1. The empirical function W = f(t), given in the graph below, comes from the Wall Street Journal, September 4, 1992. From the graph, determine the domain of the function.
A) September 1989 to August 1992 C) B) 2810 to 4090 D) Ans: A difficulty: easy section: 1.1
0 to 4200 January 1988 to January 1993
2. The curve W = f(t), given in the graph below, comes from the Wall Street Journal, September 4, 1992. W is a function.
Ans: True
difficulty: easy
section: 1.1
Page 1
Chapter 1: A Library of Functions
3. Draw a graph which accurately represents the temperature of the contents of a cup left overnight in a room. Assume the room is at 70° and the cup is originally filled with water slightly above the freezing point. Ans:
difficulty: easy
section: 1.1
4. Suppose the Long Island Railroad train from East Hampton to Manhattan leaves at 4:30 pm and takes two hours to reach Manhattan. It waits two hours at the station and then returns, arriving back in East Hampton at 10:30 pm. Draw a graph representing the distance of the train from the Farmingdale station in East Hampton as a function of time from 4:30 pm to 10:30 pm. The distance from East Hampton to Manhattan is 150 miles. Ans:
difficulty: medium
section: 1.1
Page 2
Chapter 1: A Library of Functions
5. Suppose we buy quantities x1 and x2 , respectively, of two goods. The following graph shows the budget constraint p1 x1 + p2 x2 = k , where p1 and p2 are the prices of the two goods and k is the available budget. If the budget is doubled, but prices remain the same, what is the x2 − intercept of the new budget constraint?
2k p1 Ans: B
A)
2k k C) p2 2 p1 difficulty: medium
B)
k 2 p2 section: 1.1 D)
Page 3
Chapter 1: A Library of Functions
6. Suppose we buy quantities x1 and x2 , respectively, of two goods. The following graph shows the budget constraint p1 x1 + p2 x2 = k , where p1 and p2 are the prices of the two goods and k is the available budget. If the price of the first good is doubled, but the other values are unchanged, what is the x2 − intercept of the new budget constraint?
2k p2 Ans: C
A)
k 2k C) p1 p2 difficulty: medium B)
k p1 section: 1.1 D)
7. A function is linear for x ≤ 2 and also linear for x ≥ 2. This function has the following values: f(-4) = 8; f(2) = 5; f(4) = 11. Find formula(s) (or equation(s)) which describe this function. 1 −2 x + 9, x ≤ 2 x + 4, x ≤ 2 f ( x) = 2 f ( x) = A) C) 3 x − 1, x ≥ 2 −3 x + 11, x ≥ 2 1 2 x + 6, x ≤ 2 − x + 6, x ≤ 2 f ( x) = 2 B) D) f ( x) = 1 x 11, x 2 − ≥ 3 x − 1, x≥2 3 Ans: D difficulty: medium section: 1.1
Page 4
Chapter 1: A Library of Functions
8. A pond has a population of 500 frogs. Over a ten-year period of time the number of frogs drops quickly by 20%, then increases slowly for 5 years before dropping to almost zero. Does the following graph accurately represent the number of frogs in the pond over the ten-year period of time?
Ans: no difficulty: easy
section: 1.1
9. Suppose a 40 F container of water is placed in the freezer overnight. The next morning, it is put on the counter in a 70 F room and then at the end of the day heated to the boiling point. What is the range of the function? A) From 40 F to 70 F . B) From the middle of the night until the middle of the next day. C) From the first evening until the end of the next day. D) From 32 degrees F to 212 degrees F. Ans: D difficulty: easy section: 1.1 10. A school library opened in 1980. In January, 2000 they had 25,000 books. One year later, they had 25,420 books. Assuming they acquire the same number of books at the start of each month, how many books did they have in January, 2003? Ans: 26,260 difficulty: easy section: 1.1 11. A school library opened in January of 1980. In January, 2000 they had 35,000 books. One year later, they had 35,490 books. Assuming they acquire the same number of books at the start of each month, how many books did they have in July of 1980? Ans: 25,445 difficulty: medium section: 1.1
Page 5
Chapter 1: A Library of Functions
12. A school library opened in 1980. In January, 2000 they had 25,000 books. One year later, they had 25,470 books. Assuming they acquire the same number of books at the start of each month, the linear formula for the number of books, N, in the library as a function of the number of years t the library has been open is given by N = _____ + _____ t. Part A: 15,600 Part B: 470 difficulty: easy section: 1.1 13. A school library opened in 1980. In January, 2000 they had 40,000 books. One year later, they had 40,410 books. Assume that they acquire the same number of books at the start of each month. If you graph the function with domain 1980-2010, describe the y-intercept of the graph in the context of the problem. A) The number of books the library had in 1980 B) The number of books the library will have in 2010 C) The year the library had 40,000 books D) The year the library had no books Ans: A difficulty: easy section: 1.1 14. Write a formula representing the function that says: The circumference of a circle is proportional to the diameter of the circle. Ans: C=(pi)d difficulty: easy section: 1.1 15. The illumination, I, of a candle is inversely proportional to the square of its distance, d, from the object it illuminates. Write a formula that expresses this relationship. k Ans: I = 2 D difficulty: easy section: 1.1
Page 6
Chapter 1: A Library of Functions
16. Harley Davidson (ticker symbol HOG) stock prices dropped sharply in late 2008. Series 1 in the graph below shows the actual prices at the end of each week. The trend over time is approximately linear; and the graph of a possible linear model is given by Series 2. Based on the data given, find the linear model and use it to approximate the stock's price on November 30, 2008, assuming the current trend continued.
Harley Davidson (HOG) Weekly Closing Prices September 1 - November 15, 2008
45 40 D o l l a r s
35 30 25 20
Series1
15
Series2
10 5 0 1
2
3
4
5
6
7
8
9 10 11
Week Number −2.5 x + 45 and $12.00, but answers will vary slightly. Ans: y = difficulty: easy section: 1.1 17. If f ( x) = e x + 2 , find and simplify the difference quotient e x+h A) 1 B) e − e C) h Ans: D difficulty: medium 8
6
e8 (e h − 1) D) h section: 1.1
Page 7
f (6 + h) − f (6) . h
Chapter 1: A Library of Functions
18. The following table defines three functions for 0 ≤ x ≤ 4 . The function y1 is most likely x 0 1 2 3 4
y1 4.65 7.44 11.90 19.05 30.47
y2 4.65 5.11 5.97 9.55 15.28
A) exponential B) neither Ans: A difficulty: medium
y3 4.65 3.79 2.93 2.07 1.21
C) linear section: 1.2
19. The following table defines three functions for 0 ≤ x ≤ 4 . One is linear, one is exponential, and one is neither. The equation for the linear function is y = _____ x + _____. x 0 1 2 3 4
y1 4.35 6.96 11.14 17.82 28.51
Part A: –0.76 Part B: 4.35 difficulty: medium
y2 4.35 5.11 5.97 9.55 15.28
y3 4.35 3.59 2.83 2.07 1.31
section: 1.2
20. The following table defines three functions for 0 ≤ x ≤ 4 . One is linear, one is exponential, and one is neither. The equation for the exponential function is y = ( B )x. x 0 1 2 3 4
y1 4.65 7.90 13.44 22.85 38.84
Part A: 4.65 Part B: 1.7 difficulty: medium
y2 4.65 5.11 5.97 9.55 15.28
y3 4.65 3.79 2.93 2.07 1.21
section: 1.2
Page 8
A
Chapter 1: A Library of Functions
21. The following table of data is either linear or exponential. If it is linear, give values of a y ax + b . If it is exponential, give values of a and b such that and b such that = y = a (b) x .
x y
0 0.94
a = _____; b = _____ Part A: 2.96 Part B: 0.94 difficulty: medium
0.50 2.42
1.00 3.90
1.50 5.38
section: 1.2
22. On the x − interval between E and F, the function graphed below is:
Mark all that apply. A) increasing B) decreasing C) concave up Ans: B, C difficulty: easy section: 1.2
D) concave down
23. A bar of soap starts out at 175 grams. What is the formula for the quantity S grams of soap remaining after t days if the decrease is 10 grams per day? Ans: S= 175 − 10t difficulty: easy section: 1.2 24. A bar of soap starts out at 100 grams. What is the formula for the quantity S grams of soap remaining after t days if the decrease is 15% per day? Ans: S= 100(0.85)t difficulty: easy section: 1.2
Page 9
2.00 6.86
Chapter 1: A Library of Functions
25. Give a possible formula for the function in the following figure:
Ans: y= 2(2 x ) difficulty: medium
section: 1.2
26. The following functions represent exponential growth or exponential decay. Mark the one(s) that represent exponential growth. A) P = 10(1.1)t B) Q = 4.4e0.05t C) S = 5e−0.25t D) R = 12(0.8)t Ans: A, B difficulty: easy section: 1.2 27. Joe invested $20,000 in the stock market, and Sam invested $20,000. Joe's investment increased in value by 5% per year for 10 years. Sam's investment decreased in value by 10% for 5 years and then increased by 10% for the next 5 years. At the end of the 10 years, whose investment was worth more, Joe's or Sam's? Ans: Joe's difficulty: medium section: 1.2 28. A bakery has 300 lbs of flour. If they use 10% of the available flour each day, how much do they have after 20 days? Round to the nearest pound. Ans: 36 difficulty: easy section: 1.2 29. In the book One Grain of Rice, a girl receives a reward that starts with one grain of rice on day one, two grains on day two, four on day three and eight on day four. Each day, she receives double the number of grains of rice. How many grains of rice does she receive on the 30th day? A) 900 B) 1,073,741,824 C) 4,640,650,289 D) 60 Ans: B difficulty: easy section: 1.2
Page 10
Chapter 1: A Library of Functions
30. Lisinopril is an ACE inhibitor derived from the venom of a Brazilian pit viper frequently used in the treatment of hypertension. Because of Lisinopril's relatively long half life of 12 hours, patients need to take a dose just once per day. A patient takes his first dose, 20 mg, at 6 pm on Saturday. a) How many hours does it take for the amount of Lisinopril in the patient's body to decrease to 16 mg? Round to two decimal places. b) How many milligrams remain in the patient's body right before he takes his next 20 mg at 6 pm on Sunday? Round to two decimal places Ans: a) 3.86 hours b) 5.00 milligrams difficulty: medium section: 1.2 31. One of the graphs below shows the rate of flow, R, of blood from the heart in a man who bicycles for twenty minutes, starting at t = 0 minutes. The other graph shows the pressure, p, in the artery leading to a man's lungs as a function of the rate of flow of blood from the heart. Estimate p ( R(15)) .
A) 26.5 Ans: B
B) 23 C) 17.5 D) 13 difficulty: easy section: 1.3
32. Given the function f ( x) = e− x / 3 find g ( x) and h( x) such that f ( x) = g (h( x)) 2
g ( x) = e− x and h( x) = x / 3 2
C)
g ( x) = e x and h( x) = − x 2 / 3
g ( x) = x / 3 and h( x) = e− x B) D) Ans: C difficulty: easy section: 1.3
g ( x) = − x 2 / 3 and h( x) = e x
A)
2
33. Given the function m( z ) = z 2 , find and simplify m( z – h) − m( z ) . Ans: –2zh + h 2 difficulty: easy
section: 1.3
Page 11
Chapter 1: A Library of Functions
34. Given the function= y f= ( x)
8 , give a formula for the inverse function of f(x). 5x − 8
8 + 8y 8 8 + or 5 5y 5 difficulty: medium section: 1.3 Ans:
35. The graph of y = f(x) is shown in the first figure. What graph is shown in the second figure?
y = 3 f ( x) – 4 A) y = –3 f ( x) – 4 B) Ans: A difficulty: medium
y = 3 f ( x) + 4 C) y = –3 f ( x) + 4 D) section: 1.3
Page 12
Chapter 1: A Library of Functions
36. Given the graphs of y = g ( x) and y = f ( x) in the following figure, estimate g(f(5)).
Ans: –7 difficulty: easy
section: 1.3
37. Write an equation for the graph obtained by shifting the graph of y = x3 vertically upward by 3 units, followed by vertically stretching the graph by a factor of 2 and reflecting the graph across the x − axis. Ans: y = −2( x 3 + 3) −2 x3 − 6 or y = difficulty: medium section: 1.3 38. If the graph of y = f(x) is shrunk vertically by a factor of 1/2, then shifted vertically by 4 units, then stretched vertically by a factor of 2, the resulting graph the same as the original graph. Ans: False difficulty: medium section: 1.3 39. Given the information in the table and the following conditions, find h(1) . • f(x) is even. • g(x) is odd. • h(x) = g ( x) 2 − f ( x) x -3 -2 -1 0
f(x) 7 2 –1 –2
g(x) 22 10 1 2
Ans: 2 difficulty: medium
section: 1.3
Page 13
Chapter 1: A Library of Functions
40. Is the function graphed in the following figure invertible?
Ans: no difficulty: easy
section: 1.3
41. Given the function q ( x) = x3 , which of the following is equivalent to q (2 x + a ) + q ( x) ? A) 8x3 + a3 B) 9x3 + a 3 Ans: D difficulty: easy
C) D) section: 1.3
8 x3 + 12 x 2 a + 6 xa 2 + a3 9 x3 + 12 x 2 a + 6 xa 2 + a3
42. Given the function q ( x) = x3 , which of the following is equivalent to q ( x5 ) + q ( x + a ) ? C) A) x8 + x3 + 3 x 2 a + 3 xa 2 + a 3 15 3 2 2 3 B) D) x + x + 3 x a + 3 xa + a Ans: B difficulty: easy section: 1.3
x8 + x 3 + a 3
x15 + x3 + a 3
43. Is the function f ( x) = x5 + x8 + x9 odd, even , or neither? Ans: neither difficulty: easy section: 1.3 44. The cost of shipping r kilograms of material is given by the function C = f(r) = 175+ 6r. Find a formula for the inverse function. Ans: r=(C - 175) / 6 difficulty: easy section: 1.3
) 3 x − 1 , h(g(x)) = 45. For g (= x) 3 x 2 + 7 x and h( x= Part A: 9 Part B: 21 Part C: –1 difficulty: easy section: 1.3
Page 14
A
x2 +
B
x+
C .
Chapter 1: A Library of Functions
46. The graph of a function y = f(x) is given below. Sketch a graph of it's inverse function on the same axes.
5
y
4 3 2 1 -5 -4 -3 -2 -1 -1
x 1
2
3
4
5
-2 -3 -4 -5
Ans: difficulty: medium
section: 1.3
47. The following describes a function that must be invertible: R(t) is the revenue earned by a shoe shop on the t th th day of the year. Ans: False difficulty: easy section: 1.3 48. The following describes a function that must be invertible: N(p) is the number of words on page p of a science fiction novel. Ans: False difficulty: easy section: 1.3
Page 15
Chapter 1: A Library of Functions
49. The following describes a function that must be invertible: A woman takes a tablet of 10 mg of a common allergy medicine with a half life of approximately 8 hours. L(t) gives the amount of medicine in her system at time t where 0 < t < 24. Ans: True difficulty: easy section: 1.3 50. Suppose f(x) and g(x) are both increasing functions. a) Must f(g(x)) be an increasing function? If so, explain why. If not, give an example. b) Must f(g(x)) be invertible? If so explain why. If not, give an example. Ans: Answers will vary. One example is given below. a) Yes. The function g(x) is increasing, so the inputs to f(x) are increasing. Since f is an increasing function for inputs that are increasing, f(g(x)) must also be increasing. b) Yes. Since f(g(x)) is increasing, it is one-to-one and hence invertible. difficulty: hard section: 1.3 51. A one-to-one function is described in words as follows. Take x and multiply by 4. Then add 3 and cube the result. Describe the inverse function in words. Ans: Take the cube root of x, then subtract 3, and finally divide by 4. difficulty: medium section: 1.3 52. In 1909, the Danish biochemist Sören Peter Lauritz Sörensen (1868-1939) introduced the pH function as a measure of the acidity of a chemical substance: pH = f([H+]) = − log10 [ H + ] , where [H+] is the molecular concentration of hydrogen ions (moles per liter, M). Sörensen determined that, for 0 < pH < 7, the substance is an acid; when pH = 7, the substance is neutral; and for pH > 7, the substance is a base or is said to be alkaline. The [H+] for the worst known instance of acid rain is 3.98 ×10−3 M. Find the pH of the worst known instance of acid rain. Round to 1 decimal place. Ans: 2.4 difficulty: easy section: 1.4 53. Suppose that N(t) = 100,000,000 · 2t /10 gives the population of a certain country t years after a census was taken. A historian has a collection of documents that are not dated, but do refer to the population of this country at several times. In order to help the historian date these documents, find the inverse function for the function N. Round to whole numbers. = t 14 log N − 266 Ans: difficulty: medium section: 1.4
Page 16
Chapter 1: A Library of Functions
54. Find the equation for the line L shown below.
A)
y= b( x + 1 − eb )
= y C)
b
= y D)
= y B) Ans: B
( x − 1) eb − 1 difficulty: medium
b b−e b
b
eb − e
( x − eb ) ( x − e)
section: 1.4
55. Here are some data from a Scientific American article on Old World monkeys. From the data presented, give an approximate formula for C = cranial capacity (in cm3) as a function of A = arc length of skull (in cm). Round to one decimal place.
Cranial Capacity of contemporary Old World monkeys is related to arc length of skull as shown. Ans: C = A2.3 (501) difficulty: hard section: 1.4
Page 17
Chapter 1: A Library of Functions
56. Suppose there is an initial population of 100 rabbits on Prosperity Island. Assuming that the rabbits have more than enough of everything they need to live prosperously, we might expect the population to grow exponentially. If so, find a formula for P(t), the number of rabbits on Prosperity Island at time t, given that after one year there are 140 rabbits on the island. (Assume t is in years.) Use this formula to determine how many years it will be before there are 400 rabbits on Prosperity Island. Round to 1 decimal place. Ans: P(t ) = 100(1.4)t , P (400) = 4.1 difficulty: medium section: 1.4 57. Cramped Quarters Island is a tiny island which, although able to support a limited population of rabbits, doesn't have enough space or food supplies to support unlimited exponential growth. It is suggested that if Q(t) = population of rabbits on Cramped Quarters Island at time t, then the quantity (800 - Q(t)) will be an exponentially decaying function of t. Given that there were 400 rabbits at time t = 0, and 500 rabbits one year later, find the general formula for Q(t), the population of rabbits on Cramped Quarters Island at time t and use it to determine the rabbit population after 3 years (to the nearest rabbit). Q(3) = 631 Ans: Q(t ) = 400(1.25)t , difficulty: medium section: 1.4 58. An exponentially decaying substance was weighed every hour and the results are given below. If Q = Q0e− kt gives the weight of the substance Q, at time t hours since 9:00 am, then Q0 = _____ and k = _____. Round k to 2 decimal places. Time Weight (in grams) 9am 12 10am 10.643 11am 9.440 12 noon 8.372 1 pm 7.425 Part A: 12 Part B: 0.12 difficulty: medium
section: 1.4
Page 18
Chapter 1: A Library of Functions
59. An exponentially decaying substance was weighed every hour and the results are given below. What is the approximate half-life, in hours, of the substance? Round to 1 decimal place. Time Weight (in grams) 9am 16 10am 14.623 11am 13.364 12 noon 12.214 1 pm 11.163 Ans: 7.7 hours difficulty: medium
section: 1.4
60. The number of bacteria in milk grows at a rate of 10% per day once the milk has been bottled. When the milk is put in the bottles, it has an average bacteria count of 500 million per bottle. Suppose milk cannot be safely consumed if the bacteria count is greater than 2.9 billion per bottle. How many days will the milk be safe to drink once it has been bottled? Round to the nearest integer. Ans: 18 difficulty: medium section: 1.4 61. Cramped Quarters Island is a tiny island which, although able to support a limited population of rabbits, doesn't have enough space or food supplies to support unlimited exponential growth. It is suggested that if Q(t) = population of rabbits on Cramped Quarters Island at time t, then the quantity (800 - Q(t)) will be an exponentially decaying function of t. If at time t = 0 there were 250 rabbits, and the population was increasing at an instantaneous rate of 100 rabbits per year, give the general formula for Q(t ) . Ans: Q= (t ) 800 − 250e −0.4t difficulty: medium section: 1.4 62. In 1992, the Population Crisis Committee wrote: Large cities in developing countries are growing much faster than cities in the industrialized world ever have. London, which in 1810 became the first industrial city to top 1 million, now has a population of 11 million. By contrast, Mexico City's population stood at only a million just 50 years ago and now is 20 million. Assume that the instantaneous percentage growth rates of London and Mexico City were constant over the last two centuries. How many times greater is Mexico City's percentage growth rate than London's (to one decimal place)? Ans: 4.5 difficulty: hard section: 1.4 3
63. 6eln( a ) is equivalent to 6a3 Ans: True difficulty: easy
section: 1.4
Page 19
Chapter 1: A Library of Functions
64. 8ln be is equivalent to 8b Ans: False difficulty: easy
section: 1.4
65. Solve 6 x = 2 for x. Round to 2 decimal places. Ans: 0.39 difficulty: easy section: 1.4 66. Solve 6e3 x = 11e8 x for x. Round to 3 decimal places. Ans: –0.121 difficulty: easy section: 1.4 67. What is the doubling time in years of prices which are increasing by 8% a year? Round to two decimal places. Ans: 9.01 difficulty: medium section: 1.4 68. If the size of a bacteria colony doubles in 8 hours, how many hours will it take for the number of bacteria to be 11 times the original amount? Round to 2 decimal places. Ans: 27.68 difficulty: medium section: 1.4 69. Tornados are classified in several ways. A tornado's classification on the Fujita Scale as F1 through F5 is most commonly cited. Another classification of tornados is by path = Pl 2 log( L) + 1 where L is the length of the tornado's path length, given by the formula in miles. The Binger, Oklahoma tornado of 1981 was an F4 (originally thought to be an F5) whose path was 16 miles in length. What was its Pl classification? A) P1 B) P2 C) P3 D) P4 E) P5 Ans: C difficulty: easy section: 1.4
( x) cos x + 0.5cos 5 x ? 70. What is the period of the function c= π B) π C) D) The function is not periodic A) 2π 5 Ans: A difficulty: medium section: 1.5
c( x) cos 2 x + 0.4 cos 5 x has 71. At the point x= – π , the function = A) A local maximum B) A local minimum C) Neither Ans: A difficulty: medium section: 1.5
Page 20
Chapter 1: A Library of Functions
72. At high tide, the water level is 14 feet below a certain pier. At low tide the water level is 26 feet below the pier. Assuming sinusoidal behavior, sketch a graph of y = f(t) = the water level, relative to the pier, at time t (in hours) if at t = 0 the water level is -20 feet and falling, until it reaches the first low tide at t = 3. Based on your sketch and the information provided above, the formula for f(t) can be written = f (t ) _____ sin(_____ πt ) + _____ . Part A: –6 Part B: 1/6 Part C: –20 difficulty: medium section: 1.5 73. In nature, the population of two animals, one of which preys on the other (such as foxes and rabbits) are observed to oscillate with time, and are found to be well approximated by trigonometric functions. The population of foxes is given by the graph below. What is the amplitude?
Ans: 300 difficulty: easy
section: 1.5
Page 21
Chapter 1: A Library of Functions
74. One of the functions below is a quadratic, one is a cubic, and one is a periodic function. Which one is periodic?
A) g(x) Ans: B
B) f(x) C) h(x) difficulty: medium
section: 1.5
75. Give a formula for the following sinusoidal function as a transformation of f ( x) = sin( x)
x Ans: y = 5sin 3 difficulty: medium
section: 1.5
76. Temperatures in Town A oscillate daily between 30° F at 4am and 60° F at 4pm. The temperature in Town B is consistently 10° F colder than in Town A. What does the π formula H = −15cos t + 45 represent? 12 A) Temperature in Town A, in terms of time where time is measured in hours from 4am. B) Temperature in Town B, in terms of time where time is measured in hours from 4am. C) Temperature in Town A, in terms of time where time is measured in hours from 4pm. D) Temperature in Town B, in terms of time where time is measured in hours from 4pm. Ans: A difficulty: medium section: 1.5
9 + 8cos πt 77. What is the period of the function g (t ) = Ans: 2 difficulty: easy section: 1.5
Page 22
Chapter 1: A Library of Functions
78. Consider the functions f ( x)= 5 + sin 3 x and g ( x) = 3sin x . Which has a larger vertical intercept? A) f(x) B) g(x) Ans: A difficulty: easy section: 1.5 79. Sketch a well-labeled graph of a periodic function such that : -- f(0) = 850 -- the period is 12 -- the amplitude is 550. Then write a few sentences illustrating how such a function might apply to a scorpion population. Ans: Answers will vary. The graph of f = (t ) 850 + 550sin( π6 t ) is one such periodic function. One story: At the start of a 12-month experiment about scorpion populations, there are 850 scorpions. Over the first three months, the scorpion population increases to a maximum of 1400, then returns to its initial population at mid-year. The low point of the population is after 9 months and is 300. difficulty: medium section: 1.5
) 50 + 75cos(28.274θ ) . Select all true statements about 80. Consider the function f (θ= this function from the list below. Answers are rounded to two decimal places. A) The period of the function is 0.22. D) The period of the function is 0.38. B) The amplitude of the function is 75. E) The amplitude of the function is 50. C) The vertical shift of the function is 50. F) The vertical shift of the function is 75. Ans: A, B, C difficulty: easy section: 1.5 81. Write an equation of a periodic function with amplitude 20, vertical shift 30, period 2π , and horizontal translation (phase shift) 6 units to the right. f (t ) = 30 + 20 cos(t − 6) f (t ) 30sin(6t − 20) A) D) = f (t ) = 20 + 30sin(t + 6) B) E) None of the above. f (t ) 20 cos(30t + 6) C)= Ans: A difficulty: easy section: 1.5
Page 23
Chapter 1: A Library of Functions
82. The following figure shows the graphs for x ≥ 0 of the functions y = x , y = x −2 , y = x 0 , y = x1/ 3 , and y = x5 . Which one is the graph of y = x 0 ?
Ans: D difficulty: easy
section: 1.6
83. Graphically find a solution to the equation 6 x = x 6 in the interval [−1 ≤ x ≤ 1] . Give your answer accurate to 2 decimal places. Ans: –0.79 difficulty: medium section: 1.6 84. A spherical cell takes in nutrients through its cell wall at a rate proportional to the area of the cell wall. The rate at which the cell uses nutrients is proportional to its volume. Which of the following expresses the rate at which the cell uses nutrients as a function of its radius, r. (A is a constant.) A A A) B) C) Ar 2 D) Ar 3 2 3 r r Ans: D difficulty: medium section: 1.6
Page 24
Chapter 1: A Library of Functions
85. A possible equation for the following function is y = _____ + _____ cos ( _____ πx ) . Enter fractional coefficients as decimals, accurate to 2 decimal places.
Part A: 0 Part B: 4 Part C: 0.33 difficulty: medium
section: 1.6
86. A possible equation for the following function is y = _____( _____ ) x . Enter fractions as decimals, accurate to 3 decimal places.
Part A: 5 Part B: 0.858 difficulty: medium
section: 1.6
Page 25
Chapter 1: A Library of Functions
87. A possible equation for the following function is y =_____( x − _____ )( x − _____ )( x − _____ ) . Enter fractions as decimals, accurate to 2 decimal places. The three roots should be entered in increasing order.
Part A: 0.23 Part B: –2 Part C: 3 Part D: 5 difficulty: medium
section: 1.6
88. Using the standard viewing rectangle (-10 ≤ x ≤ 10, - 10 ≤ y ≤ 10), I graphed a cubic polynomial and saw two vertical lines, as shown. There must be another root outside of the window.
Ans: True
difficulty: medium
x section: 1.6
Page 26
Chapter 1: A Library of Functions
89. Make a graphical sketch of the following function. Include a scale on the x- and y-axis.
Ans:
difficulty: medium
section: 1.6
90. Find a possible equation for the following function given a root at (0,0), a vertical asymptote at y=-1 and horizontal asymptote at x=2.
−x x−2 difficulty: medium
Ans: y =
section: 1.6
Page 27
Chapter 1: A Library of Functions
91. Coulomb's law says that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the distance between the objects. Let q1 and q2 be the charge on the two objects. Let d be the distance between the objects and F be the electrical force between them. Translate Coulomb's Law into mathematical symbols. qq kq q kd k (q2 − q1 ) 2 A) F = 1 22 B) F = C) D) F = 12 2 F = 2 2 (q1 − q2 ) d d d Ans: D difficulty: easy section: 1.6 92. Write in factored form the equation of the polynomial graphed below. All key features are shown. 11 10 9 8 7 6 5 4 3 2 1 -1 -4 -3 -2 -1 -2 -3 -4 -5 -6 -7 -8 -9
y
x 1 2 3 4 5 6 7 8 9
Ans: p ( x) = 14 ( x + 2)( x − 4)( x − 5) difficulty: medium section: 1.6
Page 28
Chapter 1: A Library of Functions
93. The solid curve below is a portion of the graph of f ( x) = 8 x 3 and the dashed curve is a portion of the graph of g ( x) = e0.25 x . The domain of both functions is all real numbers.Which of the following statements are true. Check all that apply.
5
y
4 3 2 1
-3
-2
-1
x 1
2
3
-1 -2 -3 -4 -5
A) B) C)
At x = 1, f ( x) > g ( x) . For all x > 1, f ( x) > g ( x) . lim g ( x) = 0 . x →−∞
D) There is only one value of x for which f ( x) = g ( x) . Ans: A, C, D difficulty: medium section: 1.6
Page 29
Chapter 1: A Library of Functions
94. Using base e and transformations, find a formula for the exponential function shown in the graph below.
10 9 8 7 6 5 4 3 2 1 -3
-2
-1 -1 -2 -3 -4 -5
y
x 1
2
3
Ans: The answer should be close to f ( = x) e 2 x − 3 . difficulty: medium section: 1.6
Page 30
Chapter 1: A Library of Functions
95. Delia runs the 3.1 miles from home to the park at 6 mph, jumps on her bike and returns home in 12 minutes. a) Sketch a well-labeled graph of Delia's distance from home as a function of time. b) Find the slope of each segment of the graph and interpret their meaning. c) What does it mean to say that Delia's velocity is inversely proportional to the time she takes for her run/ride?
5
distance (miles)
4 (0.516667, (0.516667, 3.1)3.1) 3 2 1 time (hours) (0.716667, 0)
(0, 0) 0.2
0.4
0.6
0.8
1
1.2
1.4
Ans: The slope of the first segment is 6 miles per hour; her velocity on the way to the park. The slope of the second segment is -15.5 miles per hour; her velocity on the way home. The faster Delia goes, the less time it takes for her to get back home. In formula form, rate = d/t.. difficulty: hard section: 1.6 96. Is the function
3π 5π 1 continuous on the interval ? ≤x≤ 4 4 sin x
Ans: no difficulty: easy 97. Is the function Ans: no difficulty: easy
section: 1.7 1 x2 − 9
continuous on the interval 0 ≤ x ≤ 4 ?
section: 1.7
98. What are the point(s) of discontinuity of the function A) 0 B) 1 C) 6 D) 12 E) 36 Ans: A, C difficulty: easy section: 1.7
Page 31
1 x( x − 6) 2
?Select all that apply.
Chapter 1: A Library of Functions
99. If f ( x) and g ( x) are both continuous functions, what can you say about f ( x) + g ( x) ? A) It cannot be determined if it is continuous or not. B) It is not continuous. C) It is continuous. Ans: C difficulty: easy section: 1.7 100. If f ( x) and g ( x) are both continuous functions, what can you say about f ( x) / 8 ? A) It is continuous. B) It is not continuous. C) It cannot be determined if it is continuous or not. Ans: A difficulty: easy section: 1.7 101. Sketch a graph of a function on the interval [-5, 5] with exactly 2 zeros and at least 2 places where the function is not continuous. Ans: Many answers possible. One possible answer:
difficulty: medium
section: 1.7
x + 1 , − 10 < x ≤ 2 102. Is the function g ( x) = 2 − x + 3 x, 2 < x < 10 Ans: no difficulty: medium section: 1.7
continuous at x = 2?
t − a, − 10 < t ≤ 3 103. For what value of a is the function f (t ) = 2 continuous? t + 10t , 3 < t < 10 A) 10 B) –36 C) 30 D) –18 Ans: B difficulty: medium section: 1.7
Page 32
Chapter 1: A Library of Functions
104. Use the following figure to give the approximate value (to the nearest integer) of lim f ( x) . If the limit does not exist, enter "DNE". x→ –4
Ans: 3 difficulty: easy
section: 1.8
(16 + h)1/ 2 − 4 . h →0 h
105. Use algebra to evaluate the exact value of lim Ans: 1/ 8 difficulty: medium
section: 1.8
x +8 exist? x→ −8 x + 8 Ans: no difficulty: medium section: 1.8
106. Does lim
107. Use algebra to evaluate the right-hand limit of f ( x) = A) 1/2 Ans: A
B) -1/2 C) 1/8 difficulty: medium
D) -1/8 section: 1.8
Page 33
x −8 as x → 8 . 2 x − 16
Chapter 1: A Library of Functions
108. Use the following figure to give the value of lim f ( x) . x→ 0 –
If the limit does not exist,
enter "DNE".
Ans: DNE difficulty: easy
section: 1.8
) 3 x − 4 , evaluate lim [ f(x) + g(x)]. If the limit does not 109. If f ( x= ) x 2 − 25 and g ( x= x→ 0
exist, enter "DNE". Ans: –29 difficulty: easy section: 1.8 110. Let f ( x) = e x and g ( x= ) x 2 − 2 . Find lim(8 f ( x) + 6 g (2 x)) . Round to two decimal x →1
places, if necessary. A) –21.75 B) 71.11 C) 20 D) 33.75 Ans: D difficulty: medium section: 1.8 111. Find, to two decimal places, lim θ →0
A) 48.00 Ans: A
8sin(6θ )
θ
B) 1.33 C) 1 D) 0 E) DNE difficulty: easy section: 1.8
Page 34
Chapter 1: A Library of Functions
4 x7 − 3x + 2 if it exists. x →∞ 11x 7 + 100 x − 5 4 7 A) 1 B) C) D) DNE E) 11 11 Ans: B difficulty: medium section: 1.8
112. Find, lim
1 if it exists. x →0 6 x 7 1 A) B) 42 C) 1 6 Ans: E difficulty: easy
−2 5
113. Find lim
1 E) DNE 7 section: 1.8
D)
114. Suppose lim f ( x) = 7 and lim g ( x) = 11 . Find lim x →0
places. A) 5.50 Ans: A
x →0
x →0
B) 1.57 C) 7.00 D) 1.27 difficulty: easy section: 1.8
7 g ( x) . Round to two decimal f ( x) + 7
E) None of the above
115. A population of bacteria is introduced into a nutrient solution in a dish. The nutrient solution is replenished by the exact same amount each day. At first the bacteria population grows very quickly. Soon, population growth must slow down, as the nutrients become more scarce given the number of bacteria. Eventually the bacteria population stabilizes. A model for the bacteria population growth is given by 5000 thousand bacteria. What is lim P (t ) ? P(t ) = t →∞ 6 + e –5t A) 6000 B) 1000 C) 833 D) 5000 E) None of the above Ans: C difficulty: medium section: 1.8 116. Could the function described by the following table of values be exponential? x 5.2 5.3 5.4 5.5 5.6 f(x) 26 28.60 31.46 34.61 38.07 Ans: yes difficulty: easy
section: 1 review
117. What is the doubling time of quantity Q growing exponentially according to the formula Q(t ) = Q0 5t ? Round to 2 decimal places. Ans: 0.43 difficulty: medium section: 1 review
Page 35
Chapter 1: A Library of Functions
118. Which function is graphed in the following figure? A. ln(e x ) + 1 B. −2 ln x 1 E. x5 + 2 x 4 − x3 + 2 x 2 + 5 x +1
Ans: C difficulty: medium
C. e − x
D.
section: 1 review
119. Give an expression for h(x) which agrees with the following table of values. x f(x) g(x) h(x) 0 –7 0 1 –3 2 5 2 1 8 2.50 3 5 18 1.67 4 9 32 1.25 5 13 50 1 Ans: h( x) = 5/x difficulty: medium
section: 1 review
120. You are offered two jobs starting on July 1st of 2008. Firm A offers you $50,000 a year to start and you can expect an annual raise of 4% every July 1st. At firm B you would start at $40,000 but can expect an annual 6% increase every July 1st. After how many years would the job at firm B first pay more than the job at firm A? Ans: 12 difficulty: medium section: 1 review 121. You have $500 invested in a bank account earning 8.5% compounded annually. How many years will it take to triple your money? Round to the nearest whole year. Ans: 13 difficulty: medium section: 1 review
Page 36
Chapter 1: A Library of Functions
122. You have $500 invested in a bank account earning 6.4% compounded monthly, earning 6.4 % interest each month. How much interest do you earn the first year? Round to the 12 nearest cent. Ans: $32.96 difficulty: medium section: 1 review 123. The elimination half-life of aspirin in plasma is estimated to be between 15 and 20 minutes. Assuming the upper estimate of 20 minutes, how long will it take a dose of 81 mg to decrease to 2/3 the original amount (54 mg) in the plasma? A) 1.4 minutes B) 9.5 minutes C) 12 minutes D) 11.7 minutes Ans: D difficulty: hard section: 1 review 124. Consider the graph of the function f(x) given below. What is lim f ( x) ? x →−2
5 (-2, 3.2) 4
y
3 (-2, 1) 2 x
1 -5 -4 -3(-2,-2-2)-1-1
1
2
3
4
5
6
-2 -3 -4 -5 -6
A) 3.2 Ans: D
B) 1 C) -2 difficulty: easy
D) DNE section: 1 review
Page 37
Chapter 1: A Library of Functions
125. The data in the table describe the percentage of fissures that developed in turbines after h hours of running time. There is moderate linear relationship (r = 0.93)between the number of hours the turbines ran and the percentage of turbines that developed fissures. Which of the formulas below best models the linear relationship? Hours Run 400 1000 1400 1800 2200 2600 3000 3400 3800 4200 4600
Number of Turbines Run 39 53 33 73 30 39 42 13 34 40 36
= f 0.00016h − 0.134 A) f 1.6h − 0.133 B) = Ans: A difficulty: medium
Number of Fissures 0 4 2 7 5 9 9 6 22 21 21
% of Turbines with Fissures 0 0.08 0.06 0.1 0.17 0.23 0.21 0.46 0.65 0.53 0.58
f = −0.002h + 0.132 C) f 0.93h − 0.13 D) = section: 1 review
Page 38
1. For any number r, let m(r) be the slope of the graph of the function y = (2.3) x at the point x = r. Estimate m(4) to 2 decimal places. Ans: 23.31 difficulty: medium section: 2.1 2. If x(V ) = V 1/ 3 is the length of the side of a cube in terms of its volume, V, calculate the average rate of change of x with respect to V over the interval 3 < V < 4 to 2 decimal places. Ans: 0.15 difficulty: easy section: 2.1 3. The length, x, of the side of a cube with volume V is given by x(V ) = V 1/ 3 . Is the average rate of change of x with respect to V increasing or decreasing as the volume V decreases? Ans: increasing difficulty: medium section: 2.1
Page 1
Chapter 2: Key Concept: The Derivative
4. If the graph of y = f(x) is shown below, arrange the following in ascending order with 1 representing the smallest value and 6 the largest. A. E.
f '( A) 1
F.
B. 0
Part A: 6 Part B: 3 Part C: 2 Part D: 4 Part E: 5 Part F: 1 difficulty: medium
f '( B)
C.
f '(C )
D. slope of AB
section: 2.1
5. The height of an object in feet above the ground is given in the following table. Compute the average velocity over the interval 1 ≤ t ≤ 3. t (sec) y (feet)
0 10
Ans: 20 difficulty: easy
1 45
2 70
3 85
section: 2.1
Page 2
4 90
5 85
6 70
Chapter 2: Key Concept: The Derivative
6. The height of an object in feet above the ground is given in the following table. If heights of the object are cut in half, how does the average velocity change, over a given interval? t (sec) y (feet)
0 10
1 45
2 70
A) It is cut in half. B) It is doubled. Ans: A difficulty: medium
3 85
4 90
5 85
6 70
C) It remains the same. D) It depends on the interval. section: 2.1
7. The height of an object in feet above the ground is given in the following table, y = f (t ) . Make a graph of f (t ) . On your graph , what does the average velocity over a the interval 0 ≤ t ≤ 3 represent? t (sec) y (feet)
0 10
1 45
2 70
3 85
4 90
5 85
6 70
A) The average height between f(0) and f(3). B) The slope of the line between the points (0, f(0)), and (3, f(3)). C) The average of the slopes of the tangent lines to the points (0, f(0)), and (3, f(3)). D) The distance between the points (0, f(0)), and (3, f(3)). Ans: B difficulty: medium section: 2.1
Page 3
Chapter 2: Key Concept: The Derivative
8. The graph of p(t), in the following figure, gives the position of a particle p at time t. List the following quantities in order, smallest to largest with 1 representing the smallest value. A. Average velocity on 1 ≤ t ≤ 3. B. Instantaneous velocity at t = 1. C. Instantaneous velocity at t = 3.
Part A: 2 Part B: 3 Part C: 1 difficulty: medium
section: 2.1
(6 + h) 2 − 36 to 2 decimal places by substituting smaller and smaller h h→0 values of h. Ans: 12 difficulty: easy section: 2.1
9. Estimate lim
sin(h 2 ) to 2 decimal places by substituting smaller and smaller values of h h h→0 (use radians). Ans: 0 difficulty: easy section: 2.1
10. Estimate lim
Page 4
Chapter 2: Key Concept: The Derivative
11. A runner planned her strategy for running a half marathon, a distance of 13.1 miles. She planned to run negative splits, faster speeds as time passed during the race. In the actual race, she ran the first 6 miles in 48 minutes, the second 4 miles in 28 minutes and the last 3.1 miles in 18 minutes. What was her average velocity over the first 6 miles? What was her average velocity over the entire race? Did she run negative splits? A) 7.50 mph for the first 5 miles, 8.36 mph for the race, No B) 8.36 mph for the first 5 miles, 7.50 mph for the race, No C) 8.12 mph for the first 5 miles, 7.35 mph for the race, Yes D) 7.35 mph for the first 5 miles, 8.12 mph for the race, No Ans: A difficulty: medium section: 2.1 12. Let f ( x) = x 2 / 3 . Use a graph to decide which one of the following statements is true. A) When x = -5, the derivative is negative; when x = 5, the derivative is positive; and as x approaches infinity, the derivative approaches 0. B) When x = -6, the derivative is positive; when x = 6, the derivative is also positive, and as x approaches infinity, the derivative approaches 0. C) When x = -7, the derivative is negative; when x = 7, the derivative is positive, and as x approaches infinity, the derivative approaches infinity. D) The derivative is positive at at all values of x. Ans: A difficulty: easy section: 2.1 13. Given the following data about a function f, estimate f '(4.75) . x f(x)
3 10
Ans: –4 difficulty: medium
3.5 8
4 7
4.5 4
5 2
5.5 0
6 -1
section: 2.2
14. Given the following data about a function f(x), the equation of the tangent line at x = 5 is approximated by x 3 3.5 4 4.5 5 5.5 6 f(x) 10 8 7 4 2 0 -1
y −= 5 –4( x − 2) A) y −= 5 –8( x − 2) B) Ans: C difficulty: medium
y −= 2 –4( x − 5) C) y −= 2 –8( x − 5) D) section: 2.2
15. For f ( x) = 2− x , estimate f '(0) to 3 decimal places. Ans: –0.693 difficulty: medium section: 2.2
Page 5
Chapter 2: Key Concept: The Derivative
16. Let f(x) = log(log(x)). Estimate f '(7) to 3 decimal places using any method. Ans: 0.032 difficulty: hard section: 2.2 17. For f ( x) = log x , estimate f ′(3) to 3 decimal places by finding the average slope over intervals containing the value x = 3. Ans: 0.145 difficulty: medium section: 2.2 18. There is a function used by statisticians, called the error function, which is written y = erf (x). Suppose you have a statistical calculator, which has a button for this function. Playing with your calculator, you discover the following: x erf(x) 1 0.29793972 0.1 0.03976165 0.01 0.00398929 0.001 0.000398942 0 0 Using this information alone, give an estimate for erf′(0), the derivative of erf at x = 0 to 4 decimal places. Ans: 0.3989 difficulty: medium section: 2.2
Page 6
Chapter 2: Key Concept: The Derivative
19. In the picture
the quantity f '(a+h) is represented by A) the slope of the line TV D) the length of the line TV. B) the area of the rectangle PQRS E) the slope of the line QU. C) the slope of the line RU. F) the length of the line QU. Ans: A difficulty: medium section: 2.2 20. Given the following table of values for a Bessel function, J 0 ( x) , estimate the derivative at x = 0.5. x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1.0 .9975 .9900 .9776 .9604 .9385 .9120 .8812 .8463 J 0 ( x) Ans: –0.242 difficulty: medium
section: 2.2
21. The data in the table report the average improvement in scores of six college freshmen who took a writing assessment before and again after they had x hours of tutoring by a tutor trained in a new method of instruction. When f(x)>0 the group showed improvement on average. x 2 3.5 5 6.5 8 9.5 11 f(x) -2 -1 0 3 7 9 10 a) Find the average change in score from 6.5 to 9.5 hours of tutoring. b) Estimate the instantaneous rate of change at 8 hours. c) Approximate the equation of the tangent line at x = 8 hours. d) Use the tangent line to estimate f(8.5). − 7 2.00( x − 8) ; d) Ans: a) 2.00 points, b) 2.00 points, but answers may vary; c) y = 8 points difficulty: medium section: 2.2
Page 7
0.9 .8075
Chapter 2: Key Concept: The Derivative
22. Use the graph of 5.5e3 x at the point (0, 5.5) to estimate f ′(0) to three decimal places. A) 16.500 B) 36.823 C) 3.500 D) 146.507 Ans: A difficulty: easy section: 2.2 23. A horticulturist conducted an experiment to determine the effects of different amounts of fertilizer on the yield of a plot of green onions. He modeled his results with the function Y ( x) = −0.5( x − 2) 2 + 5 where Y is the yield in bushels and x is the amount of fertilizer in pounds. What are Y (0.75) and Y ′(0.75) ? Give your answers to two decimal places, specify units. A) 4.22 bushels, 1.25 bushels/pound, respectively B) 4.22 bushels, 6.25 bushels/pound, respectively C) 1.25 bushels, 1.56 bushels/pound, respectively D) 1.25 bushels, 4.22 bushels/pound, respectively Ans: A difficulty: medium section: 2.2 24. Use the limit of the difference quotient to find the derivative of g ( x) = (1, 11/2). −11 A) 4 Ans: A
11 −11 C) -11 D) 2 2 difficulty: medium section: 2.2 B)
25. Could the first graph, A be the derivative of the second graph, B?
A Ans: yes difficulty: medium
B section: 2.3
Page 8
11 at the point x +1
Chapter 2: Key Concept: The Derivative
26. Could the first graph, A be the derivative of the second graph, B?
A Ans: no difficulty: medium
B section: 2.3
Page 9
Chapter 2: Key Concept: The Derivative
27. Consider the function y = f(x) graphed below. At the point x = –3, is f '( x) positive, negative, 0, or undefined?Note: f(x) is defined for -5 < x < 6, except x = 2.
Ans: positive difficulty: medium
section: 2.3
28. Estimate a formula for f '( x) for the function f ( x) = 8 x . decimal places. Ans: (2.079)8 x difficulty: hard
section: 2.3
Page 10
Round constants to 3
Chapter 2: Key Concept: The Derivative
29. Could the first graph, A be the derivative of the second graph, B?
A Ans: yes difficulty: medium
B section: 2.3
30. Find the derivative of g ( x) = 2 x 2 + 8 x − 6 at x = 4 algebraically. Ans: 24 difficulty: medium section: 2.3
Page 11
Chapter 2: Key Concept: The Derivative
31. To find the derivative of g ( x) = 2 x 2 + 5 x − 9 at x = 8 algebraically, you evaluate the following expression. 2(8 + h) 2 − 5(8 + h) − 9 − (2 ⋅ 82 + 5 ⋅ 8 − 9) A) lim h →0 h g (8 + 1) + g (8) B) h g (h) − g (8) C) lim h →∞ h D) All of the above are correct. E) None of the above is correct. Ans: A difficulty: medium section: 2.3 32. Find the derivative of m( x) = 3 x3 at x = 1 algebraically. Ans: 9 difficulty: medium section: 2.3 33. Draw the graph of a continuous function y = g(x) that satisfies the following three conditions: • g′(x) = 0 for x < 0 • g′(x) > 0 for 0 < x < 4 • g′(x) < 0 for x > 4 Ans: Answers will vary. One example:
difficulty: medium
section: 2.3
Page 12
Chapter 2: Key Concept: The Derivative
34. The graph below shows the velocity of a bug traveling along a straight line on the classroom floor. v meters/sec 4 3 2 1 t sec 1
2
3
4
5
6
7
8
9
10
11
12
-1 -2 -3 -4 -5
At what time(s) does the bug turn around? A) At 3 seconds. C) B) At 2 seconds and again at 7 seconds. D) Ans: A difficulty: easy section: 2.3
Page 13
At 4 seconds and again at 7 seconds. Never.
Chapter 2: Key Concept: The Derivative
35. The graph below shows the velocity of a bug traveling along a straight line on the classroom floor. v meters/sec 4 3 2 1 t sec 1
2
3
4
5
6
7
8
9
10
11
12
-1 -2 -3 -4 -5
When is the bug moving at a constant speed? A) Between 4 and 7 seconds. B) Whenever the velocity is linear with a positive slope. C) Whenever the velocity is linear with a negative slope. D) When the velocity is equal to zero. Ans: A difficulty: easy section: 2.3
Page 14
Chapter 2: Key Concept: The Derivative
36. he graph below shows the velocity of a bug traveling along a straight line on the classroom floor. v meters/sec 4 3 2 1 t sec 1
2
3
4
5
7
6
8
9
10
11
12
-1 -2 -3 -4 -5
Graph the bug's speed at time, t. How does it differ from the bug's velocity? 3 2 1 -1 -2 -3
speed t 1 2 3 4 5 6 7 8 9 1011
Ans: Speed is always non-negative, but has the same magnitude as the velocity. difficulty: medium section: 2.3 37. Use the definition of the derivative function to find a formula for the slope of the graph of 1 . f (t ) = 9t + 1 −9 Ans: (9t + 1) 2 difficulty: hard section: 2.3
Page 15
Chapter 2: Key Concept: The Derivative
38. What is the equation of the tangent line to the graph of f ( x) = x 3 at the point (2, 8)? y 12 x − 16 y 2x + 8 y 8x + 2 y 4 x + 64 B) = C) = D) = A) = Ans: A difficulty: easy section: 2.3 39. The definition of the derivative function is f ′( x) = Ans: False
difficulty: easy
f ( x + h) − f ( x ) h
section: 2.3
40. A runner competed in a half marathon in Anaheim, a distance of 13.1 miles. She ran the first 7 miles at a steady pace in 48 minutes, the second 3 miles at a steady pace in 28 minutes and the last 3.1 miles at a steady pace in 18 minutes. a) Sketch a well-labeled graph of her distance completed with respect to time. b) Sketch a well-labeled graph of her velocity with respect to time. 13 12 11 10 9 8 7 6 5 4 3 2 1
distance
velocity mi/min 0.4 0.3 0.2 x
10 20 30 40 50 60 70 80 90
0.1
x 10 20 30 40 50 60 70 80
Ans: Answers will vary. The graphs above give one possibility. difficulty: medium section: 2.3 41. Which of the following is NOT a way to describe the derivative of a function at a point? A) slope of the tangent line D) limit of the difference quotient B) slope of the curve E) limit of the slopes of secant lines C) y-intercept of the tangent line F) limit of the average rates of change Ans: C difficulty: easy section: 2.3 42. Suppose that f(T) is the cost to heat my house, in dollars per day, when the outside temperature is T F . If f(28) = 11.10 and f ′(28) = –0.12, approximately what is the cost to heat my house when the outside temperature is 25 F ? Ans: $11.46 difficulty: easy section: 2.4
Page 16
Chapter 2: Key Concept: The Derivative
43. To study traffic flow along a major road, the city installs a device at the edge of the road at 1:00a.m. The device counts the cars driving past, and records the total periodically. The resulting data is plotted on a graph, with time (in hours since installation) on the horizontal axis and the number of cars on the vertical axis. The graph is shown below; it is the graph of the function C(t) = Total number of cars that have passed by after t hours. When is the traffic flow greatest?
A) 2:00 am B) 3:00 am C) 4:00 am D) 5:00 am Ans: D difficulty: medium section: 2.4 44. To study traffic flow along a major road, the city installs a device at the edge of the road at 3:00a.m. The device counts the cars driving past, and records the total periodically. The resulting data is plotted on a graph, with time (in hours since installation) on the horizontal axis and the number of cars on the vertical axis. The graph is shown below; it is the graph of the function C(t) = Total number of cars that have passed by after t hours. From the graph, estimate C′(6).
A) 600 Ans: A
B) 900 C) 1200 difficulty: medium
D) 1500 section: 2.4
Page 17
Chapter 2: Key Concept: The Derivative
45. Every day the Office of Undergraduate Admissions receives inquiries from eager high school students. They keep a running account of the number of inquiries received each day, along with the total number received until that point. Below is a table of weekly figures from about the end of August to about the end of October of a recent year. Week of 8/28-9/01 9/04-9/08 9/11-9/15 9/18-9/22 9/25-9/29 10/02-10/06 10/09-10/13 10/16-10/20 10/23-10/27
Inquiries That Total for Year Week 1085 11,928 1193 13,121 1312 14,433 1443 15,876 1588 17,464 1746 19,210 1921 21,131 2113 23,244 2325 25,569
One of these columns can be interpreted as a rate of change. Which one is it? A) the first B) the second C) the third Ans: B difficulty: easy section: 2.4 46. Every day the Office of Undergraduate Admissions receives inquiries from eager high school students. They keep a running account of the number of inquiries received each day, along with the total number received until that point. Below is a table of weekly figures from about the end of August to about the end of October of a recent year. Week of 8/28-9/01 9/04-9/08 9/11-9/15 9/18-9/22 9/25-9/29 10/02-10/06 10/09-10/13 10/16-10/20 10/23-10/27
Inquiries That Total for Year Week 1085 11,928 1193 13,121 1312 14,433 1443 15,876 1588 17,464 1746 19,210 1921 21,131 2113 23,244 2325 25,569
Based on the table determine a formula that approximates the total number of inquiries received by a given week. Use your formula to estimate how many inquiries the admissions office will have received by November 24. Ans: 37,435 difficulty: medium section: 2.4
Page 18
Chapter 2: Key Concept: The Derivative
47. Let L(r) be the amount of board-feet of lumber produced from a tree of radius r (measured in inches). What does L(16) mean in practical terms? A) The amount of board-feet of lumber produced from a tree with a radius of 16 inches. B) The radius of a tree that will produce 16 board-feet of lumber. C) The rate of change of the amount of lumber with respect to radius when the radius is 16 inches (in board-feet per inch). D) The rate of change of the radius with respect to the amount of lumber produced when the amount is 16 board-feet (in inches per board-foot). Ans: A difficulty: easy section: 2.4 48. Let t(h) be the temperature in degrees Celsius at a height h (in meters) above the surface of the earth. What does t '(1200) mean in practical terms? A) The temperature in degrees Celsius at a height 1200 meters above the surface of the earth. B) The height above the surface of the earth at which the temperature is 1200 degrees Celsius. C) The rate of change of temperature with respect to height at 1200 meters above the surface of the earth (in degrees per meter). D) The rate of change of height with respect to temperature when the temperature is 1200 degrees Celsius (in meters per degree). Ans: C difficulty: easy section: 2.4 49. Let t(h) be the temperature in degrees Celsius at a height of h meters above the surface of the earth. What does h such that t(h) = 8 mean in practical terms? A) The temperature in degrees Celsius at a height 8 meters above the surface of the earth. B) The height above the surface of the earth at which the temperature is 8 degrees Celsius. C) The rate of change of temperature with respect to height at 8 meters above the surface of the earth (in degrees per meter). D) The rate of change of height with respect to temperature when the temperature is 8 degrees Celsius (in meters per degree). Ans: B difficulty: easy section: 2.4
Page 19
Chapter 2: Key Concept: The Derivative
50. Let t(h) be the temperature in degrees Celsius at a height of h meters above the surface of the earth. What does t(h) + 15 mean in practical terms? A) The temperature in degrees Celsius at a height h meters above the surface of the earth plus an additional 15 degrees B) The height above the surface of the earth at which the temperature is h degrees Celsius plus an additional 15 meters. C) The rate of change of temperature with respect to height at 15 additional meters above the surface of the earth (in degrees per meter). D) The rate of change of height with respect to temperature when the temperature is 15 additional degrees Celsius (in meters per degree). Ans: A difficulty: easy section: 2.4 51. A concert promoter estimates that the cost of printing p full color posters for a major concert is given by a function Cost = c(p) where p is the number of posters produced. a) Interpret the meaning of the statement c(450) = 5400. b) Interpret the meaning of the statement c'(450) = 11. Ans: a) It costs $5400.00 to produce 450 posters. b) When 450 posters have been produced, it costs $11.00 to produce an additional poster. difficulty: easy section: 2.4
Page 20
Chapter 2: Key Concept: The Derivative
52. The graph below gives the position of a spider moving along a straight line on the forest floor for 10 seconds. On the same axes, sketch a graph of the spider's velocity over the 10 seconds. Then write a description of the spider's movement for the 10 second period.
5
p
4 (3, 3)
(5, 3)
(9, 3) (10,
3 2 (7, 1) 1
t
(0, 0) 1
2
3
4
5
6
7
8
9
10
-1 -2 -3 -4 -5
5
p
4 (3, 3)
(0, 3)
(5, 3)
(9, 3) (10,
3 2 (7, 1)
(9, 1)
1 (0, 0) 1
(3, 0) 2
3
4
(9, 0)
(5, 0) 5
6
7
8
-1 -2 -3 -4 -5
Ans: The dashed line represents the spider's velocity. Page 21
9
10
t
Chapter 2: Key Concept: The Derivative
For the first three seconds the spider moves forward at 3 feet/sec. It stops for the next 2 seconds, turns around and goes back in the opposite direction for at a speed of 1 ft/sec for 2 seconds. It turns around again and goes forward at 1 ft/sec for the next two seconds, then stops for the final second. difficulty: medium section: 2.4 53. A typhoon is a tropical cyclone, like a hurricane, that forms in the northwestern Pacific Ocean. The wind speed of a typhoon is given by a function W = w(r) where W is measured in meters/sec., and r is measured in kilometers from the center of the typhoon. What does the statement that w'(15) > 0 tell you about the typhoon? A) At a distance of 15 kilometers from the center of the typhoon, the wind speed is increasing. B) At a distance of 15 kilometers from the center of the typhoon, the wind speed is positive. C) The wind speed of the typhoon is 15 meters per second at any distance from the center of the typhoon. Ans: A difficulty: medium section: 2.4 54. The cost in dollars to produce q bottles of a prescription skin treatment is given by the function C (q= ) 0.08q 2 + 75q + 900 . The manufacturing process is difficult and costly when large quantities are produced. The marginal cost of producing one additional dC bottle when q bottles have been produced is the derivative . dq a) Find the marginal cost function. b) Compute C(50) and explain what the number means in terms of cost and production. c) Compute C'(50) and explain what the number means in terms of cost and production. dC Ans: a) = 0.16q + 75 dq b) C(50) = $4850.00 is the cost of producing 50 bottles of the skin treatment. c) C'(50) = $83.00 per bottle of the cost of producing an additional bottle when 50 have already been produced. difficulty: medium section: 2.4
Page 22
Chapter 2: Key Concept: The Derivative
55. The graph of f ( x) is given in the following figure. What happens to f '( x) at the point x1 ?
f '( x) has an inflection point. A) f '( x) has a local minimum or maximum. B) f '( x) changes sign. C) D) none of the above Ans: C difficulty: hard section: 2.5 56. Esther is a swimmer who prides herself in having a smooth backstroke. Let s(t) be her position in an Olympic size (50-meter) pool, as a function of time (s(t) is measured in meters, t is seconds). Below we list some values of s(t) for a recent swim. Find Esther's average speed over the entire swim in meters per second. Round to 2 decimal places. t s(t)
0 0
Ans: 1.64 difficulty: medium
3.0 10
8.6 20
14.64 30
21.35 40
28.06 50
32.33 40
39.04 30
46.36. 20
54.9 10
61 0
53.9 10
60 0
section: 2.5
57. Esther is a swimmer who prides herself in having a smooth backstroke. Let s(t) be her position in an Olympic size (50-meter) pool, as a function of time (s(t) is measured in meters, t is seconds). Below we list some values of s(t), for a recent swim. Based on the data, was Esther's instantaneous speed ever greater than 3 meters/second? t s(t)
0 0
Ans: yes difficulty: medium
3.0 10
8.6 20
14.6 30
section: 2.5
Page 23
20.8 40
27.6 50
31.9 40
38.1 30
45.8 20
Chapter 2: Key Concept: The Derivative
58. The graph below represents the rate of change of a function f with respect to x; i.e., it is a graph of f ′. You are told that f (0) = 0. What can you say about f ( x) at the point x = 1.3? Mark all that apply.
f ( x) is decreasing. A) f ( x) is increasing. B) Ans: A, D difficulty: easy
C) D) section: 2.5
f ( x) is concave up. f ( x) is concave down.
59. The graph below represents the rate of change of a function f with respect to x; i.e., it is a graph of f′. You are told that f(0) = –2. For approximately what value of x other than x = 0 in the interval 0 ≤ x ≤ 2 does f ( x) = –2?
A) 0.6 Ans: C
B) 1 C) 1.4 D) 2 E) None of the above difficulty: medium section: 2.5
Page 24
Chapter 2: Key Concept: The Derivative
60. On the axes below, sketch a smooth, continuous curve (i.e., no sharp corners, no breaks) which passes through the point P(5, 6), and which clearly satisfies the following conditions: • Concave up to the left of P • Concave down to the right of P • Increasing for x > 0 • Decreasing for x < 0 • Does not pass through the origin.
5
y
4 3 2 1 -5 -4 -3 -2 -1 -1
x 1
2
3
4
5
-2 -3 -4 -5
Ans: Answers will vary. One possibility:
difficulty: easy
section: 2.5
Page 25
Chapter 2: Key Concept: The Derivative
61. One of the following graphs is of f ( x) , and the other is of f '( x) . Is f ( x) the first graph or the second graph?
Ans: second difficulty: medium
section: 2.5
62. Given the following data about a function f, estimate the rate of change of the derivative f ' at x = 4.5. . x 3 3.5 4 4.5 5 5.5 6 f(x) 10 8 7 4 2 0 -1 Ans: 4 difficulty: medium
section: 2.5
Page 26
Chapter 2: Key Concept: The Derivative
63. A function defined for all x has the following properties: • f is increasing • f is concave down • f (3) = 2 • f ′(3) = 1/2 How many zeros does f(x) have in the interval 1 ≤ x ≤ 3 ? Ans: 1 difficulty: medium section: 2.5 64. A function defined for all x has the following properties: • f is increasing • f is concave down • f (4) = 2 f '(4) = 1/ 2 • Is it possible that f '(1) = Ans: no difficulty: medium
1 ? 4
section: 2.5
65. Assume that f is a differentiable function defined on all of the real line. Is it possible that f > 0 everywhere, f ′ > 0 everywhere, and f ″ < 0 everywhere? Ans: no difficulty: medium section: 2.5 66. Assume that f and g are differentiable functions defined on all of the real line. Is it possible that f ′(x) > g′(x) for all x and f(x) < g(x) for all x? Ans: yes difficulty: medium section: 2.5 67. Assume that f and g are differentiable functions defined on all of the real line. If f ′(x) = g′(x) for all x and if f ( x0 ) = g ( x0 ) for some x0 , then must f(x) = g(x) for all x? Ans: yes difficulty: medium section: 2.5 68. Assume that f and g are differentiable functions defined on all of the real line. If f ' > 0 everywhere and f > 0 everywhere then must lim f ( x) = ∞ ? x→+ ∞
Ans: no difficulty: medium
section: 2.5
Page 27
Chapter 2: Key Concept: The Derivative
69. Suppose a function is given by a table of values as follows: x f(x)
1.1 14
1.3 17
1.5 23
1.7 25
1.9 26
2.1 27
Give your best estimate of f ''(1.9) . Ans: 0 difficulty: medium section: 2.5 70. If the Figure 1 is f ( x) , could Figure 2 be f ''( x) ?
Figure 1 Ans: no difficulty: medium
Figure 2
section: 2.5
71. The cost of mining a ton of coal is rising faster every year. Suppose C(t) is the cost of mining a ton of coal at time t. Must C ''(t) be concave up? Ans: no difficulty: medium section: 2.5 72. Let S(t) represent the number of students enrolled in school in the year t. If the number of students enrolling is increasing faster and faster, then is S '(t) positive, negative, or 0? Ans: positive difficulty: medium section: 2.5
Page 28
Chapter 2: Key Concept: The Derivative
73. A company graphs C′(t), the derivative of the number of pints of ice cream sold over the past ten years. At approximately what year was C ''(t) greatest?
Ans: 0 difficulty: medium
section: 2.5
74. A golf ball thrown directly upwards from the surface of the moon with an initial velocity of 17.00 meters per second and will attain a height of s (t ) = −0.8t 2 + 17.00t meters in t seconds. Find a formula for the velocity of the golf ball at time t. v(t ) = −1.6t + 17.00 meters/sec v(t ) = –1.36 meters/sec C) A) v(t ) = −0.8t + 8.50 meters/sec B) D) v(t ) = −16t 2 + 17.00 meters/sec Ans: A difficulty: medium section: 2.5 75. A golf ball thrown directly upwards from the surface of the moon with an initial velocity of 14.00 meters per second and will attain a height of s (t ) = −0.8t 2 + 14.00t meters in t seconds. What is the acceleration of the golf ball at time t? −1.6 meters/sec/sec −0.8t meters/sec/sec C) A) −1.6t meters/sec2 B) D) 14.00 meters/sec2 Ans: A difficulty: medium section: 2.5 76. A golf ball thrown directly upwards from the surface of the moon with an initial velocity of 20 meters per second and will attain a height of s (t ) = −0.8t 2 + 20t meters in t seconds. How fast is the golf ball going at its high point? A) 0 meters/sec B) -0.8 meters/sec C) 20 meters/sec D) -20 meters/sec Ans: A difficulty: easy section: 2.5
Page 29
Chapter 2: Key Concept: The Derivative
77. A golf ball thrown directly upwards from the surface of the moon with an initial velocity of 20 meters per second and will attain a height of s (t ) = −0.8t 2 + 20t meters in t seconds. On Earth, its height would be given by −4.9t 2 + 20t .Compare the velocity and acceleration of the golf ball on the moon after two seconds with its velocity and acceleration on Earth after two seconds. Ans: Comparisons will vary. The numerical results are: On the moon: velocity -3.2 m/s and acceleration -1.6 m/s2 On the Earth: velocity -19.6 m/s and acceleration - 9.8 m/s2 difficulty: medium
section: 2.5
78. The Chief Financial Officer of an insurance firm reports to the board of directors that the cost of claims is rising more slowly than last quarter. Let C(t) be the cost of claims. Select all statements that apply. A) C is positive. B) C is negative. C) The first derivative of C is positive. D) The first derivative of C is negative. E) The second derivative of C is positive. F) The second derivative of C is negative. Ans: A, C, F difficulty: medium section: 2.5 79. A husband and wife purchase life insurance policies. Over the next 40 years, one policy pays out when the husband dies, and the other pays out when both husband and wife die. Their life expectancy is 20 years, and the probability that both die before year t is given 1 by the function fT (t ) = 1600 t 2 . How fast is the probability that both are dead increasing in 25 years? A) 0.0313 B) 0.3906 C) 0.0500 D) 50.0006 Ans: A difficulty: hard section: 2.5
Page 30
Chapter 2: Key Concept: The Derivative
80. Sketch a graph y = f(x) that is continuous everywhere on -6 < x < 6 but not differentiable at x = -3 or x = 3. Ans: Many possible. One example:
difficulty: easy
section: 2.6
81. Sketch a graph of a continuous function f(x) with the following properties: • f″(x) < 0 for x < 4 • f″(x) > 0 for x > 4 • f″(4) is undefined Ans: Many possible. One example:
difficulty: easy
section: 2.6
82. Is the graph of f ( x)= x + 3 continuous at x = –3? Ans: yes difficulty: easy section: 2.6
Page 31
Chapter 2: Key Concept: The Derivative
83. Is the graph of f ( x) = Ans: no difficulty: easy
1 continuous at x = –9? x+9
section: 2.6
−1 ≤ r ≤ 1 1 − sin(πr / 2) . Is h(r ) differentiable at r = 84. Given the function h(r ) = 0 r < −1, r > 1 –1? Ans: no difficulty: medium section: 2.6 85. Describe two ways that a continuous function can fail to have a derivative at a point, x = a. Illustrate your description with graphs. Ans: Answers will vary but will describe two of: cusps, corners, vertical tangents. difficulty: medium section: 2.6 86. A function that has an instantaneous rate of change of 3 at a point (x, y) can fail to be continuous at that point. Ans: False difficulty: easy section: 3.6 87. Based on the graph of f(x) below: a) List all values of x for which f is NOT differentiable. b) List all values of x for which f is NOT continuous. c) List all values of x for which f '(x) = 0. 5 4 3 2 1 -5 -4 -3 -2 -1 -1
y
x 1 2 3 4 5
-2 -3 -4 -5
Ans: a) Not differentiable at x = -2.5, -1, 3, 4 b) Not continuous at x = 3, 4 c) Derivative of zero at x = -5. difficulty: easy section: 3.8
Page 32
Chapter 2: Key Concept: The Derivative
88. Let f ( x) = xsin x . Using your calculator, estimate f ′(7) to 3 decimal places. Ans: 5.605 difficulty: medium section: 2 review 89. Alone in your dim, unheated room you light one candle rather than curse the darkness. Disgusted by the mess, you walk directly away from the candle. The temperature (in F ) and illumination (in % of one candle power) decrease as your distance (in feet) from the candle increases. The table below shows this information.
distance(feet)
Temp. (° F)
0 1 2 3 4 5 6
55 54.5 53.5 52 50 47 43.5
illumination (%) 100 85 75 67 60 56 53
Does the following graph show temperature or illumination as a function of distance?
Ans: illumination difficulty: easy section: 2 review
Page 33
Chapter 2: Key Concept: The Derivative
90. Alone in your dim, unheated room you light one candle rather than curse the darkness. Disgusted by the mess, you walk directly away from the candle. The temperature (in F ) and illumination (in % of one candle power) decrease as your distance (in feet) from the candle increases. The table below shows this information.
distance(feet) 0 1 2 3 4 5 6
Temp. (° F) 56 55.5 54.5 53 51 48 44.5
illumination (%) 100 85 75 67 60 56 53
What is the average rate at which the temperature is changing (in degrees per foot) when the illumination drops from 75% to 56%? Round to 2 decimal places. Ans: 2.17 difficulty: medium section: 2 review 91. Alone in your dim, unheated room you light one candle rather than curse the darkness. Disgusted by the mess, you walk directly away from the candle. The temperature (in F ) and illumination (in % of one candle power) decrease as your distance (in feet) from the candle increases. The table below shows this information.
distance(feet)
Temp. (° F)
0 1 2 3 4 5 6
55 54.5 53.5 52 50 47 43.5
illumination (%) 100 85 75 67 60 56 53
You can still read your watch when the illumination is about 55%, so somewhere between 5 and 6 feet. Can you read your watch at 5.5 feet? A) yes B) no C) cannot tell Ans: B difficulty: medium section: 2 review
Page 34
Chapter 2: Key Concept: The Derivative
92. Alone in your dim, unheated room you light one candle rather than curse the darkness. Disgusted by the mess, you walk directly away from the candle. The temperature (in F ) and illumination (in % of one candle power) decrease as your distance (in feet) from the candle increases. The table below shows this information.
distance(feet)
Temp. (° F)
0 1 2 3 4 5 6
55 54.5 53.5 52 50 47 43.5
illumination (%) 100 85 75 67 60 56 53
Suppose you know that at 6 feet the instantaneous rate of change of the illumination is –3.5 % candle power/ft. At 7 feet, the illumination is approximately _____ % candle power. Ans: 49.5 difficulty: medium section: 2 review 93. Alone in your dim, unheated room you light one candle rather than curse the darkness. Disgusted by the mess, you walk directly away from the candle. The temperature (in F ) and illumination (in % of one candle power) decrease as your distance (in feet) from the candle increases. The table below shows this information. distance(feet)
Temp. (° F)
0 1 2 3 4 5 6
55 54.5 53.5 52 50 47 43.5
illumination (%) 100 85 75 67 60 56 53
You are cold when the temperature is below 40°F. You are in the dark when the illumination is at most 50% of one candle power. Suppose you know that at 6 feet the instantaneous rate of change of the temperature is -4.5° F/ft and the instantaneous rate of change of illumination is -3% candle power/ft. Are you in the dark before you are cold, or cold before you are dark? A) You are cold before you are in the dark. B) You are in the dark before you are cold. Ans: A difficulty: medium section: 2 review
Page 35
Chapter 2: Key Concept: The Derivative
94. Could the Function 1 be the derivative of the Function 2?
Function 1 Ans: no difficulty: medium 95. Is the function Ans: no difficulty: easy
Function 2
section: 2 review
f ( x) =
x2 2 x − 4 x−2
continuous at x = 2?
section: 2 review
96. Mark all TRUE statements. A) f ( x)= x − 3 is continuous at x = 0. B) C)
g ( x) = 3 x fails to be differentiable at x = 0. h( x)= x + 5 is not continuous at x = -5.
( x − 3) 2 is continuous for all values of x. ( x − 3) E) Any polynomial function is differentiable for all values of x. Ans: A, B, E difficulty: easy section: 2 review
D)
r ( x) =
97. In the lobby of a university mathematics building, there is a large bronze sculpture in the shape of a parabola. When the sun shines on the parabola at a certain time, its shadow falls on a mural with a coordinate plane that reveals the sculpture's height as the function − x 2 + 18 . A spider drops from its web onto the sculpture at the point (1, 17). f ( x) = What is the slope of the parabola at the point where the spider lands? A) –2 B) –20 C) 17 D) –17 E) None of the above Ans: A difficulty: easy section: 2 review 98. If f ( x) = x 4 , what is f ′(3) ? A) 81 B) 3 C) 108 Ans: C difficulty: easy
D) 12 E) 4 section: 2 review
99. The derivative of f (t ) = eπ is π eπ −1 . Ans: False difficulty: easy section: 2 review
Page 36
1. Consider the function f ( x) =x3 − 4 x 2 + 7 . The equation of the tangent line at x = 2 is y = _____ x + _____. Part A: –4 Part B: 7 difficulty: easy section: 3.1 2. Consider the function f ( x) =x3 − 4 x 2 + 4 . Estimate f (1.5) using the tangent line at x = 1. Ans: –1.5 difficulty: medium section: 3.1 3. Given f ( x) = x3 − 7 x 2 + 8 x − 3 , what is the slope of the tangent line to the curve at x = –3? Ans: 77 difficulty: easy section: 3.1 4. Given f ( x) =x3 − 9 x 2 + 24 x − 6 , at which value(s) of x does the curve have a horizontal tangent? A) 1 B) 2 C) 3 D) 4 E) 5 Ans: B, D difficulty: easy section: 3.1 5. The 12th derivative of f ( x) = x12 is 0. Ans: False difficulty: easy section: 3.1 6. Find the derivative of= y 2 x3 − A) B)
6 x2 + 6 x2 −
1 . 8x
1 8x 1
2
8x2 1 C) 6 x2 − 8 Ans: A difficulty: easy
6 x2 +
E)
None of the above
section: 3.1
Page 1
1 8
D)
Chapter 3: Short-Cuts to Differentiation
7. Find the derivative of= y 2 x + x5 . 1 x + 5x4 1 B) + 5x 4 x 1 C) − + 5x 4 x Ans: B difficulty: medium
A)
8. Find the derivative of y =
2 + 5x 4
E)
None of the above
section: 3.1
8t 3 − t 2 + 5 t2
D)
.
5 10 10 A) 7 + 3 B) 8 + 2 C) 8 − 3 D) 12t − 1 3t t t Ans: C difficulty: medium section: 3.1
E) None of the above
9. If g (t ) = 2t 3 − t 2 + 5t , then what is g ''(t ) ? Ans: 12t - 2 difficulty: easy section: 3.1 10. If g (t ) represents the position of a particle at time t seconds, then g'(t) represents the __________ of the particle at time t seconds. Ans: velocity difficulty: easy section: 3.1 11. Consider the function f ( x) = 2 x 4 − 4 x3 + 4 . Is f increasing or decreasing at the point x = 0.5? Ans: decreasing difficulty: medium section: 3.1 12. Consider the function f ( x) = 5 x 4 − 4 x3 + 4 . Is f concave up or down at the point x = –0.2? Ans: up difficulty: medium section: 3.1 13. Given a power function of the form f ( x) = ax n , find n and a so that f ′(2) = 1 and f ′(4) = 1/4. n = _____ a = _____ Part A: −1 Part B: –4 difficulty: hard section: 3.1
Page 2
Chapter 3: Short-Cuts to Differentiation
14. Given f (= x) 3 x 2 − x and g ( x) =x3 + 5 x 2 − 4 , find Ans: 3 x 2 − 8 x + 3 difficulty: medium
d [ g ( x) − 3 f ( x) ] . dx
section: 3.1
x 3/ 2 − 6 x + x1/ 2 with respect to x. x 3/ 2 –48 –8.5 3 –1 1.5 x 0.5 − 6 + 1/ 2 x −1/ 2 B) C) A) − + x 3/ 2 x2 x3/ 2 x 2 1.5 x 0.5 Ans: A difficulty: medium section: 3.1
15. Find the derivative of g ( x) =
7 . x4 7 7 28 B) 25 − 5 C) D) 25 − 3 3 4x x 4x difficulty: medium section: 3.1
D) 6x - 3
16. Differentiate = y 25 −
28 x5 Ans: A A)
E) None of the above
17. Find a formula for the slope of the tangent line to y= ( x − 9) 2 A) 2x - 18 B) 2x - 9 C) 2(x - 9)3 D) (x - 9)3/3 Ans: A difficulty: easy section: 3.1
E) none of the above
18. Given the variable θ , find r ′(θ ) when r (θ ) = 6θ 9 . A) 54θ 8 B) (6 ln 9)θ 9 C) (6 ln 9)θ 8 D) r ′(θ ) = r (θ ) Ans: A difficulty: medium section: 3.1 19. Find y′ when y = ba x b . A)
a 2
xb−2
a xb 2x ax b / 2−1 C) 2 Ans: D difficulty: medium
B)
D)
all of the above
E)
none of the above
section: 3.1
20. Find f ′(2) when f ( x) = 3 x 5 − 8 x 2 + 2 x − 1 . Ans: 210 difficulty: easy section: 3.1
Page 3
Chapter 3: Short-Cuts to Differentiation
21. A man plans to propose to a woman in romantic fashion by taking her up in an air balloon. Unfortunately, he pulls the diamond ring from his pocket and drops it over the side of the balloon's basket. The ring's position above the earth t seconds after it falls is given by the function s (t ) = −16t 2 + 1225 feet. How fast is the ring falling 3 seconds after he drops it? Ans: –96 feet/sec difficulty: easy section: 3.1 22. A man plans to propose to a woman in romantic fashion by taking her up in an air balloon. Unfortunately, he pulls the diamond ring from his pocket and drops it over the side of the balloon's basket. The ring's position above the earth t seconds after it falls is given by the function s (t ) = −16t 2 + 1325 feet. How fast is the ring falling at the instant it hits the ground? 1325 Ans: –291.20 feet/sec. difficulty: medium section: 3.1 23. On what intervals is the polynomial p ( x) = 2 x 3 − 15 x 2 + 36 x + 70. increasing? Decreasing? Ans: Increasing on (−∞, 2), (3, ∞) . Decreasing on (2, 3). difficulty: medium section: 3.1 24. On what intervals is the polynomial p ( x) = 2 x 3 − 15 x 2 + 36 x + 50. concave down? Concave up? Ans: Concave down on (−∞, 2.5) . Concave up on (2.5, ∞) . difficulty: medium section: 3.1 25. Consider the graph y = e x . What is the x-intercept of the tangent line to the graph at
( a, e a ) A) a –1 B) e a (1 – a ) C) 1 − a D) e a (a − 1) Ans: A difficulty: medium section: 3.2
Page 4
Chapter 3: Short-Cuts to Differentiation
26. If P dollars are invested at an annual rate of r%, then in t years this investment grows to F t
dF r dollars, where = . F P 1 + . Assuming P and r are constant, find dt 100 A)
r P 1 + 100
t
t
C)
r r P 1 + ln 1 + 100 100
D)
r P 1 + 100 r ln 1 + 100
t
B)
r Pt 1 + 100
Ans: C
t −1
difficulty: medium
section: 3.2
27. If P dollars are invested at an annual rate of r%, then in t years this investment grows to F t
r dollars, where = F P 1 + . If you solve this equation for P and hold F and r 100 dP be? constant, what will the sign of dt A) positive B) negative Ans: B difficulty: medium section: 3.2
28. Find the derivative of = y 4x − 4 . A) 4 x Ans: B
B) (ln 4)4 x C) (ln 4)4 x − ln 4 D) difficulty: medium section: 3.2
x 4 x −1
29. Find the derivative of f ( x) = 7e π x . B) 7 ln(π)e π x A) 7 π xe π x −1 Ans: D difficulty: medium
C) 7e π x section: 3.2
D) 7 πe π x
30. Find the derivative of= f ( x) (ln 4) x 2 + ln(7)e x . A)
2 ln(4 x) + 7 x
B) ( ln 8) x + ln(7)e x Ans: D difficulty: medium
C)
(2 ln 4) x + 7e x
D) (2 ln 4) x + ln(7)e x section: 3.2
Page 5
Chapter 3: Short-Cuts to Differentiation
31. Find the derivative of A) B)
1
3+
4/3
3x 1
3−
3x Ans: A
4/3
1 g ( x) =3 x − 3 + 3x . x
+ (ln 3)3x
C)
3+
+ (ln 3)3x
D)
3−
difficulty: medium
1 3 x 1 3 x
+ x3x −1 + x3x −1
section: 3.2
32. Find the derivative of h(= t ) t π + (π 4 )t + πt 4 . 4
A)
π 4t ( π −1) + t (π 4 )t −1 + 4πt 3
B)
π 4t ( π −1) + (π 4 )t ln(π 4 ) + 4πt 3
C)
t ( π −1) ln t + (π 4 )t ln(π 4 ) + 4πt 4 ln t
4
4
4
D) 4t π + 4t (π 3 )t −1 + 4πt 4 ln t Ans: B difficulty: medium 3
section: 3.2
33. Find the derivative of g (= t ) (1/ e)t + et + e . B) −t (1/ e)t −1 + tet −1 C) −(1/ e)t + et A) t (1/ e)t −1 + tet −1 Ans: C difficulty: medium section: 3.2
D) (1/ e)t + et
34. With a yearly inflation rate of 7%, prices are described by P = P0 (1.07)t , where P0 is the price in dollars when t = 0 and t is time in years. If P0 = 1.3, how fast (in cents/year) are prices rising when t = 19? Round to 1 decimal place. Ans: 31.8 difficulty: medium section: 3.2 35. Find the slope of the graph of g ( x)= x − 2e x at the point where it crosses the y-axis. Ans: –1 difficulty: easy section: 3.2 36. Find the equation of the line that goes through (5, 4) that is perpendicular to the line tangent to the graph of g ( x)= x − 2e x at the point where it crosses the y-axis. y = _____ x + _____. Part A: 1 Part B: –1 difficulty: medium section: 3.2
Page 6
Chapter 3: Short-Cuts to Differentiation
37. Given = y 4 x + 5 x , find y ''' . Ans: (ln 4)3 4 x + (ln 5)3 5 x difficulty: medium section: 3.2 38. Find the derivative of = y
4 3
(e − e ) . x
3
x
4
(
4 3x 1 x 4 4 x 1 x x x B) C) e 3 − e 4 e − e e − e 9 3 9 3 Ans: A difficulty: medium section: 3.2 A)
)
D)
4 3
(e − e ) x
3
x
4
39. Find the derivative of y (t ) = 3et + te3 + 3t e . A) C) 3et + e3 + 3et e −1 3 + e3 + 3t e −1 B) D) All of the above are equivalent. 3et + t + te3 Ans: A difficulty: medium section: 3.2 40. A child earns five cents from her grandfather for each dandelion she pulls out of his front yard. The child pulls out all the dandelions that are there. As the season passes, the number of dandelions in the front yard increase according to the model d (t ) = 0.0005(2t ) . After 15 days, her grandfather calls off the deal. How many dandelions does she pull on the 15th day? How fast is the number of dandelions increasing on the 15th day? Ans: 16 dandelions 11 dandelions per day difficulty: medium section: 3.2 41. On what intervals is the function f ( x= ) ax − be x : a) increasing? b) decreasing? c) concave up? d) concave down? a Ans: a) (−∞, ln ) b a b) (ln , ∞) b c) never d) all real numbers difficulty: hard section: 3.2
Page 7
Chapter 3: Short-Cuts to Differentiation
42. Prove that the function h= (t ) a (et + bt ) , where a > 1 and b > 1, is increasing for all values of t. ′(t ) a (et + (ln b)bt ) is positive for all values of t, when a and b are greater than Ans: h= 1. Since the derivative is always positive, the function is always increasing. difficulty: medium section: 3.2 43. Consider the following table of data for the function f. x
5.0
5.1
5.2
5.3
5.4
f(x)
9.1
8.7
8.2
7.6
6.9
What is the sign of f ″(5.1)? A) positive B) negative Ans: B difficulty: easy
section: 3.3
44. Consider the following table of data for the function f. x f(x)
5.0 9.2
5.1 8.8
5.2 8.3
5.3 7.7
5.4 7.0
Suppose g is a function such that g(5.1) = 9 and g′(5.1) = 3. Find h′(5.1) where h(x) = f(x)g(x). Use the right-hand estimate for f '(5.1) . Round to 2 decimal places. Ans: –18.60 difficulty: medium section: 3.3 x2 − 2 at the point at which x = 1 x +1 is y = _____ x + _____. If necessary, round to 2 decimal places. Part A: 1.25 Part B: –1.75 difficulty: medium section: 3.3
45. The equation of the tangent line to the graph of g ( x) =
46. Differentiate g (= t ) e −t + 8e−8t . A) −e −t − 64e −8t B) e −t + 64e−8t Ans: A difficulty: medium
C) te −t −1 + 8te −8t −1 D) −te −t −1 − 64te−8t −1 section: 3.3
47. Given f ( x) = e x , g ( x) = 7 x , and h(x) = g(x)f(x), find h '( x) . A) 7 x e x (ln 7 –1) B) 7 x e x (ln 7 +1) C) 7 x e x (ln 7) Ans: B difficulty: medium section: 3.3
Page 8
D)
x7 x −1 e x −1 (ln 7)
Chapter 3: Short-Cuts to Differentiation
48. Given f ( x) = e x , g ( x) = 5 x , and h(x) = g(x)/f(x), find h ''( x) . A)
5 x e – x (ln 5 +1) 2
C)
B) 5 x e – x (ln 5 –1) 2 Ans: B difficulty: medium 49. Given f ( x) =
A)
D) – x( x − 1)5 x −2 e – x −2 (ln 5) 2 section: 3.3
x3 x2 + 2 , g ( x) = , and h(x) = f(x)g(x), find h′(1). Round to 2 decimal 2x +1 3x 2
places. Ans: 0.33 difficulty: medium 50. Differentiate
5 x e – x (ln 5) 2
5x2 x3 + 1
25 x(2 − x3 )
section: 3.3 . B)
5 x( x3 − 1)
C)
5 x(2 − x3 )
( x3 + 1) 2 ( x3 + 1) 2 ( x3 + 1) 2 Ans: C difficulty: medium section: 3.3
D)
10 x 3x 2 + 1
t . 3 + et 3 + et − tet Ans: r ′(t ) = (3 + et ) 2 difficulty: Ans; easy section: 3.3
51. Differentiate r (t ) =
52. Find the slope of the line tangent to= y e x (5 − x 2 ) when x = 2. Round to two decimal places. Ans: –22.17 difficulty: easy section: 3.3 53. Determine the derivative rule for finding the derivative of the reciprocal function: − g ′( x) 1 B) C) [ g ( x)]−1 D) 2 [ g ( x)] g ′( x) Ans: A difficulty: medium section: 3.3 A)
g ′( x) 2 g ( x)
54. Use the product rule to write a proof of the constant multiple rule: Ans: Let c be any constant and f a differentiable function. Then dxd cf (= x) cf ′( x) + 0 ⋅ f (= x) cf ′(= x) c dxd f ( x) . difficulty: medium section: 3.3
Page 9
1 g ( x)
d dx
cf ( x) = c dxd f ( x) .
Chapter 3: Short-Cuts to Differentiation
55. The table below gives values for functions f and g, and their derivatives. x f g f′ g′
-1 3 1 -3 2
0 3 2 -2 3
1 1 2.5 -1.5 2
2 0 3 -1 2.5
3 1 4 1 3
d ( f ( x) g ( x) ) at x = 1. Round to 2 decimal places. dx Ans: –1.75 difficulty: easy section: 3.4 Find
56. The table below gives values for functions f and g, and their derivatives. x f g f′ g′
-1 3 1 -3 2
0 3 2 -2 3
1 1 2.5 -1.5 2
2 0 3 -1 2.5
3 1 4 1 3
d g(f(x)) at x = –1. If is cannot be computed from the information given, enter dx "cannot find". Ans: –9 difficulty: medium section: 3.4
Find
57. A table of values for a function F near x = 3 and tables of values for a function G near x = 3 and near x = 7 are given below. Estimate F'(3) using the right-hand estimate. x F(x) G(x)
2.9 6.7 5.2
3.0 7.0 5.0
3.1 7.3 4.8
x G(x)
6.9 0.95
7.0 1.00
7.1 1.05
Ans: 3 difficulty: easy
section: 3.4
Page 10
Chapter 3: Short-Cuts to Differentiation
58. A table of values for functions F and G near x = 3 is given below. If H(x) = F(x)/G(x), estimate H'(3) by using the quotient rule and then using right-hand estimates for F ' and G ' . Round to 2 decimal places. x F(x) G(x) Ans: 5.50 difficulty: medium
2.9 7.7 2.2
3.0 8 2
3.1 8.3 1.8
section: 3.4
59. A table of values for a function F near x = 3 and tables of values for a function G near x = 3 and near x = 7 are given below. If H ( x) = G ( F ( x)) , estimate H'(3) using the chain rule. Use right-hand estimates for F ' and G ' . x F(x) G(x)
2.9 6.7 5.2
3.0 7.0 5.0
3.1 7.3 4.8
x G(x)
6.9 5.85
7.0 6.00
7.1 6.15
Ans: 4.5 difficulty: medium
section: 3.4
1 2 C h , where C is the circumference of 12π the tree at the ground level and h is the height of the tree. If C is 4 feet and growing at the rate of 0.25 feet per year, and if h is 25 feet and is growing at 5 feet per year, find the rate of growth of the volume V (in ft3/yr). Round to 2 decimal places. Ans: 3.45 difficulty: medium section: 3.4
60. The volume of a certain tree is given by V =
Page 11
Chapter 3: Short-Cuts to Differentiation
61. Let f(x) and g(x) be two functions. Values of f(x), f ′(x), g(x), and g′(x) for x = 0, 1, and 2 are given in the table below. Use the information in the table to find H '(0) if H ( x) = e g(x) +π x. x 0 1 2 A) 5e 2 Ans: D
f(x) 1 -1 7
f ′(x) -1 2 3
g(x) 2 4 11
g′(x) 5 0 0.5
B) 5e + π C) e 2 + π D) 5e 2 + π difficulty: easy section: 3.4
62. Let f(x) and g(x) be two functions. Values of f(x), f ′(x), g(x), and g′(x) for x = 0, 1, and 2 are given in the table below. Use the information in the table to find H '(2) if H ( x) = [f(x)]2. x 0 1 2 Ans: 42 difficulty: easy
f(x) 1 -1 7
f ′(x) -1 2 3
g(x) 2 4 11
g′(x) 5 0 0.5
section: 3.4
63. What is the instantaneous rate of change of the function f ( x) = e− x at x = 2? Round to 3 decimal places. Ans: –0.073 difficulty: easy section: 3.4 2
64. Find the second derivative of f ( x) = e − x at x = 1.5. Round to three decimal places. A) 0.738 B) –0.105 C) 0.316 D) –1.159 Ans: A difficulty: hard section: 3.4 2
65. Find the equation of the tangent line to f ( x) = e –2 x at x = 3 and use it to approximate the value of f(3.2). Round to 5 decimal places. Ans: 0.00149 difficulty: medium section: 3.4 66. Find the equation of the tangent line to f ( x) = e –3 x at x = 3 and use it find the point where the tangent line crosses the x-axis. Round to 2 decimal places. Ans: 3.33 difficulty: medium section: 3.4
Page 12
Chapter 3: Short-Cuts to Differentiation
67. Find the derivative of h= ( x) (4 x3 + e x )3 . A)
3(4 x3 + e x ) 2 (12 x 2 + e x )
C)
B) 3(4 x3 + e x ) 2 (12 x 2 + xe x −1 ) Ans: A difficulty: medium
3(4 x3 + e x ) 2
D) 3(12 x 2 + e x ) 2 section: 3.4
68. Find the derivative of f ( x) = 54 x + 4 . D) (4 x + 4)54 x +3
A) (5ln 4)54 x+ 4 B) (4 ln 5)54 x+ 4 C) (ln 5)54 Ans: B difficulty: medium section: 3.4 69. Find the derivative of g = ( x) A)
4 x5
e + 5x e x
5
ex + ex . 1
B)
C)
2 e +e 2 e +e difficulty: medium section: 3.4 x
Ans: C
x5
e x + 5 x 4e x x
5
x5
70. If a and b are constants and g= ( x) (ax 2 + b) 2 , find g ′′′( x) . Ans: 12(ax) 2 + 4ab difficulty: medium section: 3.4 5 . (e − e − x ) 2 –10(e x + e − x ) Ans: (e x − e − x ) 3 difficulty: medium section: 3.4
71. Differentiate y =
x
3
d x 72. Find . dx 4 + 1 − x 2
Ans:
3 x 2 (4 1 − x 2 + 1)
(4 + 1 − x 2 ) 4 ( 1 − x 2 ) difficulty: hard section: 3.4
Page 13
e x + x5e x −1 5
D)
2 ex + ex
5
Chapter 3: Short-Cuts to Differentiation
73. A botanist in the field needs a quick estimate of the growth of jumping cholla cactus. In his study area, he has controlled the water and nutrients the jumping cholla receive so he can isolate the effect of sunlight on growth. He lets a function s(t) represent the average hours of sunlight per day in month t (where t = 1 is January). He lets a function g(s) represent the monthly growth of jumping cholla in centimeters at different sunlight exposures. From past experience, he quickly jots down the table below.
s s' g g'
8 13 -0.5 0 0.5
9 12 -1 0.5 0.3
10 11 -1 0.8 0.7
11 10.5 -0.5 1.5 1.5
12 10 -0.5 3 2.8
Suppose h(t ) = g ( s (t )) . a) Find and interpret meaning of the quantity g(10). b) Find and interpret the meaning of the quantity g'(10). c) Find the interpret the meaning of the quantity h(10). d) Find the interpret the meaning of the quantity h'(10). Ans: a) g(10) = 1.5. In the month when the jumping cholla receive 10 hours of sunlight, they grow by 1.5 cm. b) g'(10) = 0.7. When the jumping cholla receive 10 hours of sunlight, their growth is increasing at a rate of 0.16 cm per hour of sunlight. c) h(10) = g(s(10)) = g(11) = 1.5. In the month of October, the jumping cholla grow 1.5 centimeters. d) h'(10) = g'(s(10))s'(10) = g'(11)(-1) = -1.5. In the month of October, the cholla growth is decreasing by 1.5 cm/month. difficulty: hard section: 3.4 74. Find y′ when = y aebx (1 − ebx ) . A) abebx (1 − 2ebx ) B) abebx Ans: A difficulty: medium
C) ab 2 (ebx − 2) section: 3.4
D) abebx (−ebx )
75. Find the critical number(s) of the curve = y aebx (1 − ebx ) . ln(1/ 2) Ans: x = b difficulty: hard section: 3.4 76. Find the slope of the curve y = e0.5 x at x = 2 to the nearest whole number. Ans: 328 difficulty: medium section: 3.4 3
Page 14
Chapter 3: Short-Cuts to Differentiation
77. Find the derivative A) B)
( ( ))
d θ sin θ 2 . dθ
( ) ( ) sin (θ 2 ) + θ cos (θ 2 )
sin θ 2 + 2θ 2 cos θ 2
Ans: A
C) D)
difficulty: medium
( ) ( 2θ cos (θ 2 )
( ))
2θ sin θ 2 + θ 2θ cos θ 2
section: 3.5
78. Find a function F(x) such that F '( x= ) x 4 + sin x and F(0) = 5. x5 − cos x + 6 5 difficulty: medium
Ans:
section: 3.5
79. Which is true of the following graph?
A) g '( x) = f ( x) B) f '( x) = g ( x) Ans: B difficulty: medium section: 3.5 80. A particle moves in such a way that x(= t ) 2t 2 + 8sin t . What is the instantaneous rate of change at t = 0? Ans: 8 difficulty: medium section: 3.5 81. A particle moves in such a way that x(= t ) 2t 2 + 3sin t , where x is the horizontal distance the particle has traveled, in units, and t is time, in seconds. What is the average rate of change between t = 0 and t =
π
2
? Specify units
Ans: 5.05 units/second difficulty: medium section: 3.5
Page 15
Chapter 3: Short-Cuts to Differentiation
82. Differentiate f ( x) = x9 /10 − x10 / 9 + x −10 / 9 9 4 / 5 10 10 9 −1/10 10 1/ 9 10 −19 / 9 A) C) x − x − x −11/ 9 x − x + x 10 9 9 10 9 9 9 4 / 5 10 10 −11/ 9 9 −1/10 10 1/ 9 10 −19 / 9 B) D) x − x+ x x − x − x 10 9 9 10 9 9 Ans: B difficulty: medium section: 3.5 83. Differentiate f = ( w) w2 + 8 . 8w w 2w A) B) C) D) 2 2 w +8 w +8 Ans: C difficulty: medium section: 3.5
1 2 w2 + 8
2
84. Differentiate= f ( x) e z /( z 2 − 7) . A) e
z2
Ans: D
B)
2 ze z
2
C) z2 − 7 difficulty: medium
2
( z 2 − 7)e z − 2 ze z ( z 2 − 7) 2 section: 3.5
2
D)
2 z ( z 2 − 8)e z
2
( z 2 − 7) 2
85. Differentiate f ( x) = xe −2 x . B) e −2 x (1 − 4 x) C) e −2 x (1 + x) A) e −2 x (1 − 2 x) Ans: A difficulty: medium section: 3.5
D) −2e −2 x
86. Differentiate= f ( x) sin(3θ 6 + 1) . A)
cos(3θ 6 + 1)
C)
B) 18θ 5 cos(3θ 6 + 1) Ans: B difficulty: medium
3(ln 6)θ 6 cos(3θ 6 + 1)
D) −18θ 5 cos(3θ 6 + 1) section: 3.5
87. Find the equation of the tangent line to the curve given by f(x) = x sin x at the point x = 3π / 4 . 2 3π 3π π 2 A) y+ –1 x − = 8 2 4 4
π 2 B) y– = 8
2 3π 3π +1 x − 2 4 4
π 2 2 3π 3π C) = y– – +1 x − 8 2 4 4
π 2 2 3π 3π D) y– – –1 x − = 8 2 4 4 Ans: D difficulty: medium section: 3.5
Page 16
Chapter 3: Short-Cuts to Differentiation
88. If f ( x) = 2sin 2 x and g ( x) = −3cos 2 x , is f ′( x) = g ′( x) ? Ans: no difficulty: easy section: 3.5 89. When hyperventilating, a person breathes in and out extremely rapidly. A spirogram is a machine that draws a graph of the volume of air in a person's lungs as a function of time. During hyperventilation, the spirogram trace may be represented by V = 3 − 0.06 cos(200π t ) , where V is the volume of the lungs in liters and t is the time in minutes. What is the maximum volume of air in the lungs? Ans: 3.06 liters difficulty: hard section: 3.5 90. When hyperventilating, a person breathes in and out extremely rapidly. A spirogram is a machine that draws a graph of the volume of air in a person's lungs as a function of time. During hyperventilation, the spirogram trace may be represented by V = 3 − 0.04 cos(210π t ) , where V is the volume of the lungs in liters and t is the time in minutes. What is the period of this function in seconds? Ans: 0.571 difficulty: hard section: 3.5 91. When hyperventilating, a person breathes in and out extremely rapidly. A spirogram is a machine that draws a graph of the volume of air in a person's lungs as a function of time. During hyperventilation, the spirogram trace may be represented by V = 3 − 0.04 cos(200π t ) , where V is the volume of the lungs in liters and t is the time in minutes. Find the maximum rate (in liters/minute) of flow of air during inspiration (i.e. breathing in). This is called the peak inspiratory flow. Ans: 25.133 difficulty: hard section: 3.5 92. Find the derivative of = f ( w) sin(4 w2 ) + cos(4 w2 ) . A)
8w cos(4 w2 ) − 8w sin(4 w2 )
B) −8w cos(4 w2 ) + 8w sin(4 w2 ) Ans: A difficulty: medium
C)
4 cos(4 w2 ) − 4sin(4 w2 )
D) −4 cos(4 w2 ) + 4sin(4 w2 ) section: 3.5
93. Find the derivative of g ( x) = cos(sin(2 x)) . −2 cos(2 x) sin(sin(2 x)) −2sin(sin(2 x)) A) C) 2 cos(2 x) sin(sin(2 x)) 4sin(sin(2 x)) B) D) Ans: A difficulty: medium section: 3.5
Page 17
Chapter 3: Short-Cuts to Differentiation
94. Find the derivative of r (θ ) = 5cos(cos(cos(θ ))) A) 1.418311 −5sin(cos(cos(θ )) *(sin(cos(θ )) *(sin(θ )) B) −5sin(cos(sin(θ )) *(cos(θ )) C) sin(sin(sin(θ )) *5cos(θ ) D) Ans: B difficulty: medium section: 3.5 95. Find the equation of the tangent line to the graph of f (α ) =
π 2 4 = − 2 x − π π π 4 difficulty: hard section: 3.5
Ans: y −
tan α
α
at the point α = π / 4 .
4
96. Find the equation of the line that is "orthogonal" to the graph of the function
f (t ) = sin(t ) at the point where t = tangent line when t =
π
π
2
. In other words, find the line perpendicular to the
. 2 π −1 π Ans: y − sin = x− 2 2 cos π 2 difficulty: medium section: 3.5
y 7 x + x tan x . 97. Differentiate = x Ans: y′ = + tan x 7+ cos 2 x difficulty: easy section: 3.5 98. Find a differentiation formula for y = cot(θ ) . −1 Ans: y′ = = − csc 2 θ 2 sin θ difficulty: medium section: 3.5
Page 18
Chapter 3: Short-Cuts to Differentiation
99. If P dollars is invested at an annual interest rate of r %, then at t years this investment t
dF r grows to F dollars, where = . F P 1 + . Find dr 100 A)
r Pt 1 + 100
t −1
r 1 + 100
D)
P r r ln 1 + 1 + 100 100 100
t −1
Pt r B) 1 + 100 100 Ans: B difficulty: medium
t
C)
t
section: 3.6
100. If P dollars is invested at an annual interest rate of r %, then at t years this investment t
dF r . grows to F dollars, where = F P 1 + . Find dt 100 A)
r Pt 1 + 100
t −1
r 1 + 100
D)
P r r ln 1 + 1 + 100 100 100
t −1
Pt r 1 + 100 100 Ans: D difficulty: medium
B)
t
C)
t
section: 3.6
f ( x) 6 x ln x − 6 x . 101. Find the derivative of = A) 6 ln x − 6 B) 6 ln x + x − 6 C) 6 ln x Ans: C difficulty: easy section: 3.6
D) −6
102. Find f ′(x) if f ( x) = e6 x cos 4 x . A)
−24e6 x sin 4 x)
B) 24e6 x (cos 4 x − sin 4 x) Ans: C difficulty: easy
C)
e6 x (6 cos 4 x − 4sin 4 x)
D) section: 3.6
e6 x (6 cos 4 x − 6sin 4 x)
103. Find f ′(x) if = f ( x) sin ln x + 2 .
cos x cos ln x + 2 B) 2x x Ans: D difficulty: medium
A)
104. Differentiate A)
x cos ln x + 2 2 ln x + 2 section: 3.6
C)
cos ln x + 2 2 x ln x + 2
d 2 x 2 +8 x 7e . dx
7(4 x + 8)e 2 x +8 x 2
B) 7e 2 x +8 x Ans: A difficulty: easy 2
D)
C) D) section: 3.6
Page 19
7 ln(2 x 2 + 8 x)e 2 x +8 x 2
7(2 x 2 + 8 x)e 2 x +8 x −1 2
Chapter 3: Short-Cuts to Differentiation
105. The equation for the line tangent to the curve, y = ln x, which passes through the origin is y = _____ x + _____. Part A: 0.368 Part B: 0 difficulty: hard section: 3.6 106. Consider the equation ln x = mx where m is some constant (positive, negative, or zero). How many solutions will the equation have for m = 0.439? Ans: 0 difficulty: hard section: 3.6
(
)
d ln( x3e x ) . dx 3 3 1 3x + e x A) B) C) D) +1 x x x3e x x3e x Ans: A difficulty: medium section: 3.6
107. Find
108. Find A)
d cos θ . dθ sin θ
1
B) −
1
sin 2 θ sin 2 θ Ans: B difficulty: medium
109. Find A)
sin θ cos 2 θ D) 1 − cos θ sin 2 θ section: 3.6
C) −
d 9x 1+ 8 . dx 9 ⋅ ln 8 ⋅ 89 x
Ans: C
B)
9 x ⋅ 89 x −1
difficulty: medium
C)
9 ⋅ ln 8 ⋅ 89 x
2 1 + 89 x section: 3.6
D)
9 x ⋅ 89 x −1 2 1 + 89 x
110. Let a be a positive constant (i.e., a > 0). The equation a x = 1 + x has the solution x = 0, for all a. If a = 2.52, are there any solutions for x > 0? Ans: yes difficulty: hard section: 3.6 111. Find the derivative of f(x) = sin (7x) · sin (8x). 7 cos(8 x) sin(7 x) − 8cos(7 x) sin(8 x) A) 7 cos(8 x) sin(7 x) + 8cos(7 x) sin(8 x) B) 8cos(8 x) sin(7 x) − 7 cos(7 x) sin(8 x) C) D) 8cos(8 x) sin(7 x) + 7 cos(7 x) sin(8 x) Ans: D difficulty: medium section: 3.6
Page 20
Chapter 3: Short-Cuts to Differentiation
112. Find the derivative of f ( x) = e−(5− x ) . 2
A)
2(5 − x)e −(5− x )
2
B) −2(5 − x)e −(5− x ) Ans: A difficulty: medium 2
113. Find the derivative of f ( x) = A)
−
B)
−
Ans: B
1+ x 5 + 6 x + 7 x2
2
2
difficulty: medium
2
.
1 + 14 x + 7 x 2
(5 + 6 x + 7 x2 )
2
D) ln(−(5 − x) 2 )e −(5− x ) section: 3.6
1+ 7x
(5 + 6 x + 7 x2 )
−(5 − x) 2 e −(5− x ) −1
C)
11 + 26 x + 21x 2
C)
−
D)
1 6 + 14x
(5 + 6 x + 7 x2 )
2
section: 3.6
114. Find the derivative of f(x) = ln(sin x) A) tan x B) -cot x C) cot x D) -tan x Ans: C difficulty: medium section: 3.6 115. Differentiate h(θ ) = 5θ sin(θ 2 ) . A)
5cos(2θ )
B) 5θ cos(θ 2 ) + 5sin(θ 2 ) Ans: C difficulty: medium
2+ y 116. Differentiate f ( y ) = ln . 2− y 2− y 2− y A) B) − C) 2+ y 2+ y Ans: D difficulty: medium
10θ 2 cos(θ 2 ) + 5sin(θ 2 )
C)
D) −10θ 2 cos(θ 2 ) + sin(θ 2 ) section: 3.6
4
D)
(2 − y ) 2 section: 3.6
4 (2 + y )(2 − y )
117. What is the instantaneous rate of change of the function f ( x) = x ln x at x = 1? Round to 2 decimal places. Ans: 1.00 difficulty: medium section: 3.6
Page 21
Chapter 3: Short-Cuts to Differentiation
118. Differentiate y = x arctan(3 x) . 3x A) = + arctan(3 x) y′ (1 + (3 x) 2 ) 1 B) = + 3arctan(3 x) y′ 3 + x2 Ans: A difficulty: medium
C)
y′ = 3arctan(3 x)
D)
y′ = 1 ⋅ arctan(3 x)
section: 3.6
119. When does y = ln x increase at the same rate as y = 2 x ? 1 ln 2 A) when x = C) when x = . 2 2 B) when x = 2. D) They never increase at the same rate. Ans: A difficulty: medium section: 3.6 120. If a particle moves according to the position function s (t ) = 10 ln
3 find a formula that 3+t
gives the velocity of the particle at time t. −10 Ans: v(t ) = 3+t difficulty: medium section: 3.6 121. If a particle moves according to the position function s (t ) = 2.25ln
6 find the 6+t
acceleration of the particle at time t = 5. Ans: 41.25 difficulty: hard section: 3.6 122. At what point(s) in the interval [ 0, 2π ] , does the curve given by y = 7 ln(sin x) have horizontal tangents? Round to two decimal places. A) 0.83, 0.17 B) 1.57, 4.71 C) 0, 3.14 D) There are no horizontal tangents. Ans: B difficulty: medium section: 3.6 123. For= u 2 x 2 + 3 , find y u u + 1 and= A) B)
( 2 x2 + 4) (12 x3 + 22 x ) −1/ 2 ( 2 x2 + 3) (12 x3 + 24 x )
Ans: A
dy . dx
−1/ 2
difficulty: medium
C) D) section: 3.7
Page 22
( 2 x2 + 4) ( 4 x ) −1/ 2 ( 2 x 2 + 3) ( 4 x ) −1/ 2
Chapter 3: Short-Cuts to Differentiation
124. If p(x) = ln((x - a)(x - b)(x - c)) and a, b, c are constants, then 1 1 1 . p '( x) = + + x −a x −b x −c Ans: True difficulty: medium section: 3.7
2sin t dx , find . 1 + cos t dt 1 2 B) A) 2 + cos t 1 + cos t Ans: B difficulty: medium
125. If x =
2 cos t D) 1 + sin t section: 3.7
C)
2 cos t (1 + cos t ) 2
dy . dx 1− 2x − 2 y 3(2 x + 3) 2 3(2 x + 3) 2 − x − y B) C) A) 2x + 2 y x+ y x+ y Ans: B difficulty: medium section: 3.7
126. If ( x + y ) 2 = (2 x + 3)3 , use implicit differentiation to find
127. Suppose that xy 2 + sin y + x3 = 1 . Find
D)
6(2 x + 3) 2 ( x + y )2
dy . dx
1 3x 2 + y 2 3x 2 B) − C) − 2 xy + cos y 2 xy + cos y 2 xy + cos y Ans: C difficulty: medium section: 3.7 A) −
D) −
y2 2 xy + cos y
dy and use the result to find the tangent dx line at the point (2,0). Use this to find an approximate value for y when x = 2.02. Ans: –0.24 difficulty: hard section: 3.7
128. Suppose that xy 2 + sin y + x3 = 8 . Evaluate
129. Consider the equation y 2 − 6 cos x = 3 . Assuming this implictly defines y as a function dy of x, find . dx 3 + 6sin x 3sin x 3sin x 3 − 6sin x A) B) C) D) − y 2y 2y y Ans: D difficulty: medium section: 3.7
Page 23
Chapter 3: Short-Cuts to Differentiation
6 . Assuming this implicitly defines y as a function 130. Consider the equation x sin y + 2 y = dy of x, find . dx sin y 6sin y sin y + 2 sin y + 6 A) − B) − C) − D) − x cos y + 2 x cos y + 2 x cos y x cos y Ans: A difficulty: medium section: 3.7 131. Consider the equation ln y /(1 − y )= 0.34 x + ln 4 . Assuming this implicitly defines y as
dy . dx A) 0.34 y (1 − y ) B) 0.34 y /(1 − y ) C) 0.34 y Ans: A difficulty: medium section: 3.7 a function of x, find
D) –0.34 y (1 − y )
132. The part of the graph of sin( x 2 + y ) = x that is near (0, π) defines y as a function of x implicitly. Does the graph of this function lie above or below its tangent line at (0, π)? Ans: below difficulty: hard section: 3.7 133. If x 6 + y 6 = –27 , what is A) −
27 + x5
Ans: B
y5
B) −
dy ? dx x5
y5 difficulty: easy
C)
x −y 5
5
D) −
6x5 y5
section: 3.7
134. The equation of the lines tangent to the circle ( x − 2) 2 + ( y + 2) 2 = 20 at the upper point where x = 0 is y = _____ x + _____. Part A: 0.5 Part B: 2 difficulty: medium section: 3.7
Page 24
Chapter 3: Short-Cuts to Differentiation
135. The curve graphed below has equation 2 y 3 + y 2 − y 5 = x 4 − 2 x3 + x 2 . It is called the "bouncing wagon curve." dy a) Use implicit differentiation to find . dx b) Find the slope of the tangent line at the point (1, -1).
4
y
3
2
1
x −6
−4
−2
2
4
6
−1
−2
−3
−4
2 x(2 x − 3 x + 1) = , m 0 y (6 y + 2 − 5 y 3 ) difficulty: medium section: 3.7
Ans: y′ =
2
5 3 4 3 136. The line given by the equation y –= ( x − 2) is perpendicular to the tangent line 2 5 x2 y 2 to the ellipse + = 1 at the point x = 2. 16 25 Ans: True difficulty: medium section: 3.7
137. Find the slope of the curve x 4 − 2 xy 2 + y 5 = 13 at the point (2, 1). A) 10 B) 13 C) 16 D) 19 Ans: A difficulty: medium section: 3.7 138. Find the equation of the tangent line of x 3 + y 3= 49( x − y ) at the origin. A) y = 49x D) y - 49 = 3(x - y) B) y = x E) none of the above C) y = 3x - 49 Ans: B difficulty: easy section: 3.7
Page 25
Chapter 3: Short-Cuts to Differentiation
139. Show that the area of the triangle formed by the x-axis, the y-axis and the tangent line to the hyperbola xy = 4 is a constant for any point (a, b). What is the constant?
y 4 (a,b) 2 x 2
4
−4 4 ( x − a ) + . The x intercept of this 2 a a tangent line is x = 2a. The y-intercept is y = 8/a. Thus the area of the triangle is (1/2)xy = (1/2)(2a)(8/a) = 8. difficulty: hard section: 3.7 Ans: The equation of the tangent line is = y
11 140. Find y′ when 4 x sin y + 5 y ln x = −(4 x sin y + 5 y ) −(4 x cos y + 5) A) C) x(4 x cos y + 5ln x) (4 y cos y + 5 x) −(4 x sin y + y ) −(5 x sin y + 4 y ) B) D) x(4 x sin y + 5ln x) (4 x cos y + 5ln x) Ans: A difficulty: medium section: 3.7 141. Find the derivative of y cosh 2 (6 x) + sinh 2 (6 x) . = A) 12 cosh(6 x) + 12sinh(6 x) 24 cosh(6 x) sinh(6 x) B) Ans: B difficulty: medium
C) 12sinh 2 (6 x) + 12 cosh 2 (6 x) D) 0 section: 3.8
Page 26
Chapter 3: Short-Cuts to Differentiation
142. Differentiate y = A) B) C) D)
cosh(4 x) cosh 2 ( x)
. (Do not simplify)
−4sinh( x) cosh 2 x −4sinh(4 x) ⋅ cosh 2 ( x) − 2sinh x ⋅ cosh(4 x) (cosh x) 4 4sinh(4 x) ⋅ cosh 2 ( x) − 2sinh x ⋅ cosh( x) ⋅ cosh(4 x) (cosh x) 4 −4sinh(4 x) ⋅ cosh 2 ( x) + 2sinh x ⋅ cosh( x) ⋅ cosh(4 x) (cosh x) 4 difficulty: medium section: 3.8
Ans: C
143. Find the derivative of y = sinh ( sin(7 x) ) . A)
−7 cosh ( cos(7 x) )
B) 7 cosh ( cos(7 x) ) Ans: D difficulty: medium
C)
−7 cos(7 x) ⋅ cosh ( sin(7 x) )
D) 7 cos(7 x) ⋅ cosh ( sin(7 x) ) section: 3.8
144. Find the derivative of y = tanh(7 x) . A)
(
7 cosh 2 (7 x) + sinh 2 (7 x)
7
C)
2
cosh (7 x) B)
−
7
D)
2
cosh (7 x) Ans: A difficulty: medium
)
2
−
(
cosh (7 x)
7 cosh 2 (7 x) + sinh 2 (7 x)
)
2
cosh (7 x) section: 3.8
y 220 − 35cosh( x / 35) where x is the number of feet 145. An arch over a lake has the form= from a point on one side of the lake. What is the highest point on the arch? Ans: 185 feet high difficulty: medium section: 3.8 146. Differentiate y = sinh(ln(4 x)) . cosh(ln(4 x)) Ans: x difficulty: easy section: 3.8
Page 27
Chapter 3: Short-Cuts to Differentiation
147. Find the exact slope of the tangent line to y = sinh x when x = 1. e2 − 1 2e difficulty: easy
Ans:
section: 3.8
148. The best linear approximation to f ( x)= (1 + x)1/ 2 at x = 8 is y = _____ x + _____. Round answers to 2 decimal places. Part A: 0.17 Part B: 1.67 difficulty: medium section: 3.9 149. The graph of f(x) is shown in below. Arrange the following in ascending order by entering a "1" next to the smallest, a "2" by the next smallest, and so forth. A. f '(2) B. f (0) C. f '(−0.9) D. the number 1 E. f ''(0) F. f ''(1)
Part A: 3 Part B: 5 Part C: 6 Part D: 4 Part E: 1 Part F: 2 difficulty: medium
section: 3.9
Page 28
Chapter 3: Short-Cuts to Differentiation
150. The graph of f(x) is shown below. Suppose we want to estimate f (1.5) by using tangent line approximations at x = 1 and 2. Which tangent line yields the best approximation?
A) The one at x = 1 B) The one at x = 2 Ans: A difficulty: medium section: 3.9 151. Use local linearization to approximate the value of e0.04 . Then determine the error in your approximation to 5 decimal places. Ans: 1.04, 0.00081 difficulty: easy section: 3.9 152. Use local linearization to determine whether or not = y
1 4
x + 1 provides a good estimate
of f ( x) = x for values of x near 4. If it does not, then determine a better linear approximation. If it does provide a good approximation, use it to estimate 4.008 . Ans: The local linearization of f(x) near 4 is = y 14 x + 1 It provides an estimate of 2.002, which is close to the calculator value of 2.001999. difficulty: easy section: 3.9 153. According to the Mean Value Theorem, if f ( x) = x 2 then there exists a number c, 6 < c < 8 , such that f ′(c) = a. What is a? Ans: 14 difficulty: easy section: 3.10 154. According to the Mean Value Theorem, if f ( x) = x 2 then there exists a number c, 5 < c < 10 , such that f ′(c) = a. What is the value of c that would make the Mean Value Theorem hold? Ans: 7.5 difficulty: easy section: 3.10
Page 29
Chapter 3: Short-Cuts to Differentiation
155. What can you conclude about f if f is continuous on [0, 4], differentiable on (0, 4), and f ′(x) > 0 on (0, 4)? Mark all that apply. I.f is increasing on (0, 4) II.f is increasing on [0, 4] III. f is nondecreasing on (0, 4) IV.f is nondecreasing on [0, 4] A) I B) I, II C) I, II, III D) I, II, III, IV Ans: D difficulty: medium section: 3.10 156. The speed of a car at time t, 5 ≤ t ≤ 20, is given by f (t ) = −2(t − 10) 2 . The Mean Value Theorem guarantees that there will be a point in the interval [5, 20] when the instantaneous acceleration is equal to what number? Ans: –10 difficulty: medium section: 3.10 157. A notorious speeder leaves his house near Tucson for a popular mall in the Phoenix area -- a distance of 100 miles He arrives an hour later. A helicopter clocked him leaving home at a speed of 30 miles/hour. Another helicopter clocked him arriving at his destination going 45 miles per hour. In the mail, he receives a ticket for going 100 miles/hour. He contests the ticket in traffic court saying: "No one can prove I ever went 100 miles per hour." You are the prosecution's expert witness. What do you argue? Ans: His average speed was 100 miles/hour. By the Mean Value Theorem, there was a point in time during that one hour, when his instantaneous velocity was equal to his average velocity over the hour. Thus there was at least one point in time when he was going 100 miles per hour. difficulty: medium section: 3.10 158. If f ( x) =
x2 + 2
, then f '( x) = −
4x
. ( x − 2) 2 x −2 Ans: False difficulty: medium section: 3 review 2
159. If f ( x) = 6 x tan x , then f '(= x) Ans: True 160. If= f ( x) Ans: False
2
6x 2
+ 6 x ⋅ ln 6 ⋅ tan x .
cos x difficulty: medium section: 3 review (ln x) 2 + 6 , then f '( x) =
ln x
.
2 x (ln x) + 6 difficulty: medium section: 3 review 2
Page 30
Chapter 3: Short-Cuts to Differentiation
161. If f ( x) = (ln 6) x , then f ′(6) is A) 6 ln 6 B) (ln 6)3 C) approximately 19.297 Ans: C difficulty: medium section: 3 review
D) approximately 1.792
162. The function D(v) in the figure below gives the air resistance, or drag (in pounds), on a blunt object as a function of its velocity. (Notice that the curve rises sharply near v = 700 miles/hour, the speed of sound. This represents the “sound barrier.”) If a blunt object traveling 725 miles/hour is accelerating at the constant rate of 6000 miles/hour2, at approximately what rate (in pounds/hour) is the drag increasing at that moment? Select the closest answer.
A) 4800 lb/hr B) 8 lb/hr Ans: D difficulty: hard
C) 600 lb/hr D) 48,000 lb/hr section: 3 review
π 163. The slope of the tangent line to the curve y cos x + e y = 4 at , ln 4 is 2 ln 4 4 + ln 4 A) 0 B) 4 C) D) E) None of the above 4 4 Ans: C difficulty: medium section: 3 review
1 ( z 2 + 1) 2 is 3 z 2 + 2 − 2 . z z difficulty: medium section: 3 review
164. The derivative of Ans: True
165. The derivative of sin(θ cos θ ) is cos(θ cos θ ) ⋅ [cos θ − θ sin θ ] . Ans: True difficulty: medium section: 3 review
Page 31
Chapter 3: Short-Cuts to Differentiation
166. The derivative of Ans: False
t
+ e6 is t
1− t t
+ e6 .
e e difficulty: medium
section: 3 review
1 −3 1 x −1 − x 2 + 3 is − x 2 − ( x 2 + 3) 2 . 2 2 x difficulty: medium section: 3 review
167. The derivative of Ans: False
67 168. The derivative of arctan is ________________________. Do not simplify. x .
−67 (1 + ( 67x ) 2 ) x 2 difficulty: medium Ans:
section: 3 review
169. The derivative of cos( x 2 + 8) is −2sin( x 2 + 8) . Ans: False difficulty: easy section: 3 review 170. The derivative of Ans: True
e− x
is
−e − x (1 + 7 x 6 + x 7 )
. (1 + x 7 ) 2 1 + x7 difficulty: medium section: 3 review
6 6 1 . + 1 − x is 2 + x 2 1− x x difficulty: medium section: 3 review
171. The derivative of 6e − Ans: False
x) (7 x + sin x)e , then f '( x) =(7 + cos x) ⋅ ln(7 x + sin x) ⋅ (7 x + sin x)e . 172. If f (= Ans: False difficulty: medium section: 3 review 173. If g ( x) = e(4 x +sin x ) , then g '( x) = (4 + cos x) ⋅ e(4 x +sin x ) . Ans: True difficulty: medium section: 3 review 2 ln z − (tan 6 z ) z tan 6 z 174. If F ( z ) = , then F '( z ) = cos z . ln z (ln z ) 2 Ans: True difficulty: medium section: 3 review
6
1
−7 −7 7 −9 175. If P( w) = 7 w w 2 , then P '( w) = 7 w ⋅ ln 7 ⋅ w 2 + 7 w w 2 . 2 Ans: False difficulty: medium section: 3 review
Page 32
Chapter 3: Short-Cuts to Differentiation
( )
176. If R(θ ) = θ cos θ 6 , then R '(θ ) = Ans: False
1
( )
( )
cos θ 6 + θ ⋅ 6θ 5 sin θ 6 .
2 θ difficulty: medium section: 3 review
( )
177. The derivative of f ( x) = x5 ln x5 is f= '( x) 5 x 4 (1 + 5ln x ) . Ans: True
difficulty: medium
section: 3 review
178. The derivative of ( x 2 + 1) arctan x is 1. Ans: False difficulty: medium section: 3 review 179. The derivative of sin(cos(e6 x )) is −6e6 x ⋅ sin(e6 x ) ⋅ cos(cos(e6 x )) . Ans: True difficulty: medium section: 3 review
dy = −2sin(4 x − 2) . dx difficulty: hard section: 3 review
180. If y cos 2 (2 x − 1) , then = Ans: True
2 2 is 2 cos . x x difficulty: easy section: 3 review
181. The derivative of sin Ans: False
182. The derivative of ee is 4e 4 x +e . Ans: True difficulty: medium 4x
4x
section: 3 review
183. Let f ( x) = x 4 + 4 x + 2 . How many zeros does f have? Ans: 2 difficulty: medium section: 3 review 184. Use Newton's method to estimate the location of the point of intersection of the curves g ( x) = 1.1x and h( x) = e− x for x ≥ 0 correct to four decimal places. The point is ( _____, _____ ). Part A: 0.5333 Part B: 0.5867 difficulty: hard section: 3 review
Page 33
Chapter 3: Short-Cuts to Differentiation
185. The purpose of this problem is to find all the roots of f ( x) = 3 x3 + 2 x 2 − 4 x + 1 exactly. Do this by first using either the bisection method or Newton's method to obtain approximate roots. One of these should suggest an exact root. This is x = _____. Use this information to find exact expressions for the other roots. They are x = _____ and x = _____. (List the smaller one first) Part A: 1/3 −1 − 5 Part B: 2 −1 + 5 Part C: 2 difficulty: hard section: 3 review 186. If f ′( x= ) ln( x + 8 + e −5 x ) , find f ′(0) . Ans: -1/4 difficulty: easy section: 3 review 187. Find the derivative of = f ( x) 7esin x + sin(e7 x ) . Ans: 7(esin x cos x + e7 x cos(e7 x ) ) difficulty: medium section: 3 review
Page 34
1. Sketch a graph of a function whose f ′( x) = 0 at x=1, f ′( x) < 0 when x< 1, 0 when x> 1, Is it possible to have f ′′( x) = 0 at any value from -2<x<2? Ans: various difficulty: easy section: 4.1
f ′( x) <
2. Below is the graph of the derivative of a function f, i.e., it is a graph of y = f ′(x). Is f increasing or decreasing on the interval 2 ≤ x ≤ 3 ?
Ans: increasing difficulty: easy
section: 4.1
3. Below is the graph of the derivative of a function f, i.e., it is a graph of y = f ′(x). Which of the following values of x are local minima of f?
A) 2 B) 0 C) 4 D) 3 E) 1 Ans: A, C difficulty: easy section: 4.1
Page 1
Chapter 4: Using the Derivative
4. Below is the graph of the derivative of a function f, i.e., it is a graph of y = f ′(x). Where in the interval 0 ≤ x ≤ 4 does f achieve its global maximum?
A) 0 B) 1 C) 2 D) 3 E) 4 Ans: A difficulty: easy section: 4.1 5. Below is the graph of the derivative of a function f, i.e., it is a graph of y = f ′(x). Suppose that you are told that f (0) = 5. Estimate f(2).
A) 3 B) 3.5 C) 5 Ans: B difficulty: easy
D) 6.5 section: 4.1
Page 2
Chapter 4: Using the Derivative
6. Below is the graph of the derivative of a function f, i.e., it is a graph of y = f ′(x). Suppose that you are told that f (0) = 3. Which of the following is an exact expression for f (2) ?
A)
2
∫ f '( x) dx 3
Ans: C
2
B) 3 − ∫ f '( x) dx
difficulty: easy
0
C)
2
∫ f '( x) dx + 3 0
D)
2
∫ f '( x) dx − 3 0
section: 4.1
7. Sketch a graph of a function whose f ′( x) = 0 at x=–1, f ′( x) < 0 when x> –1, Ans: various difficulty: easy section: 4.1
f ′( x) < 0 when x< –1,
8. Determine the equation of the tangent line at x=0 and the value of f(x) at x=1.5 given f ( x) = a x for all values of a A) y=(ln a)x+1, y=1.5(ln a)+1 D) y=(ln a)x-1, y=1.5(ln a)-1 E) y=(ln a)x-1, y=1.5 B) y=(ln a)(a1.5)x-1, y=-1 C) y=(a1.5)[(ln a)x-1], y=-1 Ans: A difficulty: medium section: 4.1 9. A particle is travelling along the x-axis according to the function s (t ) = ( t-3 )( t-1 )2. When is the velocity of the particle equal to 0? Ans: t=1, t=5/3=1.667 difficulty: medium section: 4.1 10. A particle is travelling along the x-axis according to the function s (t ) = ( t-3 )( t-2 )2. Assuming t ≥ 0, when is the acceleration of the particle equal to 0? Ans: t=2.167 difficulty: medium section: 4.1
Page 3
Chapter 4: Using the Derivative
11. A particle moves along a line so that its position at any time t ≥ 0 is given by the function s(t)= -t3 + 9t2 - 18t + 10 where s is measured in meters and t is measured in seconds. A) Determine the velocity at time, t. B) Determine the acceleration at time t. C) Is the particle ever at rest? If so, when? Part A: v(t)= -3t2+18t-18 Part B: a(t)= -6t+18 Part C: Yes, t=1.2679, t=4.732 difficulty: medium section: 4.1 12. For which interval(s) is the function f ( x= ) x 5 − 16x3 decreasing? x < -4 or 4 < x D) 0 < x < 4 A) < x < B) -4 4 E) -4 < x < 4, x ≠ 0 C) -4 < x < 0 Ans: E difficulty: easy section: 4.1 13. Starting at time t = 0, water is poured at a constant rate into an empty vase (pictured below). It takes ten seconds for the vase to be filled completely to the top. Let h = f(t) be the depth of the water in the vase at time t. Is h = f(t) concave up or down on the region 3 < t < 9?
Ans: up difficulty: medium
section: 4.1
Page 4
Chapter 4: Using the Derivative
14. Starting at time t = 0, water is poured at a constant rate into an empty vase (pictured below). It takes ten seconds for the vase to be filled completely to the top. Let h = f(t) be the depth of the water in the vase at time t. For what value of h is f '(t ) the largest?
Ans: 6 difficulty: medium
section: 4.1
15. A water tank is constructed in the shape of a sphere seated atop a circular cylinder. If water is being pumped into the tank at a constant rate, let h(t ) be the height of the water as a function of time. For the interval a < t < b, which of the following is true?
A) h(t ) is concave down B) h(t ) is linear Ans: A difficulty: medium section: 4.1
Page 5
C) h(t ) is concave up
Chapter 4: Using the Derivative
16. A water tank is constructed in the shape of a sphere seated atop a circular cylinder. If water is being pumped into the tank at a constant rate, let h(t ) be the height of the water as a function of time. Which of the following is true at the point where h(t ) = b?
A) The second derivative doesn't exist C) The slope is infinite B) There is an inflection point D) None of the above Ans: B difficulty: medium section: 4.1 17. Consider a continuous function with the following properties:
• f (0) = 4 • f '( x) < 0.5 • f ''( x) < 0 for x < 0 • f '(1) = 0 . Which of the following is true? A) The graph must have a local maximum for x < 0 . B) The graph must not have a local maximum for x < 0 . C) The graph may or may not have a local maximum for x < 0 . Ans: C difficulty: medium section: 4.1
Page 6
Chapter 4: Using the Derivative
18. Consider a continuous function with the following properties:
• f (0) = 3 • f '( x) < 0.5 • f ''( x) < 0 for x < 0 • f '(1) = 0 . Which of the following are inconsistent with these four conditions? f (1) = 5 f ''(1) = 0 A) B) C) lim f ( x) = 0 x →−∞
Ans: A, B
difficulty: medium
section: 4.1
19. Given below are the graphs of two functions f(x) and g(x). Let h( x) = f ( g ( x)) . Is the point x = 1 a local minimum, a local maximum, or neither for the function h( x) ?
Ans: local minimum difficulty: easy section: 4.1
Page 7
Chapter 4: Using the Derivative
20. Given below are the graphs of two functions f(x) and g(x). Let h( x) = f ( g ( x)) . Is h( x) increasing or decreasing on the interval 3 < x < 4?
Ans: decreasing difficulty: medium
section: 4.1
Page 8
Chapter 4: Using the Derivative
21. Given below are the graphs of two functions f(x) and g(x). Graph h( x) = f ( g ( x)) on a similar set of axes.
Ans:
difficulty: medium
section: 4.1
22. Let f(x) be a function with positive values and let g = f . If f has a local maximum at x = x0 , what about g? A) g has a local maximum at x = x0 B) g has a local minimum at x = x0 C) g could have a local maximum or a local minimum at x = x0 Ans: A difficulty: medium section: 4.1
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Chapter 4: Using the Derivative
23. Let f be a function. Is it true or false that the inflection points of f are the local extrema of f ′. Ans: True difficulty: easy section: 4.1 24. Below is the graph of the rate r at which people arrive for lunch at Cafeteria Charlotte. Checkers start at 12:00 noon and can pass people through at a constant rate of 5 people/minute. Let f(t) be the length of the line (i.e. the number of people) at time t. Suppose that at 11:50 there are already 150 people lined up. Using the graph together with this information, is the time 12:34 a local minimum, a local maximum, or neither of f?
Ans: local maximum difficulty: medium
section: 4.1
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Chapter 4: Using the Derivative
25. Below is the graph of the rate r at which people arrive for lunch at Cafeteria Charlotte. Checkers start at 12:00 noon and can pass people through at a constant rate of 5 people/minute. Let f(t) be the length of the line (i.e. the number of people) at time t. Suppose that at 11:50 there are already 150 people lined up. Using the graph together with this information, is f concave up or down on the interval 11:55 < t < 12:07?
Ans: down difficulty: medium
section: 4.1
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Chapter 4: Using the Derivative
26. Below is the graph of the rate r at which people arrive for lunch at Cafeteria Charlotte. Checkers start at 12:00 noon and can pass people through at a constant rate of 5 people/minute. Let f(t) be the length of the line (i.e. the number of people) at time t. Suppose that at 11:50 there are already 150 people lined up. Using the graph together with this information, sketch a graph of f.
Ans:
difficulty: hard
section: 4.1
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Chapter 4: Using the Derivative
27. Below is the graph of the rate r at which people arrive for lunch at Cafeteria Charlotte. Checkers start at 12:00 noon and can pass people through at a constant rate of 5 people/minute. Let f(t) be the length of the line (i.e. the number of people) at time t. Suppose that at 11:50 there are already 150 people lined up. Using the graph together with this information, when is the line the longest?
Ans: 12:34 difficulty: hard
section: 4.1
28. Consider f ( x) = x 2 e − x for −1 ≤ x ≤ 3 . Is f increasing or decreasing on the interval 0 < x < 2? Ans: increasing difficulty: medium section: 4.1 29. Consider f ( x) = x 2 e − x for −1 ≤ x ≤ 3 . For which value(s) of x is f ( x) least? A) 0.586 B) –1 C) 0 D) 3 Ans: C difficulty: medium section: 4.1
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Chapter 4: Using the Derivative
30. Graph the function f ( x) = x 2 e − x for −1 ≤ x ≤ 3 . Ans:
difficulty: medium
section: 4.1
31. Given the function f ( x) =x 4 − 4 x 3 − 8 x 2 + 1 on the interval [ −5,5] , at which value(s) of x does f have a local or global maximum? A) –1 B) –5 C) –0.528 D) 4 E) 2.528 F) 0 Ans: B, F difficulty: medium section: 4.1 32. Consider the function f ( x)= x + 2 cos x , for 0 ≤ x ≤ 2π . Is f increasing or decreasing at x = 3.82? Ans: increasing difficulty: medium section: 4.1 33. Consider the function f ( x)= x + 2 cos x , for 0 ≤ x ≤ 2π . What is the largest value of f? Round to 2 decimal places. Ans: 8.28 difficulty: medium section: 4.1 34. Consider the function f ( x)= x + 2 cos x , for 0 ≤ x ≤ 2π . Which of the following values are inflection points of f? 5π 3π π π A) 0 B) C) D) E) 6 2 6 2 Ans: C, E difficulty: medium section: 4.1 35. Consider the function f ( x)= x + 2 cos x , for 0 ≤ x ≤ 2π . Where is f increasing most rapidly? 3π 5π π π A) 0 B) C) D) E) 6 2 6 2 Ans: C difficulty: medium section: 4.1
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Chapter 4: Using the Derivative
36. Consider the function f ( x)= x + 2 cos x , for 0 ≤ x ≤ 2π . Graph the function and use your graph to find how many roots there are to the equation f ( x) = 1 . Ans: 2 difficulty: medium section: 4.1 − x2 f ( x) = e b , where b is a positive constant.
37. Let points of f? A) b
b b D) − E) 0 2 2 difficulty: medium section: 4.2
B) −b
Ans: C, D
Which of the following are inflection
C)
− x2 f ( x) = e b , where b is a positive constant.
38. Let What happens to the graph of f as b decreases? A) The height of the graph increases. B) The height of the graph decreases. C) The inflection points move farther away from the y-axis. D) The inflection points move closer to the y-axis. Ans: D difficulty: medium section: 4.2 39. A normal distribution in statistics is modeled by the function − ( x − µ )2 1 2σ 2 determine where the maximum value of the function would N ( x) = e σ 2π occur. Assume x is a variable and all other terms are constants. A) –0.091 B) 0.01 C) –0.182 D) 0.008 E) 0 Ans: E difficulty: medium section: 4.2 40. Consider the two-parameter family of curves = y ax + b Classify the point – , – ab . a A) A global minimum B) A global maximum Ans: A difficulty: medium
b , with a > 0 and b > 0 . x
C) An inflection point D) None of the above section: 4.2
41. Consider the two-parameter family of curves = y ax +
b , with a > 0 and b > 0 . Is the x
graph concave up or down at the point x = –2? A) Concave down B) Concave up C) It depends on the values of a and b Ans: A difficulty: medium section: 4.2
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Chapter 4: Using the Derivative
42. Consider the function f ( x) =
1 1 + ae− x
, for a > 0 . As a increases, what happens to the
horizontal asmyptotes of f? A) They shift upward. B) They remain the same. Ans: B difficulty: medium section: 4.2
C) They shift downward.
( x) ax(1 − x) for a > 0 . For what value of x on the region 43. Consider the function f = 0 ≤ x ≤ 1 does f have a local maximum? If there is more than one value, give the smallest one. Ans: 0.5 difficulty: medium section: 4.2 ( x) ax(1 − x) for a > 0 . What is largest value of a such that 44. Consider the function f = f ( x) ≤ 3 on the region 0 ≤ x ≤ 1 ? Ans: 12 difficulty: hard section: 4.2 45. Given the table of data about the second derivative of a function f, which of the following types of a function could f be? Assume b > 0. The other constants can be positive or negative. x f ''( x)
0 1
1 –2
2 –5
3 –8
A)
aebx
D)
a cubic (i.e. ax3 + bx 2 + cx + d )
B)
e− x / b
E)
sin(bx)
2
C) a quadratic (i.e. ax 2 + bx + c ) Ans: D difficulty: medium section: 4.2
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Chapter 4: Using the Derivative
46. When light strikes a shiny surface, much of it is reflected in the direction shown. However some of it may be scattered on either side of the reflected light. If the intensity (brightness) of the scattered light at the angle θ (shown in the picture) is I, the Phong model says that = I k cos n (θ) where k and n are positive constants depending on the surface. Thus this function gives an idea of how “spread-out” the scattered light is. What effect does decreasing the parameter n have of the graph of I?
A) The graph drops less quickly C) B) The graph stretches upward D) Ans: A difficulty: hard section: 4.2
The graph shrinks downward The graph drops more quickly
47. Consider the one-parameter family of functions given by e Ax + e− Ax for A > 0 . What are the effects on the graph as the value of A is decreased? A) The global minimum is decreased. B) The global minimum is increased. C) The curvature is decreased (i.e. the curve is wider) D) The curvature is increased (i.e. the curve is narrower) Ans: C difficulty: medium section: 4.2 48. For the function y = axe −bx , choose a and b so that y has a critical point at x = 3 and a maximum value of 13. Round your answers to 3 decimal places. a = ______, Part A: 11.779 Part B: 0.333 difficulty: medium
b = _____
section: 4.2
49. Suppose f is a cubic function with critical points at x = 8 and x = 6. What is the x-coordinate of the inflection point of f? Ans: 7 difficulty: hard section: 4.2
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Chapter 4: Using the Derivative
50. Find the best possible upper and lower bounds for the function f ( x) = xe –0.8 x for x ≥ 0, i.e., find numbers A and B such that A ≤ xe –0.8x ≤ B , x ≥ 0 . The numbers A and B should be as close together as possible. Round your answers to 2 decimal points. A = _____, and B = _____. Part A: 0 Part B: 0.46 difficulty: medium section: 4.4 51. Find two positive numbers whose sum is 8 such that the sum of the cube of the first and the square of the second is a minimum. The first number is _____ and the second number is _____. Part A: 2 Part B: 6 difficulty: medium section: 4.4 52. A landscape architect plans to enclose a 3000 square-foot rectangular region in a botanical garden. She will use shrubs costing $15 per foot along three sides and fencing costing $20 per foot along the fourth side. Find the dimensions that minimize the total cost. The fenced side is _____ feet, and the other side is _____ feet. Round your answers to 1 decimal place. Part A: 50.7 Part B: 59.2 difficulty: medium section: 4.3 53. A rectangular sheet of paper is to contain 32 square inches of printed matter with 2-inch margins at top and bottom and 1-inch margins on the sides. The sheet that will use the least paper is _____ inches wide and _____ inches tall. Part A: 6 Part B: 12 difficulty: medium section: 4.3 54. The point on the parabola whose equation is y = x 2 that is nearest to the point (10,2) is has x-coordinate _____ and y-coordinate _____. Part A: 2 Part B: 4 difficulty: medium section: 4.3
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Chapter 4: Using the Derivative
55. A single cell of a bee's honey comb has the shape shown. The surface area of this cell is given by 3 − cos θ 3 A= 6hs + s 2 + 2 sin θ sin θ where h, s, θ are as shown in the picture. Keeping h and s fixed, for what angle, θ, is the surface area minimal? Round to the nearest one tenth of a degree.
Ans: 54.7 difficulty: hard
section: 4.3
56. The best possible bounds for the function y =x3 + 3 x 2 + 3 x on the interval −4 ≤ x ≤ 4 are _____ ≤ y ≤ _____. Part A: –28 Part B: 124 difficulty: medium section: 4.3
= y sin 2 x + 2 x on the interval 0 ≤ x ≤ π are 57. The best possible bounds for the function _____ ≤ y ≤ _____. Part A: 0 Part B: 2π difficulty: medium section: 4.3
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Chapter 4: Using the Derivative
58. In the function y=2 sin (x)+1.96, in the interval from 0 ≤ x ≤ π , at which value(s) of x does the function contain a global maximum? A) π B) 0 C) π D) 4 π E) 2 π 2 Ans: C difficulty: easy section: 4.3 59. In the function y=–4 sin (x)+4.96, in the interval from 0 ≤ x ≤ π , what is the global maximum value? A) 0.960 B) 8.96 C) 4.960 D) 0 E) 1.571 Ans: A difficulty: easy section: 4.3 60. For= f ( x) 4 cos 2 x − sin x and 0 ≤ x ≤ π , what is the global maximum value of f ( x) ? Ans: 4 difficulty: medium section: 4.3 61. Sketch a graph of a function with two local minima, no global maximum, but a global minimum. Ans: Answers will vary. One example:
difficulty: easy
section: 4.3
62. What does the Extreme Value Theorem allow us to conclude about f if f is continuous on [–10, 80]? Mark all that apply. A) f has a global maximum on [–10, 80]. C) f has an inflection point on [–10, 80]. B) f has a global minimum on [–10, 80]. D) None of the above Ans: A, B difficulty: easy section: 4.3 63. A rectangle is inscribed between the function y= -x2 + 49 and the x-axis. What is the maximum area of the square if the base if two vertices of the square lie on the x-axis? A) 3.5 B) 257.25 C) 428.75 D) 8.042 E) 40.459 Ans: B difficulty: hard section: 4.3
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Chapter 4: Using the Derivative
64. A window with a rectangular base is topped by a semicircle creating a Norman window. A plate of glass 24 ft2 in area. What are the dimensions that would minimize the metal frame around the window? [round to 3 decimal places] Base: _____________, Height: ________________, Radius: ___________________ Ans: Base: 2.593, Height: :2.593, Radius: 1.296 difficulty: hard section: 4.3 65. The cost C(q) (in dollars) of producing a quantity q of a certain product is shown in the graph below. Suppose that the manufacturer can sell the product for $2.50 each (regardless of how many are sold), so that the total revenue from selling a quantity q is R(q) = 2.5q. The difference π(q )= R (q ) − C (q ) is the total profit. Let q0 be the quantity that will produce the maximum profit. What is C '(q0 ) ?
Ans: 2.5 difficulty: medium
section: 4.3
66. If you want to maximize profit, you should minimize average cost. Ans: False difficulty: medium section: 4.3
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Chapter 4: Using the Derivative
67. The cost C(q) (in dollars) of producing a quantity q of a certain product is shown in the C (q) graph below. The average cost is given by a (q ) = . Graphically, a(q) is the slope q of the line between which two points?
A) (0, 0) Ans: A, B
B) (q, C (q )) C) (0, q) D) (0, C (q )) difficulty: easy section: 4.4
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Chapter 4: Using the Derivative
68. The cost C(q) (in dollars) of producing a quantity q of a certain product is shown in the C (q) graph below. The average cost is given by a (q ) = . Find on the graph the q quantity q0 where a(q) is minimal. Now suppose that the fixed costs (i.e., the costs of setting up before production starts) are doubled. Sketch the new cost function C1 (q ) on the same set of axes as the original one and let q1 be the quantity where the new a1 (q ) is minimal. Which of the following is true?
A) q0 < q1 B) q0 > q1 Ans: A difficulty: hard
C) q0 = q1 section: 4.4
D) Cannot be determined
69. Total cost and revenue are approximated by the functions C = 1200 + 3.5q and R = 6q, both in dollars. Identify the marginal cost per item. Ans: 3.5 difficulty: easy section: 4.3 70. Total cost and revenue are approximated by the functions C = 1900 + 4q and R = 6q, both in dollars. Give a formula for the profit function. Ans: 2q-1900 difficulty: easy section: 4.3 71. Write a formula for total cost as a function of quantity r when fixed costs are $30,000 and variable costs are $1,600 per item. Ans: 30,000+1,600r difficulty: easy section: 4.4
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Chapter 4: Using the Derivative
72. The revenue for selling q items is R= (q ) 600q − 2q 2 and the total cost is C(q) = 110 + 60q. Which function gives the total profit earned? A) C) −540q + 2q 2 + 110 540q − 2q 2 + 110 B) −540q + 2q 2 − 110 Ans: D difficulty: easy
D) section: 4.4
540q − 2q 2 − 110
73. The revenue for selling q items is R= (q ) 600q − 2q 2 and the total cost is C(q) = 120 + 60q. What quantity maximizes profit? Ans: 135 difficulty: easy section: 4.4 74. The function C= (r ) 10r 2 − 30 gives cost in dollars of producing r items. What is the marginal cost of increasing r by 1 item from the current production level of r = 6? Ans: 120 difficulty: easy section: 4.4 75. Find the marginal cost for q = 100 when the fixed costs in dollars are 1000, the variable costs are $190 per item, and each sells for $310. Ans: 190 difficulty: easy section: 4.4 76. Find the quantity q which maximizes profit if the total revenue, R(q), and the total cost, C(q), are given in dollars by R (q= ) 8q − 0.02q 2 C (= q ) 300 + 1.9q , where 0 ≤ q ≤ 600 units. Round to the nearest whole number. Ans: 152 difficulty: easy section: 4.4 77. A wire of length 90 is cut into two pieces. [We allow the possibility that one of the pieces has zero length.] The first piece is bent into a circle, the second into a square. A. How long should the piece of wire that is bent into a circular shape be in order to maximize the sum of the areas of the two shapes? B. How long should that piece be if you want to minimize the sum of the areas? Round both answers to two decimal places. Part A: A. 90.00 Part B: B. 39.59 difficulty: hard section: 4.5
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Chapter 4: Using the Derivative
78. A rectangular building is to cover 28,800 square feet. Zoning regulations require 20 foot frontages at the front and the rear and 10 feet of space on either side. Find the dimensions of the smallest piece of property on which the building can be constructed legally. The property is _____ feet along the front and _____ feet along the side. Part A: 140 Part B: 280 difficulty: medium section: 4.5 79. The regular air fare between Boston and San Francisco is $600. An airline flying 747s with a capacity of 480 on this route observes that they fly with an average of 400 passengers. Market research tells the airlines' managers that each $20 fare reduction would attract, on average, 20 more passengers for each flight. How should they set the fare to maximize their revenue? Ans: $520 difficulty: medium section: 4.5 80. What is the shortest distance from the point (0,1) to the curve y = e x ? You will need to use a calculator with root-finding capabilities. Give your answer to 2 decimal places. Ans: 1.41 difficulty: hard section: 4.5 81. If you throw a stone into the air at an angle of θ to the horizontal, it moves along the curve x2 = y x tan θ − 1 + tan 2 θ , 100
(
)
where y is the height of the stone above the ground, x is the horizontal distance. If the angle θ is fixed, what value of x gives the maximum height? (Your answer will θ.) tan θ 50 tan θ 1 + tan 2 θ 50 + 50 tan 2 θ A) B) C) D) 50 tan θ tan θ 50 + 50 tan 2 θ 1 + tan 2 θ Ans: A difficulty: hard section: 4.5
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Chapter 4: Using the Derivative
82. If you throw a stone into the air at an angle of θ to the horizontal, it moves along the curve x2 = y x tan θ − 1 + tan 2 θ , 160
(
)
where y is the height of the stone above the ground, x is the horizontal distance. Suppose the stone is to be thrown over a wall at a fixed horizontal distance l away from you. If you can vary θ, what is the highest wall that the stone can go over? (Your answer will contain l.) 6400 − l 2 Ans: 2k difficulty: hard section: 4.5 83. A rectangle with its base on the x-axis is inscribed in the region bounded by the curve f ( x) = x 2 , the x-axis and the line x = 1 . Find the dimensions of the rectangle with maximal area. The rectangle has base _____ and height _____. Round to 2 decimal places. Part A: 0.33 Part B: 0.44 difficulty: medium section: 4.5 84. A normal distribution in statistics is modeled by the function 1 − x2 2 determine where the maximum value of the function would occur. N ( x) = e 2π E) e –2 A) –2 B) –1 C) 0 D) –2 π Ans: C difficulty: easy section: 4.5
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Chapter 4: Using the Derivative
85. A submarine can travel 30mi/hr submerged and 60mi/hr on the surface. The submarine must stay submerged if within 200 miles of shore. Suppose that this submarine wants to meet a surface ship 200 miles off shore. The submarine leaves from a port 300 miles along the coast from the surface ship. What route of the type sketched below should the sub take to minimize its time to rendezvous? Give the value of y to 2 decimal places.
Ans: 1.85 difficulty: hard
section: 4.5
86. Daily production levels in a plant can be modeled by the function G (t ) = −3t 2 + 18t − 9 , which gives units produced t hours after the factory opened at 8am. At what time during the day is factory productivity a maximum? Answer in the form "_:_ _" (without an "am" or "pm"). Ans: 11:00 difficulty: easy section: 4.5 87. The number of plants in a terrarium is given by the function P (c) = −1.4c 2 + 3c + 15 , where c is the number of mg of plant food added to the terrarium. Find the amount of plant food that produces the highest number of plants. Round to 2 decimal places. Ans: 1.07 difficulty: easy section: 4.5
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Chapter 4: Using the Derivative
88. The function y= 0.2( x + 2) 2 − 2 x + 10 gives the population of a town (in 1000's of people) at time x where x is the number of years since 1980. When was the population a minimum? Round to the nearest year. Ans: 1983 difficulty: easy section: 4.5 89. A student is drinking a milkshake with a straw from a cylindrical cup with a radius of 5.5 cm. If the student is drinking at a rate of 4.5 cm3 per second, then the level of the milkshake dropping at a rate of _____ cm per second. Round to 2 decimal places. Ans: 0.05 difficulty: easy section: 4.6 90. Air is being blown into a spherical balloon at a rate of 70 cm3 per second. At what rate is the surface area of the balloon increasing when the radius is 10 cm? Round to 2 decimal places, and do not include units. Ans: 14.00 difficulty: medium section: 4.6 91. A fan is watching a 100-meter footrace from a seat in the bleachers 15 meters back from the midway point. The winning runner is moving approximately 8 meters per second. How fast is the distance from the fan to the winning runner changing when he is x meters into the race? 8 x − 400 meters per second Ans: x 2 + 100 x + 2725 difficulty: hard section: 4.6 92. A rectangular swimming pool is 10 meters long and 6 meters wide. It has a depth of 1 meter at the shallow end, then slopes to a depth of 1.5 meters at the deep end, as shown in the following cross section (not to scale). It is being filled with a hose at a rate of 50,000 cubic centimeters per minute. 225 minutes after the hose is turned on, the water is rising at a rate of _____ cm per second. Round to 3 decimal places.
Ans: 0.096 difficulty: hard
section: 4.6
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Chapter 4: Using the Derivative
93. A Brian's candy sugar wand is made from flavored sugar inside a straw. The straw is 210 mm long and 5 mm in diameter. The child accidentally poked a hole in the bottom, making the height of the sugar fall at a rate of 1 mm per second. The child realizes that there is a hole after 1 seconds. What was the rate of change of the volume of the sugar at this time? [V = π r 2h ] dV Ans: = –2.5 π dt difficulty: hard section: 4.6 94. A spherical lollipop has a circumference of 7.9 centimeters. A student decides to measure the rate of change of the volume of the lollipop, in cm3 per minute. The student licks the lollipop and measures the circumference every minute. The radius is decreasing at a rate of 0.18 cm/min. Determine the rate at which the volume is 4 changing when the circumference is half of it's original size. [ V = π r 3 ] 3 Ans: –0.450 difficulty: hard section: 4.6 95. A cupful of olive oil falls on the floor forming a circular puddle. Its radius is increasing at a constant rate of 0.2 cm/sec. What is the rate of increase in the area of the olive oil when its circumference measures 20 π cm? Ans: 4 π cm2/sec difficulty: medium section: 4.6 96. A bar of ice cream, with dimensions of 3 cm by 3 cm by 3 cm placed on a mesh screen on top of a cylindrical funnel that is 6 cm high and 6 cm in diameter. If the ice cream is melting at a rate of 3.3 cm3 / min into the funnel, what is the rate of change of the height 1 of the funnel when half of the ice cream has melted? V funnel = π r 2 h 3 A) 0.232 B) 0.032 C) 0.132 D) 0.284 E) 0.083 Ans: C difficulty: medium section: 4.6 97. Frank decided to ride in a hot air balloon. His brother Damien was going to videotape the lift off from a distance of 30 feet away. The hot air balloon rises to a height of 2000 feet in 19 minutes. What is the rate at which the camera's angle should be raised in order to follow the balloon? (specify units) dθ Ans: = 0.024 radians/minute dt difficulty: hard section: 4.6
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Chapter 4: Using the Derivative
98. Find the limit:
2sin x cos x x →0 x lim
Ans: 2 difficulty: easy
section: 4.7
99. Find the limit:
2e − x + 2 x x→∞ x lim
Ans: 2 difficulty: easy
section: 4.7
100. Which function dominates as x → ∞ ? x 90 e
A) Ans: A
B) x90 difficulty: medium
section: 4.7
x2 − a2 , given a=–1 x+a x→a
101. Determine the lim Ans: 0 difficulty: easy
section: 4.7 x2 − a2 , given a=5 x−a x→a
102. Determine the lim Ans: E, D difficulty: easy
section: 4.7
103. A normal distribution in statistics is modeled by the function 1 − x2 2 determine the limit of N(x) as x → - ∞ N ( x) = e 2π 1 1 B) C) 0 D) E) ∞ A) - ∞ 2π 2π Ans: C difficulty: medium section: 4.7
( )
( ) parameterize a circle.
104. The equations x = cos t 5 , y = sin t 5 Ans: True
difficulty: medium
section: 4.8
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Chapter 4: Using the Derivative
105. Which of the following diagrams represents the parametric curve = x – sin = t , y cos t , 0 ≤ t ≤ π ?
Ans: B difficulty: medium
section: 4.8
106. Find parametric equations for a line through the points, A = (–1, 1) and B = (2, 3) so that the point A corresponds to t = 0 and the point B to t = 6. The equations are x = _____ and y = _____, for −∞ < t < ∞ . Part A: –1 + t/2 Part B: 1 + t/3 difficulty: medium section: 4.8 107. A lady bug moves on the xy-plane according the the equations = x 2t (t − 6) , y= 5 − t . When does the lady bug stop moving? A) When t = 6 B) When t = 3 C) When t = 5 D) Never Ans: D difficulty: medium section: 4.8 108. A lady bug moves on the xy-plane according the the equations = x 2t (t − 8) , y= 2 − t . When is the lady bug ever moving straight up or down? A) When t = 8 B) When t = 4 C) When t = 2 D) Never Ans: B difficulty: medium section: 4.8 109. A lady bug moves on the xy-plane according the the equations = x 2t (t − 8) , y= 3 − t . Suppose that the temperature at a point (x, y) in the plane depends only on the y coordinate of the point and is equal to 4 y 2 . Find the rate of change of the temperature at the location of the lady bug at time t. Ans: 8(t − 3) difficulty: medium section: 4.8
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Chapter 4: Using the Derivative
2 2 110. The equations x = cos(πt /180), y = sin(πt /180) describe the motion of a particle π π moving on a circle. Assume x and y are in miles and t is in days. What is the radius of the circle (in miles)? Round to 2 decimal places. Ans: 0.64 difficulty: medium section: 4.8 3 3 111. The equations x =cos(πt / 90), y =sin(πt / 90) describe the motion of a particle π π moving on a circle. Assume x and y are in miles and t is in days. What is the period of the circular motion (in days)? Ans: 180 difficulty: medium section: 4.8 9 9 112. The equations x = cos(πt /180), y = sin(πt /180) describe the motion of a particle π π moving on a circle. Assume x and y are in miles and t is in days. What is the speed of the particle (in miles per day) when it passes through the point (-9/π, 0)? Round to 3 decimal places. Ans: 0.050 difficulty: hard section: 4.8 113. Which of the following parametric equations pass through the points (8, 1) and (–1, –1)? x= 8 − 9t and y = 1 − 2t x= 8 + 9t and y = 1 + 2t A) C) y –1 − 2t y –1 + 2t = x –1 − 9t and = x –1 + 9t and = B) D) = Ans: A, B difficulty: medium section: 4.8 114. The curve represented by the parametric equations x= 5 − t 3 and y= 6 + t 3 is A) A curve passing through the point (5, 6) that is concave up to the left of the point and concave down to the right of it. B) A curve passing through the point (5, 6) that is concave down to the left of the point and concave up to the right of it. C) A straight line passing through the point (5, 6) with slope -1. D) A straight line passing through the point (5, 6) with slope 1. Ans: C difficulty: medium section: 4.8 115. Motion of a particle is given by x= t 2 − 3t , y = 2t 3 − 3t where t is time in minutes. Find the instantaneous speed of the particle at t = 3. Round to 2 decimal places. Ans: 51.09 difficulty: medium section: 4.8
Page 32
Chapter 4: Using the Derivative
116. For the function f ( x) = x 2 + 2 x − 1 , there exists a value c such that 1 ≤ c ≤ 3. What is f ′(c) ? A) 6 B) 2 C) 6 D) 16 E) 7 Ans: A difficulty: easy section: 4 review 117. One fine day you take a hike up a mountain path. Using your trusty map you have determined that the path is approximately in the shape of the curve y = 3( x3 − 12 x 2 + 48 x + 36) . Here y is the elevation in feet above sea level and x is the horizontal distance in miles you have traveled, but your map only shows the path for 7 miles, horizontal distance. How many feet above sea level are you when you start your hike? Ans: 108 difficulty: easy section: 4 review 118. One fine day you take a hike up a mountain path. Using your trusty map you have determined that the path is approximately in the shape of the curve y = 3( x3 − 12 x 2 + 48 x + 36) . Here y is the elevation in feet above sea level and x is the horizontal distance in miles you have traveled, but your map only shows the path for 7 miles, horizontal distance. Where on the path is a nice flat place to stop for a picnic? A) After 2 miles D) After 5 miles B) After 3 miles E) After 6 miles C) After 4 miles Ans: C difficulty: easy section: 4 review 119. One fine day you take a hike up a mountain path. Using your trusty map you have determined that the path is approximately in the shape of the curve y = 3( x3 − 12 x 2 + 48 x + 36) . Here y is the elevation in feet above sea level and x is the horizontal distance in miles you have traveled, but your map only shows the path for 7 miles, horizontal distance. Which of the following is a good estimate of the elevation after 7.5 horizontal miles. (You do not know the shape of the path explicitly after 7 miles!) A) 381 feet B) 394.5 feet C) 408 feet D) 421.5 feet Ans: D difficulty: medium section: 4 review 120. One fine day you take a hike up a mountain path. Using your trusty map you have determined that the path is approximately in the shape of the curve y = 5( x3 − 12 x 2 + 48 x + 36) . Here y is the elevation in feet above sea level and x is the horizontal distance in miles you have traveled, but your map only shows the path for 7 miles, horizontal distance. If your friend, who is not in “good shape” followed this path for 15 miles total, in horizontal distance, the day before, it makes sense that the equation for the elevation continues to hold much beyond the 7 mile mark. Ans: False difficulty: easy section: 4 review
Page 33
Chapter 4: Using the Derivative
121. If f '' = 0 at x = 2, then the graph of f changes concavity at x = 2. A) Always true B) Sometimes true C) Never true Ans: B difficulty: easy section: 4 review 122. If f is always decreasing and concave down, then f must have at least one root. A) Always true B) Sometimes true C) Never true Ans: B difficulty: easy section: 4 review 123. Given the function f ( x) = x 2e−2 x , then at x = 1 the graph of f has A) a local maximum C) none of these B) a local minimum D) an inflection point Ans: A difficulty: medium section: 4 review 124. Graph the function f ( x) = x 2e−2 x , labeling all important points. Ans:
difficulty: medium
section: 4 review
Page 34
Chapter 4: Using the Derivative
125. Draw the graph of a polynomial of degree four that has a local minimum at (–5, –14) and inflection points at (–2, –6) and the origin. Ans:
difficulty: medium
section: 4 review
Page 35
1. Consider a sports car which accelerates from 0 ft/sec to 88 ft/sec in 5 seconds (88 ft/sec = 60mph). The car's velocity is given in the table below. t 0 1 2 3 4 5 V(t) 0 31 53 69 81 88 Find the lower bound for the distance the car travels in 5 seconds. Ans: 234 difficulty: medium section: 5.1 2. Consider a sports car which accelerates from 0 ft/sec to 88 ft/sec in 5 seconds (88 ft/sec = 60mph). The car's velocity is given in the table below. t 0 1 2 3 4 5 V(t) 0 30 52 68 80 88 In which time interval is the average acceleration smallest? A) Between t = 4 and t = 5. D) Between t = 1 and t = 2. B) Between t = 3 and t = 4. E) Between t = 0 and t = 1. C) Between t = 2 and t = 3. Ans: A difficulty: easy section: 5.1 3. The graph shown below is that of the velocity of an object (in meters/second). Find a lower estimate of the total distance traveled from t = 0 to t =5 seconds
. Ans: 70 difficulty: medium
section: 5.1
Page 1
Chapter 5: Key Concept- The Definite Integral
4. The graph shown below is that of the velocity of an object (in meters/second). At what times is the acceleration zero?
A) t = 0 Ans: C, E
B) t = 2 C) t = 4 D) t = 6 difficulty: easy section: 5.1
E) t = 8
F) t = 10
5. At time t, in seconds, the velocity, v, in miles per hour, of a car is given by v(t )= 8 + 0.8t 2 for 0 ≤ t ≤ 8 . Use ∆t = 2 to estimate the distance traveled during this time. Give the right-hand sum. Ans: 256 difficulty: medium section: 5.1 6. At time t, in seconds, the velocity, v, in miles per hour, of a car is given by v(t ) for 0 ≤ t ≤ 8 . A second car travels exactly 15 miles per hour faster than the first car. Using ∆t = 2, the right-hand estimate for the distance traveled during this time by the first car is 143 miles. What is the right-hand estimate for the distance traveled during this time by the second car? Ans: 263 miles difficulty: medium section: 5.1
Page 2
Chapter 5: Key Concept- The Definite Integral
7. If an upper estimate of the area of a region bounded by the curve in the following figure, the horizontal axis and the vertical lines x = 3 and x = -3 is 15, what is the upper estimate if the graph is shifted up one unit?
Ans: 21 difficulty: easy
section: 5.1
8. The figure below shows the graph of the velocity, v, of an object (in meters/sec.). If the graph were shifted up 4 units, how would the total distance traveled between t = 0 and t = 6 change?
A) It would increase by 4 units. D) B) It would increase by 24 units. E) C) It would remain the same. Ans: B difficulty: easy section: 5.1
Page 3
It would decrease by 4 units. It would decrease by 24 units.
Chapter 5: Key Concept- The Definite Integral
9. The figure below shows the graph of the velocity, v, of an object (in meters/sec.). If the graph were shifted up two units, what would it mean for the motion of the object?
A) The velocity at each time would be 2 m/sec greater. B) The velocity at each time would be 12 m/sec greater. C) The velocity at each time would be 20 m/sec greater. D) The velocity at each time would be 120 m/sec greater. E) The velocity at each time would be the same. Ans: A difficulty: easy section: 5.1 10. A car is observed to have the following velocities at times t = 0, 2, 4, 6: time(sec) velocity (ft/sec)
0 0
2 23
4 42
6 68
Give the lower estimate for the distance the car traveled. Ans: 130 difficulty: easy section: 5.1 11. At time t, in seconds, your velocity v, in meters/sec, is given by v(t )= 2 + 5t 2 for 0 ≤ t ≤ 6 . Use ∆t = 2 to estimate distance during this time. (Average right- and left-hand sums). Ans: 392 meters difficulty: medium section: 5.1 12. At time t, in seconds, your velocity v, in meters/sec, is given by v(t )= 4 + 4t 2 for 0 ≤ t ≤ 6 . Use ∆t = 1 to estimate distance during this time. (Average right- and left-hand sums). Ans: 316 meters difficulty: medium section: 5.1 13. At time t, in seconds, your velocity v, in meters/sec, is given by v(t )= 9 + 4t 2 for 0 ≤ t ≤ 6 . Which is more accurate? A) An estimate of the distance traveled during this time using ∆t = 1. B) An estimate of the distance traveled during this time using ∆t = 2. Ans: A difficulty: medium section: 5.1
Page 4
Chapter 5: Key Concept- The Definite Integral
14. A particle starting at the origin moves along the x-axis such that its velocity along the line can be modeled by the function v(t) = 2 - 6t units/sec. What is the exact change in position by the function from t = 0 to t = 1.333333? A) –2.667 units D) –6.333 units B) –5.333 units E) –6.833 units C) –4.333 units Ans: A difficulty: medium section: 5.1 15. Two greyhound racing dogs, A and B, start at the same time and travel in the same direction along a straight track. The figure below gives the velocity, v, of each dog as a function of time t. Which dog travels the farthest?
Ans: A difficulty: hard
section: 5.1
4
16. Estimate ∫ ln x dx to 1 decimal place, choosing a suitable ∆x. 3
Ans: 1.2 difficulty: medium
section: 5.2
Page 5
Chapter 5: Key Concept- The Definite Integral
17. Consider the region A shown in the following graph. Is the area of A more or less than 0.7?
Ans: less difficulty: medium
section: 5.2
18. Consider the region A bounded above by the graph of f ( x) = e− x , bounded below by 2
2
the graph of g (= x) e x − 1 , and bounded on the left by the y-axis. Which of the following integrals best approximates the area of the region A? (You will want to graph A to find the answer).
( − e + 1) dx B) ∫ ( e − e − 1) dx C) ∫ ( e + e − 1) dx
A)
∫0 e
1
− x2
x2
1
x2
− x2
0 1
−x
2
x
(e − e + 1) dx E) ∫ ( e − e − 1) dx F) ∫ ( e + e − 1) dx D)
2
0
Ans: D
difficulty: medium
∫0
0.7
− x2
x2
0.7
x2
− x2
0 0.7
− x2
x2
0
section: 5.2
Page 6
Chapter 5: Key Concept- The Definite Integral
19. What does the following figure represent?
The right-hand Riemann sum for the function f on the interval 0 ≤ t ≤ 12 with ∆t = 3. B) The right-hand Riemann sum for the function f on the interval 0 ≤ t ≤ 12 with ∆t = 6. C) The left-hand Riemann sum for the function f on the interval 0 ≤ t ≤ 12 with ∆t = 3. D) The left-hand Riemann sum for the function f on the interval 0 ≤ t ≤ 12 with ∆t = 6. Ans: D difficulty: easy section: 5.2 A)
20. Using the following figure, calculate the value of the right-hand Riemann sum f on the interval 0 ≤ t ≤ 12 with ∆t = 6.
Ans: 72 difficulty: medium
section: 5.2
21. Use the table to estimate ∫
50
0
f ( x) dx with n = 5 and ∆x = 10. (Average left-and
right-hand sums). x 0 f(x) 40
10 45
Ans: 3075 difficulty: medium
section: 5.2
20 55
30 60
Page 7
40 80
50 95
Chapter 5: Key Concept- The Definite Integral
22. Estimate the area of the region above the curve y = 7 cos x and below y = 7 for 0 ≤ x ≤ π/2. Round to 2 decimal places. Ans: 4.00 difficulty: medium section: 5.2 23. Estimate the area of the region between y = cos x, y = 3x, x = -π/2, and x = 0. Round to 3 decimal places. Ans: 4.701 difficulty: medium section: 5.2 24. Estimate the area of the region under the curve y = − x3 + 2 and above the x-axis for 0 ≤ x ≤ 3 2 . Round to 3 decimal places. Ans: 1.890 difficulty: medium section: 5.2
25. Calculate the area under the curve y= 3x +3 for values between [1, 4]. A) 15.75 B) 157.5 C) 993.25 D) 10.5 E) 31.5 Ans: E difficulty: easy section: 5.2 c
26. What is the value of ∫ f ( x)dx if the area of A = 15 and the area of B = –4? a
A) 5.5 Ans: D
B) 22 C) –11 difficulty: medium
D) 11 E) 19 section: 5.2
Page 8
Chapter 5: Key Concept- The Definite Integral
c
27. What is the value of ∫ f ( x) dx if the area of A = 9 and the area of B = –2? a
A) 3.5 Ans: E
B) 14 C) –7 D) 7 E) 11 difficulty: medium section: 5.2
28. The velocity and acceleration of an object are given by the graphs shown below, where v(0) = 0 . Which graph shows acceleration?
A) The one on the left. B) The one on the right. Ans: A difficulty: medium section: 5.3 29. The velocity of an object is given by v(t ) , and the acceleration is given by a (t ) . What is the relationship between the total change in v(t) over the interval 0 ≤ t ≤ 10 and a(t)? 10
A)
t ) dt a (10) − a (0) ∫0 v(=
B)
t ) dt v(10) − v(0) ∫0 a(=
10
Ans: B
difficulty: easy
t dt a (10) − v(10) ∫0 =
D)
∫0 v(t ) dt = ∫0 a(t ) dt
section: 5.3
Page 9
10
C)
10
10
Chapter 5: Key Concept- The Definite Integral
30. How can the quantity f(a)h be represented in the picture?
A) By the slope of TV B) By the length of UR Ans: C difficulty: medium
C) By the area of PQRS D) By the area of QRU section: 5.3
31. Does the quantity a+h
∫a
f ( x) dx
represent a length or an area in the picture?
A) An area B) A length Ans: A difficulty: medium
section: 5.3
Page 10
Chapter 5: Key Concept- The Definite Integral
32. Below is the graph of the rate r in arrivals/minute at which students line up for breakfast at the Cafeteria Charlotte. The first people arrive at 6:50a.m. and the line opens at 7:00a.m. Suppose that once the line is open, checkers can check peoples' meal cards at a constant rate of 20 people per minute. Use the graph and this information to find an estimate for the length of the line (i.e. the number of people) at 7:10.
A) 200 Ans: B
B) 230 C) 260 difficulty: medium
D) 290 section: 5.3
A) 1.44 Ans: D
5 π /2
π ∫0
sin t dt ?
B) 1.49 C) 1.54 D) 1.59 difficulty: easy section: 5.3
E) 1.64
33. Which of the following best approximates
34. The average value of a function g on 0 ≤ x ≤ 3 is a constant g given by 1 3 g= g ( x) dx . Which of the following must be true? 3 − 0 ∫0 A)
∫0 gg ( x) < ( g )
3
2
C)
∫0 gg ( x) = ( g )
B)
∫0 gg ( x) > ( g )
3
2
D)
None of the above
Ans: C
difficulty: medium
section: 5.3
Page 11
3
2
Chapter 5: Key Concept- The Definite Integral
35. The average value of a function g on 0 ≤ x ≤ 2 is a constant g given by 2 1 2 . Also, ( g ( x) − g ) 2 dx ≥ 0 , since ( g ( x) − g ) 2 is a square. g= g ( d ) dx ∫ ∫ 0 0 2−0 Which of the following must be true? 2
A)
2
2 g ( x) dx ≤ 2 g ( x) 2 dx ∫0 ∫0 ( )
C)
2 g ( x) dx = 2 g ( x) 2 dx ∫0 ∫0 ( )
2
2 g ( x) dx ≥ 2 g ( x) 2 dx B) D) ∫0 ∫0 ( ) Ans: A difficulty: hard section: 5.3
None of these
36. A car is moving along a straight road from A to B, starting from A at time t = 0. Below is the velocity (positive direction is from A to B) plotted against time.
How many kilometers away from A is the car at time t = 9? Ans: 8 difficulty: medium section: 5.3
Page 12
Chapter 5: Key Concept- The Definite Integral
37. A car is moving along a straight road from A to B, starting from A at time t = 0. Below are graphs of the velocity and the acceleration plotted against time (positive direction is from A to B). Which graph shows the velocity?
A) The one on the right B) The one on the left Ans: B difficulty: medium section: 5.3
Page 13
Chapter 5: Key Concept- The Definite Integral
38. A shop is open from 9am-7pm. The function r(t), graphed below, gives the rate at which customers arrive (in people/hour) at time t. Suppose that the salespeople can serve customers at a rate of 60 people per hour. When do people have to start waiting in line before being served? Answer to the nearest half-hour. You do not need to include "am" or "pm".
Ans: 12:00 difficulty: medium
section: 5.3
39. Below is the graph of the velocity, in feet per second, of a hat that is thrown up in the air from ground level. Positive velocity means upward motion. [Note that this is the graph of velocity, not distance.]
About how big is the average speed over the first 4 seconds? A) 13.1 ft/sec B) 10.8 ft/sec C) 6.8 ft/sec D) 4.4 ft/sec Ans: A difficulty: medium section: 5.3
Page 14
Chapter 5: Key Concept- The Definite Integral
40. Rashmi and Tia both go running from 7:00am to 8:00am. Both women increase their velocity throughout the hour, both beginning at a rate of 9 mi/hr. at 7:00am and running at a rate of 13 mi/hr by 8:00am. Tia's velocity increases at an increasing rate and Rashmi's velocity increases at a decreasing rate. Who has the greatest average velocity? If they had the same average velocity, enter "same". Ans: Rashmi difficulty: medium section: 5.3
) 3 x + 3 over [1, 3]. 41. Find the average value of h( x= Ans: 9 difficulty: easy section: 5.3 42. Find the average value of f ( x) = e3 x over [0, 2]. Round to the nearest whole number. Ans: 67 difficulty: medium section: 5.3 43. If E (t ) represents the energy consumed in a household in watts/month, what are the units b
of ∫ E (t )dt ? a
A) watts Ans: A
B) watts/month C) watts/month2 difficulty: eay section: 5.3
D) month
E) watts2/month2
6
44. Explain in words what ∫ v(t )dt means if v(t ) is velocity in miles/hour and t is time, in 2
hours. Ans: Total change in distance traveled, in miles, from t = 2 to t = 6 hours. difficulty: easy section: 5.3 35 1 F (t )dt ≈ 300 means if 35 − 0 ∫0 F (t ) is the temperature of the potato, in degrees Farenheit, and t is time, in minutes. Ans: The average temperature of the potato in the first 35 minutes is 300 F . difficulty: easy section: 5.3
45. A potato is cooking in an oven. Explain in words what
Page 15
Chapter 5: Key Concept- The Definite Integral
x
46. Suppose f(t) is given by the graph below. If F ( x) = ∫ f (t ) dt , what is F (4) ? 0
Ans: –3 difficulty: easy
section: 5.4
b
b
b
b
a
a
a
a
47. Suppose ∫ g ( x) dx = 6 , ∫ ( g ( x)) 2 dx = 8 , ∫ h( x) dx = 0 , and ∫ (h( x)) 2 dx = 2 . Find
∫a ( ( g ( x)) + (h( x)) ) dx . b
2
2
Ans: 10 difficulty: medium
section: 5.4
b
b
b
b
a
a 2
a
a
48. Suppose ∫ g ( x) dx = 7 , ∫ ( g ( x)) 2 dx = 8 , ∫ h( x) dx = 1 , and ∫ (h( x)) 2 dx = 4 . Find b
b
∫a g ( x) dx − ∫a ( 2h( x) ) dx . Ans: 3 difficulty: medium 49. Let ∫
10 0
section: 5.4
f ( x) dx = C . What is the average value of f(x) on the interval x = 0 to x = 10?
A) 10C
B) C
Ans: C
difficulty: easy
C)
C 10
D) 0 section: 5.4
8
8
0
−8
50. Let ∫ f ( x) dx = C . If f(x) is even, what is ∫ A) C Ans: D
B) 8C C) 16C difficulty: medium
D) 2C E) 0 section: 5.4
5
51. Evaluate the definite integral ∫ (5 + 2 x) dx . 0
Ans: 50 difficulty: medium
f ( x) dx ?
section: 5.4
Page 16
Chapter 5: Key Concept- The Definite Integral
5
52. Evaluate ∫ 3x 2 dx 1
Ans: 124 difficulty: medium 53. If f ( x)=
section: 5.4
12 12 x + 4 and g(x) = x + 1, how do ∫ f ( x) dx and ∫ g ( x) dx compare? 8 8 2
12
12
12
12
A)
∫8 f ( x) dx < ∫8 g ( x) dx
B)
∫8 f ( x) dx > ∫8 g ( x) dx
Ans: A
difficulty: easy
12
12
∫8 f ( x) dx = ∫8 g ( x) dx
C)
section: 5.4
5
8
5
1
5
8
54. If ∫ f ( x) dx = 4 and ∫ f ( x) dx = 9 , evaluate ∫ f ( x) dx . Ans: –9 difficulty: easy
section: 5.4
55. What is the value of ∫ Ans: 0 difficulty: medium
a −a
x 7 dx ?
section: 5.4
5
5
0
−5
56. If f is even and ∫ f ( x) dx = 16 , what is ∫ Ans: 32 difficulty: easy
f ( x) dx ?
section: 5.4
57. The average value of y = h(x) equals a for 0 ≤ x ≤ 5, and equals b for 5 ≤ x ≤ 15. What is the average value of h(x) for 0 ≤ x ≤ 15? a + 2b a + 2b a+b a+b A) B) C) D) 3 15 2 15 Ans: B difficulty: medium section: 5.4 5
10
10
0
5
0
58. If ∫ f ( x)dx = 11 and ∫ f ( x)dx = 16 , what is ∫ f ( x)dx ? Ans: 27 difficulty: easy
section: 5.4
7
7
−2
2
59. If ∫ f ( x)dx = 8 and f ( x) is odd, what is ∫ f ( x)dx ? Ans: 8 difficulty: medium
section: 5.4
Page 17
Chapter 5: Key Concept- The Definite Integral
1
60. If ∫ f ( x)dx = 4 and f ( x) is even, and −1
Ans: 3 difficulty: medium
∫
7
−7
7
f ( x)dx = 10 ?what is ∫ f ( x)dx ? 1
section: 5.4
61. Suppose A = the area under the curve y = e− x over the interval –5 ≤ x ≤ 5. Which of the following is/are true? 2
e− x A F (5) − F (–5) , where F ( x) = − A) = . 2x B) 1.773 < A C) 1.771 < A < 1.773 D) 1.769 < A < 1.771 Ans: C difficulty: medium section: 5 review 2
62. If r(t) represents the rate at which a country's debt is growing, then the increase in its debt between 1990 and 2000 is given by 2000 r (2000) − r (1990) A) D) r (t ) dt ∫ 1990 2000 − 1990 1 2000 r (2000) − r (1990) B) E) r '(t ) dt 10 ∫1990 1 2000 C) r (t ) dt 10 ∫1990 Ans: D difficulty: easy section: 5 review 63. The graph of f ″ is shown below.
If f is increasing at x = -1, which of the following must be true? Mark all that apply. A) f '(2) = f '(4) B) f '(4) > f '(0) C) f '(4) > 0 D) f (5) = f (6) Ans: B, C difficulty: medium section: 5 review
49π 49 − x 2 dx = 2 Ans: True difficulty: easy
64. ∫
7
−7
section: 5 review
Page 18
Chapter 5: Key Concept- The Definite Integral
65. If a function is concave up, then the left-hand Riemann sums are always less than the right-hand Riemann sums with the same subdivisions, over the same interval. Ans: False difficulty: medium section: 5 review b
66. If ∫ f ( x) = 0 , then f must have at least one zero between a and b (assume a ≠ b). a
Ans: False
difficulty: medium
67. time (hours) 3 Rate ( ml ) hr
0 20
section: 5 review
10 30
20 45
30 67.5
Find the left hand Riemann sum for n = 3 subintervals. Indicate units. Ans: 950 ml 3 difficulty: easy section: 5 review 68. time (seconds) acceleration ( m
sec 2
0
10
20
30
) 20
30
45
67.5
Find the right hand Riemann sum for n = 3 subintervals. Indicate units. Ans: 1425 meters sec difficulty: easy section: 5 review 69. Time (years) Population Growth (Millions per year)
1980 15
1984 55
1986 75
1992 135
1996 175
1998 195
1998 200
2005 270
Find the left sum for 5 subintervals, using the table above. Indicate units. Ans: 1510 million people difficulty: easy section: 5 review 70. Time (years) Balance Growth (dollars per year)
1980 20
1982 40
1989 110
1996 180
Find the right sum for 5 subintervals, using the table above. Indicate units. Ans: 4400 dollars difficulty: easy section: 5 review
Page 19
1. If f ( x) is as shown in the first graph, could the function in the second graph represent the total area (i.e., not necessarily the definite integral) between f(x) and the x-axis from 0 to a?
Ans: yes difficulty: medium
section: 6.1
Page 1
Chapter 6: Constructing Antiderivatives
2. If f is shown in the first graph, which of the two functions F could have F′ = f ?
A) The first one B) The second one C) Both of them Ans: C difficulty: easy section: 6.1
Page 2
D) Neither of them
Chapter 6: Constructing Antiderivatives
3. Could the second function be an antiderivative of the first function?
Ans: no difficulty: medium
section: 6.1
4. A young girl who aspires to be a rocket scientist launches a model rocket from the ground at time t = 0. The rocket travels straight up in the air, and the following graph shows the upward velocity of the rocket as a function of time:
Let h be the height (or vertical displacement) of the rocket, let a be the acceleration of the rocket, and let h = 0 be the ground level from which the rocket was launched. Recall that the first derivative of displacement is velocity and that the first derivative of velocity is acceleration. Is the following a graph of the acceleration or the height of the rocket as a function of time?
A) The acceleration. B) The height. Ans: A difficulty: medium section: 6.1
Page 3
Chapter 6: Constructing Antiderivatives
5. A young girl who aspires to be a rocket scientist launches a model rocket from the ground at time t = 0. The rocket travels straight up in the air, and the following graph shows the upward velocity of the rocket as a function of time:
Let v(t) be the function that gives the velocity of the rocket at time t. From the graph of the rocket's velocity, which of the following is larger? A)
t2
∫0 v(t ) dt
Ans: A
B)
t3
∫t v(t ) dt 2
difficulty: easy
section: 6.1
6. A young girl who aspires to be a rocket scientist launches a model rocket from the ground at time t = 0. The rocket travels straight up in the air, and the following graph shows the upward velocity of the rocket as a function of time:
Let v(t) be the function that gives the velocity of the rocket at time t. From the graph of t
the rocket's velocity, what does the sign of ∫ 3 v(t ) dt mean physically? 0
A) The rocket came to rest somewhere above the ground. B) The rocket came to rest somewhere below the ground. C) The rocket was accelerating when it hit the ground. D) The rocket was decelerating when it hit the ground. Ans: A difficulty: medium section: 6.1
Page 4
Chapter 6: Constructing Antiderivatives
7. You decide to take a trip down a stretch of road that runs straight east and west. The following table gives your eastward velocity (in miles per minute) measured at one-minute intervals for the first ten minutes of your trip. Time (min) 0 1 2 3 4 5 6 7 8 Velocity 0.00 0.49 0.84 1.05 1.12 1.05 0.84 0.49 0.00 (mi/min)
9 –0.63
10 –1.4
9 –0.77
10 –1.70
Does the following graph give your acceleration or your eastward distance from your starting place as a function of time?
A) Your acceleration B) Your eastward distance Ans: A difficulty: medium section: 6.1 8. You decide to take a trip down a stretch of road that runs straight east and west. The following table gives your eastward velocity (in miles per minute) measured at one-minute intervals for the first ten minutes of your trip. Time (min) 0 1 2 3 4 5 6 7 8 Velocity 0.00 0.60 1.02 1.27 1.36 1.27 1.02 0.60 0.00 (mi/min) What is your best estimate of the total eastward distance of your car from your starting position after ten minutes? Round to the nearest whole number. Ans: 6 miles difficulty: hard section: 6.1 9. Given the values of the derivative f ′(x) in the table and that f (0) = 60, estimate f (4). (Average left- and right-hand sums). x f′(x)
0 3
Ans: 120 difficulty: medium
2 15
4 27
6 39
section: 6.1
Page 5
Chapter 6: Constructing Antiderivatives
10. The graph of g (t ) is shown in the following figure. If G is an antiderivative of g such that G (0) = –6 , what is G (5) ?
Ans: 7 difficulty: medium
section: 6.1
11. The rate of change in concentration of a certain medication in a person's body, H ′(t), in micrograms per milliliter per minute, is -1 for the first 2 minutes. Then it increases at a constant rate for 2 minutes, reaching 1 at t = 4. Then it remains constant for 1 minute. Sketch H(t), assuming H(0) = 9. Ans:
difficulty: medium
section: 6.1
Page 6
Chapter 6: Constructing Antiderivatives
12. Sketch a graph of the antiderivative of g(x), given the graph g'(x) and g(0)=1.
Ans: difficulty: medium
section: 6.1
Page 7
Chapter 6: Constructing Antiderivatives
2
13. Estimate ∫ f ( x)dx for the figure given. –4
Ans: 4.283 difficulty: medium
section: 6.1
14. The graph of h(x) is given. Let H'(x)=h(x). A) Is x = 1 a critical point of H?________________ B) If x = 1 is a critical point of H, is it a local maximum or minimum?__________________
Part A: yes Part B: minimum difficulty: medium
section: 6.1
Page 8
Chapter 6: Constructing Antiderivatives
15. Find an antiderivative of 10 x +C A) 2 5(a + x 2 )5/ 2 B)
5x
. C)
+C
a2 a2 + x2 Ans: B difficulty: medium
D)
C) D)
− cos x − 7 ln x + C B) Ans: D difficulty: easy
a2 + x2
+C
1 6 x −8 x +C 6 1 6 x +8 x +C 6
section: 6.2
17. Find a function F such that F = '( x) sin x +
cos x + 7 ln x + C
10xa 2
+C
4 . x
A)
A)
a2 + x2
section: 6.2
16. Find a function F such that F '( x= ) x5 −
1 6 x +4 x +C 6 1 6 B) x −4 x +C 6 Ans: C difficulty: easy
5
7 . x C)
D) section: 6.2
cos x − 7 ln x + C − cos x + 7 ln x + C
18. Sketch the graphs of y = e x and y = ex. For which values of x is e x > ex ? Mark all that apply. A) x = 0 B) 0 < x < 1 C) x = 1 D) 1 < x < e E) x = e Ans: A, B, D, E difficulty: medium section: 6.2 19. Sketch the graphs of y = e x and y = ex. Find the average value of the difference between e x and ex on the interval between x = 0 and x = 1.5. Round to 2 decimal places. Ans: 0.28 difficulty: medium section: 6.2 20. Is x 4 sin x an antiderivative of 4 x3 cos x ? Ans: no difficulty: easy section: 6.2
Page 9
Chapter 6: Constructing Antiderivatives
21. Find the total area bounded between the curve f ( x) =x3 − 3 x 2 + 2 x and the x-axis. Round to 3 decimal places. Ans: 0.500 difficulty: medium section: 6.2 t2/3 tons per 40 week, where t is the time in weeks since the factory commenced operations. After one year of operation, how many tons of pollutant has the factory dumped into the lake? Round to 2 decimal places. Ans: 10.87 difficulty: medium section: 6.2
22. A factory is dumping pollutants into a lake continuously at the rate of
t2/3 tons per 65 week, where t is the time in weeks since the factory commenced operations. Assume that natural processes can remove up to 0.138 tons of pollutant per week from the lake and that there was no pollution in the lake when the factory commenced operations one year ago. How many tons of pollutant have now accumulated in the lake? (Note: The amount of pollutant being dumped into the lake is never negative.) Round to 2 decimal places. Ans: 0.98 difficulty: hard section: 6.2
23. A factory is dumping pollutants into a lake continuously at the rate of
24. Find an antiderivative of π + x 6 + A)
πx +
B)
πx +
Ans: A
1 πx 6
.
x7 1 − +C 7 5πx5 x7 1 + +C 7 7 πx 7 difficulty: medium
25. Find an antiderivative of
7x −
3 (7 x)3/ 2 − 6 x −1/ 2 + C 14 2 B) (7 x)3/ 2 + 12 x −1/ 2 + C 21 Ans: B difficulty: medium
7 x7 + 7 +C 7 πx
C)
πx +
D)
7 π2 x 7 + − 7 +C 2 7 πx
section: 6.2 6 x x
.
A)
C) D) section: 6.2
Page 10
3 (7 x)3/ 2 − 12 x −5/ 2 + C 14 2 5 (7 x)3/ 2 + x −5/ 2 + C 21 12
Chapter 6: Constructing Antiderivatives
26. Find an antiderivative of cos(7θ ) . A) 7 sin(7θ ) + C Ans: C
B) −7 sin(7θ ) + C
difficulty: easy
C)
1 sin(7θ ) + C 7
1 D) − sin(7θ ) + C 7
section: 6.2
27. Find an antiderivative of et + e7 . et +1 7 et B) C) et + C +e t +C + e7t + C t +1 t Ans: D difficulty: medium section: 6.2
A)
28. Find an antiderivative of 5ln x +
B)
5ln x + x ln 5 + C
Ans: A
5 x + . x 5
x2 +C 10
A)
D) et + e7t + C
difficulty: medium
C) D)
x2 +C 10 x 10 ln x + +C 5 x2 10
+ 2
section: 6.2
dV = 8t + 1 , where dt V is the volume of the ice in cubic feet, and t is the time in minutes. Use a definite integral to find how many cubic feet of ice have melted in the first 2 minutes. Ans: 18 difficulty: medium section: 6.2
29. Suppose the rate at which ice in a skating pond is melting is given by
x) 5sin x + x + 7 . Find the total area bounded by F(x), x = 0, x = π and y = 30. Suppose F (= 0. Round to 2 decimal places. Ans: 36.93 difficulty: medium section: 6.2 31. Find the exact area between the graphs of = y x3 + 9 and y =− x + 9 for 0 ≤ x ≤ 5. Ans: 168.75 difficulty: medium section: 6.2
Page 11
Chapter 6: Constructing Antiderivatives
32. Find the general antiderivative of P ( x) = A)
2 ln x + C
1 ln x + C 2 −2 −2 C) x +C 2 Ans: B difficulty: medium
1 . 2x D)
−(2 x) −2 + C
E)
ln 2x + C
B)
section: 6.2
33. Find the area above y = 3 and below f ( x)= 9 − x 2 . A) 9.7980 B) 29.3939 C) 19.5959 D) –19.5959 Ans: C difficulty: medium section: 6.2
E) 2.2020
x2 − 2 x7 dx x3 2 Ans: ln x − x 5 + C 5 difficulty: medium section: 6.2
34. Determine: ∫
35. A car is going 64 feet per second and the driver puts on the brakes, bringing the car to a stop in 4 seconds. Assume the deceleration of the car is constant while the brakes are on. The acceleration (really deceleration) of the car is _____ ft/sec. Ans: –16 difficulty: medium section: 6.3 36. A car is going 56 feet per second and the driver puts on the brakes, bringing the car to a stop in 4 seconds. Assume the deceleration of the car is constant while the brakes are on. How many feet does the car travel from the time the brakes are applied until it stops? Ans: 112 difficulty: hard section: 6.3 37. A car is going 85 feet per second and the driver puts on the brakes, bringing the car to a stop in 5 seconds. Assume the deceleration of the car is constant while the brakes are on. Suppose a second car is traveling the same speed and the brakes are twice as strong (can stop the car twice as fast) as those in the first car. How far does the second car travel before it stops? A) Half as far as the first car. D) Four times as far as the first car. B) Twice as far as the first car. E) One-fourth as far as the first car. C) The same distance as the first car. Ans: A difficulty: hard section: 6.3
Page 12
Chapter 6: Constructing Antiderivatives
38. A ball is dropped from a window 140 feet above the ground. Assume that its acceleration is a(t) = -32 ft/sec2 for t ≥ 0. Find the velocity of the ball as a function of time t. (All answers are in ft/sec.) A) −16t 2 +140 B) −32t C) −32 + 140 D) −16t 2 Ans: B difficulty: medium section: 6.3 39. A ball is dropped from a window 80 feet above the ground. Assume that its acceleration is a(t) = -32 ft/sec2 for t ≥ 0. After how many seconds does it hit the ground? Round to 2 decimal places. Ans: 2.24 difficulty: medium section: 6.3 40. On planet Janet the gravitational constant g is –11 feet per second per second: that is, for every second an object falls it picks up an extra 11 feet per second of velocity downward. A ball is thrown upward at time t = 0 at 44 feet per second. After how many seconds does the ball reach the peak of its flight? Ans: 4 difficulty: medium section: 6.3 41. On planet Janet the gravitational constant g is –10 feet per second per second: that is, for every second an object falls it picks up an extra 10 feet per second of velocity downward. A ball is thrown upward (from height zero) at time t = 0 at 40 feet per second. What is the peak height of the ball? Ans: 80 feet difficulty: medium section: 6.3 42. On planet Janet the gravitational constant g is –10 feet per second per second: that is, for every second an object falls it picks up an extra 10 feet per second of velocity downward. A ball is thrown upward at time t = 0 at 25 feet per second. On planet Nanette, g is one-fourth as great as on Janet. What is the peak height of the ball on planet Nanette? Ans: 125 feet difficulty: medium section: 6.3 43. A ball is thrown vertically upwards from the top of a 256-foot cliff with initial velocity of 96 feet per second. Find its maximum height (in ft). Ans: 400 ft difficulty: medium section: 6.3
Page 13
Chapter 6: Constructing Antiderivatives
44. Assuming the 440 feet is accurate and you neglect air resistance, determine the accuracy of the following paragraph: MY JOURNEY BENEATH THE EARTH Condensed from “A Wolverine is Eating My Leg” Tim Cahill: I am in Ellison's Cave, about to rappel down Incredible Pit, the second-deepest cave pit in the continental United States. The drop is 440 feet, about what you'd experience from the top of a 40-story building. If you took the shaft in a free fall, you'd accelerate to more than 100 miles an hour and then--about five seconds into the experience--you'd decelerate to zero. And die. A) The time (about 5 seconds) is fairly accurate, but the speed (more than 100 mph) is not. B) The speed (more than 100 mph) is fairly accurate, but the time (about 5 seconds) is not. C) Both of the numbers are fairly accurate. D) Neither of the numbers are fairly accurate Ans: C difficulty: medium section: 6.3 45. A function g is known to be linear on the interval from −∞ to 2 (inclusive) and also linear on the interval from 2 to ∞ (again inclusive). Furthermore, g(1) = 2, g(2) = 0, g(4) = 8. Another function f satisfies f (0) = 0 and f ′ = g. What is f (3) ? Ans: 6 difficulty: hard section: 6.3 46. The police observe that the skid marks made by a stopping car are 240 ft long. Assuming the car decelerated at a constant rate of 21 ft/ sec2, skidding all the way, how fast was the car going when the brakes were applied? Round to 2 decimal places. Ans: 100.40 ft/sec difficulty: hard section: 6.3 47. Find the general solution of the differential equation Ans: −4 x 2 + 3 x + C difficulty: medium
section: 6.3
48. Find the solution of the differential equation Ans: −3 x 2 + 4 x + 4 difficulty: medium
dy = −8 x + 3 . dx
section: 6.3
Page 14
dy = −6 x + 4 satisfying y (1) = 5 . dx
Chapter 6: Constructing Antiderivatives
dK = 2 − cos 5t when K (0) = –12 . dt sin 5t C) − 12 2t + 5 sin 5t D) + 12 2t + 5 section: 6.3
49. Find the solution of the initial value problems
sin 5t − 12 5 sin 5t B) 2t − + 12 5 Ans: A difficulty: medium A)
2t −
dP = −3e −t when P(0) = 15 . dt −t C) −3e + 12 D) −3e −t + 15 section: 6.3
50. Find the solution of the initial value problem A) 3e −t + 15 B) 3e −t + 12 Ans: B difficulty: medium
(1 − e−7 x ) sin 7 x is the solution to the initial value problem dy = 7 (1 − e−7 x ) ( cos 7 x ) + 7e−7 x sin 7 x , with y (0) = 2 ? dx
51. Is y=
Ans: no difficulty: medium
section: 6.3
dV t 2 = + e − t in cubic inches per minute. Determine 52. A snowball is melting at a rate dt 7 the total amount the snowball has melted in the first 9 minutes. Specify units. Ans: 35.71 cubic inches difficulty: medium section: 6.3
−32t + 18 53. A dog's bone is tossed in a yard traveling with a vertical velocity of v(t ) = feet/sec. Determine how long it takes for the bone to reach its maximum height. A) 32 B) 18 C) 1.778 D) 0.563 E) 576 Ans: D difficulty: medium section: 6.3 −32t + 2 54. A dog's bone is tossed in a yard traveling with a vertical velocity of v(t ) = feet/sec. Determine the maximum height reached by the bone. Ans: 0.06 feet difficulty: medium section: 6.3 55. For -1 ≤ x ≤ 1, define F = ( x)
x
∫−1 1 − t dt . What does F(1) represent geometrically? 2
A) The area of a quarter circle of radius 1. B) The area of a circle of radius 1. C) The area of a semicircle of radius 1. D) None of the above Ans: C difficulty: medium section: 6.4
Page 15
Chapter 6: Constructing Antiderivatives
56. For –2 ≤ x ≤ 2 , define = F ( x) 2 decimal places. Ans: 3.14 difficulty: medium
x
∫−2 4 − t dt . 2
What is the value of F (0) ? Round to
section: 6.4
F ( x) 57. For –4 ≤ x ≤ 4 , define=
x
∫−4 16 − t dt . 2
Find F′(x).
B) − 2x C) − 16 − x 2 D) difficulty: medium section: 6.4
2x A) Ans: D
16 − x 2
58. Evaluate ∫ ( x 2 + e3 x )( x3 + e3 x ) 2 / 3dx . Some of the coefficients may not be reduced.
3 3 3 x 5/ 3 (x + e ) + C 15 3 3 3 x 5/ 3 B) (x + e ) + C 5 Ans: A difficulty: medium A)
59. Evaluate ∫
A)
6e −2 w 1 − 5e −2 w
12e −2 w 1t − 10e
B)
−2 w
6e −2 w −2 w
C)
– sin(ln x) +C x
D)
section: 6.4
dw .
+C
C)
3 ln 1 − 5e −2 w + C 5
+C
D)
1 ln 1 − 5e−2 w + C 10
2t − 5e Ans: C difficulty: medium
61. Evaluate
section: 6.4
B) sin(ln x) + C
difficulty: medium
60. Evaluate ∫
D)
9 3 3 x 4 3 x 5/ 3 ( x + e )( x + e ) + C 5 3 3 3 x 4 3 x 5/ 3 ( x + e )( x + e ) + C 60
cos(ln x) dx . x
A) – sin(ln x) + C Ans: B
C)
section: 6.4
( ) ( t ) dt .
d x log11 t 21 sin ∫ dx e
( ) ( x)
Ans: log11 x 21 sin difficulty: medium
section: 6.4
Page 16
2sin(ln x) x2
+C
Chapter 6: Constructing Antiderivatives
62. Below are the graphs of (i) f ( x) , (ii)
f '( x) , and (iii)
x
∫0 f (t ) dt (not necessarily in that order).
Which one is the graph of (iii)? A) The first one. B) The third one. C) The second one. Ans: C difficulty: medium section: 6.4
Page 17
Chapter 6: Constructing Antiderivatives
63. Below are the graphs of (i) f ( x) ,
and (ii)
x
∫0 f (t ) dt (not necessarily in that order).
Which one is the graph of (ii)? A) The first one. B) The second one. Ans: B difficulty: medium section: 6.4 x
64. The function f(t) is graphed below and we define F ( x) = ∫ f (t ) dt . 0
Is F ( x) concave down for x =1/2? Ans: no difficulty: medium section: 6.4
Page 18
Chapter 6: Constructing Antiderivatives
x
x
0
0
65. Let F ( x) = ∫ sin t dt and G ( x) = ∫ sin 2 t dt . If A, B, C, D, J, and K represent positive areas as shown in the graph, what combination of these areas represent G (π / 2) on the graph?
Ans: B difficulty: medium
section: 6.4
x
x
0
0
66. Let F ( x) = ∫ sin t dt and G ( x) = ∫ sin 2 t dt . Use the graph to arrange the following in ascending order by entering a "1" next to the smallest, a "2" by the next smallest, and so forth. π A. G 2
Part A: 2 Part B: 4 Part C: 3 Part D: 1 difficulty: medium
B. F (π)
C. G (π)
section: 6.4
Page 19
D. F (2π)
Chapter 6: Constructing Antiderivatives
67. Find A) B)
( ( )
( ) ( ) sin ( x 4 ) − cos ( x 4 ) cos x 4 + sin x 4
Ans: A 68. Find
( ))
d x cos t 4 + sin t 4 dt . ∫ 0 dx
difficulty: easy
C) D)
( ) ( ) x5 sin ( x 4 ) − cos ( x 4 ) ) ( 5
sin 4 x3 − cos 4 x3
section: 6.4
d 1 (1 + t )3 dt . dx ∫x
A) (1 + x)3 Ans: B
B) −(1 + x)3
C)
difficulty: medium
1 (1 + x) 4 3 section: 6.4
D)
1 16 (1 + x) 4 − 3 3
8. 69. Write an expression for the function, f(x), with f= '( x) sin 2 x + sin x and f (π / 3) = A) B)
x
f ( x) = 8 + ∫ (sin 2 t + sin t ) dt 0
f ( x= )
Ans: C
x
∫π / 3
(sin 2 t + sin t ) dt − 8
difficulty: medium
x
C)
f ( x) = 8 + ∫ (sin 2 t + sin t ) dt
D)
f ( x) =8 −
π/3
x π + ∫ (sin 2 t + sin t ) dt 3 0
section: 6.4
70. Find the value of G(π/2) where G ′(x) = 2 sin x cos x and G(0) = 1. Ans: 2 difficulty: medium section: 6.4 x
71. For H ( x) = ∫ 6t 2 dt , find H (1) . 0
Ans: 2 difficulty: medium
section: 6.4
d x ln(t 3 )dt . ∫ 2 dx Ans: ln( x 3 ) difficulty: medium section: 6.4
72. Evaluate
d x3 sin t dt . dx ∫1 Ans: 3 x 2 sin( x3 ) difficulty: medium section: 6.4
73. Evaluate
Page 20
Chapter 6: Constructing Antiderivatives
d 0 t dt . dx ∫3x Ans: −(ln 3)9 x difficulty: medium
74. Evaluate
section: 6.4
4
d 75. Evaluate ∫ tan(t ) dt . dx sin(x) Ans: - tan(sin(x)) cos(x) difficulty: medium section: 6.4 76. At time t = 0, a bowling ball rolls off a 250-meter ledge with velocity 30 meters/sec downward. Express its height, h(t), in meters above the ground as a function of time, t, in seconds. A) C) h(t ) = −4.9t 2 − 30t + 250 h(t )= 4.9t 2 − 30t + 250 B) −4.9t 2 − 30t − 250 h(t ) = Ans: A difficulty: easy
D) section: 6.3
h(t )= 4.9t 2 − 30t − 250
77. A boulder is dropped from a 150-foot cliff. How fast is it going when it hits the ground? Round to 2 decimal places. Ans: 97.98 ft/sec difficulty: medium section: 6.3 78. A boulder is dropped from a cliff. A second boulder is dropped from a cliff that is half as high. How does the speed of the second boulder upon impact compare with that of the first? A) Ths speed of the second is approximately 0.5 times the speed of the first. B) Ths speed of the second is approximately 0.7 times the speed of the first. C) Ths speed of the second is approximately 0.25 times the speed of the first. D) Ths speed of the second is approximately 2 times the speed of the first. E) Ths speed of the second is approximately the same as the speed of the first. Ans: B difficulty: medium section: 6.3 79. On the moon the acceleration due to gravity is 5 feet/ sec2. A brick is dropped from the top of a tower on the moon and hits the ground in 16 seconds. How many feet high is the tower? Ans: 640 difficulty: medium section: 6.3
Page 21
Chapter 6: Constructing Antiderivatives
80. Suppose the acceleration due to gravity on Planet A is twice that of Planet B. A brick dropped from the top of a tower on Planet A takes 20 seconds to hit the ground. A brick dropped from the top of a tower on Planet B also takes 20 seconds to hit the ground. How does the height of the tower on Planet A compare with the height of the tower on Planet B? A) The tower on Planet A is 4 times the height of the one on Planet B. B) The tower on Planet A is twice the height of the one on Planet B. C) The tower on Planet A is half the height of the one on Planet B. D) The tower on Planet A is the same as the height of the one on Planet B. Ans: B difficulty: medium section: 6.3 81. A textbook is accidentally tossed from the top of a 32 meter bell tower with an initial velocity of 2 m . s A) Determine the equation for the velocity of the textbook as a function of time.______________________ B) Determine the equation for the height of the textbook as a function of time.________________________ C) Determine the height of the textbook at 0.9 seconds.____________________________ −9.8t + 2 Part A: v(t ) = Part B: h(t ) = −4.9t 2 + 2t + 32 Part C: 29.831 meters difficulty: medium section: 6.3
( x) sin x − cos x . 82. Find an antiderivative F(x) with F ′(x) = f (x) and F(0) = 3 when f= Ans: − cos x − sin x + 4 difficulty: easy section: 6 review 83. Use the Fundamental Theorem of Calculus to evaluate ∫ Ans: 68 difficulty: easy
2 −2
(9 x 2 − 6 x + 5) dx .
section: 6 review
84. Find the area of the region between y1 = −2( x − 4) 2 + 32 and y2 = x , accurate to 2 decimal places. Ans: 140.63 difficulty: medium section: 6 review
Page 22
Chapter 6: Constructing Antiderivatives
85. The area between y = x5/ 2 , the x-axis, and x = b is approximately 79.9. Find the value of b using the Fundamental Theorem. Round to 1 decimal place. Ans: 5.0 difficulty: medium section: 6 review 86. A boat has constant deceleration. It was initially moving at 80 mph and stopped in a distance of 300 feet. The rate of deceleration is _____ ft/ sec2. (Note: 1 mph = 22/15 ft/ sec.) Round to 2 decimal places. Ans: 22.95 difficulty: hard section: 6 review
cos 4 x dy 2 +C . = + sin 4 x is 2 ln x + 4 dx x section: 6 review
87. The general solution of the differential equation Ans: False
difficulty: medium
88. An object falls from the top of an 600 foot building. How fast is it going when it hits the ground? Round to 2 decimal places. Ans: –195.96 ft/sec difficulty: medium section: 6 review
g ( x) 7 sin 3 x − 8cos 9 x is given by 89. The general antiderivative for= −21cos 3 x − 72sin 9 x + C . Ans: False difficulty: medium section: 6 review 90. Is the antiderivative for h( x)=
1 ln x − cos ax + C ? a Ans: no difficulty: medium
a + sin ax , where a is a constant, given by x
section: 6 review
91. Is the antiderivative for j ( x) = ae x + a cos x + ax 7 , where a is a constant, given by ae x + a sin x +
ax8 +C ? 8
Ans: yes difficulty: medium
section: 6 review
92. Is the antiderivative for k ( x)= Ans: yes difficulty: medium
6 + 6e6 x given by 6 ln x + e6 x + C ? x
section: 6 review
Page 23
Chapter 6: Constructing Antiderivatives
93. If the graph below is of f ( x) where F '( x) = f ( x) , which of the equations below could represent F ( x) ?
A) 2 B) 2x C) 2 x + 3 Ans: D difficulty: medium
D) x 2 + 3 E) None of the above. section: 6 review
94. The figure shown represents the velocity of a particle over time. True or False: The area under the curve represents the particle traveling south.
Ans: False
difficulty: medium
section: 6 review
Page 24
Chapter 6: Constructing Antiderivatives
95. A particle is traveling along a straight line. (Positive velocities indicate movement to the right.) Indicate on the graph when the particle has traveled 3 units to the which.
Ans: various difficulty: medium
section: 6 review
96. A particle is traveling along a straight line. (Positive velocities indicate movement to the right.) Indicate on the graph where the particle has had a change in position of 3 units.
Ans: various difficulty: medium
section: 6 review
Page 25
Chapter 6: Constructing Antiderivatives
97. A particle is traveling along a straight line. (Positive velocities indicate movement to the right.) What is the total distance traveled (both right and left) when t = 5?
Ans: 7 units total. difficulty: medium
section: 6 review
Page 26
1. Evaluate ∫ 6x 7 dx .
3 8 B) x8 + C x +C 4 Ans: A difficulty: easy
6 8 x +C 7 section: 7.1
A)
C)
x2 − x + 5 2. Evaluate ∫ dx . x x3 x 2 A) ln x − + 5x + C 3 2 x2 B) − x + 5ln x + C 2 Ans: B difficulty: medium
D)
7 8 x +C 8
C)
2 x 10 + +C 3 x
D)
x2 10 − x+ 2 +C 2 x
section: 7.1
3. Evaluate ∫ 8e x cos(8e x ) dx . A)
sin(8e x ) e
x
+C
1 sin(8e x ) + C 9 Ans: C difficulty: medium
B)
4. Find an antiderivative of x 2 − A)
sin(8e x ) + C
D)
8e x +1 sin(8e x ) + C x +1
section: 7.1
6 8 + . x x3
x3 12 32 − − +C 3 x2 x4
x3 6 8 − + +C 3 x x2 Ans: D difficulty: easy
B)
(
x3 8 − 6 ln x + 2 + C 3 x
D)
4 x3 − 6 ln x − 2 + C 3 x
)( x − sin(3x + 9) ) .
x3 1 + cos(3 x + 9) + C 3 3
x3 1 − cos(3 x + 9) + C 3 3 Ans: A difficulty: medium
C)
section: 7.1
5. Find an antiderivative of x + sin(3 x + 9) A)
C)
B)
C)
1 x3 cos(3 x + 9) + C + 3 3x + 9
D)
x3 1 − cos(3 x + 9) + C 3 3x + 9
section: 7.1
Page 1
Chapter 7: Integration
R
sin x
∫ 9 + x 4 dx .
6. Compute
−R
A)
2 cos R
3cos(9 + R 4 )
B)
9R + R R Ans: D difficulty: easy 5
R
1
7. Compute ∫
0 (3 + x )
2
3
1
C)
D) 0
3(9 + R 4 ) section: 7.1
dx .
1 1 1 1 B) C) − 3+ R 3 3+ R 3(3 + R)3 Ans: B difficulty: medium section: 7.1 A)
( ) 4 3 2 y + 2) + C ( 8y
D)
3 R
3
−
3 (3 + R)3
3
8. Find ∫ 3 y y 2 + 2 dy A)
(
)
4 3y2 2 y +2 +C 8 Ans: C difficulty: medium
B)
5x
9. Find ∫ A)
4 − x2 10
−
(
3 4− x B)
(
10
3 4− x Ans: D
)
(
)
4 3 2 y +2 +C 8
D)
4 3 2 y +2 +C 4
section: 7.1
dx
)
+C
C)
5 4 − x2 + C
+C
D)
−5 4 − x 2 + C
2 3/ 2
)
(
C)
2 3/ 2
difficulty: medium
section: 7.1
10. Integrate ∫ cos 2 3 x sin 3 x dx . cos3 3 x cos3 3 x cos3 3 x B) − C) +C +C +C 9 3 9 Ans: B difficulty: medium section: 7.1
A)
Page 2
D) −
cos3 3 x +C 27
Chapter 7: Integration
11. Integrate ∫ (tan 7 x + cos 7 x) dx . ln cos 7 x sin 7 x − +C 7 7 ln cos 7 x sin 7 x B) − + +C 7 7 Ans: B difficulty: medium A)
C) D)
cot 7 x sin 7 x − +C 7 7 cot 7 x sin 7 x − + +C 7 7
section: 7.1 π
12. Use the Fundamental Theorem to evaluate the definite integral ∫ cos 2 x sin x dx . 0
Reduce fractions and leave them in the form "A/B". π 2 Ans: ∫ cos 2 x sin x dx = . 0 3 difficulty: medium section: 7.1 13. Calculate the area between the curve y = sin 2 θ cos θ and the x-axis between θ = 0 and θ = π / 6 . Round your answer to 2 decimal places. Ans: 0.04 difficulty: medium section: 7.1 14. Find the area between f ( x) = 4 xe − x and g(x) = x for x ≥ 0. Round to 3 decimal places. Ans: 0.807 difficulty: hard section: 7.1 2
7
1
0
0
15. Suppose ∫ f (t ) dt = a , where a is a constant. Calculate ∫ f (7t ) dt .
a D) a − 7 A) a B) 7a C) 7 Ans: C difficulty: medium section: 7.1 2
2
16. Suppose ∫ f (t ) = a , where a is a constant. Calculate ∫ 5 f (2 − t ) dt . A) −10a Ans: D
0
B) 10a C) −5a D) 5a difficulty: medium section: 7.1
0
17. Fuel pressure in the fuel tanks of the space shuttle is decreasing at a rate of r (t ) = 16e −0.1t psi per second at time t in seconds. At what rate, in psi/sec, is pressure decreasing at 10 seconds? Round to 2 decimal places. Ans: 5.89 psi/sec difficulty: easy section: 7.1
Page 3
Chapter 7: Integration
18. Fuel pressure in the fuel tanks of the space shuttle is decreasing at a rate of r (t ) = 18e−0.1t psi per second at time t in seconds. By how many total psi has the pressure decreased during the first minute? Round to 2 decimal places. Ans: 179.55 difficulty: medium section: 7.1 19. Find ∫ e −t / 4 sin4t dt . 16 −t / 4 1 e sin 4t + 4 cos 4t + C 257 4 16 −t / 4 1 B) e sin 4t − 4 cos 4t + C 257 4 1 1 C) − e −t / 4 sin 4t + 4 cos 4t + C 32 4 1 −t / 4 1 D) e sin 4t − 4 cos 4t + C 32 4 Ans: A difficulty: medium section: 7.2
A)
−
20. Find ∫ ( ln x ) dx . Hint: Integrate by parts. 2
A)
x ( ln x ) + 2 x ln x − 2 x + C 2
x ( ln x ) − 2 x ln x + 2 x + C B) Ans: B difficulty: medium 2
C)
x ( ln x ) − 2 x ln x + C 2
x ( ln x ) + 2 x ln x + C D) section: 7.2 2
21. Calculate ∫ sec 2 θ dθ 3sin θ
1 C) tan θ + C +C cos θ cos3 θ Ans: C difficulty: easy section: 7.2
A)
B) −
+C
D) tan 2 θ + C
22. Calculate ∫ ze5 z +8 dz . 2
z 5 z 2 +8 z 5 z 2 +8 1 5 z 2 +8 B) C) e e +C +C e +C 5 5 10 Ans: D difficulty: medium section: 7.2 A)
Page 4
D)
1 5 z 2 +8 +C e 10
Chapter 7: Integration
23. Calculate ∫ y sec 2 y dy . A) B)
y tan y + ln cos y + C y tan y − ln cos y + C
Ans: A
difficulty: hard
24. Calculate ∫
C)
y2 − tan y + C 2
D)
y2 tan y + C 2
section: 7.2
dt . 3+ t
2 3 + t + 6 ln 3 + t + C
(
)
(
)
C)
B)
2 3 + t − 6 ln 3 + t + C
(
)
(
)
D)
Ans: B
difficulty: medium
25. Calculate ∫ A)
dx (b + ax)8
9 a (b + ax) 7a
2 −
(
) +C
t ln 3 + t 2
) +C
section: 7.2
, where a and b are constants.
+C 9
+C (b + ax)7 Ans: C difficulty: medium B)
(
t ln 3 + t
A)
C) D)
−
1
+C 7 a (b + ax)7 1 +C 7 a (b + ax)7
section: 7.2
26. For f ( x) = x sin(4 x) , find a function F ( x) such that F ′( x) = f ( x) and F (0) = 0 . 1 1 1 1 A) C) x cos(4 x) − sin(4 x) x cos(4 x) − sin(4 x) 4 4 4 16 1 1 1 1 B) D) − x cos(4 x) + sin(4 x) − x cos(4 x) + sin(4 x) 4 16 4 4 Ans: D difficulty: medium section: 7.2 27. For f ( x) = x 2e10 x , find a function F ( x) such that F ′( x) = f ( x) and F (0) = 0 . 1 1 1 1 1 1 A) C) e10 x x 2 − x + e10 x x 2 − x − 50 500 500 50 10 10 1 1 1 e10 x x 2 − x + 50 500 10 Ans: A difficulty: medium B)
D) section: 7.2
Page 5
x3 e10 x 30
Chapter 7: Integration
(
) , find a function F ( x) such that F ′( x) = f ( x) and F (0) = 0 . 11 11 1 1 C) 8 + x3 ) − 811 8 + x3 ) ( ( 99 33
28. For f= ( x) x 2 8 + x3 A)
(
10
)
11 1 8 + x3 − 811 33 Ans: B difficulty: medium
B)
29. Integrate ∫
x4 + 5 x3
D)
(
)
11 x 8 + x3 99
section: 7.2
dx .
x2 x2 5 5 2 x5 + 5 x B) C) − +C ⋅ +C + +C 2 2 x2 5 2 2 x2 x2 Ans: A difficulty: medium section: 7.2
A)
30. Integrate ∫
x2
5
0 x3 + 1
D)
x5 + 5 x x2
+C
dx . Give an exact answer and one rounded to 3 decimal places.
x2 1 dx ln126 ≈ 1.612. ∫0 x3= 3 +1 difficulty: medium section: 7.2 5
Ans:
31. Integrate ∫ A)
( ln x )
Ans: B
4 ln x
x
dx .
4 +C B) ( ln x )5/ 4 + C C) 5 difficulty: easy section: 7.2
5/ 4
2 ( ln x )
5/ 4
x
2
+C
5
D)
ln x +C 5
32. Integrate ∫ sin(3 x)ecos(3 x ) dx . A) −ecos(3 x ) + C Ans: C
1 C) − ecos(3 x ) + C 3 section: 7.2
B) ecos(3 x ) + C
difficulty: medium
Page 6
D)
1 cos(3 x ) e +C 3
Chapter 7: Integration
1 1 33. Find ∫ − dx . x + 1 ( x + 1) 2 3 ln x + 1 − +C A) ( x + 1)3 3 ln x + 1 + +C B) ( x + 1)3 Ans: D difficulty: easy
C)
ln x + 1 −
1 +C x +1
D)
ln x + 1 +
1 +C x +1
section: 7.2
34. Find ∫ x cos 2 x dx .
1 1 x sin 2 x + cos 2 x + C 2 4 1 1 B) x sin 2 x − cos 2 x + C 2 2 Ans: A difficulty: medium
A)
x2 sin 2 x + C 8
C)
−
D)
x2 sin 2 x + C 4
section: 7.2
1 1 35. True or False: ∫ te at dt = te at − 2 e at + C , where a is a constant. a a Ans: True difficulty: medium section: 7.2 36. True or False: ∫ θ 2 cos ( aθ ) dθ = Ans: False
θ2 a
difficulty: hard
sin ( aθ ) +
2θ
a2 section: 7.2
cos ( aθ ) + C , where a is a constant.
= x ln x − x + C . 37. True or False: ∫ ln x dx Ans: True
difficulty: medium
section: 7.2
38. Use the table of antiderivatives to determine if the following statement is true. dx arctan( x − 5) + C ∫ x 2 − 10 x += 26 Ans: True difficulty: medium section: 7.3 39. Use the table of antiderivatives to determine if the following statement is true. dx = ln x + x 2 − 2 x + 2 + C ∫ 2 x − 2x + 2 Ans: False difficulty: medium section: 7.3
Page 7
Chapter 7: Integration
40. Use the table of antiderivatives to determine if the following statement is true. x2
x
x − 5 arctan +C ∫ x 2 + 5 dx = 5
Ans: True
difficulty: hard
section: 7.3
41. Use the table of antiderivatives to determine if the following statement is true. t 7 1 dt arcsin = + C ∫ 2 7 2 4 − 7t Ans: False difficulty: hard section: 7.3 42. Use the table of antiderivatives to determine if the following statement is true. 1 7x 7x = x dx e ( 7 cos8 x + 8sin 8 x ) + C ∫ e cos8 113 Ans: True difficulty: medium section: 7.3
1 43. True or False: ∫ cos(6θ ) dθ = − sin(6θ ) + C . 6 Ans: False difficulty: easy section: 7.3
( )
3e3 x 44. True or False: ∫= dx tan −1 e3 x + C . 6x 1+ e Ans: True difficulty: medium section: 7.3
t +5 45. True or False: ∫ 2 dt= ln t 2 + 10t + 75 + C . t + 10t + 75 Ans: False difficulty: medium section: 7.3 46. True or False: ∫ Ans: True
2 dx
= ln x + 1 − ln x + 3 + C . x + 4x + 3 difficulty: medium section: 7.3 2
47. True or False: ∫ sin Ans: False
sin y − y cos y + C . ( y ) dy =
difficulty: hard
section: 7.3
48. Suppose that as a storm dies down, its rainfall rate (in inches/hour) is given by 1 for 0 ≤ t ≤ 2, where t is the number of hours since the point of heaviest y= 0.09 + t 2 rainfall. What is the average rainfall rate over these two hours? Round your answer to 3 decimal places. Ans: The average rainfall rate for 0 ≤ t ≤ 2 is 2.370 inches per hour. difficulty: medium section: 7.3
Page 8
Chapter 7: Integration
−1 −4 −3 − 1 sin x cos x + sin −5 x dx . ∫ −3 −3 difficulty: medium section: 7.3
−3 49. True or False: = ∫ sin x dx
Ans: True
50. Consider the semicircle of radius 4 pictured below. Which of the following could represent the area of the semicircle? Select all that apply.
4
A)
∫−4 16 − x dx
D)
1 x x 16 − x 2 + 16 arcsin 2 4 −4
B)
1 ⋅ π ⋅ 42 2
E)
2 x 16 − x 2 −4 4
C)
4∫
4
2
1
−1
16 + x 2 dx
Ans: A, B, D 51. Find ∫
4
difficulty: medium
section: 7.4
1
dx . x + 3 x − 10 1 A) − ( ln x + 5 + ln x − 2 ) + C 10 1 B) ( ln x + 5 + ln x − 2 ) + C 7 Ans: D difficulty: medium 2
C) D)
1 ( ln x + 5 − ln x − 2 ) + C 10 1 − ( ln x + 5 − ln x − 2 ) + C 7
section: 7.4
52. Find ∫ cos3 β d β .
1 sin β − sin 3 β + C 3 1 2 B) sin β + sin 3 β + C 3 3 Ans: A difficulty: medium A)
C) D) section: 7.4
Page 9
2 sin β − sin 3 β + C 3 1 sin β + sin 3 β + C 3
Chapter 7: Integration
4x +1
53. Find ∫
dx . 4 x2 + 4 x 3 1 A) ln x + ln x + 1 + C 4 4 1 3 B) ln x + ln x + 1 + C 4 4 Ans: B difficulty: medium
3 1 ln x + ln x + 1 + C 4 4 1 ln x + ln x + 1 + C 4
C) D) section: 7.4
54. Which of the following gives the area of the circle x 2 + y 2 = 1? 1
A)
∫–1 1 − x dx
B)
4∫
2
1
–1
Ans: C
1 − x 2 dx difficulty: medium
C)
4∫
D)
2∫
1
0 1
0
1 − x 2 dx
1 − x 2 dx
section: 7.4
55. Find the area of the region bounded by y = 0 and y =
4t + 5 2t + 3t + 1 2
between t = 0 and t =
2. Round to 3 decimal places. Ans: 3.730 difficulty: medium section: 7.4 56. Find the area of the region bounded by y = decimal places. Ans: 0.464 difficulty: medium
1 x + 2x + 2 2
, x = 0, and x = 2. Round to 3
section: 7.4
57. The following are some of the values for a function known as the Gudermannian function, G(x). x G(x)
0 0
0.1 0.2 0.3 0.100 0.199 0.296
0.4 0.390
0.5 0.480 1
0.6 0.567
0.7 0.649
0.8 0.726
Use these values to approximate the value of ∫ G ( x) dx using the trapezoid rule. 0
Ans:
1
∫0 G( x) dx ≈ 0.4638.
difficulty: medium
section: 7.5
Page 10
0.9 0.798
1.0 0.866
Chapter 7: Integration
58. The following numbers are the left, right, trapezoidal, and midpoint approximations to 1
∫0 f ( x) dx , where f(x) is as shown. (Each uses the same number of subdivisions.) • • • •
0.36735 0.39896 0.36814 0.33575
Which one is the midpoint approximation? Ans: 0.36814 difficulty: medium section: 7.5
Page 11
Chapter 7: Integration
59. The following numbers are the left, right, trapezoidal, and midpoint approximations to 1
∫0 f ( x) dx , where f(x) is as shown. (Each uses the same number of subdivisions.) • • • •
0.36744 0.43401 0.38646 0.33574
1
If A < ∫ f ( x) dx < B , with B − A as small as possible, then A = ________ and B = 0
________. Part A: 0.36744 Part B: 0.38646 difficulty: medium 60. Compute ∫
3
5
2 ln x
Ans:
3
dx . Round to 3 decimal places.
5
∫2 ln x dx ≈ 5.592.
difficulty: medium 61.
Compute Ans:
section: 7.5
3
3
section: 7.5
∫0 x 16 + x dx . Round to 3 decimal places. 2
61
∫0 x 16 + x dx =3 ≈ 20.333. 2
difficulty: medium
section: 7.5
Page 12
Chapter 7: Integration
62. The table below shows the velocity v(t) of a falling object at various times (time t measured in seconds, velocity v(t) measured in meters per second). t v(t)
0 19
1 25
2 30
3 34
Due to air resistance, the object's acceleration is decreasing. What does this tell you about the shape of the graph of v(t)? A) It is concave down B) It is concave up C) Neither of the above Ans: A difficulty: medium section: 7.6 63. The table below shows the velocity v(t) of a falling object at various times (time t measured in seconds, velocity v(t) measured in meters per second). t v(t)
0 17
1 23
2 28
3 32
The distance the object fell in these three seconds lies within which interval? A) (68, 75.5) B) (75.5, 78) C) (78, 83) D) (83, 86.5) Ans: B difficulty: medium section: 7.6 3 3 x x − 1 dx . 1
∫
64. Find the exact value of
3 7/3 3 4/3 9 (2) − (2) + 7 4 28 3 7 / 3 3 4 / 3 33 B) (2) + (2) − 7 4 28 Ans: C difficulty: medium
C)
A)
D)
3 7/3 3 4/3 (2) + (2) 7 4 3 7/3 3 4/3 (2) − (2) 7 4
section: 7.5
1
65. You want to estimate ∫ cos(θ 2 ) dθ by finding values, A and B, such that 0
1
A < ∫ cos(θ 2 ) dθ < B , with B − A being as small as possible. Which method should you 0
use to find A? A) The trapezoid rule. B) The midpoint rule. Ans: A difficulty: medium
C) The left rule. D) The right rule. section: 7.5
Page 13
Chapter 7: Integration
66. Last Monday we hired a typist to work from 8am to 12 noon. His typing speed decreased between 8am and his 10am cup of coffee, and increased again afterwards, between 10am and noon. His instantaneous speed (measured in characters per second) was measured each hour and the results are given below: Time Speed
8am 6
9am 4
10am 1
11am 3
12noon 5
You want to estimate the total number of characters typed between 8am and 12 noon. Find an upper estimate using Reimann sums. Ans: 64,800 difficulty: medium section: 7.5 67. Consider the definite integral ∫
2/5
1
dx . Compute the integral using the 4 + 25 x 2 fundamental theorem of calculus and using the midpoint rule with n = 20. How far apart are your answers? A) Within 0.03 but not within 0.003 B) Within 0.003 but not within 0.0003 C) Within 0.0003 but not within 0.00003 D) Within 0.00003 but not within 0.000003 Ans: D difficulty: hard section: 7.5 0
68. Consider the definite integral ∫
π /4
−π / 4
sin 2 θ dθ . Compute the integral using the
fundamental theorem of calculus and using the trapeziod rule with n = 20. How far apart are your answers? A) Within 0.1 but not within 0.01 B) Within 0.0001 but not within 0.00001 C) Within 0.001 but not within 0.0001 D) Within 0.01 but not within 0.001 Ans: D difficulty: hard section: 7.5
Page 14
Chapter 7: Integration
69. Consider the ellipse pictured below:
The perimeter of the ellipse is given by the integral ∫
π /2
0
3 8 1 − sin 2 θ dθ . It turns out 4
3 that there is no elementary antiderivative for the function f (= θ ) 8 1 − sin 2 θ , and so 4 the integral must be evaluated numerically. A graph of the integrand f(θ) is shown below.
Calculate the right sum that approximates the definite integral with N = 4 equal divisions of the interval. Round to 4 decimal places. Ans: 8.9029 difficulty: medium section: 7.5 3 1 70. It is desired to evaluate the definite integral ∫ sin dx . As it turns out, there is no 1 x 1 antiderivative to the function f ( x) = sin , and so it will be necessary to evaluate the x integral numerically. A. Calculate the midpoint sum that approximates the definite integral with N = 4 equal divisions of the interval 1 ≤ x ≤ 3. Round to 4 decimal places. B. State whether this sum is an overestimate or an underestimate of the integral. Part A: 1.0219 Part B: underestimate difficulty: medium section: 7.5
Page 15
Chapter 7: Integration
71. A drug is being administered intravenously to a patient at a constant rate of 2 mg/hr. The following table shows the rate of change of the amount of the drug in the patient's body at half-hour intervals. Initially there is none of the drug in the patient's body. Time (hours) Rate (mg/hr)
0.0 2
0.5 1.09
1.0 0.75
1.5 0.57
2.0 0.46
2.5 0.39
3.0 0.33
3.5 0.29
4.0 0.26
Use the trapezoid rule to estimate of the total amount of the drug in the patient's body after four hours. Round to 2 decimal places. Ans: 2.51 mg difficulty: medium section: 7.5 72. Using two subdivisions, find the left approximation to ∫ (1 − e − x ) dx . Round to 4 1
0
decimal places. Ans: 0.1968 difficulty: medium
section: 7.5
73. What is shown in the following graph of ∫ (1 − e − x ) dx ? 1
0
A) The right approximation with n = 2. B) The midpoint approximation with n = 2. C) The trapezoid approximation with n = 2. D) The left approximation with n = 2. Ans: A difficulty: easy section: 7.5
Page 16
Chapter 7: Integration
74. Let S(t) be the number of daylight hours, in Cambridge, MA, on the tth day of the year. During spring (from the vernal equinox, t = 80, to the summer solstice, t = 173), the graph of S(t) is concave down. In the table below we list some values of S(t). t 80 111 142 173
S(t) 12 hours 12 minutes 13 hours 44 minutes 14 hours 50 minutes 15 hours 12 minutes
Find upper and lower bounds for the average length of the days in spring that differ by less than 10 minutes. Average these numbers, then round to the nearest 10 minutes. The average length of the days is _____ hours and _____ minutes. Part A: 14 Part B: 10 difficulty: hard section: 7.5 75. A. What is the trapezoid approximation (with n = 50 ) for ∫
0.3
1
−0.5 1 + x 4
decimal places. B. Is this an overestimate or an underestimate? Part A: 0.98789 Part B: underestimate difficulty: medium section: 7.5
Page 17
dx ? Round to 5
Chapter 7: Integration
76. Below is the graph of y = arctan x . Which of the following are true for any number of subdivisions? Select all that apply.
14
A)
LEFT(N) < ∫
B)
RIGHT(N) < ∫
arctan( x) dx < LEFT(N)
C)
TRAP(N) < ∫
arctan( x) dx < MID(N)
D)
MID(N) < ∫
E)
TRAP(N) < ∫ arctan( x) dx < MID(N)
F)
MID(N) < ∫ arctan( x) dx < TRAP(N)
arctan( x) dx < RIGHT(N)
−16 14
−16 0
−16
0
arctan( x) dx < TRAP(N)
−16 14
Ans: A, D, E
0 14
0
difficulty: medium
section: 7.5
Page 18
Chapter 7: Integration
77. Below is the graph of y = arctan x .
We have the following data: 16
For ∫ arctan( x) dx , MID(50) = -12.40537 and TRAP(50) = -12.40041. 0 0
For ∫
−10
arctan( x) dx , MID(50) = 21.36379 and TRAP(50) = 21.35097.
Using this data alone, what is the best upper bound you can give for ∫
16
−10
Ans: 8.96338 difficulty: hard
arctan( x) dx ?
section: 7.5
18
78. Evaluate ∫ arctan( x) dx “symbolically” (plug in the limits but don't evaluate). Hint: −7
Integrate by parts using A)
d 1 . arctan x = dx 1 + x2
18arctan18 − 7 arctan(−7) + ln(1 + 182 ) + ln(1 + 7 2 )
18arctan18 + 7 arctan( −7) − ln(1 + 182 ) − ln(1 + 7 2 ) 1 1 C) 18arctan18 − 7 arctan( −7) + ln(1 + 182 ) − ln(1 + 7 2 ) 2 2 1 1 D) 18arctan18 + 7 arctan( −7) − ln(1 + 182 ) + ln(1 + 7 2 ) 2 2 Ans: D difficulty: hard section: 7.2 B)
79. Consider the function f ( x)= (1 − x)1/ 3 . Does the midpoint approximation give an exact answer, an overestimate, or an underestimate? A) An overestimate B) An underestimate C) An exact answer D) Cannot tell Ans: A difficulty: easy section: 7.5
Page 19
Chapter 7: Integration
80. Suppose the points x0 , x1,..., xn are equally spaced and a = x0 < x1 < ... < xn = b . What is the formula (in terms of f and the xi 's) for the right Riemann sum approximation to b
∫a f ( x) dx ? b+ n
b
A)
∑ f ( xi ) ∆x
B)
∑
n
f ( xi ) ∆x
i =a
i= a + n
Ans: C
difficulty: medium
C)
∑ f ( xi ) ∆x i =1
n– 1
D)
∑ f ( xi ) ∆x
i =0
section: 7.5
81. Suppose the points x0 , x1,..., xn are equally spaced and a = x0 < x1 < ... < xn = b . n −1 f ( xi ) + f ( xi +1 ) Is ∑ ∆x the formula (in terms of f and the xi 's) for the midpoint 2 i =0
b
Riemann sum approximation to ∫ f ( x) dx ? a
Ans: no difficulty: medium
section: 7.5
82. True or False: For any given function, TRAP(n) is always more accurate than LEFT(n) . Ans: False difficulty: medium section: 7.5 83. True or False: The midpoint rule gives exact answers for linear functions, no matter how many subdivisions are used. Ans: True difficulty: medium section: 7.5 3
84. True or False: ∫ sin 39 ( x) dx > π . 0
Ans: False
difficulty: medium
section: 7.3
85. In the field of dynamical astronomy, integrals such as the following appear in the problem of determining trajectories of planets or spacecraft: 1 1 ∫0 1 + c cos θ dθ . In this definite integral, c is a constant that can go from 0 to +∞ depending on the shape of the orbit. Except for some special values of c, there is no fundamental antiderivative for the integrand 1 . f (θ ) = 1 + c cos θ A. What is the midpoint sum that approximates the definite integral for c = 0.5 with N = 5 equal divisions of the interval 0 ≤ θ ≤ 1? Round to 4 decimal places. B. Is this sum is an overestimate or an underestimate of the integral? Part A: 0.7052 Part B: underestimate difficulty: medium section: 7.5
Page 20
Chapter 7: Integration
86. Suppose that a computer takes 10−7 seconds to add two numbers together, and it takes 10−5 seconds to multiply two numbers together. The computer is asked to integrate the function f ( x)= 3 ⋅ x 2 from 0 to 1 using left hand sums with n divisions. As a function of n, let T(n) denote the time used by the computer to do the calculation. Compute T(n). (The computer figures x2 as x · x.) C) A) n ×10−5 + n ×10−7 2n ×10−5 + (n − 1) ×10−7 B) 2n ×10−7 + 2(n − 1) ×10−5 Ans: A difficulty: medium
D) n ×10−5 + 2n ×10−7 section: 7.5
87. Use Simpson's rule with n = 4 to approximate ∫
1
0
places. Ans: 1.2756 difficulty: medium
1 + e− x dx . Round to 4 decimal
section: 7.5
88. Suppose you used the trapezoid rule with n = 4 to approximate ∫
1
0
1 + e − x dx .
Approximately how many times more accurate would your answer be if you used n = 40? Ans: Your answer would be approximately 100 times more accurate. difficulty: medium section: 7.5 89. The table below contains numerical data for a definite integral approximated by the left-hand, midpoint, trapezoid, and Simpson's rule methods. Which method was used in the fourth column? N 1 2.4737 -44.1930 0.4737 -67.5263 3 -39.6662 -45.5098 -40.3329 -48.4316 9 -44.8687 -45.5261 -45.0909 -45.8548 27 -45.45315 -45.52630 -45.62722 -45.5629 81 -45.518171 -45.526300 -45.572862 -45.530364 A) The left-hand rule B) The midpoint rule Ans: B difficulty: medium
C) Simpson's rule D) The trapezoid rule section: 7.5
90. Compute ∫ (1 + x 2 ) −4 dx , accurate to 4 decimal places, using any method. 1
0
Ans: 0.4746 difficulty: medium
section: 7.5
Page 21
Chapter 7: Integration
4
91. Compute ∫ (1 + ln x )
−1
2
Ans: 0.971 difficulty: medium
dx , rounded to three decimal places. section: 7.6
) dx , rounded to three decimal places. 0 ( 1/ 2 4 Ans: ∫ x ⋅ ( 9 + x 2 )= dx 13 (25)3/ 2 − 9 ≈ 32.667. 0 4
92. Compute ∫ x 9 + x 2
difficulty: medium
1/ 2
section: 7.5 6
93. Find the exact value and SIMP(2) for ∫ e 2 x dx . What is your error, to the nearest 0
integer? Ans:
6 2x
∫0 e
dx =
1 e12 − 1 ≈ 81,376.8957 . 2 2
–16,651. difficulty: medium 94. Does the integral Ans: no difficulty: medium
So the error in the SIMP(2) approximation is
section: 7.5 ∞
x4
∫10 e− x + x dx converge? section: 7.6
∞
95. If ∫ x 2e − x dx converges, find its value. Otherwise, enter "DNC". 0
Ans: 2 difficulty: medium 96. If ∫
1
1
0 x1990 /1991
section: 7.6
dx converges, find its value. Otherwise, enter "DNC".
Ans: 1991 difficulty: easy
section: 7.6
1 dx converges, find its value to 3 decimal places. Otherwise, enter "DNC". 1− x Ans: 2.000 difficulty: medium section: 7.6
97. If ∫
1
98. If ∫
0
0
−1
3 dx converges, find its value. Otherwise, enter "DNC". 1+ x
Ans: 6 difficulty: medium
section: 7.6
Page 22
Chapter 7: Integration
99. If ∫
∞
100. If ∫
7
8 dx converges, find its value to 3 decimal places. Otherwise, enter "DNC". 0 1+ x Ans: DNC difficulty: medium section: 7.6
1
5 ( x − 6) 2
dx converges, find its value. Otherwise, enter "DNC".
Ans: DNC difficulty: medium
section: 7.6
∞
101. If ∫ 2 xe − x dx converges, find its value. Otherwise, enter "DNC". 0
Ans: 2 difficulty: medium 102. If ∫
section: 7.6
dx
2
converges, find its value. Otherwise, enter "DNC". 6/5 x − 1 ( ) Ans: DNC difficulty: medium section: 7.6 0
∞
103. Evaluate ∫ 7 z 6e − z dz . 7
0
Ans: 1 difficulty: medium 104. Find ∫
4
section: 7.7
1
dx to 4 decimal places. 16 − x 2 Ans: 3.1416 difficulty: medium section: 7.6 −4
105. If the rate, r, at which people get sick during an epidemic of flu is approximately r = 500te −0.5t , where r is people per day and t is time in days since the start of the epidemic, then the total number of people who get sick is 2000. How does the total number of people getting sick change if the exponent is tripled and the rate is now q = 500te −1.5t ? A) The number is one-third as large.. C) The number is three times as large. B) The number is one-ninth as large. D) The number is nine times as large. Ans: B difficulty: medium section: 7.6 a+2 b a + 2 dx dx dx 106. True or False: = + lim lim ∫−a+1 (a − x) b→a− ∫−a+1 (a − x) c→a+ ∫c (a − x) . Ans: True difficulty: easy section: 7.6
Page 23
Chapter 7: Integration
107. If ∫
∞
1
0 e9 x
dx converges, find its value to 3 decimal places. Otherwise, enter "DNC".
Ans: 0.111 difficulty: easy 108. If ∫
∞ e− x
0 1+ x
section: 7.7
dx converges, find its value to 3 decimal places. Otherwise, enter "DNC".
Ans: 0.596 difficulty: medium 109. Does ∫
∞ 7 + e− z
1
z
section: 7.7
dz converge?
Ans: no difficulty: easy
section: 7.7 ∞ sin 2 x dx . 2 0
110. Consider the integral ∫
Which if the following is true? (1 + x) A) The integral converges with an upper bound of 0. B) The integral converges with an upper bound of 1. C) The integral diverges. Ans: B difficulty: medium section: 7.7 0 ( ∞
111. Does ∫ x x3 + 2 x − 3 ∞
)
−1/ 2
dx converge or diverge?
(
Ans: The integral ∫ x x3 + 2 x − 3 0
difficulty: medium
)
−1/ 2
dx diverges.
section: 7.7
112. The following improper integral arises in the study of certain damped oscillatory motion, such as that of a pendulum that eventually comes to rest due to friction: ∞ −t
∫0 e (1 + cos t ) dt . If this integral converges, give an upper bound for it. If not, enter "DNC". ∞
Ans: Since (1 + cos t ) ≤ 2 and ∫ 2e − t dt converges to 2, then 2 is an upper bound for 0
∞ −t e (1 + cos t ) dt . 0
∫
difficulty: medium
section: 7.7
Page 24
Chapter 7: Integration
∞
1
1
x2 + 1 ∞ 1
113. If we approximate ∫
dx with ∫
1
b
1 x2 + 1
dx what value of b could we use to
estimate the value of ∫
dx with an error of less than 0.01? Of the following, x2 + 1 select the smallest value of b that will work. A) 50 B) 125 C) 250 D) 500 Ans: B difficulty: medium section: 7.7 1
114. Is the area between y =
1 x
Ans: finite difficulty: easy 115. Does ∫
∞
1
1
a2
116. Does ∫
∞
1
2
and y =
1
on (1, ∞) finite or infinite?
x5
section: 7.7
dx converge or diverge for a = –1.3?
x Ans: If a = –1.3, then the integral converges. difficulty: easy section: 7.7
1
p
dx converge or diverge for p = 0.7?
x Ans: If p = 0.7, then the integral diverges. difficulty: easy section: 7.7
1 1 117. True or False: ∫ sin 4 x cos3 x dx = sin 5 x − sin 7 x + C . 5 7 Ans: True difficulty: medium section: 7 review 118. True or False: ∫ xe − x dx = e − x ( x + 1) + C . Ans: False
difficulty: medium
119. True or False: ∫ Ans: True
7 x2 + 6
dx = 7 x − arctan x + C . x2 + 1 difficulty: medium section: 7 review
120. True or False: ∫ Ans: False
section: 7 review
t2
= dt
t 1 36 t 36 − t 2 + arcsin + C . 2 2 6
36 − t 2 difficulty: medium
section: 7 review
Page 25
Chapter 7: Integration
121. Evaluate ∫ Ans:
π /2
0 π /2
∫0
sin 4 x cos3 x dx . Leave your answer in the form "A/B". 2 . 35 section: 7 review
sin 4 x cos3 x dx =
difficulty: medium 1
2
122. True or False: ∫ 7 x3e x dx = 7 . 0
Ans: False
difficulty: medium
1, x ≤ 1 f ( x) dx = 24.5 , where f ( x) = −6 x, x > 1. difficulty: medium section: 7 review
123. True or False: ∫ Ans: True
section: 7 review
6
cos 4 x x cos 4 x sin 4 x 124. True or False: ∫ (1 + x) sin(4 x) dx = − + − +C . 4 4 16 Ans: False difficulty: medium section: 7 review
Ans: True
1 + e7t
e 6t +C. 6 et difficulty: medium section: 7 review
125. True or False: ∫
−e −t + dt =
x 4 2 x3 126. True or False: ∫ x 2 (2 x − 2) dx = − +C . 2 3 Ans: True difficulty: easy section: 7 review cos3 x sin 2 x dx = ln sin x + +C . sin x 2 difficulty: medium section: 7 review
127. True or False: ∫ Ans: False
x
128. True or False: ∫ Ans: True
dx = x tan x + ln cos x + C . cos 2 x difficulty: medium section: 7 review
129. True or False: ∫
4
2
Ans: True
e2 x
(
)
3
(
)
1
(
)
3
(
)
1
2 4 2 dx= e − 1 2 + 2 e4 − 1 2 − e2 − 1 2 − 2 e2 − 1 2 . 3 3 ex −1
difficulty: medium
section: 7 review
Page 26
Chapter 7: Integration
( )
ex 1 x 130. True or False: = ∫ 1 + e2 x dx 2 arctan e + C . Ans: False difficulty: medium section: 7 review
6
131. True or False: ∫ Ans: True
(
)
dx =6 x − 6 ln 1 + e x + C .
1+ e difficulty: medium x
section: 7 review
cos x 132. True or False: ∫ = dx ln sin x + C . sin x Ans: True difficulty: easy section: 7 review 1 1 133. True or False: ∫ = dx arcsin(5 x) + C . 2 25 1 − 25 x Ans: False difficulty: medium section: 7 review 134. True or False: ∫ x3 cos x dx = x3 sin x + x 2 cos x − x sin x − cos x + C . Ans: False
difficulty: medium
section: 7 review
t ( sin(ln t ) − cos(ln t ) ) + C . 2 difficulty: medium section: 7 review
135. True or False: ∫ sin(ln t ) dt = Ans: True
(
)
z dz z (ln z ) 2 + 2 z ln z − z + C . 136. True or False: ∫ (ln z ) 2 + 2 ln = Ans: False
difficulty: medium
section: 7 review
1 −2 x e ( −2 cos x + sin x ) + C . 5 difficulty: medium section: 7 review
137. True or False: ∫ e−2 x cos x dx = Ans: True
t dt = t − ln t + 1 + C . t +1 difficulty: medium section: 7 review
138. True or False: ∫ Ans: True
1 ( ln y − 2 + ln y + 2 ) + C . 4 4− y difficulty: medium section: 7 review
139. True or False: ∫ Ans: False
1
dy = − 2
Page 27
Chapter 7: Integration
x
1 = dx ln x 2 − 9 + C . 2 2 x −9 difficulty: easy section: 7 review
140. True or False: ∫ Ans: True
(
)
(
)
1 1 141. True or False: ∫ x 2 + 2 cos 2 x dx = x 2 + 2 sin 2 x − 1x cos 2 x − sin 2 x + C . 2 4 Ans: False difficulty: medium section: 7 review 1 1 = dx ln 6 x + 4 + C . 6x + 4 6 difficulty: easy section: 7 review
142. True or False: ∫ Ans: True
5cos x 5 cos x − 1 dx = − − ln +C . sin x 2sin 2 x 4 cos x + 1 difficulty: medium section: 7 review 5
143. True or False: ∫ Ans: False
3
1 dx = ( 4 x + 8)1/ 2 + C . 4x + 8 difficulty: easy section: 7 review
144. True or False: ∫ Ans: False
15 . 1 4 difficulty: medium section: 7 review 2
145. True or False: ∫ x3 ln x dx = ln 2 − Ans: False
3t 3 3 146. True or False: ∫ 4t dt = − te −4t − e −4t + C . 4 16 e Ans: True difficulty: medium section: 7 review 147. True or False: ∫
12
−12
Ans: True
( sin θ )5 = 0 .
θ2 difficulty: easy
section: 7 review
(
Ans: False
)
u du =u + 10 u − 50 ln u + 5 + C . u +5 difficulty: medium section: 7 review
148. True or False: ∫
3
2 149. True or False: ∫ 1 − 5 x dx = − (1 − 5 x ) 2 + C . 15 Ans: True difficulty: medium section: 7 review
Page 28
Chapter 7: Integration
7 2 x ln x + C . 2 difficulty: medium section: 7 review
150. True or False: ∫ 7 x= ln x dx Ans: False
3
7+ x 2 151. True or False: ∫ dx = 14 x + x 2 + C . 3 x Ans: True difficulty: easy section: 7 review
1 + sin(4 x) 1 1 dx = x − cos(4 x) + C . 4 4 4 difficulty: easy section: 7 review
152. True or False: ∫ Ans: False
ln(7 + x) 2 dx = ( ln(7 + x) ) + C . 7+ x difficulty: medium section: 7 review
153. True or False: ∫ Ans: False
1 154. True or False: ∫ ln dx =− x ln x + x + C . x Ans: True difficulty: medium section: 7 review
155. True or False: ∫ ( x + 4)( x − 4) dx = Ans: True
difficulty: easy
x3 − 16 x + C . 3 section: 7 review
( ln x ) + C . 5 − ln x 156. True or False: ∫ dx = 5ln x − x 2 Ans: True difficulty: medium section: 7 review 2
157. True or False: ∫ ( x − 1)e − x dx = − xe− x − e− x + C . Ans: False
difficulty: medium 1/ 7
1
0
1 − x2
158. Evaluate ∫
Ans: 0.144 difficulty: medium 4
10 x
0
x2 + 9
159. Evaluate ∫
Ans: 20 difficulty: medium
section: 7 review
dx to 3 decimal places. section: 7 review dx .
section: 7 review
Page 29
Chapter 7: Integration
8
160. True or False: ∫ ln x= dx 8ln 8 − 8 . 1
Ans: False
difficulty: medium
section: 7 review
5
1
x2 + 2 2 161. True or False: ∫ dx = x 2 + 4 x 2 + C . 5 x Ans: True difficulty: easy section: 7 review
(
)
162. True or False: ∫ sin 3 2θ + 5 cos2θ dθ = Ans: False
difficulty: medium
163. True or False: ∫ Ans: True
4 x3 + cos x
dx = ln x 4 + sin x + C .
x + sin x difficulty: easy 4
1 4 5 sin 2θ + sin 2θ + C . 4 2 section: 7 review
section: 7 review
164. True or False: ∫ u −1 ln= u du (ln u ) 2 + C . Ans: False
difficulty: medium
165. True or False: ∫ sin x ( cos x + 7 ) Ans: True
2
section: 7 review
3 cos x + 7 ) ( − +C . dx =
difficulty: easy
3 section: 7 review
xe 4 x e 4 x − +C . 4 16 difficulty: easy section: 7 review
166. True or False: ∫ xe 4 x dx = Ans: True 167. True or False:
b
If the left-hand sum, LEFT(n) , for ∫ f ( x) dx is too large for one value of n, it will be a
too large for all values of n. Ans: False difficulty: medium
section: 7 review
168. True or False: If the average value of f(x) on the interval 2 ≤ x ≤ 5 is between 0 and 1, then f is between 0 and 1 on the interval 2 ≤ x ≤ 5. Ans: False difficulty: medium section: 7 review –1
169. True or False:
∫–2 sin x dx > 3 .
Ans: False
difficulty: easy
2
section: 7 review
Page 30
Chapter 7: Integration
170. True or False: If f ′ > g′ for all a < x < b then the right-hand Riemann sum approximation of b
b
∫a f ( x) dx will have smaller error than the right-hand Riemann sum for ∫a g ( x) dx . Ans: False
difficulty: medium
section: 7 review
171. Derive the formula for the area of a circle of radius R using trigonometric substitution. R
= x R= sin θ , dx R cos θ dθ , we Ans: Area = 4 ∫ R 2 − x 2 dx . Using the substitution 0
get
π /2
π /2
π /2
1 1 2 2 4 ∫ R= − x 2 dx 4 ∫ ( R cos θ )( R cos = = θ ) dθ 4 R 2 ∫ cos θ dθ 4 R 2 cos θ sin θ += θ π R2 2 0 2 0 0 0 . difficulty: medium section: 7.4 R
1 1 x 172. Show ∫ = dx arctan + C by integrating the left hand side using the substitution 2 2 x +a a a x = a tan θ . Ans: If x = a tan θ , then 1 1 1 1 1 x a ∫ x 2 + a 2 dx = ∫ a 2 (tan 2 θ + 1) cos2 θ dθ = a ∫ dθ = a θ + C = a arctan a + C . difficulty: medium section: 7.4 5π / 2
173. The actual value of
∫e π
sin x
dx lies:
/2
A) between 7.7 and 7.8. B) between 7.8 and 7.9. Ans: C difficulty: easy
C) D) section: 7.5
between 7.9 and 8.0. between 8.0 and 8.1.
10
174. The actual value of ∫ 5ln( x 2 + 1) dx lies: 0
A) between 142.5 and 145. B) between 145 and 147.5. Ans: B difficulty: easy
C) D) section: 7.5
Page 31
between 147.5 and 150. between 150 and 152.5.
Chapter 7: Integration
a
dx
∫ (a − x)
175. Evaluate the integral
0.9
to 3 decimal places, if it converges. If not, enter
a−2
DNC. Ans: 10.718 difficulty: easy
section: 7.7
x) n dx x(ln x) n − n ∫ (ln x) n −1 dx in one 176. How can you verify the reduction formula: ∫ (ln= step? = uv − ∫ v du ) with u = (ln x) n and A) using integration by parts (that is, ∫ u dv B)
dv = dx . = uv − ∫ v du ) with u = (ln x) n −1 and using integration by parts (that is, ∫ u dv
dv = ln x dx . C) using the substitution w = ln x . Ans: A difficulty: easy section: 7 review 5
177. Evaluate exactly: ∫ xe x dx . −1
5
Ans:
dx 4e + 2e . ∫ xe = −1
5
x
−1
difficulty: easy
section: 7 review π
178. Evaluate exactly:
∫π (sin t )e
cos t
dt .
/2
π
Ans:
∫ (sin t )e π /2
difficulty: easy
cos t
1 dt = 1 − . e section: 7 review
π
179. Evaluate exactly: ∫ x 2 sin x dx . (Leave π in your answer.) 0
π
Ans:
∫ x sin x dx= π − 4 . 2
2
0
difficulty: easy
section: 7 review
Page 32
Chapter 7: Integration
3
x 2 dx . x3 + 4 0
180. Evaluate exactly: ∫
x 2 dx 1 1 31 ∫0 x3 + 4= 3 ( ln 31 − ln 4=) 3 ln 4 . difficulty: easy section: 7 review 3
Ans:
2/2
dx
∫
181. Evaluate exactly: 2/2
Ans:
∫
dx
1− x 0 difficulty: easy
2
.
1 − x2
0
π
=
. 4 section: 7 review ln 6
182. What is the value of A if
x
ln 6
A
− + ? ∫ e dx = 6 6 x
0
A) 5 B) 6 C) 7 D) None of the above. Ans: A difficulty: easy section: 7 review π /2
183. Evaluate exactly:
dθ
∫π tan θ . /6
693 A) 1000 Ans: B
B) ln 2
1 6931 D) − 2 10, 000 section: 7 review
C) ln
difficulty: easy 5π
184. Evaluate exactly:
∫π e cos 2 x dx . 2x
−5
1 5π 1 10π 1 10π B) C) e − e −5π ) e + e –10π ) e − e –10π ) ( ( ( 4 4 4 Ans: A difficulty: medium section: 7 review
A)
x 2 dx π ∫0 1 + x=2 3 − A . A) 2 B) 3 C) 4 D) 6 Ans: B difficulty: medium section: 7 review 3
185. Find the value of A if
Page 33
D)
1 10π e 4
1. A rectangular lake is 100 km long and 60 km wide. The depth of the water at any point of the surface is a tenth of the distance from the nearest shoreline. How many km3 of water does the lake contain? Round to the nearest whole number. Ans: 7200 km3. difficulty: hard section: 8.1 2. Find the area of the dumbbell shaped region bounded by the curve = y 2 x 6 (1 − x 2 ) . Round your answer to 2 decimal places. Ans: The area is approximately 0.53 square units. difficulty: medium section: 8.1 3. Focus on Engineering It's time for the School of Engineering class picture and you are the photographer! You stand at the origin with your camera and your classmates are strung out along the curve y = e− x from (0, 1) to (3, e −3 ) .
What is the formula for the average value of the distance from you to your classmates at points (x, y) on the curve? 1 3 2 −2 x 1 3 C) A) x +e dx x + e− x dx ∫ ∫ 0 0 3 3
(
B)
3
∫0 x + e
Ans: A
2
−x
dx
difficulty: medium
D) section: 8.1
Page 1
3
)
∫0 x − e 2
−2 x
dx
Chapter 8: Using the Definite Integral
4. Focus on Engineering It's time for the School of Engineering class picture and you are the photographer! You stand at the origin with your camera and your classmates are strung out along the curve y = e− x from (0, 1) to (4, e −4 ) .
Use a calculator to find the average value of the distance from you to your classmates at a point (x, y) on the curve. Round to 3 decimal places. Ans: The average distance from you and your classmates is approximately 2.099. difficulty: medium section: 8.1 5. Focus on Engineering It's time for the School of Engineering class picture and you are the photographer! You stand at the origin with your camera and your classmates are strung out along the curve y = e− x from (0, 1) to (5, e −5 ) . You focus your camera at the average value of the distance to your classmates on the curve. Who is more in focus, the person at (0, 1) or the person at (5, e −5 ) ?
A) The person at (0, 1). B) The person at (5, e−5 ) . Ans: A difficulty: medium section: 8.1
Page 2
Chapter 8: Using the Definite Integral
6. A coffee filter is in the shape of a cone, as shown below. Suppose that when it is filled with water to a height h cm, the rate at which coffee flows out the hole at the bottom is given by Volume of coffee which flows out per second = h cm3 / sec.
Approximately how many cubic centimeters are in the "slice" of coffee lying below h + ∆h and above h? A) π h3∆h B) π h∆h C) π h 2 ∆h D) π h3/ 2 ∆h Ans: C difficulty: medium section: 8.1 7. A coffee filter is in the shape of a cone, as shown below. Suppose that when it is filled with water to a height h cm, the rate at which coffee flows out the hole at the bottom is given by Volume of coffee which flows out per second = h cm3 / sec.
Suppose the coffee filter starts full. Write an integral representing the total amount of time it takes for the coffee filter to empty, and then evaluate it to find the time needed for the filter to empty. Give your answer to the nearest second. Ans: 40 difficulty: medium section: 8.1
Page 3
Chapter 8: Using the Definite Integral
8. Suppose that a new office building is being planned. The architect wants to design a building that is thick at the base and eventually tapers to a small flat roof at the top. The building is to have 10 floors, and each floor is to be 15 feet high. For purposes of air conditioning the building, an estimate of the total volume is needed. You have been hired as a consultant at an exorbitant salary to do this. You have been provided with the following information which shows how much area, in units of 100 square feet, each of the 10 floors will contain: Floor # Area
Ground 40
2 38
3 35
4 33
5 31
6 29
7 27
8 25
9 24
10 23
Earn your wage: find an approximate value for the number of cubic feet in the entire building. Ans: The building contains approximately 444,000 cubic feet. difficulty: easy section: 8.1 9. If a cylinder has radius 6 and height h (in centimeters), which of the following definite integrals represents the volume using slices? A)
h
∫0 36π dx cm
Ans: A
3
B)
h
∫0 12π dx cm
difficulty: medium
3
C)
6
∫0 6hπ dx cm
3
6 2
∫0 h π dx cm
D)
3
section: 8.1
10. A cone has base radius = 4 cm and height = 2 cm. Using horizontal slicing, which of the following definite integrals represents volume of the cone? 2
A)
∫0 6π h dh cm
B)
∫0 π (4 − 2h) dh cm
Ans: B
2
2
2
3
3
difficulty: medium
2
4
2
∫0 π (2 − h) dh cm
D)
∫0 π (4 − h) dh cm
section: 8.1
Page 4
4
C)
3
3
Roof 22
Chapter 8: Using the Definite Integral
11. Which of the following integrals determines the volume of a cone of height H and radius R? A)
R
∫ π ( R( H − h) ) dh 2
0
2
R ( H − h) dh H
∫0
D)
∫0 π ( R( H − h) ) dh
2
H B) ∫0 π R ( H − h) dh Ans: C difficulty: medium R
H
C)
H
π
2
section: 8.1 6
12. What is represented by the integral ∫ π (36 − h 2 ) dh ? 0
A) A cone with radius 6 and height 6. C) A sphere with radius 6. B) A cone with radius 3 and height 6. D) A hemisphere with radius 6. Ans: D difficulty: medium section: 8.1 2
x 13. What is represented by the integral ∫ π dx ? 0 2 A) A cone with radius 2 and height 4. C) A sphere with radius 4. B) A cone with radius 4 and height 4. D) A hemisphere with radius 4. Ans: A difficulty: medium section: 8.1 4
14. A large chocolate bar has trapezoidal cross section shown in the following figure. If you want the bar to have volume 861 cm3, what should H be?
Ans: 20.5 difficulty: easy
section: 8.1
Page 5
Chapter 8: Using the Definite Integral
x2 at a velocity of 6 2 units/second. Which of the following integrals show how far Alice has traveled when she reaches the point where x = a?
15. Alice starts at the origin and walks along the graph of y =
A) B)
a/6
∫0
a
∫0
Ans: B
1 + x 2 dx
1 + x 2 dx difficulty: medium
6a
1 + x 2 dx
a
1 + x 2 dx
C)
∫0
D)
6∫
0
section: 8.2
x2 16. Alice starts at the origin and walks along the graph of y = at a velocity of 11 2 units/second. You want to find the x-coordinate of the point Alice reaches after traveling for 2 seconds. Estimate this coordinate to 1 decimal place. Ans: After walking for 2 seconds, Alice reaches an x-coordinate of about 6.4. difficulty: hard section: 8.2
17. In a recent archaeological expedition, a scroll was discovered containing a description of a plan to build what appears to be the Tower of Babel. According to the manuscript, the tower was supposed to have a circular cross section and “go up to the heavens” (i.e., be infinitely high). A mathematician was consulted to solve some of the questions posed by the archaeologists. The mathematicians plotted half of the silhouette of the tower on a set of coordinate axes with the y-axis running through the center and discovered that it was x approximated by the curve y = −100 ln . Would such a tower have finite volume? 6 If so, calculate it to the nearest whole number. If not, enter "infinite". Ans: The tower would have a finite volume of 5655 cubic units. difficulty: medium section: 8.2 18. In a recent archaeological expedition, a scroll was discovered containing a description of a plan to build what appears to be the Tower of Babel. According to the manuscript, the tower was supposed to have a circular cross section and “go up to the heavens” (i.e., be infinitely high). A mathematician was consulted to solve some of the questions posed by the archaeologists. The mathematicians plotted half of the silhouette of the tower on a set of coordinate axes with the y-axis running through the center and discovered that it was x approximated by the curve y = −100 ln . The manuscript mentions that 2560 cubic 4 “shrims” (Babel's unit of length) of stones were available to build the tower. The base of the tower was to have radius 3 shrims. Did they have enough? Ans: yes difficulty: medium section: 8.2
Page 6
Chapter 8: Using the Definite Integral
19. Consider the region bounded by y = e x , the x-axis and the lines x = 0 and x = 2. Find the volume of the solid whose base is the given region and whose cross sections perpendicular to the x-axis are isosceles right triangles with hypotenuses lying in the region. Round to 2 decimal places. Ans: The volume of the solid is about 6.70 cubic units. difficulty: hard section: 8.2
2 3/ 2 and you also love the number 6. What is the arc length x 3 of this curve from x = 0 to x = 6? Round to 2 decimal places. Ans: The arc length is about 11.68 units. difficulty: medium section: 8.2
20. You love the function y =
2 3/ 2 and you also love the number 5. You have a gold x 3 chain which is exactly 5 feet long. As a tribute to your favorite function you want to 2 mount your chain in the shape of y = x3/ 2 from x = 0 to x = 5 on a beautiful 3 rectangular piece of rosewood. If the lower left corner is labeled with the coordinate (0, 2 0) and the upper right corner is labeled with the coordinate 5, (5)3/ 2 and a unit on 3 the x-axis and a unit on the y-axis represent the same number of feet, then the dimensions of the piece of wood are ______ feet by ______ feet. Enter the smaller dimension first, and round both numbers to 2 decimal places. Part A: 2.74 Part B: 4.08 difficulty: hard section: 8.2
21. You love the function y =
22. The circle x 2 + y 2 = a 2 is rotated around the y-axis to form a solid sphere of radius a. A plane perpendicular to the y-axis at y = a/5 cuts off a spherical cap from the sphere. What percent of the total volume of the sphere is contained in the cap? Round to 2 decimal places. Ans: The cap contains about 35.20 % of the volume of the sphere. difficulty: medium section: 8.2
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Chapter 8: Using the Definite Integral
23. Set up a Riemann sum approximating the volume of the torus (donut) obtained by rotating the circle ( x − 6) 2 + y 2 = 1 about the y-axis. 2
A)
∑ π 6 + 1 − y 2 ∆y
B)
∑ π 6 − 1 − y 2 ∆y
2
2 2 2 6 − 1 − y 2 ∆y + − + π 6 1 y C) ∑ 2 2 2 2 + − − − − π 6 1 y 6 1 y D) ∆y ∑ Ans: D difficulty: easy section: 8.2
24. Set up an integral representing the volume of the torus (donut) obtained by rotating the circle ( x − 6) 2 + y 2 = 1 about the y-axis. A)
1
2
π 6 + 1 − y 2 dy
∫−1
2
2 B) ∫−1π 6 − 1 − y dy 2 2 1 2 2 C) ∫−1π 6 + 1 − y + 6 − 1 − y dy 2 2 1 2 6 − 1 − y 2 dy + − − π 6 1 y D) ∫−1 Ans: D difficulty: easy section: 8.2 1
25. Find the volume of the solid obtained by rotating the region bounded by y = x3 , x = 0, y = 0, and y = -8 around y = 1. Round to 2 decimal places. Ans: The volume is about 420.08 cubic units. difficulty: medium section: 8.2
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Chapter 8: Using the Definite Integral
26. A plastic travel mug is made in two parts, the cup and the base. The cup part has outside shape y = x and inside shape y = 4 x cut off at x = 4 as shown in the following figure.
If the cup is 14 cm tall, and the base is 10 cm in diameter and 1 cm thick, find the volume of plastic needed to make the mug. Round to 2 decimal places. Ans: The mug requires about 419.16 cm3 of plastic. difficulty: medium section: 8.2 27. Find the length of the curve given parametrically as x = 4sin t and y = 4 cos t for 0 ≤ t ≤ 2π. Round to 2 decimal places. Ans: 25.13 difficulty: medium section: 8.2 28. A curve is given parametrically as x = 4sin t and y = 4 cos t for 0 ≤ t ≤ 2π. How does the length change if the coefficients of x and y are doubled (meaning x = 8 sin t and y = 8 cos t)? A) The length is doubled. B) The length is halved C) The length is increased by a factor of 4. D) The length remains the same. Ans: A difficulty: medium section: 8.2
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Chapter 8: Using the Definite Integral
29. Choose inequalities for r and θ which describe the following region in polar coordinates.
4 ≤ r ≤ 8 and 0 ≤ θ ≤ 2π A) C) 4 ≤ r ≤ 8 and 0 ≤ θ ≤ π B) D) Ans: B difficulty: easy section: 8.3
–8 ≤ r ≤ 8 and 0 ≤ θ ≤ 2π –8 ≤ r ≤ 8 and 0 ≤ θ ≤ π
30. Find the area inside the three-petal rose r = 3cos 3θ shown in the following figure. Round to 2 decimal places.
Ans: 7.07 difficulty: medium
section: 8.3
31. Find the area outside the limaçon r = 1 + cos θ and inside the circle r = 2. Round to 2 decimal places. Ans: 7.85 difficulty: medium section: 8.3
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Chapter 8: Using the Definite Integral
32. Find the arc length of r = 2sin θ , 0 ≤ θ ≤ π . Give an exact answer and then round to 2 decimal places. Ans: The arc length is 2π ≈ 6.28. difficulty: medium section: 8.3 33. When an oil well burns, sediment is carried up into the air by the flames and is eventually deposited on the ground. Less sediment is deposited further away from the oil well. Suppose that the density (in tons/square mile) at a distance r from the burning oil well is 9 given by . Which Riemann sum approximates the total amount of sediment 1+ r2 which is deposited within 100 miles of the well? A)
9π r 2 ∆r
∑ 1+ r2
Ans: C
B)
18π r 2 ∆r
∑ 1+ r2
difficulty: easy
C)
18π r ∆r
∑ 1+ r2
D)
9π r ∆r
∑ 1+ r2
section: 8.4
34. When an oil well burns, sediment is carried up into the air by the flames and is eventually deposited on the ground. Less sediment is deposited further away from the oil well. Suppose that the density (in tons/square mile) at a distance r from the burning oil well is 8 given by . Find and evaluate an integral which represents the total amount of 1+ r2 sediment which is deposited within 100 miles of the well. Round to 2 decimal places. Ans: 231.48 tons difficulty: medium section: 8.4 35. A straight road goes through the center of a circular city of radius 5 km. The density of the population at a distance r (in km) from the road is well approximated by D(r= ) 30 − 4r (in thousand people per km2). Find the total population of the city, to the nearest person. Ans: The total population is about 1,689,528 people. difficulty: hard section: 8.4
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Chapter 8: Using the Definite Integral
36. The globular cluster M13 is a spherical distribution of stars which orbits our galaxy. Suppose that the density of stars in the cluster is purely a function of distance r from the −5
r 3 stars/(ly)3, where r is center of the cluster and is given as ρ (r= ) 1 + 120 measured in light-years, and 0 ≤ r ≤ 120 ly. (One light-year is the distance light travels in one year; “light-year” is abbreviated as “ly”.) Set up an integral whose value is the exact number of stars in M13.
A)
120
∫0
r 3 4π r 2 1 + 120
−5
C)
dr
120
∫0
−5
B)
120
∫0
Ans: A
r 3 dr 2π r 1 + 120 difficulty: medium
D)
120
∫0
r 3 π 1 + 120
−10
r 3 4π r 1 + 120
dr −10
dr
section: 8.4
37. The globular cluster M13 is a spherical distribution of stars which orbits our galaxy. Suppose that the density of stars in the cluster is purely a function of distance r from the −5
r 3 center of the cluster and is given as ρ (r= ) 1 + stars/(ly)3, where r is 90 measured in light-years, and 0 ≤ r ≤ 90 ly. (One light-year is the distance light travels in one year; “light-year” is abbreviated as “ly”.) Find the number of stars in M13. Round to the nearest whole number. Ans: 715,694 stars difficulty: medium section: 8.4
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Chapter 8: Using the Definite Integral
38. A cylindrically-shaped mug with a 3cm radius and a 10cm height is filled with tea. You have added some sugar to the tea, which tends to settle to the bottom of the mug. It turns out that the density ρ of sugar (in gm/cm3) in the tea, as a function of the height, h, in cm, above the bottom of the mug, is given by the formula ρ(h) = 0.025(10 - h). Which of the following Riemann sums approximates the total mass of sugar (in grams) in the mug of tea?
∑ 3π (0.025)(10 − h)∆h
A)
C)
allslices
allslices
∑ 9π (0.025)(10 − h)∆h
B)
D)
∑ 9π (0.025)(10 − h)2 ∆h allslices
allslices
Ans: B
∑ 3π (0.025)(10 − h)2 ∆h
difficulty: medium
section: 8.4
39. A cylindrically-shaped mug with a 3 cm radius and a 10 cm height is filled with tea. You have added some sugar to the tea, which tends to settle to the bottom of the mug. It turns out that the density ρ of sugar (in gm/cm3) in the tea, as a function of the height, h, in cm, above the bottom of the mug, is given by the formula ρ(h) = 0.01(10 - h). Find the amount of sugar in the mug. Round to 2 decimal places.
Ans: 14.14 grams difficulty: medium
section: 8.4
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Chapter 8: Using the Definite Integral
40. After Mt. St. Helens erupted in 1980, it was found that ash was spread in decreasing density as a function of distance r from the center of the crater. Say that the density ρ of 2000 ash at a distance r (meters) from the center of the crater is given by ρ (r ) = kg/m2. 2 1+ r Find the total amount of ash within 1400 meters of the center of the crater. Round to the nearest whole number and give your answer in kg. Ans: 91,034 kg difficulty: medium section: 8.4 41. A thin strip of nutrients 20 cm long is placed in a circular petri dish of radius 10 cm, as 75 shown. The population density of bacteria in the disk after 3 hours is given by D+4 bacteria/cm2 where D is the distance (in cm) to the nutrient strip. Which of the following integrals gives the number of bacteria in the petri dish 3 hours after the nutrient strip has been introduced?
20 100 − D 2
dD
C)
300 ∫
100 − D 2 dD 0 D+4 difficulty: hard
D)
75∫
A)
150π ∫
B)
75π ∫
0
Ans: C
D+4
3
section: 8.4
Page 14
10
0
20
0
100 − D 2 dD D+4 100 − D 2 dD D+4
Chapter 8: Using the Definite Integral
42. The density of cars (in cars per mile) down a 20-mile stretch of the Massachusetts x) 500 + 100sin(π x) , where x is the Turnpike starting at a toll plaza is given by ρ (= distance in miles from the toll plaza and 0 ≤ x ≤ 20. Which of the following Riemann sums estimates the total number of cars down the 20-mile stretch? 20 20 A) ∑ ρ ( xi )∆x , where the 20-mile stretch is divided into n pieces of length n = ∆x , i =0 and xi is a point in the i th segment. n −1
B)
20
∑ n ⋅ ρ ( xi ) , where the 20-mile stretch is divided into n pieces of length n = ∆x ,
i =0
and xi is a point in the i th segment. 20
C)
20
∑ ρ (∆xi ) , where the 20-mile stretch is divided into n pieces of length n = ∆x ,
i =0
and xi is a point in the i th segment. n −1
D)
20
∑ ρ ( xi )∆x , where the 20-mile stretch is divided into n pieces of length n = ∆x ,
i =0
and xi is a point in the i th segment. Ans: D difficulty: medium section: 8.4 43. The density of cars (in cars per mile) down a 24-mile stretch of the Massachusetts x) 500 + 100sin(π x) , where x is the Turnpike starting at a toll plaza is given by ρ (= distance in miles from the toll plaza and 0 ≤ x ≤ 24. Estimate the total number of cars down the 24-mile stretch by creating a Reimann sum, converting it to an integral, and evaluating it. Ans: 12,000 cars difficulty: medium section: 8.4 44. A chlorine solution is poured over the surface of a rectangular swimming pool that is 20 meters long, 10 meters wide and 2 meters deep everywhere. Before the circulating pumps in the pool are turned on, it is discovered that the density of the chlorine solution at a height h meters above the bottom of the pool is given by ρ(h) = 100h grams/m3. In other words, the chlorine solution has distributed itself so that its density increases linearly from the bottom of the pool. Write a Riemann sum that approximates the total mass of chlorine solution in the pool, convert it into a definite integral, and evaluate it to find the total mass of the chlorine solution in the pool. Give your answer in kg. Ans: 40 kg difficulty: medium section: 8.4
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Chapter 8: Using the Definite Integral
45. Circle City is circular with a radius of 3 miles. Right in the center is a circular park with radius one mile. No one lives in the park. Elsewhere the population density is 3000(5 - r) people per square mile, where r is the distance from the center in miles. What is the total population of Circle City, to the nearest thousand? Ans: 214,000 people difficulty: medium section: 8.4 46. Circle City is circular with a radius of 3 miles. Right in the center is a circular park with radius one mile. No one lives in the park. Elsewhere the population density is 4500(5 - r) people per square mile, where r is the distance from the center in miles. What is the average population density of Circle City, to the nearest whole number? Ans: 12,750 people per square mile. difficulty: hard section: 8.4 47. A 3-gram drop of thick red paint is added to a large can of white paint. A red disk forms and spreads outward, growing lighter at the edges. Since the amount of red paint stays constant through time, the density of the red paint in the disk must vary with time. Suppose that its density p in gm/cm2 is of the form p = k (t ) f (r ) for some functions k(t) of time and f(r) of the distance to the center of the disk. Let R(t) be the radius of the disk at time t. Which of the following equations expresses the fact that there are 3 grams of red paint in the disk? R (t )
p(r )
A)
3 = k (t ) ∫
f (r )2π r dr
C)
3 = f (r ) ∫
B)
R(t ) = k (t ) ∫ f (r )2π r dr
D)
p (r , t ) = k (t ) ∫ f (r )2π r dr
0
Ans: A
3
0
difficulty: medium
0
k (t )2π t dt 3
0
section: 8.4
48. A 5-gram drop of thick red paint is added to a large can of white paint. A red disk forms and spreads outward, growing lighter at the edges. Since the amount of red paint stays constant through time, the density of the red paint in the disk must vary with time. Suppose that its density p in gm/cm2 is of the form p = k (t ) f (r ) for some functions k(t) of time and f(r) of the distance to the center of the disk. For fixed r, which of the following integrals gives the average density of red paint at a distance r from the center of the disk from 0 to T seconds? k (t ) T k (t ) T A) C) f (r ) dr f (r ) dr ∫ T 0 5T ∫0 f (r ) T f (r ) T B) D) k (t ) dt k (t ) dt ∫ T 0 5T ∫0 Ans: B difficulty: medium section: 8.4
Page 16
Chapter 8: Using the Definite Integral
1 1+ x between x = 0 and x = 4. The density of the plate x units from the y-axis is given by x2 grams/cm2. Write down a Riemann sum which approximates the total mass. Convert it to an integral, and evaluate the integral to 3 decimal places.
49. A flat metal plate is in the shape determined by the area under the graph of f ( x) =
Ans: 5.609 grams difficulty: medium
section: 8.4
50. An object is in the shape drawn below; its boundary is obtained by rotating the parabola y = 2 x 2 (for 0 ≤ x ≤ 1) around the y axis. (Units are in centimeters.) Suppose that the density of this object varies with height according to the rule ρ(y) = 12 (2 - y) grams/cm3. Which of the following Riemann sums computes (approximately) the mass in grams of this object?
2
2
A)
∑ π ⋅ y 2 ⋅12(2 − y)∆y
C)
y =0
y =0
2
B)
∑π⋅
y =0
Ans: C
2
2
y ⋅12(2 − y )∆y 4
difficulty: medium
y
∑ π ⋅ 2 ⋅12(2 − y)∆y
D)
∑ π ⋅ y ⋅12(2 − y)∆y
y =0
section: 8.4
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Chapter 8: Using the Definite Integral
51. An object is in the shape drawn below; its boundary is obtained by rotating the parabola y = 2 x 2 (for 0 ≤ x ≤ 1) around the y axis. (Units are in centimeters.) Suppose that the density of this object varies with height according to the rule ρ(y) = 8 (2 - y) grams/cm3. Compute the mass in grams of this object. Give an exact answer and then round it to 2 decimal places.
Ans: The mass of this object is difficulty: hard
16π ≈ 16.76 g. 3
section: 8.4
52. The density of a compressible liquid is 30(5 - h) kg/m3 at a height of h meters above the bottom. The liquid is put in the container as shown below (resting on the triangular side). The cross sections of the container are isosceles triangles, it has straight sides, and it looks like a triangular prism. How many kg will it hold? Round to the nearest whole number.
Ans: 900 kg difficulty: hard
section: 8.4
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Chapter 8: Using the Definite Integral
53. The Great Cone of Haverford College is a monument built by freshmen during a customs week long, long ago. It is 100 ft. high and its base has a diameter of 100 ft. It has been built from bricks (purportedly made of straw) which weigh 2 lbs/ft3. Use a definite integral to approximate the number of foot-pounds required to build the Cone. Round to the nearest million.
Ans: 13,000,000 ft-lbs difficulty: medium section: 8.5 54. The force of gravitational attraction between a thin rod of mass M and length L and a particle of mass m lying on the same line as the rod at a distance of A from one of the GmM ends is . Use this result to select an integral for the total force due to gravity A( L + A) between two thin rods, both of mass M and length L, lying along the same line and separated by a distance A. [Hint: Divide one of the rods into small pieces, each of length M dx and mass dx . Apply the formula above to each of the pieces, then form a L Riemann sum. You know the rest…]
A)
F =∫
B)
F =∫
Ans: A
A+ L A
GM 2 dx Lx( L + x)
GM 2 dx A x( L + x) difficulty: medium A+ L
C)
F =∫
D)
F =∫
A
0
GM 2 dx L( L + x)
A GM 2 dx
0
x( L + x)
section: 8.5
55. The work done lifting a 5-lb bag of sugar 4 feet off the floor is _____ ft-lbs. Ans: 20 difficulty: easy section: 8.5
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Chapter 8: Using the Definite Integral
56. If it is known to take 147 joules of work to lift a box 1.5 meters off the floor, what is the mass of the box in kg? Ans: 10 kg difficulty: easy section: 8.5 57. A bridge worker needs to pull a 14-meter uniform cable with mass 3 kg/meter up to the work platform. How many joules are needed, assuming that the cable is hanging straight down from the platform? Round to the nearest whole number. Ans: 2881 J difficulty: medium section: 8.5 58. A bridge worker needs to pull a 10-meter uniform cable with mass 3 kg/meter up to the work platform. A second worker needs to pull an identical cable (which is one-half as long) up one-half the original distance. Will the second worker do one-half the work of the first worker? Assume that each cable is hanging straight down from its platform. Ans: no difficulty: easy section: 8.5 59. A swimming pool has shape as shown in the following figure.
If the pool is 9 ft deep, how many foot-pounds does it take to pump all the water out? (Note: water weighs 62.4 pounds/ft3.) Round to the nearest whole number. Ans: 955,282 ft-lbs difficulty: medium section: 8.5
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Chapter 8: Using the Definite Integral
60. A study of the costs to produce airplanes in World War II led to the theory of “learning curves,” the idea of which is that the marginal cost per plane decreases over the duration of a production run. In other words, with experience, staff on an assembly line can produce planes with greater efficiency. The 90% learning curve describes a typical situation where the marginal cost, MC, to produce the xth plane is given by
MC ( x) = M 0 x log 2 0.9 , where M 0 = marginal cost to produce the first plane. [Note: ln x You may use the fact that log 2 x = .] If a plant produces planes with a 90% ln 2 learning curve on production costs, and the marginal cost for the first plane is $700,000, then what is the marginal cost to produce the fourth plane? Round to the nearest thousand dollars. Ans: $567,000 difficulty: easy section: 8.6 61. A study of the costs to produce airplanes in World War II led to the theory of “learning curves,” the idea of which is that the marginal cost per plane decreases over the duration of a production run. In other words, with experience, staff on an assembly line can produce planes with greater efficiency. The 90% learning curve describes a typical situation where the marginal cost, MC, to produce the xth plane is given by
MC ( x) = M 0 x log 2 0.9 , where M 0 = marginal cost to produce the first plane. [Note: ln x You may use the fact that log 2 x = .] Recall that marginal cost is related to total ln 2 cost as follows: MC ( x) = C '( x) , where C(x) = total cost to produce x units. Given this, and the formula for MC(x) with M 0 = $500,000, find a formula for C(x). What, physically, is the meaning of the constant of integration in your formula? A) The cost of producing the first plane. B) The cost if no planes are produced (the cost of setting up the plant for production). C) The marginal cost of the xth unit. D) The average cost of producing x units. Ans: B difficulty: easy section: 8.6
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Chapter 8: Using the Definite Integral
62. A study of the costs to produce airplanes in World War II led to the theory of “learning curves,” the idea of which is that the marginal cost per plane decreases over the duration of a production run. In other words, with experience, staff on an assembly line can produce planes with greater efficiency. The 90% learning curve describes a typical situation where the marginal cost, MC, to produce the xth plane is given by
MC ( x) = M 0 x log 2 0.9 , where M 0 = marginal cost to produce the first plane. [Note: ln x You may use the fact that log 2 x = .] Recall that marginal cost is related to total ln 2 cost as follows: MC ( x) = C '( x) , where C(x) = total cost to produce x units. Given this, and the formula for MC(x) with M 0 = $500,000, find a formula for C(x). If the constant for C(x) is $20 million, what is C(50), the cost of producing 50 planes, to the nearest thousand dollars? Ans: $36,267,000 difficulty: medium section: 8.6 63. Assuming 7% interest compounded continuously, arrange the following in ascending order according to present value by entering a "1" next to the smallest, a "2" by the next smallest, and so forth. A. $8000, paid today B. $8400, paid six months from now C. $8680, paid a year from now D. $8400, paid continuously over the next year Part A: 1 Part B: 3 Part C: 2 Part D: 4 difficulty: medium section: 8.6
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Chapter 8: Using the Definite Integral
64. On each of two days (January 1, 2005 and January 1, 2006), a person deposits $2500 in a savings bank. The bank deposits interest in the account at a rate of 6%, compounded 4 times per year. Choose the sum that gives the size of the bank account at the end of 2006. 4
0.06 A) 2500 1 + 4 4 0.06 B) 2500 + 2500 1 + 4 4 4 0.06 0.06 C) 2500 + 2500 1 + 1 + 4 4 4 4 0.06 0.06 D) 2500 1 + + 2500 + 2500 1 + 4 4 Ans: C difficulty: medium section: 8.6
65. On each of two days (January 1, 2005 and January 1, 2006), a person deposits $500 in a savings bank. The bank deposits interest in the account at a rate of 8%, compounded continuously. Choose the sum/integral that gives the size of the bank account at the end of 2006. A) B)
2
∫0 500e
0.08t
dt
∫0 ( 500 + 500e
Ans: D
2
0.08t
) dt
difficulty: medium
C)
500 + 500e0.08
D)
(500 + 500e0.08 ) e0.08
section: 8.6
66. By the year 2016 you will have made your first million dollars. You invest it in a new company on January 1, 2017. The new company starts to earn a profit six months later. Thus, starting July 1, 2017, you receive income from the company in a continuous stream at a constant rate of 1/2 million dollars per year. Your bank offers interest at a nominal rate of 8% per year, compounded continuously. When will you have received an income of $1 million from the company? (Do not take into account the bank's interest; this question is simply asking when the total income you have received will reach $1 million.) A) July 1, 2019 B) January 1, 2019 C) July1, 2018 D) January1, 2018 Ans: A difficulty: easy section: 8.6
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Chapter 8: Using the Definite Integral
67. By the year 2016 you will have made your first million dollars. You invest it in a new company on January 1, 2017. The new company starts to earn a profit six months later. Thus, starting July 1, 2017, you receive income from the company in a continuous stream at a constant rate of 1/2 million dollars per year. Your bank offers interest at a nominal rate of 8% per year, compounded continuously. Consider the date when will you have received an income of $1 million from the company. (Do not take into account the bank's interest; this question is simply asking when the total income you have received will reach $1 million.) Which of the following is true? A) Your investment paid off on that date. B) Your investment paid off before that date C) Your investment will not be paid off until after that date Ans: C difficulty: easy section: 8.6 68. By the year 2016 you will have made your first million dollars. You invest it in a new company on January 1, 2017. The new company starts to earn a profit six months later. Thus, starting July 1, 2017, you receive income from the company in a continuous stream at a constant rate of 1/2 million dollars per year. Your bank offers interest at a nominal rate of 6% per year, compounded continuously. Suppose T is measured in years from January 1, 2017. Compute the future value of your original investment of $1 million at time T and the future value at time T of the income that you have received by that time. Use these to determine how many years, T, it will take for your investment to have paid off. Round to 2 decimal places. Ans: 2.70 years difficulty: medium section: 8.6 69. Somebody offers to pay you money in one of the following ways: • Two $50 payments, one six months from now and one twelve months from now. • Payment in a continuous cash flow over the next year at a constant rate of $99 per year. These payments are to be deposited into a bank account that earns 10% interest compounded continuously. You want to determine which plan is preferable, i.e., which plan has a larger present value. Find the present value of the first payment plan. Then use a Riemann sum converted into a definite integral to get the exact present value of the second payment plan. Which payment plan has the larger present value? A) The second one B) The first one Ans: A difficulty: medium section: 8.6 70. You have a bank account that earns 9% nominal annual interest compounded continuously, and you want to have $90,000 in the bank account in five years so that you can buy a brand new Porsche. How much money would you have to deposit in one lump sum today so that the account balance would be $90,000 in five years? Round to the nearest dollar. Ans: $57,387 difficulty: medium section: 8.6
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Chapter 8: Using the Definite Integral
71. You have a bank account that earns 8% nominal annual interest compounded continuously, and you want to have $90,000 in the bank account in five years so that you can buy a brand new Porsche. You decide to deposit money in the account at a constant continuous rate of K dollars per year. Use a Riemann sum followed by an integral to determine the value of K (to the nearest dollar) so that you will have the $90,000 after five years. Ans: $14,639 difficulty: hard section: 8.6 72. It is estimated that in fifteen years, it will cost $200,000 to send a child to a four-year college. Suppose you want to set up an account at a bank that offers 8% nominal annual interest compounded continuously, so that 20 years from today, the account has $200,000 in it for your child's college education. Determine at what constant continuous rate K dollars per year you would need to deposit money as follows: Set up a differential equation for the rate of change of your bank balance, where B = f(t) is your bank balance at time t. Next, solve this differential equation for an initial balance of zero. Finally, use this result find K (to the nearest dollar). Ans: $4048 difficulty: hard section: 8.6 73. An insurance salesman offers you a life insurance policy with the following terms. You are to make payments at a rate of $1000 per year until age 70. If you pass away at any time, the policy will pay $150,000. Consider the payments to be made at a constant continuous rate of $1000 per year. You are 30 years old, and you have a bank account that you know will offer you 5% nominal annual interest compounded continuously for an indefinite amount of time. You could choose instead to make the $1000 payments into your bank account instead of paying for the insurance policy. If you stop these payments at age 70, just as you would stop making payments on the life insurance policy, then you would simply earn interest on the bank balance that you have accumulated to that point. If you feel that you are likely to live to be 76 years old, which is the better option? A) The insurance policy B) The bank account Ans: B difficulty: hard section: 8.6
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Chapter 8: Using the Definite Integral
74. You have $100,000 that you want to invest. Some “business men” are willing to sell you a machine for your $100,000 that prints money. You figure that every day you can print $200 with the machine, and you would deposit the $200 each day in a “special” bank account at BCCI. Your friends at BCCI will only be able to offer you 5% nominal annual interest, compounded continuously, due to the “sensitive nature” of the transaction. It would be your intention to print money each day for one year. Which of the following sums gives the value of your bank balance after one year? 365
A)
∑
0.05 ti 200(365)e 365 ∆t , with ∆t =1 day
i =1
0.05 (365−ti ) − 1 ∆t , with ∆t =1 day ∑ 200 e 365 i =1
365
B)
365
C)
∑
0.05 (365−ti ) 200e 365 ∆t , with ∆t =1 day
i =1
0.05 ti D) ∑ 200 e 365 − 1 ∆t , with ∆t =1 day i =1 Ans: C difficulty: medium section: 8.6 365
75. You have $100,000 that you want to invest. Some “business men” are willing to sell you a machine for your $100,000 that prints money. You figure that every day you can print $200 with the machine, and you would deposit the $200 each day in a “special” bank account at BCCI. Your friends at BCCI will only be able to offer you 5% nominal annual interest, compounded continuously, due to the “sensitive nature” of the transaction. It would be your intention to print money each day for one year. If you were depositing the money that you printed in a continuous stream at a constant rate of $200 per day into the bank account, then find and evaluate a definite integral that gives your balance after one year. Round to the nearest dollar. Ans: $74,856 difficulty: medium section: 8.6
Page 26
Chapter 8: Using the Definite Integral
76. AnsYou have $100,000 that you want to invest. Some “business men” are willing to sell you a machine for your $100,000 that prints money. You figure that every day you can print $310 with the machine, and you would deposit the $310 each day in a “special” bank account at BCCI. Your friends at BCCI will only be able to offer you 5% nominal annual interest, compounded continuously, due to the “sensitive nature” of the transaction. It would be your intention to print money each day for one year. If you were depositing the money that you printed in a continuous stream at a constant rate of $310 per day into the bank account, then find and evaluate a definite integral that gives your balance after one year. If you had just taken the original $100,000 and placed it in a regular bank account that compounds interest annually, then what interest rate would you have had to earn in order for this option to be more profitable than the money machine (legal concerns aside)? Round to 2 decimal places. Ans: 16.03% difficulty: hard section: 8.6 77. The capital value of an asset such as a machine is sometimes defined as the present value of all future net earnings of the asset. The actual lifetime of the asset may not be known, and since some assets last indefinitely, the capital value of the asset may be written in the form ∞
∫0 K (t )e
− rt
dt ,
where K(t) is the annual rate of earnings produced by the asset at time t, and r is the annual interest rate, compounded continuously. Find the capital value of an asset that generates income at a rate of $1000 per year, with an interest rate of 9%. Round to the nearest dollar. Ans: $11,111 difficulty: medium section: 8.6
Page 27
Chapter 8: Using the Definite Integral
78. A lightbulb company is interested in the lifespan of their lightbulbs. They have 20,000 lightbulbs burning and have collected the following information. After 2 months, 98% of the bulbs were still working. After 8 months, 80% of the bulbs were still working. We summarize all the data collected below: (Read carefully: the data was not collected at regular intervals.) # of months 2 4 8 10 12 14 18 20
% of bulbs still burning 98 92 80 64 40 20 4 0
How many bulbs out of the original 20,000 burned out during the first 4 months? Ans: 1600 difficulty: easy section: 8.7 79. A lightbulb company is interested in the lifespan of their lightbulbs. They have 20,000 lightbulbs burning and have collected the following information. After 2 months, 98% of the bulbs were still working. After 8 months, 80% of the bulbs were still working. We summarize all the data collected below: (Read carefully: the data was not collected at regular intervals.) # of months 2 4 8 10 12 14 18 20
% of bulbs still burning 98 92 80 64 40 20 4 0
Approximate the average lifespan of a lightbulb in months. Round to 2 decimal places. Ans: 10.92 months difficulty: medium section: 8.7
Page 28
Chapter 8: Using the Definite Integral
80. The probability density function f(x) shown below describes the chances that a computer circuit board will cost a manufacturer more than a certain number of dollars to produce. In this case, the cost of the circuit board, x, is measured in thousands of dollars. What is the probability that the circuit board will cost less than $2 thousand to produce?
Ans: 0 difficulty: easy
section: 8.7
81. The probability density function f(x) shown below describes the chances that a computer circuit board will cost a manufacturer more than a certain number of dollars to produce. In this case, the cost of the circuit board, x, is measured in thousands of dollars. Which of the following definite integrals give the probability that the circuit card will cost between $2 thousand and some amount $b thousand? Assume that b is between 2 and 10.
A)
b
∫2
Ans: A
f ( x) dx
B)
b
∫ (1 − f ( x)) dx 2
difficulty: easy
C)
section: 8.7
Page 29
10
∫b
f ( x) dx
D)
10
∫b (1 − f ( x)) dx
Chapter 8: Using the Definite Integral
82. The probability density function f(x) shown below describes the chances that a computer circuit board will cost a manufacturer more than a certain number of dollars to produce. In this case, the cost of the circuit board, x, is measured in thousands of dollars. Find the height of the triangle that describes the probability density function.
Ans: 0.25 difficulty: medium
section: 8.7
83. Suppose that the distribution of family sizes in the city of Boston in the year 1956 was given by: Size: # of Families:
2 13 921
3 9770
4 8955
5 5251
6 2520
≥7 2426
If this data were represented on the following histogram as a density distribution function, what would be the height of the bar located at 3? Round to 2 decimal places.
Ans: 0.23 difficulty: medium
section: 8.8
Page 30
Chapter 8: Using the Definite Integral
84. Suppose that the distribution of family sizes in the city of Boston in the year 1956 was given by: Size: # of Families:
2 13 921
3 9770
4 8955
5 5251
6 2520
≥7 2426
Using this data, find the mean of family sizes in the city of Boston in the year 1956. Assume that all of the families with 7 or more members had precisely 7 members. Round to 2 decimal places. Ans: 3.53 people difficulty: medium section: 8.8 85. Let p(t) be a probability density which is defined for 0 ≤ t ≤ 1. Could the following be the cumulative distribution function for p? (Remember that the cumulative distribution function at time t is the integral of p from 0 to t.)
Ans: yes difficulty: easy
section: 8.8
86. In the following probability density function, is the mean smaller or greater than the median?
Ans: smaller difficulty: medium
section: 8.8
Page 31
Chapter 8: Using the Definite Integral
87. A professor gives the same 100-point final exam year after year and discovers that this students' scores tend to follow the triangular probability density function f(x) pictured below:
Do the median and the mean both describe the same point on this probability density function? (All persons, places, and events in this story are fictitious. Any similarity to real persons or situations is purely coincidental.) Ans: yes difficulty: easy section: 8.8 88. A professor gives the same 100-point final exam year after year and discovers that this students' scores tend to follow the triangular probability density function f(x) pictured below:
Find the value of the height h of the triangular probability density function. (All persons, places, and events in this story are fictitious. Any similarity to real persons or situations is purely coincidental.) Ans: 0.02 difficulty: medium section: 8.8
Page 32
Chapter 8: Using the Definite Integral
89. A professor gives the same 100-point final exam year after year and discovers that this students' scores tend to follow the triangular probability density function f(x) pictured below:
What percent of the students would you expect to score above 25 points on the exam? (All persons, places, and events in this story are fictitious. Any similarity to real persons or situations is purely coincidental.) Ans: 87.5% difficulty: medium section: 8.8 90. Suppose that the distribution of people's ages in the United States is essentially constant, or uniform, from age 0 to age 60, and from there it decreases linearly until age 100. This distribution p(x) is shown below, where x is age in years, and p measures probability density. Such a probability distribution is called trapezoidal.
According to this simplified model of the distribution of people's ages in the United States, what percentage of the population is between 0 and 100 years old? Ans: 100% difficulty: easy section: 8.8
Page 33
Chapter 8: Using the Definite Integral
91. Suppose that the distribution of people's ages in the United States is essentially constant, or uniform, from age 0 to age 60, and from there it decreases linearly until age 100. This distribution p(x) is shown below, where x is age in years, and p measures probability density. Such a probability distribution is called trapezoidal.
In terms of the length of the base, b, of the trapezoidal distribution (notice that the base of the trapezoid lies along the p-axis), find the fraction of the population that is between 0 and 60 years old. A) 20b B) 60b C) 40b D) 80b Ans: B difficulty: medium section: 8.8 92. Suppose that the distribution of people's ages in the United States is essentially constant, or uniform, from age 0 to age 60, and from there it decreases linearly until age 100. This distribution p(x) is shown below, where x is age in years, and p measures probability density. Such a probability distribution is called trapezoidal.
Find the value of b. Ans: 0.0125 difficulty: medium
section: 8.8
Page 34
Chapter 8: Using the Definite Integral
93. In a hydrogen atom in the unexcited state, the probability of finding the sole electron within x meters of the nucleus is given by
−2 r x 2 a F ( x) = r e 0 dr , for 3 0 (a0 )
4
∫
x ≥ 0 , where
= a0 5.29 ×10−11 meters. What is its corresponding probability density function f(x)? −2 x
A)
−2 ) f ( x= ⋅ 2 x ⋅ e a0 3 a0 (a0 )
B)
−2 x −2 x 4 −2 2 a0 f ( x) =3 ⋅ x e + 2 xe a0 (a0 ) a0
C)
D)
4
f ( x) =
4 (a0 )3
f ( x) = −
Ans: C
12
2
x e
−2 x a0
−2 x 2 a0
x e (a0 ) 4 difficulty: easy
section: 8.8
94. In a hydrogen atom in the unexcited state, the probability of finding the sole electron within x meters of the nucleus is given by
−2 r x 2 a F ( x) = r e 0 dr , for (a0 )3 0
4
∫
x ≥ 0 , where
= a0 5.29 ×10−11 meters. Which of the following are true about the corresponding probability density function f(x)? Select all that apply. f ( x) has a local and global minimum at a0 . A) f ( x) has a local and global maximum at a0 . B) f ( x) has a local and global minimum at 0. C) f ( x) has a local and global maximum at 0. D) f ( x) → ∞ as x → ∞ . E) f ( x) → 0 as x → ∞ . F) Ans: B, C, F difficulty: medium section: 8.8
Page 35
Chapter 8: Using the Definite Integral
95. In a hydrogen atom in the unexcited state, the probability of finding the sole electron within x meters of the nucleus is given by
−2 r x 2 a F ( x) = r e 0 dr , for 3 0 (a0 )
4
∫
x ≥ 0 , where
= a0 5.29 ×10−11 meters. What is the probability that the electron will be found within a sphere of radius a0 ? Round to 3 decimal places. Ans: 0.323 difficulty: hard section: 8.8 96. In March 1995 the space shuttle carried an experiment designed by a Harvard student who studies the growth of crystals. Suppose the probability density function of the length, x cm, of a crystal grown in space is modeled by p ( x) = xe − x for x ≥ 0 . The cumulative distribution function giving the probability that a crystal has length ≤ t cm is represented by P(t). Which of the quantities (i) - (x) below best approximates the probability that a crystal has length between 4 cm and 4.01 cm? (i) p (4) (ii) p (4.01) − p (4) (iii) p (4)(0.01) p (4) (iv) 0.01 p (4.01) − p (4) (v) 0.01
(vi) P(4) (vii) P (4.01) (viii) P (4)(0.01) P (4) (ix) 0.01 P(4.01) − P(4) (x) 0.01
Ans: (iii) is best because it approximates the area underneath p ( x) between 4 and 4.01. difficulty: medium section: 8.8
Page 36
Chapter 8: Using the Definite Integral
97. In March 1995 the space shuttle carried an experiment designed by a Harvard student who studies the growth of crystals. Suppose the probability density function of the length, x cm, of a crystal grown in space is modeled by p ( x) = xe − x for x ≥ 0 . The cumulative distribution function giving the probability that a crystal has length ≤ t cm is represented by P(t). Which of the quantities (i) - (x) below represents precisely the probability that a crystal has length less than 3.01 cm? (i) p (3) (ii) p (3.01) − p (3) (iii) p (3)(0.01) p (3) (iv) 0.01 p (3.01) − p (3) (v) 0.01 Ans: (vii) difficulty: easy
(vi) P (3) (vii) P (3.01) (viii) P (3)(0.01) P(3) (ix) 0.01 P(3.01) − P(3) (x) 0.01
section: 8.8
98. In March 1995 the space shuttle carried an experiment designed by a Harvard student who studies the growth of crystals. Suppose the probability density function of the length, x cm, of a crystal grown in space is modeled by p ( x) = xe − x for x ≥ 0 . The cumulative distribution function giving the probability that a crystal has length ≤ t cm is represented by P(t). Which of the following formulas gives the cumulative distribution function P(t) for t ≥ 0? t −1 t +1 t −1 t +1 B) − t C) 1 + t D) 1 − t A) t e e e e Ans: D difficulty: medium section: 8.8 99. In March 1995 the space shuttle carried an experiment designed by a Harvard student who studies the growth of crystals. Suppose the probability density function of the length, x cm, of a crystal grown in space is modeled by p ( x) = xe − x for x ≥ 0 . The cumulative distribution function giving the probability that a crystal has length ≤ t cm is represented by P(t). What is the mean crystal length that this model predicts? Give your answer to two decimal places. Ans: 2.00 cm difficulty: hard section: 8.8
Page 37
Chapter 8: Using the Definite Integral
100. The price of crude oil in the past can be approximated by P(t ) = 40 − (t − 4) 2 , where P(t) is measured in $US/barrel, and time t is measured in months, with t = 0 on July 1, 1990. In the same time period, Saudi Arabia produced oil at a rate approximated by R(t ) = 160 + 30 arctan(t − 3) (measured in million barrels per month). Assume that the oil is sold continuously two months after its production. How many billions of dollars did Saudi Arabia get for the oil it produced in the last 6 months of 1990? Round to 1 decimal place. Ans: 29.6 difficulty: hard section: 8 review 101. Consider the 3-sided region bounded by x = 0 , y = 4 and y = x . What is the exact volume of the solid obtained when this region is rotated around the x-axis? A) 128π B) 8π C) 128 D) 8 Ans: A difficulty: easy section: 8 review 102. Consider the 3-sided region bounded by x = 0 , y = 4 and y = x . The exact volume of the solid obtained when this region is rotated around the line x = –3 can be written as Qπ . Find Q. 15 Ans: Q = 4992. difficulty: medium section: 8 review 103. Consider the 3-sided region bounded by x = 0 , y = 2 and y = x . What is the perimeter of this region? Give your answer to 3 decimal places. Ans: 2.562 difficulty: medium section: 8 review 104. The following table lists the width of the Great Pyramid of Egypt at different heights. The Pyramid has a square base and each layer is a square. height (ft) 0 100 200 300 400 side length (ft) 750 562 375 189 0 Find a good estimate (in ft3) for the volume of the Great Pyramid. Ans: 77,344,000 ft3 difficulty: easy section: 8 review
Page 38
Chapter 8: Using the Definite Integral
105. A flag is hanging over the side of a building. Assume that the flag is a right triangle that has a width (one leg) of 25 feet at the top of the building, a length (other leg) of 45 feet hanging down the side of the building, and a weight of 85 pounds. Which integral represents the work needed to lift this flag to the roof? 45 45 2(85) 5 2 85 5 25 h − h dh A) C) 25 − h dh ∫ ∫ (25)(45) 0 9 (25)(45) 0 9
85 5 2 25h − h dh ∫ (25)(45) 0 9 Ans: A difficulty: medium 45
B)
2(85) 9 2 45h − h dh ∫ (25)(45) 0 5 section: 8 review 25
D)
106. A flag is hanging over the side of a building. Assume that the flag is a right triangle that has a width (one leg) of 25 feet at the top of the building, a length (other leg) of 45 feet hanging down the side of the building, and a weight of 89 pounds. Find the amount of work (in ft-lbs) needed to lift this flag to the roof. Ans: 1335 ft-lbs difficulty: hard section: 8 review 107. Consider the region bounded by the x-axis, the line x = 1, the line x = 4, and the curve y = x 2 . Imagine that this region is the base of a solid. Each cross-section of this solid perpendicular to the x-axis is a triangle of height H . HB . What is B? The volume of this solid can be found exactly and can be written as 6 B) 15 C) 3 D) 64 A) 63 Ans: A difficulty: medium section: 8 review
Page 39
7 7 7 7 1. Find a formula for sn , n ≥ 1, for the sequence 7, − , , − , ,... 2 3 4 5 n +1 n 7(−1) 7(−1) 7(−1) n+1 7(−1) n A) sn = B) sn = C) sn = D) sn = n n +1 n n +1 Ans: A difficulty: easy section: 9.1 2. Find a recursive formula for sn , n ≥ 1, for the sequence 1, 6, 41, 286, 2001, . . . A) C) sn= 7 − sn−1 with s1 = 7 sn= 7 − sn−1 with s1 = 1 B) = D) = sn 7 sn−1 − 1 with s1 = 1 sn 7 sn−1 − 1 with s1 = 7 Ans: B difficulty: easy section: 9.1 3. Let Pn be the number of people visiting an amusement park on the nth day after it opens. What does P10 represent? A) The average number of people visiting the amusement park over the first 10 days it was open. B) The number of days it takes for 10 people to visit the amusement park. C) The number of people visiting the amusement park on the 10th day after it opens. D) The total number of people who have visited the amusement park the first 10 days it was open. Ans: C difficulty: easy section: 9.1 4. Let Pn be the number of people visiting a zoo on the nth day after it opens. What does 25
∑ Pn represent?
n =1
A) The average number of people visiting the zoo over the first 25 days it was open. B) The number of days it takes for 25 people to visit the zoo. C) The number of people visiting the zoo on the 25th day it is open. D) The total number of people who visited the zoo in the first 25 days after it opened. Ans: D difficulty: easy section: 9.1 5. Let Pn be the number of people visiting an aquarium on the nth day after it opens. What does it mean in terms of aquarium attendance if Pn > Pn+1 for all n? A) The number of people visiting the aquarium goes down each day. B) The number of people visiting the aquarium goes up each day. C) The number of people visiting the aquarium goes up some days and down some days. Ans: A difficulty: easy section: 9.1
Page 1
Chapter 9: Sequences and Series
6. Let Pn be the number of people visiting an amusement park on the nth day after it opens. Suppose= Pn 2000 − 4n . How many people visit the amusement park in its first week? A) 13888 B) 13916 C) 224 D) 14000 Ans: A difficulty: medium section: 9.1 7. Let Pn be the number of people visiting a zoo on the nth day after it opens. Suppose = Pn 2000 − 4n. You find out that the museum must close if is has fewer than 300 visitors per day. How long will it remain open? A) It will be closed on day 426. C) It will be closed on day 7. B) It will be closed on day 501. D) It will close on day 1997. Ans: A difficulty: medium section: 9.1 8. Stock prices for Abercrombie and Fitch fell steadily by an average of $0.94 per day from a high of $83.67 per share on December 24, 2007 to $70.05 on January 15, 2008. Let P1 be the price of a share of stock on December 24, 2007. Write a formula for Pn , the price of a share on the nth day after December 24. Ans: Pn =83.67 − 0.94(n − 1) difficulty: easy section: 9.1 9. Stock prices for Abercrombie and Fitch fell steadily by an average of $0.94 per day from a high of $83.67 per share on December 24, 2007 to $70.05 on January 15, 2008. Let P1 be the price of a share of stock on December 24, 2007 and Pn be the price of a share on the nth day after December 24. Write a formula for Pn +1 − Pn and then write a sentence to interpret the meaning of your formula. Ans: Pn +1 − Pn = 83.67 − 0.94n − 83.67 + 0.94(n − 1) = −0.94 , the average decrease in stock price per day. difficulty: medium section: 9.1
Page 2
Chapter 9: Sequences and Series
5
10. Consider the finite sequence An given in the graph below. Find ∑ 3 An . n =1
9
y
(1, 8)
8 7
(2, 6)
6 5
(3, 4)
4 3
(4, 2)
2 1
-1
x
(5, 0) 1
2
3
4
5
6
7 8 (6, -2)
-2 -3
(7, -4)
-4 -5
A) 60 Ans: A
B) 15 C) 75 difficulty: easy
D) 56 E) None of the above section: 9.1
11. Are the following sequences monotone? Answer "yes" or "no" to each one. n nπ A. sn = sin B. sn= (1 − n) n C. sn = n cos nπ D. sn = n +1 2 (−1) n n Part A: no Part B: no Part C: no Part D: yes Part E: no difficulty: medium
E. sn =
section: 9.1
Page 3
Chapter 9: Sequences and Series
12. Which one of the following sequences diverges to positive infinity as n → ∞?
1 n sn= n + n cos nπ Ans: D difficulty: medium A. sn = 1 +
B. sn =
1 − n2 n
C. sn =
cos nπ n
D.
section: 9.1
13. Select the appropriate word to fill in the blank: A convergent sequence is ________________. A) bounded B) monotonic C) finite Ans: A difficulty: easy section: 9.1 14. A couple puts $500,000 for their retirement into an account paying 5% annual interest. They estimate that they will need to withdraw $60,000 each year to live on. Assume that the $60,000 is withdrawn on the last day of the year. Find a recursive formula for sn , the amount of money left in the account at the end of n years, and use it to determine how many years the money will last (how many years until there is less than $60,000 in the account). Ans: 11 difficulty: medium section: 9.1 15. Compute the first 8 terms of the sequence an = 2 + (−1) n / n on plot them on a number line. To what number does it appear the sequence converges, if any? Ans: 2 difficulty: medium section: 9.1 16. Determine whether the sequence 6 + 2.5 × (−1) n converges or diverges. A) It converges. B) It diverges. Ans: B difficulty: easy section: 9.1 17. A radioactive isotope is released into the air as an industrial by-product. This isotope is not very stable due to radioactive decay. Two-thirds of the original radioactive material loses its radioactivity after each month. If 12 grams of this isotope are released into the atmosphere at the end of the first and every subsequent month, identify the closed form sum that gives the amount of the isotope in the atmosphere at the end of the nth month. A)
Sn =
Ans: A
12 − ( 13 )n ⋅12 1 − 13
B) Sn =
difficulty: easy
12 + ( 13 )n ⋅12 1 3
section: 9.2
Page 4
C) Sn =
12 ⋅ ( 13 )n 1 − 13
Chapter 9: Sequences and Series
18. A radioactive isotope is released into the air as an industrial by-product. This isotope is not very stable due to radioactive decay. Two-thirds of the original radioactive material loses its radioactivity after each month. If 15 grams of this isotope are released into the atmosphere at the end of the first and every subsequent month, how many grams of radioactive material are in the atmosphere at the end of the twelfth month? Round to 2 decimal places. Ans: 22.50 difficulty: medium section: 9.2 19. A radioactive isotope is released into the air as an industrial by-product. This isotope is not very stable due to radioactive decay. Two-thirds of the original radioactive material loses its radioactivity after each month. If 13 grams of this isotope are released into the atmosphere at the end of the first and every subsequent month and the situation goes on ad infinitum, how many grams of radioactive material are in the atmosphere at the end of each month in the long run? Ans: 19.5 difficulty: easy section: 9.2 2
100
2 2 2 20. Find the value of + + ... + 5 5 5 Ans: 0.67 difficulty: medium section: 9.2
to 2 decimal places.
21. Find the value of the infinite product e1/8 ⋅ e1/16 ⋅ e1/ 32 ⋅ e1/ 64 ⋅ ... ⋅ e1/ 2 places. Ans: 1.28 difficulty: medium section: 9.2
n+ 2
⋅ ... to 2 decimal
22. Suppose the government spends $3.5 million on highways. Some of this money is earned by the highway workers who in turn spend $1,750,000 on food, travel, and entertainment. This causes $875,000 to be spent by the people who work in the food, travel, and entertainment industries. This $875,000 causes another $437,500 to be spent; the $437,500 causes another $218,750 to be spent, and so on. (Notice that each expenditure is half the previous one.) Assuming that this process continues forever, how many million dollars in total spending is generated by the original $3.5 million expenditure (including the original $3.5 million)? Ans: 7 difficulty: medium section: 9.2 23. Does the infinite series 4 +
4 4 4 4 + + 3/ 2 + 2 + ⋅⋅⋅ converge or diverge? 3 3 3 3
Ans: It converges. difficulty: easy section: 9.2
Page 5
Chapter 9: Sequences and Series
∞
24. Find the 6th partial sum of the series ∑ ( 5 )i (to two decimal places). i =0
A) 26.12 Ans: A
3
B) ∞ C) 83.28 D) 248.66 difficulty: easy section: 9.3
25. Find the sum of the first 6 terms of the series 3 + decimal places. Ans: 8.96 difficulty: medium
3 3 3 3 + + 3/ 2 + 2 + ⋅⋅⋅ . Round to 2 2 2 2 2
section: 9.2 15
n
4 26. Find the sum of the series ∑ to 2 decimal places. n=5 3 Ans: 286.68 difficulty: medium section: 9.2 ∞
n
4 27. Find the sum ∑ . Round to 2 decimal places. n=2 5 Ans: 3.20 difficulty: medium section: 9.2 2 of the height 3 of the bounce before. Find an expression for the height to which the ball rises after it hits the floor for the nth time. 2 n 111 − n 3 2 n 2 2 A) 11 ⋅ n B) 11 C) 111 − D) 2 3 3 3 1− 3 Ans: B difficulty: medium section: 9.2
28. A ball is dropped from a height of 11 feet and bounces. Each bounce is
3 of the height 5 of the bounce before. Find the total vertical feet the ball has traveled when it hits the floor for the 4th time. Round to 1 decimal place. Ans: 60.3 difficulty: medium section: 9.2
29. A ball is dropped from a height of 18 feet and bounces. Each bounce is
Page 6
Chapter 9: Sequences and Series
30. A tennis ball is dropped from a height of 15 feet and bounces. Each bounce is
1 the 2
3 the height of the bounce 4 before, and is dropped from a height of 5 feet. Which ball bounces a greater total vertical distance? A) The tennis ball B) The superball Ans: A difficulty: hard section: 9.2 height of the bounce before. A superball has a bounce
31. Which of the following series are geometric? (1) 5 + 5a + 5a 2 + 5a 3 + ... (2) 5 + 7 a + 9a 2 + 11a 3 + ... (3) 5 + 5ak + 5a 2 k 2 + 5a 3 k 3 + ...
A) (1) and (2) B) (1) and (3) C) (2) and (3) Ans: B difficulty: easy section: 9.2
D) (1) only
E) (2) only
32. Is 3 + 6ak + 9a 2 k 2 + 12a3k 3 + ⋅⋅⋅ a geometric series? Ans: no difficulty: easy section: 9.2 33. Jamie was born in May. In August, her grandparents started a "Go to College in France" fund with $2200, earning a fixed annual interest rate of 7%. They added an additional $2200 each year in August until the last deposit in the year Jamie turned 18. Jamie estimated that she needed $90,000 to go start college in France. How much did she have in her "Go to College in France" fund? Did she have enough? 18
A)
∑ 2200(1.07) = $82233.72 , no
D)
∑ 2200(1.07)n = $74797.87 , no
E)
n
n= 0 18
B)
18
n= 0
C) $91533.97, yes Ans: A difficulty: medium
section: 9.2
Page 7
$87533.83, no
∑ 2200(1.07) n =0
n+2
= $74797.87 , no
Chapter 9: Sequences and Series
34. Consider the series: 13 + 52 + 73 + 94 + 115 + ... (a) Find a formula for the general term sn , n ≥ 1 . (b) Find the partial sums S1 , S 2 , S3 , S 4 , S5 . (c) Use your result from part (b) to predict the limit of the partial sums, lim S n . Does n →∞
this indicate that the series converges or diverges? n Ans: (a) sn = 2n + 1 (b) 1/3, 11/15, 122/105, 506/315, 7141/3465 (c) lim S n appears to be infinite. This would indicate that the series diverges. n →∞
difficulty: easy
section: 9.3
7n converge or diverge. Explain. n+8 7n Ans: It diverges because lim = 7 ≠ 0. n →∞ n + 8 difficulty: easy section: 9.3
35. Does the series ∑
∞
(ln n)3 converges or diverges. n n =1
36. Use the integral test to decide whether the series ∑ A) It converges B) It diverges Ans: B difficulty: medium section: 9.3
n+2 converges or 2 n =1 n + n ∞
37. Use the integral test, if applicable, to determine whether the series ∑
diverges. A) It converges. B) It diverges. C) The integral test does not apply. Ans: B difficulty: medium section: 9.3 ∞
5 converges or diverges. 2 n =1 1 + n
38. Use the integral test to decide whether the series ∑ Ans: Converges difficulty: medium ∞
39. Does the series ∑
n =0
Ans: converge difficulty: medium
section: 9.3 6n 2+n
6
converge or diverge?
section: 9.3
Page 8
Chapter 9: Sequences and Series
∞
40. Does the series ∑
3n 4 + 2
7 n =0 n + 2 n + 9
Ans: converges difficulty: medium
converge or diverge?
section: 9.3
1 1 1 1 41. Does 1 + + + + ⋅⋅⋅ + + ⋅⋅⋅ converge or diverge? 2 5 8 3n − 1 Ans: diverge difficulty: medium section: 9.3 ∞
∞
42. True or False: If ∑ an converges then ∑ kan converges (k ≠ 0). n =1
Ans: True
difficulty: easy
n =1
section: 9.3 ∞
43. True or False: If lim an is not zero, then ∑ an does not converge. n→∞
Ans: True
difficulty: easy
n =1
section: 9.3
1 diverges. We can form a new series from the difference n 1 1 between consecutive terms of the harmonic series obtaining ∑ [ − ] . True or n +1 n false: This series also diverges. Ans: False difficulty: medium section: 9.3
44. The harmonic series ∑
45. If ∑ ak is the sum of a series of numbers and lim ak = 0 , then the series converges. k →∞
Ans: False
difficulty: easy
section: 9.4
46. If a series of constants ∑ ak diverges, then must ∑ ak diverge? Ans: yes difficulty: easy section: 9.4 ∞
47. Use the comparison test to determine whether ∑
1
2 n n =1 5n + e
A) It converges. B) It diverges. Ans: A difficulty: medium section: 9.4
Page 9
converges.
Chapter 9: Sequences and Series
∞
(−1) n−1
n =1
8n 4
48. Use the alternating series test to decide if ∑
converges.
A) It converges. B) It diverges. Ans: A difficulty: medium section: 9.4 49. Leonard Euler found the following series to be noteworthy because it is part of a process that can be used to approximate π :
1−
1 1 1 1 + − + − ... 2 3 3 ⋅ 64 5 ⋅ 64 7 ⋅ 64 9 ⋅ 644
(a) Re-write this series in summation notation. (b) Use an appropriate test to show that the series converges.
(−1) n Ans: (a) ∑ n n = 0 (2n + 1) ⋅ 64 ∞
1 =0 and that n →∞ (2n + 1) ⋅ 64 n
(b) Using the alternating series test we find that lim 1 1 n=0, 1, 2, ... < n +1 (2n + 3) ⋅ 64 (2n + 1) ⋅ 64n difficulty: medium section: 9.4
50. Determine whether the following series converge or diverge: ∞ ∞ 12 n a) ∑ 12 b) ∑ n =1 n n =1 12n A) (a) converges and (b) diverges C) both series diverge B) (a) diverges and (b) converges D) both series converge Ans: A difficulty: easy section: 9.3 ∞
51. Estimate the error in approximating the sum of the alternating series S = ∑
(−1) n−1
n n =1 1 + 2
the sum of the first 11 terms. A) 0.000024 B) 0.00024 Ans: B difficulty: medium
C) 0.0024 D) 0.024 section: 9.4 ∞
52. What does the ratio test tell us about the series ∑
1
n =1 n
3
?
A) The series converges. B) The series diverges. C) The ratio test doesn't tell us anything about the convergence of the series. Ans: C difficulty: medium section: 9.4
Page 10
by
Chapter 9: Sequences and Series
∞
53. What does the ratio test tell us about the series ∑
n
n n =1 2
?
A) The series converges. B) The series diverges. C) The ratio test doesn't tell us anything about the convergence of the series. Ans: A difficulty: medium section: 9.4 (−0.125) n−1 ? n n =1 ∞
54. What does the ratio test tell us about the series ∑
A) The series diverges. B) The series converges. C) The ratio test doesn't tell us anything about the convergence of the series. Ans: B difficulty: medium section: 9.4 ∞
7 n5
55. Use the limit comparison test to determine whether ∑
6 2 n =1 n + n + 2 A) The series diverges. B) The series converges. Ans: A difficulty: medium section: 9.4
converges.
56. True or False: If a power series ∑ ak x k converges at x = 6 and x = 7 then it converges at x = –6. Ans: True difficulty: medium section: 9.5 57. True or False: If a power series ∑ ak x k converges at x = c then it also converges at x = -c. Ans: False difficulty: medium section: 9.5 ∞
58. Find the interval of convergence for ∑ A) −7 < x < 1 B) −1 ≤ x < 1 Ans: B difficulty: medium
n =0
xn . n+7
C) −1 ≤ x < 7 section: 9.5
D) −7 < x < 7
x x 2 x3 x 4 + + + + ⋅⋅⋅ 5 12 31 68 xn for n ≥ 1 . n3 + 4
59. Find an expression for the general term of the series A)
xn for n ≥ 1 . n+4
xn B) for n ≥ 1 . 7n − 2 Ans: C difficulty: easy
C) D) section: 9.5
Page 11
xn n 2 + 12
for n ≥ 1 .
Chapter 9: Sequences and Series
∞
60. Use the ratio test to find the radius of convergence of ∑
n( x + 2) n
n =0
2n+1
.
Ans: The radius of convergence is 2. difficulty: medium section: 9.5 61. If the series ∑ Cn x n has a radius of convergence of 4 and ∑ Dn x n has a radius of convergence of 6, what is the radius of convergence of ∑ (Cn + Dn ) x n ? Ans: 4 difficulty: easy section: 9.5
62. Find the radius of convergence of x +
3 x 2 4 x3 5 x 4 6 x5 + + + + ⋅⋅⋅ 12 18 24 30
Ans: The radius of convergence is 1. difficulty: medium section: 9.5 63. Fill in the blanks with the widest possible interval: ∞
The power series ∑
n
n =1 2
_____. Part A: 0.5 Part B: 2.5 difficulty: medium
n
(2 x − 3) n converges for all values of x between _____ and
section: 9.5
64. For what values of p does the series 1 + 9 p + (9 p ) 2 + (9 p )3 + ... + (9 p ) n + ... converge, if any? −1 1 −1 ≤ p ≤ 1 D) A) < p< 9 9 −1 1 B) E) The series diverges for all values of p. < p< 18 18 C) -1 < p < 1 Ans: A difficulty: easy section: 9.5 ( x − 2) 2 ( x − 2) 2 ( x − 2) 4 65. Find the interval of convergence of the series 1 + ( x − 2) + ... + + 2! 3! 4! Ans: (−∞, ∞) difficulty: easy section: 9.5
66. True or False: If ∑ Cn 6n is convergent, then ∑ Cn (−6) n Ans: False difficulty: medium section: 9 review
Page 12
is also convergent.
Chapter 9: Sequences and Series
67. True or False: If an > an+1 > 0 (for all n) and ∑ an converges, then ∑ (−1) n an converges. Ans: True difficulty: easy section: 9 review 6n+1 ( x − 1) n . + n 1 n =0 Ans: The radius of convergence is 1/6. difficulty: medium section: 9 review ∞
68. Find the radius of convergence of ∑
∞
1 converge? 1 sin + n n =1
69. Does ∑
Ans: no difficulty: medium ∞
70. Does ∑
section: 9 review
n 4 sin n
converge? 5 4 n + n =1 Ans: no difficulty: medium section: 9 review ∞
71. True or False: The ratio test can be used to determine whether ∑
1
n =1 n
Ans: False
difficulty: medium
2
converges.
section: 9 review
x a , a > 0 , then it converges 72. True or False: If the power series ∑ Cn x n converges for= a . 2 Ans: True for x =
difficulty: medium
section: 9 review
Page 13
1. Construct the Taylor polynomial approximation of degree 3 to the function f ( x) = arctan x about the point x = 0. Use it to approximate the value f (0.35) to 5 decimal places. How does the approximation compare to the actual value? Ans: 0.33571 The calculator value of arctan(0.35) is 0.336675. difficulty: medium section: 10.1 2. Approximate the function f ( x) = x 3e − x for values of x near 0 using the first three non-zero terms of its Taylor polynomial. x7 Ans: P6 ( x) = x3 − x 5 + 2 difficulty: medium section: 10.1 2
3. Approximate the function f ( x) = e− x to estimate the value of ∫ Ans: 0.53511 difficulty: medium
0.6 − x 2 e dx 0
2
with a Taylor polynomial of degree 6. Use this
to 5 decimal places.
section: 10.1
4. The graph of y = f(x) is given below.
Suppose we approximate f(x) near x = 17 by the second degree Taylor polynomial centered about 17, a + b( x − 17) + c( x − 17)2 . Is b positive, negative, or zero? Ans: zero difficulty: medium section: 10.1
( ) about x = 0?
5. What is the fourth degree Taylor polynomial for cos 3x 2
9 4 x 2 3 B) 1 − x 4 2 Ans: A difficulty: easy
A)
1−
C) D) section: 10.1
Page 1
9 27 3 81 4 1 − x2 + x − x 2 6 24 3 9 27 4 x + x 2 − x3 + x 2 6 24
Chapter 10: Approximating Functions Using Series
6. Suppose a function satisfies f (5) = 2 , f ′(5) = 5 , f ′′(5) = −7 , and f ′′′(5) = 12 . What is the third degree Taylor polynomial for f about x = 5? 7 A) C) 2 + 5 x − x 2 + 2 x3 2 + 5( x − 5) − 7( x − 5) 2 + 12( x − 5)3 2 7 35 B) D) 10 + 25 x − x 2 + 6 x3 2 + 5( x − 5) − ( x − 5) 2 + 2( x − 5)3 2 2 Ans: B difficulty: medium section: 10.1 7. The function g has the Taylor approximation g ( x) = c0 + c1 ( x − a ) + c2 ( x − a ) 2 and the graph given below:
Is c0 positive, negative, or zero? Ans: positive difficulty: medium section: 10.1 8. Find the Taylor polynomial of degree 3 around x = 0 for the function f ( x= ) use it to approximate Ans: 0.8371 difficulty: medium
1 − x and
0.7 . Give your answer to 4 decimal places. section: 10.1
0.9 and 0.2 using the Taylor polynomial of degree 3 around x = 0 for the function f ( x= ) 1 − x . Which approximation is more accurate?
9. Suppose you approximate
A) the approximation of 0.9 B) the approximation of Ans: A difficulty: easy section: 10.1
sin x using a Taylor approximation for sin x. x→0 4 x Ans: 1/4 difficulty: medium section: 10.1
10. Find lim
Page 2
0.2
Chapter 10: Approximating Functions Using Series
11. Estimate ∫
1.5
1
ln t dt using a 4th degree Taylor Polynomial for ln t about t = 1. Round
to 4 decimal places. Ans: 0.1078 difficulty: medium
section: 10.1 2
12. Is the Taylor polynomial of degree 6 for e – x for x near 0 given by 1+ Ans: no difficulty: easy
x 2 x 4 x6 + + ? 2! 4! 6!
section: 10.1
e x + e− x . Use 2 the Taylor polynomial for e x near 0 to find the Taylor polynomial of degree 4 for f ( x) = 4 cosh( x) .
13. The hyperbolic cosine function is defined as follows: = f ( x) cosh( x) =
4 x 2 4 x3 4 x 4 + − 2! 3! 4! 2 4 x x B) 4(1 + + ) 2 24 Ans: B difficulty: medium
A)
4−
C) D)
3x 3x 2 x 4 − + 1! 2! 24 4 x 2 4 x3 4 x 4 4 + 4x + + + 2 3! 4!
4+
section: 10.1 1.1
sin x dx . x 0.1
14. Use the first three nonzero terms of the Taylor polynomial to approximate ∫ Give your answer to 5 decimal places. A) 0.92868 B) 0.92880 C) –0.92880 D) –0.92868 Ans: B difficulty: medium section: 10.1
E) does not exist
15. Find the fourth term of the Taylor series for the function f ( x) = cos x about x = π / 3 . x − π3 ) ( A) 12 Ans: C
3
x − π3 ) ( B) –
3
12 difficulty: medium
C)
(
3 x − π3
)
3
12 section: 10.2
D) –
(
3 x − π3
)
3
12
( x) ln( x + 1) about x = 1. 16. Find the fourth term of the Taylor series for the function f = ( x − 1)3 ( x − 1)3 ( x − 1)3 A) B) C) D) 12 3 6 Ans: D difficulty: medium section: 10.2
Page 3
( x − 1)3 24
Chapter 10: Approximating Functions Using Series
17. What is the general term of the series 1 − A)
x5 x10 x15 + − + ⋅⋅⋅ ? 5! 10! 15!
(−1) k x5k for k ≥ 0 (5k )!
(−1) k +1 x5k for k ≥ 0 (5k )! Ans: A difficulty: easy
B)
C)
(−1) k x5k +5 for k ≥ 0 (5k + 5)!
D)
(−1) k +1 x5k +1 for k ≥ 0 (5k + 1)!
section: 10.2
42 43 44 + − + ⋅⋅⋅ as a Taylor series evaluated at a particular value of x 2 3 4 and find the sum to 4 decimal places. Ans: 1.6094 difficulty: medium section: 10.2
18. Recognize 4 −
53 55 57 + − + ⋅⋅⋅ as a Taylor series evaluated at a particular value of x 3! 5! 7! and find the sum to 4 decimal places. Ans: –0.9589 difficulty: medium section: 10.2
19. Recognize 5 −
20. Solve 1 + x +
x 2 x3 + + ⋅⋅⋅ = e 4 for x. 2! 3!
Ans: 4 difficulty: easy
section: 10.2
21. Solve 1 + x + x 2 + x3 + ⋅⋅⋅ = 8 for x. Round to 2 decimal places. Ans: 0.88 difficulty: medium section: 10.2 22. Suppose that you are told that the Taylor series of f ( x) = e− x about x = 0 is 2
( ) .
x 2 x 4 x6 d 2 − x2 1− + − + ⋅⋅⋅ . Find e 1! 2! 3! dx 2
Ans: –2 difficulty: medium
x =0
section: 10.2
23. Use the binomial series to find the coefficient of the x 2 term in the expansion of (1 + x) 4 . Ans: 6 difficulty: medium section: 10.2
Page 4
Chapter 10: Approximating Functions Using Series
24. Find an expression for the general term of the Taylor series for x 4 x 6 x8 x sin x = x 2 − + − ... . 3! 5! 7! n −1 2 n −1 x n +1 x 2n+2 (−1) x A) B) (−1) 2 n −1 C) (−1) n n! (2n + 1)! (2n + 1)! Ans: C difficulty: medium section: 10.2
D)
x2n 2n − 1
25. Find the first four non-zero terms of the Taylor series about zero for the function f ( x= ) 7(1 + x)1/ 2 . Leave coefficients in fraction form. 7 7 7 Ans: 7 + x − x 2 + x3 2 8 16 difficulty: medium section: 10.2 26. What is the interval of convergence of the Taylor series for the function f ( x= ) 9(1 + x)1/ 2 about zero? (Exclude any possible endpoints.) A) −1 < x < 1 B) −9 < x < 9 C) − 19 < x < 19 D) none of the above Ans: A difficulty: medium section: 10.2 x 2 x3 x 4 (−1) n −1 x n + − + ... + + ... does not converge for x = −1 . 2 3 4 n What behavior does it exhibit? It does converge for x = 1 . To what number does it appear to converge? A) At -1, the series diverges to ∞ , At 1, it appears to converge to ln(3). B) At -1 the series diverges because the terms alternate back and forth between positive and negative. At 1, it appears to converge to ln(2). C) At -1, the series diverges because some of the terms are undefined. At 1, it appears to converge to ln(3). D) At -1, the series diverges to −∞ . At 1, it appears to converge to ln(2). Ans: D difficulty: hard section: 10.2
27. The infinite series x −
28. Use the formula for the Taylor polynomial approximation to the function g ( x) = e x about 2
x0 = 0 to construct a polynomial approximation of degree 6 for f ( x) = e x . Use the 2
first four nonzero terms of this approximation to estimate the value of e(0.3) . Give your answer to 5 decimal places. Ans: 1.09417 difficulty: medium section: 10.3
Page 5
Chapter 10: Approximating Functions Using Series
29. Consider the function f ( x) = 1 − cos x . Is the Maclaurin series for f ( x) given by (−1)i +1 x 2i ∑ (2i)! ? i =1 ∞
Ans: yes difficulty: medium
section: 10.3
30. Based on the Maclaurin series for the function f ( x) = 1 − cos x , evaluate lim
1 − cos x
x→0
Ans: 1/4 difficulty: hard
2 x2
section: 10.3
31. The graph of the function f ( x) = e− x / 5 is a bell-shaped curve similar to a normal 2
∞
(−1)i x 2i
i =0
5 i (i !)
probability density function. Is ∑ Ans: yes difficulty: medium
the Maclaurin series for f ( x) ?
section: 10.3
32. The function f ( x) = e− x / 2 is part of the normal probability density function (or 2
bell-shaped curve). Find the Maclaurin series for ∫ e − x / 2 dx by first finding the 2
Maclaurin series for f ( x) and then integrating it term by term. ∞
A)
∑ (2i + 1)2 i (i !)
i =0 ∞
B)
(−1)i x 2i +1
(−1) x
i 2i
∑ (2i)2 i (i !)
∞
+C
C) D)
i =0
Ans: A
∑ (2i + 1)2 i (2i !) + C
i =0 ∞
+C
(−1)i x 2i +1
(−1)i x 2i
∑ (2i)2 i (2i !) + C
i =0
difficulty: hard
section: 10.3
33. Find the Taylor series centered at x = 0 (i.e. the Maclaurin series) for sin( x 2 ) . (−1)i x 4i +3 ∑ i =0 (4i + 1)! ∞
A)
Ans: B
∞
∞
B)
(−1)i x 4i +2 ∑ i =0 (2i +1)!
difficulty: medium
C)
section: 10.3
Page 6
(−1)i x 4i ∑ i =0 (2i )!
(−1)i x 4i +1 ∑ (4i)! i =0 ∞
D)
.
Chapter 10: Approximating Functions Using Series
34. Find the Taylor series centered at x = 0 (i.e. the Maclaurin series) for sin( x 2 ) and integrate it term by term to find ∫ 5sin( x 2 ) dx . (−1)i 5 x 2i + 2 ∑ i =0 (2i + 2) ⋅ (2i +1)!
∞
∞
A)
C)
∞
∞
B)
(−1)i 5 x 2i +1 ∑ i =0 (2i +1) ⋅ (2i +1)!
Ans: C
difficulty: medium
(−1)i 5 x 4i +3 ∑ i =0 (4i + 3) ⋅ (2i +1)!
D)
(−1)i 5 x 4i +1 ∑ i =0 (4i +1) ⋅ (2i )!
section: 10.3
35. Find the Taylor series centered at x = 0 (i.e. the Maclaurin series) for cos( x 2 ) and integrate it term by term to find ∫ cos( x 2 ) dx . Use the first two terms of this series to 1
approximate ∫ cos( x 2 ) dx . Round your answer to 4 decimal places. 0
Ans: 0.9000 difficulty: hard
section: 10.3
− ln(1 − 6 x) by substituting into the series for ln(1 + 36. Find the Taylor expansion for f ( x) = x). A)
−6 x −
B)
−6 x +
Ans: D
62 x 2 63 x3 64 x 4 − − − ⋅⋅⋅ 2 3 4
C)
6x −
62 x 2 63 x3 64 x 4 + − + ⋅⋅⋅ 2 3 4
62 x 2 63 x3 64 x 4 62 x 2 63 x3 64 x 4 D) + + + ⋅⋅⋅ − + − ⋅⋅⋅ 6x + 2 3 4 2 3 4 difficulty: medium section: 10.3
− ln(1 − 3x) by substituting into the series for ln(1 37. Find the Taylor expansion for f ( x) = + x). Plot both f(x) and its Taylor polynomials of various degrees and use the graphs to guess what the interval of convergence is. [Hint: Begin with the 3rd degree approximation. It's a good idea to use approximations as high as 10th degree!] 1 1 1 1 A) − < x < B) − < x < 1 C) −1 < x < D) −1 < x < 1 3 3 3 3 Ans: A difficulty: hard section: 10.3 38. Use the Taylor series for f ( x) = cos( x) at x = 0 to find the Taylor series for cos x = 0. A)
x 2 x3 1− x + − + ⋅⋅⋅ 2! 3!
x x 2 x3 + − + ⋅⋅⋅ 2! 4! 6! Ans: B difficulty: medium
B)
1−
C)
x x3/ 2 x 2 x− + − + ⋅⋅⋅ 2! 4! 6!
D)
x −x+
section: 10.3
Page 7
x3/ 2 x 2 − + ⋅⋅⋅ 2! 3!
( x ) at
Chapter 10: Approximating Functions Using Series
39. Find the number to which the series 1 − decimal places. Ans: –0.61727 difficulty: medium
5 25 125 + − + ⋅⋅⋅ converges. Round to 5 2! 4! 6!
section: 10.3
40. Use the derivative of the Taylor series about 0 for for
4x (1 − x) 2
1 to find the Taylor series about 0 1− x
.
A) 4 + 4 x3 + 4 x5 + 4 x 7 + ⋅⋅⋅ B) 4 + 4 x 2 + 4 x3 + 4 x 4 + ⋅⋅⋅ Ans: C difficulty: medium
C) 4 x + 8 x 2 + 12 x3 + 16 x 4 + ⋅⋅⋅ D) 4 + 8 x + 12 x 2 + 16 x3 + ⋅⋅⋅ section: 10.3
1 to find the Taylor series about 0 1− x x 1 2 3 4 5 . Use this result to find the value of + + + for + + ⋅⋅⋅ . Round 2 4 16 64 256 1024 (1 − x) to 3 decimal places. Ans: 0.444 difficulty: medium section: 10.3
41. Use the derivative of the Taylor series about 0 for
42. According to the theory of relativity, the energy, E, of a body of mass m is given as a 1 − 1 , where c is a constant, the speed of function of its= speed, v, by E mc 2 2 2 1− v / c light. Assuming v < c, expand E as a series in v/c, as far as the second non-zero term. 1 v2 1v 2 A) C) mc 1 + + ⋅⋅⋅ mc 2 1 + + ⋅⋅⋅ 2 2c 2 c 2 2 3 v4 2 1 v 3 v 2 1 v B) D) mc + + ⋅⋅⋅ mc + + ⋅⋅⋅ 2 2 8 4 c 2 c 4 c 2 c Ans: D difficulty: medium section: 10.3
Page 8
Chapter 10: Approximating Functions Using Series
43. According to the theory of relativity, the energy, E, of a body of mass m is given as a 1 − 1 , where c is a constant, the speed of function of its= speed, v, by E mc 2 2 2 1− v / c light. Assuming v < c, expand E as a series in v/c, as far as the second non-zero term. If v = 0.05c, approximate E using your expansion. Also, approximate E by the formula 1 E = mv 2 . By what percentage do your two approximations differ? 2 Ans: 0.1875% difficulty: hard section: 10.3 44. Find the Maclaurin series for f ( x) = sin(5 x) . A)
5x −
53 x3 55 x5 + − ⋅⋅⋅ 3! 5!
52 x 2 54 x 4 + − ⋅⋅⋅ 2! 4! Ans: A difficulty: medium
B)
1−
52 x 3 54 x 5 + − ⋅⋅⋅ 3! 5!
C)
x−
D)
5x −
52 x 2 53 x3 + − ⋅⋅⋅ 2! 3!
section: 10.3
45. Use the Maclaurin series for f ( x) = sin(4 x) to find the Maclaurin series for f ( x) = cos(4 x) . A)
4x −
43 x3 45 x5 + − ⋅⋅⋅ 3! 5!
42 x 2 44 x 4 + − ⋅⋅⋅ 2! 4! Ans: B difficulty: medium
B)
1−
42 x3 44 x5 + − ⋅⋅⋅ 3! 5!
C)
x−
D)
4x −
42 x 2 43 x3 + − ⋅⋅⋅ 2! 3!
section: 10.3
46. Given the fact that the Taylor series about x = 0 for e x = 1 + Taylor series about x = 0 for e x / 4 = 1 + Ans: yes difficulty: medium
x x 2 x3 + + + ⋅⋅⋅ ? 4 32 384
section: 10.3
0.3
47. Approximate
∫ 1 + x dx using the first three terms of the Taylor series about zero for 0
(1 + x) . A) 0.321 Ans: A
x x 2 x3 + + + ⋅⋅⋅ , is the 1! 2! 3!
1/ 2
B) 0.336 C) 0.3 D) 1.5 difficulty: medium section: 10.3
Page 9
Chapter 10: Approximating Functions Using Series
48. Find the first four terms of the Taylor series about x = –2 for ln(1 − x) . 1 1 1 Ans: ln(3) − ( x + 2) − ( x + 2) 2 − ( x + 2)3 3 18 81 difficulty: medium section: 10.3 –1.9
49. Estimate
∫ ln(1 − x)dx using the first two terms of the Taylor series about x = –2 for
–2
ln(1 − x) . B) 0.108195 A) 1.098612 Ans: B difficulty: medium
C) –2.08903 section: 10.3
D) 20.085537
50. a) 1 + 2 x + 4 x 2 + 8 x 3 + ... is the Maclaurin series for what function? b) What is its radius and interval of convergence (excluding possible endpoints)? c) Use the Maclaurin series to determine f ′′′(0) . 1 Ans: a) f ( x) = 1− 2x b) The radius of convergence is 12 . The interval of convergence is − 12 < x < 12 (excluding endpoints). c) 48 difficulty: medium section: 10.3 51. Since f ( x) = ln x and g ( x) = e x are inverse functions, we know that eln(1+ x ) = 1 + x for x > -1. Find the Taylor series for eln(1+ x ) using only up to the quadratic terms and show that the result is 1 + x. x2 x2 Ans: We know that e x ≈ 1 + x + and ln(1 + x) ≈ x − . Thus 2 2 2 2 x 1 x eln(1+ x ) ≈ 1 + ( x − ) + ( x − ) 2 ≈ 1 + x when you consider only quadratic terms. 2 2 2 difficulty: hard section: 10.3
( )
52. Find the 12th-degree Taylor polynomial for x sin x 2 centered at x = 0.
Suppose you
( ) for 0 < x <
use the first two non-zero terms of the polynomial to approximate x sin x 2 1. Is your approximation too big or too small? A) Too small B) Too big Ans: A difficulty: medium section: 10.4
Page 10
Chapter 10: Approximating Functions Using Series
( )
53. Find the 12th-degree Taylor polynomial for x sin x 2 centered at x = 0.
Suppose you
( ) for 0 < x < 1.
use the first two non-zero terms of the series to approximate x sin x 2 Is the magnitude of the error always less than 0.011? Ans: yes difficulty: medium section: 10.4
54. The function h(x) is a continuous differentiable function whose graph is drawn below. The accompanying table provides some information about h(x) and its derivatives.
x 0 1 2 3
h(x) 2 3.29 5.43 8.96
h′(x) 1 1.64 2.71 4.48
h″(x) 0.50 0.82 1.35 2.24
h″′(x) 0.25 0.41 0.67 1.12
Which of the following is closest to h(3.1)? 0.5 0.25 A) 3 + (0.1) + (0.1) 2 + (0.1)3 2 6 B) 3 + 3.1 + 0.5(3.1) 2 + 0.25(3.1)3 0.5 C) 3 + 3.1 + (3.1) 2 2 2.24 1.12 D) 8.96 + 4.48(3.1) + (3.1) 2 + (3.1)3 2 6 2.24 1.12 E) 8.96 + 4.48(0.1) + (0.1) 2 + (0.1)3 2 6 2 F) 8.96 + 4.48(3.1) + 2.24(3.1) + 1.12(3.1)3 Ans: E difficulty: medium section: 10.4
Page 11
Chapter 10: Approximating Functions Using Series
55. The function h(x) is a continuous differentiable function whose graph is drawn below. The accompanying table provides some information about h(x) and its derivatives.
x 0 1 2 3
h(x) 2 3.29 5.43 8.96
h′(x) 1 1.64 2.71 4.48
h″(x) 0.50 0.82 1.35 2.24
h″′(x) 0.25 0.41 0.67 1.12
h(x), h′(x), h″(x) and h″′(x) are all increasing functions. Suppose we use a tangent line approximation at zero to approximate h(0.1). Find a good upper bound for the error. A) 0.0025 B) 0.0820 C) 0.1066 D) 0.1558 Ans: B difficulty: hard section: 10.4 56. Estimate the magnitude of the error in approximating sin(1) using a third degree Taylor polynomial about x = 0. A) 0.0125 B) 0.0333 C) 0.0417 D) 0.0625 Ans: C difficulty: medium section: 10.4 1 a good bound for the maximum possible error for the nth degree Taylor (n)! x polynomial about x = 0 approximating sin on the interval [0, 1]? 2 Ans: no difficulty: medium section: 10.4
57. Is
58. Show that the Taylor series about 0 for f ( x) = 3sin( x) converges to 3sin( x) for all values of x by showing that the error En ( x) → 0 . Ans: Using the Lagrange error bound and the fact that −1 ≤ sin( x) ≤ 1 we know that 1 n +1 En ( x) ≤ x . Because the factorials dominate the power functions as x (n + 1)! increases to infinity, we can show that lim En ( x) = 0 . n →∞
difficulty: hard
section: 10.4
Page 12
Chapter 10: Approximating Functions Using Series
59. It can be shown that the Maclaurin series for e z , cos z and sin z converge for all values of z in the complex numbers, just as they do for all values of x in the real numbers. a) Write down and simplify the Maclaurin series for eix . b) Write down the Maclaurin series for cos x and i sin x c) Use the series you found in parts a) and b) to show that= eix cos x + i sin x . (This is one of several formulas called "Euler's Formula.") d) Find the value of 7(eiπ + 1) . x 2 ix 3 x 4 − + + 2! 3! 4! x2 x4 ix 3 ix 5 b) cos x =− 1 + − and i sin x =ix − + − 2! 4! 3! 5! d) 0 difficulty: hard section: 10.4
Ans: a) 1 + ix −
11 −π < x ≤ 0 . 60. Construct the third Fourier approximation to the function f ( x) = 11 0 < x ≤ π 2 33 11 11 33 11 11 A) D) − sin x − sin 3 x − sin x − sin 3 x 4 π 3π 4 2π 2π 33 11 11 33 11 11 B) E) − + sin x + sin 3 x − sin x − sin 3 x 4 π 6π 4 π π 33 11 11 33 11 11 C) F) − + sin x + sin 3 x − sin x − sin 3 x 4 2π 2π 4 2π 6π Ans: A difficulty: medium section: 10.5
−11 −π < x ≤ 0 61. Construct the third Fourier approximation to the function f ( x) = . 0< x≤π 0 22 22 11 11 D) A) 5.5 − sin x − sin 3x 5.5 − sin x − sin 3 x π 3π π 3π 22 22 11 11 B) E) −5.5 + sin x + sin 3 x −11 + sin x + sin 3 x π 3π π 6π 11 11 11 11 F) C) −5.5 + sin x + sin 3 x 11 − sin x − sin 3 x π 3π π 6π Ans: B difficulty: medium section: 10.5 −1 −π < x ≤ 0 62. Find a0 for the function f ( x) = . 1 0< x≤π A) 1 B) -1 C) 1/2 D) -1/2 E) 0 Ans: E difficulty: medium section: 10.5
Page 13
Chapter 10: Approximating Functions Using Series
−2 −π < x ≤ 0 63. Find the first harmonic of the function f ( x) = . 2 0< x≤π 8 4 8 4 B) 2 − sin x C) D) A) −2 + sin x sin x sin x π π π π Ans: C difficulty: medium section: 10.5
E) 0
−2 −π < x ≤ 0 64. Find the second harmonic of the function f ( x) = . 2 0< x≤π 4 8 4 D) A) sin 3 x −2 + sin x + sin 3 x 3π π 3π 8 4 B) E) 0 2 − sin x − sin 3 x π 3π 8 C) sin 3 x 3π Ans: E difficulty: medium section: 10.5 π −π < x ≤ 0 65. Find a0 for the function h( x) = . 0 0 < x ≤ π π π π π A) − B) C) − D) E) 0 4 4 2 2 Ans: D difficulty: medium section: 10.5
π −π < x ≤ 0 . 66. Find the first harmonic of the function h( x) = 0 0 < x ≤ π A) − sin x B) sin x C) −2sin x D) 2sin x E) 0 Ans: C difficulty: medium section: 10.5 2π −π < x ≤ 0 67. Find the second harmonic of the function h( x) = . 0 0< x≤π 4 2 4 2 A) B) − sin 3 x C) D) − sin 3 x sin 3 x sin 3 x 3 3 3 3 Ans: E difficulty: medium section: 10.5
Page 14
E) 0
Chapter 10: Approximating Functions Using Series
0 −2 < x ≤ 0 68. Find the third-degree Fourier polynomial for f (t ) = , where c is a c 0 < x ≤ 2 constant, by writing a new function, g ( x) = f (t ) , with period 2π . c 2c 2c c 2c πt 2c 3πt C) A) + sin ( t ) + sin ( 3t ) − sin − sin 2 π 3π 2 π 2 3π 2 c 2c 2c c 2c πt 2c 3πt B) D) − sin ( t ) − sin ( 3t ) + sin + sin 2 π 3π 2 π 2 3π 2 Ans: B difficulty: hard section: 10.5 69. Fill in the blanks: Fourier polynomials give good __________ approximations to a function. Taylor polynomials give good _____________ approximations to a function. Ans: global, local difficulty: easy section: 10.5 70. Which gives the better approximation of 4e0.3 , the Taylor polynomial about zero with three terms, or the Fourier polynomial with three terms? A) Taylor B) Fourier C) Both give the same approximation. Ans: A difficulty: easy section: 10.5 71. Suppose that g is the pulse train of width 0.5. What percent of the energy of g is contained in the constant term of its Fourier series? Round to one decimal place. Ans: 8.0% difficulty: hard section: 10.5 72. Find the Taylor series for sin x – cos x . x 2 x3 x 4 x5 x 2 x3 x 4 x5 C) –1+ x + – – + + ⋅⋅⋅ – + + – ⋅⋅⋅ 2! 3! 4! 5! 2! 3! 4! 5! x 2 x3 x 4 x5 x 2 x3 x 4 x5 D) B) 1 – x – –1 – x + + – – + ⋅⋅⋅ + + – – ⋅⋅⋅ 2! 3! 4! 5! 2! 3! 4! 5! Ans: C difficulty: medium section: 10 review
A)
1+ x –
73. Find the second degree Taylor polynomial approximation of A)
1 x − 1 ( x − 1) 2 − + 2 2 4
1 x − 1 ( x − 1) 2 − + − 2 3 4 Ans: A difficulty: medium
B)
C)
1 1 + x2
about x = 1.
1 x − 1 ( x − 1) 2 − + 2 4 6
1 x − 1 ( x − 1) 2 − + − 2 3 4 section: 10 review
D)
Page 15
Chapter 10: Approximating Functions Using Series
74. Use a Taylor polynomial of degree 3 for f ( x) = e 4 x to approximate the value of e0.8 . Give your answer to five decimal places. A) 23.66667 B) 1.22133 C) 2.22554 D) 2.20533 Ans: D difficulty: medium section: 10 review 75. Use the Taylor polynomials for the sine and cosine functions to find a rational function with a degree 5 numerator and no fractional coefficients that approximates the tangent function near 0. x 5 + 24 x 4 − 120 x3 30 x 4 − 6 x 2 + 120 A) C) R( x) = R ( x ) = x 4 − 6 x 2 + 120 −120 x 3 + x 5 x 5 − 20 x 3 + 120 x 120 x 5 − 24 x 3 + x B) D) R ( x ) = R( x) = 4 30 x 4 − 60 x 2 + 120 5 x − 60 x 2 + 120 Ans: B difficulty: medium section: 10 review
7 1 using a series for . 2 1+ x 1+ x b) Use the series from part a) to find the Taylor series for g ( x) = 7 arctan x . Ans: a) Use the Binomial series formula with p = -1 and then substitute x 2 for x to obtain: 7(1 − x 2 + x 4 − x 6 + ... + (−1) n x 2 n + ...), n =0,1, 2,...
76. a) Find the Taylor series for f ( x) =
b) Integrate term-by-term to obtain: 7( x −
x3 x5 x7 x 2 n +1 + − + ... + (−1) n + ...) . 3 5 7 2n + 1
Note that arctan(0) = 0 so C = 0. difficulty: medium section: 10 review 77. Medicine balls are launched from the floor to a height of six feet. They bounce, reaching x/10 the height of the previous bounce each time. The heavier the medicine ball, the smaller the value of x. Write a power series that gives the total distance that a medicine ball bounces as a function of x. What is the function that gives this Taylor polynomial? Ans: The power series is 12 + 12( 10x ) + 12( 10x ) 2 + 12( 10x )3 + ... + 12( 10x ) n + ... . This is geometric so we use the formula for the sum of an convergent geometric series to 12 120 obtain = for appropriate values of x. f ( x) = x 1 − 10 10 − x difficulty: hard section: 10 review 78. True or False: A Taylor polynomial of degree six always has six non-zero terms. Ans: False difficulty: easy section: 10 review
Page 16
P '(t ) P(t )(3 − P(t )) 1. Suppose that the function P(t) satisfies the differential equation = with the initial condition P(0) = 1. Find P″(0). Ans: 2 difficulty: medium section: 11.1 P '(t ) P(t )(6 − P(t )) 2. Suppose that the function P(t) satisfies the differential equation = with the initial condition P(0) = 4. Which of the following is a possible graph for P(t) for small t > 0?
Ans: iii difficulty: medium
section: 11.1
P '(t ) P(t )(4 − P(t )) 3. Suppose that the function P(t) satisfies the differential equation = with the initial condition P(0) = 1. Consider the behavior of the graph of P (t ) near a point t0 , where P (t0 ) = 4 (if such a point exists). Is the following graph consistent with P(t)?
Ans: no difficulty: medium
section: 11.1
Page 1
Chapter 11: Differential Equations
4. Does y = 4 cos(4t ) satisfy Ans: no difficulty: easy
d2y dt 2
+ 4y = 0?
section: 11.1
5. Select the value(s) of w for which y = e
wt
satisfies
d2y dt 2
− 49 y = 0.
A) 7 B) -7 C) 49 D) -49 E) 0 Ans: A, B difficulty: easy section: 11.1 6. Which of the following functions are solutions to the differential equation (Mark all correct answers.) = y 3cos x − sin x = y sin(3 x) + cos(3 x) B) A) Ans: C difficulty: easy section: 11.1
C)
y = ex / 3
D)
7. Which of the following functions are solutions to the differential equation (Mark all correct answers.) y sin(6 x) + cos(6 x) A)=
C)
y 6 cos x − sin x B)= Ans: D difficulty: easy
D) section: 11.1
y = e− x / 3
dy 1 = − y? dx 6
y = ex / 6 y = e− x / 6
8. Which of the following functions are solutions to the differential equation (Mark all correct answers) = y sin(5 x) + cos(5 x) = y 5cos x − sin x B) A) Ans: A difficulty: easy section: 11.1
C)
y = ex / 5
(Mark all correct answers) y sin(8 x) + cos(8 x) = y 8cos x − sin x = A) B) Ans: B difficulty: easy section: 11.1
C)
y = e x /8
dy = −25 y ? dx y = e− x / 5
D)
9. Which of the following functions are solutions to the differential equation
Page 2
dy y = ? dx 3
d2y dx 2
D)
= −y ?
y = e− x /8
Chapter 11: Differential Equations
d 2S
= −9.8 , find S if the initial velocity is 10 m/sec upward and the initial dt 2 position is 5 m above the ground.
10. Given
A) S= −4.9t 2 + 10t + 5 B) S= 4.9t 2 + 10t + 5 Ans: A difficulty: easy
C) D) section: 11.1
S= −9.8t 2 + 10t + 5 S= 9.8t 2 + 10t + 5
11. For what value(s) of n (if any) is y = enx a solution to the differential equation 1 − y′′ + y′ + 4 y = 0? 2 A) 2 B) –2 C) 4 D) -4 E) 0 Ans: B, C difficulty: medium section: 11.1 12. Is y = e x sin x a solution to
d2y dx
Ans: no difficulty: medium
2
−
dy +y= 0? dx
section: 11.1
13. Mark all the true statements about the differential equation:
d 2 y dy 0 − + By = dx 2 dx
A) It is a first order differential equation. B) It is a second order differential equation. C) Any solutions to the differential equation would be of the form x = a. D) Any solutions to the differential equation would be of the form y = f(x). Ans: B, D difficulty: easy section: 11.1
y′ 2 y − x and y (0) = 2 . Estimate y (0.05) and y (0.1) using the slope given 14. Suppose = by the differential equation. A) 4 and 3.9 D) 7.95 and 17.95 B) 5.95 and 17.75 E) 2.2 and 2.4175 C) 3.95 and 3.9 Ans: E difficulty: medium section: 11.1 15. Suppose y′ = y and y (1) = 4 . What is the best approximation of y (1.2) . A) 0.8 B) 3.6 C) 4.8 D) 16 Ans: C difficulty: easy section: 11.1
Page 3
Chapter 11: Differential Equations
16. Which of the following equations goes with (i.e. describes a solution in) the slope field shown below? (Note that the point (0, 2) is on the graph of each of the equations). A.
y 2 − 2 cos x = 2
B.
x sin y + y = 2
y C. ln = 0.71x + ln 2 1− y
Ans: C difficulty: medium
section: 11.2
Page 4
Chapter 11: Differential Equations
dy = x − y is shown below. Consider the dx solution curve to the differential equation starting at x = 0, y = 5, and ending at x = 5 and approximate the value of y when x is 5.
17. The slope field for the differential equation
A) 1.25 Ans: C
B) 2.5 C) 3.75 difficulty: medium
D) 5 section: 11.2
Page 5
Chapter 11: Differential Equations
18. Which of the following equations corresponds with the slope field shown below? I. y=' xy + 1 II. y ' = sin x III.
y ' = xe − y
IV. y=' y 2 + 1 V. y ' = sin y VI. None of them
Ans: III difficulty: medium
section: 11.2
Page 6
Chapter 11: Differential Equations
19. Which of the following equations corresponds with the slope field shown below?
dy = ( x − y)2 dx dy II. = ( x + y )2 dx dy III. = x 2 − y 2 dx IV. None of them I.
Ans: III difficulty: medium
section: 11.2
20. If the solution to the differential equation
= y ax + b , then a = _____ and b = _____. Part A: 5 Part B: -25 difficulty: easy section: 11.2
dy 1 = x − y is a straight line in the form dx 5
1 dy = x − y is a straight line. Take a 5 dx point ( x0 , y0 ) not on the line. Can a solution curve through ( x0 , y0 ) cross the line? Ans: no difficulty: medium section: 11.2
21. One of the solutions to the differential equation
Page 7
Chapter 11: Differential Equations
22. For any constant C, y = 4 x − 16 + Ce− x / 4 is a solution to the differential equation dy 1 = x − y . Find the solution through the point (0, -4). dx 4 A) C) y = 4 x − 16 + 16e− x / 4 y = 4 x − 16 + 12e− x / 4 B) y = 4 x − 16 − 4e − x / 4 Ans: C difficulty: easy 23. If a slope field for
D) section: 11.2
y = 4 x − 16 − 16e− x / 4
dy has constant slopes where y is constant, what do you know about dx
dy ? dx
dy depends on y only. dx dy B) depends on x only. dx Ans: A difficulty: medium A)
C) D)
dy must be linear in y. dx dy must be linear in x. dx
section: 11.2
dy have y = –5 as a horizontal asymptote, does it follow dt that either lim y = 5 or lim y = 5 ?
24. If all the solution curves for t →∞
Ans: no difficulty: easy
t →−∞
section: 11.2
Page 8
Chapter 11: Differential Equations
25. Sketch a slope field for the differential equation
dy − x using the points indicated on = dx y
the axes.
y
Series 1
2
1
x -2
-1
1
2
-1
-2
Ans: The slope field should be composed of short line segments at each point with slope equal to the negative of the x coordinate divided by the y coordinate. difficulty: medium section: 11.2
Page 9
Chapter 11: Differential Equations
dy = xy , sketch the solution curve in the dx fourth quadrant that goes through the point (0, -1).
26. On the slope field for the differential equation
y 4
3
2
1
x −4
−3
−2
−1
1
2
3
4
5
−1
−2
−3
−4
Ans: y 4
3
2
1
x −4
−3
−2
−1
1
2
3
4
5
−1
−2
−3
−4
difficulty: easy
section: 11.2
dy = x 2 + y . Use Euler's method with two dx subdivisions to approximate the value of y when x = 2 on the solution curve that passes through (1,4). Ans: 10.875 difficulty: medium section: 11.3
27. Consider the differential equation
Page 10
Chapter 11: Differential Equations
dy = x 2 + y . If you use Euler's method to dx approximate the value of y when x = 2 on the solution curve that passes through (1,4), is your approximation an underestimate or overestimate? A) An underestimate B) An overestimate Ans: A difficulty: medium section: 11.3
28. Consider the differential equation
dy x = − is shown below. Starting from dx y the point (0, –2) , use Euler's method with N = 3 subdivisions to approximate the value of y when x = 1. Round to 2 decimal places.
29. The slope field for the differential equation
Ans: –1.83 difficulty: medium
section: 11.3
Page 11
Chapter 11: Differential Equations
dy x = − is shown below. If you start from dx y the point (0, –2) and use Euler's method with N = 3 subdivisions to approximate the value of y when x = 1, is your answer an underestimate or overestimate?
30. The slope field for the differential equation
A) An overestimate B) An underestimate Ans: B difficulty: medium section: 11.3 31. Is the equation x 2 – y 2 = C , for C a constant, a solution to the differential equation dy x = − ? dx y Ans: no difficulty: medium section: 11.3
Page 12
Chapter 11: Differential Equations
dy 32. The slope field for the differential equation= 0.5(2 − y ) is shown below. Consider dt the solution starting from y = 0.5, t = 0, and use it to estimate the value of y when t = 1.
A) 0.5 Ans: C
B) 0.8 C) 1.1 difficulty: medium
D) 1.4 section: 11.3
dy 33. The slope field for the differential equation= 0.5(2 − y ) is shown below. Use dt Euler's method with N = 3 subdivisions of the interval 0 ≤ t ≤ 1 to approximate the value of y when t = 1, if you start at y = 1 and t = 0. Round your answer to 2 decimal places.
Ans: 1.42 difficulty: medium
section: 11.3
Page 13
Chapter 11: Differential Equations
dy 34. The slope field for the differential equation = y (2 − y ) is shown below. Use the dx slope field to estimate the solution starting from y = 3, x = 0, and use it to estimate the value of y when x = 1.
A) 1.2 Ans: D
B) 1.5 C) 1.8 difficulty: medium
D) 2.1 section: 11.3
dy 35. The slope field for the differential equation = y (2 − y ) is shown below. Use the dx Euler's method with N = 4 subdivisions of the interval 0 ≤ x ≤ 1 to approximate the value of y when x = 1, if you start at y = 1 and x = 0. Round your answer to 2 decimal places.
Ans: 1.81 difficulty: medium
section: 11.3
Page 14
Chapter 11: Differential Equations
36. Consider the solution with y(0) = 0 to the differential equation
dy 6 . Compute = dx 1 + x 2
the exact value of y(1) and then round to 4 decimal places. Ans: 4.7124 difficulty: medium section: 11.3 7 dy . If you use = dx 1 + x 2 Euler's method with 1 million steps to approximate the exact value of y (1) , will your approximation be an over- or underestimate? A) An overestimate B) An underestimate Ans: A difficulty: medium section: 11.3
37. Consider the solution with y(0) = 0 to the differential equation
dy 2 . Use Euler's = dx 1 + x 2 method with 2 steps to approximate the value of y (1) . Give your answer to 1 decimal place. Ans: 1.8 difficulty: medium section: 11.3
38. Consider the solution with y(0) = 0 to the differential equation
dy = x 2 + sin x Solve the differential equation dx analytically given that y (0) = 0 . Then approximate y (1) using Euler's method with three steps. Write a few sentences describing the error in Euler's method in this case, and what could be done to decrease the error. Ans: Euler's method gives y (1) ≈ 0.649 . The analytical solution is y (1) ≈ 0.793 . The error in Euler's method is approximately 0.144. This is an underestimate, which we also know because the second derivative evaluated at x = 1 is positive and therefore the exact solution is concave up. We could decrease the error in Euler's method by using more steps, being careful not to round too much. difficulty: medium section: 11.3
39. Consider the differential equation
Page 15
Chapter 11: Differential Equations
40. Use Euler's method with five steps to approximate a solution to the differential equation y′ = cos y given that y (0) = −1 . Give both a numerical solution by filling in the table of values, and a graphical solution by plotting your points. x 0 0.3 0.6 0.9 1.2 1.5 1.8
y
1.5 1.2 0.9 0.6 0.3 -1.8 -1.5 -1.2 -0.9 -0.6 -0.3 -0.3
x 0.3 0.6 0.9 1.2 1.5 1.8
-0.6 -0.9 -1.2 -1.5 -1.8
Ans:
Page 16
y -1
Chapter 11: Differential Equations
x 0 0.3 0.6 0.9 1.2 1.5 y
x 1
2
3
−1
2
difficulty: medium
section: 11.3
dQ 41. Solve the differential equation = 200 − 0.2Q subject to Q(0) = 1500 . dt −0.2t C)= A)= Q(t ) 500e + 1000 Q(t ) 1300e −0.2t + 200 B) = Q(t ) –500e−0.2t + 1000 Ans: A difficulty: medium
D) = Q(t ) –1300e−0.2t + 200 section: 11.4
dQ 42. Solve the differential equation = 300 − 0.2Q subject to Q(0) = 1500 and sketch dt your solution. There is a horizontal asymptote at Q = _____. Ans: 1500 difficulty: medium section: 11.4 43. Solve
dy x if y = 0 when x = 2. = dx y
2 A) x 2 + y 2 = B) y= C) y 2= x − 2 4 x2 − 4 Ans: B difficulty: medium section: 11.4
Page 17
D)
y= x − 2
y -1 -0.838 -0.637 -0.396 -0.119 +0.179
Chapter 11: Differential Equations
dy 44. Solve = y 2 + 1 if y = 1 when t = 0. dt π π A) C) y = arctan t + y = tan t + 4 4 π π B) D) y = arctan t – y = tan t – 4 4 Ans: C difficulty: medium section: 11.4 45. Solve
dy 1 if y = 4 when x = 0. = dx xy C) = y 3 x +4
A) = y
3 x + 16
B) = y
(3 x + 64)
Ans: D
1/ 3
D) = y
difficulty: medium
( 3 x + 8)
2/3
section: 11.4
dy π 46. Solve = 36 − y 2 if y = 3 when x = . 6 dx A) y = 6sin x B) y = 6 cos x C) y = 36sin x Ans: A difficulty: medium section: 11.4 47. Find the solution to the differential equation y ' =
D)
y = 36 cos x
8 satisfying y (0) = 6 . Select all 1+ y
that apply. A)
y + y 2 = 8 x + 42
y2 =8 x + 24 2 Ans: B, D difficulty: medium
B)
y y2 + =8 x + 21 2 2 (1 + y ) 2 49 D) = 8x + 2 2 section: 11.4
C)
y+
48. Find the solution to the differential equation
dy = xy satisfying y (0) = 3 . dx 2
A)
y= 3 + e
Ans: C
x2 / 2
B)
y=e
3x2 / 2
difficulty: medium
C)
y = 3e
section: 11.4
Page 18
x2 / 2
D)
ex / 2 y= 3
Chapter 11: Differential Equations
49. Find the solution to the differential equation A)
x2 2 y = arccos 2
B)
3 y = arctan ( x ) 2
Ans: D 50. Solve
difficulty: medium
dy π = x sec y if y = when x = 1. dx 6 y−
D)
x2 y = arcsin 2
section: 11.4
dN = 4 − 0.2 N , with N(0) = 0. dt
A)
N = 20e −0.2t
B)
= N 4 1 − e −0.2t
(
Ans: D
π = arccos ( x ) 6
C)
(
)
N 4 e−0.2t − 1 C)=
)
(
D) = N 20 1 − e −0.2t
difficulty: medium
)
section: 11.4
dN = 2 − 0.2 N , with N(0) = 0 and sketch the solution for t ≥ 0. There is a dt horizontal asymptote at N = _____. Ans: 10 difficulty: medium section: 11.4
51. Solve
52. Is the slope field for the differential equation Ans: no difficulty: medium
section: 11.4
53. Find the solution to the differential equation A)
dy 4 x 2 symmetric about the y-axis? = dx y
= y 2 6 x 2 + 30
2 B) 6 y= x 2 + 215 Ans: A difficulty: medium
dy 6 x passing through (1, 6) . = dx y C)
y 2 + 6 x2 = 42
D) 6 y 2 + x2 = 217 section: 11.4
Page 19
Chapter 11: Differential Equations
π dy cos 2 y with y (1) = – . = 4 dx x 2 C) = y sin −1 ln x + 2 2 D) y cos −1 ln x + = 2 section: 11.4
54. Find the solution to the differential equation A)
y = tan −1 (ln x +1)
B)
y = tan −1 (ln x –1)
Ans: B
difficulty: medium
0 with y (1) = 1 , y ( x) ≥ 0 55. Find the solution to the differential equation xy′ − (6 + x) y = for all x. A) y = x 7 e x B) y = x 7 e x −1 C) y = x 6e x −1 D) = y x 6e x + 1 − e Ans: C difficulty: medium section: 11.4 56. Is = P 250 + Ce0.2t a solution to the differential equation constant C)? Ans: no difficulty: medium 57. Is = y
section: 11.4
C − 2 cos t a solution to the differential equation
C)? Ans: no difficulty: medium
dP + 0.2 P = 50 (for some dt
dy sin t = 2 (for some constant dt y
section: 11.4
dy 15 58. Is y = − ln e− x – the solution to the differential equation = e y − x with dx 4 y (ln 4) = − ln 4 ? Ans: no difficulty: medium section: 11.4 59.
Is y=
dy a = a − by ? + Ce −bt the general solution to the differential equation b dt
Ans: yes difficulty: medium
section: 11.4
60. Is H = Cet / 2−4t the general solution to the differential equation 2
Ans: no difficulty: medium
section: 11.4
Page 20
dH = −2 H + tH ? dt
Chapter 11: Differential Equations
61. What is the minimum value of the solution to the differential equation
y (0) = 6 ? D) 6 e A) 6 B) 0 C) 6 /e Ans: C difficulty: medium section: 11.4 62. Is y= d2y 2
1 4 5 3 x − x + x 2 + C1x + C2 the general solution to the differential equation 12 2 =x 2 − 15 x + 2 ?
dx Ans: yes difficulty: medium
section: 11.4
63. Is y = cos x + C1x + C2 the general solution to the differential equation Ans: no difficulty: medium
65. Solve the differential equation using separation of variables. y=
d2y dx 2
= cos x ?
section: 11.4
64. Mark all of the differential equations that are NOT separable. dy dy dy B) = 6 x + 5 y 3 C) D) A) = 3t + 6 y = 5e xy dt dx dx Ans: A, B, C difficulty: medium section: 11.4
A)
dy = y cos x with dx
x2 y 6 y 2 2x2 + + + 12 x + C 2 2 2
B) y = xy 2 − 2 x 2 + C Ans: D difficulty: medium
dy = xy + 6 y + 2 x + 12 dx
x 5,= y 0 C) = x2 + 6 x
D)= y Ce 2 section: 11.4
Page 21
dy = 6e x + y dx
−2
Chapter 11: Differential Equations
66. Consider the Hakosalo residence in Oulu, Finland. Assume that heat is lost from the house only through windows and the rate of change of temperature in °F/hr is proportional to the difference in temperature between the outside and the inside. The 1 constant of proportionality is . Assume that it is 10°F outside constantly. On a 28 Thursday at noon the temperature inside the house was 65°F and the heat was turned off until 5 pm. Which of the following differential equations reflects the rate of change of the temperature in the house between noon and 5 pm? dy 1 dy 1 A) C) = = − ( y − 10) , 0 ≤ t ≤ 5 y − 10 , 0 ≤ t ≤ 5 dt 28 dt 28 dy 1 dy 1 D) B) = ( y − 10) , 0 ≤ t ≤ 5 = − y + 10 , 0 ≤ t ≤ 5 dt 28 dt 28 Ans: A difficulty: easy section: 11.5 67. Consider the Hakosalo residence in Oulu, Finland. Assume that heat is lost from the house only through windows and the rate of change of temperature in °F/hr is proportional to the difference in temperature between the outside and the inside. The 1 constant of proportionality is . Assume that it is 10°F outside constantly. On a 28 Thursday at noon the temperature inside the house was 65°F and the heat was turned off until 5 pm. Use differential equations to find the temperature in the house at 5 pm. Round to one decimal place, and do not include units in your answer. Ans: 56.0 difficulty: medium section: 11.5 68. Consider the Hakosalo residence in Oulu, Finland. Assume that heat is lost from the house only through windows and the rate of change of temperature in °F/hr is proportional to the difference in temperature between the outside and the inside. The 1 constant of proportionality is . Assume that it is 10° F outside constantly. On a 31 Thursday at noon the temperature inside the house was 65°F and the heat was turned off until 5 pm. At 5 pm the heat is turned on. The heater generates an amount of energy that would raise the inside temperature by 2°F per hour if there were no heat loss. Which of the following differential equations reflect what happens to the inside temperature after the heat is turned on? dy 1 52 dy 1 A) C) = − y+ = − y + 72 31 31 dt dt 31 dy 1 dy 1 72 B) D) = − y+2 = − y+ dt 31 dt 31 31 Ans: D difficulty: medium section: 11.5
Page 22
Chapter 11: Differential Equations
69. Consider the Hakosalo residence in Oulu, Finland. Assume that heat is lost from the house only through windows and the rate of change of temperature in °F/hr is proportional to the difference in temperature between the outside and the inside. The 1 constant of proportionality is . Assume that it is 10°F outside constantly. On a 26 Thursday at noon the temperature inside the house was 65°F and the heat was turned off until 5 pm. At 5 pm the heat is turned on. The heater generates an amount of energy that would raise the inside temperature by 2°F per hour if there were no heat loss. If the heat is left on indefinitely, what temperature will the inside of the house approach? Do not include units in your answer. Ans: 62 difficulty: hard section: 11.5 70. Suppose there is a new kind of savings certificate that starts out paying 1.5% annual interest and increases the interest rate by 1% each additional year that the money is left on deposit. (Assume that interest is compounded continuously and that the interest rate increases continuously.) Which of the following differential equations gives the rate of change in the balance B(t) at time t? dB dB C) A)= B(0.015 + 0.01t ) = B (0.025)t dt dt dB dB B)= B (0.015t + 0.01) D)= B (0.025 + 0.01t ) dt dt Ans: A difficulty: easy section: 11.5 71. Suppose there is a new kind of savings certificate that starts out paying 2% annual interest and increases the interest rate by 1% each additional year that the money is left on deposit. (Assume that interest is compounded continuously and that the interest rate increases continuously.) Write a differential equation that gives the rate of change in the balance B(t) at time t, and solve it assuming an initial deposit of $500. What is your equation? A)
B = 500e0.02t +0.005t
2
B) B = 500e0.02t +0.01t Ans: A difficulty: medium 2
C)
B = 500e0.02+0.01t
D) B = 500e0.03t section: 11.5
72. Suppose there is a new kind of savings certificate that starts out paying 3.5% annual interest and increases the interest rate by 1% each additional year that the money is left on deposit. (Assume that interest is compounded continuously and that the interest rate increases continuously.) Assuming an initial deposit of $1500, which of the following investment choices is better? A) Investing the money at a fixed interest rate of 5% for 4 years. B) Investing the money in the new kind of savings certificate for 4 years. Ans: B difficulty: hard section: 11.5
Page 23
Chapter 11: Differential Equations
73. The population of aphids on a rose plant increases at a rate proportional to the number present. In 5 days the population grew from 700 to 1100. Which of the following formulas give for the population of aphids at time t in days, where t = 0 is the day when there were 700 aphids? A) C) P = 700e5t P = 700e 2000t 1 11 ln t P = 700e 5 7
B) Ans: B
difficulty: medium
D) section: 11.5
1 ln 400 t 5 P = 700e
74. The population of aphids on a rose plant increases at a rate proportional to the number present. In 3 days the population grew from 600 to 1400. How many days does it take for the population to get 10 times as large? Round to 2 decimal places. Ans: 8.15 days difficulty: hard section: 11.5 75. The population of aphids on a rose plant increases at a rate proportional to the number present. In 3 days the population grew from 700 to 1400. How many aphids were there on the day before there were 700 aphids? Round to the nearest whole number. Ans: 556 aphids difficulty: hard section: 11.5 76. Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between the temperature of the object and the temperature of the surrounding air. A detective discovers a corpse in an abandoned building, and finds its temperature to be 26°C. An hour later its temperature is 18°C. Assume that the air temperature is 6°C, that normal body temperature is 37°C, and that Newton's Law of Cooling applies to the corpse. How many hours has the corpse been dead at the moment it is discovered? Round to 2 decimal places. Ans: 0.86 hours difficulty: hard section: 11.5
Page 24
Chapter 11: Differential Equations
77. Cesium 137 (Cs137) is a short-lived radioactive isotope. It decays at a rate proportional to the amount of itself present and has a half-life of 30 years (i.e., the amount of Cs137 remaining t years after A0 mg of the radioactive isotope is released is given by ln 2 − t A0 e 30 ).
As a result of its operations, a nuclear power plant releases Cs137 at a rate of 0.12 mg per year. The plant began its operations in 1990, which we will designate as t = 0. Assume there is no other source of this particular isotope. Which of the following integrals give the total amount of Cs137 T years after the plant began operations? A)
B)
∫
0.12ln 2 T − 30 t e dt 0 T
∫0
Ans: B
C)
ln 2 − t 0.12e 30 dt
D)
0.12
∫0
ln 2 − t Te 30 dt
ln 2 T − 30 t + 0.12 dt e 0
∫
difficulty: medium
section: 11.5
78. Cesium 137 (Cs137) is a short-lived radioactive isotope. It decays at a rate proportional to the amount of itself present and has a half-life of 30 years (i.e., the amount of Cs137 remaining t years after A0 mg of the radioactive isotope is released is given by ln 2 − t A0 e 30 ). As a result of its operations, a nuclear power plant releases Cs137 at a rate
of 0.12 mg per year. The plant began its operations in 1990, which we will designate as t = 0. Assume there is no other source of this particular isotope. Which of the following differential equations have a solution R(t), the amount (in mg) of Cs137 in t years? (We are assuming R(0) = 0.) ln 2
A)
− R dR = 0.12 + e 30 dt ln 2
− dR B) = 0.12 − e 30 + R dt Ans: C difficulty: medium
C)
dR ln 2 = 0.12 − R dt 30
dR ln 2 D) = 0.12 R − dt 30 section: 11.5
79. Cesium 137 (Cs137) is a short-lived radioactive isotope. It decays at a rate proportional to the amount of itself present and has a half-life of 30 years (i.e., the amount of Cs137 remaining t years after A0 mg of the radioactive isotope is released is given by ln 2 − t A0 e 30 ).
As a result of its operations, a nuclear power plant releases Cs137 at a rate of 0.1 mg per year. The plant began its operations in 1990, which we will designate as t = 0. Assume there is no other source of this particular isotope. In the long run, approximately how many mg of Cs137 will there be? Round to 2 decimal places. Ans: 4.33 mCi difficulty: medium section: 11.5
Page 25
Chapter 11: Differential Equations
80. Cesium 137 (Cs137) is a short-lived radioactive isotope. It decays at a rate proportional to the amount of itself present and has a half-life of 30 years (i.e., the amount of Cs137 remaining t years after A0 mg of the radioactive isotope is released is given by ln 2 − t A0e 30 ).
As a result of its operations, a nuclear power plant releases Cs137 at a rate of 0.14 mg per year. The plant began its operations in 1990, which we will designate as t = 0. Assume there is no other source of this particular isotope. Since Cs137 poses a great health risk, the government says that the maximum amount of Cs137 acceptable in the surrounding environment is 1 mg (spread over the surroundings). How many mg per year of the isotope can the station release and still be in compliance with the regulations? Round to 2 decimal places. Ans: 0.023 difficulty: medium section: 11.5 81. A bank account earns interest at a rate of 5% per year, compounded continuously. Money is deposited into the account in a continuous cash flow at a rate of $1000 per year. Use differential equations to find the amount of money in the bank account after 10 years, assuming an initial balance of $3000. Round to the nearest cent. Ans: $17,920.59 difficulty: medium section: 11.5 82. After a certain car is turned on, the engine block heats up according to the differential dH equation = − K ( H − 100) for K, a positive constant. Suppose that the engine block dt was 20° C when the car was started, and it was 90° after running for 15 minutes. Find K to 3 decimal places. Ans: 0.139 difficulty: medium section: 11.5 83. A diligent student has a slow leak in her bike tire, but has been too busy studying for exams to fix it. Assume that the pressure in the tire decreases at a rate proportional to the difference between the atmospheric pressure (15 lbs.) and the tire pressure. Monday at 6:00 pm she pumped up the pressure to 90 lbs. By 6:00 pm Tuesday it was down to 75 lbs. How much longer can she wait to pump up the tire if she wants to keep the pressure at a minimum of 45 lbs.? (Keep your answer in number of days from Monday at 6:00 pm, rounded to 2 decimal places.) Ans: 4.11 difficulty: medium section: 11.6
Page 26
Chapter 11: Differential Equations
84. In trying to model the response to a stimulus, psychologists use the Weber Fechner Law. This law states that the rate of change of a response, r, with respect to a stimulus, s, is inversely proportional to the stimulus. Model this law as a differential equation. What is the solution to the differential equation with the initial condition that r(s0) = r0 for some initial stimulus, s0? s r A) = C) r= r0 + k ln 0 r s0 + k ln 0 s r B)
r r0 difficulty: medium
= r s0 + k ln
Ans: D
D)
r= r0 + k ln
s s0
section: 11.6
85. The differential equation describing the motion of a 60-kg woman who has jumped into a dv k dv swimming pool is given by a= on the left hand = g − v . In this equation, a = dt m dt side is the woman's downward acceleration, v is the woman's downward velocity, g = 9.8 m/sec2 is the acceleration due to gravity, and k = 1945 kg/sec is the woman's ballistic coefficient in water. This equation assumes that the woman does not attempt to swim to the surface and just allows herself to continue sinking. The woman's terminal velocity in the water is _____ m/sec. Round to 2 decimal places. Ans: 0.30 difficulty: medium section: 11.6 86. There is a theory that says the rate at which information spreads by word of mouth is proportional to the product of the number of people who have heard the information and the number who have not. Suppose the total population is N. Which of the following dp , at which the information spreads by word differential equations describe the rate, dt of mouth? dp kp dp kp A) C) = = dt ( p − N ) dt ( N − p ) dp dp B) = kp ( N + p ) D) = kp ( N − p ) dt dt Ans: D difficulty: easy section: 11.6
Page 27
Chapter 11: Differential Equations
87. When a bacterial cell is suspended in a fluid, the concentration of a certain drug within the cell will change toward its concentration in the surrounding fluid at a rate proportional to the difference between the two concentrations. Suppose that a patient's blood is infected with these bacteria, which initially contain none of the drug. The patient is given enough of the drug to bring (and hold) its concentration in his blood to 0.0002. After two hours, the concentration within the bacterial cells is found to be 0.00003. Use differential equations to find how many hours it will be before the concentration within the bacteria reaches 0.00009. Round to 1 decimal place. Ans: 7.4 difficulty: medium section: 11.6 88. A spherical raindrop evaporates at a rate proportional to its surface area. If V = volume of the raindrop and S = surface area, which of the following is a differential equation for dV ? dt dV dV A) = kS with k a positive constant = kS with k a negative constant C) dt dt dV k dV k B) with k a negative constant D) with k a positive constant = = dt S dt S Ans: A difficulty: easy section: 11.6 89. A spherical raindrop evaporates at a rate proportional to its surface area. Let V = volume 4 3 of the raindrop and S = surface area. For a sphere, V= πr and S = 4πr 2 . If it takes 3 1 6 minutes for a spherical raindrop to evaporate to of its original volume, how many 8 minutes will it take to completely evaporate? Ans: 12 difficulty: medium section: 11.6
Page 28
Chapter 11: Differential Equations
90. A lake contains pollutants. A stream feeds clear mountain water into the lake at 13 gal/min. Polluted water is drained out of the lake at a rate of 13 gal/min by a second stream. If the volume of the lake is V gallons and time t is measured in minutes, and if it is assumed that the pollutants are spread evenly through the lake at all times, then what is the differential equation for Q(t), the quantity of pollutant in the lake at time t? dQ dQ A) F) = −13Q = 13V − Qt dt dt dQ dQ B) G) = 13Qt − V = −13t dt dt dQ 13Q dQ H) C) = − = Qe−13t dt dt V dQ dQ Qe −13t D) I) = 13 − 13Q = dt dt V dQ dQ 13Q − 13V 13Q J) E) = 13 − = dt V dt V Ans: C difficulty: medium section: 11.6 91. Suppose y is a solution to the differential equation Could the following be a graph of y?
Ans: no difficulty: easy
section: 11.7
Page 29
dy = f ( y ) and that f(y) ≥ 0 for all y. dx
Chapter 11: Differential Equations
dy = g ( y ) . The graph of g(y) is drawn below. dt Which of the following are true? (Select all that apply.)
92. Consider the differential equation
dy = g ( y) dt dy B) y = 3 is an unstable constant solution of = g ( y) dt dy C) y = 18 is a stable constant solution of = g ( y) dt dy D) y = 18 is an unstable constant solution of = g ( y) dt Ans: B, C difficulty: medium section: 11.7
A)
y = 3 is a stable constant solution of
93. The rate of change of a population for a species is given by dP = –0.05 P dt where P measures the population in thousands at time t months. The species starts with 1000 members (P = 1) at time t = 0. Assume that the differential equation holds indefinitely. Find the size of the population P when the rate of change of the population is a minimum. (Remember P(0) = 1.) If there is no minimum rate of change, enter "DNE". Ans: 1000 difficulty: medium section: 11.7
Page 30
Chapter 11: Differential Equations
94. The rate of change of a population for a species is given by dP = –0.05 P dt where P measures the population in thousands at time t months. The species starts with 1000 members (P = 1) at time t = 0. Assume that the differential equation holds indefinitely. If P = 0 is an equilibrium value of the population, is it stable or unstable? (If it is not an equilibrium value, enter "neither") Ans: stable difficulty: medium section: 11.7 95. The rate of change of a population for a species is given by dP = 0.05 dt where P measures the population in thousands at time t months. The species starts with 1000 members (P = 1) at time t = 0. Assume that the differential equation holds indefinitely. Which of the following is a graph of the population as a function of time?
Ans: i difficulty: medium
section: 11.7
Page 31
Chapter 11: Differential Equations
96. On January 1, 1879, records show that 500 of a fish called Atlantic striped bass were introduced into the San Francisco Bay. In 1899, the first year fishing for bass was allowed, 100,000 of these bass were caught, representing 10% of the population at the start of 1899. Owing to reproduction, at any moment in time the bass population is growing at a rate proportional to the population at that moment. Write a differential equation satisfied by B(t), the number of Atlantic striped bass a time t, where t is in years since January 1, 1879 and 0 ≤ t < 20 and solve it for B(t). A) C) B(t ) = 500e7.6t = B(t ) 500(1 − e0.38t ) B) B(t ) = 500e0.38t Ans: B difficulty: medium
D) = B(t ) 500(20 − e7.6t ) section: 11.7
97. On January 1, 1879, records show that 500 of a fish called Atlantic striped bass were introduced into the San Francisco Bay. In 1899, the first year fishing for bass was allowed, 100,000 of these bass were caught, representing 10% of the population at the start of 1899. Owing to reproduction, at any moment in time the bass population is growing at a rate proportional to the population at that moment. Assume that when fishing starts in 1899, the rate at which bass are caught is proportional to the square of the population with constant of proportionality 10−6 . Write a differential equation satisfied by B(t), for t > 20. dB dB A) = 7.6 B − 10−6 B 2 C) = 0.38 B − 10−6 B 2 dt dt −6 dB 10 dB 10−6 B) = 7.6 B − D) = 0.38 B − dt dt B2 B2 Ans: C difficulty: medium section: 11.7 98. On January 1, 1879, records show that 500 of a fish called Atlantic striped bass were introduced into the San Francisco Bay. In 1899, the first year fishing for bass was allowed, 100,000 of these bass were caught, representing 10% of the population at the start of 1899. Owing to reproduction, at any time the bass population is growing at a rate proportional to the population at that moment. Assume that when fishing starts in 1899, the rate at which bass are caught is proportional to the square of the population with constant of proportionality 10−9 . What happens to the bass population in the long run? A) It approaches 0 C) It approaches 7.6 ⋅108 B) It grows without bound D) It approaches 3.8 ⋅108 Ans: D difficulty: medium section: 11.7
Page 32
Chapter 11: Differential Equations
dy dx =− x + xy describe the rates of growth of = −5 y + 2 xy , dt dt two interacting species, where x is the number of species A, measured in thousands, and y is the number of species B, measured in thousands. For species A, is the interaction with species B favorable or unfavorable? Ans: favorable difficulty: easy section: 11.8
99. Suppose the equations
dx dy = −3 y + 3 xy , =− x + xy describe the rates of growth of dt dt two interacting species, where x is the number of species A, measured in thousands, and y is the number of species B, measured in thousands. Determine all the equilibrium points in the xy-phase plane. (Select all that apply.) A) (0, 0) B) (0,1) C) (1,1) D) (1,1) E) (1, 0) Ans: A, C difficulty: medium section: 11.8
100. Suppose the equations
dy dx = −4 y + 2 xy , =− x + xy describe the rates of growth of dt dt two interacting species, where x is the number of species A, measured in thousands, and y is the number of species B, measured in thousands. The slope field in the xy-phase plane is shown below. Sketch the trajectory for initial conditions of x = 1.5, y = 0.5. (In other words, there are initially 1500 of species A and 500 of species B). Is species A increasing or decreasing?
101. Suppose the equations
Ans: decreasing difficulty: medium
section: 11.8
Page 33
Chapter 11: Differential Equations
dx dy 102. Suppose the equations = 4 y − xy and = x − xy describe the rates of growth of dt dt two interacting species, where x is the number of species A, measured in thousands, and y is the number of species B, measured in thousands. Which of the following describes the nature of the interaction between these two species? A) Species A would be better off without species B, but species B needs species A in order to survive. B) Species B would be better off without species A, but species A needs species B in order to survive. C) The two populations depend on each other; each would not survive without the other. D) The two populations are competitors; each suffers in the presence of the other. Ans: D difficulty: easy section: 11.8 dx dy 103. Suppose the equations = 2 y − xy and = x − xy describe the rates of growth of dt dt two interacting species, where x is the number of species A, measured in thousands, and y is the number of species B, measured in thousands. The slope field in the xy-phase plane is shown below. Consider the trajectory for the initial conditions x = 0.5, y = 1.5. (In other words, there are initially 500 of species A and 1500 of species B.) Is the population of Species B increasing or decreasing?
Ans: increasing difficulty: medium
section: 11.8
Page 34
Chapter 11: Differential Equations
dx dy 104. Suppose the equations = 4 y − xy and = x − xy describe the rates of growth of dt dt two interacting species, where x is the number of species A, measured in thousands, and y is the number of species B, measured in thousands. Assuming that the initial conditions are 1000 of species A and 2000 of species B. What is the long run behavior of Species A? A) It dies out. B) It grows exponentially. C) It reaches an equilibrium. Ans: A difficulty: medium section: 11.8 105. On a fine spring day you stand in the Square and throw a bean bag high into the air and catch it. The following trajectory reflects the bean bag's trip.
What does B represent? A) The point where the bag is caught. B) The point where the bag reaches its highest altitude. C) The point where the bag is tossed. Ans: B difficulty: easy section: 11.8
Page 35
Chapter 11: Differential Equations
106. The following graphs show position versus time, velocity versus time, and position versus velocity (not necessarily in that order). Does the first, second, or third graph show position versus velocity?
Ans: second difficulty: medium
section: 11.8
Page 36
Chapter 11: Differential Equations
107. A fatal infectious disease is introduced into a growing population. Let S denote the number of susceptible people at time t and let I denote the number of infected people at time t. Suppose that, in the absence of the disease, the susceptible population grows at a rate proportional to itself, with constant of proportionality 0.2. People in the infected group die at a rate proportional to the infected population with constant of proportionality 0.05. The rate at which people get infected is proportional to the product of the number of susceptibles and the number of infecteds, with constant of proportionality 0.001. Which of the following systems of differential equations are satisfied by S and I? dS dI A) = 0.2 S − 0.05SI , =− 0.2 SI 0.05 I dt dt dS dI B) = 0.001S − 0.2 SI , = 0.05SI − 0.001I dt dt dS dI C) = 0.2 S − 0.001SI , = 0.001SI − 0.05 I dt dt dS dI D) =− 0.05S 0.001SI , = 0.001SI − 0.2 I dt dt Ans: C difficulty: easy section: 11.8 108. A fatal infectious disease is introduced into a growing population. Let S denote the number of susceptible people at time t and let I denote the number of infected people at time t. Suppose that, in the absence of the disease, the susceptible population grows at a rate proportional to itself, with constant of proportionality 0.2. People in the infected group die at a rate proportional to the infected population with constant of proportionality 0.15. The rate at which people get infected is proportional to the product of the number of susceptibles and the number of infecteds, with constant of proportionality 0.001. Find the equilibrium points for the system of differential equations satisfied by S and I. (Answer in the form (S,I).) (0, 0) (150, 0) E) A) (0, 200) (200,150) B) F) (0,150) (150, 200) C) G) (200, 0) D) Ans: A, G difficulty: medium section: 11.8 109. A fatal infectious disease is introduced into a growing population. Let S denote the number of susceptible people at time t and let I denote the number of infected people at time t. Suppose that, in the absence of the disease, the susceptible population grows at a rate proportional to itself, with constant of proportionality 0.3. People in the infected group die at a rate proportional to the infected population with constant of proportionality –0.2. The rate at which people get infected is proportional to the product of the number of susceptibles and the number of infecteds, with constant of proportionality 0.001. For the initial conditions S = 100 and I = 500, what happens as t starts to increase? A) S increases and I increases C) S decreases and I increases B) S increases and I decreases D) S decreases and I decreases Ans: D difficulty: medium section: 11.8
Page 37
Chapter 11: Differential Equations
110. Which system of equations could have the phase plane diagram in the following figure?
dx dy = −5 x + y, =− x 5y dt dt dx dy B) = = x − 5 y, 5x − y dt dt Ans: C difficulty: medium A)
C) D)
dx dy =− x + 5 y, =−5 x − y dt dt dx dy =− x 5 y, = −5 x − y dt dt
section: 11.9
111. Let x be the number of reptiles, y be the number of mammals, and z be the number of plants on the island of Komodo, all measured in thousands (e.g., x = 50 means 50,000 reptiles). The following differential equations give the rates of growth of reptiles, mammals, and plants on the island: dx = −0.2 x − 0.04 xy + 0.0008 xz dt dy = −0.1 y + 0.01xy dt dz = 2 z − 0.002 z 2 − 0.1xz dt
What would happen to the plants if the other two were not present? A) They would experience logistic growth. B) They would experience exponential decay. C) They would remain constant. Ans: A difficulty: medium section: 11.9
Page 38
Chapter 11: Differential Equations
112. Let x be the number of reptiles, y be the number of mammals, and z be the number of plants on the island of Komodo, all measured in thousands (e.g., x = 50 means 50,000 reptiles). The following differential equations give the rates of growth of reptiles, mammals, and plants on the island: dx = −0.2 x − 0.04 xy + 0.0008 xz dt dy = −0.1 y + 0.01xy dt dz = 2 z − 0.002 z 2 − 0.1xz dt Say that initially there are 100,000 plants (i.e., z = 100) on Komodo, and there are no reptiles or mammals. At what plant population would the number of plants be increasing the fastest? Ans: 500,000 difficulty: medium section: 11.9 113. Let x be the number of reptiles, y be the number of mammals, and z be the number of plants on the island of Komodo, all measured in thousands (e.g., x = 50 means 50,000 reptiles). The following differential equations give the rates of growth of reptiles, mammals, and plants on the island: dx = −0.6 x − 0.12 xy + 0.0024 xz dt dy = −0.1 y + 0.01xy dt dz = 6 z − 0.006 z 2 − 0.1xz dt Who is eating whom on Komodo? Mark all that apply. A) Reptiles eat plants. C) Mammals eat plants. B) Reptiles eat mammals. D) Mammals eat reptiles. Ans: A, D difficulty: easy section: 11.9
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Chapter 11: Differential Equations
114. Let x be the number of reptiles, y be the number of mammals, and z be the number of plants on the island of Komodo, all measured in thousands (e.g., x = 50 means 50,000 reptiles). The following differential equations give the rates of growth of reptiles, mammals, and plants on the island: dx = −0.2 x − 0.04 xy + 0.0008 xz dt dy = −0.1 y + 0.01xy dt dz = 2 z − 0.002 z 2 − 0.1xz dt What is the equilibrium population of plants (assuming that the population is nonzero)? Ans: 500,000 difficulty: medium section: 11.9 115. The interaction of two populations x(t) and y(t) is modeled by the system 1 dx 1 dy =1 − x − ky, =1 − y + kx , x dt y dt where k is a positive constant. What type of interaction is modeled here? A) Symbiosis B) Predator-Prey with y the predator and x the prey. C) Predator-Prey with x the predator and y the prey. D) Competition E) None of the above Ans: B difficulty: medium section: 11.9 116. The interaction of two populations x(t) and y(t) is modeled by the system 1 dx 1 dy =1 − x − ky, =1 − y + kx , x dt y dt where k is a positive constant. Do the qualitative phase plane analysis for the case k = 2. In the long run, the trajectories spiral towards the point (a, b), where a = _____ and b = _____. Part A: 0 Part B: 1 difficulty: medium section: 11.9
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Chapter 11: Differential Equations
117. The acceleration of a moving object is given by d 2x dx = x − 1 2 dt dt where x(t) is the position at time t. Which of the following is a system of first order differential equations for position x and velocity v? dx dv dx dv A) = C) = v , = x(1 − v) v , = x(v − 1) dt dt dt dt dx dv dx dv x 2 B) D) = v= , = v −1 v ,= (v − 1) dt dt dt dt 2 Ans: C difficulty: easy section: 11.9
118. The acceleration of a moving object is given by d 2x dx = x − 1 2 dt dt where x(t) is the position at time t. Do the qualitative phase plane analysis for this system (v versus x). Which of the following are true about the slope field? A) The slopes are vertical along the x-axis. B) The slopes are vertical along the v-axis. C) The slopes are horizontal along the x-axis. D) The slopes are horizontal along the v-axis. E) The slopes are vertical along the line v = 1. F) The slopes are horizontal along the line v = 1. G) There is an equilibrium point at (0, 0). H) There is an equilibrium point at (0, 1). Ans: A, D, F, G difficulty: medium section: 11.9
119. The acceleration of a moving object is given by d 2x dx = x − 1 2 dt dt where x(t) is the position at time t. Use a graphing calculator to help sketch the slope field and the trajectory for the initial values x(0) = –3, v(0) = 3. Approximate the maximum value of x to the nearest integer. If the maximum does not exist, enter "DNE". Ans: DNE difficulty: hard section: 11.9
120. The general solution of the differential equation forms? A)= s (t ) C1 cos ωt + C2 sin ωt
d 2s + ω 2s = 0 is of which of these 2 dt
C)= s (t ) C1 cos ω 2t − C2 sin ωt
B)= D) s (t ) C1 cos 2 ωt + C2 sin 2 ωt = s (t ) C= C2 sin ωt 1 cos ω t or s (t ) Ans: A difficulty: easy section: 11.8
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Chapter 11: Differential Equations
121. Show directly that s = cos ωt is a solution to Ans: −ω 2 cos ωt + difficulty: easy
k d 2s k = ω2 . + s= 0 , given that 2 m dt m
k (cos ωt ) = −ω 2 cos ωt + ω 2 cos ωt = 0. m section: 11.1
122. Find the solution to the differential equation
d 2s − 25s = 0 given the boundary dt 2
conditions s (0) = 10 and s ( π2 ) = 5 . s (t ) 5cos 25t − 10sin 25t s (t ) 5cos 25t + 25sin 5t A) = D) = s (t ) = 10 cos 5t + 5sin 5t s (t ) = 50 cos 5t B) E) s (t ) = 10 cos 5t – 5sin 5t C) Ans: B difficulty: medium section: 11.1
123. Rewrite 8cos ωt + 2sin ωt in the form A sin(ωt + ϕ ) . Give your constants as decimals rounded to two decimal places. Ans: 8.25sin(ωt + 1.33) difficulty: easy section: 11.1 124. Verify that y = –2 cos(5t ) is a solution of y′′ = −25 y . Suppose that y describes the motion of a mass on a spring. How does the motion start when t = 0? A) lowest point B) in the middle C) highest point Ans: A difficulty: easy section: 11.1 125. An artist works on a museum installation that includes a spring attached to the ceiling with a 5 kg pink quartz crystal attached to the other end. The artist sets the installation in motion by pulling the crystal down an additional 0.2 meters and then releasing it. Assuming no air resistance, set up a differential equation with initial conditions describing the motion of the crystal. Then solve your differential equation and use it to find the location of the crystal after Ans:
π seconds. 4ω
d 2s k = − s= −ω 2 s , s (0) = −0.2 , s′(0) = 0 . 2 5 dt s (t ) = −0.2 cos(ωt ) = −0.2 cos( k5 t )
π π 2 ) = −0.2 cos( ) = − ≈ −0.1414 , or about 14.14 centimeters below rest. 4ω 4 10 difficulty: hard section: 11.8 s(
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Chapter 11: Differential Equations
126. In an electric circuit, let Q(t) be the charge on a capacitor at time t. Then the current in dQ the circuit is given by I (t ) = (the rate of change of the charge). The charge and the dt dI Q current satisfy the differential equation L + = 0 . If L = 34 henry, and C = 10 farad, dt C find a formula for Q(t) with initial conditions Q(0) = 0 and I(0) = 2. Round constants to three decimal places. A) Q(t) = 36.878sin(0.054t) B) Q(t) = 36.878cos(0.054t) C) Q(t) = 0.054sin(0.054t) D) Q(t) = 36.878cos(0.054t) + 0.054sin(34t) E) Q(t) = 10cos(34t) + 10sin(34t) Ans: A difficulty: hard section: 11.8
0. 127. Show that y= (t ) C1et + C2 e − t is the general solution to y′′ − y = Ans: y′′ =C1et + C2 e − t =y . difficulty: easy section: 11.1
0. 128. Show that y (t ) = sinh(t ) is a solution to y′′ − y = = y′′ sinh( = t) y Ans: difficulty: easy section: 11.1 d2y dy + 5 − 6y = 0? 2 dx dx r 2 − 5r + 6 = 0
129. What is the characteristic equation of the differential equation
(r + 5)(r − 6) = 0 A) (r + 5)(r − 1) = 0 B) Ans: D difficulty: easy
C) D) section: 11.8
r 2 + 5r − 6 = 0
130. What is the characteristic equation of the differential equation 9
d2y dy + 2 − 27 y = 0? 2 dx dx
Ans: r 2 + 0.222r − 3 = 0 difficulty: easy section: 11.8
0. 131. Find the general solution to the differential equation y′′ + 16 y′ + 64 y = −8t Ans: y= (t ) (C1t + C2 )e difficulty: medium section: 11.8 0 describes the motion of a spring. 132. Suppose the differential equation y′′ + 16 y′ + 64 y = Which of the following terms describes the motion of the spring? A) underdamped B) critically damped C) overdamped Ans: B difficulty: easy section: 11.1
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Chapter 11: Differential Equations
133. Solve the initial value problem and then explain the motion of the spring described by this differential equation. d2y dy −10 8 2 + 64 + 96 y = 0, y (0) = 4, y′(0) = dx dx d2y dy Ans: Divide through by 8: + 8 + 12 y = 0. 2 dx dx r 2 + 8r + 12 = 0 so r1 = −6, r2 = −2 . The general solution is: = y (t ) C1e −6t + C2 e −2t . Solve to get C1 = 1 2 and C2 = 7 2 using y (0) = 4 and y′(0) = −10 . The solution is: y= ( x) 12 e −6t + 72 e −2t . The motion of the spring is overdamped. The spring is in a fluid so thick that it cannot oscillate. difficulty: medium section: 11.8 134. Solve the initial value problem and then explain the motion of the spring described by this differential equation. d2y dy − 4 + 5= = 2, y′(0) = 1 y 0, y (0) 2 dx dx Ans: Partial solution: r =α i ± β =2 ± i = y ( x) C1e 2t cos t + C2 e 2t sin t y ( x) = 2e 2t cos t – 3e 2t sin t The oscillations of the spring are underdamped. difficulty: hard section: 11.8 135. In an electrical circuit, a damping force can be provided by a resistor. In this case, the d 2Q dQ 1 differential equation L 2 + R + Q= 0 describes the charge Q on a capacitor in a dt dt C circuit with inductance L, capacitance C, and resistance R. Suppose L = 1 henry, R = 50 ohms and C = 0.0016 farads. Find a formula for the charge, Q, when Q(0) = 0 and Q′(0) = 2 . Ans: Q(= t ) (C1 + C2t )e −25t Q(t ) = 2te −25t difficulty: hard section: 11.8
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Chapter 11: Differential Equations
136. Which of the following differential equations has the solution graphed below? i.
d2y dt
=5 2
Ans: i difficulty: medium
ii.
dy = y −5 dt
iii.
dy = y (5 − y ) dt
iv.
dy = − yt dt
section: 11 review
dy 137. Is = y x 2 + x a solution to = 2( y − x 2 ) +1 ? dx Ans: yes difficulty: medium section: 11 review dy = x + 1 passing through (0, 1) the same as the solution passing dx through (0, 0), except it has been shifted one unit upward? Ans: yes difficulty: medium section: 11 review
138. Is the solution of
dP 139. Are the solutions of = kP ( L − P ) always concave down? dt Ans: no difficulty: medium section: 11 review dP = sin P . dt Is the equilibrium solution with 6 ≤ P ≤ 7 stable or unstable? Ans: unstable difficulty: medium section: 11 review
140. Consider the differential equation
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Chapter 11: Differential Equations
dP = sin P . dt What happens to the value of P (t ) as t → ∞ if P (0) = 2 ? A) It approaches 0. C) It approaches 2π . B) It approaches π . D) It approaches ∞ . Ans: B difficulty: medium section: 11 review
141. Consider the differential equation
142. Consider the differential equation
dP = sin P . dt
d 2P in terms of P and use it to decide whether the solution curve at (5, 1) is dt concave up or concave down. Ans: concave up difficulty: medium section: 11 review
Find
Page 46
1. You are in a nicely heated cabin in the winter. Deciding that it's too warm, you open a small window. Let T be the temperature in the room, t minutes after the window was opened, x feet from the window. Is T an increasing or decreasing function of x? A) Increasing B) Decreasing C) Neither Ans: A difficulty: easy section: 12.1 2. The following table gives the number f(x, y) of grape vines, in thousands, of age x in year y.
In one year a fungal disease killed most of the older grapevines, and in the following year a long freeze killed most of the young vines. Which are these years? Ans: 1982 and 1983 difficulty: easy section: 12.1
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Chapter 12: Functions of Several Variables
3. The following table gives the number f(x, y) of grape vines, in thousands, of age x in year y.
In 1986 a successful advertising campaign led to a dramatic increase in demand for premium wines. The growers followed by adding many more plants. Suppose a vine (the plant) produces the first harvestable grapes at age five, and is removed after sixteen years. How many (thousand) grape vines that bear fruit were there in the year 1986 and how many will be there in the year 1992 (assuming that no current vines die before 1992)? Enter your answers separated by a semi-colon. Ans: 11,000; 29,000 difficulty: medium section: 12.1 4. You are at (4, 2, 4) facing the yz-plane. You walk 3 units, turn right and walk for another 2 units. What are your coordinates now? Are you above or below the xy-plane? Ans: My coordinates are (1, 4, 4) and I am above the xy-plane. difficulty: easy section: 12.1
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Chapter 12: Functions of Several Variables
5. (a) Find an equation of the largest sphere that can fit inside the cubical space enclosed by the planes x = 1, x = 5, y = 2, y = 6, z = 2 and z = 6. (b) If we replace the plane z = 6 in part (a) with z = 7, what will be the new equation of the largest sphere? 2 2 2 Ans: (a) ( x − 3) + ( y − 4 ) + ( z − 4 ) = 4 (b) ( x − 3) + ( y − 4 ) + ( z − c ) = 4, difficulty: medium section: 12.1 2
2
2
4≤c≤5
6. Consider the sphere ( x +1) 2 + ( y − 0) 2 + ( z –1) 2 = 4 (a) What are the center and radius of this sphere? (b) Find an equation of the circle (if any) where the sphere intersects the plane x = –2. Ans: (a) Center (–1, 0, 1), Radius 2. (b) ( y − 0) 2 + ( z –1) 2 = 3. difficulty: medium section: 12.1 7. The points A = (4, 1, 2), B = (3, –2, 3), and C = (–2, 3, –4) are the vertices of a triangle in space. Which of the vertices is closest to the yz-plane? A) C B) A C) B Ans: A difficulty: easy section: 12.1 8. The points A = (1, 1, 1), B = (2, 4, 2), and C = (3, 2, 2) are the vertices of a triangle in space. Which of the vertices is closest to the origin? A) A B) B C) C Ans: A difficulty: easy section: 12.1 9. The points A = (–4, 5, –3), B = (–1, –3, –4), and C = (–2, 4, –4) are the vertices of a triangle in space. What is the length of the longest side of the triangle? Ans: 74 difficulty: easy section: 12.1
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Chapter 12: Functions of Several Variables
10. A certain piece of electronic surveying equipment is designed to operate in temperatures ranging from 0° C to 30° C. Its performance index, p(t, h), measured on a scale from 0 to 1, depends on both the temperature t and the humidity h of its surrounding environment. Values of the function p = f(t, h) are given in the following table. (The higher the value of p, the better the performance.)
What is the value of p(0, 25)? Ans: 0.46 difficulty: easy section: 12.1 11. A certain piece of electronic surveying equipment is designed to operate in temperatures ranging from 0° C to 30° C. Its performance index, p(t, h), measured on a scale from 0 to 1, depends on both the temperature t and the humidity h of its surrounding environment. Values of the function p = f(t, h) are given in the following table. (The higher the value of p, the better the performance.)
Describe the function p(10, h) and explain its meaning. Ans: The value of p(10, h) will first increase (as h increases from 0 to 25) then decrease (as h increases from 25 to 100). This means that when the temperature is fixed at 10° C, the equipment works best in low humidity, with optimal performance around 25% humidity. The performance will degrade severely as the humidity rises. difficulty: easy section: 12.1
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Chapter 12: Functions of Several Variables
12. Yummy Potato Chip Company has manufacturing plants in N.Y. and N.J. The cost of manufacturing depends on the quantities (in thousand of bags), q1 and q2, produced in the N.Y. and N.J. factories respectively. Suppose the cost function is given by C (q1 , q2 ) = 2q12 + q1q2 + q22 + 420 (a) Find C (10, 25) (b) By comparing the terms 2q12 and q22 in the above expression, the manager concluded that it is more expensive to produce in the N.Y. factory. Will shifting all the production to the N.J. factory minimize the production cost? Ans: (a) 1495 (b) No, the move will not minimize the production cost. To produce 100,000 bags, it is cheaper to have N.Y. produce 25,000 bags and N.J. produce 75,000 bags, rather than to have N.J. produce all 100,000 bags. The manager failed to notice from the formula that as the production in a factory increases, the cost will rise quadratically. difficulty: easy section: 12.1 13. Your monthly payment, C(s, t), on a car loan depends on the amount, s, of the loan (in thousands of dollars), and the time, t, required to pay it back (in months). What is the meaning of C(7, 48) = 250? A) If you borrow $7,000 from the bank for 48 months (4 year loan), your monthly car loan payment is $250. B) If you borrow $4,000 from the bank for 48 months (7 year loan), your monthly car loan payment is $250. C) If you borrow $250 from the bank for 48 months (4 year loan), your monthly car loan payment is $7. D) If you borrow $7 from the bank for 48 months (4 year loan), your monthly car loan payment is $250. Ans: A difficulty: easy section: 12.1 14. Your monthly payment, C(s, t), on a car loan depends on the amount, s, of the loan (in thousands of dollars), and the time, t, required to pay it back (in months). Is C an increasing or decreasing function of t? A) Decreasing B) Increasing Ans: A difficulty: easy section: 12.1
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Chapter 12: Functions of Several Variables
15. Find a possible formula for a function f(x, y) with the given values.
x
1 2 3
1 1 –1 –3
Ans: –2 x + 3 y difficulty: hard
y 2 4 2 0
3 7 5 3
section: 12.1
16. Describe in words, write equations, and give a sketch for the following set of points.
Ans:
difficulty: easy
section: 12.2
17. Describe in words the intersection of the surfaces= z Ans: A circle (of radius 1) in the plane z = 1. difficulty: medium section: 12.2
Page 6
7 − 6( x 2 + y 2 ). x 2 + y 2 and z =
Chapter 12: Functions of Several Variables
18. A spherical ball of radius four units is in a corner touching both walls and floor. What is the radius of the largest spherical ball that can be fit into the corner behind the given ball? (Hint: The smaller ball will not touch the corner point where the walls meet the floor.)
Ans: r =
( 3 − 1) ( 3 + 1)
4
difficulty: medium
section: 12.2
19. Match the graph with the function.
A)
B)
C)
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Chapter 12: Functions of Several Variables
D)
Ans: A
difficulty: easy
section: 12.2
20. Match the graph with the function.
A)
B)
C)
D)
Ans: A
difficulty: medium
section: 12.2
21. What is the slope of the contour lines of the function f(x, y) = –3 + 9 x +10 y ? 9 Ans: – 10 difficulty: easy section: 12.2
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Chapter 12: Functions of Several Variables
22. A soft drink company is interested in seeing how the demand for its products is affected by price. The company believes that the quantity, q, of soft drinks sold depends on p1 , the average price of the company's soft drinks, and p2 , the average price of competing soft drinks. Which of the graphs below is most likely to represent q as a function of p1 and p2 ? A)
B)
C)
D)
Ans: C
difficulty: medium
section: 12.2
Page 9
Chapter 12: Functions of Several Variables
23. For what values of the constant k is the intersection between the set of points y = x and the graph of f ( x, y ) = 4 x 2 – ky 2 a straight line? Ans: 4 difficulty: easy section: 12.2 24. Match the following function with the graphs below. The function z = f(x, y) giving happiness as a function of health y and money x according to the statement of a fortune cookie: 'Whoever said money cannot buy happiness does not know where to shop.' A)
B)
C)
Ans: C
difficulty: easy
section: 12.2
Page 10
Chapter 12: Functions of Several Variables
25. Match the function with the graph below.
A)
B)
Ans: B
difficulty: easy
section: 12.2
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Chapter 12: Functions of Several Variables
26. The following figure contains the graphs of the cross sections z = f(a, y) for a = -2, -1, 0, 1, 2. Which of the graphs of z = f(x, y) in A and B best fits this information?
A)
B) Ans: B
difficulty: easy
section: 12.2
Page 12
Chapter 12: Functions of Several Variables
27. The graph of the function f(x, y) is shown below.
Draw graph of cross-sections with y fixed at y = 0, and y = 1. Ans:
difficulty: easy
section: 12.2
28. Two contours of the function f(x, y) corresponding to different values of f cannot ever cross. Ans: True difficulty: easy section: 12.3 29. The contours of the function f(x, y) = 8x + 4y are all parallel lines with slope 2. A) False B) True Ans: A difficulty: easy section: 12.3
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Chapter 12: Functions of Several Variables
30. The contour diagram below shows the level curves of the difference between July and January mean temperatures in ° F.
Does this graph support or contradict the claim that the largest annual temperature variations are found on the coasts of continents? Ans: This graph supports the claim that the largest annual temperature variations are found on the coasts of continents, as level curves are very close together near the coasts of continents. difficulty: easy section: 12.3 31. Draw a possible contour diagram for the function whose graph is shown below. Label your contours with reasonable z-values.
Ans:
Page 14
Chapter 12: Functions of Several Variables
difficulty: easy
section: 12.3
32. Consider the function= z f ( x,= y ) –3 y − 2 x 2 . Suppose you are standing on the surface at the point where x = 2, y = –1. What is your altitude? Ans: –5 difficulty: easy section: 12.3 33. Consider the function = z f ( x, y= ) 3 y − 4 x 3 . Suppose you are standing on the surface at the point where x = 3, y = 1. If you start to move on the surface parallel to the y-axis in the direction of increasing y, does your height increase or decrease? Ans: Increase difficulty: easy section: 12.3 34. The diagram below shows the contour map for a circular island. Sketch the vertical cross-section of the island through the center. Your sketch should show concavity clearly.
Ans:
Page 15
Chapter 12: Functions of Several Variables
difficulty: easy
section: 12.3
35. Draw the level curves for z = 2ln(x) - ln(y). A)
B)
C)
Page 16
Chapter 12: Functions of Several Variables
D)
Ans: C
difficulty: easy
section: 12.3
36. Suppose that the temperature T of any point (x, y) is given by T ( x, y ) = 100 − x 2 − y 2 . Sketch isothermal curves (i.e. contours) for T = 100, T = 75, T = 50 and T = 0. Be sure to label each contour. What does the graph of T(x, y) look like if it is sliced by the plane x = 4? Ans:
Page 17
Chapter 12: Functions of Several Variables
The parabola = z 84 − y 2 difficulty: easy
section: 12.3
37. Which of the following is a contour diagram for f(x, y) = sin x? A)
B)
Page 18
Chapter 12: Functions of Several Variables
C)
D)
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Chapter 12: Functions of Several Variables
Ans: B
difficulty: easy
section: 12.3
38. The picture below is the contour diagram of f(x, y). The areas between contours have been shaded. Lighter shades represent higher levels, while darker shades represent lower levels.
Sketch the cross section of f(x, y) with x fixed at x = 0.5. Ans:
Page 20
Chapter 12: Functions of Several Variables
difficulty: easy
section: 12.3
39. Let f ( x, y ) =y 2 − 8 yx + 16 x 2 . Find the contour of f that passes through the point (0, 2) . y 4 x + 2 and = y 4x − 2 Ans: = difficulty: medium section: 12.3 40. Find a formula for a function f(x, y), given that its contour at level 9 has equation x 2 − 10 xy = –1. Ans: f ( x, y ) = x 2 − 10 xy + 10 difficulty: medium section: 12.3
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Chapter 12: Functions of Several Variables
41. Below is a contour diagram depicting D, the average fox population density as a function of xE , kilometers east of the western end of England, and xN , kilometers north of the same point.
Is D increasing or decreasing at the point (120, 25) in the northern direction? Ans: The function is increasing in the northern direction since as we go north the number of foxes increases. difficulty: easy section: 12.3 42. True or False: If all of the contours of a function f(x, y) are parallel lines, then the function must be linear. Ans: False difficulty: easy section: 12.4 43. True or False: If f is a linear function, then f (4, 2) − f (4,1) = f (0, 2) − f (0,1). Ans: True difficulty: easy section: 12.4 44. True or False: If f is any linear function of two variables, then f(x, y + 1) = f(x + 1, y). Ans: False difficulty: easy section: 12.4
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Chapter 12: Functions of Several Variables
45. Consider the (partial) contour diagram below for a linear function
Find an equation z = f(x, y) for the function. 7 2 4 Ans: z = + x + y 3 3 3 difficulty: medium section: 12.4 46. A plane passes through the points (1, 3, 7), (-1, 0, 6), and (2, 1, –3). Determine the equation of the plane. Ans: z =2 − 4 x + 3 y difficulty: medium section: 12.4 47. Consider the plane that passes through the points (1, 3, 10), (-1, –1, 0), and (2, 1, –1). If you were walking on this plane with no change in altitude, what would be the slope of your path in the xy-plane? 3 Ans: 4 difficulty: medium section: 12.4
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Chapter 12: Functions of Several Variables
48. Given the table of some values of a linear function complete the table: 2.5 13
y\x -1 1 3
3
3.5 15 9
8
1
Ans: 2.5 13 7 1
y\x -1 1 3 difficulty: easy
3 14 8 2
3.5 15 9 3
section: 12.4
49. Given the table of some values of a linear function determine a formula for the function.
y\x -1 1 3
2.5 –7
3
–13
3.5 –11 –15
–15
Ans: z = –4 x − 2 y + 1 difficulty: medium section: 12.4 50. Find an equation for the plane passing through (1, –5, –2) and containing the x-axis. 2 Ans: z = y 5 difficulty: medium section: 12.4
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Chapter 12: Functions of Several Variables
51. Find a formula for the linear function whose contours are shown below.
Ans: f ( x, y ) = 3 y − 2 x + 3 difficulty: easy section: 12.4 52. A linear function f(x, y) has the values f(4, 2) = 10, f(1, 2) = 4 and f(4, 1) =7. Find an equation for f. Ans: f ( x, y ) = 2 x + 3 y − 4 difficulty: medium section: 12.4 53. Determine a formula for the linear function f(x, y), such that its cross-section with y fixed at y = 1 has equation z = 4x - 1, and its contour at level 0 is the line y = (4 + 4x)/5. Ans: f ( x, y ) = 4 x − 5 y + 4 difficulty: medium section: 12.4 54. A linear function f(x,y) has cross-sections f(x,4) = 2x - 14 and f(2,y) = -4y + 6. Find an equation for f. Ans: f ( x, y ) = 2 x − 4 y + 2 difficulty: medium section: 12.4 55. Determine the equation of the plane which passes through the point (1, 3, –1), has slope 5 in the x-direction and slope –3 in the y-direction. Ans: z = 5 x − 3 y + 3 difficulty: easy section: 12.4 56. Find a formula for a linear function z = f(x, y) whose f = 0 contour is the line y = 2x + 2. Ans: z = 2 x − y + 2 difficulty: easy section: 12.4
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Chapter 12: Functions of Several Variables
57. Describe the set of points whose distance from the y-axis equals the distance from the xz-plane. Write an equation for this set in the form G(x, y, z) = 0. Can the set be described as the graph of a function of two variables? Ans: This is two cones centered around the y-axis with vertex at the origin as shown below. The equation is x 2 − y 2 + z 2 = 0. It cannot be expressed as the graph of a function of two variables, because there are two values of z for each value of x and y (except when x = ±y): z 2 = ± y 2 − x2 .
difficulty: easy
section: 12.5
58. Does the point (–5, –1, –5) lie on any of the level surfaces of f ( x, y, z ) =x 2 + 4 y 2 + z 2 ? If so, what is the equation of that level surface? Ans: Yes: x 2 + 4 y 2 + z 2 = 54 difficulty: easy section: 12.5 59. What is the domain of g ( x, y, z ) =
x2 + y 2 + z 2 ? x2 + y 2 + z 2 − 3
Describe the level surface g =5. Ans: The domain of g consists of all points in 3-space except the points that satisfy x2 + y 2 + z 2 = 3 , which lie on a sphere of radius 3 centered at the origin. 15 The level surface is a sphere x 2 + y 2 + z 2 = . 4 difficulty: easy section: 12.5
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Chapter 12: Functions of Several Variables
60. Sketch the level set of f ( x, y= ) x 2 + z 2 corresponding to f = 16 and describe it in words. Ans: The level set is a cylinder of radius 4, with its axis along the y-axis.
difficulty: easy
section: 12.5
Page 27
Chapter 12: Functions of Several Variables
61. Some level surfaces of the functions g1 ( x, y, z ), g 2 ( x, y, z ) and g3 ( x, y, z ) are shown below, respectively.
Which function takes the value 0 at (0, 0, 0)? Ans: g3 difficulty: easy section: 12.5
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Chapter 12: Functions of Several Variables
62. The level surfaces of the functions g1 ( x, y, z ), g 2 ( x, y, z ) and g3 ( x, y, z ) are shown in below respectively.
Which function is is decreasing in the positive z-direction? Ans: g3 difficulty: easy section: 12.5 63. Classify the following surface as ellipsoid, elliptical paraboloid, hyperbolic paraboloid, hyperboloid of one sheet, hyperboloid of two sheets, or cone. –9 x 2 – 4 y 2 + z 2 = 9 Ans: hyperboloid of two sheets difficulty: medium section: 12.5
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Chapter 12: Functions of Several Variables
64. Find a function f(x, y, z) whose level surface at level 4 is the graph of the function h( x, y )= 5 x 2 y + 5 y − 1 . Ans: f ( x, y, z ) =z − 5 x 2 y − 5 y + 5 . (There are other possible answers.) difficulty: medium section: 12.5 65. Given that the equation of the level surface of f(x, y, z) at level –3 is 3 x 2 + xy + z = 15 , determine f(2, 3, –3) if possible. Ans: –3 difficulty: medium section: 12.5 66. Consider the figure shown below.
Match the level curve with the function. A) f1 ( x, y, z ) = 5 x + y + z B) f 2 ( x, y , z ) = 5 x 2 + y 2 + z Ans: B difficulty: easy
C)
f 2 ( x, y , z ) = 5 x – y + z
section: 12.5
67. What value of c (if any) makes the following function continuous at (0, 0)? x4 − y 4 if ( x, y ) ≠ (0, 0) f ( x, y ) = x − y c otherwise Ans: c=0 difficulty: easy section: 12.6
68. Let f ( x, y ) =
x − 5y . Describe the level curve of f at level 2. x + 5y
1 x , provided that x ≠ 0 . 15 difficulty: easy section: 12.6 Ans: y = –
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Chapter 12: Functions of Several Variables
69. A function f(x, y) is defined for (x, y) ≠ (0, 0). Does lim ( x , y )→(0,0) f ( x, y ) exist if f has the contour diagram below?
Assume that different contours represent different values of the function. Explain your answer. Ans: The limit does not exist. The different contours meet at the origin. If it existed the limit would have to be equal to the value on each of the contours. This is impossible since the values are different. Hence the function is not continuous at the origin. difficulty: medium section: 12.6 70. Determine the values of a and b such that the following function is continuous everywhere. 5 + ax + by, y ≥ 2 f ( x, y ) = 4 + 5 x + 2 y, y < 2 3 Ans:= a 5,= b 2 difficulty: medium section: 12.6
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Chapter 12: Functions of Several Variables
71. Suppose that for all x and y the function f satisfies
f ( x, y ) − 3 > x 2 + y 2 and f ( x, y ) − 4 ≤ 5 ( ( x − 4) 2 + ( y − 2) 2 ) .
Determine, if possible, the values of the following limits. Explain your answer. Note that the limit may not exist nor be determined from the given information. (a) lim ( x , y )→(0,0) f ( x, y ) (b) lim ( x , y )→(4,2) f ( x, y ) Ans: (a) It cannot be determined from the given information. Since f ( x, y ) − 3 > x 2 + y 2 , we do not know whether |f(x, y) - 3| will approach 0 as
x 2 + y 2 approaches 0.
(b)
As (x, y) approaches (4, 2), we have 5 ( ( x − 4) 2 + ( y − 2) 2 ) approaches 0, and
hence by the inequality, f ( x, y ) − 4 will also approach zero. This implies that lim ( x , y )→(4,2) f ( x, y ) = 4 .
difficulty: medium
section: 12.6
72. Let f ( x, = y ) cos 9 − x 2 − y 2 . What is the domain of f? What are the possible values of f? Ans: Domain: x 2 + y 2 ≤ 9. The value of f can be any real number between -1 and 1. difficulty: easy section: 12R ax 2 − bx , where a, b are positive numbers. For what –7 + y values of (x, y) will the function f not be defined? Ans: y = 7 difficulty: easy section: 12R
73. Consider the function f ( x, y ) =
74. Find a linear function with f(–5, 5) = 1, whose graph is parallel to the level surfaces of g(x, y, z) = 10x + 15y + 5z. −2 x − 3 y + 6 Ans: f ( x, y ) = difficulty: medium section: 12R
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Chapter 12: Functions of Several Variables
75. Represent (if possible) the following surfaces as graphs of functions f(x, y), and as level surfaces of the form g(x, y, z) = c. (There are many possible answers.) (a) The upper half of the sphere of radius 3, centered at (5, 5, 0). (b) The lower half of the cylinder of radius 4 around the x-axis. (c) The cone = z 2 4x2 + 4 y 2 . Ans: (a) The equation of the sphere of radius 3, centered at (5, 5,0) is: ( x − 5) 2 + ( y − 5) 2 + z 2 = 9, · The upper part of this sphere is given by z= + 9 − ( x − 5) 2 − ( y − 5) 2 .
Therefore, we can choose f ( x, y )= + 9 − ( x − 5) 2 − ( y − 5) 2
and
g ( x, y,= z ) z – 9 − ( x − 5) 2 − ( y − 5) 2 .
(b) The equation of the cylinder is y 2 + z 2 =. 16 The lower part of the cylinder is given by z = − 16 − y 2 . Therefore, we can choose f ( x, y ) = − 16 − y 2
and
g ( x, y, z ) =+ z 16 − y 2 .
(c) We cannot represent the cone as the graph of a function. We can represent it as a level surface of g ( x, y, z ) =z 2 − 4 x 2 − 4 y 2 . difficulty: medium section: 12R 76. Let f(x, y, z) = ax + by + cz + d be a linear function of three variables, for constants a, b, c and d. Given that some of the cross sections of f are f(x, 1, 1) = 0 + 4x, f(0, y, –3) = 4 + 4y, and f(1, 1, z) = 6 - 2z, find a formula for f. Ans: f(x, y, z) = 4x + 4y - 2z - 2. difficulty: medium section: 12R 77. Consider the function f ( x, y ) = a x b y , for certain constants a and b. Simplify the following expression. f ( x + 4, y + 5) f ( x, y ) Ans: a 4b5 difficulty: easy
section: 12.1
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Chapter 12: Functions of Several Variables
()
78. Which of the following gives the domain and range of f ( x, y ) = ln xy ? A) domain: {( x, y ) : x, y ≠ 0} ; range: {z : z ≠ 0} B) domain: {( x, y ) : x > 0, y > 0} ; range: {z : z > 0} C) domain: {( x, y ) : xy > 0} ; range: all real numbers Ans: C difficulty: easy section: 12.1 79. Which of the following best describes the graph of g ( x,= y) A) cone centered on x-axis B) hyperbola centered at the origin C) upper half of a cone centered on the x-axis D) one branch of a hyperbola centered at the origin Ans: C difficulty: easy section: 12.2
x2 − y 2 ?
80. Describe the set of points in 3-space for which the distance from the x-axis equals the distance from the y-axis. A) two planes that intersect in the z-axis C) two planes that intersect in the y-axis B) two planes that intersect in the x-axis Ans: A difficulty: easy section: 12.2
( x, y ) cos( Ax + By ) . Find A and B if you know that the period of 81. Suppose that g= 2π g ( x, 0) is and the period of g (0, y ) is 3 . 11 2π Ans:= A 11; B . = 3 difficulty: easy section: 12.2 82. Suppose f is a function of two variables. True or False: Given a point (x,y) in the domain of f, there is a level curve of f passing through (x,y). Ans: True difficulty: easy section: 12.3 83. Suppose f is a function of two variables. True or False: Given a point (x,y) in the plane, there is a level curve of f passing through (x,y). Ans: False difficulty: easy section: 12.3 84. Suppose f is a function of two variables. True or False: The contours of f must also be functions. Ans: False difficulty: easy section: 12.3 85. Suppose f is a function of two variables. True or False: The cross-sections of f with one variable fixed must also be functions. Ans: True difficulty: easy section: 12.2
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Chapter 12: Functions of Several Variables
86. Suppose that the graph of a linear function f passes through the point (3,1,6) and that the 1 z = 9 level curve of f is the line = y x − 2 . Find a formula for f. 2 Ans: f ( x, y ) =x − 2 y + 5 difficulty: medium section: 12.4 87. Lawn King sells a push mower at price p1 and a ride-on mower at price p2. The demand function for each mower is a linear function of p1 and p2. Market research shows that if the price of the push mower is increased by $40, then the demand for it drops by 6400 units, while the demand for the ride-on mower increases by 1200 units. On the other hand, if the price of the ride-on mower is decreased by $60, then demand for it increases by 3000 units, while demand for the push mower drops by 9000 units. Currently, the company is selling 142,000 push mowers at a price of $300 and 19,000 ride-on mowers at a price of $1200. Find the two demand functions. Ans: d1 ( p1 , p2 ) = −160 p1 + 150 p2 + 10, 000 , d 2 ( p1 , p2 ) = 30 p1 − 50 p2 + 70, 000 difficulty: hard section: 12.4 88. True or False: Level surfaces of g(x,y,z) corresponding to different levels cannot intersect. Ans: True difficulty: easy section: 12.5 89. Suppose f and g are different functions of three variables. Is it possible for the level surface f = 1 and the level surface g = 1 to be the same surface? 1 Ans: Yes. For example, f ( x, y, z ) = x 2 + y 2 + z 2 and g ( x, y, z ) = 2 both x + y2 + z2 have the unit sphere as their level surface at level 1, but the functions are different. (There are other such examples.) difficulty: easy section: 12.5 90. Match the appropriate function f ( x, y, z ) with the geometric description of its level surface f = 0. (i) f ( x, y, z ) = sin x (ii) f ( x, y, z )= y + sin x (iii) f ( x, y, z ) = x 2 + y 2 + z 2 − 1 (iv) f ( x, y, z ) = x 2 + y 2 − z 2 (v) f ( x, y, z ) = 6 x 2 + 3 y 2 − z
(a) Sinusoidal curves in the xy-plane expanded in the z direction (b) A sphere of radius 1 centered at the origin. (c) A elliptical paraboloid. (d) An infinite number of planes parallel to the yz-plane. (e) Two cones with vertices at (0,0,0) centered about the z-axis.
Ans: (i)=(d), (ii)=(a), (iii)=(b), (iv)=(e), (v)=(c) difficulty: medium section: 12.5
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Chapter 12: Functions of Several Variables
( x − 2) 2 . Use the cross-sections f (2, y ) and f ( x,5) to ( x − 2) 2 + y − 5 explain why lim ( x , y )→(2,5) f ( x, y ) does not exist.
91. Consider f ( x, y ) =
Ans: Notice that f (2, = y)
= 0 if y ≠ 5 , whereas f = ( x,5)
0 y −5
( x − 2) 1 if x ≠ a . = ( x − 2)2 + 0 2
So, within any distance of (2,5) , one can always find points of the form (2,y) with y ≠ 5 and points of the form (x,5) with x ≠ 2 . In other words, no matter how close we are to (2,5) , there are points where f = 0 and points where f = 1. Therefore lim ( x , y )→(2,5) f ( x, y ) does not exist. difficulty: hard section: 12.6 92. Using the fact that y 2 ( x − 2) ≤ ( x − 2) 2 + y 2 for all x and y except ( x, y ) = (2, 0) , 2 2 ( x − 2) + y show that lim ( x , y )→(2,0)
y 2 ( x − 2) = 0. ( x − 2) 2 + y 2
Ans: We note that ( x − 2) 2 + y 2 is equal to the distance from ( x, y ) to (2, 0) . Using the given inequality, we see that, if u is any small positive number, then y 2 ( x − 2) = −0 ( x − 2) 2 + y 2
y 2 ( x − 2) ≤ ( x − 2) 2 + y 2 < u . 2 2 ( x − 2) + y
This means that the difference between ( xy− 2)( x2−+2)y 2 and 0 can be made as small as we 2
wish by choosing the distance from ( x, y ) to (2, 0) to be sufficiently small. By the definition of limit, we can conclude that lim ( x , y )→(2,0) difficulty: hard
y 2 ( x − 2) = 0. ( x − 2) 2 + y 2
section: 12.6
93. True or False: If all the level surfaces of a function f ( x, y, z ) are parallel planes, then f must be a linear function. Ans: False difficulty: easy section: 12R 94. Let f ( x, y ) = 3 . Describe the cross-sections, the contours, and the graph of f. Ans: All the cross sections of f are the same line z = 3, which is horizontal in a cross-section diagram. There is only one contour of f, which is of level 3, and contains all the points of 2-space. The graph of f is the plane z = 3, which is parallel to the xy-plane. difficulty: easy section: 12R
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Chapter 12: Functions of Several Variables
95. Let g ( x, y, z ) = 5 . Describe the level surfaces of g. Ans: There is only one level surface of g. It is of level 5 and it contains all the points of 3-space. difficulty: easy section: 12R 96. In an exam, a student wrote the following answer. "The level curve of f ( x, y ) at the point (3,–4) is x 2 − 2 xy = 34 ." How can you tell that the answer is wrong without even knowing the formula for f? Ans: The answer is wrong because the point (3,–4) does not lie on the curve x 2 − 2 xy = 34 . difficulty: easy section: 12R
Page 37
a 1. True or false: The vector − is a unit vector provided a ≠ 0 . Give a reason for a your answer. Ans: True. Dividing by a gives the unit vector parallel to a ; −a / a is a unit vector in the opposite direction. difficulty: easy section: 13.1 2. If w =− i 2 j + k , find a unit vector parallel to w. 1 2 1 Ans: i− j+ k 6 6 6 difficulty: easy section: 13.1 3. If w = 5i − j + 5k , u = 2i + a j + 5k , find the value of a making w and u perpendicular. Ans: 35 difficulty: easy section: 13.1 4. If w = 2i − j + 2k , v =i + 3 j , find a unit vector perpendicular to both w and v. 6 2 7 Ans: − i+ j+ k 89 89 89 difficulty: medium section: 13.1
5. Three men are trying to hold a ferocious lion still for the veterinarian. The lion, in the center, is wearing a collar with three ropes attached to it and each man has hold of a rope. Charlie is pulling in the direction N 60° W with a force of 380 pounds and Sam is pulling in the direction N 44° E with a force of 400 pounds. What is the direction and magnitude of the force needed on the third rope to counterbalance Sam and Charlie? Ans: The third rope must be pulled with a force of 480.47 pounds in the direction S 6.12 E. difficulty: medium section: 13.1 6. Find the missing numbers: 3 ai + b j + 3k − 3 −i + j + ck = i + 4 j − 5k .
(
) (
)
–2 7 14 = , b = ,c 3 3 3 difficulty: easy section: 13.1
Ans:= a
7. If the vector 2i + ak is parallel to 7 i + b j + 35k , find a and b. = a 10, = b 0 Ans: difficulty: easy section: 13.1
Page 1
Chapter 13: A Fundamental Tool- Vectors
8. If the vector 2i − 5 j + ak has magnitude 9, find a.
Ans: a = 52 difficulty: easy
section: 13.1
9. The vector 5i − 6 j + 6k is the displacement vector from the point (1, –1, 3) to which other point? Ans: (6, –7, 9) difficulty: easy section: 13.1 10. Find a unit vector that points in the same direction as the displacement vector from the point P(0, 2, –3 ) to the point Q(3, 5, 5). 3 3 8 Ans: PQ = i+ j+ k 82 82 82 difficulty: medium section: 13.1 11. Find a vector of length 5 that points in the same direction as the displacement vector from the point P(–1, 2, 2) to the point Q(3, 6, 5). 20 20 15 Ans: PQ = i+ j+ k 41 41 41 difficulty: medium section: 13.1 12. If u = 4i + 5 j + 3k and v = 2i − 4 j + 3k , find 5u + 4v . Ans: 28i + 9 j + 27 k difficulty: easy section: 13.1
13. If w = 2i + 5 j + ak find a value of a such that w = 30 .
Ans: 871 difficulty: easy
section: 13.1
14. If u= 2i + 2 j and v= 2i − 2 j , find the values c1 and c2 such that 18i + 6 j = c1 u + c2 v . Ans:= c1 6,= c2 3 difficulty: easy section: 13.1 15. Let v =3i + 2 j − 3k . Which of the following are parallel to v ? Select all that apply. A) 12i + 8 j − 12k D) 12i + 8 j + 12k B) E) –12i − 8 j + 12k 12i + 2 j − 3k C) 12i + 2 j − 12k Ans: A, B difficulty: easy section: 13.1
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Chapter 13: A Fundamental Tool- Vectors
16. Let v = 6i + 3 j − 6k . Considering −v as a position vector that begins at the origin, at what point of three space does its head lie? Ans: (–6, –3, 6) difficulty: easy section: 13.1
17. A bird is flying with velocity v= 8i + 3 j (measured in m/ sec ) relative to the air. The wind is blowing at a speed of 4 m/ sec parallel to the x-axis but opposing the bird's motion. Find the speed of the bird relative to the ground. Ans: 25 m/s difficulty: easy section: 13.2
18. A plane wants to go due Northeast, but a wind is blowing toward the North at 78 miles per hour. If the airspeed of the plane is 250 miles per hour, which direction should the plane head? What will be its groundspeed? Ans: 298.99 miles per hour; heading N 57.75 E difficulty: medium section: 13.2 19. An airplane wants to travel east, but there is a wind of 100 miles per hour blowing from the northwest. If the airspeed of the plane is 390 miles per hour, find the direction the plane should head (as an angle north of west) and its groundspeed. Ans: 312.83 miles per hour; heading W 10.45 N. difficulty: medium section: 13.2 20. Shortly after takeoff, a plane is climbing with velocity vector 150i + 130 j + 12k . Assume the units are kilometer per hour, that the x-axis points east, that the y-axis points north, that the z axis points up. What is the airspeed of the plane? Ans: 198.86 km/hr. difficulty: easy section: 13.2 21. Shortly after takeoff, a plane is climbing with velocity vector 140i + 130 j + 10k . Assume the units are kilometer per hour, that the x-axis points east, that the y-axis points north, that the z axis points up. If there is a wind blowing 20 km/hr to the southeast, what is the speed of the plane relative to the ground? Ans: 193.09 km/hr. difficulty: easy section: 13.2
22. A birdwatcher on the ground notices that a blue bird is heading north with a groundspeed of 28 miles/hr. The wind is blowing 25 miles/hr to the northeast. What is the airspeed of the bird? Ans: 20.47 miles/hr. difficulty: easy section: 13.2
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Chapter 13: A Fundamental Tool- Vectors
23. An object, P, is pulled by a force F1 of magnitude 16lb at an angle of 20 degrees north of east. In what direction must a force F2 of magnitude 10 lb pull to ensure that P moves due east? Ans: 33.18° south of east. difficulty: medium section: 13.2
24. Two boats leave a harbor at the same time. Boat A cruises northwest at a rate of 17 knots (nautical miles per hour). Boat B cruises north at a rate of 21 knots. (a) Find the displacement vector from boat A to boat B half an hour later. (b) For the passengers in boat A, what does the velocity of boat B appear to be? Ans: (a)= AB –6.010i + 4.490 j (b) Relative velocity –12.021i + 8.979 j difficulty: medium section: 13.2 25. Find a unit vector in the direction of the vector 0.4i − 0.8 j + 1.5k . Ans: 0.23i − 0.46 j + 0.86k difficulty: easy section: 13.2 26. True/False? The angle between 4i + 2k and i − 2 j + 5k is less than π/2. A) True B) False Ans: A difficulty: easy section: 13.3
27. True/False? The two planes z = –3x - 4y + 5 and –6x - 8y - 2z = 0 are parallel. A) False B) True Ans: B difficulty: easy section: 13.3 28. True/False? The triangle ABC with vertices A = i + 5 j , B = i − 3 j , and = C –2i + 2 j is a right triangle. A) False B) True Ans: A difficulty: easy section: 13.3 29. Suppose a is a fixed vector of length 5, b is a vector which can rotate and has length 2, and θ is the angle between them. What is the maximum value of a ⋅ b , and for what value of θ does it occur? Ans: The maximum value of a ⋅ b is 10, and it occurs when θ = 0 . difficulty: easy section: 13.3
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Chapter 13: A Fundamental Tool- Vectors
30. Suppose u = 4i + b j + 3k and v = ai + 6 j + 3k . Find values for a and b making u perpendicular to both v and j. 9 Ans: a = − ,b= 0. 4 difficulty: medium section: 13.3
0 and 31. Find the angle between the planes –4( x − 1) + 3( y + 2) + 5 z = x + 4( y − 1) + 2( z + 4) = 0. Ans: The angle between the two planes is 56.26 . difficulty: medium section: 13.3 0. Find a normal vector n to this plane. 32. Consider the plane –3 x − 5 y + 4 z = Ans: n = –3i − 5 j + 4k . difficulty: easy section: 13.3
0 has a normal n =+ 33. The plane x + 2 y + z = i 2 j + k. Let u =i + j + k . Express u as the sum of a vector v which is parallel to n and a vector w which is parallel to the plane. 1 1 1 2 4 2 Ans: The vector u= v + w , where v = i + j + k and w = i − j + k . 3 3 3 3 3 3 difficulty: easy section: 13.3
0. Find, to three decimal places, the shortest 34. Consider the plane –2 x + 2 y + 2 z = distance from the plane to the point P = (–1,1, 2). Ans: The shortest distance from P to the plane is 2.309. difficulty: medium section: 13.3 35. One of the highlights of the Calaveras Fair, besides the bullfrog jumping contest, is the watermelon seed spitting contest. The seed spitting is being done in the direction of s= 5i + j . The wind velocity during the contest is w = i + 4 j mph. The rules say that a legal wind in the direction of the seed spit must not exceed 3 mph. Mr. Samuel C. himself just spit for a record of 22 ft 1.4 inches. Will his record stand? A) Yes B) No Ans: A difficulty: easy section: 13.3
1 to the point P = (–1, –3, 4). 36. Determine the distance from the plane –2 x + 1 y + 4 z = Give your answer to 4 decimal places. Ans: The distance from the plane to P is 3.0551. difficulty: medium section: 13.3
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Chapter 13: A Fundamental Tool- Vectors
1 to the plane 2 x + y − z = 6 . Give 37. Determine the distance from the plane 2 x + y − z = your answer to 4 decimal places. (Hint: Although there are many approaches, projections come in handy.) 5 Ans: ≈ 2.0412 6 difficulty: hard section: 13.3 38. Determine a formula for the distance between a point (a, b, c) and the y-axis. Ans: a 2 + c 2 difficulty: easy
section: 13.3
39. Determine two vectors of length 7 in the xy-plane that are normal to the parabola y = x 2 at (2, 4). 7 Ans: ± 4i − j 17 difficulty: easy section: 13.3
(
)
40. Determine the equation of the plane which contains the points (1, 2, –8) and (-2, 1, –8) 2. and is perpendicular to the plane x + 3 y − 8 z = Ans: − x + 3 y + z =–3 difficulty: medium section: 13.3 41. Consider the triangle ABC where A is the point (4, 5, -3), B is the point (0, –3, 2) and C is the point (4, 1, 1). Find the cosine of angle BAC. Give your answer to 4 decimal places. Ans: 0.8971 difficulty: medium section: 13.3 42. True or False? If u , v are non-zero vectors with u ⋅ v = u v , then u is either parallel or anti-parallel to v . A) False B) True Ans: B difficulty: easy section: 13.3 43. If v = 2i + j + 4k and w =i + j − 2k , find the value of s so that v + sw is perpendicular to w. . 5 Ans: s = . 6 difficulty: easy section: 13.3
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Chapter 13: A Fundamental Tool- Vectors
44. Suppose u =+ 37 . What is the shortest i 4 j + 5k and v is a vector such that u ⋅ v = length v can be? 37 . Ans: The shortest possible length for v is 42 difficulty: easy section: 13.3
45. Suppose u =+ 41 . i 3 j + 5k . Find the vector v of shortest length with u ⋅ v = 41 Ans: The vector v is i + 3 j + 5k . 35 difficulty: easy section: 13.3
(
)
46. True/False? Let u and v be two non-zero parallel vectors. Then u ⋅ v =±1 . A) False B) True Ans: A difficulty: easy section: 13.3 47. True or false? The vector a × b is parallel to the x-axis where below (both a and b are in the xy-plane).
A) True Ans: B
B) False difficulty: easy
section: 13.4
Page 7
and
are shown
Chapter 13: A Fundamental Tool- Vectors
48. True or false? The vector a × b has a positive z component where below (both a and b are in the xy-plane).
A) False Ans: A
B) True difficulty: easy
and
are shown
section: 13.4
49. For what value(s) of a is the vector v= –50i + a j + 5k perpendicular to the plane z = 10x - 2y + 7? Ans: If a = 10, then v is perpendicular to the plane. difficulty: easy section: 13.4 50. Consider two vectors v = 5i − j + 3k and w = ai + a j − k . For what value(s) of a are v and w perpendicular? 3 Ans: The vectors v and w are perpendicular if a = . 4 difficulty: easy section: 13.4 51. Consider two vectors v= –3i − j + 3k and w = ai + a j − k . For what values of a are v and w parallel? Ans: The vectors v and w are never parallel. difficulty: easy section: 13.4
52. Find the equation of the plane parallel to 4i − j + 3k and 3i + 3 j − k , and containing the point (1, 1, 2). Ans: The equation of the plane is –8x + 13y + 15z = 35. difficulty: medium section: 13.4
53. Given the points P = (1, 2, 10), Q = (3, 5, 24), and R = (2, 5, 20), find the equation of the plane containing P, Q, and R. –2 . Ans: The equation of the plane is 4 x + 2 y − z = difficulty: medium section: 13.4
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Chapter 13: A Fundamental Tool- Vectors
54. Given the points P = (1, 2, 9), Q = (3, 5, 26), and R = (2, 5, 22), find the area of the triangle PQR. Give your answer to 4 decimal places. Ans: The area of triangle PQR is 7.6485. difficulty: medium section: 13.4 55. Given the points P = (1, 2, 16), Q = (3, 5, 35), and R = (2, 5, 33), find the (perpendicular) distance from the point R to the line through P and Q. Give your answer to 4 decimal places. Ans: The perpendicular distance from R to the line through P and Q is 0.8497. difficulty: medium section: 13.4 56. For what value(s) of a is the vector v =3i + a j + 3k parallel to the plane z = 2x − 5 y + 2 ? 3 Ans: If a = , then v is parallel to the plane. 5 difficulty: easy section: 13.4
57. Compute the area of the triangle with vertices A(0, 0, 0), B(3, 3, 0), and C(3, 6, 3). 9 3 Ans: 2 difficulty: easy section: 13.4 58. Does the point D(–3, 4, –6) lie in the same plane as the triangle with vertices A(0, 0, 0), B(5, 5, 35), C(5, 4, 33)? A) No B) Yes Ans: A difficulty: medium section: 13.4 59. Vectors a, b, c and u are shown in the picture below.
A)
<
<
B) Ans: C
<
<
difficulty: medium
C) D) section: 13.4
Page 9
<
<
<
<
Chapter 13: A Fundamental Tool- Vectors
60. Suppose that u and v are non-zero vectors, and that u is perpendicular to v . Let w =u × v × u . Which of the following statements are true? Select all that apply. A) D) w is perpendicular to u , w is parallel to v , B) E) w is perpendicular to v , w is the zero vector. C) w is parallel to u , Ans: A, D difficulty: easy section: 13.4
(
)
61. Find the volume of the parallelopiped with edges 4i − 3 j + k , i − 6 j + k , and i + j + 2k . Ans: The volume of the parallelopiped is 42 . difficulty: easy section: 13.4 62. Find the vector u in 3-space which satisfies both of the following conditions. (a) u ⋅ k = 5 (b) u × k = 5i − 2 j Ans: u = 2i + 5 j + 5k difficulty: easy section: 13.4
63. Let A(1, –2, –1), B(–1, –4, 0), C(0, –2, 1), and D(2, 0, 0) be the vertices of a parallelogram in that order. Find the area of the parallelogram to 4 decimal places. Ans: The area of the parallelogram is 5.3852. difficulty: medium section: 13.4 64. Let A(–2, 1, –1), B(–3, 2, 0), C(–2, 1, 1), and D(–1, 0, 0) be the vertices of a parallelogram in that order. Find the area of the projection of the parallelogram in the xy-plane. Ans: The area of the projection of the parallelogram in the xy-plane is 0. difficulty: medium section: 13.4 65. Let u = –4i + 2 j + 3k and v =− i 2 j + 6k . Compute u ⋅ v. 10 Ans: u ⋅ v = difficulty: easy section: 13R 66. Let u =3i + 2 j + 3k and v =− i 2 j + 4k . Compute u × v. Ans: u × v = 14i − 9 j − 8k difficulty: easy section: 13R
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Chapter 13: A Fundamental Tool- Vectors
67. The vectors v and w are shown below. Determine if the following statement is true or false.
( v – w) ⋅ j > 0 A) False Ans: A
B) True difficulty: easy
section: 13R
68. Let v = 2i − 2 j + 3k and let w be a vector of magnitude 6 at an angle of 30° to v . Find v ⋅ w . 3 51 Ans: v ⋅ w = difficulty: easy section: 13R
69. Let v = 4i − 4 j + 3k and let w be a vector of magnitude 4 at an angle of 30° to v . Find v × w . Ans: v × w = 2 41
difficulty: easy
section: 13R
70. A surveyor is standing at point A, 403 feet away from a tall building. Let i, j , k be the direction of east, north and up respectively. She observes that a vector in the direction of AC is 12 i + 5 j + 15 k , where C is the highest point of the building. What is the height of the building? Ans: The building is 465 feet high. difficulty: medium section: 13R
71. An airplane is descending in preparation to land at an airport to its northwest. Its airspeed is 640 km/hr and its altitude is decreasing a rate of 50 km/hr. Describe its velocity vector. Ans: The plane's velocity vector is −451.17i + 451.17 j − 50k , given in km/hr. difficulty: medium section: 13R
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Chapter 13: A Fundamental Tool- Vectors
20 . Find the coordinates of the points where the plane 72. Consider the plane 5 x − y + 4 z = intersects the three axes. Ans: The points of intersection are (4,0,0), (0,–20,0), and (0,0,5). difficulty: easy section: 13R 5 . Find the value of a that makes the vector 73. Consider the plane 3 x − 5 y + z = 12i − 20 j + ak perpendicular to the plane. Ans: When a = 4, then the vector is perpendicular to the plane. difficulty: easy section: 13R 3 . What is the value of a if the vector 3i − 5 j + ak is 74. Consider the plane 4 x − 3 y + z = parallel to the plane? Ans: If a = –27, then the vector is parallel to the plane. difficulty: easy section: 13R 75. Find the missing number. 5i + ___ j + 4k ⋅ i − j + k = 15
(
)(
)
Ans: The missing number is –6. difficulty: easy section: 13R 76. Find the missing numbers. 4i + ___ j + 8k × i − j += k ____ i + 4 j − 6k
(
) (
)
Ans: The missing numbers are 2 and 10, respectively. difficulty: easy section: 13R 77. Consider the points P = (1, 2, 6), Q = (3, 4, 4) and R = (-1, 2, 12). Find a vector perpendicular to PQ and PR and then use it to find the equation of the plane which contains P, Q, and R. Ans: 12i − 8 j + 4k 3x − 2 y + z = 5 difficulty: medium section: 13R 78. Let ABCDEF be a regular hexagon. Express the vectors and in terms of DE CF AB . Ans: DE = − AB and CF = −2 AB . difficulty: easy section: 13.1 79. Let ABCDEF be a regular hexagon. Express 12 AD + AB in terms of DF . Ans: 12 AD + AB = − DF . difficulty: easy section: 13.1
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Chapter 13: A Fundamental Tool- Vectors
80. Let ABCDEF be a regular hexagon. Express 12 AD − AB in terms of BE . 1 Ans: 12 AD − AB = . 2 BE difficulty: easy section: 13.1 81. True or False: If ai + bj > ci + dj , then a > b and c > d.
Ans: False
difficulty: easy
section: 13.2
82. True or False: If = v bw ≠ 0 for some constant b, then the vectors v and w are parallel or anti-parallel. Ans: True difficulty: easy section: 13.2 83. The unit sales of a company's product A in June, July, August, and September are given by a = (100, 220,100,85) , and the sales of product B in the same four months are given by b = (130, 200,320, 200) . Suppose the company makes a profit of $200 for each unit of product A and $85 for each unit of product B. Compute 200a + 85b and interpret its meaning. Ans: 200a + 85b = (31050, 61000, 47200,34000) . This means that the company's profits from the sales of products A and B in June, July, August, and September are $31050, $61000, $47200, and $34000, respectively. difficulty: easy section: 13.2 84. True or False: Suppose that A, B, C, and D are four distinct points in 3-space. Then ( AB + BC ) + CD = AD . Ans: True difficulty: easy section: 13.2
85. Find a unit vector in the same direction as 0.9i − 0.4 j + 0.9k . Round your answer to 3 decimal places. Ans: 0.675i − 0.300 j + 0.675k difficulty: easy section: 13.2 86. Suppose that A, B, C, and D are four distinct points in 3-space. Which of the following must be true? Mark all correct answers. A) C) AB + CD = AD ( AB + CD) + BC = AD B) D) AB + BC = AC ( BC + CD) + DA = BA Ans: B, C, D difficulty: easy section: 13.2 87. True or False: There are exactly two unit vectors perpendicular to the vector i + 2 j + k . Ans: False difficulty: easy section: 13.3
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Chapter 13: A Fundamental Tool- Vectors
88. Using vectors, show that if u = v , then u + v and u − v are perpendicular. Conclude that the diagonals of a rhombus are perpendicular. Ans: We can use the dot product to test whether u + v and u − v are perpendicular. We have (u + v ) ⋅ (u − v ) = u ⋅ u − u ⋅ v + v ⋅ u − v ⋅ v 2 2 = u − v −u ⋅v + u ⋅v = 0. Therefore u + v and u − v are perpendicular. We can describe the sides of a rhombus with vectors u and v , which have the same length. The diagonals of the rhombus, which are u + v and u − v , are therefore perpendicular. difficulty: medium section: 13.3
89. True or False: For any vectors u and v , u + v > u − v . Ans: False difficulty: easy section: 13.3 90. Consider the point P = ( R cos θ , R sin θ ) , which lies on the circle of radius R centered at the origin. Show that, so long as θ ≠ 0 or π radians, then the vector from (R,0) to P and the vector from (-R,0) to P are perpendicular. Ans: Let u be the vector from (R,0) to P and let v be the vector from (-R,0) to P. Then = u ( R cos θ − R )i + R sin θ j and = v ( R cos θ + R)i + R sin θ j . So we have = u ⋅ v (( R cos θ − R )i + R sin θ j ) ⋅ (( R cos θ + R )i + R sin θ j ) = R 2 cos 2 θ − R 2 + R 2 sin 2 θ = R 2 − R 2 = 0. Therefore, u and v are perpendicular. difficulty: medium section: 13.3
91. Choose values of b and c so that the following planes do not intersect. 2 x + by − cz = 24 and 2 x − 4 y = 25 Ans: The planes will intersect unless b = –4 and c = 0. difficulty: easy section: 13.3
2 , then which of the 92. Given that v and w are non-zero vectors and that v ⋅ w = following MUST be true? Select all that apply. v is perpendicular to w . A) v is parallel or anti-parallel to w . B) C) The angle θ between v and w satisfies 0 ≤ θ < π2 . D) The angle θ between v and w satisfies π2 < θ ≤ π . E) At least one of v and w is greater than 1. Ans: C, E difficulty: easy section: 13.4
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Chapter 13: A Fundamental Tool- Vectors
0 for all vectors u , 93. Given that v and w are non-zero vectors and that (v × w) ⋅ u = then which of the following MUST be true? Select all that apply. v is perpendicular to w . A) v is parallel or anti-parallel to w . B) C) The angle θ between v and w satisfies 0 ≤ θ < π2 . D) The angle θ between v and w satisfies π2 < θ ≤ π . E) At least one of v and w is greater than 1. Ans: B difficulty: easy section: 13.4
0 and v × w = i + j + k , then 94. Given that v and w are non-zero vectors and that v ⋅ w = which of the following MUST be true? Select all that apply. v is perpendicular to w . A) v is parallel or anti-parallel to w . B) C) The angle θ between v and w satisfies 0 ≤ θ < π2 . D) The angle θ between v and w satisfies π2 < θ ≤ π . E) At least one of v and w is greater than 1. Ans: A, E difficulty: easy section: 13.4 a 95. True or False: If a ≠ 0 , then is a unit vector which is parallel to a . a Ans: True difficulty: easy section: 13.4 96. If a , b ≠ 0 , then find a vector which is perpendicular to a and b . B) (a ⋅ b )(a + b ) C) a × b A) (a ⋅ b )(a − b ) Ans: C difficulty: easy section: 13.4 97. If a , b ≠ 0 , then a ⋅ (a × b ) = 0 . Why is this true? Ans: We know that a ⋅ (a × b ) = 0 , because a × b is always perpendicular to a . difficulty: easy section: 13.4 98. What is the geometric definition of v × w ? Ans: The geometric definition of v × w is v w sin θ , where θ is the angle between v and w . difficulty: easy section: 13.4
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Chapter 13: A Fundamental Tool- Vectors
99. Let A, B, C, D, E, and F be points on a plane. Explain why AB × AC × DE × DF = 0. Ans: AB × AC is perpendicular to the plane on which all the points lie. DE × DF is also perpendicular to this plane. Hence, AB × AC and DE × DF are parallel (or anti-parallel) and therefore their cross product is the zero vector. difficulty: medium section: 13.4
(
) (
)
100. Even without knowing the vector v , you can tell that the following statement must be wrong. Explain why. 4i − 9 j + 8k ×= v –4i + 4 j + 3k
(
)
(
)
Ans: The cross product of two vectors is always perpendicular to those two vectors. However, in this case, we can check that 4i − 9 j + 8k ⋅ –4i + 4 j + 3k = –28 ≠ 0 ,
(
)(
)
which means that these vectors are not perpendicular. So the statement must be wrong. difficulty: medium section: 13.4
101. True or False: (i × 78 j ) ⋅ = k 13i ⋅ ( 13 j × 6k ) Ans: True difficulty: easy section: 13.4
102. True or False: For any vectors u and v , (u + v ) ⋅ (u − v )= Ans: True difficulty: medium section: 13.4
2 2 u − v .
103. Suppose that u and v are non-zero vectors and (u + v ) × (u − v ) = 0 . Then what MUST be true about u and v ? (u + v ) ⋅ (u − v ) = 0 u and v are parallel. A) C) u and v are perpendicular. B) Ans: A difficulty: medium section: 13.4 104. Suppose that a and b are non-zero vectors. If a and b are perpendicular, then which of the following MUST be true? Select all that apply. A) a × b = b × a B) a ⋅ b = b ⋅ a C) a × b = D) a ⋅ b = 0 0 Ans: B, D difficulty: easy section: 13R 105. Suppose that a and b are non-zero vectors. If a and b are parallel, then which of the following MUST be true? Select all that apply. A) a × b = b × a B) a ⋅ b = b ⋅ a C) a × b = D) a ⋅ b = 0 0 Ans: A, B, C difficulty: easy section: 13R
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Chapter 13: A Fundamental Tool- Vectors
106. Find the missing numbers. The vector i + in the direction of the vector i − 8 j − 8k .
j+
k has magnitude 3 and is pointing
Ans: The missing numbers are -2, -2, and 4, respectively. difficulty: medium section: 13R 107. A company manufactures four grades of videotape: Normal, Extra, Super, and Supreme. Last year, the company sold a1 boxes of Normal tapes for $5.00 each, a2 boxes of Extra tapes for $5.50 each, a3 boxes of Super for $6.00 each, and a4 boxes of Supreme for $8.00 each. Let a = (a1 , a2 , a3 , a4 ) , b = (5,5.5, 6,8) , and c = (2, 2.4, 2.5,3) . (a) Compute a ⋅ b − a ⋅ c . (b) If the computation in part (a) represents the company's profit last year, then what does the vector c represent? Ans: (a) a ⋅ b − a ⋅ c = (3a1 ,3.1a2 ,3.5a3 ,5a4 ). (b) The vector c represents the production cost vector. It costs $2 per box to produce and market the Normal tapes, $2.40 per box for the Extra tapes, $2.50 per box for the Super tapes, and $3 per box for the Supreme tapes. difficulty: medium section: 13R
Page 17
∂f ∂f everywhere, then f(x, y) is a constant. = ∂x ∂y difficulty: easy section: 14.1
1. True or false? If Ans: False
2. True or False? The level curves of a function z = f(x, y) are shown below. Assume that the scales along the x and y axes are the same.
f y (Q) > f y (P)
A) False Ans: A
B) True difficulty: easy
section: 14.1
3. Suppose that the price P (in dollars) to purchase a used car is a function of C, its original ∂P cost (in dollars), and its age A (in years). So P = f(C, A). What is the sign of ? ∂C A) Positive B) Negative Ans: A difficulty: easy section: 14.1
Page 1
Chapter 14: Differentiating Functions of Several Variables
4. Given the contour diagram shown below (a) Sketch a graph of f(1, y). (b) Sketch a graph of f(x, 0).
Ans: (a)
difficulty: medium
(b)
section: 14.1
Page 2
Chapter 14: Differentiating Functions of Several Variables
5. Given the contour diagram shown below, state whether f x (0.5, 1) is positive, negative or nearly zero.
A) Almost zero B) Negative Ans: B difficulty: medium
C) Positive section: 14.1
D) Undefined
6. The figure below shows the graph of z = f(x, y) and its intersection with various planes. (The x and y-axes have the same scale.) What is the sign of f y (0, 1) ?
A) Negative B) Positive Ans: B difficulty: easy
section: 14.1
7. Suppose f ( x= , y ) (ln( x + 3 y )) xy . Use a difference quotient to estimate f x (2,1) and f y (2,1) with h = 0.01. answers to 3 decimal places. Ans: f x (2,1) = 1.885 and f y (2,1) = 4.444 . difficulty: easy section: 14.1
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Give your
Chapter 14: Differentiating Functions of Several Variables
8. Estimate the value of f y (1, 0) from the given contour diagram of f.
Ans: f y (1, 0) is approximately –3.333. difficulty: easy section: 14.1 9. The cross-sections of f when x is fixed at x = 1 and when y is fixed at y = 2 are given below. Determine, if possible, the sign of f y (1, 3).
Ans: Negative (from the first picture) difficulty: easy section: 14.1 10. The consumption of beef, C (in pounds per week per household) is given by the function C = f(I, p), where I is the household income in thousands of dollars per year, and p is the price of beef in dollars per pound. Do you expect ∂f ∂p to be positive or negative? A) Negative B) Positive Ans: A difficulty: easy section: 14.1
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Chapter 14: Differentiating Functions of Several Variables
11. The monthly mortgage payment in dollars, P, for a house is a function of three variables P = f(A, r, N), where A is the amount borrowed in dollars, r is the interest rate, and N is the number of years before the mortgage is paid off. It is given that:
Estimate the value of
∂f . ∂r (100000,7, 20)
∂f is approximately 61.15. ∂r (100000,7, 20) difficulty: easy section: 14.1 Ans:
12. The monthly mortgage payment in dollars, P, for a house is a function of three variables P = f(A, r, N), where A is the amount borrowed in dollars, r is the interest rate, and N is the number of years before the mortgage is paid off. It is given that:
∂f and interpret your answer in terms of a mortgage ∂A (100000,7, 20) payment. Select all answers that apply. A) We are currently borrowing $100,000 at 7% interest rate on a 20-year mortgage. B) The monthly payment will go up by approximately $0.007753 for each extra percentage point charged. C) The monthly payment will go up by approximately $0.007753 for each extra dollar we borrow. D) The monthly payment will go up by approximately $0.007753 for each extra year of the mortgage. E) The monthly payment will go down by approximately $0.007753 for each extra dollar we borrow. Ans: A, C difficulty: easy section: 14.1 Estimate the value of
13. True or false? There exists a function f(x, y) with fx = 2y and fy = 3x. A) False B) True Ans: A difficulty: easy section: 14.2
∂ ln( x 5 y + 5) ) . ( ∂x 5x4 5x4 y A) B) x5 y + 5 x5 y + 5 Ans: A difficulty: easy
14. Find
5x4 y D) x5 + 5 section: 14.2
C)
Page 5
x4 y x5 y + 5
Chapter 14: Differentiating Functions of Several Variables
15. Find f H if f ( H , T ) =
5H + T . (3 − H )3
15 + 10 H + 3T (3 − H )6 15 − 10 H + 3T B) (3 − H ) 4 15 + 10 H + 3T C) (3 − H ) 4 Ans: C difficulty: medium A)
16. Find Ans:
D) E)
15 − 20 H − 3T (3 − H ) 4 5 + 10 H + 3T (3 − H ) 4
section: 14.2
∂H cos(2 w + 2t ) to 2 decimal places if H ( w, t ) = e . ∂w ( π/4, π/2 ) ∂H = 2.00. ∂w ( π/4, π/2 )
difficulty: medium
section: 14.2
17. If $P is invested in a bank account earning r% interest a year, compounded continuously, the balance, $B, at the end of t years is given by rt
( P, r , t ) Pe 100 . = B f= Find ∂B / ∂P. ∂B Ans: = e0.01rt ∂P difficulty: easy section: 14.2
18. If $P is invested in a bank account earning r% interest a year, compounded continuously, the balance, $B, at the end of t years is given by rt
B f= ( P, r , t ) Pe 100 = (a) What are the units of ∂B / ∂t ? (b) What is the practical interpretation (in terms of money) of ∂B / ∂t ? Ans: (a) The units of ∂B / ∂t are dollars per year. (b) ∂B / ∂t is the rate at which B grows per year as the length of time of the investment increases. difficulty: easy section: 14.2
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Chapter 14: Differentiating Functions of Several Variables
19. The ideal gas law states that PV = RT for a fixed amount of gas, called a mole of gas, where P is the pressure (in atmospheres), V is the volume (in cubic meters), T is the temperature (in degrees Kelvin) and R is a positive constant. ∂P Find . ∂V ∂P RT Ans: =– 2 . ∂V V difficulty: easy section: 14.2 20. The ideal gas law states that PV = RT for a fixed amount of gas, called a mole of gas, where P is the pressure (in atmospheres), V is the volume (in cubic meters), T is the temperature (in degrees Kelvin) and R is a positive constant. A mole of a certain gas is at a temperature of 290° K, a pressure of 1 atmosphere, and a volume of 0.04 m3. What is ∂P / ∂T for this gas? ∂P Ans: = 0.0034 atm/degree K. ∂T difficulty: easy section: 14.2 21. Find ∂z/∂x= if z –3ln x + sin( xy 5 ). ∂z –3 Ans: = + y 5 cos( xy 5 ). ∂x x difficulty: medium section: 14.2 22. Estimate z y (3, 2) numerically if z = ( x + y )( x + y ) , x, y > 0. Ans: z y (3, 2) is approximately 8155.498. difficulty: medium section: 14.2
∂ P ln(V 5 − P ) ) . ( ∂P ∂ P Ans: . P ln(V 5 − P ) ) = ln(V 5 − P ) − 5 ( V −P ∂P difficulty: easy section: 14.2
23. Calculate the following derivative:
24. Let
f (θ , ϕ ) = sin 2 2θ cos3 2ϕ . Find
Ans: 0.612 difficulty: easy
∂f to 3 decimal places. ∂ θ = (θ π= / 6, ϕ π / 8)
section: 14.2
Page 7
Chapter 14: Differentiating Functions of Several Variables
25. Let f ( x,= y ) 3 y 2 + 2 xy. Use the appropriate partial derivative to find the slope of the cross-section at the given point. (a) The cross-section f(x, 2) at the point (3, 2). (b) The cross-section f(1, y) at the point (1, –2). Ans: (a) The desired slope is f x (3, 2) , which equals 4. (b) The desired slope is f y (1, –2) , which equals –10. difficulty: easy section: 14.2 26. The volume V = πr 2 h of a right circular cylinder is to be calculated from measured values of r and h. Suppose r is measured with an error of no more than 2.5% and h with an error of no more than 1%. Using differentials, estimate the percentage error in the calculation of V. (In general, in measuring a quantity Q, the percentage error is dQ/Q.) Ans: 6% difficulty: easy section: 14.3 27. Determine the tangent plane= to z f= ( x, y ) 3e x − 2 y at (x, y) = (2, 1). z 3.000 + 3.000( x − 2) − 6.000( y − 1) Ans: = difficulty: medium section: 14.3 28. Find an equation for the tangent plane to the graph of ( x, y ) = (2, 6π). 1 + 2( y − 6π ) . Ans: The tangent plane is z = difficulty: medium section: 14.3
f ( x, y ) = e x sin y at
29. Use the differential of f ( x, y= ) x 2 + 2 x cos 2 y to find a linear approximation of f at the point (1, π/4). Ans: z =2 + 3( x − 1) – 2( y − π/4) difficulty: medium section: 14.3 30. If a function z = g(x, y) has g(1, 2) = –5, gx(1, 2) = 4 and gy(1, 2) = 3, find the equation of the plane tangent to the surface z = g(x, y) at the point where x = 1 and y = 2. Ans: z = –15 + 4x + 3y difficulty: easy section: 14.3 x
31. Let f ( x, y= ) y 2 + 3 y ∫ sin(et )dt. Find f x (3,1) and f y (3,1) to four decimal places. 3
Ans: f x (3,1) = 2.8334 and f y (3,1) = 2.0000 . difficulty: medium section: 14.3
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Chapter 14: Differentiating Functions of Several Variables
32. The table of some values of f(x, y) is given below. Find a local linearization of f at (1 , 2).
Ans: The local linearization of f at (1 , 2) is z =11 − 5 x + 6 y . difficulty: easy section: 14.3 33. Find the differential of the function f ( x,= y) x 2 + y 2 at the point (3, 4). A point is measured to be 3 units from the y-axis with an error of ±0.01 and 4 units from the x-axis with an error of ±0.02. Approximate the error in computing its distance from the origin. = (3, 4) (3 / 5)dx + (4 / 5)dy, Approx error = 0.0220. Ans: df difficulty: hard section: 14.3 34. A function f(x, y) has linear approximation f ( x, y ) ≈ −3 + 5 x − 3 y near (2, 3) and f ( x, y ) ≈ −3 x + 2 y near (1, 2). Say whether the following statement is probably true, false, or impossible to determine: f (1.9, 2.8) is about –1.9. A) False B) True C) Impossible to determine Ans: B difficulty: easy section: 14.3 35. The equations of the tangent planes to the graph z = f(x, y) at the points (0, -2), (2, 1) are z − 3x + 2 y = 5 and 1 = 2 x + 5 y − z , respectively. Determine the value of f (0, –2). State whether the value you find is exact or an approximation. Ans: f (0, –2) is 9. The value is exact. difficulty: medium section: 14.3 36. True or false? If u is a unit vector and the level curves of f(x, y) are given below, then at point P we have fu ( P) = grad f .
Ans: False
difficulty: easy
section: 14.4
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Chapter 14: Differentiating Functions of Several Variables
37. True or false? If v is a unit vector and the level curves of f(x, y) are given below, then at point P we have f v ( P) = grad f cos θ .
A) True Ans: A
B) False difficulty: easy
section: 14.4
38. The depth of a pond at the point with coordinates (x, y) is given by h( x,= y) 5x2 + 4 y 2 . (Assume that x, y, and h are measured in feet.) If a boat at the point (–3, –5) is sailing in the direction of the vector 5i + j , then at what rate is the depth changing? 190 Ans: – ft/ft traveled 26 difficulty: medium section: 14.4 39. Suppose that f ( x, y ) = x 2 e xy . Find an equation for the tangent plane to f at the point (3, 0). z = –9 + 6 x + 27 y z = –9 + 6 x − 27 y A) C) z =9 + 6 x + 27 y z = –9 − 6 x + 27 y B) D) Ans: A difficulty: medium section: 14.4 40. For the function f ( x, y ) = x 2 y 3 find a unit vector in the direction of the steepest increase at the point (a, b) = (1, 1). 2 3 2 3 A) = C) = u i− j u i+ j 13 13 13 13 2 6 B) D) = u i+ j u= 2i + 3 j 13 13 Ans: C difficulty: easy section: 14.4 41. Find the gradient of the function f ( x, y ) = x 2 y 3 at the point (1,1) and use the result to obtain a linear approximation for 1.042 0.983. Ans: 1.02 difficulty: easy section: 14.4
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Chapter 14: Differentiating Functions of Several Variables
42. Let f ( x, y ) =x 2 + 3 y 2 − 3 x. Find the gradient vector of f at the point (–1, 2). C) A) ∇f = –5i − 12 j ∇f = 5i + 12 j B) D) ∇f = –5i + 24 j ∇f = –5i + 12 j Ans: D difficulty: easy section: 14.4 43. Let f ( x, y ) =x 2 + 5 y 2 − 5 x. What is the direction of maximum rate of change of f at (1, 1)? –3 10 Ans: = ∇f i+ j 109 109 difficulty: easy section: 14.4 44. Let f ( x, y ) =x 2 + 3 y 2 − 4 x. What is the maximum rate of change of f at (2, 1)? Ans: The maximum rate of change of f at (2, 1) is difficulty: easy section: 14.4
36 , or 6.000.
45. Given that f(2, 4) = 1.5 and f(2.1, 4.4) = 2.1, estimate the value of fu (2, 4) , where u is the unit vector in the direction of 1i + 4 j. Give your answer to four decimal places. Ans: fu (2, 4) is approximately 1.4552. difficulty: easy section: 14.4
46. Suppose that as you move away from the point (2, 0, 2), the function increases most rapidly in the direction 0.6i + 0.8 j and the rate of increase of f in this direction is 7. At what rate is f increasing as you move away from (2, 0, 2) in the direction of i + j + k ? Give your answer to 4 decimal places. Ans: 5.6580 difficulty: medium section: 14.4 47. Suppose ||∇f(a, b, c)||=19. Is it possible to choose a direction from (a, b, c) so that fu in that direction is –19? Ans: Yes: in the opposite direction to gradient. difficulty: easy section: 14.4
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Chapter 14: Differentiating Functions of Several Variables
48. Below is a contour diagram for f(x, y) which is defined and continuous everywhere. The z-values have been omitted. Explain why it is true that fi (1,1) = f − j (1,1) .
Ans: The gradient vector ∇f(1, 1) must point perpendicular to the contour at (1, 1). This means it either points in the South-East direction or in the North-West direction. In either case, the i and j components will be equal in magnitude, and different in sign. This means that ∇f (1,1) =− ai a j for some real number a ≠ 0. Since i is a unit vector, the directional derivative of f at (1, 1) in this direction is ∇f (1,1) ⋅ i =a. . Likewise, ∇f (1,1) ⋅ (− j ) =a. difficulty: medium section: 14.4 49. True or False? If grad f = grad g , then f = g . A) True B) False Ans: B difficulty: easy section: 14.5 50. True or False? If you know the directional derivative of f(x, y) in two distinct directions ∂f (i.e. not including opposite directions) at a point P then you can find ( P). ∂x Ans: True difficulty: easy section: 14.5 51. Suppose that as you move away from the point (–1, –1, –1), the function f ( x, y, z ) increases most rapidly in the direction 0.6i + 0.8 j and the rate of increase of f in this direction is 4. At what rate is f increasing as you move away from (–1, –1, –1) in the direction of i + j + k ? Give your answer to 4 decimal places. Ans: 3.2332 difficulty: medium section: 14.5 52. Find the equation of the tangent plane to the surface 3 xyz + x 3 + y 3 + z 3 = 15 at (1, –3, 2). –15 x + 33 y + 3 z = –108 A) C) 15 x + 33 y + 3 z = –108 –15 x + 33 y + 3 z = 108 –15 x – 33 y + 3 z = –108 D) B) Ans: A difficulty: medium section: 14.5
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Chapter 14: Differentiating Functions of Several Variables
53. Consider the function g ( x, y, z ) = x 2 + y 2 + z 2 . (a) Describe the level set g = 16. (b) Find a vector perpendicular to the tangent plane to the level set g = 16 at the point (–1, 2, 2). Ans: (a) Sphere of radius 4 centered at the origin.
(b) The gradient is perpendicular to the level set. Since grad g =2 xi + 2 y j + 2 zk , grad g (–1, 2, 2) = –2i + 4 j + 4k . difficulty: easy section: 14.5 54. Find the directional derivative of f ( x, = y, z ) 3 xyz 2 − 4 xz at the point (3, 3, 2), in the direction of the vector v = 2i + 3 j − k . Ans: The directional derivative of f in the direction of v is 68 or 18.1738. 14 difficulty: easy section: 14.5 55. Find the equation of the tangent plane to x 2 + 2 xy + 4 y + 6 = z 2 at the point (–4, 1, 3). 6x + 4 y + 6z = –4 6x + 4 y + 6z = 2 A) D) –6 x + 4 y + 6 z = –2 6x + 4 y + 6z = –2 B) E) 6x − 4 y + 6z = –2 C) Ans: E difficulty: medium section: 14.5
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Chapter 14: Differentiating Functions of Several Variables
56. An ant is walking along the surface which is the graph of the function f ( x, y= ) x 2 − sin( y ). (a) When the ant is at the point (1, 0, 1), what direction should it move in order to be moving on the surface in the direction of greatest ascent? (b) If the ant moves in this direction at a speed of 6 units per second, what is the rate of change of height of the ant? Ans: (a) In order to move in the direction of greatest ascent and stay on the graph of f(x, y), the ant must walk along the tangent vector to the graph at this point with the same i and j components as the gradient vector. Calculating the gradient, we get 2i − j , so the vector we are trying to find is of the form 2i − j + ak for some scalar a. To calculate a, we use the fact that any tangent vector to the surface must be perpendicular to the normal vector to the surface at this point. The normal vector at this point is 2i − j − k , so taking the dot product of these vectors, we get 4 + 1 - a = 0. Therefore the ant must move in the direction of 2i − j + 5k . (b) The velocity vector of the ant points in the direction the ant is moving which is 2i − j + 5k . The magnitude of the velocity vector is 6, so the velocity vector of 6 the ant will be 2i − j + 5k ). The rate of change of height of the ant is 30 30 simply the k component of the velocity vector, . 30 difficulty: hard section: 14.5
(
)
57. The quantity z can be expressed as a function of x and y as follows: z = f(x, y). Now x and y are themselves functions of r and θ, as follows: x = g (r , θ ) and y = h(r , θ ). Suppose you know that g(1, π/2) = –1, and h(1, π/2) = 1. In addition, you are told that ∂f ∂f ∂g π (–1,1) 1, = (–1,1) 6,= (1, ) 7, = ∂x ∂y ∂r 2 ∂g π ∂h π ∂h π (1, ) 7,= (1, ) 6,= (1, ) 4. = ∂θ 2 ∂r 2 ∂θ 2
∂z (1, π / 2). ∂r ∂z Ans: (1, π / 2) = 43. ∂r difficulty: easy section: 14.6 Find
Page 14
Chapter 14: Differentiating Functions of Several Variables
58. Let w = 3x cos y. If = x u 2 + v 2 , y = v / u , find ∂w/∂u and ∂w/∂v at the point (u , v) = (2,3) . Give your answers to 2 decimal places. ∂w ∂w Ans: = 30.03, = –18.18. ∂u ∂v difficulty: medium section: 14.6 59. Let w = 3x cos 4y. If x = e − t and y = ln t , find
dw at the point t = 3. Give your dt
answer to 2 decimal places. dw Ans: = 0.24. dt difficulty: medium section: 14.6 60. Suppose f(x, y) is a function of x and y and define g (u , v) = f (eu + cos v, eu + sin(v)). Find gu (0, 0) given that = f (0, 0) 5,= f (2,1) –1,= f x (0, 0) 2,= f x (1, 2) 1,= f x (2,1) –3, and = g (0, 0) –1, = g (2,1) –1,= f y (0, 0) 2,= f y (2,1) –3, = f (1, 2) 7. Ans: gu (0, 0) = –6. difficulty: easy section: 14.6 61. Sally is on a day hike at Mt. Baker. From 9 to 11:00 a.m. she zig-zags up z = f(x, y) where x is the number of miles due east of her starting position, y is the number of miles due north of her starting position, and z is her elevation in miles above sea level. Feeling tired, she decides to continue walking, but in such a way that her altitude remains constant from 11 a.m. to noon to settle her stomach for lunch. At 11:30 a.m., she will be passing through (2, –1, 5) where fx(2, –1) = 3 and fy(2, –1) = -2. What is the slope of her “path” in the x, y plane at this instant? (This “path” is among the level curves in the plane.) 3 Ans: 2 difficulty: easy section: 14.6 u 62. If f ( x,= y ) 2 x 2 + 2 y 2 , and = x e= , y ue w , find fw(1, 1) using the chain rule. Give your answer to 4 decimal places. Ans: f w (1,1) = 29.5562. difficulty: medium section: 14.6
63. If f (= x, y ) sin( x) + y 2 , x(u, v) = uv and y(u, v) = u + 4v. If H(u, v) = f(x(u, v), y(u, v)), what is H(0,–1)? Give your answer to 4 decimal places. Ans: 16.0000 difficulty: easy section: 14.6
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Chapter 14: Differentiating Functions of Several Variables
64. If f (= x, y ) sin( x) + y 2 , x(u, v) = uv and y(u, v) = u + 3v. If H(u, v) = f(x(u, v), y(u, v)), what is Hv(0,–2)? Give your answer to 4 decimal places. Ans: H v (0, –2) = –36.0000. difficulty: easy section: 14.6 65. Let z = ln( x 2 + 2 xy + 2 y ), x =+ t 2 3, y = t + t. Find dz/dt at t = 1 using the chain rule. Give your answer to 4 decimal places. dz Ans: = 1.0833. dt t =1 difficulty: medium section: 14.6 66. Find the following partial derivative: HP(2, 1) if H ( P, T ) =
2P . 3P + T
to 4 decimal places. Ans: H P (2,1) = 0.0408. difficulty: easy section: 14.7 67. Find the following partial derivative: fxy if f ( x, y ) = x 7 y 8 . A) D) f xy = 8 x 7 y 7 f xy = 7 x 6 y 7 B)
f xy = 7 x 6 y 8
C) f xy = 8 x 6 y 7 Ans: E difficulty: easy
E)
f xy = 56 x 6 y 7
section: 14.7
68. Consider the level curves shown for the function z = f(x, y).
Determine the sign of f yx (–1, –5). A) Positive B) Negative Ans: A difficulty: medium section: 14.7
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Give your answer
Chapter 14: Differentiating Functions of Several Variables
69. Find the quadratic approximation to the function f(x, y) = cos x cos y valid near the origin. x2 y 2 A) D) f ( x, y ) =1 − x 2 − y 2 f ( x, y ) =− 1 + 2 2 x2 y 2 x2 y 2 B) E) f ( x, y ) =+ 1 + f ( x, y ) =− 1 − 2 2 4 4 2 2 x y C) f ( x, y ) =− 1 − 2 2 Ans: C difficulty: medium section: 14.7 70. Suppose that f has continuous second order partial derivatives at (1, 2) and that f xx (1, 2) = 2, f yy (1, 2) = −1 and f xy (1, 2) = −1 . Is the following statement likely to be true or likely to be false? f y (1, 2) > f y (1, 2.01) A) True B) Neither C) False Ans: A difficulty: easy section: 14.7 71. Given that f x ( x,= y ) 3 x 2 + 2 xye xy . y ) 6 xy + y 2 e xy and f y ( x, = Suppose that f(1, 1) = 4. Find the quadratic Taylor polynomial of f(x, y) at (1, 1). Ans: 2
2
Q( x, y ) = 4 + 8.72( x − 1) + 8.44( y − 1) + 4.36( x − 1) 2 + 16.87( x − 1)( y − 1) + 8.15( y − 1) 2 difficulty: medium
section: 14.7
72. Suppose that fx(2, 1) = 2.2, fx(2.5, 1) = 1, fx(2, 1.5) = 1.8, fy(2, 1) = -0.8, fy(2.5, 1) = -1.2 and fy(2, 1.5) = -1.4. If f (2,1) = 4, estimate the value of f(1.85, 0.8) using a quadratic Taylor polynomial about (2,1). Use difference quotients to approximate all second derivatives. Ans: 3.755 difficulty: hard section: 14.7
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Chapter 14: Differentiating Functions of Several Variables
73. Using the contour diagram for f(x, y), find the sign of f yy ( P) given that fxx(P) < 0.
A) Negative B) Positive Ans: B difficulty: easy
C) Not possible to decide section: 14.7
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Chapter 14: Differentiating Functions of Several Variables
1 5
1 5
74. Consider the function g ( x, y ) = x y . (a) Find gx(x, y) and gy(x, y) for (x, y) ≠ (0, 0). (b) Use the limit definition of partial derivative to show that gx(0, 0) = 0 and gy(0, 0) = 0. (c) Are the functions gx and gy continuous at (0, 0)? Explain. (d) Is g differentiable at (0, 0)? Explain. Ans: (a) For (x, y) ≠ (0, 0), (b)
(c) No, they are not continuous at (0, 0). We can choose points very close to (0, 0) 1
that will make the fraction g x ( x, y ) =
y 5
arbitrarily large. Hence 4 5x 5 lim ( x , y )→(0,0) g x ( x, y ) does not exist and the function gx is not continuous at (0, 0). A similar argument shows that gy is not continuous at (0, 0). (d) If g is differentiable at (0, 0), then there exists a linear approximation near the origin, it would be L(x, y) = f(0, 0) + fx(0, 0)x + fy(0, 0)y = 0. Also, by the f (h, k ) − L(h, k ) definition of differentiability, lim h→0,k →0 = 0. This would give h2 + k 2 However, one can choose 1
h, k very close to (0, 0) that will make the fraction 1
Therefore lim h→0,k →0
1
h 5k 5 h2 + k 2
arbitrarily large.
1
h 5k 5
h2 + k 2 g is not differentiable at (0, 0). difficulty: medium section: 14.8
does not exist. This is a contradiction. Therefore,
75. Let f be a differentiable function with local linearization L(x, y) = –1 + 4(x - 4) - 2(y - 2) at (4, 2). Evaluate f(4, 2). Ans: f(4,2) = –1. difficulty: easy section: 14.8 76. True or False? If fx(0, 0) exists and fy(0, 0) exists, then f is differentiable at (0, 0). A) False B) True Ans: A difficulty: easy section: 14.8
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Chapter 14: Differentiating Functions of Several Variables
77. True or False? If f is non-differentiable at (0, 0), then for each direction, the directional derivative of f at (0, 0) exists. A) True B) False C) Not possible to decide Ans: B difficulty: easy section: 14.8 78. Find the directional derivative of f ( x,= y ) x 3 y + 5 xy 2 at the point (1, 1) in the direction of 3i + 5 j . 79 Ans: The directional derivative is ≈ 13.548. 34 difficulty: medium section: 14R 79. Find a unit vector perpendicular to both 6i − j + k and 8i + k . 1 2 8 A) D) − i− j+ k −i + 2 j + 8k 69 69 69 1 2 8 1 2 8 E) B) − i+ j+ k i+ j+ k 69 69 69 69 69 69 1 2 8 C) − i+ j− k 69 69 69 Ans: B difficulty: easy section: 14R
80. Find the angle between the vector c = 2i + 3 j − k and the positive z-axis. Ans: The angle is 1.84 radians. difficulty: easy section: 14R 81. The table below gives values of a function f(x, y) near x = 1, y = 2.
Estimate f x (0,1.5) . Ans: f x (0,1.5) ≈ 2 . difficulty: medium
section: 14R
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Chapter 14: Differentiating Functions of Several Variables
82. The table below gives values of a function f(x, y) near x = 1, y = 2.
Give the equation of the tangent plane to the graph z = f(x, y) at x = 1, y = 2. z = 39.5 − 3.5 x − 3 y = z 39.5 + x − y A) D) z = 27.5 + 3.5 x + 3 y z= 4 + 3.5 x − 3 y B) E) z = 39.5 + 3.5 x − 3 y C) Ans: C difficulty: medium section: 14R 83. Suppose that grad f (1, −1) = v , with v = 4. (a) What is the directional derivative of f at (1, -1) in the direction of 3v ? (b) What is the smallest value of the directional derivative of f at (1, -1) among all possible directions? Ans: (a) 4 (b) –4 difficulty: easy section: 14R
84. For the function g(x, y) with contour diagram below:
Evaluate g(–1, 1). Ans: g (–1,1) = 2 . difficulty: easy section: 14R
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Chapter 14: Differentiating Functions of Several Variables
85. For the function g(x, y) with contour diagram below:
Find the direction in which g is increasing the fastest at the point (1, 1). Ans: The required direction is −i. This vector is perpendicular to the contour curve at (1, 1) and pointing in the direction where the value of f is increasing. difficulty: easy section: 14R 86. The consumption of beef, C (in pounds per week per household) is given by the function C = f(I, p), where I is the household income in thousands of dollars per year, and p is the price of beef in dollars per pound. Explain the meaning of the statement: f p (80,3) = –1.6 , and include units in your answer. Ans: The statement f p (80,3) = –1.6 means that if the household income is $80,000 and the price of beef is $3 per pound, then the amount of beef consumed per week will decrease by approximately 1.6 pounds if the cost of beef increases by $1 per pound.. difficulty: easy section: 14.1 87. Arrange the following quantities in ascending order. (The level curves of z = f ( x, y ) are shown in the figure. Assume that the scales on the x- and y-axes are the same.) f y ( P ), f y (Q), f x ( R), − f x ( S )
Ans: f y (Q) < f x ( R) < f y ( P) < − f x ( S ) difficulty: easy section: 14.2
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Chapter 14: Differentiating Functions of Several Variables
88. In an electric circuit, two resistors (of resistance R1 and R2, respectively) are connected so that the combined resistance of the circuit, R, is given by 1 1 1 = + . R R1 R2 ∂R Find . ∂R1 R2 R1 R2 ∂R Ans:= − . ∂R1 R1 + R2 ( R1 + R2 ) 2 difficulty: easy section: 14.2 89. Let V ( x, y, z ) represent the volume of a rectangular solid of length x, width y, and height z. Find ∂V and explain its meaning in terms of how the solid changes with respect ∂x to x. Ans: ∂V = yz . As x increases by a small amount ∆x , the volume will increase by ∂x (∆x) yz . So the rate at which the volume is increasing with respect to x is yz. difficulty: easy section: 14.2 90. Let S ( x, y, z ) represent the surface area of a rectangular solid of length x, width y, and height z. Find ∂S and explain its meaning in terms of how the solid changes with ∂z respect to z. Ans: ∂S = 2 x + 2 y . As z increases by a small amount ∆z , the surface area will ∂z increase by x∆z + y∆z + x∆z + y∆z= (2 x + 2 y )∆z . So the rate at which the surface area is increasing with respect to z is 2x+2y. difficulty: easy section: 14.2 ∂g π 6, . ∂x 6 B) −2 C) π − 12 D) −π − 6 difficulty: easy section: 14.2
91. If g ( x, y ) = x 2 cos( xy ) , then find A) −12 Ans: A
∂g π 3, . ∂y 3 C) 27 D) π − 27 B) −27 difficulty: easy section: 14.2
92. If g ( x, y ) = x 2 cos( xy ) , then find A) 0 Ans: A
= 93. True or False: The tangent plane to the surface z x sin( x / y ), y ≠ 0 at any point passes through the origin. Ans: True difficulty: hard section: 14.3
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Chapter 14: Differentiating Functions of Several Variables
94. A rectangular beam, supported at its two ends, will sag when subjected to a uniform load. 4 The amount of sag is calculated from the formula: S = Cpx , where p is the load (in wh3 Newtons per meter), x is the length between supports (in meters), w is the width of the beam (in meters), h is the height of the beam (in meters), and C is a constant (depending on material and units of measurement used). Determine ∂S for a beam 3 m long, 0.1 ∂h m wide, 0.2 m high subjected to a load of 80 N/m. ∂S ∂S A) C) = 121,500,000 C. = –121,500,000 C. ∂h (80,3,0.1,0.2) ∂h (80,3,0.1,0.2) ∂S
D) = –40,500,000 C. ∂h (80,3,0.1,0.2) Ans: A difficulty: easy section: 14.3 B)
∂S
∂h (80,3,0.1,0.2)
= 40,500,000 C.
95. What is the z-coordinate of the point P (1,3, z ) if P lies on the plane which is tangent to the ellipsoid 4 x 2 + y 2 + 9 z 2 = 17 at the point (–1, –2, −1) ? Give your answer to four decimal places. Ans: The z-coordinate is –3.0000 . difficulty: medium section: 14.3 96. A ball thrown from ground level with initial speed v and at an angle α with the v 2 sin(2α ) horizontal hits the ground at a distance s = away from the starting point g (where g is the acceleration due to gravity). Calculate the differential ds. 2 α) α) A) = ds 2 v sin(2 dv + 2 v cos(2 dα g g B) = ds 2v sin(2α )dv + 2v 2 cos(2α )dα C) = ds
2 v sin(2α ) g
D) = ds Ans: A
2 v sin(2α ) g
dv + 2 v sin(2αg )cos(2α ) dα 2
α) dv − 2 v cos(2 dα g difficulty: easy section: 14.3 2
97. Let f ( x, y= ) x 2 + 2 x cos 2 y . Give a linear approximation of f at the point (2, π / 4) .
π Ans: f ( x, y ) ≈ 6 + 5( x − 2) − 2 y − . 4 difficulty: easy section: 14.3 98. True or False: If f and g have the same partial derivatives at every point in their domains, then f = g . Ans: False difficulty: easy section: 14.3
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Chapter 14: Differentiating Functions of Several Variables
99. Let f ( x,= y ) 4 x 2 − 9 y 2 . What is the equation of the tangent plane to the graph of z = f ( x, y ) at the point (5,1) ? Ans: The tangent plane is z= 40( x − 5) − 18( y − 1) + 91 . difficulty: easy section: 14.4 100. Let= u
1 2
(i + j ) and let = w
1 13
(2i + 3 j ) . Suppose that f i (1,1) = 4 2 and
f u (1,1) = 3 . Find f w (1,1) .
5 2 . 13 difficulty: medium
Ans: f w (1,1) =
section: 14.4
101. Suppose that T ( x, y ) =x 2 + 3 y 2 − x is the temperature (in degrees Celsius) at the point ( x, y ) (where x and y are in meters). If you are standing at the point (−2,1) and proceed in the direction of the point (–6, −3) , will the temperature be increasing or decreasing at the moment you begin? At what rate? Give your answer to 4 decimal places. Ans: The temperature is decreasing at a rate of 0.7071 degrees Celsius per meter. difficulty: easy section: 14.4
102. What is the angle between the unit vector u and ∇f (a, b) if fu (a= , b) –0.5 ∇f (a, b) ? A) 5π Ans: B
6
B) 2π
3 difficulty: easy
C) π
D) π 3 6 section: 14.4
103. Suppose that the maximum rate of change of f at (1, −1) is 10, and it occurs in the direction of 2i − 2 j . What is the minimum rate of change of f at (1, −1) , and in which direction does it occur? Ans: The minimum rate of change of f at (1, −1) is –10 and it occurs in the direction –2i + 2 j . difficulty: easy section: 14.4 104. Suppose that the maximum rate of change of f at (1, −1) is 20, and it occurs in the direction of 5i − 5 j . Find ∇f (1, −1) , f x (1, −1) , and f y (1, −1) . 20 100 –100 Ans: ∇f (1,= . −1) (5i − 5 j ) , f x (1, −1) = , and f y (1, −1) = 50 50 50 difficulty: easy section: 14.4
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Chapter 14: Differentiating Functions of Several Variables
105. = Let z f= ( x, y ) e( x / y ) . Find the tangent plane to f at the point (5,5) . e e Ans: The tangent plane is z = x − y + e . 5 5 difficulty: easy section: 14.4 106. = Let z f= ( x, y ) e( x / y ) . Find the instantaneous rate of change of f at the point (5,5) in the direction of the point (8,11) . Ans: The instantaneous rate of change of f at the point (5,5) in the direction of the −e point (8,11) is . 5 5 difficulty: easy section: 14.4 107. True or False? Let f ( x, y, z ) = x 2 + y 2 + z 2 . Then grad f (0, 0, 0) = 0 . Ans: True difficulty: easy section: 14.5
Let u 108. Consider the surface given as the graph of z = f ( x, y ) . =
1 (i + 5 j ) and let 26
1 21 –4 (4i − 2 j ) . Suppose that fu (1,3) = ; f v (1,3) = ; f (1,3) = −7 . Find the 20 26 20 equation of the tangent plane to the surface at the point (1,3) . Ans: The tangent plane is z =x + 4 y − 20 . difficulty: medium section: 14.5
= v
109. True or False? A partial derivative is a specific example of a directional derivative. Ans: True difficulty: easy section: 14.5 110. True or False? A gradient is a specific example of a directional derivative. Ans: False difficulty: easy section: 14.5 111. Find an equation for the tangent plane to the ellipsoid ( x − 1) 2 + 4( y − 2) 2 + ( z − 3) 2 = 17 at the point (3,1, 0) . Ans: The tangent plane is 4 x – 8 y – 6 z = 4 . difficulty: easy section: 14.5 112. Consider the surface z = xy and the point P = (–4,3, –12) . Find a vector normal to the surface at P. Ans: The vector 3i – 4 j − k is normal to the surface at P. difficulty: easy section: 14.5
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Chapter 14: Differentiating Functions of Several Variables
113. Consider the surface z = xy and the point P = (–4,3, –12) . Find an equation for the plane tangent to the surface at P. Ans: The tangent plane is z = 3 x – 4 y +12 . difficulty: easy section: 14.5 114. Suppose that the temperature at the point ( x, y, z ) is given by T ( x, y, z ) = e − z sin( xy ) . If you are at the point (1, 2π , 0) , in which direction should you go to decrease your temperature the fastest? A) 2π i + j + 0k B) 2π i − j + 0k C) 2π i + j − k D) −2π i − j + 0k E) None of the above; the gradient of T is zero at (1, 2π , 0) . Ans: A difficulty: easy section: 14.5 2
s t , y =− s t , then simplify ∂z ∂z in terms of f x and f y . 115. If z = f ( x, y ) and x =+ ∂s ∂t ∂z ∂z Ans: = ( f x + f y )( f x − f y ) = ( f x ) 2 − ( f y ) 2 . ∂s ∂t difficulty: easy section: 14.6 116. In a neighborhood clinic, the number of patient visits can be described as a function of the number of doctors, x, and the number of nurses, y, by f ( x, y ) = 1000 x 0.6 y 0.3 . With upcoming budget cuts, the clinic must reduce the number of doctors at the rate of 2 per month. Estimate the rate at which the number of nurses has to be increased in order to maintain the current patient load. Currently there are 30 doctors and 50 nurses. 20 Ans: The number of nurses must be increased at a rate of ≈ 7 nurses per month. 3 difficulty: medium section: 14.6 117. Let F (u , v, w) be a function of three variables with Fu + Fw = –1 . Suppose that G (= x, y ) F ( x, x − y, x) . Simplify Gx + G y . A) Gx + G y = B) Gx + G y = C) Gx + G y = 0. –1 . 1. Ans: B difficulty: medium section: 14.6 118. If z ( x= , y ) f ( x 2 − y 2 ) , then simplify y
∂z ∂z +x . ∂x ∂y
∂z ∂z +x = 0. ∂x ∂y difficulty: easy section: 14.6 Ans: y
Page 27
D) Gx + G y = –2 .
Chapter 14: Differentiating Functions of Several Variables
119. Let U ( x, t ) = e − ( x −ct ) . Simplify U tt − c 2U xx . 2
A)
U tt − c 2U xx = 0.
B) U tt − c 2U xx = U. Ans: A difficulty: easy
C)
U tt − c 2U xx = −U .
D) section: 14.7
U tt − c 2U xx = c 2U .
120. Let f ( x= , y ) (2e y − y 2 ) cos x . Which of the following is the quadratic Taylor approximation of f near (0,0)? A) C) Q( x, y ) =2 + 2 y − x 2 . Q( x, y ) =2 + 2 y − 2 x 2 . B) D) Q( x, y ) =2 − 2 y + 2 x 2 . Q( x, y ) =2 − 2 y + x 2 . Ans: A difficulty: easy section: 14.7 121. Suppose f = e a1x1 + a2 x2 ++ an xn , where a12 + a2 2 + + an 2 = 1 . Simplify f x1x1 + f x2 x2 + + f xn xn . A) f B) 0 C) ( x12 + x2 2 + + xn 2 ) f Ans: A difficulty: hard section: 14.7
D)
f2
122. The quadratic Taylor approximations of f ( x, y ) at the points (−1, −1), (1,1) , (2,0) and (0,2) are given respectively by: Q1 ( x, y ) =−3 + 3( x + 1) 2 + 4( y + 1) 2 , 1 3( x − 1) 2 + 4( y − 1) 2 , Q2 ( x, y ) =− Q3 ( x, y ) =−2 − 9( x − 2) − 6( x − 2) 2 − 2 y 2 , 8 + 3 x + 24( y − 2) + 22( y − 2) 2 . Q4 ( x, y ) = Determine fy (0, 2) . Ans: fy (0, 2) = 24 . difficulty: easy section: 14.7
123. The quadratic Taylor approximations of f ( x, y ) at the points (−1, −1), (1,1) , (2,0) and (0,2) are given respectively by: Q1 ( x, y ) =−3 + 3( x + 1) 2 + 4( y + 1) 2 , 1 3( x − 1) 2 + 4( y − 1) 2 , Q2 ( x, y ) =− Q3 ( x, y ) =−2 − 9( x − 2) − 6( x − 2) 2 − 2 y 2 , 8 + 3 x + 24( y − 2) + 22( y − 2) 2 . Q4 ( x, y ) = Find an approximate value of f (0.2,1.9) . Ans: f (0.2,1.9) ≈ 6.42 . difficulty: easy section: 14.7
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Chapter 14: Differentiating Functions of Several Variables
124. Suppose that the function f ( x, y ) and the linear function L( x, y ) =3 − 3 x + 4 y satisfy f ( x, y ) − L( x, y ) ≤ 5( x 2 + y 2 )3/ 2 for points ( x, y ) close to (0, 0) . Is f differentiable at (0,0)? A) Yes B) No Ans: A difficulty: medium section: 14.8 125. Suppose that the function f ( x, y ) and the linear function L( x, y ) =4 − 3 x + 5 y satisfy f ( x, y ) − L( x, y ) ≤ 5( x 2 + y 2 ) for points ( x, y ) close to (0, 0) . Find f x (0, 0) and f y (0, 0) .
f (h, k ) − L(h, k )
= 0 , then f is differentiable at (0,0), and so h2 + k 2 f x (0, 0) = –3 and f y (0, 0) = 5 . difficulty: medium section: 14.8 Ans: Since
lim
h → 0, k → 0
126. True or False: Suppose that there exists a linear function L( x, y ) near the point (0,0) such that the difference between f ( x, y ) and L( x, y ) approaches 0 as ( x, y ) approaches (0,0). Then f is differentiable at (0,0) with L( x, y ) as its local linearization. Ans: False difficulty: easy section: 14.8 − y 127. Let g ( x, = y ) ( x 2 + y 2 ) f ( x, y ) , where f ( x, y ) = − x True or False: g is differentiable at (0,0). Ans: True difficulty: medium section: 14.8
y ≤ x x ≤ y.
128. Select all the planes that have normal vectors which are perpendicular to the z-axis. 3x − 5 y − z = 2 3x + 5 y = 2 D) A) 5 x= y + 3 3x + 5 y + z = 2 B) E) 5x + 3 y = 2 y + 1 =0 C) F) Ans: B, C, D, F difficulty: easy section: 14R 129. Select all the planes that could be tangent planes to the graph of z = f ( x, y ) , where f is differentiable everywhere. 3x − 5 y − z = 2 3x + 5 y = 2 A) D) 5 x= y + 3 3x + 5 y + z = 2 B) E) 5x + 3 y = 2 y + 1 =0 C) F) Ans: A, E difficulty: medium section: 14R
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Chapter 14: Differentiating Functions of Several Variables
130. Let f ( x, y, z ) =x3 + 6 xz + 4 yz + z 2 . Find ∇f (1, −2, 6) . Ans: ∇f (1, −2, 6) = 39i + 24 j + 10k . difficulty: easy section: 14R
131. Let f ( x, y, z ) =x3 + 3 xz + 4 yz + z 2 . Find f v (1, −2, 4) , where v is a unit vector in the direction of −i + j − k . 2 Ans: f v (1, −2, 4) = . – 3 difficulty: easy section: 14R 132. Suppose the graph of z = f ( x, y ) contains the point P (2, −3,3) , and that the tangent plane at P is given by 3x + 2 y − 5 z = –15 . Find f x (2, −3) and f y (2, −3) . Ans: f x (2, −3) = 0.4 . 0.6 and f y (2, −3) = difficulty: easy section: 14R 133. True or False? grad f (a, b) is perpendicular to the graph of z = f ( x, y ) at the point (a, b, f (a, b)) . Ans: False difficulty: easy section: 14R 1 . 134. True or False? If f ( x, y ) = ln( y ) , then ∇f ( x, y ) = y Ans: False difficulty: easy section: 14R 135. Given that grad g (1,1,5) = 3i − 3 j + k , in which of the following directions does g increase the fastest? A) 7i − 5 j + 2k B) i − k C) −3i + 3 j + 2k Ans: A difficulty: easy section: 14R
136. Find the points on the ellipsoid x 2 + 2 y 2 + z 2 = 7 where the tangent plane is parallel to 0. the plane –2 x + 2 y – z = Ans: The points are (–2,1, –1) and (2, −1,1) . difficulty: medium section: 14R
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8 8 + 2. 2 x y Classify these critical points as local maxima, local minima, or saddle points. Ans: (2, 2); (-2, -2) are local minima. difficulty: hard section: 15.1
1. Find all the critical points of the function f ( x, y ) = xy +
2. Is (0, 0) a critical point of the following function?
A) No. B) Yes: (global) maximum. Ans: B difficulty: medium
C) Yes: local minimum. D) Yes: (global) minimum. section: 15.1
3. Let h( x, y ) = x 3 + y 3 + 9 xy + 6. Determine all local maxima, minima, and saddle points. Are the local extrema also global extrema? Ans: (0, 0) is a saddle point; (–3, –3) is a local maximum. There is no global extremum. difficulty: medium section: 15.1 4. Let h( x, y ) = x 3 + y 3 + 3 xy + 5. Which figure best represents the level curves of this function?
C)
A)
B) Ans: D
difficulty: easy
D) section: 15.1
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Chapter 15: Optimization- Local and Global Extrema
5. Suppose that f ( x, y ) =x 2 + 4 xy + y 2 . Find and classify the critical point(s) as local maxima, local minima, or saddle points. Ans: (0,0) is the only critical point. It is a saddle point.. difficulty: easy section: 15.1 6. Suppose that f ( x, y ) =x 2 + 3 xy + y 2 . Find an equation of the tangent plane to the graph of f at the point (2, 2). z = 20 + 10( x − 2) + 10( y − 2) z = 20 − 10( x − 2) + 10( y − 2) C) A) z = 20 − 10( x − 2) − 10( y − 2) z =20 + 10 x + 10 y D) B) Ans: A difficulty: easy section: 15.1 7. Suppose that f ( x, y ) =x 2 + 3 xy + y 2 . Find a normal vector to the tangent plane of f at the point (1, 1). Select all that apply. A) 5i + 5 j − k B) 5i + 5 j + k C) −5i − 5 j + k D) –5i + 5 j + k Ans: A, C difficulty: easy section: 15.1 8. Find all the critical points of f ( x, y ) = x 3 − 3 x + y 2 − 4 y and classify each as maximum, minimum, or saddle point. Select all possible choices. A) The point (1, 2) is a relative minimum. B) The point (-1, –2) is a saddle point. C) The point (-1, 2) is a saddle point. D) The point (1, 2) is a relative maximum. E) The point (1, -2) is a relative maximum. Ans: A, C difficulty: easy section: 15.1 9. The function f ( x, y ) = x 3 − 3 x + y 2 − 6 y has a saddle point at (–1, 12). Which of the following is a sketch of the level curves of f near this point?
A) Ans: A
difficulty: easy
B) section: 15.1
10. Find the critical points of f ( x, y ) =x3 − 9 xy + y 2 and classify each as maximum, minimum or saddle. Ans: The point (0, 0) is a saddle point. 27 243 The point , is a local minimum. 2 4 difficulty: medium section: 15.1
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Chapter 15: Optimization- Local and Global Extrema
11. The contour diagram of f is shown below. Which of the points A, B, C, D, and E appear to be critical points? Select all that apply.
A) A B) B C) C D) D E) E Ans: A, B, C difficulty: easy section: 15.1 12. The contour diagram of f is shown below.
Find and classify the critical points. Describe possible gradient vectors of f at points C, D and E. Ans: Points A, B and C are critical points. A and B are local minima and C is a saddle point. Point C is a saddle point, therefore ∇f at C is 0 . The gradient vectors ∇f at D and E are perpendicular to the level curves, pointing in the direction of increasing levels. Also, since the level curves are closer together at E than at D, we have ||∇f(E)||≥||∇f(D)||. difficulty: easy section: 15.1 13. Suppose that (0, –2) is a critical point of a smooth function f(x, y) with = f (0, –2) –2,= f xx (0, –2) –1,= f xy (0, –2) 2,= f yy (0, –2) –5. What can you conclude about the behavior of the function f near (0, –2)? Ans: f has a local maximum at (0, –2). difficulty: medium section: 15.1
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Chapter 15: Optimization- Local and Global Extrema
14. Suppose that (1, –4) is a critical point of a smooth function f(x, y) with = f (1, –4) –1,= f xx (1, –4) 2,= f xy (1, –4) 2,= f yy (1, –4) –3. Find the quadratic approximation of f at (1, –4). Ans: Q( x, y ) = –1 + 1( x − 1) 2 + 2( x − 1)( y + 4) − 1.5( y + 4) 2 difficulty: medium section: 15.1 15. Level curves of f(x, y) are shown in the figure below. (Darker shades indicate regions with lower levels.) Determine if f y (1, 0) is positive, negative or zero.
Ans: f y (1, 0) is zero, because (1, 0) is a critical point. difficulty: medium section: 15.1 16. Level curves of f(x, y) are shown in the figure below. (Darker shades indicate regions with lower levels.) Is the point (–1, 2) a local maximum, a local minimum, or a saddle point of f, or is it none of these?
A) Saddle point. B) Local minimum. C) Local maximum. these. Ans: A difficulty: easy section: 15.1
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D) None of
Chapter 15: Optimization- Local and Global Extrema
17. Let f ( x, y ) = ax 2 − 2axy + 2 y 2 − bx , where a, b are any numbers with a ≠ 2. Find the critical point of f. (Express your answer in terms of the constants a and b.) 2b b Ans: x = ,y = 2a (2 − a ) 2(2 − a ) difficulty: easy section: 15.1 18. Let f ( x, y ) = ax 2 − 2axy + 4 y 2 − bx , where a, b are any positive numbers with a ≠ 4. Find the minimum value of a such that the critical point will be a saddle point. Ans: If a > 4, then the critical point is a saddle point. difficulty: easy section: 15.1 19. The function f(x, y) has a local maximum at (–1, 1). What can you say (if anything) about the values of f xy (–1,1) ? A) Nothing. C) It cannot be greater than zero. B) It is undefined or equal to zero. D) It is equal to zero. Ans: A difficulty: easy section: 15.1 20. Consider the function f ( x, y ) = x 4 + 2 x 3 y − 6 x 2 y 2 + 7 xy 3 − y 4 + 3. Check that (0,0) is a critical point of f and classify it as a local minimum, local maximum or saddle point. Ans: The point (0,0) is a saddle point. difficulty: easy section: 15.1 21. The point (–2, 1) is a critical point of g ( x, y ) =4 x3 − 96 xy + 48 xy 2 . Classify it either as a local minimum, local maximum, or saddle point. Ans: Local maximum. difficulty: medium section: 15.1 22. The function f ( x, y ) = e − ( x − a ) −( y −b ) where a and b are constants is sometimes referred to as a "bump function" and is used to construct functions which take on maximum values at certain points. Show that f(x, y) has a maximum at (a, b). Ans: Since fx(a, b) = 0 and fy(a, b) = 0, (a, b) is a critical point of f(x, y). To determine the nature of the critical point, we need to calculate the second derivatives. We get fxx(a, b) = –2 and fyy(a, b) = –2. Calculating the mixed partial derivatives, we get fxy(a, b) = 0. Therefore, D =(−2)(−2) − 02 =4 . Since fxx(a, b) < 0, it follows that (a, b) is a local maximum of f(x, y). difficulty: easy section: 15.1 2
2
23. Find a and b so that f ( x, y ) = ax 2 + bxy + y 2 has a critical point at (1, 6). Ans: a = 36, b = –12 difficulty: medium section: 15.1
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Chapter 15: Optimization- Local and Global Extrema
24. Determine the nature of the critical points of the function f ( x, y ) = 3 x5 y + xy 5 + xy. A) f has a saddle point at (1, 0). D) f has a saddle point at (0, 0). B) f has a local maximum at (0, 0). E) f has a local minimum at (0, 0). C) f has a saddle point at (0, 1). Ans: D difficulty: easy section: 15.1 25. Without calculating the discriminant, explain using the contour diagram for f ( x, y= ) x 3 + 3 x 2 y why f has a saddle point at (0, 0).
Ans: A saddle point P of a function f(x, y) is a critical point of f with the property that any open region containing P contains points Q1 and Q2 with f(Q1) > f(P) and f(Q2) < f(P). The contour diagram of f ( x, y= ) x 3 + 3 x 2 y has three lines which cross at the point (0, 0), so they must all be the same contour (in fact they are the contour f = 0). or undefined. Since This implies that the gradient vector at this point has to be f(x, y) is differentiable, it must be , and so (0, 0) is a critical point of f(x, y). To show the point (0, 0) is a saddle, we shall show that the function is positive at (a, 0) and negative at (-a, 0) for any value a > 0. Calculating, we get f(a, 0) = a3 > 0 and f(-a, 0) = -a3 < 0. Since this holds for any value of a, it follows that any open region containing (0, 0) contains points Q1 = (a, 0) and Q2 = (-a, 0) (choosing a small enough) with f(a, 0) = a3 > 0 = f(0, 0) and f(-a, 0) = -a3 < 0 = f(0, 0). So (0, 0) is a saddle. difficulty: hard section: 15.1 26. Describe the shape of the graph of f (= x, y ) 3 x 2 + 3 y 2 − 3 x − 12 y + 18. A) Paraboloid C) Hyperbolic paraboloid B) Elliptic paraboloid D) None of the above Ans: A difficulty: easy section: 15.2
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Chapter 15: Optimization- Local and Global Extrema
27. What do the second derivatives tell you about the graph of f ( x, y ) = 7 x 2 + 1000 xy + 7 y 2 − 3 x − 12 y + 18? Select all that apply. A) Any critical point must be a saddle point. B) There is a local minimum. C) There are no local minima. D) The function has a local maximum. E) There are at least two local minima. Ans: A, C difficulty: easy section: 15.2 4 9 + . x y Determine all the local maximum, minimum and saddle points in the region x > 0, y < 0. Ans: (2, –3) is a saddle point. difficulty: medium section: 15.2
28. Consider the function f ( x, y ) = x + y +
29. The Perfect House company produces two types of bathtub, the Hydro Deluxe model and the Singing Bird model. The company noticed that demand and prices are related. In particular, for Hydro Deluxe: demand = 1900 - price of Hydro Deluxe + price of Singing Bird for Singing Bird: demand = 1450 + price of Hydro Deluxe -2(price of Singing Bird). The costs of manufacturing the Hydro Deluxe and Singing Bird are $500 and $300 per unit respectively. Determine the price of each model that gives the maximum profit. Ans: Hydro Deluxe: $2875.00 Singing Bird: $1825.00 difficulty: medium section: 15.2 30. Find a point on the surface x-yz = 14 that is closest to the origin. Ans: (1, ± 13, 13) difficulty: medium section: 15.2 31. Consider the four points A = (1, 0), B = (2, 2), C = (3, 5) and D = (4, 3) in the xy-plane. Find a and b in the line of best fit y = ax + b for these points. (The line of best fit minimizes the sum of the squares of the vertical distances from each point to the line.) = a 1.2, = b –0.5 Ans: difficulty: hard section: 15.2
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Chapter 15: Optimization- Local and Global Extrema
32. Consider the four points A = (1, 0), B = (2, 3), C = (3, 5) and D = (4, 3) in the xy-plane. Find the values of a, b and c to determine the parabola of best fit, y = ax 2 + bx + c, for these points. (The parabola of best fit minimizes the sum of the squares of the vertical distances from each point to the parabola.) = a –1.25, = b 7.35, = c –6.25 Ans: difficulty: hard section: 15.2 33. A company has two manufacturing plants which manufacture the same item. Suppose the cost function is given by C (q1 , q2 ) = 4q12 + q1q2 + q22 , where q1 and q2 are the quantities (measured in thousands) produced in each plant. The total demand q1 + q2 is related to the price, p, by p = 110 − 0.5(q1 + q2 ). How much should each plant produce in order to maximize the company's profit? Ans: The first plant should produce 4783 items and the second plant should produce 33,478 items in order to maximize profit. difficulty: medium section: 15.2 34. True or False? If C is a circle in the plane, and if f(x, y) is differentiable and is not constant when constrained to C, then there must be at least one point on C where grad f is perpendicular to C. Ans: True difficulty: easy section: 15.3 35. Find the critical points of f ( x, y ) = x 2 ye − ( x +9 y ) . Do this by setting t = −( x 2 + 9 y 2 ) and optimizing f ( x, y, t ) = x 2 yet subject to the constraint t + x 2 + 9 y 2 = 0. 2
2
1 Ans: The critical points are ±1, ± . 3 2 difficulty: medium section: 15.3
36. Find the critical point of f ( x, y ) = x 2 ye − ( x + 25 y ) . Do this by setting t = −( x 2 + 25 y 2 ) and optimizing f ( x, y, t ) = x 2 yet subject to the constraint t + x 2 + 25 y 2 = 0. What are the global maximum and minimum values of f? Give your answer to 4 decimal places. Ans: ±0.0316 difficulty: medium section: 15.3 2
Page 8
2
Chapter 15: Optimization- Local and Global Extrema
37. A zoo is designing a giant bird cage consisting of a cylinder of radius r feet and height h feet with a hemisphere on top (no bottom). The material for the hemisphere costs $20 per square foot and the material for the cylindrical sides costs $10 per square foot; the zoo has a budget of $5120. Find the values of r and h giving the birds the greatest space inside assuming the zoo stays within its budget. 8 16 Ans: = r ,= h 2= r
π
difficulty: medium
π
section: 15.3
38. Consider the diagram shown below, which shows gradient vectors of a function f(x, y).
Which is less: f(A) or f(C)? Ans: f(C) difficulty: easy section: 15.3 39. A company has $250,000 to spend on labor and raw materials. Let L be the quantity of labor and R be the quantity of raw materials. The production output P of the company is cRL (here c is a positive constant). Suppose that each unit of labor costs $6000 and the unit price of raw materials is $2000. Find the ratio of R to L that maximizes P. Ans: 3 difficulty: medium section: 15.3 40. Suppose that f ( x, y ) = 4 x 2 + 4.0000 y 2 − x. Find and classify (as local maxima, minima, or saddle points) all critical points of f. 1 Ans: , 0 is a global minimum. 8 difficulty: medium section: 15.3 41. Suppose that f ( x, y ) =x 2 + 2 y 2 − x. Find the minimum value of the function f when (x,y) is constrained to lie on or inside the triangle with vertices (0,-2), (0,1), and (1,-2). Give your answer to 4 decimal places. Ans: The minimum value of f on the triangle is –0.2237. difficulty: medium section: 15.3
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Chapter 15: Optimization- Local and Global Extrema
42. Suppose that you want to find the maximum and minimum values of f ( x, y= ) x2 + y 2 subject to the constraint x + 4y = 3. Use the method of Lagrange multipliers to find the exact location(s) of any extrema. 3 12 Ans: , 17 17 difficulty: medium section: 15.3 43. Suppose the quantity, q, of a good produced depends on the number of workers, w, and the amount of capital, k, invested and is represented by the Cobb-Douglas function 3
1
q = 6 w 4 k 4 . In addition, labor costs are $20 per worker and capital costs are $20 per unit, and the budget is $3680. Using Lagrange multipliers, find the optimum number of units of capital. Ans: 46 difficulty: medium section: 15.3
44. Determine three positive numbers x, y, z that maximize x 3 y 4 z 5 under the condition x + y + z = 17. 17 17 85 Ans: , , 4 3 12 difficulty: hard section: 15.3 45. Let f ( x, y )= x − y . The constraint g(x, y) = 3 is sketched in the picture below. In the picture, locate the point where f will have a global maximum subject to the constraint g(x, y) = 3.
Ans: The level curves of f are of the form x-y = c, or equivalently, y = x-c.
difficulty: easy
section: 15.3
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Chapter 15: Optimization- Local and Global Extrema
46. Use Lagrange multipliers to find the minimum value of 4xy on the circle x 2 + y 2 = 4. Ans: The minimum value of 4xy on the circle is –8. difficulty: medium section: 15.3 47. The Lagrange multipliers needed to find the maximum and minimum values of f ( x, y ) = 8xy on the circle x 2 + y 2 = 25 is λ = ±4 , where ∇f = λ∇g . Estimate the maximum and minimum values of 8xy subject to the constraint x 2 + y 2 = 30.25 . Ans: The maximum value will be approximately 121 and the minimum value will be approximately –121. difficulty: medium section: 15.3 48. A company manufactures a product using x, y and z units of three different raw materials. 1
1
2
The quantity produced is given by the function Q = 60 x 3 y 4 z 5 . Suppose the cost of the materials per unit is $20, $25 and $75 respectively. Find the maximum production if the budget is limited to $6000. Ans: The maximum production occurs when 50 units of each raw material are used. The maximum production is Q(50,50,50) = 60(50)59 / 60 ≈ 2811 units. difficulty: medium section: 15.3 49. A company manufactures a product using x, y and z units of three different raw materials. The quantity produced is given by The production is described by the function 1
1
2
Q = 60 x 3 y 4 z 5 . Suppose the cost of the materials per unit is $20, $15 and $24 respectively. (a) Find the cheapest way to produce 6300 units of the product. (b) Find the value of λ in ∇f = λ∇Q and interpret this value. Ans: (a) The cheapest way to produce 6300 units of the product occurs when x= y= z= 113.62 . (b) λ = 1.08. This means that it costs about $1.08 to produce one more unit of of the product. difficulty: medium section: 15.3 50. Find the maximum and minimum values of the function f ( x, y ) =x 2 + 6 xy + y 2 subject to the constraint x 2 + y 2 ≤ 32 . Ans: 128, –64 difficulty: easy section: 15.3
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Chapter 15: Optimization- Local and Global Extrema
51. Let v =3i + a j + bk be a vector in space with a, b > 0. Compute the cross product v × (3 j + k ) and then use the result and the Lagrange Multiplier method to find the values of a and b such that the magnitude of the cross product v × (3 j + k ) is the largest with v = 19. 176 −3 176 . = ,b 5 5 difficulty: medium section: 15.3
Ans: a =
52. The owner of a jewelry store has to decide how to allocate a budget of $540,000. He notices that the earnings of the company depend on investment in inventory x1 (in thousands of dollars) and expenditure x2 on advertising (in thousands of dollars) 2
1
according to the function f ( x, y ) = 6 x1 3 x2 3 . How should the owner allocate the $540,000 between inventory and advertising to maximize his profit? Ans: The owner should spend $360,000.00 on the inventory and $180,000.00 on the advertising. difficulty: easy section: 15.3 53. A coffee company sells three brands of coffee. Brand A costs p1 dollars per can, brand B costs p2 dollars per can, and brand C costs p3 dollars per can. The demand (in hundreds of cans) depends on the prices as follows: demand for brand= A 200 − p1 , demand for brand= B 300 − 2 p2 , demand for brand= C 400 − 3 p3 . The company can produce 69,000 cans. What selling prices optimize the total revenue? Ans: p1 $60.00, = = p2 $35.00, = p3 $26.67 . difficulty: medium section: 15.3 54. The following results are obtained when optimizing f(x, y) subject to the constraint g(x, y) = 35. The maximum value is f(5, 7) = 39, the Lagrange multiplier λ = 4 (when ∇f = λ∇g ) and ∇f = 2i − 5 j . What is g(5, 7)? Ans: g (5, 7) = 35 . difficulty: easy section: 15.3 55. The following results are obtained when optimizing f(x, y) subject to the constraint g(x, y) = 39. The maximum value is f(5, 7) = 42, the Lagrange multiplier λ = 3 (when ∇f = λ∇g ) and ∇f = 3i − 3 j . If the constraint condition is changed to g(x, y) = 40, what will be new maximum value of f(x, y)? Ans: The new maximum value of f is approximately 45. difficulty: easy section: 15.3
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Chapter 15: Optimization- Local and Global Extrema
56. Given that the quadratic Taylor polynomial of f at (4, 5) is p ( x, y ) = –2 + x 2 + y 2 − 4 x + 12 y, decide whether (4,5) is a critical point. If so, identify what sort of critical point it is. Ans: The point (4,5) is not a critical point.. difficulty: medium section: 15R 57. Find the saddle point of f ( x, = y ) ( x 2 + y 2 )e x / 2 . Ans: The saddle point of f is (–4,0). difficulty: easy section: 15R
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Chapter 15: Optimization- Local and Global Extrema
58. (a) Let f ( x, y ) =x 2 − 4 xy + 2 y 2 . Find the maximum and minimum values of f on the curve x 2 + 2 y 2 = 1. (b) Use the results of part (a) to find the maximum and minimum values of 2 2 1. h( x, y ) = e x − 4 xy + 2 y on the curve x 2 + 2 y 2 = Explain your work. Ans: (a) We have the objective function f ( x, y ) =x 2 − 4 xy + 2 y 2 and the constraint g ( x, y ) =x 2 + 2 y 2 =1. Using the Lagrange Multiplier method, we need to solve 2 x − 4 y= λ 2 x, − 4 x + 4 y= λ 4 y, x 2 + 2 y 2= 1. Eliminating λ from the first two equations gives x 2 = 2 y 2 . Substituting this into the third equation gives 4 y 2 = 1, so y = ±1/ 2 and x = ±1/ 2 . The critical points are
Since the curve x 2 + 2 y 2 = 1 is an ellipse without end points, we can conclude and the from the computation above that the maximum value of f is . minimum value of f is (b) If we use the Lagrange Multiplier method to optimize h(x, y), the computation will be very tedious. However, since h( x, y ) = e f ( x , y ) , we can use the results in part (a) to optimize h. Since is a maximum point of f, we have Since the exponential function is an increasing function, we have or equivalently,
This means that is a maximum point of h, and the maximum value of h is . Similarly, we can conclude that and are minimum points of h and the minimum value is . difficulty: hard section: 15R 59. Consider the function f ( x, y ) = e1− x − 2 y . Determine all local maxima, minima and saddle points of f. Does f have a global maximum? Ans: (0,0) is a local maximum and a global maximum. difficulty: medium section: 15R 2
2
Page 14
Chapter 15: Optimization- Local and Global Extrema
60. The Green Leaf Bakery makes two types of chocolate cakes, Delicious and Extra Delicious. Each Delicious requires 0.1 lb of European chocolate, while each Extra Delicious requires 0.2 lb. Currently there are only 233 lb of chocolate available each month. Suppose the profit function is given by: p ( x, y ) =151x − 0.2 x 2 + 200 y − 0.1 y 2 , where x is the number of Delicious cakes and y is the number of Extra Delicious cakes that the bakery produces each month. (a) How many of each cake should the bakery produce each month to maximize profit? (b) What is the value of λ in part (a) (if ∇p = λ∇g )? What does it mean? (c) It will cost $19.00 to get an extra pound of European chocolate. Should the bakery buy it? = = y 978.89. Ans: (a) x 372.22, (b) λ = 21.11. This means that the profit will increase by $21.11 for every extra pound of European chocolate. (c) If the bakery has to spend $19.00 to get that extra pound of chocolate, but will get back $21.11 (the value of λ) in return, it is a good deal. difficulty: medium section: 15R 61. The level curves of f(x, y) are shown in the picture below.
(a) Estimate the global maximum and minimum of f on the closed triangular region D with vertices at (-1, -1), (2, -1) and (-1, 2). (b) Find the critical point(s) of f in the interior of the region D. (c) Find the critical point(s) of f along the boundary of D. Ans: (a) The global minimum point is at (0, 0), whose value is ≈0. The global maximum point is at (2, -1) or (-1, 2), whose value is ≈8.5. (b) (0, 0) is the critical point in the interior. (c) (1/2, 1/2) is a critical point of f along the boundary. At this point the boundary is tangential to the contour of f at level 2. The points (-1/2, -1), (-1, -1/2) are also critical points along the boundary. At these two points, the boundary is tangential to the contour of f at level 3.5. difficulty: easy section: 15R 62. Let f ( x, y ) = kx 3 − 3kx + y 2 , where k ≠ 0. Find the critical points of f. Ans: The critical points of f are ( −1, 0 ) and (1, 0 ) . difficulty: easy section: 15R
Page 15
Chapter 15: Optimization- Local and Global Extrema
63. Let f ( x, y ) = kx 3 − 3kx + y 2 , where k ≠ 0. Determine the values of k (if any) for which the critical point at (–1, 0) is a local minimum. Ans: The point (–1,0) is a local minimum of f for negative values of k. difficulty: medium section: 15R 64. It can be shown that (0, 0), (0, 1 2 ), (0, −1 2 ), (1, 0), and ( −1, 0) are the critical points of the function f ( x, = y ) ( x 2 + y 2 )e − x − 4 y . Which of the following are classified correctly? Select all that apply. (0, 1 2 ) is a saddle point. A) (0,0) is a local minimum. D) (0, −1 2 ) is a local minimum. B) (1,0) is a local maximum. E) C) (-1,0) is a local maximum. Ans: A, B, C, D difficulty: medium section: 15.1 2
2
65. Let f ( x, y ) = ax 2 + bx − y 2 + cy + d , for constants a, b, c, and d, with a ≠ 0 . True or false: The constants can be chosen in such a way that f will have a local minimum. Ans: False difficulty: medium section: 15.1 66. Let f ( x, y ) = sin x + sin y + cos( x + y ) in the square S bounded by 0 ≤ x ≤ π , 0 ≤ y ≤ π . π π Then , is a critical point. What kind of critical point is it? 6 6 A) local maximum B) local minimum C) saddle point D) none of these Ans: A difficulty: easy section: 15.1 67. The point (0,0) is a critical point for the function f ( x, y ) = x 4 + y 4 . What kind of critical point is it? A) local minimum B) saddle point C) local maximum D) none of these Ans: A difficulty: easy section: 15.1 68. The temperature at each point in the first quadrant is given by T ( x,= y ) ln( xy ) −
x y − . 6 5
Find the hottest point in the first quadrant and determine its temperature. Ans: The hottest point in the first quadrant is (6,5) and its temperature is −2 + ln 30 . difficulty: easy section: 15.2 69. Find three positive numbers whose product is 11 and whose sum is a minimum. What is the minimum sum? Ans: The numbers are all equal to 3 11 . Their sum is thus 3 3 11 . difficulty: easy section: 15.2
Page 16
Chapter 15: Optimization- Local and Global Extrema
70. Find three numbers x, y, and z, such that x 2 y 3 z = 500 and x + y + z is minimal. What is this minimal sum? Ans: The numbers= are x 23= 15, y 15, = and z 13 15 . The sum is 2 15 . difficulty: medium section: 15.2 71. Suppose there are two electric generators that burn natural gas and whose efficiency declines with output. The energy output is an increasing (but concave down) function of fuel input. Specifically, say Output of generator 1 is 70 ln(1 + x) , and Output of generator 2 is 30 ln(1 + y ) , where x is the amount of fuel burned in generator 1 and y is the amount of fuel burned in generator 2. Given that T is the total amount of natural gas available, how should we use this fuel to maximize output? Your answer should be in terms of T. Ans: To maximize output, we should burn 0.7T + 0.4 units of fuel in generator 1 and 0.3T – 0.4 units of fuel in generator 2. difficulty: medium section: 15.3 72. An exam question asks students to find the maximum of f ( x, y ) on the circle g ( x, y ) = 29 , and the gradient vectors of f and g at that point. A student gave the following answer: "The maximum value of f occurs at the point (2,5) with ∇f (2,5) =− 4i 10 j and ∇g (2,5) =+ 2i 5 j ." Explain why this answer is incorrect. Ans: If the answer were correct, then the gradients of f and g would have to be parallel at the point (2,5) . But in the student's answer, grad f is not parallel to grad g at the point (2,5) . difficulty: easy section: 15.3
) ax + by . True or False: There exist values of a and b so that L( x, y ) takes 73. Let L( x, y= 1 − 3 a minimum value of 8 on the unit circle x 2 + y 2 = 1 at the point , . 2 2 Ans: False difficulty: medium section: 15.3
) ax + by . Find values of a and b so that L( x, y ) takes a maximum value of 74. Let L( x, y= 1 − 3 4 on the unit circle x 2 + y 2 = 1 at the point , . 2 2 Ans: = a 2,= b –2 3 difficulty: medium section: 15.3
Page 17
Chapter 15: Optimization- Local and Global Extrema
75. Find the maximum and minimum values of f ( x, y, z ) =x − 4 y + 6 z subject to the constraint g ( x, y, z ) = x 2 + y 2 + z 2 = 53 . Ans: The maximum and minimum values of f subject to the constraint are 53 and –53, respectively. difficulty: easy section: 15R 76. Find the maximum and minimum values of f ( x, y ) =x 2 + 7 xy + y 2 subject to the constraint x 2 + y 2 = c , with c > 0 . Your answers may depend on c. 9 Ans: The maximum and minimum values of f subject to the constraint are c and 2 –5 c , respectively. 2 difficulty: medium section: 15R 77. Let h( x, y, z ) = x 2 − 2 x sin y + z . Let f ( x, y, z ) satisfy ∇f (1, 0, 2) = i − j + 4k . Explain why the maximum of f subject to the constraint h( x, y, z ) = 3 cannot occur at the point (1, 0, 2) . Ans: By direct calculation, we have ∇h(1, 0, 2) = 2i − 2 j + k . If the maximum of f subject to the constraint h = 3 were to occur at the point (1,0,2), then grad f would have to be parallel to grad h at that point. But grad f(1,0,2) is not parallel to grad h(1,0,2). Therefore, the maximum value of f subject to the constraint h = 3 cannot occur at (1,0,2). difficulty: easy section: 15R
Page 18
1. Let f(x, y) be a positive function of x and y which is independent of x, that is, f(x, y) = g(y) for some one-variable function g. Suppose that 3
10
0
0
∫ g ( x)dx =10 and ∫ g ( x)dx =1 . Find ∫ fdA , where R is the rectangle 0 ≤ x ≤ 3, 0 ≤ y ≤ 10. R
Ans: 3 difficulty: easy
section: 16.1
2. Let R1 be the region 0 ≤ x ≤ 3, -2 ≤ y ≤ 4, and let R2 be the region 3 ≤ x ≤ 5, -2 ≤ y ≤ 4. Suppose that the average value of f over R1 is 6 and the average value over R2 is 7. Find the average value of f over R, 0 ≤ x ≤ 5, − 2 ≤ y ≤ 2 . Ans: 6.307692 difficulty: medium section: 16.1 3. Estimate ∫R f(x, y) dA using the table of values below, where R is the rectangle 0 ≤ x ≤ 4, 0 ≤y≤6
0 2 4
x
Ans: 165 difficulty: easy
0 2 6 18
y 3 3 4 14
6 4 3 12
section: 16.1
4. Upper and lower sums for a function f on a rectangle R, using n subdivisions on each side, are ( 3n 2 + 8n + 24 ) / n 2 and ( 3n 2 + 7 n + 8 ) / n 2 respectively. Evaluate ∫ fdA. R
Ans: 3 difficulty: easy
section: 16.1
5. Find a region R such that double integral ∫ ( 2 − x 2 − y 2 )dA has the largest value. R
A) x − y ≤ 2 B) x − y ≥ 2 C) x 2 + y 2 ≤ 2 Ans: C difficulty: easy section: 16.1 2
2
2
2
D)
x2 + y 2 ≥ 2
6. True or false? If f is any two-variable function, then ∫ fdA = 2 ∫ fdA , where R is the rectangle 0 ≤ x R
≤ 2, 0 ≤ y ≤ 1 and S is the square 0 ≤ x, y ≤ 1. A) False B) True Ans: A difficulty: easy section: 16.2
Page 1
S
Chapter 16: Integrating Functions of Several Variables
7. Let R be the region in the first quadrant bounded by the x- and y-axes and the line x + y = 7. Evaluate ∫ x + 2 ydA exactly and then give an answer rounded to 4 decimal places. R
(
)
4 2 2 − 1 75/ 2 or 63.2108 15 difficulty: medium section: 16.2
Ans:
5
45
1
9x
8. Consider the integral ∫ ∫
f ( x, y )dydx.
Rewrite the integral with the integration performed in the opposite order. 45
y /9
∫ ∫ ∫ ∫ f ( x, y)dydx
A)
9x 1 45 y / 9
B)
9
Ans: C
C)
f ( x, y )dxdy
D)
1
difficulty: easy
45
∫ ∫ ∫ ∫
y /9
9 1 45 9 y
9
1
f ( x, y )dxdy f ( x, y )dxdy
section: 16.2
π π sin 2 y − sin 2 y dydx by first reversing the order of integration. 9. Evaluate ∫ ∫ 0 x y Ans: 1.57 difficulty: medium section: 16.2
10. Find the volume under the graph of f ( x, y ) = xe –5 y lying over the triangle with vertices (0, 0), (2, 2) and (4, 0). 4 Ans: ( 9 + e –10 ) 25 difficulty: easy section: 16.2 11. The function f ( x, = y ) ax 2 + 5 y has an average value of 4 on the triangle with vertices at (0, 0), (0, 1) and (1, 0). Find the constant a. Ans: 14 difficulty: easy section: 16.2 12. Evaluate the iterated integral. 2
ln y
1
0
∫∫
ye 4 x dxdy
9 4 difficulty: medium Ans:
section: 16.2
Page 2
Chapter 16: Integrating Functions of Several Variables
13. Evaluate the iterated integral. e
1/ x
1
0
∫ ∫ cos(4 xy)dydx sin 4 4 difficulty: medium
Ans:
section: 16.2
14. Reverse the order of integration for the following integral. 2
8y
0
2
∫∫
y
f ( x, y )dxdy 2
A)
∫∫
B)
∫∫ ∫∫
0
x 2
x /8
4
x 2
0 x /8 4 x
C)
x2 / 8
0
Ans: E
4
x
0
8 x2
f ( x, y )dydx
D)
∫∫
f ( x, y )dxdy
E)
∫∫
4
0
f ( x, y )dydx
x 2
x /8
f ( x, y )dydx
f ( x, y )dydx
difficulty: medium
section: 16.2
15. Reverse the order of integration for the following integral. 3
4 x −12
1
x 2 −9
∫∫
f ( x, y )dydx
A)
∫∫
8
y +9
0
( y +12) / 4
B)
∫ ∫
0
y +9
0
y +9
C)
−8 ( y +12) / 4
∫ ∫
−8 ( y +12) / 4
Ans: C
f ( x, y )dxdy
D)
∫ ∫
0
( y +12) / 4
−8
y +9
f ( x, y )dydx
E)
∫ ∫
0
y +9
−8 ( y −12) / 4
f ( x, y )dxdy
f ( x, y )dxdy
f ( x, y )dxdy
difficulty: medium
section: 16.2
16. Find the volume of the region under the graph of f ( x, y ) = 1 + 2 x 2 + 5 y and above the region x 2 ≤ y, 0 ≤ y ≤ 4. 2336 Ans: 15 difficulty: medium section: 16.2
Page 3
Chapter 16: Integrating Functions of Several Variables
17. Set up but do not evaluate a (multiple) integral that gives the volume of the solid bounded above by the sphere x 2 + y 2 + z 2 = 2 and below by the paraboloid = z x2 + y 2 .
∫ ∫
1− x 2
1
∫ ∫
1− x 2
1
1− x 2
1
A)
∫
2− x2 − y 2
∫
2− x2 − y 2
∫
2− x2 − y 2
−1 − 1− x 2
B)
−1 − 1− x 2
∫ ∫
C)
−1 0
x2 + y 2
− x2 − y 2
x2 + y 2
Ans: D
dxdydz
D)
dzdydx
E)
∫ ∫
1− x 2
1
1− x 2
0
− 1− x 2
1
−1 − 1− x 2
∫∫
∫
2− x2 − y 2
∫
2− x2 − y 2
x2 + y 2
x2 + y 2
dzdydx
dzdydx
dzdydx
difficulty: easy
section: 16.3
18. Calculate the following integral exactly. (Your answer should not be a decimal approximating the true answer, but should be exactly equal to the true answer. Your answer may contain e, π, , and so on.) 1
z
y
0 0
0
∫ ∫ ∫ x y z dxdydz 2
5 5
1 405 difficulty: easy Ans:
section: 16.3
19. Calculate the following integral exactly. (Your answer should not be a decimal approximating the true answer, but should be exactly equal to the true answer. Your , and so on.) answer may contain e, π, 6
3
0
0
∫ ∫ cos 4 y sin(4 x + 5)dxdy Ans:
( cos 5 − cos17 ) sin 24
16 difficulty: medium
section: 16.3
20. Calculate the following integral exactly. (Your answer should not be a decimal approximating the true answer, but should be exactly equal to the true answer. Your answer may contain e, π, , and so on.) 4
y
3
0
∫ ∫ y e dxdy 2 xy
1 –7 − e9 + e16 ) ( 2 1 B) ( 25 − e9 + e16 ) 2 Ans: A difficulty: medium A)
C) D) section: 16.3
Page 4
1 –7 + e9 − e16 ) ( 2 1 ( –7 − e9 + e16 ) 3
Chapter 16: Integrating Functions of Several Variables
21. Find the triple integral of the function f(x, y, z) = xy sin (18yz) over the rectangular box 0 ≤ x ≤ π, 0 ≤ y ≤ 1, 0 ≤ z ≤ π/6. π2 Ans: 36 difficulty: medium section: 16.3 22. Set up (but do not evaluate) an iterated integral to compute the mass of the solid paraboloid bounded by = z x 2 + y 2 and z = 1, if the density is given by δ(x, y, z) = z2. 1
A)
Mass = ∫ ∫
B)
Mass = ∫ ∫
C)
Mass = ∫ ∫
1− x 2
−1 − 1− x 1
1− x 2
2
1
2
D)
Mass = ∫ ∫
z 2 dzdydx 2
E)
Mass = ∫ ∫
2
1
1
z 2 dzdxdy
x +y
∫
1− x 2
−1 − 1− x
Ans: D
1
x +y
0 0 1
∫
2
2
1− x 2
−1 − 1− x 1
1− x 2
0
− 1− x
2
2
∫
1
x2 + y 2
∫
1
x2 + y 2
z 2 dzdydx z 2 dzdydx
∫ z dzdydx 2
0
difficulty: medium
section: 16.3
23. Set up an iterated integral for ∫ f ( x, y, z )dV , where W is the solid region bounded W
below by the rectangle 0 ≤ x ≤ 3, 0 ≤ y ≤ 1 and above by the surface z 2 + y 2 = 1. A) B) C) D) E)
1
0
0 0
3
1
0
0 0
3
1
0
0 0
3
1
1− y 2
0
0
− 1− y 2
3
1
1− y 2
0
0
− 1− y 2
∫
W
∫
W
f ( x, y, z )dV = ∫ ∫ ∫
∫
W
f ( x, y, z )dV = ∫ ∫ ∫
∫
W
f ( x, y, z )dV = ∫ ∫ ∫
∫
f ( x, y, z )dV = ∫ ∫ ∫
W
Ans: C
1− y 2
3
f ( x, y, z )dV = ∫ ∫ ∫
1− y 2
1− y 2
difficulty: medium
f ( x, y, z )dzdxdy f ( x, y, z )dydzdx f ( x, y, z )dzdydx
f ( x, y, z )dzdydx f ( x, y, z )dydxdz
section: 16.3
24. A solid is bounded below by the triangle z = 0, x ≥ 0, y ≥ 0, x + y ≤ 1 and above by the plane z = x + 6y + 2. If the density of the solid is given by δ(x, y, z) = z, find its mass. 41 Ans: 8 difficulty: medium section: 16.3
Page 5
Chapter 16: Integrating Functions of Several Variables
25. Sketch the region of integration of the following integral and then convert the expression to polar co-ordinates (you do not have to evaluate it). 2
∫ ∫ 0
4− x2
x
x 5 y 5 dydx
Ans: The area is shown below.
Converting to polar coordinates, we get the following integral: π 2 π 4
2
∫ ∫ r cos θ sin θ dθ dr 0
11
5
5
difficulty: medium
section: 16.3
9 x2 + y 2 dxdy. Provide an exact answer. −1 1− y 2 1 + x 2 + y 2 π Ans: 9π 1 − 4 difficulty: medium section: 16.3 1
26. Evaluate ∫ ∫
0
3
27. Convert the integral ∫ ∫
–3 0
exactly. π Ans: 1 − e −9 ) ( 2 difficulty: easy
9− x2
e
(
− x2 + y 2
) dydx to polar coordinates and hence evaluate it
section: 16.4
28. Calculate the following integral: 2
y
0
0
∫ ∫ x y dxdy 5
7
Ans: 195.048 difficulty: easy
section: 16.4
Page 6
Chapter 16: Integrating Functions of Several Variables
29. Calculate the following integral: 5 ∫ r cos θ dA where R is the shaded region shown below. R
·
127 7 difficulty: easy Ans: −
section: 16.4
30. Compute the area of the flower-like region bounded by r = 6 + 3 cos (8θ). 81π Ans: 2 difficulty: medium section: 16.4 31. Evaluate exactly the integral ∫ ydA , where R is the region shown below. R
16 3 difficulty: medium
Ans:
section: 16.4
32. Evaluate the iterated integral ∫
2π
π
15066 5 difficulty: easy
6
∫ r sin θ drdθ . 4
3
Ans: −
section: 16.4
Page 7
Chapter 16: Integrating Functions of Several Variables
33. Evaluate the integral ∫ x + 3 ydA , where R is the region in the first quadrant bounded by R
2 2 the y-axis, the line y = x and the circles x 2 + y= 25, x 2 + y= 16. 61 Ans: 2+2 2 6 difficulty: easy section: 16.4
(
)
34. For the following region, decide whether to integrate using polar or Cartesian coordinates. Write an iterated integral of an arbitrary function f(x, y) over the region.
Ans: Cartesian:
difficulty: medium
section: 16.4
35. Find the volume of the solid bounded by the paraboloid = z x 2 + y 2 and the plane z = 1. 1π Ans: 2 difficulty: easy section: 16.4 36. Suppose a solid is the region in three-space in the first octant bounded by the plane x + y = 1 and the cylinder x 2 + z 2 = 1 . If the density of this solid at a point (x, y, z) is given by 3 δ ( x, y, z ) = z , find its mass. 11 Ans: 120 difficulty: medium section: 16.4 37. Consider the volume between a cone centered along the positive z-axis, with vertex at the origin and containing the point (0, 1, 1), and a sphere of radius 3 centered at the origin. Write a triple integral which represents this volume and evaluate it. Use spherical coordinates. Ans:
2π
π /4
3
0
0
0
φ 9(2 − 2)π ∫ ∫ ∫ ρ sin θ d ρ dθ d= 2
difficulty: medium
section: 16.5
Page 8
Chapter 16: Integrating Functions of Several Variables
38. A cylindrical tube of radius 2cm and length 3cm contains a gas. As the tube spins around its axis, the density of the gas increases as you get farther from the axis. The density, D, at a distance of r cm from the axis is D(r) = 1 +9 r gm/cc. Write a triple integral representing the total mass of the gas in the tube and evaluate the integral. Ans: The total mass of gas in the tube is 156π grams. difficulty: easy section: 16.5 39. Set up the three-dimensional integral ∫ ydV where R is the “ice-cream cone” enclosed R
by a sphere of radius 2 centered at the origin and the cone rectangular coordinates.
. Use
A) B) C) Ans: C
difficulty: easy
section: 16.5
40. Let R be the ice-cream cone lying inside the sphere x 2 + y 2 + z 2 = 4 and inside the cone = z
3 ( x 2 + y 2 ) . Find the center of mass of R.
3 Ans: 0, 0, ≈ (0,0,1.400) 16 − 8 3 difficulty: medium section: 16.5
41. Evaluate ∫ zdV where W is the first octant portion of the ball of radius 3 centered at the W
origin.
81π 16 difficulty: easy Ans:
section: 16.5
Page 9
Chapter 16: Integrating Functions of Several Variables
42. Choose the most appropriate coordinate system and set up a triple integral, including limits of integration, for a density function f(x, y, z) over the given region.
A) spherical coordinates B) cylindrical coordinates Ans: A difficulty: easy
C) D) section: 16.5
rectangular coordinates None of the above.
43. Find the mass of the solid cylinder x 2 + y 2 ≤ 25, 4 ≤ z ≤ 5 with density function f ( x, y , z ) = z + 4 ( x2 + y 2 ).
2725π 2 difficulty: medium Ans:
section: 16.5 16 − y 2
6
1
4
−1 − 16 − y
44. Evaluate the integral
∫∫ ∫
Ans: 16π difficulty: easy
section: 16.5 2
0
0
− 4− x2
45. Evaluate the integral ∫ ∫ coordinates. 32 Ans: π 3 difficulty: medium
∫
1 2
x + y2 2
4− x2 − y 2
− 4− x2 − y 2
dxdydz in cylindrical coordinates.
( x + y + z ) dzdydx in spherical 2
2
2 3/ 2
section: 16.5
46. Let W be the region between the cylinders x 2 + y 2 = 9 and x 2 + y 2 = 36 in the first octant and under the plane z = 1. Evaluate ∫ zdV . W
27 π 8 difficulty: medium Ans:
section: 16.5
Page 10
Chapter 16: Integrating Functions of Several Variables
47. Rewrite the integral
in spherical coordinates. You do not have to evaluate the integral. Ans:
difficulty: easy
section: 16.5
48. The region W is shown below. Write the limits of integration for ∫ f ( x, y, z )dV in W
spherical coordinates.
A)
B) C) Ans: A
difficulty: easy
section: 16.5
49. Find the condition on the non-negative constants a and b for p(x, y) to be a joint density function, where ax + by, 0 ≤ x ≤ 8, 0 ≤ y ≤ 8 p ( x, y ) = 0 otherwise 1 1 1 A) a + b = B) a + b = C) a + b < D) a=b E) 256 256 32 1 a+b = 512 Ans: A difficulty: easy section: 16.6
Page 11
Chapter 16: Integrating Functions of Several Variables
50. Let x and y have joint density function Find the probability that 0.5 ≤ x ≤ 0.6. Ans: 0.105 difficulty: easy section: 16.6 51. Let x and y have joint density function
Find the probability that x > y +0.4. Ans: 0.18 difficulty: easy section: 16.6 52. The joint density function for x, y is given by
Find the probability that (x, y) satisfies x ≥ 1.1 y. Ans: 0.6975 difficulty: easy section: 16.6 53. Jane and Mary will meet outside the library at noon. Jane's arrival time is x and Mary's arrival time is y, where x and y are measured in minutes after noon. The probability density function for the variation in x and y is 1 − x /12 − y /15 . p ( x, y ) = e e 180 After Jane arrives, she will wait up to 15 minutes for Mary, but Mary won't wait for Jane. Find the probability that they meet. Ans: 0.351 difficulty: easy section: 16.6 54. An arrow strikes a circular target at random at a point (x, y), using a coordinate system with origin at the center of the target. The probability density function for the point where the arrow strikes is given by What is the probability that the arrow strikes within 0.45 feet of the center of the target. Ans: 0.183 difficulty: easy section: 16.6
Page 12
Chapter 16: Integrating Functions of Several Variables
55. Consider the change of variables x = s + 5t, y = s - 3t. ∂ ( x, y ) Find the absolute value of the Jacobian . ∂ ( s, t ) Ans: 8 difficulty: easy section: 16.7 56. Consider the change of variables x = s + 3t, y = s - 2t. Let R be the region bounded by the lines 2x + 3y = 1, 2x + 3y = 4, x - y = -3, and x - y = 2. Find the region T in the st-plane that corresponds to region R. Use the change of variables to evaluate ∫ 2 x + 3 ydA . R
Ans: 7.5 difficulty: medium
section: 16.7
57. Let R be the region bounded between the two ellipses x2 y 2 x2 y 2 and + = 1 + = 4. 32 22 32 22 = x 3= r cos t , y 2r sin t , for r ≥ 0, 0 ≤ t ≤ 2π, to evaluate Use this change of coordinates the integral ∫ ( 4 x 2 + 9 y 2 )dA. A) 240 π Ans: D
R
B) 3240 π C) 162 π D) 1620 π difficulty: hard section: 16.6
E) 1620
58. Let R be the region in the first quadrant bounded between the circle x 2 + y 2 = 1 and the π two axes. Then ∫ ( x 2 + y 2 )dA = . R 8 Let R be the region in the first quadrant bounded between the ellipse 25 x 2 + 9 y 2 = 1 and the two axes. Use the change of variable x = s/5, y = t/3 to evaluate the integral 2 2 ∫ ( 75 x + 27 y )dA. R
1 π 40 difficulty: easy
Ans:
section: 16.7 3
0
0
− y /3
59. Evaluate the integral ∫ ∫ Ans: 1350.35 difficulty: easy
2
e y dxdy . Give your answer to two decimal places.
section: 16R
Page 13
Chapter 16: Integrating Functions of Several Variables
60. Evaluate the integral by interchanging the order of integration. 0
4
∫ ∫ e dydx . y2
–1 −4 x
Ans: 1,110,763.69 difficulty: easy section: 16R 61. Express ∫ f ( x, y )dA as an iterated integral, where R is the region in the first quadrant R
bounded between the two axes and the lines y = 5 and y = 4 - x. A) B) C)
2
∫ f ( x, y)dA = ∫ ∫ ∫ f ( x, y)dA = ∫ ∫ ∫ f ( x, y)dA = ∫ ∫
4− x
R
0 0 2 0
R
0 4− x a 4− y
R
0
Ans: C
0
f ( x, y )dydx
D)
f ( x, y )dxdy
E)
4− x
2
∫ f ( x, y)dA = ∫ ∫ f ( x, y)dydx ∫ f ( x, y)dA = ∫ ∫ f ( x, y)dxdy R
0 2
0 4− x
R
0
0
f ( x, y )dxdy
difficulty: easy
section: 16R
62. Let R be the triangular region with vertices at (0, 0), (1, 0) and (0, 1). Suppose that 1
1
0.75
0
0
0.75
= = g ( x, 0.75)dx 1.5, ∫ g ( x,1)dx 3.1, ∫ g ( x, 0)dx 1.3, ∫= 0
g ( x, 0.25)dx 1.7, ∫= 0
0.5
0.25
0.25
0
0
0
= = ∫ g ( x, 0.5)dx 1.95,= ∫ g ( x, 0.75)dx 1.5, ∫ g ( x, 0.25)dx 1.5. Find an approximate value of ∫ g ( x, y )dA. R
Ans: 1.6125 difficulty: easy
section: 16R
63. Convert the integral to polar coordinates. 1
∫ ∫
2− x2
−1 1
A) B) C)
f ( x, y )dydx
∫ ∫ ∫ ∫
3π / 4
2
π/4
1/ sin θ
3π / 4
2
3π / 4
2
0
1/ sin θ
∫ ∫ π/4
Ans: D
1
f (r cos θ , r sin θ )drdθ
D)
f (r cos θ , r sin θ )rdrdθ
E)
3π / 4
2
π/4
1/ sin θ
3π / 4
2
π/4
1/ sin θ
∫ ∫ ∫ ∫
f (r cos θ , r sin θ )rdrdθ f (r cos θ , r sin θ )rdθ dr
f (r cos θ , r sin θ )rdrdθ
difficulty: easy
64. Evaluate the integral ∫
4
∫
16 − y 2
−4 − 16 − y 2
Ans: 16π difficulty: easy
section: 16R
∫
16 − x 2 − y 2
− 16 − x 2 − y 2
1 dzdxdy in spherical coordinates. x + y2 + z2 2
section: 16R
Page 14
Chapter 16: Integrating Functions of Several Variables
65. Evaluate the integral ∫
0
0
∫ ∫
4− x2
−2 −2 − 4 − x
1024 π 9 difficulty: medium
(x + z ) 2
2
2 7/2
dzdxdy in cylindrical coordinates.
Ans:
section: 16R
66. Let W be the region between the spheres x 2 + y 2 + z 2 = 1 and x 2 + y 2 + z 2 = 4 . Given
15π , evaluate the integral ( x + y + z ) dV = ∫ ( 64 x + 36 y + 144 z ) dV , where W is the region between the ellipsoids
that ∫
2
2 1/ 2
2
W
2
2
2 1/ 2
W 2
x y2 z2 x2 y 2 z 2 and + + = 1 + + = 4. 32 42 22 32 42 22 A) 8640π B) 2160π C) 24π D) 360π Ans: A difficulty: easy section: 16R
Page 15
E) 360π 3
Chapter 16: Integrating Functions of Several Variables
67. Let W be the part of the solid sphere of radius 4, centered at the origin, that lies above the plane z = 2. Express ∫ zdV in W
(a) Cartesian (b) Cylindrical (c) Spherical coordinates. Ans: (a) Cartesian coordinates. Integrate first with respect to z from z = 2 to z=
42 − x 2 − y 2 (the equation of the sphere with radius 4, centered at the origin
is x 2 + y 2 + z 2 =), 42 then with respect to y from y = − 12 − x 2 to= y 2 2 (obtain x + y = 12 by eliminating z between the equations z = 2 and z=
12 − x 2
42 − x 2 − y 2 ), and finally with respect to x from x = − 12 to x = 12
(obtained by setting y = 0 in x 2 + y 2 =). 12 Thus,
(b) In cylindrical coordinates, the upper part of the sphere x 2 + y 2 + z 2 = 42 is changed to . Thus
c)
Spherical coordinates. The plane z = 2 becomes
so The sphere, ρ = 4, and the plane, ρ = 2/ cos ϕ, will intersect at
so hence
Therefore, we have
difficulty: hard
section: 16R
Page 16
Chapter 16: Integrating Functions of Several Variables
68. Evaluate the integral ∫ ( x 2 + y 2 ) dA , where R is the region shown below. 2
R
665 −1 tan 2 6 difficulty: medium
Ans:
section: 16R
69. The joint density function for x, y is given by 1 − x /10 − y /15 e e x ≥ 0, y ≥ 0 p ( x, y ) = 150 0 otherwise Write down an iterated integral to compute the probability that x + y ≤ 10. You do not need to do the integral. 10 10 − x 1 10 10 1 − x /10 − y /15 − x /10 − y /15 A) D) e e dydx ∫0 ∫0 10 e e dydx ∫0 ∫0 150 10 10 − x 1 10 − x 10 1 − x /10 − y /15 − x /10 − y /15 B) E) ∫0 ∫0 150 e e dxdy ∫0 ∫0 150 e e dxdy 10 10 − x 1 − x /10 − y /15 C) ∫0 ∫0 150 e e dydx Ans: C difficulty: easy section: 16R 70. Consider the integral ∫
6
∫
−6 6
Ans:
3π 4
∫ ∫ π
4
6 2
6 / sin θ
difficulty: easy
72 − x 2
f ( x, y )dydx . Convert the integral to polar coordinates.
f (r cos θ , r sin θ )rdrdθ section: 16.4
71. The function f (= x) kx 2 + 4 y has an average value of 16 on the rectangle with vertices at (0, 0),(0, 2), (2, 0) and (2, 2). Find the constant k. Ans: 9 difficulty: easy section: 16.2
Page 17
Chapter 16: Integrating Functions of Several Variables
72. The joint density function for random variables x and y is 1 ( x + y ) if 0 ≤ x ≤ 3, 0 ≤ y ≤ 2 f ( x, y ) = 21 0 otherwise Find the probability P( x ≤ 2.5, y ≥ 0.2) . Give your answer to 3 decimal places. Ans: 0.504 difficulty: easy section: 16.6 73. Find the condition on the non-negative constants a and b for p(x, y) to be a joint density function, where ax + by if 0 ≤ x ≤ 6, 0 ≤ y ≤ 6 p ( x, y ) = otherwise. 0 1 Ans: a + b = 108 difficulty: easy section: 16.6 3
0
0
−x
74. Consider the integral ∫ ∫ f ( x, y )dydx . (a) Sketch the region of integration and rewrite the integral with order of integration reversed. (b) Rewrite the integral in polar coordinates. Ans: (a) ∫
0
(b) ∫
2π
3
∫ f ( x, y)dxdy
−3 − y
π ∫
3/ cosθ
7 /4 0
difficulty: easy
f (r cos θ , r sin θ )rdrdθ
section: 16.4
75. Find the mass of the solid cylinder x 2 + y 2 ≤ 25 , 4 ≤ z ≤ 6 with density function f ( x, y, z ) =z + x 2 + y 2 . Ans: 875π difficulty: easy section: 16.5
Page 18
Chapter 16: Integrating Functions of Several Variables
76. Find the area of the part of the hyperbolic paraboloid = z y 2 − x 2 that lies between the cylinders x2 + y 2 = 1 and x 2 + y 2 =. 16 65 65 − 5 5 π 3 65 − 5 B) π 6 65 65 − 5 5 C) π 6 Ans: C difficulty: medium
A)
D) E)
65 − 5 π 3 32 65 − 4 5 π 3
77. If f and g are two continuous functions on a region R, then ∫ f ⋅ gdA= ∫ fdA ⋅ ∫ gdA . R
R
Ans: False
R
difficulty: medium
section: 16 Review
78. Consider the region in 3-space bounded by the surface f ( x, y ) = x 2 + y 2 − 1 and the plane z = k where k > 0 . Find the value of k such that the volume of this region below the xy-plane equals the volume of this region above the xy-plane. Ans: 1(1 + 2) difficulty: medium section: 16.5 79. Decide (without calculation) whether the integrals are positive, negative, or zero. Let D be the region inside of the unit circle centered at the origin, R the right half, L the left half, B the bottom half, and U the upper half of D. (A) ∫ xydA D
(B) ∫ xy 2 dA R
(C) ∫ ( x − 1)dA U
(D) ∫ cos ydA B
Part A: Zero Part B: Positive Part C: Negative Part D: Positive difficulty: easy
section: 16.1
Page 19
1. Give parameterizations for a circle of radius 2 in the plane, centered at origin, traversed anticlockwise. x cos= t , y sin t , 0 ≤ t ≤ 2π x 2 cos = t , y 2sin t , 0 ≤ t ≤ 2π A) = D)= x 4 cos = t , y 4sin t , 0 ≤ t ≤ 2π x 2 cos = t , y 2sin t , 0 ≤ t ≤ π B)= E)= = x 2 cos = t , y –2sin t , 0 ≤ t ≤ 2π C) Ans: D difficulty: easy section: 17.1 2. Give parameterizations for a circle of radius 3 in 3-space perpendicular to the y-axis centered at (4, –2, 0). x = 4 + 3cos t , y = –2, z = 3sin t , 0 ≤ t ≤ 2π A) x = 4 + 3cos t , y = –2, z = 3sin t , 0 ≤ t ≤ π B) x = 4 + 3cos t , y = 3sin t , z = –2, 0 ≤ t ≤ 2π C) x = 4 + 3cos t , z = 3sin t , 0 ≤ t ≤ 2π D) x 3cos t ,= y –2,= z 3sin t , 0 ≤ t ≤ 2π E) = Ans: A difficulty: easy section: 17.1 1 and the line with parametric equation = 3. Consider the plane 2 x + y − 8 z = r r 0 + t u. Give a value of u which makes the line parallel to the plane. (There are many possible answers.) Ans: = u i –10 j − k difficulty: easy section: 17.1 4. The equation r = 2i + 5 j + 3k + t (i − j + 2k ) parameterizes a line through the point (4, 3, 7). What is the value of t at this point? Ans: 2 difficulty: easy section: 17.1
Page 1
Chapter 17: Parameterization and Vector Fields
5. What curve, C, is traced out by the parameterization r = 2i + ( cos t ) j + ( sin t ) k for 0 ≤ t
≤ 2π? Either give a very complete verbal description or sketch the curve (or both). Ans: As r = 2i + cos t j + sin tk or x = 2, y = cos t, z = sin t, then the curve C is a circle of radius 1 centered at (2, 0, 0) in the plane x = 2.
difficulty: easy
section: 17.1
6. Let f(x, y, z) = xy + 6yz + zx. Then f(2, 2, 3) = 46. Give an equation to the tangent plane to xy + 6yz + zx = 46. Ans: 5x + 20y + 14z = 92 difficulty: easy section: 17.1 7. Which of the following equations give alternate parameterizations of the line L parameterized by A) B) C) Ans: A
difficulty: medium
section: 17.1
8. Find a parameterization for the circle of radius 4 in the xz-plane, centered at the point (3, 0, –5). Select all that apply. x= 3 + 4 cos t , y = 0, z = 5 + 4sin t A) B) r =(3i − 5k ) + 4 cos ti + sin tk
(
C) D)
)
x =3 − 4sin t , y =0, z =−5 + 4 cos t r = (5 i − 3 k ) + 4 cos t i + sin t k
(
)
x =3 + 4 cos t , y =0, z =−5 + 4sin t E) Ans: B, E difficulty: easy section: 17.1
Page 2
Chapter 17: Parameterization and Vector Fields
9. Find a parameterization for the curve y6 = x7 in the xy-plane. Select all that apply. 7/6 7 A)= D) = x t= , y t x t= , y t6 6 B) = E) = x t= , y t7 x t ,= y t6/7 6/7 C)= x t= , y t Ans: A, B, E difficulty: medium section: 17.1 10. Find a parametric equation for the line which passes through the point (5, 1, -1) and is parallel to the line (2 + 2t )i + (3 − t ) j − (3 + 2t )k . Ans: (5 + 2t )i + (1 − t ) j + (−1 + 2t )k difficulty: easy section: 19.1 11. Consider the plane x - 4y + –2z = 5 and the line x = a + bt, y = 2 + –2t, z = 2 - t. Find the value of b such that the line is perpendicular to the plane. 1 Ans: b = 2 difficulty: easy section: 17.1 12. Consider the plane x - 4y + 4z = 5 and the line x = a + bt, y = 2 + 2t, z = 5 - t. Find the values of a and b such that the line lies in the plane. Ans: a = –7 ; b = 12 difficulty: easy section: 17.1 13. Describe the similarities and differences between the following two curves. Curve 1: r (t )= (3 + 3t )i + (1 − t ) j + (3 + 4t )k , − ∞ ≤ t ≤ ∞, Curve 2 : r (t )= (3 + 3t 2 )i + (1 − t 2 ) j + (3 + 4t 2 )k , − ∞ ≤ t ≤ ∞. Ans: Curve 1 is the line with direction vector 3i − j + 4k . and passing through the point (3, 1, 3). If we replace the parameter t2 with s in the parametric equation of Curve 2, then the equation is the same as that of Curve 1. This means that Curve 2 is part of the line that points in the direction of 3i − j + 4k . and passes through the point (3, 1, 3). Curve 2 is only part of the Curve 1 because when we replace the parameter t2 with s, s can only take on values between 0 and ∞, instead of between -∞ and ∞. difficulty: medium section: 17.1
Page 3
Chapter 17: Parameterization and Vector Fields
14. Find a parameterization of a curve that looks like sin y = z when viewed from the x-axis, and looks like x = z2 when viewed from the y-axis. See the shadows drawn on the planes in the following picture.
What does the curve look like when viewed from the z-axis? Ans: We must find a curve whose points (x, y, z) satisfy the relations:
We can set y = t, and then we obtain z = sin t and x = z2 = sin2t. The parameterization is therefore: The curve will look like x = sin 2t, y = t, or equivalently, x = sin2y when viewed from the z-axis. difficulty: medium section: 17.1 15. Suppose z = f(x, y), f(1, 3) = 5 and ∇f (1,3) =+ 4 i 5 j. True or false: the vector –4i + 5 j + k is perpendicular to the graph of f(x, y) at the point (1, 3). Ans: False difficulty: easy section: 17.1
16. A child is sliding down a helical slide. Her position at time t after the start is given in feet by r= cos t i + sin t j + (12 − t )k . The ground is the xy-plane. At time t = 2π, the child leaves the slide on the tangent to the slide at that point. What is the equation of the tangent line? Ans: r = i + (12 − 2π)k + s ( j − k ) , where s is the parameter. difficulty: easy section: 17.2 17. Find parametric equations for the line through the point (1, 5, 2) and parallel to the vector 4i + 4 j − 2k in which the particle is moving with speed 24 (the parameter t represents time). Ans: x = 1 + 16 t, y = 5 + 16 t, and z = 2 - 8 t. difficulty: medium section: 17.2
Page 4
Chapter 17: Parameterization and Vector Fields
18. Find parametric equations for a line through the points, A = (–2, 5, 4) and B = (–2, 25, 9) so that the point A corresponds to t = 0 and the point B to t = 5. –2, y =+ 4t 5, z = t + 4, for -∞ < t < ∞. Ans: x = difficulty: easy section: 17.2 19. Write down a parameterization of the line through the points (2, 2, 4) and (6, 4, 2). Select all that apply. A) D) r = 2i + 2 j + 4k − t (4i + 2 j − 2k ) r = 2i + 2 j + 4k + t (4i + 2 j − 2k ) B) E) r = 2i + 2 j + 4k + t (2i + j − k ) r = 2i + 2 j + 4k + t (4i − 2 j − 2k ) C) r = 2i + 2 j + 4k + t (4i + 2 j + 2k ) Ans: A, D, E difficulty: easy section: 17.2 20. The line through the points (2, 5, 25) and (12, 7, 23) can be parameterized by r = 2i + 5 j + 25k + t (10i + 2 j − 2k ) . What value of t gives the point (42, 13, 17)? Ans: 4 difficulty: easy section: 17.2 21. The path of an object moving in xyz-space is given by ( x(t ), y (t ), = z (t )) (3t 2 , t + 1, t 3 ) . The temperature at a point (x, y, z) in space is given by f ( x, y,= z ) x 2 y − 3 z. Calculate the directional derivative of f in the direction of v at the point (12, 3, 8), where v is the velocity vector of the object.. 972 Ans: 289 289 difficulty: medium section: 17.2 22. The path of an object moving in xyz-space is given by ( x(t ), y (t ), = z (t )) (3t 2 , t + 1, t 3 ) . The temperature at a point (x, y, z) in space is given by f ( x, y,= z ) x 2 y − 3 z. d Calculate . f ( x(t ), y (t ), z (t )) dt t=2 Ans:
d f ( x(t ), y (t ), z (t )) = 972 . dt t=2
difficulty: medium
section: 17.2
23. The equation r =4 xi + 4 y j =12 cos(2πt / 360)i + 12sin(2πt / 360) j describes the motion of a particle moving on a circle. Assume x and y are in miles and t is in days. What is the speed of the particle when it passes through the point (0, 2)? 1 Ans: π miles per day. 15 difficulty: easy section: 17.2
Page 5
Chapter 17: Parameterization and Vector Fields
24. An object moves with constant velocity in 3-space. It passes through (4, 0, 1) at time t = 1 and through (13, 6, –11) at time t = 4. Find its velocity vector. Ans: v =3i + 2 j − 4k difficulty: easy section: 17.2 25. A particle moves with position vector r (t ) = ln ti + t −1 j + e − t k . Find the velocity and acceleration at time t = 8. Select all that apply. 1 1 −8 1 1 −8 A) D) v(t ) = i − j −e k a (t ) = j + e k. − i+ 8 64 64 256 1 1 −8 1 1 −8 B) E) a (t ) = j − e k. v(t ) = i + j −e k − i+ 64 256 8 64 1 1 1 1 −8 F) C) v(t ) = i − j − e −8 k a (t ) = − i− j + e k. 8 8 64 256 Ans: A, D difficulty: easy section: 17.2
26. A particle moves with position vector r (t ) = ln ti + t −1 j + e − t k . . Describe the movement of the particle as t → ∞. A) The particle will approach the positive x-axis asymptotically as t → ∞. Also, since each component of v(t ) approaches 0 as t → ∞, we expect the particle to approach the positive y-axis with slower and slower speed. B) The particle will approach the positive y-axis asymptotically as t → ∞. Also, since each component of v(t ) approaches 0 as t → ∞, we expect the particle to approach the positive x-axis with slower and slower speed. C) The particle will approach the positive x-axis asymptotically as t → ∞. Also, since each component of v(t ) approaches 0 as t → ∞, we expect the particle to approach the positive x-axis with increasing speed. D) The particle will approach the positive x-axis asymptotically as t → ∞. Also, since each component of v(t ) approaches 0 as t → ∞, we expect the particle to approach the positive x-axis with slower and slower speed. E) The particle will approach the positive x-axis asymptotically as t → ∞. Also, since each component of v(t ) approaches 0 as t → ∞, we expect the particle to approach the positive z-axis with slower and slower speed. Ans: D difficulty: easy section: 17.2 27. Find the coordinates of the point where the line tangent to the curve r (t ) =ti + t 2 j + t 3 k at the point (4, 16, 64) crosses the xy-plane. 8 16 Ans: , , 0 3 3 difficulty: medium section: 17.2
Page 6
Chapter 17: Parameterization and Vector Fields
28. A particle moves at a constant speed along a line through P = (7,–14, 13) and Q = (19, –37, 37). Find a parametric equation for the line if: The speed of the particle is 9 units per second and it is moving in the direction of PQ . Ans: r (t ) =i − 2 j + 1k + t (3i − 6 j + 6k ) difficulty: easy section: 17.2 29. A particle moves at a constant speed along a line through P = (10,–20, 22) and Q = (22, –46, 46). Find a parametric equation for the line if the particle passes through P at time t = 3 and passes through Q at time t = 7. Ans: r (t ) =− i 2 j + 4k + t (3i − 6 j + 6k ) difficulty: easy section: 17.2 30. Answer the following as “true”, “false” or “need more information”. If a particle moves with velocity v = 2ti + 3t j + 4tk , then the particle stops at the origin. Ans: Need More Information. difficulty: easy section: 17.2 31. Calculate the length of the curve y = cosh x from x = -3 to x = 3. Ans: e3 − e –3 difficulty: medium
section: 17.2
32. Find a parameterization of the curve y 4 = x5 and use it to calculate the path length of this curve from (0, 0) to (1, 1). Ans: 1.42258 difficulty: medium section: 17.2
Page 7
Chapter 17: Parameterization and Vector Fields
33. The parametric vector form of the position of a roller coaster is r (t ) =30sin(t )i + 30 cos(t ) j + 15cos(t )k . Answer the following questions about the ride. (a) The scariest point of the ride is when it is traveling fastest. For which value of t > 0 does this occur first? (b) Does the velocity vector of the roller coaster ever point directly downward? Ans: (a) We need to determine the speed of the roller coaster at a given time t. The speed is the magnitude of the velocity vector, so Speed = 900 cos 2 t + 900sin 2 t + 225sin 2 t = 900 + 225sin 2 t .
This takes its first maximum value when sin 2(t) takes it first maximum value, so t = π/2. (b) The velocity vector of the roller coaster travels directly downward if both the i and j components are simultaneously equal to 0. Since one is 30cos (t) and the other is -30sin (t), this can never happen. difficulty: easy section: 17.2 34. Let r (t ) =sin(t )i + cos(t ) j + tk and let C be the helix parameterized by r (t ). Find an expression for the outward pointing normal vector whose k component is 0 at an arbitrary point ( sin(t), cos(t), t) of C. Ans: = u (t ) sin(t )i + cos(t ) j difficulty: medium section: 17.2 35. Sketch the vector fields v = xi. A)
B)
C)
Page 8
Chapter 17: Parameterization and Vector Fields
Ans: A
difficulty: easy
section: 17.3
36. For the following vector field, identify which one of the following formulas could represent it. The scales in the x and y directions are the same. No reasons need be given.
A) Ans: C
B) difficulty: medium
C) D) section: 17.3
Page 9
Chapter 17: Parameterization and Vector Fields
37. The figure below shows the contour map of a function z = f(x, y).
Let F be the gradient vector field of f, i.e., F = gradf . Which of the vector fields show F ?
A)
B) Ans: B
C)
difficulty: easy
section: 17.3
38. Write a formula for a vector field F ( x, y ) whose vectors are parallel to the x-axis and point away from the y-axis, with magnitude inversely proportional to the cube of the distance from the x-axis. kx Ans: F ( x, y ) = 3 i y x
difficulty: easy
section: 17.3
Page 10
Chapter 17: Parameterization and Vector Fields
39. Match the vector field F ( x, y ) = x i + y j with the descriptions (a)-(d). A) A swirling in a clockwise direction. B) An attractive force field pointing toward the origin. C) A repulsive force field pointing away from the origin. D) A swirling in a counter-clockwise direction. Ans: C difficulty: easy section: 17.3 40. A vector field F ( x, y ) is shown below.
Find F (0,1). Ans: F (0,1) = –2i + j difficulty: easy section: 17.3 41. Let f(x, y) be a function that depends on only one of the variables, that is, of the form f(x, y) = g(x) or f(x, y) = g(y). Could the following picture be the gradient of f?
A) No Ans: A
B) Yes C) Not possible to say difficulty: medium section: 17.3
Page 11
Chapter 17: Parameterization and Vector Fields
42. Find an equation for a vector field with all of the following properties. • Defined at all points except (1, –5) • All vectors have length 1. • All vectors point away from the point (1, –5). x −1 y+5 Ans: F ( x, y ) = i+ j ( x − 1) 2 + ( y + 5) 2 ( x − 1) 2 + ( y + 5) 2 difficulty: easy section: 17.3 43. Suppose F ( x,= y, z ) –6 x 2 i + 12 y 3 j.
Find a function f(x, y, z) of three variables with the property that the vectors in F on a level surface of f (x, y, z) are perpendicular to the level surface of f ( x, y, z ) at each point.
Ans: f ( x, y, z ) = –2 x3 + 3 y 4 + c. difficulty: easy section: 17.3 44. Let v(t= ) 2i + 2 j be a constant velocity field. Find the flow line of v that passes through the origin at time t = 2. Ans: r (t ) = ( 2t − 4 ) i + ( 2t − 4 ) j. difficulty: easy section: 17.4
45. The vector field F ( x, y= ) xi − y j represents an ocean current. An iceberg is at the point (1, 2) at t = 0. Determine the position of the iceberg at time t = 2. Ans: r (2) = e 2 i + 2e –2 j difficulty: medium section: 17.4 46. The following equations represent a curve or a surface. Select the best geometric description. (Note: ρ, ϕ, θ are spherical coordinates; r, θ, z are cylindrical coordinates.) A) Part of a line through the origin C) B) Part of a cylinder. D) Ans: A difficulty: easy section: 17.5
Part of a cone. Disk
47. Let S be a circular cylinder of radius 0.2, such that the center of one end is at the origin and the center of the other end is at the point (5, 0, 4). Find the xyz-equation of the plane, P, containing the base of the cylinder (i.e., the plane through the origin perpendicular to the axis of the cylinder). 0 Ans: 5 x + 4 z = difficulty: easy section: 17.5
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Chapter 17: Parameterization and Vector Fields
48. Let S be a circular cylinder of radius 0.2, such that the center of one end is at the origin and the center of the other end is at the point (2, 0, 7). Find two unit vectors u and v in the plane, P, containing the base of the cylinder (i.e., the plane through the origin perpendicular to the axis of the cylinder) which are perpendicular to each other. −7i + 2k Ans: = u j= , v , other solutions are possible 53 difficulty: medium section: 17.5 49. Let S be a circular cylinder of radius 0.2, such that the center of one end is at the origin and the center of the other end is at the point (5, 0, 4). Let P be the plane containing the base of the cylinder (i.e., the plane through the origin perpendicular to the axis of the cylinder). In each case, give a parameterization ( x(θ ), y (θ ), z (θ ) ) and specify the range of values your parameters must take on. (i) the circle in which the cylinder, S, cuts the plane, P. (ii) the surface of the cylinder S. Ans: −0.8sin θ z (θ ) = $ x ^ 2 + z ^ 2] (i) y (θ ) 0.2 cos θ = 1sin θ x(θ ) = $ x ^ 2 + z ^ 2] x(θ ) =
0 ≤ θ ≤ 2π.
−0.8sin θ +t $ x ^ 2 + z ^ 2]
(ii) y (θ ) 0.2 cos θ = 1sin θ z (θ ) = +t $ x ^ 2 + z ^ 2] difficulty: medium section: 17.5
0 ≤ θ ≤ 2π, 0 ≤ t ≤ 4.
50. Let v1 = 2i − 1 j + k and v 2 = 1i + j + k Find a parametric equation for the plane through the point (1, 2, -1) and containing the vectors v1 and v 2 . Select all that apply. x =1 + 2t + 1s, y =2 − 1t + s, z =−1 + t + s. A) x =1 + 2t − 1s, y =2 − 1t − s, z =−1 + t − s. B) x =−1 + 2t + 1s, y =−2 − 1t + s, z =1 + t + s. C) x =1 − 2t + 1s, y =2 + 1t + s, z =−1 − t + s. D) Ans: A, B difficulty: easy section: 17.5
Page 13
Chapter 17: Parameterization and Vector Fields
51. Let v1 = 2i − 4 j + k and v 2 = 4i + j + k
Find a vector which is perpendicular to v1 and v 2 to find an equation of the plane through the point (1, 2, -1) and with normal vector perpendicular to both v1 and v 2 . D. Express your answer in the form Ax + By + Cz = –19 ; other solutions are possible. Ans: –5 x + 2 y + 18 z = difficulty: easy section: 17.5
52. Consider the plane r ( s, t ) = (–4 + s − 4t )i + (5 − s + 4t ) j + (6 − 4t − s )k . Does it contain the point (–7, 8, –7)? Ans: Yes difficulty: easy section: 17.5 53. Consider the plane r ( s, t )= (–4 + s − 5t )i + (5 − s + 5t ) j + (2 − 10t + s )k . Find a normal vector to the plane. 1 1 A) D) i+k i+ j+k 2 3 1 1 E) B) i− j i+ j−k 2 3 1 C) i+ j 2 Ans: C difficulty: easy section: 17.5
(
)
(
(
(
)
(
)
)
)
54. Find the parametric equation of the plane through the point (5, 2, 2) and parallel to the lines r (t ) = (1 − 2t )i + (5 + 2t ) j + (3 − 4t )k and s (t ) = (3 − 4t )i + 4t j + (4 − 2t )k . Select all that apply. x =5 − 2u − 4v, y =2 + 2u + 4v, z =2 − 4u − 2v. A) x =5 − 2u − 4v, y =2 + 2u + 4v, z =2 + 4u − 2v. B) x =5 − 2u + 4v, y =2 + 2u − 4v, z =2 − 4u + 2v. C) x= 2u + 4v, z = −2u − 4v, y = −4u − 2v. D) x =5 − 2u − 4v, y =2 + 2u + 4v, z =2 − 4u − 2v. E) Ans: A, C, E difficulty: easy section: 17.5 55. Find parametric equations for the sphere ( x − 6) 2 + ( y + 6) 2 + ( z − 12) 2 = 36. Ans: x =6 + 6sin φ cos θ , y =−6 + 6sin φ sin θ , z =12 + 6 cos φ . difficulty: easy section: 17.5
Page 14
Chapter 17: Parameterization and Vector Fields
56. Using cylindrical coordinates, find parametric equations for the cylinder x 2 + y 2 = 16. Select all that apply. x 4 cos θ= , y 4sin θ= , z z , 0 ≤ θ ≤ 2π, − ∞ < z < ∞. A) = = x 4= cos θ , y 4sin θ , 0 ≤ θ ≤ 2π, − ∞ < z < ∞. B) x 16 cos θ= , y 16sin θ= , z z , 0 ≤ θ ≤ 2π, − ∞ < z < ∞. C) = , y 4 cos θ= , z z , 0 ≤ θ ≤ 2π, − ∞ < z < ∞. x 4sin θ= D) = = x 16= cos θ , y 16sin θ , 0 ≤ θ ≤ 2π, − ∞ < z < ∞. E) Ans: A, D difficulty: easy section: 17.5 57. Find parametric equations for the cylinder y 2 + z 2 = 16. D) A) = r ( s, θ ) 4 cos θ i + 4sin θ j. r ( s, θ ) = si + 4 cos θ j + 4sin θ k . B) E) r ( s, θ ) = 4 cos θ i + 4sin θ j + sk . r ( s, θ ) = si + 4 cos θ j + 4 cos θ k . C) r ( s, θ ) = si + 16 cos θ j + 16sin θ k . Ans: D difficulty: medium section: 17.5 πt πt 58. Consider the parametric surface r ( s, t ) =s sin i + s cos j + 2tk . 2 2 Does it contain the point (0, –3, –4)? Ans: Yes difficulty: medium section: 17.5 πt πt 59. Consider the parametric surface r ( s, t ) =s sin i + s cos j + 4tk . 2 2 Does it contain the y-axis? Ans: Yes difficulty: medium section: 17.5
60. Find parametric equations for the cylinder 49 x 2 + 11= y 2 539, − 7 ≤ z ≤ 11. x 11cos t , = y 7 sin t , = z s, 0 ≤ t ≤ 2π, −7 ≤ s ≤ 11. A) = B)
= x
11 cos t , y= 7 sin t , = z t , 0 ≤ t ≤ 2π, −7 ≤ s ≤ 11.
C)
= x
7 cos t , = y 11sin t , = z s, 0 ≤ t ≤ 2π, −7 ≤ s ≤ 11.
D)
= x
11 cos t , = y 7 sin t , = z s, 0 ≤ t ≤ π, −7 ≤ s ≤ 11.
E) = x Ans: E
11 cos t , = y 7 sin t , = z s, 0 ≤ t ≤ 2π, −7 ≤ s ≤ 11. difficulty: medium section: 17.1
Page 15
Chapter 17: Parameterization and Vector Fields
61. Match the surface with its parameterization below.
A) B) Ans: A
C) difficulty: medium
D) section: 17.5
62. How many parameters are needed to parameterize a surface in 3-space? Ans: 2 difficulty: easy section: 17R 63. Consider the curve r (t ) = (2t 2 + 1)i + (t 2 − 2) j + tk . Does it pass through the point (1, –2, 0)? Ans: Yes difficulty: easy section: 17R 64. Consider the curve r (t ) = (5t 2 + 1)i + (t 2 − 5) j + tk . Find the equation of the tangent line at the point where t = 2. Ans: (21 + 20t )i + (–1 + 4t ) j + (2 + t )k . difficulty: medium section: 17R 65. Consider the curve r (t ) = (2t 2 + 1)i + (t 2 + 1) j + tk . Does the curve lie on the parametric surface x = s2 + t 2 , y = s2 , z = t? Ans: Yes difficulty: medium section: 17R 2 66. Parameterize the circle x 2 + y= 25, = z 15. Select all that apply. = x 5cos = t , y 5sin = t , z 15 = x 5cos = 2t , y 5sin = 2t , z 15 A) D) = x 5sin = t , y 5cos = t , z 15 x 25cos t , y 25sin t , z 15 = = = B) E) = x –5cos = t , y –5sin = t , z 15 C) Ans: A, B, C, D difficulty: easy section: 17R
Page 16
Chapter 17: Parameterization and Vector Fields
67. Parameterize the curve which lies on the plane 5x - 10y + z = 6 above the circle x2 + y 2 = 25. A) r (= t ) 5sin ti + 5cos t j + ( 6 − 5sin t + 10 cos t ) k B) r (= t ) 5sin ti + 5cos t j + ( 6 − 25sin t − 50 cos t ) k C) r (= t ) 5sin ti + 5cos t j + ( 6 + 25sin t + 25cos t ) k D) r (= t ) 5sin ti + 5cos t j + ( 6 − 25sin t + 50 cos t ) k E) r (= t ) 5sin ti + 5cos t j + ( 6 − 5sin t + 25cos t ) k Ans: D difficulty: medium section: 17R 68. Use spherical coordinates to parameterize the part of the sphere x 2 + y 2 + z 2 = 4 above the plane z = 1. = = ϕ cos θ , y 2sin = ϕ sin θ , z 2 cos ϕ , for 0 ≤ ϕ ≤ π/3, 0 ≤ θ ≤ 2π. Ans: x 2sin difficulty: easy section: 17R 69. Use cylindrical coordinates to parameterize the part of the plane x + y - z = 10 inside the cylinder x 2 + y 2 = 4. . Ans: x = r cos θ , y = r sin θ , z = r cos θ + r sin θ − 10, where 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2. difficulty: easy section: 17R 70. A surveyor wants to measure the height of a building. At point A = ( 733, -369, 0) on the ground, she observes that the vector AC is parallel to −3i + 2 j + 6k , where C is the highest point of the building. At point B = (418, 471, 0) she observes that the vector BC is parallel to −3i − 4 j + 12k . Given that ground level is the plane z = 0 and the units are feet, find the height of the building. Ans: 1260 feet. difficulty: medium section: 17R 71. The lines (2t + 1)i + (1 − 3t ) j + (2 + 2t )k and (2t − 3)i + (7 − 2t ) j + (−2 + at )k are perpendicular. Find a. Ans: –5 difficulty: easy section: 17R 72. The curve r (t ) = ati + (t + 4) j + (bt 2 + 6)k passes through the point (–12, 1, 51). Find a and b. Ans: a = 4, b = 5 difficulty: medium section: 17R
Page 17
Chapter 17: Parameterization and Vector Fields
73. Suppose the vector field F ( x, y ) represents an ocean current and that = r (t ) 8cos ti + 9t sin t j is a flow line of F . Find the acceleration vector of the flow line at (0, 9π/2). 9π Ans: − j 2 difficulty: medium section: 17R 74. Let S be the parametric surface r ( s,= t ) s cos ti + s sin tj + ( s 2 − s sin t )k for 0 ≤ s ≤ 2 , 0 ≤ t ≤ 2π . (a) What does the projection of S onto the xy-plane look like? (b) Show that S is part of the surface z = x 2 + y 2 − y . (c) Find a unit vector that is normal to the surface at the point (1,2,3). Ans: (a) The projection of S onto the xy-plane will be r ( s, t ) = s cos ti + s sin tj + 0k for 0 ≤ s ≤ 2 , 0 ≤ t ≤ 2π , which is a disc centered at the origin of radius 2. (b) Since = x s cos t , = y s sin t , = z s 2 − s sin t , we have
x2 + y 2 − y = ( s cos t ) 2 + ( s sin t ) 2 − s sin t =− s 2 s sin t = z. 2i + 3 j − k (c) . 14 difficulty: easy section: 17R 75. Consider the curve r (t ) = (t + 1)i + (2 − t ) j + (2t 2 − t + 3)k , 0 ≤ t ≤ 1 . (a) Find a unit vector tangent to the curve at the point (1,2,3). (b) Show that the curve lies on the surface z = x 2 + y 2 − y . i − j −k Ans: (a) 3 (b) x = t + 1, y = 2 − t , z = 2t 2 − t + 3 , so
x 2 + y 2 − y = (t + 1) 2 + (2 − t ) 2 − (2 − t ) = 2t 2 − t + 3 = z. difficulty: easy section: 17R 76. Find the parametric equations for the line of the intersection of the planes 4 x + y + 4 z = 7 and x + 4 y + 4 z = 1 . 1 – 12t , y (t ) = −1 – 12t , z (t ) = 1+ 15t Ans: x(t ) = difficulty: easy section: 17.1 77. True or False: the two planes are parallel. Plane 1: x = 2 + s + t , y = 4 + s − t , z = 1 + 2 s Plane 2: x = 2 + s + 2t , y = t , z = s − t Ans: False difficulty: easy section: 17.3
Page 18
Chapter 17: Parameterization and Vector Fields
78. Are the lines parallel? l1 : x =2t + 5, y =3t + 3, z =–4t − 2 l2 : x =5t + 1, y =2t − 4, z =11t + 7 Ans: No difficulty: easy section: 17.1 79. Do the lines intersect? l1 : x = –5t + 1, y = –7t + 1, z = 3t l2 : x =12t + 3, y = 7, z = –11t − 6 Ans: No difficulty: medium section: 17.1
80. If a particle is moving along a parameterized curve r (t ) , then the acceleration vector at any point cannot be parallel to the velocity vector at that point. Ans: False difficulty: easy section: 17R 81. If a particle moves with constant speed, the path of the particle must be a line. Ans: False difficulty: easy section: 17R 82. If the flow lines for the vector field F (r ) are all concentric circles centered at the origin, then F (r ) ⋅ r = 0 for all r . Ans: True difficulty: easy section: 17R
83. What curve, C, is traced out by the parameterization r (t ) = 2i + 4 cos tj + 4sin tk for 0 ≤ t ≤ 2π ? Give a very complete verbal description. Ans: As r (t ) = 2i + cos tj + sin tk , the curve is a circle of radius 4 centered at (2, 0,0) in the plane x=2. difficulty: easy section: 17.1
Page 19
Chapter 17: Parameterization and Vector Fields
84. Suppose that r (t ) is the parametric vector form of some curve C in 2-space. (a) Describe geometrically the curve parameterized by −r (t ) compared to the graph of C. (b) Describe geometrically the curve parameterized by r (−t ) compared to the graph of C. (c) Describe geometrically the possibility for C if r (t ) = −r (t ) . Ans: (a) For every point (x, y) on the curve, the point (-x, -y) will be on the curve parameterized by −r (t ) . Since this is true for all points on the curve C, it follows that the curve parameterized by −r (t ) is the rotation of C about the origin through π radians (or 180 degrees). (b) The curve parameterized by r (−t ) is the same curve C though the particle moves backwards along C (when compared to r (t ) ). (c) If we assume that r (t ) = −r (t ) for any value of t, then any point (x, y) on C must be equal to the point (-x, -y). Then only point in 2-space with this property is the the origin, so r (t ) = 0 . difficulty: easy section: 17.1 85. The path of an object moving in xyz-space is given by ( x(t ), y (t ),= z (t )) (4t 2 , 2t + 1, 2t 3 ) The temperature at a point (x,y,z) in space is given by f ( x, y,= z) x2 y − 2 z . (a) At time t = 1 , what is the object's velocity v ? What is its speed? (b) Calculate the directional derivative of f in the direction of v at the point (1,2,1), where v is the velocity vector you found in part (a). d (c) Calculate f ( x(t ), y (t ), z (t )) |t =1 dt (d) Explain briefly how your answers to part (a), (b) and (c) are related. Interpret them in terms of temperature. Ans: (a)Velocity : v = 8i + 2 j + 6k Speed: || v (1) ||= 104 22 (b) 104 (c) 212 d (d) f ( x(t ), y (t ), z (t )) |t= fu ⋅ speed|t=1 That is, the rate of change of temperature =1 dt at the point where the object equals the directional derivative of the temperature in the direction the object is going times the speed of the object. difficulty: easy section: 17.2
Page 20
Chapter 17: Parameterization and Vector Fields
86. In an exam, students are asked to find the arc length of the curve C parameterized by t 2 t 4 2t 3 r (t ) = − i + j + k , for −1 ≤ t ≤ 1 . 3 2 4 One student wrote the following 2 2 " || v (t )=|| (t − t 3 ) 2 + (2t 2 )= t 2 (1 + 2t + t 4= ) t 2 (1 + t 2 )= t (1 + t 2 ) 1
Thus the arc length is ∫ t (1 + t 2 )dt = 0. " −1
This answer cannot be true. (a) Which part of the student's calculation was wrong? (b) Find the correct answer. Ans: (a) Since t is between -1 and 1, t 2 (1 + t 2 ) 2 is not equal to t (1 + t 2 ) . || v (t ) ||= t 2 (1 + t 2 ) 2 =t (1 + t 2 ), when t ≥ 0 || v (t ) ||=t 2 (1 + t 2 ) 2 = −t (1 + t 2 ), when t < 0 . (b) The arc length equals to 0
1
3
. ∫ −t (1 + t ) + ∫ t (1 + t ) = 2 −1
difficulty: easy
2
2
0
section: 17.2
87. Two particles p1 and p2 are moving in the plane, with p1 moving vertically upward from initial point (0,0) and p2 moving around the unit circle x 2 + y 2 = 1 with initial point (0, 1) in the counter-clockwise direction. (a) Choose parameterizations for p1 and p2 so that the two particles collide at the point of intersection of their paths. (b) If the sum of the speeds of these two particles is greater than 3 at the time of collision, then the two particles combine to become one. Will this happen with your chosen parameterization? Ans: (a) There are many possible solutions to this question. As an example, we could (t ) 0,= y 2t / π for p1 and choose the parameterization x= = x(t ) cos( = t ), y sin(t ) for p2. With these parameterizations, the two particles collide when t = π / 2 . (b) The answer to this question depends upon the chosen parameterization from part (a). With the parameterization chosen, the speed of p1 is 2 / π and the speed of p2 is 1 when t = π / 2 . So with this chosen parameterization, the particles do not combine. difficulty: easy section: 17.2
Page 21
1. True or False? A) False Ans: A
∫ F ⋅dr
is zero if the curve is closed.
C
B) Not possible to decide C) True difficulty: easy section: 18.1
2. Given the graph of the vector field, F , shown below, list the following quantities in increasing order:
∫C1 F ⋅ d r,
Ans:
∫C2 F ⋅ d r,
∫C3 F ⋅ d r.
∫C1 F ⋅ d r < ∫C2 F ⋅ d r < ∫C3 F ⋅ d r
difficulty: easy
section: 18.1
3. Let F be the constant vector field 2i − 2 j + k . Calculate the line integral of F along a line segment L of length 9 at an angle π/3 to F . Ans: 13.5 difficulty: easy section: 18.1
4. Suppose that F (0,1)= 4i + j , F (0.5,1) = 4 j , F (1, −1) = 4i + 9 j , and F (1, 0) = 4 j . Estimate the work done by F along the line from (1, 0) to (1, 1). Ans: 2 difficulty: easy section: 18.1
5. If the length of curve C1 is longer than the length of curve C2, then ∫ F ⋅ dr ≥ ∫ F ⋅ dr. C1
A) True Ans: B
B) False difficulty: easy
section: 18.1
Page 1
C2
Chapter 18: Line Integrals
6. Let C be the curve described by r (t ) . If the angle between F (t ) and r '(t ) is less than π/2, then ∫ F ⋅ dr ≥ 0. C
A) True Ans: A
B) False difficulty: easy
section: 18.1
0 , then F is perpendicular to the curve C at every point. 7. If ∫ F ⋅ dr = C
A) It depends on the curve Ans: B difficulty: easy
B) False C) True section: 18.1
8. Let C be a segment 9 units long of the contour f(x, y) = 5. What is the work done by the gradient field of f along C? Ans: 0 difficulty: easy section: 18.1
6 , where C is the circle of radius 1, centered at the origin, 9. Suppose that ∫ F ⋅ d r = C
starting at (1, 0) and traveling counter-clockwise back to (1, 0). Determine whether the following statement is True or False.
2 12 , ∫ F ⋅dr = C1
A) True Ans: A
B) False difficulty: easy
section: 18.1
10. Consider the vector field F = 5 x j . Without using parametrization, calculate directly the line integral of F along the line from (3, 3) to (7, 3). Ans: 0 difficulty: easy section: 18.1
⋅ dr 4, 11. Suppose that ∫ F= C1
⋅ dr 3, ∫ F= ⋅ dr 16 and ∫ F ⋅ dr = –1, where C1 is ∫ F= C2
C3
C4
the line joining (0, 0) to (1, 0), C2 is the line joining (0, 0) to (3, 0), C3 is the line joining (0, 0) to (0, 1) and C4 is the line joining (0, 1) to (0, 2). Determine, if possible, the value of the line integral of F along the line from (0, 1) to (1, 0). If the value cannot be determined, say so. Ans: Cannot be determined. difficulty: easy section: 18.1
Page 2
Chapter 18: Line Integrals
12. Let C1 be the rectangular loop consisting of four line segments: from (0, 0) to (1, 0), then to (1, 2), then to (0, 2), then back to (0, 0). Suppose C2 is the triangular loop joining (0, 0) to (1, 0), then to (1, 2) then back to (0, 0), and C3 is another triangular loop joining (0, 0) to (1, 2), then to (0, 2) and then back to (0, 0). Is it true that
∫ F ⋅ dr= ∫ F ⋅ dr + ∫ F ⋅ dr C2
C3
C1
for any vector field F defined on the xy-plane? A) Not possible to decide B) Yes C) No Ans: C difficulty: easy section: 18.1
13. Calculate ∫ F ⋅ d r when F = ( y + 5 z )i + ( x + 4 z ) j + 4k and C is the line from the origin C
to the point (4, 4, 4). Ans: 104 difficulty: easy section: 18.2
14. Find ∫ F ⋅ d r where F =( x + z )i + 6 z j + 6 yk and C is the line from the point (2, 4, 4) C
to the point (0, 6, –8). Ans: –382 difficulty: easy section: 18.2
Page 3
Chapter 18: Line Integrals
15. Explain in words and symbols how to calculate the line integral ∫ F ⋅ d r given a C parameterization, r = p (t ), of the curve C. Ans: Let
and let C be a curve which connects points A and B.
where t1 and t2 denote the values of t corresponding to points A and B, respectively. difficulty: easy section: 18.2 16. Let F = x 2 + y 2 i + x 2 + y 2 j. Is the line integral of F around the unit circle traversed counterclockwise: positive, negative, or zero? A) Positive B) Negative C) Zero Ans: C difficulty: easy section: 18.2 17. Let F = x 2 + y 2 i + x 2 + y 2 j. For any θ, 0 ≤ θ ≤ 2π, let Cθ be the line segment from (0, 0) to the point (cos θ, sin θ) on the unit circle. Give examples of angles θ for which ∫ F ⋅ d r is positive. Cθ
A) 0 B) π/4 C) π/2 D) -π/4 E) 3π/4 Ans: A, B, C difficulty: medium section: 18.2
Page 4
F) 5π/4
G) π
Chapter 18: Line Integrals
18. Let= F 100i + 10 j and let Ca be the circle of radius a centered at the origin, traveled in a counter-clockwise direction.
Find ∫ F ⋅ d r. Ca
Ans: −90a 2 π difficulty: easy
section: 18.2
19. Let= F 81 yi + 9 x j and let Ca be the circle of radius a centered at the origin, traveled in a counter-clockwise direction.
Find lim a →0 ∫ F ⋅ d r. Ca
Ans: 0 difficulty: easy
section: 18.2
20. Evaluate ∫ –2 xi + (4 x + y ) j , where C is the triangular path from (0, 0) to (1, 1) to (0, –C
1) to (0, 0). Ans: - 2 difficulty: medium
section: 18.2
21. Let F be a vector field with constant magnitude F = 8. Suppose that r (t ), 0 ≤ t ≤ 5, is a parameterization of a flow line C of F . Find ∫ F ⋅ d r. C
Ans: 320 difficulty: easy
section: 18.2
22. Find, by direct computation, the line integral of yi − x j around the circle r (= t ) cos ti + sin t j , 0 ≤ t ≤ 2π. Ans: –2π difficulty: medium section: 18.2
23. Evaluate ∫ xi + ( y + 10 x) j + ( z + 9 x)k ⋅ dr , where C is the curve −C r (t ) = ti + (1 − t ) j + (t 2 + 3)k , for 0 ≤ t ≤ 1. Note that the line integral is around -C, not C. Ans: –4.5 difficulty: easy section: 18.2
Page 5
Chapter 18: Line Integrals
24. Find the work done by the force field F= (12 y + e x − 6sin x)i + 3 x 2 y j along the parabola y = 2x2 from (0, 0) to (1, 2). Ans: e + 5 + 6cos1 difficulty: easy section: 18.2 25. Let C be the circle in space with the parameterization r (= t ) cos ti + sin tk , 0 ≤ t ≤ 2π. 2 Evaluate ∫ F ⋅ d r where = F 2 ze y i + sin( x 2 + y 2 + z 2 ) j + x( y 2 − 1)k . C
Ans: –3π difficulty: medium
section: 18.2
26. Let F = ( 3 x + 2 y ) i + (3 x − 3 y ) j. Evaluate ∫ F ⋅ d r , where C is the line from (0, 0) to C
(2, 2). Ans: 10 difficulty: easy
section: 18.2
27. Let F = ( 3 x + 15 y ) i + (3 x − 3 y ) j. Evaluate ∫ F ⋅ d r , where C is parameterized by C 1 2 r (t ) = ti + 2 t j , 0 ≤ t ≤ 2. Ans: 28 difficulty: easy section: 18.2 28. Let F = ( 4 x + 5 y ) i + (4 x − 4 y ) j. I1 ∫ F ⋅ dr , where C1 is the line from (0, 0) to (2, 2). Let= C1 I 2 ∫ F ⋅ dr , where C2 is parameterized by r (t ) = ti + 12 t 2 j , 0 ≤ t ≤ 2. Let= C2
Notice that both C1 and C2 go from (0, 0) to (2, 2), but is I1 = I 2 ? Explain. Ans: No. Even though the curves C1 and C2 have the same starting and ending points, they follow different paths. Independence of parameterization applies only to different parameterizations of the same oriented path. difficulty: easy section: 18.2
Page 6
Chapter 18: Line Integrals
29. If F = 3 x 2 i + 4 z sin(4 yz ) j + 4 y sin(4 yz )k , compute ∫ F ⋅ d r where C is the curve from C
A(0, 0, 3) to B(2, 1, 5) shown below. Hint: messy computation can be avoided.
Ans: 8.5919 difficulty: easy
section: 18.3
30. True or false? There is a function f(x, y) such that grad = f –3 x 2 +6 y 2 . A) It is not possible to say B) False C) True Ans: B difficulty: easy section: 18.3
31. True or false? A) True Ans: A
∫ F ⋅ d r for any closed curve C and any gradient vector field F . C
B) False difficulty: easy
section: 18.3
32. Explain what is meant by saying a vector field is conservative. A) A vector field F is called conservative if for any two points P and Q, the line integral ∫ F ⋅ d r has the same value along any path C from P to Q lying in the C domain of F . B) A vector field F is called conservative if for any two points P and Q, the line integral ∫ F ⋅ d r has a different value along any path C from P to Q lying in the C domain of F . C) A vector field F is called conservative if for any two points P and Q, the line integral ∫ F ⋅ d r has the same value along a path C from P to Q lying in the C domain of F . D) A vector field F is called conservative if for two specific points P and Q, the line integral ∫ F ⋅ d r has the same value along any path C from P to Q lying in the C domain of F . Ans: A difficulty: easy section: 18.3
Page 7
Chapter 18: Line Integrals
33. Given that= G 3 x 2 i + 10 y 4 j find a function g so that G = grad g .
Use the function g to compute ∫ G ⋅ d r where C is a curve beginning at the point (2, 4) C
and ending at the point (0, 1). Ans: –2054 difficulty: easy section: 18.3 34. Consider the two-dimensional vector field F ( x, y ) = −4 yi + 2 x j. Write down parameterizations of the three line segments C1, C2, and C3 shown in the figure below.
F ⋅ d r , by finding Use your parameterizations to compute the line integral ∫ C3 + C2 −C1 ∫C1 F ⋅ d r , ∫C2 F ⋅ d r and ∫C3 F ⋅ d r. F ⋅dr = –6 . Ans: ∫ C3 + C2 −C1
difficulty: medium
section: 18.3
35. Is the following statement true or false? ∂P ∂Q If F= . = ( x, y ) P( x, y )i + Q( x, y ) j is a gradient vector field, then ∂x ∂y A) False B) True Ans: A difficulty: easy section: 18.3 36. Let F ( x, y ) =( 9 x 2 − 3 y 2 ) i − 6 xy j. Evaluate the line integral
∫ F ⋅ d r , where C the C
path from (0, 0) to (1, 1) that goes along the x-axis to (1, 0), and then vertically up to (1, 1). Ans: 0 difficulty: easy section: 18.3
Page 8
Chapter 18: Line Integrals
37. Let F be a conservative vector field with potential function g satisfying g(0, 0) = –5. Let C1 be the line from (0, 0) to (2, 1), C2 the path parameterized by r (t ) = (2 + 2sin t )i + cos t j , 0 ≤ t ≤ π , and C3 the path parameterized by r (t ) = (t + 2)i + (t 2 − t + 1) j , 0 ≤ t ≤ 2.
r 5, ∫ F ⋅ d= r –3, Suppose that ∫ F ⋅ d= C1
Evaluate g(2, –1). Ans: –3 difficulty: medium
C2
–3. and ∫ F ⋅ d r = C3
section: 18.3
38. Let F = ( y 2 + y sin( xy ))i + (4 xy − y + x sin( xy )) j. Is F path-independent? Ans: No difficulty: medium section: 18.3
39. Let F = grad f , where f ( x, y= ) ln(2 x 2 + 3 y 2 + 1). (a) Evaluate the line integral ∫ F ⋅ dr , where C is the line from (0, 0) to (2, –4). C (b) Do you expect the line integral of F along the parabola y = x2-4x, 0 ≤ x ≤ 2 to equal to the answer to (a)? Ans: (a) ln(57)
(b) Yes difficulty: medium
section: 18.3
40. Find a vector field F with the property that the line integral of F along the line from (0, 0) to (a, b) is 3ab 2 + 5ab, for any numbers a and b. Ans: We can choose F = grad f , where f ( x= , y ) 3 xy 2 + 5 xy . So, explicitly, F = (3 y 2 + 5 y )i + (6 xy + 5 x) j . difficulty: easy section: 18.3
41. Let C be the curve x = 2t + cos t, y = 4t, z = 2 sin t for 0 ≤ t ≤ 3π/2.
Use a potential function to evaluate ∫ F ⋅ d r exactly, where C4 3 4 F =4 x cos(4 yz )i − 4 x z sin(4 yz ) j − 4 x y sin(4 yz )k . Ans: 81π 4 − 1 difficulty: easy section: 18.3
Page 9
Chapter 18: Line Integrals
42. Consider the vector field F = ( Ax + By ) i + ( Bx + Cy ) j for certain constants A, B and C. Use the definition of the line integral to evaluate ∫ F ⋅ d r , where C is the line from (1, C
0) to (11, 1). Ans: 60 A + 11B + difficulty: easy
C 2 section: 18.3
43. Consider the vector field F = ( Ax + By ) i + ( Bx + Cy ) j for certain constants A, B and C. Show that F is path-independent by finding its potential function. Ax 2 Cy 2 Ans: A potential function for F is f ( x, y ) = . Therefore F is + Bxy + 2 2 path-independent. difficulty: easy section: 18.3 44. Consider the vector field F = ( Ax + By ) i + ( Bx + Cy ) j for certain constants A, B and C. Find the line integral ∫ F ⋅ d r , where C1 is the curve C1
π x et (t − 2) + 6t ,= y sin t , 0 ≤ t ≤ 2. = 4 C Ans: 84 A + 13B + 2 difficulty: medium section: 18.3
45. Let F = ( 9 x 2 + 6 y cos( xy ) ) i + ( 2 y + 6 x cos( xy ) ) j .
Calculate ∫ F ⋅ dr , where C is the curve shown below. C
Ans: 48 difficulty: medium
section: 18.3
Page 10
Chapter 18: Line Integrals
46. True or false? 0, where C has the parameterization If F is a path-independent field, then ∫ F ⋅ d r = C r (= t ) cos ti + sin t j , 0 < t < 3π. A) True B) False Ans: B difficulty: medium section: 18.3 47. True or false? If C1 and C2 are two curves with the same starting and ending points, then ∫ F ⋅ dr= ∫ F ⋅ dr , for any vector field F . C1
C2
A) True Ans: B
B) False difficulty: medium
section: 18.3
48. The following table gives values of a function f(x, y). The table reflects the properties of the function, which is differentiable and defined for all (x, y).
Let F = ∇f and
G = 6F.
Find ∫ G ⋅ d r if C consists of line segments connecting (1,1), (1,10), (5,6), and (9,8) in C
that order. Explain your reasoning. Ans: 228 F is a gradient field, and therefore path independent. G is also path independent so we can integrate from (1,1) to (9,8) directly. difficulty: easy section: 18.3
Page 11
Chapter 18: Line Integrals
49. The following table gives values of a function f(x, y). The table reflects the properties of the function, which is differentiable and defined for all (x, y).
Let F = ∇f and
Find ∫ H ⋅ dr C
H = 7F.
if C is the circle of radius 2 centered at (4, 3).
Ans: 0 difficulty: easy
section: 18.3
50. Is the following vector field is a gradient vector field? F = yi+ x j A) Yes B) No Ans: A difficulty: easy section: 18.4 51. Let F and G be two 2-dimensional fields, where = F 3 xi + 5 y j and = G 3 yi + 5 x j. Let C1 be the circle with center (2, 2) and radius 1 oriented counterclockwise. Let C2 be the path consisting of the straight line segments from (0, 4) to (0, 1) and from (0, 1) to (3, 1).
Find the line integral ∫ F ⋅ d r. C2
Ans: –24 difficulty: medium
Use "pi" to represent π if necessary.
section: 18.4
52. Let F ( x, y ) be the vector field= F (6 x 2 + 4 x) j. Find ∫ F ⋅ d r , where C is the line C
from (0, 0) to (4, 4). Ans: 160 difficulty: medium
section: 18.4
Page 12
Chapter 18: Line Integrals
53. Let F ( x, y ) be the vector field = F (12 x 2 + 2 x) j.
Find ∫ F ⋅ d r , using Green's theorem, if C is the unit circle traveled counterclockwise. C
Ans: 2π difficulty: medium
section: 18.4
54. Which of the two vector fields shown below is not conservative? (A)
(B)
Ans: (B) is not conservative as it rotates around the origin, so its line integral around any circle centered at the origin will be non-zero. difficulty: easy section: 18.4 2 55. Use Green's Theorem to evaluate ∫ (cos x 2 + e x + 3 y )i + (e 2sin y + 4 x + cos y 4 )k ⋅ dr , C
where C is the circle of radius
(
3, centered at ( −1, e10 , ln13) oriented in a
)
counter-clockwise direction. Ans: 3π difficulty: medium section: 18.4 56. Use Green's Theorem to calculate the circulation of = F –5 y 3 i + 4 x3 y j around the triangle with vertices (0, 0), (1, 0) and (0, 1), oriented counter-clockwise. 1 Ans: − 5 difficulty: easy section: 18.4
57. Let F =
∂F1 ∂F2 y x Check that . Is − i j . = F is ∂y ∂x 4 x 2 + 4 xy + y 2 4 x 2 + 4 xy + y 2 path-independent? ∂F1 ∂F2 Ans: Even though , we cannot conclude that F is path-independent because = ∂y ∂x there is a hole in the domain at (0,0). difficulty: easy section: 18.4
Page 13
Chapter 18: Line Integrals
58. Let F1 ( x, y ) = y 5 − 10 xy + cos y. Find a function F2 ( x, y ) , such that = F F1 i + F2 j is a gradient field. Ans: F2 ( x, y ) = 5 xy 4 − 5 x 2 − x cos y + g ( x), where g(x) is an arbitrary function of x. difficulty: easy section: 18.4
59. Let F = x j. Let C1 be the line from (0, 0) to (2, 0), C2 the line from (2, 0) to (2,-1), C3 the line from (2,-1) to (0,-1), and C4 the line from (0,-1) to (0, 0). (A) Using the definition of line integral only, without parameterizing the curves, show that the line integral of F along C = C1 + C2 + C3 + C4 is -2. That is, show −2. ∫ F ⋅ dr = C
(B) The rectangle, R, enclosed by the lines C1, C2, C3 and C4 is of area 2. So, by Green's Theorem = ∫ F ⋅ dr C
∂x
= Area of = R 2. ∫ ∂x − 0 dA R
Is something wrong? Ans: (A) Along paths C1 and C3, the vector field is perpendicular to the path. So the line integral along each of these paths is zero. Also, on path C4, the magnitude of the vector field is zero. So the line integral along this path is also zero. Finally, along C2, we have F = 2 j and the vector field points exactly opposite to the −(2)(1) = −2 . direction of the path. Therefore, ∫ F ⋅ dr = ∫ F ⋅ dr = C
C2
(B) We cannot apply Green's Theorem to the curve C. It does not satisfy the orientation condition of the Theorem since the region has to be on the left as we move around the curve. This explains the difference in sign between the two answers. If the orientation of the curve C were reversed, then Green's Theorem would apply and both answers would be equal. difficulty: medium section: 18.4 60. Determine which of the following vector fields has the property that its line integral over 2 2 any circle ( x − a ) + ( y − b ) = r 2 is zero. (A) F =2sin y i + (3 x cos y + 3cos y ) j 2 2 (B) G 3 xye x y i + x 2 e x y j = Part A: F does not have the required property. Part B: G does not have the required property. difficulty: easy section: 18.4
Page 14
Chapter 18: Line Integrals
61. Let C be the circular path which is the portion of the circle of radius 1 centered at the origin starting at (1, 0) and ending at (0,–1), oriented counterclockwise.
Determine the exact value of ∫ G ⋅ d r.
Let G = x 6 y 3 i + ( x 3 y 6 + 6 x ) j.
9 π 2 difficulty: medium
C
Ans:
section: 18.4
62. Use Green's Theorem to find the line integral of F =(– x − 4 y )i + x 2 j around the closed curve composed of the graph of y = x2n where n is a positive integer and the line y = 1. 16n Ans: 2n + 1 difficulty: medium section: 18.4 63. Find the line integral of F = ( 3 x − 2 y ) i + ( 3 x + 5 y ) j around the curve consisting of the graph of y = xn from the origin to the point (1, 1), followed by straight lines from (1, 1) to (0, 1) and from (0, 1) back to the origin. 5n Ans: n +1 difficulty: medium section: 18.4
64. Let F = −3 y i + 3 x j and let Ca be the circle x2 + y2 = a2 traversed counterclockwise.
Find ∫ F ⋅ d r. Ca
Ans: 6a 2 π difficulty: easy
section: 18R
65. Let F be the vector field shown below.
Let C be the rectangular loop from (0, 0) to (1, 0) to (1, 1) to (0, 1), then back to (0, 0).
Do you expect the line integral ∫ F ⋅ d r to be positive, negative or zero? C
A) Positive B) Negative Ans: B difficulty: medium
section: 18R
Page 15
Chapter 18: Line Integrals
66. Let F = (4 x3 − y 6 sin x)i + (6 y 5 cos x + 4 z ) j + 4 yk . Show that F is a gradient field by finding its potential function. Ans: f ( x, y, z ) = x 4 + y 6 cos x + 4 zy difficulty: easy section: 18R
67. Let= F (3 x 2 + y 4 sin x)i + (4 y 3 cos x + z ) j – yk . Is the value of the line integral of F along any loop zero? A) Yes; the function is a gradient vector field. B) No; the function is not a gradient vector field. Ans: B difficulty: easy section: 18R 68. F = (3 x 2 − y 4 sin x)i + (4 y 3 cos x + 2 z ) j + 2 yk is a gradient vector field. Is G= F + 2 z 4 i path independent? A) Yes B) It depends on the path selected C) No Ans: C difficulty: easy section: 18R 69. Let F = x3 i + ( x + sin 3 y ) j.
(a)
Find the line integral ∫ F ⋅ d r , where C1 is the line from (0, 0) to (π, 0).
(b)
Evaluate the double integral ∫ 1 dA, where R is the region enclosed by the curve y
C1
R
= sin x and the x-axis for 0 ≤ x ≤ π. What is the geometric meaning of this integral? (c) Use Green's Theorem and the result of part (a) to find ∫ F ⋅ d r , where C2 is the C2
path from (0, 0) to (π, 0) along the curve y = sin x. π4 Ans: (a) 4 (b) 2.
The area of the region is 2.
π4 +2 4 difficulty: medium
(c)
section: 18R
2 2 2 2 70. Let F = 2 + 6 xe3( x + y ) i + 6 ye3( x + y ) j. Use the curl test to check whether F is path-independent. A) Path-independent B) Not path-independent Ans: A difficulty: easy section: 18R
(
) (
)
Page 16
Chapter 18: Line Integrals
2 2 2 2 71. Let F = 2 + 8 xe 4( x + y ) i + 8 ye 4( x + y ) j. Find a potential function for F . 2 2 Ans: f ( x, y= ) 2 x + e 4( x + y ) difficulty: easy section: 18R
(
) (
)
2 2 2 2 72. Let F = 2 + 4 xe 2( x + y ) i + 4 ye 2( x + y ) j. Find the value of ∫ F ⋅ d r , where C is a path joining (0, 0) to the point (1, 2).
(
) (
)
C
Ans: 22,027.466 difficulty: easy section: 18R 73. Let f(x, y, z) be a function of three variables. Suppose that C is an oriented curve lying on the level surface f(x, y, z) = 2.
Find the line integral ∫ grad f ⋅ dr. C
Ans: 0 difficulty: easy
section: 18R
74. Let Ca be the circle x2 + y2 = a2 oriented counter-clockwise. Use Green's theorem to find ∫ ( x 3 − x 2 y )i + ( xy 2 + y 3 ) j ⋅ dr Ca
a4π 2 difficulty: easy
(
)
Ans:
section: 18R
75. Calculate the line integral of F =− ( 3 y + x)i + 2 x j along the straight line from (3, –3) to (3, 0). Ans: 18 difficulty: medium section: 18R 76. Calculate the line integral of F =− ( 9 y + x)i + 2 x j along a quarter of a circle centered at the origin, starting at (3, 0) and ending at (0, -3). 99 9 Ans: – π 4 2 difficulty: medium section: 18R
Page 17
Chapter 18: Line Integrals
77. Let F =
x 2 + y 2 i + x 2 + y 2 j.
For a fixed θ , let Cθ be the line segment from (0, 0) to the point (cos θ, sin θ) on the unit circle. Find a parameterization of Cθ and compute ∫ F ⋅ d r . (Your answer will depend on Cθ
θ .) Ans:
1
⋅ dr (cos θ + sin θ ) ∫ F= 2 Cθ
difficulty: easy
section: 18.2
78. Suppose a curve C is parameterized by r (t ) with a ≤ t ≤ b and suppose F is a vector 0. field F = (t ) r (t ) × r '(t ) for a ≤ t ≤ b . Explain why ∫ F ⋅ dr = C Ans: Since r (t ) is parametrization of C, b b F ⋅ dr = ∫C ∫a F (r (t )) ⋅ r '(t )dt = ∫a (r (t ) × r '(t )) ⋅ r '(t )dt = 0 since r (t ) × r '(t ) and r '(t ) are perpendicular. difficulty: easy section: 18.2
79. Let C be the unit circle x 2 + y 2 = 1 oriented in a counter-clockwise direction. Say whether the following statements are true or false. 0 , we can conclude that F is path-independent field. If ∫ F ⋅ dr = C
Ans: False
difficulty: easy
section: 18.3
80. Let C be the unit circle x 2 + y 2 = 1 oriented in a counter-clockwise direction. Say whether the following statements are true or false. If ∫ F ⋅ dr ≠ 0 , we can conclude that F is C
not path-independent field. Ans: True difficulty: easy
section: 18.3
81. Answer true or false, giving a reason for your answer. Let φ ( x, y ) = ln(| xy |) so that 1 1 ∇φ = i+ j. x y dr φ (1,1) − φ (−1,1) where C is given by the parametrization Then ∫ ∇φ ⋅= C
= x t ,= y 1 for −1 ≤ t ≤ 1 . Ans: The answer is false since the field is not continuous on the path C. difficulty: easy section: 18.3
Page 18
Chapter 18: Line Integrals
82. On an exam, students were asked to evaluate ∫ x 2 i + xj ⋅ dr , where C has the C
(
)
x cos= t , y sin t , 0 ≤ t ≤ π . One student wrote: parameterization= "Using Green's Theorem, ∂x ∂ ( x 2 ) π 2 x i xj dr + ⋅ = ) ∫C ( ∫D ∂x − ∂y dA= ∫D 1dA= Area of the semi-circle = 2 ." Do you agree with the student? Ans: This is wrong. We cannot use Green's Theorem because the curve C is only the upper part of the circle and does not enclose any region. difficulty: easy section: 18.4
− yi + xj 83. On an exam, students were asked to evaluate ∫ 2 ⋅ dr , where C is the circle 2 C x +y t , y r sin t , 0 ≤ t ≤ 2π . One student wrote: = centered at the origin of radius= r: x r cos "Since ∂ − y y 2 − x2 ∂ x y 2 − x2 = , = ∂y x 2 + y 2 x 2 + y 2 ∂x x 2 + y 2 x 2 + y 2
Using Green's Theorem, − yi + xj = ∫ 0 dA = 0 ." ∫C x 2 + y 2 ⋅ dr D Do you agree with the student? Ans: The answer is wrong. Since the vector field F is not defined at the origin (0,0), which is inside the region D enclosed by C, Green's Theorem cannot be applied. difficulty: easy section: 18.4 84. State the Fundamental Theorem of Calculus for Line Integrals. A) Suppose C is a piece-wise smooth oriented path with starting point P and end point Q. Then ∫ grad f ⋅ dr= f (Q) − f ( P). C
B)
Suppose C is a oriented path with starting point P and end point Q. Then ∫ grad f ⋅ dr= f (Q) − f ( P).
C)
Suppose C is a piece-wise smooth oriented path with starting point P and end point Q. If f is a function whose gradient is continuous on the path C, then ∫ grad f ⋅ dr= f (Q) − f ( P).
D)
Suppose C is a piece-wise smooth oriented path with starting point P and end point Q. If f is a function whose gradient is continuous on the path C, then ∫ grad f ⋅ dr= f ( P) − f (Q).
C
C
C
Ans: C
difficulty: easy
section: 18.3
Page 19
1. True or false?
∫ F ⋅ d A is a vector. S
A) False Ans: A
B) True difficulty: easy
section: 19.1
2. A circular disk, S, of radius 2 and centered on an axis, is perpendicular to the y-axis at y = –6 with normal in the direction of decreasing y. Consider the vector field F = xi + y j + ( z + x)k . Is the flux integral ∫ F ⋅ dA positive, S
negative or zero? A) Positive B) Zero C) Negative Ans: A difficulty: easy section: 19.1 3. Compute the flux integral of the vector field G = −6 yi + 2 x j + 6 xyzk through the square 0 ≤ y ≤ 2, 0 ≤ z ≤ 2, in the yz-plane, oriented so that the normal vector points in the direction of the x-axis. Ans: –24 difficulty: easy section: 19.1 4. What is the flux of the vector field F =−5i + 6 j + 2k through a circle in the xy-plane of radius 2 oriented upward with center at the origin? Ans: 8π difficulty: easy section: 19.1
5. Let S be the sphere of radius 4 centered at the origin, oriented outward. Find n , the unit normal vector to S in the direction of orientation. x y z A) n= i+ j+ k 4 4 4 B) n =xi + y j + zk x2 y2 z2 C) n= i+ j+ k x2 + y 2 + z 2 x2 + y 2 + z 2 x2 + y 2 + z 2 x y z D) n= i+ j+ k 16 16 16 Ans: A difficulty: easy section: 19.1 6. Let S be the sphere of radius 3 centered at the origin, oriented outward. Suppose F is normal to n at every point of S. Find the flux of F out of S. Ans: 0 difficulty: easy section: 19.1
Page 1
Chapter 19: Flux Integrals and Divergence
7. Let S be the sphere of radius 6 centered at the origin, oriented outward. Let G be a vector field such that G =3 xi + 3 y j + 3 zk at every point of S. Find the flux of G out of S. Ans: 1296π difficulty: easy section: 19.1 8. Let G be the constant vector field ai + b j + ck . 0 for any surface S lying on the plane Find a condition on a, b and c such that ∫ F ⋅ dA = S
–5x + 3y - 2z = 1. 0 Ans: –5a + 3b − 2c = difficulty: easy section: 19.1 9. Let S be an oriented surface with surface area 6. Suppose F = ai + b j + ck is a constant vector field with magnitude 3. If the angle between F and n is π/6 at each point of the surface S, determine the value of the flux integral ∫ F ⋅ dA . S
Ans: 9 3 difficulty: easy
section: 19.1
10. Let F = –2i − 5 j + 4k , and let S1 be a horizontal rectangle with corners at (0,0,1), (0,2,1), (3,0,1) and (3,2,1), oriented upward; S2 a rectangle parallel to the xz-plane, with corners at (1,3,1), (2,3,1), (1,3,5) and (2,3,5), oriented in the positive y-direction, and S3 a rectangle parallel to the yz-plane, with corners at (1,2,1), (1,4,1), (1,2,5) and (1,4,5), oriented in the negative x-direction.
Arrange ∫ F ⋅ dA , ∫ F ⋅ dA and ∫ F ⋅ dA in ascending order. S3 S S 2 1 Ans: ∫ F ⋅ dA < ∫ F ⋅ dA < ∫ F ⋅ dA S2
S3
S1
difficulty: medium
section: 19.1
11. True or False? If the surface area of surface S1 is larger than the surface area of surface S2, then ∫ F ⋅ d A ≥ ∫ F ⋅ d A. . S1
S2
A) False B) True Ans: A difficulty: easy
section: 19.1
12. True or False? Let n be the unit normal vector of S. If the angle between F and n is less than π/2 at each point of the surface, then ∫ F ⋅ dA ≥ 0. S
A) False Ans: B
B) True difficulty: easy
section: 19.1
Page 2
Chapter 19: Flux Integrals and Divergence
0 , then F is perpendicular to the surface S at every 13. True or False? If ∫ F ⋅ dA = S
point. A) True B) False Ans: B difficulty: easy
section: 19.1
14. Calculate the flux of F =3i + 5 j − 6k through a surface of area 3 lying in the plane 2x + 5y + 5z = 10, oriented away from the origin. 1 Ans: 54 18 difficulty: medium section: 19.1 15. Let F = axi + by j + czk , where a, b and c are constants. Suppose that the flux of F through a surface of area 3 lying in the plane y = 3, oriented in the positive y-direction, is 45. Find the flux of F through a surface of area 4 lying in the plane y = 3, oriented in the negative y-direction. Ans: –60 difficulty: medium section: 19.1
16. Suppose S is a disk of radius 2 in the plane x + z = 0 centered at (0, 0, 0) oriented “upward”. Write down an area vector for the surface S. i+k Ans: A = 4π 2 difficulty: easy section: 19.1
(
)
17. Suppose S is a disk of radius 4 in the plane x + z = 0 centered at (0, 0, 0) oriented “upward”. Calculate the flux of F = 4i + 5 j − 5k through S. Ans: –8π 2 difficulty: easy
section: 19.1
18. Suppose S is a disk of radius 2 in the plane x + z = 0 centered at (0, 0, 0) oriented “upward”. Calculate the flux of G = ( e –2 xy z secy ) j through S. Ans: 0 difficulty: easy
section: 19.1
Page 3
Chapter 19: Flux Integrals and Divergence
19. Explain why it is impossible to give the Möbius strip a continuous orientation. Ans: A continuous orientation of a surface is a choice of normal vector at every point of the surface with the property that a “small change” in distance on the surface results in a “small change” in the normal vector on the surface. Assume that the Möbius strip M is orientable. Fix a point P on M and let C be a closed curve based at P on M which wraps once around M. Since the orientation is continuous on M, as we move along C, the chosen normal vector at any point cannot “jump” to the other side of the strip. However, once we return to P, the normal vector points in the other direction to the original one. This is absurd, and so the assumption that M was orientable must be incorrect. difficulty: hard section: 19.1 20. Compute the flux of the vector field F =( x − 4 y )i + (4 y − 2 z ) j + 5 xk through the surface S, where S is the part of the plane z = x + 2y above the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3, oriented upward. Ans: 84 difficulty: medium section: 19.2 21. Compute the flux of the vector field F =( x − 3 y )i + (4 y − 2 z ) j + 4 xk through the surface S, where S is the part of the plane z = x + 2y above the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3, oriented downward. Ans: –69 difficulty: medium section: 19.2 22. Compute the flux of the vector field F =( x − 4 y )i + (4 y − 2 z ) j + 6 xk through the surface S, where S is the part of the plane z = x + 2y above the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3, oriented upward. What is the answer if the plane is oriented downward? Ans: 90,–90 difficulty: medium section: 19.2 23. Compute the flux of the vector field F = yi + ( x + y ) j + zk through the surface S that is the part of the surface = z x 2 − y 2 above the disk x 2 + y 2 ≤ 1 , oriented in the positive z-direction. π Ans: 2 difficulty: medium section: 19.2
Page 4
Chapter 19: Flux Integrals and Divergence
24. Suppose the surface S is the part of the surface x = g(y, z), for points (y, z) belonging to a region R in the yz-plane. If S is oriented in the positive x-direction, what will be the formula for computing the flux of F through S? Ans: The normal vector of the surface x = g(y, z) oriented in the positive x-direction is given by ∇( x − g ( y, z )) =i − g y j − g z k . Therefore, one would expect = F ⋅ d A F ( g ( y , z ), y , z ) ⋅ i − g j − g k dydz. y z ∫ ∫ S
difficulty: medium
R
(
)
section: 19.2
25. Compute the flux of F = xz 2 i + yz 2 j + 4 z 5 k through the cylindrical surface 2 x 2 + y= 16, 0 ≤ z ≤ 5, oriented away from the z-axis. 4000 Ans: π 3 difficulty: medium section: 19.2
26. A greenhouse is in the shape of the graph z = 36 − x 2 − y 2 with the floor at z = 0. Suppose the temperature around the greenhouse is given by T = 3 x 2 + 3 y 2 + ( z − 6) 2 . Let H = −∇T be the heat flux density field. Calculate the total heat flux outward across the boundary wall of the greenhouse. Ans: –8640π difficulty: medium section: 19.2 27. (a) Compute the flux of the vector field F = yi − x j + 5( x 2 + y 2 + z 2 ) zk through Sa, the sphere of radius a, x 2 + y 2 + z 2 =, a 2 oriented outward. (b)
Find lim a →0 ∫ F ⋅ d A .
Ans: (a)
Sa
20 5 aπ 3
(b) 0 difficulty: medium
section: 19.2
28. Let S be the part of the sphere x 2 + y 2 + z 2 = 25 with x ≥ 0, y ≥ 0, z ≥ 0, oriented outward. Evaluate ∫ xzi + 4 yz j ⋅ dA . S
3125 π 16 difficulty: medium
(
)
Ans:
section: 19.2
Page 5
Chapter 19: Flux Integrals and Divergence
29. Evaluate ∫ F ⋅ dA , where F= xi + 6 y j and S is the part of the cylinder x 2 + y 2 = 4 S
with x ≥ 0, y ≥ 0, 0 ≤ z ≤ 2, oriented toward the z-axis. Ans: –14π difficulty: easy section: 19.2 30. Let F= xi + y j . Write down an iterated integral that computes the flux of F through S, where S is the part of the surface = z x 2 + y 2 below the plane z = 16, oriented downward.
∫ ∫
B)
−∫
C)
16 − x 2
4
A)
−4 − 16 − x 4
∫
2 ( x 2 + y 2 ) dydx 2
D)
2 ( x 2 + y 2 ) dxdy
E)
16 − x 2
−4 − 16 − x
4
16 − x 2
0
− 16 − x 2
∫∫
Ans: A
2
−∫
4
∫
16 − x 2
−4 − 16 − x 2
16 − x 2
4
∫∫ 0
0
( x + y ) dydx 2
2
( x + y ) dydx 2
2
2 ( x 2 + y 2 ) dydx
difficulty: easy
section: 19.2
31. Let F= yi + x j . Write down an iterated integral that computes the flux of F through S, where S is the part of the cylinder x 2 + y 2 = 9 with x ≥ 0, y ≥ 0, 0 ≤ z ≤ 4, bounded between the planes y = 0 and y = x, oriented outward. 4
A)
−2 ∫ ∫
B)
2∫ ∫
C)
2∫ ∫
0
4
π/4
0 π/2
0 0 4 π/4 0
0
Ans: C
4
9 cos θ sin θ dθ dz
D)
−2 ∫ ∫
9 cos θ sin θ dθ dz
E)
2∫ ∫
0
π/4
4
0 π/4
0
0
9 cos θ sin θ dzdθ 3cos θ sin θ dθ dz
9 cos θ sin θ dθ dz
difficulty: easy
section: 19.2
32. Let F = xzi + yz j + 9 x 2 y 2 k . Calculate the flux of F through the surface oriented upward and given by z = f(x, y) = xy, over the region in the xy-plane bounded by the curves y = x 2 and y = x 3 between the origin and the point (1, 1). 7 Ans: 108 difficulty: easy section: 19.2
= –4 ∫ G ⋅ dA . 33. True or False? If F = –4G and S is an oriented surface, then ∫ F ⋅ dA S
Ans: True
difficulty: easy
S
section: 19.2
34. True or False? If F is a constant vector field and S1 and S2 are oriented rectangles with areas 1 and 2 respectively, then ∫ F ⋅ dA= 2 ∫ F ⋅ dA . S2
Ans: False
difficulty: easy
section: 19.2
Page 6
S1
Chapter 19: Flux Integrals and Divergence
35. If F is a constant vector field and S1 and S2 are oriented rectangles with areas 3 and 15 respectively, both orientated with the vector i, is it true that ∫ F ⋅ dA= 5∫ F ⋅ dA ? S2
A) True Ans: B
B) False difficulty: easy
S1
section: 19.2
36. Find the flux of = F ( x 2 + z 2 ) j through the disk of radius 5 in the xz-plane, centered at the origin, and oriented upward. Give an exact answer. 625 Ans: π 2 difficulty: easy section: 19.2
37. Let S be the spherical region of radius R with π/3 ≤ ϕ ≤ 2π/3 and π/3 ≤ θ ≤ 2π/3. Find the 40π. value of R so that ∫ r ⋅dA = S
Give your answer to two decimal places. Ans: 4.93 difficulty: medium section: 19.2 38. Let C be the portion of the cylinder x 2 + y 2 = R 2 of fixed radius R with π/3 ≤ θ ≤ 2π/3 and -a ≤ z ≤ a oriented outward for some positive number a. Let S be the portion of the sphere x 2 + y 2 + z 2 = R 2 with π/3 ≤ θ ≤ 2π/3 and π/3 ≤ ϕ ≤ 2π/3 oriented outward. Determine the value of a for which the flux of r through each of these surfaces is equal in magnitude but opposite in sign for any choice of R. Ans: –0.5 difficulty: easy section: 19.2 39. Calculate ∫ 5 xi + 3 y j ⋅ dA where C is a cylinder of radius R with 0 ≤ z ≤ 1. C
A) 8πR 4 Ans: C
(
)
B) 2πR 3 C) 8πR 2 D) 8πR 3 difficulty: easy section: 19.2
Page 7
E) 8R 3
Chapter 19: Flux Integrals and Divergence
40. Let F = ai + b j + ck be a constant vector field and S be an oriented surface. Show that ∫ F ⋅ dA ≤ a 2 + b 2 + c 2 ( Area of S ) . S Ans: Let n be the unit normal vector on S.
)
(
F ⋅ ∆A = F ⋅ n∆A = F ⋅ n cos θ∆A ≤
∫ F ⋅ dA ≤ ∫ S
difficulty: hard
S
( a + b + c ) (1)(1)∆A 2
2
2
( a + b + c ) dA ≤ ( a + b + c ) ∫ dA= ( a + b + c ) ( Area of S ) . 2
2
2
2
2
2
2
2
2
S
section: 19R
41. Find the flux of F =xi + y j + 8 zk over the sphere Sa, x 2 + y 2 + z 2 =, a 2 oriented outward, with a > 0. 40 3 40 2 40 3 A) B) 40a 3 C) D) aπ a π a 3 3 3 Ans: A difficulty: easy section: 19R 42. Let F be a constant vector field with F = ai + b j + ck , where a, b, c are constants satisfying the condition a 2 + b 2 + c 2 = 1 . Let S be a surface lying on the plane x + 4y - 5z = 10 oriented upward. If the surface area of S is 10, what is the smallest possible value of ∫ F ⋅ dA , and what S
are the corresponding values of a, b, c? 1 4 5 Ans: –10; = a = ,b = ,c – 42 42 42 difficulty: medium section: 19R 43. Calculate the flux of F = 4i + 2 j − 5 xk through a disk of radius 5 in the plane x = 3, oriented away from the origin. Ans: 100π difficulty: easy section: 19R
44. Calculate the flux of F =5i + 5 j − 2 xk through the plane rectangle y = 1, 0 ≤ x ≤ 1, 0 ≤ z ≤ 3, oriented in the negative y-direction. Ans: –15 difficulty: easy section: 19R
Page 8
Chapter 19: Flux Integrals and Divergence
45. Calculate the flux of F =5i + 2 j − 2 xk through the plane rectangle z= 1, 0 ≤ x ≤ 5, 0 ≤ y ≤ 2, oriented downward. Ans: 50 difficulty: easy section: 19R 46. Calculate the flux of = F ( x 2 + z 2 ) yi , through the plane rectangle z = 3, 0 ≤ x ≤ 2, 0 ≤ y ≤ 5, oriented in the positive z-direction. Ans: 0 difficulty: easy section: 19R 47. Calculate the flux of = F ( x 2 + z 2 ) y j , through the plane rectangle y = 4, 0 ≤ x ≤ 4, 0 ≤ z ≤ 5, oriented in the positive y-direction. 3280 Ans: 3 difficulty: easy section: 19R
48. Calculate the flux of = F ( x 2 + z 2 ) y j through the disk x 2 + z 2 ≤ 25 on the plane y = 5 , oriented in the positive y-direction. 3125 Ans: π 2 difficulty: easy section: 19R 2 49. Let S be the cylinder x 2 + y= a 2 , 0 ≤ z ≤ h . Which of the following are outward pointing normal vectors to the cylinder? Select all that apply. x y A) D) i+ 2 j xi + yj a2 a E) B) xi + yj − k − yi + xj x y C) F) i+ j yi − xj a a Ans: A, C, D difficulty: easy section: 19.1
50. Let S be the cylinder x 2 + y 2= 4, 0 ≤ z ≤ 6 . Find Q if
x
y
Qπ . ∫ x + y i + x + y j + k ⋅ dA = 2
2
2
Ans: Q = 24. difficulty: medium S
2
section: 19.1
Page 9
Chapter 19: Flux Integrals and Divergence
51. Let S be the portion of the parabolic bowl = z f ( x, y= ) x 2 + y 2 above x 2 + y 2 ≤ 1 oriented upward. The following vector fields have i and j components which are parallel to the contours of f. Rank them from smallest to largest (1, 2 and 3, with 1 = smallest, etc.) according to the flux of each one through S. A) F = − yi + x j B) F = − yi + x j – k C) F = − yi + x j + k Part A: 2 Part B: 1 Part C: 3 difficulty: easy section: 19.2 52. Let S be the portion of the parabolic bowl = z f ( x, y= ) x 2 + y 2 above x 2 + y 2 ≤ 1 oriented upward. The following vector field has i and j components which are perpendicular to the contours of f. Which best describes its flux through S? F= xi + y j A) positive B) negative Ans: B difficulty: easy
C) zero section: 19.2
53. Let S be the portion of the parabolic bowl = ) x 2 + y 2 above x 2 + y 2 ≤ 1 z f ( x, y= oriented upward. The following vector field has i and j components which are perpendicular to the contours of f. Which best describes its flux through S? F= − xi − y j + k A) positive B) negative Ans: A difficulty: easy
C) zero section: 19.2
54. Suppose that S is the surface which is a portion of the graph of a smooth function z = f ( x, y ) over a region R in the xy-plane, oriented upward. Consider the vector field F = f x ( x, y )i + f y ( x, y ) j + g ( x, y , z ) k . 0. Find g ( x, y, z ) so that ∫ F ⋅ dA = S
Ans: g= ( x, y , z ) difficulty: hard
( f ( x, y ) + f ( x, y ) ) . 2
x
2
y
section: 19.2
Page 10
Chapter 19: Flux Integrals and Divergence
55. Let F =
xi + yj
(x + y ) 2
2 3/ 2
.
2 True or False: The flux of F through any cylinder C (with x 2 + y= R 2 , 0 ≤ z ≤ 1 ) does not depend on the radius R. Ans: True difficulty: medium section: 19.2 3r 5r 56. The flux of F = 5 − 3 , r ≠ 0 out of the sphere of radius R centered at the origin can r r
be written as
π
( A − BR 2 ) . Find A and B. R2 Part A: 12 Part B: 20 difficulty: easy section: 19R 57. Suppose T is the triangle with vertices (1, 0, 0), (0, 2, 0) and (0, 0, 2) oriented upward. Calculate the flux of F e8 x + 4 y + 4 z i − 5 j through T exactly, and then give an answer = rounded to 3 decimal places. Ans: The flux of F through T is 2e8 − 5 ≈ 5956.916. difficulty: medium section: 19.1
Page 11
1. Let H =5 xi − 2 xy j + xz 2 k . What is div H ? Ans: 5 - 2 x + 2 xz difficulty: easy section: 20.1 2. Let H = –3 xi − 2 xy j + xz 2 k . , What is ∫ H ⋅ dA , where S is the cube with corners at (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, S
1), (1, 1, 0), (1,0,1), (0,1,1), and (1, 1, 1)? Ans: –3.5 difficulty: easy section: 20.1 3. Consider the two-dimensional fluid flow given by a a F = x 2 + y 2 xi + x 2 + y 2 y j.
(
)
(
)
where a is a constant. (We allow a to be negative, so F may or may not be defined at (0, 0).) (a) Is the fluid flowing away from the origin, toward it, or neither? (b) Calculate the divergence of F . Simplify your answer. (c) For what values of a is div F positive? Zero? Negative? (d) What does your answer to (c) mean in terms of flow? How does this fit in with your answer to (a)? Ans: (a) The fluid is flowing away from the origin. (b)
2 ( x 2 + y 2 ) (1 + a ) a
a > −1, a = −1, a < −1 (d) If div F > 0, it means that the flow rate is increasing as it is away from origin, i.e. the vectors in the vector field will be longer and longer when we move away from origin. If div F = 0, it means that the flow is constant everywhere. If div F < 0, then the flow rate is decreasing when we move away from origin. There is no contradiction between (c) and (a), no matter what div F , is, the flow is always from origin. difficulty: hard section: 20.1 (c)
Page 1
Chapter 20: The Curl and Stokes' Theorem
4.
2 If F =5 ye x i + 4 xye y j + 4 z cos( xy )k , find div F . 2 A) 10 xye x i + ( 4 xe y + 4 xye y ) j + 4 cos( xy )k
B)
10 xye x + 4 xe y + 4 xye y − 4 cos( xy )
C)
10 xye x + 4 xye y + 4 cos( xy )
D)
10 xye x + 4 xe y + 4 xye y + 4 cos( xy ) 2 5 xye x i + ( 4 xe y + 4 xye y ) j + 4 cos( xy )k
E)
2
2
Ans: D
2
difficulty: easy
section: 20.1
5. Let F be a smooth vector field. The flux of F out of a small sphere of radius r centered at the point (1, π, 2π) is 12πr3. Find divF at (1, π, 2π) and use the result to estimate the flux of F out of the small cube of side 0.025 centered around the point (1, π, 2π). A) 0.000141 B) 0.005625 C) 0.000125 D) 0.00625 E) 0.25 Ans: A difficulty: easy section: 20.1
6. Let F =( x 2 + 5 y )i + axy j + z 4 k . Given that div F (2, 2, –1) = 10, find the value of a. Ans: 5 difficulty: easy section: 20.1 7. Let F be a smooth velocity vector field describing the flow of a fluid. Suppose that div F (1, 2, −1) = 10.
Will there be an inflow or outflow of fluid at the point (1, 2,-1)? Ans: Since div F (1, 2, −1) = 10 > 0 there will an outflow of fluid at that point. difficulty: easy section: 20.1 8. Let F be a smooth velocity vector field describing the flow of a fluid. Suppose that div F (1, 2, −1) = –4. Estimate the value of ∫ F ⋅ dA, where S is a sphere of radius 0.25 centered at (1, 2,-1) S
oriented outward. Give your answer to 4 decimal places. Ans: –0.2618 difficulty: easy section: 20.1
Page 2
Chapter 20: The Curl and Stokes' Theorem
9. Let g ( x, y, z ) = 2 x 2 + z 2 y + 3 x cos( y ). Compute ∇g and div (∇g). Select all that apply. A) ∇g= ( 4 x + 3cos( y ) ) i + ( z 2 − 3 x sin( y ) ) j + 2 yzk B) ∇g= ( 4 x + 3cos( y ) ) i + ( z 2 + 3 x sin( y ) ) j + 2 yzk C) ∇g= ( 4 x + 3cos( y ) ) i − ( z 2 − 3 x sin( y ) ) j + 2 yzk
div∇g = 4 + 2 y − 3 x cos( y ) D) div∇g = 4 − 2 y − 3 x cos( y ) E) div∇g = 4 x + 2 y − 3 x cos( y ) F) Ans: A, D difficulty: medium
section: 20.1
p 10. Let F = r r , where p is a positive constant. Is there a value of p such that F is a divergence free vector field? Ans: No difficulty: easy section: 20.1 11. Let a = a1 i + a2 i + a3 i be a constant vector and f(x, y, z) be a smooth function. Which statement is true? A) If div f a is a divergence free vector field then ∇f is parallel to a. B) If div f a is not a divergence free vector field then ∇f is perpendicular to a. C) If div f a is a divergence free vector field then ∇f is perpendicular to a. D) If div f a is not a divergence free vector field then ∇f is parallel to a. E) If div f a is a divergence free vector field then ∇f is constant. Ans: C difficulty: medium section: 20.1 12. (a) Is F = F1 ( y, z )i + F2 ( x, z ) j + F3 ( x, y )k a divergence free vector field? (b) Do all divergence free vector fields have the form of the vector field in (a)? 0 for any closed (c) If F has the form given in (a) can we conclude that ∫ F ⋅ dA = S
surface S? Ans: (a) Yes (b) No (c) No difficulty: easy
section: 20.1
13. True or false? 4V . If div F = 4 for all x, y, z and if S is a surface enclosing a volume V, then ∫ F ⋅ dV = S
A) Not possible to decide Ans: C difficulty: easy
B) True C) False section: 20.2
Page 3
Chapter 20: The Curl and Stokes' Theorem
14. An oceanographic vessel suspends a paraboloid-shaped net below the ocean at depth of 1200 feet, held open at the top by a circular metal ring of radius 20 feet, with bottom 90 feet below the ring and just touching the ocean floor. Set up coordinates with the origin at the point where the net touches the ocean floor and with z measured upward.
If the water is flowing with velocity 2 F =2 xzi − (1290 + xe −3 x ) j + z (1290 − z )k
write down an iterated integral for the flux of water through the net (oriented from i Include the limits of integration but do not evaluate.
A)
9 2 y 9 2 9 2 2 2 2 x −3 x 2 2 x x y i 1290 xe j x y 1290 x y k i+ + + + − + − + ⋅ ( ) ( ) ( ) ∫−20 ∫− 400− y2 40 40 40 2 2 20
(
400 − y 2
)
B) 9 2 9 2 9 2 2 2 2 −3 x 2 2 1290 1290 x x y i xe j x y x y k + + + − + − + ( ) ( ) ( ) dxdy ∫−20 ∫− 400− y2 40 40 40 20
(
400 − y 2
)
C)
9 2 9 2 9 x y 2 x ( x + y 2 ) i − 1290 + xe −3 x j + ( x 2 + y 2 ) 1290 − ( x 2 + y 2 ) k ⋅ i + −20 − 400 − y 40 40 2 2 40 20
∫ ∫
)
(
400 − y 2
2
D)
9 2 y 9 2 9 2 2 2 2 x −3 x 2 2 x x + y i − 1290 + xe j + x + y 1290 − x + y k ⋅ i+ j ( ) ( ( ) ) ∫−20 ∫− 20− y 40 40 40 2 2 20
20 − y 2
2
(
)
(
)
E) 2 9 2 9 9 2 x ( x + y 2 ) − 1290 + xe −3 x + ( x 2 + y 2 ) 1290 − ( x 2 + y 2 ) dxdy −20 − 400 − y 40 40 40
20
∫ ∫ Ans: C
400 − y 2
2
difficulty: easy
section: 20.2
Page 4
Chapter 20: The Curl and Stokes' Theorem
15. An oceanographic vessel suspends a paraboloid-shaped net below the ocean at depth of 750 feet, held open at the top by a circular metal ring of radius 30 feet, with bottom 70 feet below the ring and just touching the ocean floor. Set up coordinates with the origin at the point where the net touches the ocean floor and with z measured upward.
If the water is flowing with velocity 2 F = 2 xzi − (820 + xe −3 x ) j + z (820 − z )k
use the Divergence Theorem to find the flux of water through the net (oriented from
Ans: –21420000π difficulty: easy section: 20.2 16. State the Divergence Theorem. A) If W is a solid region whose boundary S is a piecewise smooth surface, then F ⋅ dA = div ∫ ∫ F dV . S
B)
W
If W is a solid region whose boundary S is a smooth surface oriented outward, then F ⋅ dA = div ∫ ∫ F dV . S
W
C)
If W is a solid region whose boundary S is a piecewise smooth surface oriented outward, and if F is a smooth vector field on an open region containing W and S, div then ∫ F ⋅ dA = ∫ F dV .
D)
If W is a solid region whose boundary S is a piecewise smooth surface oriented ⋅ dA ∫ div F ⋅ dV . outward, then ∫ F =
E)
If S is a solid region whose boundary W is a piecewise smooth surface oriented div outward, then ∫ F ⋅ dA = ∫ F dV .
S
W
S
W
S
Ans: C
W
difficulty: easy
section: 20.2
17. Evaluate the flux integral ∫ 6 x3 i + ( 6 y 3 + cos z ) j + ( 7 z 3 + 6 y 2 + 42 x 3 ) k ⋅ dA , where S is S
2
(
)
2
the surface of the cylinder x + y = 1 bounded by the planes z = -1, z = 1 (including the ends of the cylinder). Ans: 32π difficulty: easy section: 20.2
Page 5
Chapter 20: The Curl and Stokes' Theorem
18. Let F = ( e y + 6 x) ) i + ( 6 y + z cos x 6 ) j + ( 5 z + 6 x ) k . Calculate div F .
Ans: 17 difficulty: easy
section: 20.2
19. Let F = ( e y + 5 x) ) i + ( 4 y + z cos x 5 ) j + ( 2 z + 5 x ) k . Calculate the flux ∫ F ⋅ dA , where S is the sphere (x-2)2 + (y-3)2 + z2 = 4 oriented S
inward.
352 π 3 difficulty: easy Ans: −
section: 20.2
20. Use the Divergence Theorem to find the flux of the vector field F = ( y 2 x − 3 z 2 y ) i + ( z 2 y + zx3 ) j + ( x 2 z + y x ) k through the sphere x2 + y2 + z2 = 1 . 4 Ans: π 5 difficulty: easy section: 20.2 21. Suppose G is a vector field with the property that G = 10 zk at every point of the surface r = 0.5. If div G = c, where c is a constant, find c. Ans: 10 difficulty: easy section: 20.2 22. Suppose G is a vector field with the property that G = xi at every point of the surface r = 0.8. If div G is not a constant, estimate the value of div G (0, 0, 0). Ans: 1 difficulty: medium section: 20.2
23. Let F=–
x
( x + 4 y + 3z ) 2
2
2
i– 3/ 2
y
( x + 4 y + 3z ) 2
2
2
Is F a divergence free vector field? Ans: Everywhere except the origin . difficulty: medium section: 20.2
Page 6
j– 3/ 2
z
( x + 4 y + 3z ) 2
2
2
k . 3/ 2
Chapter 20: The Curl and Stokes' Theorem
24. Let F=
x
( x + 2 y + 4z )
i+ 3/ 2
y
j+ 3/ 2
z
k . 3/ 2
( x + 2 y + 4z ) ( x + 2 y + 4z ) Use the Divergence Theorem to calculate ∫ F ⋅ dA, where S is the sphere of radius 3 a 2
2
2
2
2
2
2
2
2
S
centered at the point ( 4 a, 0, 0). Ans: 0 difficulty: medium section: 20.2 25. Let F=
x
( x + 5 y + 2z ) 2
2
2
i+ 3/ 2
y
( x + 5 y + 2z ) 2
2
2
j+ 3/ 2
z
( x + 5 y + 2z ) 2
2
2
k . 3/ 2
Let S1 be the sphere of radius 5 centered at the origin, oriented outward and let S2 be the sphere of radius 2 centered at the origin, also oriented outward.
Do you expect the value of ∫ F ⋅ dA, to be larger, smaller or the same as ∫ F ⋅ dA ? S1
S2
Ans: The same. difficulty: medium
section: 20.2
26. Use the Divergence Theorem to find the flux of the vector field F = ( 4 x 3 + 4 yz ) i + ( 4 y 3 + 8 xz ) j + ( 4 z 3 + 16 yx ) k
through the cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. Ans: 12 difficulty: easy section: 20.2 27. Let F = g ( x, z ) j , where g ( x, z ) is a smooth function of x and z. (a) Use the Divergence Theorem to find the flux of F out of the cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. (b) Let Si, i = 1, …, 6 be the six faces of the cube in part (a), oriented outward. Find ∫ F ⋅ dA, for i = 1, …, 6, and hence find the flux of F out of the cube 0 ≤ x Si
≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. Compare your answers for (a) and (b). A) The answer in (a) is bigger than (b). C) B) The answer in (a) is smaller than (b). Ans: C difficulty: easy section: 20.2
Page 7
They are the same.
Chapter 20: The Curl and Stokes' Theorem
28. If div F = –7 everywhere and S is a smooth surface (oriented outward) enclosing a volume W of size V find ∫ F ⋅ dA . S
A) 7V Ans: C
B) –7 C) –7V D) –7W difficulty: easy section: 20.2
E) 0
29. Show that the vector field F 3 xe x i y 3 z 5 e x j 3z 5 e x k is a divergence free vector field. Use this result to calculate F dA where S is the open surface which is the graph of S
f ( x, y ) = − x − y + 4 with f(x, y) ≥ 0. Ans: –20π difficulty: medium section: 20.2 2
2
30. Suppose that F is a vector field defined everywhere with constant negative divergence C. Decide if the following statement is true and explain your answer. F dA 0 for every oriented surface S. S
Ans:
difficulty: hard
section: 20.2
31. Suppose that F is a vector field defined everywhere with constant negative divergence C. Decide if the following statement is true. There is no oriented surface S for which F dA 0. S
A) False Ans: A
B) True difficulty: easy
section: 20.2
32. Calculate curl H , where H 9 xi 8 xy j xz 5 k . A) C) curl H z 5 j 8 yk curl H z 5 i 8 yk B) D) curl H z 5 j 8 yk curl H z 5 j 8 yk Ans: B difficulty: easy section: 20.1
Page 8
Chapter 20: The Curl and Stokes' Theorem
33. What is meant by curl where is a vector field? Is curl F F F a vector or a scalar? A) curl F is a scalar which tells you how a vector field is rotating. Its direction gives the axis of rotation; its magnitude gives the strength of rotation about the axis. B) curl F is a vector which tells you how a scalar field is rotating. Its direction gives the axis of rotation; its magnitude gives the strength of rotation about the axis. C) curl F is a vector which tells you how a vector field is rotating. Its direction gives the axis of rotation; its magnitude gives the strength of rotation about the axis. D) curl F is a vector which tells you how a vector field is rotating. Its magnitude gives the axis of rotation; its direction gives the strength of rotation about the axis. E) curl F is a scalar which tells you how a scalar field is rotating. Its magnitude gives the strength of rotation about the axis. Ans: C difficulty: easy section: 20.1 34. How is curl F defined? Select all that apply. A) curl F is a vector in the direction n for which circ n ( x, y, z ) is the greatest that has magnitude equal to the circulation density of F around that direction. B) If F F1 i F2 j F3 k , then F F F F F F curl F 3 2 i 1 3 j 2 1 k . y z z x x y C) curl F is a vector in the direction n for which the circulation is greatest and has magnitude equal to the circulation of F around that direction. D) If F F1 i F2 j F3 k , then F F F F F F curl F 3 2 i 1 3 j 2 1 k . y z z x x y E) If F F1 i F2 j F3 k , then F F F F F F curl F 3 2 i 3 1 j 2 1 k . y z x z x y
Ans: A, D
difficulty: hard
section: 20.1
35. Find curl 3 yi 3 x j ( x5 y 5 z 5 )k . Ans: 5 y 4 i 5 x 4 j 6k difficulty: easy section: 20.1
Page 9
Chapter 20: The Curl and Stokes' Theorem
36. The figure below shows a vector field of the form F F1 i F2 j. Assume F1 and F2 depend only on x and y and not on z. Answer (a)-(d). No reasons need be given.
F d r positive, negative or zero? C (b) Is div F at P is positive, negative or zero? (c) Is the z component of curl F is positive, negative or zero? (d) Could F be a gradient field? Ans: (a) zero (b) positive (c) zero (d) yes difficulty: medium section: 20.1
(a) Is
37. Let F
y 4 x cos z i z j z 4 xy k. Find the divergence of F . 2
3
2
Ans: div F 4 cos z 2 z. difficulty: easy section: 20.1
38. Let F
y 4 x cos z i z j z 4 xy k. Find the curl of F . Ans: curl F 4 x 3 z i 4 y 4 x sin z j 2 yk . 2
3
2
2
difficulty: medium
section: 20.1
39. Let G be a smooth vector field with curl G (0, 0, 0) –3i 4 j 4k . Estimate the circulation around a circle of radius 0.01 in the yz plane, oriented counterclockwise when viewed from the positive x-axis. Ans: –0.0003π difficulty: hard section: 20.1
Page 10
Chapter 20: The Curl and Stokes' Theorem
40. Suppose that G d r 5a 3 4a 2 , where Ca is the circle Ca r (t ) a cos t + 5 i a sin t 4 j + 20k , for any a > 0. Does knowing this tell you anything about curl G (5, 4, 20) ? Is so, what? If not, why not? 4 Ans: The k -component of curl G (5, 4, 20) is . π difficulty: medium section: 20.1
41. Below is a picture of the vector field F (= x, y, z ) F1 ( x, y )i + F2 ( x, y ) j from above.
Match the value of curl F (1,1, 0) with one of the following values: 0, k , − k , k / 2, − k / 2 Ans: k difficulty: easy section: 20.1
42. Let = F ( –2 x 2 − axz ) i + ( bx 2 + 16 ) j + ( x + cy ) k be a smooth vector field with curl F (1, –4, 4 ) = –3i + 3 j + 8k . Determine the values of a, b and c. Ans: a = –4, b = 4, c = –3. difficulty: easy section: 20.1
Page 11
Chapter 20: The Curl and Stokes' Theorem
43. Suppose that curl F (1, 2,1) =3i − 2 j + –5k , curl F (0, 2,1) =6i + 2 j + 5k and curl F (1,3, −1) = –3i + 2 j + 10k . Estimate the following line integrals. (a) ∫ F ⋅ d r , where C1 is given by C1 r (t ) = i + ( 3 + 0.1cos t ) j + ( 0.1sin t − 1) k , 0 ≤ t ≤ 2π. (b) ∫ F ⋅ d r , where C2 is given by= r (t ) 0.1sin ti + 2 j + (1 + 0.1cos t ) k , 0 ≤ t ≤ 2π. C2 (c) ∫ F ⋅ d r , where C3 is given by C3 r (t ) = (1 + 0.1cos t ) i + ( 2 + 0.1sin t ) j + k , 0 ≤ t ≤ 2π.
Ans: (a) –0.03π (b) 0.02π (c) –0.05π difficulty: medium
section: 20.1
44. You want to build a windmill at the origin that maximizes the circulation of the wind. The wind vector field at any point (x, y, z) in your coordinate world is given by F y 5 z i x 2 z j 2 y x k . (a) In which direction should you face the windmill to get maximum use from the wind? (b) What will be the strength of the circulation of the wind when you face it in this direction? Ans: (a) curl F 4i 6 j
52 (b) difficulty: easy
section: 20.1
45. Suppose that curl F is not zero. True or false? If curl F is parallel to the x -axis for all x, y, and z and if C is a circle in the xy-plane, then the circulation of F around C is zero. A) True B) False Ans: A difficulty: easy section: 20.2
46. Let F 2 xi z j ( y 2 x)k .
Use Stokes' Theorem to find F dr , where C is a circle in the xy-plane of radius 4 , C
centered at 3,1, 0 , oriented counter-clockwise when viewed from above. Ans: 0 difficulty: easy section: 20.2
Page 12
Chapter 20: The Curl and Stokes' Theorem
47. Let F 5 xi z j ( y 5 x)k .
Use Stokes' Theorem to find F dr , where C is a circle in the xz-plane of radius 6 , C
centered at 6, 0, 2 , oriented clockwise when viewed from the positive y-axis. Ans: 180π difficulty: medium section: 20.2 48. Let F –3 xzi 4( y x) j 4 xk .
(a) By direct computation, find the circulation of F around the circle of radius a, r (t ) a cos t i a sin t j , for 0 ≤ t ≤ 2π. (b) Use this result to find the k component of curl F (0, 0, 0). Ans: (a) –4a 2 π (b) –4 difficulty: medium section: 20.2
49. Let F –2 xzi 5( y x) j 5 xk .
(a) Use Stokes' Theorem to find the circulation of F around the circle of radius a, r (t ) a cos ti a sin t j , for 0 ≤ t ≤ 2π. (b) Use this result to find the k component of curl F (0, 0, 0). Ans: (a) –5a 2 π (b) –5 difficulty: medium section: 20.2
50. State Stokes' Theorem. A) If S is a smooth oriented surface with smooth, oriented boundary C, then dr ∫ curl F ⋅ dA. ∫ F ⋅= C
S
B)
If S is a smooth oriented surface with piecewise smooth, oriented boundary C, then F ⋅ = dr curl ∫ ∫ F ⋅ dA.
C)
If S is a smooth oriented surface with piecewise smooth, oriented boundary C, then F ⋅ dr = curl ∫ ∫ FdV .
S
C
C
S
D)
If S is a smooth oriented surface with piecewise smooth, oriented boundary C, and if F is a smooth vector field on an open region containing S and C, then F ⋅ = dr curl ∫ ∫ F ⋅ dA.
E)
If S is a smooth oriented surface with piecewise smooth, oriented boundary C, then ∫ F ⋅ dr = ∫ div FdV .
C
C
Ans: D
S
S
difficulty: easy
section: 20.2
Page 13
Chapter 20: The Curl and Stokes' Theorem
51. Let F 4 z 3 yi 4 x 4 z 3 x j x 2 12 z 2 xy k . (a) Evaluate the line integral F d r , where C is the circle x 2 y 2 1, on the
C
xy-plane, oriented in a counter-clockwise direction when viewed from above. (b) Without any computation, explain why the answer in part (a) is also equal to the flux integral curl F dA, where S1 is lower hemisphere x 2 y 2 z 2 1, z 0, oriented S1
inward. Ans: (a) The orientation of C is determined form the orientation of S (the disk with boundary C) according to the right hand rule and curl F 2 x j 4k By Stokes' Theorem, curl 2 4 F d r F d A x j k kdA 4Area of S 4π. C
S
S
(b) The circle C is also the boundary curve of the lower hemisphere S1. The orientations of C and S1 will satisfy the right hand rule. By Stokes' Theorem, d r curl F d A. F C
S
In this case, the computation of flux integral through a hemisphere is tedious, but it should give the same answer as part (a). difficulty: medium section: 20.2 52. True or false? If curl F 0, then by Stokes' Theorem the line integral F dr is equal to zero, where C
C is the curve y = x , for 0 ≤ x ≤ 2. Ans: False. We cannot apply Stokes' Theorem here because the curve is not the boundary of any surface. difficulty: easy section: 20.2 2
Page 14
Chapter 20: The Curl and Stokes' Theorem
53. The trumpet surface, S, is given parametrically by r ( s, t ) s 2 cos ti s j s 2 sin tk , 0 s 5, 0 t 2π.
If S has outward pointing normal, use an appropriate line integral to calculate curl zi cos( x z ) j xk dA. S
Ans: –1250π difficulty: medium
section: 20.2
54. (a) The vector field F 2 xi 4 y j azk has the property that the flux of F through any closed surface is 0. What is the value of the constant a? (b) The vector field G ay bz i (2 x 4 z ) j 4 x cy k has the property that the circulation of G around any closed curve is 0. What are the values of the constants a, b and c? Ans: (a) a –6 a 2, b 4, c 4 (b) difficulty: easy section: 20.2 55. Let = F 4 zi − 4 xk . Let C be the circle of radius a parameterized by x = a cos t, y = 0, z = a sin t, 0 ≤ t ≤ 2π and let S be the disk in the xz-plane enclosed by C, oriented in the positive y-direction. (a) Evaluate directly ∫ F ⋅ d r. C (b) Evaluate directly ∫ curl F ⋅ d A. S
(c) Do these results contradict Stokes' Theorem? Ans: (a) –8a 2 π (b) 8a 2 π (c) They do not contradict Stokes' Theorem. The orientations of C and S do not satisfy the right hand rule. If we keep the orientation of C, then S has to be oriented in the negative y-direction in order to satisfy the right hand rule. As a result, the answers in parts (a) and (b) are different by a negative sign. difficulty: medium section: 20.2
Page 15
Chapter 20: The Curl and Stokes' Theorem
56. Let F be a vector field such that div F =− 4 2 x2 − 2 y 2 − 2z 2 . (a) Is F a curl field? (b) Use spherical coordinates to evaluate ∫ div FdV , where W is the solid ball of radius W
R centered at the origin. (c) Use the result of part (b) to find the radius of a sphere centered at the origin, such that the flux of F out of this sphere is zero. Ans: (a) Since div F ≠ 0, it fails the Divergence Test, and hence cannot be a curl field. (b) In spherical coordinates 2π π R 2 2 div = FdV ∫ ∫ ∫ ∫ ( 4 − 2 ρ ) ρ sin φ d ρ dφ dθ W
0 0 0
4 3 R ( 20 − 6 R 2 ) 15 (c) In order for the flux integral to be zero, we need 20-6R 2 to be zero; this means 10 that R = . 3 difficulty: medium section: 20.3 =
57. Use the curl test to determine whether the following vector field is a gradient field. F =y 3 z 2 i + ( 4 y 2 xz 2 + 4 yz 3 ) j + ( 2 xy 3 z + 3 y 2 z 2 ) k . A) Only at the origin B) No C) Yes Ans: B difficulty: easy section: 20.3 58. Given that grad f = 4 xyzi + ( ax 2 z + 10 yz ) j + ( bx 2 y + cy 2 ) k , use the curl test to find the values of the constants a, b and c. a 2,= b 2,= c 5 Ans: = difficulty: easy section: 20.3 59. Given curl F = 10 xyzi + ( ay 2 z + 2byz ) j − 4 z 2 k , find the values of the constants a and b, without knowing the expression of F . = a –5, = b 4 Ans: difficulty: easy section: 20.3 60. Suppose that the flux of a smooth vector field F out of a sphere of radius r centered at the origin is ar3 + br4, where a and b are constants. Calculate div F dV , where W is W
the solid 1 x y z 16. Ans: 63a 255b difficulty: medium section: 20R 2
2
2
Page 16
Chapter 20: The Curl and Stokes' Theorem
61. Suppose that the flux of a smooth vector field F out of a sphere of radius r centered at the origin is ar br 2 , where a and b are constants. If F curl G, for a smooth vector field G, find the values of a and b. Ans: a b 0 difficulty: easy section: 20R
62. Suppose a vector field G is always perpendicular to the normal vector at each point of a surface S. What is the value of G dA ? S
Ans: 0 difficulty: easy
section: 20R
0. Let F be a vector 63. Let P be a plane through the origin with equation ax 4 y 3 z field with curl F 3i 5 j 2k . Suppose F dr 0 for any closed curve on the C
0. Using Stokes' Theorem, determine the value of a. plane ax 4 y 3 z 26 Ans: 3 difficulty: medium section: 20R
64. For the following integral, say whether Stokes' Theorem, the Divergence Theorem, or neither applies. 2 6 yi 9 z j 2k dr , where C is a closed loop in space. C
A) Stokes' Theorem B) Divergence Theorem. Ans: A difficulty: easy section: 20R
C) Neither
65. For the following integral, say whether Stokes' Theorem, the Divergence Theorem, or neither applies. 3 –5 xi z j 4 y k dA, where S is the sphere of radius 4 centered at the origin S
oriented outward. A) Stokes' Theorem B) Divergence Theorem. Ans: B difficulty: easy section: 20R
C) Neither
66. For the following integral, say whether Stokes' Theorem, the Divergence Theorem, or neither applies. 3 curl 2 xi z j 2 y k dA, where S is a triangular plane in space oriented S
downward. A) Stokes' Theorem B) Divergence Theorem. Ans: A difficulty: easy section: 20R
Page 17
C) Neither
Chapter 20: The Curl and Stokes' Theorem
67. For the following integral, say whether Stokes' Theorem, the Divergence Theorem, or neither applies. 3 2 5 xi z j y k dr , where S is a triangular plane in space oriented upward. C
A) Stokes' Theorem B) Divergence Theorem. Ans: C difficulty: easy section: 20R
C) Neither
68. Using either Stokes' theorem or the Divergence theorem (whichever is appropriate), evaluate the following: C –4 xi 3 y j 3zk dr , where C is a closed loop parameterized by r (t ) cos t sin t i cos3 t j sin 4 tk . Ans: 0 difficulty: medium section: 20R
69. Using either Stokes' theorem or the Divergence theorem (whichever is appropriate), evaluate dA, x yz i y z j z x k 2 cos 5 sin 2 3 S
where S is the sphere of radius 2 oriented outward and centered at the point ( 2, e −25 ,π 2 ) .
320 π 3 difficulty: medium
Ans:
section: 20R
70. Let F = –4 zk . (a) Compute div F (0, 0, 0).
(b) By direct computation, find the flux of F through a cube with edge length l, centered at the origin and edges parallel to the axes. (b) Explain how your answers in parts (b) are related to that of part (a). Ans: (a) div F (0, 0, 0) = –4. (b) –4l 3 (c) By the geometric definition of divergence, Flux of F out of the cube –4l 3 div F (0, 0, 0) lim –4. = = = l →0 Volume of the cube l3 difficulty: easy section: 20R
Page 18
Chapter 20: The Curl and Stokes' Theorem
71. Let S be the surface of the upper part of the cylinder 4x2 + z2 = 1, z ≥ 0, between the planes y = -1, y = 1, with an upward-pointing normal. (a) Evaluate the flux integral ∫ –3 xy 2 zi + 7 cos z j + 3k ⋅ dA. S
(
)
(b) Consider W, the solid region described by -1 ≤ y ≤ 1, 4x2 + z2 ≤ 1, z ≥ 0. Evaluate ∫ div –3 xy 2 zi + 7 cos z j + 3k dV . W
(
)
Does this contradict the Divergence Theorem? Explain. 20 Ans: (a) 3 –2 (b) 3 This does not contradict the Divergence Theorem. We cannot apply the Divergence Theorem in this case because S is only the curved part of the cylinder, we also need the two ends and a base to enclose the solid W. difficulty: medium section: 20R 72. A physicist is asked to find the flux of a magnetic field F = 4k through the curved surface of a right circular cone. The cone is of height 4 and has a circular base of radius 2. He says immediately that, by the Divergence Theorem, the answer is 16π. How did he know? Ans: The physicist noticed that the curved surface together with the circular base would enclose a solid region, so one could apply the Divergence Theorem in this case. Since div F 0, Flux out of curved surface + Flux through circular base (oriented downward) = 0, or equivalently,
Flux out pf the curved surface=-Flux out of circular base, oriented downward = Flux out of circular base, oriented upward = (4)(22 π) 16π. difficulty: easy
section: 20R
73. Let S be the boundary surface of a solid region W with outward-pointing normal. Using an appropriate theorem, change the following flux integral into volume integral over W. 2 4 cos 3 sin 5 x + z i + y + x j − zk ⋅ d A ( ) ( ) ∫ S
(
Ans:
)
∫ 2dV W
difficulty: easy
section: 20R
Page 19
Chapter 20: The Curl and Stokes' Theorem
74. Suppose ∫ F ⋅ d A = S
∫ ( 4 xi − 6 y j + 2 zk ) ⋅ d A for any closedsurface S in space with S
outward-pointing normal. What does this tell you about div F ? Ans: div F = 0 difficulty: easy section: 20R
75. On an exam, students are asked to find the line integral of F = 2 xyi + yz j + y 2 k over the curve C which is the boundary of the upper hemisphere x 2 + y 2 + z 2= 22 , z ≥ 0, oriented in a counter-clockwise direction when viewed from above. One student wrote: “ curl F = −2 x j + 2k . By Stokes' Theorem ∫C F ⋅ dr = ∫S −2x j + 2k ⋅ dA, where S is the hemisphere. Since div −2 x j + 2k = 0, by the Divergence Theorem = ∫ 0dV = 0, ∫S −2 x j + 2k ⋅ dA W 0. ” where W is the solid hemisphere. Hence we have ∫ F ⋅ dr =
(
(
(
)
)
)
C
This answer is wrong. Which part of the student's argument is wrong? Select all that apply. A) The student has been careless with the orientations of the curve and surface. B) The student has to be careful with the orientations of the curve and surface. However, Stokes' Theorem has been applied correctly. C) The student has used Divergence Theorem incorrectly. The upper hemisphere S does not include the bottom, hence it does not enclose any region in space, and we cannot apply the Divergence Theorem to this surface. D) The student has to be careful with the orientations of the curve and surface. However, the Divergence Theorem has been applied correctly. E) The student has the correct orientations of the curve and surface but the student has used Divergence Theorem for the hemisphere incorrectly. The upper hemisphere S does not include the bottom, hence it does not enclose any region in space, and we cannot apply the Divergence Theorem to this surface. Ans: A, C difficulty: medium section: 20R 76. Calculate the divergence of the following vector fields at the point (1,1,0). A) F = 5 x 2 yi + 2 xyz j + (cos z )k B) G= (2 y − 3 x)i + (1 − cos z ) j + x 2 yk C) H (cos z )i + 2 xyz j + x 2 yk = Part A: ∇ ⋅ F (1,1, 0) = 10. Part B: ∇ ⋅ G (1,1, 0) = –3. Part C: ∇ ⋅ H (1,1, 0) =0. difficulty: easy section: 20.1
Page 20
Chapter 20: The Curl and Stokes' Theorem
77. Suppose that F is a smooth vector field, defined everywhere. True or False: It is possible that ∫ F ⋅ dA = 3r 2 + 2r , where S is a sphere of radius r S
centered at the origin. Ans: False difficulty: medium
section: 20.1
78. Which of the following vector fields have zero divergence everywhere they are defined? Select all that apply. A) E) xi − yj xi + yj + zk B) F) xzi + yzj − z 2 k xi + yj xi + yj + zk C) G) yi + xj − zk x2 + y 2 + z 2 D) yi + zj + xk Ans: D, E, F difficulty: easy section: 20.1 79. Which of the following vector fields have positive divergence everywhere they are defined? Select all that apply. E) A) xi − yj xi + yj + zk F) B) xzi + yzj − z 2 k xi + yj xi + yj + zk C) G) yi + xj − zk x2 + y 2 + z 2 D) yi + zj + xk Ans: A, B, G difficulty: easy section: 20.1 80. Let G be a smooth vector field with div G = 3 at every point in space and let S1 and S2 be spheres of radius r, oriented outward, centered at (0,0,0) and at (1,2,1), respectively. True or False: ∫ G ⋅ dA = ∫ G ⋅ dA . S1
Ans: True
S2
difficulty: easy
section: 20.2
81. Let G be a smooth vector field with div G = 3 at every point in space. Find the exponent p in the following: p The flux of G out of any sphere of radius r1 r1 = The flux of G out of any sphere of radius r2 r2 Ans: p = 3 . difficulty: easy section: 20.2
Page 21
Chapter 20: The Curl and Stokes' Theorem
82. Let a = a1i + a2 j + a3k be a nonzero constant vector and let r = xi + yj + zk . Suppose S is the sphere of radius one centered at the origin. There are two (related) reasons why a × r ⋅ dA = 0 . Select them both. ( ) ∫ S
A)
S
B)
0 because div ( a × r ) = 0. ∫ ( a × r ) ⋅ dA = S
C)
0 because a × r = 0. ∫ ( a × r ) ⋅ dA =
0 because r is parallel to the dA element everywhere on S and so ∫ ( a × r ) ⋅ dA = S
D)
a × r is perpendicular to dA on S. a × r ⋅ dA = 0 because a × r is a constant vector field. ) ( ∫ S
E)
0 because ∫ ( a × r ) ⋅ dA = Q > 0 and ∫ ( a × r ) ⋅ dA = −Q , where H is ∫ ( a × r ) ⋅ dA = S
L
H
the upper unit hemisphere and L is the lower unit hemisphere. Ans: B, C difficulty: medium section: 20.2 83. Suppose that f ( x, y, z ) is defined and differentiable everywhere and satisfies the ∂f ∂f ∂f differential equation x + y + z = 0 . Let F = f ( x, y, z )r , where ∂x ∂y ∂z r = xi + yj + zk . Suppose that S is a closed surface and W is its interior. Find Q in the following equation: Q ∫ f ( x, y, z ) dV . ∫ F ⋅ dA = S
W
Ans: Q = 3. difficulty: easy
section: 20.2
84. Let S be the closed surface which is the portion of the sphere x 2 + y 2 + z 2 = 4 with −2 ≤ z ≤ 1 topped by the disk in the plane z = 1 , oriented outward. Then the flux of r = xi + yj + zk through S is: A) 9π B) 24π C) 27π D) 18π E) 36π Ans: C difficulty: medium section: 20.2 85. Let G ( x, y, z ) = − yi + 2 x j + xyzk . Calculate curl G . Ans: curl = G xzi − yz j + 3k . difficulty: easy section: 20.1
Page 22
Chapter 20: The Curl and Stokes' Theorem
86. Let G ( x, y, z ) = − yi + x j + 3 xyzk . Calculate div G . Ans: div G = 3 xy . difficulty: easy section: 20.1
87. Which of the following vector fields has the following properties: 1) the largest value of its circulation density at (1,2,1) is 50 2) the largest value of its circulation density at (1,2,1) occurs around the direction 3 4 i+ j 5 5 A) D) F1 = − y 2 xi − z 3 j + z 2 k F4 =10 − y 2 xi − z 3 j + z 2 k 1 E) B) F2 = − y 2 xi − z 3 j + z 2 k F5 = 100 − y 2 xi − z 3 j + z 2 k 100 1 C) F3 = − y 2 xi − z 3 j + z 2 k 10 Ans: D difficulty: easy section: 20.1
(
(
(
)
)
(
)
Page 23
)
Chapter 20: The Curl and Stokes' Theorem
88. Suppose W consists of the interior of two intersecting cylinders of radius 2. One cylinder is centered on the y-axis and extends from y = -5 to y = 5. The other is centered on the x-axis and extends from x = -5 to x = 5. Let S be the entire surface of W except for one circular end of one cylinder, namely the circular end centered at (0,5,0). The boundary of S is therefore the circle x 2 + z 2 = 4, y = 5 ; the surface S is oriented outward. Let F = (3 x 2 + 3 z 2 ) j = curl ( z 3i + y 3 j − x 3 k ) . Which of the following integrals is equal to ∫ F ⋅ dA ? Select all that apply. S
A)
− ∫ F ⋅ dA , where D is the disk described by x 2 + z 2 ≤ 4, y = 5 , oriented outward
B)
(away from the origin) 2 2 5 , oriented outward ∫ F ⋅ dA , where D is the disk described by x + z ≤ 4, y =
C)
(away from the origin) − ∫ F ⋅ dA , where D is the disk described by x 2 + z 2 ≤ 4, y = 5 , oriented inward
D
D
D
D)
(toward the origin) 2 2 5 , oriented inward ∫ F ⋅ dA , where D is the disk described by x + z ≤ 4, y =
E)
(toward the origin) 3 3 3 + ( z i y j − x k ) ⋅ dr , where C is the circle x 2 + z 2 = 4, y = 5 oriented clockwise ∫
D
C
F)
when viewed from y > 5 on the positive y-axis − ∫ ( z 3i + y 3 j − x3 k ) ⋅ dr , where C is the circle x 2 + z 2 = 4, y = 5 oriented C
clockwise when viewed from y > 5 on the positive y-axis Ans: A, D, E difficulty: medium section: 20.2 89. Suppose W consists of the interior of two intersecting cylinders of radius 2. One cylinder is centered on the y-axis and extends from y = -5 to y = 5. The other is centered on the x-axis and extends from x = -5 to x = 5. Let S be the entire surface of W except for one circular end of one cylinder, namely the circular end centered at (0,5,0). The boundary of S is therefore the circle x 2 + z 2 = 4, y = 5 ; the surface S is oriented outward. Let F = (3 x 2 + 3 z 2 ) j = curl ( z 3i + y 3 j − x 3 k ) . Qπ . Find Q. Then ∫ F ⋅ dA = S
Ans: Q = −24 . difficulty: medium
section: 20.2
Page 24
Chapter 20: The Curl and Stokes' Theorem
90. Suppose that F is a curl field. Which of the following must also be curl fields? Select all that apply. A) F + xi + yj + zk B) F + yi + zj + xk C) F + yzi + xzj + xyk D) F + (cos z )i + ze x j + k E) F +i + j +k x F) F + ln(1 + y 2 + z 2 )i + arctan j + x 2 y 4 k z G) F + xyzi + xyzj + xyzk Ans: B, C, D, E, F difficulty: easy section: 20.3
91. Let S be the triangle with vertices at (12,0,0), (0,6,0) and (0,0,6), oriented outward (away 12 . Let C be the oriented from the origin). So S lies in the plane x + 2 y + 2 z = 2 y z 2x boundary of S. Evaluate ∫ − i − j − k ⋅ dr . 3 3 3 C 2 y z 2x Ans: ∫ − i − j − k ⋅ dr = 54 . 3 3 3 C difficulty: hard section: 20R
Page 25
1. Consider the change of variables x = s + 4t, y = s - 5t. ∂ ( x, y ) Find the absolute value of the Jacobian . ∂ ( s, t ) Ans: 9 difficulty: easy section: 21.2 2. Consider the change of variables x = s + 3t, y = s - 2t. Let R be the region bounded by the lines 2x + 3y = 1, 2x + 3y = 4, x - y = -3, and x - y = 2. Find the region T in the st-plane that corresponds to region R. Use the change of variables to evaluate ∫ 2 x + 3 ydA . R
Ans: 7.5 difficulty: medium
section: 21.2
3. Let R be the region in the first quadrant bounded between the circle x 2 + y 2 = 1 and the π two axes. Then ∫ ( x 2 + y 2 )dA = . R 8 Let R be the region in the first quadrant bounded between the ellipse 25 x 2 + 9 y 2 = 1 and the two axes. Use the change of variable x = s/5, y = t/3 to evaluate the integral 2 2 ∫ ( 75 x + 27 y )dA. R
1 π 40 difficulty: easy Ans:
section: 21.2
4. The following equations represent a curve or a surface. Select the best geometric description. (Note: ρ, ϕ, θ are spherical coordinates; r, θ, z are cylindrical coordinates.) A) Part of a line through the origin C) Part of a cylinder. B) Disk D) Part of a cone. Ans: A difficulty: easy section: 21.1 5. Let S be a circular cylinder of radius 0.2, such that the center of one end is at the origin and the center of the other end is at the point (2, 0, 7). Find the xyz-equation of the plane, P, containing the base of the cylinder (i.e., the plane through the origin perpendicular to the axis of the cylinder). 0 Ans: 2 x + 7 z = difficulty: easy section: 21.1
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Chapter 21: Parameters, Coordinates, & Integrals
6. Let S be a circular cylinder of radius 0.2, such that the center of one end is at the origin and the center of the other end is at the point (4, 0, 7). Find two unit vectors u and v in the plane, P, containing the base of the cylinder (i.e., the plane through the origin perpendicular to the axis of the cylinder) which are perpendicular to each other. −7i + 4k Ans: = u j= , v , other solutions are possible 65 difficulty: medium section: 21.1 7. Let S be a circular cylinder of radius 0.2, such that the center of one end is at the origin and the center of the other end is at the point (5, 0, 4). Let P be the plane containing the base of the cylinder (i.e., the plane through the origin perpendicular to the axis of the cylinder). In each case, give a parameterization ( x(θ ), y (θ ), z (θ ) ) and specify the range of values your parameters must take on. (i) the circle in which the cylinder, S, cuts the plane, P. (ii) the surface of the cylinder S. −0.8sin θ z (θ ) = $ x ^ 2 + z ^ 2] Ans: (i) y (θ ) 0.2 cos θ = 1sin θ x(θ ) = $ x ^ 2 + z ^ 2] x(θ ) =
0 ≤ θ ≤ 2π.
−0.8sin θ +t $ x ^ 2 + z ^ 2]
(ii) y (θ ) 0.2 cos θ = 1sin θ z (θ ) = +t $ x ^ 2 + z ^ 2] difficulty: medium section: 21.1
0 ≤ θ ≤ 2π, 0 ≤ t ≤ 4.
8. Let v1 = 2i − 5 j + k and v 2 = 5i + j + k Find a parametric equation for the plane through the point (1, 2, -1) and containing the vectors v1 and v 2 . Select all that apply. x =1 + 2t + 5s, y =2 − 5t + s, z =−1 + t + s. A) x =1 + 2t − 5s, y =2 − 5t − s, z =−1 + t − s. B) x =−1 + 2t + 5s, y =−2 − 5t + s, z =1 + t + s. C) x =1 − 2t + 5s, y =2 + 5t + s, z =−1 − t + s. D) Ans: A, B difficulty: easy section: 21.1
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Chapter 21: Parameters, Coordinates, & Integrals
9. Let v1 = 2i − 5 j + k and v 2 = 5i + j + k
Find a vector which is perpendicular to v1 and v 2 to find an equation of the plane through the point (1, 2, -1) and with normal vector perpendicular to both v1 and v 2 . D. Express your answer in the form Ax + By + Cz = –27 ; other solutions are possible. Ans: –6 x + 3 y + 27 z = difficulty: easy section: 21.1
10. Consider the plane r ( s, t ) = (2 + s − 4t )i + (5 − s + 4t ) j + (6 − 4t − s )k . Does it contain the point (1, 6, –1)? Ans: Yes difficulty: easy section: 21.1 11. Consider the plane r ( s, t ) = (3 + s − 5t )i + (5 − s + 5t ) j + (5 − 10t + s )k . Find a normal vector to the plane. 1 1 A) D) i+k i+ j+k 2 3 1 1 E) B) i− j i+ j−k 2 3 1 C) i+ j 2 Ans: C difficulty: easy section: 21.1
(
)
(
(
(
)
(
)
)
)
12. Find the parametric equation of the plane through the point (–4, 2, 4) and parallel to the lines r (t ) = (1 − 2t )i + (5 + 2t ) j + (3 − 4t )k and s (t ) = (3 − 4t )i + 4t j + (4 − 2t )k . Select all that apply. x =–4 − 2u − 4v, y =2 + 2u + 4v, z =4 − 4u − 2v. A) x =–4 − 2u − 4v, y =2 + 2u + 4v, z =4 + 4u − 2v. B) x =–4 − 2u + 4v, y =2 + 2u − 4v, z =4 − 4u + 2v. C) x= 2u + 4v, z = −2u − 4v, y = −4u − 2v. D) x =–4 − 2u − 4v, y =2 + 2u + 4v, z =4 − 4u − 2v. E) Ans: A, C, E difficulty: easy section: 21.1 13. Find parametric equations for the sphere ( x − 4) 2 + ( y + 4) 2 + ( z − 8) 2 = 16. Ans: x =4 + 4sin φ cos θ , y =−4 + 4sin φ sin θ , z =8 + 4 cos φ . difficulty: easy section: 21.1
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Chapter 21: Parameters, Coordinates, & Integrals
14. Using cylindrical coordinates, find parametric equations for the cylinder x 2 + y 2 = 9. Select all that apply. x 3cos θ= , y 3sin θ= , z z , 0 ≤ θ ≤ 2π, − ∞ < z < ∞. A) = = x 3cos = θ , y 3sin θ , 0 ≤ θ ≤ 2π, − ∞ < z < ∞. B) x 9 cos θ= , y 9sin θ= , z z , 0 ≤ θ ≤ 2π, − ∞ < z < ∞. C) = x 3sin θ= , y 3cos θ= , z z , 0 ≤ θ ≤ 2π, − ∞ < z < ∞. D) = = x 9= cos θ , y 9sin θ , 0 ≤ θ ≤ 2π, − ∞ < z < ∞. E) Ans: A, D difficulty: easy section: 21.1 15. Find parametric equations for the cylinder y 2 + z 2 = 49. D) A) = r ( s, θ ) 7 cos θ i + 7 sin θ j. r ( s, θ ) = si + 7 cos θ j + 7 sin θ k . B) E) r ( s, θ ) = 7 cos θ i + 7 sin θ j + sk . r ( s, θ ) = si + 7 cos θ j + 7 cos θ k . C) r ( s, θ ) = si + 49 cos θ j + 49sin θ k . Ans: D difficulty: medium section: 21.1 πt πt 16. Consider the parametric surface r ( s, t ) =s sin i + s cos j + 4tk . 2 2 Does it contain the point (0, –2, 0)? Ans: Yes difficulty: medium section: 21.1 πt πt 17. Consider the parametric surface r ( s, t ) =s sin i + s cos j + 4tk . 2 2 Does it contain the x-axis? Ans: No difficulty: medium section: 21.1
18. Find parametric equations for the cylinder 49 x 2 + 5= y 2 245, − 7 ≤ z ≤ 5. x 5cos t , = y 7 sin t , = z s, 0 ≤ t ≤ 2π, −7 ≤ s ≤ 5. A) = B)
= x
5 cos t , y= 7 sin t , = z t , 0 ≤ t ≤ 2π, −7 ≤ s ≤ 5.
C)
= x
7 cos t , = y 5sin t , = z s, 0 ≤ t ≤ 2π, −7 ≤ s ≤ 5.
D)
= x
5 cos t , = y 7 sin t , = z s, 0 ≤ t ≤ π, −7 ≤ s ≤ 5.
E) = x Ans: E
5 cos t , = y 7 sin t , = z s, 0 ≤ t ≤ 2π, −7 ≤ s ≤ 5. difficulty: medium section: 21.1
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Chapter 21: Parameters, Coordinates, & Integrals
19. Let F =xi + 5 y j + zk and S be parametric surface r ( s, t ) = s i + t j + (4 s + t − 4)k , 0 ≤ s ≤ 1, 0 ≤ t ≤ 2, oriented upward. Use the formula for a flux integral over a parametric surface to find ∫ F ⋅d A. . S
Ans: –16 difficulty: medium learning objective: LO 19.3.1 Compute the divergence of a vector field. section: 21.3 20. Compute ∫ zi + x j + yk ⋅ dA , where S is oriented in the positive j direction and S
(
)
given, for 0 ≤ s ≤ 1, 0 ≤ t ≤ 2, by x = s, y = t2, z =8 t. Ans: 0 difficulty: medium learning objective: LO 19.3.1 Compute the divergence of a vector field. section: 21.3 21. Compute the flux of the vector field F= yi + 2 z j over the surface S, which is oriented upward and given, for 0 ≤ s ≤ 1, 0 ≤ t ≤ 2 by x = s + t2, y = s2 , z = 4t. Ans: 28 difficulty: medium learning objective: LO 19.3.1 Compute the divergence of a vector field. section: 21.3 22. Let F = x(1 + z )i + 2 y (1 + z ) j. Find the flux of F across the parametric surface S given by x = s cos t, y = s sin t, z = s, for 1 ≤ s ≤ 2, 0 ≤ t ≤ 2π, oriented downward. 73 Ans: π 4 difficulty: medium learning objective: LO 19.3.1 Compute the divergence of a vector field. section: 21.3
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Chapter 21: Parameters, Coordinates, & Integrals
23. Let F = x(1 + z )i + 7 y (1 + z ) j. Show that the parametric surface S given by x = s cos t, y = s sin t, z = s, for 1 ≤ s ≤ 2, 0 ≤ t ≤ 2π, oriented downward can also be written as the
surface z=
x2 + y 2 , 1 ≤ z ≤ 2 .
Which of the following iterated integrals calculates the flux of F across S? Select all that apply.
( x2 +7 y2 )1+ x2 + y2 dxdy
A)
∫
B)
∫ ∫ r (1 + r )drdθ ∫ ∫ r (1 + r )rdθ dr ∫ ∫ r (1 + r )(cos θ + b sin θ )drdθ
C) D) E)
T
x2 + y 2
2
2
0 1 2π 2
2
0 1 2π 2
2
2π
0
2
2
1
−∫
T
( x2 + y2 )1+ x2 + y2 dxdy x2 + y 2
Ans: A, D difficulty: easy learning objective: LO 19.3.1 Compute the divergence of a vector field. section: 21.3
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