) dx + f y ( x, y) dy

1 -1/2 y x e dx + x1/2e y dy 2

ey dx + 2 x

xe y dy.

66.

To approximate 4.06e0.04 , let x = 4, dx = 0.06, y = 0, and dy = 0.04. Therefore, e0 (0.06) + 4e0 (0.04) 2 4 1 dz = (0.06) + 2(0.04) 4 dz = 0.095.

ò y (7 x + 11y ) 2

3 -1/2

3 2 (7 x + 11y 3 )1/2 1 33 2 [(7 x + 297)1/2 - (7 x + 11)1/2 ] = 33

dz =

Chapter 9 Review 67.

2é

739

ù ( x 2 y 2 + 5 x) dx úú dy 0 úû

2 é

ò ò ê ê 0 êë

2æ

æ 64 y 3 ÷ö ç = çç + 40 y ÷÷÷ çè 9 ÷ø 0

1 10-7 y (e - e 6-7 y ) 14 1

1 -4 (e - e-8 - e3 + e-1) 14

64 (8) + 40(2) 9 512 720 = + 9 9 1232 = 9 =

68.

2æ

ö çç 64 y 2 + 40 ÷÷ dy ÷ø ç è 3 0

ò1 ò

ö çç 1 x3 y 2 + 5 x 2 ÷÷ dx ç 2 ÷ø 0 0 è3

ù (e2 x-7 y ) dx úú dy 3 úû 2 1 10-7 y (e - e6-7 y ) dy 2 1

ê ê êë

71.

e3 + e-8 - e-4 - e-1 14 4

dxdy = y 2

ò2 ò

ò ò (2x + 6 y + y )dydx 2

ù ö ê çç10 x + 75 + 125 ÷÷ - 0 ú dx êç ú 3 ø÷ 0 ëè û 3 æ ö çç10 x + 350 ÷÷ dx ÷ ç 3 ø 0 è

72.

ò1 ò

ò ò

ê ê 3 ëê 2

= =

3/2

73.

- (12 + 3 y )3/2 ] dy

4 1 1 2 = ⋅ ⋅ ⋅ [(30 + 3 y )5/2 - (12 + 3 y )5/2 ] 3 3 9 5 2 5/2 5/2 5/2 5/2 = [(42) - (24) - (39) + (21) ] 135

70.

ò e

2 x -7 y

ò1 ln 2 dy

( x 2 + 2 y 2 )dx dy; 0 £ x £ 5, 0 £ y £ 2 òò R ( x 2 + 2 y 2 ) dx dy òò R = = =

1 2 x -7 y e 2 3

1 10-7 y - e6- 7 y e 2

ò0 ò0 (x2 + 2 y 2 ) dx dy 2 æ

ö çç 1 x3 + 2 xy 2 ÷÷ dy ÷ø ç 0 è3 0

ù ö ê çç 125 + 10 y 2 ÷÷ - 0 ú dy ÷ êç ú ø 0 ëè 3 û

2 éæ

(

ò1 ln x 1 dy

= ln 2

ò 9 [(30 + 3 y)

= 2 ln 2 - ln 2

1 (6 x + 3 y )3/2 dx 9 3 2 4

ù 6 x + 3 y dx úú dy ûú

= y ln 2 1

= (45 + 350) - 0 = 395

69.

ò2 êêë y (4 - 2) úúû dy

dxdy = x 1

æ 350 ÷ö = çç 5 x 2 + x÷ çè 3 ÷ø 0 4é

4 é

3 éæ

4 = 2 ln 2 = 2 ln 2 or ln 4

3 æ

ö çç 1 x ÷÷ dy ÷ ç 2 è y ø÷

= 2 ln y

ö çç 2 xy + 3 y 2 + 1 y 3 ÷÷ dx ç 3 ÷ø 0 0 è

4æ

740

Chapter 9 MULTIVARIABLE CALCULUS 4 (1024 - 32(4 2) - 243 + 1) 15 4 = (782 - 128 2) 15 4 = (782 - 85/2 ) 15

2 æ

ö çç 125 + 10 y 2 ÷÷ dy ç è ø÷ 3 0

æ 125 10 3 ö÷ y+ y ÷÷ = çç çè 3 ø0 3 250 80 + = 110 3 3

74.

2 x + y dx dy; 1 £ x £ 3, 2 £ y £ 5 òò R

76.

òò 2x + y dx dy R

1/2

2 (2 x + y)3/2 dx 1 3 2

2 [(2 x + 5)3/2 - (2 x + 2)3/2 ] dx 2 3

2 [(2 x + 5)5/2 - (2 x + 2)5/2 ] 15 1

2 (115/2 - 85/2 - 75/2 + 45/2 ) 15 2 = (115/2 - 85/2 - 75/2 + 32) 15

77.

òò y + x dx dy; 0 £ x £ 7, 1 £ y £ 9 R

òò y + x dx dy =

ò ò 0

= =

1 y2 + x e dx 0 2 0

1 2 [e - e - e + 1] 2

e2 - 2e + 1 2

z = x + 8 y + 4; 0 £ x £ 3, 1 £ y £ 2 V =

ò0 ò1 (x + 8 y + 4) dy dx 3

ò0

( xy + 4 y 2 + 4 y)

2 2 = ⋅ [(9 + x)5/2 - (1 + x)5/2 ] 3 5 0

2 1

ò0 [(2x + 16 + 8) - ( x + 4 + 4)] dx 3

ò0 (x + 16) dx 3

2 [(9 + x)3/2 - (1 + x)3/2 ] dx 3 0

ù 2 ê ( y + x)3/2 ú dx úû 0 êë 3 1

ò0 ò0 ye y xdy dx

1 1+ x [e - ex ] 2 0

7é

y + x dy dx

ò0 2 [e1 x - e x ] dx

75.

òò R

ye y + x dx dy

ò ò (2x + y) dydx 1

òò R

ye y + x dx dy; 0 £ x £ 1, 0 £ y £ 1

æ1 ö = çç x 2 + 16 x ÷÷÷ çè 2 ø0 æ9 ö 105 = çç + 48 ÷÷÷ - 0 = çè 2 ø 2

4 [(16)5/2 - (8)5/2 - (9)5/2 + (1)5/2 ] 15 4 5 [4 - (2 2)5 - 35 + 1] = 15 =

Chapter 9 Review 78.

741

z = x 2 + y 2 ; 3 £ x £ 5, 2 £ y £ 4

81.

( x 2 + y 2 ) dy dx òò R

V =

ò3 ò2

5 é

ò3

ò 2 (x - x ) dx 2

1 2

64 8ù ê 4x2 + - 2 x 2 - ú dx êë 3 3 úû

ò ( x - x ) dx 5

1 æç x6 x8 ö÷÷ = çç ÷ 2 çè 6 8 ÷÷ø

250 280 + - 18 - 56 3 3 308 = 3

1 3

5 é

5æ

1æ 3

æ 3 ö ö çç 2 x 2 + 56 ÷÷ dx = çç 2 x + 56 x ÷÷÷ ç ç 3 ø÷ 3 ÷ø÷ 3 è èç 3 3

ò òx x y dy dx ò

ò3 êêëê x y + 3 úúûú 2 dx

ö çç x 2 ÷÷ y ÷÷ dx çç 0 è 2 ø÷ 2

( x 2 + y 2 ) dy dx 3ù

1æ1 1ö 1 1 1 = çç - ÷÷÷ = ⋅ = ç 2è 6 8ø 2 24 48

79.

òò 0

82.

xy dy dx

= =

2ö çç xy ÷÷ ÷÷ ç 0 çè 2 ø÷

1æ

dx 0

ò0 2x3dx

83.

2x2

ò (2x - 2) dx 4

òò

1 y2 + 1

dx dy

ò y + 1 dy 0

ò òx y + 1 dy dx =

1 2 y dx 2 2

Change the order of integration. 0

y dy dx

ò òx y + 1 dy dx 0

æ1 ö 1 = çç x 4 ÷÷÷ = çè 2 ø 2 0

ò1 ò2

1æ1 1ö 1 = çç - ÷÷÷ = ç 2è 2 3ø 12

ò0 2 (4x2 - 0) dx

ò 2 ( y - y ) dy

1 æç y 2 y 3 ö÷÷ = çç ÷ 2 çè 2 3 ÷÷ø

dy y

80.

x 0 2

ò ò y x dx dy = ò

1 2

ù 1 ê 1 (2 y) ú dy (0) ê 2 ú y 2 + 1 úû 0 êë y + 1

1é

æ2 ö = çç x5 - 2 x ÷÷÷ çè 5 ø1

æ 64 ö æ2 ö = çç - 4 ÷÷ - çç - 2 ÷÷ ÷ø èç 5 ø÷ èç 5

= ln ( y 2 + 1)

52 = 5

= ln 2 - ln 1 = ln 2 - 0 = ln 2

ò y + 1 dy 0

1 0

742 84.

Chapter 9 MULTIVARIABLE CALCULUS 8

x /2

ò ò

86.

y 2 + 4 dy dx

òò (2 - x - y ) dy dx; 0 £ x £ 1, 2

Change the order of integration. 8

x /2 4

ò ò =

ò ò

ò ò (2 - x - y ) dy dx =

ò ( y + 4) x 4

y 2 + 4 dx dy

1/2

x2 £ y £ x

y 2 + 4 dy dx

2y 0

dy =

[( y 2 + 4)1/2 (2 y)] - ( y 2 + 4)(0)] dy

ò ( y + 4) (2 y) dy 2

1/2

( y + 4)

3/2

3 2

òò R

ò0 ò y = = = =

87.

(a)

x x 6 ö÷÷ çç 3 2 4 çç 2 x - x - 3 - 2 x + x + 3 ÷÷ dx 0 è ø÷

1æ

x6 ö÷÷ 4 x3 ç 2 4 ççç 2 x - 2 x - 3 + x + 3 ÷÷ dx 0 è ø÷

1æ

2 1 1 1 26 - + + = 3 3 5 21 105 x

C (10,5) = 4(10)2 + 5(5) 2 - 4(10)(5) + 10 = 400 + 125 - 200 + 10 = 325 + 10 » 328.16

The cost is about $328.16. (b) C (15,10) = 4(15) 2 + 5(10)2 - 4(15)(10) + 15 = 800 + 15 » 803.87

(2 x + 3 y) dx dy 2

2- y

The cost is about $803.87. (c)

= 1600 + 2000 - 1600 + = 2000 +

88.

ò0 (4 + 2 y - 6 y2 ) dy

= (4 y + y 2 - 2 y 3 ) = 4 +1- 2 = 3

20 » 2004.47

The cost is about $2004.47.

(4 - 4 y + y 2 - y 2 + 6 y - 6 y 2 ) dy

C (20, 20) = 4(20)2 + 5(20)2 - 4(20)(20) +

[(2 - y) 2 - y 2 + 3 y(2 - y - y)] dy

ò0

C ( x, y) = 4 x 2 + 5 y 2 - 4 xy +

y 3 ö÷÷ çç 2 2 y x y ÷ dx ç 3 ÷÷ø 2 0 çè

= 900 + 500 - 600 + 15

ò0 (x + 3xy) y ò0

1æ

= 1-

(2 x + 3 y) dx dy; 0 £ y £ 1,

2- y

y £ x £ 2- y 1

æ x4 x5 x7 ö÷÷ 2 x3 ç = çç x 2 + + ÷ çè 3 3 5 21 ø÷÷ 0

2 2 = (42 + 4)3/2 - (02 + 4)3/2 3 3 2 = (203/2 - 43/2 ) 3 2 = (20 20 - 8) 3 2 = (40 5 - 8) 3 16 = (5 5 - 1) 3

85.

c( x, y) = 2 x + y 2 + 4 xy + 25

(a)

cx = 2 + 4 y cx (640, 6) = 2 + 4(6) = 26

For an additional 1 MB of memory, the approximate change in cost is $26.

Chapter 9 Review (b)

743

c y = 2 y + 4x

91.

c y (640, 6) = 2(6) + 4(640) = 2572

For an additional hour of labor, the approximate change in cost is $2572. 89.

z = x0.7 y 0.3

g ( x, y) = 2 x + 4 y - 80

F ( x, y) = xy 3 -  (2 x + 4 y - 80)

Fx ( x, y,  ) = y 3 - 2 Fy ( x, y,  ) = 3xy 2 - 4

(a) The marginal productivity of labor is ¶z 0.7 y 0.3 = 0.7 x0.7-1 y 0.3 = . ¶x x 0.3

F ( x, y,  ) = -2 x - 4 y + 80

3xy 2 - 4 = 0 -2 x - 4 y + 80 = 0

(a) Minimize c( x, y) 2

y 3 - 2 = 0

(b) The marginal productivity of capital is ¶z 0.3x0.7 = 0.3x0.7 y 0.3-1 = . ¶y y 0.7 90.

Maximize f ( x, y) = xy 3 subject to 2 x + 4 y = 80.

Use the first and second equations to express 4 and eliminate 4. 2 y3 = 4

= x + 5 y + 4 xy - 70 x - 164 y + 1800

3xy 2 = 4 2 y3 = 3xy 2

cx = 2 x + 4 y - 70

If y equals 0, the utility will have a minimum value of 0. So we can assume

c y = 10 y + 4 x - 164

y ¹ 0 and divide by y 2 to find that 2 y = 3x. Substitute 6x for 4y in the last equation. -2 x - 6 x + 80 = 0 3 x = 10, y = x = 15. 2 f (10,15) = 33, 750.

2 x + 4 y - 70 = 0 4 x + 10 y - 164 = 0 -4 x - 8 y + 140 = 0 4 x + 10 y - 164 = 0 2y -

24 = 0 y = 12

4 x + 10(12) - 164 = 0 4 x = 44 x = 11

Extremum at (11,12)

The maximum utility is 33,750, obtained by purchasing 10 units of x and 15 units of y. 92.

Maximize f ( x, y) = x5 y 2 subject to 10 x + 6 y = 42 or 5 x + 3 y = 21.

cxx = 2, c yy = 10, cxy = 4

g ( x, y) = 5 x + 3 y - 21

For (11, 12), D = (2)(10) - 16 = 4 > 0 and cxx (11, 12) = 2 > 0.

F ( x, y) = x5 y 2 -  (5x + 3 y - 21)

Fx ( x, y,  ) = 5x 4 y 2 - 5 Fy ( x, y,  ) = 2 x5 y - 3

There is a relative minimum at (11, 12). (b)

F ( x, y,  ) = -5 x - 3 y + 21

c(11,12) = (11)2 + 5(12)2 + 4(11)(12) - 70(11) - 164(12) + 1800 = 121 + 720 + 528 - 770 - 1968 + 1800 = $431

5x 4 y 2 - 5 = 0 2 x5 y - 3 = 0 -5x - 3 y + 21 = 0

744

Chapter 9 MULTIVARIABLE CALCULUS 5.

Solve the first two equations for  and eliminate .

94.

x4 y 2 =  2 5 x y = 3 2 x 4 y 2 = x5 y 3

1 2 r h 3 r = 2 cm, h = 8 cm dr = 0.21 cm, dh = 0.21 cm V =

dV = =

If either x or y equals 0, the utility will have a minimum value of 0. So we can assume

xy ¹ 0 and divide by x y to find that y = (2/3) x. Substitute for y in the last equation. æ2 ö -5x - 3çç x ÷÷÷ + 21 = 0 çè 3 ø 7 x = 21 2 (3) = 2. 3 f (3, 2) = 972.

95.

96.

æ 200 x y xy /20 ö÷÷ dC = ççç 2 e + ÷ dx çè x + y 20 ø÷÷ æ 100 x xy /20 ö÷÷ e + ççç 2 + ÷ dy çè x + y 20 ø÷÷

Costs decrease by $243.82.

 3

[2(2)(8)(0.21) + 4(0.21)] (6.72 + 0.84) (7.56)

4 3  r , r = 2 ft, 3 1 ft dr = 1 in = 12

1 2 r h 3 r = 2.9 cm, h = 11.4 cm dr = dh = 0.2 cm V =

dV = = = =

æ 100 15 (9)(15)/20 ö÷ ÷÷ (-1) e + ççç 2 + 20 è 15 + 9 ø÷ 1450 3 27/4 e 117 10 = -243.82

V =

dC (15,9) æ 200(15) 9 (15)(9)/20 ö÷ ÷÷ (1) = ççç 2 + e 20 è 15 + 9 ø÷



(2rh dr + r 2dh)

æ 1ö dV = 4 r 2dr = 4 (2) 2 çç ÷÷÷ » 4.19 ft 3 çè 12 ø

The maximum utility is 927, obtained by purchasing 3 units of x and 2 units of y. C ( x, y) = 100 ln ( x 2 + y) + e xy /20 x = 15, y = 9, dx = 1, dy = -1

» 7.92 cm3

x = 3, y =

93.



 3



 3



(2rh dr + r 2dh) [2(2.9)(11.4)(0.2) + (2.9)2 (0.2)] (13.224 + 1.682) (14.906)

» 15.6 cm3

97.

P( x, y) = 0.01(-x 2 + 3xy + 160 x - 5 y 2 + 200 y + 2600) with x + y = 280.

(a)

y = 280 - x

P( x) = 0.01[-x 2 + 3x(280 - x) + 160 x - 5(280 - x)2 + 200(280 - x) + 2600] = 0.01(-x 2 + 840 x - 3x 2 + 160 x - 392, 000 + 2800 x - 5 x 2 + 56, 000 - 200 x + 2600)

Chapter 9 Review

745

P( x) = 0.01(-9 x 2 + 3600 x - 333, 400) P¢( x) = 0.01(-18x + 3600) 0.01(-18 x + 3600) = 0 - 18x = -3600 x = 200

If x < 200, P¢( x) > 0, and if x > 200, P′( x) < 0. Therefore, P is maximum when x = 200. If x = 200, y = 80. P(200,80)

P(200,80) = 266, as in part (a) Thus, $200 spent on fertilizer and $80 spent on seed will produce a maximum profit of $ 266 per acre.

(c)

Maximize P( x, y) = 0.01(-x 2 + 3xy + 160 x - 5 y 2 + 200 y + 2600)

subject to x + y = 280. 1.

g ( x, y) = x + y - 280

F ( x, y,  ) = 0.01(-x 2 + 3xy + 160 x - 5 y 2

= 0.01[-200 + 3(200)(80) + 160(200) - 5(80)2 + 200(80) + 2600] = 0.01(26, 600) = 266

+ 200 y + 2600) -  ( x + y - 280)

Fy = 0.01(3x - 10 y + 200) - 

Thus, $200 spent on fertilizer and $80 spent on seed will produce a maximum profit of $266 per acre. (b)

P( x, y) = 0.01(-x 2 + 3xy + 160 x - 5 y 2 + 200 y + 2600) Px ( x, y) = 0.01(-2 x + 3 y + 160) Py ( x, y) = 0.01(3x - 10 y + 200) 0.01(-2 x + 3 y + 160) = 0 0.01(3x - 10 y + 200) = 0

These equations simplify to -2 x + 3 y = -160 3x - 10 y = -200. Solve this system. -6 x + 9 y = -480 6 x - 20 y = -400 - 11y = -880 y = 80

Fx = 0.01(-2 x + 3 y + 160) - 

F = -( x + y - 280)

0.01(-2 x + 3 y + 160) -  = 0 (1) 0.01(3x - 10 y + 200) -  = 0 (2) x + y - 280 = 0 (3)

Equations (1) and (2) give 0.01(-2 x + 3 y + 160) = 0.01(3x - 10 y + 200) - 2 x + 3 y + 160 = 3x - 10 y + 200 - 5x + 13 y = 40. Multiplying equation (3) by 5 gives 5x + 5 y - 1400 = 0. - 5 x + 13 y =

5 x + 5 y = 1400 18 y = 1440

If y = 80, 3x - 10(80) = -200 3x = 600 x = 200.

y = 80

If y = 80, 5x + 5(80) = 1400 5x = 1000

Pxx ( x, y) = 0.01(-2) = -0.02

x = 200.

Pyy ( x, y) = 0.01(-10) = -0.1 Pxy ( x, y) = 0

Thus, P(200,80) is a maximum. As before, P(200,80) = 266.

For (200,80), D = (-0.02)(-0.1) - 02 = 0.002 > 0, and Pxx < 0, so there is a relative maximum at (200,80).

Thus, $200 spent on fertilizer and $80 spent on seed will produce a maximum profit of $266 per acre.

746

Chapter 9 MULTIVARIABLE CALCULUS

98. The average production level will be given by the integral 1 P( x, y ) dx dy, 200

òò R

where R is the region defined by 40 £ x £ 50 and 20 £ y £ 40, which has area (10)(20) = 200. Compute the integral as an iterated integral: 1 200

= 2

0.3 0.7

æ çç è

0.7 çç

÷ö dx ÷÷÷ dy ÷ø

0.3

æ 1.3 - 401.3 ÷ö 0.7 çç 50 ÷

çç è

1.3

1.3 ö æ 1.3 ç 50 - 40 ÷÷ = 2 çç ÷÷ 1.3 çè ø÷

V =  r 2h r = 0.7, h = 2.7 dr = dh = .1 dV = 2 rh dr +  r 2dh = 2 (0.7)(2.7)(0.1) +  (0.7) 2 (0.1) » 1.341

The possible error is 1.341 cm3.

ò ò 400x y dx dy 400 = 200

100. Assume that blood vessels are cylindrical.

101. T ( A, W , S ) = -18.37 - 0.09 A + 0.34W + 0.25S (a)

T (65,85,180) = -18.37 - 0.09(65) + 0.34(85) + 0.25(180)

÷÷ dy ø÷

= 49.68

y 0.7 dy

The total body water is 49.68 liters.

1.3 öæ 1.7 1.7 ö æ 1.3 ç 50 - 40 ÷÷çç 40 - 20 ÷÷ = 2 çç ÷÷ç ÷÷ çè 1.3 1.7 ÷ç øè ø÷

= 13, 492.701 » 13, 493 units

99.

The average weekly cost will be given by the integral 1 C ( x, y ) dx dy, 2500

òò R

where R is the region defined the inequalities 100 £ x £ 150 and 50 £ y £ 100, which has area (50)(50) = 2500. Compute the integral as an iterated integral: 1 2500

100

150

100

ò ò ( 0.03x + 6 y + 2xy + 10 ) dx dy 2

First compute the inner integral: 150

ò ( 0.03x + 6 y + 2xy + 10 ) dx 2

100

(

= 0.01x 3 + 6 xy + x 2 y + 10 x

150

) 100

= 24,250 + 12,800 y The outer integral is then: 1 2500

100

( 24, 250 + 12,800 y ) dy

100 1 24, 250 y + 6400 y 2 2500 50

(

)

49, 212,500 = 19,685 2500 The average weekly cost for the two products is $19,685. =

Chapter 9 Review (b)

747

TA ( A, M , S ) = -0.09

The approximate change in total body water if age is increased by 1 yr and mass and height are held constant is -0.09 liter. TM ( A, M , S ) = 0.34

The approximate change in total body water if mass is increased by 1 kg and height are held constant is 0.34 liter. TS ( A, M , S ) = 0.25

The approximate change in total body water if height is increased by 1 cm and age and mass are held constant is 0.25 liter. 102.

L(w, t ) = (0.00082t + 0.0955)e(ln w +10.49)/2.842

(a)

L(450, 4) = [0.00082(4) + 0.0955] e(ln(450) +10.49) / 2.842 » 33.982

The length is about 33.98 cm. (b)

Lw (w, t ) = (0.00082t + 0.0955) e(ln w+10.49)/2.842 ⋅

1 2.842w

Lw (450, 7) » 0.02723

The approximate change in the length of a trout if its mass increases from 450 to 451 g while age is held constant at 7 yr is 0.027 cm. Lt (w, t ) = 0.00082e(ln w +10.49)/2.842 Lt (450, 7) » 0.2821

The approximate change in the length of a trout if its age increases from 7 to 8 yr while mass is held constant at 450 g is 0.28 cm. 103. (a) (b)

f (60,1900) » 50

In 1900, 50% of those born 60 years earlier are still alive. f (70, 2000) » 75 In 2000, 75% of those born 70 years earlier are still alive.

(c)

f x (60,1900) » -1.25

In 1900, the percent of those born 60 years earlier who are still alive was dropping at a rate of 1.25 percent per additional year of life. (d)

f x (70, 2000) » -2

In 2000, the percent of those born 70 years earlier who are still alive was dropping at a rate of 2 percent per additional year of life. 104.

f (a, b) =

(a)

1 b 4a 2 - b 2 4

1 (2) 4(3)2 - 22 4 1 = 32 = 2 2 » 2.828 2 The area of the bottom of the planter is a approximately 2.828 ft2. f (3, 2) =

748

Chapter 9 MULTIVARIABLE CALCULUS (b)

Since x = 0 is not a solution of equation (3), then 4 - x = 0 4  = . x

1 b 4a 2 - b 2 4 1 1 dA = b ⋅ (4a 2 - b 2 )-1/2 (8a)da 4 2 é1 1 + ê b ⋅ (4a 2 - b 2 )-1/2 (-2b) êë 4 2 1 + (4a 2 - b 2 ) 1/2 ùú db û 4 ab dA = da 2 4a - b 2 ö÷ -b 2 1 æç + çç + 4a 2 - b 2 ÷÷÷ db ÷ø 4 ççè 4a 2 - b 2 A=

Substituting into equation (1) gives æ4ö 4 x + 4 y - 2 xy çç ÷÷÷ = 0 çè x ø or

4x + 4 y - 8 y = 0 x = y.

Substituting x = y into equation (3) gives x 2 y - 125 = 0 y 3 = 125 y = 5.

If a = 3, b = 2, da = 0, and db = 0.5, dA =

æ 1 çç -2 2 + çç 4 çç 4(3) 2 - 22 è

ö÷ ÷ 4(3) 2 - 22 ÷÷ (0.5) ÷÷ ø÷

dA » 0.6187.

The approximate effect on the area is an increase of 0.6187 ft2. 105. Let x be the length of each of the square faces of the box and y be the length of the box.

Therefore, x = y = 5. The dimensions are 5 inches by 5 inches by 5 inches. 106. Maximize f ( x, y) = xy, subject to 2 x + y = 400.

g ( x, y) = 2 x + y - 400

F ( x, y,  ) = xy -  (2 x + y - 400)

Fx = y - 2 Fy = x -  F = -(2 x + y - 400)

y - 2 = 0 x- = 0 2 x + y - 400 = 0

 =

Since the volume must be 125, the constraint is 125 = x 2 y. f ( x, y) = 2 x 2 + 4 xy is the surface area of the box.

g ( x) = x 2 y - 125

F ( x, y,  ) = 2 x 2 + 4 xy -  ( x 2 y - 125)

Fx ( x, y,  ) = 4 x + 4 y - 2 xy Fy ( x, y,  ) = 4 x - x 2 F ( x, y,  ) = -( x 2 y - 125)

4 x + 4 y - 2 xy = 0

Substituting into 2 x + y - 400, we have 2 x + 2 x - 400 = 0, so x = 100, y = 200. Dimensions are 100 feet by 200 feet for maximum area of 20,000 ft2.

(1)

4x - x  = 0

( 2)

x 2 y - 125 = 0

(3)

y ,  = x 2 y = x 2 y = 2x

Factoring equation (2) gives x(4 - x ) = 0 x = 0 or 4 - x = 0.

Extended Application

749

Extended Application: Using Multivariable Fitting to Create a Response Surface Design 1.

The general cubic function of two variables has the form

With this more stringent requirement, the allowable regions in the three contour plots do not overlap. The blanching efficiency would need to be allowed to be less that 93%.

The lower area of overlap suggests a processing temperature between 145C and 155C with treatment times from 50 to 70 seconds. This region represents processing nuts at higher temperatures but for a shorter time.

G( x, y) = Ax3 + By3 + Cx 2 y + Dxy 2 + Ex 2 + Fy 2 + Gxy + Hx + Ty + J

which has 10 terms. 2.

The maximum appears to be close to orange = 56, banana = 48.

G( x, y) = -0.00202 x 2 - 0.00163 y 2 + 0.000194 xy + 0.21380 x + 0.14768 y - 2.36204 Gx ( x, y) = -0.00404 x + 0.000194 y + 0.21380 G y ( x, y) = -0.00326 y + 0.000194 x + 0.14768 -0.00404 x + 0.000194 y = -0.21380 -0.00326 y + 0.000194 x = -0.14768

The solution to the system is x » 55.254, y » 48.589. 4.

Gxx ( x, y) = -0.00404 G yy ( x, y) = -0.00326 Gxy ( x, y) = 0.000194 D = (-0.00404)(-0.00326) - (0.000194) 2 » 0.0000131

Since D > 0 and Gxx = -0.00404 < 0, G(55.254, 48.589) is a relative maximum of G( x, y). 5.

The test results include random errors or “noise,” which may explain the 7.2 rating. Also, the function that best fits all of the data points is not necessarily above all of the points.

The two flavor surfaces have saddle points near the middle of the domain. For overall flavor, the maxima are at the edges of the domain, either 400 minutes at about 130C or 20 minutes at about 160C.

Chapter 10

DIFFERENTIAL EQUATIONS 10.1 Solutions of Elementary and Separable Differential Equations Your Turn 1 Find all solutions of

y =

dy = 12 x5 + dx

x + e5 x .

ò (12x + x + e )dx 5

= 2 x6 + 2 x3/2 + 1 e5 x + C 3 5

Your Turn 4 Find the goat population in 5 years if the reserve can support 6000 goats, the growth rate is 15%, and there are currently 1200 goats in the area.

The general solution will be y = N - Me-kt as in Example 5, where N is the maximum population, k is the growth rate constant, and M is a constant to be determined using the initial population. For this problem, N = 6000 and k = 20 = 15% = 0.15.

Solve for M. 1200 = 6000 - Me(-0.15)(0) M = 6000 - 1200 M = 4800

Your Turn 2

dy - 12 x3 = 6 x 2 , dx

Find the particular solution of y(2) = 60.

The model is y = 6000 - 4800e-0.15t .

First solve for dy /dx. dy = 12 x3 + 6 x 2 dx

Now find y(5). y(5) = 1200 - 4800e-0.15(5)

y = 3x 4 + 2 x3 + C Use the initial condition to find the value of C. y (2) = 60

= 6000 - 4800e-0.75 » 3733

The goat population in 5 years will be 3733.

60 = 3(2)4 + 2(2)3 + C 60 = 48 + 16 + C 60 = 64 + C

10.1 Warmup Exercises

-4 = C

The particular solution is y = 3x 4 + 2 x3 - 4. W1. Integrate the sum term by term. Your Turn 3

ò ( e2 + 4x + x1/ 2 + x-1 ) dx x

Find the general solution of

dy 1. = x + 5 dx y

Separate the variables.

1 2x 2 e + 2 x 2 + x 3/ 2 + ln x + C 2 3

y 5 dy = ( x 2 + 1) dx

W2. Use the substitution u = 2 x3, du = 6 x 2dx.

Now integrate. 5

y dy = ( x + 1) dx

ò y dy = ò (x + 1) dx ò y dy = ò x dx + ò 1dx 5

1 y 6 = 1 x3 + x + C 6 3

x 2e2 x dx =

1 u

ò 6 e du

1 u e +C 6 3 1 = e2 x + C 6 =

751

752

Chapter 10 DIFFERENTIAL EQUATIONS

W3. Use the substitution u = x 2 + 1, du = 2 x dx. 4x

10.1 Exercises 1.

ò x + 1 dx = ò u du 2

equation dy/dx = g(x) is y =

= 2ln u + C

(

False. The general solution of the differential

)

= 2ln x 2 + 1 + C

True

False. A differential equation of the form dy/dx = p(x)/q(y) is separable.

False. A solution y(t) to the logistic equation levels off as t goes to infinity.

W4. Use integration by parts with u = 3x, du = 3 dx; dv = e5x dx, v =

1 5x e . 5

ò 3xe dx = ò u dv = uv ò v du 3 3 e dx = xe 5 5ò 5x

dy = -4 x + 6 x 2 dx y =

dy = 4e-3x dx y =

ò 4e-3xdx

-4e-3x +C 3

W5. Use tabular integration.

ò x e- dx 3x

Choose x as the part to be differentiated and

e-3x as the part to be integrated. D

e-3x

ò (-4x + 6x2 ) dx

= -2 x 2 + 2 x 3 + C

3 3 5x e +C = xe5x 5 25

dy =0 dx dy . Solve for dx 4 x3 - 2

dy = 2 x3 dx

y = 2

1 - e-3x 3

1 -3 x e 9

+ 0

ò x dx 3

æ x 4 ö÷ ç = 2çç ÷÷÷ + C çè 4 ø÷

ò g (x) dx.

1 -3 x e 27

1 2 2 -3 x - x 2e-3x - xe-3x e +C 3 9 27

3x 2 - 3

x4 +C 2

dy = 2 dx dy 3x 2 - 2 = dx 3 2 = x2 3 æ 2 2ö y = ççç x - ÷÷÷ dx è 3ø

x3 2x +C 3 3

Section 10.1 9.

753

dy = x2 dx Separate the variables and take antiderivatives. y

13.

ò ydy =ò x dx 2

y2 x3 = +K 2 3 2 y 2 = x3 + 2 K 3 2 y 2 = x3 + C 3

10.

2 x3 - x2 + C 3

y3 y2 x2 = +C 3 2 2 2 y 3 - 3 y 2 = 3x 2 + C

15.

ln | y | = x 2 + C e

dy y = ,x > 0 dx x dy dx = dx x ln | y | = ln x + C1

2 x3 +C 2 x2

y = eln x ⋅ eC1

y = e x + C

y = Celn x

y = e x ⋅ eC

y = Cx

16. dy = x2 y dx

12.

1 dy = x 2dx y 1 dy = x 2dx y

ln | y | =

eln| y | = eln x + C1

y = ke x

dy = x dx

ò ( y - y)dy = ò x dx

dy = 2 xy dx dy = 2 xdx y

ln | y | =

( y 2 - y) dy = x dx

y x 1 = - x2 + C 2 3 2

ln | y |

( y 2 - y)

14.

ò y dy =ò (x - x) dx

y = ke x - x

11.

3x3 2x2 +C 3 2

eln | y | = e x - x + C æ 3 2ö y =  çç e x - x ÷÷ eC è ø

y dy = ( x 2 - x) dx

y2 =

ln | y | =

dy = x2 - x dx

dy = 3x 2 y - 2 xy dx dy = y(3x 2 - 2 x) dx dy (3x 2 - 2 x) dx = y

y = Me

x3 /3

x-1 + C1 -1

| y | = e-1/x + C1

| y | = e x /3+C = e x /3 ⋅ eC

y = eC e x /3

ln | y | =

x3 +C 3 3

dy y = 2 dx x 1 1 dy = 2 dx y x 1 dy = x-2dx y

= e-1/x ⋅ eC1 y = eC1 e-1/x y = Ce-1/x

754

Chapter 10 DIFFERENTIAL EQUATIONS dy y2 + 6 = 2y dx

17. 2y

21.

dy = dx

y +6 2y

dy + 3x 2 = 2 x dx dy = 2 x = 3x 2 dx y =

ò y + 6 dy = ò dx 2

y = x 2 - x3 + C

ln |( y 2 + 6)| = x + C

Since y = 5 when x = 0,

Since y 2 + 6 is always greater than 0 we can

5 = 0-0+C

write this as ln ( y 2 + 6) = x + C.

C = 5.

Thus,

dy ey = dx y

18. y 2

ey y

dy = dx

2x2 3x3 +C 2 3

y = x 2 - x3 + 5.

22.

ò e dy = ò dx ò -2 ye dy = ò -2dx

dy = 4 x3 - 3x 2 + x; y = 0 when x = 1. dx y =

-y2

ò (4x - 3x + x) dx 3

= x 4 - x3 +

e- y = -2 x + C

y-2dy =

0 = 1-1+ -

e2 x dx

1 2x e +C 2 1 1 - = e2 x + C 2 y -1 y = 1 e2 x + C 2

y = x 4 - x3 +

23.

x2 1 2 2

dy = 4 xe-x dx dy = 2 xe-x dx

Use the table of integrals or integrate by parts.

dy ex = y dx e

y = 2(-x - 1)e-x + C

e y dy = e x dx

Since y = 42 when x = 0,

ò e dy = ò e dx y

1 +C 2

1 =C 2

- y-1 =

20.

x2 +C 2

Substitute.

dy = y 2e 2 x dx

19.

42 = 2(0 - 1)(1) + C 42 = -2 + C

e y = ex + C y = ln (e x + C )

C = 44

Thus, y = -2 xe-x - 2e-x + 44.

Section 10.1 24.

755

dy 8 = x 2e3x ; y = when x = 0. dx 9 dy 3x = xe dx y =

Let

27.

ò xe dx dv = e3x dx

e x + 3x + C = eln| y |

du = dx

v = e . 3

y = (e x + 3x )(eC )

y = x e3x 3

e3x dx 3

y = ke x + 3x

Since y = 1 when x = 0.

3x y = x e3x - e + C 3 9 8 = 0- 1 +C 9 9 C =1

1 = ke0 + 0 k = 1. 2

So y = e x + 3x .

y = x e3x - e + 1 3 9

28.

dy x2 + 5 = ; y = 11 when x = 0. dx 2y - 1

3 dy = x ; y = 5 when x = 0. dx y

25.

y dy =

(2 y - 1) dy = ( x 2 + 5) dx

ò (2 y - 1) dy = ò (x + 5)dx 2

x3 dx

x3 + 5x + C 3 121 - 11 = C C = 110 y2 - y =

4 y2 = x +C 2 4 y 2 = 1 x 4 + 2C 2 1 2 y = x4 + k 2

y2 - y =

Since y = 5 when x = 0, 25 = 0 + k k = 25. 1 2 4 So y = x + 25. 2 26.

2 x2 + 3x + C = ln | y | 2

u = x

dy ; y = 1 when x = 0. dx dy (2 x + 3)dx = y

(2 x + 3) y =

29.

dy 2x + 1 = ; y = 4 when x = 0. dx y-3

ò ( y - 3) dy = ò (2x + 1)dx

dy - y x = 0; y = e-2 when x = 1. dx 1 x dy = 2 dx y x 1 dy = x-3/2dx y

x3 + 5 x + 110 3

y2 2x2 - 3y = +x+C 2 2

Since y = 4 when x = 0, 16 - 12 = 0 + 0 + C 2 C = -4.

ln | y | = -2 x-1/2 + C -1/2

| y| = e-2 x

-1/2

y = Me-2 x

So, y2 - 3 y = x 2 + x - 4. 2

e-2 = Me-2 M =1 -1/2

y = e-2 x

756 30.

Chapter 10 DIFFERENTIAL EQUATIONS dy = y; y = -1 when x = 1. dx 1 dy = 1 dx y x2 See Exercise 16. x2

33.

dy = ( y - 1)2 e x -1; y = 2 when x = 1. dx dy = e x -1 dx 2 ( y - 1)

ò ( y - 1)-2 dy = ò e x-1dx

y = Me-1/x

( y - 1)-1 = e x -1 + C -1 - 1 = e x -1 + C y -1 -( y - 1) = x -11 +C e 1 - y + 1 = x -1 +C e

-1 = Me-1 M = -e y = -e1-1/x = -e(-1/x) +1

31.

dy y2 = ; y = 3 when x = e. dx x

ò y dy = ò x -2

= y 1 - x -11 e +C

- y-1 = ln | x | + C

x -1 y = e x -1 + C - x -11 e +C e +C

- 1 = ln | x | + C y -1 y = ln |x | + C

x -1

C -1 y = e x-+ 1 e +C y = 2, when x = 1.

Since y = 3 when x = e, 3=

-1 ln e + C

e0 + C C 2= 1+C 2 + 2C = C C = -2

-1 1+C 3 + 3C = -1 3=

3C = -4 C = -4. 3

y =

-1 -3 = . So y = 4 ln | x | - 3 3 ln | x | - 4

32.

dy = x1/2 y 2 ; y = 9 when x = 4. dx

e x-1 - 2

dy = ( x + 2)2 e y ; y = 0 when x = 1. dx

e- y dy = ( x + 2)2 dx -y

( x + 2)3 +C 3 -1 = 9 + C C = -10

-e- y =

x1/2dx

- y-1 = 2 x 2/3 + C 3 1 - = 2 (4)3/2 + C 9 3 C = 49 9 -1 = 2 x3/2 - 49 y 3 9 -1 = 6 x y

34.

e x -1 - 3

ò e dy = ò (x + 2) dx

1 dy = x1/2dx y2 y-2dy =

e0 + C - 1

-e- y =

( x + 2)3 - 10 3

e- y = 10 -

3/2

- 49 9 9 y = 3/2 6 x - 49

( x + 2)3 3

é ( x + 2)3 ùú - y = ln êê 10 ú 3 ë û é ( x + 2)3 ùú y = -ln êê 10 ú 3 ë û

Section 10.1 35.

757

dy = y( y 2 - 1) dx

The equilibrium points are solutions of y( y 2 - 1) = 0, that is, y = -1, y = 0, and y = 1.

Checking the sign of dy / dx in the four intervals into which these points divide the y-axis we get the following diagram.

Thus 0 is a stable equilibrium point and 3 is an unstable equilibrium point. 38.

dy = (ln y - 1)(5 - y) dx

The equilibrium points are solutions of (ln y - 1)(5 - y) = 0, that is, y = e and y = 5. Checking the sign of dy / dx in the 3 intervals into which these points divide the y-axis we get the following diagram.

Thus –1 and 1 are unstable equilibrium points and 0 is a stable equilibrium point. 36.

dy = (4 - y 2 )( y + 1) dx

The equilibrium points are solutions of (4 - y 2 )( y + 1) = 0, that is, y = -2, y = -1, and y = 2.

Checking the sign of dy / dx in the four intervals into which these points divide the y-axis we get the following diagram.

Thus 5 is a stable equilibrium point and e is an unstable equilibrium point. 39.

dy k = ( N - y) y dx N N dy (a) = k dx ( N - y) y

Since Thus –1 is an unstable equilibrium point and -2 and 2 are stable equilibrium points. 37.

1 1 N + = , y N -y ( N - y) y

ò y + ò N - y = k dx

dy = (e y - 1)( y - 3) dx

y = kx + C N -y y = Cekx . N -y

The equilibrium points are solutions of y

(e - 1)( y - 3) = 0, that is, y = 0 and y = 3.

For 0 < y < N , Cekx > 0.

Checking the sign of dy / dx in the three intervals into which these points divide the y-axis we get the following diagram.

For 0 < N < y, Cekx < 0.

Solve for y. y =

Cekx N 1 + Ce

N 1 + C -1e-kx

758

Chapter 10 DIFFERENTIAL EQUATIONS y =

N 1 + be

(b)

1 + be-kx0 = 2 and b = e kx0 .

N 1 - be-kx

For 0 < y < N ; t = 0, y = y0.

41.

(a)

0 < y0 < N implies that y0 > 0, N > 0,

and N - y0 > 0.

N y0 = = 0 1+ b 1 + be

Therefore,

Solve for b.

N - y0 > 0. y0

1 + be-kx

Let x0 be the time when z = 1/2, then

Let -b = C -1 < 0 for 0 < N < y. y =

Letting b = e-C , z =

-kx

N - y0 y0

Also, e-kx > 0 for all x, which implies that

1 + be-kx > 1. (c) For 0 < N < y; t = 0, y = y0.

y0 =

1 - be Solve for b. b=

40.

y ( x) =

(1)

N 1- b

N 1 + be-kx

1 + be-k x > 1. y ( x) =

(2)

y0 - N y0

N 1 + be-kx

Combining statements (1) and (2), we have 0<

N 1 + be-kx

Note 1 - z > 0 also.

(b)

1 1 1 + = 1- z (1 - z ) z z é 1 1ù ê + ú = dz = k dx êë 1 - z z úû

(c)

y ¢( x) = =

Solve for z.

1 + e kx + C 1

1 + be-kx

< N

N = N 1 + b(0)

Note that as x  -¥, 1 + be-kx becomes infinitely large. Therefore, the horizontal asymptotes are y = N and y = 0.

z = e kx + C 1- z

lim

1ù

-ln(1 - z ) + ln z = kx + C æ z ÷ö ln çç = kx + C çè 1 - z ÷÷ø

lim

x ¥ 1 + be-kx x -¥ 1 + be-kx

ò êêë 1 - z + z úúû dz = ò k dx

ekx + C

or 0 < y( x) < N for all x.

Observe that

z = e kx + C - ze kx + C

= y ( x) =

1 dz = k dx (1 - z ) z

> 0 since N > 0

and 1 + be-kx > 0.

dz = k (1 - z ) z, 0 < z < 1 dx

< N since

(1 + be-kx )(0) - N (-kbe-kx ) (1 + be-kx )2 Nkbe-kx (1 + be-kx )2

> 0 for all x.

Therefore, y(x) is an increasing function. (d) To find y ¢¢( x), apply the quotient rule to

find the derivation of y ¢( x). The numerator of y ¢¢( x), is

1 + e-kx -C

Section 10.1

759

y¢¢( x) = (1 + be-kx )2 (-Nk 2be-kx )

42.

- Nkbe-kx [-2kbe-kx (1 + be-kx )] = -Nk 2be-kx (1 - be-kx )(1 + be-kx ),

y =

N -kx

1 - be 0 < N < y0 and

and the denominator is [(1 + be-kt )2 ]2 = (1 + be-kx )4.

-Nk 2be-kx (1 - be-kx ) -kx 3

(1 + be

)

y0 - N < 1 and b < 1. y0

y ¢¢( x) = 0 when

(b)

lim e-kx = 0 (assuming k > 0), so

x ¥

k - kbe-kx = 0

x ¥ 1 - be-kx

When x =

lim e-kx = ¥, so

( 1b )

lim

-1

( 1b ) k

1 + be N 1 + be

ln(1/b)

1 - be-kx = 0

1 + be(-ln b) N

1+b

( ) 1 b

1 = be-kx 1 = e-kx b

N . 2

( lnkb , N2 ) is a point of inflection.

b-1 = e-kx

ln b-1 = ln e-kx

(e) To locate the maximum of dx , we must

- ln b = -kx ln b = x. k

consider, from part (d), d æç dy ö÷ -Nkbe-kx (k - kbe-kx ) = . ÷ ç dx çè dx ÷ø (1 + be-kx )3 ln b and k ln b , y ¢¢( x) < 0 for x > k

Since y ¢¢( x) > 0 for x <

we know that x = dy

= 0.

Thus, y = 0 is a horizontal asymptote to the curve as x  -¥.

ln b . k

Therefore,

x -¥ 1 - be-kx

ln b , k

y =

N = N. 1- 0

x -¥

k ln

Thus, y = N is a horizontal asymptote to the curve as x  ¥.

æ1ö -kx = ln çç ÷÷÷ çè b ø

lim

be-kx = 1 1 e-kx = b

x =

y0 - N . y0

(a) Since 0 < N < y0 , 0 < y0 - N and 0 < y0 , so b > 0. Also y0 - N < y0 , so

Thus, y¢¢( x ) =

ln b , where k

for all x ¹

maximum of dx .

ln b k

locates a relative

(d)

y = N (1 - be-kx )-1, x ¹

ln b k

We assume that k > 0. dy = -N (1 - be-kx )-2 (-be-kx ⋅ (-k )) dx =

-kbNe-kx (1 - be-kx ) 2

760

Chapter 10 DIFFERENTIAL EQUATIONS e-kx > 0 for all x, (1 - be-kx )2 > 0 for all ln b x ¹ , k > 0, b > 0, and N > 0. k

44.

E =-

p dq ⋅ q dp

If E =

4 p2 , q2

Therefore, dx < 0, and y is decreasing on

(

ln b ,¥ k

(e) To find

)

(

ln b

4 p2

)

and on -¥, k .

ò 4 p dp = -ò q dq

change the sign of b. Thus, d y dx 2

-kx

k bNe

-kx

(1 + be

(1 - be-kx )3

)

1 2 p 2 = - q 2 + C1 2

Multiplying by 2, we have

e-kx > 0 and b > 0, so 1 + be-kx > 0

4 p 2 = -q 2 + 2C2

for all x. k 2 > 0 and N > 0, so the numerator is always positive.

q 2 = -4 p 2 + 2C2.

(1 - be-kx )3 > 0, if and only if

Let C = 2C2. Then q 2 = -4 p 2 + C

1 - be-kx > 0 1 > be-kx

q =  -4q 2 + C.

b-1 > e-kx

Since q cannot be negative,

ln b-1 > ln e-kx

q =

-ln b > -kx ln b > x. k

45.

( lnkb , ¥ ) and ln b concave downward on ( -¥, k ). Thus, y is concave upward on

43.

Let y = sales, t = time, and k = -0.15. (a)

p dq ⋅ q dp

q 4 p dp = -q dq.

d2y , see Exercise 33(d) and just dx 2

dy = -0.15 y dt

E =-

p dq with p > 0 and q > 0 ⋅ q dp

If E = 2, p dq ⋅ q dp 2 1 dp = - dq p q 2 1 dp = dq p q 2 ln p = - ln q + K 2 =-

ò -0.15t

(b) From Example 5, y = Me

-4 p 2 + C .

ln p 2 + ln q = K

0.25M = Me-0.15t

ln ( p 2q) = K

0.25 = e-0.15t

p 2q = e K

-0.15t = ln 0.25

p 2q = C C q = 2. p

t =

ln 0.25 » 9.2 -0.15

Sales will become 25% of their original value in approximately 9.2 yr.

Section 10.1

761

46. (a)

48.

dy = -0.03 y dt dy = -0.03 y dt y ln | y | = -0.03t + C

eln | y | = e-0.03t +C

(b)

y = 32.350(1.2143)t

y = Me-0.03t

Since y = 6 when t = 0, 6 = Me0 M = 6.

So y = 6e-0.03t . If t = 10, (c) y =

y = 6e-0.03(10) » 4.4

4081 1 + 116.8e-0.2381t

After 10 minutes, about 4.4 cc of dye will be present. 49.

(a)

dI = 0.088(2.4 - I ) dW

Separate the variables and take antiderivatives. (d) As t gets very large, the value of the function in (c) approaches 6252, so 6252 million is the limiting number of internet users predicted by the model. 47.

ln |2.4 - I | = -0.088W - k |2.4 - I | = e-0.088W - k = e-k e-0.088W I - 2.4 = Ce-0.088W , where C = e-k .

I = 2.4 + Ce-0.088W

Since I (0) = 1, then

eln A = eit +C A = Meit When t = 0, A = 5000. Therefore, M = 5000. Find i so that A = 20,000 when t = 24. 4 = e24i ln 4 = 24i i = ln 4 24 2 = ln 2 24 ln = 2 12 The answer is d.

- ln |2.4 - I | = 0.088W + k

Solve for I.

dA = Ai dt dA = i dt A dA = i dt A ln A = it + C

20, 000 = 5000e24i

ò 2.4 - I = ò 0.088 dW

1 = 2.4 + Ce0 C = 1 - 2.4 = -1.4.

Therefore, I = 2.4 - 1.4e-0.088W . (b) Note that as W gets larger and larger e-0.088W approaches 0, so lim I = lim (2.4 - 1.4e-0.088W )

W ¥

= 2.4 - 1.4(0) = 2.4,

so I approaches 2.4.

762

Chapter 10 DIFFERENTIAL EQUATIONS -200 ln | C - 17.5w | = t + k ln | C - 17.5w | = 0.005t - 0.005k

dy = 0.02(4000 - y) dx

50. dy

ò 4000 - y = ò 0.02 dx

| C - 17.5w | = e-0.005t -0.005k | C - 17.5w | = e-0.005t ⋅ e-0.005k

-ln |4000 - y | = 0.02 x + C ln |4000 - y | = -0.02 x + C

C - 17.5w = e-0.005M e-0.005t -17.5w = -C + e-0.005M e-0.005t

|4000 - y | = e-0.02 x +C |4000 - y | = e-0.02 x ⋅ eC

w =

4000 - y = Me-0.02 x

(e) Since w = w0 when t = 0,

y = 4000 - Me-0.02 x

Since y = 320 when x = 0, 320 = 4000 - Me0 M = 3680.

If x = 10,

dw = k (C - 17.5w) dt

C being constant implies that the calorie intake per day is constant. pounds/day = k (calories/day) pounds/day = k calories/day

The units of k are pounds/calorie. (c) Since 3500 calories is equivalent to 1 pound, 1 and k = 3500

dw 1 = (C - 17.5w); w = w0 3500 dt when t = 0.

3500 -17.5

C e-0.005M =17.5 17.5

æ C ö C - çç - w0 ÷÷÷ e-0.005t ç ø 17.5 è 17.5 æ C C ÷ö -0.005t w= . + çç w0 ÷e 17.5 çè 17.5 ø÷

52.

(a)

1 dw (2500 - 17.5w); wt = 180 = 3500 dt when t = 0.

3500 dw = dt 2500 - 17.5w 3500 dw = dt 2500 - 17.5w -17.5 3500 dw = dt -17.5 2500 - 17.5w -200 ln |2500 - 17.5w | = t + C1

ò ò

3500 dw = dt C - 17.5w -17.5 dw = dt C - 17.5w

ò ò

ln |2500 - 17.5w | = -0.005t + C2

dw 1 (C - 17.5w). = 3500 dt

(d)

w0 -

At the end of 10 years, there will be about 987 fish.

(b)

C e-0.005M (1) 17.5 17.5

Therefore,

y = 4000 - 3680e-0.02(10) » 987.

(a)

w0 =

e-0.005M C = - w0. 17.5 17.5

So y = 4000 - 3680e-0.02 x .

51.

C e-0.005M -0.005t e 17.5 17.5

|2500 - 17.5w | = e-0.005t + C2 |2500 - 17.5w | = eC2 e-0.005t 2500 - 17.5w =  eC2 e-0.005t -17.5w = -2500 + C3e-0.005t w = 143 + C4e-0.005t

Since w = 180 when t = 0, 180 = 143 + C4 (1) C4 = 37 w = 143 + 37e-0.005t .

Section 10.1 (b)

763 lim w = lim (143 + 37e-0.005t )

t ¥

54. (a)

t ¥

15,000

= 143 + 37(0) = 143

The asymptote is w = 143.

300

y =

(b)

1383 1 + 1.867e-0.05615t

y = 1383 (1 + 1.867e–0.05615t ) 2000

According to the model, the value 143 will never be attained. 145 = 143 + 37e-0.005t

(c)

110

2 = 37e-0.005t

(c)

2 37 æ 2 ö -0.005t = ln çç ÷÷÷ çè 37 ø e-0.005t =

t =

According to the model, the limiting population for China is 1383 million.

(d) 2000

( )

2 ln 37

-0.005 t » 583.55 0

About 584 days (about 19.5 months) would be required to reach a weight of 145. 53.

110

y =

(a)

1959 1 + 4.800e-0.03415t

y = 1959 (1 + 4.800e–0.03415t) 2000

(b)

y =

25,538

110

According to the logistic model, the limiting population for India is 1959 million.

1 + 110.28e-0.01819t

(d) As t gets very large, the value of the function in (b) approaches 25,538, so this is the limiting number of deaths predicted by the model.

dy = ky dt First separate the variables and integrate. dy = k dt y dy = k dt y ln | y | = kt + C. Solve for y.

| y | = ekt +C1 = eC1 ekt y = Cekt , where C = eC1 .

764

Chapter 10 DIFFERENTIAL EQUATIONS Since y(16) = 57.5, so 57.5 = Ce16k and

dP = -4 DRP 2 dR dP = -4 DR dR P2

57.

since y(60) = 111.2, so 111.2 = Ce60k . Solve for k. 111.2 = e60k 57.5 e16k 111.2 = e 44k 57.5 æ ö 44k = ln çç 111.2 ÷÷÷ è 57.5 ø k =

(

ln 111.2 57.5 44

ò P dP = -4 Dò R dR 2

1 = 2 DR 2 + C P 1 P = C + 2 DR 2 Since P(0) = 1, 1 1= and thus C = 1. C+0 1 P( R ) = 1 + 2 DR 2

) » 0.01499,

then y = Ce0.01499t . Since y (16) = 57.5 , solve for C. 57.5 = Ce0.01499(16) C = 57.5e-0.01499(16) » 45.2

Thus, y = 45.2e0.01499t 56.

dy = ky dt First separate the variables and integrate. dy = k dt y dy = k dt y ln | y | = kt + C. Solve for y.

58.

(a) A calculator with a logistic regression function determined that y =

11.74 1 + (1.423 ´ 1022 )e-0.02554t

best fits the data. (b) From the graph, the function from part (a) seems to fit the data from 1927 on very well. For the year 1804, the function does not fit the data very well.

| y | = ekt + C1 = eC1 ekt y = Ce kt , where C = eC1 .

Since y(16) = 18.3, so 18.3 = Ce16k and since y(60) = 36.8, so 36.8 = Ce60k . Solve for k. 36.8 = e60k 18.3 e16k 36.8 = e 44k 18.3 æ ö 44k = ln çç 36.8 ÷÷÷ è 18.3 ø k =

(

36.8 ln 18.3

(c) After subtracting 0.99 from the y-values in the list, a calculator with a logistic regression function determines that y =

9.803 1 + (2.612 ´ 1029 )e-0.03391t

best fits the data. (d)

) » 0.01588,

0.01588t

then y = Ce . Since y (16) = 18.3 , solve for C. 18.3 = Ce0.01588(16) -0.01588(16)

C = 18.3e

Thus, y = 14.2e

0.01588t

» 14.2

From the graph, the function in part (c) does seem to fit the data better than the graph found in part (b).

Section 10.1

765

(e) As x gets larger and larger e-0.03391x approaches 0 so that y approaches

60.

(a)

p dq ⋅ = k q dp dq -1 = dp q p ln | q | = -k ln | p | + C

9.803 = 9.803. 1+ 0

If you add back the 0.99 that was subtracted from the y-values, the result is approximately 10.79 billion. (f ) For the function found in part (c), as x gets smaller and approaches negative infinity, the denominator of this logistic function approaches infinity so that y approaches 0. After adding back the 0.99 that was subtracted earlier, this would imply that the limiting value for the world population as you go further and further back in time is 0.99 billion. This does not seem reasonable though because the world population was not always more than 990 million. 59.

ln | q | = ln | p |-k + C -k

q = Celn | p | q =

pk As p approaches 0, q becomes infinitely large.

(b)

p + P dq ⋅ = -k q dp p+P 1 -k dq = dp q q ln | q | = -k ln | p + P | + C

ln | q | = -k ln | p + P | + k ln | P ln | q | = ln |

-kq dq = dp 1 + kp

ò q = ò 1 + kp dp (c)

ln | q | = - ln |1 + kp | + C eln | q | = e- ln |1+ kp |+C 1 (eC ) 1 + kp 1 q = 1 + kp

q =

(c)

q(1) =

k 1 = æç 0.1 ö÷ ÷ çè 1 + 0.1 ø÷ 2

C =0

k 1 = æç 1 ö÷ ÷ ç è 11 ø 2 æ ö ln ( 0.5 ) = k ln çç 1 ÷÷÷ è 11 ø

1 2

( ) k = ln 0.5 » 0.289 æ ö ln çç 1 ÷÷ è 11 ø

æ 0.1 ö÷0.289 Thus, q = çç ÷ çè p + 0.1 ÷÷ø

k =1 1 Thus, q = 1+ p

(d)

p = 0.20 1 » 0.83 1 + 0.2 Thus, the consumers will pay about 83%. q =

1 , P = 0.10 2

q(0) = 1

1 1 = 2 1+ k 1+ k = 2

(d)

P p+P

æ P ö÷k q = çç ÷ çè p + P ÷÷ø

-k

q(0) = 1

C = k ln | P

(1 + kp) dq = -kq dp

(b)

ln1 = -k| P | + C

dq = -k[ p dq + q dp] dq = -kp dq - kp dp

(a)

p = 0.20 0.289 æ 0.1 ÷ö q = çç » 0.73 ÷ çè 0.2 + 0.1 ø÷ Thus, the consumers will pay about 73%.

766 61.

Chapter 10 DIFFERENTIAL EQUATIONS (c) When t = 5,

dy = 7.5e-0.3 y , y = 0 when x = 0. dx

y = e0.8(5)

e0.3 y dy = 7.5 dx

òe

0.3 y

dy =

= e4 » 55.

ò 7.5 dx

(d) When t = 10,

e0.3 y = 7.5 x + C 0.3

y = e0.8(10) = e8 » 3000.

e0.3 y = 2.25 x + C 1= 0+C =C e0.3 y = 2.25 x + 1 0.3 y = ln (2.25 x + 1) y =

ln (2.25x + 1) 0.3

When x = 8, y =

62.

(a)

ln [2.25(8) + 1] » 10 items. 0.3

Let t = 0 be the time it started snowing. If h is the height of the snow and if the rate of snowfall is constant, dh = k1, where k1 is a constant. dt dh = k1 and h = 0 when t = 0. dt dh = k1 dt

ò dh = ò k dt 1

dy = ky dt dy = k dt y ln | y | = kt + C

63.

h = k1t + C1

Since h = 0 and t = 0, 0 = k1(0) + C1. Thus, C1 = 0 and h = k1t. Since the snowplow removes a constant volume of snow per hour and the volume is proportional to the height of the snow, the rate of travel of the snowplow is inversely proportional to the height of the snow.

eln | y | = e k t + C y =  (e kt )(eC ) y = Mekt

If y = 1 when t = 0 and y = 5 when t = 2, we have the system of equations 1 = Me k (0)

dx k2 = , where k2 is a constant. dt h When t = T , x = 0.

When t = T + 1, x = 2.

5 = Me 2k .

When t = T + 2, x = 3.

1 = M (1)

Since dt = kh and h = k1t ,

M =1

Substitute.

dy k = 2 dt k1t

5 = (1)e2k e 2k = 5

dx k 1 = 2 ⋅ . dt k1 t

2k ln e = ln 5 ln 5 2 » 0.8

Let k3 = k2 . Then

k =

(b) If k = 0.8 and M = 1, y = e0.8t .

When t = 3, y = e0.8(3) = e2.4 » 11.

dx 1 = k3 dt t 1 dx = k3 dt t 1 dx = k3 dt t x = k3 ln t + C2.

Section 10.1

767

Since x = 0, when t = T ,

64.

(a)

0 = k3 ln T + C2 (b)

C2 = -k3 ln T . Thus,

dy = -0.03 y dt dy = - 0.03 dt y ln | y | = -0.03t + C

eln | y | = e-0.03+C

x = k3 ln t - k3 ln T x = k3 (ln t - ln T )

y =  e-0.03t ⋅ eC

æt ö x = k3 ln çç ÷÷÷. çè T ø

y = Me-0.03t

Since x = 2, when t = T + 1, æ T + 1 ö÷ 2 = k3 ln çç . çè T ÷÷ø

75 = Me0 M = 75.

(1)

So y = 75e-0.03t . (d) At t = 10,

Since x = 3 when t = T + 2,

y = 75e-0.03(10) » 56.

æ T + 2 ÷ö 3 = k3 ln çç . çè T ÷÷ø

(2)

After 10 months, about 56 g are left.

We want to solve for T, so we divide equation (1) by equation (2).

66.

lim T = lim (Ce-kt + TM )

( ) 2 = 3 k3 ln ( T T+ 2 ) k3 ln T T+1

t ¥

= lim (Ce-kt ) + TM t ¥

= TM (The exponential term has limit 0 since k > 0.)

2 ln (T + 1) - ln T = 3 ln (T + 2) - ln T 2 ln (T + 2) - 2 ln T = 3ln (T + 1) - 3 ln T ln (T + 2)2 - ln T 2 - ln (T + 1)3 + ln T 3 = 0 ln

If T = Ce-kt + TM ,

(T + 2)2T 3 2

T (T + 1)

T (T + 2) 2 (T + 1)

67.

Use the formula from Exercise 66: T = Ce-kt + TM .

(a) We know that TM = 68, T (0) = 98.6 and T (1) = 90. 98.6 = (C )(1) + 68

C = 30.6 T = 30.6e-kt + 68

T (T 2 + 4T + 4) = T 3 + 3T 2 + 3T + 1 T 3 + 4T 2 + 4T = T 3 + 3T 2 + 3T + 1 T2 + T -1 = 0

90 = 30.6e-k (1) + 68 30.6e-kt = 22 22 30.6 æ 22 ö÷ -k = ln çç çè 30.6 ÷÷ø

e-k =

-1  1 + 4 T = 2

T = -1-2 5 is negative and is not a possible

æ 22 ö÷ k = - ln çç çè 30.6 ÷÷ø

solution. Thus, T = -1+2 5 » 0.618 hr.

» 0.33

Therefore, T = 30.6e-0.33t + 68.

0.618 hr » 37 min and 5sec

Now, 37 min and 5 sec before 8:00 A.M. is 7:22:55 A.M. Thus, it started snowing at 7:22:55 A.M.

When t = 2, T = 30.6e-0.33(2) + 68 = 83.8. After two hours the temperature of the body will be 83.8°F.

768

Chapter 10 DIFFERENTIAL EQUATIONS (b)

75 = 30.6e-0.33t + 68 7 = 30.6e-0.33t

88.6e-0.24t = 30

7 30.6 æ 7 ö÷ -0.33t = ln çç çè 30.6 ÷÷ø

30 88.6 æ 30 ö÷ -0.24t = ln çç çè 88.6 ÷÷ø

e-0.33t =

(

7 ln 30.6 -0.33 » 4.5

t =

e-0.24t =

)

(

30 ln 88.6 -0.24 t » 4.5

t =

The temperature of the body will be 75°F in approximately 4.5 hours. (c)

40 = 88.6e-0.24t + 10

(e)

)

The body will reach a temperature of 40°F in 4.5 hours.

68.01 = 30.6e-0.33t + 68 0.01 = 30.6e-0.33t

69.

0.01 e = 30.6 æ 0.01 ö÷ -0.33t = ln çç çè 30.6 ÷ø÷

(a)

-0.33t

t =

(

0.01 ln 30.6

)

-0.33 » 24.3

(b)

The temperature of the body will be within 0.01° of the surrounding air in approximately 24.3 hours. 68.

y =

250, 724 1 + 9.9928e-0.23631t

(c)

Use the formula from Exercise 66: T = Ce-kt + TM .

(a) In this problem, T0 = 10, C = 88.6, and k = 0.24.

Therefore,

(d) The limiting size is 250,724.

T = 88.6e-0.24t + 10.

(b)

T = 88.6e-0.24(4) + 10 T » 43.9. The temperature of the body will be 43.9°F after 4 hours.

Section 10.2

769

10.2 Linear First-Order Differential Equations Your Turn 1 Give the general solution of dy x - y - x 2e x = 0, x > 0. dx

Multiply all terms by the integrating factor, express the left side as the derivative of a product, and integrate on both sides. 2 dy 2 ex + 2 xe x y = x dx

(

)

Dx e x y = x 2

ex y =

Step 1: Put the equation in the required from dy + P( x) y = Q( x). dx

e x y = 1 x2 + C 2 2

1 x2 + C

dy æç 1 ÷ö + ç - ÷ y = xe x dx çè x ÷ø

y = 2

æ 1ö

ò çççè - x ÷÷ø÷dx

= e-ln x 1 = x

(0)2 + C 2(1) C =6

-1

integrating factor x . dy - ( x-2 ) y = e x dx

Step 4: Write the left side as the derivative of a product with respect to x.

The particular solution is y =

10.2

Warmup Exercises

W1.

ò x dx = 4 ln x + C

W2.

ò e- dt = -10e-

Dx ( x-1 y) = e x

Step 5: Integrate on both sides.

ò Dx (x-1y) dx = ò exdx -1

0.1t

The integrating factor is

W4.

Use the substitution u = t 3 - 3t , du = (3t 2 - 3) dt.

ò (t - 1)e - dt 1 1 = e du = e + C ò 3 3 t 3 3t

2 xdx I ( x) = e ò

1 eu du 2 1 = eu + C 2 1 3x 2 = e +C 2

3xe3x dx =

y = xe x + Cx

dy + (2 x) y = xe-x dx

2e x

Step 6: Multiply both sides by x to solve for y.

Your Turn 2 Solve the initial value problem 2 dy + 2 xy - xe-x = 0, y(0) = 3. dx

x2 + 6

W3. Use the substitution u = 3x 2 , du = 6 x dx.

y = e +C

Step 3: Multiply each term in the equation by the

( x-1)

2 = x +2 C 2e x To find the particular solution, substitute 0 for x and 3 for y.

Step 2: Find the integrating factor I(x). I ( x) = e

ò xdx

1 t 3 -3t e +C 3

770

Chapter 10 DIFFERENTIAL EQUATIONS

10.2

Exercises

True

False. A differential equation of the form dy/dx + P(x) y = Q(x) is linear.

True

dy + 3y = 6 dx

dy + 2 xy = 4 x dx

I ( x) = e ò ex

= ex

= e3 x

y = 2 + Ce-x

dy + 3e3x y = 6e3x dx 3x

Dx (e y ) = 6e

e3 x y =

I ( x) = e ò = e2 x 2 dy 2 2 + 4 xe2 x y = 4 xe2 x e2 x dx 2 æ 2 ö Dx çç e 2 x y ÷÷ = 4 xe2 x è ø 4 x dx

ò 6e dx 3x

= 2e3x + C

e2 x y =

y = 2 + Ce-3x

4 xe2 x dx 2

= e2 x + C

dy + 5 y = 12 dx

y = 1 + Ce-2 x

I ( x) = e ò

5 dx

= e5 x

dy + 5e5 x y = 12e5 x dx e5 x y =

dy - y - x = 0; x > 0 dx dy 1 - y =1 dx x

Dx (e5 x y) = 12e5 x

I ( x) = e ò

- 1/x dx

ò 12e xdx 5

12 5 x = e +C 5 12 y = + Ce-5 x 5

dy + 4 xy = 4 x dx

Integrate both sides.

e5 x

4 xe x dx

= 2e x + C

Multiply each term by e3x . e3x

2 2 dy + 2 xe x y = 4 xe x dx 2 æ 2 ö Dx çç e x y ÷÷ = 4 xe x è ø

ex y =

I ( x) = e3 ò

2 x dx

= e- ln x =

1 x

1 dy 1 1 - 2y = x dx x x æ1 ö 1 Dx çç y ÷÷÷ = çè x ø x 1 y = dx x x y = ln x + C x y = x ln x + Cx

Section 10.2 10.

771

dy + 2 xy - x 2 = 0; x > 0 dx

ex y =

e2 x

2 dx

= e2 x

dy + 2e 2 x y = xe2 x dx

13.

Dx (e2 x y) = xe2 x

ò xe2xdx

e2 x y =

dy + 2 y = x 2 + 6 x; x > 0 dx dy 2 + y = x+6 dx x

I ( x) = e ò = e2 ln x = x 2 dy + 2 xy = x3 + 6 x 2 x2 dx 2/x dx

Integration by parts: dv = e 2 x dx

Let u = x du = dx

v =

e2 x . 2

Dx ( x 2 y) = x3 + 6 x 2 x2 y =

xe2 x e2 x xe2 x e2 x +C dx = 2 2 2 4 1 x y = - + Ce-2 x 2 4

e2 x y =

11.

dy - 2 xy - x = 0 dx dy x - xy = 2 dx

I ( x) = e ò -

e-x /2

xdx

14. 2

= e-x /2

dy 3 + 6 xy + x = 0 dx dy x + 2 xy = 3 dx I ( x) = e ò e

2 x dx

2 dy x 2 + 2 xe x y = - e x dx 3 x 2 x2 Dx (e y) = - e x 3

ò (x + 6x ) dx 3

x4 + 2 x3 + C 4

y =

x2 C + 2x + 2 4 x

dy - x 2 y = x3 - 4 x3; x > 0 dx dy 1 - y = x-4 dx x -1/x dx

= e- ln x =

1 x

1 dy 1 4 ⋅ - 2 y = 1x dx x x æ y ö÷ 4 Dx çç ÷÷ = 1 çè x ø x æ ö y çç1 - 4 ÷÷ dx = ÷ ç è x xø = x - 4 ln x + C

y = x 2 - 4 x ln x + Cx

15.

= ex

I ( x) = e ò

2 2 dy x - xe-x /2 y = e-x /2 2 dx 2 2 x Dx (e-x /2 y) = e-x /2 2 2 x -x 2 /2 e-x /2 y = e dx 2 -1 -x 2 /2 = +C e 2 2 1 y = - + Ce x /2 2

12.

1 2 = - ex + C 6 2 1 y = - + Ce-x 6

dy + 2y = x dx I ( x) = e ò

ò - 3 e dx

dy = x3; x > 0 dx dy y - = -x 2 dx x

y-x

I ( x) = e ò

- 1/x dx

= e- ln x = x-1

772

Chapter 10 DIFFERENTIAL EQUATIONS 1 dy y - 2 = -x x dx x æ 1 ö÷ Dx çç y ÷÷ = -x çè x ø y = x = y =

17.

dy + y = 4e x ; y = 50 when x = 0. dx I ( x) = e ò ex

-x dx

= ex

dy + ye x = 4e2 x dx Dx (e x y) = 4e2 x

-x 2 +C 2

ex y =

-x + Cx 2

ò 4e2xdx

= 2e2 x + C

y = 2e x + Ce-x 2 xy + x3 = x

16. x

dy dx

Since y = 50 when x = 0,

dy - 2 xy = x3 dx dy - 2 y = x2 dx I ( x) = e ò

-2 x dy

-2 dx

-2 x

- 2e

50 = 2e0 + Ce0 50 = 2 + C C = 48.

Therefore,

= e-2 x

y = 2e x + 48e-x .

2 -2 x

y = x e

Dx (e-2 x y) = x 2e-2 x

18.

ò x e- dx

e-2 x y =

I ( x) = e ò

Integration by parts: u = x2

dv = e-2 x dx

du = 2 x dx

v =

-x 2e-2 x y = + 2

ò xe

Let

-2 x

du = dx

e4 x

e-2 x . -2 -2 x

2 -2 x

-x e 2

v = -

ò 9e9xdx + C

= e9 x + C

y = e5 x + Ce-4 x 25 = 1 + C C = 24 -2 x

ò 2

y = e5 x + 24e-4 x

-x 2e-2 x xe-2 x e-2 x +C 2 2 4

y =-

= e4 x

dy + 4e 4 x y = 9e5 xe4 x dx

e4 x y =

e-2 x . -2

-2 x

4 dx

Dx (e 4 x y) = 9e9 x

dv = e-2 x dx

Let u = x

e-2 x y =

dy + 4 y = 9e5 x ; y = 25 when x = 0. dx

19.

dy - 2 xy - 4 x = 0; y = 20 when x = 1. dx

1 x2 x - - + Ce2 x 2 2 4

dy - 2 xy - 4 x = 0 dx dy - 2 xy = 4 x dx I ( x) = e ò

- 2 x dx

e-x

= e-x

2 2 dy - 2 xe-x y = 4 xe-x dx

Section 10.2

773 2

Dx (e-x y) = 4 xe-x

21.

e-x y =

ò 4xe x dx

- 2

dy + 5 y = x2 dx dy 5 + y = x dx x

e-x y = -2e-x + C y = -2 + Ce x

dy + 5 y = x 2 ; y = 12 when x = 2. dx

Since y = 20 when x = 1,

I ( x) = e ò

20 = -2 + Ce1

5/x dx

22 = Ce

= e5ln x = x5

dy + 5x 4 y = x6 dx

22 . e

Dx ( x5 y) = x6 x5 y =

ò x dx

22 æ x 2 ö÷ çe ÷ y = -2 + ø e çè

x5 y =

x7 +C 7

( x 2 -1)

y =

x2 C + 5 7 x

C =

Therefore,

= -2 + 22e

20.

Since y = 12, when x = 2,

dy - 3 y + 2 = 0; y = 8 when x = 1. dx

4 C + 7 32 80 C = 7 32 2560 . C = 7 12 =

dy 3 2 - y =dx x x I ( x) = e ò

-3/x dx -3

= eln x

= e-3 ln x

= x-3

1 dy 3 2 - 4 y =- 4 3 dx x x x æ 1 ö÷ 2 Dx çç 3 y ÷ = - 4 çè x ÷ø x 1 -2 x-4dx y = 3 x

2 x-3 +C 3 2 y = + Cx3 3 2 8= +C 3

Therefore, y =

22.

22 3

y =

2 22 x3 + 3 3

x2 2560 . + 7 7 x5

dy - 4 xy = 5x; y = 10 when x = 1. dx dy 5x - 2 xy = dx 2

C =

= e-x I ( x) = e ò 2 dy 2 5 x -x 2 - 2 xe-x y = e-x e dx 2 2 ö 5 x -x 2 æ Dx çç e-x y ÷÷ = e è ø 2 2 5 x -x 2 e-x y = e dx 2 2 5 = - e-x + C 4

-2 x dx

774

Chapter 10 DIFFERENTIAL EQUATIONS 2 5 + Ce x 4 5 10 = - + Ce 4 45 C = 4e 5 45 x 2 y =- + e 4 4e 5 45 x 2 -1 - + e or 4 4

y =-

23.

24.

3 dy + 3x 2 y - 2 xe-x ; y = 100 when x = 0. dx

I ( x) = e ò

ex y =

= eò

(1/x) dx + dx

= xe xe x

⋅e

Dx ( xe x y) = 3e x

ò 3e dx x

xe x y = 3e x + C C 3 y = + x x xe

Since y = 50 when x = 4, 3 C + 4 4 4e C

197 = 4 4 4e

C = 197e4.

Therefore, 4

3 197e + x xe x 3 197 4- x = + e x x

y =

dA = 0.05 A - 50 dt

3 + 197e 4- x . x

eò = e-0.05t . Multiply both sides of the differential equation by the integrating factor: -0.05t dA + (-0.05) A(t )e-0.05t = -50e-0.05t e dt The left side is now the derivative of -0.05 dt

dy + (1 + x)e x y = 3e x dx

50 =

(b) Rearrange in standard linear form: dA + (-0.05) A(t ) = -50 dt The integrating factor will be

xe x y =

y = x 2e-x + 1000e-x

= e(ln x) + x =e

ò 2xdx

y = x 2e-x + Ce-x 1000 = 0 + C

25. (a)

ln x

= x2 + C

dy æç 1 + x ö÷ 3 +ç ÷÷ y = ç dx è x ø x (1+ x) dx /x

= ex

3 3 3 dy + 3x 2e x y = 2 xe x ⋅ e-x dx æ 3 ö Dx çç e x y ÷÷ = 2 x è ø

dy + (1 + x) y = 3; y = 50 when x = 4 dx

I ( x) = e ò

3 x 2dx

e-0.05t A(t ), so integrating both sides with respect to t yields e-0.05t A(t ) = 1000e-0.05t + C or, multiplying through by e0.05t , A(t ) = 1000 + Ce0.05t . Since A(0) = 2000, C = 1000 and

A(t ) = 1000 + 1000e0.05t . (c)

A(1) = 2051.27 or $2051.27 A(5) = 2284.03 or $2284.03 A(10) = 2648.72 or $2648.72

(d) The effective yield after n years is A(n) - 2000 . 2000 ⋅ n 51.27 For 1 year: = 2.56% 2000 284.03 For 5 years: = 2.84% 2000 ⋅ 5 648.72 For 10 years: = 3.24% 2000 ⋅ 10

Section 10.2

775

(e) If Shauntel invests $1000, then the value of C will be 0 and A(t ) = 1000 for all t.

26.

(b) Let z (0) = y1 . 0

1 p + Kce0 p + Kc = = y0 c c

dA dA = 0.05 A - 50 so = dt dt 0.05 A - 50 Integrate both sides: dA = dt 0.05 A - 50 200 ln(0.05 A - 50) = t + C t +C ln(0.05 A - 50) = 200 Exponentiate on both sides:

c = p + Kc y0

ln(0.05 A-50)

= e

K = y =

(t + C ) / 200

0.05 A - 50 = e0.05t C ¢

A = 1000 + 1000e

27.

dy Let y = 1z and dx = - z2¢ . z

æ 1 ö æ1ö z¢ - 2 = c çç ÷÷÷ - p çç 2 ÷÷ ç ç èzø è z ÷ø z z ¢ = -cz + p

z ¢ + cz = p I ( x) = e ò Dx (ecx ⋅ z ) =

c dx

p cx e +K c p z = + Ke-cx c

ecx ⋅ z =

-cx

p + Kce c

) ce

-cx

cy0

py0 + (c - py0 )e-cx

cy0 py0 - 0

c p

ò 1 Kò -

1 dG = dt a - KG 1 dG = dt a - KG -K dG = dt a - KG

ò ò

1 ln| a - KG | = t + C1 K ln| a - KG | = -Kt + C2 | a - KG | = e-Kt +C2

Therefore, y =

(

dG = a - KG dt dG = (a - KG) dt

28.

= ecx

ò pecxdx

c c - py0 cy0

æ ö÷ cy0 ç ÷÷ lim y = lim çç x ¥ x ¥ çè py0 + (c - py0 )e-cx ø÷÷

(c)

dy = cy - py 2 dx

(a)

c - py0 cy0

From part (a)

A = 1000 + 20e0.05t C ¢ Since A(0) = 2000. C ¢ = 20 and 0.05t

c - py0 c - p = y0 y0

Kc =

= eC2 e-Kt

. p + Kce-cx

a - KG = ke-Kt - KG = -a + ke-Kt

Dividing by -K , we get G =

a k + Ce-Kt where C = . K -K

776 29.

Chapter 10 DIFFERENTIAL EQUATIONS

I = eò e

æ ö÷ çç ÷÷ + + dN  r (  b v ) ç ÷÷ - e-rt rN = e-rt çç e-rt é ù ÷÷÷ çç dt v çè  êêë  - r 1 + b + γ úúû ÷÷ø  r ( + b + v) Dt ( Ne-rt ) = e-rt é ù v  ê  - r 1 + b+ γ ú ëê ûú

dC = -kC + D(t ) dt dC + kC = D(t ) dt

(a)

kt dC

k dt

(

= ekt

ekt C =

(

+ e kC = e D(t )

Ne-rt = -

Dt (e C ) = e D(t ) ekt D(t )d (t )

ò0 e D( y)dy kt

C (t ) = e-kt

ò0 e D( y)dy + C2 ky

If C (0) = 0,

C (0) = e0

ò0

Therefore, eky D( y)dy.

ò0

= De-kt

(

)

v b+ γ

)

 +b+v  +b+v  = ⋅    -r 1+

v b+ γ

)

(

)

-rdt

(

v b+ γ

)

v b+ γ

)

ö÷ ÷÷ ÷÷ ÷÷ ÷ø

Replace this in the anti-derivative previously found.

dN  r ( + b + v) = rN é ù dt  ê  - r 1 + b +v γ ú êë úû  r ( + b + v) dN - rN = é ù dt  ê  - r 1 + b +v γ ú ëê ûú

(

çè

N (t ) =

(

+ Cer (0)

 +b+v  +b+v  ⋅    -r 1+

 + b + v ççç  = çç1  çç  -r 1+

D(1 - e-kt ) C (t ) = k

Integrating factor: I = e ò

)

ò0 e dy

æ1ö = De-kt çç ÷÷÷ ekt - ek (0) çè k ø

30.

)

 +b+v  ⋅   -r 1+

C =

eky D dy t

(

(b) Let D( y) = D, a constant. C (t ) = e-kt

 r ( + b + v)

(

N (0) =

ò0

-rt

Use the initial condition N (0) = ( + b + γ)/ .

0 = 0 + C2 t

(

e dt ò )

é ù  ê  - r 1 + b +v γ ú ëê ûú

(

eky D( y)dy + C2

C (t ) = e-kt

 r ( + b + v)

)

æ 1 ÷ö -rt ç- ÷e +C é ù ççè r ÷ø v  ê  - r 1 + b+ γ ú ëê ûú  ( + b + v) = e-rt + C é ù v  ê  - r 1 + b+ γ ú êë úû  +b+v  ⋅ + Cert N (t ) = é ù  v ê  - r 1 + b+ γ ú êë úû =-

)

 +b+v  ⋅   -r 1+

(

v b+γ

)

 + b + v ççç  + çç1  çç  -r 1+ çè

Now use the substitution

= e-rt .

(

v b+ γ

)

ö÷ ÷÷ rt ÷÷ e ÷÷ ø÷

Section 10.2

777

æ v ö÷ ÷. R =  - r çç 1 + çè b + γ ÷÷ø

 +b+v  ⋅ N (t ) =   - r 1+

(

32.

v b+ γ

dy = ky + f (t ), k = 0.02 dt

)

dy = 0.02 y + e-t dt

æ ö÷ ÷÷ rt  + b + v ççç  ÷÷ e + ⋅ ç1 çç ÷÷ v   1 r + b + γ ø÷ èç  +b+v   + b + v çæ  ö rt = ⋅ + çç 1 - ÷÷÷ e è   R Rø

(

dy - 0.02 y = e-t dt

)

I (t ) = e ò

ye . Integrate both sides with respect to t.



e( +  )t + C

+  y = + C ⋅ e-( +  )t + 

+

C = y0 y =

1.02t

y = -0.98e-t + Ce-0.02t

Since y = 10,000 when t = 0, 10,000 = -0.98e0 + Ce0 = -0.98 + C C = 10,000.98.

Therefore, y = -0.98e-t + 10,000.98e0.02t .

33.

dy = 0.02 y + et ; y = 10,000 when t = 0. dt dy - 0.02 y = et dt I (t ) = e ò e-0.02t

-0.02dt

= e-0.02t

dy - 0.02e-0.02t y = e-0.02t ⋅ et dt

Since y(0) = y0 , y0 =

ò e-

= -0.98e-1.02t + C

( +  ) dt

( +  )t

= e-0.02t

ye-0.02t =

31. (a) The differential equation is dy =  (1 - y ) -  y. dt Rearrange in standard linear form: dy + ( +  ) y =  dt The integrating factor is eò = e( +  )t . Multiply through by the integrating factor: dy ( +  )t + ( +  ) ye( +  )t =  e( +  )t e dt The left side is now the derivative of

-0.02dt

Dt ( y ⋅ e-0.02t ) = e-0.02t e-t

 +b+v [ + ( R -  )e rt ] = R  +b+v [( R -  )e rt +  ] = R

ye( +  )t =

f (t ) = e-t

Dt (e-0.02t y) = e0.98t e-0.02t y =



+

æ  ÷ö -( +  )t ÷e + ççç y0 + è  +  ÷÷ø



(b) As t  ¥ the second term in the expression for y above goes to 0 because the exponential factor has limit 0, so the limiting value of y is  /( +  ).

òe

0.98t

e0.98t +C 0.98

et + Ce0.02t 0.98 1 10, 000 = +C 0.98 C » 9999 y =

y »

et + 9999e0.02t 0.98

= 1.02et + 9999e0.02t

778 34.

Chapter 10 DIFFERENTIAL EQUATIONS f (t ) = t dy = ky + f (t ) dt

du = -dt dy = ky + t dt

dy - ky = t dt I (t ) = e ò

- kdt

e-kt

Dt (e

e-0.02t y =

te-0.02t e-0.02t + +C 0.02 0.0004

10,000 = 2500 + C C = 7500

-kt

y = 50t + 2500 + 7500e0.02t

ò te dt -kt

36.

æ kt + 1 ö÷ -kt +C e-kt y = -çç ÷e çè k 2 ÷ø

dT = -k (T - TM ) dt dT = -kT + kTM dt dT + kT = kTM dt

æ kt + 1 ö÷ y = -çç ÷ + Cekt çè k 2 ÷ø

I (t ) = e ò

At k = 0.02, y = 10,000, t = 0,

kdt

= ekt

Multiply both sides by ekt.

æ 0 + 1 ö÷ 0 10, 000 = -çç ÷ + Ce ççè (0.02) 2 ÷ø÷

e kt

dT + kektT = kTM e kt dt

æ 1 ö÷ 10, 000 = -ççç ÷ + C (1) è 0.0004 ø÷

Dt (Te kt ) = kTM e kt Tekt =

12,500 = C.

ò kTM ektdt

Tekt = TM ekt + C

Therefore,

T = TM + Ce-kt

æ 0.02t + 1 ö÷ + 12,500e0.02t y = -çç çè 0.0004 ÷÷ø

T = Ce-kt + TM

= -50t - 2500 + 12,500e0.02t . dy = 0.02 y - t; y = 10, 000 when t = 0. dt

10.3 Euler’s Method

dy - 0.02 y = -t dt

e-0.02t

e-0.02t dt -0.02

y = 50t + 2500 + Ce0.02t

Use the table or integration by parts.

I (t ) = e ò

e-0.02t -0.02

te-0.02t 0.02

y) = te

e-kt y =

v =

e-0.02t =

= e-kt

dy - ke-kt y = te-kt dt -kt

35.

dv = e-0.02t dt

Let u = -t

-0.02

Your Turn 1 Use Euler’s method to approximate the solution

= e-0.02t

of dy /dx - x 2 y 2 = 1, y(0) = 2, for the interval [0, 1] with h = 0.2.

dy - 0.02e-0.02t y = -te-0.02t dt

First rewrite the equation with the derivative on the left.

Dt (e-0.02t y) = -te-0.02t e-0.02t y =

Integration by parts:

ò -te

-0.02t

dy = 1 + x2 y 2 dx

g ( x, y) = 1 + x 2 y 2

Section 10.3

779

i 0

0.2

2.2

1.1936

0.4

2.43872

1.95157684

0.6

2.82903537 3.88123880

0.8

3.60528313 9.31876252

1.0

5.46903563

10.3 Exercises

xi 0 0.1 0.2 0.3 0.4 0.5

True

False. Numerical method can be used on differential equations that are linear or separable.

True

yi 2 2.4 2.977 3.867 5.372 8.273

y(0.5) » 8.273. dy = xy + 4; y(0) = 0, h = 0.1. Find y (0.5). dx

g ( x, y) = xy + 4 x0 = 0; y0 = 0 g ( x0 , y0 ) = 0 + 4 = 4 x1 = 0.1; y1 = 0 + 4(0.1) = 0.4

Note: In each step of the calculation shown in this section, all digits should be kept in your calculator as you proceed through Euler’s method. Do not round intermediate results. 5.

These results are tabulated as follows.

g ( xi , yi )

dy = x 2 + y 2 ; y(0) = 2, h = 0.1. Find y (0.5). dx g ( x, y) = x 2 + y 2

g ( x1, y1) = (0.1)(0.4) + 4 = 4.04 x2 = 0.2; y2 = 0.4 + 4.04(0.1) = 0.804 g ( x2 , y2 ) = (0.2)(0.804) + 4 » 4.161 x3 = 0.3; y3 = 0.804 + 4.161(0.1)

x0 = 0; y0 = 2

» 1.220 g ( x3, y3 ) = (0.3)(1.220) + 4

g ( x0 , y0 ) = 0 + 4 = 4

» 4.366

x1 = 0.1; y1 = 2 + 4(0.1) = 2.4 2

g ( x1, y1) = (0.1) + (2.4)

x4 = 0.4; y4 = 1.220 + 4.366(0.1)

» 1.657 g ( x4 , y4 ) = (0.4)(1.657) + 4

= 5.77

» 4.663

x2 = 0.2; y2 = 2.4 + 5.77(0.1) = 2.977

x5 = 0.5; y5 = 1.657 + 4.663(0.1)

g ( x2 , y2 ) = (0.2) + (2.977)

» 8.903

» 2.123

These results are tabulated as follows.

x3 = 0.3; y3 = 2.977 + 8.903(0.1) » 3.867 2

g ( x3, y3 ) = (0.3) + (3.867)

xi 0 0.1 0.2 0.3 0.4 0.5

» 15.046 x4 = 0.4; y4 = 3.867 + 15.046(0.1) » 5.372 g ( x4 , y4 ) = (0.4)2 + (5.372)2

» 29.016

x5 = 0.5; y5 = 5.372 + 29.016(0.1) » 8.273

0 0.4 0.804 1.220 1.657 2.123

y(0.5) » 2.123

Use Euler’s method as outlined as in the solutions for Exercises 5 and 6 in the following exercises. The results are tabulated.

780 7.

Chapter 10 DIFFERENTIAL EQUATIONS dy = 1 + y; y(0) = 2, h = 0.1; find y(0.6). dx xi 0 0.1 0.2 0.3 0.4 0.5 0.6

11.

yi 2 2.3 2.63 2.993 3.3923 3.8315 4.31465

dy = 2 x 1 + y 2 ; y(1) = 0, h = 0.1; dx find y(1.5). xi 1 1.1 1.2 1.3 1.4 1.5

y(0.6) » 4.315

y(1.5) » 6.191.

dy = x + y 2 ; y(0) = 0, h = 0.1; find y(0.6). dx xi 0 0.1 0.2 0.3

12.

0 0.0 0.01 0.030

1 1.1 1.2 1.3 1.4 1.5

0.100 0.151

y ; y(0) = 1, h = 0.1; find y(0.4). xi 0 0.1 0.2 0.3 0.4

1 1.1 1.215 1.345 1.491

dy y = 1 + ; y(1) = 0, h = 0.1; find y(1.4). dx x xi 1 1.1 1.2 1.3 1.4 y(1.4) » 0.452.

yi 0 0.1 0.209 0.327 0.452

13.

dy = -4 + x; y(0) = 1, h = 0.1, find y(0.4). dx xi yi 0 1 0.1 0.6 0.2 0.21 -0.17 0.3 -0.540 0.4 y(0.4) » -0.540

y(0.4) » 1.491

10.

1 1.309 1.636 1.988 2.368 2.783

y(1.5) » 2.783.

y(0.6) » 0.151. dy = x+ dx

dy = e- y + e x ; y(1) = 1; h = 0.1; find y(1.5). dx xi yi

0.0601

0.4 0.5 0.6

yi 2 2.447 3.029 3.794 4.815 6.191

Exact solution: dy = -4 + x dx

y = -4 x + x + C 2

At y(0) = 1, 1 = -4(0) + 0 + C 2 C =1 Therefore, 2

y = -4 x + x + 1 2 (0.4)2 y(0.4) = -4(0.4) + +1 2 = -0.520.

Section 10.3 14.

781

dy = 4 x + 3; y(1) = 0, h = 0.1; find y(1.5). dx xi yi 1 0 1.1 0.7 1.2 1.44 1.3 2.22 1.4 3.04 1.5 3.9

16.

y(1.5) » 3.900.

y(1.4) » 3.053.

Exact solution: dy = 4x + 3 dx

Exact solution: dy 3 = dx x y = 3ln x + C

y = 2 x 2 + 3x + C

At y(1) = 2, 2 = 3ln1 + C = C.

At y(1) = 0, 0 = 2(1)2 + 3(1) + C

Therefore,

C = -5.

y = 3ln x + 2 y(1.4) = 3ln1.4 + 2 » 3.009.

Therefore, y = 2 x 2 + 3x - 5 y(1.5) = 2(1.5) 2 + 3(1.5) - 5 = 4.000.

15.

dy 3 = ; y(1) = 2, h = 0.1; find y(1.4). dx x xi yi 1 2 1.1 2.3 1.2 2.573 1.3 2.823 1.4 3.053

dy = x3; y(0) = 4, h = 0.1, find y(0.4). dx xi

0 0.1 0.2 0.3 0.4 0.5

4 4 4.0001 4.0009 4.0036 4.010

17.

dy = 2 xy; y(1) = 1, h = 0.1, find y(1.6). dx g ( x, y) = 2 xy xi 1 1.1 1.2 1.3 1.4 1.5 1.6

1 1.2 1.464 1.815 2.287 2.927 3.806

y(1.6) » 3.806

y(0.5) » 4.010

Exact solution:

Exact solution: dy = x3 dx 4 y = x +C 4

dy = 2 x dx y dy 2 x dx = y

ln | y | = x 2 + C

At y(0) = 4,

y | = ex + C

4= 0 +C 4 C = 4.

y = ke x

Therefore, 4

At y(1) = 1,

y = x +4 4 (0.5) 4 y(0.5) = + 4 » 4.016. 4

1 = ke1 = ke 1 k = . e

782

Chapter 10 DIFFERENTIAL EQUATIONS Exact solution:

Therefore, y =

1 æ x 2 ö÷ çe ÷ ø e çè

= e x -1 (1.6)2 -1

y(1.6) = e = 4.759.

18.

ln | y | = e x + c x

| y | = e e + c = ec e e

At y(0) = 2, 2 = kee = ke, so k =

so y(0.4) = 2ee

20.

dy = x 2dx y dy = x 2dx y

x3 +C 3

At y(0) = 1,

0.4

-1

» 3.271.

dy 2x = ; y(0) = 3, h = 0.1; find y(0.6). dx y

Exact solution:

0 0.1 0.2 0.3 0.4 0.5 0.6

3 3 3.007 3.020 3.040 3.066 3.099

Therefore, y(0.6) » 3.099. ln 1 = 0 + C 0 = C.

Exact solution: ydy = 2 x dx

Therefore,

ò y dy = ò 2x dx

x3 ln y = 3

y 3

y = y( x) = e x /3. y(0.6) = e

19.

x 2 ex e = 2ee -1, e

y =

y(0.6) » 1.056.

ln y =

2 . e

Therefore,

1 1 1.001 1.005 1.014 1.030 1.056

y = ke x , where k = ec .

dy = x 2 y; y(0) = 1, h = 0.1; find y(0.6). dx xi 0 0.1 0.2 0.3 0.4 0.5 0.6

dy = e x dx y dy = ex + c y

(0.6)3 /3

xi 0 0.1 0.2 0.3 0.4

So, y(0.4) » 3.112.

y2 = x2 + C 2

» 1.075.

dy = ye x ; y(0) = 2, h = 0.1, find y(0.4). dx yi

2 2.2 2.443 2.742 3.112

dy = x dx

At y(0) = 3, 9 = 0 + C = C. 2 y2 9 = x2 + 2 2 y2 = 2x2 + 9 y = y(0.6) =

2 x 2 + 9, assuming y > 0. 2(0.6)2 + 9 » 3.118.

Section 10.3 21.

783

dy + y = 2e x ; y(0) = 100, h = 0.1. dx Find y(0.3). xi 0 0.1 0.2 0.3

23.

dy = ye x ; y(0) = 2, h = 0.05, find y(0.4). dx xi yi

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

yi 100 90.2 81.401 73.505

y(0.3) » 73.505.

Exact solution: I ( x) = e ò ex

= ex

dy + e x y = 2e xe x dx Dx (e x y) = 2e2 x ex y =

So, y(0.4) » 3.186. Exact solution: y =

2e2 x dx + C

= e2 x + C

y(0.3) = e0.3 + 99e-0.3 » 74.691 dy - 2 y = e2 x ; y(0) = 10, h = 0.1. Find y(0.4). dx xi yi 0 10 0.1 12.1 0.2 14.642 0.3 17.720 0.4 21.446

24.

Exact solution:

e-2 x

= e-2 x

dy - 2e-2 x y = e-2 xe2 x dx

Dx (e-2 x y) = 1 e-2 x y =

ò dx + C

-1

» 3.271.

dy 2x = ; y(0) = 3, h = 0.05; find y(0.6). dx y

y(0.4) » 21.446. -2dx

0.4

Error: 3.186 – 3.271 = –0.085 Percentage error: 3.271 - 3.186 » 0.026 = 2.6% 3.271

y = e x + 99e-x

I ( x) = e

x 2 ex e = 2ee -1, e so

y(0.4) = 2ee

y = e x + Ce-x 100 = 1 + C C = 99

22.

2 2.1 2.21 2.333 2.468 2.619 2.787 2.975 3.186

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6

3.000 3.000 3.002 3.005 3.010 3.017 3.025 3.035 3.046 3.060 3.074 3.090 3.108

Therefore, y(0.6) » 3.108.

= x+C

Exact solution:

y = ( x + C )e2 x 10 = C

y = y(0.6) =

y = ( x + 10)e2 x y(0.4) = (0.4 + 10)e2(0.4) » 23.146

2 x 2 + 9, assuming y > 0. 2(0.6)2 + 9 » 3.118.

784

Chapter 10 DIFFERENTIAL EQUATIONS Error: 3.108 – 3.118 = –0.01 Percentage error: 3.118 - 3.108 » 0.003 = 0.3% 3.118

25.

27.

xi 0 0.2 0.4 0.6 0.8 1.0

dy = 3 x , y(0) = 0 dx

Using the program for Euler’s method in the Graphing Calculator Manual, the following values are obtained: xi

0 0.2 0.4 0.6 0.8 1.0 xi

yi - f ( xi )

0 0 0 0.08772053 0.11696071 0.22104189 0.26432197 0.37954470 0.43300850 0.55699066 0.61867206 0.75000000

0 0.08772053 0.10408118 0.11522273 0.12398216 0.13132794

( yi - f ( xi )) / f ( xi ) ´ 100 0 100 47.09 30.36 22.26 17.51

0 0.2 0.4 0.6 0.8 1.0

26.

f ( xi )

yi - f ( xi )

0 0.2 0.4 0.6 0.8 1.0

1 1.2 1.44 1.728 2.0736 2.48832

1 1.2214028 1.4918247 1.8221188 2.2255409 2.7182818

0 0.0214028 0.0518247 0.0941188 0.1519409 0.2299618

0 0.2 0.4 0.6 0.8 1.0

( yi - f ( xi )) / f ( xi ) ´ 100 0 1.75 3.47 5.17 6.83 8.46

f ( xi )

0 0.8 1.44 1.952 2.3616 2.68928

0 0.725077 1.3187198 1.8047535 2.2026841 2.5284822

28.

dy = x - 2 xy, y(0) = 1 dx Using the program in the Graphing Calculator Manual, the following values are obtained: xi 0 0.2 0.4 0.6 0.8 1.0

f ( xi ) 1 1 1 0.9803947 0.96 0.9260719 0.8864 0.8488382 0.793664 0.7636462 0.699692 0.6839397

0 0.2 0.4 0.6 0.8 1.0 29.

yi - f ( xi ) 0 0.07492 0.12128 0.14725 0.15892 0.16080

( yi - f ( xi )) / f ( xi ) ´ 100 0 10.33 9.20 8.16 7.21 6.36

0 0.2 0.4 0.6 0.8 1.0

Using the program for Euler’s method in the Graphing Calculator Manual, the following values are obtained: yi

dy = y, y(0) = 1 dx

dy = 4 - y, y(0) = 0 dx Using the program for Euler’s method in the Graphing Calculator Manual, the following values are obtained:

( yi - f ( xi )) / f ( xi ) ´ 100 0 2.00 3.66 4.43 3.93 2.30

dy = 3 x ; y(0) = 0 dx y = 34 x 4/3 See Exercise 25.

yi - f ( xi ) 0 0.01961 0.03393 0.03756 0.03002 0.01575

Section 10.3 30.

785

dy = y; y(0) = 1 dx y = ex

dy = y2 dx 1

See Exercise 26.

y2 1

dy = dx

ò y2 dy = ò dx -

1 = x+C y

When x = 0, y = 1.

31.

dy = 4 - y; y (0) = 0 dx

1 - = 0+C 1 C = -1

y = 4(1 - e-x ) See Exercise 27.

1 = x -1 y -1 = ( x - 1) y -1 x -1 1 y = 1- x y =

32.

y =

33.

As x approaches 1 from the left, y approaches ¥.

y ¢ = x - 2 xy; y(0) = 1 2 1 (1 + e-x ) See Exercise 28. 2

0 0.2 0.4 0.6 0.8 1.0

dy = ky(125 - y); k = 0.002 dt dy = 0.002 y(125 - y) dt dy = 0.25 y - 0.002 y 2 dt

y0 = 20; g (t , y) = 0.25 y - 0.002 y 2 g (t0 , y0 ) = 0.25(20) - 0.002(20)2

= 4.2

1 1.2 1.488 1.9308288 2.676448771 4.109124376

t1 = 1; y1 = 20 + (4.2)(1) = 24.2 g (t1, y1) = 0.25(24.2) - 0.002(24.2) 2 = 4.87872 t2 = 2; y2 = 24.2 + (4.87872)(1) = 29.0787 g (t2 , y2 ) = 0.25(29.0787) - 0.002(29.0787) 2

Thus, y(1.0) » 4.109. (b)

(a)

(b) If t0 = 0 corresponds to 2015, then t4 = 5 corresponds to 2020.

dy = y 2 ; y(0) = 1 dx (a) xi

34.

dy = y 2 ; y = 1 when x = 0 dx

= 5.57857 t3 = 3; y3 = 29.07870 + (5.57857)(1) = 34.6573 g (t3, y3 ) = 0.25(34.6573) - 0.002(34.6573) 2 = 6.26206

786

Chapter 10 DIFFERENTIAL EQUATIONS

t4 = 4; y4 = 34.6573 + (6.26206)(1)

t4 = 2; y4 = 12.9008 + 1.9282(0.5)

= 40.9193

= 13.8649

g (t4 , y4 ) = 0.25(40.9193) - 0.002(40.9193)

g (t4 , y4 ) = 2 - 0.02 13.8649 = 1.9255

= 6.88105

t5 = 5, y5 = 40.9193 + (6.88105)(1)

t5 = 2.5; y5 = 13.8649 + 1.9255(0.5)

= 47.8004

= 14.8277

About 48 firms are bankrupt in 2020. 35.

g (t5 , y5 ) = 2 - 0.02 14.8277 = 1.9230

Let y = the number of algae (in thousands) at time.

t6 = 3, y6 = 14.8277 + 1.9230(0.5)

y £ 500; y(0) = 5

(a)

= 15.7892

dy = 0.01y(500 - y) dt

g (t6 , y6 ) = 2 - 0.02 15.7892 = 1.9205

= 5 y - 0.01y 2

t7 = 3.5, y7 = 15, 7892 + 1.9205(0.5)

(b) Find y (2); h = 0.5.

= 16.7495

g (t7 , y7 ) = 2 - 0.02 16.7495

0 0.5 1 1.5 2

5 17.375 59.303 189.976 484.462

= 1.9181 t8 = 4, y8 = 16.7495 + 1.9181(0.5) = 17.7086 g (t8 , y8 ) = 2 - 0.02 17.7086 = 1.9158

Therefore, y (2) » 484, so about 484,000 algae are present when t = 2. 36.

t9 = 4.5; y9 = 17.7086 + 1.9158(0.5) = 18.6665

dy = 0.02(100 - y1/2 ) = 2 - 0.02 y dt t0 = 0; y0 = 10; h = 0.5

g (t9 , y9 ) = 2 - 0.02 18.6665 = 1.9136 t10 = 5; y10 = 18.6665 + 1.9136(0.5)

g (t , y) = 2 - 0.02 y

= 19.6233

g (t0 , y0 ) = 2 - 0.02 10 = 1.9368

There will be about 20 species.

t1 = 0.5; y1 = 10 + 1.9368(0.5) = 10.9684 g (t1, y1) = 2 - 0.02 10.9684 = 1.9338

37.

dy = 0.05 y - 0.1y1/2 ; y(0) = 60, h = 1; dt find y(6).

t2 = 1, y2 = 10.9684 + 1.9338(0.5) = 11.9353 g (t2 , y2 ) = 2 - 0.02 11.9353 = 1.9309 t3 = 1.5; y3 = 11.9353 + 1.9309(0.5) = 12.9008 g (t3, y3 ) = 2 - 0.02 12.9008 = 1.9282

0 1 2 3 4 5 6

60 62.22541 64.54785 66.97182 69.50205 72.14347 74.90127

Therefore, y (6) » 75, so about 75 insects are present after 6 weeks.

Section 10.3 38.

787

dy = - y + 0.02 y 2 + 0.003 y 3; for [0, 4] dt h = 1, t0 = 0, y0 = 15 2

g (t , y) = - y + 0.02 y + 0.003 y

40.

of the text, store dt for the function variable Y1 with k = 0.5 and m = 4.

That is, Y1 should equal 0.5( P - P 2 )1.5. Store 5 for H (use keystrokes 5  H ), to -h = -5 for T (-5  T ), and p0 = 0.1 for P(0.1  P). Next enter the keystrokes T + H  T : P + Y1H  P. Each time the ENTER key is pressed, the subsequent values for ti will be stored into T and the corresponding values for pi will appear on the screen. This summarized in the table below.

g (t0 , y0 ) = -15 + 0.02(15)2 + 0.003(15)3 = -0.375 t1 = 1; y1 = 15 + (-0.375)(1) = 14.625 g (t1, y1) = -14.625 + 0.02(14.625) 2 + 0.003(14.625)3 = -0.963 t2 = 2; y2 = 14.625 + (-0.963)(1)

ti 0 5 10 15 20 25 30

= 13.662 g (t2 , y2 ) = -13.662 + 0.02(13.662) 2 + 0.003(13.662)2 = -2.279 t3 = 3; y3 = 13.662 + (-2.279)(1) = 11.383 g (t3, y3 ) = -11.383 + 0.02(11.383)2 + 0.003(11.383) = -4.367

= 7.016 thousand There will be about 7000 whales. 39.

dW = -0.01189W + 0.92389W 0.016 dt W (0) = 3.65; h = 1; find W (5)

Using the program for Euler’s method in The Graphing Calculator Manual, the following values are obtained: t4

0 1 2 3 4 5

3.6500000 4.5498301 5.4422927 6.3268604 7.2032009 8.0710987

0.1 0.1675 0.297678 0.536660 0.846644 0.963605 0.980024

(b) By continuing to press the ENTER key, it appears that the values for pi are approaching 1.

t4 = 4; y4 = 11.383 + (-4.367)(1)

(a) Using Method 2 described after Example 1

41.

dN = 0.02(500 - N ) N 1/2 ; N (0) = 2, h = 0.5; dt Find N (3). ti

0 0.5 1 1.5 2 2.5 3

2 9.043 23.806 47.041 78.108 115.394 156.709

Therefore, N (3) » 157, so about 157 people have heard the rumor.

The weight of a goat at 5 weeks is about 8.07 kg.

788 42.

Chapter 10 DIFFERENTIAL EQUATIONS dT = -0.24(T - 10); T (0) = 98.6, h = 0.1; dt Find T (1). ti

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

98.6 96.4736 94.3982336 92.37267599 90.39573177 88.46623421 86.58304459 84.74505152 82.95117028 81.20034219 79.49153398

Find A when t = 21. A = -24,000 + 30,000e0.05(21) A » 61,729.53 When Michael is 21 the account will be worth $61,729.53.

Your Turn 2

Solve the equation given in Example 2 with p = 4, q = 1, r = 3, and s = 5, given that there was a time

when x = 1 and y = 1.

Therefore, T (1) » 79.5, so after an hour, the temperature is about 79.5°F, which is close to the solution of 79.7°F found in Section 10.1.

dy y(4 - x) = dx x(5 y - 3) 5y - 3 4-x dy = dx y x 5y - 3 dy = y 5 y - 3ln y = 4ln x - x + C

4-x dx x

Use the condition y = 1 when x = 1 to find C. 5(1) - 3ln(1) = 4ln(1) - 1 + C 6=C

10.4 Applications of Differential Equations

The equation relating x and y is x + 5 y - 4 ln x - 3 ln y = 6.

Your Turn 1

For this problem, r = 0.05 and D = 1200 so the differential equation is dA = 0.05 A + 1200 dt

Separate variables and integrate on both sides. dA = dt 0.05 A + 1200 1 ln(0.05 A + 1200) = t + C 0.05 ln(0.05 A + 1200) = 0.05t + K 0.05 A + 1200 = e0.05t + K 0.05 A = -1200 + M e0.05t A = -24, 000 +

M 0.05t e 0.05

Use the fact that A(0) = 6000. 6000 = -24, 000 +

Your Turn 3

Using the notation of Example 3, N - y0 b= y0 50,000 - 80 80 = 624. =

Thus the formula for y is y =

640 =

A = -24, 000 + 30, 000e

50,000 1 + 624e-k (15)

49,360 399, 360 1 æç 399, 360 ö÷ k = ln ç ÷ 15 çè 49,360 ÷÷ø

e-15k =

1500 = M

0.05t

640 + 399,360e-15k = 50,000

M 0.05(0) e 0.05

1500 0.05t e 0.05

1 + 624e-kt

Use the fact that y (15) = 640 to find k.

30, 000(0.05) = M A = -24, 000 +

50,000

k » 0.13938

Therefore, y =

50, 000 1 + 624e-0.13938t

Section 10.4

789

When t = 25,

10.4 Warmup Exercises 50,000

y =

1 + 624e-0.13938(25) = 2483. About 2483 people are infected 25 days into the epidemic.

W1.

dy x2 = 2 dx y

Separate the variables: y 2dy = x 2dx

Your Turn 4

Integrate on both sides:

Suppose that a tank initially contains 500 liters of a solution of water and 5 kg of salt, and that pure water flows in at a rate of 6 L/min and solution flows out at a rate of 4 L/min. These last two numbers indicate that the volume of solution increases at the rate of 2 L/min, so at time t the volume V (t ) = 500 + 2t. How many kilograms of salt remain after 20 minutes? We will assume that the tank is large enough so that it does not overflow during the 20 minutes of interest. Following the argument in Example 4 we can write the differential equation for this process as

y3 x3 = + C so 3 3 y3 = x3 + D y = 3 x3 + D Since y(0) = 2, D = 8 and y = 3 x 3 + 8.

W2.

dy = y dx

dy 4y =dt V (t ) dy 4y =dt 500 + 2t

Separate the variables:

with initial condition y (0) = 5, where y is the amount of salt in the tank in kilograms. Separate the variables and integrate on both sides.

Integrate on both sides:

ò ò

dy dx = 2 y x

4y dy =dy 500 + 2t dt 1 4 dy = dt y 500 + 2t 1 4 dy = dt y 500 + 2t 1 1 dy = -2 2dt y 500 + 2t ln y = -2ln(500 + 2t ) + C

ò ò

Use the initial condition y(0) = 5. ln 5 = -2 ln[500 + 2(0)] + C C = ln 5 + 2 ln 500

y = De-1/ x with D positive or negative Since y (1) = e-2 , D = e-1 and y = e-(1/ x)-1.

W3.

dy + xy - x 2 = 0 dx The equation will hold for x = 0 as long as the derivative of y is defined there, so assume x ¹ 0. Then dy + y = x dx x

1 dx

= ex.

dy + y ⋅ ex = x ⋅ ex dx Integrate on both sides, integrating by parts on the right. ex

ln y = -2 ln(500 + 2t ) + ln(5 ⋅ 5002 ) 5 ⋅ 5002

ye x = ( x - 1) e x + C

(500 + 2t )2

y = x - 1 + Ce-x

When t = 20, y =

1 +C x

The integrating factor is e ò

C = ln(5 ⋅ 5002 )

y =

ln y = -

Since y (0) = 3, D = 4 and y = x - 1 + 4e-x .

5 ⋅ 5002

(500 + 40)2 » 4.29. After twenty minutes, the tank contains 540 L of solution and 4.29 kg of salt.

(Note that y and its derivative are continuous at x  0, so our solution is well-behaved there.)

790

Chapter 10 DIFFERENTIAL EQUATIONS

W4.

When t = 10 yr,

dy - 4 xy = 10 x dx

A = 55,000e0.06(10) - 50,000

Separate the variables by rewriting the

= $50, 216.53.

equation: dy = (2 y + x) ⋅ x dx dy dx = 2y + 5 x

1 ln 2 y + 5 = ln x + C 2 2

2 y + 5 = e x ⋅ D (D positive or negative)

30, 000 =

-3000 + 3300e0.06t 0.06

æ 16 ö 1 ln çç ÷÷ 0.06 çè 11 ø÷ » 6.2 yr

t =

5 + e x ⋅ E (E positive or negative) 2 2

Since y(0) =

-3000 + 3300e0.06t 0.06

16 = e0.06t 11 æ 16 ö ln çç ÷÷÷ = 0.06t çè 11 ø

Integrate on both sides:

y =-

2 1 5 , E = 8 and y = - + 3e x . 2 2

dA = rA + D; r = 0.01; D = $50,000; dt t = 0, A = 0

10.4 Exercises

dA = 0.1A + 50, 000 dt

True

False. If y represents the number of individuals affected by an epidemic at time t, the infection

dA = dt 0.1A + 50, 000

rate is at a maximum when d 2 y / dt 2 = 0. 5.

1 ln |0.1A + 50, 000| = t + C 0.1 ln |0.1A + 50, 000| = 0.1t + C 0.1A + 50, 000 = e0.1t + C

dA = rA + D; r = 0.06, D = $5000 dt dA = 0.06 A + 3000 dt dA dt = 0.06 A + 3000 ln |0.06 A + 3000| = 0.06t + C

M 0.1t 50, 000 e 0.1 0.1

When t = 0, A = 0. M 50,000 (1) 0.1 0.1 M = 50,000 A=

0.06 A + 3000 = Me0.06t M 0.06t 3000 e 0.06 0.06

A = 16.66667Me0.06t - 50, 000 t = 0, A = 5000 5000 = 16.66667Me0 - 50,000 M = 3300 A = 16.66667(3300)e0.06t - 50,000 = 55,000e0.06t - 50,000

|0.06 A + 3000| = e0.06t +C

= Me0.1t

50,000 0.1t 50,000 e 0.1 0.1

= 500,000(e0.1t - 1)

Find t when A = $500,000. 500, 000 = 500, 000(e0.1t - 1) e0.1t = 2 1 ln 2 0.1 » 6.9 yr It will take about 6.9 years to accumulate $500,000.

t =

Section 10.4

791 (c) If A = 0, then

dA = 0.1A + D dt

20,000 - 12,000e0.06t = 0.

1 dA = dt 0.1A + D ln(0.1A + D) = t +C 0.1 D A= (-1 + e0.1t ) 0.1 When t = 3, A = 500,000. D 500, 000 = -1 + e0.1(3) 0.1 D (-1 + e0.3 ) = 0.1 500, 000(0.1) D = » $142,914.80 -1 + e0.3

(

(a)

12,000e0.06t = 20,000 5 e0.06t = 3 æ5ö 0.06t = ln çç ÷÷÷ , so çè 3 ø

æ5ö 1 ln çç ÷÷÷ » 8.51. 0.06 çè 3 ø Therefore, the account will be completely depleted in 8.51 years. t =

)

dy = 4 y - 2 xy dt = y(4 - 2 x)

11. (a)

dA = rA + D; r = 0.06; D = $1200; dt

dx = -3x + 2 xy dt = x(-3 + 2 y)

A(0) = 8000 dA = 0.06 A - 1200 dt

dy dt = y (4 - 2 x) = dx dx x(-3 + 2 y)

(b) To solve for A, first separate the variables.

dA = dt 0.06 A - 1200

Integrate.

æ 3ö çç 2 - ÷÷÷ dy = çè y ÷ø

ò ççèç x - 2 ÷÷ø÷ dx

0.06 A = 1200 + Me0.06t , so A = 20, 000 + C1e0.06t , M C1 = . Where 0.06

Solve for C1. Since A(0) = 8000, then 8000 = 20,000 + C1e0 , so

æ4

When x = 1, y = 1, 2 - 3 ln1 = 4ln1 - 2 + C 4 = C.

|0.06 A - 1200| = e0.06t + C = eC e0.06t , where M = e

æ 4 - 2 x ö÷ ÷ dx x ÷ø

2 y - 3ln y = 4 ln x - 2 x + C

Solve for A. ln|0.06 A - 1200| = 0.06t + C , where C = 0.06k .

0.06 A - 1200 = Me

ò çççè

dA dt = 0.06 A - 1200 1 ln |0.06 A - 1200| = t + k 0.06

0.06t

æ 2 y - 3 ö÷ ÷ dy = y ÷ø÷

ò çççè

Therefore, 2 y - 3ln y - 4 ln x + 2 x = 4 is an equation relating x and y in this case.

(b)

0 = 4 y - 2 xy = - y(-4 + 2 x) 0 = -3x + 2 xy = x(-3 + 2 y)

0 = - y(-4 + 2 x) and

0 = x(-3 + 2 y )

y =0

x = 0 or

-4 + 2 x = 0 x = 2

C1 = -12,000. Therefore, A = 20,000 - 12,000e0.06t

Since A(2) = 20, 000 - 12, 000e0.06(2) » 6470.04, the account will have $6470.04 left after two years.

and

- 3 + 2y = 0 3 and y = 2

792 12.

Chapter 10 DIFFERENTIAL EQUATIONS Now substitute k  0.25.

dx dy = -4 x + 4 xy, = -3 y + 2 xy dt dt

5000 - 1 + e0.25t 24,995,000

-3 y + 2 xy = -4 x + 4 xy y(-3 + 2 x) = x(-4 + 4 y) -4 + 4 y -3 + 2 x dy = dx y x æ 4 ö æ 3 ö çç - + 4 ÷÷ dy = çç - + 2 ÷÷ dx çè y ÷ø çè x ÷÷ø

4999 + e0.25t

(b) t = 30 y =

(c)

24,995,000 4999 + e0.25(30)

4999 + e0.25(50)

ln ( N - 1) k ln(5000 - 1) = 0.25 = 34.

tm =

4 = -3ln 5 + 10 + C

-6 + 3ln 5 = C 3ln x - 4 ln y - 2 x + 4 y = 3ln 5 - 6 -4 x + 4 xy = 0

The maximum infection rate will occur on the 34th day.

x(-4 + 4 y ) = 0 -3 y + 2 xy = 0 y(-3 + 2 x) = 0

Let N = population size, y = number infected.

14.

The solution x = 0, y = 0 may be feasible. If not, x = 32 and y = 1 .

(a)

N -y = N -

(a) Let y  the number of individuals infected. The differential equation is

1 + ( N - 1)e-kt

The number of individuals uninfected at time t is y = N =

1 + ( N - 1)e-aNt Example 4

N + N ( N - 1)e-aNt - N 1 + ( N - 1)e-aNt N ( N - 1)e-aNt

= aNt + ( N - 1) e N ( N - 1) = N - 1 + e a Nt N ( N - 1) = with k = aN N - 1 + e kt

The solution is Equation 7 in Example 3: N

1 + ( N - 1)e-aNt N ( N - 1)

æ dy yö = k çç1 - ÷÷÷ y. ç è dt Nø

y =

= 91

(d) From Example 3,

(a) When x = 5, y = 1. so

13.

= 3672

t = 50 24,995,000

-4 ln y + 4 y = -3 ln x + 2 x + C

(b)

5000(5000 - 1)

y =

dy dt = dx dx dt

(b)

N = 100, a = 0.01 y =

1 + ( N - 1)e-kt

N + N ( N - 1)e-kt - N

1 + ( N - 1)e-kt N ( N - 1) = . N - 1 + ekt

100 1 + 99e-t

and 100 - y =

9900 99 + et

Section 10.4

793 (e)

100

lim y = lim

t ¥ 1 + 99e-t

t ¥

100 = 100 people 1 + 99(0)

Therefore, y =

(c)

100

lim (100 - y) = 100 - lim y

= 100(1 + 99e-t )-1

-t

1 + 99e

t ¥

= 100 - 100

dy = -100(1 + 99e-t )-2 (-99e-t ) dt = 9900 

= 0 people.

In general,

e-t (1 + 99e-t )

lim y = N and lim ( N - y) = 0.

t ¥

d y = 9900 dt 

-t

15. -t

-t

(1 + 99e )  (-e ) - e

(a) The differential equation is

-t

 2(1 + 99e )  (-99e )

dy = a( N - y) y. dt

-t 4

(1 + 99e ) -t

9900(-e )(1 - 99e-t )

y0 = 100; y = 400 when t = 10, N = 20, 000.

-t 3

(1 + 99e )

-9900e-t

< 0 for all t, (1 + 99e-t )3 changes sign.

Since

d2y dt

Thus, 1 - 99e-t changes sign: 1 - 99e-t > 0 1 1 > t 99 e

The solution is y =

where b =

1 + be-kt N - y0 y0

and k = aN .

-100 b = 20,000 = 199; k = 20, 000a. 100

Therefore, y =

Therefore, y is concave upward if t > ln 99 and concave downward if t < ln 99.

20, 000

y = 400 when t = 10. 400 =

1 + 99

( ) 1 99

= 50.

1 + 199e-20,000(10)a

19, 600 400(199) = 0.2462312

e-200,000a =

a =

The second derivative of 100 - y merely d2y dt

20, 000

400 + 400(199)e-200,000a = 20, 000

Therefore, (ln 99,50) » (4.6 days, 50 people) is a point of inflection.

has the opposite sign as

1 + 199e-20,000 at

If t = ln 99, 100

Since y0 = 100 and N = 20, 000,

et > 99 t > ln 99.

y =

= 7 ´ 10-6

above, so

100 - y has the same point of inflection

as y. (d) At the point of inflection, the number of infected people is the same as the number of uninfected people.

ln (0.2462312) -200, 000

k = 20,000a = 20,000(7 ´ 10-6 ) = 0.14

Therefore, y =

20, 000 1 + 199e-0.14t

20, 0000.14t e0.14t + 199

794

Chapter 10 DIFFERENTIAL EQUATIONS (b) Half the community is y = 10, 000. Find t for y = 10, 000. 10, 000 =

So 0.1977 -0.02t e + 13.7970 -0.02

ln y =

20, 000

» -9.885e-0.02t + 13.7970.

1 + 19e-0.14t

10, 000 + 10, 000(199)e-0.14t = 20, 000 10, 000 = 0.005 e-0.14t = 10, 000(199) ln(0.005) t = -0.14 = 37.77

Thus, -0.02 t

y » e-9.885e

-0.02 t

= e13.7970e-9.885e

-0.02 t

» 982,000e-9.89e

Half the community will be infected in about 38 days.

(b) Find t when y =

16.

(a)

+13.7970

1 (10, 000) = 5000. 2

dy = kye-at dt

ln 5000 = -9.89e-0.02t + 13.7970

a = 0.02, so

e-0.02t =

ln 5000 - 13.7970 -9.89 é ln 5000 - 13.7970 ù 1 ú t = ln ê úû -0.02 êë -9.89 » 31 days

dy = kye-0.02t . dt y = 50 when t = 0; y = 300 when t = 10.

1 dy = ke-0.02t dt y k e-0.02t + C ln y = -0.02

17.

(a)

dy = -ay + b( f - y)Y dt a = 1, b = 1, f = 0.5, Y = 0.01; y = 0.02 when t = 0. dy = - y + 1(0.5 - y)(0.01) dt = -1.010 y + 0.005

To find k and C, solve the system k +C -0.02 k ln 300 = e-0.2 + C. -0.02 ln 50 =

(1) ( 2)

Subtract equation (1) from equation (2). k (e-0.2 - 1) -0.02 æ 300 ö÷ k = ln 6 = ln çç (1 - e-0.2 ) çè 50 ÷÷ø 0.02 0.02 ln 6 » 0.1977 k = 1 - e-0.2

ln 300 - ln 50 =

ò -1.010 y + 0.005 = ò dt 1 ln | -1.010 y + 0.005| = t + C2 -1.010 ln | -1.010 y + 0.005| = -1.010t + C1 | -1.010 y + 0.005| = e-1.010t +C1 = eC1 e-1.010t -1.010 y + 0.005 = Ce-1.010t y = 0.005 - 0.990Ce-1.010t

Therefore, k 0.02 0.1977 » ln50 + 0.02 » 13.7970.

C = ln50 +

Since y = 0.02 when t = 0, 0.02 = 0.005 - 0.990Ce0 -0.990C = 0.015. Therefore, y = 0.005 + 0.015e-1.010t .

Section 10.4 (b)

795

dY = - AY + B( F - Y ) y dt

497 -1000a e = 10 3 27 e-1000a = 497

A = 1, B = 1, y = 0.1, F = 0.03; Y = 0.01 when t = 0.

ln 497 æ 27 ö÷ 1 27 = ln çç ÷ 1000 çè 497 ø÷ 1000 500a » 1.456

dY = -Y + 1(0.03 - Y ) (0.1) dt = -1.1Y + 0.003

a =-

The number of people who have heard the rumor in t days is given by

dY = dt -1.1Y - 0.003 -

1 ln | -1.1Y + 0.003| = t + C2 1.1 ln | -1.1Y + 0.003| = -1.1t + C1

y =

| -1.1Y + 0.003| = e-1.1t + C1

= eC1 e-1.1t

When t = 5,

500 1 + 497 e-1.456t 3 1500 3 + 497e-1.456t

-1.1t

-1.1Y + 0.003 = Ce Y =

C -1.1t 0.003 e -1.1 -1.1

Since Y = 0.01 when t = 0, -0.909C = 0.00727.

19.

(a)

dy = a ( N - y) y dt y0 = 3; y = 12 when t = 3; N = 45

Therefore, Y = 0.00727e-1.1t + 0.00273.

18.

From equations (4) and (5) in Section 1 of this chapter, the solution to dy = a( N - y ) y dt

From equations (4) and (5) in Section 1 of this chapter, the solution is

N 1 + be-aNt

1 + be k = aN .

y = 50 =

where b =

N - y0 and y0

45 - 3 = 14; k = 45a. 3 45 y = . 1 + 14e-45at

y = 12 when t = 3, so

N - y0 . y0

12 =

If y0 = 3, N = 500, and y = 50 when t = 2, find y. b=

-kt

Since y0 = 3 and N = 45,

where y = number of people who have heard the information and b=

y =

is y =

» 449,

3 + 497e-1.456(5)

so in 5 days, about 449 people have heard the rumor.

= 0.909Ce-1.1t + 0.00273

0.01 = -0.909Ce0 + 0.00273

1500

y =

33 11 = 168 56 11 -135a = ln = -1.627 56 -45a = -0.542.

e-135a =

1 + 497 e-500at 3 1 + 497 e-1000a 3

1 + 14e-135a

12 + 168e-135a = 45

500 - 3 497 = , so 3 3 500 500

Therefore,

y »

45 1 + 14e-0.54t

796

Chapter 10 DIFFERENTIAL EQUATIONS Since y = 5 when t = 0,

(b) When y = 30, 30 = -0.54t

30 + 450e

5 = Ce-10k

1 + 14e-0.54t

C = 5e10k .

= 45

Since y = 15 when t = 3,

15 1 e-0.54t = = 450 30 1 1 t == 6.30. ln 0.54 30

15 = Ce-10ke

Solve the system C = 5e10k C = 15e7.41k .

y0 = 5, N = 100, and when t = 1, y = 20.

5e10k = 15e7.41k

100 - 5 = 19, so 5 100

e10k = 3e7.41k

b= y =

Take natural logarithms on both sides. 10k ln e = ln 3 + 7.41k

1 + 19e-100at 100

20 =

2.59k = ln 3 1 k = ln 3 = 0.424 2.59

1 + 19e-100a

1 + 19e-100a = 5

C = 5e10(0.424) = 347

e-100a =

4 19

1 4 a == ln 100 19 100a » 1.558

Therefore,

-0.1t

( )

-0.1t

30 = 347e-4.24e

100 -1.558t

1 + 19e

-0.1t

e-4.24e

1 ln 0.0865 4.24 = 0.5773 -0.1t ln e = ln 0.5773

100

t = -10 ln 0.5773

dy = kye-at ; a = 0.1; y = 5 when t = 0; dt y = 15 when t = 3. dy

ò y = kò e

-01t

= 5.493.

-0.1t

+ C1

= eC1 e-10ke

-0.1t

| y | = e-10ke

30 employees have heard the rumor in about 5.5 days. 22.

Let y = amount (in pounds) of salt at time t. At t = 0, y = 20 lb and V = 100 gallons. (a)

ln | y | = -10ke-0.1t + C1

y = Ce-10ke

30 = 0.0865 347

e-0.1t = -

so after 3 days, about 85 people have heard the news. (a)

-4.24e-0.1t ln e = ln 0.0865

1 + 19e-1.558(3) » 85,

21.

(b) If y = 30,

100

When t = 3, y =

-0.1t

y = 347e-10(0.424)e = 347e-4.24e

19 4

The number of people who have heard the news in t days is given by y =

= Ce-7.41k

C = 15e7.41k .

In about 6 days, 30 employees have heard the rumor. 20.

-0.3

Rate of Salt In = (2 gal/min) ⋅ (2 lb/gal) = 4 lb/min.

-0.1t

Section 10.4

797 dy 2y = 6; y(0) = 20 lb dt V dV = (rate of liquid in) dt - (rate of liquid out)

Rate of Salt Out æy ö = (2 gal/min) ⋅ çç lb/gal ÷÷÷ çè V ø =

2y (lb/min) V

= 3 gal/min - 2 gal/min

rate of liquid in

= 1 gal/min

= rate of liquid out

dV =1 dt V = t + C1

= 2 gal/min

So dV = 0 and V = 100 at all times t. dt

When t = 0, V = 100. Thus,

Therefore, 2y 200 - y dy lb/min = 4= dt 100 50 dy dt = . 200 - y 50

C1 = 100. V = t + 100.

Therefore,

t +C 50 - ln180 = C

- ln(200 - y ) =

dy 2y = 6dt t + 100 dy 2 y = 6. + dt t + 100 I (t ) = e ò

t - ln 180 50 ln(200 - y ) = ln 180 - 0.02t

- ln(200 - y ) =

= e2 ln | t + 100| = (t + 100) 2

200 - y = eln 180-0.02t = 180e-0.02t y = 200 - 180e-0.02t

dy (t + 100) 2 + 2 y(t + 100) = 6(t + 100) 2 dt Dt [ y(t + 100)2 ] = 6(t + 100) 2

(b) One hour later, t = 60 min,

y(t + 100)2 = 6

y = 200 - 180e0.02(60) » 146 lb.

(c)

y = 2(t + 100) +

x ¥

As t  ¥, y increases, approaching 200 lb.

dy = (rate of salt in) dt - (rate of salt out)

y = 2(t + 100) -

rate of salt in = (3 gal/min)(2 lb/gal)

æ 2y ö lb/min ÷÷÷ = çç çè V ø

(t + 100)2

1002 C = -1,800, 000.

Let y = the amount of salt present at time t.

= 6 lb/min æy ö rate of salt out = çç lb/gal ÷÷÷ (2 gal/min) çè V ø

Since t = 0 when y = 20, 20 = 2(100) +

(a)

ò (t + 100) dt

y(t + 100)2 = 2(t + 100)3 + C

lim (200 - 180e-0.02t ) = 200 - 180(0) = 200 lb

23.

2 dt /(t +100)

1,800, 000 (t + 100)2

2(t + 100)3 - 1,800, 000 (t + 100)2

(b) t = 1hr = 60 min y =

2(160)3 - 1,800, 000

= 249.69 (160)2 After 1 hr, about 250 lb of salt are present.

798 24.

Chapter 10 DIFFERENTIAL EQUATIONS

(

(a) Change rate of salt in to (1 gal/min) 1 lb/gal = 2 lb/min and rate of liquid in

2y lb/min V dV 2y = - ; y(0) = 20 dt V dV = (rate of liquid in) dt - (rate of liquid out)

to 1 gal/min, so dV = 1 - 2 = -1 gal/min. dt

V = -t + C and V = 100 when t = 0, so V = 100 - t.

Therefore, dy 2y = 2dt 100 - t dy 2y + = 2. dt 100 - t I ( x) = e ò

= 2 gal/min - 2 gal/ min = 0 dV =0 dt V = C1

2 / (100-t ) dt

= e-2 ln (100-t ) =

When t = 0, V = 100, so C1 = 100.

Therefore,

(100 - t )2

1 (100 - t ) 2

dy 2y == -0.02 y 100 dt dy = -0.02 dt y ln | y | = -0.02t + C1

y =

ò (100 - t) dt

ò 2(100 - t)

-2

+ dt

= 2(100 - t )-1 + C

| y | = e-0.02t +C1 = eC1 e-0.02t = Ce-0.02t

y = 2(100 - t ) + C (100 - t )2

Since t = 0 when y = 20,

20 = 2(100) + C (100)2 C = -0.018

20 = Ce0

y = 2(100 - t ) - 0.018(100 - t )2

C = 20. y = 20e-0.02t

(b) When t = 60 min,

(b) t = 1 hr = 60 min

y = 2(100 - 60) - 0.018(100 - 60)2

y = 20e-0.02(60) = 6.024 After 1 hr, about 6 lb of salt are present.

» 51 lb.

100 since y = 20 + 1.6t - 0.018t 2. Thus, y increases from t = 0 to t = 400 9 to t = 100. and decreases from t = 400 9 Let y = the amount of salt present at time t minutes. dy = (rate of salt in) - (rate of salt out) (a) dt rate of salt in = 0

is a parabola opening downward with vertex and t-intercepts of - 100 and at t = 400 9 9

25.

)

rate of salt out = V lb/gal (2 gal/min)

26.

Let y = amount (in grams) of chemical at time t. At t = 0, y = 5g and V = 100 liters. (a) rate of chemical in = 0 rate of chemical out

= (1 liter/min) V g/liter =

y g/min V

rate of liquid In = 2 liters/min; rate of liquid out = 1 liter/min; so

)

Section 10.4

799 Therefore,

dV = 1 liter/min. dt

dy y = 0.5 dt t + 100

Therefore,

dy 1 + ⋅ y = 0.5 dt t + 100

V = t + C and 100 = C , so V = t + 100. dy -y = dt t + 100 dy -dt = y t + 100

I (t ) = e ò = t + 100

dt /(t +100)

dy (t + 100) + y = 0.5(t + 100) dt Dx (t + 100) y = 0.5(t + 100)

ln y = - ln (t + 100) + C ln 5 = - ln 100 + C ln 5 + ln 100 = C C = ln (500)

(t + 100) y =

ln y = - ln (t + 100) + ln(500) æ 500 ö÷ = ln çç çè t + 100 ÷÷ø y =

y = 0.25(t + 100) +

5 = 0.25(100) +

500 » 3.8 g. 130

C 100

500 = 2500 + C C = -2000

Therefore,

Let y = amount of the chemical at time t. (a)

C t + 100

t = 0, y = 5

(b) When t = 30,

27.

ò 0.5(t + 100) dt

(t + 100) y = 0.25(t + 100)2 + C

500 t + 100

y =

= eln|t +100|

dy = (rate of chemical in) dt - (rate of chemical out)

y = 0.25(t + 100) + =

rate of chemical in = 0.5 g/min rate of chemical out æy ö = çç g/liter ÷÷÷ (1 liter/min) çè V ø

0.25(130)2 - 2000 130 = 17.115,

y =

After 30 min, about 17.1g of chemical are present.

= 2 liter/min - 1 liter/min = 1 liter/min dV =1 dt V = t + C1

0.25(t + 100)2 - 2000 . t + 100

(b) When t = 30 min,

= (2 liters/min) (0.25 g/liter)

y g/liter V dy y = 0.5 - ; y(0) = 5 dt V dV = (rate of liquid in) dt - (rate of liquid out)

-2000 t + 100

28.

Let y = amount (in pounds) of soap at time t. At t = 0, y = 4 pounds and V = 200 gallons. rate of soap in = 0; rate of soap out æy ö = (8 gal/min) çç lb/gal ÷÷÷ çè V ø 8y lb/min V rate of liquid in =

= rate of liquid out = 8 gal/min,

800

Chapter 10 DIFFERENTIAL EQUATIONS so

dV = 0, and V = 200 for all t. dt dy 8y =dt 200 1 1 dy = - dt 25 y t +C ln y = 25 ln 4 = C ln y = -

11.

True

12.

True

17.

Since you cannot separate the variables, that is, rewrite the equation in the form g ( y )dy = f ( x)dx where g is a function of y alone and f is a function of x alone, then the equation is not separable. Since you cannot

t + ln 4 25

rewrite the equation in the form dx + P( x) y

y = 4e-0.04t

= Q( x), then the equation is not a linear firstorder differential equation. Therefore, the equation is neither linear nor separable.

If y = 1, 1 = 4e-0.04t e-0.04t =

dy = 2x + y dx

1 4

18.

dy + y 2 = xy 2 dx

Since the equation can be rewritten in the form

æ1ö 1 ln çç ÷÷÷ 0.04 çè 4 ø = 25 ln 4 » 34.7 min .

t =-

dy y2

= ( x - 1)dx, then the equation is separable,

but since it cannot be rewritten in the form dy + P( x) y = Q( x), then the equation is not dx

Chapter 10 Review Exercises 1. 2.

19.

True

linear. dy 1 + ln x x = dx y Since you can rewrite the equation in the form y dy =

False: No; y = e satisfies the given differential equation.

1+ ln x x dx, then the equation x

is separable, but since it cannot be rewritten dy

in the form dx + P( x) y = Q( x), then the

True

False: No; the term y (dy /dx) makes the equation nonlinear.

equation is not linear. 20.

dy = xy + e x dx

False: Many (most) differential equations are neither separable nor linear.

Since the equation can be rewritten in the form

True

Since the equation cannot be rewritten in the form g ( y)dy = f ( x)dx, then it is not separable.

False: There is no way to separate the variables in this equation.

True

False: The integrating factor is x5.

10.

False: Euler’s method finds a numerical approximation to a particular solution over some interval.

dy + (-x) y = e x , then the equation is linear. dx

21.

dy + x = xy dx

Since you can rewrite the equation in the form dy + (-x) y = -x, then the equation is linear. dx

Since the equation can be rewritten in the form dy = x dx, then it is also separable. Therefore, y -1

it is both linear and separable.

Chapter 10 Review 22.

801

x dy = 4 + x3/2 y dx

29.

Since the equation can be rewritten in the form dy + dx

4 + x3/2 x

) y = 0, then the equation is

y2 3x 2 = + x + C1 2 2

linear. Since it can be rewritten in the form

y 2 = 3x 2 + 2 x + C

3/2 dy = 4 +xx dx, then it is also separable. y

Therefore, it is both linear and separable. 23.

30.

dy + y = e x (1 + y ) dx æ 1 - e x ö÷ dy ex ç + y çç , ÷÷÷ = dx x çè x ÷ø

y2 x2 - y = ex + +C 2 2

31.

then the equation is linear. Since the equation cannot be rewritten in the form g ( y)dy = f ( x)dx, then the equation is not separable.

dy 2y + 1 = dx x dy dy = 2y + 1 x 1 æç 2 dy ö÷ dx ÷÷ = çç ÷ 2 è 2y + 1ø x

1 ln|2 y + 1| = ln | x | + C1 2

dy = x2 + y2 dx

ln |2 y + 1|1/2 = ln | x | + ln k Let ln k = C1

Since the equation cannot be rewritten in dy

either form dx + P( x) y = Q( x) or the form

ln |2 y + 1|1/2 = ln k | x |

g ( y)dy = f ( x)dx, then the equation is neither nor separable.

25.

dy ex + x = dx y -1 ( y - 1) dy = (e x + x) dx

Since the equation can be rewritten in the form

24.

dy 3x + 1 = dx y y dy = (3x + 1) dx

|2 y + 1|1/2 = k| x | 2 y + 1 = k 2 x2 2 y + 1 = Cx 2

dy = 3x 2 + 6 x dx

2 y = Cx 2 - 1

dy = (3x + 6 x)dx 3

y =

y = x + 3x + C

26.

dy = 4 x3 + 6 x5 dx y = x 4 + x6 + C

27.

dy = 4e2 x dx

dy 3- y = dx ex

32.

1 dy = e-x dx 3- y - ln |3 - y | = -e-x + C ln |3 - y | = e-x - C 3 - y = kee

dy = 4e2 x dx y = 2e2 x + C

28.

1 dy = 3x + 2 dx 1 y = ln |3x + 2| + C 3

Cx 2 - 1 2

-x

y = 3 - kee or

-x -x

y = 3 + Mee

802

Chapter 10 DIFFERENTIAL EQUATIONS dy + y = x dx

33.

36.

I ( x) = e ò ex

= ex

dy + e x y = xe x dx x

Dx (e y) = xe

dy + 2 y - e2 x = 0 dx dy 2 1 + y = e2 x dx x x I ( x) = e ò

Integrate both sides, integrating the right side by parts.

2dx /x

= e 2 ln x = x 2

dy + 2 xy = xe2 x dx Dx ( x 2 y) = xe2 x

e x y = xe x - e x + C

ò xe dx x 1 = ò xe dx = 2 e - 4 e + C 2x

x2 y =

y = x - 1 + Ce-x

34.

dy x4 + 3x3 y = 1 dx dy 3 1 + y = 4 dx x x

y = 3

I ( x) = e

ò x dx

37.

= x3

dy 1 + 3x 2 y = dx x 1 Dx ( x3 y) = x Integrate both sides.

e2 x e2 x C - 2 + 2 2x 4x x

dy = x 2 - 6 x; y = 3 when x = 0. dx x3 - 3x 2 + C 3 When x = 0, y = 3. y =

3= 0-0+C C =3

x3 y = ln x + C

35.

ln x + C x3

y =

dy + y = 2x2 dx 1 2x dy + y = ln x dx x ln x x ln x

38.

x3 - 3x 2 + 3 3

dy = 5(e-x - 1); y = 17 when x = 0. dx dy = 5(e-x - 1) dx

I ( x) = e

dy = ( x 2 - 6 x)dx

y =

y = 5(-e-x - x) + C

ò x ln x dx

= -5e-x - 5x + C

1 æ1 ö ò ln x çççè x dx ÷÷÷ø

17 = -5e0 - 0 + C 17 = -5 + C

= eln(ln x)

22 = C

= ln x

Multiply each term by the integrating factor, express the left side as the derivative of a product and integrate on both sides. dy 1 + y = 2x dx x Dx[(ln x) y] = 2 x

ln x

(ln x) y = x 2 + C y =

x2 + C ln x

y = -5e-x - 5x + 22

Chapter 10 Review 39.

803

dy = ( x + 2)3 e y ; y = 0 when x = 0. dx

42.

dy = xy; y = 4 when x = 1. dx

e- y dy = ( x + 2)3 dx 1 ( x + 2) 4 + C 4 1 -1 = (2)4 + C 4 C = -5

-e- y =

1 ( x + 2) 4 4 é ù 1 y = - ln ê 5 - ( x + 2) 4 ú êë úû 4

3 2

+ C1

2 3/2 x + C1 3

| y | = e(2/3) x

3/ 2

+ C1

| y | = eC1 e(2/3) x

3/ 2

y = eC1 e(2/3) x

dy (3 - 2 x) y = ; y = 5 when x = 0. dx dy (3 - 2 x)dx = y

y = Ce(2/3) x

3/ 2

3/2

Since y = 4 when x = 1, 4 = Ce(2/3)(1) 4 C = 2/3 e C » 2.054

3x - x 2 + C = ln | y | 2

e3x-x + C = y 2

e3x-x ⋅ eC = y 2

Me3x - x = y

3/2

y = 2.054e(2/3) x .

Me0 = 5 M =5

41.

x 3/2

ln| y | =

Notice that x < 4 20 - 2.

y = 5e3x - x

ln| y | =

e- y = 5 -

40.

1 x dy = dx y x 1 dy = x1/2 dx y

43.

dy - e x y = x 2 - 1; y = 42 when x = 0. dx dy - y = ( x 2 - 1)e-x dx

dy 1 - 2x ; y = 16 when x = 0. = dx y+3

I ( x) = e ò

-1 dx

ò (x - 1)e ⋅ e dx - x = ò (x - 1)e dx

( y + 3) dy = (1 - 2 x) dx

e-x y =

y2 + 3 y = x - x2 + C 2 162 + 3(16) = 0 + C 2 176 = C

= e-x

-x

Integration by parts: u = x2 - 1

Let

y + 3 y = x - x 2 + 176 2 y 2 + 6 y = 2 x - 2 x 2 + 352

du = 2 x dx e-x y = Let

( x 2 - 1) -2 x e + 2

u = x du = dx

dv = e-2 x dx 1 v = - e-2 x . 2

ò xe- xdx 2

dv = e-2 x dx 1 v = - e-2 x . 2

804

Chapter 10 DIFFERENTIAL EQUATIONS

e-x y = =-

( x 2 - 1) -2 x x e - e-2 x + 2 2

1 -2 x e dx 2

45.

( x - 1) -2 x 1 x e - e-2 x - e-2 x + C 2 2 4

dy - 2 x 2 y + 3x 2 = 0 dx y = 15 when x = 0.

dy - 2 xy = -3x dx

( x 2 - 1) -x x 1 y =e - e-x - e-x + Ce x 2 2 4

I ( x) = e ò

1 1 -0- +C 2 4 C = 41.75

Dx (e-x y) = -3xe-x

x ( x 2 - 1) -x 1 e - e-x - e-x + 41.75e x 2 2 4 é x2 ù x 1 = e-x êê - + úú + 41.75e x 2 4ú êë 2 û

e-x y =

3x 2 dx

= ex

ex y =

3 + Ce0 2 27 C = . 2

15 =

Therefore, 2

3 27e x . y = + 2 2

3 3

x 2e x dx

1 x3 e +C 3 3 1 y = + Ce-x 3 =

Since x = 0 when y = 2,

46.

3 dy + 4 xy - e2 x = 0; y = e2 dx when x = 1.

3 dy 4 1 + y = 2 e2 x dx x x

I ( x) = e ò

4/x dx

= e4 ln x = x 4 3 1 x4 y = x 4 ⋅ 2 e2 x dx x

ò = ò x e x dx

1 + Ce0 3 5 C = . 3 2=

2 2 3

1 2 x3 e +C 6 1 1 ⋅ e2 = e2 + C 6 5 C = e2 6 =

Therefore, y =

-3xe-x dx

Since x = 0 when y = 15,

3 3 dy + x 2e x y = x 2e x dx

D x ( e x y ) = x 2e x

3 -x 2 e +C 2 2 3 y = + Ce x 2

dy + 3x 2 y = x 2 ; y = 2 when x = 0. dx I ( x) = e ò

-x 2e-x xe-x e-x + + 41.75e x 2 2 4

y =-

44.

= e-x

e-x y - 2 xe-x y = -3xe-x

42 =

-2 x dx

3 1 5 + e-x . 3 3

y =

e2 x + 5e2 6x 4

Chapter 10 Review 47.

805

dy = (1 - e y )( y - 2) dx

g ( x2 , y2 ) = 0.4 +

The equilibrium points are solutions of

x3 = 0.6; y3 = 1.4067 + 1.1109(0.2)

1 = 1.1109 1.4067

(1 - e )( y - 2) = 0, that is, y = 8 and y = 2.

= 1.6289

Checking the sign of dy / dx in the three intervals into which these points divide the y-axis we get the following diagram.

g ( x3 , y3 ) = 0.6 +

1 1.6289 = 1.2139

x 4 = 0.8; y4 = 1.6289 + 1.2139(0.2) = 1.8717 g ( x4 , y4 ) = 0.8 +

1 = 1.3343 1.8717

x5 = 1.0; y5 = 1.8719 + 1.3343(0.2) y5 = 2.13855

Thus 0 is an unstable equilibrium point and 2 is a stable equilibrium point. 48.

dy = ( y - 8)( y 2 - 1) dx

y (1) » 2.138

51.

The equilibrium points are solutions of 2

dy = e x + y; y(0) = 1, h = 0.2; find y (0.6). dx g ( x, y) = e x + y

( y - 8)( y - 1) = 0, that is, y = 1 and y = 8.

x0 = 0; y0 = 1

Checking the sign of dy / dx in the four intervals into which these points divide the y-axis we get the following diagram.

x1 = 0.2; y1 = y0 + g ( x0 , y0 )h

g ( x0 , y0 ) = e0 + 1 = 2 = 1 + 2(0.2) = 1.4

g ( x1, y1) = e.2 + 1.4 » 2.6214 x2 = 0.4; y2 = y1 + g ( x1, y1)h = 1.4 + 2.6214(0.2) » 1.9243 g ( x2 , y2 ) = e0.4 + 1.9243 » 3.4161 x3 = 0.6; y3 = y2 + g ( x2 , y2 )h = 1.9243 + 3.4161(0.2) » 2.6075

Thus 1 is a stable equilibrium point and -1 and 8 are unstable equilibrium points. 50.

xi 0 0.2 0.4 0.6

dy = x + y-1; y(0) = 1; h = 0.2; dx 1 y x0 = 0; y0 = 1 g ( x, y) = x +

1 =1 1 x1 = 0.2; y1 = 1 + 1(0.2) = 1.2 g ( x0 , y0 ) = 0 +

g ( x1, y1) = 0.2 +

So, y (0.6) » 2.608.

1 = 1.0333 1.2

x2 = 0.4; y2 = 1.2 + 1.0333(0.2) = 1.4067

1 1.4 1.9243 2.608

806 52.

Chapter 10 DIFFERENTIAL EQUATIONS dy x = + 4; y(0) = 0; h = 0.1, dx 2 x +4 2 x0 = 0; y0 = 0 g ( x, y) =

0 +4= 4 2 x1 = 0.1; y1 = 0 + 4(0.1) = 0.4 g ( x0 , y0 ) =

55.

dy = 6e0.3x dx dy = 6e0.3x dx

0.1 + 4 = 4.05 2 x2 = 0.2; y2 = 0.4 + 4.05(0.1) g ( x1, y1) =

y = 20e0.3x + C

When x = 0, y = 0

= 0.805 g ( x2 , y2 ) =

(a)

0.2 + 4 = 4.1 2

0 = 20e0 + C C = -20

x3 = 0.3; y3 = 0.805 + 4.1 (0.1)

y = 20e0.3x - 20.

= 1.215

When x = 6, y = 20e1.8 - 20

Solving the differential equation gives

» 100.99.

dy x = +4 dx 2 æx ö dy = çç + 4 ÷÷÷ dx çè 2 ø

Sales are $10,099.

y =

(b) When x = 12,

y = 20e3.6 - 20

x2 + 4 x + C. 4

» 711.96.

Sales are $71,196.

Since x = 0 when y = 0, C = 0. x2 + 4x 4 y( x3 ) = y(0.3) y =

53.

56.

dy = 0.1(150 - y); y = 15 dx when x = 0; y £ 150

0.32 + 4(0.3) = 1.223 4 y3 - y( x3 ) = 1.215 - 1.223 = -0.008

1 dy = 0.1 dx 150 - y - ln (150 - y) = 0.1x + C - ln 135 = C

dy = 3+ dx

y, y(0) = 0, h = 0.2, find y (1).

- ln (150 - y) = 0.1x - ln 135

xi 0

yi 0

0.2

0.6

0.4

1.354919

0.6

2.187722

0.8

3.083541

4.034741

ln (150 - y) = ln 135 - 0.1x 150 - y = 135e-0.1x y = 150 - 135e-0.1x

(a) If x = 10, y = 150 y - 135e-0.1(10) » 100 items.

Therefore, y (1) » 4.035.

Chapter 10 Review (b)

807 dA = 0.05 A - 20, 000 dt

100 = 150 - 135e-0.1t 50 = 135e-0.1t 50 10 e-0.1t = = 135 27 æ 10 ÷ö -0.1t = ln çç ÷÷ çè 27 ø

ò 0.05 A - 20, 000 = ò dt 1 ln| 0.05 A - 20, 000| = t + C2 0.05 ln |0.05 A - 20, 000| = 0.05t + C1

æ 10 ö t = -10ln çç ÷÷÷ çè 27 ø

0.05 A - 20, 000 = Ce0.05t A = 20Ce0.05t + 400, 000

» 9.83 It will take 10 days.

57.

Since t = 0 when A = 300, 000, 300,000 = 20C (1) + 400,000

A = 300,000 when t = 0; r = 0.05,

C = -5000.

D = -20,000

(a)

dA = 0.05 A - 20, 000 dt

(b)

1 dA = dt 0.05 A - 20,000

A = 20(-5000)e0.05t + 400,000 A = 400,000 - 100,000e0.05t .

Find t for A = 0. 0 = 400, 000 - 100, 000e0.05t 400, 000 e0.05t = = 4 100, 000

1 ln |0.05 A - 20, 000| = t + C 0.05 ln |0.05 A - 20, 000| = 0.05t + k

ln e0.05t = ln 4

ln |0.05(300, 000) - 20, 000| = k

t = 20 ln 4 = 27.73

k = ln | -5000| = ln 5000

It will take about 27.7 yr.

ln |0.05 A - 20, 000| = 0.0.5t + ln 5000 |0.05 A - 20, 000| = 5000e0.05t

59.

Since 0.05 A < 20, 000, |0.05 A - 20, 000| = 20, 000 - 0.05 A 20, 000 - 0.05 A = 5000e A=

y = -2 ln (1 + 5t ) + C

0.05t

50 = -2 ln 1 + C

1 (20, 000 - 5000e0.05t ) 0.05 = 5

= 100, 000(4 - e0.05t ).

When t = 10,

(

-10 dy = ; y = 50 when t = 0. dt 1 + 5t

A = 100, 000 4 - e0.05(10)

)

» $235,127.87.

=C y = 50 - 2 ln (1 + 5t )

(a) If t = 24, y = 50 - 2ln [1 + 5t (24)] » 40 insects.

(b) If y = 0, 50 = 2 ln(1 + 5t )

58.

dA = rA - D; t = 0, A = $300,000; dt r = 0.05; D = $20,000

1 + 5t = e 25 t =

e25 - 1 5

» 1.44 ´ 1010 hours » 6 ´ 108 days » 1.6 million years.

808 60.

Chapter 10 DIFFERENTIAL EQUATIONS dy = ky; k = 0.1, t = 0, y = 120 dt

62.

dy = k dt y

dy = rate of smoke in - rate of smoke out dt

rate of smoke in = 0 æ yö 1200 y rate of smoke out = çç ÷÷÷ (1200) = çè V ø V

ln | y | = kt + C1 | y | = e kt +C1

120 = Me0

dy -1200 y = dt V dV = 1200 - 1200 = 0 dt

M = 120

V (t ) = C1; at t = 0, V = 15,000.

y = Me kt y = Me0.1t

C1 = 15,000

y = 120e0.1t

V (t ) = 15,000

Let t = 6 and find y. y = 120e

dy -1200 y = = -0.08 y dt 15,000 dy = -0.08 dt y ln y = -0.08t + C

0.6

» 219

After 6 weeks, about 219 are present. 61.

y = ke-0.08t

dx = 0.2 x - 0.5 xy dt dy = -0.3 y + 0.4 xy dt

At t = 0, y = 20. 20 = ke0

k = 20

dy dt = dx dt dt

Therefore,

-0.3 y + 0.4 xy 0.2 x - 0.5xy y(-0.3 + 0.4 x) = x(0.2 - 0.5 y) -0.3 + 0.4 x 0.2 - 0.5 y dy = dx y x æ ö çç 0.2 - 0.5 ÷÷ dy = æç -0.3 + 0.4 ÷÷ö dx çç ÷÷ ÷ø çè y è x ø

y = 20e-0.08t .

At y = 5, 5 = 20e-0.08t .25 = e-0.08t - ln (0.25) 0.08 = 17.3 min.

t =

0.2 ln y - 0.5 y = -0.3 ln x + 0.4 x + C 0.3 ln x + 0.2 ln y - 0.4 x - 0.5 y = C

Both growth rates are 0 if 0.2 x - 0.5 xy = 0 and -0.3 y + 0.4 xy = 0.

63.

Let y = the amount in parts per million (ppm) of smoke at time t. When t = 0, y = 20 ppm, V = 15,000 ft 3.

If x ¹ 0 and y ¹ 0, we have

rate of smoke in = 5 ppm,

0.2 - 0.5 y = 0 and -0.3 + 0.4 x = 0, so

rate of smoke out

x = 34 units and y = 52 units.

æy ö = (1200 ft 3 /min) çç ppm/ft 3 ÷÷÷ çè V ø Rate of air in = rate of air out,

so V = 15,000ft 3 for all t.

Chapter 10 Review

809

dy 1200 y 2y = 5= 5dt 15,000 25 125 - 2 y = 25 dt 1 dy = 125 - 2 y 25 t 1 - ln (125 - 2 y) = +C 2 25 1 - ln (125 - 2(20)) = C 2 1 C = - ln 85 2

300 = =

a =

b

N  y0 , k  aN , and t is in weeks. y0

The number of individuals uninfected at time t is y = N -

1 + ( N - 1)e-aNt N ( N - 1) = . N - 1 + eaNt

y = =

700(699) 699 + e700at 489,300 699 + e700at

489,300 699 + e1.140t

(b) At t = 7 weeks, 489,300

y =

699 + e1.140(7) » 135 people.

65.

y =

1 + be-kt i = 1, 2,3.

; y = yi when x = xi ,

t1, t2 , t3 are equally spaced: t3 = 2t2 - t1, so t1 + t3 = 2t2 , t + t3 or t2 = 1 . 2

Show N =

1 + 1 - 2 y1 y3 y2 . 1 - 1 y1 y3 y22

Let

Substitute N = 700.

ln 932

Therefore,

Which is negative. It is impossible to reduce y to 10 ppm.

The solution is N y = , where 1 + ( N - 1)e-kt

4200a

» 0.00163 4200 k = 700a » 700(0.00163) » 1.140

1 t 1 - ln [125 - 2(10)] = - ln 85 2 25 2 2t ln 105 = ln 85 25 25 t = [ln 85 - ln 105], 2

dy = a( N - y) y. dt

699 + e

e 4200a = 932

y =

(a) The differential equation for y, the number of individuals infected, is

700(6) a

699 + e 489, 300

209, 700 + 300e 4200a = 489, 300

If y = 10,

64.

489, 300

1 1 2 + y1 y3 y2

(

)

2 1 + be-kt2 1 + be-kt1 1 + be-kt3 = + N N N 1 -kt3 -kt1 1 + be = + 1 + be - 2 - 2be-kt2 N b é -kt1 = + e-kt3 - 2e-kt2 ùú êe û Në

(

Substitute t = 6, y = 300.

)

810

Chapter 10 DIFFERENTIAL EQUATIONS Let

N - y0 326.347 - 39.8 = y0 39.8

b = B =

1 1 - 2 y1y3 y2

» 7.20.

1 + be-kt )( 1 + be-kt ) ( =

If 1920 corresponds to t = 5 decades, then

(

1 + be-kt2

105.7 =

1 + 7.23e-5k 326.3 1 + 7.20e-5k = 105.7 ö 1 æç 326.3 - 1÷÷÷ e-5k = çç ø 7.20 è 105.7

)

N 1 é = 2 ê 1 + be-kt1 + e-kt3 + b2e-k (t1 + t3 ) N ë -1 - 2be-kt2 - b2e-2kt2 ùú û b é -kt1 -kt3 -k (2t2 ) = 2 êe + be + be N ë -2e-kt2 - be-2kt2 ùú û b é -kt1 -kt3 = 2 êe +e - 2e-kt2 ùú ë û N

öù 1 é 1 æç 326.3 so k = - ln ê - 1÷÷÷ ú çç ê ø úû 5 ë 7.20 è 105.7 » 0.248.

(b)

326

y =

1 + 7.20e-0.248t

In 2010, t = 14. If t = 14,

Clearly, BA = N .

326

y =

1 + 7.20e-0.248(14)

Hence, N =

66.

N =

1 + 1 - 2 y1 y3 y2 1 - 1 y1 y3 y22

(a)

y0 = 39.8,

1 + 7.20e-0.248(16)

» 287 million.

In 2050, t = 18, so 326

y =

68.

(a)

1 + 7.20e-0.248(18)

» 301 million.

300

The points suggest the lower portion of a logistic growth curve, so yes, a logistic function seems appropriate. (b) A calculator with a logistic regression function determines that

From Exercise 66, 1 + 1 - 2 39.8 203.3 105.7 1 1 (39.8)(203.3) (105.7)2

326

y =

(b) Using the formula for N with y1 = 23.2, y2 = 76.0, and y3 = 150.7, N » 207. So, the upper bound that the U.S population was approaching during these years was approximately 207 million. (c) Using the formula for N with y1 = 39.8, y2 = 105.7, and y3 = 203.3, N » 326. So, the upper bound that the U.S. population was approaching during these years was approximately 326 million.

N =

» 266 million

The predicted population is 266 million which is less than the table value of 308.7 million.

1 + 1 - 2 y1 y3 y2 . 1 - 1 2 y1 y3 y2

(a) Using the formula for N with y1 = 5.3, y2 = 23.2, and y3 = 76.0, N » 185. So, the upper bound that the U.S. population was approaching during these years was approximately 185 million.

67.

326.3

= 326.347 » 326.

y =

487 1 + 58.1e-0.208t

best fits the data.

Chapter 10 Review

811

(c)

70.

y 487/(1 58.1e 0.208t )

(a)

300

The solution to the differential equation is N y = , 1 + be-kx

dy = a( N - y) y; N = 200; dx y0 = 10; y = 35 when x = 3.

N - y0 and k = aN . y0

where b = The model does produce appropriate y-values for the given t-values. (d) As t gets larger and larger, e 0.208t approaches 0, so y approaches

y =

487 487 = = 487. 1 + 58.1 ⋅ 0 1

35 =

Exponential: y = Mekt Limited growth: y = N - ( N - y0 )e-kt N 1 + be-kt

1 + 19e-200a(3)

1 æ 33 ö÷ k = -200a = - ln çç ÷ » 0.465. 3 çè 133 ø÷

k > 0, N > y0 , b > 0

All three models increase for all t, as one can see by looking at the behavior of the

Therefore,

exponential (either ekt or e-kt ).

y =

200 1 + 19e-0.465 x

(b) For x = 5 days,

(b) The second derivatives are: Exponential: y ¢¢ = Mk 2ekt

y =

Limited growth: y ¢¢ = -( N - y0 )k e

(

Nbk 2e-kt be-kt - 1

)

In 5 days, about 70 people have heard the rumor.

-kt 3

(1 + be ) For the exponential, y ¢¢ is always positive and the model is concave upward everywhere. For the limited growth, y ¢¢ is always negative, so the model is concave downward everywhere. The logistic second derivative is positive for ln b be-kt > 1, or ln b - kt > 0, or t < , k so the model is concave upward for t < (ln b) / k . For t > (ln b) / k the second derivative is negative and the model is concave downward. There is an inflection point at t = (ln b) / k.

200

1 + 19e-0.465(5) y » 69.98

2 -kt

Logistic: y ¢¢ =

200

35 + 665e-600a = 200 165 33 e-600a = = 665 133 æ 33 ÷ö -600a = ln çç çè 133 ÷÷ø

69. The models are

(a)

1 + 19e-200ax

Since y = 35 when x = 3,

Therefore, according to this model, the U.S. population has a limiting size of 487 million.

Logistic: y =

200 - 10 = 19; k = 200a 10 200

71.

(a)

dx = 1 - kx dt Separate the variables and integrate. dx

ò 1 - kx =ò dt -

1 ln |1 - kx | = t + C1 k

Solve for x. ln|1 - kx | = -kt - kC1

|1 - kx | = e-kC1 e-kt

812

Chapter 10 DIFFERENTIAL EQUATIONS Therefore,

1 - kx = Me-kt , where M = e-kC1 . 1 1 x = - (Me-kt - 1) = + Ce-kt , k k

T = -260ekt + 300.

At T (1) = 150,

M where C = - . k

150 = -260ek + 300 -150 = -260ek 15 = ek 26 æ 15 ö ln çç ÷÷÷ = k ln e çè 26 ø

(b) Write the linear first-order differential equation in the linear form dx + kx = 1. dt

The integrating factor is I (t ) = e ò

k dt

= ekt .

æ 15 ö k = ln çç ÷÷÷ çè 26 ø

Multiply both sides of the differential equation by I(t).

k = -0.55.

dx kt e + kxekt = ekt dt

Therefore, T = -260e-.055t + 300.

Replace the left side of this equation by Dt ( xekt ) = kt

Dt ( xe ) = e

T = -260e-0.55(2) + 300

= 213.

Integrate both sides with respect to t. 1 xekt = ekt dt = ekt + C k

Solve for x. x =

At t = 2,

dx kt e + xkekt . dt

1 + Ce-kt . k

73.

dT = k (T - TF ); TF = 300; T = 40 dt when t = 0. T = 150 when t = 1.

From Exercise 65 in Section 1 of this chapter, the solution to the differential equation is

and larger, Ce-kt approaches 0,

(

1 + Ce-kt x ¥ k

so lim 72.

) = 1k .

dT = k (T - TF ) dt dT = k (T - 300) dt T (0) = 40 T (1) = 150 dT = k dt T - 300 ln | T - 300| = kt + C1 | T - 300| = ekt +C1 T - 300 = Cekt T = Cekt + 300

Since T (0) = 40, 40 = Ce0 + 300 C = -260.

T = Cekt + TM

where C is a constant (-k has been replaced by k in this exercise.) Here, T = Cekt + 300. 40 = C + 300 C = -260 T = 300 - 260ekt 150 = 300 - 260ek 15 ek = 26 æ 15 ö k = ln çç ÷÷÷ » -0.55 çè 26 ø T = 300 - 260e-0.55t 250 = 300 - 260e-0.55t 5 e-0.55t = 26 æ 5 ö 1 ln çç ÷÷÷ » 3 hr t =0.55 çè 26 ø

Extended Application 74.

813 dv = G 2 - K 2v 2 dt

(a)

(b)

dv = (G 2 - K 2v 2 ) dt 1 G 2 - K 2v 2 1

G e2GKt - 1 lim 2GKt K t ¥ e t ¥ +1 G G lim v = ⋅1 = K K t ¥ lim v =

dv = dt

ò G - K v dv = ò dt 1 1 = ò dt K ò ( ) -v 2

2 2

G K

A falling object in the presence of air resistance has a limiting velocity, G . K

(c)

Use entry 7 from the table of integrals. 1 K

⋅ 2

G +v 1 K = t + C1 ln G -v 2G K K

G + Kv 1 = t + C1 ln G - Kv 2GK ln

æ G e2GKt - 1 ö÷ ç ÷÷ lim v = lim çç ⋅ 2GKt t ¥ t ¥ çè K e + 1 ÷ø÷

G k G 88 = k G k = 88

lim v =

t ¥

G2 88 32 Gk = 88 Gk =

G + Kv = 2GKt + C2 G - Kv

+ Kv Since v < G , Kv < G and G is K G - Kv

64 88 2Gk » 0.727 2Gk =

positive. æ G + Kv ÷ö = 2GK (t ) + C2 ln çç çè G - Kv ÷÷ø

v =

When t = 0, v = 0, so

G e2Gkt - 1 k e2Gkt + 1

e0.727t - 1 v = 88 0.727t +1 e

æ G + 0 ÷ö = 2GK (0) + C2 ln çç çè G - 0 ÷÷ø ln1 = C2 C2 = 0.

Thus, æ G + Kv ÷ö = 2GKt ln çç çè G - Kv ø÷÷

Extended Application: Pollution of the Great Lakes

G + Kv = e2GKt G - Kv G + Kv = Ge2GKt - Kve2GKt

t =

1 æç PL (0) ö÷ ÷÷ ln ç k ççè PL (t ) ø÷

(a)

1 P (t ) = 2.6; L = 0.4 k PL (0)

Kv + Kve2GKt = Ge2GKt - G vK (e2GKt + 1) = G(e2GKt - 1) v = v =

G(e2GKt - 1)

æ 1 ö÷ = 2.4 yr t = 2.6 ln çç çè 0.4 ÷÷ø

K (e2GKt + 1) G e2GKt - 1 ⋅ K e2GKt + 1

(b)

1 P (t ) = 2.6; L = 0.3 k PL (0) æ 1 ö÷ = 3.1 yr t = 2.6 ln çç çè 0.3 ø÷÷

814 2.

Chapter 10 DIFFERENTIAL EQUATIONS t =

(a)

1 æç PL (0) ÷ö ln ç ÷÷ k ççè PL (t ) ÷ø

1 P (t ) = 30.8; L = 0.4 k PL (0)

òe

ò P (x)e dx. 0

(k - p) x

1 P (t ) = 30.8; L = 0.3 k PL (0)

(a) For Lake Michigan, k » 0.0325. Graph 1 (0.0325e-0.02 x - 0.02e-0.0325 x ) y1 = 0.0125

1 æ P (0) ö t = ln ççç L ÷÷÷ k çè PL (t ) ÷ø

(a)

PL (0)e- px ekx dx = PL (0)

é 1 ù = PL (0) ê e(k - p) x ú êk - p ú ë û0 1 = PL (0) éê e(k - p)t - 1ùú ë û k- p

æ 1 ö÷ = 37.1 yr t = 30.8 ln çç çè 0.3 ÷÷ø

æ 1 ö÷ = 28.2 yr t = 30.8 ln çç çè 0.4 ÷÷ø

(b)

Substitute Pi ( x) = PL (0)e- px in

and y2 = 0.50, and find the intersection point:

1 P (t ) = 189; L = 0.4 k PL (0) æ 1 ö÷ = 173 yr t = 189 ln çç çè 0.4 ÷ø÷

(b)

1 P (t ) = 189; L = 0.3 k PL (0) æ 1 ÷ö = 228 yr t = 189 ln çç çè 0.3 ÷÷ø

It will take about 67 years to reduce pollution to 50% of its current value. This compares with 21 years assuming pollution-free inflow.

(a) ekt PL (t ) = PL (0) + k ekx

ò P (x)e dx 0

dPL + kekt PL (t ) = kPi (t )ekt dt dPL + kPL (t ) = kPi (t ) dt

(b) With p = 0, the inflow is constant at the same pollution level as the lake. The ratio PL (t ) 1 = (kept - pe-kt ) PL (0) k- p

(b) Substituting t = 0 in both sides of t é ù ê ú PL (t ) = e-kt ê PL (0) + k Pi ( x)ekx dx ú ê ú êë úû 0

becomes 1 (ke0 - 0) = 1, k -0

which means that the lake pollution level is constant, as would be expected.

produces PL (0) on both sides. On the right side, e-kt becomes e0 = 1, and the integral goes to zero (because lower limit = upper limit). 1 » 0.0325. (c) Since 30.8 = 1k , k = 30.8 flow rate

This represents Vr = lake volume , which means that every year, the total flow is 3.25% of the lake’s volume. That is the percentage replaced. The higher k is, the lower 1k is. So the lake with the biggest turnover is Lake Erie

Chapter 11

PROBABILITY AND CALCULUS Your Turn 3

11.1 Continuous Probability Models Your Turn 1

ò 2xe

P(0 £ X £ 1) =

f ( x) =

2 x2

on [1, 2]

f ( x) dx =

2 ö æ = çç -e-x ÷÷ è ø0

1 = - - (-1) e 1 = 1 - » 0.6321 e

ò x 1

dx 2

2 x 1

= -1 - (-2) = 1. Thus f ( x) is a probability density function for the interval [1, 2]. P(3/2 £ X £ 2) =

Your Turn 4

The cumulative distribution function from Example 5 is 2

F ( x) = -e-x + 1. The probability that there is a bird’s nest within 1 km of the given point is F (1) = -

3/2 x

dx 2

2 x 3/2

11.1

æ 2 ö÷ = -1 - çç çè 3/2 ÷÷ø

W1.

= -1 + =

Warmup Exercises 4

ò (12x + 9 x ) dx 2

(

4 3

= 4 x3 + 6 x3 / 2

1 3

2æ

6 ö

ò çççè x - x ÷÷÷ø dx 4

The function f ( x) = kx3 will be nonnegative on the interval [0, 4] for any positive k.

= 4 ⋅ 43 + 6 ⋅ 43 / 2 = 304

Your Turn 2

kx3 dx =

1 + 1 » 0.6321. e

W2.

Condition 1 holds because f ( x) ³ 0 on [1, 2]. Condition 2 holds because

-x 2

(

= 4 ln x + 2 x-3

æ 1ö = çç 4 ln 2 + ÷÷÷ - ( 0 + 2 ) çè 4ø

k 4 4 x 0 4

= 4 ln 2 -

k = 44 4 = 64k

Thus f ( x) will be a probability density function on [0, 4] when 64k = 1, or k = 1/64.

W3.

7 » 1.023 4

ò 20e xdx 5

= 4e5 x

1 0

= 4e5 - 4 » 589.7

815

816

Chapter 11 PROBABILITY AND CALCULUS

Show that condition 2 holds.

11.1

Exercises

False. The probability of an event can never be larger than 1.

False. The probability density function must be greater than or equal to zero.

True

ò f (x) dx

1 2 x ; [1, 4] 21

f ( x) =

Since x 2 ³ 0, f ( x) ³ 0 on [1, 4]. 4 1 1 æç x3 ö÷÷ çç ÷ x 2 dx = 21 1 21 çè 3 ÷÷ø

1 1 x - ; [2, 5] 9 18 Show that condition 1 holds. Since 2 £ x £ 5,

10.

5æ

x3 3 2 x dx = 98 3 98

1ö

ò çèçç 9 x - 18 ÷÷ø÷ dx = 9 ò çççè x - 2 ÷÷ø÷ dx 2

3 2 x ; [3, 5] 98

f ( x) =

Since x 2 ³ 0, f ( x) ³ 0 on [3, 5].

Show that condition 2 holds. 1ö

Yes, f ( x) is a probability density function.

Hence, f ( x) ³ 0 on [2, 5].

1 æç 64 1ö = - ÷÷÷ = 1 çç è 21 3 3ø

2 1 5 £ x£ 9 9 9 1 1 1 1 £ x£ . 6 9 18 2

1 æç x 2 1 ö÷ = çç - x ÷÷÷ 9 çè 2 2 ø÷

1 (8 + 1) 9 =1

Yes, f ( x) is a probability density function.

5 3

Yes, f ( x) is a probability density function. 11.

1 1 x - ; [3, 4] 3 6 Show that condition 1 holds. Since 3 £ x £ 4,

1 = (125 - 27) 98 =1

ö 1 æ 25 5 4 = çç - - + 1÷÷÷ ø 9 çè 2 2 2

1 = (16 - 4 - 9 + 3) 6 =1

f ( x) =

5æ

Yes, f ( x) is a probability density function.

True

False. P(a £ X £ b) =

æ x2 1 1ö x ö÷ ç - ÷÷÷ ççç x - ÷÷÷ dx = çç çè 6 6ø 6 ÷ø 3 è3 4æ

f ( x) = 4 x3; [0, 3] 3

æ x 4 ö÷ ç x dx = 4çç ÷÷÷ 4 çè 4 ÷ø 0 0

æ 81 ö = 4çç - 0 ÷÷÷ çè 4 ø

f ( x) =

= 81 ¹ 1

No, f ( x) is not a probability density function.

1 4 x£ 3 3 5 1 1 7 £ x- £ . 6 3 6 6 1£

Hence, f ( x) ³ 0 on [3, 4].

Section 11.1 12.

817

f ( x) =

x3 ; [0, 3] 81

17.

x3 x4 dx = 81 324

f ( x) = kx1/2 ; [1, 4]

kx1/2dx =

2 k (8 - 1) 3 14 = k 3 =

1 4 ¹1 =

If 14 k = 1, 3

No, f ( x) is not a probability density function.

13.

f ( x) =

3 . 14

k =

x ; [-2, 2] 16

Notice that f ( x) = 34 x1/2 ³ 0 for all x in [1, 4].

2 1 1 æç x3 ö÷÷ ç ÷ x 2 dx = 16 -2 16 ççè 3 ÷ø÷

18.

-2

f ( x) = kx3/2 ; [4, 9]

1 æç 8 8ö çç + ÷÷÷ 16 è 3 3ø 1 = ¹1 3

2k (243 - 32) 5 422k = 5 =

f ( x) = 2 x 2 ; [-1, 1] 1

ò-1

2 x 2 dx =

2 3 x 3 -1

422k = 1, k = 5 . 5 422

5 x3/2 ³ 0 for all x in Notice that f ( x) = 422

[4, 9].

2 = (1 + 1) 3 4 = ¹1 3

19.

No, f ( x) is not a probability density function. 15.

f ( x) = kx 2 ; [0, 5] x3 kx dx = k 3 0

æ 125 ö æ 125 ö÷ = k çç - 0 ÷÷ = k çç ÷ èç 3 ø÷ èç 3 ø÷

5  0. 90

( )

If k 124 = 1, 3

So f ( x) < 0 for at least one x-value in [-1, 1].

k =

No, f ( x) is not a probability density function.

f ( x) =

5 5 f ( x) = x 2 ; [-1, 1] 3 90

Let x = 0. Then f ( x) = f (0) = -

16.

2k 5/2 x 5 4

kx3/2 dx =

No, f ( x) is not a probability density function. 14.

2 3/2 kx 3 1

3 x 2 ³ 0 for all x in Notice that f ( x) = 125

3 2 12 45 x x+ ; [0, 4] 13 13 52

Let x = 2. Then f ( x) = f (2) = -

3 . 125

[0, 5].

3 < 0. 52

So f ( x) < 0 for at least one x-value in [0, 4]. No, f ( x) is not a probability density function.

818 20.

Chapter 11 PROBABILITY AND CALCULUS f ( x) = kx 2 ; [-1, 2] 2

-1

24. 2

k 3 x 3 -1

kx 2dx =

f ( x) = kx3; [2, 4]

k (8 + 1) 3 = 3k = 1

k (256 - 16) 4 = 60k = 1

1 3

k =

k 4 x 4 2

kx3 dx =

k =

1 60

1 x 3 ³ 0 for all x in [2, 4]. Notice that f ( x) = 60

Notice that f ( x) = 13 x 2 ³ 0 for all x in [-1, 2].

25. 21.

f ( x) = kx; [0, 3] x2 kx dx = k 2 0 3

distribution function is

F ( x) =

æ9 ö = k çç - 0 ÷÷÷ çè 2 ø =

9 k 2

k 2 k 5k x = (9 - 4) = =1 2 2 2 2

2 5

k =

Notice that f ( x) = 52 x ³ 0 for all x in [2, 3]. 23.

f ( x) = kx ; [1, 5]

kx3 dx = k

x4 4

For the probability density function 1 1 f ( x ) = x - on [3, 4], the cumulative 3 6 distribution function is F ( x) = =

òa f (t) dt xæ

ö çç 1 t - 1 ÷÷ dt ç 6 ÷ø 3 è3

1 2 [( x - x) - (9 - 3)] 6 1 = ( x 2 - x - 6), 3 £ x £ 4. 6 =

æ 625 1ö = k çç - ÷÷÷ çè 4 4ø = 156k If 156k = 1, k =

æ1 1 ö = çç t 2 - t ÷÷÷ çè 6 6 ø3

ö çç 1 t - 1 ÷÷ dt ÷ ç 18 ø a è9

1 [( x 2 - x) - (4 - 2)] 18 1 2 ( x - x - 2), 2 £ x £ 5. = 18

26. f ( x) = kx; [2, 3] kx dx =

xæ

Notice that f ( x) = 92 x ³ 0 for all x in [0, 3].

òa f (t) dt x

2 k = . 9

22.

æ1 1 ö = çç t 2 - t ÷÷÷ çè 18 18 ø 2

9 k = 1, 2

For the probability density function 1 on [2, 5], the cumulative f ( x) = 19 x - 18

1 . 156

1 x ³ 0 for all x in [1, 5]. Notice that f ( x) = 156

Section 11.1 27.

819 2

x For the probability density function f ( x) = 21

35.

f ( x) =

on [1, 4], the cumulative distribution function is t dt 21 1

F ( x) =

t = 63

1 2 ¥ a

æ -1 ö = lim ççç + 1÷÷÷ a ¥ çè 1 + a ø÷ = 0 +1= 1

Since x ³ 0, f ( x) ³ 0. f ( x) is a probability density function.

(a)

P(0 £ X £ 2)

(b)

F ( x) =

31.

ò (1 + x)

-3/2

= -4

1 -1/2

» 0.2071 P( X ³ 5) =

1 2

ò (1 + x)

-3/2

1 a ¥ 2

= lim

5 3/2 t dt 4 422

-1/2

(c)

ò (1 + x)

-3/2

a ¥

-1/2

= lim [-(1 + a)

1 ( x5/2 - 32), 4 £ x £ 9. 211

The total area under the graph of a probability density function always equals 1.

= lim [-(1 + x)-1/2]

5 2 ⋅ t 5/2 422 5 4

1 3/2 ( x - 1), 1 £ x £ 4. 7

1 2

= -(1 + x)-1/2

3 2 3/2 ⋅ t 14 3 1

[4, 9], the cumulative distribution is

P(1 £ X £ 3) =

5 . For the The value of k was found to be 422 5 x3/2 on probability density function f ( x) = 422

» 0.4226

3 1/2 t dt 14 1

30.

-3/2

= -3

ò (1 + x) -1/2

3 x1/2 on probability density function f ( x) = 14

[1, 4], the cumulative distribution function is

1 2

= -(1 + x)-1/2

3 . For the The value of k was found to be 14

a ¥

3 2 t dt 3 98

F ( x) =

3/2

= lim [-(1 + a)-1/2 + 1]

1 3 ( x - 27), 3 £ x £ 5. = 98

29.

ò (1 + x)- dx a

distribution function is

t3 = 98

= lim

3 x 2 on [3, 5], the cumulative f ( x) = 98

æ -2 ö÷ 1 (1 + x)-1/2 çç ÷ ç è 1 ÷ø 0 a ¥ 2

For the probability density function

F ( x) =

3/2

= lim

1 3 ( x - 1), 1 £ x £ 4. 63

ò (1 + x)-

28.

1 2

x 2

1 (1 + x)-3/2 ; [0, ¥) 2

a ¥

a 5

+ 6-1/2 ]

æ -1 ö = lim çç + 6-1/2 ÷÷÷ ÷ø ç a ¥ è 1 + a » 0 + 0.4082 = 0.4082

820 36.

Chapter 11 PROBABILITY AND CALCULUS f ( x) = e-x ; [0, ¥)

37. b

ò e dx

e-x dx = lim

b ¥

f ( x) =

-x

1 -x / 2 e ; [0, ¥) 2 1 2

bö æ = lim çç -e-x ÷÷÷ ç 0 ÷ø b ¥ è æ 1 ö = lim çç1 - b ÷÷÷ = 1 ç è b ¥ e ø

P(0 £ X £ 1) =

(b)

P(1 £ X £ 2) =

(c)

P( X £ 2) =

= 0 +1 =1

1 » 0.6321 e e-x dx

f ( x) > 0 for all x. f ( x) is a probability density function.

(a)

1 2

P(0 £ X £ 1) =

- /2

1 0

+1 e x /2 » 0.3935

(b)

1 2

P(1 £ X £ 3) =

1 e

ò e x dx

-1

= -e-x /2

ò e dx

= 1-

æ -1 ö = lim çç a/2 + 1÷÷÷ ç ø a ¥ è e

-x

= - e-x

a ¥

1 1 = - 2 » 0.2325 e e 2

- /2

= lim - e-x/2

= -e

ò e x dx a

-x

1 æç -2 -x/2 ö÷ e ÷÷ ç ø0 a ¥ 2 çè 1

= 1-

= lim

ò e dx

= -e-x

- /2

1 a ¥ 2

f ( x) is a probability density function.

(a)

ò e x dx = lim

f ( x) ³ 0 for all x.

» 0.8647 2

ò e x dx - /2

= -e-x /2

Notice that

-1

P( X £ 2) = P(0 £ X £ 1) + P(1 £ X £ 2).

3 1

3/2

1/2

e e » 0.3834

(c)

P( X ³ 2) =

1 2

ò e

-x /2

1 a ¥ 2

= lim

ò e

-x /2

= lim (-e-x /2 ) a ¥

a 2

æ -1 1ö = lim çç a /2 + ÷÷÷ ç eø a ¥ è e » 0.3679

Section 11.1 38.

821 20

f ( x) =

( x + 20) ¥

Alternatively,

; [0, ¥)

P( X ³ 5) = 1 - P(0 £ X £ 5) = 1 - (0.0476 + 0.1524)

ò (x + 20) dx

= 1 - 0.2 = 0.8.

= lim

b ¥

39.

20( x + 20)-2 dx

= lim -20( x + 20)-1 b ¥

b 0

First, note that f ( x) > 0 for x > 0. Next,

æ 20 ÷ö = lim çç1 ÷=1 b + 20 ÷ø b ¥ çè

ò0 f (x)dx

f ( x) ³ 0 for all x.

Therefore f ( x) is a probability density function. (a)

P(0 £ X £ 1) =

ò 20(x + 20) dx -2

= -20( x + 20)-1

(b)

P(1 £ X £ 5) =

1 » 0.0476 21 5

2 3

a x 16 dx + lim dx 3 12 a ¥ 0 2 3x

a æ x 4 ö÷ æ 8 ö ç = çç ÷÷÷ + lim çç - 2 ÷÷÷ çè 48 ÷ø a ¥ çè 3x ø 2 0

æ 8 ö æ 8 öù æ1 ö é = çç - 0 ÷÷ + ê lim çç - 2 ÷÷÷ - çç - ÷÷÷ ú èç 3 ø÷ ëê a ¥ èç 3a ø çè 12 ÷ú øû 1 2 = + 3 3 = 1.

æ 1 1 ö÷ = -20 çç çè 21 20 ÷÷ø =

ìï x3 ïï if 0 £ x £ 2 ï f ( x) = ïí 12 ïï 16 ïï 3 if x > 2 ïî 3x

ò 20(x + 20) dx -2

Therefore, f ( x) is a probability density function.

= -20( x + 20)-1

(a)

P(0 £ X £ 2) =

(c)

P( X ³ 5) =

æ x 4 ÷ö ç = çç ÷÷÷ çè 48 ÷ø 0

16 » 0.1524 105

ò 20( x + 20) dx -2

(b)

b ¥ ò5

= lim

b ¥

1 » 0.3333 3

P( X ³ 2) = P( X > 2)

20( x + 20)-2 dx

= lim [-20( x + 20)-1]

ò f (x)dx 0

æ 1 1 ö÷ = -20 çç ÷ çè 25 21 ÷ø =

ò 3x dx 2

= lim

a ¥

é æ 1 1 ö÷ ùú = lim ê -20 çç ÷ èç b + 20 25 ø÷ úû b ¥ êë 4 = = 0.8 5

ò 3x dx 2

æ 8 ö = lim çç - 2 ÷÷ a ¥ çè 3x ÷ø

æ 8 ö æ 8 ÷ö = lim çç - 2 ÷÷ - çç ÷ ç ç ÷ è 3 ⋅ 22 ÷ø a ¥ è 3a ø æ 2ö = 0 - çç - ÷÷÷ çè 3 ø =

2 » 0.6667 3

822

Chapter 11 PROBABILITY AND CALCULUS P(1 £ X £ 3)

(c)

= æ x 4 ö÷ ç = çç ÷÷÷ çè 48 ø÷

(c)

2 3

x dx + 12 1

æ 8 ö + çç - 2 ÷÷÷ çè 3x ø

ò 3x 2

P(0 £ X £ 2)

40.

139 » 0.9653 144

41.

f (t ) =

(a)

1 -t /2 ; [0, ¥) e 2

P(0 £ T £ 12) =

1 2

f ( x) dx

20 x 4 dx + lim 9 a ¥ 0 1

ò 9x 1

dx 5

(b)

(c)

1 (-2)e-s /2 2 0

= -(e-t /2 - 1) = 1 - e-t /2 , t ³ 0.

(d)

F (6) = 1 - e-6/2 = 1 - e-3 » 0.9502

dx a ¥ ò1 9 x5

æ 5 ö = lim çç - 4 ÷÷÷ a ¥ èç 9 x ø

The probability is 0.9502.

æ 5 ö æ 5ö = lim çç - 4 ÷÷ - çç - ÷÷÷ ç èç 9 ø a ¥ è 9a ÷ø =

ò0 2 e-s/2 ds

P( X ³ 1) = P( X > 1) = lim

F (t ) =

4 » 0.4444 9

20 12

1 = 10 + 6 e e » 0.0024

æ 4 x5 ö÷ ç ÷÷ = çç çè 9 ø÷÷

-1

20 x 4 dx 9 0

= -e-t /2

4 5 = + 9 9 = 1.

P(0 £ X £ 1) =

1 20 -t /2 e dt 2 12

P(12 £ T £ 20) =

æ æ4 ö é 5 ö æ 5 öù = çç - 0 ÷÷÷ + ê lim çç - 4 ÷÷÷ - çç - ÷÷÷ ú çè 9 ø êë a ¥ çè 9a ø çè 9 ÷ú øû

Therefore, f( x) is a probability density function.

t /2

+1 e6 » 0.9975

a æ 4 x5 ö÷ æ ö ç ÷÷ + lim çç - 5 ÷÷ = çç ÷ çè 9 ø÷÷ a ¥ èç 9 x 4 ø 1 0

(b)

ò e- dt

-1

(a)

= -e-t /2

æ4 ö æ 5 5ö = çç - 0 ÷÷ + çç + ÷÷ ÷ èç 9 ø èç 144 9 ÷ø

ì ï 20 x 4 ï ï if 0 £ x £ 1 ï 9 f ( x) = ï í ï 20 ï if x > 1 ï ï 5 ï 9 î x First, note that f ( x) > 0 for all x > 0. Next,

ò 9x dx

295 » 0.6829 432

2 æ 4 x5 ö÷ æ ö ç ÷÷ + çç - 5 ÷÷ = çç çè 9 ÷÷ø èç 9 x 4 ÷ø 1 0

æ1 1 ö÷ çæ 8 2ö = çç + ç+ ÷÷ ÷ ÷ èç 3 48 ø çè 27 3 ÷ø =

20 x 4 dx + 9 0

5 » 0.5556 9

Section 11.1 42.

f (t ) =

(a)

823 1 æç 3 ö÷ ÷; [4, 9] çç 1 + 11 è t ÷ø

P(6 £ T £ 9) =

ò0

-1

9 1 (t + 6t1/2 ) 6 11 1 (9 + 18 - 6 - 6 6) = 11 1 (21 - 6 6) = 11 » 0.5730

(b)

P(4 £ T £ 5) =

f ( x) =

F ( x) =

1 (t + 6t1/2 ) 11 4

ò 11 (1 + 3t

-1/2

)dt

1 (t + 6t1/2 ) 11 4

1 (7 + 6 7 - 4 - 12) 11 1 = (6 7 - 9) » 0.6250 11 t

1 æç 3 ö÷ ÷ ds çç 1 + s ø÷ 4 11 è

» 0.47

The correct answer choice is c. 44.

We are told that the payment is the loss minus the deductible. Therefore, P(payment < 0.5) = P( X - C < 0.5) = P( X < C + 0.5)

1 (s + 6 s ) 11 4

1 [(t + 6 t ) - (4 + 6 4)] 11 1 = (t + 6 t - 16), 4 £ t £ 9. 11 =

1 (e) F (8) = (8 + 6 8 - 16) 11 » 0.8155 The probability is 0.8155.

43.

25 (10 + t )-1 2 0

25 [(10 + x)-1 - 10-1] 2 25 æ 1 1ö = - çç - ÷÷÷ ç è 2 10 + x 10 ø 25 æç 1 1 ö÷ = ÷ ç ç 2 è 10 10 + x ÷ø 25 æç 1 1 ö÷ F (6) = ÷ çç è 2 10 106 ø÷

F (t ) =

(d)

ò0 2 (10 + t) 2 dt

1 (5 + 6 5 - 4 - 12) 11 1 = (6 5 - 11) 11 » 0.2197

P(4 £ T £ 7) =

25 (10 + x)-2 , 0 £ x £ 40 2

So the probability distribution function is

ò4 11 (1 + 3t-1/2 )dt

(c)

2 x 25

2 k = 1, then k = 25 . Therefore If 25 2

0 -1 -1

= -k (50 - 10 ) æ 1 1 ö = -k çç - ÷ çè 50 10 ÷ø÷

1 (1 + 3t-1/2 )dt 6 11

k (10 + x)-2 dx = -k (10 + x)-1

C + 0.5

ò0

= x2

2 x dx

C + 0.5

= (C + 0.5)2 - 0 = C 2 + C + 0.25.

We are given that the probability is 0.64. C 2 + C + 0.25 = 0.64

If f ( x) is proportional to (10 + x)-2 , then, for some value of k, f ( x) = k (10 + x)-2 on [0, 40]. Find k. We know the total probability must equal 1.

C 2 + C - 0.39 = 0 (C + 1.3)(C - 0.3) = 0 C = -1.3 or C = 0.3

824

Chapter 11 PROBABILITY AND CALCULUS The solution C = -1.3 is extraneous since 0 < C. Thus, C = 0.3. The correct answer choice is b.

45.

f ( x) =

(a)

1 2 x

47.

(a)

P(0 £ X £ 150) =

; [1, 4]

P(3 £ X £ 4) =

1 2

ò3

(b)

200

1.185 ⋅ 10-9 x 4.5222e-0.049846 x dx

100

» 0.4901

= 2 - 31/2 » 0.2679

48.

p( x, t ) =

e-x /(4 Dt )

-u 2 /(4 Dt ) du

ò0 e

1 ö

ò1 çççè 2 x ÷÷ø÷÷ dx

P(1 £ X £ 2) =

ò0 1.185 ⋅ 10-9 x4.5222e-0.049846xdx

P(100 £ x £ 200) =

x-1/2 dx

1 (2) x1/2 2 3 2æ

150

» 0.8131

4æ

ö çç 1 ÷÷ dx ÷ 3 çè 2 x ø÷

(b)

f ( x) = 1.185 ⋅ 10-9 x 4.5222 - 0.049846 x

(a) Let t = 12, L = 6, and D = 38.3.

1 (2) x1/2 2 1

1/2

= 2

(c)

p( x, 12) =

3æ

= 31/2 - 21/2 = 0.3178 f (t ) =

(a)

ò1 (ln 20) t dt 5

(b)

1 ln t ln 20 1

ln 5 » 0.5372 ln 20

P(10 £ T £ 20) =

1 dt ln20 t 10

1 = ln t ln20 10 = 1-

2 1 e-x /1838.4 5.9611

P(0 £ X £ 3)

ln10 » 0.2314 ln20

Evaluate this (and other) integrals using the integration feature on a graphing calculator.

1 ; [1, 20] (ln 20) t

P(1 £ T £ 5) =

ò0 e-u /1838.4 du

p( x, 12) »

1 = (2) x1/2 2 2

46.

- 1 = 0.4142

ö çç 1 ÷÷ dx ÷ ç 2 è 2 x ÷ø

P(2 £ X £ 3) =

e-x /1838.4

ò0 p(x, 12)dx 3

ò0 5.9611 e-x /1838.4 dx 2

» 0.5024

The probability that a flea beetle will be recaptured within 3 meters of the release point is about 0.5024. (b) P(1 £ X £ 5) = »

ò1 p(x, 12)dx 5

ò1 5.9611 e

-x 2 /1838.4

» 0.6673

The probability that a flea beetle will be recaptured between 1 and 5 meters of the release point is about 0.6673.

Section 11.1 49.

825

(a)

50.

f (t ) =

(a)

8 7(t - 2)2

P(3 £ T £ 4) =

N (t ) = -0.00008621t 4 + 0.01632t 3 - 1.1280t 2 + 32.66t - 280.8

(c)

(b)

71.5

ò14.4 N (t) dt » 1995, as found using a

P(5 £ T £ 10) =

1 N (t ) 1995 1 = (-0.00008621t 4 1995

S (t ) =

+ 0.01632t - 1.1280t + 32.66t - 280.8).

P(35 £ age < 50) =

(d) Use a calculator to compute the integrals needed in (d).

51.

f ( x) =

(a)

P(3 £ X £ 5) =

» 0.3131; P(18 £ age < 30) =

ò18 S (t) dt

» 0.2844; P(40 £ age) =

ò (t - 2) dt -2

4 » 0.5714 7 8 7

ò (t - 2) dt -2

5 » 0.2381 21

5.5 - x ; [0, 5] 15

ò 35 S (t) dt

10 8 = - (t - 2)-1 5 7 8 æç 1 1 ö÷ = ç - ÷÷ 7 çè 8 3ø 8æ 5 ö = - çç - ÷÷÷ 7 çè 24 ø

calculator. Thus the density function corresponding to the quartic fit is

8 7

4 8 = - (t - 2)-1 3 7 ö÷ 8 æç 1 = - ç - 1 ÷÷ ø 7 çè 2 æ ö 8 1 = - çç - ÷÷÷ 7 çè 2 ø

A polynomial function could fit the data. (b)

; [3, 10]

71.5

ò 40 S (t) dt

» 0.4354;

5.5 - x dx 15 3

æ 5.5 1 x 2 ÷÷ö ç x= çç ⋅ ÷ 15 2 ÷÷ø çè 15 3 æ 5.5 1 52 ö÷÷ ç = çç ⋅5⋅ ÷ 15 2 ø÷÷ çè 15 æ 1 32 ö÷÷ ç 5.5 - çç ⋅3⋅ ÷ 15 2 ÷÷ø çè 15 = 0.2

826

Chapter 11 PROBABILITY AND CALCULUS (b)

5.5 - x dx 15 0

P(0 £ X £ 2) =

f (t ) =

53. 2

(a)

æ 5.5 1 x 2 ö÷÷ ç = çç x⋅ ÷ çè 15 15 2 ÷÷ø 0

(b)

P(T < 365) =

-b /3650.1

= lim (-e b ¥

(b)

P(T > 960) =

= lim

b ¥

P(16 £ T £ 25) = =-

ò e-t

/960

(b)

ò16 f (t) dt

4.045 -0.532 25 t 16 0.532

P(35 £ T £ 80) = =-

= lim (-e-t /960 )

ò 35 f (t) dt

4.045 -0.532 80 t 35 0.532

(

» 0.4081

960

= lim (-e-b /960 + e-1)

» -7.6034 80-0.532 - 35-0.532

960

b ¥

)

» 0.3676

» -7.6034(25-0.532 - 16-0.532 )

1 -t /960 e dt 960 960

7300 -7300/3650.1

4.045 f (t ) = 1.532 t

365

= -e » 0.3163

= 0 + e-7300/3650.1 » 0.1353

1 -t /960 dt e 960

0 -365/960

1 e-t /3650.1 dt 3650.1 7300

b ¥

(a)

= -e-t /960

= lim (-e-t /3650.1)

54.

(a)

1 e-t /3650.1 dt 3650.1 7300

b ¥

= 0.6

365

= lim

æ 5.5 1 42 ÷÷ö ç = çç ⋅4⋅ ÷ çè 15 15 2 ÷÷ø æ 5.5 1 12 ö÷÷ ç - çç ⋅1⋅ ÷ çè 15 15 2 ÷÷ø

1 -t /960 e 960

+ e-365/3650.1

P(T > 7300)

f (t ) =

1095

= -e » 0.1640

æ 5.5 1 x 2 ö÷÷ ç = çç x⋅ ÷ çè 15 15 2 ÷÷ø 1

52.

ò365 3650.1 e-t /3650.1 dt 365 -1095/3650.1

5.5 - x dx 15 1

P(1 £ X £ 4) =

1095

= (-e-t /3650.1)

= 0.6

(c)

P(365 < T < 1095) =

æ 5.5 1 22 ÷÷ö ç = -çç ⋅2⋅ ÷ 15 2 ÷÷ø çè 15 æ 5.5 1 02 ö÷÷ ç - çç ⋅0⋅ ÷ 15 2 ÷÷ø çè 15

1 e-t /3650.1 3650.1

(c)

P(21 £ T £ 30) =

b ¥

= 0 + e-1 » 0.3679

ò 21 f (t) dt

4.045 -0.532 30 t 21 0.532

» -7.6034(30-0.532 - 21-0.532 ) » 0.2601

)

Section 11.1 (d)

827

F (t ) =

ò16

(e)

f ( s) ds

= 1.8838(0.0887) = 0.1671

4.045 -0.532 s 16 0.532 4.045 -0.532 =- 16-0.532 ) (t 0.532 =-

= 1.7395 - 7.6034t

-0.532

The probability is 0.1671. 56.

f (t ) = 3t-4 ; [1, ¥)

16 £ t £ 80

(e)

F (21) = 1.8838(0.5982 - e-0.03211⋅ 21)

(a)

P(1 £ T £ 2) =

F (21) = 1.7395 - (7.6034)(21-0.532 )

P(16 £ T £ 25)

= =

ò16 f (t)dt

(b)

P(3 £ T £ 5) =

ò16 0.06049e 0.03211 dt -

(c)

P(T ³ 3) =

- e-0.03211⋅16 )

ò35 0.06049e-0.03211tdt

b ¥

1 1 » 0.0290 27 125 -4

ò3 3t 4 dt -

æ 1 1 ö = lim çç - 3 ÷÷÷ b ¥ èç 27 b ø

84 35

57.

ò21 0.06049e-0.03211t dt

» -1.88384 (e-0.03211t )

= lim (-t-3)

» 0.4853 P(21 £ T £ 30) =

ò 3t dt b ¥

» -1.88384 (e-0.03211t )

(c)

= lim

P(35 £ T £ 84) =

» 0.2829

(b)

ò3 3t 4 dt

16 -0.03211⋅ 25

= -1.88384(e

= -t-3

0.06049 -0.03211t 25 = (e ) 16 -0.03211 » -1.88384 (e-0.03211t )

1 = 18 7 = = 0.875 8

f (t ) = 0.06049e-0.03211t ; [16, 84]

(a)

ò1 3t 4 dt

= -t-3

» 0.2343

55.

1 » 0.0370 27

(a)

30 21

» 0.2409

(d)

F (t ) =

A polynomial function could fit the data.

0.06049e-0.03211s ds

= 0.06049 ⋅

1 e-0.03211s 16 -0.03211

(b) T (t ) = -2.067t 3 + 78.97t 2 - 704.6t + 4633

= -1.8838(e-0.03211t - e-0.03211⋅16 ) = 1.8838(0.5982 - e-0.03211t ), 16 £ t £ 84

828

Chapter 11 PROBABILITY AND CALCULUS (c)

ò0

T (t ) dt » 100, 716, as found using a

calculator. Thus the density function corresponding to the cubic fit is

2æ

8 -1 64 -2 128 -3 ö÷ x + x ÷÷ dx ççç x ø 9 27 1 è3

  0.0706  0.2657

P(4 pm £ time < 5:30 pm) 17.5

ò16 S (t) dt

Your Turn 2 Use the alternative formula to find the variance of the random variable X with probability density function f ( x) = 45 for x ³ 1. First find the mean . x

» 0.0785;

 =

Expected Value and Variance of Continuous Random Variables

Your Turn 1 Find the expected value and the variance of the random variable X with probability density function f ( x) = 83 on [1, 2].

ò1 x 3x3 dx 8 -

ò1 3 x 2 dx

= Var(X ) =

4 3

8 æç 8 ö÷ - ç- ÷ 6 çè 3 ÷ø

4 x-4 dx

1 b

ò 4x dx -4

4 3

x 2 f ( x) dx -  2

8 = - x-1 3 1

æ 4 ö x çç 5 ÷÷÷ dx çè x ø

bö æ ÷ ç 4 = lim çç - x-3 ÷÷÷ ç b ¥ çè 3 1 ø÷÷ æ 4 æ 4 öö = lim çç - 3 - çç - ÷÷ ÷÷÷ ç ç è 3 ø÷ ø÷ b ¥ è 3b

ò1 x f (x) dx 2

b ¥

= lim

8 16 » 0.0706 ln 2 3 9

» 0.0800;

 =

æ8 32 16 ö÷ æç 64 64 ÷ö = çç ln 2 + ÷÷ - çç ÷ çè 3 9 27 ø è 9 27 ÷ø

P(12 am £ time < 2 am)

11.2

æ8 64 64 ÷ö = çç ln( x) + ÷÷ èç 3 9x 27 x 2 ø

(d) Use a calculator to compute the integrals needed in (d).

ò 0 S (t) dt

16 ö 8

ò çççè x - 3 x + 9 ÷÷÷ø 3x dx 1

1 T (t ) 100, 716 1 = (-2.067t 3 + 78.97t 2 100, 716 - 704.6t + 4633).

S (t ) =

2æ

ö çç x - 4 ÷÷ 8 dx ç 3 ÷ø 3x3 1 è

Var( x) =

= 2

æ 4 ö2 4 x 2 5 dx - çç ÷÷÷ çè 3 ø x

ò 2x dx - 9 -3

bö æ 16 = 2 lim çç -x-2 ÷÷÷ ç ÷ 1ø 9 b ¥ è

æ ö 16 = 2 çç 1 - lim b-2 ÷÷÷ è ø 9 b ¥ = 2-

16 2 = 9 9

Section 11.2

829

Your Turn 3 Find the median m for the probability density function

f ( x) =

1 ; [3, 7] 4

f ( x) = 45 for x ³ 1. x

E( X ) =  =

1 dx = 5 2 1 x m

æ 1 ö÷ çç - ÷ çè x 4 ø÷

= 1

1 1 æç x 2 ö÷ x dx = çç ÷÷÷ 4 èç 2 ø÷ 3 4 7

49 9 = 8 8 =5

1 2

Var( X ) =

1 - 4 - (-1) = 2 m

2æ 1 ö

ò (x - 5) çççè 4 ÷÷÷ø dx 3

1 ( x - 5)3 = ⋅ 4 3

m4 = 2 m = 4 2 » 1.1892

7 3

8 8 + 12 12 4 = » 1.33 3 =

11.2

Warmup Exercises

W1.

P(1 £ X £ 2) =

 »

ò f (x) dx 1

1 ö

ò x dx = ççèç - x ÷÷÷ø 3

4 3 » 1.15

2 1

1 3 = - +1= 4 4

P(1 £ X £ 2) =

Var(X )

f ( x) =

1 ; [0, 10] 10

E( X ) =  =

f ( x) dx

æ 1 ö 5 3/ 2 x dx = çç x5 / 2 ÷÷÷ ç è 31 ø1 62

1 5/ 2 4 2 -1 -1 = » 0.1502 2 31 31

(

)

Exercises

True

False. The standard deviation of a continuous random variable X must be non-negative, since the square root of a non-negative number is non-negative. False. E ( X ) =  =

Var( X ) =

ò xf (x) dx a

x2 20

æ 1 ö x çç ÷÷÷ dx çè 10 ø

=5 0 10

True

æ1 ö

ò0 (x - 5)2 çççè 10 ÷÷ø÷ dx

1 ( x - 5)3 = ⋅ 10 3

11.2

10 0

25 25 = + 6 6 25 = » 8.33 3

 »

Var( X ) » 2.89

830 7.

Chapter 11 PROBABILITY AND CALCULUS x 1 - ; [2, 6] 8 4 6 æ x 1ö  = x çç - ÷÷÷ dx ç 4ø 2 è8

æ 2 x3 x 4 ö÷÷ 1 ç = çç ÷÷ 2 ÷ø 9 çè 3 0

f ( x) =

2 1 1 1 - - = » 0.06. 3 2 9 18  = Var( X ) » 0.24 =

x ö÷÷ çx çç 8 - 4 ÷÷ dx ÷ø 2 çè

6æ 2

æ x3 x 2 ö÷÷ ç = çç ÷ çè 24 8 ÷÷ø 2

æ 216 36 ÷ö çæ 8 4ö = çç - ÷÷ ÷÷ - çç çè 24 8 ø è 24 8 ÷ø

 =

208 -4 24 26 = -4 3 14 = » 4.67 3 =

6æ 3

Var( X ) = =

æ x4 x3 ÷÷ö 196 ç = çç ÷ 12 ÷÷ø 9 çè 32 2

2 x(1 - x)dx =

æ 2 x3 ÷÷ö ç = çç x 2 ÷ çè 3 ÷÷ø

10.

(2 x - 2 x )dx

2 1 = 1 - = » 0.33 3 3

Use the alternative formula to find

æ 1 ö2 2 x 2 (1 - x)dx - çç ÷÷÷ çè 3 ø 0

ò (2x - 2x )dx - 9 0

3/2

64 64 1 2 289 - + 3 5 3 5 36 » 0.57

f ( x) =

 =

Var( X ) » 0.76

0.89 » 0.94

289

ò1 ( x - x )dx - 36

 »

Var( X ) »

Var( X ) =

f ( x) = 2(1 - x); [0, 1]

 =

» 0.89.

æ 17 ö2 x 2 (1 - x-1/2 )dx - çç ÷÷÷ çè 6 ø 1 4

æ x3 2 x5/2 ö÷÷ 289 ç = çç ÷÷ çè 3 5 ÷ø 36 1

æ 1296 216 ÷ö = çç ÷ çè 32 12 ÷ø æ 16 8 ö 196 - çç - ÷çè 32 12 ÷÷ø 9

ò1 ( x - x1/2 )dx

16 16 1 2 - + 2 3 2 3 17 = » 2.83 6

 =

x 2 ö÷÷ 196 çç x ç 8 - 4 ÷÷ dx - 9 ÷ ç 2 è ø

ò1 x(1 - x-1/2 )dx 4

æx æ 14 ö2 1ö x 2 çç - ÷÷ dx - çç ÷÷ çè 3 ø÷ èç 8 4 ÷ø 2

æ x2 2 x3/2 ÷÷ö ç = çç ÷ çè 2 3 ÷÷ø 1

Use the alternative formula to find Var( X ) =

1 ; [1, 4] x

f ( x) = 1 -

1 æç 3 ö÷ ÷; [4, 9] çç1 + 11 è x ÷÷ø 9

x æ

3 ö

ò4 11 çççè1 + x ÷÷÷ø dx ò4 11 ( x + 3x1/2 )dx

ö÷ 1 æç x 2 çç = + 2 x 3/2 ÷÷÷ 11 çè 2 ÷ø

ö 1 æç 81 = + 54 - 8 - 16 ÷÷÷ çç ø 11 è 2 141 = » 6.41 22

Section 11.2

831

Var( X ) = =

9 2æ

x ç 3 ö÷ ÷ dx -  2 çç 1 + x ÷ø 4 11 è

12.

f ( x) = 3x-4 ; [1, ¥)

 =

1 2 ( x + 3x 3/2 )dx -  2 4 11

æ3 3 ö 3 = lim çç - 2 ÷÷÷ = = 1.5 2 ø b ¥ èç 2 2b

» 2.09

Var( X ) =

Var( X ) » 1.45

ò1 x(4x-5)dx

= lim

a ¥

15.

ò1 x (4x a ¥

9 4

9 3 = = 0.75 4 4 Var( X ) » 0.87

 =

4 = » 1.33 3

= lim

= 3-

æ -4 4ö = lim çç 3 + ÷÷÷ ç 3ø a ¥ è 3a

-5

æ 3ö 9 = lim çç 3 - ÷÷÷ ç bø 4 b ¥ è

æ 4 x-3 ö÷ ç ÷÷ = lim çç a ¥ çè -3 ÷ø÷ 1

Var( X ) =

b ¥

-4

æ3ö )dx - çç ÷÷÷ çè 2 ø

= lim (-3x-1)

ò1 4x dx

ò1 x (3x

-4

= lim

3x-2 dx ò 4 b ¥ 1

f ( x) = 4 x-5; [1, ¥)

 =

ò1 3x-3 dx

æ 3x-2 ö÷ ç ÷÷ = lim çç 2 ÷ø÷ b ¥ çè 1

æ 141 ö÷2 - çç çè 22 ø÷÷

11.

b ¥

1 æç 1458 64 192 ö÷ ÷ ç 243 + 11 çè 5 3 5 ø÷

 »

ò1 x(3x-4 )dx

= lim

ö÷ 1 æç x 3 6 çç = + x5/2 ÷÷÷ -  2 11 çè 3 5 ø÷ 4 =

æ4ö )dx - çç ÷÷÷ çè 3 ø

ò1 4x

-3

æ 4 x-2 ö÷ ç ÷÷ = lim çç a ¥ çè -2 ÷ø÷

(a)

x ; [0, 9] 18

x x

ò0 18 dx 9 3/ 2

ò0 18 dx

2 x5/2 = 90

16 9

x5/2 = 45

243 27 = = = 5.40 45 5

æ -2 ö 16 = lim çç 2 + 2 ÷÷÷ 9 ø a ¥ çè a 16 2 = 2= » 0.22 9 9 2 » 0.47  = Var( X ) = 9

E( X ) =  =

16 dx 9

f ( x) =

(b)

Var( X ) =

9 2

æ 27 ö2 dx - çç ÷÷÷ çè 5 ø

æ 27 ö2 x dx - çç ÷÷÷ çè 5 ø 0 18

9 5/2

x 7/2 = 63 =

æ 27 ö2 - çç ÷÷÷ çè 5 ø

2 2187 æç 27 ö÷ - ç ÷÷ » 5.55 çè 5 ø 63

832

Chapter 11 PROBABILITY AND CALCULUS (c)

 =

(d)

P(5.40 < X £ 9) = =

Var( X ) » 2.36

(d)

x3/2 27

5.4

(e)

7.76

3.04 3/2

17.

f ( x) =

(a) 16.

f ( x) =

(a)

8 2/3

(b)

Var( X ) =

æ x-1/3 ö÷ æ ö2 ç ÷÷ dx - çç 16 ÷÷ x 2 çç èç 5 ÷ø çè 6 ø÷÷ 0

ò ò

8 5/3

8/3

x 16

= 16 -

(c)

 =

6 8

ò 4 x dx - 25 5

64 25

8 64 = 3 25 8 = » 0.11 75

16 = = 3.2 5

(b)

x6 = 24

x5/3 = 10

1 4 x5 x dx = 20 0 4

E( X ) =  =

0.8

32 8 = = = 1.6 20 5

æ x-1/3 ö÷ ç ÷÷ dx x çç ÷ 0 çè 6 ø÷

5.6

1 3 x ; [0, 2] 4

x-1/3 ; [0, 8] 6

 =

1 = [(5.6)2/3 - (0.8) 2/3 ] » 0.5729 4

7.76 3.043/2 27 27 » 0.6043

x 2/3 = 4

x dx 3.04 18

x3/2 27

5.6 -1/3

0.8

7.76 1/2

3.2

(3.2)2/3 » 0.4571 4 P(3.2 - 2.4 < X < 3.2 + 2.4) = P(0.8 < X < 5.6)

P(5.40 - 2.36 £ X £ 5.40 + 2.36) =

= 1-

27 (5.4)1.5 = 27 27 » 0.5352

(e)

ò 6 dx

x 2/3 = 4

x1/2 dx 5.4 18 9

8 -1/3

P( X >  ) =

dx -

256 25

(c)

 =

Var( X )

8 75 » 0.3266 =

» 0.33

256 25

256 144 = = 5.76 25 25

Var( X ) = 2.4

2 0

Section 11.2 (d)

833 x3 dx 8/5 4

P ( 8/5 < X £ 2 ) =

x4 16

(d)

P( X >  ) =

2 8/ 5

x x4 dx = 16 1.2734 4

1.9266 0.1.2734 16 16 » 0.6967 =

18.

f ( x) =

(a)

1.9266

3/4

P(0.75 - 0.4873 < X < 0.75 + 0.4873) = P(0.2627 < X < 1.2373)

1.2734

1.2373

3 (4 - x 2 )dx 0.2627 16

x 3 ö÷÷ 3 æç çç 4 x = ÷ 16 çè 3 ÷÷ø

1.2373

0.2627

3 éê (1.2373)3 (4)(1.2373) = ê 16 êë 3 (0.2627)3 ùú - (4)(0.2627) + ú 3 úû » 0.6137

ò 16 x(4 - x )dx 0

(e) Use a four-place value for the standard deviation.

3 (4 - x 2 ); [0, 2] 16

 =

3 æç 8 9 ö÷ = ÷ çç 8 - - 3 + 16 è 3 64 ÷ø 475 = 1024 » 0.4639

(e) Use a four-place value for the standard deviation. P ( 1.6 - 0.3266 £ X £ 1.6 + 0.3266 ) 1.9266 3

ò 16 (4 - x )dx

3 æç x 3 ö÷÷ çç 4 x = ÷ 16 çè 3 ÷÷ø

256 = 1625 369 = » 0.5904 625

ò 16 (4 x - x )dx 0

æ3 3 4 ö÷ = çç x 2 x ÷ çè 8 64 ÷ø 0 =

(b)

19.

3 = 0.75 4

Var( X ) = =

f ( x) =

(a)

1 ; [3, 7] 4

m = median:

æ3ö 3 2 x (4 - x 2 )dx - çç ÷÷÷ çè 4 ø 0 16

m 3 1 - = 4 4 2 m-3 = 2

3 æç 4 x 3 x 5 ö÷÷ 9 çç = ÷÷ 16 çè 3 5 ÷ø 16 0

3 æç 32 32 ÷ö 9 = ÷÷ çç 16 è 3 5 ø 16 4 9 19 = = 5 16 80 » 0.24

(c)

 =

1 1 x = 4 3 2

ò 16 (4 x - x )dx - 16 2

ò 4 dx = 2 3

m=5

(b)

E ( X ) =  = 5 (from Exercise 1) P( X = 5) =

Var( X ) » 0.4873

» 0.49

ò 4 dx = 0 5

834 20.

Chapter 11 PROBABILITY AND CALCULUS

(a)

If you do the integration on a calculator using rounded values for the limits you may get a slightly different answer, such as 0.0553.

1 ; [0, 10] 10

f ( x) =

1 x m dx = = 10 10 0 10

22.

1 = when m = 5. 2

(b)

f ( x) = 2(1 - x); [0, 1]

(a)

2(1 - x )dx = (2 x - x 2 )

E ( X ) =  = 5 (from Exercise 2)

= x 1 - ; [2, 6] 8 4

f ( x) =

(a)

2- 2 2 » 0.2929

m =

1ö

ò ççèç 8 - 4 ÷÷ø÷ dx = 2

(Reject m = 2 +2 2 , which is not in the

æ x2 x ö÷÷ ç ççç 16 - 4 ÷÷ ÷ø è

= 2

interval [0, 1].)

1 2

(b)

m2 m 1 1 1 - - + = 16 4 4 2 2 m2 - 4m - 4 = 0 m=

4  16 + 16(1) 2

Reject 4-2 32 since it is not in [2, 6]. 4 + 32 m = 2 = 2 + 2 2 » 4.8284 E( X ) =  =

(

14 (from Exercise 3) 3

14 £ X £ 2 + 2 3

E( X ) =  =

)

(2 - 2 ) / 2

= (2 x - x 2 )

If you do the integration on a calculator using rounded values for the limits you may get a slightly different answer, such as 0.0553.

2+ 2 2

(2 + 2 2 ) - 2 + 2 2 =

1/ 3

1 » 0.0556 18

æ x2 x ÷ö ç = çç - ÷÷÷ 4 ÷ø çè 16 14 / 3

(2 - 2 ) / 2

æ 2 - 2 ÷ö æ 2 - 2 ö÷2 æ 1 ö æ 1 ö2 ÷÷ + çç ÷÷ = 2 çç ÷÷÷ - çç ÷÷÷ - 2 ççç ÷ø ççè ÷ø èç 3 ø èç 3 ø çè 2 2

ö ç x - 1 ÷÷ dx ççè 8 4 ø÷

2(1 - x)dx

1/ 3

2+ 2 2 æ

14 / 3

1 (from Exercise 4) 3

æ2- 2 1 ö÷ < X < ÷÷ P ççç çè 2 3 ÷ø

m 2 - 4m - 4 + 8 = 8

(b)

1 2

2m 2 - 4m + 1 = 0

m = median: mæ

= 2m - m 2

P(5 £ X £ 5) = P( X = 5) = 0

21.

(14/3) 14/3 + 16 4

1 » 0.0556 18

Section 11.2 23.

835

f ( x) = 4 x-5; [1, ¥)

25.

m = median:

(a)

ò 4x- dx = 2 5

4 x-4 -4

1 2

= 1

ì ï x3 ï ï ï 12 if 0 £ x £ 2 f ( x) = ï í ï 16 ï ï ï 3x3 if x > 2 ï î Expected value: ¥

ò x f (x)dx

E( X ) =  =

1 2 1 1 1- 4 = 2 m

-m-4 + 1 =

a æ æ x3 ö÷ 16 ö ç x çç ÷÷÷ dx + lim x çç 3 ÷÷÷ dx ç ç è 3x ø a ¥ 2 0 è 12 ø÷

E( X ) =  =

4 (from Exercise 7) 3

(

) ò1.189 4x-5 dx

P 1.19 £ X £ 43 »

»-

æ8 ö é æ 16 ö æ 16 ö ù = çç - 0 ÷÷ + ê lim çç - ÷÷ - çç - ÷÷ ú ÷ø ê a ¥ çè 3a ø÷ çè 6 ÷ø ú çè 15 ë û

1.333

» -x-4

1.189

Var( X ) =

= lim

a ¥

ò 3x- dx 2

1.5

= =

( 3 2 )3

2 æ 16 ö 3

ò x çççè 3x ÷÷÷ø dx 2

ò 3x dx 2

Since the limit diverges, neither the variance nor the standard deviation exists.

16 16 = lim ln | a | - ln |2| 3 a ¥ 3

P( 3 2 < X < 1.5)

= -x - 3 3

2 ö çç 16 ÷÷ dx - æçç 16 ö÷÷ çè 5 ÷ø ç 3÷ 2 è 3x ø ¥æ

2 æ 16 ö 3

a ¥

E ( X ) =  = 1.5 (from Exercise 8)

= lim

» 1.2599 » 1.260

1.5

æ x3 ö÷ ç x 2 çç ÷÷÷ dx + çè 12 ÷ø 0

m3 = 2

ò x çççè 3x ÷÷÷ø dx

1 = 1- 3 = 2 m

(b)

m =

ò x f (x)dx - 

Examine the second integral.

3x-4 dx = -x-3

f ( x) = 3x-4 ; [1, ¥)

16 = 3.2 5

Variance:

1.333

+ 1.3334 1.1894 » 0.1836

a æ x5 ö÷ æ 16 ö ç = çç ÷÷÷ + lim çç - ÷÷ çè 60 ÷ø a ¥ çè 3x ÷ø 2 0

m = 4 2 » 1.189

(a)

m4 = 2

24.

a x4 16 dx + lim dx 2 12 a ¥ 0 2 3x

2m 4 - 2 = m 4

(b)

26.

ìï 20 x 4 ïï if 0 £ x £ 1 ï f ( x) = ïí 9 ïï 20 ïï 5 if x > 1 ïî 9 x

(1.5)3

1 1 » 0.2037 2 (1.5)3

836

Chapter 11 PROBABILITY AND CALCULUS Expected value:

E( X ) =  = = =

27.

x f ( x) dx

0 1

a 20 x 4 20 dx + lim x 5 dx 9 a ¥ 1 9x

0 1

20 x dx + lim 9 a ¥ 0

ì -x for -2 £ x £ 0 ï | x| = ï í ï ï î x for 0 £ x £ 4

ò 9x dx 1

ì | x| ï ï for -2 £ x £ 4 ï f ( x) = í 10 ï ï ï î 0 otherwise First, note that

The expected value is

æ 20 x 6 ö÷ æ ö ç ÷÷ + lim çç - 20 ÷÷ = çç ÷ çè 54 ÷÷ø a ¥ èç 27 x3 ø 1 0

E( X ) =  =

æ æ 10 ö é 20 ÷ö æç 20 ÷ö ùú = çç - 0 ÷÷ + ê lim çç ÷ - ç- ÷ èç 27 ø÷ êë a ¥ èç 27a3 ÷ø èç 27 ÷ø÷úû 10 20 = + 27 27 10 = » 1.111 9

= =

ò x f ( x)dx -  0 1

+ lim

a ¥

æ 10 ö÷2 ç dx ç çè 9 ø÷÷ 9 x5 20

ò-2

-x dx + 10

x3 + 30

-2

ò0 x ⋅ 10 dx 4 2

ò0 10 dx

4 0

28.

56 28 = 30 15

The correct answer choice is d. 1 f (t ) = ; [1, 900] 58 t (a)

E (T ) =

900

ò1 t ⋅ 58 t dt

900 1 t1/2 dt 58 1

1 3/2 t 87 1

æ 20 x7 ö÷ æ 10 ö 100 ç = çç ÷÷ + lim ççç - 2 ÷÷÷ çè 63 ÷÷ø 81 a ¥ è 9 x ø 1 0

900

1 (9003/2 - 13/2 ) 87 » 310.3 hr

æ 10 ö æ 20 ö é = çç - 0 ÷÷÷ + ê lim çç - 2 ÷÷ ê çè 63 ç ø ë a ¥ è 9a ÷ø

æ 10 ö ù 100 -çç - ÷÷÷ ú çè 9 ø ú 81 û

(b)

900

Var(T ) =

ò1 t 2 58 t dt - (310.3)2

900 1 t 3/2 dt - (310.3)2 58 1

1 5/2 t 145 1

20 10 100 + 63 9 81 110 = » 0.1940 567 =

900

Standard deviation:

 =

x⋅

x2 dx + -2 10

a 20 x 6 20 dx + lim dx 9 ¥ a x3 9 0 1 100 81

æ ö -8 ö÷ æç 64 = -çç 0 - 0 ÷÷ ÷+ç ÷ø èç 30 ÷ø èç 30

20 x 4 dx 9 a

| x|

ò-2 x ⋅ 10 dx

x3 =30

Variance: Var( X ) =

Var( X ) » 0.4405

1 (9005/2 - 15/2 ) - (310.3)2 145 = 71, 300.11 =

 =

- (310.3)2

71, 300.11 » 267.0 hr

Section 11.2 (c)

837

P(310.3 + 267 < T £ 900) =

900

1 dt 58 t 577.3

f (t ) =

(a)

900

30.

2 1/2 t 58 577.3

1 -t /2 ; [0, ¥) e 2

E (T ) =  =

1 2

1 1 dt = 2 58 t

1 1 t = 29 2 1

t /2

= - lim (te a ¥

)

- 2 lim (e-t /2 ) a ¥

æ a 0 ö = - lim çç a /2 - 0 ÷÷÷ a ¥ çè e e ø æ 1 1 ö - 2 lim çç a /2 - 0 ÷÷÷ ç a ¥ è e e ø = (0 - 0) - 2(0 - 1)

(b)

Var(T ) =

1 2

ò t e-t dt - 2 2

Use integration by parts twice. 1 2

1 (1 + 3t-1/2 ) dt 141/ 22 11

9 1 (t + 6t1/2 ) 141/ 22 11 é æ 141 ÷ö1/ 2 ùú 1 ê 141 çç = + 9 18 6 ê çè 22 ÷÷ø úú 11 ê 22 ë û » 0.4910

ò e- dt

E (T ) = 2 mo

141 yr 22 » 6.409 yr.

(c)

= -(0) + 2

(a) From Exercise 6,  =

æ 141 ö÷ P çç T > ÷ çè 22 ÷ø

a 1 lim (-2te-t /2 ) 0 2 a ¥

-t /2

1 æç 3 ö÷ ÷; [4, 9] çç 1 + 11 è t ÷ø

 = 1.447 yr.

a ¥

The median life of the bulbs is 240.3 hours.

(b)

a 1 lim -2e-t /2 dt 2 a ¥ 0

+ lim

1 1 ( m - 1) = 29 2 29 m -1 = 2 31 m = 2 961 = 240.3 m= 4

f (t ) =

a 1 lim (-2te-t /2 ) 0 2 a ¥

29.

Use integration by parts.

ò f (t)dt = . a

t /2

a 1 lim te-t /2 dt 2 a ¥ 0

(d) The median life of the bulbs is the value of

ò t ⋅ 2 e- dt 0

2 (9001/2 - 577.31/2 ) » 0.2059 58

m such that

ò t e-t dt 2

1 (-2t 2e-t /2 + 4 te-t /2 dt ) 2 1 = [-2t 2e-t /2 + 4(-2t e-t /2 2

+2 =

ò e-t dt)] /2

1 (-2t 2e-t /2 - 8te-t /2 - 16e-t /2 ) 2

= -t 2e-t /2 - 4te-t /2 - 8e-t /2

838

Chapter 11 PROBABILITY AND CALCULUS Var(T ) = lim (-t 2e-t /2 - 4te-t /2 - 8e-t /2 ) a ¥

2 -a /2

= lim (-a e a ¥

-a /2

- 4ae

 = -4

0 -a /2

- 8e

0.6 2

4 = 2 mo

= -e

-t /2

dt = - e

-t /2

2.5(0.6)2.5

x3.5

2.5 0.6 x

3.5

æ 1 ö÷ 1 + 5(0.6)2.5 çç çè -2.5 ÷÷ø x 2.5 2

(d) The median life is the value of m such that 1

æ 1 5 1 ö÷ = - (0.6)2.5 çç 1.5 ÷ ç 3 è2 0.61.5 ø÷

1 -t /2 1 e dt = 2 0 2 m 1 e-t /2 = 0 2 1 -e-m /2 - (-e0 ) = 2 1 -m /2 -e +1 = 2

æ 1 ö - 2(0.6)2.5 çç 0 - 2.5 ÷÷÷ çè 2 ø

òa f (t)dt = 2 . ò

-e-m /2 =

» 0.8357 + 0.0986 » 0.93

The correct answer choice is c. 32.

Using the hint, we have ìï y for 1 < y < 10 benefit paid = ïí ïïî10 for y > 10

1 2

æ1ö m = -2 ln çç ÷÷÷ » 1.386 çè 2 ø

The median life of the parts is 1.386 months. 31.

æ 1 ÷ö 1 2 = 2.5(0.6)2.5 çç çè -1.5 ÷÷ø x1.5 0.6

+ 1 » 0.6321

2.5

ò e 0 -1

+ 5(0.6)2.5

P(0 < T < 2)

ò 2 ⋅ f ( x)dx

x3.5

= 2.5(0.6)2.5

Var(T ) = 4 mo

1 = 2

2.5(0.6)

= -0 - 0 - 0 + 0 + 0 + 8 - 4

(c)

0.6

+ 0 + 0 + 8e0 ) - 4

 =

x ⋅ f ( x)dx +

Therefore, the expected value of the benefit paid will be E (Y ) = =

Using the hint, we have ïì x for 0.6 < x < 2 loss not paid = ïí ïïî 2 for x > 2

Therefore, the mean of the manufacturer’s annual losses not paid will be

y ⋅ f ( y)dy +

1 10

ò 10 ⋅ f ( y)dy 10 ¥

ò y y + dy +ò 10 y dy 3

= 2

dy + 20

æ 1ö = 2 çç - ÷÷÷ çè y ÷ø

æ 1 ö÷ + 20 ççç - 2 ÷÷ çè 2 y ÷ø÷ 1 10

æ 1 ö æ 1 ö÷ = 2 çç - + 1÷÷ + 20 çç 0 + ÷ ÷ø çè 10 çè 200 ø÷ =

9 1 + = 1.9. 5 10

The correct choice is d.

Section 11.2 33.

839

Since the probability density function is -4

proportional to (1 + x)

ò0 ce-0.004xdx = 1

, we have f ( x)

= k (1 + x)-4 , 0 < x < ¥. To determine k,

solve the equation ò0 k f ( x)dx = 1. ¥

ò k (1 + x) dx = 1 -4

æ 1ö =1 k çç - ÷÷÷ (1 + x)-3 çè 3 ø 0

k - (0 - 1) = 1 3 k =1 3 k =3 -4

Thus, f ( x) = 3(1 + x)

1 e-0.004 x =1 -0.004 0 c (e-¥ - e0 ) = 1 0.004 c (0 - 1) = 1 0.004 c =1 0.004 c = 0.004

Thus, f ( x) = 0.004e-0.004 x , x ³ 0. Now solve for m. m

ò 0.004e

,0 < x < ¥.

-0.004 x

dx =

1 2

æ 1 ö÷ -0.004 x 1 = e 0.004 çç çè -0.004 ÷÷ø 2 0

The expected monthly claims are

1 2 1 e-0.004m = 2

ò (1 + x) dx

x ⋅ 3(1 + x)-4 dx = 3

-(e-0.004m - 1) =

The antiderivative can be found using the substitution u = 1 - x. u -1

ò (1 + x) dx = ò u 4

1 2u

The correct answer choice is c.

3u 3

35.

Resubstitute u = 1 + x. ¥

(a)

æ ö÷ 1 1 ÷÷ = 3ççç + 2 3(1 + x)3 ÷ø÷ èç 2(1 + x )

1 ; [1, 20] (ln 20)t

 =

é æ 1 æ1ö 1 öù 1 = 3 ê 0 - çç - + ÷÷ ú = 3çç ÷÷ = ÷ ÷ ç 2 ç ê ú è ø è ø 3 û 6 2 ë

The correct answer choice is c. 34.

f (t ) =

ò (1 + x) dx

æ1ö 1 ln çç ÷÷÷ -0.004 çè 2 ø m » 173 m=

æ 1 1 ö çç - 4 ÷÷÷ du 3 èç u u ø

æ1ö -0.004m = ln çç ÷÷÷ çè 2 ø

The median benefit is the value of m such that

ò1 t ⋅ (ln 20)t dt 20

ò1 ln 20 dt

t ln 20 1

19 » 6.3424 » 6.342 seconds ln 20

òa f ( x)dx= 12 . First, we need to determine the value of c. Since f is a probability density function,

840 (b)

Chapter 11 PROBABILITY AND CALCULUS

ò1

Var(T ) = =

ò1

t2 ⋅

1 dt -  2 (ln 20)t

(a)

t dt -  2 ln 20

t2 = 2 ln 20

- (6.3424)

(b)

(t - 6.366) 2 0.07599t1.43e-t / 2.62 » 16.6677

The standard deviation is about 16.6677 days, or 4.08 days.

 »

26.3687 » 5.135 sec

(c)

P(6.3424 - 5.1350 < T < 6.3424 + 5.1350) = P(1.2074 < T < 11.4774) 11.4774

ln t ln 20 1.2074

ò5 0.07599t1.43e / 2.62 » 0.3903 -t

The probability that the time from onset to infection is between 5 and 10 days is

1 dt (ln 20)t

ò1.2074

0.07599t 2.43e-t / 2.62 » 6.366

The mean is about 6.37 days.

399 = - (6.3424) 2 2 ln 20 » 26.3687

(c)

36. Compute the integrals required with a calculator, using an upper limit of 40.

about 0.3903.

11.4774

37.

1 (ln 11.4774 - ln 1.2074) ln 20 » 0.7517 =

f ( x) =

(a)

If you do the integration on a calculator using differently rounded values for the limits you may get a slightly different answer, such as 0.7518. (d)

 =

ò1 x ⋅ 2 x dx ò1

4 1/2

x3/2 dx = 2 3

4 1

1 = (8 - 1) 3 7 = » 2.333 cm 3

that òa f (t ) dt = 1 . 1 1 dt = (ln 20) 2 t 1

; [1, 4]

2 x

The median clotting time is the value of m such

(b)

Var( X ) =

x2 ⋅

æ 7 ö2 dx - çç ÷÷÷ çè 3 ø 2 x

4 3/2

ò1

1 1 ln t = ln 20 2 1

ò1 2 dx - 9

1 1 (ln m - 0) = ln 20 2 ln 20 ln m = 2

x5/2 5

m = eln 20/2 » 4.472

49 9

1 49 = (32 - 1) 5 9 » 0.7556

 =

Var( X )

» 0.8692 cm

Section 11.2 (c)

841

P( X > 2.33 + 2(0.87)) = P( X > 4.07)

40.

æ 1 öù 8 éê 1 1 - çç - ÷÷÷ ú = ç ê ú 7 ë m - 2 è 1 øû 2

41.

S (t ) =

e-x /(4 Dt ) L

ò0

1 7 9 = -1 = 16 16 m-2 16 m-2 = 9 34 m = » 3.778 9

The median time is about 3.778 min.

p( x, t ) =

8 æç 1 ö÷ 1 ÷ = çç 7 è t - 2 ÷ø 3 2

1 dx = 2 1 2 x m 1 x = 1 2 1 m -1 = 2 3 m = 2 9 m = = 2.25 4 The median petal length is 2.25 cm.

such that òa f ( x)dx = 1 .

38.

1 (-0.00008621t 4 + 0.01632t 3 1995 - 1.128t 2 + 32.66t - 280.8)

2 e-u /(4 Dt ) du

for t in the interval [14.4, 71.5]. Evaluate the required integrals with a calculator.

Letting t = 12, L = 6, and D = 38.3, 2

p( x, 12) =

e-x (4⋅38.3⋅12) 6

 =

-u /(4⋅38.3⋅12) du

ò0 e

e-x /1838.4 -u 2 /(1838.4) du

ò0 e

Var( X ) =

2 1 e-x /1838.4. 5.9611 The integral in the denominator was evaluated using the integration feature on the calculator.

E( X ) =

» 13.76 years

xe-x /1838.4dx

42.

f (t ) =

2 6 æ 1ö 1 = ⋅ çç - ÷÷÷ ⋅ 1838.4e-x 5.9611 çè 2 ø 0 » 2.990 The expected recapture distance is 2.990 m.

39.

f ( x) = 1.185 ⋅ 10-9 x 4.5222e-0.049846 x

E( X ) =

1000

t 2 S (t ) dt - (39.05) 2

14.4

 » 189.3

ò x ⋅ p( x, 12)dx 6

71.5

= 189.3

1 5.9611

t S (t ) dt

14.4

» 1714.2 - 1524.9

71.5

» 39.05 years

ò 7(t - 2) dt = 2

(d) The median petal length is the value of m

ò 7(t - 2) dt = 2 . 3

=0 The probability is 0 since two standard deviations falls out of the given interval [1, 4].

The median m satisfies

1 -t /960 e ; [0, ¥) 960

E (T ) =

ò 960 e

-t/960

= lim

b ¥

ò 960 e

-t/960

Integrating by parts, choose dv = e-t /960dt and u =

x f ( x) dx

Using the integration function on our calculator. E ( X ) » 110.80

The expected size is about 111.

t . 960

Then v = -960e-t /960 and du =

1 dt. 960

842

Chapter 11 PROBABILITY AND CALCULUS So,

Var(T ) =

ò u dv = uv - ò v du b

b t -t /960 e dt = -te-t /960 0 0 960

= -te-t /960

b 0

ò0 -e-t /960dt

- 960e-t /960

= -(t + 960)e-t /960

43. b ¥

ò 960 e

-t /960

f ( x) =

5.5 - x ; [0, 5] 15

(a)

 = =

= 960 days

æ 5.5 - x ö

ò xçççè 15 ÷÷÷ø dx 0 5æ

1 2ö

5.5

ò çççè 15 x - 15 x ÷÷ø÷ dx 0

t e-t/960dt - 9602 0 960 b ù 1 éê 2 -t/960 ú 2 = lim t e dt ú - 960 960 êëê b ¥ 0 ûú

9602 = 960 days

Var(T ) =

b ¥

Var(T ) =

1 [ lim [-960e-b /960 (b 2 + 2(960)b 960 b ¥

 =

= lim [-(b + 960)e-b /960 + 960]

= 9602

Thus, b

ù t 2e-t /960dt úú - 9602 0 ûú b

+ 2(9602 )) + 2(9603 )]] - 9602 1 = ⋅ 2(9603 ) - 9602 960

= -(b + 960)e-b /960 + 960

E (T ) = lim

1 êé lim 960 êëê b ¥

æ 5.5 ö 1 = çç ⋅ 25 ⋅ 125 ÷÷÷ - 0 çè 30 ø 45

» 1.806

Use the column method for integration by parts. Let u = t 2 and dv = e-t /960dt.

5.5 2 1 3 x x 30 45 0

(b)

Var( X ) =

æ 5.5 - x ÷ö 2 x 2 çç ÷÷ dx -  ç è ø 15 0

5æ

-t 960

-960e-t /960

9602 e-t /960

æ 5.5 3 1 4 ö÷ = çç x x ÷ - 2 çè 45 60 ø÷ 0

-9603 e-t /960

5.5 1 ⋅ 125 ⋅ 625 - 0 -  2 45 60 » 1.60108

ò0 t 2e-t /960dt = -960t e

 = 2 -t /960

- 2t (960 e

+ 2(-9603 e-t /960 )

(c)

)

= -960e-t /960 (t 2 + 2(960)t + 2(9602 ))

+ 2(9602 )) + 2(9603 )

Var( X ) » 1.265

P( X £  -  )

= P( X £ 1.806 - 1.265) = P( X £ 0.541)

= -960e-b /960 (b 2 + 2(960)b

2 -t /960

5.5

ò0 çççè 15 x2 - 15 x3 ÷÷÷ø dx -  2

b 0

0.541

5.5 - x dx 15 0.541

æ 5.5 1 2 ö÷ = çç xx ÷ çè 15 30 ÷ø 0

æ 5.5 ö 1 = çç (0.541) (0.541) 2 - 0 ÷÷÷ çè 15 ø 30 » 0.1886

Section 11.2 44.

843

ln 2 T (a) Find the constant b that will make A(t) a probability density function. A(t ) = A0ekt where k = -

1= 1=

ò ò

(a)

bA(t ) dt

b ( A0e

-(t ln 2) / T

) dt

Ae 1=b 0 -(ln 2) / T 0

ln 2 =b A0T Thus, the probability that a radioactive particle decays after time t is given by f (t ) = bA(t ) =

ln 2 ( A0e-(t ln 2) / T ) A0T

ln 2 -(t ln 2) / T e T

ln 2 -(t ln 2/T ) ö÷ ÷dt ø÷

ò te 0

-(t ln 2/T )

ò16 4.045t 0.532 dt

4.045 0.468 t 0.468 16

4.045 0.468 - 160.468 ) (80 0.468 » 35.56 years

(b)

ò ò

ò16 t 2 f (t) dt - (35.555)2

æ 4.045 ö t 2 çç 1.532 ÷÷ dt - (35.555)2 ç ÷ø èt 16 80

4.045t 0.468 dt - (35.555) 2

16 80

4.045 1.468 (t ) - (35.555)2 1.468 16

ò t çççè T e ¥

4.045 ö

ò16 tçççè t1.532 ÷÷÷ø dt

4.045 1.468 (80 - 161.468 ) - (35.555)2 1.468 » 288.141

ò t f (t)dt

ln 2 éê = T êë

80 æ

ò16 t f (t) dt

Var(T ) =

(b) Calculate the expected life of a particle. Factor out the constant (ln 2/T), use integration by parts, and then distribute the constant (ln 2/T) before evaluating.

0 ¥ æ

æ T ö÷ 1 = bA0 ççç 0 + ÷ è ln 2 ø÷

=

for t in [16, 80].

t1.532

 =

4.045

f (t ) =

45.

 »

288.141 » 16.98 years

ù dt úú û

¥ é æ T ö÷ ln 2 ê æç T ö÷ -(t ln 2/T ) = - ççç ÷÷ e ÷ ê -t çèç ø è ln 2 ø÷ T ê ln 2 0 ë

ù ú e-(t ln 2/T ) dt ú úû

2 é ù ln 2 ê æç T ö÷ -(t ln 2/T ) çæ T ö÷ -(t ln 2/T ) ú = + çç ÷÷ e ÷÷ e ê -t ççè ú è ln 2 ø T êë ln 2 ø úû

P(T < 18.57) =

é T -(t ln 2/T ) ù ú = ê -te-(t ln 2/T ) + e êë úû ln 2 0

ò ò

18.580

f (t ) dt

16 18.580

4.045

t1.532

dt 18.580

T ln 2 » 1.44T

4.045 (18.580-0.532 - 16-0.532 ) 0.532 » 0.1330 =-

Thus, the mean life span of a muon is 1.44T.

4.045 -0.532 (t ) 0.532 16

844

Chapter 11 PROBABILITY AND CALCULUS

(d) The median age m satisfies m

1.532

4.045

ò t

1.532

4.045

dt =

1 . 2

1 2

dt =

ò16 t 2(0.06049e-0.03211t )dt -  2 » -1.88384(t 2 + 62.28589t + 1939.76619) e-0.03211t

4.045 -0.532 m 1 = (t ) 16 0.532 2 4.045 -0.532 1 - 16-0.532 ) = (m 0.532 2

m-0.532 = -

- 38.5122

» 308.305

 =

Var(T ) =

308.290 » 17.558

s » 17.56 years

0.532 æç 1 ö÷ ç ÷ 4.045 çè 2 ÷ø

+ 16-0.532

m-0.532 = 0.16301

20.954

0.06049e-0.03211t dt

e-0.532ln m = 0.16301

= -1.88384 e-0.03211t

-0.532 ln m = ln 0.16301

» 0.1657

-1.8139 -0.532 ln m = 3.4096 ln m =

20.954 16

If you do the integration on a calculator using differently rounded values for the limits you may get a slightly different answer, such as 0.1656.

m = e3.4096 m » 30.35 years

(d) To find the median age, find the value of m m

46.

such that òa f (t )dt = 12 .

f (t ) = 0.06049e-0.03211t ; [16, 84]

ò 0.06049e

(a) Expected value: E (T ) =  =

ò16

ò16 t(0.06049e-0.03211t )dt

(

)

0.06049 -0.03211m 1 e - e-0.03211(16) = 2 -0.03211 0.03211 2 ⋅ 0.06049 0.03211 -0.03211m -0.51376 e = e 2 ⋅ 0.06049 e-0.03211m - e-0.51376 = -

t (0.06049e-0.03211t )dt 84

16 -0.03211⋅84

= -1.88384[(84 + 31.14295)e

- (16 + 31.14295)e-0.03211⋅16

æ 0.03211 ö÷ -0.03211m = ln çç e-0.51376 ÷ çè 2 ⋅ 0.06049 ÷ø æ 1 0.03211 ö÷ ln çç e-0.51376 ÷ ç è -0.03211 2 ⋅ 0.06049 ø÷ m » 34.26 m=

» 38.512 » 38.51

The expected value is 38.51 years.

The median age is 34.26 years.

Var(T )

1 2

» -1.88384(t + 31.14295) e-0.03211t

(b)

dt =

0.06049 -0.03211t 1 e = 2 -0.03211 16

Use integration by parts. 84

-0.03211t

ò16 t 2(0.06049e-0.03211t )dt -  2

Use integration by parts twice.

Section 11.3 47.

845 A temperature one standard deviation below the mean is 5 3 34.5 » 30.16987. A temperature one standard 2 5 3 deviation above the mean is 34.5 + » 38.83013. 2

f (t ) = 3t-4 for t in [1, ¥). ¥

ò t(3t ) dt

 =

-4

= lim

b ¥

ò 3t dt -3

P(  -  £ T £  +  ) =

bö æ ÷ ç 3 = lim çç - t-2 ÷÷÷ ç b ¥ çè 2 1 ø÷÷

æ 3 æ 3 öö = lim çç - b-3 - çç - ÷÷÷ ÷÷÷ èç 2 ø ø÷ b ¥ èç 2

48.

The expected length of a phone call is 1.5 minutes. 1 (-2.067t 3 + 78.97t 2 S (t ) = 100, 716 - 704.6t + 4633) for t in [0, 24]

 = =

ò 30.16987 15 dx

This integral is 1/15 times the difference of the limits, but this difference is just twice the standard deviation, so the probability is 5 3 2 3 ⋅ = » 0.5774 2 15 3

3 2

38.83013

Your Turn 2 f (t ) =

(a) t S (t ) dt

1 -t / 25 for t ³ 0 e 25

The probability that a randomly selected battery has a useful life less than 100 hours is

1 100, 716

(-2.067t 4 + 78.97t 3

100

ò 0 25 e /25 dt

1 (-25e-t /25 ) 25 0

100

- 704.6t 2 + 4633t )dt

Use a calculator to evaluate the integral.  » 1.22

= -(e-100/25 - e0 ) = 1 - e-4

The expected time of day at which a fatal accident will occur is about 1:22 A.M.

11.3 Special Probability Density Functions

» 0.9817.

(b)

Your Turn 1 42 - 27 = 15, so the uniform distribution for the 1 for t in maximum daily temperature T is f (t ) = 15

1 -t

P(T £ 100) =

(c)

1 = 25 1/25 1 = 25  = 1/25

 =

P(T > 40) =

[27, 42].

1 -t

ò 40 25 e /25 dt b

1 (-25e-t /25 ) 25 b ¥ 40

= lim

27 + 42  = 2 69 = 2 = 34.5

= lim (-e-b /25 + e-40/25 ) b ¥ -8/5

= e

The expected maximum daily temperature is 34.5C. 1 (42 - 27) 12 15 = 12

 =

5 3 » 4.33 2

» 0.2019

846

Chapter 11 PROBABILITY AND CALCULUS

Your Turn 3 (a)

The z-score for an age of 79 is

b ¥



79 - 75 16 = 0.25.

P( X > 79) = P( z > 0.25) = 1 - P( z £ 0.25) = 1 - 0.5987

W2.

Compute the z-scores for 67 and 83. 67 - 75 = -0.5 16 83 - 75 = 0.5 z = 16 z =

éæ b ù 1 1ö + lim ê çç - - ÷÷÷ e-2b ú ú 4 4ø b ¥ êë èç 2 û

1 4

x 2e-3x dx

òxe

2 -3 x

= 0.6915 - 0.3085 = 0.3830

(We find the value 0.6915 in the row for 0.5 and the column for 0.00 in the normal table, and the value 0.3085 in the row for -0.5 and the column for 0.00.)

e-3x 1 - e-3x 3 1 -3 x e 9 1 - e-3x 27

(

)

1 9 x 2 + 6 x + 2 e-3x + C 27

x 2e-3x dx = lim

b ¥

ò xe

2 -3 x

é 1 ùb 9 x 2 + 6 x + 2 e-3x ú = lim ê úû 0 b ¥ êë 27

(

Warmup Exercises

ò0

P(67 £ X £ 83) = P(-0.5 £ z £ 0.5)

-2 x

Use tabular integration. D I

(We find the value 0.5987 in the row for 0.2 and the column for 0.05 in the table giving the area under the normal curve.)

W1.

ò xe

= 0.4013

11.3

éæ x ùb 1ö = lim ê çç - - ÷÷÷ e-2 x ú ú 4ø b ¥ êë çè 2 û0

(b)

xe-2 x dx = lim

79 - 

z =

xe-2 x dx

)

2 27

Using integration by parts with dv = e-2 x and u = x, so that

11.3

Exercises

1 v = - e-2 x and du = dx, 2

True

ò xe-2xdx

True

False. The graph of a normal distribution with a larger value of  has more values being farther from the mean than a normal distribution with a smaller value of  .

x 1 = - e-2 x + e-2 x dx 2 2 æ x x 1 1ö = - e-2 x - e-2 x = ççç - - ÷÷÷ e-2 x + C è 2 2 4 4ø

Section 11.3 5.

847

f ( x) =

5 for x in [3, 4.4] 7

(b)  =

This is a uniform distribution: a = 3, b = 4.4.

(c)

P(0.25 < T < 0.25 + 0.25) = P(0.25 < T < 0.5)

1 1 (4.4 + 3) = (7.4) 2 2 = 3.7 cm 1 (b)  = (4.4 - 3) 12 1 (1.4) = 12 » 0.4041 cm

(a)

(c)

 =

f (t ) = 0.05e-0.05t for [0, ¥)

 =

1 = 20 yr 0.05

(b)  =

1 = 20 yr 0.05

(a)

5 x 7 3.7

(c)

P(20 < X < 20 + 20)

f ( x) = 4for x in [2.75, 3]

= P(20 < x < 40)

This is a uniform distribution: a = 2.75, b = 3. (a)

1  = (3 + 2.75) 2 1 = (5.75) 2 = 2.875 hundred dollars, or $287.50 1 (3 - 2.75) 12 1 (0.25) = 12 » 0.0722 hundred dollars, or $7.22

P(2.875 < X < 2.875 + 0.0722) = P(2.875 < X < 2.9472) =

2.9472

-1

20 -2

-e

» 0.2325

f (t ) =

e-t /3 for t in [0, ¥) 3

This is an exponential distribution: a = 13 . (a)

 = 1 = 3 days 3

(c)

P(3 < T < 3 + 3) = P(3 < T < 6)

» 0.2888

f (t ) = 4e-4t for t in [0, ¥)

6 -t /3

ò3 3 dt

= e-t /3

This is an exponential distribution: a = 4. (a)

ò20 0.05e-0.05t dt

1 (b)  = 1 = 3 days

ò2.875 4 dx

2.9472 = 4 x 2.875

= -e-0.05t

(b)  =

(c)

0.25

This is a exponential distribution.

5 dx 7

» 0.2886

0.5

1 = - -2 + -1 e e » 0.2325

3.7 4.1041

ò0.25 4e-4t dt 1

= P(3.7 < X < 4.1041) 4.1041

0.5

= -e-4t

P(3.7 < X < 3.7 + 0.4041)

1 = 0.25 year 4

1  = = 0.25 year 4

6 3

1 = - -2 + -1 e e » 0.2325

848 10.

Chapter 11 PROBABILITY AND CALCULUS f ( x) = 0.1e-0.1x for [0, ¥)

15.

Since 10% = 0.10, the z-score that corresponds to the area of 0.10 to the left of z is -1.28.

16.

Since 2% = 0.02, the z-score that corresponds to the area of 0.02 to the left of z is -2.05.

17.

18% of the total area to the right of z means 1 - 0.18 of the total area is to the left of z.

This is an exponential distribution.

 =

1 = 10 m 0.1

(b)  =

1 = 10 m 0.1

(a)

(c)

P(10 < X < 10 + 10) = P(10 < X < 20) =

ò10

1 - 0.18 = 0.82

The closest z-score that corresponds to the area of 0.82 is 0.92

0.1e-0.1x dx

= -e-0.1x

20 10 -2

= e-1 - e

18. » 0.2325

1 - 0.22 = 0.78

In Exercise 11–18, use the table in the Appendix for areas under the normal curve. 11.

22% of the total area to the right of z means 1 - 0.22 of the total area to the left of z.

z = 3.50

Area to the left of z = 3.50 is 0.9998. Given mean  = z - 0, so area to left of  is 0.5.

The closest z-score that corresponds to the area of 0.78 is 0.77. 22.

For the uniform distribution, f ( x) = b-1 a for x in [a, b]. If m is the median,

Area between  and z is 0.9998 - 0.5 = 0.4998.

Therefore, this area represents 49.98% of total area under normal curve. 12.

1 2

òa b - a dx = 2 1 b-a

z = 1.68

Area between mean z = 0 and z = 1.68 is 0.9535 - 0.5000 = 0.4535.

P( X £ m) =

òa dx = 2

Multiply both sides by b - a. m

Percent of area = 45.35%

ò dx = 2 b - 2 a a

13.

Between z = 1.28 and z = 2.05 Area to left of z = 2.05 is 0.9798 and area to left of z = 1.28 is 0.8997. 0.9798 - 0.8997 = 0.0801

Percent of total area = 8.01% 14.

1 1 b- a 2 2 1 1 m-a = b- a 2 2 1 1 m= b+ a 2 2 b+a m= 2 m

Area between z = -2.13 and z = -0.04 is 0.4840 - 0.0166 = 0.4674.

Percent of area = 46.74% Using the TI-84 Plus C command normalcdf ( 2.13, 0.04) you will get an answer of 46.75%.

xa =

Section 11.3 23.

849

Let m be the median of the exponential -ax

distribution f ( x) = ae m

for [0, ¥).

du = dx and v = -xe-ax -

ò ae- dx = 0.5 ax

ò x2ae-axdx

= 0.5

-e-am + 1 = 0.5

æ ö 1 = x çç -xe-ax - e-ax ÷÷÷ çè ø a

0.5 = e-am -am = ln 0.5 ln 0.5 m=a

24.

f ( x) = ae-ax for [0, ¥) with a > 0. ¥

ò xae

-ax

1 -ax xe a 1 1 1 - xe-ax - 2 e-ax - 2 e-ax a a a

We first consider ò xae-ax dx using integration by parts. u = x

and dv = ae-ax dx; -ax

du = dx and

v = -e

ò xae dx = -xe

ò (-e

-ax

) dx

1 = -xe - e-ax + C a ax + 1 =+C ae ax -ax

 =

ò0

xae-ax dx = lim

b ¥

ò0 xae dx -ax

æ ax + 1 ö÷ æ ab + 1 1ö = lim çç + ÷÷ ÷ = lim çç ax ab ÷ ç ç a ø÷ ø0 b ¥ è b ¥ è ae ae 1  = a

Var (X ) =

Var( X ) =

a 2 x 2 + 2ax + 2 a 2eax ¥

ò0 x2ae-axdx - a2

= lim

b ¥

ò0 x2ae-axdx - a2

= lim -

a 2 x 2 + 2ax + 2

b ¥

a 2e ax

ò0 x ae 2

-ax

æ 1 ö2 dx - çç ÷÷÷ çè a ø

1 - 2 a 0

é a 2 x 2 + 2ax + 2 2 ù 1 = lim êê + 2 úú - 2 ab 2 b ¥ ê a e a úû a ë 2 1 = 2 - 2, a a

since eab grows more rapidly than a 2b 2 + 2ab + 2. Var ( X ) =

since eab grows more rapidly than ab. ¥

-ax

= - x 2 e-ax -

Let

1 -ax xe a 1 + xe-axdx + e-axdx a 1 = - x 2 e-ax - xe-ax a ö 1æ 1 1 + çç -xe-ax - e-ax ÷÷÷ - 2 e-ax ø a a çè a

1 2 -am = - ln 2 ln 2 m= a

 =

ò çççè - xe-ax - a e-ax ÷÷÷ødx

= - x 2 e-ax -

-am = ln

1 -ax e . a

from the previous integration by parts.

- e-ax

Let u = x and dv = xae-ax dx

Thus,

We first consider ò x 2 ae-ax dx using integration by parts.

 =

1 . a

850 25.

Chapter 11 PROBABILITY AND CALCULUS The area that is to the left of x is A=

dt =  du.

2

(t -  )2

1 2 e 2 dt. -¥  2

(t -  )

Let u =

( x -  )2



( x - )2 =  2 ( x -  ) =  x =   .

. Then du = 1 dt and

If x <  -  , f ¢¢( x) > 0.

If t = x, u =

x-

If  -  < x <  +  , f ¢¢( x) < 0. If x >  +  , f ¢¢( x) > 0.

= z.



Therefore, there are points of inflection when x =  -  and when x =  +  .

As t  -¥, u  -¥. Therefore,

 

2 1 e (-1/2)u  du -¥  2

-¥

1 -u 2 /2 e du 2

1 -u 2 /2 e du. 2

In Exercise 27, use Simpson’s rule with n = 140 or the integration feature on a graphing calculator to approximate the integrals. Answers may very slightly from those given here depending on the method that is used.

27.

This is the area to the left of z for the standard normal curve. f ( x) =

26.

(b)

2 2 1 e-( x-  ) /2  2

é 1 (x - ) ù ú f ¢( x ) = e-( x-  ) /2 ê -2 êë  2 2 2 úû -( x -  ) f ¢( x ) = 2 2 3  2 e( x-  ) /2 2

(c)

28.

2 2 ( x-) ö æ ( x -  )2 /2 2 ÷÷ çç e (-1) + ( x -  )e( x -  ) /2 ç  2 ÷÷ = 3 çç ÷÷ 2 2 ÷÷  2 çç e( x -  ) / è ø÷

2 2æ ( x -  )2 ö÷÷ ç -e( x -  ) /2 çç 1 ÷ çè  2 ÷÷ø

f ¢¢( x) = -

 3 2 e( x -  ) / 1-

 3 2 e

( x -  )2

2 ( x -  )2 /2 2

If f ¢¢( x) = 0, then 1 -

2

ò0 0.5xe 0.5 dx » 1.99999 -

ò 0.5x e

2 -0.5 x

dx = 7.999998

The exponential distribution with a = 0.5 has 1 1  = = 2 and  = = 2 . The answer 0.5 0.5 in part (b) of Exercise 23 approximates the mean very closely. If we approximated  using the integrals in parts (b) and (c) of Exercise 27 we would get

Use Simpson’s rule with n = 40 and limits of -6 and 6 to approximate the mean and standard deviation of a normal probability distribution. (a)

( x -  )2

ò0 0.5e 0.5 dx » 1.00000

8.00000 - 1.999992 » 2.00001, again very close to the exact value.

29.

f ¢¢( x)

f ¢¢( x) =

(a)

x -x 2 /2 e dx -¥ 2 »

x -x 2 /2 e dx -6 2

Section 11.3 (b)

851

F ( x ) = P ( X £ x)

x 2 -x 2 / 2 e dx -¥ 2

x 2 -x 2 / 2 » e dx - 6 2  » 0.9999999224 (Simpson's rule)

(a)

 = =

 »

0 ¥

1 t b-a a

1 ( x - a) b-a x-a = , a £ x £ b. b-a =

1.5

32.

1.5 -4 x 1.5

The probability density function for the uniform distribution is f ( x) = ae-ax for x in [0, ¥).

1.5

x(6 x0.5e-4 x )dx

ò 6x e 0 3

ò b - a dt

= 6 x0.5e-4 x

Use f ( x) = abxb-1e-ax with a = 4 and b = 1.5. f ( x) = (4)(1.5) x1.5-1e-4 x

ò f (t) dt a

 » 0.9999999251 (calculator) 30.

The cumulative distribution function for f is F ( x) = P(T £ x)

ò f (t)dt

1.5

6 x1.5e-4 x dx

Using the integration feature on a graphing calculator,

0 x

ò ae dt

= -e-at

 » 0.3583.

-ax

= -e 2

(b)  »

ò (x - 0.3583) 6x e 2

0.5 -4 x 1.5

x 0

+ -e-a(0)

= 1 - e-ax , x ³ 0. dx

Using the integration feature on a graphing calculator,

 2 » 0.05916  » 0.2432. 31.

-at

33.

For a uniform distribution, 1 for [a, b]. f ( x) = b-a Thus, we have f ( x) =

The probability density function for the uniform 1 for x in [a, b]. distribution is f ( x) = ba

for [10, 85].

The cumulative distribution function for f is

(a)

1 1 = 85 - 10 75

1 1 (10 + 85) = (95) 2 2 = 47.5 thousands

 =

Therefore, the agent sells $47,500 in insurance.

852

Chapter 11 PROBABILITY AND CALCULUS (b)

P(50 < X < 85) =

(a)

1 dx 75 50

x 75 50

= P( z < -0.73) = 0.2327 Using the TI-84 Plus C command normaldcf ( 1E99, 32, 32.8, 1.1) we get an answer of 0.2335.

85 50 75 75 35 = = 0.4667 75 =

(b) 34.

We have an exponential distribution with mean  = 5.

f ( x) = 0.2e-0.2 x for [0, ¥)

(b)

P(2 < X < 6) =

ò 0.2e

-0.2 x

= -e-0.2 x

6 2 -1.2

37.

= e-0.4 - e » 0.3691

35.

Probability = 0.1587 (b) Within 1.2 standard deviations of the mean is the area between z = -1.2 and z = 1.2.

Therefore, f ( x) = 0.235e-0.235 x on [0, ¥).

Area to left of z = 1.2 = 0.8849

P( X > 10)

Area to the left of z = -1.2 = 0.1151

0.8849 - 0.1151 = 0.7698

ò10 0.235e-0.235xdx a ¥

2.7 - 2.5 =1 0.2

1 - 0.8413 = 0.1587.

1 = 4.25 a a = 0.235.

= lim

 = 2.5,  = 0.2, x = 2.7

Area to the right of z = 1 is

 =

(a)

(a) Since we have exponential distribution with  = 4.25,

(b)

P( X > 33) æ x - 32.8 33 - 32.8 ö÷ = P çç < ÷÷ çè 1.1 ø 1.1 = P( z > 0.18) = 1 - P( z £ 0.18) = 1 - 0.5714 = 0.4286 Using the TI-84 Plus C command normaldcf (33, 1E99, 32.8, 1.1) we get an answer of 0.4279.

1 =5 a a = 0.2

 =

(a)

P( X < 32) æ x - 32.8 32 - 32.8 ö÷ = P çç < ÷÷ çè 1.1 ø 1.1

ò10 0.235e

Probability = 0.7698

-0.235 x

= lim (-e-0.235 x ) a ¥

Using the TI-84 Plus C command normalcdf (-1.2, 1.2) you will get an answer of 0.7699.

a 10

1 = lim (- 0.235a + 2.35 ) a ¥ e e 1 = 2.35 = 0.0954 e

36. We have a normal distribution with  = 32.8,  = 1.1.

Let x = number of ounces of juice.

38.

We have a normal distribution, with  = 54.40,  = 13.50. 1 2 P( z < -a) = 0.25

P(-a < z < a) =

Since the closest value to 0.25 is 0.2514, we use z = -0.67.

Section 11.3

853

-0.67 < z < 0.67

500 - D +

x - 54.40 < 0.67 13.50 45.36 < x < 63.45

-0.67 <

1 D 2 - D + 500 = 125 2000 1 D 2 - D + 375 = 0 2000

Therefore, P(45.36 < X < 63.45) = 12 , and the middle 50% of the customers spend between $45.36 and $63.45. If we use the TI-84 Plus C command invNorm(0.25, 54.40, 13.50) we directly find the value b such that that P( X < b) = 0.25, which is b = 45.29. The smallest amount spent by the middle 50% is thus $45.29, and the largest amount is 54.40 + (54.40 - 45.29) = $63.51. 39.

D 2 - 2000 D + 750,000 = 0 ( D - 1500)( D - 500) = 0 D = 1500 or D = 500

We reject D = 1500 since it is not in [0, 1000]. Therefore, D = 500. The correct answer choice is c. 40.

If X has a uniform distribution on [0, 1000], then 1 for x in its density function is f ( x) = 1000

1000

ò0

x⋅

We are given P(0 £ X £ 50) = 0.3 so that 60

1 dx 1000

ò ae-axdx = 0.3 0

1000

Let the random variable X be the number of days that elapse between the beginning of a calendar year and the moment a high-risk driver is involved in an accident. Then it has exponential distribution f ( x) = ae-ax for x ³ 0.

[0, 1000]. The expected payment with no deductible is E( X ) =

-e-ax

1 1 ⋅ x2 1000 2 0

50 0

e-50a = 0.7 -50a = ln 0.7

Now, let the deductible be D. According to the hint,

a =-

x£ D

E( X ) =

ò0

1 dx + 0⋅ 1000

1000

òD

The portion expected to be involved in an accident during the first 80 days is 80

ò 0.007133e

1 dx ( x - D) ⋅ 1000

-0.007133x

1000

= -e-0.007133x

0 -0.007133(80)

= -e

éæ 1 ö æ1 öù 1 = ⋅ ê çç 10002 - 1000D ÷÷÷ - çç D 2 - D 2 ÷÷÷ ú ø èç 2 ø úû 1000 êë èç 2 1 D 2. 2000

» 0.43

The correct answer choice is c. 41.

For this amount to be 25% of the amount with no deductible, we must have

æ1 ö 1 = 0+ ⋅ çç x 2 - Dx ÷÷÷ ç øD 1000 è 2

= 500 - D +

1 ln 0.7 » 0.007133 50

Thus, f ( x) = 0.007133e-0.007133x for x ³ 0.

x > D.

The expected payment with the deductible is therefore D

= 0.3

-e-50a + 1 = 0.3

1 = ⋅ (10002 - 0) 2000 = 500.

ìï 0 for payment = ï í ïïî x - D for

1 ⋅ D 2 = 0.25 ⋅ 500 2000

Let the random variable X be the lifetime of the printer in a years. Then it has exponential distribution f ( x) = ae-ax for x ³ 0. If the mean is 2 years, then 1a = 2, or a = 12 and the function is f ( x) = 12 e-x /2 for x ³ 0.

854

Chapter 11 PROBABILITY AND CALCULUS We wish to find P(0 £ X £ 1) and P(1 £ X £ 2). P(0 £ X £ 1) =

P(5 £ X £ ¥) =

» 0.4204

1 -x /2

The correct answer choice is d. 43.

= -e

For a uniform distribution, 1 for x in [a, b]. b-a 1 1 f ( x) = = for x in [20, 36] 36 - 20 16

-1

= 0+e

ò 2e

= -e-x /2

5 -0.1733(5)

= -e-1/2 + 1

P(1 £ X £ 2) =

-0.1733 x

= -e-0.1733x

1 -x /2 e dx 0 2

= -e-x /2

ò 0.1733e 5

f ( x) =

2 1

+ e-1/2

1 1 (20 + 36) = (56) 2 2 = 28 days

If 100 printers are sold, then (1 - e-1/2 + 1)(100)

(a)

 =

will fail in the first year, and (e-1/2 - e-1/2 )(100) will fail in the second year. The manufacturer pays a full refund on those failing first year and one-half refund on those failing during the second year.

(b)

P(30 < X £ 36)

Refunds = (1 - e-1/2 + 1)(100)($200)

1 1 dx = x 16 16 30

36 30

1 = (36 - 30) 16 = 0.375

+ (e-1/2 - e-1)(100)($100) » $10, 255.90

The correct answer choice is d. 44. 42.

f ( x) = ae-ax for [0, ¥). Since a = 2,

Let the random variable X be the life time to failure. Then it has exponential distribution -ax

f ( x) = ae

f ( x) = 2e-2 x for [0, 1].

for x ³ 0.

(a) The expected proportion is

If the mean is 4 hours, then 4

ò ae 0

-ax

For an exponential distribution,

 =

1 dx = 2 4

1 0 2 1 -e-4a + 1 = 2 1 -e-4a = 2 -e-ax

(b)

1 = 0.5. 2

P(0 < X < 13 ) =

1/3

2e-2 x dx

= -e-2 x

1/3 0

1 2 1 1 a = - ln » 0.1733 4 2

= - 2/3 + 1 e » 0.4866

-4a = ln

45.

We have an exponential distribution, with a = 1. f (t ) = e-t , [0, ¥)

So the exponential function for X is f ( x) = 0.1733e-0.1733x for x ³ 0. The probability that the component will work without failing for at least 5 hours is,

(a)

 =

1 = 1 hr 1

Section 11.3 (b)

855

P(T < 30 min)

0.5

(b)

P( X ³ 60)

e-t dt

= lim

b ¥

1 - e-0.5 » 0.3935 46.



48.

Since  = (a)

1 -x /28.9 e dx 25 28.9

= lim

b ¥

ò 28.9 e

-x /28.9

(

= lim -e-b /28.9 + e-25/28.9 b ¥ -25/28.9

» 0.4210

(b)

P( X < 20) =

20 0

1 -x /28.9 e dx 28.9

= 0.90

= (-e-x /28.9 )

-e-0.04t + 1 = 0.90

-20/28.9

= -e » 0.4994

0.10 = -e-0.04t

20 0

-0.04t = ln 0.10 ln 0.10 -0.04 t » 57.56 t =

The longest time within which the predator will be 90% certain of finding a prey is approximately 58 min.

)

dx = 0.90

b ö æ ç 1 -x /28.9 ÷÷ = lim çç ÷÷ e b ¥ ççè 28.9 25 ÷÷ø

(a) We must find t such that P( X £ t ) = 0.90.

-e

1 1 . = 28.9, a = a 28.9

P( X ³ 25) =

This, f ( x) = 0.04e-0.04 x .

f ( x) = ae-ax for x in [0, ¥).

1 , a 1 = 0.04. a = 25

-0.04 x

For an exponential distribution,

Since  = 25 and  =

-0.04 x

The probability that the predator will have to spend more than one hour looking for a prey is approximately 0.0907.

for [0, ¥]

ò 0.04e

-0.04 x

» 0.0907

The smallest height is 3.065 ft, and the largest height is 3.335 ft.

ò 0.04e

= 0 + e-2.4

x = 3.2  0.135 x = 3.065,3.335

f ( x) = ae

= lim éê -e-0.04b + e-0.04(60) ùú û b ¥ ë

x-

x - 3.2 0.675 = 0.2 0.135 = x - 3.2

47.

b ¥

If we wish to find the middle 50%, or 0.50, we want to find the value of z corresponding to a probability of 0.25 in the standard normal distribution table. This is about 0.675. Since the desired area is symmetric about the mean, we use 0.675. z =

-0.04 x

= lim (e-0.04 x )

 = 3.2 ft,  = 0.2 ft

-ax

ò 0.04e 60

0.5

= -e-t

49.

For an exponential distribution, f ( x) = ae-ax for x in [0, ¥).

Since  =

1 1 . = 14.5, a = a 14.5

856

Chapter 11 PROBABILITY AND CALCULUS (a)

P( X ³ 20) =

1 -x /14.5 e dx 14.5 20

= lim

b ¥

53.

We have an exponential distribution, with a = 0.229. So f (t ) = 0.229e-0.229t , for [0, ¥).

1 -x /14.5 e dx 20 14.5

(a) The life expectancy is

b ö æ ç 1 -x /14.5 ÷÷ = lim çç e ÷÷ b ¥ ççè 14.5 20 ÷÷ø

 =

1 1 = » 4.37 millennia. a 0.229

The standard deviation is

= lim (-e-b /14.5 + e-20/14.5 ) b ¥ -20/14.5

 =

1 1 = » 4.37 millennia. a 0.229

» 0.2518

(b)

P(10 £ X £ 20) =

(b)

1 -x /14.5 e dx 10 14.5

-x /14.5

= (-e

)

P(T ³ 2) =

= 1-

54.

1 1 = for [32, 44] 44 - 32 12 1 (32 + 44) = 38 inches 2

(a)

 =

(b)

P(38 < X < 40) =

ò 12 dx 38 40

(b) We have a normal distribution with  = 0.6 and  = 0.3.

x 12 38

æ 6 - 0.6 ÷ö P( X ³ 6) = P çç z ³ ÷ çè 0.3 ÷ø

1 » 0.1667 6

55.

For an exponential distribution, f ( x) = ae-ax for [0, ¥).

Yes, by today’s standards there is sufficient evidence to conclude that Andrew Jackson suffered from mercury poisoning.

-0.229t

Uniform distribution on [32, 44] f ( x) =

No, this is insufficient evidence. Using the TI-84 Plus C command normaldcf(6.0, 1E99, 6.9, 4.6) we get an answer of 0.5776.

Since  =

Use the TI-84 Plus C command invNorm. invNorm(0.85, 52, 8) = 60.291. The 85th percentile is a speed of 60.29 mph.

52.

ò 0.229e

= e-0.458 » 0.6325

» 0.5793

51.

= 1 + éê e-0.229(2) - 1 ùú ë û

= P ( z ³ -0.19565 )

»0

2ö æ = 1 + çç e-0.229t ÷÷÷ 0 ø÷ èç

(a) We have a normal distribution with  = 6.9 and  = 4.6.

= P( z ³ 18)

-0.229t

æ 6 - 6.9 ö÷ P( X > 6) = P çç z ³ ÷ çè 4.6 ÷ø

ò 0.229e 2

= -e-20/14.5 + e-10/14.5 » 0.2500

50.

1 1 = 8, a = . a 8

Chapter 11 Review (a)

857 (b) The probability that the time for a goal is 499 minutes or more is

1 -x /8 e dx 10 8

P( X ³ 10) =

1 -x /8 = 1e dx 0 8 10 ö æ = 1 + çç e-x /8 ÷÷÷ çè 0 ÷ø

P( X ³ 499) =

= e-x / 90

1 -x /8 e dx 0 8

= -e-x /8

We have a normal distribution with  = 0 and  = 13.861. The probability that the margin of victory over the point spread is greater than -3 is æ -3 - 0 ö÷ P( X > -3) = P çç z ³ ÷ çè 13.861 ÷ø

-2/8

= -e +1 » 0.2212

56.

= P( z ³ -0.2164) = P( z £ 0.2164).

We have an exponential distribution,

Using the normal table in the book with z = 0.22, we have the probability is about 0.5871. Using the TI-84 Plus C command normaldcf( -3 , 1E99, 0, 13.861) we get the answer 0.5857.

f ( x) = ae-ax for [0, ¥). Since 1 , f ( x) = 1 e-x / 609.5. a = 609.5 609.5

(a) The expected number of days is  = 1a = 609.5. The standard deviation is

 = 1a = 609.5. (b)

P( x > 365) =

1 e-x /609.5dx 609.5 365

Chapter 11 Review Exercises 1.

True

1 e-x /609.5dx 609.5 0 æ -x /609.5 365 ÷ö ÷÷ = 1 + çç e çè 0 ÷ø

True

= 1 + (e-365/609.5 - 1)

False: A density function is always nonnegative.

= e-365/609.5 » 0.5494 We have an exponential distribution f ( x)

False: If the random variable takes on negative values the expectation may also be negative.

True

False: The normal distribution is symmetrical; the exponential distribution has a long tail to the right.

10.

False: The expected value is 0 and the standard deviation is 1.

= 1-

57.

= 0+e » 0.0039.

58.

499 -499 / 90

= e-10/8 » 0.2865 P( X < 2) =

1 -x / 90

ò 90e 499

= 1 + [e-10/8 - 1]

(b)

365

1 , f ( x) = = ae-ax for x ³ 0. Since a = 90 1 e-x /90 for x ³ 0. 90

(a) The probability that the time for a goal is no more than 71 minutes is P(0 < X < 71) =

1 -x /90 e dx 90

= -e-x /90 -71/90

= -e » 0.5457.

71 0

858 11.

13.

Chapter 11 PROBABILITY AND CALCULUS In a probability function, the y-values (or function values) represent probabilities.

6.5

f ( x) ³ 0 for all x in [4, 6.5].

Therefore f( x) is a probability density function.

òa f (x)dx = 1.

19.

f ( x) = k x 2 ; [1, 4]

In a probability density function, the probability that X equals a specific value, P( X = c), is zero. f ( x) =

k x 2dx =

2 3/2 x 3

20.

f ( x) = k x ; [4, 9]

k x dx =

1 (2 x + 4); [1, 4] 27

2 k (27 - 8) 3 38 k = 3

Since f( x) is a probability density function, 38 k =1 3 3 k = . 38

Since 1 £ x £ 4, f ( x) ³ 0. Therefore, f ( x) is a probability density function. 21. f ( x) = 0.7e-0.7 x ; [0, ¥)

ò 0.7e

-0.7 x

dx = - e-0.7 x

2 3/2 kx 3 4

1/2

1 1 2 (2 x + 4)dx = ( x + 4 x) 27 27 1 1 1 (32 - 5) = 27 =1

17.

ò k x dx

21k = 1 1 k = . 21

f ( x) is not a probability density function.

f ( x) =

Since f( x) is a probability density function, 9

2 = (27 - 8) 3 38 = ¹1 3

16.

k x3 3

= 21k

x1/2dx =

x ; [4, 9] 9

6.5

0.4 dx = 0.4 x 4

= 0.4(6.5 - 4) = 1

f ( x) ³ 0 for all x in the interval [a, b];

(2)

15.

f ( x) = 0.4; [4, 6.5]

A probability density function f for [a, b] must satisfy the following two conditions: (1)

14.

18.

f ( x) =

(a)

1 for [10, 20] 10

P(10 £ X £ 12)

= lim (-e-0.7b ) + e0 b ¥

æ 1 ö = lim çç - 0.7b ÷÷ + 1 ç ÷ø b ¥ è e = 0 +1= 1

ò 10 dx 10 12

x 10 10

1 = 0.2 5

f ( x) ³ 0 for all x in [0, ¥).

Chapter 11 Review (b)

859

(

P 31 £ X £ 20 2

)

The distribution that is tallest or most peaked has the smallest standard deviation. This is the distribution pictured in graph (b).

25.

f ( x) =

1 dx 31/ 2 10

x 10 31/2

31 = 220 9 = = 0.45 20

(c)

24.

(a)

Var( X ) =

1 ; [2, 5] x -1

ò [1 - (x - 1)

-1/2

= [ x - 2( x - 1)

] dx

]

ò 9 (x - 2x )dx - 16 3

2 æ 625 250 16 ö÷ = çç -4 + ÷ - 16 9 çè 4 3 3 ø÷ = 0.5

(c)

P(2 £ X £ 4)

ò2 [1 - (x - 1)

-1/2

= [ x - 2( x - 1)1/2 ]

(d) ]dx

4 2

» 0.5359 P(3 £ X £ 4) 4

ò3 [1 - (x - 1)-1/2 ]dx

= [ x - 2( x - 1)1/2 ]

 = m

0.5 » 0.7071

ò 9 (x - 2)dx = 2 2

1 1 (m - 2) 2 = 9 2 2

= 4-2 3-2+2

4 3

= 4-2 3-3+2 2 » 0.3643

23.

» 0.8284

2 æç x 4 2 x3 ö÷÷ = çç ÷ - 16 9 èç 4 3 ÷ø÷

= 5 - 2(2) - 3 + 2 2

(c)

2x2 ( x - 2)dx - (4)2 9 2 5

1/2

16.2

x = 0.54 10 10.8

(b)

ò2 9 (x2 - 2x)dx

P(3 £ X £ 5) =

ö 2 æ 125 8 = çç - 25 - + 4 ÷÷÷ = 4 ç ø 9è 3 3

f ( x) = 1 -

(a)

ò2 9 (x - 2)dx

16.2

(b) 22.

ö÷ 2 æç x3 = çç - x 2 ÷÷÷ 9 çè 3 ÷ø

1 dx 10.8 10

 = =

P(10.8 £ X £ 16.2) =

2 ( x - 2); [2, 5] 9

1 1 [(m - 2)2 - 0] = 9 2 9 2 m - 4m + 4 = 2 1 2 m - 4m - = 0 2 43 2 m= 2 » -0.121, 4.121

We reject -0.121 since it is not in [2, 5]. So, m = 4.121

If we consider the probabilities as weights, the expected value or mean of a probability distribution represents the point at which the distribution balances.

860

Chapter 11 PROBABILITY AND CALCULUS (e)

27.

2 1 (t - 2)dt = (t - 2) 2 9 2 9 2

f ( x) = 5x-6 ; [1, ¥)

(a)

1 = [( x - 2)2 - 0] 9

f ( x) =

b ¥

E( X ) =  =

æ1ö x2 x çç ÷÷÷ dx = ç 10 4 è5ø

(b)

Var( X ) =

 = m

1 ( x - 6.5)3 15 4

(c)

Var( X ) » 1.443

(d)

ò 5 dx = 2

 »

ò1

æ 5 ö2 x 2 ⋅ 5x-6 dx - çç ÷÷÷ çè 4 ø b

25 16

5 25 5 = » 0.1042 3 16 48

Var( X ) » 0.3227

ò1 5x-6dx = 2 m

1 2 1 -m-5 + 1 = 2 1 -5 m = 2 -x-5

1 1 x = 5 4 2 1 1 (m - 4) = 5 2 5 m-4 = 2 13 = 6.5 m= 2 x

5æ 1 ö 25 = lim çç1 - 3 ÷÷÷ ç b ¥ 3 è b ø 16

1 (2.53 + 2.53) = 15 » 2.083

F ( x) =

5 x-4 b ¥ -4

5x-3 b ¥ -3

(e)

m5 = 2 m = 5 2 » 1.149

(e)

ò 5 dt

ò1 5t-6dt = -t-5 1

= -x-5 + 1

1 = 1- 5, x ³ 1 x

1 t 5 4

x-4 , 4 £ x £ 9. 5

5 x-5 dx = lim

= lim

ò 5x- dx

5x-4 dx 16 b ¥ ò1

= lim

1 ( x - 6.5)2 dx 4 5

Var( X ) =

5æ 1 ö 5 = lim çç1 - 4 ÷÷ = = 1.25 ç ÷ 4 b ¥ 4 è b ø

81 16 65 = = = 6.5 10 10 10

(b)

x ⋅ 5 x-6 dx =

= lim

1 ; [4, 9] 5

(a)

( x - 2) 2 = ,2 £ x £ 5 9

26.

 =

Chapter 11 Review 28.

f ( x) =

(a)

861 We reject 65.5941 since it is not in [1, 9]. So, m » 4.406.

1 æç 3 ö÷ ÷; [1, 9] çç 1 + 20 è x ÷ø 9

é 1 ù x ê (1 + 4 x-1/2 ) ú dx ê úû 1 ë 20

E( X ) =  =

(e)

1 [( x + 6 x ) - (1 + 6)] 20 1 = ( x + 6 x - 7), 1 £ x £ 9 20

(b) Var( X ) = =

x (1 + 3x-1/2 )dx - (4.6)2 1 20

29.

f ( x) = 4 x - 3x 2 ; [0, 1]

(a)

4æ 2

ò x(4x - 3x )dx ò (4x - 3x )dx

(b)

Var( X ) =

» 5.4933 » 5.493

1 æ

æ 7 ö2 x 2 (4 x - 3x 2 )dx - çç ÷÷÷ çè 12 ø 0 1

æ 7 ö2 (4 x3 - 3x 4 )dx - çç ÷÷÷ çè 12 ø 0

3 ö

æ æ 7 ÷ö2 3x5 ö÷÷ ç çç ÷ = çç x 4 ÷ çè èç 12 ÷ø 5 ÷÷ø 0

ò1 20 çççè1 + x ÷ø÷÷ dx = 2

2 3 æç 7 ö÷ - ç ÷÷ 5 çè 12 ø » 0.0597

= 1-

1 1 ( x + 6 x1/2 ) = 20 2 1 1 3 1/ 2 7 1 m+ m = 20 10 20 2 m + 6 m - 17 = 0

To solve this equation, let u =

 = Var( X ) » 2.3438  » 2.344 m

4 3 7 = - = 3 4 12 » 0.5833

æ 729 729 1 3 ö÷ = çç + ÷ çè 60 50 60 50 ÷ø

(d)

æ 4 x3 3x 4 ö÷÷ ç = çç ÷ 4 ø÷÷ èç 3

(c)

0 1 0

æ x3 3 5/2 ö÷÷ ç = çç + x ÷÷ - (4.6)2 50 çè 60 ø÷ 1

- (4.6)2

 = =

3 3/2 ö÷÷ çç x + x ÷÷ dx - (4.6)2 ç 20 1 èç 20 ø÷

1 (t + 6t1/2 ) 20 1

9 2

3 ö

ö 1 æç 81 1 + 54 - - 2 ÷÷÷ çç ø 20 è 2 2 92 = = 4.6 20

1 æ

ò 20 çççè1 + t ÷÷÷ø dt 1

9 1 ( x + 3x1/2 ) dx = 20 1

ö÷ 1 æç x 2 çç = + 2 x 3/2 ÷÷÷ 20 çè 2 ø÷

m so u 2 = m.

 »

Var( X )

» 0.2444

(c)

(

7 P 0 £ X £ 12

7 /12

)

(4 x - 3x3 )dx

u 2 + 6u - 17 = 0 -6  36 + 68 2 = -3  26

u =

m = u 2 » 4.4059, 65.5941

= (2 x 2 - x3 )

æ 7 ö æ 7 ö3 = 2 çç ÷÷ - çç ÷÷ çè 12 ÷ø çè 12 ÷ø » 0.4821

7/12

862 (d)

Chapter 11 PROBABILITY AND CALCULUS P(  -  £ X £  +  ) » P(0.3389 £ x £ 0.8277)

0.8277

31.

(4 x - 3x 2 )dx

f ( x) = 0.01e-0.01x for [0, ¥) is an exponential distribution.

(a)

 =

0.3389 0.8277

= (2 x 2 - x 3 )

(b)  =

0.3389

(c)

= 2(0.8277)2 - (0.8277)3

1 = 100 0.01 1 = 100 0.01 P(100 - 100 < X < 100 + 100)

= P(0 < X < 200)

- 2(0.3389)2 + (0.3389)3 » 0.6123

200

0.01e-0.01x dx

30.

= -e-0.01x

f ( x) = 4 x - 3x ; [0, 1] 1

-2

ò0 x(4x - 3x )dx

 =

= 1- e

32.

ò0 (4x - 3x )dx

f ( x) =

(a)

æ 3 3x 4 ÷÷ö ç 4x = çç ÷ 4 ÷÷ø çè 3 0

 =

= 2m 2 - m3 =

5 112

1 2

(b)

)dx

æ 95 ö2 5x 2 (1 - x-3/2 ) dx - çç ÷÷÷ çè 7 ø 112

ò1

5 112

ö÷ 5 æçç x 3 2 - x 3/2 ÷÷÷ ç ÷ø 112 çè 3 3

This equation has no rational roots, but trial and error used with synthetic division reveals that m » -0.4516, 0.5970, and 1.855. The only one of these in [0, 1] is 0.5970.

)

æ 95 ö2 ( x 2 - x1/2 ) dx - çç ÷÷÷ çè 7 ø 1

æ 95 ö2 - çç ÷÷÷ çè 7 ø

7/12 0.5970

6560 147  = Var( X ) » 6.6803 =

7/12

æ 7 ö2 = 2(0.5970)2 - (0.5970)3 - 2 çç ÷÷÷ çè 12 ø æ 7 ö3 + çç ÷÷÷ » 0.0180 çè 12 ø

æ 95 ö2 - çç ÷÷÷ èç 7 ø

5 æçç 253 250 1 2 ö÷÷ = + ÷ ç 112 çè 3 3 3 3 ÷ø÷

(4 x - 3x 2 )dx

= 2 x 2 - x3

-1/2

Var( X ) =

2m3 - 4m 2 + 1 = 0.

7 < X < 0.5970 P 12

ò (x - x

Therefore,

)dx

0.5970

-3/2

ö 5 æç 625 1 - 10 - + 2 ÷÷÷ ç ø 112 çè 2 2 95 = » 13.5714 7 » 13.6

1 . 2

ò0 (4x - 3x2 )dx = (2x2 - x3) 0

ò 112 (1 - x

ö÷ 5 æç x 2 çç = - 2 x1/2 ÷÷÷ 112 çè 2 ø÷

(

» 0.8647

5 (1 - x-3/2 ); [1, 25] 112

Find m such that

(4 x - 3x 2 )dx =

4 3 7 = - = » 0.5833 3 4 12

200

 » 6.7

Chapter 11 Review (c)

863

P(  -  £ X £  +  ) = P(6.8911 £ X £ 20.2517) =

20.2517

6.8911

39.

P( z < a) = 0.48 for a = -0.05

5 (1 - x-3/2 )dx 112

Thus, 52% of the area lies to the right of z = -0.05.

20.2517

5 = ( x + 2 x-1/2 ) 112 6.8911

æ ö 2 ÷÷÷ ççç 20.2517 + 5 ç 20.2517 ÷÷÷ = ç ÷÷ 2 112 ççç - 6.8911 ÷÷÷ çç è 6.8911 ø » 0.5823

If you do the integration on a calculator using differently rounded values for  and you may get a slightly different answer, such as 0.5840. For Exercises 33–40, use the table in the Appendix for the areas under the normal curve. 33.

52% of area is to the right implies that 48% is to the left.

40.

We want to find the z-score for 21% of the area under the normal curve to the left of z. We note that 21% < 50%, so z must be negative. The zscore for the value in the table nearest 0.21 is z » -0.81.

41.

f ( x) = 0.05 for [10, 30]

(a) This is a uniform distribution. (b) The domain of f is [10, 30]. The range of f is {0.05}. (c)

Area to the left of z = -0.43 is 0.3336. Percent of area is 33.36%.

34.

Area to right of z = 1.62 is 1 - 0.9474 = 0.0526 or 5.26%.

35.

Area between z = -1.17 and z = -0.09 is 0.4641 - 0.1210 = 0.3431.

(d) For a uniform distribution,

 =

Percent of area is 34.31%. 36.

Area to left of z = 1.28 is 0.8997. Area to left of z = -1.39 is equivalent to area to right of z = 1.39: 1 - 0.9177 = 0.0823.

Area between is 0.8997 - 0.0823 = 0.8174 or 81.74%. Using the TI-84 Plus C command normalcdf  1.39,1.28 you will get an answer of 0.8175 or 81.75%. 37.

The region up to 1.2 standard deviations below the mean is the region to the left of z = -1.2. The area is 0.1151, so the percent of area is 11.51%.

38.

 = 2.5,  = 0, z = 0 + 2.5

1 (b + a) and 2

Var( X ) =

b 2 - 2ab + a 2 . 12

Thus,

 =

1 1 (30 + 10) = (40) = 20 2 2 302 - 2(10)(30) + 102 12 400 = . 12

Var( X ) =

Area to left of z = 2.5 is 0.9938 or 99.38%.

 =

400 » 5.77 12

864

Chapter 11 PROBABILITY AND CALCULUS (e)

P(  -  £ X £  +  ) = P(20 - 5.77 £ X £ 20 + 5.77)

-x 2 =

= P(14.23 £ X £ 25.77) =

25.77

( )

1 2

and

ò14.23 0.05 dx

= 0.05(25.77 - 14.23)



25.77 = 0.05 x 14.23

2

1 2

f ( x) is a normal distribution with  = 0

» 0.577

42.

-( x - 0) 2

and  =

f ( x) = e-x for [0, ¥)

1 . 2

(b) The domain of f is (-¥, ¥).

(

ù The range of f is 0, 1 ú .  û

-1x

f ( x) = 1e

(a) This is an exponential distribution with a = 1.

(c)

(b) The domain of f is [0, ¥). The range of f is (0, 1]. (c) (d) For this normal distribution,  = 0 and

 =

1 . 2

P(  -  £ X £  +  )

(e)

= 2 P(0 £ X £  +  )

(

(d) For an exponential distribution,  = 1a and

 = 1a .

If x =

Thus

 = (e)

1 1 = 1 and  = = 1. 1 1

= 2 P(0 £ z £ 1.00) = 2(0.3413) » 0.6826

ò e dx -x

-x

= -e

44.

f ( x) =

-2

= -e

xe-x /2 for x in [0, ¥) 4

P(0 £ X £ ¥) =

(a)

» 0.8647

1 -0 2 = 1.00. 1 2

1 , z = 2

P(  -  £ X £  +  )

= P(0 £ X £ 2) 2

Thus,

P(  -  £ X £  +  ) = P(1 - 1 £ X £ 1 + 1)

)

= 2P 0 £ X £ 1

xe-x /2 dx 4

-x /2

43.

f ( x) =

e-x



For all x ³ 0, e > 0 so that f ( x) ³ 0 for x in [0, ¥).

for (-¥, ¥)

(a) Since the exponent of e in f ( x) may be written

Evaluate

ò xe- dx using integration by x /2

parts. and

dv = e-x /2dx

Then du = dx and

v = -2e-x /2

Let

u = x

Chapter 11 Review 1 4

ò xe

-x /2

865 dx

ö 1 æç -x /2 + 2e-x /2dx ÷÷÷ çè -2 xe ç ø 4 1 = -2 xe-x /2 - 4e-x /2 4 1 = - xe-x /2 - e-x /2 2

(

)

òx

1 é 1/2 -x /2 ê 2x e 2 ë

1 é 1/2 -x /2 + ê 2x e 2 ë

æ 1 ö¥ = çç - xe-x /2 - e-x /2 ÷÷÷ çè 2 ø0

æ 1 öb = lim çç - xe-x /2 - e-x /2 ÷÷÷ ø0 b ¥ çè 2

(b)

æ 1 ö = lim çç - be-b /2 - e-b /2 ÷÷÷ ç ø b ¥ è 2 lim be-b /2 = 0. Therefore,

(b)

-1/2 -x /2

1 é 1/2 -x /2 b ê 2x e + 0 2 ëê

ò x e

-0

1/2 -x /2

dx » 0.4962

Therefore, P(0 £ X £ 1) 1 1 x-1/2e-x /2dx 2 0 1 » (1.2131 + 0.4962) 2 » 0.6819.

5 -3/2 e 2 » 0.4422

= 1-

x-1/2e-x /2 for x in (0, ¥) 2

(c)

(a) Using integration by parts: and dv = x

Using Simpson’s rule with n = 12 to evaluate the improper integral, we have

- (0 - 1)

u = e

» 1.2131.

æ 3 ö = çç - e-3/2 - e-3/2 ÷÷÷ çè 2 ø

-x /2

ù x1/2e-x /2dx úú . 0 úû b

P(0 < X £ 1) 1 é 1/2 -x /2 1 ê 2x e = 0 2 êë

= 2e

xe-x /2 dx 4 0

0 -1/2

Let

ò x

dx = 0 - (-1) = 1.

P(0 £ X £ 3) =

f ( x) =

2 x1/2e-x /2

æ 1 ö = çç - xe-x /2 - e-x /2 ÷÷÷ çè 2 ø0

45.

ù æ 1ö 2 x1/2 çç - ÷÷÷ e-x /2dx ú ú èç 2 ø û ù x1/2e-x /2dx ú úû

ù x1/2e-x /2dx úú 0 úû Notice that

b ¥

1 2

(0 - 1)

-x /2

P(0 < X £ b)

-x /2

-1/2 -x /2

Thus,

Therefore, ¥

1 2

-1/2

Then du = - 12 e-x /2 and v = 2 x1/2.

P(0 < X £ 10) =

1 é 1/2 -x /2 10 ê 2x e 2 ë 0 +

ù x1/2e-x /2dx úú úû

866

Chapter 11 PROBABILITY AND CALCULUS First, 2 x1/2e-x /2

(b)

10 0 -5

= 2 10e

 = =

-0

» 0.0426.

ò x e

1/2 -x /2

3 4

ò (x - 16x + 65x)dx

8 9

3 æ 6561 5265 = çç - 3888 + 4 çè 4 2

dx » 2.3928

- 1024 +

Therefore,

ö 8192 - 2080 ÷÷÷ ø 3

137 16 » 8.56 The expected value of the price is $8.56.

P(0 £ X £ 1)

10 1 x-1/ 2e-x / 2dx 2 0 1 » (0.0426 + 2.3928) 2 » 0.9716.

ò x(x - 16x + 65)dx

3 æç x 4 16 3 65 2 ö÷÷ = çç x + x ÷÷ 4 çè 4 3 2 ø÷

Using Simpson’s rule with n = 12 to evaluate the improper integral, we have 10

3 4

(d) Since f ( x) is a probability density function, the limit as b  ¥ should be 1. The previous results do support this conclusion.

æ 137 ö÷2 x 2 ( x 2 - 16 x + 65) dx - çç çè 16 ÷÷ø 8

3 4

æ 137 ö÷2 ( x 4 - 16 x 3 + 65x 2 ) dx - çç çè 16 ÷÷ø 8

46.

(a)

3 f ( x) = ( x 2 - 16 x + 65) for [8, 9] 4

P(8 £ X £ 8.50) =

8.5

3 2 ( x - 16 x + 65)dx 4

ö÷ 3 æç x3 = çç - 8 x 2 + 65 x ÷÷÷ 4 çè 3 ø÷

8.5

3 é 8.53 = êê - 8(8.5)2 + 65(8.5) 4ê 3 ë ù 83 + 8(8)2 - 65(8) úú 3 úû » 0.4063

æ 137 ö÷2 3 æç x 5 65 3 ö÷÷ = çç - 4 x4 + x ÷÷ - çç çè 16 ÷ø÷ 4 çè 5 3 ÷ø 8

3 æ 59, 045 = çç - 26, 244 + 15,795 4 çè 5 32,768 33, 280 ö÷ + 16, 384 ÷ 5 3 ÷ø 18,769 256 107 = 1280 107 » 0.29  = 1280 -

The standard deviation of the price is $0.29.

Chapter 11 Review 47.

f (t ) =

(a)

867 5 (1 - t -3/2 ); [1, 25] 112

48.

P(No repairs in years 1-3) = P(First repair needed in years 4-25) =

ò 112 (1 - t

-3/2

1 -x /6 e for [0, ¥) is an exponential 6 distribution. f ( x) =

(a)

 = 1 = 6 outlets 6

) dt

(b)  = 1 = 6 outlets

5 (t + 2t -1/2 ) = 112 4

(c)

ù 5 é 2 ê 25 + - 4 - 1ú ê úû 112 ë 5 51 = » 0.9107 56

P( X > 6) =

25 5 (t - t -1/2 ) dt 112 1

= lim - e-x / 6 b ¥

49.

(a)

ö 5 æç 625 1 - 10 - + 2 ÷÷÷ ç ø 112 çè 2 2 95 = 7 » 13.57 The expected value for the number of years before the machine requires repairs is 13.57 years. =

1 » 0.3679 e

 =8

f ( x) =

1 -x /8 e for [0, ¥) 8

(b)

Expected number =  = 8 repairs

(c)

 =  = 8 repairs

(d)

P(5 £ X £ 10) =

æ 95 ö 5 Var = t 2 (1 - t -3/2 ) dt - çç ÷÷÷ çè 7 ø 112 1 25 æ 95 ö2 5 = (t 2 - t1/2 )dt - çç ÷÷÷ çè 7 ø 112 1

æ 95 ö2 - çç ÷÷ çè 7 ø÷

5 æç 15,625 250 1 2 ö 9025 - + ÷÷÷ ç 112 çè 3 3 3 3ø 49 6560 = 147 6560  = 147 » 6.68 =

1 -x /8

ò 8e

= -e-x /8

1 =8 a 1 a = 8

æ 1 ö = lim çç e-1 - b /6 ÷÷÷ ç è b ¥ e ø

5 æçç t 3 2 3/2 ö÷÷ = çç - t ÷÷ 112 çè 3 3 ø÷

e-x /6 dx b ¥ ò6 6

25 5 t (1 - t -3/2 ) dt 112 1

= lim

 =

ö 5 æç t 2 1/2 ÷÷ = ççç - 2t ÷÷ 112 è 2 ÷ø

1 - /6

ò 6 e x dx 6

(b)

10 5

= -e-10/8 + e-5/8 » 0.2488

50.

 = 46.2,  = 15.8, x = 60 z =

x-



60 - 46.2 » 0.8734 15.8

0.8734 is the z-score for the area of about 0.8078 (from the table).

P( X ³ 60) » P( z ³ 0.8734)

The standard deviation of the number of years before repairs are required is 6.68 years.

» 1 - 0.8088 = 0.1912

868 51.

Chapter 11 PROBABILITY AND CALCULUS Let the random variable X be the number of printers. We have an exponential distribution

52.

f (t ) = ae-at for t in [0, ¥). Since

f ( x) =

(a)

 = 10, a = 1 = 0.1 so that f (t ) = 0.1e-0.1t , t ³ 0.

We need to determine the portion of the printers sold in the first year and in the second and third year. In other words, we need to calculate P(0 £ X £ 1) and P(1 £ X £ 3). 1

ò 0.1e

P(0 £ X £ 1) =

-0.1t

= -e

 =

8 ln x 7 1

8 (ln 8 - ln 1) » 2.3765 g 7

(b) Var(X ) =

-0.1t

ò 7 x (x ) dx - 2.3765 -2

8 x - 2.37652 7 1

64 8 - - 2.37652 7 7 » 2.35225

-0.1

ò 0.1e

= -e

ò1 7 x-1 dx

= 1 - e-0.1

-0.3

1 0

= -e-0.1t

ò1 7 x-2 (x) dx

= -e-0.1 + e0

P(1 £ X £ 3) =

-0.1t

8 -2 x for [1, 8] 7

 =

1 -0.1

(c)

- e-0.3

Since

Var( X ) » 1.534 g

P (u    X     ) = P(0.844 £ X £ 3.91) = P(1 £ X £ 3.91)

ìx for 0 £ x £ 1 ï ï ï ï payment = í 0.5x for 1 £ x £ 3 ï ï for x > 3 ïï î0

+ (e-0.1 - e-0.3 )(0.5x) + 0 = x - 1.5 xe-0.1 - 0.5xe-0.3.

= 1000.

-8 -1 x 7 1

-8 æç 1 ö÷ 8 ÷+ ç 7 çè 3.91 ø÷ 7 » 0.8506

E ( X ) = (1 - e-0.1)( x)

E ( X ) = x - 1.5xe-0.1 - 0.5 xe-0.3

8 -2 x dx 7 3.91

the expected payment will be

To determine the level x must be set for this to be 1000, solve

3.91

53.

f ( x) = 0.01e-0.01x for [0, ¥) is an exponential distribution.

P(0 £ X £ 100)

Using our calculators, we find x » 5644. The correct answer choice is d.

100

0.01e-0.01x dx

= - e-0.01x 1 e » 0.6321

= 1-

100 0

Chapter 11 Review 54.

f ( x) =

869

1 ; for [a, b] b-a

(b)

P( X £  ) =

1 1 f ( x) = = 30 - 2 28

3 19,696

40.07

( x 2 + x) dx

40.07 3 æç x3 x 2 ö÷÷ çç = + ÷ 19, 696 çè 3 2 ÷÷ø

This is uniform distribution for a = 2, b = 30.

(a)

(b)

1 E ( X ) =  = (b + a ) 2 1 = (30 + 2) = 16 in. 2 P(20 < X £ 30) =

3 æç (40.07)3 (40.07)2 çç = + 19, 696 çè 3 2 (38)3 (38) 2 ö÷÷ ÷ 3 2 ÷÷ø

ò20 28 dx

» 0.4928 The probability of a body temperature below the mean is 0.4928.

1 x 28 20

30 20 28 28 10 = » 0.3571 28 =

55.

f ( x) =

(a)

56.

z =

3 19,696

ò x(x + x) dx

3 19, 696

ò38 ( x3 + x2 ) dx

x-



9 - 7.8 » 1.09 1.1

P( X > 9) » P( z > 1.09) = 1 - 0.8621 » 0.1379

3 ( x 2 + x) for x in [38, 42] 19,696

 =

 = 7.8 lb,  = 1.1 lb, x = 9 lb

Using the TI-84 Plus C command we calculate 1 - normalcdf (9, 7.8, 1.1) = 0.1377.

38 42

3 æç x 4 x3 ö÷÷ çç = + ÷ 19, 696 çè 4 3 ÷÷ø 38

57.

P( X < 1.9) æ x - 2.2 1.9 - 2.2 ö÷ = P çç < ÷ çè 0.4 0.4 ÷ø

3 æç (42) 4 (42)3 ç = + ç 19, 696 çè 4 3 (38)4 (38)3 ö÷÷ ÷ 4 3 ÷÷ø » 40.07 The expected body temperature of the species is 40.07C.

Normal distribution,  = 2.2 g,  = 0.4 g, X = tension

= P ( z < -0.75 ) » 0.2266.

58.

For an exponential distribution, f ( x) = ae-ax for x in [0, ¥).

Since  =

1 1 . = 17.0, a = a 17.0

870

Chapter 11 PROBABILITY AND CALCULUS (a) P( X ³ 15) =

1 -x /17.0 e dx 17.0 15

ò b

= lim

b ¥

60.

ò 17.0 e

-x /17.0

b ö æ ç 1 -x /17.0 ÷÷ e = lim çç ÷÷ b ¥ ççè 17.0 15 ÷÷ø

(

-b /17.0

-15/17.0

= lim -e

b ¥ -15/17.0

A polynomial function could fit the data. Your answers to (b) through (f) may differ slightly from the answers given here, depending on how many places you keep in each of the coefficients of N and S.

)

= e » 0.4138

(b)

P( X < 10) =

(b)

- 38.50574t 2 + 1090.5874t - 4695.1476

1 -x /17.0 e dx 17.0 0

-10/17.0

= -e » 0.4447

10 0

+ e0

For an exponential distribution, -ax

(c)

f ( x) = ae

for x in [0, ¥).

Since  =

1 1 . = 32.5, a = 32.5 a

(a)

P( X ³ 40) =

b ¥

ò5.2 N (t) dt » 197,023, as found using S (t ) =

1 -x /32.5 e dx 32.5 40

88.9

a calculator. Thus the density function corresponding to the quartic fit is

= lim

1 (-0.002154802t 4 197,023 + 0.493396t 3 - 38.50574t 2 + 1090.5874t - 4695.1476).

1 -x /32.5 e dx 32.5 40

Use a calculator to compute the integrals needed in (d), (e), and (f ) .

æ b ö ç 1 -x /32.5 ÷÷ ÷÷ = lim çç e ÷ b ¥ççè 32.5 40 ÷ø

(

= lim -e-b /32.5 + e-40/32.5 b ¥ -40/32.5

= e

)

(d)

P(age < 25) =

ò 5.2 S(t) dt » 0.3971; 1096 + 5729 18, 488 » 0.3692

actual relative frequency =

» 0.2921

(b)

N (t ) = -0.002154802t 4 + 0.493396t 3

= (-e-x /17.0 )

59.

(a)

P(30 £ X < 50) =

P(45 £ age < 65)

ò 30 32.5 e-x/32.5 dx

= (-e-x /32.5 ) -50/32.5

= -e » 0.1826

S (t ) dt » 0.1320;

+ e-30/32.5

2207 + 1011 18, 488 » 0.1741

actual relative frequency =

P(75 £ age) =

88.9 75

S (t ) dt » 0.0265;

Extended Applcation

871 274 + 32 18, 488 » 0.0166

actual relative frequency =

(e)

 =

88.9

Var =

» 267.59  = 267.59 » 16.36 The standard deviation of the distribution of age from death by assault is 16.36 years.

61.

= 1 - (-e-2 + 1) = e-2 » 0.135

For the exponential schedule, P(9 < t < 10) =

ò 10 e-t dt -1

» 0.039

For the uniform schedule, P(9 < t < 10)

So the expected value is  = 1a = 3650.1 days.

Normal distribution,  = 40,  = 13, X = “take”

æ X - 40 50 - 40 ÷ö P( X > 50) = P çç > ÷ çè 13 13 ÷ø

= 1-

1 t 10 9

9 = 0.1 10

For the exponential model, P(8 < t < 12) =

Using the TI-84 Plus C command we calculate 1 - normalcdf (50, 40, 13) = 0.2209.

ò 10 e-t dt /10

= -e-t /10

12 8

-12/10

= -e

Extended Application: Exponential Waiting Times

If the density function is continuoues, the probability of any event of the form ( X = c) is 0, so it does not matter whether we use strict or inclusive inequalities. We already know that the probability of a gap of 30 minutes or longer is about 0.05. Over 40 trips, the expected number of waits that are 30 minutes or longer is 40(0.05) = 2 times.

ò 10 dt

The standard deviation is  = 1a = 3650.1 days.

= 1 - 0.7794 = 0.2206

+ e-9/10

= -e

= 1 - P ( Z £ 0.77 )

= -e-t /10

1 f (t ) = e-t /3650.1 3650.1

= P(Z > 0.77)

/10

This is an exponential distribution with 1 . a = 3650.1

62.

/10

20 ö æ = 1 - çç -e-t /10 ÷÷÷ çè 0 ÷ø

t 2 S (t ) dt - (31.80)2

5.2

ò 10 e-t dt 0

tS (t ) dt » 31.80

5.2

/10

= 1-

88.9

ò 10 e-t dt 20

The expected age at which a person will die by assault is 31.80 years. (f)

P(t > 20) =

+ e-8/10

» 0.148

For the uniform model, all interarrival times are 10 minutes, so they all meet the 2 criterion. 6.

The trick used at the HotBits site is the following: they measure times between successive pairs of decay events; if the first wait is shorter than the the second, the generator omits a 0, and if the first wait is a longer than the second, it omits a 1. For more details, see http://www.fourmilab.ch/ hotbits/how/html.

Chapter 12

SEQUENCES AND SERIES Your Turn 5

12.1 Geometric Sequences

Here a = 81 and r = 1/3. The index of summation i runs from 0 to 4, so n = 5.

Your Turn 1

å0

If = 1, an = 3(1) - 6 = 3 - 6 = -3. If = 2, an = 3(2) - 6 = 6 - 6 = 0.

æ 1 öi 81çç ÷÷÷ = S5 çè 3 ø 5 é ù 81 ê ( 13 ) - 1 ú êë úû = 1 -1 3

If = 3, an = 3(3) - 6 = 9 - 6 = 3. If = 4, an = 3(4) - 6 = 12 - 6 = 6.

Your Turn 2 Verify that r = -3. 6 18 -54 - = = = -3 2 -6 18

81( - 242 243 )

- 242 243

- 23

a7 = 2(-3)7-1 = 2(-3)6 = 2(729) = 1458

- 23

= 121

12.1

Exercises

False. The sequence a(n) = 3n + 1, for n ³ 1, is not a geometric sequence because there is no common ratio.

Your Turn 3

True

Since the machine loses 10% of its value each year, at the end of each year its value is 0.9 times its value at the end of the year before. Thus its values form a geometric sequence with r = 0.9 and a1 = 10,000. Its value at the end of 10 years is its value at n = 11.

False. If a geometric sequence has a common ratio that is negative, the consecutive terms will alternate in sign.

True

a1 = 2, r = 3, n = 4

a11 = a0r11-1

Since n = 4, we must find a1, a2, a3, and a4 with a = a1 = 2.

= 10,000(0.9)10 = 3486.78

a1 = 2

The value of the machine at the end of its tenth year is $3486.78.

a2 = 2(3) 2-1 = 2(3)1 = 2(3) = 6

Your Turn 4

a3 = 2(3)3-1 = 2(3)2 = 2(9) = 18

Since -8/2 = -4, this is a geometric sequence with a1 = 2 and r = -4. 2[(-4)7 - 1] -4 - 1 2(-16,384 - 1) = -5 = 6554

a4 = 2(3) 4-1 = 2(3)3 = 2(27) = 54

The first four terms of this geometric sequence are 2, 6, 18, and 54.

S7 =

873

874 6.

Chapter 12 SEQUENCES AND SERIES a1 = 4, r = 2, n = 5

6 = a (2)3-1

Since n = 5, we must find a1, a2, a3, a4, and a5 with a = a1 = 4.

6 = a (2)2 6 = 4a 3 3 = = a 2 2

a1 = 4 a2 = 4(2)2-1 = 4(2)1 = 4(2) = 8

3 2 3 3 3 a2 = (2)2-1 = (2)1 = (2) = 3 2 2 2 3 3-1 3 2 3 a3 = (2) = (2) = (4) = 6 2 2 2 3 4-1 3 3 3 a4 = (2) = (2) = (8) = 12 2 2 2 3 5-1 3 4 3 a5 = (2) = (2) = (16) = 24 2 2 2

a3 = 4(2)3-1 = 4(2)2 = 4(4) = 16

a1 =

a4 = 4(2)4-1 = 4(2)3 = 4(8) = 32 a5 = 4(2)5-1 = 4(2) 4 = 4(16) = 64

The first five terms of this geometric sequence are 4, 8, 16, 32, and 64. 7.

a1 = 12 , r = 4, n = 4

Since n = 4, we must find a1, a2, a3, and a4 , with a = a1 = 12 . 1 2 1 2-1 1 1 = (4)1 = (4) = 2 a2 = (4) 2 2 2 1 3-1 1 2 1 = (4) = (16) = 8 a3 = (4) 2 2 2 1 4-1 1 3 1 = (4) = (64) = 32 a4 = (4) 2 2 2

The first five terms of this geometric sequence are 32 , 3, 6, 12, and 24.

a1 =

10.

a2 = 9, a3 = 3, n = 4

Since n = 4, we must find a1, a2 , a3, and a4 a

with r = a3 = 93 = 13 . To find a, use a2 = 9, 2

r = 13 , and n = 2 in the formula. æ 1 ö2-1 9 = a çç ÷÷÷ çè 3 ø

The first four terms of this geometric sequence are 12 , 2, 8, and 32. 8.

a1 =

æ 1 ö1 9 = a çç ÷÷÷ çè 3 ø æ1ö 9 = a çç ÷÷÷ çè 3 ø

2 , r = 6, n = 3 3

Since n = 3, we must find a1, a2, and a3 with a = a1 = 23 .

27 = a

2 3 2 2 2 a2 = (6) 2-1 = (6)1 = (6) = 4 3 3 3 2 3-1 2 2 2 = (6) = (36) = 24 a3 = (6) 3 3 3 a1 =

a1 = 27 æ 1 ö2-1 æ 1 ö1 æ1ö a2 = 27 çç ÷÷ = 27 çç ÷÷ = 27 çç ÷÷ = 9 çè 3 ÷ø çè 3 ÷ø çè 3 ÷ø æ 1 ö3-1 æ 1 ö2 æ1ö a3 = 27 çç ÷÷ = 27 çç ÷÷ = 27 çç ÷÷ = 3 ÷ ÷ çè 9 ø÷ èç 3 ø èç 3 ø

The first three terms of this geometric sequence are 23 , 4, and 24.

æ 1 ö4-1 æ 1 ö3 æ 1 ö a4 = 27 çç ÷÷÷ = 27 çç ÷÷÷ = 27 çç ÷÷÷ = 1 çè 27 ø èç 3 ø èç 3 ø

a3 = 6, a4 = 12, n = 5

The first four terms of this geometric sequence are 27, 9, 3, and 1.

Since n = 5, we must find a1, a2, a3, a4 and a5 a

with r = a4 = 12 = 2. To find a, use a3 = 6, 6 3

r = 2, and n = 3 in the formula.

Section 12.1 11.

875

a1 = 4, r = 3

æ 1 ö5-1 æ 1 ö4 æ1 ö 3 = 24 çç ÷÷ = 24 çç ÷÷ = a5 = 24 çç ÷÷ çè 2 ÷ø çè 2 ÷ø çè 16 ÷ø 2

Since we want a5 , use n = 5 in the formula with a = a1 = 4 and r = 3.

æ 1 ön-1 24 an = 24 çç ÷÷÷ or n-1 çè 2 ø 2

a5 = 4(3)5-1 = 4(3)4 = 4(81) = 324 an = 4(3) n-1 12.

16.

a1 = 8, r = 4

a2 = 2, r =

To find a, use a3 = 2, r = 13 , and n = 3 in

Since we want a5 , use n = 5 in the formula with a = a1 = 8 and r = 4.

the formula. æ 1 ö3-1 a 2 = a çç ÷÷÷ = çè 3 ø 9

a5 = 8(4)5-1 = 8(4)4

a = 18

= 8(256) = 2048 an = 8(4)

13.

1 3

n -1

Since we want a5, use n = 5 in the formula with a = 18 and r = 13 .

a1 = -3, r = -5

æ 1 ö5-1 æ 1 ö4 æ1ö 2 = 18 çç ÷÷ = 18 çç ÷÷ = a5 = 18 çç ÷÷ ÷ ÷ èç 3 ø èç 3 ø èç 81 ø÷ 9

Since we want a5, use n = 5 in the formula with a = a1 = -3 and r = -5.

æ 1 ön-1 18 an = 18 çç ÷÷÷ or n-1 çè 3 ø 3

a5 = -3(-5)5-1 = -3(-5) 4 = -3(625) = -1875 an = -3(-5)n-1

14.

17.

a4 = 64, r = -4

To find a, use a4 = 64, r = -4, and n = 4 in the formula.

a1 = -4, r = -2

64 = a (-4)4-1

Since we want a5, use n = 5 in the formula with a = a1 = -4 and r = -2.

64 = a (-4)3 64 = a (-64) -1 = a

a5 = -4(-2)5-1 = -4(-2) 4 = -4(16) = -64

15.

an = -4(-2) n-1

Since we want a5 , use n = 5 in the formula with a = -1 and r = -4 .

1 2

a5 = -1(-4)5-1 = -1(-4)4

a2 = 12, r =

= -1(256) = -256

To find a, use a2 = 12, r = 12 , and n = 2 in

an = -(-4)n-1

the formula. 18.

æ 1 ö2-1 a 12 = a çç ÷÷÷ = çè 2 ø 2

a4 = 81, r = -3

To find a, use a4 = 81, r = -3, and n = 4 in the formula.

a = 24

Since we want a5, use n = 5 in the formula with a = 24 and r = 12 .

81 = a (-3)4-1 81 = a (-3)3 81 = a (-27) -3 = a

876

Chapter 12 SEQUENCES AND SERIES Since we want a5, use n = 5 in the formula with a = -3 and r = -3.

26.

a5 = -3(-3)5-1 = -3(-3) 4 = -3(81) = -243 an = -3(-3)

19.

n -1

1+ (n-1)

= (-3)

7 - 12 7 4

n -1

( )

16 64 256 = = = 4 4 16 64

an = 74 - 13

27.

an = 4(4) n-1 = 4n.

7 - 108 7 36

1 3

3, 6, 12, 24, 

Since a = a1 = 3 and r = 63 = 2,

6, 12, 24, 48, . . .

3(25 - 1) 2 -1 3(32 - 1) = 1 = 93.

S5 =

12 24 48 r = = = = 2 6 12 24

Since r = 2 and a = a1 = 6, an = 6 (2)n-1. 21.

7 36 7 - 12

Since r = - 13 and a = a1 = 74 ,

Since r = 4 and a = a1 = 4,

20.

r =

4, 16, 64, 256, . . .

r =

7 7 7 7  ,- , 4 12 36 108

3 3 , , 3, 6, 12,  4 2

The sum of the first five terms of this geometric sequence is 93.

3 6 12 r = 23 = 3 = = = 2 3 6 4 2

28.

Since a = a1 = 5 and r = 20 = 4, 5

Since r = 2 and a = a1 = 34 , an = 34 (2) n-1. 22.

5, 20, 80, 320, 

5(45 - 1) 4 -1 5(1024 - 1) = 3 = 1705.

S5 =

-7, -5, -3, -1, 1, 3, . . . -5 = 5 and -3 = 3 , the ratio is not Since 7 7 5 -5

constant, so the sequence is not geometric. 23.

The sum of the first five terms of this geometric sequence is 1705.

4, 8, -16, 32, 64, -128, 

Since 84 = 2 and -816 = -2, the ratio is not constant, so the sequence is not geometric. 24.

29.

3 12, -6, 3, - ,  2 6 = -1, Since a = a1 = 12 and r = 12 2

6, 8, 10, 12, 14, 

5 é ù 12 ê ( - 12 ) - 1ú êë úû S5 = 1 (- 2 ) - 1

= 54 , the ratio is not Since 86 = 43 and 10 8

constant, so the sequence is not geometric. 25.

5 5 5 5 - , ,- , , 8 12 18 27

r =

5 12 - 85

5 - 18 5 12

5 27 5 - 18

Since r = - 23 and a = a1 = - 85 ,

(

an = - 85 - 23

n -1

)

2 3

1 - 1ù 12 éê - 32 ë ûú - 32

33 . 4

The sum of the first five terms of this geometric . sequence is 33 4

Section 12.1 30.

18, -3,

877 1 1 ,- , 2 12

-2.772 [(-1.335)5 - 1] -1.335 - 1 -2.772 (-4.2404 - 1) » -2.335 » -6.221.

S5 =

3 = -1, Since a = a1 = 18 and r = 18 6

5 é ù 18 ê - 16 - 1ú êë úû S5 = 1 -6 -1

(

)

1 - 1ù 18 éê - 7776 ë ûú - 76

1111 . 72

The sum of the first five terms of this sequence is about -6.221.

35.

For  8(2)i , use the formula with a = 8, i=0

r = 2, and n = 8. 8(28 - 1) 2 -1 8(256 - 1) = 1 = 2040

The sum of the first five terms of this geometric sequence is 1111 . 72 31.

S8 =

a1 = 3, r = -2

Since a = a1 = 3, 3[(-2)5 - 1] 3(-32 - 1) = S5 = . -2 -1 -3

36.

i=0

r = 3, and n = 7.

The sum of the first five terms of this geometric sequence is 33. 32.

For  4(3)i , use the formula with a = 4,

4(37 - 1) 3 -1 4(2187 - 1) = 2 = 4372

S7 =

a1 = -5, r = 4

Since a = a1 = -5, -5(45 - 1) 4 -1 -5(1024 - 1) = 3 = -1705.

S5 =

37.

3 (4)i , use the formula with a = 3 , 2 i =0 2

For 

r = 4, and n = 9. 3 (49 - 1)

The sum of the first five terms of this sequence is -1705. 33.

S9 = 2

a1 = 6.324, r = 2.598

3 262,143 = 2

Since a = a1 = 6.324, 6.324(2.5985 - 1) 2.598 - 1 6.324(118.3575 - 1) » 1.598 » 464.4.

S5 =

38.

3 (2)i , use the formula with a = 3 , 4 i =0 4

For 

r = 2, and n = 10.

The sum of the first five terms of this sequence is about 464.4. 34.

4 -1

3 (262,144 - 1) 2

a1 = -2.772, r = -1.335

3 (210 - 1)

S5 = 4 =

Since a = a1 = -2.772,

2 -1

3 (1024 - 1) 4

3069 = 4

878 39.

Chapter 12 SEQUENCES AND SERIES 4

3 (-3)i , use the formula with a = 3 , 4 i =0 4

For 

43.

r = -3, and n = 5. S5 = =

3 [(-3)5 - 1] 4

-3 - 1 3 (-243 - 1) 4

-4

183 = 4

40.

3 (-2)i , use the formula with a = 3 , 2 i =0 2

For 

r = -2, and n = 6. 3 [(-2)6 - 1]

a6 = 12, 000 (0.80)6-1 = 12, 000 (0.80)5

S5 = 2 =

-2 - 1 3 (64 - 1) 2

= 12, 000 (0.32768) = 3932.16

Therefore, the value of the machine at the end of the fifth year is about $3932.

-3 63 =2

41.

(b) To find the value of the machine at the end of the eighth year, we are looking for a9 in the geometric sequence.

( )

For  64 12 , use the formula with n = 9, i =0 r = 12 , and a = 64.

a9 = 12, 000 (0.80)9-1 = 12, 000 (0.80)8 = 12, 000 (0.16777216) » 2013.27

9 é ù 64 ê 12 - 1ú êë úû S9 = 1 -1 2

Therefore, the value of the machine at the end of the eighth year is about $2013.

( )

= =

44.

( 5121 - 1) - 12

511 4

8 æ 1 öi 511 Therefore,  64 çç ÷÷÷ = . ç è ø 2 4 0 i=

42.

( )

For    , use the formula with n = 7, i =0 r =  , and a = 81.

7 é ù 81 ê 23 - 1ú 128 - 1 81 2187 êë úû 2059 S7 = = = 2 -1 1 9 -3 3

( )

(

i =0

)

If the machine loses 30% of its value, it maintains 70% of its value. With an initial value of $200,000, its value at the end of the first year will be 70% of $200,000, and its value at the end of each subsequent year will be 70% of its value at the end of the previous year. Thus, the end-ofyear values form a geometric sequence with r = 0.70. If we let a1 = 200,000, then a2 represents the value of the machine at the end of the first year, a3 is the value at the end of the second year, and so on. To find the value at the end of sixth year we are looking for a7 in the geometric sequence with n = 7, r = 0.70, and a = a1 = 200,000. a7 = 200,000(0.70)7-1 = 200,000(0.70)6 = 200,000(0.117649) = 23,529.8

. ( ) = 2059 9

Therefore,  81 23

(a) A machine that loses 20% of its value maintains 80% of its value. If the initial value of the machine is $12,000, then its value at the end of the first year will be 80% of $12,000, and its value at the end of each subsequent year will be 80% of its value at the end of the previous year. This means that the end-of-year values form a geometric sequence with r = 0.80. If we let a1 = 12, 000, then a2 will represent the value at the end of the first year, a3 will represent the value at the end of the second year, and so on. Thus, to find the value of the machine at the end of the fifth year, we are looking for a6 in the geometric sequence with n = 6, r = 0.80 and a = a1 = 12, 000.

The value of the machine at the end of the sixth year will be $23,530.

Section 12.1 45.

879

(a) The first year’s salary is $30,000. First year: $30,000.

47.

The salary for the second year is increased by 5%, or 100% + 5%, which is 1.05 times the first year’s salary. Second year: 1.05(30,000) = $31,500

a31 = 1(2)31-1 = 230

Similarly, the salary for the third year is 1.05 times the second year’s salary. Third year: 1.05(31,500) = $33,075

= 1,073,741,824

To determine the total amount saved during January, we want to use the formula to find Sn with n = 31, r = 2, and a = a1 = 1.

(b) The common ratio is r = 1.05. Also, a1 = 30, 000. Then the general term, an , is an = a1r n-1

1(231 - 1) 2,147, 483,648 - 1 = 2 -1 1 = 2,147, 483,647

S31 =

an = 30, 000(1.05)n-1

The savings on January 31 would be $230 or $1,073,741,824 and for the month of January

a10 = 30, 000(1.05)10-1 » 46,539.85 The salary in the 10th year is $46,539.85.

For the 40th year, n = 40, we find a40. a40 = 30, 000(1.05)

would be $231 - $1 or $2,147, 483,647. 48.

40-1

» 201,142.53 The salary in the 40th year is $201,142.53.

(d) To find the total amount earned over 40 years, use the formula to find S40. 30, 000(1.0540 - 1) 1.05 - 1 = 3, 623,993.23 The total amount earned over 40 years is $3,623,993.23. S40 =

46.

a1 = 4, 000, 000. To determine the total amount

produced in 8 years, use the formula to find Sn with n = 8, r = 34 , and a = a1 = 4, 000, 000.

8 é ù 4, 000, 000 ê 34 - 1ú êë úû

( )

3 -1 4

(

6561 - 1 4, 000, 000 65,536

If we let a1 represent the initial bacteria population, then the population at the end of the first hour will be a1 + 5% (a1) , or 105% (a1). The populations form a geometric sequence with r = 105% = 1.05. To determine the percent increase in the population after 7 hours, we want to use the formula to find an with n = 8, r = 1.05, and a = a1. a6 = a(1.05)8-1 » 1.40710a

Since a = a1, a8 » 1.40710a1. This means that the population at the end of 7 hours is about 141% of the initial population.

The yearly incomes produced by the well form a geometric sequence with r = 34 and

S8 =

The amounts saved form a geometric sequence with a1 = 1 and r = 2. To determine the amount saved on January 31, we want to use the formula to find an with n = 31, r = 2, and a = a1 = 1.

After five hours, the population has increased by about 41%. 49.

If 1015 molecules are present initially, then 1 (1015 ) molecules will be present after 3 years. 2

The numbers of molecules present at the end of each 3-year period form a geometric sequence with r = 12 and a1 = 1015. To determine the

)

- 14

» 14,398,193

The total amount of income produced by the well in eight years is $14,398,193.

number of molecules unchanged after 15 years, we want the number of molecules after five 3-year periods. If we let a1 = 1015 , then a2 represents the number of molecules after the first 3-year period, a3 is the number present after the second 3-year period, and so on. To find the number of molecules after five 3-year periods,

880

Chapter 12 SEQUENCES AND SERIES we are looking for a6. Use the formula to find an

(b)

with n = 6, r = 12 , and a = a1 = 1015.

The stack would be about 9.007 ´ 1012 inches thick, or about 142 million miles thick.

æ 1 ö6-1 a6 = 1015 çç ÷÷÷ çè 2 ø æ 1 ö5 = 1015 çç ÷÷÷ çè 2 ø æ 1 ö = 1015 çç ÷÷÷ çè 32 ø

52.

= 1015 (0.03125) = 3.125 ´ 1013

After fifteen years, 3.125 ´ 1013 molecules will be unchanged. 50.

The number of rotations form a geometric sequence with r = 34 and a1 = 400. If we let a1 = 400, then a2 represents the number of rotations at the end of the first minute, a3 is the number of rotations at the end of the second minute, and so on. To determine the number of rotations in the fifth minute, we want a6. Use the formula to find an with n = 6, r = 34 , and a = a1 = 400. æ 3 ö6-1 æ 3 ö5 a6 = 400çç ÷÷÷ = 400çç ÷÷÷ çè 4 ø èç 4 ø æ 243 ö÷ = 400çç = 94.921875 çè 1024 ÷÷ø

In the fifth minute after the rider’s feet are removed from the pedals, the wheel will rotate about 95 times. 51.

(a) The thicknesses of the paper form a geometric sequence with r = 2 and a1 = 0.008. If a1 = 0.008, then a2 represents the thickness after the first fold, a3 is the thickness after the second fold, and so on. To determine the thickness after 12 folds, use the formula to find an with n = 13, r = 2, and a = a1 = 0.008. a13 = 0.008(2)13-1

a51 = 0.008(2)50 » 9.007 ´ 1012

(a) Initially, there are 64 teams that play a1 = 32 games in round one. After each round, the number of teams (and the number of games) decreases by 12 . At the end of the

first round, there are 32 teams left to play a2 = 16 games in round two. At the end of the second round, there will be 16 teams left to play a3 = 8 games in round 3, and so on. Note that a1 = 25 , a2 = 24 , a3 = 23, so that by the sixth round there will be 2 teams left playing a6 = 20 = 1 game to decide the champion. Therefore, to determine the total number of games, we need to sum the terms of the geometric sequence a1 = 25 , a2 = 24 , a3 = 23, a 4 = 22 , a5 = 21, a6 = 1 or 1 + 2 + 22 + 23 + 24 + 25.

(b) To find the sum from part (a), use the formula to find Sn with n = 6, r = 12 ,

and a = a1 = 25. 6 é ù 25 ê 12 - 1ú êë úû S6 = 1 -1 2

( )

( 12 - 32 ) = 63, - 12

so 63 games must be played to produce the champion. (c) To generalize the methods of (a) and (b) for 2n teams rather than 64 = 26 teams, first

note that 2n teams play 2n-1 games in the initial round. In part (a), note that it required 6 rounds to determine a champion when 26 teams are initially involved. Similarly, for 2n teams it will require n rounds. So, the total number of games is given by

= 0.008(2)12 = 0.008(4096) = 32.768

Sn =

a (r n - 1) r -1

Section 12.1

881

with a = a1 = 2n-1, and r = 12 , which is the desired sum 2n-1 + 2n-2 + ⋅ ⋅ ⋅ + 22 + 21 + 20. n é ù 2n-1 ê ( 12 ) - 1ú n -1 1 ëê ûú = 2 - 2 Sn = 1 -1 - 12 2

æ1 ö = -2çç - 2n-1 ÷÷÷ çè 2 ø

required to determine a champion. For 3n players, n rounds are required. To determine the total number of games in such a tournament, we know that 3n players will play 3n-1 games in the first round, and for each succeeding round, 13 fewer games will be played. Therefore, the number of games played in each round forms a

= -1 + 2 n

geometric sequence with a1 = a = 3n-1

= 2 - 1.

53.

and r = 13 .

(a) In round one, there are 81 players playing = 27 games that produce 27 winners in 81 3

The total number of games played is then

to play in round two. In round two, there will be 27 = 9 games that produce 9 winners 3

Sn =

These games will determine the 3 winners that will play in the championship in round four. Therefore, the total number of games is the sum 27 + 9 + 3 + 1 = 40, or 2

ö 3æ1 = - çç - 3n-1 ÷÷÷ ç ø 2è 3 =

3 + 3 + 3 +3 , which is the sum of

a3 = 31, a4 = 30.

(b) Notice that the ratio for the geometric 2

sequence in part (a) is r = 33 = 13 . The 3

number of players and the number of games decreases by 13 from one round to the next. The total number of games therefore is the sum Sn with n = 4, r = 13 , and a = a1 = 27, or 4 é ù 27 ê 13 - 1ú êë úû S5 = 1 -1 3

( )

1 - 27

= 3

1 -1 3

- 23

the geometric sequence a1 = 33, a2 = 32 ,

( )

1 - 3n -1 ( ) 3 =

to play 93 = 3 games in round three.

æ 1 n ö÷ ç 3n-1 ççè 3 - 1÷÷ø

1 (-1 + 3n ), 2

or Sn =

3n - 1 = 3n-1 + ⋅ ⋅ ⋅ + 32 + 31 + 1. 2

(d) For a tournament where t n players are initially present, if the total number of games played is a sum of a geometric sequence as in parts (a) and (c), then it must be that t n players will play t n-1 games in the first round. This means the number of winners and the number of games played in subsequent rounds will decrease by a factor of 1t . We also know from parts (a) and (c) that a tournament that begins with t n players will require n rounds to produce a champion. Therefore, the total number of games played is the sum t n-1 + t n-2 + ⋅ ⋅ ⋅ + t 2 + t + 1,

- 23

3 æ 1 - 81 ö÷ = - çç ÷ 2 çè 3 ÷ø -80 -2 = 40, =

882

Chapter 12 SEQUENCES AND SERIES Your Turn 3

which is equal to

Sn =

æ 1 ç t n-1 ççè t

The only difference from Example 3 is that the interest rate per period is now 0.025/4. There are still (4)(3) = 12 compounding periods.

( ) - 1÷÷ø÷

1 -1 t

12 é ù ê ( 1 + 0.025 - 1ú ) 4 ú 2400 = R ê 0.025 ê ú ê ú 4 ë û 2400 » R(12.421246)

( 1t - t n-1 ) = 1-t t

t æç 1 n -1 ö÷ ÷ çç - t è ø÷ 1- t t

1 - tn = 1-t

R »

2400 » 193.22 12.421246

Melissa must make twelve deposits of $193.22.

tn - 1 . = t -1

Your Turn 4

12.2 Annuities: An Application of Sequences

We need to find the present value of an annuity of payments of $2500 made at the end of each year for 12 years, with interest at 4.5% compounded annually. é 1 - (1 + i)-n ù ú P = R êê ú i êë úû é 1 - (1.045)-12 ù ú = 2500 êê ú 0.045 êë úû » (2500)(9.118581) » 22,796.45

Your Turn 1

This is an ordinary annuity with R = 22, 000, n = 7, and i = 0.05. é (1.05)7 - 1 ù ú S = 22,000 êê ú 0.05 êë úû = (22,000)(8.142008453125) » 179,124.19

The amount accumulated at the end of 12 years will be $22,796.45.

Erin will have $179,124.19 on deposit.

Your Turn 5

Your Turn 2

The present value is $11,000, the interest rate is 0.06/12 per period, or 0.005, and there will be 48 payments.

Since the interest is compounded monthly for 5 years, the number of periods n is (5)(12) = 60. The annual interest is 6%, so the interest per monthly compounding period is 0.06/12. The initial deposit is $125. 60 é ù ê 1 + 0.06 - 1ú 12 ú S = 125 êê ú 0.06 ê ú 12 ëê ûú » (125)(69.770031) » 8721.25

(

)

The amount of the annuity is $8721.25.

é 1 - (1 + i )-n ù ú P = R êê ú i ëê ûú é 1 - (1 + 0.005)-48 ù ú 11,000 = R êê ú 0.005 ëê ûú 11,000 » R(42.580318) R »

11,000 » 258.34 42.580318

Each payment will be $258.34.

Section 12.2

883

Your Turn 6

The present value is $256,000 - $32,000 = $224,000. The monthly interest rate is 0.049/12 and there are (12)(30) = 360 periods. é 1 - (1 + i ) ú P = R êê ú i ëê ûú -360 ù é ê 1 - ( 1 + 0.049 ú ) 12 ú 224,000 = R ê 0.049 ê ú ê ú 12 ë û 224,000 » R(188.420888)

R = 1500, i = 0.04, n = 12

S = R ⋅ sni = 1500 ⋅ s 12 0.04 é (1.04)12 - 1 ù ú = 1500 êê ú 0.04 ëê ûú

-n ù

» 1500 (15.02580546) » 22,538.71

The amount is $22,538.71. 7.

224,000 188.420888 » 1188.83

R »

R = 9000, i = 0.06, n = 18

S = 9000 ⋅ s 18 0.06 é (1.06)18 - 1 ù ú = 9000 êê ú 0.06 êë úû » 9000 (30.90565255) » 278,150.87

Each monthly payment will be $1188.83.

12.2

Exercises

True 8.

True

False. Amortizing a loan is when both the principal and interest are paid by a sequence of equal period payments. The concept of depositing a lump sum P today, at an interest rate period i, in order to have a particular sum of money in the future describes an annuity.

False. It is possible for the interest paid on a loan to be more than the amount of principal borrowed. The total amount of interest paid on a loan depends on the interest rate and the number of payments.

S = 80, 000 ⋅ s 24 0.07 é (1.07) 24 - 1 ù ú = 80, 000 êê ú 0.07 ëê ûú » 80, 000 (58.17667076) » 4, 654,133.66

The amount is $4,654,133.66. 9.

é (1.055)30 - 1 ù ú = 11,500 êê ú 0.055 êë úû » 11,500 (72.43547797)

S = R⋅ sn i

» 833, 008.00.

= 120 ⋅ s 10 0.05

» 1509.35

The amount is $1509.35.

R = 11,500, i = 0.055, n = 30 S = 11,500 ⋅ s 30 0.055

R = 120, i = 0.05, n = 10

é (1.05)10 - 1 ù ú = 120 êê ú 0.05 êë úû » 120(12.57789254)

The amount is $278,150.87. R = 80,000, i = 0.07, n = 24

The amount is $833,008.00 10.

R = 13, 400, i = 0.045, n = 25

S = 13, 400 ⋅ s 25 0.045 é (1.045)25 - 1 ù ú = 13, 400 êê ú 0.045 ëê ûú » 13, 400 (44.56521015) » 597,173.82

884 11.

Chapter 12 SEQUENCES AND SERIES R = 10,500

15.

Interest is earned semiannually for 7 years, so i = 0.10 = 0.05 and n = 2 ⋅ 7 = 14 periods. 2 é (1.05)14 - 1 ù ú S = 10,500 êê ú 0.05 êë úû » 10,500(19.59863199) » 205,785.6359

10, 000 = R ⋅ s 12 0.08 10, 000 = R (18.97712646) R » 526.95

The periodic payment should be $526.95. 16.

The amount is $205,785.64. 12.

This describes an ordinary annuity with S = 10, 000, i = 0.08, and n = 12 periods.

This describes an ordinary annuity with S = 80,000, i = 0.06 = 0.03, and 2 n = 2 ⋅ 9 = 18 periods.

R = 4200

S = R ⋅ sni

Interest is earned semiannually for 11 years, so i = 0.06 = 0.03 and n = 2 ⋅ 11 = 22 periods. 2

R =

é (1.03) 22 - 1 ù ú S = 4200 êê ú 0.03 êë úû » 4200(30.5367803) » 128, 254.48

80, 000 é (1.03)18 -1 ù ê 0.03 ú êë úû 80, 000 » 23.41443537 » 3416.70

The amount is $128,254.48. 13.

R = 1800

The periodic payment should be $3416.70.

Interest is earned quarterly for 12 years, so i = 0.08 = 0.02 and n = 4 ⋅ 12 = 48 periods. 4

17.

)

50, 000 = R ⋅ s 32 0.03 50, 000 » R(52.50275852) R » 952.33

The amount is $142,836.33.

The periodic payment should be $952.33.

R = 5300

18.

Interest is earned quarterly for 9 years, so i = 0.04 = 0.01 and n = 4 ⋅ 9 = 36 periods. 4 é (1.01)36 - 1 ù ú S = 5300 êê ú 0.01 êë úû » 5300(43.07687836)

The amount is $228,307.46.

(

n = 8 ⋅ 4 = 32 periods.

» 142,836.33

» 228,307.46

This describes an ordinary annuity with S = 50,000, i = 0.03 = 12% , and 4

é (1.02)48 - 1 ù ú S = 1800 êê ú 0.02 ëê ûú » 1800(79.35351927)

14.

80, 000 s 18 0.03

This describes an ordinary annuity with S = 8000, i = 0.04/12, and n = (12)(5) = 60 periods. 8,000 = R ⋅ s 60 (0.04 /12) 60 é ù ê ( 1 + 0.04 - 1ú ) 12 ú 8000 = R ê 0.04 ê ú ê ú 12 ë û 8000 » R(66.298978)

R » 120.67

The periodic payment should be $120.67.

Section 12.2 19.

885

R = 5000, i = 0.06, and n = 11 payments.

24.

n = 4 ⋅ 15 = 60 periods.

é 1 - (1.06)-11 ù ú = 7.886874577, a 11 0.06= êê ú 0.06 ëê ûú

é 1 - (1.01)-60 ù ú P = 9800 êê ú 0.01 êë úû » 9800(44.95503841)

so P » 5000(7.886874577) » 39, 434.37,

» 440,559.38

or $39,434.37. 20.

The present value is $440,559.38.

R = 1280, i = 0.07, and n = 9 periods.

25.

P = 1280 ⋅ a 9 0.07 é 1 - (1.07)-9 ù ú = 1280 êê ú 0.07 ëê ûú » 1280(6.515232249) » 8339.50

The present value is $8339.50. 21.

R = 1400, i = 0.06 = 0.03, and 2

26.

n = 2 ⋅ 8 = 16 periods. é 1 - (1.03)-16 ù ú P = 1400 êê ú 0.03 êë úû » 1400(12.56110203) » 17,585.54

The present value is $17,585.54. 22.

R = 960, i = 0.05 = 0.025, and 2

27.

n = 2 ⋅ 16 = 32 periods. é 1 - (1.025)-32 ù ú P = 960 êê ú 0.025 ëê ûú » 960(21.84917796) » 20,975.21

The present value is $20,975.21. 28. 23.

R = 9800, i = 0.04 = 0.01, and 4

R = 50,000, i = 0.08 = 0.02, and 4 n = 10 ⋅ 4 = 40 payments. é 1 - (1.02)-40 ù ú a40 0.02 = êê ú 0.02 ëê ûú » 27.35547924,

é 1 - (1.04)-15 ù ú » 11.11838743, so a 15 0.04= êê ú 0.04 ëê ûú P » 10,000(11.11838743) » 111,183.87, or $111,183.87. A lump sum deposit of $111,183.87 today at 4% compounded annually will yield the same total after 15 years as deposits of $10,000 at the end of each year for 15 years at 4% compounded annually. é 1 - (1.05)-15 ù ú » 10.37965804, so a 15 0.005= êê ú 0.05 ëê ûú P » 10,000(10.37965804) » 103,796.58, or $103,796.58.

A lump sum deposit of $103,796.58 today at 5% compounded annually will yield the same total after 15 years as deposits of $10,000 at the end of each year for 15 years at 5% compounded annually. é 1 - (1.06)-15 ù ú » 9.712248988, so a 15 0.06= êê ú 0.06 êë úû P » 10,000(9.712248988) » 97,122.49, or $97,122.49. A lump sum deposit of $97,122.49 today at 6% compounded annually will yield the same total after 15 years as deposits of $10,000 at the end of each year for 15 years at 6% compounded annually. é 1 - (1.08)-15 ù ú = 8.559478688, so a 15 0.08= êê ú 0.08 êë úû P » 10,000(8.559478688) » 85,594.79, or $85,594.79. A lump sum deposit of $85,594.79 today at 8% compounded annually will yield the same total after 15 years as deposits of $10,000 at the end of each year for 15 years at 8% compounded annually.

so P » 50, 000 (27.35547924) » 1,367, 773.96,

or $1,367,773.96.

29.

$2500 is the present value of this annuity of R = 4% dollars, with 6 periods, and i = 16% 4 = 0.04 per period.

886

Chapter 12 SEQUENCES AND SERIES P = R ⋅ an i

33.

2500 = R ⋅ a6 0.04 R =

$55,000 is the present value of this annuity of R = 0.005, and n = 36 dollars, with i = 0.06 12 periods.

2500 a6 0.04

P = R ⋅ ani

2500 5.242136857 » 476.90

R =

55, 000 32.87101624 » 1673.21 »

Each payment is $476.90. 30.

$1000 is the present value of this annuity of R dollars, with 9 periods, and i = 8% = 0.08 per period.

Each payment is $1673.21. 34.

P = R ⋅ an i

$6800 is the present value of this annuity of R = 0.01, and n = 24 periods. dollars, i = 0.12 12

1000 = R ⋅ a 9 0.08

P = R ⋅ an i

1000 a 9 0.08

R =

1000 6.246887911 » 160.08

R =

Each payment is $320.10.

Each payment is $160.08. $90,000 is the present value of this annuity of R dollars, with 12 periods, and i = 8% = 0.08 per period. P = R ⋅ an i

35.

(a) Sarah’s payments form an ordinary annuity with R = 12,000, n = 9, and i = 0.05. The amount of this annuity is S = 12, 000s 9 0.05 é (1.05)9 - 1 ù ú = 12, 000 êê ú 0.05 êë úû » (12, 000)(11.026564) » 132,318.77,

90, 000 = R ⋅ a 12 0.08 R =

90, 000 a 12 0.08

90, 000 7.536078017 » 11,942.55 »

Each payment is $11,942.55. 32.

$41,000 is the present value of this annuity of R dollars, with n = 10 periods, and i = 12% = 6% = 0.06 per period. 2 P = R ⋅ an i 41, 000 = R ⋅ a 10 0.06 R =

41,000 a 10 0.06

41, 000 7.36008705 » 5570.59 »

6800 a24 0.01

6800 21.24338726 » 320.10

31.

55, 000 a 36 0.005

or $132,318.77. (b) Using her brother-in-law’s bank gives her an annuity with R = 12,000, n = 9, and i = 0.03. The amount of this annuity is S = 12, 000s 9 0.03 é (1.03)9 - 1 ù ú S = 12, 000 êê ú 0.03 êë úû » (12, 000)(10.159106) » 121,909.27,

or $121,909.27. (c) The amount Sarah would lose using the second option is 132,318.77 - 121,909.27 = 10,409.50, or $10,409.50.

Section 12.2 36.

887

Yvette’s payments form an ordinary annuity with R = 100, n = (12)(8) = 96, and i = 0.06/12 = 0.005.

24, 000 = Rs 20 0.0125 é (1.0125) 20 - 1 ù ú = R êê ú 0.0125 ëê ûú

(a) Yvette’s payments total (96)(100) or $9600.

» R(22.562979)

(b) The final amount in the annuity will be

R »

S = 100s 95 0.005

The required payments at the end of each quarter are $1063.69.

é (1.005)95 - 1 ù ú 100 êê ú 0.005 êë úû » (100)(122.828542) » 12, 282.85,

39.

or $12,282.85. (c) The interest Yvette earned is 12, 282.85 - 9600.00 = 2682.85, or $2682.25. 37.

Harv’s payments will form an ordinary annuity with final amount 12,000, R unknown, n = (2)(4) = 8, and i = 0.04 / 2 = 0.02. 12, 000 = Rs 8 0.02 é (1.02)8 - 1 ù ú = R êê ú 0.02 ëê ûú

(a) Isaac’s payments will form an ordinary annuity with final amount 20,000, R unknown, n = (4)(8) = 32, and i = 0.06/4 = 0.0015.

» R(8.582969) R »

12,000 » 1398.12 8.582969

The required payments at the end of each quarter are $1398.12.

20, 000 = Rs 32 0.015 é (1.05)32 - 1 ù ú = R êê ú 0.015 êë úû » R(40.688288) 20, 000 R » 40.688288 » 491.54

40.

(a) The firm’s payments will form an ordinary annuity with final amount 40,000, R unknown, n = 7, and i = 0.06. 40,000 = Rs 7 0.06 é (1.06)7 - 1 ù ú = R êê ú 0.06 êë úû » R(8.393838)

The required payments at the end of each quarter are $491.54. (b) In this scenario, i = 0.04/4 = 0.01. 20,000 = Rs 32 0.01 é (1.01)32 - 1 ù ú = R êê ú 0.01 êë úû » R(37.494068) 20, 000 » 533.42 R » 37.494068

R »

40,000 » 4765.40 8.393838

The required payments at the end of each quarter are $4765.40. (b) The firm’s payments will form an ordinary annuity with final amount 40,000, R unknown, n = 7, and i = 0.08. 40,000 = Rs 7 0.08 é (1.08)7 - 1 ù ú = R êê ú 0.08 ëê ûú » R(8.922803) 40,000 R » » 4482.90 8.922803

The required payments at the end of each quarter are $533.42. 38.

24,000 » 1063.69 22.562979

Mariah’s payments will form an ordinary annuity with final amount 24,000, R unknown, n = (4)(5) = 20, and i = 0.05/4 = 0.00125.

The required payments at the end of each quarter are $4482.90.

888 41.

Chapter 12 SEQUENCES AND SERIES Interest of 6% = 3% is earned semiannually. 2 In 65 - 40 = 25 years, there are 25 ⋅ 2 = 50 semiannual periods. Since é (1.03)50 - 1 ù ú s 50 0.03= êê ú 0.03 êë úû » 112.7968673,

the $1000 semiannual deposits will produce a total of S = 1000(112.7968673) » 112, 796.87,

(a) The total amount of interest paid is (60, 000)(0.08)(7) = 33, 600, or $33, 600. This total is divided into 7 ⋅ 4 = 28 equal quarterly interest payments. Since 33,600 = 1200, each quarterly interest 28

payment will be $1200. (b) This ordinary annuity will amount to $60.000 in 7 years at 6% compounded semiannually. Thus, S = 60, 000, n = 7 ⋅ 2 = 14, and i = 6% = 3% 2

= 0.03, so 60,000 = R ⋅ s 14 0.03

or $112,796.87. 42.

45.

Interest of 8% = 4% is earned semiannually. 2 In 65 - 40 = 25 years, there are 25 ⋅ 2 = 50 semiannual periods. Since é (1.04)50 - 1 ù ú » 152.6670837, s 50 0.04= êê ú 0.04 êë úû

R = »

S = 1000(152.6670837) » 152,667.08,

or $152,667.08. Interest of 10% = 5% is earned semiannually. 2 In 65 - 40 = 25 years, there are 25 ⋅ 2 = 50 semiannual periods. Since é (1.05)50 - 1 ù ú » 209.3479957, s 50 0.05= êê ú 0.05 êë úû

the $1000 semiannual deposits will produce a total of S » 1000(209.3479957) » 209,348.00, or $209,348.00. 44.

60,000 » 3511.58, 17.08632416

or $3511.58.

the $1000 semiannual deposits will produce a total of

43.

60,000 s 14 0.03

Interest of 12% = 6% is earned semiannually. 2 In 65 - 40 = 25 years, there are 25 ⋅ 2 = 50 semiannual periods. Since é (1.06)50 - 1 ù ú » 290.3359046, s 50 0.06= êê ú 0.06 êë úû

the $1000 semiannual deposits will produce a total of S » 1000(290.3359046) » 290,335.90,

or $290,335.90.

Section 12.2

889

Amount of Deposit $3511.58 $3511.58 $3511.58 $3511.58 $3511.58 $3511.58 $3511.58 $3511.58 $3511.58 $3511.58 $3511.58 $3511.58 $3511.58 $3511.58

Interest Earned $0 (3511.58) (0.03) = $105.35 (7128.51) (0.03) = $213.86 (10,853.95) (0.03) = $325.62 (14,691.15) (0.03) = $440.73 (18,643.46) (0.03) = $559.30 (22,714.34) (0.03) = $681.43 (26,907.35) (0.03) = $807.22 (31,226.15) (0.03) = $936.78 (35,674.51) (0.030) = $1070.24 (40,256.33) (0.03) = $1207.69 (44,975.60) (0.03) = $1349.27 (49,836.45) (0.03) = $1495.09 (54,843.12) (0.03) = $1645.29

Total in Account $3511.58 3511.58 + 3511.58 + 105.35 = $7128.51 7128.51 + 3511.58 + 213.86 = $10,853.95 10,853.95 + 3511.58 + 325.62 = $14.691.15 14,691.15 + 3511.58 + 440.73 = $18,643.46 18,643.46 + 3511.58 + 559.30 = $22,714.34 22,714.34 + 3511.58 + 681.43 = $26,907.35 26,907.35 + 3511.58 + 807.22 = $31,226.15 31,226.15 + 3511.58 + 936.78 = $35,674.51 35,674.51 + 3511.58 + 1070.24 = $40,256.33 40,256.33 + 3511.58 + 1207.69 = $44,975.60 44,975.60 + 3511.58 + 1349.27 = $49,836.45 49,836.45 + 3511.58 + 1495.09 = $54,843.12 54,843.12 + 3511.58 + 1645.29 = $59,999.99

So that the final total in the account is $60,000, add $0.01 to the last amount of deposit. Thus, line 14 of the table will be: 14

46.

$3511.59

(54,843.12)(0.03) = $1645.29 54,843.12 + 3511.59 + 1645.29 = $60, 000.00

(a) The total amount of interest paid is (4000)(0.06)(5) = $1200. This total is divided into 5 ⋅ 2 = 10 equal

= 120, each semiannual interest payment will be $120. semiannual interest payments. Since 1200 10 (b) This ordinary annuity will amount to $4000 in 5 years at 8% compounded annually. Thus, S = 4000, n = 5, and i = 0.08, so 4000 = R ⋅ s 5 0.08 R =

4000 4000 » s 5 0.08 5.86660096

» 681.83.

or $681.83. (c) Payment Number 1 2 3 4 5

Amount of Deposit $681.83 $681.83 $681.83 $681.83 $681.83

Interest Earned $0 (681.83) (0.08) = $54.55 (1418.21) (0.08) = $113.46 (2213.50) (0.08) = $177.08 (3072.41) (0.08) = $245.79

Total in Account $681.83 681.83 + 681.83 + 54.55 = $1418.21 1418.21 + 681.83 + 113.46 = $2213.50 2213.50 + 681.83 + 177.08 = $3072.41 3072.41 + 681.83 + 245.79 = $4000.03

So that the final total in the account is $4000, subtract $0.03 from the last amount of deposit. Thus, line 5 of the table will be: 5

$681.80

(3072.41) (0.08) = $245.79 3072.41 + 681.80 + 245.79 = $4000.00

890 47.

Chapter 12 SEQUENCES AND SERIES $150,000 is the future value of an annuity over 79 yr compounded quarterly. So, there are 79(4) = 316 payment periods. (a)

48.

First, we want to find the amount of the annuity with R = 1000, n = 8, and i = 0.06. é (1.06)8 - 1 ù ú. S = 1000 êê ú 0.06 ëê ûú

The interest per quarter is 5.25% = 1.3125%. 4 Thus, S = 150, 000, n = 316, i = 0.013125, and we must find the quarterly payment R in the formula

The number in brackets, s 8 0.06, is 9.897467909, so that

é (1 + i)n - 1 ù ú S = R êê ú i êë úû é (1.013125)316 - 1 ù ú 150, 000 = R êê ú 0.013125 êë úû R » 32.4923796

S » 1000(9.897467909) » 9897.47,

or $9897.47. Next, we want to find the lump sum that must be deposited today at 5% compounded annually for 8 years to provide $9897.47. A = 9897.47, i = 0.05, and n = 8, so

She would have to put $32.49 into her savings at the end of every three months.

9897.47 = P(1.05)8 9897.47 P = » 6699.00, (1.05)8

(b) For a 2% interest rate, the interest per = 0.5%. Thus, S = 150, 000, quarter is 2% 4

n = 316, i = 0.005, and we must find the quarterly payment R in the formula é (1 + i)n - 1 ù ú S = R êê ú i êë úû é (1.005)316 - 1 ù ú 150, 000 = R êê ú 0.005 êë úû R » 195.5222794

or $6699.00. 49.

é 1 - (1.08)-9 ù ú » 6.246887911, so a9 0.08 = êê ú 0.08 êë úû P » 2000 (6.246887911) » 12, 493.78,

She would have to put $195.52 into her savings at the end of every three months. For a 7% interest rate, the interest per quarter = 1.75%. Thus, S = 150, 000, is 7% 4

or $12,493.78. A lump sum deposit of $12,493.78 today at 8% compounded annually will yield the same total after 9 years deposits of $2000 at the end of each year for 9 years at 8% compounded annually.

n = 316, i = 0.0175, and we must find the quarterly payment R in the formula é (1 + i) n - 1 ù ú S = R êê ú i ëê ûú é (1.0175)316 - 1 ù ú 150, 000 = R êê ú 0.0175 ëê ûú

R » 10.9663932

She would have to put $10.97 into her savings at the end of every three months.

We want to find the present value of an annuity of $2000 per year for 9 years at 8% compounded annually.

50.

We want to find the present value of an annuity of $50,000 per year for 20 years at 6% compounded annually. é 1 - (1.06) -20 ù ú a 20 0.06= ê êë úû 0.06 » 11.46992122, so P » 50, 000(11.46992122) » 573, 496.06,

or $573,496.06. A lump sum deposit of $573, 496.06 today at 6% compounded annually will yield the same total after 20 years as deposits of $50,000 at the end of each year for 20 years at 6% compounded annually.

Section 12.2 51.

891

For parts (a) and (b), if $1 million is divided into 20 equal payments, each payment is $50,000. (a)

i = 0.05, n = 20 é 1 - (1 + i)-n ù ú P = R êê ú i êë úû é 1 - (1 + 0.05)-20 ù ú = 50, 000 êê ú 0.05 êë úû » 623,110.52

The present value is $623,110.52. (b)

i = 0.09, n = 20 é 1 - (1 + i)-n ù ú P = R êê ú i êë úû é 1 - (1 + 0.09)-20 ù ú = 50,000 êê ú 0.09 êë úû » 456, 427.28

The present value is $456,427.28. For parts (c) and (d), if $1 million is divided into 25 equal payments, each payment is $40,000. (c)

i = 0.05, n = 25 é 1 - (1 + i)-n ù ú P = R êê ú i ëê ûú é 1 - (1 + 0.05)-25 ù ú = 40,000 êê ú 0.05 ëê ûú » 563,757.78

The present value is $563,7573.78. (d)

i = 0.09, n = 25 é 1 - (1 + i )-n ù ú P = R êê ú i ëê ûú é 1 - (1 + 0.09)-25 ù ú = 40,000 êê ú 0.09 ëê ûú » 392,903.18

The present value is $392,903.18. 52.

= 0.0075 per month and of n = 4 ⋅ 12 = 48 $22,000 is the present value of this annuity with interest i = 0.09 12 monthly payments. S = R⋅an i R =

22, 000 22, 000 » » 547.47 a 48 0.075 40.18478189

The amount of each payment is $547.47. Since 48 payments of $547.47 each are made, 48($547.47) = $26, 278.56 will be paid in total. This means that $26, 278.56 - $22, 000 = $4278.56 is the total amount of interest Susan will pay. Copyright © 2022 Pearson Education, Inc.

892 53.

Chapter 12 SEQUENCES AND SERIES The present value, P, is 249,560, i = 0.0775 » 0.0064583333, and n = 12 ⋅ 25 = 300. 12

é 1 - (1 + 0.0064583333)-300 ù ú » R éê 1 - 0.1449639356 ùú » R éê 0.8550360644 ùú 249,560 = R ⋅ a 300 0.0064583333= R êê ú 0.0064583333 ëê 0.0064583333 ûú ëê 0.0064583333 ûú ëê ûú or R » 1885.00.

Monthly payments of $1885.00 will be required to amortize the loan. Use the formula for the unpaid balance of a loan with R = 1885, i » 0.0064583333, n = 12 ⋅ 15 = 300, and x = 12 ⋅ 5 = 60. -240 ù é 1 - (1 + i)-(n- x) ù é ú » 1885 ê 1 - (1 + 0.0064583333) ú » 229, 612.44 y = R êê ú ê ú i 0.0064583333 ëê ûú ëê ûú

The unpaid balance after 5 years is $229,612.44. 54.

The present value, P, is 170,892, i = 0.0811 » 0.0067583333, and n = 12 ⋅ 30 = 360. 12

é 1 - (1 + 0.0067583333)-360 ù ú » R éê 1 - 0.0884944586 ùú » R éê 0.9115055414 ùú 170,892 = R ⋅ a 360 0.0067583333= R êê ú êë 0.0067583333 úû êë 0.0067583333 úû 0.0067583333 êë úû R » 1267.07.

Monthly payments of $1267.07 will be required to amortize the loan. Use the formula for the unpaid balance of a loan with R = 1267.07, i » 0.0067583333, n = 12 ⋅ 30 = 360, and x = 12 ⋅ 5 = 60. -300 ù é 1 - (1 + i)-(n- x) ù é ú » 1267.07 ê 1 - (1 + 0.0067583333) ú » 162, 628.91 y = R êê ú ê ú i 0.0067583333 êë úû êë úû

The unpaid balance after 5 years is $162,628.91. 55.

The present value, P, is 353,700, i = 0.0795 = 0.006625, and n = 12 ⋅ 30 = 360 periods. 12 P = R⋅an i R =

é 1 - (1.006625)-360 ù ú » 353, 700 » 2583.01 = 353, 700 êê ú a 360 0.006625 0.006625 136.9334083 êë úû 353, 700

Monthly payments of $2583.01 will be required to amortize the loan. To find the unpaid balance, use the formula for the unpaid balance of a loan with R = 2583.01, i = 0.006625, n = 360, and x = 12 ⋅ 5 = 60. -300 ù é 1 - (1 + i)-(n- x) ù é ú = 2583.01 ê 1 - (1 + 0.006625) ú » 2583.01(130.1224488) » 336,107.59 y = R ⋅ êê ú ê ú i 0.006625 ëê ûú ëê ûú

The unpaid balance after 5 years is $336,107.59.

Section 12.2 56.

893

The present value, P is 196,511, i = 0.0757 » 0.0063083333, and n = 12 ⋅ 25 = 300. 12 é 1 - (1 + 0.0063083333)-300 ù ú 196,511 = R ⋅ a 300 0.0063083333 = R êê ú 0.0063083333 ëê ûú é 1 - 0.1515930387 ù é 0.8484069613 ù ú = Rê ú » Rê êë 0.0063083333 úû êë 0.0063083333 úû R » 1461.16.

Monthly payments of $1461.16 will be required to amortize the loan. Use the formula for the unpaid balance of a loan with R = 1461.16, i » 0.0063083333, n = 12 ⋅ 25 = 300, and x = 12 ⋅ 5 = 60. -240 ù é 1 - (1 + i)-(n- x) ù é ú » 1461.16 ê 1 - (1 + 0.0063083333) ú » 180, 417.11 y = R êê ú ê ú i 0.0063083333 êë úû êë úû

The unpaid balance after 5 years is $180,417.11. 57.

(a) $25,000 is the present value of this annuity of 8 years with interest of 6% per year. a 8 0.06= 6.209793811, so

25, 000 = R(6.209793811) R » 4025.90

Annual withdrawals of $4025.90 each will be needed. (b) $25,000 is the present value of this annuity of 12 years with interest of 6% per year. a 12 0.06= 8.38384394, so

25, 000 = R(8.38384394) R » 2981.93

Annual withdrawals of $2981.93 each will be needed. 58.

(a) $150,000 is the present value of this annuity of 5 ⋅ 2 = 10 periods with interest of 6% = 3% = 0.03 2

per year. a 10 0.03» 8.530202837, so 150, 000 = R (8.530202837) R » 17,584.58

Withdrawals of $17,584.58 at the end of each six-month period are needed. (b) $150,000 is the present value of this annuity of 7 ⋅ 2 = 14 periods with interest 6% = 3% = 0.03 per year. 2 a 14 0.03» 11.29607314, so 150, 000 = R (11.29607314) R » 13, 278.95

Withdrawals of $13,278.95 at the end of each six-month period are needed.

894

Chapter 12 SEQUENCES AND SERIES First, find the amount of each payment. $4000 is the present value of an annuity of R dollars, with 4 periods, and i = 0.08 per period.

59.

P = R⋅ an i 4000 = R ⋅ a4 0.08 R =

4000 4000 » » 1207.68 3.31212684 a 4 0.08

Each payment is $1207.68. Payment Amount of Number Payment 0

Interest for Period

Portion to Principal

Principal at End of Period

—

$4000.00

—

$1207.68

(4000)(0.08)(1) = $320.00 1207.68 - 320.00 = $887.68

4000.00 - 887.68 = $3112.32

$1207.68

(3112.32) (0.08)(1) = $248.99 1207.68 - 248.99 = $958.69

3112.32 - 958.69 = $2153.63

$1207.68

(2153.63) (0.08)(1) = $172.29 1207.68 - 172.29 = $1035.39 2153.63 - 1035.39 = $1118.24

$1207.68

(1118.24) (0.08)(1) = $89.46

1207.68 -

89.46 = $1118.22 1118.24 - 1118.22 = $0.02

So that the final principal is zero, add $0.02 to the last payment. Thus, line 4 of the table will be: 4

$1207.70

(1118.24) (0.08) (1) = $89.46 1207.70 - 89.46 = $1118.24 1118.24 - 1118.24 = $0

First, find the amount of each payment. $72,000 is the present value of an annuity of R dollars, with 9 periods, and i = 9.5% = 4.75% = 0.0475 per period. 2

60.

P = R⋅an i 72, 000 = R ⋅ a 9 0.0475 R =

72, 000 72, 000 » » 10, 017.22 a 9 0.0475 7.187624181

Each payment is $10,017.22. Payment Amount of Number Payment

Interest for Period

Portion to Principal

Principal at End of Period

—

$72,000,00

—

$10,017.22

$10,017.22 65,402.78(0.0475) = $3106.63 10,017.22 - 3106.63 = $6910.59 65,402.78 - 6910.59 = $58,492.19

$10,017.22 58,492.19(0.0475) = $2778.38 10,017.22 - 2778.38 = $7238.84 58,492.19 - 7238.84 = $51,253.35

$10,017.22 51,253.35(0.0475) = $2434.53 10,017.22 - 2434.53 = $7582.69 51,253.35 - 7582.69 = $43,670.66

$10,017.22 43,670.66(0.0475) = $2074.36 10,017.22 - 2074.36 = $7942.86 43,670.66 - 7942.86 = $35,727.80

$10,017.22 35,727.80(0.0475) = $1697.07 10,017.22 - 1697.07 = $8320.15 35,727.80 - 8320.15 = $27,407.65

$10,017.22 27,407.65(0.0475) = $1301.86 10,017.22 - 1301.86 = $8715.36 27,407.65 - 8715.36 = $18,692.29

$10,017.22 18,692.29(0.0475) = $887.88

$10,017.22

72,000(0.0475) = $3420.00 10,017.22 - 3420.00 = $6597.22 72,000.00 - 6597.22 = $65,402.78

9562.95(0.0475) = $454.24

10,017.22 - 887.88 = $9129.34 18,692.29 - 9129.34 = $9562.95 10,017.22 - 454.24 = $9562.98

9562.95 - 9562.98 = –$0.03

So that the final principal is zero, subtract $0.03 from the last payment. Thus, line 9 of the table will be: 9

$10, 017.19

9562.95(0.0475) = $454.24 10, 017.19 - 454.24 = $9562.95 9562.95 - 9562.95 = $0

Section 12.2

895

The firm’s total cost is 8(1048) = $8384.00. After making a down payment of $1200, a balance of 8384 - 1200 = $7184 is owed. To amortize this balance, first find the amount of each payment. $7184 is the present value of an annuity of R dollars, with 4 ⋅ 12 = 48 periods, and i = 10.5% = .875% = 0.00875 12

61.

per period. P = R⋅ ani 7184 = R ⋅ a 48 0.00875 R =

7184 7184 » » 183.93 a 48 0.00875 39.05734359

Each payment is $183.93. Payment Number

Amount of Payment

Interest for Period

Portion to Principal

Principal at End of Period

—

— 183.93–62.86 = $121.07

$7184.00 7184.00–121.07 = $7062.93

183.93–61.80 = $122.13

7062.93–122.13 = $6940.80

183.93–60.73 = $123.20

6940.80–123.20 = $6817.60

183.93–59.65 = $124.28

6817.60–124.28 = $6693.32

183.93–58.57 = $125.36

6693.32–125.36 = $6567.96

183.93–57.47 = $126.46

6567.96–126.46 = $6441.50

62.

$183.93

( ) = $62.86 (7062.93) (0.105) ( 1 ) = $61.80 12 (6940.80) (0.105) ( 1 ) = $60.73 12 (6817.60) (0.105) ( 1 ) = $59.65 12 (6693.32) (0.105) ( 1 ) = $58.57 12 (6567.96) (0.105) ( 1 ) = $57.47 12 (7184.00)(0.105) 1 12

Aaliyah’s total cost is 14, 000 + 7200 = $21, 200.00. After making a down payment of $1200, she owes a balance of 21, 200 - 1200 = $20,000. To amortize this balance, first find the amount of each payment. $20,000 is the = 4% = 0.04 per period. present value of an annuity of R dollars, with 5 ⋅ 2 = 10 periods, and i = 8% 2 P = R⋅ an i 20, 000 = R ⋅ a 10 0.04 R =

20, 000 a 10 0.04

20, 000 8.110895779 » 2465.82 »

Each payment is $2465.82.

896

Chapter 12 SEQUENCES AND SERIES

Payment Amount of Number Payment

Interest for Period

Portion to Principal

Principal at End of Period

—

$20,000.00

—

$2465.82

(20,000)(0.08)

( 12 ) = $800.00 2465.82 - 800.00 = $1665.82 20,000.00 - 1665.82 = $18,334.18

$2465.82

(18,334.18) (0.08)

( 12 ) = $733.37 2465.82 - 733.37 = $1732.45 18,334.18 - 1732.45 = $16,601.73

$2465.82

(16,601.73) (0.08) 1

$2465.82 (14,799.98) (0.08)

( 12 ) = $592.00 2465.82 - 592.00 = $1873.82 14,799,.98 - 1873.82 = $12,926.16

$2465.82 (12,926.16) (0.08)

( 12 ) = $517.05 2465.82 - 517.05 = $1948.77 12,926.16 - 1948.77 = $10,977.39

$2465.82 (10,977.39) (0.08)

( 12 ) = $439.10 2465.82 - 439.10 = $2026.72 10,977.39 - 2026.72 = $8950.67

$2465.82

(8950.67) (0.08) 1

$2465.82

(6842.88) (0.08) 1

$2465.82

(4650.78) (0.08) 1

$2465.82

(2370.99) (0.08) 1

( 2 ) = $664.07 2465.82 - 664.07 = $1801.75 16,601.73 - 1801.75 = $14,799.98

( 2 ) = $358.03 2465.82 - 358.03 = $2107.79

8950.67 - 2107.79 = $6842.88

( 2 ) = $273.72 2465.82 - 273.72 = $2192.10

6842.88 - 2192.10 = $4650.78

( 2 ) = $186.03 2465.82 - 186.03 = $2279.79

4650.78 - 2279.79 = $2370.99

( 2 ) = $94.84

2370.99 - 2370.98 = $0.01

2465.82 - 94.84 = $2370.98

So that the final principal is zero, add $0.01 to the last payment. Thus, line 10 of the table will be: 10

$2465.83

( )

(2370.99)(0.08) 12 = $94.84 2465.83 - 94.84 = $2370.99 2370.99 - 2370.99 = $0

12.3 Taylor Polynomials at 0 Your Turn 1 1 1 1 1 1 (-0.15) + (-0.15)2 + (-0.15)3 + (-0.15)4 + (-0.15)5 1! 2! 3! 4! 5! » 0.860708

P5 (-0.15) = 1 +

Your Turn 2

The derivatives of f ( x) = x + 4. Derivative

x + 4 are exactly as shown in Example 2 except that x + 1 is replaced everywhere by

Value at 0 1/2

f ( x) = ( x + 4) 1 f (1) ( x) = 2( x + 4)1/2 -1 f (2) ( x) = 4( x + 4)3/2 3 f (3) ( x) = 8( x + 4)5/2 -15 f (4) ( x) = 16( x + 4)7/2

f (0) = 2 1 f (1) (0) = 4 f (2) (0) = -

1 32

f (3) (0) =

3 256

f (4) (0) =

-15 2048

Section 12.3

897

f (1) (0) f (2) (0) f (3) (0) f (4) (0) + + + 1! 2! 3! 4! 1/4 -1/32 3/256 -15/2048 = 2+ + + + 1! 2! 3! 4! 1 1 2 1 3 5 = 2+ xx + x x4 4 64 512 16,384

P4 ( x) = f (0) +

Your Turn 3 4.05 we evaluate f (0.05) with f ( x) =

To approximate

x + 4. Using the approximation from Your Turn 2,

4.05 = f (0.05) 1 1 1 5 (0.05) (0.05)2 + (0.05)3 (0.05)4 4 64 512 16,384 » 2.0124612. » 2+

12.3 Warmup Exercises W1.

f ( x) =

2 x + 5 = (2 x + 5)1/ 2

1 (2 x + 5)-1/ 2 (2) = (2 x + 5)-1/ 2 2 1 = 2x + 5 æ 1 ö÷ f ¢¢( x ) = çç - ÷÷ (2 x + 5)-3/ 2 (2) çè 2 ø

f ¢( x ) =

= -(2 x + 5)-3/ 2 æ 3ö f ¢¢¢( x ) = çç - ÷÷÷ éê -(2 x + 5)-5/ 2 ùú (2) çè 2 ø ë û = 3(2 x + 5)-5/ 2

W2.

f ( x) =

1 = ( x + 2)-1 x+2

f ¢( x ) = -( x + 2)-2 f ¢¢( x ) = (-2) éê -( x + 2)-3 ùú = 2( x + 2)-3 ë û 4 é ù f ¢¢¢( x ) = (-3) ê 2( x + 2) ú = -6( x + 2)-4 ë û

W3.

f ( x ) = e3x f ¢( x ) = 3e3x

( ) f ¢¢¢( x ) = 3( 9e3x ) = 27e3x f ¢¢( x ) = 3 33x = 9e3x

898

Chapter 12 SEQUENCES AND SERIES f ( x) = ln(1 + 2 x) 1 2 f ¢( x) = (2) = = 2(1 + 2 x)-1 1 + 2x 1 + 2x é ¢¢ f ( x ) = (-1) ê 2(1 + 2 x )-2 ùú (2) ë û

W4.

= -4(1 + 2 x )-2 f ¢¢¢( x ) = (-2) éê -4(1 + 2 x )-3 ùú (2) ë û = 16(1 + 2 x )-3

12.3 Exercises 1. False. Only functions that can be differentiated n times at 0 can be approximated at 0 by a Taylor polynomial of degree n. 2. False. A Taylor polynomial of degree 4 is the sum of the first 4 derivatives and the original function. 3. True n

f (n) (0) i x i! i =0

4. False. Pn ( x) = å Derivative

Value at 0

f ( x) = e-2 x f

(1)

f (0) = 1

-2 x

( x) = -2e

(1)

(0) = -2

f (2) ( x) = 4e-2 x

f (2) (0) = 4

f (3) ( x) = -8e-2 x

f (3) (0) = -8

f (4) ( x) = 16e-2 x

f (4) (0) = 16

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! -2 -8 3 16 4 4 =1+ x + x2 + x + x 1! 2! 3! 4! 4 2 = 1 - 2 x + 2 x 2 - x3 + x 4 3 3

P4 ( x) = f (0) +

Derivative

Value at 0

f ( x) = e

f (0) = 1

(1)

(0) = 3

(2)

(0) = 9

(3)

(0) = 27

(4)

(0) = 81

( x) = 3e

( x) = 9e

( x) = 27e ( x) = 81e

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 3 9 27 3 81 4 = 1 + x + x2 + x + x 1! 2! 3! 4! 9 9 27 4 = 1 + 3x + x 2 + x3 + x 2 2 8

P4 ( x) = f (0) +

Section 12.3

899

Derivative

Value at 0

f ( x) = e

x +1

f (0) = e

(1)

( x) = e

x +1

(1)

(0) = e

(2)

( x) = e

x +1

(2)

(0) = e

f (3) ( x) = e x +1

f (3) (0) = e

f (4) ( x) = e x +1

f (4) (0) = e f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! e e e e = e + x + x 2 + x3 + x 4 1! 2! 3! 4! e 2 e 3 e 4 = e + ex + x + x + x 2 6 24

P4 ( x) = f (0) +

Derivative

Value at 0

f ( x) = e-x f

(1)

-x

( x) = -e

f (0) = 1 f

(1)

(0) = -1

f (2) ( x) = e-x

f (2) (0) = 1

f (3) ( x) = -e-x

f (3) (0) = -1

f (4) ( x) = e-x

f (4) (0) = 1 f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! -1 1 -1 3 1 =1+ x + x2 + x + x4 1! 2! 3! 4! 1 2 1 3 1 4 = 1- x + x - x + x 2 6 24

P4 ( x) = f (0) +

Derivative f ( x) = f (1) ( x) =

Value at 0 1/2

x + 9 = ( x + 9)

1 1 ( x + 9)-1/2 = 2 2( x + 9)1/2

1 1 f (2) ( x) = - ( x + 9)-3/2 = 4 4( x + 9)3/2 f (3) ( x) =

3 3 ( x + 9)-5/2 = 8 8( x + 9)5/2

f (4) ( x) = -

15 15 ( x + 9)-7/2 = 16 16( x + 9)7/2

f (0) = 3 1 f (1) (0) = 6 f (2) (0) = f (3) (0) =

1 108

1 648

f (4) (0) = -

5 11, 664

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 5 1 1 - 11,664 - 1 x4 = 3 + 6 x + 108 x 2 + 648 x3 + 1! 2! 3! 4! 1 1 2 1 3 5 x + x x4 = 3+ x6 216 3888 279,936

P4 ( x) = f (0) +

900

Chapter 12 SEQUENCES AND SERIES

10.

Derivative f ( x) = f (1) ( x ) =

Value at 0 1/2

x + 16 = ( x + 16)

f (0) = 4

1 1 ( x + 16)-1/2 = 2 2( x + 16)1/2

f (1) (0) =

1 1 f (2) ( x ) = - ( x + 16)-3/2 = 4 4( x + 16)3/2 f (3) ( x ) =

3 3 ( x + 16)-5/2 = 8 8( x + 16)5/2

f (4) ( x ) = -

f (2) (0) = f (3) (0) =

15 15 ( x + 16)-7 /2 = 16 16( x + 16)7/2

1 8 1 256

3 8192

f (4) (0) = -

15 262,144

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 15 3 1 - 262,144 - 1 = 4 + 8 x + 256 x 2 + 8192 x3 + x4 1! 2! 3! 4! 1 1 2 1 5 = 4+ xx + x3 x4 8 512 16,384 2, 097,152

P4 ( x) = f (0) +

11.

Derivative f ( x) =

Value at 0 1/3

x - 1 = ( x - 1)

f (0) = -1 f (1) (0) =

1 3

2 2 f (2) ( x) = - ( x - 1)-5/3 = 9 9( x - 1)5/3

f (2) (0) =

2 9

10 10 ( x - 1)-8/3 = 27 27( x - 1)8/3

f (3) (0) =

10 27

f (4) (0) =

80 81

f (1) ( x) =

f (3) ( x) =

1 1 ( x - 1)-2/3 = 3 3( x - 1)3/2

f (4) ( x) = -

80 80 ( x - 1)-11/3 = 81 81( x - 1)11/3

P4 ( x) = f (0) +

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4!

= -1 + 3 x + 9 x 2 + 27 x3 + 81 x 4 1! 2! 3! 4! 1 1 2 5 3 10 4 = -1 + x + x + x + x 3 9 81 243

12.

Derivative

Value at 0 1/3

f ( x) = 3 x + 8 = ( x + 8) f (1) ( x) =

1 1 ( x + 8)-2/3 = 3 3( x + 8) 2/3

2 2 f (2) ( x) = - ( x + 8)-5/3 = 9 9( x + 8)5/3 f (3) ( x) =

10 10 ( x + 8)-8/3 = 27 27( x + 8)8/3

f (4) ( x) = -

80 80 ( x + 8)-11/3 = 81 81( x + 8)11/3

f (0) = 2 f (1) (0) =

1 12

f (2) (0) = f (3) (0) =

1 144

5 3456

f (4) (0) = -

5 10,368

Section 12.3

901 f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 5 5 1 1 - 10,368 - 144 2 3 3456 12 x+ x + x + x4 = 2+ 1! 2! 3! 4! 1 1 2 5 5 3 xx + x x4 = 2+ 12 288 20, 736 248,832

P4 ( x) = f (0) +

13.

Derivative

Value at 0

f ( x) = 4 x + 1 = ( x + 1)1/4 1 1 f (1) ( x) = ( x + 1)-3/4 = 4 4( x + 1)3/4 f (2) ( x) = f (3) ( x) =

3 3 ( x + 1)-7/4 = 16 16( x + 1)7/4

21 21 ( x + 1)-11/4 = 64 64( x + 1)11/4

f (4) ( x) = -

231 231 ( x + 1)-15/ 4 = 256 256( x + 1)15/4

f (0) = 1 1 f (1) (0) = 4 f (2) (0) = f (3) (0) =

3 16

21 64

f (4) (0) = -

231 256

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 21 1 -3 - 231 = 1 + 4 x + 16 x 2 + 64 x3 + 256 x 4 1! 2! 3! 4! 1 3 2 7 3 77 4 x + x x =1+ x4 32 128 2048

P4 ( x) = f (0) +

14.

Derivative

Value at 0 1/4

f ( x) = 4 x + 16 = ( x + 16) f (1) ( x) =

1 1 ( x + 16)-3/4 = 4 4( x + 16)3/4

f (2) ( x) = f (3) ( x) =

3 3 ( x + 16)-7/4 = 16 16( x + 16)7/4

21 21 ( x + 16)-11/4 = 64 64( x + 16)11/4

f (4) ( x) = -

231 231 ( x + 16)-15/4 = 256 256( x + 16)15/4

f (0) = 2 f (1) (0) =

1 32

f (2) (0) = f (3) (0) =

3 2048

21 131, 072

f (4) (0) = -

231 8,388, 608

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 231 21 3 1 - 8,388,608 - 2048 131,072 3 2 32 x+ x + x + x4 = 2+ 1! 2! 3! 4! 1 3 7 77 xx2 + x3 x4 = 2+ 32 4096 262,144 67,108,864

P4 ( x) = f (0) +

902 15.

Chapter 12 SEQUENCES AND SERIES Derivative

Value at 0

f ( x) = ln (1 - x)

f (0) = 0

1 1 = = ( x - 1)-1 f (1) ( x) = 1- x x -1 1 f (2) ( x) = -( x - 1)-2 = ( x - 1)2 2 f (3) ( x) = 2( x - 1)-3 = ( x - 1)3 6 f (4) ( x) = -6( x - 1)-4 = ( x - 1)4

f (1) (0) = -1 f (2) (0) = -1 f (3) (0) = -2 f (4) (0) = -6

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! -1 -1 2 -2 3 -6 4 =0+ x+ x + x + x 1! 2! 3! 4! 1 1 1 = -x - x 2 - x 3 - x 4 2 3 4

P4 ( x) = f (0) +

16.

Derivative f ( x) = ln (1 + 2 x)

Value at 0 f (0) = 0

2 1 + 2x 4 f (2) ( x) = (1 + 2 x)2 16 f (3) ( x) = (1 + 2 x)3 96 f (4) ( x) = (1 + 2 x) 4

f (1) (0) = 2

f (1) ( x) =

f (2) (0) = -4 f (3) (0) = 16 f (4) (0) = -96 f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 2 -4 2 16 3 -96 4 x + x + x =0+ x+ 1! 2! 3! 4! 8 = 2 x - 2 x 2 + x3 - 4 x 4 3

P4 ( x) = f (0) +

Derivative

17.

Value at 0

f ( x) = ln (1 + 2 x ) 4x f (1) ( x) = 1 + 2x2 f (2) ( x) = f (3) ( x) = f (4) ( x) =

f (0) = 0 f (1) (0) = 0

4 - 8x 2

f (2) (0) = 4

(1 + 2 x 2 ) 2 32 x3 - 48x

f (3) (0) = 0

(1 + 2 x 2 )3 -192 x 4 + 576 x 2 - 48 2 4

(1 + 2 x )

f (4) (0) = -48

Section 12.3

903 f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! -48 4 0 4 0 = 0 + x + x 2 + x3 + x 1! 2! 3! 4!

P4 ( x) = f (0) +

= 2x2 - 2x4

Derivative

18.

Value at 0

f ( x) = ln (1 - x ) f (1) ( x) = f (2) ( x) = f (3) ( x) = f (4) ( x) =

f (0) = 0

-3 x 2

3x 2

1- x

= 3

f (1) (0) = 0

x -1

-3 x - 6 x 3

( x - 1)

f (2) (0) = 0

6 x 6 + 42 x3 + 6

f (3) (0) = -6

( x3 - 1)3 -18 x8 - 288x5 - 180 x 2

f (4) (0) = 0

( x3 - 1)4

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 0 0 0 -6 3 = 0 + x + x2 + x + x4 1! 2! 3! 4!

P4 ( x) = f (0) +

= -x 3

19.

Derivative

Value at 0

-x

f (0) = 0

f ( x) = xe

f (1) ( x) = -xe-x + e-x

f (1) (0) = 1

f (2) ( x) = xe-x - 2e-x

f (2) (0) = -2

f (3) ( x) = -xe-x + 3e-x

f (3) (0) = 3

f (4) ( x) = xe-x - 4e-x

f (4) (0) = -4 f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 1 3 -2 2 -4 4 =0+ x+ x + x3 + x 1! 2! 3! 4! 1 1 = x - x 2 + x3 - x 4 2 6

P4 ( x) = f (0) +

20.

Derivative 2 x

f ( x) = x e

Value at 0 f (0) = 0

f (1) ( x) = x 2e x + 2 xe x

f (1) (0) = 0

f (2) ( x) = x 2e x + 4 xe x + 2e x

f (2) (0) = 2

f (3) ( x) = x 2e x + 6 xe x + 6e x

f (3) (0) = 6

f (4) ( x) = x 2e x + 8xe x + 12e x

f (4) (0) = 12

904

Chapter 12 SEQUENCES AND SERIES f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 0 2 6 12 4 = 0 + x + x 2 + x3 + x 1! 2! 3! 4! 1 = x 2 + x3 + x 4 2

P4 ( x) = f (0) +

21.

Derivative

Value at 0

3/2

f (0) = 27

f ( x) = (9 - x)

3 f (1) ( x) = - (9 - x)1/2 2 3 3 (2) f ( x) = (9 - x)-1/2 = 4 4(9 - x)1/2

f (1) (0) = f (2) (0) =

1 4

9 2

f (3) ( x) =

3 3 (9 - x)-3/2 = 8 8(9 - x)3/2

f (3) (0) =

1 72

f (4) ( x) =

9 9 (9 - x)-5/2 = 16 16(9 - x)5/2

f (4) (0) =

1 432

P4 ( x) = f (0) +

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4!

- 92

x + 4 x 2 + 72 x3 + 432 x 4 1! 2! 3! 4! 9 1 2 1 3 1 = 27 - x + x + x + x4 2 8 432 10,368 = 27 +

Derivative

22.

Value at 0

3/2

f (0) = 1

f (1) ( x) =

3 (1 + x)1/2 2 3 3 f (2) ( x) = (1 + x)-1/2 = 4 4(1 + x)1/2

3 2 3 f (2) (0) = 4

3 3 f (3) ( x) = - (1 + x)-3/2 = 8 8(1 + x)3/2

f (3) (0) = -

f ( x) = (1 + x)

f (4) ( x) =

9 9 (1 + x)-5/2 = 16 16(1 + x)5/2

f (1) (0) =

f (4) (0) =

3 8

9 16

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 9 3 3 -3 = 1 + 2 x + 4 x 2 + 8 x3 + 16 x 4 1! 2! 3! 4! 3 3 2 1 3 3 4 =1+ x + x x + x 2 8 16 128

P4 ( x) = f (0) +

Section 12.3 23.

905

Derivative 1 f ( x) = = (1 + x)-1 1+ x f (1) ( x) = -(1 + x)-2 = f (2) ( x) = 2(1 + x)-3 =

Value at 0 f (0) = 1 1

(1 + x) 2

(1 + x)3 6 f (3) ( x) = -6(1 + x)-4 = (1 + x)4 24 f (4) ( x) = 24(1 + x)-5 = (1 + x)5

f (1) (0) = -1 f (2) (0) = 2 f (3) (0) = -6 f (4) (0) = 24

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! -1 -6 3 2 24 4 =1+ x + x2 + x + x 1! 2! 3! 4!

P4 ( x) = f (0) +

= 1 - x + x 2 - x3 + x 4

24.

Derivative 1 f ( x) = = ( x - 1)-1 x -1 f (1) ( x) = -( x - 1)-2 = f (2) ( x) = 2( x - 1)-3 =

Value at 0 f (0) = -1 1

( x - 1)2 2

( x - 1)

f (3) ( x) = -6( x - 1)-4 = f (4) ( x) = 24( x - 1)-5 =

6 ( x - 1)4 24

( x - 1)

f (1) (0) = -1 f (2) (0) = -2 f (3) (0) = -6 f (4) (0) = -24

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! -1 -2 2 -6 3 -24 4 = -1 + x+ x + x + x 1! 2! 3! 4!

P4 ( x) = f (0) +

= -1 - x - x 2 - x 3 - x 4

For Exercises 25–38 each approximation can be determined using the Taylor polynomials from Exercises 5–19, respectively. 25.

Using the result of Exercise 5, with f ( x) = e-2 x and P4 ( x) = 1 - 2 x + 2 x 2 - 43 x3 + 23 x 4 , we can approximate e-0.04 by evaluating f (0.01) = e-2(0.02) = e-0.04. Using P4 ( x) from Exercise 5 with x = 0.02 gives 4 2 P4 (0.02) = 1 - 2(0.02) + 2(0.02)2 - (0.02)3 + (0.02) 4 3 3 » 0.096078944. To four decimal places, P4 (0.02) approximates the value of e-0.04 as 0.9608. Copyright © 2022 Pearson Education, Inc.

906 26.

Chapter 12 SEQUENCES AND SERIES Using the result of Exercise 6, with f ( x) = e3x and P4 ( x) = 1 + 3x + 92 x 2 + 92 x3 + 27 x 4 , we can 8 approximate e0.06 by evaluating f (0.02) = e3(0.02) = e0.06. Using P4 ( x) from Exercise 6 with x = 0.02 gives P4 (0.02) = 1 + 3 (0.02) +

9 9 27 (0.02) 2 + (0.02)3 + (0.02) 4 2 2 8

» 1.06183654

To four decimal places, P4 (0.02) approximates the value of e0.06 as 1.0618. 27.

e x 4 , we can Using the result of Exercise 7, with f ( x) = e x +1 and P4 ( x) = e + ex + 2e x 2 + 6e x3 + 24

approximate e1.02 by evaluating f (0.02) = e0.02 +1 = e1.02. Using P4 ( x) from Exercise 7 with x = 0.02 gives P4 (0.02) = e + e(0.02) +

e e e (0.02)2 + (0.02)3 + (0.02)4 2 6 24

» 2.773194764.

To four decimal places, P4 (0.02) approximates the value of e1.02 as 2.7732. 28.

1 x 4 , we can Using the result of Exercise 8, with f ( x) = e-x and P4 ( x) = 1 - x + 12 x 2 - 16 x3 + 24

approximate e-0.07 by evaluating f (0.07) = e-(0.07) = e-0.07 . Using P4 ( x) from Exercise 8 with x = 0.07 gives 1 1 1 (0.07)2 - (0.07)3 + (0.07) 4 2 6 24 » 0.9323938338.

P4 (0.07) = 1 - 0.07 +

To four decimal places, P4 (0.07) approximates the value of e-0.07 as 0.9324. 29.

Using the result of Exercise 9, with f ( x) =

5 1 x 2 + 1 x3 x + 9 and P4 ( x) = 3 + 16 x - 216 x 4 , we 3888 279,936

can approximate 8.92 by evaluating f (-0.08) = x = -0.08 gives

-0.08 + 9 =

8.92. Using P4 ( x) from Exercise 9 with

1 1 1 5 (-0.08) (-0.08) 2 + (-0.08)3 (-0.08) 4 6 216 3888 279,936 » 2.986636905.

P4 (-0.08) = 3 +

To four decimal places, P4 (-0.08) approximates the value of 30.

Using the result of Exercise 10, with f ( x) =

8.92 as 2.9866.

x + 16 and

5 1 x2 + 1 x3 P4 ( x) = 4 + 18 x - 512 x 4 , we can approximate 16,384 2,097,152

f (0.3) =

16.3 by evaluating

0.3 + 16 = 16.3. Using P4 ( x) from Exercise 10 with x = 0.3 gives

1 1 1 5 (0.3) (0.3)2 + (0.3)3 (0.3)4 8 512 16,384 2, 097,152 » 4.037325847.

P4 (0.3) = 4 +

To four decimal places, P4 (0.3) approximates the value of 16.3 as 4.0373.

Section 12.3 31.

907

5 x3 + 10 x 4 , we can Using the result of Exercise 11, with f ( x) = 3 x - 1 and P4 ( x) = -1 + 13 x + 19 x 2 + 81 243

approximate 3 -1.05 by evaluating f (-0.05) = 3 -0.05 - 1 = 3 -1.05. Using P4 ( x) from Exercise 11 with x = -0.05 gives 1 1 5 10 (-0.05) + (-0.05)2 + (-0.05)3 + (-0.05) 4 3 9 81 243 » -1.016396348.

P4 (-0.05) = -1 +

To four decimal places, P4 (-0.05) approximates the value of 32.

3 -1.05

as -1.0164.

Using the result of Exercise 12, with f ( x) = 3 x + 8 and 5 x3 5 1 x - 1 x2 + P4 ( x) = 2 + 12 x 4 , we can approximate 3 7.91 by evaluating 288 20,736 248,832

f (-0.09) = 3 -0.09 + 8 =

7.91. Using P4 ( x) from Exercise 12 with x = -0.09 gives

1 1 5 5 (-0.09) (-0.09) 2 + (-0.09)3 (-0.09) 4 12 288 20, 736 248,832 » 1.992471698.

P4 (-0.09) = 2 +

To four decimal places, P4 (-0.09) approximates the value of 3 7.91 as 1.9925. 33.

Using the result of Exercise 13, with f ( x) =

3 x 2 + 7 x3 - 77 x 4 , x + 1 and P4 ( x) = 1 + 14 x - 32 128 2048

we can approximate 4 1.06 by evaluating f (0.06) = 4 0.06 + 1 = 4 1.06. Using P4 ( x) from Exercise 13 with x = 0.06 gives 1 3 7 77 (0.06) (0.06)2 + (0.06)3 (0.06)4 4 32 128 2048 » 1.014673825.

P4 (0.06) = 1 +

To four decimal places, P4 (0.06) approximates the value of 4 1.06 as 1.0147. 34.

Using the result of Exercise 14, with f ( x) =

x + 16 and

7 77 1 x - 3 x2 + P4 ( x) = 2 + 32 x3 - 67,108,864 x 4 , we can approximate 4 15.88 4096 262,144

by calculating

f (-0.12) = 4 -0.12 + 16 = 4 15.88. Using P4 ( x) from Exercise 14 with x = -0.12 gives 1 3 7 77 (-0.12) (-0.12)2 + (-0.12)3 (-0.12) 4 32 4096 262,144 67,108,864 » 1.996239407.

P4 (-0.12) = 2 +

To four decimal places, P4 (-0.12) approximates the value of 4 15.88 as 1.9962. 35.

Using the result of Exercise 15, with f ( x) = ln(1 - x) and P4 ( x) = -x - 12 x 2 - 13 x3 - 14 x 4 , we can approximate ln 0.97 by evaluating f (0.03) = ln (1 - 0.03) = ln 0.97. Using P4 ( x) from Exercise 15 with x = 0.03 gives 1 1 1 (0.03)2 - (0.03)3 - (0.03)4 2 3 4 » -0.0304592025.

P4 (0.03) = -(0.03) -

To four decimal places, P4 (0.03) approximates the value of ln 0.97 as -0.0305.

908 36.

Chapter 12 SEQUENCES AND SERIES Using the result of Exercise 16, with f ( x) = ln (1 + 2 x) and P4 ( x) = 2 x - 2 x 2 + 83 x3 - 4 x 4 , we can approximate ln 1.06 by evaluating f (0.03) = ln(1 + 2(0.03)) = ln(1 + 0.06) = ln1.06. Using P4 ( x) from Exercise 16 with x = 0.03 gives P4 (0.03) = 2(0.03) - 2(0.03) 2 +

8 (0.03)3 - 4(0.03)4 3

» 0.05826876.

To four decimal places, P4 (0.03) approximates the value of ln1.06 as 0.0583. 37.

Using the result of Exercise 17, with f ( x) = ln(1 + 2 x 2 ) and P4 ( x) = 2 x 2 - 2 x 4 , we can compute ln 1.008 by evaluating f

2ù

( ) = ln êêëê1 + 2( ) úúûú = ln[1 + 2(0.004)] = ln1.008. Using P4 (x) with x = 5010 gives 10 50

10 50

æ 10 ö÷ æ ö2 æ ö4 ÷÷ = 2 çç 10 ÷÷÷ - 4 çç 10 ÷÷÷ P4 ççç ç ç èç 50 ÷ø èç 50 ø÷ èç 50 ø÷ = 0.007968

To four decimal places, P4 38.

( ) approximates ln1.008 as 0.0080. 10 50

Using the result of Exercise 18, with f ( x) = ln(1 - x3 ) and P4 ( x) = -x3 , we can approximate ln 0.992 by evaluating f (0.2) = ln (1 - 0.23 ) = ln 0.992. Using P4 ( x) from Exercise 18 with x = 0.2 gives P4 (.1) = -(0.2)3 » -0.008.

To four decimal places, P4 (0.2) approximates the value of ln 0.992 as -0.0080.

39.

P3 ( x) = f (0) +

40.

P4 ( x) = f (0) +

f (1) (0) f (2) (0) 2 f (3) (0) 3 6 12 2 24 3 x+ x + x = 3+ x+ x + x = 3 + 6 x + 6 x 2 + 4 x3 1! 2! 3! 1 2 6

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 1 2 6 24 4 = 1 + x + x 2 + x3 + x 1 2 6 24 = 1 + x + x 2 + x3 + x 4

For a polynomial of degree n with f (n) (0) = n ! and 0! = 1, f (1) (0) f (2) (0) 2 f (n-1) (0) n-1 f (n) (0) n + x+ x ++ x x 1! 2! (n - 1)! n! 1! 2! (n - 1)! n-1 n! = 1 + x + x2 +  + + xn x 1! 2! (n - 1)! n!

Pn ( x) = f (0) +

= 1 + x + x 2 +  + x n-1 + x n

Section 12.3 41.

909

(a) Derivative

f ( x) = ( a + x ) f

(1)

(2)

Value at 0

1 n

f (0) = a n 1

1 -1 1 ( x) = ( a + x ) n n

1 -3 öæ 1 ö 1 æç 1 çç - 1÷÷÷ççç - 2 ÷÷÷ (a + x) n øè n ø nè n

f (4) ( x) =

1 -4 öæ 1 öæ 1 ö 1 æç 1 çç - 1÷÷÷ççç - 2 ÷÷÷ççç - 3 ÷÷÷ (a + x) n øè n øè n ø nè n

1/n

1 æç 1 - n öæ 1 - 2n ö÷ 1n -3 (1 - n)(1 - 2n)a n = ÷÷÷ççç ÷÷ a çç n è n øè n ø n 3a 3 1 æ 1 - n ÷öæç 1 - 2n ÷öæç 1 - 3n ö÷ 1n -4 f (4) (0) = çç ÷ç ÷ç ÷a n çè n ÷øèç n ÷øèç n ø÷ f (3) (0) =

(1- n)a a1/n 2 2 + na x + n a

1 æ 1 - n ÷ö 1n -2 (1 - n)a n (0) = çç = ÷÷ a ç nè n ø n 2a 2 1

f (3) ( x) =

P4 ( x) = a

(2)

1 1 -1 an (0) = a n = n na 1

1 -2 ö 1æ1 ( x) = çç - 1÷÷ (a + x) n ÷ ç ø nè n

1/n

(1)

(1 - n)(1 - 2n)(1 - 3n) a n n 4a 4

(1- n)(1- 2n)a1/n (1- n)(1- 2n)(1-3n)a1/n 3 3 3 n a n4a 4 + x + x4

2 1/n

3 1/n

4 1/n

xa x a (1 - n) x a (1 - n)(1 - 2n) x a (1 - n)(1 - 2n)(1 - 3n) + + + 2 2 3 3 na 2!n a 3!n a 4!n 4a 4 xa1/n éê x(1 - n) x 2 (1 - n)(1 - 2n) x3 (1 - n)(1 - 2n)(1 - 3n) ùú = a1/n + 1 + + + ú na êê 2!na 3!n 2a 2 4!n3a3 ë ûú = a1/n +

For values of x that are small compared with a, the terms after 1 in the brackets get closer and closer to zero. For example, consider the term

x(1- n) 2!na

1 a. Then where x = 10

x(1- n) - n . For even the smallest value of = 120 2!na n

1 . This term, and the terms which follow will have little impact on our n, n = 2, the value of this term is - 40 1/n

approximation. Thus the value of the expression is approximately 1, so P4 ( x) » a1/n + xana . Thus, if x is 1/n

small compared with a, the Taylor polynomial P4 ( x) approximates (a + x)1/n as a1/n + xana . (b) Using the result of part (a), we can approximate 3 66 by evaluating f (2) given f ( x) = (64 + x)1/3 since f (2) = (64 + 2)1/3 = 661/3 = 3 66. Using P4 ( x) from part (a) with x = 2, a = 64, and n = 3 gives (64 + 2)1/3 = 641/3 +

2 ⋅ 641/3 3 ⋅ 64

8 192 = 4.0416 = 4+

To five decimal places, P4 ( x) » (64 + x)1/ 3 approximates the value of 3 66 as 4.04167. To five decimal places, a calculator gives an approximation of 4.04124.

910 42.

Chapter 12 SEQUENCES AND SERIES Derivative

(a)

f (r ) = (1 + r )

Value at 0 D

f (0) = 1

f (1) (r ) = D (1 + r ) D -1

(1)

(0) = D

D r = 1 + rD 1!

P1(r ) = 1 +

So the Taylor polynomial of degree 1 for the function f (r ) = (1 + r ) D is P1(r ) = 1 + rD. If r is small, then (1 + r ) D » 1 + rD, and so V (1 + r ) D » V (1 + rD). Therefore, for small r, the two formulas for S give approximately equal values. (b) Using S » V (1 + rD) with V = 1000, r = 0.1, and D = 3.2 gives S » 1000[1 + (0.1)(3.2)] = 1320.

Using S » V (1 + r ) D with V = 1000, r = 0.1, and D = 3.2 gives S » 1000(1 + 0.1)3.2 » 1357. The two approximations for S are 1320 and 1357 which are relatively close. 43.

(a)

Derivative f (N ) = e f

(1)

(2)

Value at 0

N

(N ) = e

f (0) = 1

N

2 N

(N ) =  e

P2 ( N ) = f (0) + =1+

 1!

(1)

(0) = 

(2)

(0) =  2

f (1) (0) f (2) (0) 2 N + N 1! 2!

N +

2 2!

N2 = 1 + N +

2 2

The Taylor polynomial of degree 2 at N = 0 for e N is P2 ( N ) = 1 +  N + 2 N 2. 2

(b) Substituting P2 ( N ) = 1 +  N + 2 N 2 for e N in the original equation gives: 2

1 +  N + 2 N 2



-N =



æ  2 2 ö÷÷ çç 1 + +  N N ÷÷ -  N = 1 +  k çç 2 è ø÷

2 2

N 2 = k N2 =



N =



Since N is a positive quantity, use the positive square root. So, N =

2k .



Section 12.3

911

44. Derivative P ( x) = P (1) ( x) = P (2) ( x) =

Value at 0

20 + x 50 + x

P(0) = 0.4

x 2 + 100 x - 20 (50 + x)

P (1) (0) = -0.008

(50 + x) 2 (2 x + 100) - ( x 2 + 100 x - 20)[2(50 + x)]

P2 ( x) = P(0) +

(50 + x)

P (2) (0) = 0.04032

P (1) (0) P (2) (0) 2 -0.008 0.04032 2 x+ x = 0.4 + x+ x = 0.4 - 0.008 x + 0.02016 x 2 1! 2! 1 2

P2 (0.3) = 0.4 - 0.008(0.3) + (0.02016) (0.3)2 = 0.3994144

The Taylor polynomial P2 (0.3) approximates P(0.3) as 0.3994144. The approximate profit when 300 tons of apples are sold is 0.399 thousand dollars, or $399. Finding P(0.3) by direct substitution gives: P(0.3) =

20 + (0.3) 2 » 0.3994035785 50 + (0.3)

Direct substitution also estimates the profit as 0.399 thousand dollars, or $399. 45.

Derivative P( x) = ln (100 + 3x) 3 P (1) ( x) = 100 + 3x 9 P (2) ( x) = (100 + 3x)2

Value at 0 P(0) = 4.605 P (1) (0) = 0.03 P (2) (0) = -0.0009

P (1) (0) P (2) (0) 2 x+ x 1! 2! 0.03 -0.0009 2 x+ x = 4.605 + 1 2

P2 ( x) = P(0) +

= 4.605 + 0.03x - 0.00045x 2 P2 (0.6) = 4.605 + 0.03(0.6) - 0.00045(0.6) 2 = 4.622838

The Taylor polynomial P2 (0.6) approximates P(0.6) as 4.622838. The approximate profit is 4.623 thousand dollars, or $4623. Finding P(0.6) by direct substitution gives: P(0.6) = ln100 + 3(0.6) = ln101.8 = 4.623010104

Direct substitution also estimates the profit as 4.623 thousand dollars, or $4623.

912 46.

Chapter 12 SEQUENCES AND SERIES Derivative

Value at 0

-x /50

C (0) = 1

C ( x) = e

1 C (1) ( x) = - e-x /50 50 1 -x /50 (2) C ( x) = e 2500

1 50 1 (2) C (0) = 2500 C (1) (0) = -

C (1) (0) C (2) (0) 2 1 1 x+ x = 1 + 50 x + 2500 x 2 = 1 x+ x2 1! 2! 1! 2! 50 5000 1 1 P2 (5) = 1 (5) + (5)2 = 0.905 50 5000

P2 ( x) = C (0) +

The Taylor polynomial P2 (5) approximates C (5) as 0.905. The approximate cost is 0.905 dollar, or about $0.91. Finding C (5) by direct substitution gives: C (5) = e-5 / 50 » 0.904837418

Direct substitution estimates the profit as 0.905 dollar, or about $0.90. 47.

Derivative æ x ö÷ R( x) = 500 ln çç 4 + ÷ çè 50 ÷ø 500 200 + x 500 R (2) ( x) = (200 + x)2 R (1) ( x) =

P2 ( x) = R(0) +

Value at 0 R(0) = 500(1.3863) = 693.15 R (1) (0) = 2.5 R (2) (0) = -0.0125

R (1) (0) R (2) (0) 2 2.5 -0.0125 2 x+ x = 693.15 + x+ x = 693.15 + 2.5 x - 0.00625 x 2 1! 2! 1 2

P2 (10) = 693.15 + 2.5(10) - 0.00625(10)2 = 717.525

The Taylor polynomial P2 (10) approximates R(10) as 717.525. Thus, the approximate revenue is $718. Finding R(10) by direct substitution gives: æ 10 ö÷ R(10) = 500 ln çç 4 + ÷ » 717.5422626 çè 50 ÷ø

Direct substitution estimates the revenue as $718. 48.

Derivative 6x A( x) = 1 + 10 x 6 A(1) ( x) = (1 + 10 x) 2 120 A(2) ( x) = (1 + 10 x)3 P2 ( x) = A(0) +

Value at 0 A(0) = 0 A(1) (0) = 6 A(2) (0) = -120

-120 2 A(1) (0) A(2) (0) 2 6 x+ x =0+ x+ x = 6 x - 60 x 2 1! 2! 1 2

P2 (0.05) = 6(0.05) - 60(0.05)2 = 0.15

Section 12.3

913

The Taylor polynomial P2 (0.05) approximates A(0.05) as 0.15. The approximate amount of the drug in the bloodstream after 0.05 minutes is 0.15 milliliters. Finding A(0.05) by direct substitution gives: A(0.05) =

6(0.05) = 0.2 1 + 10(0.05)

Direct substitution indicates that 0.2 milliliter of the drug remain in the bloodstream after 0.05 minutes. 49.

Derivative

Value at 0

P(k ) = 1 - e-2k P (1) (k ) = 2e-2k P1(k ) = P(0) +

P(0) = 1 - e0 = 0 P (1) (0) = 2

P (1) (0) 2 k = 0 + k = 2k 1! 1!

So, if k is small, that is, close to 0, then P(k ) » P1(k ) = 2k.

50.

(a) The Taylor polynomial of degree 1 for 1 + x was found in Example 2. 1+ x » 1+

1 x 2

Therefore, æ ö÷ æ ö÷ çç 2 æç ç R 2 ö÷÷ R2 æ 2 ö÷ ÷ 2 V = k1 çç z + R - z ÷÷ = k1 çç z çç1 + 2 ÷÷ - z ÷÷ = k1 ççç z 1 + 2 - z ÷÷÷. ÷ ÷ è ø ç z ø÷ z èç èç ø÷ èç ø÷÷

Now use the Taylor polynomial to approximate the radical expression, letting x be replaced by R 2 /z 2. æ é ö÷ ù 2æ ö æ ö÷ æ R 2 ö÷ ç R2 z 1 æç R 2 ö÷ ç ç ÷÷ = k1R çç 1 ÷÷ V » k1 ççç z êê 1 + çç 2 ÷÷÷ úú - z ÷÷÷ = k1 çç z + 2 - z ÷÷÷ = k1 çç ÷ ÷ 2 çè z ÷ø ú 2 èç z ø÷ ÷ø çè çè êë 2z èç 2 z ÷ø ø÷ û

If z is much larger than R, the expression is dominated more by the z-factor and less so by the R 2. Let k1R 2 /2 = k2. Then V »

k2 . z

æ ö÷ æ ö÷ çç 2 çæ ç æ 2 ö÷ z 2 ÷ö÷ z2 ÷ 2 ÷ (b) V = k1 çç z + R - z ÷÷ = k1 çç R çç 1 + 2 ÷÷ - z ÷÷ = k1 çç R 1 + 2 - z ÷÷÷ ç ÷ è ø çç ÷ R ø÷ R èç çè ø÷ è ø÷

As before, use the Taylor polynomial of degree 1 for 1 + x to approximate the radical expression. This time let x be replaced by z 2/R 2. æ é ö÷ ù 2 ö æ ö÷ æ ç Rz 2 1 æç z 2 ÷ö ç ÷÷ = k1 çç R - z + z ÷÷÷ V » k1 ççç R êê 1 + çç 2 ÷÷÷ úú - z ÷÷÷ = k1 çç R + z ç ÷÷ 2 çè R ÷ø ú 2R ø÷÷ ÷÷ø çè çè êë 2R2 èç ø û

This is the desired result.

914

Chapter 12 SEQUENCES AND SERIES Turtle’s distance from the starting line at any number of hours t after 2:00 P.M. is d1 = 5400 + (t )(60)(15) = 5400 + 900t feet.

12.4 Infinite Series Your Turn 1

The sequence is 1,

1 1 1 1 , , , , ... . 4 9 16 25

Rabbit’s distance from the starting line at time t is d 2 = (t )(60)(45) = 2700t feet. d1 = d 2

S1 = 1 1 5 = 4 4 1 1 49 S3 = 1 + + = 4 9 36 1 1 1 205 S4 = 1 + + + = 4 9 16 144 1 1 1 1 5269 S5 = 1 + + + + = 4 9 16 25 3600

5400 + 900t = 2700t

S2 = 1 +

5400 = 1800t t =3

Rabbit catches up with Turtle 3 hours after 2 P.M., that is, at 5 P.M.. (b)

Your Turn 2 (a)

This is a geometric series with a = a1 = 2 and r = 1/3. Since r is in (-1, 1), the series converges and has sum a 2 = 1- r 1 - (1/3) 2 = 2/3 = 3.

a 120 = 1- r 1 - (1/3)

2 + 6 + 18 + 54 +  is geometric series with a = a1 = 2 and r = 3.

(b)

(c)

120 2/3

= 180,

This is a geometric series with a = a1 = 4 and r = -1/4. Since r is in (-1, 1), the series converges and has sum a 5 = 1- r 1 - (-1/4) 5 = 5/4 = 4.

that is, 180 minutes or 3 hours after 2 P.M., or 5 P.M., as found in (a).

12.4 Exercises

This is a geometric series with a = a1 = 2 and r = 1.01. Since r is not in (-1, 1), the series diverges.

Your Turn 3 (a)

Solve using a geometric series. At 2 P.M. Turtle has traveled 5400 ft. Traveling at 45 feet per minute, it will take Rabbit 5400/45 = 120 minutes to make up this distance. During this time, Turtle will have traveled a further (120)(15) = 1800 feet, and it will take Rabbit 1800/45 = 40 minutes to make up this distance. The total number of minutes it takes Rabbit to make up all these distances is 120 + 40 + 40/3 + . This is a geometric series with a = a1 = 120 and r = 1/3. Since r is in (-1, 1), the series converges and has sum

Solve using algebra. Let t be the number of hours since 2:00 P.M. Between 8:00 A.M. and 2:00 P.M., Turtle has traveled (6)(60)(15) = 5400 feet.

False. It is possible for the sum of an infinite number of positive values to equal a finite number.

True

False. Some infinite geometric series diverge.

20 + 10 + 5 + 52 +  is a geometric series

with a = a1 = 20 and r = 12 . Since r is in (-1, 1), the series converges and has sum 20 20 a = = 1 = 40. 1 1- r 1- 2 2

Section 12.4 6.

915

1 + 0.8 + 0.64 + 0.512 +  is a geometric

13.

series with a = a1 = 1 and r = 0.8. Since r

the series converges and has sum

is in (–1, 1), the series converges and has sum a 1 1 = = = 5. 1- r 1 - 0.8 0.2

2 + 6 + 18 + 54 +  is geometric series with a = a1 = 2 and r = 3. Since r > 1, the series diverges.

3 + 6 + 12 + 24 +  is a geometric series with a = a1 = 3 and r = 2. Since r > 1, the series diverges.

27 + 9 + 3 + 1 +  is a geometric series with a = a1 = 27 and r = 13 . Since r is in (-1, 1),

14.

15.

a 3 = 1- r 1- -2

( 3)

16.

= 53 = 2

1+ 3

1 . 5

1 +  is a geometric series (1.01)2

1 1 a = = 0.01 = 101. 1 1- r 1 - 1.01 1.01

17.

e - 1 + 1e - 12 +  is a geometric series with e

a = a1 = e and r = - 1e . Since r is in (-1, 1),

the series converges and the sum a e e = = 1 1- r 1 + 1e 1 - (- )

44 + 22 + 11 +  is a geometric series with a = a1 = 44 and r = 12 . Since r is in (-1, 1),

e = e +1 =

the series converges and has sum a 44 44 = = 1 = 88. 1- r 1 - 12 2

1 3

(-1, 1), the series converges and has sum

a 100 100 1000 = = 9 = . 1 1- r 9 1 - 10 10

12.

1 + 1 + 1.01

1 . Since r is in with a = a1 = 1 and r = 1.01

is in (-1, 1), the series converges and has sum

1 - 2 + 4 - 8 +  is a geometric series 3 9 27 81 with a = a1 = 13 and r = - 23 . Since r is in

(-1, 1), the series converges and has sum

a 64 64 256 = = 3 = . 1 1- r 3 1- 4 100 + 10 + 1 +  is a geometric series 1 . Since r with a = a1 = 100 and r = 10

8 a = 5 1 = 15 = . 1- r 5 1- 2 2

in (-1, 1), the series converges and has sum

11.

4 + 2 + 1 +  is a geometric series with 5 5 5 a = a1 = 54 and r = 12 . Since r is in (-1, 1),

the series converges and has sum

a 27 27 81 = = 2 = . 1 1- r 2 1- 3 3 64 + 16 + 4 + 1 +  is a geometric series with a = a1 = 64 and r = 14 . Since r is

a 5 = 4 1 = 14 = . 1- r 2 1- 2 2

the series converges and has sum

10.

5 + 5 + 5 +  is a geometric series with 4 8 16 a = a1 = 54 and r = 12 . Since r is in (-1, 1),

18.

e2 . e +1

e + e2 + e3 + e4 +  is a geometric series with a = a1 = e and r = e. Since r > 1, the series diverges.

916 19.

Chapter 12 SEQUENCES AND SERIES S1 = a1 =

1 =1 1

1 3 = 2 2 1 1 11 S3 = a1 + a2 + a3 = 1 + + = 2 3 6 1 1 1 25 S4 = a1 + a2 + a3 + a4 = 1 + + + = 2 3 4 12 1 1 1 1 137 S5 = a1 + a2 + a3 + a4 + a5 = 1 + + + + = 2 3 4 5 60 S2 = a1 + a2 = 1 +

20.

21.

1 1 = 1+1 2 1 1 1 1 5 S2 = a1 + a2 = + = + = 1+1 2 +1 2 3 6 1 1 1 1 1 1 13 S3 = a1 + a2 + a3 = + + = + + = 1+1 2 +1 3+1 2 3 4 12 1 1 1 1 1 1 1 1 77 S4 = a1 + a2 + a3 + a4 = + + + = + + + = 1+1 2 +1 3+1 4 +1 2 3 4 5 60 1 1 1 1 1 1 1 1 1 1 29 S5 = a1 + a2 + a3 + a4 + a5 = + + + + = + + + + = 1+1 2 +1 3+1 4 +1 5 +1 2 3 4 5 6 20 S1 = a1 =

S1 = a1 =

1 1 = 2 (1) + 5 7

S2 = a1 + a2 =

1 1 1 1 16 + = + = 2 (1) + 5 2 (2) + 5 7 9 63

S3 = a1 + a2 + a3 =

1 1 1 1 1 1 239 + + = + + = 2 (1) + 5 2 (2) + 5 2 (3) + 5 7 9 11 693

S4 = a1 + a2 + a3 + a4 =

1 1 1 1 1 1 1 1 3800 + + + = + + + = 2 (1) + 5 2 (2) + 5 2 (3) + 5 2 (4) + 5 7 9 11 13 9009

S5 = a1 + a 2 + a3 + a4 + a5 =

1 1 1 1 1 + + + + 2 (3) + 5 2 (4) + 5 2 (5) + 5 2 (1) + 5 2 (2) + 5

1 1 1 1 1 22, 003 + + + + = 7 9 11 13 15 45, 045

Section 12.4 22.

917

S1 = a1 =

1 1 = 3(1) - 1 2

S2 = a1 + a2 =

1 1 1 1 7 + + + = 3(1) - 1 3(2) - 1 2 5 10

1 1 1 1 1 1 33 + + + + + = 3(1) - 1 3(2) - 1 3(3) - 1 2 5 8 40 1 1 1 1 1 1 1 1 403 + + + = + + + = S4 = a1 + a2 + a3 + a4 = 3(1) - 1 3(2) - 1 3(3) - 1 3(4) - 1 2 5 8 11 440 S3 = a1 + a2 + a3 =

S5 = a1 + a2 + a3 + a4 + a5 = =

23.

1 1 1 1 1 + + + + 3(1) - 1 3(2) - 1 3(3) - 1 3(4) - 1 3(5) - 1

1 1 1 1 1 3041 + + + + = 2 5 8 11 14 3080

S1 = a1 =

1 1 = (1 + 1) (1 + 2) 6

S2 = a1 + a2 =

1 1 1 1 1 + = + = (1 + 1) (1 + 2) (2 + 1) (2 + 2) 6 12 4

S3 = a1 + a2 + a3 =

1 1 1 1 1 1 3 + + = + + = (1 + 1) (1 + 2) (2 + 1) (2 + 2) (3 + 1) (3 + 2) 6 12 20 10

S4 = a1 + a2 + a3 + a4 =

1 1 1 1 + + + (1 + 1) (1 + 2) (2 + 1) (2 + 2) (3 + 1) (3 + 2) (4 + 1) (4 + 2)

1 1 1 1 1 = + + + 3 6 12 20 30 S5 = a1 + a2 + a3 + a4 + a5 =

24.

1 1 1 1 1 + + + + (1 + 1) (1 + 2) (2 + 1) (2 + 2) (3 + 1) (3 + 2) (4 + 1) (4 + 2) (5 + 1) (5 + 2)

1 1 1 1 1 5 + + + + = 6 12 20 30 42 14

S1 = a1 =

1 1 = (1 + 3)[2 (1) + 1] 12

1 1 1 1 37 + = + = (1 + 3)[2 (1) + 1] (2 + 3)[2 (2) + 1] 12 25 300 1 1 1 S3 = a1 + a2 + a3 = + + (1 + 3)[2 (1) + 1] (2 + 3)[2 (2) + 1] (3 + 3)[2 (3) + 1] 1 1 1 103 = + + = 12 25 42 700

S2 = a1 + a2 =

1 1 1 1 + + + (1 + 3)[2(1) + 1] (2 + 3)[2(2) + 1] (3 + 3)[2(3) + 1] (4 + 3)[2(4) + 1] 1 1 1 1 1027 = + + + = 12 25 42 63 6300 S5 = a1 + a2 + a3 + a4 + a5

S4 = a1 + a2 + a3 + a4 =

1 1 1 1 1 + + + + (1 + 3)[2(1) + 1] (2 + 3)[2(2) + 1] (3 + 3)[2(3) + 1] (4 + 3)[2(4) + 1] (5 + 3)[2(5) + 1] 1 1 1 24,169 1 1 + + = = + + 42 63 88 138,600 12 25 =

918 25.

Chapter 12 SEQUENCES AND SERIES The infinite geometric series æ ö æ ö2 æ ö3 0.2 + 0.2 çç 1 ÷÷÷ + 0.2 çç 1 ÷÷÷ + 0.2 çç 1 ÷÷÷ +  è 10 ø è 10 ø è 10 ø

is a geometric series with a = a1 = 0.2 and r = 1/10. Since r is in (–1, 1), the series converges and has sum a = 0.2 1- r 1 - (1/10) = 0.2 = 2/9. 0.9 26.

The infinite geometric series æ ö æ ö2 æ ö3 0.18 + 0.18 çç 1 ÷÷÷ + 0.18 çç 1 ÷÷÷ + 0.18 çç 1 ÷÷÷ +  è 100 ø è 100 ø è 100 ø

is a geometric series with a = a1 = 0.18 and r = 1/100. Since r is in (-1, 1), the series converges and has sum a = 0.18 1- r 1 - (1 / 100) = 0.18 = 18 / 99 = 2 / 11. 0.99

27.

(a) It follows from Viète’s formula that 2



» =

2 ⋅ 2 2

2+ 2

⋅

2+ 2

( 2 + 2 )( 2 + 2 + 2 )

» 0.641

2 » 3.12. Thus,  » 0.641

It follows from Liebniz’s formula that  » 1 - 1 + 1 - 1 » 0.724. 4 3 5 7 Thus,  » 4 (0.724) » 2.90. Therefore, approximating  by multiplying the first three terms of Viète’s formula together is more accurate than approximating  by adding the first four terms of Liebniz’s formula together. æ æ (-1) N -1 öö (b) Using the graphing calculator with Y1 = 4 * sum çç seq ççç 2 N -1 , N , 1, X ÷÷÷ ÷÷÷ and the table function gives the è øø èç

following values for Y1 for 35 £ X £ 41.

Thus, 38 terms of the second formula must be added together to produce the same accuracy as the product of the first three terms of the first formula. 28.

(a) With a = a1 = 1000 and r = 0.1, the production manager should order a = 1000 = 1000 = 10, 000 » 1111 units. 1- r 1 - 0.1 0.9 9 (b)

0.9 x = 1000 x = 1000 = 0.9

10, 000 » 1111 units 9

Section 12.4 29.

919

(a) The total expenditure is the sum of an infinite geometric series with a = a1 = 200 and r = 0.9. The sum of this convergent series is 200 a = 1- r 1 - 0.9 200 = 0.1 = 2000. The total expenditure, including the government’s original $200 payout, will be $2000. (b) Since the original payout was $200, the multiplier is $2000/$200 = 10.

30.

(a) The total withholdings is the sum of an infinite geometric series with a = a1 = 1000 and r = 0.25. The sum of this convergent series is 1000 a = 1- r 1 - 0.25 1000 = 0.75 = 1333.33. The total withholdings will be $1333.33. (b) Since the original amount was $4000, the effective tax rate is $1333.33 / $4000 = 0.33 = 33%.

31.

(a) Consider the finite sequence C (1 + r )-1, C (1 + r )-2 , C (1 + 5)-3, , C (1 + r )-n .

This is a finite geometric sequence having first term a = a1 = C (1 + r )-1 and common ratio w = (1 + r )-1. Thus, P = Sn = =

a(wn - 1) w -1

C (1 + r )-1[(1 + r )-n - 1]

(1 + r )-1 - 1 (1 + r )-1[(1 + r )-n - 1] (1 + r )-1[1 - (1 + r )]

(1 + r )-n - 1 -r

(1 + r )-n - 1 (1 + r ) n ⋅ -r (1 + r ) n

(1 + r )n - 1 r (1 + r )n

(b) The present value over an infinite amount of time is given by the infinite geometric series P = C (1 + r )-1 + C (1 + r )-2 + C (1 + r )-3 + 

where a = a1 = C (1 + r )-1 and the common ratio w = (1 + r )-1. This series converges to a C (1 + r )-1 C (1 + r )-1 C = = = 1 -1 1- w r 1 - (1 + r ) (1 + r ) [(1 + r ) - 1]

Here, P = Cr .

920 32.

Chapter 12 SEQUENCES AND SERIES Let Sn be the total medical malpractice payments the company pays after n years. If each year’s payments are 20% less than those of the previous year, then they are 80% of the previous year’s payments. Thus, S0 = 60 S1 = 60 + 60(0.8) S2 = 60 + 60(0.8) + 60(0.8)2  Sn = 60 + 60(0.8) +  + 60(0.8n .

The total medical malpractice payments that the company pays in all years is, therefore, S = 60 + 60(0.8) +  + 60(0.8)n + 

This is a geometric series with a = 60 and r = 0.8. S =

a 60 60 = = = 300 1- r 1 - 0.8 0.2

The correct answer choice is d. 33.

Let p0 = p, the probability that the policyholder files no claims during the period. Then, because pn +1 =

( 15 ) pn , we have that

p0 = p p1 =

1 1 p0 = p 5 5

p2 =

æ 1 ö2 1 1 1 p1 = ⋅ p = çç ÷÷÷ p çè 5 ø 5 5 5

 Pn =

n -1 æ 1 ön 1 1 æ1ö pn-1 = ⋅ çç ÷÷ p = çç ÷÷ p. èç 5 ÷ø 5 5 èç 5 ÷ø

The total probability is the sum of the probabilities of the policyholder filing no claim, 1 claim, 2 claims, …, n claims, …, and so on. æ1ö æ 1 ö2 æ 1 ön S = p + p1 + p2 +  + pn +  = p + p çç ÷÷÷ + p çç ÷÷÷ +  + p çç ÷÷÷ +  èç 5 ø èç 5 ø èç 5 ø

This is a geometric series with a = p and r = 15 . S =

a p p 5 = = 4 = p 1- r 4 1 - 15 5

And, since the sum of all probabilities must equal 1, 5 p =1 4 4 p = . 5

The probability that a policyholder files more than one claim is æ 1 ö 6 6 æ4ö 24 pn>1 = 1 - ( p0 + p1) = 1 - çç p + p ÷÷ = 1 - p = 1 - çç ÷÷ = 1 = 0.04. ÷ çè 5 ø 5 5 çè 5 ÷ø 25

Section 12.4 34.

921

( )

( ) , after

The height that the ball returns to after the first bounce is given by 10 34 , after the second bounce 10 34 3

( ) , and so on. Thus, the distance that the ball travels before it comes to rest is given by

the third bounce 10 34

æ3ö æ 3 ö2 æ 3 ö3 10 + 2(10) çç ÷÷ + 2(10) çç ÷÷ + 2(10) çç ÷÷ + . çè 4 ÷ø èç 4 ø÷ èç 4 ÷ø

( )

( ) + 2(10)( 34 ) +  is an infinite geometric series with a1 = a = 2(10)( 34 ) = 15 and

2(10) 34 + 2(10) 34

r = 34 . This series converges to a 15 15 = = 1 = 60. 1- r 1- 3 4 4

So, the total distance traveled by the ball is 10 + 60 = 70 meters. 35.

( )

( ) , in

The number of times the wheel rotates in the second minute is given by 400 34 , in the third minute 400 34 3

( ) , and so on. Thus, the total number of rotations of the wheel before coming to a

the fourth minute 400 34

complete stop is given by æ3ö æ 3 ö2 æ 3 ö3 400 + 400 çç ÷÷ + 400 çç ÷÷ + 400 çç ÷÷ + . èç 4 ÷ø èç 4 ÷ø èç 4 ÷ø

This is an infinite geometric series with a = a1 = 400 and r = 34 . This series converges to a 400 400 = = 1 = 1600. 1- r 1 - 34 4 The wheel makes 1600 rotations before coming to a complete stop. 36.

On its second swing, the pendulum bob will swing through an arc 40(0.8) centimeters long, on its third swing 40(0.8) 2 centimeters long, on its fourth swing 40(0.8)3 centimeters long, and so on. Thus, the distance it will swing altogether before coming to a complete stop is given by 40 + 40(0.8) + 40(0.8) 2 + 40(0.8)3 +  

This is an infinite geometric series with a1 = a = 40 and r = 0.8. This series converges to a 40 40 = = = 200. 1- r 1 - 0.8 0.2

So, the total distance altogether before coming to a complete stop is 200 centimeters. 37.

The second triangle has sides 1 meter in length, the third triangle has sides 12 meter in length, the fourth triangle has sides 14 meter in length, and so on. Thus, the total perimeter of all the triangles given by æ1ö æ 1 ö2 3(2) + 3(1) + 3çç ÷÷ + 3çç ÷÷ +  èç 2 ø÷ èç 2 ÷ø

is an infinite geometric series with a = a1 = 3(2) = 6 and r = 12 . This series converges to a 6 6 = = 1 = 12. 1- r 1 - 12 2

922 38.

Chapter 12 SEQUENCES AND SERIES The first triangle has sides 2 meters in length, the second triangle has sides 1 meter in length, the third triangle has sides 12 meter in length, the fourth triangle has sides 14 meter in length, and so on. Thus, the height of the first triangle is

3 meters, the height of the second triangle is

the height of the fourth triangle is

3 8

3 2

meter, the height of the third triangle is

3 4

meter,

meter, and so on. The total area of the triangles, disregarding the overlaps, is

given by 1 1 æ 3 ö÷÷ 1 æç 1 ö÷æç 3 ÷÷ö 1 æç 1 ö÷çæ 3 ÷÷ö (2)( 3) + (1) ççç ÷ + ç ÷ç ÷ + ç ÷ç ÷+⋅⋅⋅ = 2 2 èç 2 ø÷ 2 çè 2 ÷øèçç 4 ÷ø 2 çè 4 ÷øççè 8 ÷ø

This is an infinite geometric series with a = a1 = This series converges to

3 a = = 1- r 1 - 14

3 3 4

3 3 3 + + + ⋅ ⋅ ⋅. 4 16 64

3 and r = 14 .

4 3 . 3

The total area of the triangles, disregarding the overlaps, is given by 4 3 3 square meters. 39.

(a)

Solve using algebra. Let t be the number of hours since 2:00 pm. Between noon and 2 pm the first train has traveled (2)(100) = 200 miles. Its distance from the station at any time t is d1 = 200 + 100t. At time t, the second train has traveled d 2 = 125t miles.

meter in this 1 second since the tortoise’s rate is 1 meter per second. After Achilles reaches the tortoise’s starting point, the tortoise has traveled 1 meter. Since Achilles travels 10 meters per second, it takes 1 second to reach this new point. Achilles 10 1 meter in this The tortoise has traveled 10 1 second since the tortoise’s rate is 10

d1 = d 2

1 meter per second. The time that it takes Achilles to reach the tortoise is given by

200 + 100t = 125t 200 = 25t t =8

The trains are the same distance from the station 8 hours after 2 pm, that is, at 10 pm. (b) Solve using a geometric series. At 2 pm the first train has traveled 200 miles, and it takes the second train 200/125 hours to make up this distance. During this time the first train travels an additional (100)(200/125) miles, which takes the second train (100)(200/125)/125 hours, to make up. The total of the second train’s catch-up times is a geometric series with a = a1 = 200/125 and r = 100/125 . The sum of this series is

This is an infinite geometric series with 1 . This series a = a1 = 1 and r = 10 converges to

40.

(a) The tortoise’s starting point is 10 meters. Since Achilles runs 10 meters per second, it takes him 1 second to reach the tortoise’s starting point. The tortoise has traveled 1

a 1 10 = = . 1 1- r 1 - 10 9

sec to catch the tortoise It takes Achilles 10 9 (b) Using distance = rate ´ time, Achilles’ distance is given by 10t meters in t seconds and the tortoise’s distance is given by t + 10 meters in t seconds. The distances are equal when 10t = t + 10

a 200/125 8/5 = = = 8, 1- r 1 - (100/125) 1/5

or 8 hours, as found in (a). Again the two trains meet 8 hours after 2 pm, at 10 pm.

1 1 + + ⋅ 10 100

t =

10 9

It takes Achilles 10 seconds to catch the 9 tortoise. (c) The error is the assumption that an infinite series must have an infinite sum.

Section 12.5

923

Since the cyclists travel at 10 miles per hour and start 20 miles apart, they will meet in one hour. Since the fly flies at 15 miles per hour for 1 hour, it must travel 15 miles.

41.

Now find the fly’s total travel by summing each piece of its trip. It leaves the first bike and will meet the second bike when 15t = 20 - 10t or t = 4/5 hours. The fly has traveled 15(4/5) = 12 miles and the second bike has traveled 8 miles toward it, so they meet at a point 12 miles from the first bike’s original starting point. During this same 4/5 hour, the first bike has also traveled 8 miles, so when the fly turns around, the first bike is 4 miles away. Now repeat the calculation: 15t = 4 - 10t or t = 4/25 hours. During this time the fly travels 15(4/25) miles. The fly’s travel segments form a geometric series with first term a = a1 = 12 and r = 1/5. This series has sum 12 a = 1- r 1 - (1/5) 12 = = 15, 4/5

or 15 miles, as found using the argument based on the total travel time. 42. (a) To win the next point, A has to win on B’s serve and then on her own serve. This can happen

immediately with probability x 2 , after one trade of serve with probability ( x )(1 - x ) x 2 , after two trades of serve, probability [ x(1 - x )]x 2 , and so on. Note that each trade of serve happens when A wins on B’s serve (probability x) and then loses on her own serve (probability 1 - x ). So the probability that A wins the next point is x2



[ x(1  x)]n .  n0 

(b) Since 0  x(1  x)  1,

 [x(1  x)] is a n

n0

convergent geometric series with sum

1 . 1  [ x(1  x)]

Thus the probability in (a) simplifies to

x2 1  x  x2

(d) The only difference from the computation in (a) is that A already has the serve, so the factor of x can be omitted x from each of the summands in (a), giving a probability of . 1 - x + x2

12.5

Taylor Series

Your Turn 1 (a) Use the Taylor series for ln(1 + x) with property (2). ln(1 + x) = x -

x2 x3 x4 (-1)n +1 x n + ++ + 2 3 4 n

-7 ln(1 + x) = -7 x +

7x2 7 x3 7x4 7(-1)n x n + ++ + 2 3 4 n

for all x in (-1, 1]

924

Chapter 12 SEQUENCES AND SERIES

(b) Use the Taylor series for 1-1 x with property (3). 1 = 1 + x + x 2 + x3 +  + x n +  1- x x2 = x 2 + x3 + x 4 + x5 +  + x n +  1- x

for all x in (1, 1) Your Turn 2 (a) Use the Taylor series for ln(1 + x ) together with composition. ln(1 + x ) = x -

x2 x3 x4 (-1)n +1 x n + ++ + 2 3 4 n

ln(1 + 2 x 2 ) = (2 x 2 ) -

(2 x 2 )2 (2 x 2 )3 (2 x 2 )4 (-1)n +1(2 x 2 )n + ++ + 2 3 4 n

= 2 x2 - 2x4 +

8x6 (-1)n +12n x 2n - 4 x8 +  + + 3 n

The interval of convergence of the original series is -1 < x £ -1. 1 , then -1 < 0 £ 2 x 2 £ 1, so the interval of convergence is now é - 1 , 1 ù . ê ú 2 2û 2 ë

If - 1 £ x £ 2

(b) Use the series for 1-1 x together with composition. g ( x) =

3 4-x

3/4 æ x ö2 1 - çç ÷÷÷ çè 2 ø

3 æç x ö÷ 1 f ç ÷÷ where f ( x) = ç 4 è2ø 1- x

1 = 1 + x + x 2 + x3 +  + x n +  1- x n 2 3 2 2ù 2ù 2ù é é é 3 3 æç x ö÷ 3 ê çæ x ö÷ ú 3 ê æç x ö÷ ú 3 ê çæ x ÷ö ú = + ç ÷÷ + ê ç ÷÷ ú + ê ç ÷÷ ú +  + ê ç ÷÷ ú +  4 4 èç 2 ø 4 ê çè 2 ø ú 4 ê èç 2 ø ú 4 ê çè 2 ø ú æ x ö÷2 ë û ë û ë û 1 - çç ÷÷ çè 2 ø

3/4

3 3x 2 3x 4 3x 6 3 x 2n + + + +  + n +1 +  4 16 64 256 4

The interval of convergence of the original series is -1 < x < -1. When -2 < x < 2, -1 < 0 £

( 2x ) < 1 so the interval of convergence of the new series is (-2, 2).

Your Turn 3 2 For an interest rate of 3.5%, the doubling time is ln(1+ln0.035) » 20.149. Since this interest rate is less than 5%, we can use the 70 rule of 70 to estimate this doubling time; the estimate is 100(0.035) = 20 years.

Section 12.5

925

2 For an interest rate of 8%, the doubling time is ln(1ln » 9.006. Since this interest rate is more than 5%, we can use the rule + 0.08) 72 of 72 to estimate this doubling time; the estimate is 100(0.08) = 9 years.

12.5

Exercises

False. The interval of convergence for a Taylor series depends on the function.

True

Use the Taylor series for 1-1 x with property (2). 1 = 1 + x + x2 + x3 +  + xn +  1- x 6 = 6 + 6x + 6x2 + 6x3 +  + 6xn +  1- x

for all x in (-1, 1) 6.

Use the Taylor series for 1-1 x with property (2). 1 = 1 + x + x2 + x3 +  + xn +  1- x -3 = -3 - 3x - 3x 2 - 3x 3 -  - 3x n -  1- x

for all x in (-1, 1) 7.

Use the Taylor series for e x with property (3). ex = 1 +

x x2 x3 xn + + ++ + n! 1! 2! 3!

x 2e x = x 2 +

x3 x4 x5 xn+2 + + ++ + n! 1! 2! 3!

for x in ( -¥, ¥) 8.

Use the Taylor series for e x with property (3). ex = 1 +

x x2 x3 xn + + ++ + n! 1! 2! 3!

x5e x = x5 +

x6 x7 x8 x n +5 + + ++ + n! 1! 2! 3!

for x in (-¥, ¥) 9.

This function most nearly matches 1-1 x . To get 1 in the denominator, instead of 2, divide the numerator and denominator by 2. 5

5 = 2 x 2-x 1- 2

926

Chapter 12 SEQUENCES AND SERIES Thus, we can find the Taylor series for

5 2

1 - 2x

by starting with the Taylor series for 1-1 x , multiplying each term by 52 ,

and replacing each x with 2x . 5

5 = 2 x 2-x 1- 2 =

n 2 3 5 5æ x ö 5æ x ö 5æ x ö 5æ x ö ⋅ 1 + çç ÷÷ + çç ÷÷ + çç ÷÷ +  + çç ÷÷ +  2 2 çè 2 ÷ø 2 èç 2 ø÷ 2 èç 2 ø÷ 2 èç 2 ø÷

5 5x 5x 2 5 x3 5xn + + + +  + n +1 +  2 4 8 16 2

The Taylor series for 1-1 x is valid when -1 < x < 1. Replacing x with 2x gives -1 <

x < 1 or -2 < x < 2. 2

The interval of convergence of the new series is (-2, 2). 10.

This function most nearly matches 1-1 x . To get 1 in the denominator, instead of 4, divide the numerator and denominator by 4. - 34 -3 = 4-x 1 - 4x

Thus, we can find the Taylor series for

- 34

1 - 4x

by starting with the Taylor series for 1-1 x , multiplying each term by

- 34 , and replacing each x with 4x . - 34 -3 = 4-x 1 - 4x æ 3ö æ 3 öæ x ö æ 3 öæ x ö2 æ 3 öæ x ö3 æ 3 öæ x ön = çç - ÷÷ ⋅ 1 + çç - ÷÷çç ÷÷ + çç - ÷÷çç ÷÷ + çç - ÷÷çç ÷÷ +  + çç - ÷÷çç ÷÷ +  ÷ç 4 ÷ø çè 4 ÷øèç 4 ÷ø çè 4 ÷øèç 4 ÷ø çè 4 øè çè 4 ÷øèç 4 ÷ø èç 4 ø÷ =-

3 3x 3x 2 3x 3 3x n -  - n +1 -  4 16 64 256 4

The Taylor series for 1-1 x is valid when -1 < x < 1. Replacing x with 4x gives -1 <

x < 1 or -4 < x < 4. 4

The interval of convergence of the new series is (-4, 4).

11.

8x 8 = x⋅ 1 + 3x 1 - (-3x)

Use the Taylor series for 1-1 x , multiply each term by 8, and replace x with -3x. Also, use property (3) with k = 1.

Section 12.5

927 8x 8 = x⋅ 1 + 3x 1 - (-3x) = x ⋅ 8 ⋅ 1 + x ⋅ 8(-3x) + x ⋅ 8(-3x) 2 + x ⋅ 8(-3x)3 +  + x ⋅ 8(-3x) n +  = 8x - 24 x 2 + 72 x3 - 216 x 4 +  + (-1)n ⋅ 8 ⋅ 3n x n +1 + 

The Taylor series for 1-1 x is valid when -1 < x < 1. Replacing x with -3x gives -1 < -3x < 1 or

(

1 1 > x>- . 3 3

)

The interval of convergence of the new series is - 13 , 13 . 12.

7x 7 = x⋅ 1 + 2x 1 - (-2 x)

Use the Taylor series for 1-1 x , multiply each term by 7, and replace x with -2 x. Also, use property (3) with k = 1. 7x 7 = x⋅ 1 + 2x 1 - (-2 x) = x ⋅ 7 ⋅ 1 + x ⋅ 7(-2 x) + x ⋅ 7(-2 x)2 + x ⋅ 7(-2 x)3 +  + x ⋅ 7(-2 x) n +  = 7 x - 14 x 2 + 28 x3 - 56 x 4 +  + (-1) n ⋅ 7 ⋅ 2n x n +1 + 

The Taylor series for 1-1 x is valid when -1 < x < 1. Replacing x with -2 x gives -1 < -2 x < 1 or

(

1 1 > x>- . 2 2

)

The interval of convergence of the new series is - 12 , 12 .

13.

1 x2 = x2 ⋅ 4 x 4-x 1- 4

Use the Taylor series for 1-1 x , multiply each term by 14 , and replace x with 4x . Also, use property (3) with k = 2. 1 x2 = x2 ⋅ 4 x 4-x 1- 4

= x2 ⋅ =

2 3 n 1 1æ x ö 1æ x ö 1æ x ö 1æ x ö ⋅ 1 + x 2 ⋅ çç ÷÷ + x 2 ⋅ çç ÷÷ + x 2 ⋅ çç ÷÷ +  + x 2 ⋅ çç ÷÷ +  4 4 çè 4 ÷ø 4 èç 4 ø÷ 4 èç 4 ø÷ 4 èç 4 ø÷

x2 x3 x4 x5 xn+2 + + + +  + n +1 +  4 16 64 256 4

The Taylor series for 1-1 x is valid when -1 < x < 1. Replacing x with 4x gives -1 <

x < 1 or -4 < x < 4 ⋅ 4

The interval of convergence of the new series is (-4, 4).

928 14.

Chapter 12 SEQUENCES AND SERIES 9x4 9 = x4 ⋅ 1- x 1- x

Use the Taylor series for 1-1 x , and multiply each term by 9. Also, use property (3) with k = 4. 9x4 9 = x4 ⋅ 1- x 1- x = x 4 ⋅ 9 ⋅ 1 + x 4 ⋅ 9 ( x) + x 4 ⋅ 9 ( x) 2 + x 4 ⋅ 9 ( x) 3 +  + x 4 ⋅ 9 ( x ) n +  = 9 x 4 + 9 x5 + 9 x 6 + 9 x 7 +  + 9 x n + 4 +  4

The Taylor series for 19-x x has the same interval of convergence, (-1, 1), as the Taylor series for 1-1 x .

15.

We find the Taylor series of ln (1 + 4 x) by starting with the Taylor series for ln (1 + x) and replacing each x with 4x. ln (1 + 4 x) = 4 x -

(4 x) 2 (4 x)3 (4 x) 4 (-1)n (4 x) n +1 + ++ + 2 3 4 n +1

= 4 x - 8x 2 +

64 3 (-1)n 4n +1 x n +1 x - 64 x 4 +  + + 3 n +1

The Taylor series for ln (1 + x) is valid when -1 < x £ 1. Replacing x with 4x gives -1 < 4 x £ 1 or -

1 1 < x£ . 4 4

(

The interval of convergence of the new series is - 14 , 14 ùú . û 16.

(

)

We find the Taylor series of ln 1 - 2x by starting with the Taylor series for ln(1 + x) and replacing each x with - 2x .

(

- 2x æ xö ln 1 - 2x = çç - ÷÷÷ çè 2 ø 2

(

)

n +1

) + ( - 2x ) - ( - 2x ) +  + (-1)n ( - 2x ) 3

n +1

+

x x2 x3 x4 x n +1 -- 2 8 24 64 (n + 1)2n +1

The Taylor series for ln (1 + x) is valid when -1 < x £ 1. Replacing x with - 2x gives -1 < -

x £ 1 or 2 > x ³ -2. 2

The interval of convergence of the new series is [-2, 2). 17.

We find the Taylor series for e 4x by starting with the Taylor series for e x and replacing each x with 4 x 2. 2

e 4 x = 1 + (4 x 2 ) +

1 1 1 (4 x 2 )2 + (4 x 2 )3 +  + (4 x 2 ) n +  n! 2! 3!

= 1 + 4 x 2 + 8x 4 +

32 6 4n x 2n + x ++ n! 3

Section 12.5 18.

929 2

We find the Taylor series for e-3x by starting with the Taylor series for e x and replacing each x with -3x 2. -3 x 2

= 1 + (-3x 2 ) + = 1 - 3x 2 +

1 1 1 (-3x 2 ) 2 + (-3x 2 )3 +  + (-3x 2 ) n +  n! 2! 3!

9 4 9 (-1)n 3n x 2n x - x6 +  + + n! 2 2

The Taylor series for e-3x has the same interval of convergence, (-¥, ¥), as the Taylor series for e x . 19.

Use the Taylor series for e x and replace x with -x. Also, use property (3) with k = 3. x3e-x = x3 ⋅ 1 + x3 (-x) + x3 ⋅ = x3 - x 4 +

1 1 1 (-x) 2 + x3 ⋅ (-x)3 +  + x3 ⋅ (-x)n +  n! 2! 3!

1 5 1 (-1)n x n +3 x - x6 +  + + n! 2 6

The Taylor series for x3e-x has the same interval of convergence, (-¥, ¥), as the Taylor series for e x . 20.

f ( x ) = x 4e 2 x

Use the Taylor series for e x and property (3) with k = 4. x 4e2 x = x 4 ⋅ 1 + x 4 ⋅ (2 x) + x 4 ⋅ = x 4 + 2 x5 + 2 x 6 +

1 1 1 (2 x)2 + x 4 ⋅ (2 x)3 +  + x 4 ⋅ (2 x)n +  n! 2! 3!

4 7 2n x n + 4 x ++ + n! 3

The Taylor series for x 4e 2x has the same interval of convergence, (-¥, ¥), as the Taylor series for e x . 21.

2 1+ x

2 1 - (-x 2 )

Use the Taylor series for 1-1 x , multiply each term by 2, and replace x with -x 2. 2 1+ x

2 1 - (-x 2 )

= 2 ⋅ 1 + 2(-x 2 ) + 2(-x 2 )2 + 2(-x 2 )3 +  + 2(-x 2 )n +  = 2 - 2 x 2 + 2 x 4 - 2 x6 +  + (-1) n ⋅ 2 ⋅ x 2n + 

The Taylor series for 1-1 x is valid when -1 < x < 1. Replacing x with -x 2 gives -1 < -x 2 < 1 or 1 > x 2 > -1.

The inequality is satisfied by any x in the interval (-1, 1). 22.

6 3+ x

( ) 2

1 - - x3

930

Chapter 12 SEQUENCES AND SERIES 2

Use the Taylor series for 1-1 x , multiply each term by 2, and replace x with - x3 . 6 3+ x

( ) 2

1 - - x3

æ x 2 ö÷ æ x 2 ö÷2 æ x 2 ÷ö3 æ x 2 ö÷n çç çç çç ç ÷ ÷ ÷ = 2 ⋅ 1 + 2 ç - ÷÷ + 2 ç - ÷÷ + 2 ç - ÷÷ +  + 2 çç - ÷÷÷ +  çè 3 ÷ø çè 3 ø÷ çè 3 ÷ø çè 3 ø÷ = 2-

2 2 2 2 6 (-1)n ⋅ 2 ⋅ x 2n x + x4 x ++ + 3 9 27 3n 2

The Taylor for 1-1 x is valid when -1 < x < 1. Replacing x with - x3 gives -1 < -

x2 < 1 or 3 > x 2 > -3. 3

The inequality is satisfied by any x in the interval (- 3, 3).

23.

e x + e-x 1 1 = e x + e-x 2 2 2

We find the Taylor series for 12 e x by starting with the Taylor series for e x and multiplying each term by 12 . 1 x 1 1 1 1 1 1 1 1 e = ⋅ 1 + ⋅ x + ⋅ x 2 + ⋅ x3 +  + ⋅ x n +  2 2 2 2 2! 2 3! 2 n! 1 1 1 2 1 3 1 n = + x+ x + x ++ x + 2 2 4 12 2n !

We find the Taylor series for 12 e-x by starting with the Taylor series for e x , multiplying each term by 12 , and replacing x with -x. 1 -x 1 1 1 1 1 1 1 1 e = ⋅ 1 + (-x) + ⋅ (-x)2 + ⋅ (-x)3 +  + ⋅ (-x)n +  2 2 2 2 2! 2 3! 2 n! =

1 1 1 1 3 (-1)n n - x + x2 x ++ x + 2 2 4 12 2n !

Use property (1) with f ( x) = 12 e x and g ( x) = 12 e-x . e x + e-x 2 1 1 = e x + e-x 2 2 æ 1 æ1 æ1 æ 1 1ö æ1 1ö 1ö 1 ö (-1) n ÷ö÷ n ç = çç + ÷÷ + çç - ÷÷ x + çç + ÷÷ x 2 + çç - ÷÷ x3 +  + çç + ÷x +  çè 2 èç 4 èç 12 12 ø÷ 2 ÷ø èç 2 2 ø÷ 4 ø÷ 2n ! ÷÷ø çè 2n ! 1 2 1 4 1 6 1 + (-1)n n x + x + x ++ x + 2 24 720 2n ! 1 1 4 1 6 1 2n = 1 + x2 + x + x ++ x + 2 24 720 (2n)! =1+

The new series has the same interval of convergence, (-¥, ¥), as the Taylor series for e x .

Section 12.5 24.

931

e x - e-x 1 1 = e x - e-x 2 2 2

We find the Taylor series for 12 e x by starting with the Taylor series for e x and multiplying each term by 12 . 1 x 1 1 1 1 1 1 1 1 e = ⋅ 1 + ⋅ x + ⋅ x2 + ⋅ x3 +  + ⋅ xn +  2 2 2 2 2! 2 3! 2 n! 1 1 1 2 1 3 1 n = + x+ x + x ++ x + 2 2 4 12 2n !

We find the Taylor series for - 12 e-x by starting with the Taylor series for e x , multiplying each term by - 12 , and replacing x with -x. æ 1ö æ 1ö 1 æ 1ö1 æ 1ö 1 1 1 - e-x = - ⋅ 1 + çç - ÷÷ (-x) + çç - ÷÷ (-x) 2 + çç - ÷÷ (-x)3 +  + çç - ÷÷ (-x)n +  ÷ ÷ ÷ ç ç ç è ø è ø è ø èç 2 ø÷ n ! 2 2 2 2 2! 2 3! =-

1 1 1 1 3 (-1) n +1 n + x - x2 + x ++ x + 2 2 4 12 2n !

Use property (1) with f ( x) = 12 e x and g ( x) = - 12 e-x . e x - e-x 2 1 1 = e x - e-x 2 2 æ 1 æ1 æ1 æ1 1ö æ1 1ö 1ö 1 ö÷ 3 (-1) n +1 ö÷÷ n ç = çç - ÷÷÷ + çç + ÷÷÷ x + çç - ÷÷÷ x 2 + çç + x +  + çç + ÷ ÷x +  ÷ èç 2 èç 4 èç 12 2 ø çè 2 2ø 4ø 12 ø 2n ! ÷ø÷ çè 2n ! 1 3 1 5 1 7 1 + (-1) n +1 n x + x + x ++ x + 6 120 5040 2n ! 1 1 5 1 7 1 x + x3 + x + x ++ x 2n +1 +  6 120 5040 (2n + 1)!

= x+

The new series has the same interval of convergence, (-¥, ¥), as the Taylor series for e x . 25.

Use the Taylor series for ln(1 + x) and replace x with 2 x 4. ln (1 + 2 x 4 ) = 2 x 4 -

(2 x 4 )2 (2 x 4 )3 (2 x 4 )4 (-1)n (2 x 4 )n +1 + ++ + 2 3 4 n +1

= 2 x 4 - 2 x8 +

8 12 (-1) n ⋅ 2n +1 x 4n + 4 x - 4 x16 +  + + 3 n +1

The Taylor series for ln(1 + x) is valid when -1 < x £ 1. Replacing x with 2x 4 gives -1 < 2 x 4 £ 1 or -

1 1 < x4 £ . 2 2

é ù The inequality is satisfied by any x in the interval ê - 41 , 41 ú . 2 2 ë û

26.

ln (1 - 5 x 2 ) = ln [1 + (-5x 2 )]

932

Chapter 12 SEQUENCES AND SERIES Use the Taylor series for ln (1 + x) and replace x with -5x 2. ln (1 - 5 x 2 ) = (-5 x 2 ) = -5 x 2 -

(-5 x 2 ) 2 (-5 x 2 )3 (-5 x 2 ) 4 (-1)n (-5 x 2 ) n +1 + ++ + 2 3 4 n +1

25 4 125 6 625 8 5n +1 x 2n + 2 x x x -- 2 3 4 n +1

The Taylor series for ln (1 + x) is valid when -1 < x £ 1. Replacing x with -5x 2 gives -1 < -5 x 2 £ 1 or

(

1 1 > x2 ³ - ⋅ 5 5

)

The inequality is satisfied by any x in the interval - 1 , 1 .

27.

1+ x 1 x = + 1- x 1- x 1- x

The Taylor series for 1-1 x is given as 1 = 1 + x + x 2 + x3 +  + x n +  1- x

We find the Taylor series for 1-x x by starting with the Taylor series for 1-1 x and using property (3) with k = 1. x 1 = x⋅ 1- x 1- x = x ⋅ 1 + x ⋅ x + x ⋅ x 2 + x ⋅ x3 +  + x ⋅ x n +  = x + x 2 + x3 + x 4 +  + x n +1 + 

Use property (1) with f ( x) = 1-1 x and g ( x) = 1-x x . 1+ x 1 x = + 1- x 1- x 1- x = 1 + (1 + 1) x + (1 + 1) x 2 + (1 + 1) x3 +  + (1 + 1) x n +  = 1 + 2 x + 2 x 2 + 2 x3 +  + 2 x n + 

28.

æ 1 + x ÷ö ln çç = ln(1 + x) - ln(1 - x) çè 1 - x ø÷÷

The Taylor series for ln(1 + x) is given as ln (1 + x) = x -

x2 x3 x4 (-1) n x n +1 + ++ + 2 3 4 n +1

We find the Taylor series for -ln (1 - x) by starting with the Taylor series for ln (1 + x) multiplying each term by -1, and replacing x with -x. -ln (1 - x) = -1(-x) - (-1) = x+

(-x)2 (-x)3 (-x)4 (-1) n (-x)n +1 + (-1) - (-1) +  + (-1) + 2 3 4 n +1

x2 x3 x4 x n +1 + + ++ + 2 3 4 n +1

Section 12.5

933

Use property (1) with f ( x) = ln(1 + x) and g ( x) = - ln (1 - x). æ 1 + x ö÷ ln çç = ln (1 + x) - ln (1 - x) çè 1 - x ÷÷ø æ (-1) n æ 1 æ1 æ 1 1ö 1ö 1ö 1 ö÷÷ n +1 ç = (1 + 1) x + çç - + ÷÷ x 2 + çç + ÷÷ x3 + çç - + ÷÷ x 4 +  + çç + + ÷x çè 4 çè n + 1 èç 2 èç 3 2 ø÷ 3 ø÷ 4 ÷ø n + 1 ÷÷ø

29.

= 2x +

2 3 2 2 (-1)n + 1 n +1 x + x5 + x 7 +  + x + 3 5 7 n +1

= 2x +

2 3 2 2 2 x 2n +1 x + x5 + x 7 +  + + 3 5 7 2n + 1

The Taylor series for e x is given as ex = 1 + x +

1 2 1 1 1 1 x + x3 + x 4 + x5 +  + x n +  n! 2! 3! 4! 5!

x2 x3 x4 x5 xn + + + ++ + n! 2 6 24 120 ù x 2 éê x x2 x3 x n-2 ú =1+ x + + + + + + +   1 ú n! 2 êê 3 12 60 2 ë ûú =1+ x +

n-2

x + x ++ x Compare the series 3x + 12 +  , for n ³ 3, with the series n! 60 2

x + x 2 + x3 +  + x n - 2 +  , for n ³ 3. The second series is an infinite geometric series that is larger than 3 12 48 3⋅4n -3

or equal to the first series term by term. The geometric series has a = 3x and r = 4x , and sums to 124-x3x . Thus, for x > 0, 1+ x +

ù 2 ù x2 x 2 éê x x2 x3 x n-2 4x ú < 1 + x + x éê 1 + ú   1 , <1+ x + + + + + + + ú ê n! ê ú x 2 2 ê 3 12 60 2 12 3 ú ë û 2 ë û

1+ x +

x2 x2 é 4x ù ê1 + ú. < ex < 1 + x + 2 2 êë 12 - 3x úû

2 2 For values of x sufficiently close to 0, 124-x3x approaches 0, and 1 + x + x2 éê 1 + 124-x3x ùú » 1 + x + x2 ⋅ ë û 2

Thus, e x » 1 + x + x2 . For x < 0, a similar argument can be made. 30.

Use the Taylor series for e x and replace x with -x to find the Taylor series for e-x . e-x = 1 + (-x) +

1 1 1 1 1 (-x) 2 + (-x)3 + (-x)4 + (-x)5 +  + (-x)n +  2! 3! 4! 5! n!

(-1) n x n x2 x3 x4 x5 + ++ + 2 6 24 120 n! ù (-1)n x n-2 x 2 éê x x2 x3 ú 1 = 1- x + + + + +   ú n! 2 êê 3 12 60 úû 2 ë ù x n-2 æ x ö÷ x 2 éê xæ x ö x3 æç xö ú  1 - çç1 - ÷÷ 1 - ÷÷ -  - n ! çç1 = 1- x + ÷ ç ê ú èç ø÷ + 2 ê 3 èç 4 ø÷ 60 èç 6 ø÷ n 1 û 2 ë

= 1- x +

934

Chapter 12 SEQUENCES AND SERIES

(

)

(

)

x 1- x - Compare the series - 3x 1 - 4x - 60 6

(

)

(

)

(

x n-2 n! 2

(1 - n+x 1 ) -  with the series

)

x3 1 - x -  - x n - 2 1 - x - . The second series is an infinite geometric series that is - 3x 1 - 4x - 48 n -3 4 4 3⋅ 4

(

)

x , and less than or equal to the first series term by term. The geometric series has a = - 3x 1 - 4x and r = 16 2

sums to - 16 x -4 x2 . Thus, for 0 < x < 1, 48-3 x

1- x +

ù x 2 éê 16 x - 4 x 2 ùú x2 é x æç x ö÷ x3 æç x ö÷ x n-2 æç x ö÷ ê < + 1 1 x 1 1 1  ÷ ÷ ÷÷ -  úú ç ç çç1 ê ú 2 ! n ÷ ÷ ç ç ê è 2 ê 2 ë 3è 4 ø 60 è 6ø n + 1ø 48 - 3x úû û 2 ë

< 1- x +

x2 , 2

or 1- x +

x 2 éê 16 x - 4 x 2 ùú x2 -x 1 e 1 x . < < + ú 2 êê 2 48 - 3x 2 ûú ë 2

For values of x sufficiently close to 0, 16 x- 4 x2 approaches 0, and 48-3 x

1- x +

x 2 éê 16 x - 4 x 2 ùú x2 11 x . » + ê ú 2 2 ê 2 48 3 x ú ë û

Thus, e-x » 1 - x + x2 . For x < 0, a similar argument can be made. 31.

The Taylor series for ex is given as ex = 1 + x +

1 2 1 1 1 1 x + x3 + x 4 + x5 +  + x n +  2! 3! 4! 5! n!

For x > 0, n1! x n > 0 when n ³ 2. Thus, 1 2 1 1 1 1 x + x3 + x 4 + x5 +  + x n +  > 0 n! 2! 3! 4! 5!

since each term on the left-hand side is positive. Adding 1 + x to both sides we have 1+ x +

1 2 1 1 1 1 x + x3 + x 4 + x5 +  + x n +  > 1 + x n! 2! 3! 4! 5!

or e x > 1 + x. For x = 0, e x = e0 = 1 = 1 + 0 = 1 + x. b

| x|

For -1 £ x < 0, when b is an even positive integer, xb > 0. Also, when a ³ 3, a < 1, or 1 + ax > 0. Thus, xb b

(1 + ax ) > 0 for b an even positive integer and a ³ 3, since both factors are positive. Thus x 2 æç xö x 4 æç xö x 6 æç xö x 2n æç x ö÷ ÷ +  > 0, çç 1 + ÷÷÷ + çç 1 + ÷÷÷ + çç 1 + ÷÷÷ +  + ç1 + 2! è 3ø 4! è 5ø 6! è 7ø (2n)! èç 2n + 1 ø÷

Section 12.5

935 1 2 1 1 1 1 x + x3 + x 4 + x5 +  + x n +  n! 2! 3! 4! 5! 2æ 4æ 2n æ ö ö x ç x x ç x x ç x ö÷ =1+ x + ÷+ çç1 + ÷÷÷ + çç1 + ÷÷÷ +  + çç1 + 2! è 3ø 4! è 5ø (2n)! è 2n + 1 ø÷

ex = 1 + x +

We have already shown the series to be positive from the third term on, so e x > 1 + x. For x < -1, 1 + x < 0 and e x > 0. Thus e x > 1 + x. 32.

The Taylor series for e-x is found from the Taylor series for e x by replacing x with -x. x2 x3 x4 x5 (-1) n x n + ++ + 2! 3! 4! 5! n!

e-x = 1 - x +

For x £ 0, all the terms after the first two are greater than or equal to 0, so e-x ³ 1 - x. For 0 < x £ 1, take the terms in pairs beginning with the third term. For example, x 2 - x3 = x 2 2! 3! 2!

(1 - 3x ). If 0 < x £ 1, then 0 < 3x £ 13 , so 1 - 3x > 0 and x2! (1 - 3x ) > 0. 2

2 n +1

(

)

A similar argument works for every pair of terms of the form (2x n !) - (2xn +1)! = (2x n)! 1 - 2nx+1 . Thus e-x ³ 1 - x. Finally, if x > 1, e-x is positive and 1 - x is negative, so again e-x ³ 1 - x. 33.

The area is given by

ò0

1 3

e x dx.

Find the Taylor series for e x by using the Taylor series for e x and replacing x with x 2. 2

e x = 1 + (x2 ) +

1 2 2 1 1 1 ( x ) + ( x 2 )3 + ( x 2 ) 4 +  + ( x 2 ) n +  n! 2! 3! 4!

Using the first five terms of this series gives 1 3

1 3

2 4ù

ò e dx » ò êêë1 + x + 2! (x ) dx + 3! (x ) + 4! (x ) úúû dx x2

2 2

2 3

1 4

1 6

1 8ö

ò çççè1 + x + 2 x + 6 x + 24 x ÷÷÷ø dx 0

æ 1 1 5 1 7 1 9 ö÷ 3 = çç x + x3 + x + x + x ÷ çè 3 10 42 216 ø÷ 0 1 1 1 1 1 + + + + -0 3 81 2430 91,854 4, 251,528 » 0.3461.

34.

The area is given by 1 2

ò 1 - x dx. 0

936

Chapter 12 SEQUENCES AND SERIES by using the Taylor series for 1-1 x and replacing x with x3.

1 1- x3

Find the Taylor series for 1

1 - x3

= 1 + ( x 3 ) + ( x 3 ) 2 + ( x 3 )3 + ( x 3 ) 4 +  + ( x 3 ) n +  = 1 + x3 + x 6 + x9 + x12 +  + x3n + 

Using the first five terms of this series gives 1 2

ò 1 - x dx » ò 3

1 2

æ 1 1 1 10 1 13 ö÷ x + x ÷÷ (1 + x + x + x + x )dx = çç x + x 4 + x 7 + çè ø 4 7 10 13 0

1 1 1 1 1 = + + + + - 0 » 0.5168. 2 64 896 10.240 106, 496

35.

The area is given by 1 3

ò 1 - x dx. 1 4

Find the Taylor series for

1 1- x

by using the Taylor series for 1-1 x and replacing x with

1 = 1+ 1- x

x +

x ++

x +

Using the first five terms of this series gives

1 3

1 dx » 1 1x 4

1 3

ò (1 + x + x + x 1 4

3/ 2

3 æ 2 x2 2 x3 ö÷÷ ç + x ) dx = çç x + x3/ 2 + + x5/ 2 + ÷÷ 3 2 5 3 ø÷ 1 çè

3/ 2 5/ 2 æ ö ç1 2æ1ö 1 2æ1ö 1 ÷÷ çæ 1 1 1 1 1 ö÷ = çç + çç ÷÷ + + çç ÷÷ + ÷-ç + + + + ÷ ÷ ÷ ÷ ç ç ç çç 3 3è 3ø 18 5è 3ø 81 ø÷÷ è 4 12 32 80 192 ø÷ è

» 0.1729.

36.

The area is given by 1

ò0 e x dx. Find the Taylor series for e x by using the Taylor series for e x and replacing x with e x =1+

x +

x x x x + + ++ + 2! 3! 4! n!

Using the first five terms of this series gives 1

ò0 e

ç dx » ççç1 + 0è

=1+

37.

1æ

æ x x3/ 2 x 2 ö÷÷ 2 3/ 2 x2 x5/ 2 x3 ö÷÷ ç x + + + + + + ÷÷ dx = çç x + x ÷ 2 6 24 ø÷ 3 4 15 72 ø÷÷ èç

2 1 1 1 + + + - 0 » 1.9972 3 4 15 72

The area is given by

ò0

0.4

1 -x 2 /2 e dx = 2

1 2

ò0

0.4

e-x /2dx.

1 2

Section 12.5

937 2

Find the Taylor series for e-x /2 by using the Taylor series for e x and replacing x with - x2 . 2

e-x /2 = 1 -

1 2 1 4 1 6 1 8 (-1) n 2n x + x x x  x + + + + n !2n 2 2!22 3!23 4!24

Using the first five terms of this series gives 1 2

0.4

e-x /2dx »

1 2

0.4 æ

1 2 1 4 1 6 1 8 ÷ö x x + x ÷÷ dx ççç1 - x + 2 3 2 è 2!2 3!2 4!24 ø

0 0.4 æ

ö çç1 - 1 x 2 + 1 x 4 - 1 x 6 + 1 x8 ÷÷ dx çè 2 8 48 384 ø÷ 0.4

1 æç 1 3 1 5 1 7 1 9 ö÷ x x + x ÷ çç x - x + è 6 40 336 3456 ø÷ 0 2

ö 1 1 æç 1 1 1 3 (0.4)9 - 0 ÷÷÷ (0.4)5 (0.4)7 + çç 0.4 - (0.4) + ø 3456 6 40 336 2 è 1 (0.389585) 2(3.1416)

= »

» 0.1554.

38.

The area is given by 0.6

1 -x 2 /2 e dx = 2

ò0

1 2

0.6

ò0 e x /2dx. - 2

Find the Taylor series for e-x /2 by using the Taylor series for e x and replacing x with - x2 . 2

e-x /2 = 1 -

1 2 1 4 1 6 1 8 (-1) n 2n x + x x + x ++ x + 2 3 4 2 n !2n 2!2 3!2 4!2

Using the first five terms of this series gives 1 2

0.6

ò0

0.6 æ

e-x /2dx »

1 2

1 æç 1 3 1 5 1 7 1 9 ö÷ x x + x ÷÷ çç x - x + è ø0 6 40 336 3456 2

ò0 çççè1 - 2 x2 + 2!22 x4 - 3!23 x6 + 4!24 x8 ø÷÷÷ dx 1

ò0 çççè1 - 2 x2 + 8 x4 - 48 x6 + 384 x8 ÷÷÷ø dx 0.6

= »

ö 1 1 æç 1 1 1 3 5 (0.6)9 - 0 ÷÷÷ (0.6)7 + çç 0.6 - (0.6) + (0.6) è ø 3456 6 40 336 2 1 (0.565864) 2(3.1416)

» 0.2257.

39.

The doubling time n for a quantity that increases at an annual rate r is given by n=

ln2 ln2 = » 14.94. ln(1 + r ) ln0.0475

It will take about 14.94 years. According to the Rule of 70, the doubling time is given by

938

Chapter 12 SEQUENCES AND SERIES 70 70 = 100r 100(0.0475) = 14.74.

Doubling time »

It will take about 14.74 years, a difference of 0.2 years, or about 10 weeks. 40.

The doubling time n for a quantity that increases at an annual rate r is given by n =

ln 2 ln 2 = » 11.896. ln (1 + r ) ln1.06

It will take about 11.9 years. According to the Rule of 72, the doubling time is given by 72 72 = 100r 100(0.06) = 12.

Doubling time »

It will take about 12 years, a difference of 0.1 years, or about 5 weeks.

41.

(a)

f ( x) =

x=0

 xe-

å x! x=0 0 -

 e

1e- 1!

= e- +  e- +

 2e- 2!

 3e- 3!

++

 ne-

 ne- n!

+

æ ö÷ 2 3 n ç = e- çç1 +  + + ++ +  ÷÷÷ n! 2! 3! ÷ø çè = e- ⋅ e = e0 =1 ¥

(b)

å x=0

x f ( x) =

 xe-

x=0 0 -

 e

x! +1

1e- 1!

= 0 +  e- +  2e- +

 2e- 2!

 3e-

++

++ n

 ne-

(n - 1)! æ ö÷ 2  n-1 ç =  e- çç1 +  + + + +  ÷÷÷ çè 2! (n - 1)! ø÷ 2!

 ne-

+

=  e- ⋅ e =  e0 =

+

Section 12.5

939

(c) Since 5.3 is the expected value from part b we have  = 5.3. P( x < 4) = P( x = 0) + P( x = 1) + P( x = 2) + P( x = 3) 5.30 e-5.3 5.31e-5.3 5.32 e-5.3 5.33 e-5.3 + + + 0! 1! 2! 3! æ ö 28.09 148.877 = e-5.3 çç1 + 5.3 + + ÷÷÷ çè ø 2 6 =

» 0.2254

42.

(a)

x =1

å f (x) = å (1 - p) x p = (1 - p) -1

1-1

p + (1 - p)2-1 p + (1 - p)3-1 p + (1 - p) 4-1 p +  + (1 - p)n-1 p + 

= p + p(1 - p) + p(1 - p)2 + p(1 - p)3 +  + p(1 - p)n-1 + 

This is an infinite geometric series with a = p and r = (1 - p). Thus the series sums to a p p p = = = = 1. 1- r 1 - (1 - p) 1-1+ p p ¥

(b) Let g ( z ) = p

å z x = p(z + z + z + z + + z n + ). 2

x =1

Then g ¢( z ) = p(1 + 2 z + 3z 2 + 4 z 3 +  + nz n-1 + ) and g ¢(1 - p) = p[1 + 2(1 - p) + 3(1 - p)2 + 4(1 - p)3 +  + n (1 - p) n-1 + ] = p [1(1 - p)1-1 + 2(1 - p) 2-1 + 3(1 - p)3-1 + 4(1 - p) 4-1 +  + n (1 - p) n-1 + ] ¥

= p

x(1 - p) x -1 =

x =1

x(1 - p) x -1 p =

x =1

å xf ( x). x =1

Using the Taylor series for 1-1 x , write g ( z ) as ¥

g ( z) = p

å z x = p(z + z + z + z +  + z n + ) = pz(1 + z + z + z +  + z n + ) 2

-1

x =1

= pz ⋅

Then

1 zp = . 1- z 1- z

g ¢( z ) = ¥

Thus,

(1 - z ) p - zp(-1) (1 - z )

p (1 - z )

, and

g ¢(1 - p) =

p 2

[1 - (1 - p)]

p p

1 . p

å x f (x) = g ¢(1 - p) = p . x =1

251 . The expected value is 1 = 882 » 3.514. (c) The probability p of meeting a foreign-born player is p = 882 p 251

(d)

P(meeting in first three trials) = f (1) + f (2) + f (3) 1-1 2-1 3-1 æ 251 ö÷ æç 251 ö÷ æç 251 ö÷ æç 251 ö÷ æç 251 ö÷ æç 251 ö÷ = çç1 + ç1 + ç1 ÷ ÷ ÷ ÷ ÷ ÷ ç ç ç èç 882 ø÷ èç 882 ø÷ èç 882 ø÷ èç 882 ø÷ èç 882 ø÷ èç 882 ø÷

» 0.6338.

940

Chapter 12 SEQUENCES AND SERIES (a) The probability, p, of popping a 6 is p = 16 . From Exercise 42b, the expected value is given by

43.

å x f ( x) = p = x =1

1 1 6

= 6.

(b) For x ³ 4, find ¥

f ( x) = 1 -

x=4

å f ( x) x =1 3

= 1-

å (1 - p) x p x =1 3

= 1-

å x =1

-1

æ öx -1 æ ö çç1 - 1 ÷÷ çç 1 ÷÷ çè 6 ÷ø çè 6 ÷ø

1-1 2-1 3-1 éæ ù 1ö æ1ö æ 1ö æ1ö æ 1ö æ1ö = 1 - êê çç1 - ÷÷÷ çç ÷÷÷ + çç1 - ÷÷÷ çç ÷÷÷ + çç1 - ÷÷÷ çç ÷÷÷ úú 6 ø çè 6 ø çè 6 ø çè 6 ø çè 6 ø çè 6 ø ú êë çè û » 1 - 0.4213 = 0.5787.

12.6

Newton’s Method

Your Turn 1 f ( x) = 2 x3 - 5x 2 + 6 x - 10

First note that f (1) = -7 and f (3) = 17, so f has a zero on [1, 3]. Use c1 = 2 as a first guess. For Newton’s method we need to know f ¢, which is f ¢( x) = 6 x 2 - 10 x + 6. f (c1) f ¢(c1)

c2 = c1 = 2-

-2 10

= 2.2 c3 = c2 -

f (c2 ) f ¢(c2 )

0.296 13.04 » 2.177301 f (c3 ) c4 = c3 f ¢(c3 ) = 2.2 -

» 2.177301 -

0.004207 12.670828

» 2.176969

Section 12.6

941

Your Turn 2

Let c1 = 1.

We find 3 15 by finding a zero of the function f ( x) = x3 - 15. f ¢( x) = 3x 2. Use 2.5 as the initial guess c1. c2 = c1 -

f (c1) f ¢(c1)

c2 = c1 -

-1 f (c1) = 1= 1.1429 f ¢(c1) 7

c3 = c2 -

f (c2 ) 0.1020 = 1.1429 = 1.1308 f ¢(c2 ) 8.4286

c4 = c3 -

f (c3 ) 0.0007 = 1.1308 = 1.1307 f ¢c3 8.3075

0.625 18.75 » 2.466667

= 2.5 -

c3 = c2 -

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = 1.13.

f (c2 ) f ¢(c2 )

= 2.466667 -

f ( x) = 2 x 2 - 8 x + 3 f ¢( x) = 4 x - 8

0.008302 18.25334

f (3) = -3 < 0 and

» 2.466212

f (4) = 3 > 0 so a

solution exists in (3, 4). Let c1 = 3.

3 15 » 2.466. In fact, c is correct to six decimal places. 3

12.6

Warmup Exercises

W1.

1/ 3

f ( x) = 6 x + 3 ln x æ 1 ÷ö 3 f ¢( x) = çç ÷÷ 6 x-2 / 3 + çè 3 ø x

(

= 2 x-2 / 3 +

)

3 x

-3 f (c1) = 3= 3.75 f ¢(c1) 4

c3 = c2 -

f (c2 ) 1.125 = 3.75 = 3.5893 f ¢(c2 ) 7

c4 = c3 -

f (c3 ) 0.0517 = 3.5893 = 3.5812 f ¢(c3 ) 6.3571

c5 = c4 -

f (c4 ) 0.0001 = 3.5812 = 3.5811 f ¢(c4 ) 6.3246

Subsequent approximations will agree with c4 and c5 to the nearest hundredth. Thus, x = 3.58.

W2. Use the product rule. f ( x) = x 2e-3x

(

c2 = c1 -

)

(

f ¢( x) = x 2 -3e-3x + (2 x) e-3x

(

)

= -3x 2 + 2 x e-3x

f ¢( x) = 6 x 2 - 12 x - 1 f (3) = -1 < 0 and f (4) = 30 > 0 so a solution exists in (3, 4).

12.6

Exercises

True

False. When using Newton’s method to find the zeros of a differentiable function, the first step is to find the derivative of the function.

False. Newton’s method finds the zeros of a differentiable function on a closed interval [a, b] so that f (a) and f (b) are of opposite sign, one positive and one negative.

True

f ( x) = 5 x 2 - 3 x - 3

Let c1 = 3. c2 = c1 -

-1 f (c1) = 3= 3.0588 f ¢(c1) 17

c3 = c2 -

f (c2 ) 0.0419 = 3.0588 = 3.0565 f ¢(c2 ) 18.4325

Subsequent approximations will agree with c2 and c3 to the nearest hundredth. Thus, x = 3.06. 8.

f ( x ) = -x 3 + 4 x 2 - 5 x + 4 f ¢ ( x ) = -3 x 2 + 8 x - 5

f ¢( x) = 10 x - 3 f (1) = -1 < 0 and f (2) = 11 > 0 so a

solution exists in (1, 2).

f ( x) = 2 x 3 - 6 x 2 - x + 2

f (1) = 2 > 0 and

solution exists in (2, 3). Let c1 = 2.

f (3) = -2 < 0 so a

942

Chapter 12 SEQUENCES AND SERIES

c2 = c1 -

f (c1) 2 = 2= 4 -1 f ¢(c1)

c2 = c1 -

f (c1) 6 = 2= 1.7273 f ¢(c1) 22

c3 = c2 -

-16 f (c2 ) = 4= 3.2381 -21 f ¢(c2 )

c3 = c2 -

f (c2 ) 1.3324 = 1.7273 = 1.6210 f ¢(c2 ) 12.5298

c4 = c3 -

f (c3 ) -4.2017 = 3.2381 = 2.8399 -10.5510 f ¢(c3 )

c4 = c3 -

f (c3 ) 0.1734 = 1.6210 = 1.6024 f ¢(c3 ) 9.3217

c5 = c4 -

-0.8430 f (c4 ) = 2.8399 = 2.7097 -6.4756 f ¢(c4 )

c5 = c4 -

f (c4 ) 0.0050 = 1.6024 = 1.6018 f ¢(c4 ) 8.7882

c6 = c5 -

f (c5 ) -0.0744 = 2.7097 = 2.6958 -5.3497 f ¢(c5 )

Subsequent approximations will agree with c5 and c6 to the nearest hundredth. Thus, x = 2.70. 9.

f ( x ) = -3 x + 5 x + 3 x + 2 f ¢( x) = -9 x 2 + 10 x + 3 f (2) = 4 > 0 and

f (3) = -25 < 0 so a

solution exists in (2, 3).

11.

f ( x) = 2 x 4 - 2 x 3 - 3 x 2 - 5 x - 8 f ¢( x ) = 8 x 3 - 6 x 2 - 6 x - 5

In the interval [-2, -1]: f (-2) = 38 > 0 and f (-1) = -2 < 0 so a solution exists in (-2, -1). Let c1 = -2.

Let c1 = 2. c2 = c1 -

f (c1) 4 = 2= 2.3077 -13 f ¢(c1)

c3 = c2 -

-1.3182 f (c2 ) = 2.3077 = 2.2474 -21.8521 f ¢(c2 )

c4 = c3 -

f (c3 ) -0.0567 = 2.2474 = 2.2445 -19.9824 f ¢(c3 )

c5 = c4 -

-0.0001 f (c4 ) = -2.2445 = 2.2445 -19.8960 f ¢(c4 )

Subsequent approximations will agree with c4 and c5 to the nearest hundredth. Thus, x = 2.24. 10.

Subsequent approximations will agree with c4 and c5 to the nearest hundredth. Thus, x = 1.60.

f ( x) = 4 x 3 - 5 x 2 - 6 x + 6

c2 = c1 -

f (c1) 38 = -2 = -1.5309 f ¢(c1) -81

c3 = c2 -

f (c2 ) 10.7834 = -1.5309 = -1.2513 f ¢(c2 ) -38.5773

c4 = c3 -

f (c3 ) 2.3817 = -1.2513 = -1.1458 f ¢(c3 ) -22.5622

c5 = c4 -

f (c4 ) 0.2457 = -1.1458 = -1.1322 f ¢(c4 ) -18.0356

c6 = c5 -

f (c5 ) 0.0036 = -1.1319 = -1.1322 -17.5069 f ¢(c5 )

Subsequent approximations will agree with c5 and c6 to the nearest hundredth. Thus, x = -1.13 is a solution in [-2, -1].

f ¢( x) = 12 x 2 - 10 x - 6

In the interval [2, 3]: f (2) = -14 < 0 and f (3) = 58 > 0 so a solution exists in (2, 3).

f (1) = -1 < 0 and f (2) = 6 > 0 so a solution exists in (1, 2).

Let c1 = 2.

Let c1 = 1.

c2 = c1 -

-14 f (c1) = 2= 2.6087 f ¢(c1) 23

c2 = c1 -

-1 f (c1) = 1= 0.75 -4 f ¢(c1)

c3 = c2 -

f (c2 ) 15.6588 = 2.6087 = 2.4143 f ¢(c2 ) 80.5396

c3 = c2 -

f (c2 ) 0.375 = 0.75 = 0.8056 -6.75 f ¢(c2 )

c4 = c3 -

f (c3 ) 2.2460 = 2.4143 = 2.3756 f ¢(c3 ) 58.1188

c5 = c4 -

f (c4 ) 0.0774 = 2.3756 = 2.3742 f ¢(c4 ) 54.1413

c6 = c5 -

f (c5 ) 0.0001 = 2.3742 = 2.3742 f ¢(c5 ) 53.9972

The approximations clearly are not leading to a solution in (1, 2). (There is another solution to the equation in [0, 1], x = 0.81.) Therefore, start again, this time letting c1 = 2.

Section 12.6

943

Subsequent approximations will agree with c5 and c6 to the nearest hundredth. Thus, x = 2.37 is a solution in [2, 3]. 12.

13.

f ¢( x ) =

f (0) = 4 > 0

so a solution exists in (-3,0).

f ¢( x) = 12 x + 12 x - 12 x - 2

In the interval [-3, -2]: f (-3) = 75 > 0 and f (-2) = -16 < 0 so a solution exists in (-3, -2).

4 -2/3 x - 4x 3

f (-3) = -19.7690 < 0 and

f ( x) = 3x 4 + 4 x3 - 6 x 2 - 2 x - 12 3

f ( x) = 4 x1/3 - 2 x 2 + 4

Let c1 = -3. c2 = c1 -

-19.7690 f (c1) = -3 = -1.4361 f ¢(c1) 12.6410

c3 = c2 -

-4.6378 f (c2 ) = -1.4361 = -0.7533 f ¢(c2 ) 6.7920

c4 = c3 -

f (c3 ) -0.7744 = -0.7533 = -0.5858 f ¢(c3 ) 4.6236

f (c2 ) 18.2141 = -2.5879 = -2.4030 -98.5428 f ¢(c2 )

c5 = c4 -

f (c4 ) -0.0332 = -0.5858 = -0.5780 4.2477 f ¢(c4 )

f (c3 ) 2.6873 = -2.4030 = -2.3633 c4 = c3 -65.2105 f ¢(c3 )

c6 = c5 -

f (c5 ) -0.0001 = -0.5780 - 0.5780 f ¢(c5 ) 4.2335

Let c1 = -3. f (c1) 75 = -3 = -2.5879 c2 = c1 -182 f ¢(c1) c3 = c2 -

c5 = c4 -

Subsequent approximations will agree with c5 and c6 to the nearest hundredth. Thus, x = -0.58.

f (c4 ) 0.0005 = -2.3633 = -2.3633 -65.0118 f ¢(c4 )

Subsequent approximations will agree with c4 and c5 to the nearest hundredth. Thus, x = -2.36 is a solution in [-3, - 2].

14.

f ¢( x ) =

In the interval [1, 2]: f (1) = -13 < 0 and f (2) = 40 > 0 so a solution exists in (1, 2).

-13 f (c1) = 1= 2.3 f ¢(c1) 10

f (c2 ) 84.2803 = 2.3 = 1.8315 c3 = c2 f ¢(c2 ) 179.884 c4 = c3 c5 = c4 -

f (c3 ) 22.5408 = 1.8315 = 1.5810 f ¢(c3 ) 89.9975 f (c4 ) 4.3913 = 1.5810 = 1.5032 f ¢(c4 ) 56.4444

f (3) = -8.2310 < 0 so

a solution exists in (0, 3). Notice that f ¢(0) is undefined. Therefore, instead of choosing 0 for c1, let c1 = 3. c2 = c1 -

-8.2310 f (c1) = 3= 2.2754 -11.3590 f ¢(c1)

c3 = c2 -

-1.0938 f (c2 ) = 2.2754 = 2.1441 -8.3309 f ¢(c2 )

c4 = c3 -

f (c3 ) -0.0364 = 2.1441 = 2.1394 -7.7745 f ¢(c3 )

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = 2.14.

f (c5 ) 0.3400 = 1.5032 c6 = c5 = 1.4961 f ¢(c5 ) 47.8367

Subsequent approximations will agree with c5 and c6 to the nearest hundredth. Thus, x = 1.50 is a solution in [1, 2].

4 -2/3 x - 4x 3

f (0) = 4 > 0 and

Let c1 = 1. c2 = c1 -

f ( x) = 4 x1/3 - 2 x 2 + 4

15.

f ( x) = e x + x - 2 f ¢( x) = e x + 1 f (0) = -1 < 0 and solution exists in (0, 3).

Let c1 = 0.

f (3) = 21.085537 so a

944

Chapter 12 SEQUENCES AND SERIES

c2 = c1 -

-1 f (c1) = 0= 0.5 f ¢(c1) 2

c3 = c2 -

f (c2 ) 0.14872 = 0.5 = 0.4439 f ¢(c2 ) 2.6487

c4 = c3 -

f (c3 ) 0.00267 = 0.4439 = 0.4429 f ¢(c3 ) 2.5588

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = 1.25. 18.

f ( x) = x 2e-x + x 2 - 2 f ¢( x) = -x 2e-x + 2 xe-x + 2 x f (-3) = 187.76983 > 0 and f (0) = -2 < 0

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = 0.44.

so a solution exists in (-3, 0). Let c1 = -3.

16.

f ( x) = e 2 x + 3 x - 4

c2 = c1 -

187.77 f (c1) = -3 = -2.3890 f ¢(c1) -307.3

c3 = c2 -

65.932 f (c2 ) = -2.3890 = -1.8354 f ¢(c2 ) -119.1

c4 = c3 -

f (c3 ) 22.482 = -1.8354 = -1.3650 f ¢(c3 ) -47.79

c5 = c4 -

7.1591 f (c4 ) = -1.3650 = -1.0195 f ¢(c4 ) -20.72

f (c2 ) 1.1201 = 0.6 = 0.4838 f ¢(c2 ) 9.6402

c6 = c5 -

f (c5 ) 1.9203 = -0.08378 = -1.0195 -10.57 f ¢(c5 )

f (c3 ) 0.08302 = 0.4838 = 0.4738 c4 = c3 f ¢(c3 ) 8.2632

c7 = c6 -

f (c6 ) 0.32421 = -0.8378 = -0.7926 -7.171 f ¢(c6 )

c8 = c7 -

f (c7 ) 0.01602 = -0.7926 = -0.7901 -6.475 f ¢(c7 )

f ¢( x) = 2e2 x + 3 f (0) = -3 < 0 and

f (3) = 408.42879 > 0

so a solution exists in (0, 3). Let c1 = 0. -3 f (c1) = 0= 0.6 c2 = c1 f ¢(c1) 5 c3 = c2 -

f (c4 ) 9.1 ⋅ 10-4 = 0.4738 = 0.4737 c5 = c4 f ¢(c4 ) 8.159

Subsequent approximations will agree with c7 and c8 to the nearest hundredth. Thus, x = -0.79.

Subsequent approximations will agree with c4 and c5 to the nearest hundredth. Thus, x = 0.47. 19. 17.

2 -x

f ( x) = x e

+x -2

2 -x

f ¢( x ) = - x e

f (0) = -2 and

-x

+ 2 xe

f ( x) = ln x + x - 2 f ¢( x ) =

+ 2x

f (3) = 7.4480836 so a

solution exists in (0, 3).

1 +1 x

f (1) = -1 < 0 and

f (4) = 3.3862944 > 0

so a solution exists in (1, 4). Let c1 = 1.

Let c1 = 0. f (c1) -2 c2 = c1 = 0f ¢(c1) 0

Since c2 is undefined, let c1 = 1. c2 = c1 -

f (c1) -0.6321 = 1= 1.2670 f ¢(c1) 2.3679

c3 = c2 -

f (c2 ) 0.05746 = 1.2670 = 1.2464 f ¢(c2 ) 2.7956

c4 = c4 -

f (c3 ) 2.1 ⋅ 10-4 = 1.2464 = 1.2463 f ¢(c3 ) 2.7629

c2 = c1 -

f (c1) -1 = 1= 1.5 2 f ¢(c1)

c3 = c2 -

-0.0945 f (c2 ) = 1.5 = 1.5567 1.6667 f ¢(c2 )

c4 = c3 -

f (c3 ) -7 ⋅ 10-4 = 1.5567 = 1.5571 1.6424 f ¢(c3 )

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = 1.56.

Section 12.6 20.

945

f ( x) = 2 ln x + x - 3

23.

11 is a solution of x 2 - 11 = 0.

2 +1 x f (1) = -2 < 0 and f (4) = 3.7725887 > 0 so

f ( x) = x 2 - 11

f ¢( x ) =

f ¢( x ) = 2 x

a solution exists in (1, 4).

Since 3 < 11 < 4, let c1 = 3.

Let c1 = 1. c2 = c1 -

f (c1) -2 = 1= 1.6667 3 f ¢(c1)

c3 = c2 -

-0.3116 f (c2 ) = 1.6667 = 1.8083 2.2 f ¢(c2 )

c4 = c3 -

f (c3 ) -0.0069 = 1.8083 = 1.8116 2.106 f ¢(c3 )

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = 1.81. 21.

-2 = 3.333 6 0.10889 = 3.317 c3 = 3.333 6.666 0.00249 = 3.317 c4 = 3.317 6.634 c2 = 3 -

Since c3 = c4 = 3.317, to the nearest thousandth, 11 = 3.317. 24.

15 is a solution of x 2 - 15 = 0. f ( x) = x 2 - 15

2 is a solution of x 2 - 2 = 0.

f ¢( x ) = 2 x

f ( x) = x 2 - 2

Since 3 < 15 < 4, let c1 = 3.

f ¢( x ) = 2 x

Since 1 <

2 < 2, let c1 = 1.

-6 = 4 6 1 c3 = 4 - = 3.875 8 0.01563 = 3.873 c4 = 3.875 7.75 c2 = 3 -

-1 c2 = 1 = 1.5 2 0.25 c3 = 1.5 = 1.417 3 0.00789 c4 = 1.417 = 1.414 2.834

c5 = 3.873 -

-6 ⋅ 10-4 c5 = 1.414 = 1.414 2.828

Since c4 = c5 = 3.873, to the nearest

Since c4 = c5 = 1.414, to the nearest

22.

thousandth, 15 = 3.873.

2 = 1.414.

thousandth,

3 is a solution of x 2 - 3 = 0.

25.

250 is a solution of x 2 - 250 = 0. f ( x) = x 2 - 250

f ( x) = x 2 - 3 f ¢( x ) = 2 x

Since 1 <

f ¢( x ) = 2 x

Since 15 <

3 < 2, let c1 = 1.

-2 = 2 2 1 c3 = 2 - = 1.75 4 0.0625 = 1.732 c4 = 1.75 3.5 c2 = 1 -

c5 = 1.732 -

-2 ⋅ 10-4 = 1.732 3.464

250 < 16, let c1 = 15.

-25 = 15.833 30 0.68389 = 15.811 c3 = 15.833 31.666 -0.0123 = 15.811 c4 = 15.811 31.622 c2 = 15 -

Since c3 = c4 = 15.811, to the nearest thousandth,

Since c4 = c5 = 1.732, to the nearest thousandth,

1.3 ⋅ 10-4 = 3.873 7.746

250 = 15.811.

946 26.

Chapter 12 SEQUENCES AND SERIES 300 is a solution of x 2 - 300 = 0.

Since 17 <

29.

f ( x) = x3 - 100

f ¢( x) = 2 x

f ¢( x ) = 3 x 2

Since 4 < 3 100 < 5, let c1 = 4.

300 < 18, let c1 = 17.

-36 = 4.75 48 7.1719 = 4.644 c3 = 4.75 67.688 0.15592 = 4.642 c4 = 4.644 64.7 0.02658 = 4.642 c5 = 4.642 64.644

-11 = 17.324 34 0.12098 = 17.321 c3 = 17.324 34.648 0.01704 = 17.321 c4 = 17.321 34.642

c2 = 4 -

Since c3 = c4 = 17.321, to the nearest

27.

100 is a solution of x3 - 100 = 0.

f ( x) = x 2 - 300

c2 = 17 -

thousandth,

300 = 17.321.

Since c4 = c5 = 4.642, to the nearest thousandth, 3 100 = 4.642.

3 9 is a solution of x3 - 9 = 0.

30.

f ( x) = x 3 - 9

3 121 is a solution of x3 - 121 = 0.

f ( x) = x3 - 121

f ¢( x ) = 3 x 2

Since 2 < 3 9 < 3, let c1 = 2.

Since 4 < 3 121 < 5, let c1 = 4.

-1 = 2.083 12 0.03791 = 2.080 c3 = 2.083 13.017 -0.0011 = 2.080 c4 = 2.080 12.979 c2 = 2 -

-57 = 5.188 48 18.637 = 4.957 c3 = 5.188 80.746 0.80266 = 4.946 c4 = 4.957 73.716 -0.0064 = 4.946 c5 = 4.946 73.389 c2 = 4 -

Since c3 = c4 = 2.080, to the nearest thousandth, 3 9 = 2.080. 28.

Since c4 = c5 = 4.946, to the nearest

3 15 is a solution of x3 - 15 = 0.

thousandth, 3 121 = 4.946.

f ( x) = x3 - 15 f ¢( x) = 3x 2

31.

Since 2 < 3 15 < 3, let c1 = 2.

f ( x) = x3 - 3x 2 - 18x + 4 f ¢( x) = 3x 2 - 6 x - 18

-7 = 2.583 12 2.2335 = 2.471 c3 = 2.583 20.016 0.08753 = 2.466 c4 = 2.471 18.318 -0.0039 = 2.466 c5 = 2.466 18.243 c2 = 2 -

Since c4 = c5 = 2.466, to the nearest

To find critical points, solve f ¢( x) = 3x 2 - 6 x - 18 = 0. f  ( x) = 6 x - 6 f ¢(-2) = 6 > 0 and

f ¢(-1) = -9 < 0 so

a solution exists in (-2, - 1). Let c1 = -2.

thousandth, 3 15 = 2.466.

Section 12.6

947 f ¢(0) = -6 < 0 and f ¢(1) = 15 > 0 so a

6 = -1.67 -18 0.3867 c3 = -1.67 = -1.65 -16.02 0.0675 c4 = -1.65 = -1.65 -15.9 c2 = -2 -

solution exists in (0, 1). Let c1 = 0. -6 = 0.33 18 0.2667 = 0.32 c3 = 0.33 19.98 0.0672 = 0.32 c4 = 0.32 19.92 c2 = 0 -

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = -1.65. Since f (-1.65) = -15.9 < 0, the graph has a relative maximum at x = -1.65.

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = 0.32. Since f (0.32) = 19.92 > 0, the graph has a relative minimum at x = 0.32.

f ¢(3) = -9 < 0 and f ¢(4) = 6 > 0 so a

solution exists in (3, 4). Let c1 = 3. -9 = 3.75 12 1.6875 = 3.65 c3 = 3.75 16.5 0.0675 = 3.65 c4 = 3.65 15.9 c2 = 3 -

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = 3.65. Since f (3.65) = 15.9 > 0, the graph has a relative minimum at x = 3.65. 32.

33.

f ( x) = x 4 - 3 x 3 + 6 x - 1 f ¢( x ) = 4 x 3 - 9 x 2 + 6

To find critical points, solve f ¢( x) = 4 x3 - 9 x 2 + 6 = 0. f ( x) = 12 x 2 - 18x f ¢(-1) = -7 < 0 and

f ¢(0) = 6 so a

solution exists in (-1, 0). Let c1 = -1.

f ( x) = x 3 + 9 x 2 - 6 x + 4

-7 = -0.77 30 -1.162 = -0.71 c3 = -0.77 20.975 0.03146 = -0.71 c4 = -0.71 18.829

f ¢( x) = 3x 2 + 18 x - 6

c2 = -1 -

To find critical points, solve f ¢( x) = 3x 2 + 18x - 6 = 0. f ( x) = 6 x + 18 f ¢(-7) = 15 > 0 and f ¢(-6) = -6 < 0 so a solution exists in (-7, - 6).

Let c1 = -7. 15 = -6.38 -24 1.2732 c3 = -6.38 = -6.32 -20.28 0.0672 c4 = -6.32 = -6.32 -19.92 c2 = -7 -

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = -6.32. Since f (-6.32) = -19.92 < 0, the graph has a relative maximum at x = -6.32.

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = -0.71. Since f (-0.71) = 18.829 > 0, the graph has a relative minimum at x = -0.71. f ¢(1) = 1 > 0 and f ¢(1.5) = -0.75 < 0 so a

solution exists in (1, 1.5). Let c1 = 1. 1 = 1.17 -6 0.08635 c3 = 1.17 = 1.19 -4.633 -0.00043 c4 = 1.19 = 1.19 -4.427 c2 = 1 -

948

Chapter 12 SEQUENCES AND SERIES Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = 1.19. Since f (1.19) = -4.427 < 0, the graph has a relative maximum at x = 1.19. f ¢(1.5) = -0.75 < 0 and f ¢(2) = 2 > 0 so a

solution exists in (1.5, 2).

Subsequent approximations will agree with c3 and c4 to the nearest hundredth. Thus, x = 0.75. Since f (0.75) = 15.75 > 0, the graph has a relative minimum at x = 0.75. 35.

f ¢( x) =

Let c1 = 1.5. c2 = 1.5 -

-0.75 0

-1 f (c1) = 0=3 f ¢(c1) 0.3333

c3 = c2 -

f (c2 ) 1.2599 = 3= -3 f ¢(c2 ) 0.2100

c4 = c3 -

f (c3 ) -1.5874 = -3 =9 f ¢(c3 ) 0.1323

c5 = c4 -

f (c4 ) 2 = 9= -15 f ¢(c4 ) 0.0833

because the derivative, f ¢( x) = 13 ( x - 1)-2/3 , is undefined at x = 1; the function has a vertical tangent line there.

f ( x) = x + 2 x - 5 x + 2 f ¢( x ) = 4 x 3 + 6 x 2 - 5

36.

To find critical points, solve f ¢( x) = 4 x3 + 6 x 2 - 5 = 0.

Break-even occurs when R( x) = C ( x), so solve R( x) - C ( x) = 0. f ( x) = R( x) - C ( x) = 10 x 2/3 - (2 x - 9)

f ( x) = 12 x 2 + 12 x

= 10 x 2/3 - 2 x + 9

f ¢(0) = -5 < 0 and f ¢(1) = 5 > 0 so a

solution exists in (0, 1).

f ¢( x ) =

20 -1/3 x -2 3

Graphing the function f ( x) on a graphing calculator will help locate a value for c1 . With a little trial-and-error, you should see that f ( x) intersects the x-axis around 138.

Let c1 = 0. -5 0

Since c2 is undefined, let c1 = 0.5. -3 = 0.83 9 1.4205 = 0.75 c3 = 0.83 18.227 0.0625 = 0.75 c4 = 0.75 15.75 c2 = 0.5 -

c2 = c1 -

Not only are successive approximations alternating in sign but they are moving further and further apart. Thus, the approximations are not approaching any specific value. The method fails in this case

Subsequent approximations will agree with c5 and c6 to the nearest hundredth. Thus, x = 1.77. Since f (1.77) = 5.7348 > 0, the graph has a relative minimum at x = 1.77.

c2 = 0 -

f (2) = 1 > 0 so a

Let c1 = 0.

-0.656 = 1.94 c2 = 1.6 1.92 1.3331 = 1.81 c3 = 1.94 10.243 0.23406 = 1.78 c4 = 1.81 6.7332 0.04341 = 1.77 c5 = 1.78 5.9808 -0.0152 = 1.77 c6 = 1.77 5.7348

34.

1 ( x - 1)-2/3 3

f (0) = -1 < 0 and solution exists in (0, 2).

Since c2 is undefined, let c1 = 1.6.

f ( x) = ( x - 1)1/3

f (138) = 0.0460 > 0 and f (139) = -0.6655 < 0 so a solution exists in (138, 139). Let c1 = 138.

Section 12.6

949

c2 = c1 -

f (c1) 0.0460 = 138 = 138.06 -0.7099 f ¢(c1)

c3 = c2 -

f (c2 ) 0.0034 = 138.06 = 138.06 -0.7101 f ¢(c2 )

Subsequent approximations will agree with c2 and c3 to the nearest hundredth. Thus, x = 138.06. The break-even point for the product is 138.06 units.

37.

The process should be used at least until the savings produced is equal to the increased costs incurred, costs incurred, or when S ( x) = C ( x) . Therefore, solve S ( x) - C ( x). f ( x ) = S ( x ) - C ( x) = ( x3 + 5 x 2 + 9) - ( x 2 + 40 x + 20) = x3 + 4 x 2 - 40 x - 11 f ¢( x) = 3x 2 + 8 x - 40 f ¢( x) = 3x 2 + 8 x - 40 f (4) = -43 < 0 and f (5) = 14 > 0 so a solution exists in (4, 5).

Let c1 = 4. -43 = 5.08 40 20.122 c3 = 5.08 = 4.82 78.059 1.1098 c4 = 4.82 = 4.80 68.257 -0.248 c5 = 4.80 = 4.80 67.52 c2 = 4 -

Subsequent approximations will agree with c4 and c5 to the nearest hundredth. Thus, x = 4.80. The process should be used for at least 4.80 years.

38.

(a)

f (i) = f ¢(i) = = =

1 - (1 + i)-n P i M i [n(1 + i)-n-1] - [1 - (1 + i)-n ] i2 ni (1 + i)-n-1 - 1 + (1 + i)-n i2 -1 + ni (1 + i)-n-1 + (1 + i)-n i2

950

Chapter 12 SEQUENCES AND SERIES (b)

é 1 - (1 + i)-n f (i) P ùú éê -1 + ni (1 + i)-n-1 + (1 + i)-n ùú = êê ¸ ú f ¢(i) i M úú êê i2 êë úû û ë

= =

(c)

M - M (1 + i)-n - Pi i2 ⋅ Mi -1 + ni(1 + i)-n-1 + (1 + i)-n Mi - Mi (1 + i)-n - Pi 2 M [-1 + ni (1 + i)-n-1 + (1 + i)-n ]

197(0.01) - 197(0.01)(1 + 0.01)-24 - 4000(0.01) 2 0.018494729 = 0.01 -24-1 -24 ù é 4.982020913 197 ê -1 + (24)(0.01)(1 + 0.01) + (1 + 0.01) úû ë = 0.013712295

i2 = 0.01 -

(d) i3 = 0.01371229 -

197(0.01371229) - 197(0.01371229)(1 + 0.01371229)-24 - 4000(0.01371229) 2

197[-1 + (24)(0.01371229)(1 + 0.01371229)-24-1 + (1 + 0.01371229)-24 ] 0.0010602245 = 0.01371229 -8.803254706 = 0.01383273

39.

i2 = 0.02 -

57(0.02) - 57(0.02)(1 + 0.02)-12 - 600(0.02)2

57[-1 + (12)(0.02)(1 + 0.02)]-12-1 + (1 + 0.02)-12 ] 0.0011177798 = 0.02 -1.480804049 = 0.02075485

i3 = 0.02075485 = 0.02075485 -

57(0.02075485) - 57(0.02075485)(1 + 0.02075485)-12 - 600(0.02075485)2 57[-1 + (12)(0.02075485)(1 + 0.02075485)-12-1 + (1 + 0.02075485)-12 ] 4.0638916 ⋅ 10-6 -1.583924743

= 0.02075742

40.

i2 = 0.01 -

337(0.01) - 337(0.01)(1 + 0.01)-60 - 15, 000(0.01) 2

337[-1 + (60)(0.01)(1 + 0.01)-60-1 + (1 + 0.01)-60 ] 0.0149847943 = 0.01 -41.29955632 = 0.01036283

i3 = 0.01036283 = 0.01036283 -

337(0.01036283) - 337(0.01036283)(1 + 0.01036283)-60 - 15, 000(0.01036283) 2 337[-1 + (60)(0.01036283)(1 + 0.01036283)-60-1 + (1 + 0.01036283)-60 ] 1.128666375 ⋅ 10-4 -43.73087288

= 0.01036541

Section 12.7

12.7

951 Your Turn 3

L’Hospital’s Rule

x3 . x  0 ln( x + 1)

Your Turn 1

Find lim 3 x 4

Find lim

3x - 12 . x+4-2

First note that lim 3x - 12 = 0 and

First note that lim x3 = 0 and lim ln( x + 1) = 0, so

lim 3 x + 4 - 2 = 0, so the conditions of l’Hospital’s

the conditions of l’Hospital’s rule are satisfied. Differentiate the numerator and denominator.

x 4

x0

x 4

rule are satisfied. Differentiate the numerator and denominator. For f ( x) = 3x - 12, f ¢( x) = 3.

For g ( x) = 3 x + 4 - 2, g ¢ x  

x0

For f ( x) = x3 , f ¢( x) = 3x 2. For g ( x) = ln( x + 1), g ¢( x) =

 ( x + 4)-2/3. 

f ¢( x ) 3x 2 = lim 1 x  0 g ¢( x) x0 x +1 lim

f ¢( x ) 3 = lim x  4 g ¢( x ) x  4 (1/3)( x + 4)-2/3

= lim [3x 2 ( x + 1)]

lim

x0

(1/3)(8)-2/3 = 36

By l’Hospital’s rule, lim 3 x 4

By l’Hospital’s rule, lim ln (xx +1) = 0. x0

3x - 12 = 36. x+4-2

Your Turn 4

Find lim Your Turn 2

Find lim

e3x - 92 x 2 - 3x - 1 x3

x0

e x -1 - 1

x 1 x 2 - 2 x + 1

First note that lim e x 1

x0

and lim x = 0, so the conditions of l’Hospital’s rule are 2

- 1 = 0 and lim x - 2 x + 1 x 1

= 0, so the conditions of l’Hospital’s rule are satisfied. Differentiate the numerator and denominator.

For g ( x) = x 2 - 2 x + 1, g ¢( x) = 2 x - 2. f ¢( x) e x -1 lim = lim x 1 g ¢( x) x 1 2 x - 2 The numerator has limit l but the denominator has limit 0, so the limit of the quotient does not exist. e x -1 - 1

x 1 x 2 - 2 x + 1

x0

satisfied. Differentiate the numerator and denominator. For f ( x) = e3x -

9 2 x - 3x - 1, 2

f ¢( x) = 3e3x - 9 x - 3.

For f ( x) = e x -1 - 1, f ¢( x) = e x -1.

By l’Hospital’s rule, lim

First note that lim e3x - 92 x 2 - 3x - 1 = 1 - 1 = 0,

x -1

1 . x +1

For g ( x) = x3 , g ¢( x) = 3x 2. f ¢( x ) 3e3x - 9 x - 3 = lim x  0 g ¢( x) x0 3x 2 lim

Both the numerator and denominator have limit equal to 0, so we apply l’Hospital’s rule again with f ( x) = 3e3x - 9 x - 3 and

does not exist.

g ( x) = 3x 2 .

f ¢( x) = 9e3x - 9 g ¢( x ) = 6 x

We see that both f ¢ and g ¢ still have limit 0 as x  0, so we need one more application of l’Hospital’s rule, now with f ( x) = 9e3x - 9 and g ( x) = 6 x.

952

Chapter 12 SEQUENCES AND SERIES f ¢( x) = 27e3x g ¢( x ) = 6

Thus xlim 0

e3x - 92 x 2 - 3x - 1 x

= lim

x0

27e 6

12.7

Warmup Exercises

W1.

3x 2 + 4 x + 5 -1/ 2 ö æ1 ÷÷ f ¢( x) = ( 6 x + 4 )çç 3x 2 + 4 x + 5 ÷ø÷ èç 2 f ( x) =

(

27 6 9 = . 2 =

Your Turn 5

f ( x) = ( ln(5 x + 3) )

d ln(5 x + 3) dx 5 = (2) ( ln(5x + 3) ) 5x + 3 10 ln(5 x + 3) = 5x + 3

x  0+

Write this as lim

ln(3x) 1 x2

Now both numerator and denominator become infinite in magnitude as x  0, so we can apply l’Hospital’s rule, differentiating the numerator and the denominator. ln(3x )

lim

x  0+

1 x2

= lim

x  0+ -2/x

Your Turn 6 x ¥ e

and lim

ln x

x ¥ x 2

True

False. L’Hospital’s rule gives a quicker way to decide whether a quotient with the indeterminate form 0/0 has a limit.

False. In order to use L’Hospital’s rule, first be sure that lim [ f ( x) / g ( x)] leads to the x a

indeterminate form 0/0 or ¥ / ¥.

x  0+

True

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives

3x 2 + 2 x - 1 3(1) 2 + 2(1) - 1 = = 4. 2x - 1 2(1) - 1 x 1 lim

Each expression is a quotient in which both the numerator and denominator tend to infinity as x  ¥, so we can apply l’Hospital’s rule. lim

ln x

x ¥ e

lim

ln x

x ¥ x 2

= lim

1/x

x ¥ e x

= lim

1/x

x ¥ 2 x

Exercises

Thus lim x 2 ln(3x) = 0.

ln x

12.7

3/x

3 = lim - x 2 = 0 + 2 x0

Find lim

3x 2 + 4 x + 5

f ¢( x) = (2) ( ln(5x + 3) )

Find lim x 2 ln(3x).

x  0+

3x + 2

W2.

)

= lim

x ¥ xe x

= lim

x ¥ 2 x 2

By l’Hospital’s rule,

= 0.

lim

x 1

= 0.

x3 + x 2 - x - 1 x2 - x

= 4.

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives 3x 2 + 2 x - 11 3(3)2 + 2(3) - 11 22 lim . = = 2x - 3 2(3) - 3 3 x3 By l’Hospital’s rule, lim

x3

x3 + x 2 - 11x - 3 2

x - 3x

22 . 3

Section 12.7 7.

lim

953

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives 5x 4 - 6 x 2 + 8x

x  0 40 x 4 - 4 x + 5

5(0)4 - 6(0)2 + 8(0) 40(0) 4 - 4(0) + 5

= 0.

By l’Hospital’s rule, lim

x5 - 2 x3 + 4 x 2

x  0 8 x5 - 2 x 2 + 5 x

12.

= 0.

48x5 + 12 x3 - 9

x  0 63x 6 - 8 x3 + 3x 2

which does not exist.

8 x 6 + 3x 4 - 9 x

x  0 9 x 7 - 2 x 4 + x3

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives 2e 2 x 2e 2(0) = = -2. 10(0) - 1 x  0 10 x - 1

By l’Hospital’s rule, lim

13.

does not exist.

lim

e x + xe x ex

e 0 + 0 ⋅ e0 e0

= 1.

xe x lim x = 1. x 0 e - 1

14.

By l’Hospital’s rule, ln ( x - 1) = 1. x 2 x - 2 lim

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives e-x - xe-x

lim

x 0

10.

By l’Hospital’s rule,

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives 1

= -2.

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives

x 0

1 lim x -1 = = 1. 2 -1 x 2 1

e2 x - 1

x 0 5x2 - x

By l’Hospital’s rule, lim

does not exist.

lim

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives lim

x 0

By l’Hospital’s rule, lim

ex - 1

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives

e0 - 0 ⋅ e 0

2(0)

1 . 4

By l’Hospital’s rule, xe-x

lim

x  0 2e 2 x - 2

1 . 4

1 lim x +1 = = 1. 0 +1 x 0 1

15.

However,

By l’Hospital’s rule, ln ( x + 1) lim = 1. x x 0

11.

x  0 2 x3 + 9 x 2 - 11x

16.

which does not exist. 3

does not exist.

lim e x = 1 and l’Hospital’s rule does not apply.

x0

However lim

x 0 4 x

lim

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives lim

lim e x = 1 and l’Hospital’s rule does not apply.

x0

x  0 8 x5 - 3x 4

does not exist.

954 17.

Chapter 12 SEQUENCES AND SERIES The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives lim

x0

1 2(2 + x)1/2

21.

2 or . = = 1/2 4 2 2 2(2)

2/3

lim 3x x 8 1

x0

18.

2+x - 2 1 2 = or . 4 x 2 2

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives lim

x0

1 2(9 + x)1/ 2

lim

x 8

22.

19.

1/ 2

lim

20.

1 1/ 2

2⋅4

lim

3x -2

x-8

3x -3

x  27 x - 27

23.

1 . 4

1 . 12

1 . 27

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives 9 x8 + 24 x 7 + 20 x 4 9 + 24 + 20 = = 53. 1 1 x 1 lim

By l’Hospital’s rule,

x -2 1 = . 4 x-4

x 9 + 3 x8 + 4 x 5 - 8 = 53. x -1 x 1

lim

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives 1 1/ 2 lim 2 x x 9 1

1 . 12

2/3 1 1 lim 3x . = = 27 x 27 1 3 ⋅ 27 2/3

By l’Hospital’s rule, x 4

By l’Hospital’s rule,

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives lim 2 x x 4 1

3⋅8

2/3

1 = . 1/ 2 6 2(9 + 0)

9+ x -3 1 = . 6 x

x0

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives

By l’Hospital’s rule, lim

By l’Hospital’s rule,

By l’Hospital’s rule, lim

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives

1 = = . 6 2 ⋅ 91/ 2

By l’Hospital’s rule, x -3 1 lim = . 6 x9 x - 9

24.

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives 7 x 6 - 30 x5 + 25 x 4 = 448 - 960 + 400 1 x 2 = -112. lim

By l’Hospital’s rule, x 7 - 5x 6 + 5x5 + 32 = -112. x-2 x 2 lim

Section 12.7 25.

955

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives

29.

e x - e-x lim = e0 - e0 = 0. 1 x0

lim

e +e x x0

-2

= 0.

lim

e +1

lim

x  0 -e-x - 1

2x 2 9(1+ x)5/3

x0

0 . 0

e +1 -e0 - 1

lim

2 - 1. -2

e -1+ x

Find lim

1 9(1 + 0)

5/3

1 . 9

2e5 x - 25x 2 - 10 x - 2 5 x3

x0

= -1.

x  0 e-x - 1 - x

x  0 (1 + x)5/3

1 + 13 x - (1 + x)1/3

x0

30.

By l’Hospital’s rule,

By l’Hospital’s rule, lim

2⋅0

= lim =

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives x

1 1 3 3(1+ 0) 2/3

Using l’Hospital’s rule a second time, gives -x

lim

26.

1 1 3 3(1+ x)2/3

x0

By l’Hospital’s rule, x

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives

lim 2e5 x - 25 x 2 - 10 x - 2 = 2 - 2 = 0, and

x0

27.

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives lim

x3

x ( x + 7)1/2 2

= lim =

1 = . 2 1/2 8 2(3 + 7)

By l’Hospital’s rule, x2 + 7 - 4

lim

x2 - 9

x3

28.

lim

x5

Both derivatives still have limit equal to 0, so we apply l’Hospital’s rule again with

f ¢( x) = 50e5 x - 50 g ¢( x) = 30 x

We see that both f ¢ and g ¢ still have limit 0 as x  0, so we need one more application of l’Hospital’s rule, now with f ( x) = 50e5 x - 50 and g ( x) = 30 x. f ¢( x) = 250e5 x g ¢( x) = 30

x  5 2( x 2 + 11)1/2

1 2

1/2

2(5 + 11)

x 2 + 11 - 6 2

x - 25

1 . 12

Thus lim

By l’Hospital’s rule,

x5

For g ( x) = 5 x3, g ¢( x) = 15 x 2.

= lim =

lim

f ¢( x) = 10e5 x - 50 x - 10.

f ( x) = 10e5 x - 50 x - 10 and g ( x) = 15x 2.

1 = . 8

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives x ( x 2 +11)1/2

Differentiate the numerator and denominator. For f ( x) = 2e5 x - 25 x 2 - 10 x - 2,

x  3 2( x 2 + 7)1/2

lim 5 x3 = 0, so l’Hospital’s rule applies.

x0

2e5 x - 25x 2 - 10 x - 2

x0

1 . 12

5 x3

250e5 x x  0 30 250 25 = = . 30 3 = lim

956 31.

35.

1 1 + 2(1+ x)1/ 2 2(1- x)1/2

lim

æ ö÷ 1 1 = lim ççç + ÷÷ x  0 çè 2(1 + x)1/2 2(1 - x)1/2 ÷÷ø 1

x ln ( x + 1) + 5x+ +1

1/2

2(1 + 0)

1 1/2

2(1 - 0)

= lim =

= 1.

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives

36.

2⋅3

1 1/2

2⋅3

= lim

lim

33.

lim

x0

-e-x (1 - x) -(1 - 0)ln (1 - 0) - 7 + 0

-e0 (1 - 0)

34.

37.

lim x 2 + 5 x + 9 = 15 and l’Hospital’s rule

x 1

(ln x)2 1 x2

Now both the numerator and denominator become infinite and l’Hospital’s rule applies to the limit of the form ¥/¥. Differentiate the numerator and denominator. lim

does not apply. However

= 7.

We have a limit of the form 0 ´ ¥. Rewrite the expression.

x 1

lim

e-x - 1

x0

does not apply. However,

x0

(7 - x)ln (1 - x)

x 2 (ln x) 2 =

x 2 - 5x + 4 does not exist. x

By l’Hospital’s rule, lim

x 2 - 5 x + 4 = 2 and l’Hospital’s rule

lim

-e-x -(1 - x)ln (1 - x) - 7 + x

x0

1 3 or . 3 3

3- x - 3+ x 1 3 =or . x 3 3

= 5.

-x -ln(1 - x) - 71x

x0

By l’ Hospital’s rule,

x0

= 5.

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives lim

1 æ ö÷ 1 -1 = lim ççç ÷÷ x  0 çè 2(3 - x)1/2 2(3 + x)1/2 ÷ø÷ 1/2

ex - 1

x0

-1

(5 + x)ln ( x + 1)

lim

-1 -1 2(3- x)1/2 2(3+ x)1/2

e0 (0 + 1)

By l’Hospital’s rule,

1+ x - 1- x lim = 1. x x0

lim

e x ( x + 1) (0 + 1)ln (0 + 1) + 5 + 0

x0

By l’Hospital’s rule,

32.

ex ( x + 1) ln ( x + 1) + 5 + x

x0

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives

x  0+

x2 + 5x + 9 does not exist. x -1

(ln x )2 1 x2

2(ln x ) x -2 x  0+ x3 2

= lim

= lim - x ln x x  0+

This problem is similar to what we started with so we handle it in the same manner.

Section 12.7

957 - ln x

lim - x 2 ln x = lim

x0

x  0+ e

Therefore, by l’Hospital’s rule, lim x ln(e x - 1) = 0.

x  0+

lim x 2 (ln x)2 = 0.

x  0+

40.

We have a limit of the form 0 ´ ¥. Rewrite the expression. e1/x

lim

x  0+

1 x

This problem is similar to what we started with so we handle it in the same manner.

( )

2(ln x) = lim x x x ¥ x ¥ 1 2 = lim =0 x ¥ x lim

x -1 x2

x  0+

= lim e1/x = ¥ x  0+

Therefore, by l’Hospital’s rule,

Therefore, by l’Hospital’s rule, 1/x

lim xe

x  0+

(ln x)2 = 0. x x ¥ lim

= ¥.

41. 39.

We have a limit of the form 0 ´ ¥. Rewrite the expression. x ln(e x - 1) =

lim

x0

1 x

Notice that

The limit in the numerator is ¥, as in the limit in the denominator, so l’Hospital’s rule applies. Taking derivatives separately in the numerator and the denominator gives

(ln e x - 1)

1 x

e (e x -1) = lim -1 x  0+ x2

x (2 x ) = lim x ¥ ln(ln x) x ¥ 1 ( x ln x) lim

Now both the numerator and denominator become infinite and l’Hospital’s rule applies to the limit of the form ¥/¥. Differentiate the numerator and denominator. ln(e x - 1)

2(ln x) x

1 2(ln x) = lim x x ¥

e1/x -21

= lim

The limit in the numerator is ¥, as in the limit in the denominator, so l’Hospital’s rule applies. Taking derivatives separately in the numerator and the denominator gives (ln x)2 = lim lim x x ¥ x ¥

Now both the numerator and denominator become infinite and l’Hospital’s rule applies to the limit of the form ¥/¥. Differentiate the numerator and the denominator. e1/x

-1

= 0 ⋅1 = 0

Therefore, by l’Hospital’s rule,

xe1/x =

æ 1 ö÷ = lim -x 2 ⋅ çç1 + x ÷ + ç è x0 e - 1 ÷ø

x = lim =0 + 2 x0

38.

-x 2 e x

lim

1 x0 x2 -1 = lim -x2 x  0+ 3 x 2

= lim

x0

-x 2e x x

e -1

= lim

x ¥

x ln x = ¥. 2

By l’Hospital’s rule, lim

x ¥ ln(ln x )

ex = 1 + 1 . e x -1 e x -1

= ¥ (does not exist).

958 42.

Chapter 12 SEQUENCES AND SERIES The limit in the numerator is ¥, as is the limit in the denominator, so l’Hospital’s rule applies. Taking derivatives separately in the numerator and the denominator gives x

lim

x ¥

lim

x  0+ 5(e

e x

This problem is similar to what we started with so we handle it in the same manner. In fact, perform the same steps five times. e x (2 x ) = lim 5/2 3/2 x ¥

e x

= lim

x ¥

e x (2 x )

60 x

= lim

x ¥

e x (2 x )

360

= lim

x ¥

720

e x (2 x ) = lim 1440 x ¥ (2 x )

44.

e x

x ¥

43.

e x x3

e x x ¥ 720 x e x x ¥ 1440 x

x0

e x =¥ x ¥ 1440

x0

=1+

1 4e

x -1

3(4e x - 1)

1 4e x . lim 3 x  0+ (4e x - 1)

æ 1 1 ÷÷ö ⋅ lim ççç1 ÷÷ + 3 x  0 çè 4e x - 1 ø

1 1 ⋅1 = 3 3

Therefore, by l’Hospital’s rule, ln (4e x - 1) 1 = . + 3 3 x x0 lim

45.

We have a limit of the form ¥ ⋅ 0. Rewrite the expression as a quotient.

x5e-0.001x =

5(e + 1)

1 = 1- x . e +1 e +1 x

x -1

4e x

lim

= lim

4e x

Notice that

= lim

4e x

x  0+ 3(4e x - 1)

= lim

ln(e x + 1) (e x +1) lim = lim 5x 5 x  0+ x  0+

2e x x (4e x -1) 3 (2 x )

= lim

= ¥ (does not exist).

The limit in the numerator is ¥, as is the limit in the denominator, so l’Hospital’s rule applies. Taking derivatives separately in the numerator and the denominator gives

e x x ¥ 360 x

The limit in the numerator is ¥, as is the limit in the denominator, so l’Hospital’s rule applies. Taking derivatives separately in the numerator and the denominator gives

Notice that

1 1 ⋅1 = 5 5

= lim

Therefore, by l’Hospital’s rule, lim

ln(4e x - 1) lim = lim 3 x x  0+ x  0+

x ¥ 120 x3/2

e (2 x )

æ 1 1 ö÷ ÷÷ ⋅ lim çç1 - x + ç 5 x0 è e + 1 ÷ø

ln (e x + 1) 1 = . + 5 x 5 x0

e x

= lim

e x (2 x ) = lim x ¥ 180 x1/2

lim

x ¥ 30 x 2

15 x

1 ex . lim x 5 x  0+ e + 1

Therefore, by l’Hospital’s rule,

x ¥ 6 x5/2

x ¥ 6 x

+ 1)

e x (2 x ) = lim 2 x ¥

= lim

lim

x5 e0.001x

The limit in the numerator is ¥, as is the limit in the denominator, so l’Hospital’s rule applies. Taking derivatives separately in the numerator and the denominator gives

Section 12.7 lim

959 x5

x ¥ e

5x 4

= lim 0.001x

x ¥ 0.001e0.001x

47.

This problem is similar to what we started with so we handle it in the same manner. In fact, continue the process four more times. 5x4

lim

x ¥ 0.001e

Rewrite over a common denominator. æ e x - 1 - x ö÷ ç ÷÷ lim çç x  0 çè ÷÷ø x2

20 x3

= lim 0.001x

x ¥ (0.001) 2 e0.001x

60 x

= lim

The numerator and denominator have limit 0, so l’Hospital’s rule applies. Differentiate the numerator and denominator.

x ¥ (0.001)3 e0.001x

æ e x - 1 - x ÷ö æ x ö ç ÷÷ = lim çç e - 1 ÷÷÷ lim çç ç ÷÷ x  0 èç x  0 çè 2 x ÷÷ø x2 ø

120 x

= lim

x ¥ (0.001) 4 e0.001x

120

= lim

The numerator and denominator still have limit 0, so differentiate the numerator and denominator again.

x ¥ (0.001)5 e0.001x

æ e x - 1 ö÷ æ xö ç ÷÷ = lim çç e ÷÷÷ = 1 lim çç ç 2 x  0 çè 2 x ÷÷ø x  0 çè 2 ø÷÷

Therefore, by l’Hospital’s rule, lim x5e-0.001x = 0.

x ¥

46.

Thus æ ex 1 1 ö÷ 1 ç lim çç 2 - 2 - ÷÷÷ = . x ø÷ 2 x  0 èç x x

The limit in the numerator is ¥, as is the limit in the denominator, so l’Hospital’s rule applies. Taking derivatives separately in the numerator and the denominator gives 3x 2 x ¥ x ln x x ¥ 2 x ln x + x x(3x) = lim x ¥ x(2ln x + 1) 3x = lim x +1 2ln x ¥ lim

x3 + 1 2

= lim

This problem is similar to what we started with so we handle it in the same manner. lim

x ¥ 2ln x + 1

= lim

x ¥ 2 x

= lim

x ¥ 2

=¥

Therefore, by l’Hospital’s rule, lim

x3 + 1

x ¥ x 2 ln x

æ ex 1 1 ö÷ ç Find lim çç 2 - 2 - ÷÷÷. x ø÷ x  0 çè x x

= ¥ (does not exist).

48.

æ 12e x 12 12 6 ö÷ ç Find lim çç 3 - 3 - 2 - ÷÷÷. x ÷ø x  0 çè x x x

Rewrite over a common denominator. æ 12e x - 12 - 12 x - 6 x 2 ö÷ ç ÷÷ lim çç ÷÷ x  0 çè x3 ø

The numerator and denominator have limit 0, so l’Hospital’s rule applies. Differentiate the numerator and denominator. æ 12e x - 12 - 12 x - 6 x 2 ö÷ ç ÷÷ lim çç x  0 çè ÷÷ø x3 æ 12e x - 12 - 12 x ö÷ ç ÷÷ = lim çç x  0 çè ÷÷ø 3x 2

The numerator and denominator still have limit 0, so differentiate the numerator and denominator again. x æ 12e x - 12 - 12 x ö÷ æ ö ç ÷÷ = lim çç 12e - 12 ÷÷÷ lim çç ç ÷÷ ÷÷ 6x x  0 èç x  0 èç 3x 2 ø ø

960

Chapter 12 SEQUENCES AND SERIES The numerator and denominator still have limit 0, so differentiate the numerator and denominator again. æ 12e x - 12 ö÷ æ 12e x ö÷ 12 ç ç = 2 lim çç ÷÷÷ = lim çç ÷÷÷ = 6x 6 x  0 èç x  0 èç 6 ø÷ ø÷

Thus æ 12e x 12 12 6 ö÷ ç lim çç 3 - 3 - 2 - ÷÷÷ = 2. x ÷ø x  0 çè x x x

49.

æ x 1 ÷ö Find lim çç ÷. ln x ÷ø x 1 çè x - 1

Rewrite over a common denominator:

æ x ln x - x + 1 ö÷ ÷÷. lim ççç x 1 è ( x - 1) ln( x) ø÷

The numerator and denominator have limit 0, so l’Hospital’s rule applies. Differentiate the numerator and denominator. æ ö÷ çç æ x ln x - x + 1 ÷ö æ ö ÷÷÷ ln x x ln x ç ç ÷ = lim çç lim ç ÷÷ = lim çç ÷÷÷ ÷ ç ç ÷÷ x 1 è ( x - 1) ln( x ) ÷ø x 1 çç x - 1 x 1 è x - 1 + x ln x ø çè x + ln x ø÷÷

The numerator and denominator still have limit 0, so differentiate the numerator and denominator again. æ ö÷ æ 1 + ln x ö÷ x ln x 1 lim çç ÷÷ = lim ççç ÷= ç 2 x 1 è x - 1 + x ln x ø x 1 è 1 + 1 + ln x ÷ø

Thus æ lim çç

x 1 çè x - 1

50.

1 ö÷ 1 ÷= . ln x ÷ø 2

æ2 ln(1 + 2 x) ö÷ Find lim çç ÷÷. ø x  0 çè x x2 æ 2 x - ln(1 + 2 x) ö÷ Rewrite over a common denominator: lim çç ÷÷ ø x  0 çè x2

The numerator and denominator have limit 0, so l’Hospital’s rule applies. Differentiate the numerator and denominator. æ 2 - 2 ÷ö æ 2 x - ln(1 + 2 x ) ö÷ çç 1+ 2 x ÷ = lim çç lim ÷÷ ÷ çç ÷ø 2 ÷ø 2 x x  0 èç x  0 x èç æ ö æ 2 ÷ö 2x ÷÷ = lim çç = lim çç ÷= 2 ÷ ç ç x  0 è x(1 + 2 x ) ø÷ x  0 è 1 + 2 x ø÷

Thus æ2 ln(1 + 2 x) ö÷ lim çç ÷÷ = 2. ç ø x0 è x x2

Chapter 12 Review 51.

52.

961

lim x 2 + 3 ¹ 0, so l’Hospital’s rule does not apply.

x0

First, verify that the form is one for which l’Hospital’s rule applies. æ ö 3 3 lim çç 2a3 x - x 4 - a a 2 x ÷÷÷ = 2a 4 - a 4 - a a3 = a 2 - a 2 = 0 ø æ ö 4 4 lim ç a - ax3 ÷÷÷ = a - a 4 = a - a = 0 ø x  a çè

x  a çè

The limit in the numerator is 0, as is the limit in the denominator, so l’Hospital’s rule applies. Taking derivatives separately in the numerator and the denominator gives lim

x a

( 2a3 x - x 4 - a 3 a 2 x) 4

a - ax3

= lim

x a

= = =

a3 - 2 x3 - a5/3 33 x 2a 3 x - x 4 2 -3ax 4(ax3 )3/4

4(a 4 )3/4 (a5/3 a 4 + 6a11/3 - 3a11/3 ) 9a(a8/3 ) a 4 4a3 (a11/3 + 6a11/3 - 3a11/3 ) 9a11/3a 2 4a3 (4a11/3 )

9a17/3 16a = 9

Therefore, by l’Hospital’s rule, 3

lim

x a

2a3 - x - x 4 - a a 2 x a - 4 ax3

16a . 9

55. (b) Since s(b) and 1 - b are both 0 at b = 1, we can apply s(b) l’Hospital’s rule and evaluate lim as b 1 1 - b d s(b) s¢(1) db = = -s¢(1). d -1 (1 - b) db

962

Chapter 12 SEQUENCES AND SERIES æ 1 ön-1 an = 128çç ÷÷÷ çè 2 ø

Chapter 12 Review Exercises 1. 2.

True

a (r 5 - 1) S5 = 1 r -1

False: The amounts are a constant arithmetic sequence.

True =

128[(1/2)5 - 1] (1/2) - 1

(

31 128 - 32

)

- 12

True

False: In general the fifth derivatives will be different at 0.

True

False: It converges as long as -1 < r < 1.

True

æ 1 ön-1 an = 27 çç ÷÷÷ çè 3 ø

False: It converges only for x in (-1, 1].

S5 =

10. 11.

15.

æ 1 ö3 = 27 çç ÷÷÷ çè 3 ø

False: Newton’s method may fail to converge by alternating between two values, neither of which is a zero. False: The rule applies to limits of quotients, not derivatives of quotients.

13.

a4 = a1 ⋅ r 3 = 5(-2)3 = -40 an = 5(-2) S5 =

a4 = a1 ⋅ r 3

12.

n -1

a1(r 5 - 1) r -1

5[(-2)5 - 1] = (-2) - 1 5(-33) = -3 = 55

14.

= (8)(31) = 248

a4 = a1 ⋅ r 3 æ 1 ö3 = 128çç ÷÷÷ çè 2 ø

a1(r 5 - 1) r -1 27[(1/3)5 - 1] (1/3) - 1

(

27 - 242 243

)

2 3 242 = 6 121 = 3

16.

a4 = a1 ⋅ r 3 = 2(-5)3 = -250 an = 2(-5) n-1 a (r 5 - 1) S5 = 1 r -1 2[(-5)5 - 1] (-5) - 1 2(-3126) = -6 = 1042 =

= 16

Chapter 12 Review

17.

963

Derivative

Value at 0

2- x

f (0) = e2

f ( x) = e

f (1) ( x) = -e2- x

f (1) (0) = -e2

f (2) ( x) = e 2- x

f (2) (0) = e2

f (3) ( x) = -e 2- x

f (3) (0) = -e2

f (4) ( x) = e 2- x

f (4) (0) = e2

P4 ( x) = f (0) + = e2 +

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4!

-e 2 e2 2 -e 2 3 e2 4 x+ x + x + x 1! 2! 3! 4!

= e2 - e2 x +

e2 2 e2 3 e2 4 x x + x 2 6 24

Derivative

18. f

Value at 0

f ( x) = 5e

f (0) = 5

(1)

( x) = 10e

(0) = 10

f (2) ( x) = 20e 2 x

f (2) (0) = 20

f (3) ( x) = 40e 2 x

f (3) (0) = 40

f (4) ( x) = 80e 2 x

f (4) (0) = 80

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 10 20 2 40 3 80 4 20 3 10 4 =5+ x+ x + x + x = 5 + 10 x + 10 x 2 + x + x 1! 2 6 24 3 3

P4 ( x) = f (0) +

19.

f ( x) = f (1) ( x) =

Derivative

Value at 0

x + 1 = ( x + 1)1/2

f (0) = 1

1 1 ( x + 1)-1/2 = 2 2( x + 1)1/2

1 1 f (2) ( x) = - ( x + 1)-3/2 = 4 4( x + 1)3/2 f (3) ( x) =

3 3 ( x + 1)-5/2 = 8 8( x + 1)5/2

f (4) ( x) = -

15 15 ( x + 1)-7/2 = 16 16( x + 1)7 /2

f (1) (0) =

1 2

f (2) (0) = f (3) (0) =

1 4

3 8

f (4) (0) = -

15 16

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 3 1 - 15 -1 = 1 + 2 x + 4 x 2 + 8 x3 + 16 x 4 1 2 6 24 1 1 2 1 3 5 4 x x =1+ x- x + 2 8 16 128

P4 ( x) = f (0) +

964 20.

Chapter 12 SEQUENCES AND SERIES Derivative

Value at 0

f ( x) = 3 x + 27 = ( x + 27)1/3

f (0) = 3 1 f (1) (0) = 27

1 1 f (1) ( x) = ( x + 27)-2/3 = 3 3( x + 27) 2/3 2 2 f (2) ( x) = - ( x + 27)-5/3 = 9 9( x + 27)5/3 f (3) ( x) =

f (2) (0) = -

10 10 ( x + 27)-8/3 = 27 27( x + 27)8/3

f (4) ( x) = -

f (3) (0) =

80 80 ( x + 27)-11/3 = 81 81( x + 27)11/3

2 2187

10 177,147

f (4) (0) = -

80 14,348,907

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 10 80 1 2 - 14,348,907 - 2187 177,147 3 2 27 = 3+ x+ x + x + x4 1 2 6 24 1 1 5 10 2 3 = 3+ xx + x x4 27 2187 531, 441 43, 046, 721

P4 ( x) = f (0) +

21.

Derivative f ( x) = ln (2 - x)

Value at 0 f (0) = ln 2

1 = -(2 - x)-1 2-x 1 f (2) ( x) = -(2 - x)-2 = (2 - x) 2 2 f (3) ( x) = -2(2 - x)-3 = (2 - x)3 6 f (4) ( x) = -6(2 - x)-4 = (2 - x) 4 f (1) ( x) = -

1 2 1 (2) f (0) = 4 f (1) (0) = -

f (3) (0) = -

1 4

f (4) (0) = -

3 8

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! -3 -1 -1 -1 = ln 2 + 2 x + 4 x 2 + 4 x3 + 8 x 4 1 2 6 24 1 1 2 1 3 1 4 = ln 2 - x - x x x 2 8 24 64

P4 ( x) = f (0) +

22.

Derivative

Value at 0

f ( x) = ln (3 + 2 x) 2 = 2(3 + 2 x)-1 f (1) ( x) = 3 + 2x

f (0) = ln 3 2 f (1) (0) = 3 4 f (2) (0) = 9

f (2) ( x) = -4(3 + 2 x)-2 = f (3) ( x) = 16(3 + 2 x)-3 =

4 (3 + 2 x)2 16 3

(3 + 2 x) 96 f (4) ( x) = -96(3 + 2 x)-4 = (3 + 2 x) 4

f (3) (0) =

16 27

f (4) (0) = -

32 27

Chapter 12 Review

965

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 16 2 -4 - 32 = ln 3 + 3 x + 9 x 2 + 27 x3 + 27 x 4 1 2 6 24 2 2 2 8 3 4 4 x x = ln 3 + x - x + 3 9 81 81

P4 ( x) = f (0) +

23.

Derivative

Value at 0

2/3

f (0) = 1

f ( x) = (1 + x) f (1) ( x) =

2 2 (1 + x)-1/3 = 3 3(1 + x)1/3

f (1) (0) =

2 2 f (2) ( x) = - (1 + x)-4/3 = 9 9(1 + x) 4/3 f (3) ( x) =

f (2) (0) = -

8 8 (1 + x)-7/3 = 27 27(1 + x)7/3

f (4) ( x) = -

2 3

f (3) (0) =

56 56 (1 + x)-10/3 = 81 81(1 + x)10/3

2 9

8 27

f (4) (0) = -

56 81

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 8 2 -2 - 56 = 1 + 3 x + 9 x 2 + 27 x3 + 81 x 4 1 2 6 24 2 1 2 4 3 7 4 =1+ x- x + x x 3 9 81 243

P4 ( x) = f (0) +

24.

Derivative

Value at 0

3/2

f ( x) = (4 + x) 3 f (1) ( x) = (4 + x)1/2 2 3 (2) f ( x) = (4 + x)-1/2 4 3 f (3) ( x) = - (4 + x)-3/2 8 9 (4 + x)-5/2 f (4) ( x) = 16 P4 ( x) = f (0) +

f (0) = 8 f (1) (0) = 3 f (2) (0) =

3 8

3 64 9 f (4) (0) = 512 f (3) (0) = -

f (1) (0) f (2) (0) 2 f (3) (0) 3 f (4) (0) 4 x+ x + x + x 1! 2! 3! 4! 3

-3

3 x + 8 x 2 + 64 x3 + 512 x 4 1 2 6 24 3 2 1 3 3 = 8 + 3x + x x + x4 16 128 4096 =8+

966 25.

Chapter 12 SEQUENCES AND SERIES 2

e x 4 , we can Using the result of Exercise 17, with f ( x) = e2- x and P4 ( x) = e2 - e2 x + e2 x 2 - e6 x3 + 24

approximate e1.93 by evaluating f (0.07) = e2-0.07 = e2-0.07 = e1.93. Using P4 ( x) from Exercise 17 with x = 0.07 gives P4 (0.07) = e 2 - e2 (0.07) +

e2 e2 e2 (0.07)2 (0.07)3 + (0.07) 4 2 6 24

» 6.88951034388.

To four decimal places, P4 (0.07) approximates the value of e1.93 as 6.8895. 26.

Using the result of Exercise 18, with f ( x) = 5e 2 x and P4 ( x) = 5 + 10 x + 10 x 2 + 20 x3 + 10 x 4 , we can 3 3 approximate 5e0.04 by evaluating f (0.02) = 5e2(0.02) = 5e0.04. Using P4 ( x) from Exercise 18 with x = 0.02 gives P4 (0.02) = 5 + 10 (0.02) + 10 (0.02)2 +

20 10 (0.02)3 + (0.02)4 3 3

» 5.20405386667.

To four decimal places, P4 (0.02) approximates the value of 5e0.04 as 5.2041. 27.

Using the result of Exercise 19, with f ( x) =

1 x3 - 5 x 4 , we can x + 1 and P4 ( x) = 1 + 12 x - 18 x 2 + 16 128

approximate 1.03 by evaluating f (0.03) = x = 0.03 gives

0.03 + 1 = 1.03. Using P4 ( x) from Exercise 19 with

1 1 1 5 (0.03) - (0.03) 2 + (0.03)3 (0.03)4 2 8 16 128 » 1.01488915586.

P4 (0.03) = 1 +

To four decimal places, P4 (0.03) approximates the value of 1.03 as 1.0149. 28.

5 1 x - 1 x2 + Using the result of Exercise 20, with f ( x) = 3 x + 27 and P4 ( x) = 3 + 27 x3 531,441 2187 10 - 43,046,721 x 4 , we can approximate 3 26.94 by evaluating f (-0.06) = 3 -0.06 + 27 = 3 26.94.

Using P4 ( x) from Exercise 20 with x = -0.06 gives 1 1 5 10 (-0.06) (-0.06)2 + (-0.06)3 (-0.06) 4 27 2187 531, 441 43, 046, 721 » 2.99777612965.

P4 (-0.06) = 3 +

To four decimal places, P4 (-0.06) approximates the value of 3 26.94 as 2.9978. 29.

1 x3 - 1 x 4 , we Using the result of Exercise 21, with f ( x) = ln (2 - x) and P4 ( x) = ln 2 - 12 x - 18 x 2 - 24 64

can approximate ln 2.05 by evaluating f (-0.05) = ln (2 - (-0.05)) = ln 2.05. Using P4 ( x) from Exercise 21 with x = -0.05 gives 1 1 1 1 (-0.05) - (-0.05)2 (-0.05)3 (-0.05) 4 2 8 24 64 » 0.717842610677 (using ln 2 = 0.69315).

P4 (-0.05) = ln 2 -

To four decimal places, P4 (-0.05) approximates the value of ln 2.05 as 0.7178.

Chapter 12 Review 30.

967

8 x3 - 4 x 4 , we Using the result of Exercise 22, with f ( x) = ln(3 + 2 x) and P4 ( x) = ln 3 + 23 x - 92 x 2 + 81 81

can approximate ln 3.06 by evaluating f (0.03) = f (3 + 2 (0.03)) = f (3 + 0.06) = f (3.06). Using P4 ( x) from Exercise 22 with x = 0.03 gives 2 2 8 4 (0.03) - (0.03)2 + (0.03)3 - (0.03) 4 3 9 81 81 = 1.118412627 (using ln 3 = 1.1184).

P4 (0.03) = ln 3 +

To four decimal places, P4 (0.03) approximates the value of ln 3.06 as 1.1184. 31.

4 x3 - 7 x 4 , Using the result of Exercise 23, with f ( x) = (1 + x)2/3 and P4 ( x) = 1 + 23 x - 19 x 2 + 81 243

we can approximate (0.92)2 / 3 by evaluating f (-0.08) = (1 + (-0.08))2/3 = (0.92)2/3 . Using P4 ( x) from Exercise 23 with x = -0.08 gives 2 1 4 7 (-0.08) - (-0.08)2 + (-0.08)3 (-0.08) 4 3 9 81 243 » 0.945929091687.

P4 (-0.08) = 1 +

To four decimal places, P4 (-0.08) approximates the value of (0.92)2/3 as 0.9459. 32.

3 x 2 - 1 x3 + 3 x 4 , Using the result of Exercise 24, with f ( x) = (4 + x)3/2 and P4 ( x) = 8 + 3x + 16 128 4096

we can approximate 4.023/2 by evaluating f (0.02) = (4 + 0.02)3/2 = 4.023/2. Using P4 ( x) from Exercise 24 with x = 0.02 gives 3 1 3 (0.02) 2 (0.02)3 + (0.02)4 16 128 4096 » 8.06007493762.

P4 (0.02) = 8 + 3(0.02) +

To four decimal places, P4 (0.02) approximates the value of 4.023/ 2 as 8.0601. 33.

9 - 6 + 4 - 83 +  is a geometric series with a = a1 = 9 and r = - 23 . Since r is in (-1, 1), the series

converges and has sum a 9 9 9 27 = = = 5 = . 1- r 5 1 - (- 23 ) 1 + 23 3 34.

2 + 1.4 + .98 + .686 +  is a geometric series with a = a1 = 2 and r = .7. Since r is in (-1, 1), the series converges and has sum a 2 2 20 = = = . 1- r 1 - .7 .3 3

35.

3 + 9 + 27 + 81 +  is a geometric series with a = a1 = 3 and r = 3. Since r > 1, the series diverges.

36.

4 + 4.8 + 5.76 + 6.912 +  is a geometric series with a = a1 = 4 and r = 1.2. Since r > 1, the series diverges.

968 37.

Chapter 12 SEQUENCES AND SERIES 2 - 2 + 2 - 2 +  is a geometric series with a = a = 2 1 5 25 125 625 5

and r = - 15 . Since r is in (-1, 1), the series

converges and has sum 2

a 5 = 1- r 1 - -1

( 5)

38.

2 5

= 56 = 1

1+ 5

1 . 3

1 +  is a geometric series with a = a = 36 and r = 1 . Since r is in (-1, 1), the series 36 + 3 + 14 + 48 1 12

converges and has sum a 36 36 432 . = = 11 = 1 1- r 11 1 - 12 12

39.

40.

1 =1 2(1) - 1 1 1 1 4 S2 = a1 + a2 = + = 1+ = 2(1) - 1 2(2) - 1 3 3 1 1 1 1 1 23 S3 = a1 + a2 + a3 = + + =1+ + = 2(1) - 1 2(2) - 1 2(3) - 1 3 5 15 1 1 1 1 S4 = a1 + a2 + a3 + a4 = + + + 2(1) - 1 2(2) - 1 2(3) - 1 2(4) - 1 1 1 1 176 =1+ + + = 3 5 7 105 1 1 1 1 1 S5 = a1 + a2 + a3 + a4 + a5 = + + + + 2(1) - 1 2(2) - 1 2(3) - 1 2(4) - 1 2(5) - 1 1 1 1 1 563 1+ + + + = 3 5 7 9 315 S1 = a1 =

1 1 = (1 + 2)(1 + 3) 12 1 1 1 1 2 S2 = a1 + a2 = + = + = (1 + 2)(1 + 3) (2 + 2)(2 + 3) 12 20 15 1 1 1 1 1 1 1 S3 = a1 + a2 + a3 = + + = + + = (1 + 2)(1 + 3) (2 + 2)(2 + 3) (3 + 2)(3 + 3) 12 20 30 6 1 1 1 1 S4 = a1 + a2 + a3 + a4 = + + + (1 + 2)(1 + 3) (2 + 2)(2 + 3) (3 + 2)(3 + 3) (4 + 2)(4 + 3) 1 1 1 1 4 = + + + = 12 20 30 42 21 S5 = a1 + a2 + a3 + a4 + a5 S1 = a1 =

1 1 1 1 1 + + + + (1 + 2)(1 + 3) (2 + 2)(2 + 3) (3 + 2)(3 + 3) (4 + 2)(4 + 3) (5 + 2)(5 + 3) 1 1 1 1 1 5 = + + + + = 12 20 30 42 56 24 =

Chapter 12 Review 41.

969

This function most nearly matches 1-1 x . To get 1 in the denominator, instead of 3, divide the numerator and denominator by 3. 4

4 = 3 x 3- x 1- 3

Thus, we can find the Taylor series for

4 3

1- 3x

by starting with the Taylor series for 1-1 x , multiplying each term by

4 , and replacing x with x . 3 3 4

4 = 3 x 3- x 1- 3 2

4 4 æxö 4 æxö 4 æxö 4 æxö ⋅ 1 + çç ÷÷ + çç ÷÷ + çç ÷÷ +  + çç ÷÷ +  ÷ ÷ ÷ ç ç ç è ø è ø è ø 3 3 3 3 3 3 3 3 èç 3 ø÷

4 4x 4x2 4 x3 4xn + + + +  + n +1 +  3 9 27 81 3

The Taylor series for 1-1 x is valid when -1 < x < 1. Replacing x with 3x gives -1 < 3x < 1 or - 3 < x < 3. The interval of convergence of the new series is (-3, 3).

42.

2x 2 = x⋅ 1 + 3x 1 - (-3x) Use the Taylor series for 1-1 x , multiply each term by 2, and replace x with -3x. Also, use property (3) with k = 1. 2x 2 = x⋅ = x ⋅ 2 ⋅ 1 + x ⋅ 2(-3x) + x ⋅ 2(-3x) 2 + x ⋅ 2(-3x)3 +  + x ⋅ 2(-3x) n +  1 + 3x 1 - (-3x) = 2 x - 6 x 2 + 18x3 - 54 x 4 +  + (-1)n ⋅ 2 ⋅ 3n ⋅ x n +1 + 

The Taylor series for 1-1 x is valid when -1 < x < 1 . Replacing x with -3x gives -1 < -3x < 1 or

(

1 1 > x>- . 3 3

)

The interval of convergence of the new series is - 13 , 13 .

43.

x2 1 = x2 ⋅ 1 - (-x) x +1

Use the Taylor series for 1-1 x and replace x with -x. Also, use property (3) with k = 2. x2 1 = x2 ⋅ x +1 1 - (-x) = x 2 ⋅ 1 + x 2 (-x) + x 2 (-x) 2 + x 2 (-x)3 +  + x 2 (-x) n +  = x 2 - x3 + x 4 - x5 +  + (-1) n x n + 2 + 

970

Chapter 12 SEQUENCES AND SERIES The Taylor series for 1-1 x is valid when -1 < x < 1 . Replacing x with -x gives -1 < -x < 1 or 1 > x > -1.

The interval of convergence of the new series is (-1, 1). 3

44.

3x3 = x3 ⋅ 2 x 2-x 1- 2

Use the Taylor series for 1-1 x , multiply each term by 32 , and replace x with 2x . Also, use property (3) with k = 3. 2 3 n 3 3x3 3 3æ x ö 3æ x ö 3æ x ö 3æ x ö = x3 ⋅ 2 x = x3 ⋅ ⋅ 1 + x3 ⋅ çç ÷÷ + x3 ⋅ çç ÷÷ + x3 ⋅ çç ÷÷ +  + x3 ⋅ çç ÷÷ +  2-x 2 2 èç 2 ÷ø 2 çè 2 ÷ø 2 çè 2 ÷ø 2 çè 2 ÷ø 12

3x 3 3x 4 3x5 3x 6 3x n + 3 = + + + +  + n +1 +  2 4 8 16 2

The Taylor series for 1-1 x is valid when -1 < x < 1 . Replacing x with 2x gives -1 <

x < 1 or - 2 < x < 2. 2

The interval of convergence of the new series is (-2, 2). 45.

We find the Taylor series for ln (1 - 2 x) by starting with the Taylor series for ln (1 + x) and replacing each x with -2 x. ln (1 - 2 x) = -2 x -

(-2 x)2 (-2 x)3 (-2 x)4 (-1)n (-2 x) n +1 + ++ + 2 3 4 n +1 8 3 2n +1 x n +1 x - 4x4 -  - 3 n +1

= -2 x - 2 x 2 -

The Taylor series for ln (1 + x) is valid when -1 < x £ 1 . Replacing x with -2x gives 1 1 > x³- . 2 2

-1 < -2 x £ 1 or

)

The interval of convergence of the new series is éê - 12 , 12 . ë 46.

(

)

We find the Taylor series for ln 1 + 13 x by starting with the Taylor series for ln(1 + x) and replacing each x with 13 x. æ 1 ö 1 ln çç1 + x ÷÷÷ = x çè 3 ø 3 =

n +1

( 13 x ) + ( 13 x ) - ( 13 x ) +  + (-1)n ( 13 x ) 2

n +1

+

1 1 2 1 3 1 4 (-1)n x n +1 xx + x x +  + n +1 + 3 18 81 324 3 (n + 1)

The Taylor series for ln (1 + x) is valid when -1 < x £ 1 . Replacing x with 13 x gives -1 <

1 x £ 1 or - 3 < x £ 3. 3

The interval of convergence of the new series is (-3, 3].

Chapter 12 Review

47.

971

We find the Taylor series for e-2x by starting with the Taylor series for e x and replacing each x with -2 x 2. 2

e-2 x = 1 + (-2 x 2 ) +

1 1 1 (-2 x 2 )2 + (-2 x 2 )3 +  + (-2 x 2 )n +  2! 3! n!

= 1 - 2x2 + 2x4 -

4 6 (-1)n ⋅ 2n x 2n + x ++ 3 n!

The Taylor series for e-2 x has the same interval of convergence, (-¥, ¥), as the Taylor series for e x . 48.

We find the Taylor series for e-5 x by starting with the Taylor series for e x and replacing each x with -5x. e-5 x = 1 + (-5x) +

= 1 - 5x +

1 1 1 (-5 x) 2 + (-5x)3 +  + (-5x)n +  2! 3! n!

25 2 125 3 (-1)n 5n x n + x x ++ 2 6 n!

The Taylor series for e-5x has the same interval of convergence, (-¥, ¥), as the Taylor series for e x . 49.

Use the Taylor series for e x , multiply each term by 2, and replace x with -3x. Also, use property (3) with k = 3. 2 x3e-3x = 2 x3 ⋅ 1 + 2 x3 (-3x) + 2 x3 ⋅

1 1 1 (-3x) 2 + 2 x3 ⋅ (-3x)3 +  + 2 x3 ⋅ (-3x) n +  2! 3! n!

= 2 x3 - 6 x 4 + 9 x5 - 9 x 6 +  +

(-1)n ⋅ 2 ⋅ 3n x n +3 + n!

The Taylor series for 2 x3e-3x has the same interval of convergence, (-¥, ¥) , as the Taylor series for e x . 50.

Use the Taylor series for e x and replace x with -x . Also, use property (3) with k = 6 . x6e-x = x6 ⋅ 1 + x6 ⋅ (-x) + x6 ⋅ = x6 - x7 +

1 1 1 (-x)2 + x6 ⋅ (-x)3 +  + x 6 ⋅ (-x) n +  2! 3! n!

(-1)n x n + 6 x8 x9 ++ + 2 6 n!

The Taylor series for x6e-x has the same interval of convergence, (-¥, ¥), as the Taylor series for e x . 51.

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives, 3x 2 - 2 x - 1 3(2)2 - 2(2) - 1 7 = = . 2x 2(2) 4 x 2 lim

By l’Hospital’s rule, lim

x 2

x3 - x 2 - x - 2 2

x -4

7 . 4

972 52.

By l’Hospital’s rule, x3 - 4 x 2 + 6 x = 2. 3x x0 lim

53.

lim x3 - 3x 2 + 4 x - 1 = -221 and l’Hospital’s rule does not apply. However,

x -5

x3 - 3x 2 + 4 x - 1

lim

x 2 - 25

x -5

54.

does not exist.

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives, 3

3 lim 3x +1 = = 3. 3(0) + 1 x0 1

By l’Hospital’s rule, lim

x0

55.

ln (3x + 1) = 3. x

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives, lim

5e x

x  0 3x

- 16 x + 7

5e0 2

3(0) - 16(0) + 7

5 . 7

By l’Hospital’s rule, lim

5e x - 5

x  0 x3 - 8 x 2 + 7 x

56.

5 . 7

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives, lim

x0

1 2(5 + x)1/2

1 1/2

2(5 + 0)

1 2 5

5 . 10

By l’Hospital’s rule, lim

x0

5+x - 5 1 5 or . = x 10 2 5

Chapter 12 Review 57.

973

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives, lim

-e2 x - 2 xe 2 x 2e2 x

x0

-e 2(0) - 2(0)e2(0)

2e 2(0)

1 =- . 2

By l’Hospital’s rule,

-xe2 x

lim

x  0 e2 x - 1

58.

1 =- . 2

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives, lim

1 2 x1/2

x16

1 1/2

2(16)

1 . 8

By l’Hospital’s rule, x -4 1 = . 8 x 16 x - 16 lim

59.

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives, lim

2 - 12 (1 + x)-1/2 3x 2

x0

= lim

4(1 + x)1/2 - 1

x  0 3x 2 (2(1 + x)1/2 )

3 which does not exist. 0

By l’Hospital’s rule, lim

1 + 2 x - (1 + x)1/2 x3

x0

60.

does not exist.

The limit in the numerator is 0, as is the limit in the denominator, so that l’Hospital’s rule applies. Taking derivatives separately in the numerator and denominator gives,

lim

1 1 + 2(5+ x )1/2 2(5- x )1/2

x0

æ ö÷ 1 1 1 ÷ + lim ççç ÷ 2 x  0 èç 2(5 + x )1/2 2(5 - x)1/2 ÷ø÷

ù 1 éê 1 1 5 ú = 1 or + ê 1/2 ú 2 ëê 2(5 + 0)1/2 10 2 5 2(5 - 0) ûú

By l’Hospital’s rule, lim

x0

5+ x - 5-x 1 = 2x 2 5

5 . 10

974 61.

Chapter 12 SEQUENCES AND SERIES We have a limit of the form ¥ ⋅ 0. Rewrite the expression. x 2e- x =

x2 e x

Now both the numerator and denominator become infinite and l’Hospital’s rule applies to the limit of the form ¥/¥. Differentiate the numerator and the denominator. x2 2x = lim x ¥ e x x ¥ e x

4 x3/2 x ¥ e x

= lim

lim

(2 x )

This problem is similar to what we started with so we handle it in the same manner. In fact, continue the process three more times. lim

4 x3/2

x ¥ e x

= lim

6 x1/2

= lim

x ¥ e x (2 x )

12 x

= lim

x ¥ e x (2 x )

x ¥ e x

= lim

x ¥

24 x e

= lim

12 x

x ¥ e x (2 x )

= lim

x ¥ e x

= 0

Therefore, by l’Hospital’s rule, lim x 2e- x = 0

x ¥

62.

The limit in the numerator is ¥, as is the limit in the denominator, so l’Hospital’s rule applies. Taking derivatives separately in the numerator and the denominator gives x

lim

x ¥ ln( x3 + 1)

1 (2 x )

= lim

x ¥

3x 2 ( x3 +1)

= lim

x3 + 1

x ¥ 6 x5/2

This problem is similar to what we started with so we handle it in the same manner. In fact, continue the process three more times. 3x 2 6x = lim = lim = lim 1/2 5/2 3/2 45 x ¥ 6 x x ¥ 15 x x ¥ x x ¥ lim

x3 + 1

6 45 (4 x1/ 2 )

8 x1/2 =¥ x ¥ 15

= lim

Therefore, by l’Hospital’s rule, lim

x ¥ ln( x

63.

x 3

+ 1)

= ¥.

æ e3 x 1 3 ö÷ ç lim çç 2 - 2 - ÷÷÷. x ÷ø x  0 çè x x

Rewrite over a common denominator. æ 3x ö ç e - 1 - 3x ÷÷ lim çç ÷ 2 ÷ x  0 çè ÷ø x

The numerator and denominator have limit 0, so l’Hospital’s rule applies. Differentiate the numerator and denominator. æ e3x - 1 - 3x ÷ö æ 3x ö ç ÷÷ = lim çç 3e - 3 ÷÷÷ lim çç ç 2x ÷ ÷÷ x  0 èç x  0 çè ÷ø x2 ø

Chapter 12 Review

975 The numerator and denominator have limit 0, so l’Hospital’s rule applies. Differentiate the numerator and denominator.

The numerator and denominator still have limit 0, so differentiate the numerator and denominator again. æ 3e3x - 3 ö÷ æ 3x ö ç ÷÷ = lim çç 9e ÷÷÷ = 9 lim çç ç 2 x ÷ø÷ 2 x  0 èç x  0 çè 2 ÷÷ø

æ -4 + 4 ö÷ æ ln(1 - 4 x) + 4 x ö÷ çç 1-4 x ÷÷ = lim çç lim ÷ ç ÷ø ÷ 2x x  0 çè x  0 ççè ÷ø x2 æ ö÷ -16 x ÷÷ = lim çç ç x  0 è (1 - 4 x)(2 x) ÷ø

Thus æ e3 x 1 3 ö÷ 9 ç lim çç 2 - 2 - ÷÷÷ = . x ÷ø 2 x  0 çè x x

64.

æ -8 ö÷ = lim çç ÷ = -8 x  0 çè 1 - 4 x ÷ø

æ 2 2 1 2e x ö÷ ç Find lim çç 3 + 2 + - 3 ÷÷÷. x x  0 çè x x x ø÷

Thus

æ ln(1 - 4 x) 4ö lim çç + ÷÷÷ = -8. 2 ç xø è x0 x

Rewrite over a common denominator. 2 xö æ ç 2 + 2 x + x - 2e ÷÷ lim çç ÷÷ x  0 èç x3 ø÷

66.

Rewrite over a common denominator.

The numerator and denominator have limit 0, so l’Hospital’s rule applies. Differentiate the numerator and denominator.

æ x + 1ö lim çç 3 ÷÷÷ ø x  0 çè x

2 xö xö æ æ ç 2 + 2 x + x - 2e ÷÷ ç 2 + 2 x - 2e ÷÷ lim çç ÷ = lim çç 3 2 ÷ ÷÷÷ x  0 çè x  0 çè ÷ø x 3x ø

The numerator and denominator still have limit 0, so differentiate the numerator and denominator again. xö æ 2 + 2 x - 2e x ö÷ æ ç ÷÷ = lim çç 2 - 2e ÷÷÷ lim çç ç ÷ 6 x ÷ø÷ x  0 çè x  0 çè 3x 2 ø÷

The numerator and denominator still have limit 0, so differentiate the numerator and denominator again. xö æ 2 - 2e x ö÷ æ 1 -2 ç ç -2e ÷÷ lim çç =÷÷÷ = lim çç ÷÷ = 6 x ÷ø 6 3 x  0 çè x  0 çè 6 ÷ø

Thus æ 2 2 1 2e x ö÷ 1 ç lim çç 3 + 2 + - 3 ÷÷÷ = - . x 3 x  0 çè x x x ÷ø

65.

æ ln(1 - 4 x) 4 ÷ö Find lim çç + ÷. x ø÷ x  0 çè x2

Rewrite over a common denominator.

æ ln(1 - 4 x) + 4 x ö÷ lim çç ÷÷ ø x  0 çè x2

æ 1 1 ö Find lim çç 3 + 2 ÷÷÷. x  0 çè x x ø

The numerator has limit 1 and the denominator has limit 0, so this limit does not exist. 67.

f ( x) = x3 - 8x 2 + 18x - 12 f ¢( x) = 3x 2 - 16 x + 18 f (4) = -4 < 0 and

f (5) = 3 > 0 so a

solution exists in (4, 5). Let c1 = 4. c2 = c1 -

-4 f (c1) = 4=6 f ¢(c1) 2

c3 = c2 -

f (c2 ) 24 = 6= 5.2 f ¢(c2 ) 30

c4 = c3 -

f (c3 ) 5.888 = 5.2 = 4.8302 f ¢(c3 ) 15.92

c5 = c4 -

f (c4 ) 0.98953 = 4.8302 = 4.7378 f ¢(c4 ) 10.709

c6 = c5 -

f (c5 ) 0.05462 = 4.7378 = 4.7321 f ¢(c5 ) 9.5354

c7 = c6 -

f (c6 ) 4.7 ⋅ 10-4 = 4.7321 = 4.7321 f ¢(c6 ) 9.4647

Subsequent approximations will agree with c6 and c7 to the nearest hundredth. Thus, x = 4.73.

976 68.

Chapter 12 SEQUENCES AND SERIES f ( x) = 3 x 3 - 4 x 2 - 4 x - 7 f ¢( x ) = 9 x 2 - 8 x - 4 f (2) = -7 < 0 and f (3) = 26 > 0 so a solution exists in (2, 3).

Let c1 = 2.

c2 = c1 -

-78 f (c1) = 3= 5.1667 f ¢(c1) 36

c3 = c2 -

f (c2 ) 384.31 = 5.1667 = 4.3527 f ¢(c2 ) 472.11

c4 = c3 -

f (c3 ) 95.883 = 4.3527 = 3.9689 f ¢(c3 ) 249.83

c2 = c1 -

-7 f (c1) = 2= 2.4375 f ¢(c1) 16

c5 = c4 -

f (c4 ) 15.586 = 3.9689 = 3.8779 f ¢(c4 ) 171.202

c3 = c2 -

f (c2 ) 2.9309 = 2.4375 = 2.3397 f ¢(c2 ) 29.973

c6 = c5 -

f (c5 ) 0.75897 = 3.8779 = 3.8730 f ¢(c5 ) 154.8

c4 = c3 -

f (c3 ) 0.16835 = 2.3397 = 2.3334 f ¢(c3 ) 26.55

c7 = c6 -

f (c6 ) 0.00256 = 3.8730 = 3.8730 f ¢(c6 ) 153.94

c5 = c4 -

f (c4 ) 0.00176 = 2.3334 = 2.3333 f ¢(c4 ) 26.336

Subsequent approximations will agree with c4 and c5 to the nearest hundredth. Thus, x = 2.33. 69.

Subsequent approximations will agree with c6 and c7 to the nearest hundredth. Thus, x = 3.87. 71.

37.6 is a solution of x 2 - 37.6 = 0. f ( x) = x 2 - 37.6 f ¢( x ) = 2 x

f ( x) = x 4 + 3x3 - 4 x 2 - 21x - 21 f ¢( x) = 4 x3 + 9 x 2 - 8 x - 21

Since 6 <

f (2) = -39 < 0 and f (3) = 42 > 0 so a solution exists in (2, 3).

Let c1 = 2. c2 = c1 -

-39 f (c1) = 2= 3.2581 f ¢(c1) 31

c3 = c2 -

f (c2 ) 84.558 = 3.2581 = 2.8055 f ¢(c2 ) 186.81

c2 = c1 -

-1.6 f (c1) = 6= 6.1333 f ¢(c1) 12

c3 = c2 -

f (c2 ) 0.01737 = 6.1333 = 6.1319 f ¢(c2 ) 12.2666

c4 = c3 -

f (c3 ) 0.0002 = 6.1319 = 6.1319 f ¢(c3 ) 12.2638

Since c4 = c3 = 6.1319, to the nearest

f (c3 ) 16.796 = 2.8055 = 2.6604 c4 = c3 f ¢(c3 ) 115.72 c5 = c4 -

f (c4 ) 1.4037 = 2.6604 = 2.6459 f ¢(c4 ) 96.735

thousandth, 72.

70.

f ( x) = x + x - 14 x - 15 x - 15 f ¢( x) = 4 x3 + 3x 2 - 28x - 15 f (3) = -78 < 0 and f (4) = 21 > 0 so a solution exists in (3, 4).

Let c1 = 3.

37.6 = 6.132.

51.7 is a solution of x 2 - 51.7 = 0. f ( x) = x 2 - 51.7 f ¢( x) = 2 x

f (c5 ) 0.01411 c6 = c5 = 2.6459 = 2.6458 f ¢(c5 ) 94.933

Subsequent approximations will agree with c5 and c6 to the nearest hundredth. Thus, x = 2.65.

37.6 < 7, let c1 = 6.

Since 7 <

51.7 < 8, let c1 = 7.

c2 = c1 -

-2.7 f (c1) =7= 7.1929 f ¢(c1) 14

c3 = c2 -

f (c2 ) 0.03781 = 7.1929 = 7.1903 f ¢(c2 ) 14.3858

c4 = c3 -

f (c3 ) 0.0004 = 7.1903 = 7.1903 f ¢(c3 ) 14.3806

Since c4 = c3 = 7.1903, to the nearest thousandth,

51.7 = 7.190.

Chapter 12 Review 73.

3 94.7

977

is a solution of x3 - 94.7 = 0.

76.

f ( x) = x3 - 94.7

n = 4 ⋅ 2 = 8, and i = 8% = 4% = 0.04, 2

f ¢( x ) = 3 x 2

Since 4 < 3 94.7 < 5, let c1 = 4 c2 = c1 -

f (c1) -30.7 = 4= 4.6396 48 f ¢(c1)

c3 = c2 -

5.1715 f (c2 ) = 4.6396 = 4.5595 64.5777 f ¢(c2 )

c4 = c3 -

f (c3 ) 0.08763 = 4.5595 = 4.5581 62.3671 f ¢(c3 )

c5 = c4 -

0.0003 f (c4 ) = 4.5581 = 4.5581 62.3288 f ¢(c4 )

5000 = R ⋅ s 8 0.04 R =

77.

this annuity is é (1.0235)36 - 1 ù S = 491 êê . úú 0.0235 êë úû

The number is brackets, s 36 0.0235, is

4 102.6 is a solution of x 4 - 102.6 = 0.

55.64299673, so that

S = 491(55.64299673) » 27,320.71.

f ( x) = x - 102.6 f ¢( x ) = 4 x 3

or $27,320.71

Since 3 < 4 102.6 < 4, let c1 = 3. c2 = c1 -

f (c1) -21.6 = 3= 3.2 108 f ¢(c1)

c3 = c2 -

2.2576 f (c2 ) = 3.2 = 3.1828 131.072 f ¢(c2 )

c4 = c3 -

f (c3 ) 0.02127 = 3.1828 = 3.1826 128.9698 f ¢(c3 )

78.

The yearly incomes produced by the mine form a geometric sequence with r = 118% = 1.18 and a1 = 750,000. To determine the total amount produced in 8 years, use the formula to find Sn with n = 8, r = 1.18, and a = a1 = 750,000. 8

750, 000 [(1.18) - 1] 1.18 - 1 750, 000 (3.75886 - 1) = » 11, 495, 247 0.18

The payments form an ordinary annuity with R = 1526.38, n = 5 ⋅ 2 = 10, and i = 7.6% 2 = 3.8% = 0.038. The amount of this annuity is é (1.038)10 - 1 ù S = 1526.38 êê . úú 0.038 êë úû

The number in brackets, s 10 0.038, is 11.89534558, so that

Rounded to the nearest thousandth, c3 , c4 , and subsequent approximations agree. Thus, 4 102.6 = 3.183.

S8 =

The payments form an ordinary annuity with R = 491, n = 9 ⋅ 4 = 36, and i = 9.4% = 2.35% = 0.0235. The amount of 4

thousandth, 3 94.7 = 4.558.

75.

5000 5000 = » 542.64 9.21423 s 8 0.04

or $542.64.

Since c3 = c4 = 4.5581, to the nearest

74.

This ordinary annuity will amount to $5000 in 4 years at 8% compounded semiannually. Thus, S = 5000,

S = 1526.38(11.89534558) » 18,156.82

or $18,156.82. 79.

$20,000 is the present value of an annuity of R dollars, with 9 periods, and i = 8.9% = 0.089 per period. P = R ⋅ an i 20, 000 = R ⋅ a 9 0.089 R =

20, 000 20, 000 = 6.019696915 a 9 0.089

» 3322.43

Each payment is $3322.43.

The total amount of income produced by the mine in five years is $11,495,247.

978 80.

Chapter 12 SEQUENCES AND SERIES $49,275 is the present value of an annuity of R dollars, with 48 period, and i = 12.2% = 1.016% 12

83.

= 0.01016 per period.

P = R ⋅ an i

ln 2 ln 2 = » 21.67. ln(1 + r ) ln1.0325

It will take about 21.67 years. According to the Rule of 70, the doubling time is given by

49, 275 = R ⋅ a 48 0.01016 R =

The doubling time n for a quantity that increases at an annual rate r is given by

49, 275 a 48 0.01016

Doubling time »

49, 275 37.83272858 » 1302.44

70 100(0.0325) = 21.54. =

Each payment is $1302.44. 81.

The present value, P, is 156,890, i = 0.0774 = 0.00645, and n = 12 ⋅ 25 = 300. 12

It will take about 21.54 years, a difference of 0.13 year, or about 7 weeks. 84.

156,890 = R ⋅ a 300 0.00645 é 1 - (1 + 0.00645)-300 ù ú = R êê ú 0.00645 ëê ûú é 1 - 0.1453244675 ù ú = Rê êë úû 0.00645

The doubling time n for a quantity that increases at an annual rate r is given by n =

ln 2 ln 2 = » 8.04. ln(1 + r ) ln1.09

It will take about 8.04 years. According to the Rule of 72, the doubling time is given by

é 0.8546755325 ù ú = Rê ëê 0.00645 ûú

Doubling time »

R » 1184.01

The present value, P, is 177,110, i = 0.0845 12

85.

86.

é 1 - (1 + 0.0070416)-360 ù ú = R êê ú 0.0070416 ëê ûú é 1 - 0.0799689882 ù ú = Rê êë úû 0.0070416 é 0.9200310118 ù ú = Rê ëê 0.0070416 ûú

2 hours amounts to six doubling times, so after two hours the number of bacteria will be 1000(26 ) = 64, 000 bacteria.

= 0.0070416, and n = 12 ⋅ 30 = 360.

177,110 = R ⋅ a 360 0.0070416

70 72 = = 8. 100r 100(0.09)

It will take about 8 years, a difference of 0.04 year, or about 2 weeks.

Monthly payments of $1184.01 will be required to amortize the loan. 82.

70 100r

Since the number of crimes decreases by 8% each year, each year it will be 0.92 times the number reported the previous year. Since number of crimes was 22,700 in the year before the neighborhood program started, the number of crimes n years after program began will be 22, 700(0.92) n . The number after five years will

be 22,700(0.92)5 » 14,961.

R » 1355.55

Monthly payments of $1355.55 will be required to amortize the loan.

Extended Application

979

Extended Application: Living Assistance and Subsidized Housing 1.

The distances proposed by Westville, measured from Westville toward Eastville, are half of the corresponding changes proposed by Westville (since Westville chose first), so the series of distances will be 3 + 34 + 1.5 + ⋅ ⋅ ⋅ or 4

Solve the following system: S = 0.3R 0.2(100 + S ) = R

(1 + 14 + 161 + ⋅ ⋅ ⋅ ) with a sum of 3/(3/4) = 4

Substitute for S in the second equation.

miles from Westville, the same location as found by summing Eastville’s proposals.

0.2(100 + 0.3R) = R 20 + 0.06R = R R =

20 » 21.28 0.94

4 = 6, and Westville’s changes form the 1 - 13

Then S = 0.3R æ 20 ö÷ = 0.3çç çè 0.94 ÷÷ø

same series, so in this case the chosen location for the pool is half way between the two towns.

» 6.38

Thus the rent will be 300 + R = $321.28, and the stipend will be 1000 + 100 + S = $1106.38, as found using infinite series. 2.

Under this scenario, Eastville’s proposed changes from the series 4 + 43 + 94 +  which has sum

Under this scenario, the sequence of stipends will be

100(0.06n - 1) 100 = (1 - 0.06n ). 0.06 - 1 0.94 S2 = 106

Sn =

1000 - 50 - 50 (0.3) (0.2) - 50[(0.3)(0.2)]2 - ⋅ ⋅ ⋅.

After the first term, the amount subtracted is a geometric series with a1 = 50 and r = 0.06. 50 50 Its sum is = » 53.19. Thus the 1 - 0.06 0.94 final stipend will be 1000 - 53.19 = 946.81, or $946.81.

S3 = 106.36 S4 = 106.3816

Thus, including the first term of 1000, 3 terms give the value to the nearest dollar, four terms to the nearest dime and 5 terms to the nearest penny.

The rents will be 300 - 10 - 10(0.06)2 -10(0.06)2 ⋅ ⋅ ⋅ .

After the first term, the amount subtracted is a geometric series with a1 = 10 and r - 0.096.

The initial $1000 doesn’t affect our answers to this questions, so we will look at the terms after the first term, which are a geometric series with a1 = 100 and r = 0.06. The sum of the first n term is

Experimentation shows that you need to sum about 50 terms to get within 0.01 of ln 2.

10 » 10.64. Thus the final Its sum is 1-10 = 0.94 0.06

rent will be 300 - 10.64 = 289.36, or $289.36. 3.

The distances proposed by Eastville, measured from Eastville toward Westville, are 6 + 32 + 1.5 + ⋅ ⋅ ⋅ Which we can write as 2

(

)

1 +  . In parentheses is a 6 1 + 14 + 16

geometric series with sum

1 1- 14

= 43 , so the total

changes proposed by Eastville would locate the fire station 6/(3/4) = 8 miles from Eastville.

Chapter 13

THE TRIGONOMETRIC FUNCTIONS 13.1

Definitions of the Trigonometric Functions

Your Turn 1 (a)

Your Turn 3 Sketch the angle and label the sides of an appropriate right triangle. y

Since 1 =  /180 radians, æ  ö÷ 210 = 210 çç radians çè 180 ÷÷ø 7 radians. = 6

(b)

 x y

0 r

Since 1 radian = 180/ , 3 3 æç 180 ö÷ radians = ÷ ç 4 4 çè  ÷ø = 3(45) = 135

Your Turn 2 Sketch the triangle formed by joining the origin, the point (9, 40), and the point (9, 0) on the x-axis. y (9, 40) 41 40

The triangle formed in the third quadrant is a 30--90 triangle, so we label the sides as in Example 4, except that here in the third quadrant both x and y are negative. sin

7 -1 1 = =6 2 2

sec

7 2 2 3 = =6 3 - 3

cos

cot

7 - 3 = = 6 -1

7 2 = = -2 6 -1

tan

7 -1 3 = = 6 3 - 3

csc

7 - 3 3 = =6 2 2

Your Turn 4

x=9 y = 40 r = 41

 0

x= 3 y = 1 r=2

(a)

cos 6 » 0.9945

(b)

sec 4 » -1.5299

Your Turn 5 To find r , use r =

x + y . Here x = 9, and

y = 40, so

r =

92 + 402

= 1681 = 41. y 40 = sin  = r 41 r 41 = sec  = x 9 x 9 = cot  = y 40

y 40 = x 9 x 9 = cos  = r 41 r 41 = csc  = y 40 tan  =

The cosine function is negative in quadrants II and III. In quadrant II we draw a triangle with an angle whose cosine is - 2 /2, which we recognize as a 45-45-90° degree triangle (Figure (a)). Thus the angle corresponding to the terminal side has measure  - ( /4) = 3 /4. Drawing the same triangle in quadrant III (Figure (b)) we see that the angle corresponding to the terminal side is  + ( /4) = 5 /4. Thus the equation cos θ = - 2 /2 has two solutions between 0 and 2 , namely 3 /4 and 5 /4.

981

982

Chapter 13 THE TRIGONOMETRIC FUNCTIONS

12.

æ  ö÷ 17 = 510 = 510çç çè 180 ÷÷ø 6

13.

5 5 æç 180 ö÷ = ÷ = 225 ç 4 4 çè  ø÷

14.

2 2 æç 180 ö÷ = ÷ = 120 ç 3 3 çè  ÷ø

15.

16.

(b)

17.

8 8 æç 180 ö÷ = ÷ = 288 ç 5 5 çè  ÷ø

Exercises

18.

5 5 æç 180 ö÷ = ÷ = 100 ç 9 9 çè  ø÷

19.

7 7 æç 180 ö÷ = ÷ = 105 ç 12 12 çè  ÷ø

20.

æ 180 ö÷ = 900 5 = 5 çç çè  ÷÷ø

y 2

 4 1

0 (a) y 

1 0

4 1

13.1

False. An angle of 180° or π radians represents half of one complete rotation around a circle.

True

æ  ö÷  = 60 = 60çç çè 180 ÷ø÷ 3

21. 6.

æ  ö÷  = 90 = 90çç çè 180 ÷÷ø 2

æ  ö÷ 5 150 = 150çç = çè 180 ÷÷ø 6

æ  ö÷ 3 135 = 135çç = çè 180 ÷÷ø 4

æ  ö÷ 3 270 = 270çç = çè 180 ÷ø÷ 2

10.

æ  ö÷ 16 320 = 320çç = çè 180 ÷÷ø 9

11.

æ  ö÷ 11 495 = 495çç = çè 180 ÷÷ø 4

13 13 æç 180 ö÷ =÷ = -390 ç 6 6 çè  ÷ø

 æ 180 ö÷ = - çç ÷ = -45 4 4 çè  ÷ø



Let  = the angle with terminal side through (-3, 4). Then x = -3, y = 4, and x2 + y 2 =

r = =

25 = 5.

y 4 = r 5 3 x cos  = =r 5 y 4 tan  = =x 3 sin  =

22.

(-3)2 + (4) 2

x 3 =y 4 5 r sec  = =x 3 r 5 csc  = = y 4 cot  =

Let  = the angle with terminal side through (-12, -5). Then x = -12, y = -5, and r =

x 2 + y 2 = 144 + 25

= 169 = 13.

Section 13.1

983

y 5 =r 13 x 12 =cos  = r 13 y 5 = tan  = x 12 sin  =

23.

27.

r =

x2 + y 2 =

625 = 25.

24 y =25 r 7 x = cos  = 25 r 24 y =tan  = 7 x

y < 0, so the sign is -. r x cos  = < 0, so the sign is -. r y tan  = > 0, so the sign is +. x x > 0, so the sign is +. cot  = y r sec  = < 0, so the sign is -. x r < 0, so the sign is -. csc  = y

sin  =

49 + 576

7 x =24 y 25 r = sec  = 7 x 25 r =csc  = 24 y cot  =

28.

x2 + y 2 =

625 = 25.

y 3 = r 5 x 4 = cos  = r 5 y 3 = tan  = x 4

sin  =

y < 0, so the sign is -. r x cos  = > 0, so the sign is +. r y tan  = < 0, so the sign is -. x x < 0, so the sign is -. cot  = y r sec  = > 0, so the sign is +. x r < 0, so the sign is -. csc  = y

sin  =

400 + 225

x 4 = y 3 r 5 = sec  = x 4 r 5 = csc  = y 3

cot  =

25.

In quadrant I, all six trigonometric functions are positive, so their sign is +.

26.

In quadrant II, x < 0 and y > 0. Furthermore, r > 0. y > 0, so the sign is +. r x cos  = < 0, so the sign is -. r y tan  = < 0, so the sign is -. x x < 0, so the sign is -. cot  = y r sec  = < 0, so the sign is -. x r > 0, so the sign is +. csc  = y

sin  =

In quadrant IV, x > 0 and y < 0. Also, r > 0.

Let  = the angle with terminal side through (20,15). Then x = 20, y = 15, and r =

In quadrant III, x < 0 and y < 0. Furthermore, r > 0.

Let  = the angle with terminal side through (7, -24). Then x = 7, y = -24, and

sin  =

24.

x 12 = y 5 r 13 =sec  = x 12 r 13 =csc  = y 5 cot  =

29.

When an angle  of 30 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = ( 3, 1). Then r =

x2 + y 2 =

3 + 1 = 2.

y 1 3 = = x 3 3 x cot  = = 3 y r csc  = = 2 y tan  =

984 30.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS When an angle  of 45 is drawn in standard position, ( x, y) = (1, 1) is one point on its terminal side. Then r = 1+1 = sin  =

34.

r =

3 + 1 = 2. 1 y sin  = = 2 r x =- 3 cot  = y

y 1 2 = = r 2 2

x 1 2 = = r 2 2 r sec  = = 2 x r csc  = = 2 y

cos  =

31.

35.

y 3 = 2 r 1 3 x cot  = = = 3 y 3

x2 + y 2 =

3 + 1 = 2.

x 3 =r 2 r 2 2 3 sec  = = =x 3 - 3

cos  =

sin  =

32.

When an angle  of 210 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (- 3, -1) . Then r =

x 2 + y 2 = 1 + 3 = 2.

csc  =

2 2 3 r ==3 x 3

sec  =

When an angle  of 60 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (1, 3). Then r =

When an angle  of 150 is drawn in standard position, ( x, y) = (- 3, 1) is one point on its terminal side. Then

36.

2 2 3 r = = 3 y 3

When an angle  of 120 is drawn in standard position, ( x, y) = (-1, 3) is one point on its terminal side. Then

When an angle  of 240 is drawn in standard position, ( x, y) = (-1, - 3) is one point on its terminal side. tan  =

y = x

cot  =

x -1 3 = = y 3 - 3

r = 1 + 3 = 2. x 1 =r 2 x 1 3 cot  = ==y 3 3 r sec  = = -2 x

37.

cos  =

33.

one choice of a point on its terminal side is ( x, y ) = (1, 3). Then

r =

x + y

= 1+1 =

y = -1 x x cot  = = -1 y

x 2 + y 2 = 1 + 3 = 2.

r =

sin

When an angle  of 135 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (-1, 1). Then 2

When an angle of 3 is drawn in standard position,

38.

 3

y 3 = r 2

When an angle of 6 is drawn in standard position, ( x, y) = ( 3, 1) is one point on its terminal side. Then

tan  =

cos

r =

3 + 1 = 2.



x 3 = r 2

Section 13.1 39.

985

When an angle of 4 is drawn in standard position,

45.

one choice of a point on its terminal side is ( x, y) = (1,1). tan

40.

 4

 3

x2 + y 2 =

csc



r = sin

47.

48.

0 + 1 = 1.

When an angle of 3 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (-1,0). Then r =

x + y

cos 3 =

5 r 2 = = =- 2 4 -1 x

When an angle of 5 is drawn in standard position, ( x, y) = (-1,0) is one point on its terminal side. Then r = 1 + 0 = 1. x cos 5 = = -1 r

49.

When an angle of - 34 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (-1, -1). Then

= 1 = 1.

æ 3 ö y -1 cot çç - ÷÷÷ = = =1 çè 4 ø x -1

x = -1 r

When an angle of  is drawn in standard position, ( x, y) = (-1,0) is one point on its terminal side. Then r = 1 + 0 = 1. r sec  = = -1 x

x2 + y 2 = 1 + 1 =

sec

3 y -1 = = = -1 2 r 1

When an angle of 54 is drawn in standard

r =

is drawn in standard

x2 + y2 =

y 5 1 = = undefined. 2 0 x

position, one choice of a point on its terminal side is ( x, y) = (-1, -1). Then

3 + 1 = 2.

position, one choice of a point on its terminal side is ( x, y) = (0, -1). Then

7 -1 2 y = = =4 2 r 2

When an angle of 52 is drawn in standard

tan

r 2 = = 2 y 1

When an angle of 32

position, one choice of a point on its terminal side is ( x, y) = (0,1). Then

When an angle of 6 is drawn in standard

r =

44.

46.

x 1 3 = = y 3 3

position, one choice of a point on its terminal side is ( x, y) = ( 3,1). Then

43.

sin

When an angle of 3 is drawn in standard position,

cot

42.

x2 + y 2 = 1 + 1 =

r =

( x, y) = (1, 3) is one point on its terminal side.

41.

position, one choice of a point on its terminal side is ( x, y) = (1, -1). Then

y =1 x

When an angle of 74 is drawn in standard

50.

When an angle of - 56 is drawn in standard position, one choice of a point on its terminal side is ( x, y ) = (- 3, -1). Then æ 5 ö -1 y 3 tan çç - ÷÷÷ = = = çè 6 ø x 3 - 3

986 51.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS When an angle of - 76 is drawn in standard

The two solutions of sec θ = -2/ 3 between 0 and 2 are 5 /6 and 7 /6.

position, one choice of a point on its terminal side is ( x, y) = (- 3,1). Then r =

x2 + y 2 =

3 + 1 = 2.

58.

æ 7 ö÷ y 1 sin çç = = çè 6 ÷ø÷ r 2

52.

r =

3 + 1 = 2.

æ ö 3 cos çç - ÷÷÷ = çè 6 ø 2

53.

54.

55.

solutions of sec θ =  /4 and 7 /4.

When an angle of - 6 is drawn in standard position, ( x, y) = ( 3, -1) is one point on its terminal side. Then

The cosine function is positive in quadrants I and IV. We know that cos ( /3) = 1/2, so the solution in quadrant I is  /3. The solution in quadrant IV is 2 - ( /3) = 5 /3. The two solutions of cos θ = 1/2 between 0 and 2 are  /3 and 5 /3. The sine function is negative in quadrants III and IV. We know that sin ( /6) = 1/2. The solution in quadrant III is  + ( /6) = 7 /6. The solution in quadrant IV is 2 - ( /6) = 11 /6. The two solutions of sin θ = -1/2 between 0 and 2 are 7 /6 and 11 /6. The tangent function is negative in quadrants II and IV. We know that tan ( /4) = 1. The solution in quadrant II is  - ( /4) = 3 /4. The solution in quadrant IV is 2 - ( /4) = 7 /4. The two solutions of tan θ = -1

The tangent function is positive in quadrants I and III. We know that tan ( / 3) = 3, so the solution in quadrant I is  /3. The solution in quadrant III is  + ( /3) = 4 /3. The two solutions of tan θ =  /3 and 4 /3.

2 between 0 and 2 are

59.

sin 39 » 0.6293

60.

cos 67 » 0.3907

61.

tan 123 » -1.5399

62.

tan 54 » 1.3764

63.

sin 0.3638 » 0.3558

64.

tan 1.0123 » 1.6004

65.

cos 1.2353 » 0.3292

66.

sin 1.5359 » 0.9994

67.

f ( x) = cos(3x) is of the form f ( x) = a cos(bx) where a = 1 and b = 3.

Thus, a = 1 and T = 2b = 23 . 68.

between 0 and 2 are 3 /4 and 7 /4. 56.

The secant function is positive in quadrants I and IV. We know that sec ( /4) = 2 , so the solution in quadrant I is  /4. The solution in quadrant IV is 2 - ( /4) = 7 /4. The two

69.

1 h( x ) = - sin(4 x ) is of the form 2 h( x ) = a sin(bx + c ) where a = - 1 , b = 4 2 and c = 0. Thus  a = 1/2 and T = 2 = 2 4 = 1/2. b

(

)

g (t ) = -2 sin 4 t + 2 is of the form g (t ) = a sin (bt + c) where a = -2, b = 4 ,

3 between 0 and 2 are

and c = 2. Thus, a = 2 and T = 2b = 2 = 8. 4

57.

The secant function is negative in quadrants II and III. We know that sec ( /6) = 2/ 3, so the solution in quadrant II is  - ( /6) = 5 /6. The solution in quadrant III is  + ( /6) = 7 /6.

70.

s( x) = 3sin(880 t - 7) is of the form s( x) = a sin(bt + c) where a = 3, b = 880 ,

Section 13.1

987

(

and c = -7. Thus, a = 3 and 2 = 1 . T = 2b = 880  440

71.

the left relative to the graph of g ( x) = sin ( 12 x ).

The graph of y = 2 cos x is similar to the graph of y = cos x except that it has twice the amplitude. (That is, its height is twice as great.)

76. 72.

)

y = sin 12 x +  + 2 is shifted 2 units to

(

)

y = 2 cos 3x - 4 + 1 has amplitude a = 2,  , period T = 2b = 23 , phase shift -3/4 = - 12

The graph of y = 2 sin x is similar to the graph of y = sin x except that it has twice the amplitude. (That is, its height is twice as great.)

and vertical shift d = 1. Thus, the graph of

(

)

y = 2cos 3x - 4 + 1 is similar to the graph

of f ( x) = cos x except that it has 2 times the amplitude, a third of the period, and is shifted up 1 unit vertically. Also, y = 2 cos

( 3x - 4 ) + 1

 units to the right relative to the is shifted 12

graph of graph of g ( x) = cos x. 73.

The graph of y = - 12 cos x is similar to the graph of y = cos x except that it has half the amplitude and is reflected about the x-axis.

77. 74.

75.

The graph of y = - sin x is similar to the graph of y = sin x except that it is reflected about the x-axis.

The graph of y = -3tan x is similar to the graph of y = tan x except that it is reflected about the x-axis and each ordinate value is three times larger in absolute value. Note that the

(

)

(

)

points - 4 ,3 and 4 , -3 lie on the graph.

y = 4sin ( 12 x +  ) + 2 has amplitude a = 4,

period T = 2b = 21 = 4 , phase shift 2

c =  1 b

= 2 , and vertical shift d = 2. Thus,

the graph of y = 4sin (

1x + 2

) + 2 is similar

to the graph of f ( x ) = sin x except that it has 4 times the amplitude, twice the period, and is shifted up 2 units vertically. Also,

78.

The graph of y = 12 tan x is similar to the graph of y = tan x except that the y-values of points on the graph are one-half the y-values of points on the graph of y = tan x.

988

Chapter 13 THE TRIGONOMETRIC FUNCTIONS é æ ö ù S (2) = 500 + 500 ê cos çç ÷÷÷ (2) ú ê èç 6 ø úû ë = 500 + 500cos



3 æ 1 ÷ö = 500 + 500 çç ÷÷ çè 2 ø = 750 snowblowers.

(a)

Since the three angles  are equal and their sum is 180°, each angle  is 60.

(b) The base angle on the left is still 60. The bisector is perpendicular to the base, so the other base angle is 90. The angle formed by bisecting the original vertex angle  is 30. (c) Two sides of the triangle on the left are given in the diagram: the hypotenuse is 2, and the base is half of the original base of 2, or 1. The Pythagorean Theorem gives the length of the remaining side (the vertical

bisector) as 80.

(a)

2 -1 =

The Pythagorean Theorem gives the length of the hypotenuse as 12 + 12 =

S (t ) = 500 + 500cos



t 6 (a) November corresponds to t = 0.

é æ ö ù S (3) = 500 + 500 ê cos çç ÷÷÷ (3) ú ê çè 6 ø ú ë û = 500 + 500 cos

= 500 snowblowers.

(d) May corresponds to t = 6.

Therefore, é æ ö ù S (6) = 500 + 500 ê cos çç ÷÷÷ (6) ú çè 6 ø ú ê ë û = 500 + 500 cos  = 500 + 500(-1) = 0 snowblowers.

(e) August corresponds to t = 9.

Therefore, é æ ö ù S (9) = 500 + 500 ê cos çç ÷÷÷ (9) ú ê èç 6 ø úû ë 3 = 500 + 500 cos 2 = 500 + 500(0) = 500 snowblowers.

Therefore, é æ ö ù S (0) = 500 + 500 ê cos çç ÷÷÷ (0) ú çè 6 ø ú ê ë û = 500 + 500 cos 0



= 500 + 500(0)

(b) The angle at the lower left is a right angle with measure 90. Since the sum of all three angles is 180, the measures of the remaining two angles sum to 90. Since these two angles are the base angles of an isosceles triangle they are equal, and thus each has measure 45. 81.

Therefore,

(f) Use the ordered pairs obtained in parts (a)-(e) to plot the graph.

= 1000 snowblowers.

(b) January corresponds to t = 2.

Therefore,

Section 13.1 82.

989

(a) It is reasonable to assume that electrical consumption is periodic across the seasons of the year, with higher use in summer and winter and lower use in spring and fall.

The formula predicts that the number of consultations was 98.75% of the daily mean. 84.

(a) Since the amplitude is 2 and the period is 0.350, 2 b 1 b = 0.350 2 2 = b. 0.350

0.350 =

(b) C (t ) = 80.63sin(0.8998t + 1.154) + 416.1

Therefore, the equation is y = 2 sin(2 t /0.350), where t is the time in seconds. æ 2 t ö÷ 2 = 2 sin çç çè 0.350 ÷÷ø

(b)

(c)

T =

æ 2 t ö÷ 1 = sin çç çè 0.350 ø÷÷

2 2 = »7 b 0.8998

 2

83.

(a)

The period is

2

 ( 14.77 )

(c)

= 29.54

(t - 6) y = 100 + 1.8cos éê 14.77 ùú reaches a ë û (t - 6) ù é maximum value when cos ê 14.77 ú = 1 ë û which occurs when

t = 2 æ 2 (2) ö÷ y = 2 sin çç » -1.95. çè 0.350 ÷÷ø

The position of the object after 2 seconds is -1.95.

There is a lunar cycle every 29.54 days. (b)

2 t 0.350

0.350 = t 4 0.0875 = t The image reaches its maximum amplitude after 0.0875 seconds.

The period is about 7 months, which seems reasonable since it is about half a year. (d) C (10) » 362.5. The estimated October consumption is 362.5 trillion BTUs, just a bit less than the actual value given in the table.

85. (a) T 37.29 0.46 cos(2p(t 16.37)/24) 38

t-6 = 0 t = 6

Six days from February 9, 2020 is February 15, 2020. é (6 - 6) ù ú y = 100 + 1.8cos ê êë 14.77 úû = 101.8

There is a percent increase of 1.8 percent. (c) On February 26, t = 17. é (17 - 6) ù ú y = 100 + 1.8cos ê ëê 14.77 ûú

T 36.91 0.32 cos(2p(t 14.92)/24)

(b) The cosine function in the expression for the body temperature T of patients will have a maximum where k = t. For the patients without Alzheimer’s this will be when t = 14.92. Since 0.92 ´ 60 » 55, this time is 2:55 P.M.

» 98.75

990

Chapter 13 THE TRIGONOMETRIC FUNCTIONS (c) For the patients with Alzheimer’s the maximum temperature will occur when t = k = 16.37. Since 0.37 ´ 60 » 22, this time is 4:22 P.M.

86.

89.

sin 

Solving c1 = sin 1 for c2 gives c2 =

P(t ) = 7(1 - cos 2 t )(t + 10) + 100e0.2t

c1 = 3 ⋅ 108 , 1 = 46, and  2 = 31 so

(a) Since January 1 of the base year corresponds to t = 0, the pollution level is

c2 =

P(0) = 7(1 - cos 0)(0 + 10) + 100e0

» 2.1 ´ 108 m/sec.

90.

P(0.5) = 7(1 - cos  )(0.5 + 10) + 100e0.1 = 7(2) (10.5) + 100e0.1 = 258.

P(1) = 7(1 - cos 2 )(1 + 10) + 100e0.2 » 122.

3 ⋅ 108 (sin 31) sin 46

= 214, 796,150

= 7(0)(10) + 100 = 100.

(b) Since July 1 of the base year corresponds to t = 0.5, the pollution level is

c1 sin  2 . sin 1

Since each square represents 30 and the sine wave repeats itself every 8 squares, the period of the sine wave is 8 ´ 30 = 240.

91.

On the horizontal scale, one whole period clearly spans four square, so 4 ⋅ 30° = 120° is the period.

92.

(a)

(d) Since July 1 of the following year corresponds to t = 1.5, the pollution level is P(1.5) = 7(1 - cos 3 ) (1.5 + 10) + 100e0.3 = 7(2) (11.5) + 100e0.3 » 296.

87.

(b) Since the sine function is zero for multiples of  , we can determine the values (s) of t where P = 0 by setting 880 t = n , where n is an integer, and solving for t. After some algebraic manipulations, t =

88.

sin 

Solving c1 = sin 1 for c2 gives c2 =

c1 sin  2 . sin 1

When  = 39,  2 = 28, and c1 = 3 ´ 108 , c2 =

(3 ´ 108 )(sin 28) sin 39°

» 2.2 ´ 108 m/sec.

n 880

and P = 0 when n =  , -2, - 1, 0, 1, 2,  . However, only values of n = 0, n = 1, or n = 2 produce values of t that lie in the interval [0, 0 .003]. Thus, 1 » 0.0011, P = 0 when t = 0, t = 880 1 » 0.0023. These values check and t = 440

with the graph in part (a). 2 = 1 . Therefore, (c) The period is T = 880  440

the frequency is 440 cycles per second.

Section 13.1 93.

991

ætö T (t ) = 60 - 30 cos çç ÷÷÷ çè 2 ø

(a)

37(-1) + 25 = -12F.

t = 1 represents February, so the maximum afternoon temperature in February is

(f) The period is 22 = 365. 365

95.

(a)

1 T (0) = 60 - 30cos » 34F. 2

(b)

t = 3 represents April, so the maximum afternoon temperature in April is T (3) = 60 - 30 cos

(c)

3 » 58F. 2

t = 8 represents September, so the maximum afternoon temperature in September is T (8) = 60 - 30 cos 4 » 80F.

(d)

(b) The function s(t ), derived by a TI-84 Plus using the sine regression function under the STATCALC menu, is given by s(t ) = 94.1528sin(0.0166t - 1.2162) + 347.2588.

t = 6 represents July, so the maximum afternoon temperature in July is T (6) = 60 - 30 cos3 » 90F.

(e)

Yes; because of the cyclical nature of the days of the year, it is reasonable to assume that the times of the sunset are periodic.

(c)

s(60) = 94.1528sin[0.0166(60) - 1.2162] + 347.2588

t = 11 represents December, so the maximum afternoon temperature in December is T (11) = 60 - 30 cos

11 » 39F. 2

» 327 minutes = 5:27 P.M. s(120) = 94.1528sin[0.0166(120) -1.2162] + 347.2588 » 413 minutes + 60 minutes

94.

é 2 ù T ( x ) = 37 sin ê (t - 101) ú + 25 êë 365 úû

(a)

é 2 ù T (74) = 37 sin ê (-27) ú + 25 êë 365 úû » 8F

(daylight savings)

= 7:53 P.M. s(240) = 94.1528sin[0.0166(240) -1.2162] + 347.2588 » 382 minutes + 60 minutes (daylight savings)

é 2 ù (b) T (121) = 37 sin ê (20) ú + 25 êë 365 úû » 37F

(c)

é 2 ù (149) ú + 25 T (250) = 37 sin ê êë 365 úû » 45F

é 2 ù (d) T (325) = 37 sin ê (224) ú + 25 êë 365 úû » 1F

(e) The maximum and minimum values of the sine function are 1 and -1, respectively. Thus, the maximum value of T is

= 7:22 P.M.

(d) The following graph shows s(t ) and y = 360 (corresponding to a sunset at 6:00 P.M.). These graphs first intersect on day 81. However because of daylight savings time, to find the second value we find where the graphs of s(t ) and y = 360 - 60 = 300 intersect. These graphs intersect on day 295. Thus, the sun sets at approximately 6:00 P.M. on the 81st and 295th days of the year.

37(1) + 25 = 62F

and the minimum value of T is

992

96.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS

w1 =

u 2 (tan  ) and L + u (tan  )

w2 =

u 2 (tan  ) L - u(tan  )

1 , L = 0.00625, and u = 6, then If  = 30

w1 =

w2 =

(

1  62 tan 30

(

)

1  0.00625 + 6 tan 30 1 °) 62 (tan 30

(

1  0.00625 - 6 tan 30

)

» 2.2 and

)

» 7.6.

The near and far limits of the depth of field when the object being photographed is 6 m from the camera, are 2.2 m and 7.6 m, respectively. 97.

Let h = the height of the building. h 65 h = 65 tan 42.8 » 60.2

tan 42.8 =

The height of the building is approximately 60.2 meters. 98.

Let x be the distance to the opposite side of the canyon. Then

The coordinates of C can be found by solving the right triangle BCF. Those coordinates are (1 + cos 72,sin 72), or about (1.309, 0.951). The y-coordinate of D can be found by solving the right triangle ADG, noting that the measure of angle DAG is also 72. That gives the coordinates of D to be (0.5, 0.5 tan 72), or about (0.5, 1.539). Similarly, the coordinate of E are (- cos 72, sin 72), or about (-0.309, 0.951). The x-coordinate of D is half-way between the x-coordinates of E and C, namely at 0.5. 101. We need to find the values of t for which 3.5 £ h(t ) £ 4.

The following graphs show where

(

)

h(t ) = sin t - 2 + 4 intersects the horizontal

lines y = 3.5 and y = 4.

105 x 105 x = tan 27

tan 27 =

» 206

The distance to the opposite side of the canyon is approximately 206 ft. 99.

Let  = the average angle with the horizontal. tan  =

26 5280

Using the TAN-1 key on the calculator, æ 26 ö÷ ÷ » 0.28. è 5280 ÷ø

Thus, 3.5 £ h(t ) £ 4 when t is in the interval [4.6, 6.3], to the nearest tenth.

 = TAN-1 ççç

100. Label the vertices as shown. Since the five exterior angles have equal measure and the sum of their measures is 360, the measure of each exterior angle is 72.

Section 13.2

13.2

993 dy = 1 ⋅ tan 2 x + x ⋅ Dx (tan 2 x) dx

Derivatives of Trigonometric Functions

= 1 ⋅ tan 2 x + x ⋅ (2 tan x) ⋅ Dx (tan x) = 1 ⋅ tan 2 x + x ⋅ (2 tan x) ⋅ sec2 x

Your Turn 1

= 2 x tan x sec2 x + tan 2 x.

Find the derivative of y = 5sin (3x 4 ).

Your Turn 5

By the chain rule,

Find the derivative of y = sec2 ( x ).

dy = 5cos(3x 4 ) ⋅ Dx (3x 4 ) dx

By two applications of the chain rule,

= 60 x3 cos(3x 4 ).

dy = [2sec( x )] ⋅ Dx[sec( x )] dx = [2sec( x )] ⋅ [sec( x ) tan( x )] ⋅ Dx ( x )

Your Turn 2

Find the derivative of y = 2sin 3 ( x ).

= [2sec( x )] ⋅ [sec( x ) tan( x )] ⋅

This derivative requires two applications of the chain rule. dy = 6sin 2 ( x ) ⋅ Dx[sin( x )] dx = 6sin 2 ( x ) ⋅ cos( x ) ⋅ Dx ( x ) = 6sin 2 ( x ) ⋅ cos( x ) ⋅ =

1 2 x

sec ( x ) tan( x ) . x

Your Turn 6

Find the derivative of f ( x) = sin(cos x) when x =  /2. Use the chain rule.

3sin ( x ) ⋅ cos( x ) x

f ¢( x) = cos(cos x) ⋅ (- sin x) æ ö æ ö æ ö f ¢ çç ÷÷÷ = cos çç cos ÷÷÷ ⋅ çç - sin ÷÷÷ çè èç 2 ø 2 ø çè 2ø

Your Turn 3

= (cos 0) ⋅ (-1) = (1)(-1)

Find the derivative of y = x cos ( x ). By the product rule and two applications of the chain rule, dy = 1 ⋅ cos( x 2 ) + x ⋅ Dx[cos( x 2 )] dx = cos( x 2 ) + x ⋅ [- sin( x 2 )] ⋅ Dx ( x 2 ) = cos( x 2 ) + x ⋅ [- sin( x 2 )] ⋅ (2 x) = -2 x 2 sin( x 2 ) + cos( x 2 ).

Your Turn 4

= -1

13.2

Warmup Exercises

W1. y = x ln 2 x Use the product rule. dy d d = x ( ln 2 x ) + ( ln 2 x ) x dx dx dx æ 2 ö = ( x )çç ÷÷÷ + ( ln 2 x )(1) çè 2 x ø = 1 + ln 2 x

Find the derivative of y = x tan 2 ( x). By the product rule and two applications of the chain rule,

W2. y =

ex x2

Use the quotient rule.

( )( x2 ) - ( e x )( 2x ) 2 ( x2 )

ex dy = dx =

e x (1 - 2 x)

994

Chapter 13 THE TRIGONOMETRIC FUNCTIONS

W3. y = (1 + 2 x)5 Use the chain rule. dy d = 5(1 + 2 x)4 (1 + 2 x) dx dx

= 10(1 + 2 x) 4

= 108 sec2 (9 x + 1)

W4. y = e5 x - 3x Use the chain rule. 2 dy d = e5 x - 3x 5 x 2 - 3x dx dx

(

)

= (10 x - 3) e5 x - 3x

(

W5. y = ln x 2 + 1

)

(

= 56 x sin (7 x 2 - 4)

) 9.

y = -4 cos (7 x 2 - 4) dy = [4 sin (7 x 2 - 4)] ⋅ Dx (7 x 2 - 4) dx

= [4sin (7 x 2 - 4)] ⋅ 14 x

Use the chain rule.

1 dy d = x2 + 1 2 dx dx x +1

y = 12 tan (9 x + 1) dy = [12 sec2 (9 x + 1) ⋅ Dx (9 x + 1) dx = [12 sec (9 x + 1)] ⋅ 9

x +1

y = cos4 x dy = [4(cos x)3 ] Dx (cos x) dx

= (4 cos3 x)(- sin x) = -4 sin x cos3 x

13.2

Exercises

False. lim =

True. Since sin 2 x + cos 2 x = 1, then

x0

sin x =1 x

10.

= -45 sin 4 x cos x

Dx (sin 2 x + cos 2 x) = Dx (1) = 0.

True

11.

1 sin 8 x 2 1 dy = (cos 8 x) ⋅ Dx (8 x) dx 2 1 = (cos 8 x) ⋅ 8 2 = 4 cos 8 x y =

y = - cos 2 x + cos

y = tan8 x dy = 8(tan x)7 ⋅ Dx (tan x) dx

= 8 tan 7 x sec2 x

12.

y = 3cot 5 x dy = 15(cot x)4 ⋅ Dx (cot x) dx

= -15 cot 4 x csc2 x



6 dy = (sin 2 x) ⋅ Dx (2 x) + 0 dx = (sin 2 x) ⋅ 2 = 2 sin 2 x

y = -9 sin 5 x dy = -45(sin x) 4 ⋅ Dx (sin x) dx

13.

y = -6 x ⋅ sin 2 x dy = -6 x ⋅ Dx (sin 2 x ) + sin 2 x ⋅ Dx (-6 x) dx = -6 x(cos 2 x) ⋅ Dx (2 x) + (sin 2 x) (-6)

= -6 x(cos 2 x) ⋅ 2 - 6 sin 2 x = -12 x cos 2 x - 6sin 2 x

Section 13.2 14.

995

y = 2 x ⋅ sec 4 x dy = 2 x ⋅ Dx (sec 4 x) + sec 4 x ⋅ Dx (2 x) dx = 2 x(sec 4 x tan 4 x) ⋅ Dx (4 x) + (sec 4 x)(2)

20.

dy = -2ecot x ⋅ Dx (cot x) dx = -2ecot x ⋅ (- csc 2 x)

= 2 x(sec 4 x tan 4 x) ⋅ 4 + 2sec 4 x = 8x sec 4 x tan 4 x + 2sec 4 x

15.

y = dy = dx

= =

16.

y = dy = dx

17.

csc x x x ⋅ Dx (csc x) - (csc x) ⋅ Dx x

= (2csc2 x)ecot x

21.

= cos (ln 3x 4 ) ⋅

x -( x csc x cot x + csc x)

= cos (ln 3x 4 )

tan x x -1 ( x - 1) ⋅ Dx (tan x) - (tan x) Dx ( x - 1)

22.

( x - 1) sec x - tan x

y = sin e4 x dy = cos e4 x ⋅ Dx (e4 x ) dx

23.

19.

y =e dy = ecos x ⋅ Dx (cos x) dx

= ecos x ⋅ (- sin x) = (- sin x)ecos x

Dx (2 x3 ) 2 x3

y = ln |sin x 2 | D (sin x 2 ) (cos x 2 ) ⋅ Dx ( x 2 ) dy = x = dx sin x 2 sin x 2 =

y = cos 4e dy = (-sin 4e2 x ) ⋅ Dx (4e2 x ) dx

cos x

y = cos (ln | 2 x3 |)

3 = - sin (ln |2 x3 |) x

= -8e2 x sin 4e2 x

12 x3

= - sin (ln |2 x3 |) ⋅

= 4e4 x cos e4 x

18.

3x 4

dy = [- sin (ln |2 x3 |)] ⋅ Dx (ln | 2 x3 |) dx

( x - 1)2

= (cos e4 x ) ⋅ e4 x ⋅ 4

Dx (3x 4 )

3x 4 4 = cos (ln 3x 4 ) ⋅ x 4 = cos (ln 3x 4 ) x

= (cos e4 x ) ⋅ e4 x ⋅ Dx (4 x)

y = sin (ln 3x 4 ) dy = [cos (ln 3x 4 )] ⋅ Dx (ln 3x 4 ) dx

x2 -x csc x cot x - csc x

( x - 1) 2

y = -2ecot x

(cos x 2 ) ⋅ 2 x sin x 2

2 x cos x 2 sin x 2

or 2 x cot x 2

24.

y = ln | tan 2 x | dy 1 = ⋅ Dx (tan 2 x) 2 dx tan x 1 = ⋅ (2 tan x) ⋅ Dx (tan x) tan 2 x =

2 sec 2 x tan x

996

Chapter 13 THE TRIGONOMETRIC FUNCTIONS

25. y = dy = dx

= = =

2sin x 3 - 2sin x (3 - 2sin x) Dx (2sin x) - (2sin x) ⋅ Dx (3 - 2sin x) (3 - 2sin x)2

28.

æ cos 4 x ö÷1/2 cos 4 x = çç ÷ çè cos x ÷ø cos x

y =

-1/2 æ cos 4 x ö÷ 1 æ cos 4 x ö÷ dy ÷ = ççç ⋅ Dx ççç ÷ ÷÷ 2 çè cos x ÷ø dx èç cos x ø÷

-1/2

(3 - 2sin x) ⋅ 2Dx (sin x) - (2sin x) ⋅ [-2 Dx (sin x)] (3 - 2sin x) 2

1 æç cos 4 x ö÷ ÷÷ ç 2 ççè cos x ÷ø

1 æç cos x ÷ö ÷÷ ç 2 çèç cos 4 x ÷ø

1/2

6 cos x - 4sin x cos x + 4sin x cos x (3 - 2sin x) 2

é (cos x) ⋅ D (cos 4 x) - (cos 4 x) ⋅ D (cos x) ù x x ú ⋅ êê ú cos2 x ë û

æ -4 cos x sin 4 x + cos 4 x sin x ÷ö ÷÷ ⋅ ççç ÷ø èç cos2 x

-4 cos x sin 4 x + cos 4 x sin x

6 cos x

2 cos3/2 x cos1/2 4 x

(3 - 2sin x) 2

29. 26. y = dy = dx

3cos x 5 - cos x (5 - cos x) ⋅ D(3cos x) - (3cos x) ⋅ Dx (5 - cos x)

æ1 ö æ1 ö = 3sec 2 çç x ÷÷ ⋅ Dx çç x ÷÷ çè 4 ÷ø çè 4 ÷ø

(5 - cos x) 2

+ 4(- csc2 2 x) ⋅ Dx (2 x) - 5(- csc x cot x)

(5 - cos x)(-3sin x) - (3cos x) ⋅ (sin x)

(5 - cos x) 2

+ e-2 x ⋅ Dx (-2 x)

-15sin x

æ1 ö 1 = 3sec 2 çç x ÷÷÷ ⋅ - (4 csc 2 2 x) ⋅ 2 çè 4 ø 4

(5 - cos x) 2

+ 5csc x cot x + e-2 x ⋅ (-2) æ1 ö 3 = sec 2 çç x ÷÷÷ - 8 csc 2 2 x çè 4 ø 4

1/2

y =

27.

æ sin x ö÷ sin x = çç çè sin 3x ÷÷ø sin 3x

1/2 æ sin x ö÷ dy 1 æç sin x ö÷ ÷ ÷ = çç ⋅ Dx ççç ÷ çè sin 3x ÷÷ø dx 2 çè sin 3x ÷ø

+ 5csc x cot x - 2e-2 x

é (sin 3x) ⋅ Dx (sin x)

ù ú - (sin x) ⋅ Dx (sin 3x) úú 1 çæ sin 3x ÷ö ê ÷ = çç ⋅ê ÷ ú 2 çè sin x ÷ø (sin 3x)2 ê ú ê ú ë û é (sin 3x)(cos x) ù ú -1/2 ê æ ö ê ú (3 x ) ⋅ (sin x )(cos 3 x ) D 1 ç sin 3x ÷ x ú ÷÷ = çç ⋅ê 2 ê ú 2 çè sin x ø÷ sin 3x ê ú ê ú ë û -1/2 êê

= = =

1(sin 3x)1/2

⋅ 2(sin x)1/2 1/2

(sin 3x)

æ1 ö y = 3tan çç x ÷÷÷ + 4cot 2 x - 5csc x + e-2 x çè 4 ø æ1 ö dy = 3 tan çç x ÷÷÷ + 4 cot 2 x - 5csc x + e-2 x çè 4 ø dx

sin 3x cos x - 3 sin x cos 3x sin 2 3x

30.

y = (sin 3x + cot( x3 ))8

dy = 8[sin 3x + cot( x3 )]7 ⋅ Dx [sin 3x + cot( x3 )] dx = 8[sin 3x + cot( x3 )]7 ⋅[cos3x ⋅ Dx (3x) - csc 2 ( x3 ) ⋅ Dx ( x3 )] = 8[sin 3x + cot( x3 )]7[3cos3x - 3x 2 csc 2 ( x3 )]

31.

y = sin x; x = 0

Let f ( x) = sin x. Then f ¢( x) = cos x, so

(sin 3x cos x - 3 sin x cos 3x) 2(sin x)1/2 (sin 2 3x)

sin 3x [sin 3x cos x - 3 sin x cos 3x] 2 sin x (sin 2 3x)

f ¢(0) = cos 0 = 1.

The slope of the tangent line to the graph of y = sin x at x = 0 is 1.

Section 13.2 32.

997

y = sin x; x =



æ ö æ ö f ¢ çç ÷÷ = - csc2 çç ÷÷ = -( 2)2 = -2. çè 4 ÷ø çè 4 ÷ø

Let f ( x) = sin x.

The slope of the tangent line to the graph of y = cot x at x = 4 is -2.

Then f ¢( x) = cos x and æ ö 2 . f ¢ çç ÷÷÷ = çè 4 ø 2

37.

The slope of the tangent line at x = 4 is 33.

y = cos x; x = -

æ cos x ÷ö Dx (cos x) = Dx ççç ÷÷ è sin x ø÷

2 . 2

5 6

Let f ( x) = cos x.

Then f ¢( x) = -sin x, so

38.

sin 2 x + cos 2 x sin 2 x 1 sin 2 x

= Dx[(cos x)-1] = -1(cos x)-2 (- sin x)

Then f ¢( x) = - sin x and

sin x

cos 2 x 1 sin x = ⋅ cos x cos x

æ ö 2 . f ¢ çç - ÷÷÷ = çè 4 ø 2 The slope of the tangent line at x = - 4 is

2 . 2

= sec x tan x

39. y = tan x; x = 0

Since csc x = sin1 x = (sin x)-1, æ 1 ö÷ ÷÷ = Dx (sin x)-1 Dx (csc x) = Dx ççç è sin x ÷ø

Let f ( x) = tan x.

= -1 (sin x)-2 cos x

Then f ¢( x) = sec2 x, so 1 cos 2 0

= 1.

 4

cos x

sin 2 x 1 cos x =⋅ sin x sin x

The slope of the tangent line to the graph of y = tan x at x = 0 is 1.

y = cot x; x =

sin 2 x

æ 1 ö÷ ÷ Dx (sec x) = Dx çç çè cos x ÷÷ø

Let f ( x) = cos x.

36.

- sin 2 x - cos 2 x

= - csc 2 x.

π y = cos x; x = 4

f ¢(0) = sec 2 0 =

sin 2 x

The slope of the tangent line to the graph of y = cos x at x = - 56 is 12 .

35.

(sin x)(- sin x) - (cos x)(cos x)

æ 5 ö æ 5 ö 1 f ¢ çç - ÷÷÷ = - sin çç - ÷÷÷ = . èç 6 ø èç 6 ø 2

34.

cos x

Since cot x = sin x , by using the quotient rule,

= - csc x cot x.

41.

y = sin x

dy = cos x dx

Let f ( x) = cot x. Then f ¢( x) = - csc2 x, so

(a) The critical numbers are the zeros of cos x, that is, n / 2 where n is an odd integer.

998

Chapter 13 THE TRIGONOMETRIC FUNCTIONS (b) The sine function is increasing where the cosine is positive, that is, on the intervals ( n , (n + 1/2) ) È ( (n + 3/2) , (n + 2) )

45.

f ( x ) = cos( x 3 )

( )

where n is an even integer.

f ¢( x ) = - sin( x 3 ) 3x 2

f ¢¢( x ) = - sin( x 3 )(6 x ) + (- cos( x 3 ))(3x 2 )(3x 2 )

(c) The sine function is decreasing where the cosine is negative, that is, on the intervals ( (n + 1/2) , (n + 3/2) ) where n is an even

= -9 x 4 cos( x 3 ) - 6 x sin( x 3 ) f ¢¢(0) = 0

integer. 42.

y = tan x

Use the chain rule and the product rule.

f ¢¢(2) = -144 cos 8 - 12sin 8

dy = sec 2 x dx

46.

Use the chain rule and the product rule. f ( x ) = cos3 x

(a)

The derivative is never 0; the critical numbers are the numbers where the tangent is undefined, that is, n /2 for n an odd integer.

(b)

The tangent function is increasing between every pair of critical points, that is, on intervals ( n /2, (n + 2) /2 ) for n an even

( ) f ¢¢( x ) = ( 3cos2 x ) (- cos x )

f ¢( x ) = 3cos2 x ( - sin x )

+ (6 cos x )(- sin x )(- sin x ) 2

= 6sin x cos x - 3cos3 x f ¢¢(0) = -3

integer.

f ¢¢(2) = 6sin 2 2 cos 2 - 3cos3 2

44.

The relative maxima of y = sin  x have value 1 and occur at the values of x for which  x = (n + 1/2) where n is an even integer, that is, at x = n + 1/2 for n an even integer. The relative minima of y = sin  x have value -1 and occur at the values of x for which  x = (n - 1/2) where n is an even integer, that is, at x = n - 1/2 where n is an even integer. The relative maxima of y = cos 2 x have value 1 and occur at the values of x for which 2 x = 2n where n is an integer, that is, at x = n for n an integer. The relative minima of y = cos 2 x have value -1 and occur at the values of x for which 2 x = (2n + 1) where n is an integer, that is, at x = (n + 1/2) where n is an integer.

47.

f ( x) = sin x f ¢( x) = cos x f ¢¢( x) = - sin x f ¢¢¢( x) = - cos x f (4) ( x) = sin x f (4n) ( x) = sin x

48.

f ( x) = cos x f ¢( x) = - sin x f ¢¢( x) = - cos x f ¢¢¢( x) = sin x f (4) ( x) = cos x f (4n) ( x) = cos x

49.

f ( x) = sin 2 x f ¢( x) = 2 cos 2 x and f ¢¢( x) = -4 sin 2 x

The graph of f will be concave upward where sin 2 x < 0, that is, for (2n - 1) < 2 x < 2n , where n is an integer; x will then be in one of the intervals ( (n - 1/2) , n ) with n an integer. The graph will be concave downward on the intervals ( n , (n + 1/2) ) with n an integer. The Copyright © 2022 Pearson Education, Inc.

Section 13.2

999

inflection points will occur when x is a point separating the intervals of opposite concavity, that is, for x = n /2. The inflection points are therefore (n /2, 0) for integer n. 50.

52.

f ¢¢( x) = - sin x

Since f ¢( x) is positive except at the isolated points x = (2n + 1) for integer n, the function f is increasing everywhere. The graph has a single intercept at (0, 0), which is both an x-intercept and a y-intercept. The concavity is given by the sign of - sin x; the graph will be concave downward on the intervals ( 2n , (2n + 1) ) for integer n and concave

f ( x) = tan  x f ¢( x) =  sec2  x f ¢¢( x) = 2 2 sec2 x tan  x sin  x = 2 2 cos3  x The graph of f will be concave upward where sine and cosine have the same sign, that is, when  x is in the first or third quadrant. This occurs in the intervals n <  x < (n + 1/2) or for x in (n, n + 1/2) for n an integer. The graph will be concave downward on the intervals (n + 1/2, n) for n an integer. Inflection points don’t occur at all points separating the intervals of opposite concavity, but only at those where tan  x is defined, which have the form x = n. So the inflection points are (n, 0).

51.

f ( x) = x + sin x f ¢( x) = 1 + cos x

upward on the intervals ( (2n + 1) , (2n + 2) ) for integer n. Inflection points occur at the x-values separating regions of opposite concavity, that is, where sin x = 0, for example at x = - , x = 0, x =  , and x = 2 . The y-values at these inflection points are equal to the corresponding x-values.

f(x) 8

f ( x) = x + cos x f ¢( x) = 1 - sin x

(2p, 2p) 4

Since f ¢( x) is positive except at the isolated points x = (2n + 1/2) for integer n, the function f is increasing everywhere. The graph has a single y-intercept at (0, 1) and a single x-intercept at approximately (-0.739, 0), found by solving x + cos x = 0 with a calculator. The concavity is given by the sign of - cos x; the graph will be concave downward on the intervals ( (2n - 1/2) , (2n + 1/2) ) for integer n and concave upward on the intervals ( (2n + 1/2) , (2n + 3/2) ) for integer n. Inflection points occur at the x-values separating regions of opposite concavity, that is, where cos x = 0, for example at x = - /2, x =  /2, and x = 3 /2. The corresponding y-values are - /2,  /2, and 3 /2.

–10

–8 –6 –4 p

冸– 2 , – 2 冹

f(x) = x + cos x

冸

3p , 3p 2 2

冹

冸

p, p 2 2

冹

(p, p)

(0, 0)

f ¢¢( x) = - cos x

f(x) 8 6 4 (0, 1) 2

f(x) = x + sin x

–8 –4 (–p, –p)

8 x

–4 –8

53.

x - sin x on [0,  ] 2 1 f ¢( x) = - cos x 2 The only critical number in [0,  ] is  /3 , since f ( x) =

f ¢( /3) = 0 and f ¢ is negative in the second quadrant. Thus extrema could only be located at x = 0, x =  /3, and x =  . f (0) = 0 f ( /3) =

 -3 3 6

» -0.342

f ( ) =  /2

So there is an absolute maximum of  /2 at x =  and an absolute minimum of

(  - 3 3 )/6 at x =  /3.

10 x

–4 –6 –8 –10

1000 54.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS f ( x) = tan x - 2 x on [0,  /3]

æ 1 dy æ 2 öö - sin [  (1/2)(1) ] ( ) çç + (1) çç - ÷÷ ÷÷÷ çè 2 dt èç  ø÷ ø÷ æ 2ö dy + 2 çç - ÷÷÷ + 2(1) = 0 çè  ø dt

f ¢( x) = sec2 x - 2 sec2 ( /4) =

( 2 ) = 2, and the only critical

number in [0,  /3] is  /4. Thus extrema could only be located at x = 0, x =  /4, and x =  /3. f (0) = 0 f ( /4) = 1 -  /2 » -0.571

55.

58.

sin( xy) = x Differentiate both sides with respect to x. d d x sin( xy) = dx dx d cos( xy) xy = 1 dx æ dy ö cos( xy) çç x + y ÷÷÷ = 1 çè dx ø Solve for dy / dx. æ dy ö cos( xy ) çç x + y ÷÷÷ = 1 çè dx ø

(

)

dy 1 = - 2 = - cos 2 y dx sec y

57.

cos( xy) + 2 x + y = 2 Differentiate with respect to t. æ dy dx ö dx dy - sin( xy)( ) çç x + y ÷÷÷ + 2 + 2y = 0 çè dt dt ø dt dt Substitute the given values for dx / dt , x and y.

dy = 0 dt

sin( x + y) + (1 + x)2 + (2 + y)2 = 5 Differentiate with respect to t. æ dx dy ö÷ cos( x + y) çç + ÷ çè dt dt ÷ø dx dy + 2(2 + y) = 0 dt dt

+ 2(1 + 0)(-10) + 2(2 + 0) dy dy - 20 + 4 = 0 dt dt dy dy 5 = 30, so = 6 dt dt

-10 +

59.

tan y + x = 4 Differentiate both sides with respect to x. d d tan y + x = 4 dx dx æ dy ö sec 2 y çç ÷÷÷ + 1 = 0 çè dx ø



Substitute the given values for dx / dt , x and y. æ dy ö÷ cos(0 + 0) çç -10 + ÷ çè dt ø÷

56.

+2-

+ 2(1 + x)

ö 1æ 1 dy = çç - y ÷÷÷ ÷ø ç dx x è cos( xy ) sec xy y x x æ dy ö + y ÷÷÷ = 1 cos( xy ) çç x çè dx ø

2 dt

æ ö çç 2 -  ÷÷ dy = 4 - 2 çè 2 ø÷ dt  4 -2 dy 8 - 4  = =  dt (4 -  )  22

f ( /3) = 3 - 2 /3 » -0.362 So there is an absolute maximum of 0 at x = 0 and an absolute minimum of 1 -  /2 at x =  /4.

 dy

d sin x = cos x dx sin(0 + h) » sin 0 + h ( cos 0 )

sin(0.03) » 0 + (0.03)(1) = 0.03 To four places, sin(0.03) = 0.0300. The absolute difference to four places is 0.

60.

d cos x = - sin x dx cos(0 + h) » cos 0 + h ( - sin 0 )

cos(-0.002) » 1 + (-0.002)(0) = 1 To four places, cos(-0.002) = 1.0000. The absolute difference to four places is 0.

dy = 0 dt

Section 13.2 61.

1001

R(t ) = 120 cos 2 t + 150

(a)

R¢(t ) = 120 ⋅ (- sin 2 t )(2 ) + 0 = -240 sin 2 t

1 (for 1 of a year). (b) Replace t with 12 12

æ 1ö æ 1 ö R¢ çç ÷÷ = -240 sin 2 çç ÷÷ çè 12 ÷ø çè 12 ÷ø = -240 sin



63.

y =

æ 1ö cos 3 çç t - ÷÷÷ ç è 8 3ø



(a) The graph should resemble the graph of y = cos x with following difference: The

6 æ 1 ö÷ = -240 çç ÷÷ çè 2 ø

maximum and minimum values of y are 8

= -120

2 3

and - 8 . The period of the graph will be

R¢(t ) for August 1 is -$120 per year.

= 23 units. The graph will be shifted

horizontally 13 units to the right.

6 = 1 of a year (c) January 1 is 6 months, or 12 2

for July 1. Replace t with 12 . æ1ö æ1ö R¢ çç ÷÷÷ = -240 sin 2 çç ÷÷÷ èç 2 ø èç 2 ø = -240 sin  = -240 (0) =0

(b) velocity =

R¢(t ) for January 1 is $0 per year. 11 of a year from July 1. Replace (d) June 1, is 12 11 . t with 12

æ 11 ö æ 11 ö R¢ çç ÷÷ = -240 sin 2 çç ÷÷ çè 12 ÷ø çè 12 ÷ø 11 6 æ 1 ö÷ = -240 çç - ÷÷ çè 2 ø = -240 sin

= 120 R¢(t ) for June 1 is $120 per year.

62.

dy dt

é æ 1 öù = Dt ê cos 3 çç t - ÷÷÷ ú ê8 èç 3 ø úû ë é æ  1 öù = Dt ê cos 3 çç t - ÷÷÷ ú ê èç 8 3 ø úû ë é æ æ é 1 öù 1 öù = ê - sin 3 çç t - ÷÷÷ ú Dt ê 3 çç t - ÷÷÷ ú çè ê çè 8 êë 3 ø úû 3 ø úû ë æ é 1 öù = ê - sin 3 çç t - ÷÷÷ ú ⋅ 3 ç è 8 êë 3 ø úû é æ 3 2 1 öù sin ê 3 çç t - ÷÷÷ ú =ê èç 8 3 ø úû ë

acceleration =

We are given that C (t ) = 80.63 sin(0.8998t + 1.154) + 416.1. (a) C ¢(t ) = (0.8998)(80.63) cos(0.8998t + 1.154) = 72.55 cos(0.8998t + 1.154) (b) C ¢(3) » -54.93 In March, the amount of electricity consumed is decreasing by about 54.9 billion BTUs per month.

d2y dt 2

1002

Chapter 13 THE TRIGONOMETRIC FUNCTIONS é -3 2 sin 3 t - 1 ù ( 3)ú ê = Dt ê ú 8 ê ú êë úû 2 é æ 1 öù -3 Dt ê sin 3 çç t - ÷÷÷ ú = ç ê è 8 3 ø úû ë

æ4ö æ4 9 3 1ö cos 3 çç - ÷÷ a çç ÷÷ = çè 3 ÷ø ç è3 8 3 ø÷ =-

é æ æ 1 öù 1 öù -3 2 êé cos 3 çç t - ÷÷÷ ú Dt ê 3 çç t - ÷÷÷ ú ê çè èç 8 êë 3 ø úû 3 ø úû ë

æ 1 öù -3 2 êé cos 3 çç t - ÷÷÷ ú ⋅ 3 èç 8 êë 3 ø úû

(c)

9 3 9 3 (-1) = >0 8 8 æ4ö æ4  1ö y çç ÷÷ = cos 3 çç - ÷÷ ÷ èç 3 ø èç 3 8 3 ø÷ =-

é æ 9 1 öù cos ê 3 çç t - ÷÷÷ ú ç ê è 8 3 ø úû ë d2y dt

æ 9 3 1ö cos 3 çç t - ÷÷÷ çè 8 3ø é æ 1 öù + 9 2 ê cos 3 çç t - ÷÷÷ ú ê8 èç 3 ø úû ë æ 9 3 1ö cos 3 çç t - ÷÷÷ =çè 8 3ø

(-1) = -

 8

9 3 cos (4 ) 8

9 3 ⋅1 8

9 3 <0 8 æ5ö æ5 1ö  y çç ÷÷ = cos 3 çç - ÷÷ çè 3 ÷ø ç è3 8 3 ÷ø =-

==-

æ 9 3 1ö cos 3 çç t - ÷÷÷ ç è 8 3ø 9 3 cos 2 8

9 3 ⋅1 8

9 3 <0 8 æ  1ö y(1) = cos 3 çç t - ÷÷÷ ç è 8 3ø

æ5ö æ5 9 3 1ö a çç ÷÷ = cos 3 çç - ÷÷ çè 3 ÷ø ç è3 8 3 ø÷

æ 9 3 1ö cos 3 çç t - ÷÷÷ çè 8 3ø

a(1) = -



arm is moving counterclockwise.

(d)

cos (3 )

counter clockwise and the arm makes an angle of - 8 radians from the vertical. The



Therefore, at t = 43 seconds, the force is

+ 9 2 y

9 3 cos (3 ) 8

 8



cos (4 ) ⋅1



Therefore, at t = 53 seconds, the answer corresponds to t = 1 second. So the arm is moving clockwise and makes an angle of 8π

cos 2 ⋅1 =



from the vertical.



8 8 Therefore, at t = 1 second, the force is clockwise and the arm makes an angle of 8

radians forward from the vertical. The arm is moving clockwise.

64.

(a) The following is a table of values for y =

1 sin π(t - 1). 5

0.5 1.0 1.5 2.0

-0.2 0

0.2 0

2.5 2.0 -0.2 0

Section 13.2

1003 At t = 2.5 sec, a(2.5) = =-

(b)

dy dt é1 ù = ê cos  (t - 1) ú ⋅ Dx [ (t - 1)] êë 5 úû

v(t ) =



a(t ) = v¢(t ) =

(c)



2

a(3.5) = -

sin [ (t - 1)]

1 + ( 2 ) sin  (t - 1) 5 =0

a(1.5) = ===-

2 5

y =

sin

65.



> 0,

2 5

sin  (2.5) sin

5 2

⋅1 = -

2 5

< 0,

1 5 1 1 = ⋅1 = . sin 5 2 5 5

Thus, at t = 3.5, acceleration is negative, the arm is moving clockwise, and the arm is at an angle of 15 radian from vertical.

sin  (t - 1) sin  (0.5)

2

(-1) =

and

(d) Since the constant of proportionality is positive, the force and acceleration are in the same direction. At t = 1.5 sec, a(t ) = -

2

3 2

At t = 3.5 sec,

sin  (t - 1)

+  2y = 2

sin

Thus, at t = 2.5, acceleration is positive, the arm is moving counterclockwise, and the arm is at angle of - 15 radian from vertical.

d2y

2

sin  (1.5)

1 3 1 1 = (-1) = - . sin 5 2 5 5

y =

dt é  ù = ê - sin  (t - 1) ú ⋅ Dx [ (t - 1)] ëê 5 ûú 2

and

cos  (t - 1)

2

L(t ) = 0.022t 2 + 0.55t + 316 + 3.5 sin(2 t ) (a) L(t) 0.022t 2 0.55t 316 3.5sin (2pt)

360

⋅1 < 0,

and 1 1 y = sin  (t - 1) = sin  (0.5) 5 5 1 1 1  = sin = ⋅1 = . 5 2 5 5

Thus, at t = 1.5, acceleration is negative, the arm is moving clockwise, and the arm is at an angle of 15 radian from vertical.

310

(b) L(25) = 343.5 parts per million L(35.5) = 363.25 parts per million L(50.2) = 402.38 parts per million (c) L¢(t ) = 0.044t + 0.55 + 7.0 cos(2 t ) L¢(50.2) = 9.55 parts per million per year At the beginning of 2010, the level of carbon dioxide was increasing at 9.55 parts per million per year.

1004 66.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS (a) Using the calculator to graph

(e)

C (t ) = 0.04t + 0.6t + 330 + 7.5sin(2 t ) in a [0, 25] by [320,380] viewing window, gives the following graph.

(b)

C (25) = 0.04(25) 2 + 0.6(25) + 330

(f) Set f ¢(t ) = 2000 cos(t )e2sin(t ) equal to zero to find critical numbers. 2000 cos(t )e2sin(t ) = 0

+ 7.5sin[2 (25)]

cos(t )e2sin(t ) = 0

= 370 parts per million C (35.5) = 0.04(35.5)2 + 0.6(35.5) + 330 + 7.5sin[2 (35.5)] = 401.71 parts per million C (50.2) = 0.04(50.2) 2 + 0.6(50.2) + 330 + 7.5sin[2 (50.2)] » 468.05 parts per million

+ 15 cos[2 (50.2)] » 19.18 parts per million per year. The level of carbon dioxide will be increasing at the beginning of 2010 at 19.18 parts per million.

(a)

for all t

We will use the second derivative test. f ¢¢(t ) = -2000sin(t )e2sin(t ) + 4000 cos 2 (t )e2sin(t ) = 2000e 2sin t (- sin t + 2 cos 2 t )

when - sin t + 2cos 2 t > 0, when sin t < 2 cos 2 t. Also, f ¢¢(t ) < 0 when - sin t + 2cos 2 t < 0, when

sin t > 2 cos 2 t. Since

(

sin 2 + n

) < 0 = 2 cos2 ( 2 + n ) for

n any odd integer, and

(

sin 2 + n

f (t ) = 1000e2 sin(t )

) > 0 = 2cos2 ( 2 + n )

for n any even integer, f has a maximum at t = 2 + n for n any even integer and f

f (0.2) = 1000e 2 sin(0.2) » 1488

(b)

e2sin(t ) ¹ 0,

Since 2000e2sin t > 0 for all t, f ¢¢(t ) > 0

C ¢(50.2) = 0.08(50.2) + 0.6

67.



+ n 2 for n any integer t =

or e 2sin(t ) = 0

cos(t ) = 0

has a minimum at t = 2 + 2 n for n any

f (t ) = 1000e2 sin(t )

odd integer. This is equivalent to f having a maximum for t = 2 + 2 n for n any

f (1) = 1000e 2 sin(1) » 5381

(d) Since f ¢(t ) = 2000 cos(t )e2sin(t ) , f ¢(0.2) is given by f (0.2) = 2000 cos(0.2)e2 sin(0.2)

integer and f having a minimum for t = 32 + 2 n for n any integer. To find the maximum and minimum values we will

( ) and f ( 32 ) respectively.

find f 2

( ) » 7389

( ) = 1000e

2sin 2

( 32 ) = 1000e

2sin 32

f 2

» 2916.

( ) » 135

Section 13.2 68.

1005 can be used to find a more precise value. The maximum velocity occurs at 4.944 radians.

s( ) = 2.625cos  + 2.625(15 + cos 2  )1/2

(a) The position of the piston when  = 2 is

given by

69.

(a) Using the graphing calculator to graph

æ ö s çç ÷÷÷ = 2.625(0) + 2.625(15 + 0)1/2 çè 2 ø

P(t ) = 0.003sin(220 t ) + 0.003 sin(660 t ) 3

= 2.625 15 » 10.17 in.

(b)

+ 0.003 sin(1100 t ) 5

æ1ö ds = 2.625(- sin  ) + 2.625çç ÷÷÷ çè 2 ø d

+ 0.003 sin(1540 t ) 7

in a [0,0.01] by [-0.004,0.004] viewing window, gives the following graph.

(15 + cos 2  )-1/2 ⋅ 2 cos  (- sin  ) = -2.625sin  -

2.625sin  cos 

æ ç = -2.625sin  çç1 + çç è

15 + cos 2  ö÷ ÷÷÷ 15 + cos 2  ÷ø÷ cos 

(c) One way to find the maximum velocity of the piston is to take the second derivative of s( ) and then find the values of  where the second derivative is equal to zero. Using the second derivative, we can determine which critical numbers maximize ds /d . Because the second derivative is so complicated, we will employ a graphing calculator to find the critical number(s). The figures below show the calculatorgenerated graphs of the functions ds /d

(b) Using the graphing calculator to find P( x) and evaluate P(0.002) gives .151. However, the slope of the graph of P( x) at x = 0.002 appears to be approximately -1. Thus, we calculate P¢( x) by hand and use the

calculator to evaluate P¢(0.002). Since P¢(t ) = 0.66 cos(220 t ) + 0.66 cos(660 t ) + 0.66 cos(1100 t ) + 0.66 cos(1540 t ), P¢(0.002) is given by

and d 2 s /d 2 , respectively.

P¢(0.002) = 0.66 cos[220 (0.002)] + 0.66 cos[660 (0.002)] + 0.66 cos[1100 (0.002)] + 0.66 cos[1540 (0.002)] » -1.05 pounds per square foot per second. At 0.002 seconds the pressure is decreasing at a rate of about 1.05 pounds per square foot per second. 70.

(a)

æ ö u( x, t ) = T0 + A0e-ax cos çç t - ax ÷÷÷ çè 6 ø æ ö = 16 + 11e-0.00706 x cos çç t - 0.00706 x ÷÷÷ çè 6 ø

The amplitude of  ( x, t ) is given by 11e-0.00706 x . We need to find where 11e-0.00706 x £ 1.

Although there are two critical numbers, the graph of ds /d shows that only one critical number is associated with maximum velocity. This occurs when  » 5. The equation solver on the graphing calculator

1006

Chapter 13 THE TRIGONOMETRIC FUNCTIONS satisfies the heat equation. (In our case w =  /6. ) ¶u = A0e-ax [- sin ( wt - ax )]( w) ¶t

11e-0.00706 x £ 1 1 11 æ1ö -0.00706 x £ ln çç ÷÷÷ çè 11 ø e-0.00706 x £

= -wA0e-ax sin ( wt - ax) ¶u = -aA0e-ax cos ( wt - ax) ¶x

1 ln ( 11 )

x£

+ A0e-ax [- sin ( wt - ax)](-a )

-0.00706 » 340 cm

= -aA0e-ax ⋅ [cos ( wt - ax )

The amplitude is at most 1C at a minimum depth of about 340 centimeters.

- sin ( wt - ax )] ¶ 2u

(b) We wish to find x for which 14 £ u ( x, t ) £ 18. Since æ ö -1 £ cos çç t - 0.00706 x ÷÷÷ £ 1, we have çè 6 ø

= a 2 A0e-ax [cos ( wt - ax) - sin ( wt - ax)]

16 - 11e-0.00706 x £ u( x, t ) £ 16 + 11e-0.00706 x .

= a 2 A0e-ax [cos ( wt - ax ) - sin ( wt - ax )]

¶x 2 - aA0e-x [a sin ( wt - ax ) + a cos ( wt - ax)] - a 2 A0e-ax [sin ( wt - ax ) + cos ( wt - ax )]

For 14 £ u ( x, t ) £ 18, we need to find

= -2a 2 A0e-ax sin( wt - ax)

where 16 - 11e-0.00706 x = 14 and

Now a =

16 + 11e-0.00706 x = 18.

w 2k

These two conditions are equivalent to -0.00706 x

11e

w 2k w

a2 =

= 2

k =

2 11 æ2ö -0.00706 x = ln çç ÷÷÷ çè 11 ø e-0.00706 x =

x =

( ) » 242 cm 2 11

¶ 2u ¶x 2

6 months when =6

0.00706 x = 

 0.00706

» 445 cm

w 2a 2

[-2a 2 A0e-ax sin( wt - ax)] ¶u . ¶t

Thus,

¶u ¶ 2u = k 2. ¶t ¶x 71.

x =

= -wA0e-ax sin( wt - ax) =

-0.00706

2a 2

It follows that

A minimum depth of about 242 centimeters will keep the wine at a temperature between 14C and 18C.



implies that

(a)

y = x tan  -

16 x 2 2

sec2 

V æ  ö÷ 16(40)2 æ ö = 40 tan çç ÷÷ sec2 çç ÷÷÷ 2 èç 4 ø èç 4 ø 44 = 40(1) -

A depth of about 445 centimeters gives a ground temperature prediction of winter when it is summer and vice versa. (d) We show that the more general function u( x, t ) = T0 + A0e-ax cos ( wt - ax)

16(40) 2 442

(2) » 13.55 feet

Section 13.2 (b)

1007

y = x tan  0 = x tan  -

16 x 2 V

sec2  2

x =

16 x 2

= 88 and  =

16 x V2

sec2 

(for x ¹ 0)

sec  = tan 

s( ) = 2.625cos  + 2.625(15 + cos 2  )1/2 ds ds d = ⋅ dt d dt d = éê -2.625sin  + (15 + cos2  )-1/2 ⋅ D (15 + cos2  ) ùú ⋅ ë û dt é ù 1.3125 ê ú d = ê -2.625sin  + ⋅ (-2sin  cos ) ú ⋅ 2 ê ú dt 15 + cos  ë û æç ö÷ cos ÷÷ d = -2.625sin  ççç 1 + ÷ 2 ÷÷ dt çè 15 + cos  ø

V 2 tan  ⋅ 16 sec2 

V2 ⋅ sin  cos  16

V2 ⋅ 2sin  cos  32

V2 sin(2 ) 32

(b) With  = 4.944 and d = 505,168.1 radians , we have hour dt

V2 x = sin(2 ) 32 é æ  öù 442 sin ê 2 çç ÷÷÷ ú = ê èç 3 ø ú 32 ë û » 52.39 feet

(d)

72. (a)

x =

(c)

16 x

dx V2 = cos(2 ) ⋅ D (2 ) d 32

ds = -2.625sin(4.944) dt é ù cos(4.944) ê ú ´ ê1 + ú (505,168.1) 2 ê 15 + cos (4.944) úúû êë inches hour inches 1 foot 1 mile = 1,367, 018.749 ´ ´ hour 12 inches 5280 feet » 1,367, 018.749

V2 = cos(2 ) 16 Find critical values.

» 21.6 miles per hour

V2 cos(2 ) = 0 16 cos(2 ) = 0 2 =

 2

73. + n , for n any integer

s(t ) = sin t + 2cos t

v(t ) = s¢(t ) = cos t - 2sin t

n , for n any integer 2

(a)

v(0) = 1 - 2(0) = 1

Since 0 <  < 2 , ddx = 0 for  = 4 .

(b)

æ 2 ö÷ æ ö 2 ÷÷ = - 2 - 2 ççç v çç ÷÷÷ = çè 4 ø çè 2 ø÷ 2 2

 =



. 4 é æ  öù 882 = sin ê 2 çç ÷÷÷ ú = 242 feet ç ê ú 32 ë è 4 øû

sec2  2

V æ ö 16 x 0 = x çç tan  - 2 sec2  ÷÷ çè ÷ ø V 0 = tan  -

V2 sin(2 ) when V 32

Furthermore, d x2 = - V8 sin(2 ) which is

» -0.7071

d

less than zero at  =  . Therefore, x is 4

maximized when  =  .

(c)

(e) Since 60 miles per hour is 88 feet per second, evaluate

æ 3 ö v çç ÷÷÷ = 0 - 2(-1) = 2 çè 2 ø a(t ) = v¢(t ) = - sin t - 2 cos t

(d)

a(0) = 0 - 2(1) = -2

1008

Chapter 13 THE TRIGONOMETRIC FUNCTIONS

(e)

Since

æ 2 ÷ö æ ö 2 3 2 a çç ÷÷÷ = - 2ççç ÷÷ = çè 4 ø ÷ çè 2 ø 2 2

dx = 600, dt 600 d = dt 60sec2  10 = . sec 2 

» -2.1213

(f) 74.

(a)

a (  ) = -0 - (-2) = 2 x 50 Differentiate both sides with respect to time, t. tan  =

(a) When the car is at the point on the road closest to the camera  = 0. Thus, d 10 = dt (sec 0)2

æ x ö Dt (tan  ) = Dt çç ÷÷÷ çè 50 ø sec 2  ⋅

d 1 dx = ⋅ dt 50 dt

= 10 radians/min 12 d 10 radians 1 rev = ⋅ dt min 2 radians

Since the light rotates twice per minute, d dt

2(2 radians) = 4 1 min

radians per

minute. When the light beam and shoreline are at right angles,  = 0 and sec  = 1. Thus,

rev/min

10 10 d = = 2 dt ( 2)2 sec 4

(

1 dx ⋅ 50 dt

This is one-half the previous value, so d 1 5 rev/min = ⋅ dt 2  5 rev/min. = 2

76.

dx = 400 . dt

The beam is moving along the shoreline at 400 m/min. From the figure, we see that

)

= 5 radians/min

(b) When the beam hits the shoreline 50 m from the point on the shoreline closest to the lighthouse,  = 4 and sec  = 2. Thus,

Let x be the length of the ladder and let y be the distance from the wall to the bottom of the ladder. Then y+2 x x cos  = y + 2 cos  =

and and

y 9 y = 9 cot .

cot  =

Thus, x cos  = 9 cot  + 2 9 cot  + 2 x = cos  x = 9 csc  + 2 sec  .

x tan  = 60 60 tan  = x.

Differentiating with respect to time, t, gives 60sec 2  ⋅



the car has traveled 60 feet. Thus, tan  = 60 and  = 4 . 60

The beam is moving along the shoreline at 200 m/min.

75.

1 of a minute later, and (b) Six seconds later is 10

1 dx ⋅ (1)2 (4 ) = 50 dt dx = 200 . dt

( 2)2 (4 ) =

d dx = dt dt

This expression gives the length of the ladder as a function of . Find the minimum value of this function.

d dt = . dt 60sec 2 

Section 13.2

1009

dx = -9 csc  cot  + 2 sec  tan  d æ 1 öæ ÷÷çç cos  ö÷÷ + 2æçç 1 öæ ÷÷çç sin  ö÷÷ = -9ççç ÷ ÷ ÷÷ç cos  ÷÷ ç ç ÷ sin  ø÷ è sin  øè è cos  øè ø =

-9 cos 

dx d

sin 2 

2 sin  cos 2 

= 0, then 2sin  cos 2 

9 cos  sin 2 

10 + 5 cot  = 10 sec  + 5 csc  cos 

dx = 10 sec  tan  - 5 csc  cot  d æ 1 öæ ÷÷çç sin  ÷÷ö - 5æçç 1 öæ ÷÷çç cos  ö÷÷ = 10 ççç ÷ ÷ ÷ ÷ ç ç è cos  ÷øè cos  ÷ø è sin  ÷øèç sin  ø÷ 10 sin  cos 

10sin 

cos 

5 cos  sin 2 

5cos  sin 2 

10sin 3  = 5cos3  sin 3 

1 2 cos  1 tan 3  = 2

9 tan  = 3 2  » 1.02619 radians.

If  < 1.02619, ddx < 0.

1 2  » 0.6709 radian.

tan  = 3

If  > 1.02619, ddx > 0. Therefore, there is a minimum when  = 1.02619. If  = 1.02619,

If  < 0.6709 radian, ddx < 0. If  > 0.6709 radian, ddx > 0.

x » 14.383.

The minimum length of the ladder is approximately 14.38 ft. 77.

If ddx = 0, then

9 2 cos  9 tan 3  = 2 3

x =

2sin 3  = 9 cos3  sin 3 

Thus,

Let x represent the length of the ladder.

Thus, x will be minimum when  = 0.6709. If  = 0.6709, then x =

10 + 5 cot 0.6709 » 20.81. cos 0.6709

The length of the longest possible ladder is approximately 20.81 feet.

tan  =

5 y

10 + y x 10 + y x = cos 

and cos  =

y = 5cot 

and

1010

Chapter 13 THE TRIGONOMETRIC FUNCTIONS

13.3

Integrals of Trigonometric Functions

Find

ò x cos (3x) dx.

Find

Your Turn 1 (a)

Your Turn 3

æxö

ò sin çççè 2 ÷÷÷ø dx.

Let u = x and dv = cos (3x ) dx. Then du = dx and v =

If u = x /2, then du = (1/2) dx. æxö

æ1

ò x cos(3x) dx = ò u dv = uv - v du ò

ò sin çççè 2 ÷÷÷ø dx = 2ò sin xçççè 2 dx ÷÷÷ø = 2 sin u du ò

x 1 sin (3x) sin (3x) dx 3 3 ö x 1æ 1 = sin (3x) - çç - cos(3x) ÷÷÷ + C ç ø 3 3è 3

= 2(- cos u) + C æxö = -2cos çç ÷÷÷ + C çè 2 ø

Find

(b)

ò 6(sec x) tan x dx. 2

If u = tan x, then du = sec 2 x dx. 2

= 4u 3/2 + C

 4 

Your Turn 2

Find

tan( x ) dx. x

1 x sin (3x) + cos(3x) + C 3 9



y  sec



( 3)

2 x

4 x

 æxö æ x öæ 1 ö sec2 çç ÷÷ dx = 3 sec2 çç ÷÷çç dx ÷÷ ÷ ÷ç 3 ø÷ ç è ø èç 3 øè 3 - -



x , then du =

If u =

y 4

= 4 (tan x)

Find the area under the curve y = sec2 ( x /3) between x = - and x =  .

æ2 ö = 6 çç u 3/2 ÷÷÷ + C çè 3 ø 3/2

Your Turn 4

ò 6(sec x) tan x dx = 6ò tan x (sec x dx) =6 ò u du 2

1 sin 3x. 3

1 2 x

tan( x ) dx = 2 x

æxö = 3 tan çç ÷÷÷ çè 3 ø -

dx.

æ 1

ò tan ( x )çççè 2 x dx ø÷÷÷ = 2 tan u du ò = -2 ln |cos u | + C = -2 ln |cos( x )| + C

æ æ ö æ  öö = 3ççç tan çç ÷÷ - tan çç - ÷÷ ÷÷÷ ÷ ç çè 3 ÷ø ø÷ è è3ø = 3[ 3 - (- 3)] =6 3

Section 13.3

1011

13.3 Warmup Exercises

W1. Let u = x 2 , then du = 2 x dx. 2 1 xe x dx = eu du 2 1 = eu + C 2 1 x2 = e +C 2



(sin 2 x + cos 2 x)dx =



=

ò cos3x dx Let u = 3x, so du = 3 dx or 13 du = dx.

ò cos3x dx = ò cos u ⋅ 3 du 1 = cos u du 3ò 1 sin u + C 3 1 = sin 3x + C 3 =

W3. Let dv = e x dx and u = x + 1. Then v = e x and du = dx.

ò (x + 1)e dx = ò u dv = uv ò v du



ò 1 dx

= x

W2. Let u = x 2 + 6, then du = 2 x dx. x 1 1 dx = du 2 u 2 x +6 1 = ln u + C 2 1 = ln( x 2 + 6) + C 2

True. Recall sin 2 x + cos 2 x = 1.

= ( x + 1)e x -

Let u = 5 x, so du = 5 dx or 15 du = dx.

e x dx

ò sin 5x dx = ò sin u ⋅ 5 du 1 = sin u du 5ò

= ( x + 1)e - e x + C = xe x + C

W4. Let dv = 8x dx and u = ln x. 1 Then v = 4 x 2 and du = dx. x

ò 8 ln x dx = ò u dv = uv ò v du = 4 x 2 ln x ò (4x2 )(1/ x) dx

ò sin 5x dx

1 = - cos u + C 5 1 = - cos5 x + C 5

= 4 x 2 ln x - 2 x 2 + C

13.3 Exercises

ò (3cos x - 4sin x)dx = ò 3cos x dx - ò 4sin x dx = 3 cos x dx - 4 sin x dx ò ò = 3sin x + 4cos x + C

False.

ò tan x dx = - ln cos x + C

True.

sin x dx = - cos x + C

ò (9sin x + 8cos x)dx = ò 9sin x dx + ò 8cos x dx = 9 sin x dx + 8 cos x dx ò ò = -9cos x + 8sin x + C

1012 9.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS

ò x sin x dx 2

13.

ò sin x cos x dx 7

Let u = sin x.

Let u = x 2.

Then du = cos x dx.

Then du = 2 x dx 1 = du = x dx. 2

ò sin x(cos x) dx = ò u du 7

1 8 u +C 8 1 = sin8 x + C 8 =

1 sin u ⋅ du 2

ò x sin x dx = ò 1 = sin u du 2ò 2

1 cos u + C 2 1 = - cos x 2 + C 2 =-

14.

ò sin x cos x dx 4

Let u = sin x, so du = cos x dx.

ò sin x cos x dx = ò u du 4

10.

ò 2x cos x dx 2

1 5 u +C 5 1 = sin 5 x + C 5 =

Let u = x 2 , so that du = 2 x dx.

ò 2x cos x dx = ò cos u du 2

15.

= sin u + C = sin x 2 + C

11.

Let u = cos x, so du = -sin x dx or -du = sin x dx.

ò 3 sec 3x dx 2

ò 3 cos x (sin x) dx = ò 3 u (-du) = -3 u du ò

Let u = 3x, so du = 3 dx or 13 du = dx. -

ò 3 sec 3x dx = -ò 3sec u ⋅ 3 du = - sec u du ò 2

ò 3 cos x (sin x) dx

1/2

æ2 ö = -3çç u 3/2 ÷÷÷ + C çè 3 ø

= - tan u + C = - tan 3x + C

= -2u 3/2 + C = -2(cos x)3/2 + C

12.

ò 2 csc 8x dx 2

16.

Let u = 8 x, so that du = 8 dx. 1

Let u = sin x, so that du = cos x dx.

ò 2 csc 8x dx = - 4 ò csc 8x(8 dx) 1 (- csc u ) du = 4ò

cos x

ò sin x dx

cos x

ò sin x dx = ò sin x cos x dx = ò u du = 2u + C

1 cot u + C 4 1 = cot 8x + C 4 =

-1/2

= 2 sin1/2 x + C

1/2

Section 13.3

1013

ò ( x + 2) sin (x + 2) dx 1 = sin ( x + 2) ⋅ 5( x + 2) dx 5ò 1 sin u du = 5ò 1 cos u + C =5ò 4

17.

sin x dx 1 + cos x

Let u = 1 + cos x. du = - sin x

Then

-du = - sin x dx.

sin x

ò 1 + cos x dx = ò u (-du) = -ò u du = - ln |1 + cos x | + C

21. cos x

ò 1 - sin x dx

ò tan 3 x dx Let u = 13 x, so du = 13 dx or 3 du = dx. 1

ò tan 3 x dx = ò (tan u) ⋅ 3 du = 3 tan u du ò

Let u = 1 - sin x, that du = - cos x dx. cos x

ò 1 - sin x dx = - (1 - sin x) (- cos x) dx ò = - u du ò

= -3ln |cos u | + C

-1

= -3ln cos

-1

= - ln | u | +C = - ln |1 - sin x | + C

22.

19.

æ 3x ö

1 du = x 7 dx. 8

ò 2x cos x dx = ò 2(cos u) ⋅ 8 du 1 = cos u du 4ò 8

1 sin u + C 4 1 = sin x8 + C 4

ò (x + 2) sin ( x + 2) dx 4

Let u = ( x + 2) , so that du = 5( x + 2)4 dx.

æ 3x ö æ 3

8 = - ln |sin u | +C 3 æ 3x ö 8 = - ln sin çç - ÷÷÷ + C çè 8 ø 3

20.

æ 3x ö

ò cot ççèç - 8 ÷ø÷÷ dx ò cot çççè - 8 ÷÷÷ø dx = - 3 ò cot çççè - 8 ÷÷÷ø çççè - 8 dx ÷÷÷ø 8 =cot u du 3ò

Let u = x8 , so du = 8x 7 dx or

1 x +C 3

Let u = - 38x , so that du = - 83 dx. Then

ò 2x cos x dx 7

1 = - cos ( x + 2)5 + C 5

= - ln | u | + C

18.

23.

ò x5 cot x6 dx Let u = x 6 , so du = 6 x5dx or 16 du = x5dx. 1

ò x5 cot x6dx = ò (cot u) ⋅ 6 du 1 cot u du = 6ò 1 ln |sin u | +C 6 1 = ln |sin x 6 | + C 6 =

1014

24.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS æ x ö2 x tan çç ÷÷÷ dx çè 4 ø 4

28.

æ x ö2 x tan çç ÷÷÷ dx çè 4 ø 4

29.

= -2 ln |cos u | + C

ò -6x cos 5x dx

= - cos u + C = - cos e x + C

ò e-x tan e-x dx

30.

Let u = e-x , so that du = -e-x dx.

ò 7x sin 5x dx Let u = 7 x and dv = sin 5 x dx.

ò e- tan e- dx = -ò tan u du x

Then du = 7 dx and v = - 15 cos 5x.

= ln |cos u | + C = ln |cos e-x | + C

ò 7x sin 5x dx æ 1 ö = 7 x çç - cos 5 x ÷÷÷ çè 5 ø

æ 1

ò çççè - 5 cos 5x) ø÷÷÷ ⋅ 7dx 7 7 = - x cos 5 x + cos 5 x du 5 5ò

ò e csc e cot e dx x

Let u = e x , so that du = e x dx.

ò e csc e cot e dx = ò csc e cot e (e dx) = ò csc u cot u du x

æ1

ò çççè 5 sin 5x) ÷÷÷ø (-6 dx)

6 6 sin 5x dx = - sin 5 x + 5 5 6 6 1 = - x sin 5 x + ⋅ (- cos 5 x) + C 5 5 5 6 6 cos 5x + C = - x sin 5x 5 25

ò e sin e dx = ò sin u du x

1 sec x5 + C 5

æ1 ö = (-6 x) çç sin 5x ÷÷ ÷ø çè 5

Then du = e x dx.

ò -6x cos 5x dx

e x sin e x dx

Then du = -6 dx and v = 15 sin 5 x.

Let u = e x .

27.

Let u = -6 x and dv = cos 5x dx.

æ x ö2 = -2 ln cos çç ÷÷÷ + C çè 4 ø

ò = 2 tan u du ò

26.

æ x ö2 x tan çç ÷÷÷ dx çè 4 ø 8

= 2

ò x sec x tan x dx = 5 ò sec x tan x (5x dx) 1 = sec u tan u du 5ò 1 = sec u + C 5ò

æ x öæ 1 ö x du = 2çç ÷÷÷ çç dx ÷÷÷ = dx. çè 4 øèç 4 ø 8

25.

Let u = x5 , so that du = 5 x 4dx

æ x ö2 Let u = çç ÷÷÷ , so that çè 4 ø

ò x sec x tan x dx

7 7 1 = - x cos 5 x + ⋅ sin 5 x + C 5 5 5 7 7 = - x cos 5 x + sin 5x + C 5 25

= - csc u + C = - csc e x + C

Section 13.3 31.

1015

ò 4x sin x dx

ò -6x cos 8x dx 2

3 = - x 2 sin 8 x 4 3é 1 + ê - x cos 8 x 2 êë 8

Let u = 4 x and dv = sin x dx. Then du = 4 dx and v = - cos x.

ò 4x sin x dx ò (-cos x) ⋅ 4 dx = -4 x cos x + 4 cos x du ò

= -4 x cos x + 4sin x + C

ò -11x cos x dx Let u = -11x and dv = cos x dx. Then du = -11 dx and v = sin x.

ò -11x cos x dx

34.

ò (sin x)(-11dx) = -11x sin x + 11 sin x dx ò = -11x sin x -

ò 10x sin 2 x dx 2

Let u = 10 x 2 and dv = sin 12 x dx Then du = 20 x dx and v = -2 cos

= -11x sin x - 11cos x + C

ò -6x2 cos 8x dx Let u = -6 x 2 and dv = cos 8 x dx. Then du = -12 x dx and v = 18 sin 8x.

æ1 ö çç sin 8x ÷÷ (-12 x dx) ÷ø çè 8

3 3 = - x 2 sin 8x + 4 2

ò x sin 8x dx

ò x sin 8x dx, let u = x

and dv = sin 8x dx.

Then du = dx and

æ 1 ö = (10 x 2 ) çç -2 cos x ÷÷÷ çè 2 ø æ

1 ö

ò ççèç -2cos 2 x ÷÷ø÷ (20x dx) 1 1 = -20 x cos x + 40 x cos x dx ò 2 2 -

Let u = x and dv = cos 12 x dx.

æ1 ö = (-6 x 2 ) çç sin 8x ÷÷÷ çè 8 ø

-6 x 2 cos 8 x dx

1 x. 2

ò 10x sin 2 x dx 2

33.

3 3 = - x 2 sin 8 x x cos 8 x 4 16 3 cos 8 x dx + 16 3 3 = - x 2 sin 8 x x cos 8 x 4 16 3 1 + ⋅ sin 8 x + C 16 8 3 3 = - x 2 sin 8 x x cos 8 x 4 16 3 + sin 8 x + C 128

= 4 x(- cos x) -

32.

æ 1

v = - 18 cos 8x.

Then du = dx and v = 2sin 12 x. 1

ò 10x sin 2 x dx 2

= -20 x 2 cos

1 x 2

æ 1 + 40 çç 2 x sin x çè 2

2 sin

1 1 x + 80 x sin x 2 2 1 + 160 cos x + C 2

= -20 x 2 cos

ò çççè - 8 cos 8x ÷÷÷ø dx úúû

ö 1 x dx ÷÷÷ ø 2

1016

35.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS

 /4

sin x dx = - cos x 0 = - cos



39.

37.

2 /3

cos x dx = sin x  /2

- ( - cos 0 )

4 2 =+1 2 2 = 12

36.

ò

2 /3

= sin

3 -1 2

40.

 2 - sin 3 2

 /4

ò sin x dx = - cos x  /6

ò  cos x dx = sin x 

0 - /2

- /2

= - cos

æ ö = sin 0 - sin çç - ÷÷÷ çè 2 ø

= 0 - (-1) =1

 /6

æ ö - çç - cos ÷÷÷ ç 4 è 6ø



2 3 + 2 2 3- 2 2

 /6

ò0 tan x dx = -ln |c os x | 0 = - ln cos

 6

41. - ( - ln |cos 0| )

3 + ln1 2 3 = - ln +0 2 3 = - ln 2

Use the fnInt function on the graphing calculator to enter fnInt (e ^ (-x) sin ( x), x, 0, b), for successively larger values of b, which returns a value of 0.5 for sufficiently large enough b. Thus, ¥

an estimate of ò 0 e-x sin x dx is 0.5.

= - ln

42.

Use the fnInt function on the graphing calculator to enter fnInt (e ^ (-x) cos x, x, 0, b), for successively larger values of b, which returns a value of 0.5 for sufficiently large enough b. Thus, ¥

38.

an estimate of ò 0 e-x cos x dx is 0.5.

 /2

ò cot x dx /4

43.

 /2

= ln |sin x |  /4 = ln sin

 2

= ln 1 - ln = - ln

- ln sin

 4

f ( x) = sin x on [0, 3 /2] f is positive to the left of x =  and negative to the right of this point, so we compute the area as the sum of two integrals:  æ 3 / 2 ö÷ Area = sin x dx + ççç - sin x dx ÷÷÷ ÷ø çè  0

2 2

= ( - cos x )

 0

- ( - cos x )

3 / 2 

= (1 - (-1) ) - ( 0 - 1)

æ 2 ö÷-1 1 or ln ççç ÷÷÷ = ln 2 = ln 2 2 èç 2 ø

= 3

44.

f ( x) = tan x on [- /4,  /3] f is negative to the left of x = 0 and positive to the right of this point, so we compute the areas as

Section 13.3

1017

the sum of two integrals: Area =

47.  /3

ò  - tan x dx + ò - /4

Sketch the graphs:

tan x dx

= - ( ln sec x )

(

0 - / 4

+ ( ln sec x )

 /3 0

)

= - ln 1 - ln 2 + ( ln 2 - ln 1) =

45.

1 3 ln 2 + ln 2 = ln 2 » 1.040 2 2 Area =

Sketch the graphs:

 /4

(tan x - sin x ) dx

= ( ln sec x + cos x )

 /4 0

æ 2 ö÷÷ = ççç ln 2 + ÷ - ( ln1 + 1) çè 2 ÷ø =

Area =

 /4

(cos x - sin x) dx

48.

1 2 -1 ln 2 + 2 2

Sketch the graphs;

= ( sin x + cos x )

 /4 0

æ 2 2 ö÷÷ = ççç + ÷ - ( 0 + 1) çè 2 2 ø÷ =

46.

2 -1

Since the graph is symmetric around x =  /2 , we can double the result of adding the integral from 0 to  /6 of (1 - sin x) - sin x to the integral from  /6 to  /2 of sin x - (1 - sin x).

Sketch the graphs:

é  /6 ù ê ú (1 - 2sin x) dx ê ú 0 ú Area = 2 êê ú  /2 ê ú + (2sin x - 1) dx ú ê êë úû  /6  /6 é ù ê ( x + 2 cos x ) ú 0 ê ú = 2ê ú / 2  ê ú x x + + 2 cos ( ) ê  / 6 úû ë éæ  ù ö ê çç + 3 ÷÷ - ( 0 + 2 ) ú ÷ ê èç 6 ú ø ê ú = 2ê ú æ ö æ ö   ê çç - + 2 ÷÷ - çç - - 3 ÷÷ ú + ê ÷ø èç 6 çè 2 ø÷ úûú ëê

Area =

 /4

(sec2 x - sin 2 x ) dx

 /4

æ ö 1 = çç tan x + cos 2 x ÷÷÷ çè ø0 2 æ 1ö = ( 1 + 0 ) - çç 0 + ÷÷÷ çè 2ø =

1 2

= 4 3-4-

 3

1018

49.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS æ ö S (t ) = 500 + 500 cos çç t ÷÷÷ çè 6 ø

Total sales over a year’s time are approximated by the area under the graph of S during any twelve-month period due to periodicity of S. Total sales »

ò S (t) dt 0 12 é

æ öù ê 500 + 500 cos çç  t ÷÷ ú dt ê èç 6 ÷ø úû 0 ë 12 12 æ ö 500 dt + 500 cos çç t ÷÷÷ dt çè 6 ø 0 0

çè

= 6000 +

50.

(f) The function in the model has a period of 2 T = b 2 = 0.3860 » 16.3 months.

One might expect a period of 12 months, but longer-term trends might make this annual period hard to see.

æ ö ç6 12 = 500t 0 + 500 çç sin çç t ÷÷÷ ç çè 6 ø = 6000 +

(e) Integrating C(t) from 0 to 12 using a calculator gives the estimate 1020 trillion BTUs; the sum of the values in the table is 1002 trillion BTUs.

3000

 3000



12 ö

÷÷ ÷÷÷ 0 ÷ø

51.

æ ö T (t ) = 50 + 50 cos çç t ÷÷÷ çè 6 ø

Since T is periodic, the number of animals passing the checkpoint is equal to the area under the curve for any 12-month period. Let t vary from 0 to 12.

(sin 2 - sin 0) (0 - 0) = 6000

Total =

12 é

æ öù ê 50 + 50 cos çç  t ÷÷ ú dt çè 6 ø÷ ú ê ë û

(a) It is not clear whether petroleum consumption is likely to be periodic or not.

ò 50 dt 0

æ  öæ 

= 50t

+ 0

æ ö sin çç t ÷÷÷ çè 6 ø  0

300

= (600 - 0) +

300



(0 - 0)

= 600 (in hundreds) The total number of animals is 60,000.

(b) C (t ) = 75.02sin(0.3860t + 2.004) + 107.0 52.

V (t ) = 170 sin (120πt ) T

V 2 (t ) dt ò 0 Root mean square = T

(a) The period is T = (c) C (4) » 77.3, so our estimate for April is 77.3 trillion BTUs; the actual value is 70 trillion BTUs. (d) C ¢(t ) = 0.386(75.02) cos(0.386t + 2.004) = 38.9577 cos(0.3860t + 2.004)

we find that C ¢(4) » -26.6, so in April the rate of consumption is decreasing at about 26.6 trillion BTUs per month.

50 cos çç t ÷÷ çç dt ÷÷ çè 6 ÷øèç 6 ÷ø  ò

2π 1 = sec. 120π 60

Chapter 13 Review

1019 55.

(b) T

The amount of water in the tank is k

ò sec t dt = tan t

V 2 (t ) dt

1/60

= tan k - tan 0

[170 sin (120  t 2 )]2 dt

= tan k (“tank”).

= 1702 = 1702

1/60

0 1/60

1702 2

sin 2 (120 t ) dt

56.

1 [1 - cos (240  t )] dt 2

1/60

[1 - cos (240  t ) dt

1702 ⋅0 2 1702 = 120 Root mean square 1702 120 1 60

2 = 170 » 120.21 2

Thus, 120 volts is the root mean square value for V(t). 53.

Chapter 13

ò0

365

ò0

N (t )dt =

[183.549sin(0.0172t

- 1.329) + 728.124]dt é 183.549 cos(0.0172t = êêë 0.0172 -1.329) + 728.124 ]

365

False; The period of cosine is 2 .

True

False; There’s no reason to suppose that the Dow is periodic.

False; cos(a + b) = cos a cos b - sin a sin b

True

False; Dx tan( x 2 ) = 2 x sec2 ( x 2 )

True

False; The secant function has no absolute maximum or minimum

False; Since the sine function is negative on half of this interval, the true area is ò

» 265,819.0192 minutes » 4430 hours.

The amount of water in the sink is k

ò cos t dt = sin t 0

= sin k (“sink”).

|sin x | dx.

False; Use substitution, with u = 5 + sin x and du = cos x dx.

14.

The exact value for the trigonometric functions can be determined for any integer multiple of 6 or 4 .

k 0

= sin k - sin 0

2

10.

The result is relatively close to the actual value. 54.

Review Exercises

The total amount of daylight is given by 365

ò -sin t = cos t + C

Then a formula for cost is C (t ) = cos t (“cost”).

1/ 60

æ 1702 é 1 1 1 ÷ö ùú ê sin çç 240 ÷ ç ê è 2 ë 60 240 60 ÷ø úû

C (t ) =

When t = 0, cos 0 + C = 1 1+C = 1 C = 0.

ù 1702 é 1 êt sin (240 t ) ú = úû 2 êë 240 0 =

k 0

15.

æ  ö÷ 90  90 = 90 çç = = çè 180 ÷÷ø 180 2

1020

Chapter 13 THE TRIGONOMETRIC FUNCTIONS

16.

æ  ö÷ 8 160 = 160 çç = çè 180 ÷÷ø 9

17.

æ  ö÷ 5 225 = 225 çç = çè 180 ÷ø÷ 4

26.

When an angle of 150 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (- 3,1). Then r =

x2 + y 2 =

3 + 1 = 2,

so 18.

æ  ö÷ 3 270 = 270 çç = çè 180 ÷÷ø 2

19.

æ 180 ö÷ 5 = 5 çç = 900 çè  ÷÷ø

20.

3 3 æç 180 ö÷ = ÷ = 135 ç 4 4 çè  ÷ø

21.

9 9 æç 180 ö÷ = ÷ = 81 ç 20 20 çè  ÷ø

22.

3 3 æç 180 ö÷ = ÷ = 54 ç 10 10 çè  ÷ø

sec 150 =

27.

r 2 2 3 . = =3 x - 3

When an angle of 120 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (-1, 3). Then r =

x 2 + y 2 = 1 + 3 = 2,

23.

28.

When an angle 60 is drawn in standard position, one choice of a point on its terminal side is ( x, y ) = (1, 3). Then r =

x 2 + y 2 = 1 + 3 = 2,

24.

29.

25.

y 3 . = r 2

y = - 3. x

x2 + y 2 = 1 + 1 =

When an angle of 6 is drawn in standard position,

r =

x2 + y 2 =

sin

 6

y 1 = . r 2

When an angle of 73 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (1, 3). Then r =

x 2 + y 2 = 1 + 3 = 2,

cos cos(-45) =

3 + 1 = 2,

30.

When an angle of -45 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (1, -1). Then r =

x 1 3 ==. y 3 3

one choice of a point on its terminal side is ( x, y ) = ( 3,1). Then

When an angle of 120 is drawn in standard position, ( x, y) = (-1, 3) is one point on its terminal sides, so tan120 =

When an angle of 300 is drawn in standard position, ( x, y) = (1, - 3) is one point on its terminal side, so cot 300 =

so sin 60 =

r 2 2 3 = = . y 3 3

csc 120 =

x 1 2 . = = r 2 2

7 x 1 = =- . 3 r 2

Chapter 13 Review 31.

1021

When an angle of 53 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (1, - 3). Then x 2 + y 2 = 1 + 3 = 2,

r =

so sec

32.

5 r 2 = = = 2. x 3 1

39.

When an angle of 73 is drawn in standard position, ( x, y) = (1, 3) is one point on its terminal side. Then r = 1 + 3 = 2,

A sample calculation: csc

When x = 4 ,

7 r 2 2 3 = = = . 3 y 3 3

33.

tan 115 » -2.1445

34.

sin (-123) » -0.8387

35.

sin 2.3581 » 0.7058

36.

cos 0.8215 » 0.6811

y = - tan

40. 37.

The graph of y = tan x appears in Figure 18 in Section 1 in this chapter. The difference between the graph of y = tan x and y = - tan x is that the y-values of points on the graph of y = - tan x are the opposites of the y-values of the corresponding points on the graph of y = tan x.

 4

= -1.

The graph of y = - 23 sin x is similar to the graph of y = sin x except that it has two-thirds the amplitude and is reflected about the x-axis.

The graph of y = cos x appears in Figure 17 in Section 1 of this chapter. To get y = 4 cos x, each value of y in y = cos x must be multiplied by 4. This gives a graph going through (0, 4), ( , -4) and (2 , 4).

41.

Because the derivative of y = sin x is y ¢ = cos x, the slope of y = sin x varies from -1 to 1.

38.

The graph of y = 12 tan x is similar to the graph of y = tan x except that each ordinate value is multiplied by a factor of 12 . Note that

(

)

(

the points 4 , 12 and - 4 , - 12 graph.

) lie on the

42.

f ( x) = x + 2 cos x f ¢( x) = 1 - 2sin x f ¢( x) = 0 when sin x = 1/2. In the interval [0,  ] this happens at x =  /6 and x = 5 /6.

1022 43.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS y = 2 tan 5x dy = 2 sec 2 5 x ⋅ Dx (5 x) dx

æ1 ö y = cot çç x 4 ÷÷÷ çè 2 ø

50.

æ1 ö æ1 ö dy = - csc çç x 4 ÷÷÷ ⋅ Dx çç x 4 ÷÷÷ ç çè 2 ø è2 ø dx æ1 ö = - csc 2 çç x 4 ÷÷÷ ⋅ 2 x3 çè 2 ø

= 10 sec 5x

44.

y = -4sin 7 x

æ1 ö = -2 x3 csc 2 çç x 4 ÷÷÷ çè 2 ø

dy = -4(cos 7 x) ⋅ Dx (7 x) dx = -4(cos 7 x)7 = -28 cos 7 x

45.

y = e-2 x sin x dy = e-2 x ⋅ Dx (sin x) + sin x ⋅ Dx (e-2 x ) dx

51.

y = cot(6 - 3x 2 )

= e-2 x (cos x)

dy = [- csc 2 (6 - 3x 2 )] ⋅ Dx (6 - 3x 2 ) dx 2

+ (sin x)(e-2 x ) ⋅ Dx (-2 x)

= 6 x csc (6 - 3x )

46.

= e-2 x (cos x) + (sin x)(e-2 x )(-2) = e-2 x (cos x - 2sin x)

y = tan(4 x + 3) dy = sec 2 (4 x 2 + 3) ⋅ Dx (4 x 2 + 3) dx

y = x 2 csc x

52.

dy = x 2 Dx (csc x) + csc x Dx ( x 2 ) dx

= sec 2 (4 x 2 + 3) ⋅ (8x) = 8x sec 2 (4 x 2 + 3)

47.

= x 2 (- csc x cot x) + csc x(2 x) = -x 2 csc x cot x + 2 x csc x

y = 2sin (4 x ) dy = [8sin 3 (4 x 2 )] ⋅ Dx[sin (4 x 2 )] dx = 8sin 3 (4 x 2 ) ⋅ cos (4 x 2 ) ⋅ Dx (4 x 2 ) 3

= 64 x sin (4 x ) cos (4 x )

48.

53. y =

cos 2 x 1 - cos x

dy (1 - cos x)(-2 cos x sin x) - (cos 2 x)(sin x) = dx (1 - cos x)2 =

y = 2cos5 x

-2 cos x sin x + cos 2 x sin x (1 - cos x) 2

dy = 2Dx (cos x)5 dx

sin x - 1 sin x + 1

54.

= 10(cos x)4 (- sin x)

dy (sin x + 1)(cos x) - (sin x - 1)(cos x) = dx (sin x + 1) 2

= -10 sin x cos x

49.

y =

= [2 ⋅ 5(cos x) 4 ] ⋅ Dx (cos x)

y = cos (1 + x 2 ) dy = [- sin (1 + x 2 )] ⋅ Dx (1 + x 2 ) dx

sin x cos x + cos x - sin x cos x + cos x (sin x + 1)2 2 cos x (sin x + 1)2

= -2 x sin (1 + x 2 )

Chapter 13 Review 55.

y =

1023

tan x 1+ x

61.

The relative maxima of f ( x) = 2 cos( x) have value 2 and occur when  x = 2n , that is, when x = 2n for integer n. The relative minima have value -2 and occur when  x = (2n + 1) , that is, when x = 2n + 1 for integer n.

62.

The relative maxima of f ( x) = -5 sin 3x have 3 value 5 and occur when 3x = + 2n , that is, 2  2n + for integer n. The relative when x = 2 3 minima have value 5 and occur when  2n  3x = + 2n , that is, when x = + 2 6 3 for integer n.

dy (1 + x)(sec2 x) - (tan x)(1) = dx (1 + x)2 =

56.

y =

sec2 x + x sec2 x - tan x (1 + x) 2

6-x sec x

(sec x) ⋅ Dx (6 - x) - (6 - x) ⋅ Dx (sec x) dy = dx sec 2 x (sec x) ⋅ (-1) - (6 - x) ⋅ (sec x tan x) = sec2 x sec x[-1 - (6 - x) tan x] = sec 2 x 1 + (6 - x) tan x = - sec x

57.

63.

f ¢( x) = 7 sec 2 7 x f ¢¢( x) = (7)(14)(sec 7 x)(sec 7 x tan 7 x) = 98 sec 2 7 x tan 7 x

y = ln |5sin x |

f ¢¢(1) = 98 sec 2 7 tan 7

dy 1 = ⋅ Dx (5sin x) dx 5sin x

f ¢¢(-3) = 98 sec 2 (-21) tan(-21)

cos x sin x or cot x

= -98 sec 2 21 tan 21

58.

64.

60.

f ( x) = x cos 3x f ¢( x) = x ( -3 sin 3x ) + cos 3x

y = ln |cos x |

f ¢¢( x) = x ( -9 cos 3x ) - 3 sin 3x - 3 sin 3x

D (cos x) dy = x dx cos x - sin x = = - tan x cos x

59.

f ( x) = tan 7 x

= -9 x cos 3x - 6 sin 3x f ¢¢(1) = -9 cos 3 - 6 sin 3

f ¢¢(-3) = (-9)(-3) cos(-9) - 6 sin(-9) = 27 cos 9 + 6 sin 9

f ( x) = tan(2 x) is decreasing on its whole domain, so there are no intervals where it is increasing. It is decreasing wherever defined, that is, on intervals of the form æ  ö çç - + n ,  + n ÷÷ for integer n. çè 4 2 4 2 ø÷ f ( x) = - sin x is increasing on intervals of the æ ö 3 form çç + 2n , + 2n ÷÷÷ for integer n, çè 2 ø 2

65.

f ( x) = x - sin x f ¢( x) = 1 - cos x f ¢¢( x) = sin x

Since f ¢( x) is positive except at the isolated points x = 2n for integer n, the function f is increasing everywhere. The graph has a single intercept at (0, 0), which is both an x-intercept and a y-intercept. The concavity is given by the sign of sin x; the graph will be concave upward on the intervals ( 2n , (2n + 1) ) for integer n and

so f ( x) = -3 sin 4 x is increasing on the intervals æ ö çç + n , 3 + n ÷÷ for integer n, and is çè 8 2 8 2 ÷ø

concave downward on the intervals ( (2n + 1) , (2n + 2) ) for integer n.

decreasing on the complementary intervals æ  ö çç - + n ,  + n ÷÷ for integer n. çè 8 2 8 2 ø÷

Inflection points occur at the x-values separating regions of opposite concavity, that is, where sin x = 0, for example at x = - , x = 0,

1024

Chapter 13 THE TRIGONOMETRIC FUNCTIONS x =  , and x = 2 . The y-values at these inflection points are equal to the corresponding x-values. f(x) 8 6 f(x) = x – sin x 4 2

f (0) = 1 f ( /2) =  /2 f ( ) =  - 1 Thus there is an absolute maximum of  - 1 at x =  and an absolute minimum of 1 at x = 0.

(2p, 2p) (p, p)

–3p –2p –p–2 (–p, –p) –4 (–2p, –2p) –6 –8 (–3p, –3p) –10

68.

2p 3p x

The only value in [0,  ] for which f ¢( x) = 0 is x = 3 /4, so this is the only critical number, and extrema could only be located at 0, 3 /4, and  . f (0) = 0

f ( x) = x - cos x f ¢( x) = 1 + sin x

66.

f (3 /4) = 1 +

f ¢¢( x) = cos x

Since f ¢( x) is positive except at the isolated points x = (2n - 1/2) for integer n, the function f is increasing everywhere. The graph has a single y-intercept at (0, -1) and a single x-intercept at approximately (0.739, 0), found by solving x - cos x = 0 with a calculator. The concavity is given by the sign of cos x; the graph will be concave upward on the intervals ( (2n - 1/2) , (2n + 1/2) ) for

integer n. Inflection points occur at the x-values separating regions of opposite concavity, that is, where cos x = 0, for example at x = - /2, x =  /2, and x = 3 /2. The corresponding y-values are - /2,  /2, and 3 /2.

8 6 4 p p 冸– 2 , – 2 冹 2

Thus there is an absolute maximum of 1 + 2 at x = 3 /4 and an absolute minimum of 0 at x = 0. 69.

–3p –2p –p – 3p, – 3p 2 2

冸

冹

冸– 5p2 , – 5p2 冹

67.

冸

70.

tan( xy) + x3 = y Differentiate with respect to x. tan( xy) + x3 = y æ dy ö dy sec2 ( xy) çç y + x ÷÷÷ + 3x 2 = çè dx ø dx

(

y sec2 ( xy) + 3x 2 = 1 - x sec2 ( xy)

冸5p2 , 5p2 冹

dy y sec2 ( xy) + 3x 2 = dx 1 - x sec2 ( xy)

冹 2p 3p x

71.

–4 –6 –8

f ( x) = cos x + x on [0,  ] f ¢( x) = - sin x + 1

The only value in [0,  ] for which f ¢( x) = 0 is x =  /2, so this is the only critical number, and extrema could only be located at 0,  /2, and  .

dy dx

dy 1 - sin( x + y) = dx 2 y + sin( x + y)

p, p 2 2

x + cos( x + y ) = y 2 Differentiate with respect to x. æ dy ö÷ dy 1 - sin( x + y) çç1 + ÷ = 2y çè dx ÷ø dx

1 - sin( x + y) = ( 2 y + sin( x + y) )

f(x) = x – cos x

冸3p2 , 3p2 冹

f ( ) = 2

integer n and concave downward on the intervals ( (2n + 1/2) , (2n + 3/2) ) for

f ( x) = sin x - cos x + 1 on [0,  ] f ¢( x) = cos x + sin x

y = sin x Differentiate with respect to t. æ dx ö dy = (cos x) çç ÷÷÷ çè dt ø dt

Substitute the given values. æ æ  öö dy 1 = çç cos çç ÷÷ ÷÷÷ ( -1) = ÷ ç ÷ ç è 3 øø dt 2 è

) dy dx

Chapter 13 Review 72.

1025

y = tan x Differentiate with respect to t. æ dx ö dy = sec2 x çç ÷÷÷ èç dt ø dt

(

76.

)

Let u = 7 x, so du = 7dx or 17 du = dx.

Substitute the given values. æ æ  öö dy = çç sec2 çç ÷÷ ÷÷÷ ( 3 ) = (2)(3) = 6 çè 4 ÷ø ÷ø ç dt è

73.

ò tan 7x dx 1

ò tan 7x dx = ò (tan u) ⋅ 7 du 1 = tan u du 7ò 1 ln |cos u | + C 7 1 = - ln |cos 7 x | + C 7

ò cos 5x dx

Let u = 5x, so du = 5dx or 15 du = dx. 1

ò cos 5x dx = ò cos u ⋅ 5 du 1 = cos u du 5ò 1 sin u + C 5 1 = sin 5 x + C 5

77.

= -4 cot x + C

74.

78.

79.

Let u = 2 x.

æ1

ò 5x sec 2x2 tan 2x2 dx 5 = sec 2 x 2 tan 2 x 2 (4 x dx) 4ò 5 = sec u tan u du 4ò

ò sin 2x dx = ò (sin u)çççè 2 du ÷÷÷ø 1 sin u du = 2ò

5 sec u + C 4 5 = sec 2 x 2 + C 4

1 (- cos u) + C 2 1 = - cos 2 x + C 2

ò sec2 5x dx

80.

ò sec2 5x dx = 5 ò (sec2 5x) (5 dx) 1 = sec2 u du 5ò 1 tan u + C 5 1 = tan 5 x + C 5

ò x2 sin 4x3 dx Let u = 4 x3 , so du = 12 x 2 dx or

Let u = 5x, so that du = 5 dx.

ò 5x sec 2x2 tan 2x2 dx Let u = 2 x 2 , so that du = 4 x dx.

du = 2 dx 1 du = dx. 2

75.

ò 8sec2 x dx = 8ò sec2 x dx = 8 tan x + C

ò sin 2x dx Then

ò 4 csc2 x dx = -4ò - csc2 x dx

1 du = x 2 dx. 12

ò x2 sin 4x2 dx = ò (sin u) ⋅ 12 du 1 = sin u du 12 ò 1 cos u + C 12 1 = - cos 4 x3 + C 12 =-

1026 81.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS

ò cos8 x sin x dx

84.

Let u = cos x, so du = - sin x dx.

Let u = 11x 2. Then du = 22 x dx.

ò cos8 x sin x dx = -ò u8 du

ò x tan 11x2 dx 1 = (tan 11x 2 ) ⋅ (22 x dx) 22 ò 1 = tan u du 22 ò

1 = - u9 + C 9 1 = - cos9 x + C 9

82.

ò x tan 11x2 dx

1 (- ln |cos u |) + C 22 1 =ln |cos 11x 2 | + C 22

ò cos x sin x dx

Let u = cos x. Then

du = - sin x dx -du = sin x dx.

ò cos x sin x dx = ò u (-du) = - u1/2 du ò

85.

ò (cos x)-4/3 sin x dx Let, u = cos x so that du = - sin x dx.

ò (cos x) sin x dx = - (cos x)-4/3 (sin x) dx ò = - u-4/3 du ò -4/3

2 = - u 3/2 + C 3 2 = - (cos x)3/2 + C 3

83.

= 3u-1/3 + C

x 2 cot 8 x3 dx

= 3(cos x)-1/3 + C

Let u = 8 x3, so that du = 24 x3 dx.

ò x cot 8x dx 1 = (cot 8 x3 ) (24 x 2 ) dx 24 ò 1 = cot u du 24 ò 2

86.

ò sec2 5x tan 5x dx Let u = tan 5 x, so du = 5 sec 2 5 x dx.

ò sec2 5x tan 5x dx 1 u du = 5ò

1 ln |sin u | + C 24 1 = ln |sin 8 x3 | + C 24 =

1 u2 ⋅ +C 5 2 1 2 u +C = 10 1 tan 2 5x + C = 10 =

87.

 /2

ò0 cos x dx = sin x 0

= sin 2 - sin 0 = 1- 0 = 1

Chapter 13 Review 88.

1027

2 /3

ò- -sin x dx = cos x -

ò x2 cos 2x dx

= cos 23 - cos (- )

1 1 = - - (-1) = 2 2

89.

2

1 x2 x sin 2 x + cos 2 x - sin 2 x + C 2 2 4

93. (a) Residential usage of natural gas is probably periodic.

(10 + 10 cos x) dx

ò0

2

10 dx +

= 10(2 ) + 10

2

ò0 10 cos x dx 2

ò0 cos x dx 2

= 20 + 10(sin x ) 0

= 20 + 10(sin 2 - sin 0)

(b)

C (t ) = 35, 072sin(0.32954t + 2.3160)

= 20 + 10(0 - 0)

+ 35, 706.

= 20

90.

 /3

ò0 (3 - 3 sin x) dx ò0

 /3

3 dx -

 /3

= 3x 0

91.

 /3

ò0 3 sin x dx  /3

+ 3 cos x 0

æ3 ö 3 = (  - 0) + çç - 3 ÷÷÷ =  çè 2 ø 2

Let dv = sin x dx and u = x + 2.

C ¢(11) » 10,886. In November the consumption is increasing at 10,886 million cubic feet per month.

Then v = - cos x and du = dx.

ò (x + 2) sin x dx = ò u dv = uv ò v du = -( x + 2) cos x + ò cos x dx = -( x + 2) cos x + sin x + C

92.

ò x2 cos 2x dx Use column integration. D x

2x 2 0

I +

cos 2 x

1 sin 2 x 2 1 + - cos 2 x 4 -

(d) Integrating C (t ) from 0 to 12 using a calculator gives the estimate 249,882 million cubic feet; the sum of the values in the table is 238,391 million cubic feet. (e) The function in the model has a period of 2 T = b 2 = 0.32954 » 19.07 months. One might expect a period of 12 months, but longer-term trends might make this annual period hard to see.

1 - sin 2 x 8

1028 94.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS P(t ) = 90 + 15 sin 144 t

R2 = k ⋅

(f)

The maximum possible value of sin  is 1, while the minimum possible value is -1. Replacing  with 144 t gives -1 £ sin144 t £ 1,

L2 r24

where R2 is the resistance along

DC. (g) R = R1 + R2 L L = k 14 + k 24 r1 r2

90 + 15(-1) £ P(t ) £ 90 + 15(1) 75 £ P(t ) £ 105.

æL L ö÷ = k ççç 14 + 24 ÷÷÷ çè r1 r2 ø÷

Therefore, the minimum value of P(t ) is 75 and the maximum value of P(t ) is 105.

æL L ö÷ ç (h) R = k çç 14 + 24 ÷÷÷ çè r1 r2 ÷ø æ L - s cot  ö÷ s ç ÷÷ = k çç 0 + 4 4 çè r1 (sin  )r2 ÷÷ø =

95.

(a) The sketch shows  in standard position.

From the diagram, we see that s sin  = . L2 (b)

sin  =

L2 sin  = s s L2 = sin 

(d)

r14

ks 4

r2 sin 

(i) R¢ = D R æ L sk s k ö÷÷ = D ççç k 04 - 4 cot  + 4 ⋅ ÷ sin  ÷ø÷ r1 r2 èç r1 æL 1 ö÷÷ s s = kD ççç 04 - 4 cot  + 4 ⋅ ÷ sin  ÷ø÷ r1 r2 èç r1

s L2

cot  =

k ( L0 - s cot  )

L0 - L1 . s

é æ L ö÷ æ s ö÷ = k êê D ççç 04 ÷÷÷ - D ççç 4 cot  ÷÷÷ êë èç r1 ø÷ èç r1 ø÷ æ s 1 ö÷÷ ùú + D ççç 4 ⋅ ÷ çè r2 sin  ÷÷ø úú û é æ 1 ö÷ ù s s ÷÷ úú = k êê 0 - 4 D (cot  ) + 4 D ççç ÷ø ú  sin è r r 1 2 ëê û é s ù s æ - cos  ö÷ ú = k êê - 4 (- csc 2  ) + 4 çç ÷ú r2 çè sin 2  ÷ø ûú ëê r1 æ s 1 cos  ö÷÷ s ç = k çç 4 ⋅ ÷ çè r1 sin 2  r24 sin 2  ÷ø÷

L0 - L1 s s cot  = L0 - L1 cot  =

L1 + s cot  = L0 L1 = L0 - s cot 

(e) The length of AD is L1, and the radius of that

section of blood vessel is r1, so the general equation L k = 4 r is similar to L R1 = k 14 r1 for that particular segment of the blood vessel.

= =

æ 1 cos  ö÷÷ çç ÷ ç sin 2  çè r14 r2 4 ø÷÷ ks

ks csc2 

(j)

r14

ks cos  r2 4 sin 2 

Using part (h) gives ks csc2  r14

ks cos  r2 4 sin 2 

= 0.

Chapter 13 Review

1029

(k) If the left side of the equation in the solution to part (h) is multiplied

(b)

by sins  , we get

Use the graphing calculator to find a trigonometric function. y = 12.5631 sin (0.548067t - 2.35326) + 50.2793

sin 2  ks æç 1 cos  ö÷÷ çç ⋅ ÷ 2 4 s r2 4 ÷÷ø sin  çè r1

æ 1 cos  ÷÷ö ç = k çç 4 ÷ çè r1 r24 ÷÷ø

This gives the equation k r14

(l)

k cos  r24

= 0.

Using part (k) gives

(d) T =

æ 1 cos  ö÷÷ ç k çç 4 ÷=0 çè r1 r2 4 ÷÷ø

1 r14

cos  r24 1 r14

The period is about 11.5 months. One might expect a period of 12 months, but longer-term trends might make this annual period hard to see.

=0 k ¹0 =

cos  =

cos  r24

97.

r2 4 r14

2 2 = » 11.46426 b 0.548067

(a) y = x tan  = 39 tan

 24

16 x 2 V -

sec2  + h

16(39)2 732

sec2

 24

» 9.5

(n)

If r1 = 1 and r2 = 14 , then

Yes, the ball will make it over the net since the height of the ball is about 9.5 feet when x is 39 feet.

æ 1 ö4 æ 1 ö4 cos  = çç 14 ÷÷÷ = çç ÷÷÷ èç ø èç 4 ø

(b) Entering

1 = » 0.0039, 256

Y1 = 39 tan x -

from which we get

 » 90. (o) If r1 = 1.4 and r2 = 0.8, then cos  =

Thus,

(0.8) 4 (1.4)

» 0.1066.

 » 84.

Y2 =

16(39)2 442

sec2 x + 9 and

é 442 sin x cos x ù ê ú ê ú ê + 442 cos2 x tan 2 x + 576 sec2 x ú 2 êë úû 44

into the graphing calculator and using the table function, indicates that the tennis ball will clear the net and travel between 39 and 60 feet for 0.18 £ x £ 0.41 or 0.18 £  £ 0.41 in radians. In degrees, 10.3 £  £ 23.5.

96.

(a) Enter the data into a graphing calculator and plot.

1030

Chapter 13 THE TRIGONOMETRIC FUNCTIONS At t = 7,

é ù dy 12.5 = cos ê (7 + 1.2) ú êë 6 úû dt 6 é ù 12.5 cos ê (8.2) ú = êë 6 úû 6 » -2.66 MCF/month

In July, the monthly gas usage is decreasing by 2.66 MCF per month.

feet » 0.995 feet . Note that 57.01 radian degree

The distance the tennis ball travels will increase by approximately 1 foot by increasing the angle of the tennis racket by one degree. 98.

(d) Use the integration function on a calculator to

compute

ò0 y(t ) dt » 176.4. We estimate

the total gas usage for the year at 176.4 MCF.

(a) 99.

s(t ) = A cos( Bt + C ) s ¢(t ) = - A sin( Bt + C ) ⋅ Dt ( Bt + C ) = - A sin( Bt + C ) ⋅ B = - AB sin( Bt + C ) s ¢¢(t ) = - AB cos( Bt + C ) ⋅ Dt ( Bt + C ) = - AB cos( Bt + C ) ⋅ B

(b) For February:

= -B 2 A cos( Bt + C )

é ù y = 12.5sin ê (2 + 1.2) ú + 14.7 êë 6 úû é ù = 12.5sin ê (3.2) ú + 14.7 êë 6 úû » 27.13 MCF

= -B 2 s(t )

100.

Refer to the figure below.

For July: é ù y = 12.5sin ê (7 + 1.2) ú + 14.7 êë 6 úû é ù = 12.5sin ê (8.2) ú + 14.7 êë 6 úû » 3.28 MCF

(c)

é ù y = 12.5sin ê (t + 1.2) ú + 14.7 êë 6 úû

dy dt é ù éæ  öù = 12.5cos ê (t + 1.2) ú Dt ê çç (t + 1.2) ÷÷÷ ú êë 6 úû êë çè 6 ø úû é ù  = 12.5cos ê (t + 1.2) ú ⋅ êë 6 úû 6 é ù 12.5 cos ê (t + 1.2) ú = ê ú 6 ë6 û

Let A be the area of the triangle. A=

1 bh 2

In this triangle, h b and cos  = 6 6 h = 6 sin  and b = 6 cos  .

sin  =

7-x (7 - x )2 + 52

x x 2 + 32

Extended Application

1031

Thus,

(d)

1 (6 cos  )(6 sin  ) 2 = 18 sin  cos 

= k ln |sec  + tan  | - k ln |sec 0 + tan 0| = k ln |sec  + tan  | - k ln |1 + 0| = k ln |sec  + tan  | -k ln 1 = k ln |sec  + tan  | -0 = k ln |sec  + tan  |

dA = 9 cos (2 ) ⋅ 2 = 18 cos (2 ). d

If ddA = 0, 18cos (2 ) = 0 cos 2 = 0

Since D(3403¢) = 7,



7 = k ln |sec 3403¢ + tan 3403¢ | 7 k = ln |sec 3403¢ + tan 3403¢ | » 11.0635.



If  < 4 , ddA > 0. 

D(4045¢) » 11.0635ln sec 4045¢ + tan 4045¢

< 0. If  > 4 , ddA 

» 8.63

Thus, A is maximum when  = 4 or 45. 101. (a)

New York City should be placed approximately 8.63 inches from the equator.

d ln |sec x + tan x | dx

(e)

D(2546¢) » 11.0635 ln |sec 2546¢ + tan 2546¢ | » 5.15

sec x tan x + sec2 x = sec x + tan x =

ò0 sec x dx 

A = 9 sin (2 )

 =



= k ln | sec x + tan x | 0

= 9(2 sin  cos  )

2 =

D( ) = k

sec x(tan x + sec x) (sec x + tan x)

Miami should be placed approximately 5.15 inches from the equator.

= sec x

(b)

d (- ln |sec x - tan x |) dx 2

sec x tan x - sec x sec x - tan x sec x(tan x - sec x) =(sec x - tan x) = -(- sec x) = sec x =-

Extended Application: The Shortest Time and the Cheapest Path 1.

Since both rays are coming from point A, but meet the interface at two different points, they cannot be parallel. If the two rays were parallel, the alternate interior angle of the labeled right angle would also measure 90 implying that the two angles  1 are equal. As Dx becomes closer to zero the rays become more nearly parallel.

The segment labeled Dp is only an approximation to the change in length of the ray from point A. The actual change in length would be found by measuring the length of the original ray from point A to the crossing point along the new ray which would leave the change in length left over. As Dx becomes closer to zero the two rays become more nearly parallel and Dp becomes a better approximation.

1032 3.

Chapter 13 THE TRIGONOMETRIC FUNCTIONS Speed of light in air = 1.6 Speed of light in plastic

Speed of light in plastic =

Speed of light in air 1.6

The percentage of the speed of light in 1.6-index 1 = 0.625 or 62.5% . plastic is given by 1.6 4.

The cheapest route would be where the route goes almost perpendicularly across the swamp, making x1 almost zero.

If construction over land was actually more expensive than construction over the swamp, we would solve k

7-x 2

(7 - x) + 5

x 2

x + 32

where k represents how many more times expensive building over land is than building over the swamp. Letting k = 2 in the above equation, we get The solver on the calculator gives x » 4.68 miles. The angle  1 has tangent equal to 4.68/3 which means that  1 is about 57.3. For the straight line route,  1 has tangent equal to 7/8 which means that  1 is about 41.2. The angle between the north-south line which passes A and the line from A to the point where the road emerges from the swamp is also  1, the larger  1 corresponds to the route which lies to the west of the other route. Thus, the route which minimizes cost lies to the west of the straight line route. 6.

When you see the sun begin to set, the center of the sun looks to be 12 (0.53) = 0.265 above the horizon, but is actually 0.265 - 0.57 = -0.305 below the horizon. Since the radius of the sun’s disk is only 0.265, this means the sun is already below the horizon.