APPLIED CALCULUS 7TH EDITION BY DEBORA HUGHES, PATTI FRAZER LOCK, DANIEL E FLATH CHAPTER 1_10 AND P by QuentinAxel

TEST BANK

1.1 SOLUTIONS

CHAPTER ONE Solutions for Section 1.1 1. (a) The story in (a) matches Graph (IV), in which the person forgot her books and had to return home. (b) The story in (b) matches Graph (II), the fat tire story. Note the long period of time during which the distance from home did not change (the horizontal part). (c) The story in (c) matches Graph (III), in which the person started calmly but sped up later. The frst graph (I) does not match any of the given stories. In this picture, the person keeps going away from home, but his speed decreases as time passes. So a story for this might be: I started walking to school at a good pace, but since I stayed up all night studying calculus, I got more and more tired the farther I walked. 2. The height is going down as time goes on. A possible graph is shown in Figure 1.1. The graph is decreasing. height

time

Figure 1.1

3. The amount of carbon dioxide is going up as time goes on. A possible graph is shown in Figure 1.2. The graph is increasing. CO2

time

Figure 1.2

Chapter One /SOLUTIONS 4. The number of air conditioning units sold is going up as temperature goes up. A possible graph is shown in Figure 1.3. The graph is increasing. AC units

temperature

Figure 1.3

5. The noise level is going down as distance goes up. A possible graph is shown in Figure 1.4. The graph is decreasing. noise level

distance

Figure 1.4

6. If we let t represent the number of years since 1900, then the population increased between t = 0 and t = 40, stayed approximately constant between t = 40 and t = 50, and decreased for t ≥ 50. Figure 1.5 shows one possible graph. Many other answers are also possible. population

years 80 100 120 since 1900

Figure 1.5

1.1 SOLUTIONS

7. Amount of grass G = f (r) increases as the amount of rainfall r increases, so f (r) is an increasing function. 8. We are given information about how atmospheric pressure P = f (ℎ) behaves when the altitude ℎ decreases: as altitude ℎ decreases the atmospheric pressure P increases. This means that as altitude ℎ increases the atmospheric pressure P decreases. Therefore, P = f (ℎ) is a decreasing function. 9. We are given information about how battery capacity C = f (T ) behaves when air temperature T decreases: as T decreases, battery capacity C also decreases. This means that increasing temperature T increases battery capacity C. Therefore, C = f (T ) is an increasing function. 10. Time T = f (m) increases as m increases, so f (m) is increasing. 11. The attendance A = f (P ) decreases as the price P increases, so f (P ) is a decreasing function. 12. The cost of manufacturing C = f (v) increases as the number of vehicles manufactured v increases, so f (v) is an increasing function. 13. We are given information about how commuting time, T = f (c) behaves as the number of cars on the road c decreases: as c decreases, commuting time T also decreases. This means that increasing the number of cars on the road c increases commuting time T . Therefore, T = f (c) is an increasing function. 14. The statement f (4) = 20 tells us that W = 20 when t = 4. In other words, in 2019, Argentina produced 20 million metric tons of wheat. 15. (a) If we consider the equation C = 4T − 160 simply as a mathematical relationship between two variables C and T , any T value is possible. However, if we think of it as a relationship between cricket chirps and temperature, then C cannot be less than 0. Since C = 0 leads to 0 = 4T − 160, and so T = 40 F, we see that T cannot be less than 40 F. In addition, we are told that the function is not defned for temperatures above 134 . Thus, for the function C = f (T ) we have Domain = All T values between 40 F and 134 F = All T values with 40 ≤ T ≤ 134 = [40, 134].

(b) Again, if we consider C = 4T − 160 simply as a mathematical relationship, its range is all real C values. However, when thinking of the meaning of C = f (T ) for crickets, we see that the function predicts cricket chirps per minute between 0 (at T = 40 F) and 376 (at T = 134 F). Hence, Range = All C values from 0 to 376 = All C values with 0 ≤ C ≤ 376 = [0, 376].

16. (a) The statement f (19) = 415 means that C = 415 when t = 19. In other words, in the year 2019, the concentration of carbon dioxide in the atmosphere was 415 ppm. (b) The expression f (22) represents the concentration of carbon dioxide in the year 2022. 17. (a) At p = 0, we see r = 8. At p = 3, we see r = 7. (b) When p = 2, we see r = 10. Thus, f (2) = 10. 18. Substituting x = 5 into f (x) = 2x + 3 gives f (5) = 2(5) + 3 = 10 + 3 = 13. 19. Substituting x = 5 into f (x) = 10x − x2 gives f (5) = 10(5) − (5)2 = 50 − 25 = 25. 20. We want the y-coordinate of the graph at the point where its x-coordinate is 5. Looking at the graph, we see that the y-coordinate of this point is 3. Thus f (5) = 3.

Chapter One /SOLUTIONS

21. Looking at the graph, we see that the point on the graph with an x-coordinate of 5 has a y-coordinate of 2. Thus f (5) = 2. 22. In the table, we must fnd the value of f (x) when x = 5. Looking at the table, we see that when x = 5 we have f (5) = 4.1 23. (a) We are asked for the value of y when x is zero. That is, we are asked for f (0). Plugging in we get f (0) = (0)2 + 2 = 0 + 2 = 2. (b) Substituting we get f (3) = (3)2 + 2 = 9 + 2 = 11. (c) Asking what values of x give a y-value of 11 is the same as solving y = 11 = x2 + 2 x2 = 9

√ x = ± 9 = ±3.

We can also solve this problem graphically. Looking at Figure 1.6, we see that the graph of f (x) intersects the line y = 11 at x = 3 and x = −3. Thus, when x equals 3 or x equals −3 we have f (x) = 11. y f (x)

11 x −6 −3

Figure 1.6 (d) No. No matter what, x2 is greater than or equal to 0, so y = x2 + 2 is greater than or equal to 2. 24. The year 2018 was 2 years before 2020 so 2018 corresponds to t = 2. Thus, an expression that represents the statement is: f (2) = 7.088.

25. The year 2020 was 0 years before 2020 so 2020 corresponds to t = 0. Thus, an expression that represents the statement is: f (0) meters. 26. The year 1949 was 2020 − 1949 = 71 years before 2020 so 1949 corresponds to t = 71. Similarly, we see that the year 2000 corresponds to t = 20. Thus, an expression that represents the statement is: f (71) = f (20).

27. The year 2018 was 2 years before 2020 so 2018 corresponds to t = 2. Similarly, t = 3 corresponds to the year 2017. Thus, f (3) and f (2) are the average annual sea level values, in meters, in 2017 and 2018, respectively. Because 11 millimeters is the same as 0.011 meters, the average sea level in 2018, f (2), is 0.011 less than the sea level in 2017 which is f (3). An expression that represents the statement is: f (2) = f (3) − 0.011. Note that there are other possible equivalent expressions, such as: f (2) − f (3) = −0.011.

1.1 SOLUTIONS

28. (a) Since the potato is getting hotter, the temperature is increasing. The temperature of the potato eventually levels o˙ at the temperature of the oven, so the best answer is graph (III). (b) The vertical intercept represents the temperature of the potato at time t = 0 or just before it is put in the oven. 29. (a) The graph shows that the yield increases as more kg of seeds are planted. Moreover the yield initially increases very rapidly but then levels o˙ at about 2.5 tons, meaning there is little beneft from continuing to increase the quantity of seed. (b) We see that at 80 kg the graph is almost a horizontal line, meaning that any additional seed planted will not impact the crop yield, so it is not worth the cost of planting additional seed. 30. (a) f (30) = 10 means that the value of f at t = 30 was 10. In other words, the temperature at time t = 30 minutes was 10 C. So, 30 minutes after the object was placed outside, it had cooled to 10 C. (b) The intercept a measures the value of f (t) when t = 0. In other words, when the object was initially put outside, it had a temperature of a C. The intercept b measures the value of t when f (t) = 0. In other words, at time b the object’s temperature is 0 C. 31. (a) Since N = f (ℎ), the statement f (500) = 100, means that if ℎ = 500, then N = 100. This tells us that at an elevation of 500 feet above sea level, there are 100 species of bats. (b) The vertical intercept k is on the N-axis, so it represents the value of N when ℎ = 0. The intercept, k, is the number of bat species at sea level. The horizontal intercept, c, represents the value of ℎ when N = 0. The intercept, c, is the lowest elevation above which no bats are found. 32. The number of species of algae is low when there are few snails or lots of snails. The greatest number of species of algae (about 10) occurs when the number of snails is at a medium level (around 125 snails per square meter.) The graph supports the statement that diversity peaks at intermediate predation levels, when there are an intermediate number of snails. 33. (a) From the graph, we estimate f (3) = 0.14. This means that after 3 hours, the level of nicotine is about 0.14 mg. (b) About 4 hours. (c) The vertical intercept is 0.4. It represents the level of nicotine in the blood right after the cigarette is smoked. (d) A horizontal intercept would represent the value of t when N = 0, or the number of hours until all nicotine is gone from the body. 34. (a) The original deposit is the balance, B, when t = 0, which is the vertical intercept. The original deposit was $1000. (b) It appears that f (10) 2200. The balance in the account after 10 years is about $2200. (c) When B = 5000, it appears that t 20. It takes about 20 years for the balance in the account to reach $5000. 35. (a) The statement f (10) = 0.2 means that C = 0.2 when t = 10. In other words, CFC consumption was 0.2 million tons in 1997. (b) The vertical intercept is the value of C when t = 0. It represents CFC consumption in 1987. (c) The horizontal intercept is the value of t when C = 0. It represents the year when CFC consumption is expected to be zero. 36. (a) The graph shows maximum range is about 78 miles and it occurs at approximately 65 F. (b) The graph is decreasing for temperatures above 65 F, so above 65 F, every increase in the temperature will reduce the range. 37. (a) We see that the defcit is decreasing during this time period. (b) The statement f (4) = 485 means that the defcit was 485 billion dollars in 2014. (c) We can estimate the location of 4 on the horizontal axis, since the graph covers 2011–2015, corresponding to 1 ≤ t ≤ 5. The value 485 on the vertical axis is then the corresponding y-coordinate. These values, and a dot at the corresponding point, are shown in Figure 1.7. billion $

D = f (t) 485

Figure 1.7

t (years since 2010)

Chapter One /SOLUTIONS (d) A vertical intercept represents the value of D when t = 0, which is the US defcit (in billions of dollars) in the year 2010. (e) A horizontal intercept represents the value of t when D = 0, or the number of years after 2010 when the defcit is zero: the horizontal intercept occurs at a moment when the defcit is zero. 38. See Figure 1.8. heart rate

time administration of drug

Figure 1.8

39. Figure 1.9 shows one possible graph of gas mileage increasing to a high at a speed of 45 mph and then decreasing again. Other graphs are also possible. miles per gallon

speed (mph)

Figure 1.9

40. One possible graph is shown in Figure 1.10. Since the patient has none of the drug in the body before the injection, the vertical intercept is 0. The peak concentration is labeled on the concentration (vertical) axis and the time until peak concentration is labeled on the time (horizontal) axis. See Figure 1.10. concentration Peak concentration

time Time of peak concentration

Figure 1.10

41. (a) We want a graph of expected return as a function of risk, so expected return is on the vertical axis and risk is on the horizontal axis. The return increases as the risk increases, so the graph might look like Figure 1.11. However, the graph may not be a straight line; many other answers are also possible.

1.1 SOLUTIONS

expected return

risk

Figure 1.11 (b) High expected return and low risk is in the top left corner. See Figure 1.11. 42. (a) The statement f (1000) = 3500 means that when a = 1000, we have S = 3500. In other words, when $1000 is spent on advertising, the number of sales per month is 3500. (b) Graph I, because we expect that as advertising expenditures go up, sales will go up (not down). (c) The vertical intercept represents the value of S when a = 0, or the sales per month if no money is spent on advertising. 43. (a) In 1970, fertilizer use in the US was about 13 million tons, in India about 2 million tons, and in the former Soviet Union about 10 million tons. (b) Fertilizer use in the US rose steadily between 1950 and 1980 and has stayed relatively constant at about 18 million tons since then. Fertilizer use in India has risen steadily throughout this 50-year period. Fertilizer use in the former Soviet Union rose rapidly between 1950 and 1985 and then declined very rapidly between 1985 and 2000. 44. Since P (t) is the total cases, it can’t decrease. So (I) is the graph of the function N(t), and (II) is the graph of P (t). 45. (a) P (100) is the total number of confrmed cases of the disease up to day 100, which is the last day of the outbreak, so it is the total number of people who contract the disease during the outbreak. Therefore P (100) = 20,000. N(100) is the number of new cases on day 100. Since the outbreak is over on day 100 there are no new cases, so N(100) = 0. (b) Since the outbreak is over on day 100, there are no new cases after day 100, so P (t) = P (100) for t > 100. 46. (a) On January 8, 2021, there were 1419 new Covid-19 cases reported in Iowa. (b) On January 9, 2021, there were 898 fewer Covid-19 cases reported than on January 8, 2021. (c) On January 10 there were 581 more cases reported than on January 9. 47. (a) On February 3, 2021, there were 2296 new Covid-19 cases reported in Arizona. (b) On February 4, 2021, there were 2121 more Covid-19 cases reported than on February 3, 2021. (c) On February 5 there were 591 fewer cases reported than on February 4. (I) The incidence of cancer increase with age, but the rate of increase slows down slightly. The graph is nearly linear. This type of cancer is closely related to the aging process. (II) In this case a peak is reached at about age 55, after which the incidence decreases. (III) This type of cancer has an increased incidence until the age of about 48, then a slight decrease, followed by a gradual increase. (IV) In this case the incidence rises steeply until the age of 30, after which it levels out completely. (V) This type of cancer is relatively frequent in young children, and its incidence increases gradually from about the age of 20. (VI) This type of cancer is not age-related – all age-groups are equally vulnerable, although the overall incidence is low (assuming each graph has the same vertical scale). (b) Graph (V) shows a relatively high incidence rate for children. Leukemia behaves in this way. (c) Graph (III) could represent cancer in women with menopause as a signifcant factor. Breast cancer is a possibility here. (d) Graph (I) shows a cancer which might be caused by toxins building up in the body. Lung cancer is a good example of this.

48. (a)

49. (a) Since 2014 corresponds to t = 0, the average annual sea level in Aberdeen in 2014 was 7.071 meters. (b) Looking at the table, we see that the average annual sea level was 7.083 twenty fve years before 2014, or in the year 1989. Similar reasoning shows that the average sea level was 6.985 meters 100 years before 2014, or in 1914. (c) Because 1889 was 125 years before 2014, we see that the sea level value corresponding to the year 1889 is 6.900 (this is the sea level value corresponding to t = 125). Similar reasoning yields Table 1.1.

Chapter One /SOLUTIONS Table 1.1 Year

1889

1914

1939

1964

1989

2014

6.900

6.985

6.964

6.990

7.083

7.071

50. (a) When the temperature is 14 F the graph is at 60%, so on average an electrical vehicle will achieve 60% of its advertised range. (b) The temperatures that give 80% of advertised range are approximately 32 F and 104 F. (c) If the vehicle has a shorter range than the advertised range, then the percentage of the advertised range achieved will be less than 100%. All temperatures where the percentage is less than 100% correspond to temperatures where the vehicle will not achieve its advertised range. From the graph we see that these correspond to temperatures below 50 F and above approximately 90 F. 51. See Figure 1.12. driving speed

time

Figure 1.12 52. See Figure 1.13. distance driven

time

Figure 1.13 53. See Figure 1.14. distance from exit

time

Figure 1.14 54. See Figure 1.15. distance between cars

distance driven

Figure 1.15

1.2 SOLUTIONS

Solutions for Section 1.2 1. The slope is (3 − 2)∕(2 − 0) = 1∕2. So the equation of the line is y = (1∕2)x + 2. 2. The slope is (1 − 0)∕(1 − 0) = 1, so the equation of the line is y = x. 3. The slope is 3−1 2 1 = = . 2 − (−2) 4 2 Now we know that y = (1∕2)x + b. Using the point (−2, 1), we have 1 = −2∕2 + b, which yields b = 2. Thus, the equation of the line is y = (1∕2)x + 2. Slope =

4. The slope is −1 − 5 −6 = = 3. 2−4 −2 Substituting m = 3 and the point (4, 5) into the equation of the line, y = b + mx, we have Slope =

5 = b+3⋅4 5 = b + 12 b = −7. The equation of the line is y = −7 + 3x. 5. Rewriting the equation as 12 2 x+ 7 7 shows that the line has slope −12∕7 and vertical intercept 2∕7. y=−

6. Rewriting the equation as 3 y=4− x 2 shows that the line has slope −3∕2 and vertical intercept 4. 7. Rewriting the equation of the line as 12 4 x− 6 6 2 y = 2x − , 3

we see that the line has slope 2 and vertical intercept −2∕3. s 8. Rewriting the equation of the line as −2 x−2 4 1 y = x + 2, 2

−y =

we see the line has slope 1∕2 and vertical intercept 2. 9. The slope is positive for lines l1 and l2 and negative for lines l3 and l4 . The y-intercept is positive for lines l1 and l3 and negative for lines l2 and l4 . Thus, the lines match up as follows: (a) l1 (b) l3 (c) l2 (d) l4 10. (a) Lines l2 and l3 are parallel and thus have the same slope. Of these two, line l2 has a larger y-intercept. (b) Lines l1 and l3 have the same y-intercept. Of these two, line l1 has a larger slope. 11. (a) The slope is constant at 750 people per year so this is a linear function. The vertical intercept (at t = 0) is 28,600. We have P = 28,600 + 750t.

Chapter One /SOLUTIONS (b) The population in 2020, when t = 5, is predicted to be P = 28,600 + 750 ⋅ 5 = 32,350. (c) We use P = 35,000 and solve for t: P = 28,600 + 750t 35,000 = 28,600 + 750t 6,400 = 750t t = 8.53. The population will reach 35,000 people in about 8.53 years, or during the year 2023.

12. This is a linear function with vertical intercept 35 and slope 0.20. The formula for the monthly charge is C = 35 + 0.20m.

13. (a) The frst company’s price for a day’s rental with m miles on it is C1 (m) = 40 + 0.15m. Its competitor’s price for a day’s rental with m miles on it is C2 (m) = 50 + 0.10m. (b) See Figure 1.16. C (cost in dollars) C1 (m) = 40 + 0.15m

150 100

C2 (m) = 50 + 0.10m

50 0 200

400

600

800

m (miles)

Figure 1.16

(c) To fnd which company is cheaper, we need to determine where the two lines intersect. We let C1 = C2 , and thus 40 + 0.15m = 50 + 0.10m 0.05m = 10 m = 200. If you are going more than 200 miles a day, the competitor is cheaper. If you are going less than 200 miles a day, the frst company is cheaper. 14. This is a linear function with two parts to the formula: For 0 ≤ d ≤ 2, we pay only the $25, so P = 25. For 2 < d, the $5 charge is only on the data above 2 gigabytes, that is d − 2 gigabytes: P = 25 + 5(d − 2).

15. (a) The vertical intercept appears to be about 300 miles. When this trip started (at t = 0) the person was 300 miles from home. (b) The slope appears to be about (550 − 300)∕5 = 50 miles per hour. This person is moving away from home at a rate of about 50 miles per hour. (c) We have D = 300 + 50t.

1.2 SOLUTIONS

16. (a) On the interval from 0 to 1 the value of y decreases by 2. On the interval from 1 to 2 the value of y decreases by 2. And on the interval from 2 to 3 the value of y decreases by 2. Thus, the function has a constant rate of change and it could therefore be linear. (b) On the interval from 15 to 20 the value of s increases by 10. On the interval from 20 to 25 the value of s increases by 10. And on the interval from 25 to 30 the value of s increases by 10. Thus, the function has a constant rate of change and could be linear. (c) On the interval from 1 to 2 the value of w increases by 5. On the interval from 2 to 3 the value of w increases by 8. Thus, we see that the slope of the function is not constant and so the function is not linear. 17. For the function given by table (a), we know that the slope is slope =

27 − 25 = −2. 0−1

We also know that at x = 0 we have y = 27. Thus we know that the vertical intercept is 27. The formula for the function is y = −2x + 27. For the function in table (b), we know that the slope is slope =

72 − 62 10 = = 2. 20 − 15 5

Thus, we know that the function will take on the form s = 2t + b. Substituting in the coordinates (15, 62) we get s = 2t + b 62 = 2(15) + b = 30 + b 32 = b. Thus, a formula for the function would be s = 2t + 32. 18. No. Linear growth corresponds to the same increase in revenue every year. Facebook’s ad revenue growth was faster than linear. 19. (a) The vertical intercept is the value of S when t = 0, so it is S = 124 mph. This tells us that according to the linear model, the top speed of the fastest car in 1949 is S = 124 mph. (b) The slope is 2.25 mph/year. This tells us that according to the linear model, the top speed of the fastest car is increasing by 2.25 mph each year. 20. (a) For the frst function, from t = 0 to t = 5, the slope is $15.5 million dollars per quarter. This tells us that for this period, Zoom’s revenue increased, on average, by $15.5 million per quarter. For the second function, from t = 6 onward in 2020, the slope is $22.1 million dollars per quarter. This tells us that for this period, Zoom’s revenue increased, on average, by $22.1 million per quarter. (b) The sharp change in the rate at which Zoom earned revenue corresponds to the start of the Covid-19 pandemic in March 2020. With schools and businesses moving online, many people were spending much more time on Zoom. 21. (a) The slope is −1.22 billion dollars per year. McDonald’s revenue is decreasing at a rate of 1.22 billion dollars per year. (b) The vertical intercept is 25.44 billion dollars. In 2015, McDonald’s revenue was 25.44 billion dollars. (c) Substituting t = 6, we have R = 25.44 − 1.22 ⋅ 6 = 18.12 billion dollars. (d) We substitute R = 17 and solve for t: 25.44 − 1.22t = 17 −1.22t = −8.44 8.44 t= 1.22

.92 years.

The function predicts that annual revenue will fall to 17 billion dollars in 2022.

Chapter One /SOLUTIONS

22. (a) Reading coordinates from the graph, we see that rainfall r = 100 mm corresponds to about Q = 600 kg∕hectare, and r = 600 mm corresponds to about Q = 5800 kg∕hectare. Using a di˙erence quotient we have ΔQ 5800 − 600 = = 10.4 kg∕hectare per mm. Δr 600 − 100 (b) Every additional 1 mm of annual rainfall corresponds to an additional 10.4 kg of grass per hectare. (c) Using the slope, we see that the equation has the form Slope =

Q = b + 10.4r. Substituting r = 100 and Q = 600 we can solve for b. b + 10.4(100) = 600 b = −440. The equation of the line is Q = −440 + 10.4r. 23. (a) Reading coordinates from the graph, we see that rainfall r = 100 mm corresponds to about Q = 300 kg∕hectare, and r = 600 mm corresponds to about Q = 3200 kg∕hectare. Using a di˙erence quotient we have ΔQ 3200 − 300 = = 5.8 kg∕hectare per mm. Δr 600 − 100 (b) Every additional 1 mm of annual rainfall corresponds to an additional 5.8 kg of grass per hectare. (c) Using the slope, we see that the equation has the form Slope =

Q = b + 5.8r. Substituting r = 100 and Q = 300 we can solve for b. b + 5.8(100) = 300 b = −280. The equation of the line is Q = −280 + 5.8r. 24. The di˙erence quotient ΔQ∕Δr equals the slope of the line and represents the increase in the quantity of grass per millimeter of rainfall. We see from the graph that the slope of the line for 1939 is larger than the slope of the line for 1997. Thus, each additional 1 mm of rainfall in 1939 led to a larger increase in the quantity of grass than in 1997. (i) The slope −0.0566 = ΔA∕Δt has units seconds/year. This tells us that the time it takes the fastest car to accelerate from 0–60 mph drops by 0.0566 seconds or about 1∕20 of a second every year. (ii) The vertical intercept has the same units as A, and is therefore seconds. It corresponds to the value of A when t = 0, which is the prediction of the model for the fastest car in 1955. It tells us that the model predicts the fastest car in 1955 took 6 seconds to accelerate from 0 to 60 miles per hour. (b) We solve for t in A(t) = 1:

25. (a)

−0.0566t + 6 = 1 −5 −0.0566 = 88.339 years.

Thus, we expect the 0–60 mph time to reach 1 second a little more than 88 years after 1955, so in 2044. 26. We know W = 22.42 when t = 2001 and W = 2.17 when t = 2019. We use these two points to fnd the slope: ΔW 2.17 − 22.42 −20.25 = = = −1.125 million metric tons/year. Δt 2019 − 2001 18 We substitute the point W = 22.42 and t = 2001 and the slope m = −1.125 into the point-slope form to fnd the equation: m=

W − W0 = m(t − t0 ) W − 22.42 = −1.125(t − 2001) W − 22.42 = −1.125t + 2251.125 W = −1.125t + 2273.545. The equation of the line is W = −1.125t + 2273.545.

1.2 SOLUTIONS

27. (a) We select two points on the line. Using the values (0, 12.5) and (10, 4) we have Slope =

million barrels per day 4 − 12.5 = −0.85 . 10 − 0 year

(b) With t in years since 2006, and f (t) as the quantity of net imports in millions of barrels per day, we have f (t) = 12.5 − 0.85t. (c) We have 12.5 − 0.85t = 0 t = 14.71. The model predicts US will stop being a net importer of oil 14.71 years after 2006, so in the year 2006 + 14.71 = 2020.71. This agrees with the actual date net US imports of oil dropped below 0 million barrels per day. 28. (a) We know that the function for q in terms of p will take on the form q = mp + b. We know that the slope will represent the change in q over the corresponding change in p. Thus m = slope =

4−3 1 1 = =− . 12 − 15 −3 3

Thus, the function will take on the form 1 q = − p + b. 3 Substituting the values q = 3, p = 15, we get 1 3 = − (15) + b 3 3 = −5 + b b = 8. Thus, the formula for q in terms of p is 1 q = − p + 8. 3 (b) We know that the function for p in terms of q will take on the form p = mq + b. We know that the slope will represent the change in p over the corresponding change in q. Thus m = slope =

12 − 15 = −3. 4−3

Thus, the function will take on the form p = −3q + b. Substituting the values q = 3, p = 15 again, we get 15 = (−3)(3) + b 15 = −9 + b b = 24. Thus, a formula for p in terms of q is p = −3q + 24. 29. (a) The vertical intercept appears to be about 495. In 2015, world milk production was about 495 million tons of milk. (b) World milk production appears to be about 495 at t = 0 and about 525 at t = 4. We estimate Slope =

525 − 495 = 7.5 million tons/year. 4−0

World milk production increased at a rate of about 7.5 million tons per year during this 5-year period. (c) We have M = 495 + 7.5t.

Chapter One /SOLUTIONS

30. (a) Let x be the average minimum daily temperature ( C), and let y be the date the marmot is frst sighted. Reading coordinates from the graph, we see that temperature x = 12 C corresponds to date y = 137 days after Jan 1, and x = 22 C corresponds to y = 109 days. Using a di˙erence quotient, we have Slope =

days Δy 109 − 137 = = −2.8 . C Δx 22 − 12

(b) The slope is negative. An increase in the temperatures corresponds to an earlier sighting of a marmot. Marmots come out of hibernation earlier in years with warmer average daily minimum temperature. (c) We have Δy = Slope × Δx = (−2.8)(6) = 16.8 days. If temperatures are 6 C higher, then marmots come out of hibernations about 17 days earlier. (d) Using the slope, we see that the equation has the form y = b − 2.8x. Substituting x = 12 and y = 137 we solve for b. b − 2.8(12) = 137 b = 171. The equation of the line is y = 171 − 2.8x. 31. (a) If the frst date of bare ground is 140, then, according to the fgure, the frst bluebell fower is sighted about day 150, that is 150 − 140 = 10 days later. (b) Let x be the frst date of bare ground, and let y be the date the frst bluebell fower is sighted. Reading coordinates from the graph, we see that date x = 130 days after Jan 1 corresponds to date y = 142 days after Jan 1, and x = 170 days corresponds to y = 173 days. Using a di˙erence quotient we have Slope =

Δy 173 − 142 = = 0.775 days per day. Δx 170 − 130

(c) The slope is positive, so an increase in the x-variable corresponds to an increase in the y-variable. These variables represents days in the year, and larger values indicate days that are later in the year. Thus if bare ground frst occurs later in the year, then bluebells frst fower later in the year. Positive slope means that bluebells fower later when the snow cover lasts longer. (d) Using the slope, we see that the equation has the form y = b + 0.775x. Substituting x = 130 and y = 142 we can solve for b. b + 0.775(130) = 142 b = 41.25. The equation of the line is y = 41.25 + 0.775x. 32. (a) We have m =

256 = 14.222 cm per hour. When the snow started, there were 100 cm on the ground, so 18 f (t) = 100 + 14.222t.

(b) The domain of f is 0 ≤ t ≤ 18 hours. The range is 100 ≤ f (t) ≤ 356 cm. 33. (a) Substituting into the function g(t): g(10) = 300 + 1.4 ⋅ (10) = 314 ppm g(20) = 300 + 1.4 ⋅ (20) = 328 ppm g(100) = 300 + 1.4 ⋅ (100) = 440 ppm

1.2 SOLUTIONS

(b) Substituting into ℎ(t): ℎ(10) = 300 + 1.5 ⋅ (10) = 315 ppm ℎ(20) = 300 + 1.5 ⋅ (20) = 330 ppm ℎ(100) = 300 + 1.5 ⋅ (100) = 450 ppm (c) They di˙er in slope, 1.4 vs. 1.5 ppm per year. This creates a larger di˙erence between g(t) and ℎ(t) for larger values of t: For t = 10, the di˙erence is ℎ(10) − g(10) = 315 − 314 = 1 ppm. For t = 100, the di˙erence is ℎ(100) − g(100) = 450 − 440 = 10 ppm. 34. (a) A linear function has the form P = b + mt where t is in years since 2015. The slope is Slope =

ΔP = −0.6. Δt

The function takes the value 13.5 in the year 2015 (when t = 0). Thus, the formula is P = 13.5 − 0.6t. (b) In 2019, we have t = 4. We predict P = 13.5 − 0.6 ⋅ 4 = 11.1, that is, there are 11.1% below the poverty level in 2019. (c) The di˙erence is 11.1 − 10.5 = 0.6%; the prediction is too high. 35. (a) We want P = b + mt, where t is in years since 2005. The slope is Slope = m =

ΔP 2170 − 2048 = = 8.71429 Δt 14 − 0

Since P = 2048 when t = 0, the vertical intercept is 2048 so P = 2048 + 8.71429t. (b) The slope tells us that world grain production has been increasing at a rate of about 8.7 million tons per year. (c) The vertical intercept tells us that world grain production was 2048 million tons in 2005 (when t = 0). (d) We substitute t = 15 to fnd P = 2048 + 8.71429t ⋅ 15 = 2178.714 million tons of grain. (e) Substituting P = 2250 and solving for t, we have 2250 = 2048 + 8.71429t 202 = 8.71429t t = 23.180. Grain production is predicted to reach 2250 million tons during the year 2028. 36. (a) Let G be the increase in average global temperature above the benchmark average between 1951 and 1980 as a function of year, t. The slope of the line joining the two points (2010, 0.72) and (2020, 1.02) is m=

ΔG 1.02 − 0.72 0.3 = = = 0.03 C/year. Δt 2020 − 2010 10

We use the point-slope form to fnd the equation of the line. We substitute the point (2010, 0.72) and the slope m = 0.03 into the equation: G − G0 = m(t − t0 ) G − 0.72 = 0.03(t − 2010) G − 0.72 = 0.03t − 60.3 G = 0.03t − 59.58. The equation of the line is G = 0.03t − 59.58. (b) To calculate the global average temperature increase over the benchmark predicted for the year 2025, we substitute t = 2025 into the equation of the line and calculate: G = 0.03(2025) − 59.58 = 1.17. The formula predicts that in the year 2025, the average global temperature will have increased by 1.17 C over the benchmark.

Chapter One /SOLUTIONS

37. (a) We are given two points: S = 0.9 when t = 0 and S = 18.84 when t = 13. We use these two points to fnd the slope. Since S = f (t), we know the slope is ΔS∕Δt: ΔS 18.84 − 0.9 = = 1.38. Δt 13 − 0

The vertical intercept (when t = 0) is 0.9, so the linear equation is S = 0.9 + 1.38t. (b) The slope is 1.38 million vinyl records per year. Sales of vinyl records were increasing at a rate of 1.38 million vinyl records a year. The vertical intercept is 0.9 million vinyl records and represents the sales in the year 2006. (c) In the year 2025, we have t = 19 and we predict Sales = 0.9 + 1.38 ⋅ 19 = 27.12 million vinyl records. 38. (a) Since the initial value is $25,000 and the slope is −2000, the value of the vehicle at time t is V (t) = 25000 − 2000t. The cost of repairs has initial value 0 and slope 1500, so C(t) = 1500t. Figure 1.17 shows the graphs of these two functions. dollars 25000 V (t)

20000

C(t)

15000 10000 5000 0 2

t (years)

Figure 1.17

(b) The vehicle value equals the repair cost, when V (t) = C(t). 25000 − 2000t = 1500t 25000 = 3500t 7.143 = t. Thus, replace during the eighth year (because the frst year ends at t = 1). Since 0.143 years is 0.143(12) = 1.7 months, replacement should take place in the second month of the eighth year. (c) Since 6% of the orginal value is 1500, the vehicle should be replaced when V (t) = 1500. 25000 − 2000t = 1500 25000 = 1500 + 2000t 23500 = 2000t 11.750 = t. Thus, in the twelfth year (because the frst year ends at t = 1). Since 0.75 years is 9 months, replacement should take place at the end of the ninth month of the twelfth year.

1.2 SOLUTIONS

39. (a) It appears that P is a linear function of d since P decreases by 10 on each interval as d increases by 20. (b) The slope of the line is ΔP −10 m= = = −0.5. Δd 20 We use this slope and the frst point in the table to fnd the equation of the line. We have P = 100 − 0.5d. (c) The slope is −0.5 %/ft. The percent found goes down by 0.5% for every additional foot separating the rescuers. (d) The vertical intercept is 100%. If the distance separating the rescuers is zero, the percent found is 100%. This makes sense: if the rescuers walk with no distance between them (touching shoulders), they will fnd everyone. The horizontal intercept is the value of d when P = 0. Substituting P = 0, we have P = 100 − 0.5d 0 = 100 − 0.5d d = 200. The horizontal intercept is 200 feet. This is the separation distance between rescuers at which, according to the model, no one is found. This is unreasonable, because even when the searchers are far apart, the search will sometimes be successful. This suggests that at some point, the linear relationship ceases to hold. We have extrapolated too far beyond the given data. 40. (a) We fnd the slope m and intercept b in the linear equation C = b + mw. To fnd the slope m, we use m=

ΔC 1098.32 − 694.55 = = 134.59 dollars per ton. Δw 8−5

We substitute to fnd b: C = b + mw 694.55 = b + (134.59)(5) b = 21.6 dollars. The linear formula is C = 21.6 + 134.59w. (b) The slope is 134.59 dollars per ton. Each additional ton of waste collected costs $134.59. (c) The intercept is $21.60. The fat fee for waste collection is $21.60. This is the amount charged even if there is no waste. 41. (a) Since the di˙erences in P -values are constant (ΔP = 147) for equally spaced ℎ-values (Δℎ = 2), P could be a linear function of ℎ. (b) To fnd a formula for P = f (ℎ), we frst fnd the slope: ΔP 147 = = 73.5 torr per meter. Δt 2 Thus, the equation of the line has the form P = 73.5ℎ + b, where b is the vertical intercept. To fnd b, we choose a point from the table: for example, when ℎ = 6, we have P = 1201. Substituting these values, we have P = 73.5ℎ + b 1201 = 73.5(6) + b b = 1201 − 73.5(6) = 760. Therefore, P = f (ℎ) = 73.5ℎ + 760. The slope is 73.5 torr per meter which means that the pressure increases by 73.5 torr for each additional meter of depth below the surface of the lake. The vertical intercept is 760 torr, which means that the pressure at the surface of the lake is 760 torr. (c) The pressure at the surface of the lake is f (0) = 760 torr. Since we want the value of ℎ when the pressure is twice that at the surface, we solve f (ℎ) = 2 ⋅ 760: 2 ⋅ 760 = f (ℎ) 1520 = 73.5ℎ + 760 73.5ℎ = 760 760 ℎ= = 10.34. 73.5

Chapter One /SOLUTIONS

42. (a) We are given two points: N = 34 when l = 11 and N = 26 when l = 44. We use these two points to fnd the slope. Since N = f (l), we know the slope is ΔN∕Δl: m=

ΔN 26 − 34 = = −0.2424. Δl 44 − 11

We use the point l = 11, N = 34 and the slope m = −0.2424 to fnd the vertical intercept: N = b + ml 34 = b − 0.2424 ⋅ 11 34 = b − 2.67 b = 36.67. The equation of the line is N = 36.67 − 0.2424l. (b) The slope is −0.2424 species per degree latitude. In other words, the number of coastal dune plant species decreases by about 0.2424 for every additional degree South in latitude. The vertical intercept is 36.67 species, which represents the number of coastal dune plant species at the equator, if the model (and Australia) extended that far. (c) See Figure 1.18. number of species 40 30 20 10 l S 10

Figure 1.18

43. (a) This could be a linear function because w increases by 3.768 as ℎ increases by 4. (b) We fnd the slope m and the intercept b in the linear equation w = b + mℎ. We frst fnd the slope m using the frst two points in the table. Since we want w to be a function of ℎ, we take m=

Δw 80.032 − 76.264 = = 0.942. Δℎ 176 − 172

Substituting the frst point and the slope m = 0.942 into the linear equation w = b + mℎ, we have 76.264 = b + (0.942)(172), so b = −85.76. The linear function is w = 0.942ℎ − 85.76. The slope, m = 0.942, is in units of kg per cm. (c) We fnd the slope and intercept in the linear function ℎ = b + mw using m = Δℎ∕Δw to obtain the linear function ℎ = 1.0616w + 91.04. Alternatively, we could solve the linear equation found in part (b) for ℎ. The slope, m = 1.0616, has units cm per kg. 44. Both formulas are linear, with slope −1, which means that in one year there is a decrease of one beat per minute, so (c) is correct. 45. If M is the male’s maximum heart rate and F is the female’s, then M = 220 − a F = 226 − a, so F is 6 larger than M. Thus, female’s maximum heart rate exceeds the male’s if they are the same age.

1.2 SOLUTIONS

46. If f is the age of the female, and m the age of the male, then if they have the same MHR, 226 − f = 220 − m, so f − m = 6. Thus the female is 6 years older than the male. 47. (a) If old and new formulas give the same MHR for females, we have 226 − a = 206.9 − 0.67a, so a = 57.88 years. Thus, a woman’s MHR is the same under both methods when she is approximately 58 years old. For males we need to solve the equation 220 − a = 206.9 − 0.67a, so a = 39.7 years. Thus, a man’s MHR is the same under both methods when he is approximately 40 years old. (b) The old formula starts at either 226 or 220 and decreases. The new formula starts at 206.9 and decreases slower than the old formula. Thus, the new formula predicts a lower MHR for young people and a higher MHR for older people, so the answer is (ii). (c) Under the old formula, Heart rate reached = 0.85(220 − 65) = 131.75 beats∕minute, whereas under the new formula, Heart rate reached = 0.85(206.9 − 0.67 ⋅ 65) = 138.85 beats∕minute. The di˙erence is 138.85 − 131.75 = 7.1 beats∕minute more under the new formula. 48. If the MHR is linear with age then the rate of change, in beats per minute per year, is approximately the same at any age. Between 0 and 21 years, 2 Rate of change = − = −0.095 beats per minute per year, 21 whereas between 0 and 33 years, Rate of change = −

5 = −0.152 beats per minute per year. 33

These are not approximately the same, so are not consistent with MHR being approximately linear with age. 49. If the MHR is linear with age then the slope, in beats per minute per year, would be approximately the same at any age. Between 0 and 21 years, 6 Rate of change = − = −0.286 beats per minute per year, 21 whereas between 0 and 33 years, Rate of change = −

11 = −0.333 beats per minute per year. 33

These are not approximately the same, so they are not consistent with MHR being approximately linear with age. 50. (a) We have y = 3.5 when u = 0, which occurs when the unemployment rate does not change. When the unemployment rate is constant, Okun’a law states that US production increases by 3.5% annually. (b) When unemployment rises from 5% to 8% we have u = 3 and therefore y = 3.5 − 2u = 3.5 − 2 ⋅ 3 = −2.5. National production for the year decreases by 2.5%. (c) If annual production does not change, then y = 0. Hence 0 = 3.5 − 2u and so u = 1.75. There is no change in annual production if the unemployment rate goes up 1.75%. (d) The coeÿcient −2 is the slope Δy∕Δu. Every 1% increase in the unemployment rate during the course of a year results in an additional 2% decrease in annual production for the year.

Chapter One /SOLUTIONS

51. (a) The average hourly wage for men is AU$29.40 and the height premium is 3%, so Slope of men’s line =

(3%) AU$29.40 = 0.0882 AU$/cm. 10 cm

Since the wage is AU$29.40 for height x = 178, we have Men’s wage = 0.0882(x − 178) + 29.40 so Men’s wage = 29.40 + 0.0882x − 0.0882 ⋅ 178 Men’s wage = 13.70 + 0.0882x. (b) The average hourly wage for women is AU$24.78 and the height premium is 2%, so Slope of women’s line =

(2%) AU$24.78 = 0.0496 AU$/cm. 10 cm

Since the wage is AU$24.78 for height y = 164, we have Women’s wage = 0.0496(y − 164) + 24.78 so Women’s wage = 24.78 + 0.0496y − 0.0496 ⋅ 164 Women’s wage = 16.65 + 0.0496y. (c) For x = 178, we are given that Men’s average wage = AU$29.40. For y = 178, we fnd Women’s average wage = 16.65 + 0.0496 ⋅ 178 = AU$25.48. Thus the wage di˙erence is 29.40 − 25.48 =AU$3.92. (d) The line representing men’s wages has vertical intercept 13.70 and slope 0.0082; the line representing women’s wages has vertical intercept 16.65 and slope 0.0496. Since the vertical intercept of men’s wages is lower but the slope is larger, the two lines intersect. Thus, the model predicts there is a height making the two average hourly wages equal. For men and women of the same height, x = y. If the average hourly wages are the same Men’s wage = Women’s wage 13.70 + 0.0882x = 16.65 + 0.0496x 16.65 − 13.70 x= = 76.425 cm. 0.0882 − 0.0496 However, this height 76.425 cm, or about 30 inches, or 2 foot 6 inches, is an unrealistic height. Thus for all realistic heights, men’s average hourly wage is predicted to be larger than women’s average hourly wage.

Solutions for Section 1.3 1. The graph shows a concave down function. 2. The graph shows a concave up function. 3. The graph is concave up. 4. This graph is neither concave up or down. 5. As t increases w decreases, so the function is decreasing. The rate at which w is decreasing is itself decreasing: as t goes from 0 to 4, w decreases by 42, but as t goes from 4 to 8, w decreases by 36. Thus, the function is concave up.

1.3 SOLUTIONS

6. One possible answer is shown in Figure 1.19. y

Figure 1.19

7. The function is increasing and concave up between D and E, and between H and I. It is increasing and concave down between A and B, and between E and F . It is decreasing and concave up between C and D, and between G and H. Finally, it is decreasing and concave down between B and C, and between F and G 8. The average rate of change R between x = 1 and x = 3 is f (3) − f (1) 3−1 18 − 2 = 2 16 = 2 = 8.

9. The average rate of change R between x = −2 and x = 1 is: R=

f (1) − f (−2) 3(1)2 + 4 − (3(−2)2 + 4) 7 − 16 = = = −3. 1 − (−2) 1+2 3

Slope = −3

10 f (x) x −2

−1

10. When t = 0, we have B = 1000(1.02)0 = 1000. When t = 5, we have B = 1000(1.02)5 = 1104.08. We have Average rate of change =

ΔB 1104.08 − 1000 = = 20.82 dollars/year. Δt 5−0

During this fve year period, the amount of money in the account increased at an average rate of $20.82 per year. 11. In February of 2019, there were 14.4 million square kilometers of Arctic Sea covered by ice. 12. In February of 2009, there were 14.9 million square kilometers of Arctic Sea covered by ice. 13. Over the 10 year period from February 2009 to February 2019, the area of Arctic Sea covered by ice, shrunk, on average, by 40,000 square kilometers per year. 14. On February 1st, 2019, there were about 14.11 million square kilometers of Arctic Sea covered by ice. 15. Over the period between February 1 and March 1, the area of Arctic Sea covered by ice, grew, on average, by about 15,000 square kilometers per day.

Chapter One /SOLUTIONS

16. (a) Between 2017 and 2019, Change in net sales = Net sales in 2019 − Net sales in 2017 = 16.38 − 15.86 = 0.52 billion dollars. (b) Between 2015 and 2019, Average rate of change

in net sales

Change in net sales Change in time

Net sales in 2019 − Net sales in 2015 2019 − 2015 16.38 − 15.80 = 2019 − 2015 = 0.145 billion dollars per year. =

This means that the Gap’s net sales increased on average by 0.145 billion (145 million) dollars per year between 2015 and 2019. (c) The average rate of change was negative from 2015 to 2016, and from 2018 to 2019, when sales decreased. 17. (a) Between 2006 and 2018 Change in number of bicyclists = Bicyclists in 2018 − Bicyclists in 2006 = 47.88 − 39.69

(in millions)

= 8.19 million bicyclists. (b) The average rate of change R is the change in the number (from part (a)) divided by the change in time. 47.88 − 39.69 2018 − 2006 8.19 = 12 = 0.6825 million bicyclists per year.

This means that number of bicyclists has increased on average by 682,500 per year between 2006 and 2018. 18. (a) The average rate of change is the change in attendance divided by the change in time. Between 2015 and 2019, Average rate of change =

16.67 − 17.26 = −0.1475 million people per year 2019 − 2015 = −147,500 people per year.

(b) For each of the years from 2015–2019, the annual change in the average attendance, in million people per year, was: 2015 to 2016 ∶ 17.79 − 17.26 = 0.53 2016 to 2017 ∶ −0.53 2017 to 2018 ∶ −0.16 2018 to 2019 ∶ −0.43. (c) We average the four yearly rates of change: −0.53 + 0.53 − 0.16 − 0.43 4 0.59 =− = −0.1475, which is the same as part (a). 4

Average of the four fgures in part (b) =

19. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t) gives the distance of the particle from a point, we fnd the values of s(3) = 72 and s(10) = 144. Using these values, we fnd Δs(t) s(10) − s(3) 144 − 72 72 Average velocity = = = = = 10.286 cm∕sec. Δt 10 − 3 7 7

1.3 SOLUTIONS

20. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t) gives the distance of the particle from a point, we read o˙ the graph that s(0) = 1 and s(3) = 4. Thus, Average velocity =

Δs(t) s(3) − s(0) 4−1 = = = 1 meter∕sec. Δt 3−0 3

21. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t) gives the distance of the particle from a point, we read o˙ the graph that s(1) = 2 and s(3) = 6. Thus, Average velocity =

Δs(t) s(3) − s(1) 6−2 = = = 2 meters∕sec. Δt 3−1 2

22. (a) Average velocity = (10 − 5)∕1 = 5 ft/sec. (b) 0 ft/sec, because the distance is not changing. 23. (a) Average velocity = (−5 − 5)∕1 = −10 ft/sec. (b) Average velocity = (5 − (−5))∕1 = 10 ft/sec. 24. (a) Since f (20) − f (0) 3.6 − 0 = = 0.18, 20 − 0 20 the average velocity of the plane is about 0.18 thousand feet per second, or 180 feet per second. (b) Since f (40) − f (20) 5.6 − 3.6 Average velocity = = = 0.1, 40 − 20 40 − 20 the average velocity of the plane is about 0.1 thousand feet per second, or 100 feet per second. Average velocity =

25. (a) The value of imports was higher in 2020 than in 2005. The 2005 fgure is about 1.7 and the 2020 fgure is about 2.3, making the 2020 fgure 0.6 trillion dollars higher than the 2005 fgure. (b) The average rate of change R is the change in values divided by the change in time. 2.3 − 1.7 2020 − 2005 0.6 = 15 = 0.04 trillion dollars per year.

Between 2005 and 2020, the value of US imports has increased by an average of about 40 billion dollars per year. 26. (a) During the years 1997 through 2016 the use of the internet increased dramatically, replacing earlier newspaper readership which declined. Hence graph A records newspaper ad revenue and graph B gives internet ad revenue. (b) We compute average rates of change using di˙erence quotients. Newspaper ad revenues were about 50 billion dollars in 1997, corresponding to t = 0, and about 15 billion dollars in 2016, corresponding to t = 19. So Average rate newspapers =

15 − 50 = −1.842. 2016 − 1997

The average rate of change is negative because newspaper revenue decreased. Newspaper ad revenue decreased at an average rate of about 1.8 billion dollars per year. Internet ad revenues were about 1 billion dollars in 1997 and 60 billion dollars in 2016. So Average rate internet =

60 − 1 = 3.105. 2016 − 1997

The average rate of change is positive because internet ad revenue increased. Internet ad revenue increased at an average rate of about 3.1 billion dollars per year. 27. (a)

(i) The average velocity of the car over the interval 0 ≤ t ≤ 0.5 is 10 − 0 = 20 miles per hour. 0.5 − 0 (ii) The average velocity over the interval 0.5 ≤ t ≤ 1 is 40 − 10 = 60 miles per hour. 1 − 0.5

Chapter One /SOLUTIONS (iii) The average velocity over the interval 3 ≤ t ≤ 3.5 is 25 − 65 = −80 miles per hour. 3.5 − 3 (iv) The average velocity over the interval 3.5 ≤ t ≤ 4 is 0 − 25 = −50 miles per hour. 4 − 3.5 (b) The average velocity over the time interval 0 ≤ t ≤ 5 is given by 0−0 = 0 miles per hour. 5−0 This means that, after 5 hours, the car has returned home, the same position it started at when t = 0.

28. (a) We have 7.34 − 6.51 = 0.083 billion people per year. 2015 − 2005 90.8 − 59.7 Average rate of change of A = = 3.1 million cars per year. 2015 − 2005 2168 − 91 Average rate of change of C = = 0.444 billion subscribers per year. 2015 − 2005 (b) (i) The number of people is increasing faster since the population is increasing at 83 million per year and car production is increasing at about 3 million per year. (ii) The number of cell phone subscribers is increasing faster since the population is increasing at 83 million per year and the number of cell phone subscribers is increasing at about 444 million per year. Average rate of change of P =

29. (a) Between 2007 and 2018, we have Change in sales = Sales in 2018 − Sales in 2007 = 64.66 − 39.47 = 25.19 billion dollars. (b) Over the same period, we have Average rate of change in sales between 2007 and 2018

Change in sales Change in time

Sales in 2018 − Sales in 2007 2018 − 2007 64.66 − 39.47 = 2018 − 2007 = 2.29 billion dollars per year. =

This means that Pepsico’s sales increased on average by 2.29 billion dollars per year between 2007 and 2018. 30. (a) Negative. Rain forests are being continually destroyed to make way for housing, industry and other uses. (b) Positive. Virtually every country has a population which is increasing, so the world’s population must also be increasing overall. (c) Negative. Since cell phones became widely available in the 1990s, the number of pay phones has dropped every year to almost zero today. (d) Negative. As time passes and more sand is eroded, the height of the sand dune decreases. (e) Positive. As time passes the price of just about everything tends to increase. 31. (a) The average rate of change R is the di˙erence in amounts divided by the change in time. 468 − 728 2019 − 2006 −260 = 13 = −20 million pounds/yr.

This means that in the years between 2006 and 2019, the production of tobacco decreased at a rate of approximately 20 million pounds per year.

1.3 SOLUTIONS

(b) To have a positive rate of change, the production has to increase during one of these 1 year intervals. Looking at the data, we can see that between 2006 and 2007, production of tobacco increased by 60 million pounds. Thus, the average rate of change is positive between 2003 and 2004. Likewise, the annual average rate of change is positive between 2007 and 2008, between 2008 and 2009, between 2011 and 2012, between 2013 and 2014, and between 2016 and 2017. Nevertheless, the average rate of change over the period 2006 to 2019 is negative. 32. (a) This function is increasing and the graph is concave down. (b) From the graph, we estimate that when t = 5, we have L = 70 and when t = 15, we have L = 130. Thus, ΔL 130 − 70 = = 6 cm/year. Δt 15 − 5 During this ten year period, the sturgeon’s length increased at an average rate of 6 centimeters per year. Average rate of change of length =

33. Between 2000 and 2020: 160.74 − 142.58 20 18.16 = 20 = 0.908 million people/year.

Average rate of change =

This means that, between 2000 and 2020, the labor force increased by an average of 908,000 workers per year. Between 2000 and 2010: 153.89 − 142.58 Average rate of change = 10 11.31 = 10 = 1.131 million people/year. This means that, between 2000 and 2010, the labor force increased by an average of 1,131,000 workers per year. Between 2010 and 2020: 160.74 − 153.89 Average rate of change = 10 6.85 = 10 = 0.685 million people/year. This means that, between 2010 and 2020, the labor force increased by an average of about 685,000 workers per year. 34. Between 1950 and 2016, we have Average rate of change =

Change in marine catch 90.91 − 18.71 = = 1.09 million tons/year. Change in years 2016 − 1950

Between 1950 and 2016, marine catch increased at an average rate of 1.09 million tons each year. 35. (a) The change is given by Change between 2013 and 2019 = Revenues in 2019 − Revenues in 2013 = 137.2 − 138.79 = −1.59 billion dollars. (b) The rate of change is given by Average rate of change

between 2013 and 2019

Change in revenues Change in time

Revenues in 2019 − Revenues in 2013 2019 − 2013 137.2 − 138.79 = 2019 − 2013 = −0.265 billion dollars per year.

This means that General Motors’ revenues decreased on average by 0.265 billion dollars per year, or 265 million dollars per year, between 2013 and 2019.

Chapter One /SOLUTIONS (c) To have a positive rate of change, the revenue has to increase during one of these intervals. Looking at the data, we can see that between 2015 and 2016, revenues increased by 13.45 billion dollars. Thus the average rate of change is positive between 2015 and 2016. Likewise, the annual average rate of change is positive between 2017 and 2018.

36. We have

88.4 − 104.1 −15.7 = = −1.744 million households/year. 2019 − 2010 9 During this 9-year period, the number of US households with cable television decreased, on average, by 1.74 million households per year. Average rate of change =

37. (a) As time passes after a cigarette is smoked, the nicotine level in the body decreases, so the average rate of change is negative. The slope of a secant line between t = 0 and t = 3 is negative. (b) The nicotine level at t = 0 is N = 0.4 and the nicotine level at t = 3 is about N = 0.14. We have Average rate of change =

f (3) − f (0) 0.14 − 0.4 = = −0.087 mg/hour. 3−0 3−0

During this three hour period, nicotine decreased at an average rate of 0.087 mg of nicotine per hour. 38. Between 1804 and 1927, the world’s population increased 1 billion people in 123 years, for an average rate of change of 1∕123 billion people per year. We convert this to people per minute: 1, 000, 000, 000 1 people/year ⋅ years/minute = 15.468 people/minute. 123 60 ⋅ 24 ⋅ 365 Between 1804 and 1927, the population of the world increased at an average rate of 15 people per minute. Similarly, we fnd the following: Between 1927 and 1960, the increase was 58 people per minute. Between 1960 and 1974, the increase was 136 people per minute. Between 1974 and 1987, the increase was 146 people per minute. Between 1987 and 1999, the increase was 159 people per minute. Between 1999 and 2012, the increase was 146 people per minute. 39. (a)

(i) Between the 6th and 8th minutes the concentration changes from 0.336 to 0.298 so we have Average rate of change =

0.298 − 0.336 = −0.019 (mg/ml)/min. 8−6

(ii) Between the 8th and 10th minutes we have Average rate of change =

0.266 − 0.298 = −0.016 (mg/ml)/min. 10 − 8

(b) The signs are negative because the concentration is decreasing. The magnitude of the decrease is decreasing because, as the concentration falls, the rate at which it falls decreases. 40. (a) The muscle contracts at 30 cm/sec when it is not pulling any load at all. (b) If the load is 4 kg, the contraction velocity is zero. Thus, the maximum load the muscle can move is 4 kg. 41. (a) The contraction velocity for 1 kg is about 13 cm/sec and for 3 kg it is about 2 cm/sec. The change is 2 − 13 = −11 cm/sec. (b) Using the answer to part (a), we have Average rate of change =

2 − 13 = −5.5 (cm/sec)/kg. 3−1

42. (a) Between 2005 and 2020 Change in sales = Sales in 2020 − Sales in 2005 = 77.87 − 38.8 = 39.07 billion dollars. (b) Between 2005 and 2020 Average rate of change in sales

Change in sales Change in time

1.3 SOLUTIONS

Sales in 2020 − Sales in 2005 2020 − 2005 77.87 − 38.3 = 15 = 2.60 billion dollars per year. =

This means that Intel’s sales increased on average by 2.60 billion dollars per year between 2005 and 2020. (i) f (2001) = 272 (ii) f (2019) = 607 (b) The average yearly increase is the rate of change.

43. (a)

Yearly increase =

f (2019) − f (2001) 607 − 272 335 = = = 18.611 billionaires per year. 2019 − 2001 18 18

So on average, there were between 18 and 19 more billionaires in the US each year. (c) Since we assume the rate of increase remains constant, we use a linear function with slope 335∕18 = 18.611 billionaires per year. The equation is 335 f (t) = b + t 18 where f (2001) = 272, so 335 272 = b + (2001) 18 b = −36,968.833. Thus, f (t) = 18.611t − 36,968.833. 44. (a) See Figure 1.20. (b) The average velocity between two times can be represented by the slope of the line joining these points on the position curve. In Figure 1.21, the average velocity between t = 0 and t = 3 is equal to the slope of l1 , while the average velocity between t = 3 and t = 6 is equal to the slope of l2 . We can see that the slope of l2 is greater than the slope of l1 and so the average velocity between t = 3 and t = 6 is greater. s (ft)

s (ft)

✛l2

(6, s(6))

(8, s(8))

Slope =

Average velocity from t = 3 to t = 6

✛

Average velocity Slope = from t = 2 to t = 8

✛

Slope between t = 0 and t = 3

(3, s(3)) (0, s(0))

(2, s(2)) 1

t (sec)

Figure 1.20

t (sec)

Figure 1.21

(c) Since the average velocity can be represented as the slope of the line between the two values given, we just have to look at the slope of the line passing through (6, s(6)) and (9, s(9)). We notice that the distance traveled at t = 6 is greater than the distance traveled at t = 9. Thus, the line between those two points will have negative slope, and thus, the average velocity will be negative. 45. (a) If the lizard were running faster and faster, the graph would be concave up. In fact, the graph looks linear; it is not concave up or concave down. The lizard is running at a relatively constant velocity during this short time interval. (b) During this 0.8 seconds, the lizard goes about 2.1 meters, so the average velocity is about 2.1∕0.8 = 2.6 meters per second. 46. We know that a function is linear if its slope is constant. A function is concave up if the slope increases as t gets larger. And a function is concave down if the slope decreases as t gets larger. Looking at the points t = 10, t = 20 and t = 30 we see that the slope of F (t) decreases since F (20) − F (10) 22 − 15 = = 0.7 20 − 10 10

Chapter One /SOLUTIONS while

F (30) − F (20) 28 − 22 = = 0.6. 30 − 20 10 Looking at the points t = 10, t = 20 and t = 30 we see that the slope of G(t) is constant G(20) − G(10) 18 − 15 = = 0.3 20 − 10 10 and

G(30) − G(20) 21 − 18 = = 0.3. 30 − 20 10 Also note that the slope of G(t) is constant everywhere. Looking at the points t = 10, t = 20 and t = 30 we see that the slope of H(t) increases since H(20) − H(10) 17 − 15 = = 0.2 20 − 10 10 while

H(30) − H(20) 20 − 17 = = 0.3. 30 − 20 10 Thus F (t) is concave down, G(t) is linear, and H(t) is concave up.

47. Between 0 and 21 years, Average rate of change = −

9 = −0.429 beats per minute per year, 21

whereas between 0 and 33 years, Average rate of change = −

26 = −0.788 beats per minute per year. 33

Because the average rate of change is negative, a decreasing function is suggested. Also, as age increases, the average rate of change decreases, suggesting the graph of the function is concave down. (Since the average rate of change is negative and increasing in absolute value, this rate is decreasing.) 48. Velocity is the slope of the graph of distance against time, so we draw a graph with a slope that starts out small and increases (so the graph gets steeper). Then the slope decreases until the car stops. When the car is stopped, the distance stays constant and the slope is zero. See Figure 1.22. distance

time

Figure 1.22

49. (a) Using a rate of change of 2.25 miles per hour per year, we calculate that the top speed in 2016 is 2.25 mph less than that in 2017, so 2016 Top speed = 277.9 − 2.25 = 275.65 mph. In addition, we calculate 2020 Top speed = 3 ⋅ 2.25 + 277.9 = 284.65 mph. (b) The top speed increased 316.1 − 277.9 = 38.2 mph over three years so Average rate of change =

38.2 = 12.7 mph per year. 3

(c) The di˙erence may have been due to disruptive technological advances, which caused the rate to increase. It may also be due to the fact that the 2.25 mph per year rate is an average. Some years may be higher than 2.25 mph and some may be lower, even though the average is 2.25 mph per year.

1.3 SOLUTIONS

(d) The top speed in 2020 is already at 316.1 mph, so half the speed of sound requires a gain of 67.9 mph. If the rate of change is 2.25 mph per year 67.9 Time for 67.9 mph gain = = 30.178 years. 2.25 Thus, top speed is expected to reach 384 mph by 2051 if the rate of change is 2.25 mph per year. If the rate is 12.7 mph per year as in part (b), then Time for 67.9 mph gain =

67.9 = 5.35 years 12.7

which tells us top speed is expected to reach 384 mph by 2026. 50. We have Relative change =

15,000 − 12,000 = 0.25. 12,000

The quantity B increases by 25%. 51. We have Relative change =

450 − 400 = 0.125. 400

Relative change =

0.05 − 0.3 = −0.833. 0.3

The quantity S increases by 12.5%. 52. We have The quantity W decreases by 83.3%. 53. We have Relative change =

47 − 50 = −0.06. 50

The quantity R decreases by 6%. 54. The changes are −86.7 in 1931 and −4485.4 in 2008 (they are negative because the Dow Jones Industrial average fell in both years), but the relative changes are Relative change, 1931, = − Relative change, 2008, = −

86.7 = −52.7% 164.6

4485.4 = −33.8%. 13,261.8

The 1931 relative change is bigger in magnitude. 55. The changes are 2.0 million for 1800-1810 and 28.0 million for 1950-1960. The relative changes are 2.0 = 38.5% 5.2

28.0 = 18.5%. 151.3

and

The relative change is bigger for 1800-1810. On the other hand, the change in 1950-1960 represents the baby boom that has had various important consequences for the US in the last 50 years so many would consider the implications of that change greater. 56. The changes are 5 students for the small class and 20 students for the larger class. The relative changes are 5 = 100% 5

and

20 = 66.667%. 30

The relative change for the small class is bigger than the relative change for the large class, but the extra 20 students added to a class of 30 has a potentially larger e˙ect on the class. 57. The changes are 400,000 and 500,000. The relative changes are 400,000 = 400% 100,000

and

500,000 = 2.5%. 20,000,000

The relative change for the frst sales is much larger and would have a much larger e˙ect on the company.

Chapter One /SOLUTIONS

58. (a) We have Relative change =

2000 − 1000 = 1 = 100%. 1000

(b) We have Relative change =

1000 − 2000 = −0.5 = −50%. 2000

1,001,000 − 1,000,000 = 0.001 = 0.1%. 1,000,000

260.64 − 298.97 = −0.128 = −12.8%. 298.97

The Dow Jones average dropped by 12.8%. 60. We have Relative change =

Change in cost 55 − 50 = = 0.1. Initial cost 50

The cost to mail a letter increased by 10%. 61. Let f (t) be the CPI in January of year t. Since the infation rate is the relative rate of change with Δt = 1, we have Infation rate =

f (t + 1) − f (t) 1 Δf 1 f (t + 1) − f (t) = = f Δt f (t) 1 f (t)

we have 237.017 − 236.736 = 0.0012 = 0.12% per year 236.736 240.007 − 237.017 2015 infation = = 0.0126 = 1.26% per year 237.017 245.120 − 240.007 2016 infation = = 0.0213 = 2.13% per year 240.007 251.107 − 245.120 2017 infation = = 0.0244 = 2.44% per year 245.120 255.657 − 251.107 2018 infation = = 0.0181 = 1.81% per year 251.107 2014 infation =

Rounding to two decimal places, we have Table 1.2 Year

2014

2015

2016

2017

2018

Infation (% per year)

0.12%

1.26%

2.13%

2.44%

1.81%

62. (a) Let f (t) be the GDP at time t. We are using Δt = 0.25, so Relative growth rate =

(Δf (t))∕Δt (f (t + 0.25) − f (t))∕0.25 ≈ f (t) f (t)

we have (14.81 − 14.67)∕0.25 = 0.038 = 3.8% per year 14.67 (14.84 − 14.81)∕0.25 Relative growth rate at 0.25 ≈ = 0.008 = 0.8% per year 14.81 (14.55 − 14.84)∕0.25 Relative growth rate at 0.5 ≈ = −0.078 = −7.8% per year 14.84 (14.38 − 14.55)∕0.25 Relative growth rate at 0.75 ≈ = −0.047 = −4.7% per year. 14.55 Relative growth rate at 0 ≈

We have

1.3 SOLUTIONS

Table 1.3 t (years since 2008)

0.25

0.5

0.75

GDP growth rate (% per year)

3.8

0.8

−7.8

−4.7

(b) Yes, since the relative growth rate was negative in the last two quarters of 2008, the GDP decreased then, and the economy was in recession. 63. (a) We know P (37) is the total number of Covid-19 cases confrmed up to and including day 37 in Minnesota. Similarly, P (36) is the total number of cases confrmed up to and including day 36. Thus the di˙erence P (37) − P (36) is the number of new cases confrmed on day 37 (April 11). This means that P (37) − P (36) = N(37). There were 80 new cases on day t = 37 (April 11). (b) We have P (37) − P (36) = 0.074 P (36) P (37) − P (36) = 0.074P (36) P (37) = 1.074P (36). There are 7.4% more total cases by day 37 than by day 36. In other words, the total number of cases increased by 7.4% from day 36 to day 37 (so, from April 10 to April 11). (c) The number P (37) − P (36) is the number of new cases confrmed on day 37. Thus, (P (37) − P (36))∕P (37) gives the fraction of confrmed cases by day 37 that were frst confrmed on day 37. Hence 6.9% of the total cases confrmed by day 37 are new cases on that day. (d) The sum P (35) + N(36) + N(37) gives the total number of cases up to day 35, plus the number of new cases on day 36, plus the number of new cases on day 37. The total, 1164, is thus the total number of Covid-19 cases up to and including day 37 (April 11). Note that P (35) + N(36) + N(37) = P (37). 64. (a) The quantity P (102) is the total number of Covid-19 cases confrmed up to and including day 102 (June 15) in Maryland, and P (95) is the total number of cases confrmed up to and including day 95 (June 8). The di˙erence P (102) − P (95) is thus the number of cases that were confrmed during the 7 days from t = 96 to t = 102 (from June 9 to June 15). In other words, there were 4081 cases confrmed in the week ending on day 102 (June 15). (b) This expression is the average rate of change of P (t) between t = 95 and t = 102. Thus, the total number of cases was growing on average by 583 cases each day between t = 95 and t = 102. In other words, there were on average 583 new reported cases each day between t = 95 and t = 102. (c) This is the relative rate of change of P (t) between t = 95 and t = 102. Thus, it says that there were 7.2% more total cases by day 102 than by day 95. 65. (a) The largest time interval was 2015–2017 since the percentage growth rate increased each year from 2015 to 2016 to 2017. This means that from 2015 to 2017 the US consumption of hydroelectric power grew relatively more with each successive year. (b) The largest time interval was 2012–2015 since the percentage growth rates were negative for each of these four consecutive years. This means that the amount of hydroelectric power consumed by the US industrial sector steadily decreased during this period. 66. (a) We have Relative change in price of candy =

Change in price 1.25 − 1 = = 0.25. Initial price 1

The price of the candy has gone up by 25%. (b) We have Relative change in quantity of sold candy =

Change in quantity 2440 − 2765 = = −0.12. Initial quantity 2765

The number of candy sold dropped down by 12%. (c) We have | Relative change in quantity | | −0.12 | | | | | | Relative change in price | = | 0.25 | = 0.48. | | | | So for a 1% increase in a $1 candy the quantity sold drops by 0.48%.

Chapter One /SOLUTIONS

67. (a) The largest time interval was 2016–2018 since the percentage growth rate increased each year from 0.33 to 6.20 to 10.10 from 2016 to 2018. This means US exports of biofuels were growing faster and faster between 2016 and 2018. (Note that the percentage growth rate was a decreasing function of time over 2012–2014 and in 2015–2016.) (b) The largest time interval was 2015–2018 since the percentage growth rates were positive for each of these consecutive years. This means that the amount of biofuels exported from the US steadily increased during the four years from 2015 to 2018, after a large fall in 2014. 68. (a) The largest time interval was 2008–2010 since the percentage growth rate decreased each year from 2008 to 2009 to 2010. This means that from 2008 to 2010 the US price per watt of a solar panel fell relatively more with each successive year. (b) The largest time interval was 2009–2010 since the percentage growth rates were negative for each of these two consecutive years. This means that the US price per watt of a solar panel decreased during these two years, after an increase in the previous year.

Solutions for Section 1.4 1. (a) The company makes a proft when the revenue is greater than the cost of production. Looking at the fgure in the problem, we see that this occurs whenever more than roughly 335 items are produced and sold. (b) If q = 600, revenue ≈ 2400 and cost ≈ 1750 so proft is about $650. 2. (a) The fxed costs are the price of producing zero units, or C(0), which is the vertical intercept. Thus, the fxed costs are roughly $75. The marginal cost is the slope of the line. We know that C(0) = 75 and looking at the graph we can also tell that C(30) = 300. Thus, the slope or the marginal cost is Marginal cost =

300 − 75 225 = = 7.50 dollars per unit. 30 − 0 30

(b) Looking at the graph it seems that C(10) ≈ 150. Alternatively, using what we know from parts (a) and (b) we know that the cost function is C(q) = 7.5q + 75. Thus, C(10) = 7.5(10) + 75 = 75 + 75 = $150. The total cost of producing 10 items is $150. 3. See Figure 1.23 $

✛

R (c) (ii) loses money

✲

✛ (a) Fixed costs

✲

✲ (c) (i) makes profit q

✒ (b) Break-even quantity

Figure 1.23

1.4 SOLUTIONS

4. We know that the fxed cost is the cost that the company would have to pay if no items were produced. Thus Fixed cost = C(0) = $5000. We know that the cost function is linear and we know that the slope of the function is exactly the marginal cost. Thus Slope = Marginal cost =

5020 − 5000 20 = = 4 dollars per unit produced. 5−0 5

Thus, the marginal cost is 4 dollars per unit produced. We know that since C(q) is linear C(q) = m ⋅ q + b, where m is the slope and b is the value of C(0), the vertical intercept. Or in other words, m is equal to the marginal cost and b is equal to the fxed cost. Thus C(q) = 4q + 5000. 5. The 5.7 represents the fxed cost of $5.7 million. The 0.002 represents the variable cost per unit in millions of dollars. Thus, each additional unit costs an additional $0.002 million = $2000 to produce. 6. (a) A company with little or no fxed costs would be one that does not need much start-up capital and whose costs are mainly on a per unit basis. An example of such a company is a consulting company, whose major expense is the time of its consultants. Such a company would have little fxed costs to worry about. (b) A company with marginal cost of zero (or almost) would be one that can produce a product with little or no additional costs per unit. An example is a computer software company. The major expense of such a company is software development, a fxed cost. Additional copies of its software can be very easily made. Thus, its marginal costs are rather small. 7. (a) The statement f (12) = 60 says that when p = 12, we have q = 60. When the price is $12, we expect to sell 60 units. (b) Decreasing, because as price increases, we expect less to be sold. 8. (a) The cost of producing 500 units is C(500) = 6000 + 10(500) = 6000 + 5000 = $11,000. The revenue the company makes by selling 500 units is R(500) = 12(500) = $6000. Thus, the cost of making 500 units is greater than the money the company will make by selling the 500 units, so the company does not make a proft. The cost of producing 5000 units is C(5000) = 6000 + 10(5000) = 6000 + 50000 = $56,000. The revenue the company makes by selling 5000 units is R(5000) = 12(5000) = $60,000. Thus, the cost of making 5000 units is less than the money the company will make by selling the 5000 units, so the company does make a proft. (b) The break-even point is the number of units that the company has to produce so that in selling those units, it makes as much as it spent on producing them. That is, we are looking for q such that C(q) = R(q). Solving we get C(q) = R(q) 6000 + 10q = 12q 2q = 6000 q = 3000.

Chapter One /SOLUTIONS Thus, if the company produces and sells 3000 units, it will break even. Graphically, the break-even point, which occurs at (3000, $36,000), is the point at which the graphs of the cost and the revenue functions intersect. (See Figure 1.24.) p($) R(q) C(q) 50000 40000 (3000, 36,000)

30000 20000 10000

q 1000 2000 3000 4000 5000

Figure 1.24 9. The 5500 tells us the quantity that would be demanded if the price were $0. The 100 tells us the change in the quantity demanded if the price increases by $1. Thus, a price increase of $1 causes demand to drop by 100 units. 10. Since the price p is on the vertical axis and the quantity q is on the horizontal axis we would like a formula which expresses p in terms of q. Thus 75p + 50q = 300 75p = 300 − 50q 2 p = 4 − q. 3 p 4 3 2 1 q 1

Figure 1.25 From the graph in Figure 1.25 we see that the vertical intercept occurs at p = 4 dollars and the horizontal intercept occurs at q = 6. This tells us that at a price of $4 or more nobody would buy the product. On the other hand even if the product were given out for free, no more than 6 units would be demanded. 11. The cost function C(q) = b + mq satisfes C(0) = 500, so b = 500, and MC = m = 6. So C(q) = 500 + 6q. The revenue function is R(q) = 12q, so the proft function is (q) = R(q) − C(q) = 12q − 500 − 6q = 6q − 500. 12. The cost function C(q) = b + mq satisfes C(0) = 35,000, so b = 35,000, and MC = m = 10. So C(q) = 35,000 + 10q. The revenue function is R(q) = 15q, so the proft function is (q) = R(q) − C(q) = 15q − 35,000 − 10q = 5q − 35,000.

1.4 SOLUTIONS

13. The cost function C(q) = b + mq satisfes C(0) = 5000, so b = 5000, and MC = m = 15. So C(q) = 5000 + 15q. The revenue function is R(q) = 60q, so the proft function is (q) = R(q) − C(q) = 60q − 5000 − 15q = 45q − 5000. 14. The cost function C(q) = b + mq satisfes C(0) = 0, so b = 0. There are 32 ounces in a quart, so 160 in 5 quarts, or 20 cups at 8 ounces per cup. Thus each cup costs the operator 20 cents = 0.20 dollars, so MC = m = 0.20. So C(q) = 0.20q. The revenue function is R(q) = 0.25q. So the proft function is (q) = R(q) − C(q) = 0.25q − 0.20q = 0.05q. 15. (a) We know that the cost function will be of the form C(q) = b + m ⋅ q where m is the slope of the graph and b is the vertical intercept. We also know that the fxed cost is the vertical intercept and the variable cost is the slope. Thus, we have C(q) = 5000 + 30q. We know that the revenue function will take on the form R(q) = pq where p is the price charged per unit. In our case the company sells the chairs at $50 a piece so R(q) = 50q. (b) Marginal cost is $30 per chair. Marginal revenue is $50 per chair. (c) p ($) 25000

R(q)

20000

C(q)

15000 (250, 12,500)

10000 5000

q 100

300

500

Figure 1.26 (d) We know that the break-even point is the number of chairs that the company has to sell so that the revenue will equal the cost of producing these chairs. In other words, we are looking for q such that C(q) = R(q). Solving we get C(q) = R(q) 5000 + 30q = 50q 20q = 5000 q = 250. Thus, the break-even point is 250 chairs and $12,500. Graphically, it is the point in Figure 1.26 where the cost function intersects the revenue function.

Chapter One /SOLUTIONS

16. (a) Regardless of the number of rides taken, it costs $23 to get in. In addition, it costs $6.00 per ride, so the function R(n) is R(n) = 23 + 6n. (b) Substituting in the values n = 2 and n = 8 into the formula for R(n), we get R(2) = 23 + 6⋅ = 23 + 12 = $35. This means that admission and 2 rides costs $35. R(8) = 23 + 6 ⋅ 8 = 23 + 48 = $71. This means that admission and 8 rides costs $71. 17. (a) A 20-page hardcover book is $30 and the price of each additional page is $1.50. Thus, the price of a hardcover book with q additional pages is P1 (q) = 30 + 1.50q. For the softcover book, the price is $25 for a 20-page book and $2 for each additional page. Thus, the price of a softcover book with q additional pages is P2 (q) = 25 + 2q. (b) At 20 additional pages, the price of a hardcover book is P1 (20) = 30 + 1.50(20) = 30 + 30 = $60. At 20 additional pages, the price of a softcover book is P2 (20) = 25 + 2(20) = 25 + 40 = $65. Thus, for 20 additional pages, the hardcover book is cheaper. (c) We fnd the point q at which P1 (q) = P2 (q). Solving we get P1 (q) = P2 (q) 30 + 1.50q = 25 + 2q 5 = 0.50q q = 10. Thus, the price of the two formats is the same for 10 additional pages. We evaluate each price at q = 10 to check that they are equal: P1 (q) = 30 + 1.5(10) = $45 P2 (q) = 20 + 2(10) = $45. 18. (a) The fxed cost is the cost that would have to be paid even if nothing was produced. That is, the fxed cost is C(0) = 4000 + 2(0) = $4000. (b) We know that for every additional item produced, the company must pay another $2. Thus, the marginal cost is $2 per chair. (c) We know that the revenue takes on the form R(q) = p ⋅ q where q is the quantity produced and p is the price the company is charging for an item. Thus in our case, p= So the company is charging $10 per item.

R(q) 10q = = $10. q q

1.4 SOLUTIONS (d)

$ R(q)

C(q) 5000

q q0 = 500

Figure 1.27 We know that the company will make a proft for q > q0 since the line R(q) lies above the line C(q) in that region, so revenues are greater than costs. (e) Looking at Figure 1.27 we see that the two graphs intersect at the point where q0 = 500. Thus, the company will break even if it produces 500 units. Algebraically, we know that the company will break even for q0 such that the cost function is equal to the revenue function at q0 . That is, when C(q0 ) = R(q0 ). Solving we get C(q0 ) = R(q0 ) 4000 + 2q0 = 10q0 8q0 = 4000 q0 = 500. To fnd the total revenue, we evaluate and fnd: C(500) = R(500) = $5000. 19. (a) We know that the function for the cost of running the theater is of the form C = b + mq where q is the number of customers, m is the variable cost and b is the fxed cost. Thus, the function for the cost is C = 5000 + 11q. We know that the revenue function is of the form R = pq where p is the average revenue per customer per customer. Thus, the revenue function is R = 16q. The theater makes a proft when the revenue is greater than the cost, that is when R > C. Substituting R = 16q and C = 11q + 5000, we get R>C 16q > 11q + 5000 5q > 5000 q > 1000. Thus, the theater makes a proft when it has more than 1000 customers.

Chapter One /SOLUTIONS (b) The graph of the two functions is shown in Figure 1.28. $ 22,000

15,000 10,000 5000 q 1000

2000

Figure 1.28 20. (a) We have C(q) = 6000 + 2q and R(q) = 5q and so (q) = R(q) − C(q) = 5q − (6000 + 2q) = −6000 + 3q. (b) See Figure 1.29. We fnd the break-even point, q0 , by setting the revenue equal to the cost and solving for q: Revenue = Cost 5q = 6000 + 2q 3q = 6000 q = 2000. The break-even point is q0 = 2000 puzzles. Notice that this is the same answer we get if we set the proft function equal to zero. p($) R(q) = 15q

20,000

C(q) = 6000 + 2q

q 5000

Figure 1.29: Cost and revenue functions for the jigsaw puzzle company 21. The price at any quantity q is the slope of the revenue function at that value of q. To identify the graphs, look at how the slope changes as we move along the line to the right in the direction of increasing q. Graph (I) corresponds to (c), price increasing with quantity. Graph (II) corresponds to (b), price increasing for small quantities and decreasing for large quantities. Graph (III) corresponds to (d), as the quantity increases, the price decreases. Graph (IV) corresponds to (a), constant price because the curve is a line. 22. (a) The cost function is of the form C(q) = b + m ⋅ q where m is the variable cost and b is the fxed cost. Since the variable cost is $20 and the fxed cost is $650,000, we get C(q) = 650,000 + 20q.

1.4 SOLUTIONS

The revenue function is of the form R(q) = pq where p is the price that the company is charging the buyer for one pair. In our case the company charges $70 a pair so we get R(q) = 70q. The proft function is the di˙erence between revenue and cost, so (q) = R(q) − C(q) = 70q − (650,000 + 20q) = 70q − 650,000 − 20q = 50q − 650,000. (b) Marginal cost is $20 per pair. Marginal revenue is $70 per pair. Marginal proft is $50 per pair. (c) We are asked for the number of pairs of shoes that need to be produced and sold so that the proft is larger than zero. That is, we are trying to fnd q such that (q) > 0. Solving we get (q) > 0 50q − 650,000 > 0 50q > 650,000 q > 13,000. Thus, if the company produces and sells more than 13,000 pairs of shoes, it will make a proft. 23. (a) We know that the function for the value of the robot at time t will be of the form V (t) = m ⋅ t + b. We know that at time t = 0 the value of the robot is $15,000. Thus the vertical intercept b is b = 15,000. We know that m is the slope of the line. Also at time t = 10 the value is $0. Thus m=

0 − 15,000 −15,000 = = −1500. 10 − 0 10

Thus we get V (t) = −1500t + 15,000 dollars. (b) The value of the robot in three years is V (3) = −1500(3) + 15,000 = −4500 + 15,000 = $10,500. 24. (a) We know that the function for the value of the tractor will be of the form V (t) = m ⋅ t + b where m is the slope and b is the vertical intercept. We know that the vertical intercept is simply the value of the function at time t = 0, which is $50,000. Thus, b = $50,000. Since we know the value of the tractor at time t = 20 we know that the slope is m=

V (20) − V (0) 10,000 − 50,000 −40,000 = = = −2000. 20 − 0 20 20

Thus we get V (t) = −2000t + 50,000 dollars.

Chapter One /SOLUTIONS (b)

$ 50,000

10,000

t (years)

Figure 1.30 (c) Looking at Figure 1.30 we see that the vertical intercept occurs at the point (0, 50,000) and the horizontal intercept occurs at (25, 0). The vertical intercept tells us the value of the tractor at time t = 0, namely, when it was brand new. The horizontal intercept tells us at what time t the value of the tractor will be $0. Thus the tractor is worth $50,000 when it is new, and it is worth nothing after 25 years. 25. (a) The slope of the linear function is 25,000 − 100,000 ΔV = = −5000. Δt 15 − 0 The vertical intercept is V = 100,000 so the formula is Slope =

V = 100,000 − 5000t. (b) At t = 5, we have V = 100,000 − 5000 ⋅ 5 = 75,000. In 2020, the bus is worth $75,000. (c) The vertical intercept is 100,000 and represents the value of the bus in dollars in 2015. The horizontal intercept is the value of t when V = 0. Solving 0 = 100,000 − 5000t for t, we fnd t = 20. The bus is worth nothing when t = 20 years, that is, in the year 2035. (d) The domain of the function is 0 ≤ t ≤ 20. The bus cannot have negative value. 26. We know that at the point where the price is $5 per scoop the quantity must be 240. Thus we can fll in the graph as follows: p (price per scoop) 10

Demand

q (quantity)

120 240 360 480

Figure 1.31 (a) Looking at Figure 1.31 we see that when the price per scoop $2.50, the quantity given by the demand curve is roughly 360 scoops. (b) Looking at Figure 1.31 we see that when the price per scoop is $7.50, the quantity given by the demand curve is roughly 120 scoops. 27. (a) Since cost is less than revenue for quantities in the table between 20 and 60 units, production appears to be proftable between these values. (b) Proft = Revenue − Cost is show in Table 1.4. The maximum proft is obtained at a production level of about 40 units. Table 1.4 Quantity

Cost ($)

120

400

600

780

1000

1320

1800

2500

3400

Revenue ($)

300

600

900

1200

1500

1800

2100

2400

Proft

−120

−100

120

200

180

−400

−1000

1.4 SOLUTIONS

28. (a) We know that as the price per unit increases, the quantity supplied increases, while the quantity demanded decreases. So Table 1.30 of the text is the demand curve (since as the price increases the quantity decreases), while Table 1.31 of the text is the supply curve (since as the price increases the quantity increases.) (b) Looking at the demand curve data in Table 1.30 we see that a price of $155 gives a quantity of roughly 14. (c) Looking at the supply curve data in Table 1.31 we see that a price of $155 gives a quantity of roughly 24. (d) Since supply exceeds demand at a price of $155, the shift would be to a lower price. (e) Looking at the demand curve data in Table 1.30 we see that if the price is less than or equal to $143 the consumers would buy at least 20 items. (f) Looking at the data for the supply curve (Table 1.31) we see that if the price is greater than or equal to $110 the supplier will produce at least 20 items. 29. (a) We know that the equilibrium point is the point where the supply and demand curves intersect. Looking at the fgure in the problem, we see that the price at which they intersect is $10 per unit and the corresponding quantity is 3000 units. (b) We know that the supply curve climbs upward while the demand curve slopes downward. Thus we see from the fgure that at the price of $12 per unit the suppliers will be willing to produce 3500 units while the consumers will be ready to buy 2500 units. Thus we see that when the price is above the equilibrium point, more items would be produced than the consumers will be willing to buy. Thus the producers end up wasting money by producing that which will not be bought, so the producers are better o˙ lowering the price. (c) Looking at the point on the rising curve where the price is $8 per unit, we see that the suppliers will be willing to produce 2500 units, whereas looking at the point on the downward sloping curve where the price is $8 per unit, we see that the consumers will be willing to buy 3500 units. Thus we see that when the price is less than the equilibrium price, the consumers are willing to buy more products than the suppliers would make and the suppliers can thus make more money by producing more units and raising the price. 30. (a) We know that the cost function will be of the form C = b + mq where m is the variable cost and b is the fxed cost. In this case this gives C = 7000 + 10q. We know that the revenue function is of the form R = pq where p is the price per shirt. Thus in this case we have R = 25q. (b) We are given q = 4000 − 40p. We are asked to fnd the demand when the price is $25. Substituting p = 25 into the equation, we get q = 4000 − 40(25) = 4000 − 1000 = 3000. Given this demand we know that the cost of producing q = 3000 shirts is C = 7000 + 10 ⋅ 3000 = 7000 + 30000 = $37,000. The revenue from selling q = 3000 shirts is R = 25 ⋅ 3000 = $75,000. Thus the proft is (25) = R − C or in other words (25) = 75,000 − 37,000 = $38,000. (c) Since we know that q = 4000 − 40p, C = 7000 + 10q,

Chapter One /SOLUTIONS and R = pq, we can write C = 70001 0q = 7000 + 10(4000 − 40p) = 7000 + 40,000 − 400p = 47,000 − 400p and R = pq = p(4000 − 40p) = 4000p − 40p2 . We also know that the proft is the di˙erence between the revenue and the cost so (p) = R − C = 4000p − 40p2 − (47,000 − 400p) = −40p2 + 3600p − 47,000. (d) Looking at Figure 1.32 we see that the maximum proft occurs when the company charges about $55 per shirt. At this price, the proft is about $74,000. 75000 (p)

50000 25000

p 20

100

−25000 −47000

Figure 1.32

31. We know that a formula for passengers versus price will take the form N = mp + b where N is the number of passengers on the boat when the price of a tour is p dollars. We know two points on the line thus we know that the slope is 650 − 500 150 slope = = = −30. 20 − 25 −5 Thus the function will look like N = −30p + b. Plugging in the point (20, 650) we get N = −30p + b 650 = (−30)(20) + b = −600 + b b = 1250. Thus a formula for the number of passengers as a function of tour price is N = −30p + 1250.

32. (a) If we think of q as a linear function of p, then q is the dependent variable, p is the independent variable, and the slope m = Δq∕Δp. We can use any two points to fnd the slope. If we use the frst two points, we get Slope = m =

Δq 460 − 500 −40 = = = −20. Δp 18 − 16 2

The units are the units of q over the units of p, or tons per dollar. The slope tells us that, for every dollar increase in price, the number of tons sold every month will decrease by 20.

1.4 SOLUTIONS

To write q as a linear function of p, we need to fnd the vertical intercept, b. Since q is a linear function of p, we have q = b + mp. We know that m = −20 and we can use any of the points in the table, such as p = 16, q = 500, to fnd b. Substituting gives q = b + mp 500 = b + (−20)(16) 500 = b − 320 820 = b. Therefore, the vertical intercept is 820 and the equation of the line is q = 820 − 20p. (b) If we now consider p as a linear function of q, we have Δp 18 − 16 2 1 Slope = m = = = =− = −0.05. Δq 460 − 500 −40 20 The units of the slope are dollars per ton. The slope tells us that, if we want to sell one more ton of the product every month, we should reduce the price by $0.05. Since p is a linear function of q, we have p = b + mq and m = −0.05. To fnd b, we substitute any point from the table such as p = 16, q = 500 into this equation: p = b + mq 16 = b + (−0.05)(500) 16 = b − 25 41 = b. The equation of the line is p = 41 − 0.05q. Alternatively, notice that we could have taken our answer to part (a), that is q = 820 − 20p, and solved for p. 33. (a) The quantity demanded at a price of $100 is q = 120,000 − 500(100) = 70,000 units. The quantity supplied at a price of $100 is q = 1000(100) = 100,000 units. At a price of $100, the supply is larger than the demand, so some goods remain unsold and we expect the market to push prices down. (b) The supply and demand curves are shown in Figure 1.33. The equilibrium price is about p∗ = $80 and the equilibrium quantity is about q ∗ = 80,000 units. The market will push prices downward from $100, toward the equilibrium price of $80. This agrees with the conclusion to part (a) which says that prices will drop. p ($/unit) 240

S p∗ = 80 D q ∗ = 80,000

120,000

q (units)

Figure 1.33: Demand and supply curves for a product 34. One possible demand curve is shown in Figure 1.34. Many other answers are possible. P ($/oz) 35 25 15 5 155

165

185

Figure 1.34

205

Q (million ozs)

Chapter One /SOLUTIONS

35. One possible supply curve is shown in Figure 1.35. Many other answers are possible. P (milion $/oz) 35 25 15 5 160

180

200

220

240

Q (million ozs)

Figure 1.35

36. (a) The total amount spent on compact cars is 20,000c and the total amount spent on SUV’s is 30,000s, so the budget constraint is 20,000c + 30,000s = 720,000. (b) When c = 0, we have 30,000s = 720,000 which gives s = 24 SUVs. If the company does not replace any compact cars, it can replace 24 SUVs. When s = 0, we have 20,000c = 720,000 which gives c = 36 compact cars. If the company does not replace any SUVs, it can replace 36 compact cars. 37. (a) We know that the total amount the company will spend on raw materials will be price for raw materials = $100 ⋅ m where m is the number of units of raw materials the company will buy. We know that the total amount the company will spend on paying employees will be total employee expenditure = $25,000 ⋅ r where r is the number of employees the company will hire. Since the total amount the company spends is $500,000, we get 25,000r + 100m = 500,000. (b) Solving for m we get 25,000r + 100m = 500,000 100m = 500,000 − 25,000r m = 5000 − 250r. (c) Solving for r we get 25,000r + 100m = 500,000 25,000r = 500,000 − 100m 100 r = 20 − m 25,000 1 = 20 − m. 250 38. (a) The amount spent on books will be Amount for books = $80 ⋅ b where b is the number of books bought. The amount of money spent on outings is money spent on outings = $20 ⋅ s where s is the number of social outings. Since we want to spend all of the $2000 budget we end up with 80b + 20s = 2000.

1.4 SOLUTIONS (b)

15 5 s 25

100

Figure 1.36 (c) Looking at Figure 1.36 we see that the vertical intercept occurs at the point (0, 25) and the horizontal intercept occurs at (100, 0). The vertical intercept tells us how many books we would be able to buy if we wanted to spend all of the budget on books. That is, we could buy at most 25 books. The horizontal intercept tells how many social outings we could a˙ord if we wanted to spend all of the budget on outings. That is, we would be able to go on at most 100 outings. 39. (a) If the bakery owner decreases the price, the customers want to buy more. Thus, the slope of d(q) is negative. If the owner increases the price, she is willing to make more cakes. Thus the slope of s(q) is positive. (b) To determine whether an ordered pair (q, p) is a solution to the inequality, we substitute the values of q and p into the inequality and see whether the resulting statement is true. Substituting the two ordered pairs gives (60, 18):

18 ≤ 20 − 60∕20, or 18 ≤ 17.

This is false, so (60, 18) is not a solution.

(120, 12):

12 ≤ 20 − 120∕20, or 12 ≤ 14.

This is true, so (120, 12) is a solution.

The pair (60, 18) is not a solution to the inequality p ≤ 20 − q∕20. This means that the price $18 is higher than the unit price at which customers would be willing to buy a total of 60 cakes. So customers are not willing to buy 60 cakes at $18. The pair (120, 12) is a solution, meaning that $12 is not more than the price at which customers would be willing to buy 120 cakes. Thus customers are willing to buy a total of 120 (and more) cakes at $12. Each solution (q, p) represents a quantity of cakes q that customers would be willing to buy at the unit price p. (c) In order to be a solution to both of the given inequalities, a point (q, p) must lie on or below the line p = 20 − q∕20 and on or above the line p = 11 + q∕40. Thus, the solution set of the given system of inequalities is the region shaded in Figure 1.37. A point (q, p) in this region represents a quantity q of cakes that customers would be willing to buy, and that the bakery-owner would be willing to make and sell, at the price p. p 25 20

d(q) = 20 − q∕20 (120, 14)

15 10

s(q) = 11 + q∕40

5 q 40

120

160

200

Figure 1.37: Possible cake sales at di˙erent prices and quantities (d) To fnd the rightmost point of this region, we need to fnd the intersection point of the lines p = 20 − q∕20 and p = 11 + q∕40. At this point, p is equal to both 20 − q∕20 and 11 + q∕40, so these two expressions are equal to each other: q q 20 − = 11 + 20 40 q q 9= + 20 40 3q 9= 40 q = 120.

Chapter One /SOLUTIONS Therefore, q = 120 is the maximum number of cakes that can be sold at a price at which customers are willing to buy them all, and the owner of the bakery is willing to make them all. The price at this point is p = 20 − 120∕20 = 14 dollars. (In economics, this price is called the equilibrium price, since at this point there is no incentive for the owner of the bakery to raise or lower the price of the cakes.)

40. (a) See Figure 1.38. (b) If the slope of the supply curve increases then the supply curve will intersect the demand curve sooner, resulting in a higher equilibrium price p1 and lower equilibrium quantity q1 . Intuitively, this makes sense since if the slope of the supply curve increases. The amount produced at a given price decreases. See Figure 1.39. p

p New supply

Supply

Old supply

p1 p0

p0 Demand q0

Demand

q1 q0

Figure 1.38

Figure 1.39

(c) When the slope of the demand curve becomes more negative, the demand function will decrease more rapidly and will intersect the supply curve at a lower value of q1 . This will also result in a lower value of p1 and so the equilibrium price p1 and equilibrium quantity q1 will decrease. This follows our intuition, since if demand for a product lessens, the price and quantity purchased of the product will go down. See Figure 1.40. p Supply

p0 p1 Old demand q1 q0

New demand

Figure 1.40

41. The quantity demanded for a product is q = 4 when the price is p = 50. The demand for the product is q = 3.9 when the price is p = 51. The relative change in demand is 3.9 − 4 = −0.025 4 and the relative change in price is 51 − 50 = 0.02. 50 Hence | Relative change in demand | | −0.025 | | | | | | Relative change in price | = | 0.02 | = 1.25. | | | | A 1% increase in the price resulted in a 1.25% decrease in the demand.

1.4 SOLUTIONS

42. The original demand equation, q = 100 − 5p, tells us that H Quantity demanded = 100 − 5

I Amount per unit

paid by consumers

The consumers pay p + 2 dollars per unit because they pay the price p plus $2 tax. Thus, the new demand equation is q = 100 − 5(p + 2) = 90 − 5p. See Figure 1.41. p 20 18 Demand without tax: q = 100 − 5p

Demand with tax∶ q = 90 − 5p q 90 100

Figure 1.41 43. The original supply equation, q = 4p − 20, tells us that H Quantity supplied = 4

I Amount per unit received by suppliers

− 20.

The suppliers receive only p −2 dollars per unit because $2 goes to the government as taxes. Thus, the new supply equation is q = 4(p − 2) − 20 = 4p − 28. See Figure 1.42.

Supply with tax ∶ q = 4p − 28 Supply without tax ∶ q = 4p − 20

7 5 q(quantity)

Figure 1.42

Chapter One /SOLUTIONS

44. Before the tax is imposed, the equilibrium is found by solving the equations q = 0.5p − 25 and q = 165 − 0.5p. Setting the values of q equal, we have 0.5p − 25 = 165 − 0.5p p = 190 dollars Substituting into one of the equations for q, we fnd q = 0.5(190) − 25 = 70 units. Thus, the pre-tax equilibrium is p = $190,

q = 70 units.

The original supply equation, q = 0.5p − 25, tells us that H Quantity supplied = 0.5

I Amount per unit received by suppliers

− 25.

When the tax is imposed, the suppliers receive only p − 8 dollars per unit because $8 goes to the government as taxes. Thus, the new supply curve is q = 0.5(p − 8) − 25 = 0.5p − 29. The demand curve is still q = 165 − 0.5p. To fnd the equilibrium, we solve the equations q = 0.5p − 29 and q = 165 − 0.5p. Setting the values of q equal, we have 0.5p − 29 = 165 − 0.5p p = 194 dollars Substituting into one of the equations for q, we fnd that q = 0.5(194) − 29 = 68 units. Thus, the post-tax equilibrium is p = $194, q = 68 units. 45. (a) The equilibrium price and quantity occur when demand equals supply. If we graph these functions on the same axes, we get Figure 1.43. p Supply: q = 10p − 500 (0, 125)

(500, 100)

Demand: q = 2500 − 20p (2500, 0) q

(−500, 0)

Figure 1.43

We can see from the graph in Figure 1.43 that the supply and demand curves intersect at the point (500, 100). The equilibrium price is $100 and the equilibrium quantity is 500. This answer can also be obtained algebraically, by setting the values of q equal and solving 2500 − 20p = 10p − 500 3000 = 30p p = 100. q = 2500 − 20(100) = 500.

1.4 SOLUTIONS

(b) The $6 tax imposed on suppliers has the e˙ect of moving the supply curve upward by $6. The original supply equation, q = 10p − 500, tells us that H I Amount received per Quantity supplied = 10 − 500. unit by suppliers The suppliers receive only p − 6 dollars per unit because $6 goes to the government as taxes. Thus, the new supply equation is q = 10(p − 6) − 500 = 10p − 560. The demand equation q = 2500 − 20p is unchanged. We can fnd the new equilibrium price and quantity by graphing the new supply equation with the demand equation. p, price

(0, 125)

✛ New Supply Curve ✛ Old Supply Curve ✛ ▼

(500, 100)

(460, 102)

Demand (2500, 0) q

(−500, 0)

Figure 1.44: Specifc tax shifts supply curve From Figure 1.44, we see that the new equilibrium price (including tax) is $102 and the new equilibrium quantity is 460 units. We can also obtain these results algebraically: 2500 − 20p = 10p − 560 3060 = 30p p = 102, so q = 10(102) − 560 = 460. (c) The tax paid by the consumer is $2, since the new equilibrium price of $102 is $2 more than the old equilibrium price of $100. Since the tax is $6, the producer pays $4 of the tax and receives $102 − $6 = $96 per item after taxes. (d) The tax received by the government per unit product is $6. Thus, the total revenue received by the government is equal to the tax per unit times the number of units sold, which is just the equilibrium quantity. Thus, Government revenue = Tax ⋅ Quantity = 6 ⋅ 460 = $2760. 46. (a) The original demand equation, q = 100 − 2p, tells us that H Quantity demanded = 100 − 2

I Amount per unit

paid by consumers

The consumers pay p + 0.05p = 1.05p dollars per unit because they pay the price p plus 5% tax. Thus, the new demand equation is q = 100 − 2(1.05p) = 100 − 2.1p. The supply equation remains the same. (b) To fnd the new equilibrium price and quantity, we fnd the intersection of the demand curve and the supply curve q = 3p − 50. 100 − 2.1p = 3p − 50 150 = 5.1p p = $29.41.

Chapter One /SOLUTIONS The equilibrium price is $29.41. The equilibrium quantity q is given by q = 3(29.41) − 50 = 88.23 − 50 = 38.23 units. (c) Since the pre-tax price was $30 and the suppliers’ new price is $29.41 per unit, Tax paid by supplier = $30 − $29.41 = $0.59. The consumers’ new price is 1.05p = 1.05(29.41) = $30.88 per unit and the pre-tax price was $30, so Tax paid by consumer = 30.88 − 30 = $0.88. The total tax paid per unit by suppliers and consumers together is 0.59 + 0.88 = $1.47 per unit. (d) The government receives $1.47 per unit on 38.23 units. The total tax collected is (1.47)(38.23) = $56.20.

47. (a) The original supply equation, q = 3p − 50, tells us that H Quantity supplied = 3

I Amount received per unit by suppliers

− 50.

The suppliers receive only p − 0.05p = 0.95p dollars per unit because 5% of the price goes to the government. Thus, the new supply equation is q = 3(0.95p) − 50 = 2.85p − 50. The demand equation, q = 100 − 2p is unchanged. (b) At the equilibrium, supply equals demand: 2.85p − 50 = 100 − 2p 4.85p = 150 p = $30.93. The equilibrium price is $30.93. The equilibrium quantity q is the quantity demanded at the equilibrium price: q = 100 − 2(30.93) = 38.14 units. (c) Since the pretax price was $30 and consumers’ new price is $30.93, Tax paid by consumers = 30.93 − 30 = $0.93 The supplier keeps 0.95p = 0.95(30.93) = $29.38 per unit, so Tax paid by suppliers = 30 − 29.38 = $0.62. The total tax per unit paid by consumers and suppliers together is 0.93 + 0.62 = $1.55 per unit. (d) The government receives $1.55 per unit on 38.14 units. The total tax collected is (1.55)(38.14) = $59.12.

Solutions for Section 1.5 1. (a) Town (i) has the largest percent growth rate, at 12%. (b) Town (ii) has the largest initial population, at 1000. (c) Yes, town (iv) is decreasing in size, since the decay factor is 0.9, which is less than 1. 2. (a) Initial amount = 100; exponential growth; growth rate = 7% = 0.07. (b) Initial amount = 5.3; exponential growth; growth rate = 5.4% = 0.054. (c) Initial amount = 3500; exponential decay; decay rate = −7% = −0.07. (d) Initial amount = 12; exponential decay; decay rate = −12% = −0.12.

1.5 SOLUTIONS

3. Graph (I) and (II) are of increasing functions, therefore the growth factor for both functions is greater than 1. Since graph (I) increases faster than graph (II), graph (I) corresponds to the larger growth factor. Therefore graph (I) matches Q = 50(1.4)t and graph (II) is Q = 50(1.2)t . Similarly, since graphs (III) and (IV) are both of decreasing functions, they have growth factors between 0 and 1. Since graph (IV) decreases faster than graph (III), graph (IV) has the smaller decay factor. Thus graph (III) is Q = 50(0.8)t and graph (IV) is Q = 50(0.6)t . 4. (a) The city’s population is increasing at 5% per year. It is growing exponentially. Thus, it must be either I or II. We notice that part (b) has a higher rate of growth and thus we choose the exponential curve that grows more slowly, II. (b) Using the same reasoning as part (a), we know that the graph of the population is an exponential growth curve. Comparing it to the previous city, part (b) has a faster rate of growth and thus we choose the faster growing exponential curve, I. (c) The population is growing by a constant number of people each year. Thus, the graph of the population of the city is a line. Since the population is growing, we know that the line must have positive slope. Thus, the city’s population is best described by graph III. (d) The city’s population does not change. That means that the population is a constant. It is thus a line with 0 slope, or graph V. The remaining curves are IV and VI. Curve IV shows a population decreasing at an exponential rate. Comparison of the slope of curve IV at the points where it intersects curves I and II shows that the slope of IV is less steep than the slope of I or II. From this we know that the city’s population is decreasing at a rate less than I or II’s is increasing. It is decreasing at a rate of perhaps 4% per year. Graph VI shows a population decreasing linearly—by the same number of people each year. By noticing that graph VI is more downward sloping than III is upward sloping, we can determine that the annual population decrease of a city described by VI is greater than the annual population increase of a city described by graph III. Thus, the population of city VI is decreasing by more than 5000 people per year. 5. (a) Since the y-intercept of each graph is above the x-axis, we know that a, c, and p must all be positive. (b) Since the graph of y = abx is increasing and goes through (1, 1), the y-intercept, a, must be less than 1. Since the y-intercept is positive, we know that 0 < a < 1. Since y = cd x and y = pq x are both decreasing, both d and q must be between 0 and 1. (c) The graphs of y = cd x and y = pq x have the same y-intercepts, therefore we have c = p. (d) Since the point (1, 1) lies on the graph of y = abx , we know that 1 = ab1 . Therefore, a and b are reciprocals of each other. Similarly, p and q are reciprocals of each other. 6. This looks like an exponential function y = Cat . The y-intercept is 500 so we have y = 500at . We use the point (3, 2000) to fnd a: y = 500at 2000 = 500a3 4 = a3 a = 41∕3 = 1.59. The formula is y = 500(1.59)t . 7. This looks like an exponential decay function y = Cat . The y-intercept is 30 so we have y = 30at . We use the point (25, 6) to fnd a: y = 30at 6 = 30a25 0.2 = a25 a = (0.2)1∕25 = 0.94. The formula is y = 30(0.94)t . 8. We look for an equation of the form y = y0 ax since the graph looks exponential. The points (0, 3) and (2, 12) are on the graph, so 3 = y0 a0 = y0 and 12 = y0 ⋅ a2 = 3 ⋅ a2 , x

Since a > 0, our equation is y = 3(2 ).

giving

a = ±2.

Chapter One /SOLUTIONS 9. We look for an equation of the form y = y0 ax since the graph looks exponential. The points (0, 18) and (2, 8) are on the graph, so 18 = y0 a0 = y0 and

u 8 = y0 ⋅ a2 = 18 ⋅ a2 ,

giving

y = 18

x 2 . 3

Since a > 0, our equation is

a=±

4 . 9

10. (a) If G is increasing at a constant percent rate, G is an exponential function of t. The formula is G = 678.97(1.0109)t . (b) If G is increasing at a constant rate, G is a linear function of t. The formula is G = 678.97 + 7t. 11. (a) The function is linear with initial population of 1000 and slope of 50, so P = 1000 + 50t. (b) This function is exponential with initial population of 1000 and growth rate of 5%, so P = 1000(1.05)t . 12. (a) Since the price is decreasing at a constant absolute rate, the price of the product is a linear function of t. In t days, the product will cost 80 − 4t dollars. (b) Since the price is decreasing at a constant relative rate, the price of the product is an exponential function of t. In t days, the product will cost 80(0.95)t . 13. (a) This is a linear function with slope −2 grams per day and intercept 30 grams. The function is Q = 30 − 2t, and the graph is shown in Figure 1.45. Q (grams)

Q (grams)

30 Q = 30 − 2t

Figure 1.45

Q = 30(0.88)t

t (days)

Figure 1.46

(b) Since the quantity is decreasing by a constant percent change, this is an exponential function with base 1−0.12 = 0.88. The function is Q = 30(0.88)t , and the graph is shown in Figure 1.46. 14. (a) We see from the base 1.0103 that the percent rate of growth is 1.03% per year. (b) The initial population (in the year 2020) is 7.68 billion people. When t = 5, we have P = 7.68(1.0103)5 = 8.084. The predicted world population in the year 2025 is 8.084 billion people. (c) We have ΔP 8.084 − 7.68 Average rate of change = = = 0.081 billion people per year. Δt 2025 − 2020 The population of the world is projected to increase by about 81 million people per year from 2020 to 2025. 15. (a) Since the percent rate of change is constant, A is an exponential function of t. The initial quantity is 50 and the base is 1 − 0.06 = 0.94. The formula is A = 50(0.94)t . (b) When t = 24, we have A = 50(0.94)24 = 11.33. After one day, 11.33 mg of quinine remains. (c) See Figure 1.47. The graph is decreasing and concave up, as we would expect with an exponential decay function. (d) In Figure 1.47, it appears that when A = 5, we have t ≈ 37. It takes about 37 hours until 5 mg remains.

1.5 SOLUTIONS

A (mg) 50

5 20

t (hours)

Figure 1.47

16. This function is exponential with initial value 258.8 and growth rate of 2.5%, so CPI = 258.8(1.025)t . 17. Since the decay rate is 0.33%, the base of the exponential function is 1 − 0.0033 = 0.9967. If P represents the percent of forested land in 2010 remaining t years after 2010, when t = 0 we have P = 100. The formula is P = 100(0.9967)t . The percent remaining in 2030 will be P = 100(0.9967)20 = 93.603. About 93.603% of the land will be covered in forests 20 years later if the rate continues. 18. (a) The total number of confrmed cases can be modeled by P = 2979(1.051)t . (b) May 10, 2020 corresponds to t = 26, so the expected number of confrmed cases in Colombia is 2979(1.051)26 = 10,858. (c) We compute 11,063 − 10,858 = 0.01853 = 1.853%, 11,063 which tells us that the value predicted by the model is 1.853% less than the actual value. 19. See Figure 1.48. The graph is decreasing and concave up. It looks like an exponential decay function with y-intercept 100. y 100

y = 100e−0.4x

Figure 1.48

20. (a) See Table 1.5. Table 1.5 x

2.72

7.39

20.09

(b) See Figure 1.49. The function y = ex is an exponential growth function.

Chapter One /SOLUTIONS y 20 y 1

y = ex y = e−x

10 0.5 5

x 1

Figure 1.49

Figure 1.50

e−x

0.37

0.14

0.05

(d) See Figure 1.50. The function y = e−x is an exponential decay function. 21. The values of f (x) given seem to increase by a factor of 1.4 for each increase of 1 in x, so we expect an exponential function with base 1.4. To assure that f (0) = 4.30, we multiply by the constant, obtaining f (x) = 4.30(1.4)x .

22. Each increase of 1 in t seems to cause g(t) to decrease by a factor of 0.8, so we expect an exponential function with base 0.8. To make our solution agree with the data at t = 0, we need a coeÿcient of 5.50, so our completed equation is g(t) = 5.50(0.8)t .

23. Table A and Table B could represent linear functions of x. Table A could represent the constant linear function y = 2.2 because all y values are the same. Table B could represent a linear function of x with slope equal to 11∕4. This is because x values that di˙er by 4 have corresponding y values that di˙er by 11, and x values that di˙er by 8 have corresponding y values that di˙er by 22. In Table C, y decreases and then increases as x increases, so the table cannot represent a linear function. Table D does not show a constant rate of change, so it cannot represent a linear function. 24. Table D is the only table that could represent an exponential function of x. This is because, in Table D, the ratio of y values is the same for all equally spaced x values. Thus, the y values in the table have a constant percent rate of decrease: 9 4.5 2.25 = = = 0.5. 18 9 4.5 Table A represents a constant function of x, so it cannot represent an exponential function. In Table B, the ratio between y values corresponding to equally spaced x values is not the same. In Table C, y decreases and then increases as x increases. So neither Table B nor Table C can represent exponential functions. 25. (a) See Figure 1.51. (b) No. The points in the plot do not lie even approximately on a straight line, so a linear model is a poor choice. (c) No. The plot shows that H is a decreasing function of Y that might be leveling o˙ to an asymptote at H = 0. These are features of an exponential decay function, but their presence does not show that H is an exponential function of Y . We can do a more precise check by dividing each value of H by the previous year’s H. Houses in 2011 13 million = = 0.710 Houses in 2010 18.3 million

1.5 SOLUTIONS

Houses in 2012 7.8 million = = 0.600 Houses in 2011 13 million Houses in 2013 3.9 million = = 0.500 Houses in 2012 7.8 million Houses in 2014 1 million = = 0.256 Houses in 2013 3.9 million Houses in 2015 0.5 million = = 0.500. Houses in 2014 1 million If H were an exponential function of Y , the ratios would be approximately constant. Since they are not approximately constant, we see that an exponential model is a poor choice. H 20 15 10 5 Y 2010 2011 2012 2013 2014 2015

Figure 1.51 26. (a) We divide the two equations to solve for a: P0 a3 75 = 50 P0 a2 a = 1.5 Now that we know the value of a, we can use either of the equations to solve for P0 . Using the frst equation, we have: P0 a3 = 75 P0 (1.53 ) = 75 75 = 22.222 P0 = 1.53 We see that a = 1.5 and P0 = 22.222. (b) The initial quantity is 22.222 and the quantity is growing at a rate of 50% per unit time. 27. (a) We divide the two equations to solve for a: P0 a4 18 = 20 P0 a3 a = 0.9 Now that we know the value of a, we can use either of the equations to solve for P0 . Using the frst equation, we have: P0 a4 = 18 P0 (0.94 ) = 18 18 P0 = = 27.435 0.94 We see that a = 0.9 and P0 = 27.435. (b) The initial quantity is 27.435 and the quantity is decaying at a rate of 10% per unit time. 28. From the statements given, we have the equations: P0 a5 = 320 and P0 a3 = 500. (a) We divide the two equations to solve for a: P0 a5 320 = 500 P0 a3 a2 = 0.64 a = 0.641∕2 = 0.8

Chapter One /SOLUTIONS Now that we know the value of a, we can use either of the equations to solve for P0 . Using the frst equation, we have: P0 a5 = 320 P0 (0.85 ) = 320 320 = 976.563 P0 = 0.85 We see that a = 0.8 and P0 = 976.563. (b) The initial quantity is 976.563 and the quantity is decaying at a rate of 20% per unit time.

29. From the statements given, we have the equations: P0 a3 = 1600 and P0 a1 = 1000. (a) We divide the two equations to solve for a: P0 a3 1600 = 1000 P0 a1 a2 = 1.6 a = 1.61∕2 = 1.2649111 ≈ 1.265. Now that we know the value of a, we can use either of the equations to solve for P0 . From the frst equation, we have: P0 a3 = 1600 P0 (1.26491113 ) = 1600 1600 = 790.569. P0 = 1.26491113 We see that a = 1.265 and P0 = 790.569. (b) The initial quantity is 790.569 and the quantity is growing at a rate of 26.5% per unit time. 30. If the population increases exponentially, the formula for the population at time t can be written in the form P (t) = P0 ekt where P0 is the population at time t = 0 and k is the continuous growth rate; we measure time in years. If we let 2000 be the initial time t = 0, we have P0 = 6.081, so P (t) = 6.081ekt . In 2018, time t = 18, so we have P (18) = 7.504. Thus we get 7.504 = 6.081e18k 7.504 e18k = 6.081 7.504 ln e18k = ln 6.081 ln(7.504∕6.081) k= = 0.0117. 18 Thus, the formula for the population, assuming exponential growth, is P (t) = 6.081e0.0117t billion. Using our formula for the year 2020, time t = 20, we get P (20) = 6.081e0.0117(20) = 6.081e0.2340 = 7.68 billion. Thus, we see that we were not too far o˙ the mark when we approximated the population by an exponential function. 31. (a) The revenue is modeled by the exponential function R = 1.9(1.545)t where t is years since 2010. In 2017 the revenue was R = 1.9(1.545)7 = 39.926 billion dollars. (b) If growth continued exponential at the same rate, then in 2020 the revenue would be R = 1.9(1.545)10 = 147.244 billion dollars. This is much more than the actual revenue, which was $80.9 billion. The growth did not continue at 54.5% per year.

1.5 SOLUTIONS

32. (a) Since the population was 2.7 million in 2010, and t is measured since 2010 so t = 0 in 2010, we have N0 = 2.7. Then N = 3.0 when t = 8, so 3 = 2.7a8 3 1∕8 a= = 1.0133. 2.7 Thus, N = 2.7(1.0133)t . (b) In 2000, we have t = −10, so

0 N = 2.7

3 2.7

1∕8 1−10 = 2.367 million.

33. Use an exponential function of the from P = P0 at for the number of passengers. At t = 0, the number of passengers is P0 = 190,205, so P = 190,205at . After t = 5 years, the number of passengers is 174,989, so 174,989 = 190,205a5 174,989 a5 = = 0.920002 190,205 a = (0.920002)1∕5 = 0.983. Then P = 190,205(0.932)t , which means the annual percentage decrease over this period is 1 − 0.983 = 0.017 or 1.7%. 34. Since we are looking for an annual percent increase, we use an exponential function. If V represents the value of the company, and t represent the number of years since it was started, we have V = Cat . The initial value of the company was $4000, so V = 4000at . To fnd a, we use the fact that V = 14,000,000 when t = 40: V = 4000at 14,000,000 = 4000a40 3500 = a40 a = (3500)1∕40 = 1.226. The value of the company increased at an average rate of 22.6% per year. 35. Let y represent the number of zebra mussels t years since 2014. (a) The slope of the line is (3186 − 2700)∕1 = 486 and the vertical intercept is 2700. We have y = 2700 + 486t. The slope of the line, 486 zebra mussels per year, represents the rate at which the zebra mussel population is growing. (b) When t = 0, we have y = 2700, so the exponential function is in the form y = 2700at . When t = 1 we have y = 3186, and we substitute to get 3186 = 2700a1 so a = 1.18. The exponential function is y = 2700(1.18)t and the zebra mussel population is growing at 18% per year. 36. (a) We have (1.03)120 = 34.711, so infections in Arizona grew by a factor of about 35. Similarly, (1.0175)120 = 8.019, so infections in West Virginia grew by a factor of about 8. (b) Note that 35 is about 4 times 8. So, although the growth rates di˙er by a factor of two, the four month growth di˙ers by a factor of more that 4. 37. (a) The slope of the line is ΔW 591.1 − 371.4 = = 54.925. Δt 2018 − 2014 The initial value (when t = 0) is 371.4, so the linear formula is Slope =

W = 371.4 + 54.925t. The rate of growth of wind generating capacity is 54.925 gigawatts per year.

Chapter One /SOLUTIONS (b) If the function is exponential, we have W = W0 at . The initial value (when t = 0) is 371.4, so we have W = 371.4at . We use the fact that W = 591.1 when t = 4 to fnd a: 591.1 = 371.4a4 1.591546 = a4 a = 1.5915461∕4 = 1.123194. The exponential formula is W = 371.4(1.123194)t . The percent rate of growth of wind generating capacity is 12.319% per year. (Alternatively, if we used the form W = W0 ekt , the formula would be W = 371.4e0.11618t , giving a continuous rate of 11.618% per year.) (c) See Figure 1.52. (d) The linear function, 371.4+54.925t, gives a predicted generating capacity in 2019, when t = 5, of 371.4+54.925⋅5 = 646.025 gigawatts, 4.775 less than the real power generating capacity of 650.8 gigawatts. The exponential model predicts the capacity in 2019 to be W = 371.4(1.123194)5 = 663.920, which is about 13.12 gigawatts more than the real fgure, so the linear model is better. W (GW) W = 371.4(1.123194)t 700 W = 371.4 + 54.925t (4, 591.1) 500 371.4 300 1

t (years 6 since 2014)

Figure 1.52 38. (a) We see that f cannot be a linear function, since f (x) increases by di˙erent amounts (24 − 16 = 8 and 36 − 24 = 12) as x increases by one. Could f be an exponential function? We look at the ratios of successive f (x) values: 24 36 54 81 = 1.5 = 1.5 = 1.5 = 1.5. 16 24 36 54 Since the ratios are all equal to 1.5, this table of values could correspond to an exponential function with a base of 1.5. Since f (0) = 16, a formula for f (x) is f (x) = 16(1.5)x . Check by substituting x = 0, 1, 2, 3, 4 into this formula; you get the values given for f (x). (b) As x increases by one, g(x) increases by 6 (from 14 to 20), then 4 (from 20 to 24), so g is not linear. We check to see if g could be exponential: 20 24 = 1.43 and = 1.2. 14 20 Since these ratios (1.43 and 1.2) are di˙erent, g is not exponential. (c) For ℎ, notice that as x increases by one, the value of ℎ(x) increases by 1.2 each time. So ℎ could be a linear function with a slope of 1.2. Since ℎ(0) = 5.3, a formula for ℎ(x) is ℎ(x) = 5.3 + 1.2x. 39. (a) A linear function must change by exactly the same amount whenever x changes by some fxed quantity. While ℎ(x) decreases by 3 whenever x increases by 1, f (x) and g(x) fail this test, since both change by di˙erent amounts between x = −2 and x = −1 and between x = −1 and x = 0. So the only possible linear function is ℎ(x), so it will be given by a formula of the type: ℎ(x) = mx + b. As noted, m = −3. Since the y-intercept of ℎ is 31, the formula for ℎ(x) is ℎ(x) = 31 − 3x. (b) An exponential function must grow by exactly the same factor whenever x changes by some fxed quantity. Here, g(x) increases by a factor of 1.5 whenever x increases by 1. Since the y-intercept of g(x) is 36, g(x) has the formula g(x) = 36(1.5)x . The other two functions are not exponential; ℎ(x) is not because it is a linear function, and f (x) is not because it both increases and decreases.

1.5 SOLUTIONS 40. (a) In this case we know that f (1) − f (0) = 12.7 − 10.5 = 2.2 while f (2) − f (1) = 18.9 − 12.7 = 6.2. Thus, the function described by these data is not a linear one. Next we check if this function is exponential.

while

f (1) 12.7 = f (0) 10.5

.21

f (2) 18.9 = f (1) 12.7

.49

thus f (x) is not an exponential function either. (b) In this case we know that s(0) − s(−1) = 30.12 − 50.2 = −20.08 while s(1) − s(0) = 18.072 − 30.12 = −12.048. Thus, the function described by these data is not a linear one. Next we check if this function is exponential. s(0) 30.12 = = 0.6, s(−1) 50.2 s(1) 18.072 = = 0.6, s(0) 30.12 and

s(2) 10.8432 = = 0.6. s(1) 18.072 Thus, s(t) is an exponential function. We know that s(t) will be of the form s(t) = P0 at where P0 is the initial value and a = 0.6 is the base. We know that P0 = s(0) = 30.12. Thus, s(t) = 30.12at . Since a = 0.6, we have s(t) = 30.12(0.6)t . (c) In this case we know that g(2) − g(0) 24 − 27 −3 = = = −1.5, 2−0 2 2 g(4) − g(2) 21 − 24 −3 = = = −1.5, 4−2 2 2 and

g(6) − g(4) 18 − 21 −3 = = = −1.5. 6−4 2 2 Thus, g(u) is a linear function. We know that g(u) = m ⋅ u + b where m is the slope and b is the vertical intercept, or the value of the function at zero. So b = g(0) = 27 and from the above calculations we know that m = −1.5. Thus, g(u) = −1.5u + 27.

Chapter One /SOLUTIONS

41. Assuming that, on average, the minimum wage has increased exponentially from its value of $1.60 in 1968, we have m = 1.60bt , where m is the minimum wage t years after 1968. Since t = 2021 − 1968 = 53 in 2021, we have m = 1.60(1.0392)53 = $12.28. Thus, a $15 minimum wage would be growth faster than the infation rate. 42. (a) Using Q = Q0 (1 − r)t for loss, we have Q = 10,000(1 − 0.1)10 = 10,000(0.9)10 = 3486.78. The investment was worth $3486.78 after 10 years. (b) Measuring time from the moment at which the stock begins to gain value and letting Q0 = 3486.78, the value after t years is Q = 3486.78(1 + 0.1)t = 3486.78(1.1)t . We can estimate the value of t when Q = 10,000 by tracing along a graph of Q, giving t ≈ 11. It will take about 11 years to get the investment back to $10,000. 43. (a) We look at the ratios of successive total cases to estimate the daily growth rate. We have Ratio of total number of cases, March 11 and March 12 =

813 = 1.3416. 606

This means that the total number of cases grew by about 34% between March 11 and March 12. Similarly, Ratio of total number of cases, March 12 and March 13 =

1086 = 1.3358. 813

This means that the total number of cases grew by about 34% between March 12 and March 13 as well. (b) Since we estimate the daily growth rate to be 34%, an exponential function that models the number of confrmed cases t days after March 10, 2020 is f (t) = 444(1.34)t . (c) March 27 corresponds to t = 17, so (rounding) Expected number of cases on March 27 = f (17) = 444(1.34)17 = 64,293.126 ≈ 64,293. This estimate is close to the total number of confrmed cases on March 27 (about 10% error). (d) April 30 corresponds to t = 51, so Expected number of cases on April 30 = f (51) = 444(1.34)51 = 1,348,117,315 ≈ 1348 million. This is more than 1300 times larger than the total number of cases, 988,487. In fact, it is over four times the reported population of the US1 in 2020 of about 330 million. 44. (a) We know w = 1248 when t = 1 and w = 25,827 when t = 24. One algorithm for exponential regression gives w = 1093.965(1.1408)t . (b) The annual growth rate is 0.1408 = 14.08%. (c) The model predicts that in 2000 the number of cases would be w = 1093.965(1.1408)20 ≈ 15,247. If the number of cases had doubled between 2000 and 2004, there would have been 15,247 ⋅ 2 = 30,494 cases in 2004. However, there were 25,827 cases in 2004. Thus the model does not confrm this report. The reason is that the number of cases has been increasing at an average annual rate of 14.08% between 1980 and 2004, but at a higher rate between 2000 and 2004. 1 Data from https://www.census.gov/popclock , accessed on June 27, 2020

1.5 SOLUTIONS

45. Since the price dropped 20% every year, the price P per kilowatt-hour, where t is in years since 2013 and P0 is the price per kilowatt-hour in 2013, is given by P (t) = P0 (0.8)t . When the price is half the price in 2013, it is P0 ∕2, so we solve P0 = P0 (0.8)t 2 1 = (0.8)t 2 1 ln = ln(0.8)t 2 1 ln = t ln(0.8) 2 ln(1∕2) t= = 3.106. ln(0.8) It took about three years for the price to go down by 50%. 46. Direct calculation reveals that each 1000 foot increase in altitude results in a longer takeo˙ roll by a factor of about 1.096. Since the value of d when ℎ = 0 (sea level) is d = 670, we are led to the formula d = 670(1.096)ℎ∕1000 , where d is the takeo˙ roll, in feet, and ℎ is the airport’s elevation, in feet. Alternatively, we can write d = d0 a ℎ , where d0 is the sea level value of d, d0 = 670. In addition, when ℎ = 1000, d = 734, so 734 = 670a1000 . Solving for a gives

734 670

1∕1000 = 1.00009124,

so d = 670(1.00009124)ℎ . 47. (a) Let P represent the total number of confrmed cases in Africa with P0 = 1 at time t = 0. We model the total number of cases by P = at . Since P = 100,000 when t = 98, we have 100,000 = a98 a = 100,0001∕98 a = 1.125. So P = 1.125t and the number of confrmed cases, P , was growing at a 12.5% daily rate. (b) After 116 days, the predicted number of cases is much higher than the 200,000 cases: P = (100,0001∕98 )116 = 828,642.773. 48. (a) Since the annual growth factor from 2012 to 2013 was 1 + (58.95∕100) = 1.5895 and 899(1.5895) = 1428.961, the US consumed about 1429 million gallons of biodiesel in 2013. Since the annual growth factor from 2013 to 2014 was 1 + (−0.0084) = 0.9916 and 1428.961(0.9916) = 1416.96, the US consumed approximately 1417 million gallons of biodiesel in 2014. (b) Completing the table of annual consumption of biodiesel and plotting the data gives Figure 1.53. Table 1.7 Year

2012

2013

2014

2015

2016

2017

2018

Consumption of biodiesel (mn gal)

899

1429

1417

1494

2085

1985

1896

Chapter One /SOLUTIONS consumption of biodiesel (mn gal) 2500 2000 1500 1000 year

500 2012

2014

2016

2018

Figure 1.53

49. (a) False, because the annual percent growth is not constant over this interval. (b) (i) The US consumption of biodiesel grew by more that 58% in 2013. (ii) The US consumption of biodiesel fell by 4.8% in 2017. 50. (a) Since the annual growth factor from 2013 to 2014 was 1 − 0.0342 = 0.9658 and 2.686 ⋅ 0.9658 = 2.5941, the US generated approximately 2.6 quadrillion BTUs of hydroelectric power in 2014. Since the annual growth factor from 2014 to 2015 was 1 − 0.0397 = 0.9603 and 2.5941(1 − 0.0397) = 2.4911, the US generated about 2.5 quadrillion BTUs of hydroelectric power in 2015. (b) Completing the table of annual consumption of hydroelectric power and plotting the data gives Figure 1.54.

Year Generation of hydro. power (quadrillion BTU)

2013

2014

2015

2016

2017

2018

2.686

2.594

2.491

2.678

3.004

2.918

(c) The largest increase in the US generation of hydroelectric power occurred in 2017. In this year, the US generation of hydroelectric power increased by about 326 trillion BTUs to 3.004 quadrillion BTUs, up from 2.678 quadrillion BTUs in 2016. consumption of hydro. power (quadillion BTU) 3.5 3 2.5 year 2014

2016

2018

Figure 1.54

51. (a) From Figure 1.81 in the text, we can see the approximate percent growth for each year over the previous year:

Year

2013

2014

2015

2016

2017

2018

% growth over previous yr

Since the annual growth factor from 2013 to 2014 was 1 + 0.08 = 1.08 and 1601(1 + 0.08) = 1729.08, the US consumed approximately 1729 trillion BTUs of wind power energy in 2014. Since the annual growth factor from 2014 to 2015 was 1 + 0.03 = 1.03 and 1729.08(1 + 0.03) = 1780.95, the US consumed about 1781 trillion BTUs of wind power energy in 2015.

1.6 SOLUTIONS

(b) Completing the table of annual consumption of wind power and plotting the data gives Figure 1.55. Year

2013

2014

2015

2016

2017

2018

Consumption of wind power (trillion BTU)

1601

1729

1781

2102

2354

2542

(c) The largest increase in the US consumption of wind power energy occurred in 2016. In this year the US consumption of wind power energy rose by about 321 trillion BTUs to 2102 trillion BTUs, up from 1781 trillion BTUs in 2015. consumption of wind power energy (trillion BTU) 2500 2250 2000 1750 1500 year 2014

2016

2018

Figure 1.55 52. (a) The US consumption of wind power energy increased by at least 10% in 2013, 2016 and in 2017, relative to the previous year. Consumption did not decrease during the time period shown because all the annual percent growth values are positive, indicating a steady increase in the US consumption of wind power energy between 2013 and 2018. (b) True. From the fgure we see that consumption increased by about 18% during 2016, by about 12% during 2017, and by about 12% during 2018. This means that x(1 + 0.18) units of wind power energy were consumed in 2016 if x had been consumed in 2015. Similarly, x(1 + 0.18)(1 + 0.12) units of wind power energy were consumed in 2017 if x had been consumed in 2015, and x(1 + 0.18)(1 + 0.12)(1 + 0.08) units of wind power energy were consumed in 2018 if x had been consumed in 2015. Since x(1 + 0.18)(1 + 0.12)(1 + 0.08) = x(1.4273) = x(1 + 0.4273), wind power consumption was about 42.73% greater in 2018 than in 2015. Thus, consumption of wind power energy grew by about 42.73% from the beginning of 2015 to the end of 2018. Note that adding these three percentages would be only 38%, however there is a multiplier e˙ect which makes the change more than 40%.

Solutions for Section 1.6 1. Taking natural logs of both sides we get ln 10 = ln(2t ). This gives t ln 2 = ln 10 or in other words

ln 10 ln 2

.3219

2. Taking natural logs of both sides we get ln(5t ) = ln 7. This gives t ln 5 = ln 7 or in other words t=

ln 7 ln 5

.209.

Chapter One /SOLUTIONS 3. Taking natural logs of both sides we get ln 2 = ln(1.02t ). This gives t ln 1.02 = ln 2 or in other words

ln 2 ln 1.02

.003.

4. Taking natural logs of both sides we get ln 130 = ln(10t ). This gives t ln 10 = ln 130 or in other words

ln 130 ln 10

.1139.

5. Taking natural logs of both sides we get ln(et ) = ln 10 which gives t = ln 10

2.3026.

6. Dividing both sides by 25 we get 4 = 1.5t . Taking natural logs of both side gives ln(1.5t ) = ln 4. This gives t ln 1.5 = ln 4 or in other words

ln 4 ln 1.5

.419.

7. Dividing both sides by 10 we get 5 = 3t . Taking natural logs of both sides gives ln(3t ) = ln 5. This gives t ln 3 = ln 5 or in other words t=

ln 5 ln 3

.465.

8. Dividing both sides by 2 we get 2.5 = et . Taking the natural log of both sides gives ln(et ) = ln 2.5. This gives t = ln 2.

.9163.

1.6 SOLUTIONS 9. Taking natural logs of both sides we get ln(e3t ) = ln 100. This gives 3t = ln 100 or in other words t=

ln 100 3

.535.

10. Dividing both sides by 6 gives e0.5t =

10 5 = . 6 3

Taking natural logs of both sides we get ln(e0.5t ) = ln This gives 0.5t = ln

5 . 3

5 = ln 5 − ln 3 3

or in other words t = 2(ln 5 − ln 3)

1.0217.

11. We divide both sides by 100 and then take the natural logarithm of both sides and solve for t: 40 = 100e−0.03t 0.4 = e−0.03t ln(0.4) = ln(e−0.03t ) ln(0.4) = −0.03t ln(0.4) t= = 30.54. −0.03 12. Taking natural logs of both sides (and assuming a > 0, b > 0, and b ≠ 1) gives ln(bt ) = ln a. This gives t ln b = ln a or in other words t=

ln a . ln b

13. Dividing both sides by P we get B = ert . P Taking the natural log of both sides gives ln(ert ) = ln This gives

rt = ln

B P

B . P

= ln B − ln P .

Dividing by r gives t=

ln B − ln P . r

Chapter One /SOLUTIONS

14. Dividing both sides by P we get 2 = e0.3t . Taking the natural log of both sides gives ln(e0.3t ) = ln 2. This gives 0.3t = ln 2 or in other words t=

ln 2 0.3

.3105.

15. Taking natural logs of both sides we get ln(7 ⋅ 3t ) = ln(5 ⋅ 2t ) which gives ln 7 + ln(3t ) = ln 5 + ln(2t ) or in other words ln 7 − ln 5 = ln(2t ) − ln(3t ). This gives ln 7 − ln 5 = t ln 2 − t ln 3 = t(ln 2 − ln 3). Thus, we get t=

ln 7 − ln 5 ln 2 − ln 3

−0.8298.

16. Taking natural logs of both sides we get ln(5e3t ) = ln(8e2t ). This gives ln 5 + ln(e3t ) = ln 8 + ln(e2t ) or in other words ln(e3t ) − ln(e2t ) = ln 8 − ln 5. This gives 3t − 2t = ln 8 − ln 5 or in other words t = ln 8 − ln 5

0.47.

17. Taking natural logs of both sides, we get ln(Ae2t ) = ln(Bet ) so ln A + ln(e2t ) = ln B + ln(et ) or in other words ln A + 2t ln e = ln B + t ln e. Since ln e = 1, this gives ln A + 2t = ln B + t. Thus, combining ts on the left side of the equation, we get t = ln B − ln A = ln(B∕A).

1.6 SOLUTIONS 18. We cannot take the log of 0. Moving the 5 to the other side of the equation, we have 2et = 5. Then we take natural logs of both sides, giving ln(2et ) = ln 5 so ln 2 + ln(et ) = ln 5 or in other words ln 2 + t ln e = ln 5. Since ln e = 1, this gives ln 2 + t = ln 5. Thus, we get t = ln 5 − ln 2 = ln(5∕2)

0.9163.

19. We cannot take the log of 0. Moving the 3et to the other side of the equation, we have 3et = 7. Then we take natural logs of both sides, giving ln(3et ) = ln 7 so ln 3 + ln(et ) = ln 7 or in other words ln 3 + t ln e = ln 7. Since ln e = 1, this gives ln 3 + t = ln 7. Thus, we get t = ln 7 − ln 3 = ln(7∕3)

0.8473.

20. Since we cannot take the log of 0, we move Qe−t to the other side of the equation P e4t = Qe−t . Then, taking natural logs of both sides, we get ln(P e4t ) = ln(Qe−t ) so ln P + ln(e4t ) = ln Q + ln(e−t ) or in other words ln P + 4t ln e = ln Q − t ln e. Since ln e = 1, this gives ln P + 4t = ln Q − t. Thus, combining ts on the left side of the equation, we get 5t = ln Q − ln P ln(Q∕P ) 1 . t = (ln Q − ln P ) = 5 5

Chapter One /SOLUTIONS

21. Initial quantity = 5; growth rate = 0.07 = 7%. 22. Initial quantity = 7.7; growth rate = −0.08 = −8% (decay). 23. Initial quantity = 15; growth rate = −0.06 = −6% (continuous decay). 24. Initial quantity = 3.2; growth rate = 0.03 = 3% (continuous). t

25. Since e0.25t = e0.25 (1.2840)t , we have P = 15(1.2840)t . This is exponential growth since 0.25 is positive. We can also see that this is growth because 1.2840 > 1. 26. Since e−0.5t = (e−0.5 )t (0.6065)t , we have P = 2(0.6065)t . This is exponential decay since −0.5 is negative. We can also see that this is decay because 0.6065 < 1. 27. P = P0 (e0.2 )t = P0 (1.2214)t . Exponential growth because 0.2 > 0 or 1.2214 > 1. 28. P = 7(e− )t = 7(0.0432)t . Exponential decay because − < 0 or 0.0432 < 1. 29. Since we want (1.5)t = ekt = (ek )t , so 1.5 = ek , and k = ln 1.5 = 0.4055. Thus, P = 15e0.4055t . Since 0.4055 is positive, this is exponential growth. 30. We want 1.7t = ekt so 1.7 = ek and k = ln 1.7 = 0.5306. Thus P = 10e0.5306t . 31. We want 0.9t = ekt so 0.9 = ek and k = ln 0.9 = −0.1054. Thus P = 174e−0.1054t . 32. Since we want (0.55)t = ekt = (ek )t , so 0.55 = ek , and k = ln 0.55 = −0.5978. Thus P = 4e−0.5978t . Since −0.5978 is negative, this represents exponential decay. 33. We use the information to create two equations: P0 e3k = 140 and P0 e1k = 100. (a) We divide the two equations to solve for k: P0 e3k 140 = 100 P0 ek e2k = 1.4 2k = ln 1.4 ln 1.4 k= = 0.168 2 Now that we know the value of k, we can use either of the equations to solve for P0 . Using the frst equation, we have: P0 e3k = 140 P0 e3(0.168) = 140 140 P0 = 0.504 = 84.575 e We see that a = 0.168 and P0 = 84.575. (b) The initial quantity is 84.575 and the quantity is growing at a continuous rate of 16.8% per unit time. 34. We use the information to create two equations: P0 e4k = 40 and P0 e3k = 50. (a) We divide the two equations to solve for k: P0 e4k 40 = 50 P0 e3k ek = 0.8 k = ln 0.8 = −0.223144. Now that we know the value of k, we can use either of the equations to solve for P0 . Using the frst equation, we have: P0 e4k = 40 4(−0.223144)

P0 e

= 40

P0 =

40 = 97.656. e−0.892576

We see that k = −0.22314 and P0 = 97.656. (b) The initial quantity is 97.656 and the quantity is decaying at a continuous rate of 22.314% per unit time.

1.6 SOLUTIONS

35. (a) The continuous percent growth rate is 6%. (b) We want P0 at = 100e0.06t , so we have P0 = 100 and a = e0.06 = 1.0618. The corresponding function is P = 100(1.0618)t . The annual percent growth rate is 6.18%. An annual rate of 6.18% is equivalent to a continuous rate of 6%. 36. (a) Since 0.88 = 1 − 0.12, the annual percent decay rate is 12%. (b) We want P0 ekt = 25(0.88)t , so we have P0 = 25 and ek = 0.88. Since ek = 0.88, we have k = ln(0.88) = −0.128. The corresponding function is P = 25e−0.128t . The continuous percent decay rate is 12.8%. A continuous rate of 12.8% is equivalent to an annual rate of 12%. 37. We fnd a with at = e0.08t . Thus, a = e0.08 = 1.0833. The corresponding annual percent growth rate is 8.33%. 38. We fnd k with ekt = (1.10)t . We have ek = 1.10 so k = ln(1.10) = 0.0953. The corresponding continuous percent growth rate is 9.53%. 39. (a) Town D is growing fastest, since the rate of growth k = 0.12 is the largest. (b) Town C is largest now, since P0 = 1200 is the largest. (c) Town B is decreasing in size, since k is negative (k = −0.02). 40. If production continues to decay exponentially at a continuous rate of 9.1% per year, the production f (t) at time t years after 2008 is f (t) = 84e−0.091t . In 2025, production is predicted to be f (17) = 84e−0.091(17) = 17.882 million barrels per day.

41. (a) (i) P = 1000(1.05)t ; (ii) P = 1000e0.05t (b) (i) 1629; (ii) 1649 42. (a) P = 5.97(1.0099)t (b) Since P = 5.97ekt = 5.97(1.0099)t , we have ekt = (1.0099)t kt = t ln(1.0099) k = 0.00985. Thus, P = 5.97e0.00985t . (c) The annual growth rate is 0.99%, while the continuous growth rate is 0.985%. Thus the growth rates are not equal, though for small growth rates (such as these), they are close. The annual growth rate is larger. 43. We fnd k with ekt = (1.032)t . We have ek = 1.032 so k = ln(1.032) = 0.0315. An equivalent formula for gross world product as a function of time is W = 123.3e0.0315t . Notice that the annual growth rate of 3.2% is equivalent to a continuous growth rate of 3.15%. 44. We fnd k with ekt = (1.0103)t . We have ek = 1.0103 so k = ln(1.0103) = 0.01025. An equivalent formula for the population of the world as a function of time is P = 7.68e0.01025t . Notice that the annual growth rate of 1.03% is equivalent to a continuous growth rate of 1.025%. 45. (a) Substituting t = 0.5 in P (t) gives P (0.5) = 1000e−0.5(0.5) = 1000e−0.25 ≈ 1000(0.7788) = 778.8 ≈ 779 trout.

Chapter One /SOLUTIONS Substituting t = 1 in P (t) gives P (1) = 1000e−0.5 1000(0.6065) = 606.5

607 trout.

(b) Substituting t = 3 in P (t) gives P (3) = 1000e−0.5(3) = 1000e−1.5 1000(0.2231) = 223.1

223 trout.

This tells us that after 3 years the fsh population has gone down to about 22% of the initial population. (c) We are asked to fnd the value of t such that P (t) = 100. That is, we are asked to fnd the value of t such that 100 = 1000e−0.5t . Solving this gives 100 = 1000e−0.5t 0.1 = e−0.5t ln 0.1 = ln e−0.5t = −0.5t ln 0.1 t= −0.5 .6. Thus, after roughly 4.6 years, the trout population will have decreased to 100. (d) A graph of the population is given in Figure 1.56. Trout 1000

500

P (t) = 1000e−0.5t

t 2

Figure 1.56 Looking at the graph, we see that as time goes on there are fewer and fewer trout in the pond. A possible cause for this may be the presence of predators, such as fshermen.The rate at which the fsh die decreases due to the fact that there are fewer fsh in the pond, which means that there are fewer fsh that have to try to survive. Notice that the survival chances of any given fsh remains the same, it is just the overall mortality that is decreasing. 46. (a) Since the percent rate of growth is constant and given as a continuous rate, we use the exponential function S = 7.986e0.025t . (b) In 2022, we have t = 3 and S = 7.986e0.025⋅3 = 8.608 billion dollars. (c) Using the graph in Figure 1.57, we see that S = 10 at approximately t = 9. Solving 7.986e0.025t = 10, we have 10 7.986 10 0.025t = ln 7.986 ln(10∕7.986) t= = 8.996 years. 0.025 e0.025t =

1.6 SOLUTIONS

In other words, assuming this growth rate continues, annual net sales are predicted to pass 10 billion dollars in 2028. S (billion dollars) 15 (9, 10)

10 5

t (years 10 since 2019)

Figure 1.57

47. (a) The current revenue is R(0) = 5e−0 = 5 million dollars. In two years time the revenue will be R(2) = 5e−0.15(2) = 3.704 million dollars. (b) To fnd the time t at which the revenue has fallen to $2.7 million we solve the equation 5e−0.15t = 2.7. Divide both sides by 5 giving e−0.15t = 0.54. Now take natural logs of both sides: −0.15t = −0.6162, so t = 4.108. The revenue will have fallen to $2.7 million early in the ffth year. 48. (a) We use the formula P = P0 ekt , so we have

P = 50,000e0.045t ,

where t is the number of years since 2014. See Figure 1.58. (b) The year 2020 is when t = 6, so we have P = 50,000e0.045(6) = 65,498.2. We can predict that the population will be about 65,500 in the year 2020. (c) We see in Figure 1.58 that P = 100,000 approximately when t = 15, so the doubling time is about 15 years. population P = 50,000e0.045t

200,000 150,000 100,000 50,000 10

t (years since 2014)

Figure 1.58: The doubling time is about 15 years To fnd the doubling time more precisely, we use logarithms. Solving for the value of t for which P = 100,000: 100,000 = 50,000e0.045t 2 = e0.045t ln 2 = 0.045t ln 2 t= = 15.403. 0.045 The doubling time is 15.403 years. 49. (a) Since the percent increase in deaths during a year is constant for constant increase in pollution, the number of deaths per year is an exponential function of the quantity of pollution. If Q0 is the number of deaths per year without pollution, then the number of deaths per year, Q, when the quantity of pollution is x micrograms per cubic meter of air is Q = Q0 (1.0033)x .

Chapter One /SOLUTIONS (b) We want to fnd the value of x making Q = 2Q0 , that is, Q0 (1.0033)x = 2Q0 . Dividing by Q0 and then taking natural logs yields ln ((1.0033)x ) = x ln 1.0033 = ln 2, so

ln 2 = 210.391. ln 1.0033 When there are 210.391 micrograms of pollutants per cubic meter of air, respiratory deaths per year are double what they would be in the absence of air pollution. x=

50. If C0 is the concentration of NO2 on the road, then the concentration x meters from the road is C = C0 e−0.0254x . We want to fnd the value of x making C = C0 ∕2, that is, C0 e−0.0254x =

C0 . 2

Dividing by C0 and then taking natural logs yields ln e−0.0254x = −0.0254x = ln

1 = −0.6931, 2

so x = 27 meters. At 27 meters from the road the concentration of NO2 in the air is half the concentration on the road. 51. (a) B(t) = B0 e0.067t (b) P (t) = P0 e0.033t (c) If the initial price is $50, then B(t) = 50e0.067t P (t) = 50e0.033t . We want the value of t such that B(t) = 2P (t) 50e0.067t = 2 ⋅ 50e0.033t e0.067t = e0.034t = 2 e0.033t ln 2 t= = 20.387 years . 0.034 Thus, when t = 20.387 the price of the textbook was predicted to be double what it would have been had the price risen by infation only. This occurred in the year 2000. 52. The population of China, C, in billions, is given by C = 1.379(1.0041)t where t is time in years from 2017, and the population of India, I, in billions, is given by I = 1.282(1.0117)t . The two populations will be equal when C = I, thus, we must solve the equation: 1.379(1.0041)t = 1.282(1.0117)t for t, which leads to

(1.0117)t 1.0117 t 1.379 = = . 1.282 (1.0041)t 1.0041

Taking logs on both sides, we get t ln so

1.0117 1.379 = ln , 1.0041 1.282

ln (1.379∕1.282) = 9.67 years. ln (1.0117∕1.0041) This model predicts the population of India will exceed that of China in 2026. t=

1.6 SOLUTIONS

53. If t is time in decades, then the number of vehicles, V , in millions, is given by V = 246(1.155)t . For time t in decades, the number of people, P , in millions, is given by P = 308.7(1.097)t . There is an average of one vehicle per person when

V = 1, or V = P . Thus, we solve for t in the equation: P

246(1.155)t = 308.7(1.097)t , which leads to

1.155 1.097

t =

(1.155)t 308.7 = 246 (1.097)t

Taking logs on both sides, we get t ln so

1.155 308.7 = ln , 1.097 246

ln (308.7∕246) = 4.41 decades. ln (1.155∕1.097) This model predicts one vehicle per person in 2054. t=

54. (a) Using the properties of the natural logarithm, we have ln P = ln(P0 ekt ) = ln P0 + ln(ekt ) = ln P0 + kt Thus, y = ln P is of the form y = b + mt, so it is a line. (b) The slope of y = ln P = ln P0 + kt is k. (c) The vertical intercept of y = ln P = ln P0 + kt is ln P0 . 55. (a) We see in Figure 1.85 that ln(P (t)) is a straight line. From Problem 54 we know the slope of this line is the continuous growth rate k and the vertical intercept is ln(P0 ). We use the points (t, ln(P )) = (10, 13.75) and (t, ln(P )) = (60, 14.2) to estimate the slope: k=

14.2 − 13.75 = 0.009. 60 − 10

This tells us that daily new cases in Florida were growing approximately exponentially with a continuous growth rate k = 0.009 = 0.9% per day. (b) We estimate the vertical intercept to be 13.65, so 13.65 = ln(P0 ) which gives P0 = e13.65 = 847,460.916 ≈ 850,000 cases. 56. (a) If the graph of ln(P ) is a straight line with slope m and intercept b, then ln(P ) = b + mt, so ln(P ) = b + mt P = eb+mt = eb emt P = Cemt

where C = eb is a constant.

This tells us that the slope m of the line is the continuous growth rate of P . On the frst straight segment in Figure 1.86, we estimate Slope = m1 =

3 = 0.3, 10

so the number of confrmed cases was growing at a continuous rate of about 30% per day before day t = 10. On the second straight segment in Figure 1.86, we estimate Slope = m2 =

10 − 8 = 0.057, 70 − 35

so the number of confrmed cases was growing at a continuous rate of about 5.7% per day after day t = 10. (b) On March 27, 2020, the South African government imposed a strict lockdown.2 This date corresponds to t = 9. 2 https://en.wikipedia.org /wiki/COVID-19_pandemic_in_South_Africa, accessed on January 19, 2021.

Chapter One /SOLUTIONS

Solutions for Section 1.7 1. At the doubling time, t = 10, we have p = 2p0 . Thus p0 e10k = 2p0 e10k = 2 10k = ln 2 1 k= ln 2 = 0.0693. 10 The function p = p0 e0.0693t has doubling time equal to 10. 2. At the doubling time, t = 0.4, the we have p = 2p0 . Thus p0 e0.4k = 2p0 e0.4k = 2 0.4k = ln 2 1 k= ln 2 = 1.733. 0.4 The function p = p0 e1.733t has doubling time equal to 0.4. 3. In both cases the initial deposit was $20. Compounding continuously earns more interest than compounding annually at the same interest rate. Therefore, curve A corresponds to the account which compounds interest continuously and curve B corresponds to the account which compounds interest annually. We know that this is the case because curve A is higher than curve B over the interval, implying that bank account A is growing faster, and thus is earning more money over the same time period. 4. Since y(0) = Ce0 = C we have that C = 2. Similarly, substituting x = 1 gives y(1) = 2e so 2e = 1. Rearranging gives e = 1∕2. Taking logarithms we get

= ln(1∕2) = − ln 2 = −0.693. Finally,

y(2) = 2e2(− ln 2) = 2e−2 ln 2 =

1 . 2

5. (a) If the interest is added only once a year (i.e. annually), then at the end of a year we have 1.015x where x is the amount we had at the beginning of the year. After two years, we’ll have 1.015(1.015x) and after eight years, we’ll have (1.015)8 x. Since we started with $1000, after eight years we’ll have (1.015)8 (1000) ≈ $1126.49. (b) If an initial deposit of $P0 is compounded continuously at interest rate r then the account will have P = P0 ert after t years. In this case, P = P0 ert = 1000e(0.015)(8) ≈ $1127.50. 6. We use the equation B = P ert , where P is the initial principal, t is the time in years since the deposit is made to the account, r is the annual interest rate and B is the balance. After t = 5 years, we will have B = (10,000)e(0.08)5 = (10,000)e0.4 ≈ $14,918.25.

7. We use the equation B = P ert . We want to have a balance of B = $20,000 in t = 6 years, with an annual interest rate of 10%. 20,000 = P e(0.1)6 20,000 P = e0.6 ≈ $10,976.23.

1.7 SOLUTIONS

8. We know that the formula for the balance in an account after t years is P (t) = P0 ert where P0 is the initial deposit and r is the nominal rate. In our case the initial deposit is $15,000 that is P0 = 15,000 and the nominal rate is r = 0.015. Thus we get P (t) = 15,000e0.015t . We are asked to fnd t such that P (t) = 20,000, that is we are asked to solve 20,000 = 15,000e0.015t 20,000 4 e0.015t = = 15,000 3 ln e0.015t = ln(4∕3) 0.015t = ln 4 − ln 3 ln 4 − ln 3 t= 0.015

.18.

Thus after roughly 19.18 years there will be $20,000 in the account. 9. The formula which models compounding interest continuously for t years is P0 ert where P0 is the initial deposit and r is the interest rate. So if we want to double our money while getting 1.25% interest, we want to solve the following for t: P0 e0.0125t = 2P0 e0.0125t = 2 0.0125t = ln(2) ln 2 t= 0.0125

.5 years

Alternatively, by the Rule of 70, we have Doubling time

70 = 56 years. 1.25

10. We know that the formula for the total process is P (t) = P0 at where P0 is the initial size of the process and a = 1 + 0.07 = 1.07. We are asked to fnd the doubling time. Thus, we solve 2P0 = P0 (1.07)t 1.07t = 2 ln(1.07t ) = ln 2 t ln 1.07 = ln 2 ln 2 t= ln 1.07

.24.

The doubling time is roughly 10.24 years. Alternatively we could have used the “rule of 70” to estimate that the doubling time is approximately 70 = 10 years. 7

Chapter One /SOLUTIONS

11. (a) The concentration decreases from 100 to 50 during the time from t = 0 to t = 5, a period of 5 years. (b) The concentration decreases from 50 to 25 during the time from t = 5 to t = 10, a period of 5 years. (c) The decay is exponential if the half-life of the concentration is the same beginning at any concentration level. We only checked at two levels, 100 and 50, but for those the half-lives were both 5 years. This is some evidence that the decay might be exponential. Also, if we look at any other 5-year period, we can check that the concentration appears to decreases by half. (d) Since the initial value of concentration is 100, an exponential model is of the form C = 100ekt . We can evaluate k by using the value C = 25 at t = 10. 25 = 100e10k 0.25 = e10k ln(0.25) = 10k k = −0.139. An exponential model is given by C = 100e−0.139t . Alternative solution. Using the half-life T = 5 years, we have C = 100 ⋅ 2−t∕T = 100 ⋅ 2−t∕5 . 12. Every 2 hours, the amount of nicotine remaining is reduced by 1∕2. Thus, after 4 hours, the amount is 1∕2 of the amount present after 2 hours. Table 1.8 t (hours)

Nicotine (mg)

0.4

0.2

0.1

0.05

0.025

0.0125

From the table it appears that it will take just over 6 hours for the amount of nicotine to reduce to 0.04 mg. 13. (a) We have W = 591e0.092t . (b) We substitute W = 1000 and solve for t: 1000 = 591e0.092t 1000 = e0.092t 591 1000 ln = 0.092t 591 ln 1000 591 t= = 5.72. 0.092 Wind capacity is expected to pass 1000 gigawatts in 2024. 14. Let f (t) be oil consumption t years after 2015. Since consumption is exponential, and f (0) = 11.986 million barrels per day, we have f (t) = 11.986ekt . In 2018, when t = 3, consumption was 13.525, so f (3) = 13.525 11.986e3t = 13.525 13.525 3t = ln 11.986 k = 0.040267. Our model predicts Chinese oil consumption in 2025, when t = 10, to be f (10) = 11.986e0.040267(10) = 17.929 million barrels per day.

1.7 SOLUTIONS

15. The doubling time for N(t) is the time t such that N(t) = 2N(0). Since N(t) = N0 e0.054t and N(0) = N0 we have N0 e0.054t = 2N0 e0.054t = 2 0.054t = ln 2 1 t= ln 2 = 12.84 0.054 During the 5 years from January 2014 to January 2019, the doubling time for the number of Wikipedia articles was about 12.84 years. 16. (a) If the CD pays 9% interest over a 10-year period, then r = 0.09 and t = 10. We must fnd the initial amount P0 if the balance after 10 years is P = $12,000. Since the compounding is annual, we use P = P0 (1 + r)t 12,000 = P0 (1.09)10 , and solve for P0 :

12,000 12,000 = 5068.93. 2.36736 (1.09)10 The initial deposit must be $5068.93 if interest is compounded annually. (b) On the other hand, if the CD interest is compounded continuously, we have P0 =

P = P0 ert 12,000 = P0 e(0.09)(10) . Solving for P0 gives

12,000 12,000 12,000 = 4878.84. 2.45960 e(0.09)(10) e0.9 The initial deposit must be $4878.84 if the interest is compounded continuously. Notice that to achieve the same result, continuous compounding requires a smaller initial investment than annual compounding. This is to be expected, since continuous compounding yields more money. P0 =

17. (a) Since sales of electronic devices doubled in 12 years, sales in 1997 were 438∕2 = 219 million devices. Because part (b) asks for the annual percentage growth rate (not a continuous growth rate), we use an exponential function of the form S(t) = S0 bt , where b is the annual growth factor. We have 438 = 219b12 , so 2 = b12 and b = 21∕12 = 1.05946. Thus, S(t) = 219(1.05946)t . (b) The annual growth rate was 5.946%. 18. (a) Let P represent the population of the world (in billions), and let t represent the number of years since 2017. Then we have P = 7.41(1.011)t . (b) According to this formula, the population of the world in the year 2025 (at t = 8) will be P = 7.41(1.011)8 = 8.09 billion people. (c) The graph is shown in Figure 1.59. The population of the world has doubled when P = 14.82; we see on the graph that this occurs at approximately t = 63.4. Under these assumptions, the doubling time of the world’s population is about 63.4 years. P (billion people) 14.82 7.41

63.4

Figure 1.59

t (years)

Chapter One /SOLUTIONS

19. (a) Since the initial amount of ca˙eine is 100 mg and the exponential decay rate is −0.17, we have A = 100e−0.17t . (b) See Figure 1.60. We estimate the half-life by estimating t when the ca˙eine is reduced by half (so A = 50); this occurs at approximately t = 4 hours. A (mg) 100

t (hours)

Figure 1.60 (c) We want to fnd the value of t when A = 50: 50 = 100e−0.17t 0.5 = e−0.17t ln 0.5 = −0.17t t = 4.077. The half-life of ca˙eine is about 4.077 hours. This agrees with what we saw in Figure 1.60. 20. (a) Since P (t) is an exponential function, it will be of the form P (t) = P0 at , where P0 is the initial population and a is the base. P0 = 200, and a 5% growth rate means a = 1.05. Thus, we get P (t) = 200(1.05)t . (b) The graph is shown in Figure 1.61. P P = 200(1.05)t 800

531

200 t 10

Figure 1.61 (c) Evaluating gives that P (10) = 200(1.05)10 326. (d) From the graph we see that the population is 400 at about t = 15, so the doubling time appears to be about 15 years. 21. (a) The pressure P at 6194 meters is given in terms of the pressure P0 at sea level to be P = P0 e−0.00012ℎ = P0 e(−0.00012)6194 = P0 e−0.74328 .4756P0

or about 47.6% of sea level pressure.

(b) At ℎ = 12,000 meters, we have P = P0 e−0.00012ℎ = P0 e(−0.00012)12,000 = P0 e−1.44 .2369P0

or about 23.7% of sea level pressure.

1.7 SOLUTIONS

22. We use P = P0 ekt . Since the half-life is 3 days, we can fnd k: P = P0 ekt 0.50P0 = P0 ek(3) 0.50 = e3k ln(0.50) = 3k ln(0.50) k= 3

−0.231.

After one day, P = P0 e−0.231(1) = P0 (0.79). About 79% of the original dose is in the body one day later. After one week (at t = 7), we have P = P0 e−0.231(7) = P0 (0.20). After one week, 20% of the dose is still in the body. 23. If the rate of increase is r% then each year the output will increase by 1 + r∕100 so that after fve years the level of output is r 5 20,000 1 + . 100 To achieve a fnal output of 30,000, the value of r must satisfy the equation r 5 20,000 1 + = 30,000. 100 Dividing both sides by 20,000 gives

r 5 = 1.5, 100 r so (1 + 100 ) = 1.51∕5 = 1.0845. Solving for r gives an annual growth rate of 8.45%. 1+

24. (a) Using the formula for exponential decay, A = A0 e−kt , with A0 = 10.32. Since the half-life is 12 days, when t = 12 = 5.16. Substituting this into the formula, we have we have A = 10.32 2 5.16 = 10.32e−12k 0.5 = e−12k ,

and, taking ln of both sides

ln 0.5 = −12k ln(0.5) k=− 12

.057762.

The full equation is A = 10.32e−0.057762t . (b) We want to solve for t when A = 1. Substituting into the equation from part (a) yields 1 = 10.32e−0.057762t 1 10.32

.096899 = e−0.057762t ln 0.096899 = −0.057762t −2.33408 t = 40.41 days. −0.057762

25. (a) Since P (t) is an exponential function, it will be of the form P (t) = P0 at . We have P0 = 1, since 100% is present at time t = 0, and a = 0.976, because each year 97.6% of the contaminant remains. Thus, P (t) = (0.976)t . (b) The graph is shown in Figure 1.62.

Chapter One /SOLUTIONS P 100%

50%

t 50

100

Figure 1.62 (c) The half-life is about 29 years, since (0.976)29 0.5. (d) At time t = 100 there is about 9% remaining, since (0.976)100

0.09.

26. (a) We use an exponential function in the form H = H0 e . Since t is in years since 1985, the initial value of H is 2.5 and we have H = 2.5ekt . We use the fact that H = 36.7 when t = 30 to fnd k: H = 2.5ekt 36.7 = 2.5ek⋅30 14.68 = e30k ln(14.68) = 30k ln(14.68) k= = 0.0895. 30 The formula is H = 2.5e0.0895t . See Figure 1.63. living with HIV (millions) 40 30 20 10 year 1985

1995

2005

2015

Figure 1.63: HIV infections worldwide (b) The annual continuous percent change is an increase of approximately 8.95% per year. 27. (a) At t = 0, the population is 1000. The population doubles (reaches 2000) at about t = 4, so the population doubled in about 4 years. (b) At t = 3, the population is about 1700. The population reaches 3400 at about t = 7. The population doubled in about 4 years. (c) No matter when you start, the population doubles in 4 years. 28. We have P0 = 500, so P = 500ekt . We can use the fact that P = 1,500 when t = 2 to fnd k: 1,500 = 500ekt 3 = e2k ln 3 = 2k ln 3 k= 2

.5493.

The size of the population is given by P = 500e0.5493t . At t = 5, we have P = 500e0.5493(5)

7794.

1.7 SOLUTIONS

29. We know that a function modeling the quantity of substance at time t will be an exponential decay function satisfying the fact that at t = 10 hours, 4% of the substance has decayed, or in other words 96% of the substance is left. Thus, a formula for the quantity of substance left at time t is Q(t) = Q0 ekt , where Q0 is the initial quantity. We solve for k: 0.96Q0 = Q0 ek⋅10 0.96 = ek⋅10 ln 0.96 = 10k ln 0.96 k= .004. 10 We are asked to fnd the time t at which the quantity of material will be half of the initial quantity. Thus, we are asked to solve 0.5Q0 = Q0 e−0.004t 0.5 = e−0.004t ln 0.5 = −0.004t ln 0.5 t= −0.004 Thus, the half life of the material is roughly 173 hours.

173.

30. We use Q = Q0 ekt and let t = 0 represent 8 am. Then Q0 = 100, so Q = 100ekt . For the husband, the half-life is four hours. We solve for k: 50 = 100ek⋅4 0.5 = e4k ln 0.5 = 4k ln 0.5 k= 4

.173.

We have Q = 100e−0.173t . At 10 pm, we have t = 14 so the amount of ca˙eine for the husband is Q = 100e−0.173(14) = 8.87 mg

mg.

For the pregnant woman, the half-life is 10. Solving for k as above, we obtain ln 0.5 = −0.069. 10 At 10 pm, the amount of ca˙eine for the pregnant woman is k=

Q = 100e−0.069(14)

38 mg.

31. (a) Since the population of koalas is growing exponentially, we have P = P0 ekt . Since the initial population is 5000, we have P0 = 5000, so P = 5000ekt . We use the fact that P = 27,000 in 2005 (when t = 9) to fnd k: 27,000 = 5000ek(9) Divide both sides by 5000: 27,000 = e9k 5000 0 1 27,000 ln = 9k 5000

Take logs of both sides:

Divide both sides by 9: ln(27,000∕5000) = 0.187. 9 The koala population increased at a continuous rate of 18.7% per year. k=

Chapter One /SOLUTIONS (b) The population, for t in years since 1996, is given by P = 5000e0.187t . (c) In the year 2020, we have t = 24, and using the exact value of k = (ln(27∕5))∕9, we have P = 5000e24 ln(27∕5)∕9 = 448,766 koalas. Thus, if growth continues at the same rate, there will be nearly half a million koalas in 2020.

32. (a) A daily growth rate of 3.4% corresponds to a daily growth factor of 1.034. If this is constant through December 31, 2020, we predict Total number of cases on December 31, 2020 = P (60) = 337,245 ⋅ (1.034)60 = 2,507,135.723. That is, around 2.5 million cases. (b) We want to fnd a new growth factor a so that 2,507,135.723 2 2,507,135.723 60 a = = 3.717 2(337,245) a = (3.717)1∕60 = 1.022.

337,245 ⋅ a60 =

If the daily growth factor is reduced to 1.022 for the period from November 1 to December 31, 2020, then the number of total cases by December 31, 2020 will be half of what continuation with the growth factor of 1.034 predicts. In other words, if the growth rate can be cut from 3.4% per day to 2.2% per day, then the number of total cases by December 31, 2020 would be cut in half. 33. Let t = 0 be April 14. The total number of confrmed cases can be modeled by P = 2979(1.051)t . To fnd when P = 50 million cases, we set up the equation and solve for t. 2979(1.051)t = 50,000,000 (1.051)t = 16,784.15576 ln 16,784.15576 t= ln 1.051 t = 195.573 days Thus, the total number of cases would have reached the population of Colombia in about 196 days; that is, six and a half months. 34. With t in days since April 14 and P0 the total number of cases on April 14, the total number of cases at time t is given by P = P0 (1.051)t , (a) The doubling time can be found by setting P equal to 2P0 and solving for t: 2 ⋅ P0 = P0 (1.051)t 2 = (1.051)t ln 2 = t ln 1.051 ln 2 t= = 13.935 days. ln 1.051 Thus, at this rate, it takes about two weeks for the number of confrmed cases to double. (b) We set up an equation similar to in part (a) but with 10 instead of 2, and solve for t: 10 ⋅ P0 = P0 (1.051)t 10 = (1.051)t ln 10 = t ln 1.051 ln 10 t= = 46.29 days. ln 1.051 Thus, at this rate, it takes about 1.5 months for the number of cases to increase by a factor of 10. (Note, this is less than 5 times 2 weeks.)

1.7 SOLUTIONS

35. (a) If the daily growth rate is 33.33%, the daily growth factor is 1 + 0.3333 = 1.3333. We see that Growth over 3 days = (1.3333)3 = 2.37 = 1 + 1.37. So a 33.33% daily growth corresponds to a 137% increase over 3 days, which is larger than a 100% increase. (b) We want a so that 2P0 = P0 a3 . We take the cube root: 2P0 = P0 a3 2 = a3 1∕3

a = 1.2599. Thus, P = P0 at = P0 (1.2599)t , and the daily growth rate is about 26%. 36. Since P doubles every 10 days we know P is exponential, so P = P0 at for some a. The doubling time of 10 days tells us that 2P0 = P0 a10 . We fnd a by taking 10th roots: 2P0 = P0 a10 2 = a10 1∕10

a = 1.0718. Thus, P = P0 (1.0718)t . To fnd the time it takes to multiply by a factor of 10 we set up the equation 10P0 = P and solve for t: 10P0 = P0 (1.0718)t 10 = (1.0718)t ln 10 = ln((1.0718)t ) ln 10 = t ln(1.0718) t = ln 10∕ ln(1.0718) t = 33.2074 This tells us that in t = 33.207 days, the number of cases of the new variant P will have multiplied by 10. 37. (a) Assuming the US population grows exponentially, we have population P (t) = 282.2ekt at time t years after 2000. Using the 2018 population, which corresponds to t = 18, we have 327.2 = 282.2e18k ln(327.2∕282.2) k= = 0.0082198. 18 We want to fnd the time t in which 350 = 282.2e0.0082198t ln(350∕282.2) t= = 26.195 years. 0.0082198 This model predicts the population to go over 350 million 26.195 years after 2000, in the year 2026. (b) Evaluate P = 282.2e0.0082198t for t = 20 to fnd P = 332.6 million people. 38. We know that the world population is an exponential function over time. Thus the function for world population will be of the form P (t) = P0 at where P0 is the initial population, t is measured in years after 2020 and a is the growth factor. We know that the initial population is the population in the year 2020 so P0 = 7.68 billion.

Chapter One /SOLUTIONS We are also told that in the year 2037 the population will be 9 billion, that is P (17) = 9 billion. Solving for a we get 9 = 7.68a17 9 = a17 7.68 9 1∕17 =a 7.68 a .009 Since the growth factor is 1.009, we know the annual growth rate is projected to be 0.9%.

39. We assume exponential decay and solve for k using the half-life: e−k(5730) = 0.5

k = 1.21 ⋅ 10−4 .

= 0.995,

ln 0.995 = 41.43 years. −1.21 ⋅ 10−4

Now fnd t, the age of the painting: e−1.21⋅10

−4 t

Since Vermeer died in 1675, the painting is a fake. 40. (a) The 1.109 represents the number of food parcels, in millions, when t = 0. Approximately 1,109,000 food parcels were supplied in 2015. (b) The continuous growth rate is 0.12, or 12%, per year. (c) Since e0.12t = (e0.12 )t = 1.127t , we see that the annual growth factor is 1.127. Thus, the annual growth rate is 1.127−1 = 0.127 = 12.7% per year. (d) Since the annual growth factor is only 1.127, the number of food parcels does not double each year. The doubling time is more than one year. 41. (a) We assume f (t) = Ae−kt , where A is the initial population, so A = 100,000. When t = 110, there were 3200 tigers, so 3200 = 100,000e−k⋅110 Solving for k gives e−k⋅110 = k=− so

3200 = 0.032 100,000

1 ln(0.032) = 0.0313 = 3.13% 110 f (t) = 100,000e−0.0313t .

(b) In 2000, the predicted number of tigers was f (100) = 100,000e−0.0313(100) = 4372. In 2010, we know the number of tigers was 3200. The predicted percent reduction is 3200 − 4372 = −0.268 = −26.8%. 4372 Thus the actual decrease is larger than the predicted decrease. 42. For an exponentially growing population, p = p0 ekt , that doubles in n days, we fnd the continuous growth rate k in terms of n. Since p = 2p0 when t = n, we have 2p0 = p0 ekn ekn = 2 kn = ln 2 1 k = ln 2 n p = p0 e(ln 2)t∕n .

1.7 SOLUTIONS

Doubling time for the phytoplankton population, P , is 0.5 days, and for the foraminifera population, F , it is 5 days. If they both initially have the same population C, then after t days we have P = Ce(ln 2)t∕0.5 = Ce1.3863t F = Ce(ln 2)t∕5 = Ce0.13863t . (a) We have P = 2F Ce1.3863t = 2Ce0.13863t e1.3863t = 2e0.13863t 1.3863t = ln 2 + 0.13863t 1.24767t = ln 2 t = 0.556 days. The phytoplankton population is double the foraminifera population after 0.556 days, a little over 13 hours. (b) We have P = 1000F Ce1.3863t = 1000Ce0.13863t e1.3863t = 1000e0.13863t 1.3863t = ln 1000 + 0.13863t 1.24767t = ln 1000 t = 5.537 days. The phytoplankton population is 1000 times the foraminifera population after 5.537 days, about fve and a half days. 43. Let P (t) be the world population in billions t years after 2010. (a) Assuming exponential growth, we have P (t) = 6.9ekt . In 2050, we have t = 40 and we expect the population then to be 9 billion, so 9 = 6.9ek⋅40 . Solving for k, we have ek⋅40 =

9 6.9

1 9 ln = 0.00664 = 0.664% per year. 40 6.9 (b) The “Day of 8 Billion” should occur when 8 = 6.9e0.00664t . k=

Solving for t gives 8 6.9 ln(8∕6.9) t= = 22.277 years. 0.00664

e0.00664t =

So the “Day of 8 Billion” should be 22.277 years after the end of 2010. This is 22 years and 0.277 ⋅ 365 so 101 days into 2032. That is, April 11, 2032. This solution does not take leap years into account.

101 days;

44. (a) From the table, we can see that the number of E85 vehicles increased by 393,553−302,341 = 91,212 between 2012 to 2017. Thus the percent increase is 91,212∕302,341 = 0.302 = 30.2%. Alternatively, we see that 393,553∕30,2341 = 1.302, again showing an increase of 30.2%.

Chapter One /SOLUTIONS (b) To fnd A, we solve when t = 0: Aer⋅0 = 302,341 A ⋅ 1 = 302,341 A = 302,341. Since t is the number of years since 2012, we know that t = 5 in 2017 and get: 393,553 = 302,341e5r . Solving for r, we get 393,553 = e5r 302,341 0 1 393,553 ln = 5r 302,341 r = 0.052732. Substituting t = 1, 2, 3, 4 into 302,341e(0.052732)t , we fnd the remaining table values:

Year

2012

2013

2014

2015

2016

2017

E85 vehicles

302,341

318,712

335,969

354,161

373,338

393,553

(c) The year 2011 is the year t = −1. Therefore, If N is the number of E85-powered vehicles in 2011, we have N = 302,341e0.052732(−1) = 286,811. 45. We have Future value = 10,000e0.03⋅8 = $12,712.49. 46. We have Future value = 20,000e0.038⋅15 = $35,365.34. 47. We have Future value = Present value ⋅ e0.04⋅5 8000 = Present value ⋅ e0.04⋅5 8000 Present value = 0.04⋅5 = $6549.85. e 48. We have Future value = Present value ⋅ e0.032⋅10 20,000 = Present value ⋅ e0.032⋅10 20,000 Present value = 0.032⋅10 = $14,522.98. e

1.7 SOLUTIONS

49. (a) The total present value for each of the two choices are in the following table. Choice 1 is the preferred choice since it has the larger present value.

Choice 1

Choice 2

Year

Payment

Present value

Payment

Present value

1500

1900

3000

3000∕(1.05) = 2857.14

2500

2500∕(1.05) = 2380.95

Total

4357.14

Total

4280.95

(b) The di˙erence between the choices is an extra $400 now ($1900 in Choice 2 instead of $1500 in Choice 1) versus an extra $500 one year from now ($3000 in Choice 1 instead of $2500 in Choice 2). Since 400 × 1.25 = 500, Choice 2 is better at interest rates above 25%. 50. (a)

(i) We know that the formula for the account balance at time t is given by P (t) = P0 ert (since the account is being compounded continuously), where P0 is the initial deposit and r is the annual rate. In our case the initial deposit is P0 = 24 and the annual rate is r = 0.05. Thus, the formula for the balance is P (t) = 24e0.05t . We are asked for the amount of money in the account in the year 2012, that is we are asked for the amount of money in the account 386 years after the initial deposit. This gives amount of money in the account in the year 2012 = P (386) = 24e0.05(386) = 24e19.3 ≈ 24(240,925,906) = 5,782,217,428. Thus, in the year 2012, there would be about 5.78 billion dollars in the account. (ii) Here the annual rate is r = 0.07. Thus, the formula for the balance is P (t) = 24e0.07t . Again, we want P (386). This gives Amount of money in the account in the year 2012 = P (386) = 24e0.07(386) = 24e27.02 ≈ 24(332,531,537,200) = 13,027,111,872,500.

Thus, in the year 2012, there would be about 13.02 trillion dollars in the account. (b) Here the annual rate is r = 0.06. Thus, the formula for the balance is P (t) = 24e0.06t . We are asked to solve for t such that P (t) = 1,000,000,000.

Chapter One /SOLUTIONS Solving we get 1,000,000,000 = B(t) = 24e0.06t 1,000,000,000 = e0.06t 24 e0.06t 41,666,666.7 ln e0.06t

ln 41,666,666.7

0.06t

ln 41,666,666.7 ln 41,666,666.7 0.06

292.4.

Thus, there would be one million dollars in the account 292.4 years after the initial deposit, that is roughly in 1871. 51. (a) Using the Rule of 70, we get that the doubling time of the investment is 70 = 8.75. 8 That is, it would take about 8.75 years for the investment to double. (b) We know that the formula for the balance at the end of t years is B(t) = P at where P is the initial investment and a is 1+(interest per year). We are asked to solve for the doubling time, which amounts to asking for the time t at which B(t) = 2P . Solving, we get 2P = P (1 + 0.08)t 2 = 1.08t ln 2 = ln 1.08t t ln 1.08 = ln 2 ln 2 t= ln 1.08

.01.

Thus, the actual doubling time is 9.01 years. And so our estimation by the “rule of 70” was o˙ by a quarter of a year. 52. You should choose the payment which gives you the highest present value. The immediate lump-sum payment of $2800 obviously has a present value of exactly $2800, since you are getting it now. We can calculate the present value of the installment plan as: PV = 1000e−0.03(0) + 1000e−0.03(1) + 1000e−0.03(2) $2912.21. Since the installment payments o˙er a higher present value, you should accept the installment option. 53. (a) Option 1 is the best option since money received now can earn interest. (b) In Option 1, the entire $2000 earns 5% interest for 1 year, so we have: Future value of Option 1 = 2000e0.05⋅1 = $2102.54. In Option 2, the payment in one year does not earn interest, but the payment made now earns 5% interest for one year. We have Future value of Option 2 = 1000 + 1000e0.05⋅1 = 1000 + 1051.27 = $2051.27. Since Option 3 is paid all in the future, we have Future value of Option 3 = $2000. As we expected, Option 1 has the highest future value.

1.7 SOLUTIONS

(c) In Option 1, the entire $2000 is paid now, so we have: Present value of Option 1 = $2000. In Option 2, the payment now has a present value of $1000 but the payment in one year has a lower present value. We have Present value of Option 2 = 1000 + 1000e−0.05⋅1 = 1000 + 951.23 = $1951.23. Since Option 3 is paid all in the future, we have Present value of Option 3 = 2000e−0.05⋅1 = $1902.46. Again, we see that Option 1 has the highest value. Alternatively, we could have computed the present values directly from the future values found in part (b). 54. (a) The following table gives the present value of the cash fows. The total present value of the cash fows is $116,224.95.

Year

Payment

Present value

50,000

50,000∕(1.075) = 46, 511.63

40,000

40,000∕(1.075)2 = 34, 613.30

25,000

25,000∕(1.075)3 = 20, 124.01

20,000

20,000∕(1.075)4 = 14, 976.01

Total

116, 224.95

(b) The present value of the cash fows, $116,224.95, is larger than the price of the machine, $97,000, so the recommendation is to buy the machine. 55. (a) For Myers, the present value of the frst year’s payment is equal to the payment, $17 million. The payment the next year is discounted by 4%, so it is 2∕(1.04) = $1.923 million. Continuing, we get: Present value, Myers contract = 17 +

2 3 7.4 20 20 + + + + = 61.810 million dollars. 1.04 (1.04)2 (1.04)3 (1.04)4 (1.04)5

For Fowler, we calculate similarly: Present value, Fowler contract = 16.5 +

16.5 16.5 7.4 1.8 + + + = 55.738 million dollars. 1.04 (1.04)2 (1.04)3 (1.04)4

We see that Myers’s contract is about 6 million dollars more valuable in present value. (b) We perform the same calculations using a discount rate of 13%: Present value, Myers contract = 17 +

2 3 7.4 20 20 + + + + = 49.369 million dollars. 1.13 (1.13)2 (1.13)3 (1.13)4 (1.13)5

For Fowler, we calculate similarly: Present value, Fowler contract = 16.5 +

16.5 16.5 7.4 1.8 + + + = 50.256 million dollars. 1.13 (1.13)2 (1.13)3 (1.13)4

If investment returns are very high, Fowler’s contract becomes more valuable because his payments are not as far in the future. 56. In e˙ect, your friend is o˙ering to give you $17,000 now in return for the $19,000 lottery payment one year from now. Since 19,000∕17,000 = 1.11764 ⋯, your friend is charging you 11.7% interest per year, compounded annually. You can expect to get more by taking out a loan as long as the interest rate is less than 11.7%. In particular, if you take out a loan, you have the frst lottery check of $19,000 plus the amount you can borrow to be paid back by a single payment of $19,000 at the end of the year. At 8.25% interest, compounded annually, the present value of 19,000 one year from now is 19,000∕(1.0825) = 17,551.96. Therefore the amount you can borrow is the total of the frst lottery payment and the loan amount, that is, 19,000 + 17,551.96 = 36,551.96. So you do better by taking out a one-year loan at 8.25% per year, compounded annually, than by accepting your friend’s o˙er.

Chapter One /SOLUTIONS

57. The following table contains the present value of each of the payments, though it does not take into account the resale value if you buy the machine.

Buy Year

Lease Payment

Present Value

Payment

Present Value

12000

2650

580

580∕(1.0775) = 538.28

2650

2650∕(1.0775) = 2459.40

464

464(1.0775)2 = 399.65

2650

2650∕(1.0775)2 = 2282.50

290

290∕(1.0775)3 = 231.82

2650

2650∕(1.0775)3 = 2118.33

Total

13,169.75

Total

9510.23

Now we consider the $5000 resale. Present value of resale =

5000 = 3996.85. (1.0775)3

The net present value associated with buying the machine is the present value of the payments minus the present value of the resale price, which is Present value of buying = 13,169.75 − 3996.85 = 9172.90 Since the present value of the expenses associated with buying ($9172.90) is smaller than the present value of leasing ($9510.23), you should buy the machine. 58. The following table contains the present value of each of the expenses. Since the total present value of the expenses, $352.01, is less than the price of the extended warranty, it is not worth purchasing the extended warranty.

Present value of expenses Year

Expense

Present value

150

150∕(1.05)2 = 136.05

250

250∕(1.05)3 = 215.96

Total

352.01

59. Let P be the required present value in 2020 in millions of dollars. Then with an interest rate of r per year, compounded annually over a time period of t years, a payment of $P grows to a future value of $B where B = P (1 + r)t . We have r = 0.02 over t = 10 years with a required future value of B = $222 million, giving P (1 + 0.02)10 = 222. Solving for P gives P =

222 ≈ $182.117 million. 1.0210

60. Let P be the required present value in 2020 in millions of dollars. Then with an interest rate of r per year, compounded continuously over a time period of t years, a payment of $P grows to a future value of $B where B = P ert . We have r = 0.015 over t = 30 years with a required future value of B = $650 million, giving P e(0.015)30 = 650. Solving for P gives P =

650 ≈ $414.458 million. e0.45

1.7 SOLUTIONS

61. The future value of $150 million invested in 2020 at an interest rate r compounded annually for 10 years is 150(1 + r)10 so we must solve 150(1 + r)10 = 222. Solving, we get 150(1 + r)10 = 222 222 (1 + r)10 = 150 1 222 10 1+r = 150 0 11 222 10 r= 150

.03998

To achieve the required payment would require an interest rate of about 4%. 62. The future value of $300 million invested in 2020 at an interest rate r compounded continuously for 30 years is 300e30r so we must solve 300e30r = 650. Solving, we get 300e30r = 650 650 e30r = 300 650 30r = ln 300 1 650 r= ln 30 300

.0258

To achieve the required payment would require an interest rate of about 2.6% 63. Let P , be the present value in 2020 in billions of the $19.5 billion cost of damage attributed to climate change in 2075. Since the interest rate is 2% over the period of 55 years from 2020 to 2075 we get P (1.02)55 = 19.5 Solving for P gives P =

19.5 = 6.562. 1.0255

The present value required is $6.562 billion. 64. (a) We want the present value of $1.4 billion in 2020. Since 2070 is 50 years in the future and the interest rate is 1% = 0.01, we have 1.4 Present value in 2020 = = 0.851 billion dollars. (1 + 0.01)50 (b) To fnd out how much of the $1.4 billion the $0.28 billion will cover, we fnd the future value of the $0.28 billion: Future value in 2070 = 0.28(1.01)50 = 0.460 billion dollars. So Amount not covered = 1.4 − 0.460 = 0.94 billion dollars. 65. The interest rate is 1.5% = 0.015. Thus, the present value of a beach nourishing t years in the future is 12∕(1 + 0.015)t = 12(1.015)−t . The beach nourishings are 1, 6, 11, 16 years in the future, so Present value = 12(1.015)−1 + 12(1.015)−6 + 12(1.015)−11 + 12(1.015)−16 = 42.441 million dollars.

Chapter One /SOLUTIONS

66. (a) We fnd the future value of $7 million in 2011. Since this is one year in the future and the continuous interest rate is 2% = 0.02, we have Future value in 2011 = 7e0.02⋅1 = 7.141 million dollars. Thus, 7.141 − 2.5 = 4.641 million dollars remain. (b) To fnd out how many foods are covered by the $7 million, we can calculate the present value of the money remaining after each food or we can use future values. The foods are 1, 6, 11, . . . years after 2010. We see how many foods occur before the present value remaining is negative—at that point the food wall has paid o˙: Present value after 2011 food = 7 − 2.5e−0.02⋅1 = 4.550 million dollars. Present value after 2016 food = 7 − 2.5e−0.02⋅1 − 2.5e−0.02⋅6 = 2.332 million dollars. Present value after 2021 food = 7 − 2.5e−0.02⋅1 − 2.5e−0.02⋅6 − 2.5e−0.02⋅11 = 0.326 million dollars. Present value after 2026 food = 7 − 2.5e−0.02⋅1 − 2.5e−0.02⋅6 − 2.5e−0.02⋅11 − 2.5e−0.02⋅16 = −1.489 million dollars. Thus the food wall has paid o˙ after 4 foods. Alternatively we can use future value: Future value after 2011 food = 7e0.02⋅1 − 2.5 = 4.6414 million dollars. Five years later, at the 2016 food, we have Future value after 2016 food = 4.6414e0.02⋅5 − 2.5 = 2.63 million dollars. And then for the 2021 and 2026 foods: Future value after 2021 food = 2.63e0.02⋅5 − 2.5 = 0.41 million dollars. Future value after 2026 food = 0.41e0.02⋅5 − 2.5 = −2 million dollars. As we saw with the present value calculation, the food wall pays o˙ after 4 foods. 67. The interest rate is 1.2% = 0.012. The present value of the repairs is $150,000, since they are paid at the time of the purchase. Then 2400 Present value of the frst $2400 = = 2400(1.012)−1 1.012 2400 Present value of the second $2400 = = 2400(1.012)−2 . 1.0122 Similarly for the third and fourth payments. Thus the total present value is Present value = 150,000 + 2400(1.012)−1 + 2400(1.012)−2 + 2400(1.012)−3 + 2400(1.012)−4 = 159,318.770 dollars. 68. The interest rate is 1% = 0.01. Suppose they take in $P each year for 5 years. Then, over the fve years, Present value of earnings = P e−0.01⋅1 + P e−0.01⋅2 + P e−0.01⋅3 + P e−0.01⋅4 + P e−0.01⋅5 � = P e−0.01⋅1 + e−0.01⋅2 + e−0.01⋅3 + e−0.01⋅4 + e−0.01⋅5 = 4.85271P dollars. To pay o˙ the $700,000, the present value must satisfy 4.85271P = 700,000 so P =

700,000 = 144,249 dollars each year. 4.85271

Solutions for Section 1.8 1. (a) We have f (g(x)) = f (3x + 2) = 5(3x + 2) − 1 = 15x + 9. (b) We have g(f (x)) = g(5x − 1) = 3(5x − 1) + 2 = 15x − 1. (c) We have f (f (x)) = f (5x − 1) = 5(5x − 1) − 1 = 25x − 6.

1.8 SOLUTIONS 2. (a) We have f (g(x)) = f (x2 + 8) = (x2 + 8) − 2 = x2 + 6. (b) We have g(f (x)) = g(x − 2) = (x − 2)2 + 8 = x2 − 4x + 12. (c) We have f (f (x)) = f (x − 2) = (x − 2) − 2 = x − 4. 3. (a) We have f (g(x)) = f (e2x = 3e2x . (b) We have g(f (x)) = g(3x) = e2(3x) = e6x . (c) We have f (f (x)) = f (3x) = 3(3x) = 9x. 4. Notice that f (2) = 4 and g(2) = 5. (a) f (2) + g(2) = 4 + 5 = 9 (b) f (2) ⋅ g(2) = 4 ⋅ 5 = 20 (c) f (g(2)) = f (5) = 52 = 25 (d) g(f (2)) = g(4) = 3(4) − 1 = 11. 5. (a) g(2 + ℎ) = (2 + ℎ)2 + 2(2 + ℎ) + 3 = 4 + 4ℎ + ℎ2 + 4 + 2ℎ + 3 = ℎ2 + 6ℎ + 11. (b) g(2) = 22 + 2(2) + 3 = 4 + 4 + 3 = 11, which agrees with what we get by substituting ℎ = 0 into (a). (c) g(2 + ℎ) − g(2) = (ℎ2 + 6ℎ + 11) − (11) = ℎ2 + 6ℎ. 6. (a) f (t + 1) = (t + 1)2 + 1 = t2 + 2t + 1 + 1 = t2 + 2t + 2. (b) f (t2 + 1) = (t2 + 1)2 + 1 = t4 + 2t2 + 1 + 1 = t4 + 2t2 + 2. (c) f (2) = 22 + 1 = 5. (d) 2f (t) = 2(t2 + 1) = 2t2 + 2. 2 (e) (f (t))2 + 1 = t2 + 1 + 1 = t4 + 2t2 + 1 + 1 = t4 + 2t2 + 2. 7. (a) f (g(1)) = f (1 + 1) = f (2) = 22 = 4 (b) g(f (1)) = g(12 ) = g(1) = 1 + 1 = 2 (c) f (g(x)) = f (x + 1) = (x + 1)2 (d) g(f (x)) = g(x2 ) = x2 + 1 (e) f (t)g(t) = t2 (t + 1) 5 8. (a) f (g(1)) = f 2 ) = f (1) = 1 + 4 2 (b) g(f (1)) = g 1 + = g( 5) = ( 5) = 5 (c) f (g(x)) = f (x2 ) = x2 + 4 (d) g(f (x)) = g( x + 4) = ( x + 4)2 = x + 4 (e) f (t)g(t) = ( t + 4)t2 = t2 t + 4 9. (a) f (g(1)) = f (12 ) = f (1) = e1 = e (b) g(f (1)) = g(e1 ) = g(e) = e2 2 (c) f (g(x)) = f (x2 ) = ex x x 2 (d) g(f (x)) = g(e ) = (e ) = e2x (e) f (t)g(t) = et t2 10. (a) f (g(1)) = f (3 ⋅ 1 + 4) = f (7) =

1 7

(b) g(f (1)) = g(1∕1) = g(1) = 7 1 (c) f (g(x)) = f (3x + 4) = 3x + 4 1 1 3 (d) g(f (x)) = g =3 +4 = +4 x x x 1 4 (e) f (t)g(t) = (3t + 4) = 3 + t t 11. (a) We see in the tables that g(1) = 6 and f (6) = 5, so f (g(1)) = f (6) = 5. (b) We see in the tables that f (1) = 5 and g(5) = 2, so g(f (1)) = g(5) = 2. (c) We see in the tables that g(4) = 3 and f (3) = 3, so f (g(4)) = f (3) = 3. (d) We see in the tables that f (4) = 3 and g(3) = 4, so g(f (4)) = g(3) = 4. (e) We see in the tables that g(6) = 1 and f (1) = 5, so f (g(6)) = f (1) = 5. (f) We see in the tables that f (6) = 5 and g(5) = 2, so g(f (6)) = g(5) = 2. 12. (a) We see in the tables that g(0) = 2 and f (2) = 3, so f (g(0)) = f (2) = 3. (b) We see in the tables that g(1) = 3 and f (3) = 4, so f (g(1)) = f (3) = 4. (c) We see in the tables that g(2) = 5 and f (5) = 11, so f (g(2)) = f (5) = 11. (d) We see in the tables that f (2) = 3 and g(3) = 8, so g(f (2)) = g(3) = 8. (e) We see in the tables that f (3) = 4 and g(4) = 12, so g(f (3)) = g(4) = 12.

Chapter One /SOLUTIONS

13. (a)

(d)

(b)

(c)

f (x − 2)

5g(x)

f (x) + 3

(e)

(f)

g(x − 3)

f (x) + g(x)

−8

−4

−1

−2

−5

−9

−f (x) + 2

14. (a) Let u = 5t2 − 2. We now have y = u6 . (b) Let u = −0.6t. We now have P = 12eu . (c) Let u = q 3 + 1. We now have C = 12 ln u. 15. (a) Let u = 3x − 1. We now have y = 2u .√ (b) Let u = 5t2 + 10. We now have P = u. (c) Let u = 3r + 4. We now have w = 2 ln u. 16. m(z + 1) − m(z) = (z + 1)2 − z2 = 2z + 1. 17. m(z + ℎ) − m(z) = (z + ℎ)2 − z2 = 2zℎ + ℎ2 . 18. m(z) − m(z − ℎ) = z2 − (z − ℎ)2 = 2zℎ − ℎ2 . 19. m(z + ℎ) − m(z − ℎ) = (z + ℎ)2 − (z − ℎ)2 = z2 + 2ℎz + ℎ2 − (z2 − 2ℎz + ℎ2 ) = 4ℎz. 20. We see in the graph of g(x) that g(1) = 2 so we have f (g(1)) = f (2) = 4. 21. We see in the graph of f (x) that f (1) = 5 so we have g(f (1)) = g(5) = 6. 22. We see in the graph of g(x) that g(4) = 5 so we have f (g(4)) = f (5) = 1. 23. We see in the graph of f (x) that f (4) = 2 so we have g(f (4)) = g(2) = 3. 24. We see in the graph of f (x) that f (2) = 4 so we have f (f (2)) = f (4) = 2. 25. We see in the graph of g(x) that g(2) = 3 so we have g(g(2)) = g(3) = 4. 26. f (g(1)) = f (2) ≈ 0.4. 27. g(f (2)) ≈ g(0.4) ≈ 1.1. 28. f (f (1)) ≈ f (−0.4) ≈ −0.9. 29. We see in the graph of g(x) that g(3) ≈ −2.5 so we have f (g(3)) = f (−2.5) ≈ 0. 30. To fnd f (g(−3)), we notice that g(−3) = 3 and f (3) = 0 so f (g(−3)) = f (3) = 0. We fnd the other entries in the table similarly. x

−3

−2

−1

f (g(x))

g(f (x))

−2

−3

−2

31. The tree has B = y − 1 branches on average and each branch has n = 2B 2 − B = 2(y − 1)2 − (y − 1) leaves on average. Therefore Average number of leaves = Bn = (y − 1)(2(y − 1)2 − (y − 1)) = 2(y − 1)3 − (y − 1)2 .

1.8 SOLUTIONS

32. We have v(10) = 65 but the graph of u only enables us to evaluate u(x) for 0 ≤ x ≤ 50. There is not enough information to evaluate u(v(10)). 33. We have approximately v(40) = 15 and u(15) = 18 so u(v(40)) = 18. 34. We have approximately u(10) = 13 and v(13) = 60 so v(u(10)) = 60. 35. We have u(40) = 60 but the graph of v only enables us to evaluate v(x) for 0 ≤ x ≤ 50. There is not enough information to evaluate v(u(40)). 36. 2H(x)

(a)

H(x) + 1

(b) 2

1 x

(c)

H(x + 1)

(d)

−1

−H(x)

−1 (e)

H(−x) 1

37. The graph of y = f (x) is shifted vertically upward by 1 unit. See Figure 1.64. y 3

x −3

−2

−1

Figure 1.64: Graph of y = f (x) + 1

38. The graph of y = f (x) is shifted horizontally to the right by 2 units. See Figure 1.65.

Chapter One /SOLUTIONS y 3

x −1

Figure 1.65: Graph of y = f (x − 2)

39. The graph of y = f (x) is stretched vertically 3 units. See Figure 1.66. y 4

x −3

−2

−1

Figure 1.66: Graph of y = 3f (x)

40. The graph of y = f (x) is shifted horizontally to the left by 1 unit and is shifted vertically down by 2 units. See Figure 1.67. y 3 2 1 x −4

−3

−2

−1

−1 −2 −3

Figure 1.67: Graph of y = f (x + 1) − 2

1.8 SOLUTIONS

41. The graph of y = f (x) is refected over the x-axis and shifted vertically up by 3 units. See Figure 1.68. y 4

x −3

−2

−1

Figure 1.68: Graph of y = −f (x) + 3

42. The graph of y = f (x) is refected over the x-axis, stretched by 2 units, and shifted horizontally to the right by 1 units. See Figure 1.69. y 3 2 1 x −2

−1

−1 −2 −3

Figure 1.69: Graph of y = −2f (x − 1)

98 43.

Chapter One /SOLUTIONS y

(a)

(b) 4

−2

(e)

x −2

−2

−4

(f) 4

x 2

−4 y

(e)

−4

(c)

−4

x −2

−2

(b) 4

(d)

−4

(a)

(f)

−2

−4 y

−4

44.

−2

(d)

(c)

−2

−4

x −2

−4

1.8 SOLUTIONS 45.

(a)

(b) 4

x −2

−2

−4 y

(e)

(f)

−2

−4

−2

−4

y 4 2 x −4

−2

Figure 1.70: Graph of y = f (x) + 2

47. See Figure 1.71. y 4 3 2 1 x −5

−3

−1

−4

46. See Figure 1.70.

Figure 1.71: Graph of y = 2f (x)

48. See Figure 1.72.

−4 y

−2

−4 (d)

(c)

100

Chapter One /SOLUTIONS y

x −4

−2

Figure 1.72: Graph of y = f (x − 1)

49. See Figure 1.73. −5

−1 y

−3

−2 −4 −6

Figure 1.73: Graph of y = −3f (x)

50. See Figure 1.74. y y = 2f (x) − 1 2 x −4

−2

Figure 1.74

51. See Figure 1.75. y 3 2 1 −4 −5

−3

−2

x −1

Figure 1.75: Graph of 2 − f (x)

52. The graph shows the concentration leveling o˙ to a. (a) The value of a represents the value at which the concentration levels o˙. (b) We see a is positive. (c) At t = 0, we have f (0) = a − bek⋅0 = a − b. Since the graph goes through the origin, f (0) = 0. Thus a − b = 0 so a = b. (d) Since the concentration of f levels o˙ to a and f (t) = a − bekt , the term bekt must decay to zero. To make the term bekt decay to zero, k must be negative.

1.8 SOLUTIONS

101

53. (a) Since the IV line is started 4 hours later, the drug saturation curve is as shown in Figure 1.76. (b) The function g is a horizontal shift of f . (c) Since g is f shifted 4 hours to the right, we have g(t) = f (t − 4) for t ≥ 4. mg 7.2

hours since 16 8 am

Figure 1.76 54. (a) The equation is y = 2x2 + 1. Note that its graph is narrower than the graph of y = x2 which appears in gray. See Figure 1.77. y = 2x2 + 1

y = 2(x2 + 1)

y = x2

2 x

Figure 1.77

Figure 1.78

(b) y = 2(x2 + 1) moves the graph up one unit and then stretches it by a factor of two. See Figure 1.78. (c) No, the graphs are not the same. Since 2(x2 + 1) = (2x2 + 1) + 1, the second graph is always one unit higher than the frst. 55. The volume of the balloon t minutes after infation began is: g(f (t)) ft3 . 56. The volume of the balloon if its radius were twice as big is: g(2r) ft3 . 57. The time elapsed is: f −1 (30) min. 58. The time elapsed is: f −1 (g −1 (10,000)) min. 59. (a) We fnd an exponential model in the form c(t) = c0 at . Using c(0) = 25 and c(1) = 21.8, we have c(t) = 25at 21.8 = 25a1 = 25a. Thus

21.8 = 0.872 25 c(t) = 25(0.872)t .

Alternately, we could use c(t) = c0 ekt . Then we have 21.8 = 25ek⋅1 = 25ek , giving 21.8 k = ln = ln(0.872) = −0.136966 25 c(t) = 25e−0.136966t . t

(b) We fnd t so that 25(0.872) = 10, giving 10 25 10 t ln(0.872) = ln 25 ln(10∕25) t= = 6.6899. ln(0.872) (0.872)t =

102

Chapter One /SOLUTIONS Alternatively, solving 25e−0.136966t = 10 gives t=−

ln(10∕25) = 6.6899. 0.136966

Using either method shows that it takes 6.690 years for the cyanide concentration to fall to 10 ppm. (c) If D(t) = c(2t), then � t D(t) = 25(0.872)2t = 25 0.8722 = 25(0.7604)t . Alternatively, D(t) = 25e−0.136966(2t) = 25e−0.273932t . (d) Starting three years earlier, but with t measured from the original time, we have E(t) = c(t + 3), so E(t) = 25(0.872)t+3 = 25(0.872)3 (0.872)t = 16.576(0.872)t . Alternatively, E(t) = 25e−0.136966(t+3) = 25e−0.136966(3) e−0.136966t = 16.576e−0.136966t . 60. (a) The patients in quarantine on day t are all the cases that were confrmed up to and including day t but who have not yet left quarantine. Since people stay for exactly 10 days, all cases confrmed before day t − 10 leave. Thus the number in quarantine is the di˙erence Q(t) = P (t) − P (t − 10). (b) Of the confrmed cases P (t) up to and including day t, only rP (t) have entered quarantine. The rP (t − s) who were in quarantine before day t − s leave by day t. Thus, the number in quarantine on day t is the di˙erence r ⋅ (P (t) − P (t − s)) or rP (t) − rP (t − s). 61. We have ℎ(t + 1) = People spending night t in the hospital = People spending night (t − 1) in the hospital +(People checking in on day t) − (People checking out on day t). We investigate the three terms of the sum on the right hand side separately. By defnition, ℎ(t) = People spending night (t − 1) in the hospital. There were N(t) new cases on day t. Therefore 0.14N(t) = People checking in to the hospital on day t. The people who check out of the hospital on day t are exactly those who checked in s days earlier. They are the ones who were confrmed on day t − s. Therefore 0.14N(t − s) = People checking out on day t. Putting it all together gives ℎ(t + 1) = ℎ(t) + 0.14N(t) − 0.14N(t − s). 62. Using the defnitions given: (a) P (t − 1) is the number of cases confrmed up to and including day t − 1. (b) P (t) = P (t − 1) + N(t). The number of cases confrmed up to and including day t is the number confrmed up to and including day t − 1, that is P (t − 1), plus the number who are diagnosed on day t, that is N(t). (c) N(t − 1) is the number diagnosed on day t − 1 (yesterday, if t is today) and N(t − 2) is the number diagnosed two days before day t (2 days ago, if t is today). 63. Using the defnitions given: (a) The infectious people on day t is the number of cases minus those who have recovered and those who have passed away: A(t) = P (t) − R(t). (b) The active cases are those that were diagnosed yesterday and for the 13 days before: A(t) = N(t) + N(t − 1) + … + N(t − 13).

1.9 SOLUTIONS

103

Solutions for Section 1.9 1. y = (1∕5)x; k = 1∕5, p = 1. 2. y = 5x1∕2 ; k = 5, p = 1∕2. 3. y = 8x−1 ; k = 8, p = −1. 4. y = 3x−2 ; k = 3, p = −2. 5. Not a power function. 6. y = 38 x−1 ; k = 38 , p = −1. 7. y = 9x10 ; k = 9, p = 10. 8. y = 25 x−1∕2 ; k = 25 , p = −1∕2. 9. Not a power function. 10. y = 0.2x2 ; k = 0.2, p = 2. 11. y = 53 ⋅ x3 = 125x3 ; k = 125, p = 3. 12. Not a power function because of the +4. 13. For some constant k, we have S = kℎ2 . 14. We know that E is proportional to v3 , so E = kv3 , for some constant k. d 15. If distance is d, then v = . t k 16. For some constant k, we have F = 2 . d 17. (a) This is a graph of y = x3 shifted to the right 2 units and up 1 unit. A possible formula is y = (x − 2)3 + 1. (b) This is a graph of y = −x2 shifted to the left 3 units and down 2 units. A possible formula is y = −(x + 3)2 − 2. 18. Since S = kM 2∕3 we have 18,600 = k(702∕3 ) and so k = 1095. We have S = 1095M

2∕3

. If M = 60, then S = 1095(602∕3 ) = 16,782 cm2 .

19. If we let N represent the number of species of lizard on an island and A represent the area of the island, we have N = kA1∕4 . A graph is shown in Figure 1.79. The function is increasing and concave down. Larger islands have more species of lizards. For small islands, the number of species increases rapidly as the size of the island increases. For larger islands, the increase is much less. species of lizard

area

Figure 1.79

104

Chapter One /SOLUTIONS

20. Let M = blood mass and B = body mass. Then M = k ⋅ B. Using the fact that M = 150 when B = 3000, we have M = k⋅B 150 = k ⋅ 3000 k = 150∕3000 = 0.05. We have M = 0.05B. For a human with B = 70, we have M = 0.05(70) = 3.5 kilograms of blood. 21. (a) Since daily calorie consumption, C, is proportional to the 0.75 power of weight, W , we have C = kW 0.75 . (b) Since the exponent for this power function is positive and less than one, the graph is increasing and concave down. See Figure 1.80. (c) We use the information about the human to fnd the constant of proportionality k. With W in pounds we have 1800 = k(150)0.75 1800 k= = 42.0. 1500.75 The function is C = 42.0W 0.75 . For a horse weighing 700 lbs, the daily calorie requirement is C = 42.0 ⋅ 7000.75 = 5,715.745 calories. For a rabbit weighing 9 lbs, the daily calorie requirement is C = 42.0 ⋅ 90.75 = 218.238 calories. (d) Because the exponent is less than one, the graph is concave down. A mouse has a faster metabolism and needs to consume more calories per pound of weight. C (Calories)

W (weight)

Figure 1.80 22. If y = k ⋅ x3 , then we have y∕x3 = k with k a constant. In other words, the ratio Weight y = x3 (Length)3 should all be approximately equal (and the ratio will be the constant of proportionality k). For the frst fsh, we have y 332 = = 0.0088. (33.5)3 x3 If we check all 11 data points, we get the following values of the ratio y∕x3 : 0.0088, 0.0088, 0.0087, 0.0086, 0.0086, 0.0088, 0.0087, 0.0086, 0.0087, 0.0088, 0.0088.These numbers are indeed approximately constant with an average of about 0.0087. Thus, k .0087 and the allometric equation y = 0.0087x3 fts this data well. 23. Since N is inversely proportional to the square of L, we have N=

k . L2

As L increases, N decreases, so there are more species at small lengths.

1.9 SOLUTIONS

105

24. The specifc heat is larger when the atomic weight is smaller, so we test to see if the two are inversely proportional to each other by calculating their product. This method works because if y is inversely proportional to x, then y=

k x

for some constant k,

yx = k.

Table 1.9 shows that the product is approximately 6 in each case, so we conjecture that sw

, or s

6∕w.

Table 1.9 Element

Lithium

Magnesium

Aluminum

Iron

Silver

Lead

Mercury

6.9

24.3

27.0

55.8

107.9

207.2

200.6

s (cal/deg-gm)

0.92

0.25

0.21

0.11

0.056

0.031

.033

6.3

6.1

5.7

6.1

6.0

6.4

6.6

25. (a) T is proportional to the fourth root of B, and so 4

T = k B = kB 1∕4 . (b) 148 = k ⋅ (5230)1∕4 and so k = 148∕(5230)1∕4 = 17.4. (c) Since T = 17.4B 1∕4 , for a human with B = 70 we have T = 17.4(70)1∕4 = 50.3 seconds It takes about 50 seconds for all the blood in the body to circulate and return to the heart. 26. (a) Since P is inversely proportional to R, we have 1 P =k . R (b) If k = 300,000, we have P = 300,000∕R. If R = 1, the population of the city is 300,000 people, since P = 300,000∕1. The population of the second largest city is about 150,000 since P = 300,000∕2. The population of the third largest city is about 100,000 since P = 300,000∕3. (c) If k = 6 million, the population of the largest city is about 6 million, the population of the second largest city is 6 million divided by 2, which is 3 million, and the population of the third largest city is 6 million divided by 3, which is 2 million. (d) The constant k represents the population of the largest city. 27. (a) Since N is proportional to the 0.77 power of P , we have N = kP 0.77 . (b) Since PA = 10 ⋅ PB , we have NA = k(PA )0.77 = k(10PB )0.77 = k(10)0.77 (PB )0.77 = (10)0.77 (k(PB )0.77 ) = (10)0.77 ⋅ (NB ) = 5.888 ⋅ NB . City A has about 5.888 times as many gas stations as city B. (c) The number of gas stations per person is kP 0.77 N k = = 0.23 = kP −0.23 . P P P This is a power function with negative exponent, so is decreasing. Therefore, the town of 10,000 people has more gas stations per person. 28. (a) We know that the demand function is of the form q = mp + b where m is the slope and b is the vertical intercept. We know that the slope is m=

q(30) − q(25) 460 − 500 −40 = = = −8. 30 − 25 5 5

106

Chapter One /SOLUTIONS Thus, we get q = −8p + b. Substituting in the point (30, 460) we get 460 = −8(30) + b = −240 + b, so that b = 700. Thus, the demand function is q = −8p + 700. (b) We know that the revenue is given by R = pq where p is the price and q is the demand. Thus, we get R = p(−8p + 700) = −8p2 + 700p. (c) Figure 1.81 shows the revenue as a function of price. R 15000 10000 5000 p −5000

100 R = −8p2 + 700p

−10000

Figure 1.81

Looking at the graph, we see that maximal revenue is attained when the price charged for the product is roughly $44. At this price the revenue is roughly $15,300. 29. (a) We know that q = 3000 − 20p, C = 10,000 + 35q, and R = pq. Thus, we get C = 10,000 + 35q = 10,000 + 35(3000 − 20p) = 10,000 + 105,000 − 700p = 115,000 − 700p and R = pq = p(3000 − 20p) = 3000p − 20p2 . (b) The graph of the cost and revenue functions are shown in Figure 1.82.

1.10 SOLUTIONS

107

$ 120,000 C 80,000

40,000

R 92 50

100

p 150

Figure 1.82

(c) It makes sense that the revenue function looks like a parabola, since if the price is too low, although a lot of people will buy the product, each unit will sell for a pittance and thus the total revenue will be low. Similarly, if the price is too high, although each sale will generate a lot of money, few units will be sold because consumers will be repelled by the high prices. In this case as well, total revenue will be low. Thus, it makes sense that the graph should rise to a maximum proft level and then drop. (d) The club makes a proft whenever the revenue curve is above the cost curve in Figure 1.82. Thus, the club makes a proft when it charges roughly between $40 and $145. (e) We know that the maximal proft occurs when the di˙erence between the revenue and the cost is the greatest. Looking at Figure 1.82, we see that this occurs when the club charges roughly $92.

Solutions for Section 1.10 y

1. 3

x 3 2

3 4

−3

Figure 1.83 See Figure 1.83. The amplitude is 3; the period is 2 . y

2. 4

−4

Figure 1.84 See Figure 1.84. The amplitude is 4; the period is .

108

Chapter One /SOLUTIONS y

3. 3

3 4

3 2

−3

Figure 1.85 See Figure 1.85. The amplitude is 3; the period is . y

4. 3

−3

Figure 1.86 See Figure 1.86. The amplitude is 3; the period is . y

5. 6 4 2

Figure 1.87 See Figure 1.87. The amplitude is 1; the period is . y

6. 4

−4

Figure 1.88 See Figure 1.88. The amplitude is 4; the period is 4 . 7. (a) The graph looks like a periodic function because it appears to repeat itself, reaching approximately the same minimum and maximum each year. (b) The maximum occurs in the 2nd quarter and the minimum occurs in the 4th quarter. It seems reasonable that people would drink more beer in the summer and less in the winter. Thus, production would be highest the quarter just before summer and lowest the quarter just before winter. (c) The period is 4 quarters or 1 year. Amplitude =

max − min 2

51 − 41 2

million barrels

1.10 SOLUTIONS

109

8. It makes sense that sunscreen sales would be lowest in the winter (say from December through February) and highest in the summer (say from June to August). It also makes sense that sunscreen sales would depend almost completely on time of year, so the function should be periodic with a period of 1 year. sunscreen sales

months 12

Figure 1.89

9. (a) The function appears to vary between 5 and −5, and so the amplitude is 5. (b) The function begins to repeat itself at x = 8, and so the period is 8. (c) The function is at its highest point at x = 0, so we use a cosine function. It is centered at 0 and has amplitude 5, so we have f (x) = 5 cos(Bx). Since the period is 8, we have 8 = 2 ∕B and B = ∕4. The formula is f (x) = 5 cos x . 4 10. (a) Since C represents the vertical shift, C = 27.5 C, the average temperature. (b) The amplitude, A, is the distance above and below the midline at C, which represents the temperature variation above and below the average. Since the data rises 36.7 − 27.5 = 9.2 C above 27.5 C, and falls 27.5 − 18.4 = 9.1 C. We take A to be the average 9.2 + 9.1 A= = 9.15 C. 2 (c) Since the period is 12 months, 2 12 = B so B = ∕6. (d) The function is H(n) = 9.15 cos n + 27.5. 6 See Figure 1.90. temperature ( C) 36.7 27.5 18.4

n (months)

Figure 1.90

11. Since the volume of the function varies between 2 and 4 liters, the amplitude is 1 and the graph is centered at 3. Since the period is 3, we have 2 2 3= so B = . B 3 The correct formula is (b). This function is graphed in Figure 1.91 and we see that it has the right characteristics.

110

Chapter One /SOLUTIONS t) y = 3 + sin( 2 3

4 3 2

t 3

Figure 1.91

12. If we graph the data, it certainly looks like there’s a pattern from t = 20 to t = 45 which begins to repeat itself after t = 45. And the fact that f (20) = f (45) helps to reinforce this theory. f (t) 3 2 1 20

t 30

Figure 1.92 Since it looks like the graph repeats after 45, the period is (45 − 20) = 25. The amplitude = 12 (max − min) =

1 (2.3 − 1.4) = 0.45. Assuming periodicity, 2

f (15) = f (15 + 25) = f (40) = 2.0 f (75) = f (75 − 25) = f (50) = 1.4 f (135) = f (135 − 25 − 25 − 25) = f (60) = 2.3.

13. Since the temperature oscillates over time, we use a trigonometric function. The function is at its lowest point at t = 0, so we use a negative cosine curve. The amplitude is about (26 − (−6))∕2 = 16 and the midline is halfway between −6 and 26, at 10 C. We have H = −16 cos(Bt) + 10. The period is 12 months so the coeÿcient B = 2 ∕12 = ∕6. Thus, the formula is H = −16 cos t + 10. 6 14. The levels of both hormones certainly look periodic. In each case, the period seems to be about 28 days. Estrogen appears to peak around the 12th day. Progesterone appears to peak from the 17th through 21st day. (Note: the days given in the answer are approximate so your answer may di˙er slightly.) 15. Let b be the brightness of the star, and t be the time measured in days from a when the star is at its peak brightness. Because of the periodic behavior of the star, b(t) = A cos(Bt) + C. The oscillations have amplitude A = 0.35, shift C = 4.0,and period 5.4, so 5.4B = 2 and B = 2 ∕5.4. This gives 2 b(t) = 0.35 cos t + 4. 5.4 16. (a) See Figure 1.93. (b) The number of species oscillates between a high of 28 and a low of 10. The amplitude is (28 − 10)∕2 = 9. The period is 12 months or one year. (c) The graph is at its maximum at t = 0 so we use a cosine function. The amplitude is 9 and the graph is shifted up 19 units, so we have N = 19 + 9 cos(Bt). The period is 12 so B ⋅ 12 = 2 and we have B = ∕6. The formula is N = 19 + 9 cos t . 6

1.10 SOLUTIONS

111

bird species 30 20 10

months

Figure 1.93

17. The graph looks like a sine function of amplitude 7 and period 10. If the equation is of the form y = 7 sin(kt), then 2 ∕k = 10, so k = ∕5. Therefore, the equation is y = 7 sin

t . 5

18. The graph looks like an upside-down sine function with amplitude (90 − 10)∕2 = 40 and period . Since the function is oscillating about the line x = 50, the equation is x = 50 − 40 sin(2t). x . 19. This graph is a cosine curve with period 6 and amplitude 5, so it is given by f (x) = 5 cos 3 x . 20. This graph is a sine curve with period 8 and amplitude 2, so it is given by f (x) = 2 sin 4 21. This graph is an inverted sine curve with amplitude 4 and period , so it is given by f (x) = −4 sin(2x). 22. The graph is a sine curve which has been shifted up by 2, so f (x) = (sin x) + 2. 23. This graph is an inverted cosine curve with amplitude 8 and period 20 , so it is given by f (x) = −8 cos 24. This graph has period 5, amplitude 1 and no vertical shift or horizontal shift from sin x, so it is given by 2 f (x) = sin x . 5 25. This graph has period 6, amplitude 5 and no vertical or horizontal shift, so it is given by 2 x . f (x) = 5 sin x = 5 sin 6 3 26. The graph is a cosine curve with period 2 ∕5 and amplitude 2, so it is given by f (x) = 2 cos(5x). 27. This graph has period 8, amplitude 3, and a vertical shift of 3 with no horizontal shift. It is given by 2 x . f (x) = 3 + 3 sin x = 3 + 3 sin 8 4 28. This can be represented by a sine function of amplitude 3 and period 18. Thus, x . f (x) = 3 sin 9

x . 10

112

Chapter One /SOLUTIONS

29. The graph is a stretched sine function, so it starts at the midline, y = 0, when x = 0. The amplitude is k, so the sinusoidal graph oscillates between a minimum of −k and a maximum of k. Since k is positive, the function begins by increasing. The period is 2 . See Figure 1.94. y 2

k 2 −2

k −k

−2

Figure 1.94

30. The graph is a stretched cosine function refected about the x-axis, so it starts at its minimum when x = 0. The amplitude is k, so the sinusoidal graph oscillates between a minimum of −k and a maximum of k. The period is 2 . See Figure 1.95. y 2

k −2

k −k

−2

Figure 1.95

31. The graph is a stretched cosine function shifted vertically, so it starts at its maximum when x = 0. We begin by fnding the midline, which is k, since the sine function is shifted up by k. The amplitude is k, so the sinusoidal graph oscillates between a minimum of k − k = 0 and a maximum of k + k = 2k. The period is 2 . See Figure 1.96. y 2

k x −2

−k

−2

Figure 1.96

1.10 SOLUTIONS

113

32. The graph is a stretched sine function shifted vertically, so it starts at its midline when x = 0. We begin by fnding the midline, which is −k, since the sine function is shifted up by −k (down by k). The amplitude is k, so the sinusoidal graph oscillates between a minimum of −k − k = −2k and a maximum of −k + k = 0. Since k is positive, the function begins by increasing. The period is 2 . See Figure 1.97. y 2

k −2

k −k

−2

Figure 1.97

33. (a) D = the average depth of the water. (b) A = the amplitude = 15∕2 = 7.5. (c) Period = 12.4 hours. Thus (B)(12.4) = 2 so B = 2 ∕12. (d) C is the time of a high tide.

.507.

34. The depth function is a vertical shift of a function of the form A sin(Bt). The vertical shift is A = (5.5 + 8.5)∕2 = 7. The amplitude is A = 8.5 − 7 = 1.5. The period is 6 = 2 ∕B so B = 2 ∕6 = ∕3. Thus, we have t . Depth = 7 + 1.5 sin 3 35. We use a cosine of the form H = A cos(Bt) + C and choose B so that the period is 24 hours, so 2 ∕B = 24 giving B = ∕12. The temperature oscillates around an average value of 60 F, so C = 60. The amplitude of the oscillation is 20 F. To arrange that the temperature be at its lowest when t = 0, we take A negative, so A = −20. Thus A = 60 − 20 cos t . 12 36. (a)

g(t)

18 16 14 12 10

t 5 10 15 20 25

Figure 1.98 From the graph in Figure 1.98, the period appears to be about 12, and the table tells us that g(0) = 14 = g(12) = g(24), which supports this guess. Amplitude

max − min 2

19 − 11 2

. (b) To estimate g(34), notice that g(34) = g(24 + 10) = g(10) = 11 (assuming the function is periodic with period 12). Likewise, g(60) = g(24 + 12 + 12 + 12) = 14.

114

Chapter One /SOLUTIONS

37. (a) The monthly mean CO2 increased about 12 ppm between December 2013 and December 2018. This is because the black curve shows that the December 2013 monthly mean was about 398 ppm, while the December 2018 monthly mean was about 410 ppm. The di˙erence between these two values, 410 − 398 = 12, gives the overall increase. (b) The average rate of increase is given by 410 − 398 1 = ppm/month. 60 − 0 5 This tells us that the slope of a linear equation approximating the black curve is 1∕5. Since the vertical intercept is about 398, a possible equation for the approximately linear black curve is Average monthly increase of monthly mean =

1 t + 398, 5

where t is measured in months since December 2013. (c) The period of the seasonal CO2 variation is about 12 months since this is approximately the time it takes for the function given by the blue curve to complete a full cycle. The amplitude is about 3.5 since, looking at the blue curve, the average distance between consecutive maximum and minimum values is about 7 ppm. So a possible sinusoidal function for the seasonal CO2 cycle is y = 3.5 sin t . 6 1 t and g(t) = t + 398, we have (d) Taking f (t) = 3.5 sin 6 5 1 ℎ(t) = 3.5 sin t + t + 398. 6 5 See Figure 1.99. ppm 415 410 405 400 395

t (months since Dec 2013)

Figure 1.99

Solutions for Chapter 1 Review 1. Since t represents the number of years since 2020, we see that f (5) represents the population of the city in 2025. In 2025, the city’s population is 8 million. 2. Dan runs 5 kilometers in 23 minutes. 3. If there are no workers, there is no productivity, so the graph goes through the origin. At frst, as the number of workers increases, productivity also increases. As a result, the curve goes up initially. At a certain point the curve reaches its highest level, after which it goes downward; in other words, as the number of workers increases beyond that point, productivity decreases. This might, for example, be due either to the ineÿciency inherent in large organizations or simply to workers getting in each other’s way as too many are crammed on the same line. Many other reasons are possible. 4. See Figure 1.100. temperature

sunrise

noon

Figure 1.100

sunset

time

SOLUTIONS to Review Problems For Chapter One

115

5. The contamination is probably greatest right at the tank, at a depth of 6 meters. Contamination probably goes down as the distance from the tank increases. If we assume that the gas spreads both up and down, the graph might look that the one shown in Figure 1.101. contamination

depth below 12 ground (meters)

Figure 1.101

6. See Figure 1.102. Distance from Kalamazoo 155 120

start in Chicago

arrive in Kalamazoo

Time arrive in Detroit

Figure 1.102

7. (a) The hottest oven results in the largest fnal temperature, so the answer is (I). (b) The initial temperature is represented by the vertical intercept, and the curve with the lowest vertical intercept is (IV). (c) The initial temperature is represented by the vertical intercept, and the two curves with the same vertical intercepts are (II) and (III). (d) The loaf represented by curve (III) heats up much faster than the loaf represented by curve (II). One possible reason for this is that the curve shown in (III) is for a much smaller loaf. 8. The equation of the line is of the form y = b + mx where m is the slope given by Slope =

Rise 3 − (−1) = =2 Run 2−0

and b is the vertical intercept. Since y = −1 when x = 0 the vertical intercept is b = −1. Therefore, the equation of the line passing through (0, −1) and (2, 3) is y = −1 + 2x. Note that when x = 0 this gives y = −1 and when x = 2 it gives y = 3 as required. 9. The equation of the line is of the form y = b + mx where m is the slope given by Slope =

Rise 2−3 1 = =− Run 2 − (−1) 3

so the line is of the form y = b − x∕3. For the line to pass through (−1, 3) we need 3 = b + 1∕3 so b = 8∕3. Therefore, the equation of the line passing through (−1, 3) and (2, 2) is y=

8 1 − x. 3 3

Note that when x = −1 this gives y = 3 and when x = 2 it gives y = 2 as required.

116

Chapter One /SOLUTIONS

10. The equation of the line is of the form y = b + mx where m is the slope given by Rise 0 = =0 Run 2−0 so the line is y = 0 ⋅ x + b = b. Since the line passes through the point (0, 2) we have b = 2 and the equation of the line is y = 2. Slope =

11. If the equation of the line were of the form y = b + mx where m is the slope, then the slope would be given by Rise 1 = Run 0 so the line cannot be of this form. Instead we have the special case of a vertical line of the form x = −1. Slope =

12. (a) is (V), because slope is positive, vertical intercept is negative (b) is (IV), because slope is negative, vertical intercept is positive (c) is (I), because slope is 0, vertical intercept is positive (d) is (VI), because slope and vertical intercept are both negative (e) is (II), because slope and vertical intercept are both positive (f) is (III), because slope is positive, vertical intercept is 0 13. Given that the function is linear, choose any two points, for example (5.2, 27.8) and (5.3, 29.2). Then 29.2 − 27.8 1.4 = = 14. 5.3 − 5.2 0.1 Using the point-slope formula, with the point (5.2, 27.8), we get the equation Slope =

y − 27.8 = 14(x − 5.2) which is equivalent to y = 14x − 45. Δ$ 102 − 69 14. (a) Charge per cubic foot = = = $0.055/cu. ft. Δ cu. ft. 1600 − 1000 Alternatively, if we let c = cost, w = cubic feet of water, b = fxed charge, and m = cost/cubic foot, we obtain c = b + mw. Substituting the information given in the problem, we have 69 = b + 1000m 102 = b + 1600m. Subtracting the frst equation from the second yields 33 = 600m, so m = 0.055. (b) The equation is c = b + 0.055w, so 69 = b + 0.055(1000), which yields b = 14. Thus the equation is c = 14 + 0.055w. (c) We need to solve the equation 201 = 14 + 0.055w, which yields w = 3400. It costs $201 to use 3400 cubic feet of water. 15. (a) We fnd the slope m and intercept b in the linear equation S = b + mt. To fnd the slope m, we use ΔS 66 − 113 = = −0.94. Δt 50 − 0 When t = 0, we have S = 113, so the intercept b is 113. The linear formula is m=

S = 113 − 0.94t. (b) We use the formula S = 113 − 0.94t. When S = 20, we have 20 = 113 − 0.94t and so t = 98.9. If this linear model were correct, the average male sperm count would drop below the fertility level during the year 2038. 16. (a) Substituting x = 1 gives f (1) = 3(1) − 5 = 3 − 5 = −2. (b) We substitute x = 5: y = 3(5) − 5 = 15 − 5 = 10. (c) We substitute y = 4 and solve for x: 4 = 3x − 5 9 = 3x x = 3. (d) Average rate of change =

f (4) − f (2) 7−1 6 = = = 3. 4−2 2 2

SOLUTIONS to Review Problems For Chapter One

117

17. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t) gives the position of the particle, we fnd the values of s(3) = 12 ⋅ 3 − 32 = 27 and s(1) = 12 ⋅ 1 − 12 = 11. Using these values, we fnd Δs(t) s(3) − s(1) 27 − 11 Average velocity = = = = 8 mm∕sec. Δt 3−1 2 18. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t) gives the position of the particle, we fnd the values of s(3) = ln 3 and s(1) = ln 1. Using these values, we fnd Average velocity =

Δs(t) s(3) − s(1) ln 3 − ln 1 ln 3 = = = = 0.549 mm∕sec. Δt 3−1 2 2

19. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t) gives the position of the particle, we fnd the values of s(3) = 11 and s(1) = 3. Using these values, we fnd Δs(t) s(3) − s(1) 11 − 3 = = = 4 mm∕sec. Δt 3−1 2

Average velocity =

20. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t) gives the position of the particle, we fnd the values of s(3) = 4 and s(1) = 4. Using these values, we fnd Δs(t) s(3) − s(1) 4−4 = = = 0 mm∕sec. Δt 3−1 2 Though the particle moves, its average velocity over the interval is zero, since it is at the same position at t = 1 and t = 3. Average velocity =

21. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t) gives the position of the particle, we fnd the values on the graph of s(3) = 2 and s(1) = 3. Using these values, we fnd Average velocity =

Δs(t) s(3) − s(1) 2−3 1 = = = − mm∕sec. Δt 3−1 2 2

22. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t) gives the position of the particle, we fnd the values on the graph of s(3) = 2 and s(1) = 2. Using these values, we fnd Δs(t) s(3) − s(1) 2−2 = = = 0 mm∕sec. Δt 3−1 2 Though the particle moves, its average velocity over the interval is zero, since it is at the same position at t = 1 and t = 3. Average velocity =

23. (a) More fertilizer increases the yield until about 40 lbs.; then it is too much and ruins crops, lowering yield. (b) The vertical intercept is at Y = 200. If there is no fertilizer, then the yield is 200 bushels. (c) The horizontal intercept is at a = 80. If you use 80 lbs. of fertilizer, then you will grow no apples at all. (d) The range is the set of values of Y attainable over the domain 0 ≤ a ≤ 80. Looking at the graph, we can see that Y goes as high as 550 and as low as 0. So the range is 0 ≤ Y ≤ 550. (e) Looking at the graph, we can see that Y is decreasing near a = 60. (f) Looking at the graph, we can see that Y is concave down everywhere, so it is certainly concave down near a = 40. 24. (a) Advertising is generally cheaper in bulk; spending more money will give better and better marginal results initially, hence concave up. (Spending $5,000 could give you a big newspaper ad reaching 200,000 people; spending $100,000 could give you a series of TV spots reaching 50,000,000 people.) See Figure 1.103. But after a certain point, you may “saturate” the market, and the increase in revenue slows down – hence concave down. (b) The temperature of a hot object decreases at a rate proportional to the di˙erence between its temperature and the temperature of the air around it. Thus, the temperature of a very hot object decreases more quickly than a cooler object. The graph is decreasing and concave up. See Figure 1.104 (We are assuming that the co˙ee is all at the same temperature.) temperature

revenue

advertising

Figure 1.103

time

Figure 1.104

118

Chapter One /SOLUTIONS

25. (a) This is the graph of a linear function, which increases at a constant rate, and thus corresponds to k(t), which increases by 0.3 over each interval of 1. (b) This graph is concave down, so it corresponds to a function whose increases are getting smaller, as is the case with ℎ(t), whose increases are 10, 9, 8, 7, and 6. (c) This graph is concave up, so it corresponds to a function whose increases are getting bigger, as is the case with g(t), whose increases are 1, 2, 3, 4, and 5. 26. (a) See Figure 1.105. customers

time

Figure 1.105

(b) “The rate at which new people try it” is the rate of change of the total number of people who have tried the product. Thus, the statement of the problem is telling you that the graph is concave down—the slope is positive but decreasing, as the graph shows. 27. (a) The death rate from stomach cancer has decreased fairly steadily. For lung and bronchus cancer the death rate increased fairly steadily until around 1990 and has decreased since. Colorectum cancer death rates increased slightly until about 1945, leveled o˙ until about 1980 and has decreased since. Prostate cancer deaths increased until about 1945 and then stayed level until a small jump between 1990 and 1995. Since 1995, prostate cancer death rates have decreased. Death rates from liver and intrahepatic bile duct decreased slightly until about 1975 and have increased since. Deaths from leukemia increased slightly until around 1980 and then decreased slightly. Deaths from pancreatic cancer increased until around 1980 and then leveled o˙. (b) Lung and bronchus cancer deaths have had the greatest average rate of change. Average rate of change =

55 − 3 2012 − 1930 additional deaths per 100,000 .634 per year between 1930 and 2012.

3 − 43 2012 − 1930 −0.488 or 0.488

fewer deaths per 100,000 per year between 1930 and 2012.

= 1. 28. For y = x, average rate of change = 10−0 10−0

For y = x2 , average rate of change = 100−0 = 10. 10−0

= 100. For y = x3 , average rate of change = 1000−0 10−0

For y = x4 , average rate of change = 10000−0 = 1000. 10−0 So y = x4 has the largest average rate of change. For y = x, the line is the same as the original function.

SOLUTIONS to Review Problems For Chapter One y

119

100 y=x y = x2

x 10

10 y

1000

10000 y = x3

y = x4

29. Each of these questions can also be answered by considering the slope of the line joining the two relevant points. (a) The average rate of change is positive if the volume of water is increasing with time and negative if the volume of water is decreasing. (i) Since volume is rising from 500 to 1000 from t = 0 to t = 5, the average rate of change is positive. (ii) We can see that the volume at t = 10 is greater than the volume at t = 0. Thus, the average rate of change is positive. (iii) We can see that the volume at t = 15 is lower than the volume at t = 0. Thus, the average rate of change is negative. (iv) We can see that the volume at t = 20 is greater than the volume at t = 0. Thus, the average rate of change is positive. (b) (i) The secant line between t = 0 and t = 5 is steeper than the secant line between t = 0 and t = 10, so the slope of the secant line is greater on 0 ≤ t ≤ 5. Since average rate of change is represented graphically by the slope of a secant line, the rate of change in the interval 0 ≤ t ≤ 5 is greater than that in the interval 0 ≤ t ≤ 10. (ii) The slope of the secant line between t = 0 and t = 20 is greater than the slope of the secant line between t = 0 and t = 10, so the rate of change is larger for 0 ≤ t ≤ 20. (c) The average rate of change in the interval 0 ≤ t ≤ 10 is about 750 − 500 250 = = 25 cubic meters per week 10 10 This tells us that for the frst ten weeks, the volume of water is growing at an average rate of about 25 cubic meters per week. 30. (a) We know that the fxed cost for this company is the amount of money it takes to produce zero units, or simply the vertical intercept of the graph. Thus, the fxed cost = $1000. We know that the marginal cost is the price the company has to pay for each additional unit, or in other words, the slope of the graph. We know that C(0) = 1000 and looking at the graph we also see that C(200) = 4000. Thus the slope of the line, or the marginal cost, is 4000 − 1000 3000 = = $15 per unit. 200 − 0 200 (b) C(q) gives the price that the company will have to pay for the production of q units. Thus if marginal cost =

C(100) = 2500 we know that it will cost the company $2500 to produce 100 items.

120

Chapter One /SOLUTIONS

31. We know that the cost function will take on the form C(q) = b + m ⋅ q where m is the variable cost and b is the fxed cost. We know that the company has a fxed cost of $350,000 and that it costs the company $400 to feed a student. That is, $400 is the variable cost. Thus C(q) = 350,000 + 400q. We know that the revenue function will be of the form R(q) = pq where p is the price that the company charges a student. Since the company intends to charge $800 per student we have R(q) = 800q. We know that the proft is simply the di˙erence between the revenue and the cost. Thus (q) = 800q − (350,000 + 400q) = 800q − 350,000 − 400q = 400q − 350,000. We are asked to fnd the number of students that must sign up for the plan in order for the company to make money. That is, we are asked to fnd the number of students q such that (q) > 0. Solving we get 400q − 350,000 > 0 400q > 350,000 q > 875. Thus, if more than 875 students sign up, the company will make a proft. 32. We are looking for a linear function y = f (x) that, given a time x in years, gives a value y in dollars for the value of the refrigerator. We know that when x = 0, that is, when the refrigerator is new, y = 950, and when x = 7, the refrigerator is worthless, so y = 0. Thus (0, 950) and (7, 0) are on the line that we are looking for. The slope is then given by m=

950 −7

It is negative, indicating that the value decreases as time passes. Having found the slope, we can take the point (7, 0) and use the point-slope formula: y − y1 = m(x − x1 ). So, 950 (x − 7) 7 950 y=− x + 950. 7

y−0 = −

33. Generally manufacturers will produce more when prices are higher. Therefore, the frst curve is a supply curve. Consumers consume less when prices are higher. Therefore, the second curve is a demand curve. 34. (a) We know that the equilibrium point is the point where the supply and demand curves intersect. They appear to intersect at a price of $250 per unit, so the corresponding quantity is 750 units. (b) We know that the supply curve climbs upward while the demand curve slopes downward. At the price of $300 per unit the suppliers will be willing to produce 875 units while the consumers will be ready to buy 625 units. Thus, we see that when the price is above the equilibrium point, more items would be produced than the consumers will be willing to buy. Thus, the producers end up wasting money by producing that which will not be bought, so the producers are better o˙ lowering the price. (c) Looking at the point on the rising curve where the price is $200 per unit, we see that the suppliers will be willing to produce 625 units, whereas looking at the point on the downward sloping curve where the price is $200 per unit, we see that the consumers will be willing to buy 875 units. Thus, we see that when the price is less than the equilibrium price, the consumers are willing to buy more products than the suppliers would make and the suppliers can thus make more money by producing more units and raising the price.

SOLUTIONS to Review Problems For Chapter One

121

35. This is a line with slope −3∕7 and y-intercept 3, so a possible formula is 3 y = − x + 3. 7 36. This is a line, so we use y = mx + b. The slope is a positive 2∕5 = 0.4 and the y-intercept is 2. The formula is y = 0.4x + 2. 37. Starting with the general exponential equation y = Aekx , we frst fnd that for (0, 1) to be on the graph, we must have A = 1. Then to make (3, 4) lie on the graph, we require 4 = e3k ln 4 = 3k ln 4 k= 3

.4621.

Thus the equation is y = e0.4621x . Alternatively, we can use the form y = ax , in which case we fnd y = (1.5874)x . 38. Since this function has a y-intercept at (0, 2), we expect it to have the form y = 2ekx . Again, we fnd k by forcing the other point to lie on the graph: 1 = 2e2k 1 = e2k 2 1 ln = 2k 2 ln( 12 ) k= 2

.34657.

This value is negative, which makes sense since the graph shows exponential decay. The fnal equation, then, is y = 2e−0.34657x . Alternatively, we can use the form y = 2ax , in which case we fnd y = 2(0.707)x . 39. This looks like an exponential function. The y-intercept is 3 and we use the form y = 3ekt . We substitute the point (5, 9) to solve for k: 9 = 3ek5 3 = e5k ln 3 = 5k k = 0.2197. A possible formula is y = 3e0.2197t . Alternatively, we can use the form y = 3at , in which case we fnd y = 3(1.2457)t . 40. z = 1 − cos 41. We look frst at f (x). As x increases by 1, f (x) decreases by 5, then 6, then 7. The rate of change is not constant so the function is not linear. To see if it is exponential, we check ratios of successive terms: 20 = 0.8, 25

14 = 0.7, 20

7 = 0.5. 14

Since the ratios are not constant, this function is not exponential either. What about g(x)? As x increases by 1, g(x) decreases by 3.2 each time, so this function is linear with slope −3.2. The vertical intercept (when x = 0) is 30.8 so the formula is g(x) = 30.8 − 3.2x.

122

Chapter One /SOLUTIONS Now consider ℎ(x). As x increases by 1, ℎ(x) decreases by 6000, then 3600, then 2160. The rate of change is not constant so this function is not linear. To see if it is exponential, we check ratios of successive terms: 9000 = 0.6, 15,000

5400 = 0.6, 9000

3240 = 0.6. 5400

Since the ratios are constant, this is an exponential function with base 0.6. The initial value (when x = 0) is 15, 000 so the formula is ℎ(x) = 15,000(0.6)x . 42. (a) The slope of the line is ΔC 36.44 − 35.24 = = 0.24 billion tons per year. Δt 2019 − 2014 The initial value (when t = 0) is 35.24, so the linear formula is Slope =

C = 35.24 + 0.24t. The annual rate of increase in carbon dioxide emission is 0.24 billion tons per year. (b) If the function is exponential, we have C = C0 at . The initial value (when t = 0) is 35.24, so we have C = 35.24at . We use the fact that C = 36.44 when t = 5 to fnd a: 36.44 = 35.24a5 a5 = 1.034 a = (1.034)1∕5 = 1.0067. The exponential formula is C = 35.24(1.0067)t . Carbon dioxide emission is increasing 0.67% per year. (Alternately, if we used the form C = C0 ekt , the formula would be C = 35.24e0.0067t .) 43. The population has increased by a factor of 48,000,000∕40,000,000 = 1.2 in 10 years. Thus we have the formula P = 40,000,000(1.2)t∕10 , and t∕10 gives the number of 10-year periods that have passed since 2005. In 2005, t∕10 = 0, so we have P = 40,000,000. In 2015, t∕10 = 1, so P = 40,000,000(1.2) = 48,000,000. In 2020, t∕10 = 1.5, so P = 40,000,000(1.2)1.5 ,581,000. To fnd the doubling time, solve 80,000,000 = 40,000,000(1.2)t∕10 , to get t = 38.02 years. 44. Taking logs of both sides ln 3x = x ln 3 = ln 11 ln 11 x= = 2.2. ln 3 45. Isolating the exponential term 20 = 50(1.04)x 20 = (1.04)x . 50 Taking logs of both sides 2 = ln(1.04)x 5 2 ln = x ln(1.04) 5 ln(2∕5) x= = −23.4. ln(1.04)

SOLUTIONS to Review Problems For Chapter One

123

46. We use logarithms: ln(e5x ) = ln(100) 5x = ln(100) ln(100) x= = 0.921. 5 47. We divide both sides by 25 and then use logarithms: 10 = 0.4 25 3x ln(e ) = ln(0.4) e3x =

3x = ln(0.4) ln(0.4) x= = −0.305. 3 48. Since e0.08 = 1.0833, and e−0.3 = 0.741, we have P = e0.08t = e0.08

= (1.0833)t and

−0.3 t

Q = e−0.3t = e

= (0.741)t .

49. (a) The continuous percent growth rate is 15%. (b) We want P0 at = 10e0.15t , so we have P0 = 10 and a = e0.15 = 1.162. The corresponding function is P = 10(1.162)t . (c) Since the base in the answer to part (b) is 1.162, the annual percent growth rate is 16.2%. This annual rate is equivalent to the continuous growth rate of 15%. (d) When we sketch the graphs of P = 10e0.15t and P = 10(1.162)t on the same axes, we only see one graph. These two exponential formulas are two ways of representing the same function, so the graphs are the same. See Figure 1.106. P

P = 10e0.15t and P = 10(1.162)t

t 5

Figure 1.106

50. We know that the formula for the balance after t years in an account which is compounded continuously is P (t) = P0 ert where P0 is the initial deposit and r is the interest rate. In our case we are told that the rate is r = 0.08, so we have P (t) = P0 e0.08t . We are asked to fnd the value for P0 such that after three years the balance would be $10,000. That is, we are asked for P0 such that 10,000 = P0 e0.08(3) .

124

Chapter One /SOLUTIONS Solving we get 10,000 = P0 e0.08(3) = P0 e0.24

P0 (1.27125) 10,000 1.27125 7866.28.

Thus, $7866.28 should be deposited into this account so that after three years the balance would be $10,000. 51. (a) We want to fnd t such that 0.15Q0 = Q0 e−0.000121t , ln 0.15 so 0.15 = e−0.000121t , meaning that ln 0.15 = −0.000121t, or t = ,678.7 years. −0.000121 (b) Let T be the half-life of carbon-14. Then 0.5Q0 = Q0 e−0.000121T , ln 0.5 so 0.5 = e−0.000121T , or T = = 5728.489 5730 years. −0.000121 52. We use P = P0 ekt . Since 200 grams are present initially, we have P = 200ekt . To fnd k, we use the fact that the half-life is 8 years: P = 200ekt 100 = 200ek(8) 0.5 = e8k ln(0.5) = 8k ln(0.5) k= 8

.087.

We have P = 200e−0.087t . The amount remaining after 12 years is P = 200e−0.087(12) = 70.4 grams. To fnd the time until 10% remains, we use P = 0.10P0 = 0.10(200) = 20: 20 = 200e−0.087t 0.10 = e−0.087t ln(0.10) = −0.087t ln(0.10) t= = 26.5 years. −0.087 53. Given the doubling time of 5 hours, we can solve for the bacteria’s growth rate; 2P0 = P0 ek5 ln 2 k= . 5 So the growth of the bacteria population is given by: P = P0 eln(2)t∕5 . We want to fnd t such that 3P0 = P0 eln(2)t∕5 . Therefore we divide both sides by P0 and apply ln. We get t=

5 ln(3) = 7.925 hours. ln(2)

SOLUTIONS to Review Problems For Chapter One

125

54. If the pressure at sea level is P0 , the pressure P at altitude ℎ is given by 0.4 ℎ∕100 , P = P0 1 − 100 since we want the pressure to be multiplied by a factor of (1 − 0.4∕100) = 0.996 for each 100 feet we go up to make it decrease by 0.4% over that interval. At Mexico City ℎ = 7340, so the pressure is P = P0 (0.996)7340∕100 ≈ 0.745P0 . So the pressure is reduced from P0 to approximately 0.745P0 , a decrease of 25.5%. 55. Let A represent the revenue (in billions of dollars) at Apple t years since 2016. Since A = 216 when t = 0 and we want the continuous growth rate, we write A = 216ekt . We use the information from 2020, that A = 275 when t = 4, to fnd k: 275 = 216ek⋅4 1.27315 = e4k ln(1.27315) = 4k k = 0.06037. We have A = 216e0.06037t , which represents a continuous growth rate of 6.037% per year. 56. Marine catch, M (in millions), is increasing exponentially, so M = M0 ekt . If we let t be the number of years since 2007, we have M0 = 140.7 and M = 140.7ekt . Since M = 158 when t = 5, we can solve for k: 158 = 140.7ek(5) 158∕140.7 = e5k ln(158∕140.7) = 5k ln(158∕140.7) k= = 0.0232. 5 Marine catch has increased at a continuous rate of 2.32% per year. Since M = 140.7e0.0232t in the year 2020, we have t = 13 and M = 140.7e0.0232⋅13 = 190.2 million tons.

57. The following table contains the present value of each of the expenses. Since the total present value of the repairs, $195.72, is less than the cost of the service contract, you should not buy the service contract.

Present value of repairs Year

Repairs

Present Value

30∕(1.05) = 28.57

70∕(1.05)2 = 63.49

120

120∕(1.05)3 = 103.66

Total

195.72

126

Chapter One /SOLUTIONS

3 + 1. = 58. (a) g(ℎ(x)) = g x3 3 (b) ℎ(g(x)) = ℎ( x) = ( x) + 1 = x3∕2 + 1. (c) ℎ(ℎ(x)) = ℎ(x3 + 1) = (x3 + 1)3 + 1. (d) g(x) + 1 = x + 1.

(e) g(x + 1) =

x + 1.

59. (a) We know that f (x) = 2x + 3 and g(x) = ln x. Thus, g(f (x)) = g(2x + 3) = ln(2x + 3). (b) We know that f (x) = 2x + 3 and g(x) = ln x. Thus, f (g(x)) = f (ln x) = 2 ln x + 3. (c) We know that f (x) = 2x + 3. Thus, f (f (x)) = f (2x + 3) = 2(2x + 3) + 3 = 4x + 6 + 3 = 4x + 9. 60. (a) f (g(x)) = f (x + 3) = 2(x + 3)2 = 2(x2 + 6x + 9) = 2x2 + 12x + 18. (b) g(f (x)) = g(2x2 ) = 2x2 + 3. (c) f (f (x)) = f (2x2 ) = 2(2x2 )2 = 8x4 . 61. (a) We have f (g(x)) = f (5x2 ) = 2(5x2 ) + 3 = 10x2 + 3. (b) We have g(f (x)) = g(2x + 3) = 5(2x + 3)2 = 5(4x2 + 12x + 9) = 20x2 + 60x + 45. (c) We have f (f (x)) = f (2x + 3) = 2(2x + 3) + 3 = 4x + 9. 62. (a) We have f (g(x)) = f (ln x) = (ln x)2 + 1. (b) We have g(f (x)) = g(x2 + 1) = ln(x2 + 1). (c) We have f (f (x)) = f (x2 + 1) = (x2 + 1)2 + 1 = x4 + 2x2 + 2. 63. For 5f (x), the values are 5 times as large. See Figure 1.107. 10 5 x −4 −2

Figure 1.107 64. For f (x + 5), the graph is shifted 5 units to the left. See Figure 1.108. 10 5 x −4 −2

Figure 1.108 65. For f (x) + 5, the graph is shifted 5 upward. See Figure 1.109. 10 5 x −4 −2

Figure 1.109

SOLUTIONS to Review Problems For Chapter One 66.

(a)

(b)

f (x − 1)

f (x) + 2

−1 1 x 1 (c)

4 (d)

−f (x)

3f (x)

x 1 x

−1

−3

67.

(a)

(b)

f (x − 1) x

f (x) + 2

−1 1 x 1 (c)

4 1

(d)

−f (x) x

3f (x)

1 −1

x 1

−3

68.

(a)

(b)

2 f (x − 1) 1

f (x) + 2

1 x −1 (c)

3 (d)

x −1

3f (x) 4

−f (x) −2

x −1

127

128 69.

Chapter One /SOLUTIONS (a)

(b)

4 2

f (x) + 2

f (x − 1)

x x 1

(c)

(d)

x 1 −1

−f (x)

−2

3f (x) −3

x 1

70. This graph is the graph of m(t) shifted upward by two units. See Figure 1.110. 4 n(t)

t −4

−4

Figure 1.110

71. This graph is the graph of m(t) shifted to the right by one unit. See Figure 1.111. y 4

p(t) t −4

−4

Figure 1.111

72. This graph is the graph of m(t) shifted to the left by 1.5 units. See Figure 1.112.

SOLUTIONS to Review Problems For Chapter One

129

k(t)

−4

t 4

−4

Figure 1.112

73. This graph is the graph of m(t) shifted to the right by 0.5 units and downward by 2.5 units. See Figure 1.113. 4

t −4

6 w(t) −4

Figure 1.113

74. (a) First note that C is negative, because if C = 0 then the quantity of pollution Q = f (t) = A + BeCt does not change over time and if C > 0 then Q = f (t) approaches +∞ or −∞ as t approaches ∞, which is not realistic. Since C < 0, the quantity Q = f (t) approaches A + B(0) = A as t increases. Since Q can not be negative or 0, we have A > 0. Finally, since Q = f (t) is a decreasing function we have B > 0. (b) Initially we have Q = f (0) = A + BeC(0) = A + B. (c) We saw in the solution to part (a) that Q = f (t) approaches A as t approaches ∞, so the legal limit, which is the goal of the clean-up project, is A. 75. This looks like the graph of y = −x2 shifted up 2 units and to the left 3 units. One possible formula is y = −(x + 3)2 + 2. Other answers are possible. You can check your answer by graphing it using a calculator or computer. 76. This looks like the graph of y = x2 shifted down 5 units and to the right 2 units. One possible formula is y = (x − 2)2 − 5. Other answers are possible. You can check your answer by graphing it using a calculator or computer. 77. If p is proportional to t, then p = kt for some fxed constant k. From the values t = 10, p = 25, we have 25 = k(10), so k = 2.5. To see if p is proportional to t, we must see if p = 2.5t gives all the values in the table. However, when we check the values t = 20, p = 60, we see that 60 ≠ 2.5(20). Thus, p is not proportional to t. 78. Substituting w = 65 and ℎ = 160, we have (a) s = 0.01(650.25 )(1600.75 ) = 1.3 m2 . (b) We substitute s = 1.5 and ℎ = 180 and solve for w: 1.5 = 0.01w0.25 (1800.75 ). We have w0.25 =

1.5 = 3.05. 0.01(1800.75 )

Since w0.25 = w1∕4 , we take the fourth power of both sides, giving w = 86.8 kg.

130

Chapter One /SOLUTIONS (c) We substitute w = 70 and solve for ℎ in terms of s: s = 0.01(700.25 )ℎ0.75 , so ℎ0.75 =

s . 0.01(700.25 )

Since ℎ0.75 = ℎ3∕4 , we take the 4∕3 power of each side, giving 0 ℎ=

s 0.01(700.25 )

14∕3 =

s4∕3 (0.014∕3 )(701∕3 )

so ℎ = 112.6s4∕3 . 79. The period is 2 ∕3, because when t varies from 0 to 2 ∕3, the quantity 3t varies from 0 to 2 . The amplitude is 7, since the value of the function oscillates between −7 and 7. 80. The period is 2 ∕(1∕4) = 8 , because when u varies from 0 to 8 , the quantity u∕4 varies from 0 to 2 . The amplitude is 3, since the function oscillates between 2 and 8. 81. The period is 2 ∕ = 2, since when t increases from 0 to 2, the value of t increases from 0 to 2 . The amplitude is 0.1, since the function oscillates between 1.9 and 2.1. 82. (a) It makes sense that temperature would be a function of time of day. Assuming similar weather for all the days of the experiment, time of day would probably be the over-riding factor on temperature. Thus, the temperature pattern should repeat itself each day. (b) The maximum seems to occur at 19 hours or 7 pm. It makes sense that the river would be the hottest toward the end of daylight hours since the sun will have been beating down on it all day. The minimum occurs at around 9 am which also makes sense because the sun has not been up long enough to warm the river much by then. 32 − 28 (c) The period is one day. The amplitude is approximately = 2◦ C 2 83. (a) The period looks to be about 12 months. This means that the number of mumps cases oscillates and repeats itself approximately once a year. Amplitude =

max − min 2

11, 000 − 2000 = 4500 cases 2

This means that the minimum and maximum number of cases of mumps are within 9000(= 4500 ⋅ 2) cases of each other. (b) Assuming cyclical behavior, the number of cases in 30 months will be the same as the number of cases in 6 months which is about 2000 cases. (30 months equals 2 years and six months). The number of cases in 45 months equals the number of cases in 3 years and 9 months which assuming cyclical behavior is the same as the number of cases in 9 months. This is about 2000 as well. 84.

population 900 800 700

Jan.1

July 1

Jan. 1

t (months)

Figure 1.114 x + 2. 4 86. The graph is an inverted sine curve with amplitude 1 and period 2 , shifted up by 2, so it is given by f (x) = 2 − sin x.

85. This graph is the same as in Problem 20 of Section 1.10 but shifted up by 2, so it is given by f (x) = 2 sin

STRENGTHEN YOUR UNDERSTANDING

131

STRENGTHEN YOUR UNDERSTANDING 1. False, the domain is the set of inputs of a function. 2. False, f (10) is the value of the car when it is 10 years old. 3. True, the point (2, 4) is on the graph since f (2) = 22 = 4. 4. False, since we include the points 3 and 4, the set of numbers is written [3, 4]. 5. True. Plugging in r = 0 to fnd the intercept on the D axis gives D = 10. 6. False, a function could be given by a graph or table, or it could be described in words. 7. True, plugging in x = 3, we have f (3) = 32 + 2(3) + 1 = 9 + 6 + 1 = 16. 8. True, consider the function f (x) = x2 − 1. This has horizontal intercepts at x = ±1. 9. False, if a graph has two vertical intercepts, then it cannot correspond to a function, since the function would have two values at zero. 10. True. The vertical intercept is the value of C when q = 0, or the cost to produce no items. 11. True, the slope of a linear function is given by rise over run, where rise is the di˙erence in the function values and run is the di˙erence in the corresponding inputs. 12. False, the linear function m(x) has slope 3. 13. True, any constant function is a linear function with slope zero. 14. True. We frst check that the slope is −1, and then make sure that the point (2, 5) is on the graph. Since −(2) + 7 = 5, the line y = −x + 7 passes through the point (2, 5). 15. True, for every increase in s by 2, the function ℎ(s) decreases by 4, so ℎ(s) could correspond to a linear function with slope −2. 16. False, any two linear functions which have the same slope and di˙erent y-intercepts will not intersect, since they are parallel. 17. False, consider the linear function y = x − 1. The slope of this function is 1, and the y-intercept is −1. 18. False, the slope of a linear function f (x) is given by 3−2 = 15 . 9−4

f (x2 )−f (x1 ) . Plugging in 4 for x1 and 9 for x2 , we see that the slope is x2 −x1

19. False, the units of the slope are acres per meter. 20. True, since a slope of 3 for a linear function y = f (x) means increasing the x coordinate by 1 causes the y coordinate to increase by 3. 21. False. Since the slope of this linear function is negative, the function is decreasing. 22. True, this is just the defnition of the average rate of change. 23. False, the average rate of change is given by C(b)−C(a) . The units of the numerator are dollars, and the units of the denomib−a nator are students, so the units of the average rate of change is in dollars per student. 2

−(−1) . Substituting b = 2 and a = −1, we get 22−(−1) = 4−1 = 24. True, the average rate of change is given by the formula s(b)−s(a) b−a 2+1 3 = 1 > 0. 3

25. True, consider f (x) =

26. False, if Q(r) was concave up, then the average rate of change would be increasing. However, the average rate of change = 5, and the average rate of change from r = 2 to r = 3 is 17−15 = 2. from r = 1 to r = 2 is 15−10 2−1 3−2 27. False, the velocity includes the direction of the particle, whereas the speed does not. 28. True, the rate of change formula is the same as the formula for the slope of the secant line. 29. True, for any values t1 < t2 ,

s(t2 )−s(t1 ) t2 −t1

3t2 +2−(3t1 +2) t2 −t1

3(t2 −t1 ) t2 −t1

= 3.

30. False, consider the function f (x) = 1∕x for x > 0. f (x) is concave up, but it is a decreasing function. 31. True. This is the defnition of relative change. 32. False. Percent change does not have units. 33. True. When quantity increases by 100 from 500 to 600, the cost increases by 15%, which is 0.15 ⋅ 1000 = 150 dollars. Cost goes from $1000 to $1150. 34. False. Relative change can be positive (if the quantity is increasing) or negative (if the quantity is decreasing.)

132

Chapter One /SOLUTIONS

35. False. Relative rate of change is given as a percent per hour. 36. False, proft is the di˙erence between revenue and cost. 37. True, this is the defnition of revenue. 38. False, although the cost function depends on the quantity produced q, there can also be fxed costs of production which are present even when nothing is being produced. 39. False, as q increases C(q) should increase, since producing more units will cost more than producing less. 40. True, since the more units sold, the more revenue is produced. 41. False. It is possible that supply is greater than demand. 42. True. 43. True, both marginal cost and marginal revenue have units dollars per item. 44. True, since the additional proft in selling one more item is the additional revenue from that item minus the additional cost to produce the item. 45. False. Imposing a sales tax shifts the curves which can change where they intersect. 46. True, this is a function of the form P0 at which is the defnition of an exponential function. 47. True. 48. False, an exponential function can be increasing or decreasing. 49. False, the percent growth rate is 3%. 2

= 35(1∕3) = 1∕3. 50. True, substituting x = 2 and x = 1, we have Q(2) Q(1) 35(1∕3) 51. True, substituting s = 0, we have R(0) = 16. 52. True, the frst two decimal digits of e are 2.7. 53. True. 54. True. 55. False, Q(r) has constant di˙erences in successive values so it is linear. It does not have constant ratios of successive values so it is not exponential. 56. False, ln(x) is only defned for x > 0. 57. True, since e0 = 1. 58. False, for example, ln(2) = ln(1 + 1) ≠ ln(1) + ln(1) = 0 + 0 = 0. 59. True, since ln x = c means ec = x. 60. True, since as x increases, the exponent e needs to be raised to in order to equal x also increases. 61. True. We take ln of both sides and bring the exponent t down. 62. True, an exponential growth function is one which has the form P = P0 ekt for k a positive constant. 63. True, when b > 0, eln b = b and ln(eb ) = b. 64. False, since if B = 1, then ln(2B) = ln(2) ≠ 2 ln(B) = 2 ln(1) = 0. 65. True, by the properties of logarithms. 66. True, the doubling time depends only on the rate of growth, which in this case is the same. 67. False. It will take two half-lives, or 12 years, for the quantity to decay to 25 mg. 68. False, the half-life only depends on the rate of decay, which is the same in this case (k = −0.6). So, they have the same half-life. 69. False, if an initial investment Q is compounded annually with interest rate r for t years, then the value of the investment is Q(1 + r)t . So, we should have 1000(1.03)t as the value of the account. 70. False, if the rate of growth is larger, the doubling time will be shorter. 71. True, the doubling time is the value of t which gives double the initial value for P . Since the initial value is 5, we fnd the value of t when P = 10. 72. True, since a deposit made today (present value) will always grow with interest over time (future value). 73. True, since we have a continuous rate of growth, we use the formula P0 ekt . In our case, P0 = 1000, k = .03, and t = 5, giving 1000e.15 .

PROJECTS FOR CHAPTER ONE

133

74. True, since the longer the initial payment is invested, the more interest will be accumulated. 75. False, since less money needs to be invested today to grow to $1000 ten years from now than to grow to $1000 fve years from now. 76. False, since f (x + 5) shifts the graph of f to the left by 5, while f (x) + 5 shifts the graph of f up by 5. 77. True, since f (x + k) is just a shift of the graph of f (x) k units to the left, this shift will not a˙ect the increasing nature of the function f . So, f (x + k) is increasing. 78. True, since −2g(t) fips and stretches the concave up graph of g(t). 79. True, crossing the x-axis at x = 1 means that f (1) = 0. If we multiply both sides by 5, we see that 5f (1) = 0, which means that 5f (x) crosses the x-axis at x = 1. 80. True, since the graph of f (x + 1) is the graph of f (x) shifted to the left by 1. 81. False. We have g(3 + ℎ) = (3 + ℎ)2 = 9 + 6ℎ + ℎ2 so g(3 + ℎ) ≠ 9 + ℎ2 for many values of ℎ. 82. False, f (g(t)) = (t + 1)2 = t2 + 2t + 1 and g(f (t)) = t2 + 1 which are clearly not equal. 83. True, we have f (g(x)) = f (ln x) = (ln x)3 − 5. 84. True, we can check that g(u(x)) = g(3x2 + 2) = (3x2 + 2)3 = ℎ(x). 85. False, since f (x + ℎ) = (x + ℎ)2 − 1 = x2 + 2xℎ + ℎ2 − 1, we see that f (x + ℎ) − f (x) = 2xℎ + ℎ2 which is not equal to ℎ2 . 86. True. 87. False, if A is inversely proportional to B, then A = k∕B for some non-zero constant k. 88. True, a power function has the form g(x) = k ⋅ xp . 89. False, this is an exponential function. 90. False. It can be written as the power function ℎ(x) = 3x−1∕2 . 91. False. It can be written as a power function in the form g(x) = (3∕2)x−2 92. True, since f (x) = (3∕2)x1∕2 = 1.5x1∕2 . 93. True. 94. True. 95. True. If p is proportional to q, then p = kq for nonzero constant k, so p∕q = k, a constant. 96. False, the amplitude is the coeÿcient of the trig function, which in this case is 3. 97. True, for a trig function of the form cos(kx) + b, the period is given by 2 ∕k. Since k = 1 in this case, the period is 2 . 98. False, if we substitute t = 0, then the frst expression is not defned, whereas the second expression is 3∕5. Thus, they cannot be equal. 99. True. It is shifted ∕2 to the left. 100. False, for a trig function of the form A cos(kx) + b, the period is given by 2 ∕k. Since k = 5 in this case, the period is 2 ∕5. 101. False, since the period of y = sin(2t) is and the period of sin(t) is 2 , the period is half. 102. True, since both have amplitude of 5. The second function has a vertical shift of 8. 103. False, one way to see that the graphs are di˙erent is that (sin x)2 is never negative (as it is a square). On the other hand, y = sin(x2 ) is both positive and negative. 104. False, the function sin(x) oscillates between −1 and 1. 105. True.

PROJECTS FOR CHAPTER ONE 1. (a) Compounding daily (continuously), P = P0 ert = $450,000e(0.06)(213) $1.5977 ⋅ 1011 . This amounts to approximately $160 billion.

134

Chapter One /SOLUTIONS (b) Compounding yearly, A = $450,000 (1 + 0.06)213 = $450,000(1.06)213 $450,000(245,555.29) $1.10499882 ⋅ 1011 . This is only about $110.5 billion. (c) We frst wish to fnd the interest that will accrue during 1990. For 1990, the principal is $1.105 ⋅ 1011 . At 6% annual interest, during 1990 the money will earn 0.06 ⋅ $1.105 ⋅ 1011 = $6.63 ⋅ 109 . The number of seconds in a year is 0 10 1 days hours mins secs 365 24 60 60 = 31,536,000 sec. year day hour min Thus, over 1990, interest is accumulating at the rate of $6.63 ⋅ 109 31,536,000 sec

$210.24 /sec.

2. (a) Since the population center is moving west at 40 miles per 10 years, or 4 miles per year, if we start in 2020, when the center is in Hartville, its distance d west of Hartville t years after 2020 is given by d = 4t. (b) It moved over 1000 miles over the 230 years from 1790 to 2020, so its average speed was greater than 1000∕230 = 4.35 miles/year, somewhat bigger than its present rate. (c) According to the function in (a), after 500 years, the population center would be 2000 miles west of Hartville, in the Pacifc Ocean, which is impossible. 3. Assuming that Tr decays exponentially, we have Tr = Ae−kt . We fnd the values of A and k from the data in the table in the problem. We know that: Ae−k⋅4 = 37 so

since Tr = 37 at t = 4

Ae−k⋅19.5 = 13 since Tr = 13 at t = 19.5 Ae−k⋅4 37 = dividing 13 Ae−k⋅19.5 37 e15.5k = 13 1 37 k= ln = 0.06748. 15.5 13

Having found k, we can now fnd A: Ae−0.06748(4) = 37

since Tr = 37 at t = 4

A = 48.5 ng∕ml. Thus, the concentration of tryptase t hours after surgery is modeled by the exponential decay function Tr = 48.5e−0.06748t ng∕ml. The greatest concentration is 48.5 ng∕ml, which occurs at t = 0 when the patient leaves surgery. We can diagnose anaphylaxis since this peak level is above 45 ng∕ml.

2.1 SOLUTIONS

135

CHAPTER TWO Solutions for Section 2.1 1. Recall that f (x) gives the y-coordinate at a point while f ¨ (x) gives the slope at a point. We see that (a) f (−2) is negative (b) f (0) is positive (c) f (4) is negative (d) f (5) is zero (e) f ¨ (−2) is positive (f) f ¨ (2) is negative (g) f ¨ (4) is zero (h) f ¨ (5) is positive 2. Recall that f (x) gives the y-coordinate at a point while f ¨ (x) gives the slope at a point. (a) We see that f (1) is higher on the graph than f (2) so f (1) is larger. (b) We see that f (6) is higher on the graph than f (5) so f (6) is larger. (c) We see that f (3) is higher on the graph than f (4) so f (3) is larger. (d) We see that f ¨ (−2) is positive while f ¨ (2) is negative so f ¨ (−2) is larger. (e) We see that f ¨ (4) is approximately zero while f ¨ (5) is positive, so f ¨ (5) is larger. (f) We see that f ¨ (−1) and f ¨ (0) are both positive, but that the graph is steeper at x = −1, so the slope is greater there and f ¨ (−1) is larger. 3. Recall that f (x) gives the y-coordinate at a point while f ¨ (x) gives the slope at a point. In this case, we are determining whether the y-coordinate at x = 2 is positive. We see that this is true only for the functions shown in A and B. 4. Recall that f (x) gives the y-coordinate at a point while f ¨ (x) gives the slope at a point. In this case, we are determining whether the y-coordinate at x = 4 is negative. We see that this is true only for the functions shown in C and D. 5. Recall that f (x) gives the y-coordinate at a point while f ¨ (x) gives the slope at a point. In this case, we are determining whether the slope at x = 2 is positive. We see that this is true only for the functions shown in A and B. 6. Recall that f (x) gives the y-coordinate at a point while f ¨ (x) gives the slope at a point. In this case, we are determining whether the slope at x = 4 is negative. We see that this is true only for the functions shown in A and D. 7. (a) We have s = f (t) where s is the height at time t and we are told that the height is s = 96 at time t = 2, so we have f (2) = 96. (b) We know that the derivative f ¨ (t) gives the instantaneous rate of change at time t, and we are told that the rate of change at t = 2 is 16. We have f ¨ (2) = 16. 8. (a) We have w = f (t) where w is the quantity at time t and we are told that the quantity is w = 50 at time t = 20, so we have f (20) = 50. (b) We know that the derivative f ¨ (t) gives the instantaneous rate of change at time t, and we are told that the rate of change at t = 20 is 3. We have f ¨ (20) = 3. 9. (a) The average rate of change is the slope of the secant line in Figure 2.1, which shows that this slope is positive. (b) The instantaneous rate of change is the slope of the graph at x = 3, which we see from Figure 2.2 is negative.

136

Chapter Two /SOLUTIONS f (x)

Slope = Average rate of change of f

Slope = f ¨ (3) = Rate of change of f at 3

x 7

Figure 2.1

Figure 2.2

10. We use the interval x = 1 to x = 1.01: f (1.01) − f (1) 4.05583 − 4 41.01 − 41 = = = 5.583. 1.01 − 1 0.01 0.01

g¨

For greater accuracy, we can use the smaller interval x = 1 to x = 1.001: g¨

f (1.001) − f (1) 4.005549 − 4 41.001 − 41 = = = 5.549. 1.001 − 1 0.001 0.001

11. (a) Figure 2.3 shows that for t = 2 the function g(t) = (0.8)t is decreasing. Therefore, g ¨ (2) is negative.

2 g(t) = (0.8)t

t 1

Figure 2.3 (b) To estimate g ¨ (2) we take the small interval between t = 2 and t = 2.001 to the right of t = 2. g¨

g(2.001) − g(2) 0.6399 − 0.64 −0.0001 0.82.001 − 0.82 = = = = −0.1 2.001 − 2 0.001 0.001 0.001

12. The slope is positive at A and D; negative at C and F . The slope is most positive at A; most negative at F . 13.

Slope

−3

−1

1∕2

Point

14. (a) f ¨ (−1), f ¨ (0) are positive; f ¨ (1) is zero; f ¨ (2), f ¨ (3) are negative. (b) In ascending order, the derivatives are: f ¨ (3) < f ¨ (2) < f ¨ (1) < f ¨ (0) < f ¨ (−1). 15. Using the interval 1 ≤ x ≤ 1.001, we estimate f (1.001) − f (1) 3.0033 − 3.0000 = = 3.3 0.001 0.001 x The graph of f (x) = 3 is concave up so we expect our estimate to be greater than f ¨ (1). f¨

f (x)

2.1 SOLUTIONS

137

16. (a) The function N = f (t) is decreasing when t = 1950. Therefore, f ¨ (1950) is negative. That means that the number of farms in the US was decreasing in 1950. (b) The function N = f (t) is decreasing in 1960 as well as in 1980 but it is decreasing faster in 1960 than in 1980. Therefore, f ¨ (1960) is more negative than f ¨ (1980). (c) Since the graph of N = f (t) has a horizontal tangent line at t = 2000 we have f ¨ (2000) = 0. That means that the number of farms in the US stayed approximately constant in 2000. 17. (a) For the interval 0 ≤ t ≤ 0.2, we have H I Average velocity 0 ≤ t ≤ 0.2 (b) For the interval 0.2 ≤ t ≤ 0.4, we have H I Average velocity 0.2 ≤ t ≤ 0.4

s(0.2) − s(0) 0.5 = = 2.5 ft/sec. 0.2 − 0 0.2

s(0.4) − s(0.2) 1.3 = = 6.5 ft/sec. 0.4 − 0.2 0.2

(c) To estimate the instantaneous velocity at t = 0.2, we can average the average velocities found on the left and the right of t = 0.2. So a reasonable estimate of the velocity at t = 0.2 is 12 (6.5 + 2.5) = 4.5 ft/sec. 18. (a) Let s = f (t). (i) We wish to fnd the average velocity between t = 1 and t = 1.1. We have Average velocity =

f (1.1) − f (1) 7.84 − 7 = = 8.4 m/sec. 1.1 − 1 0.1

(ii) We have Average velocity =

f (1.01) − f (1) 7.0804 − 7 = = 8.04 m/sec. 1.01 − 1 0.01

(iii) We have f (1.001) − f (1) 7.008004 − 7 = = 8.004 m/sec. 1.001 − 1 0.001 (b) We see in part (a) that as we choose a smaller and smaller interval around t = 1 the average velocity appears to be getting closer and closer to 8, so we estimate the instantaneous velocity at t = 1 to be 8 m/sec. Average velocity =

19. (a) The average rate of change of a function over an interval is represented graphically as the slope of the secant line to its graph over the interval. See Figure 2.4. Segment AB is the secant line to the graph in the interval from x = 0 to x = 3 and segment BC is the secant line to the graph in the interval from x = 3 to x = 5. We can easily see that slope of AB > slope of BC. Therefore, the average rate of change between x = 0 and x = 3 is greater than the average rate of change between x = 3 and x = 5. y

4 y 3

This slope is larger

❄

2 1 1

A x

x 1

Figure 2.4

Figure 2.5

(b) We can see from the graph in Figure 2.5 that the function is increasing faster at x = 1 than at x = 4. Therefore, the instantaneous rate of change at x = 1 is greater than the instantaneous rate of change at x = 4. (c) The units of rate of change are obtained by dividing units of cost by units of product: thousands of dollars/kilogram.

138

Chapter Two /SOLUTIONS

20. (a) The average velocity between t = 2 and t = 5 is s(5) − s(2) Distance 25 − 4 21 = = = = 7 ft/sec. Time 5−2 3 3 (b) Using an interval of size 0.1, we have H I Instantaneous velocity at t = 2 Using an interval of size 0.01, we have H I Instantaneous velocity at t = 2

≈

s(2.1) − s(2) 4.41 − 4 = = 4.1. 2.1 − 2 0.1

s(2.01) − s(2) 4.0401 − 4 = = 4.01. 2.01 − 2 0.01

From this we guess that the instantaneous velocity at t = 2 is about 4 ft/sec. 21. First, draw a line joining the origin to any point in the frst quadrant labeled (5, f (5)). This line represents the distance as a function of time if the particle had moved at a constant speed, the average velocity. The average velocity is the slope of the line between (0, 0) and (5, f (5)). Since the particle’s velocity varies, it must sometimes be more than and sometimes less than the average velocity. The curve we want has instantaneous velocity at exactly two points equal to the average velocity. Thus, we want a curve whose endpoints are the origin and (5, f (5)) and with tangents at exactly two points that are parallel the line. To achieve this, the curve must be sometimes above, and sometimes below, the line. For example, we can draw a smooth curve that is above the line for the frst half of the interval from 0 to 5 and below the line for the second half of this interval. One possibility is in Figure 2.6. Other answers are possible. s f (t)

t 1

Figure 2.6

22. (a) We have 7.68 − 6.53 = 0.082 billion people per year. 2020 − 2006 (b) To estimate f ¨ (2018), we use the rate of change formula on an interval containing 2018: Rate of change of population =

f ¨ (2018) ≈

7.68 − 7.50 = 0.09 billion people per year. 2020 − 2018

23. (a) The size of the tumor when t = 0 months is S(0) = 20 = 1 cubic millimeter. The size of the tumor when t = 6 months is S(6) = 26 = 64 cubic millimeters. The total change in the size of the tumor is S(6) − S(0) = 64 − 1 = 63 mm3 . (b) The average rate of change in the size of the tumor during the frst six months is: Average rate of change =

S(6) − S(0) 64 − 1 63 = = = 10.5 cubic millimeters/month. 6−0 6 6

t (months)

6.001

6.01

6.1

S (cubic millimeters)

64.0444

64.4452

68.5935

2.1 SOLUTIONS

139

68.5935 − 64 4.5935 = = 45.935 6.1 − 6 0.1 64.4452 − 64 0.4452 Average rate of change = = = 44.52 6.01 − 6 0.01 64.0444 − 64 0.0444 Average rate of change = = = 44.4 6.001 − 6 0.001 Average rate of change =

We can continue taking smaller intervals but the value of the average rate will not change much. Therefore, we can say that a good estimate of the growing rate of the tumor at t = 6 months is about 44.4 cubic millimeters/month. 24. (a) Since the values of P are always decreasing we see that f ¨ (2) is negative. The values of N are always increasing, so g ¨ (4) is positive. The number of pay TV subscriptions in the US is decreasing at t = 2 (in the year 2016). The number of households with Netfix subscriptions in the US is increasing at t = 4 (in the year 2018). (b) We estimate f ¨ (2) using the di˙erence quotient for the interval to the right of t = 2, as follows: f¨

ΔP 90.3 − 97.7 −7.4 = = = −3.7. Δt 4−2 2

The fact that f ¨ (2) = −3.7 tells us that the number of pay TV subscriptions in the US was decreasing at a rate of 3.7 million per year when t = 2 (in the year 2016). Similarly, ΔN 203.7 − 139.3 64.4 g¨ = = = 32.2. Δt 6−4 2 The fact that g ¨ (4) = 32.2 tells us that the number Netfix subscriptions in the US was increasing at a rate of 32.2 million per year when t = 4 (in the year 2018). 25. (a) Since 80.8 percent live in the city in 2010 and 35.1 percent in 1890, we have Average rate of change =

80.8 − 35.1 45.7 = = 0.381. 2010 − 1890 120

Thus the average rate of change is 0.381 percent/year. (This rate is said to be 0.381 percentage points per year, to distinguish it from a percent change.) (b) By looking at 2000 and 2010 we see that Average rate of change =

80.8 − 79.1 1.7 = = 0.17 2010 − 2000 10

which gives an average rate of change of 0.17 percent/year between 2000 and 2010. (That is, 0.17 percentage points per year.) However, the defnition of an urban area was changed for the 2000 data, so this estimate should be used with care. We could also look at 1990 and 2000 and see that Average rate of change =

79.1 − 75.3 3.8 = = 0.38 2000 − 1990 10

giving an average rate of change of 0.38 percent/year between 1990 and 2000. We see that the rate of change in the year 2000 is somewhere between 0.17 and 0.38 percent/year. (c) By looking at 1830 and 1860 we see that Average rate of change =

19.8 − 9.0 10.8 = = 0.36, 1860 − 1830 30

giving an average rate of change of 0.36 percent/year between 1830 and 1860. (That is, 0.36 percentage points per year.) Alternatively, we can look at 1800 and 1830 and see that Average rate of change =

9.0 − 6.0 3.0 = = 0.10, 1830 − 1800 30

giving an average rate of change of 0.10 percent/year between 1800 and 1830. We see that the rate of change at the year 1830 is somewhere between 0.10 and 0.36. This tells us that in the year 1830 the percent of the population in urban areas is increasing at a rate somewhere between 0.10 percent/year and 0.36 percent/year.

140

Chapter Two /SOLUTIONS

26. (a) The value 6.08 is the population (in billions of people) of the world in the year 2000, and is not a rate of change. The answer is (iii). (b) The value 7.68 is the population (in billions of people) of the world in the year 2020, and is not a rate of change. The answer is (iii). (c) The value 0.08 is the average rate of change in the world’s population, in billions of people per year, during the period from the year 2000 to the year 2020. The answer is (i). (d) The value 0.079 is the instantaneous rate of change in the world’s population, in billions of people per year, right at the start of the year 2000. The answer is (ii). (e) The value 0.081 is the instantaneous rate of change in the world’s population, in billions of people per year, right at the start of the year 2020. The answer is (ii). 27. (a) The average velocity between t = 2 and t = 3 is given by f (3) − f (2) 56 − 136 = = −80 ft/sec. 3−2 1 (b) We use the average rate of change formula in each case: (i) If ℎ = 0.1, the average rate of change is f (2.1) − f (2) 129.44 − 136 = = −65.6 ft/sec. 2.1 − 2 0.1 (ii) If ℎ = 0.01, the average rate of change is f (2.01) − f (2) 135.3584 − 136 = = −64.16 ft/sec. 2.01 − 2 0.01 (iii) If ℎ = 0.001, the average rate of change is f (2.001) − f (2) 135.935984 − 136 = = −64.016 ft/sec. 2.001 − 2 0.001 (c) The values in part (b) appear to be getting closer to −64 so we estimate that the instantaneous velocity at time t = 2 is about −64 ft/sec. (d) We see that f (2) = 136 ft and we estimated in part (c) that f ¨ (2) is approximately −64 ft/sec. 28. f (x) = x2

9 8 7

✲

g(x) = x2 + 3

6 5 4 3 2 1 x

−4

−3

−2

−1 −1

Figure 2.7 (a) The tangent line to the graph of f (x) = x2 at x = 0 coincides with the x-axis and therefore is horizontal (slope = 0). The tangent line to the graph of g(x) = x2 + 3 at x = 0 is the dashed line indicated in the fgure and it also has a slope equal to zero. Therefore both tangent lines at x = 0 are parallel. We see in Figure 2.7 that the tangent lines at x = 1 appear parallel, and the tangent lines at x = 2 appear parallel. The slopes of the tangent lines at any value x = a will be equal. (b) Adding a constant shifts the graph vertically, but does not change the slope of the curve.

2.1 SOLUTIONS

141

29. (a) The velocity of the car represented by f (t) at time t is equal to the slope of f (t) at that point. Therefore, since the slope of f (t) starts large and positive and then gets less steep as t increases, this car starts out fast and then slows down. Likewise, since the slope of g(t) starts small and positive and then gets steeper as t increases, this car starts out slowly and then speeds up. (b) The average velocity of the car represented by f (t) for a ≤ t ≤ b is the slope of the line joining the points on the graph of f (t) corresponding to t = a and t = b. Therefore, we are looking for an interval a ≤ t ≤ b on which the average slopes of g(t) and f (t) are the same. The interval from t = 0 to t = 1 is one such interval. Other answers are possible. (c) Item (i) is true. The cars are traveling at the same instantaneous velocity when the slopes of both curves are the same. This appears to occur near the end of the frst half minute, or between t = 0 and t = 1∕2. This also means that (iv) is not true. Item (ii) is not true. For times between t = 1∕2 and t = 1, the slopes of f (t) appear to be less than the slopes of g(t), so the car represented by g(t) is going faster. Item (iii) is not true. At t = 1 minute, the cars are at the same position, but the slopes are di˙erent, so the velocities are di˙erent. 30. The coordinates of A are (4, 25). See Figure 2.8. The coordinates of B and C are obtained using the slope of the tangent line. Since f ¨ (4) = 1.5, the slope is 1.5 From A to B, Δx = 0.2, so Δy = 1.5(0.2) = 0.3. Thus, at C we have y = 25 + 0.3 = 25.3. The coordinates of B are (4.2, 25.3). From A to C, Δx = −0.1, so Δy = 1.5(−0.1) = −0.15. Thus, at C we have y = 25 − 0.15 = 24.85. The coordinates of C are (3.9, 24.85). Tangent line B 1.5(0.2) = 0.3 0.2

A = (4, 25)

0.15 C 0.1

Figure 2.8 31. (a) Since the point A = (7, 3) is on the graph of f , we have f (7) = 3. (b) The slope of the tangent line touching the curve at x = 7 is given by Slope =

Rise 3.8 − 3 0.8 = = = 4. Run 7.2 − 7 0.2

Thus, f ¨ (7) = 4. 32. We have Δy ≈ f ¨ (100)(Δx). Taking Δx = 1 we have Change in y = f (101) − f (100) ≈ (0.4)(1) = 0.4. 33. We have Δy ≈ f ¨ (12)(Δx). Taking Δx = 0.2 we have Change in y = f (12.2) − f (12) ≈ (30)(0.2) = 6. 34. We have Δy ≈ g ¨ (250)(Δx). Taking Δx = 1.5 we have Change in y = g(251.5) − g(250) ≈ (−0.5)(1.5) = −0.75.

142

Chapter Two /SOLUTIONS

35. We have p¨ (400)(Δx).

Δy Taking Δx = −2 we have

Change in y = p(398) − p(400)

(2)(−2) = −4.

36. The answers to parts (a)–(d) are shown in Figure 2.9. Slope= f ¨ (3)

❄

✻ ✻ ✛ ❄

✻

f (4) − f (2)

f (x) (2) Slope = f (5)−f 5−2

f (4)

❄ 1

Figure 2.9 37. (a) Since f is increasing, f (4) > f (3). (b) From Figure 2.10, it appears that f (2) − f (1) > f (3) − f (2). f (2) − f (1) (c) The quantity represents the slope of the secant line connecting the points on the graph at x = 1 and 2−1 f (3) − f (1) . x = 2. This is greater than the slope of the secant line connecting the points at x = 1 and x = 3 which is 3−1 (d) The function is steeper at x = 1 than at x = 4 so f ¨ (1) > f ¨ (4). f (x)

✻ ❄f (3) − f (2) ✻f (2) − f (1) ✻✻ ❄ slope =

slope =

f (3) − f (1) 3−1

f (2) − f (1) 2−1

x 1

Figure 2.10 38. How are the slopes of secant lines and tangent lines related to the values of the derivative and the average rate of change of a function? 39. Using a di˙erence quotient with ℎ = 0.001, say, we fnd 1.001 ln(1.001) − 1 ln(1) = 1.0005 1.001 − 1 2.001 ln(2.001) − 2 ln(2) = 1.6934 f¨ 2.001 − 2 The fact that f ¨ is larger at x = 2 than at x = 1 suggests that f is concave up between x = 1 and x = 2. f¨

2.1 SOLUTIONS

143

40. We want to approximate P ¨ (0) and P ¨ (1). Since for small ℎ P ¨ (0)

P (ℎ) − P (0) ℎ

and

P ¨ (1)

P (1 + ℎ) − P (1) , ℎ

if we take ℎ = 0.01, we get 1.434(1.0043)0.01 − 1.434 = 0.00615 billion∕year 0.01 = 6.15 million people∕year in 2019, 1.434(1.0043)1.01 − 1.434(1.0043)1 = 0.00618 billion∕year 0.01 = 6.18 million people∕year in 2020 P¨

P¨

41. The quantity f (0) represents the population on October 17, 2006, so f (0) = 300 million. The quantity f ¨ (0) represents the rate of change of the population (in millions per year). Since 1 person 1∕106 million people = = 2.867 million people∕year, 11 seconds 11∕(60 ⋅ 60 ⋅ 24 ⋅ 365) years so we have f ¨ (0) = 2.867. 42. With the defnitions given, we have: (a) Since P (t + 1) is the total number of cases up to and including day t + 1 and P (t) is the total number of cases up to and including day t, we see that P (t + 1) − P (t) is the number of new cases on day t + 1, so P (t + 1) − P (t) = N(t + 1). (b) Comparing P (t + 1) − P (t) 1

and

P (t + Δt) − P (t) ΔP = , Δt Δt

we see that Δt = 1 day. (c) The derivative P ¨ is the limit of the di˙erence quotient (P (t + Δt) − P (t))∕Δt as Δt gets very small. Since 1 day is the smallest time interval in the data, we take Δt = 1: P (t + 1) − P (t) 1

P ¨ (t).

But we also know P (t + 1) − P (t) = N(t + 1), so P ¨ (t

P (t + 1) − P (t) = N(t + 1). 1

43. (a) f ¨ (t) is negative or zero, because the average number of hours worked a week has been decreasing or constant over time. g ¨ (t) is positive, because hourly wage has been increasing. ℎ¨ (t) is positive, because average weekly earnings has been increasing. (b) We use a di˙erence quotient to the right for our estimates. (i) 36.0 − 37.0 = −0.2 hours/year f ¨ (1970) 1975 − 1970 34.3 − 34.3 f ¨ (1995) = 0 hours/year. 2000 − 1995 In 1970, the average number of hours worked by a production worker in a week was decreasing at the rate of 0.2 hours per year. In 1995, the number of hours was not changing. (ii) 4.73 − 3.40 g ¨ (1970) = $0.27 per year 1975 − 1970 14.00 − 11.64 g ¨ (1995) = $0.47 per year. 2000 − 1995 In 1970, the hourly wage was increasing at a rate of $0.27 per year. In 1995, the hourly wage was increasing at a rate of $0.47 per year.

144

Chapter Two /SOLUTIONS (iii) 170.28 − 125.80 = $8.90 per year 1975 − 1970 480.41 − 399.53 = $16.18 per year. ℎ¨ (1995) ≈ 2000 − 1995 In 1970, average weekly earnings were increasing at a rate of $8.90 a year. In 1995, weekly earnings were increasing at a rate of $16.18 a year. ℎ¨ (1970) ≈

Solutions for Section 2.2 1. Estimating the slope of the lines in Figure 2.11, we fnd that f ¨ (−2) ≈ 1.0, f ¨ (−1) ≈ 0.3, f ¨ (0) ≈ −0.5, and f ¨ (2) ≈ −1. 4

x −4

−2

Figure 2.11 2. We visualize tangent lines on the graph at the given points, and estimate the slopes of the tangent lines. We fnd that f ¨ (1) ≈ −2, f ¨ (2) ≈ −1, f ¨ (3) ≈ 0, f ¨ (4) ≈ 1, and f ¨ (2) ≈ 2. 3. The graph is that of the line y = −2x + 2. The slope, and hence the derivative, is −2. See Figure 2.12. 4

x −4

−4

Figure 2.12 4. See Figure 2.13. 4

x −4

−4

Figure 2.13

2.2 SOLUTIONS

145

5. See Figure 2.14. 4

−4

Figure 2.14

6. The slope of this curve is approximately −1 at x = −4 and at x = 4, approximately 0 at x = −2.5 and x = 1.5, and approximately 1 at x = 0. See Figure 2.15. 4

x −4

−4

Figure 2.15

7. See Figure 2.16. 4

−4

Figure 2.16

8. See Figure 2.17.

146

Chapter Two /SOLUTIONS 4

−4

Figure 2.17 9. (a) x3

(b) x4

(d) x3

10. For x = 0, 5, 10, and 15, we use the interval to the right to estimate the derivative. For x = 20, we use the interval to the left. For x = 0, we have f (5) − f (0) 70 − 100 −30 = = = −6. f¨ 5−0 5−0 5 Similarly, we fnd the other estimates in Table 2.1. Table 2.1 x

f ¨ (x)

−6

−3

−1.8

−1.2

11. The value of R¨ (0) is the derivative of the function R(x) = 100(1.1)x at x = 0. This is the same as the rate of change of R(x) at x = 0. We estimate this by computing the average rate of change over intervals near x = 0. If we use the intervals −0.001 ≤ x ≤ 0 and 0 ≤ x ≤ 0.001, we see that: 0 1 100(1.1)0 − 100(1.1)−0.001 Average rate of change 100 − 99.990469 = = = 9.531, 0 − (−0.001) 0.001 on − 0.001 ≤ x ≤ 0 0 1 100(1.1)0.001 − 100(1.1)0 Average rate of change 100.009531 − 100 = = = 9.531. 0.001 − 0 0.001 on 0 ≤ x ≤ 0.001 It appears that the rate of change of R(x) at x = 0 is 9.531, so we estimate R¨ (0) = 9.531. 12. See Figure 2.18.

f ¨ (x) x

Figure 2.18 13. This is a line with slope 1, so the derivative is the constant function f ¨ (x) = 1. The graph is the horizontal line y = 1. See Figure 2.19.

f ¨ (x)

x −3

Figure 2.19

2.2 SOLUTIONS

147

14. See Figure 2.20. f ¨ (x)

Figure 2.20

15. See Figure 2.21.

1 −1

x 2

f ¨ (x)

Figure 2.21

16. See Figure 2.22. x

f ¨ (x)

Figure 2.22

17. See Figure 2.23.

f ¨ (x)

Figure 2.23

18. The function is decreasing for x < −2 and x > 2, and increasing for −2 < x < 2. The matching derivative must be negative (below the x-axis) for x < −2 and x > 2, positive (above the x-axis) for −2 < x < 2, and zero (on the x-axis) for x = −2 and x = 2. The matching derivative is in graph VIII.

148

Chapter Two /SOLUTIONS

19. The function is a line with negative slope, so f ¨ (x) is a negative constant, and the graph of f ¨ (x) is a horizontal line below the x-axis. The matching derivative is in graph IV. 20. The function is increasing for x < 2 and decreasing for x > 2. The corresponding derivative is positive (above the x-axis) for x < 2, negative (below the x-axis) for x > 2, and zero at x = 2. The matching derivative is in graph II. 21. The function is increasing for x < −2 and decreasing for x > −2. The corresponding derivative is positive (above the x-axis) for x < −2, negative (below the x-axis) for x > −2, and zero at x = −2. The derivatives in graphs VI and VII both satisfy these requirements. To decide which is correct, consider what happens as x gets large. The graph of f (x) approaches an asymptote, gets more and more horizontal, and the slope gets closer and closer to zero. The derivative in graph VI meets this requirement and is the correct answer. 22. Since f ¨ (x) > 0 for all x, we know that f is increasing for all x, which rules out choices (II) and (IV). Since f ¨ (x) reaches its highest value when x = 0, we know that f is the steepest when x = 0. Thus, the answer is (III). 23. Since f ¨ (x) is a constant function, the slope of f is constant; that is, f is linear. Therefore, our answer is (V). 24. Since f ¨ (x) > 0 for all x, we know that f is increasing for all x, which rules out choices (II) and (IV). Since f ¨ (x) achieves its smallest value when x = 0, we know that f is the least steep at x = 0. Therefore, our answer is (I). 25. Since f ¨ (x) < 0 when x < 0 and f ¨ (x) > 0 when x > 0, we know that f is decreasing when x < 0 and increasing when x > 0. Thus, the choices are either (II) or (IV). Since the magnitude of f ¨ (x) increases as x moves away from 0, the graph of f becomes steeper as x moves away from 0. Thus, the answer is (II). 26. The graph is increasing for 0 < t < 15 and is decreasing for 15 < t < 30. One possible graph is shown in Figure 2.24. The units on the horizontal axis are years and the units on the vertical axis are people. people

f (t)

years 15

Figure 2.24 The derivative is positive for 0 < t < 15 and negative for 15 < t < 30. Two possible graphs are shown in Figure 2.25. The units on the horizontal axes are years and the units on the vertical axes are people per year. people/year

people/year f ¨ (t)

f ¨ (t) years

years

Figure 2.25

27. The value of g(x) is increasing at a decreasing rate for 2.7 < x < 4.2 and increasing at an increasing rate for x > 4.2. Δy 7.4 − 6.0 = = 2.8 Δx 5.2 − 4.7 Δy 9.0 − 7.4 = = 3.2 Δx 5.7 − 5.2 Thus g ¨ (x) should be close to 3 near x = 5.2.

between x = 4.7 and x = 5.2 between x = 5.2 and x = 5.7

2.2 SOLUTIONS

149

28. Notice that the x-values in the table are equally spaced 3 units apart, so we expect f ¨ (x) to be greatest on the interval where f (x) has the largest increase. This occurs between x = −9 and x = −6, where f (x) increases by 0.07. So the greatest value of f ¨ (x) likely occurs between x = −9 and x = −6. To fnd the least value of f ¨ (x), we look for the interval where f (x) has the largest decrease. This occurs between x = 6 and x = 9, where f (x) decreases by 0.06. So the least value of f ¨ (x) on the interval −12 ≤ x ≤ 9 likely occurs between x = 6 and x = 9, possibly at x = 9. 29. (a) The values in the table are decreasing, so f ¨ (t) appears to be negative. This means that Arctic sea ice is shrinking. (b) Since t measures years since 2000, we see that t = 17 corresponds to 2017. We see in the table that f (17) = 4.82 million km2 . We estimate f ¨ (17) using the interval to the right, between t = 17 (2017) and t = 18 (2018): f (18) − f (17) 4.79 − 4.82 = = −0.03 mn km2 /year. 18 − 17 1

f¨

In 2017, Arctic sea ice covered 4.82 million square kilometers and was shrinking at a rate of 0.03 million (30,000) square kilometers per year. (c) Since t measures years since 2000, we see that t = 20 corresponds to 2020. We see in the table that f (20) = 3.92 million km2 . Since we do not have data for 2021, we estimate f ¨ (20) using the interval to the left, between t = 19 (2019) and t = 20 (2020): f (20) − f (19) 3.92 − 4.36 f¨ = = −0.44 mn km2 /year. 20 − 19 1 In 2020, Arctic sea ice covered 3.92 million square kilometers and was shrinking at a rate of 0.44 million (440,000) square kilometers per year. 30. (a) The values in the table are increasing, so f ¨ (t) appears to be positive. This means that Amazon was increasing the size of its workforce throughout this four year period. (b) Since t measures years since 2015, we see that t = 3 corresponds to 2018. We see in the table that f (3) = 647.5 thousand employees. We estimate f ¨ (3) using the interval to the right, between t = 3 (2018) and t = 4 (2019): f¨

f (4) − f (3) 798 − 647.5 = = 150.5 thousand employees/year. 4−3 1

In 2018, Amazon employed 647.5 thousand employees and was hiring new employees at a rate of 150.5 thousand a year. (c) The year with the greatest increase of employees was 2020, when t = 5, so we guess that f ¨ (5) is largest. We estimate f ¨ (5) using the interval to the left, between t = 4 (2019) and t = 5 (2020): f¨

f (5) − f (4) 1298 − 798 = = 500 thousand employees/year. 5−4 1

This signifcant increase in the number of employees was in response to the Covid-19 pandemic when many more people started to shop online. 31. (a) The values in the table are increasing, from January to May, so f ¨ (t) appears to be positive then. From May to September, the values are decreasing, so f ¨ (t) appears to be negative then. This means that the unemployment rate was going up until around May, 2020 and then started going back down. (In fact, unemployment peaked in April, 2020.) (b) Since t measures months since the beginning of 2020, we see that t = 3 corresponds to March, 2020. We see in the table that f (3) = 4.4%. We estimate f ¨ (3) using the interval to the right, between t = 3 (March) and t = 5 (May): f¨

f (5) − f (3) 13.3 − 4.4 = = 4.45 %/month. 5−3 2

In March, 2020, the US unemployment rate was 4.4% and was increasing at a rate of 4.45 percent per year. (c) Since t measures months since the beginning of 2020, we see that t = 9 corresponds to September, 2020. We see in the table that f (9) = 7.8%. Since we do not have data for November, 2020, we estimate f ¨ (9) using the interval to the left, between t = 7 (July) and t = 9 (September): f¨

f (9) − f (7) 7.8 − 10.2 = = −1.2 %/month. 9−7 2

In September, 2020, the US unemployment rate was 7.8% and was being reduced at a rate of 1.2 percent per month.

150

Chapter Two /SOLUTIONS

32. Since f ¨ (x) > 0 for x < −1, f (x) is increasing on this interval. Since f ¨ (x) < 0 for x > −1, f (x) is decreasing on this interval. Since f ¨ (x) = 0 at x = −1, the tangent to f (x) is horizontal at x = −1. One possible shape for y = f (x) is shown in Figure 2.26.

x −1

Figure 2.26 33. Since f ¨ (x) > 0 for 1 < x < 3, we see that f (x) is increasing on this interval. Since f ¨ (x) < 0 for x < 1 and for x > 3, we see that f (x) is decreasing on these intervals. Since f ¨ (x) = 0 for x = 1 and x = 3, the tangent to f (x) will be horizontal at these x’s. One of many possible shapes of y = f (x) is shown in Figure 2.27. y

x 1

Figure 2.27 34. (a) Graph II (b) Graph I (c) Graph III f (1.1) − f (1) ln(1.1) − ln(1) 35. (a) f ¨ = 0.1 0.1 f (2.1) − f (2) ln(2.1) − ln(2) f¨ = 0.1 0.1 f (3.1) − f (3) ln(3.1) − ln(3) f¨ = 0.1 0.1 f (4.1) − f (4) ln(4.1) − ln(4) ¨ f = 0.1 0.1 f (5.1) − f (5) ln(5.1) − ln(5) ¨ f = 0.1 0.1 (b) It looks like the derivative of ln(x) is 1∕x.

.95. .49. .33. .25. .20.

36. Table 2.2 x

f (x)

1.998

2.6587

2.998

8.9820

3.998

21.3013

1.999

2.6627

2.999

8.9910

3.999

21.3173

2.000

2.6667

3.000

9.0000

4.000

21.3333

2.001

2.6707

3.001

9.0090

4.001

21.3493

2.002

2.6747

3.002

9.0180

4.002

21.3653

2.3 SOLUTIONS

151

Near 2, the values of f (x) seem to be increasing by 0.004 for each increase of 0.001 in x, so the derivative appears 0.004 to be 0.001 = 4. Near 3, the values of f (x) are increasing by 0.009 for each step of 0.001, so the derivative appears to be 9. Near 4, f (x) increases by 0.016 for each step of 0.001, so the derivative appears to be 16. The pattern seems to be, then, that at a point x, the derivative of f (x) = 31 x3 is f ¨ (x) = x2 . 37. (a) The only graph in which the slope is 1 for all x is Graph (III). (b) The only graph in which the slope is positive for all x is Graph (III). (c) Graphs where the slope is 1 at x = 2 are Graphs (III) and (IV). (d) Graphs where the slope is 2 at x = 1 are Graphs (II) and (IV). 38. We use the fact that f ¨ positive means the slope is positive so f is increasing while f ¨ negative means the slope is negative so f is decreasing. (a) The description tells us that f ¨ is negative to the left of x = 2 and positive to the right of x = 2. The appropriate graph of f (x) is one that is decreasing to the left of 2 and increasing to the right of 2, so the best match is graph J. The appropriate graph of f ¨ (x) is one that is negative (below the x-axis) to the left of 2 and positive (above the x-axis) to the right of 2, so the best match is graph S. (b) The description tells us that f ¨ is positive to the left of x = 2 and negative to the right of x = 2. The appropriate graph of f (x) is one that is increasing to the left of 2 and decreasing to the right of 2, so the best match is graph I. The appropriate graph of f ¨ (x) is one that is positive (above the x-axis) to the left of 2 and negative (below the x-axis) to the right of 2, so the best match is graph U. (c) The description tells us that f ¨ is positive to the left of x = 4 and negative to the right of x = 4. The appropriate graph of f (x) is one that is increasing to the left of 4 and decreasing to the right of 4, so the best match is graph H. The appropriate graph of f ¨ (x) is one that is positive (above the x-axis) to the left of 4 and negative (below the x-axis) to the right of 4, so the best match is graph T. (d) The description tells us that f ¨ is negative to the left of x = 4 and positive to the right of x = 4. The appropriate graph of f (x) is one that is decreasing to the left of 4 and increasing to the right of 4, so the best match is graph K. The appropriate graph of f ¨ (x) is one that is negative (below the x-axis) to the left of 4 and positive (above the x-axis) to the right of 4, so the best match is graph R. 39. (a) t = 3 (b) t = 9 (c) t = 14 (d)

V ¨ (t)

18 t

−2

Solutions for Section 2.3 1. In Leibniz notation the derivative is dD∕dt and the units are feet per minute. This is because D is a function of t. 2. In Leibniz notation the derivative is dC∕dW and the units are dollars per pound. This is because C is a function of W . 3. In Leibniz notation the derivative is dN∕dD and the units are gallons per mile. This is because N is a function of D. 4. In Leibniz notation the derivative is dP ∕dH and the units are dollars per hour. This is because P is a function of H. 5. Since Q = g(t) gives the amount of water in the reservoir at time t, this means g ¨ (t) gives the rate at which the amount changes. Thus, at time t = 15, the rate is given by g ¨ (15) = 13, which is expression (iv). 6. (a) The statement f (5) = 18 means that when 5 milliliters of catalyst are present, the reaction will take 18 minutes. Thus, the units for 5 are ml while the units for 18 are minutes. (b) As in part (a), 5 is measured in ml. Since f ¨ tells how fast T changes per unit a, we have f ¨ measured in minutes/ml. If the amount of catalyst increases by 1 ml (from 5 to 6 ml), the reaction time decreases by about 3 minutes. 7. (a) If the price is $150, then 2000 items will be sold. (b) If the price goes up from $150 by $1 per item, about 25 fewer items will be sold. Equivalently, if the price is decreased from $150 by $1 per item, about 25 more items will be sold.

152

Chapter Two /SOLUTIONS

8. (a) The 12 represents the weight of the chemical; therefore, its units are pounds. The 5 represents the cost of the chemical; therefore, its units are dollars. The statement f (12) = 5 means that when the weight of the chemical is 12 pounds, the cost is 5 dollars. (b) We expect the derivative to be positive since we expect the cost of the chemical to increase when the weight bought increases. (c) Again, 12 is the weight of the chemical in pounds. The units of the 0.4 are dollars/pound since it is the rate of change of the cost as a function of the weight of the chemical bought. The statement f ¨ (12) = 0.4 means that the cost is increasing at a rate of 0.4 dollars per pound when the weight is 12 pounds, or that an additional pound will cost about an extra 40 cents. 9. (a) On April 30, 2020 there were 45,798 total Covid-19 cases in California and the number was growing at a rate of about 1538 cases per day. (b) Taking the cases on April 30 (t = 30) and adding the approximate number of cases per day for 5 days we have P (35)

45,798 + 5(1538) = 53,488 cases.

(c) The estimate is lower than the actual number of cases by 394 cases. This means that the increase in cases per day on April 30 must have been lower than during the following 5 days since using it to project into the future gave us an underestimate. 10. We use the interval from t = 6 to t = 8 to estimate f ¨ (6), f (8) − f (6) 500 − 300 = = 100. 8−6 2 100 gigawatts per year. In 2016, the world solar energy output was increasing at a rate of about 100 f¨

Thus, we have f ¨ (6) gigawatts per year.

11. (a) The output units of P (t) are cases and the input units are days. Therefore, the units of P ¨ (t) are cases per day. (b) Since P ¨ (t) is the instantaneous rate of change of total cases with respect to time, in days, it tells us how fast the total number of cases is changing at time t, in cases per day. (c) We know that N(t) is the number of new cases on day t and N(t + 1) is the number of new cases on day t + 1. We can interpret N(t) as the change in total cases between day t − 1 and day t and N(t + 1) as the change in total cases between day t and day t + 1. They are both average rates of change of P (t) for a 1-day time interval, and therefore they are good approximations of the instantaneous rate of change P ¨ (t). 12. (a) Covid-19 cases per day. (b) Drawing the tangent line at t = 30, we estimate the slope with the points (30, 500) and (40, 1000) (see Figure 2.28), giving 1000 − 500 P¨ = 50 cases per day. 40 − 30 Similarly, we estimate the slope of the tangent at t = 50 using the points (50, 2000) and (58, 3000), giving 3000 − 2000 = 125 cases per day. 58 − 50 This means that the rate of change of P , or the additional (new) cases per day, has more than doubled in 20 days. P ¨ (50)

P (total cases) 5000 4000 3000 2000 1000 10

t (days)

Figure 2.28 13. (a) Note that f ¨ (10) and f ¨ (30) give us the slope of the curve with units of tons of wheat per kg of seed. This means that f ¨ (10) tells us by approximately how much the yield changes when the amount of seed used increases from 10 kg by 1 kg. When 30 kg of seed are used, f ¨ (30) tells us by approximately how much a 1-kg increase in the amount of seed used changes the yield. (b) We see that f ¨ (10) and f ¨ (30) are both positive, so planting an additional kg of seed will increase the yield at both values of s. We also see that f ¨ (30) > f ¨ (10) so one additional kilogram planted will have a higher impact on the crop yield if you have already planted 30 kg.

2.3 SOLUTIONS

153

14. (a) This means that investing the $1000 at 2% would yield $1221 after 10 years. (b) Writing g ¨ (r) as dB∕dr, we see that the units of dB∕dr are dollars per percent (interest). We can interpret dB as | the extra money earned if interest rate is increased by dr percent. Therefore g ¨ (2) = dB means that the dr r=2 balance, at 2% interest, would increase by about $122 if the interest rate were increased by 1%. In other words, g(3) g(2) + 122 = 1221 + 122 = 1343. 15. (a) The statement f (200) = 1300 means that it costs $1300 to produce 200 gallons of the chemical. (b) The statement f ¨ (200) = 6 means that when the number of gallons produced is 200, costs are increasing at a rate of $6 per gallon. In other words, it costs about $6 to produce the next (the 201st ) gallon of the chemical. 16. (a) The turbine generates the most power when the wind speed is 10 meters per second. Maximum power is approximately 54,000 watts. (b) Since p¨ (7) is positive, the function p is increasing for wind speeds near 7 meters per second: for wind speeds near 7 m∕sec, the turbine generates more power from faster winds. Since p¨ (15) is negative, the function p is decreasing for wind speeds near 15 meters per second: for wind speeds near 15 m∕sec, the turbine generates less power from faster winds. (c) The statement p(7) = 17, 500 means that the turbine generates 17,500 watts when the wind is blowing at a speed of 7 meters per second. The statement p¨ (7) = 8000 means that when the wind speed is around 7 meters per second, a 1 m∕sec increase in wind speed corresponds to about an 8000 watt increase in power. (d) The statement p(15) = 50, 000 means means that the turbine generates 50,000 watts when the wind is blowing at a speed of 15 meters per second. The statement p¨ (15) = −800 means that when the wind speed is around 15 meters per second, a 1 m∕sec increase in wind speed corresponds to about an 800 watt decrease in power. 17. (a) The units of compliance are units of volume per units of pressure, or liters per centimeter of water. (b) The increase in volume for a 5 cm reduction in pressure is largest between 10 and 15 cm. Thus, the compliance appears maximum between 10 and 15 cm of pressure reduction. The derivative is given by the slope, so 0.70 − 0.49 = 0.042 liters per centimeter. 15 − 10

Compliance

(c) When the lung is nearly full, it cannot expand much more to accommodate more air. 18. (a) The derivative f ¨ (r) is a function of the rainfall, r. Hence the units of 1500 are the units of rainfall, r, millimeters. (b) Since f ¨ (r) = dw∕dr, we have Units of f ¨ (r) =

Units of w mm (of leaf width) = . Units of r mm (of rain)

Δw

= (0.0218)(200) = 4.36 mm. The average leaf width in the rainier forest is about 4.4 mm greater. 19. Let p be the rating points earned by the CBS Evening News, let R be the revenue earned in millions of dollars, and let R = f (p). When p = 4.3, Rate of change of revenue

$5.5 million = 55 million dollars/point. 0.1 point

Thus f ¨ (4.3)

55.

20. We have W ¨ (t) = 0.6∕10 = 0.06 pounds per day. 21. (a) The yam is cooling o˙ so T is decreasing and f ¨ (t) is negative. (b) Since f (t) is measured in degrees Fahrenheit and t is measured in minutes, df ∕dt must be measured in units of F∕min. 22. The units of f ¨ (x) are feet/mile. The derivative, f ¨ (x), represents the rate of change of elevation with distance from the source, so if the river is fowing downhill everywhere, the elevation is always decreasing and f ¨ (x) is always negative. (In fact, there may be some stretches where the elevation is more or less constant, so f ¨ (x) = 0.)

154

Chapter Two /SOLUTIONS

23. (a) The units of lapse rate are the same as for the derivative dT ∕dz, namely (units of T )∕(units of z) = C∕km. (b) Since the lapse rate is 6.5, the derivative of T with respect to z is dT ∕dz = −6.5 C∕km. The air temperature drops about 6.5 for one more kilometer you go up. 24. (a) The derivative has units of people/second, so we fnd the rate of births, deaths, and migrations per second and combine them. 1 people per second 8 1 Death rate = people per second 12 1 Migration rate = people per second 33 Birth rate =

Thus

1 1 1 − + = 0.072 people/second. 8 12 33 In other words, the population is increasing at 0.072 people per second. (b) From the answer to part (a), we see that it took 1∕0.072 = 13.89 14 seconds to add one person. f ¨ (0) = Rate of change of population =

25. (a) The derivative f ¨ (t) appears to be positive for most of the period 2014-2018, because according to the table, gold production is increasing. (b) The derivative (or rate of change) appears to be greatest between 2016 and 2017. (c) We have 3300 − 3230 f ¨ (2018) = 70 metric tons/year. 2018 − 2017 In 2018, gold production was increasing at a rate of approximately 70 metric tons per year. (d) In 2018, gold production was 3300 metric tons and was increasing at a rate of 70 metric tons each year. Therefore, in 2019 (one year later), we have f (2019) 3300 + 70 = 3370 metric tons, and in 2023 (5 years later), we have f (2023)

3300 + 70(5) = 3650 metric tons.

We estimate that gold production in 2019 is 3370 metric tons and gold production in 2023 is 3650 metric tons. 26. (a) An additional dollar per year of government purchases increases national output for the year by about $0.60. The derivative is called a fscal policy multiplier because if government purchases increase by x dollars per year, then national output increases by about 0.60x dollars per year. (b) An additional tax dollar collected per year decreases national output for the year by about $0.26. The derivative is called a fscal policy multiplier because if government tax revenues increase by x dollars per year, then national output decreases by about 0.26x dollars per year. 27. (a) Positive, since weight increases as the child gets older. (b) f (8) = 45 tells us that when the child is 8 years old, the child weighs 45 pounds. (c) The units of f ¨ (a) are lbs/year. The value of f ¨ (a) gives approximate increase in weight for a 1 year increase in age. (d) f ¨ (8) = 4 tells us that an 8-year-old child weighs about 4 more pounds after the next year. (e) As a increases, f ¨ (a) will decrease since the rate of growth slows down as the child grows up. 28. Let f (t) be the age of onset of Alzheimer’s, in weeks, of a typical man who retires at age t years. The derivative f ¨ (t) has units weeks∕year. The study reports that each additional year of employment is associated with about a six week later age of onset, that is, f ¨ (t . 29. (a) Since f ¨ (c) is negative, the function T = f (c) is decreasing: on average, pelican eggshells are thinner if the PCB concentration, c, in the eggshell is higher. (b) The statement f (200) = 0.28 means that the thickness of pelican eggshells is 0.28 mm when the concentration of PCBs in the eggshell is 200 parts per million (ppm). The statement f ¨ (200) = −0.0005 means that when the PCB concentration is 200 ppm, a 1 ppm increase in the concentration typically corresponds to about a 0.0005 mm decrease in eggshell thickness. 30. Since N(t) is approximately the derivative of P (t), the slope of the graph of P (t) approximates N(t). (a) The slope of the graph of P (t) is greater between about t = 20 and t = 60 (for the last third of March and all of April, 2020) than at nearby points, so N(t) has a peak here. This was the frst wave.

2.3 SOLUTIONS

155

Similarly, P (t) climbs steeply between about t = 260 and t = 350 (from about the last third of November, 2020 until mid-February, 2021), so N(t) also has a peak there. This was the second wave. (b) The graph of P (t) is steeper around between t = 260 and t = 350, so the second wave peak was higher. These conclusions are refected in the graph of N(t) in Figure 2.29. Massachusetts, new cases N 6000 4000 2000 0 100

200

t (days)

300

Figure 2.29 N(t), meaning the derivative of P (t) is approximately N(t), and N(t) is positive or zero (new cases 31. (a) Since P ¨ (t are never negative), P (t) cannot decrease. In other words, P (t) is the total number of cases that have been confrmed, which can only increase as new cases are added. (b) If N(t) = 0 or small, the graph of P (t) is horizontal, or sloping very gently. There appear to be two such intervals, one from t = 0 to about t = 30, and one from about t = 100 to about t = 210. (c) Peaks in N(t) occur where P (t) is sloping upward. There are two peaks, the smaller one between about t = 50 and t = 80, the larger one between about t = 250 and t = 330. Figure 2.30 shows the actual data for N(t), daily new confrmed cases on day t in Connecticut. Connecticut, new cases N 3000 2000 1000 0 100

200

300

t (days)

Figure 2.30 32. The number of daily cases, N, is approximately equal to the derivative of the total cases, P , so N changes according to the slope of P . We can use this to identify the corresponding graph of daily Covid-19 cases for New York. From Figure 2.44 we see that there were two rapid growth periods for P around t = 25, and t = 275. These correspond to high values for the daily new cases around these times. Thus, the correct graph is (III) since no other graph has high values for N at these two times. 33. The number of daily cases, N, is approximately equal to the derivative of the total cases, P , so N changes according to the slope of P . We can use this to identify the corresponding graph of daily Covid-19 cases for Louisiana. From Figure 2.46 we see that there were three rapid growth periods for P around t = 25, t = 125 and t = 275. These correspond to high values for the daily new cases around these times. Thus, the correct graph is (II) since no other graph has high values for N at these three times. 34. Since we do not have information beyond t = 25, we will assume that the function will continue to change at the same rate. Therefore, f (26)

f (25) + f ¨ (25) = 3.6 + (−0.2) = 3.4.

Since 30 = 25 + 5, then f (30)

f (25) + f ¨ (25)(5) = 3.6 + (−0.2)(5) = 2.6.

156

Chapter Two /SOLUTIONS

35. Using Δx = 1, we can say that f (21) = f (20) + change in f (x) f (20) + f ¨ (20)Δx = 68 + (−3)(1) = 65. Similarly, using Δx = −1, f (19) = f (20) + change in f (x) f (20) + f ¨ (20)Δx = 68 + (−3)(−1) = 71. Using Δx = 5, we can write f (25) = f (20) + change in f (x) f (20) + (−3)(5) = 68 − 15 = 53.

36. We have Δx = 4.02 − 4 = 0.02. From the tangent line approximation we get: Δy

f ¨ (4)Δx = 7(0.02) = 0.14.

Thus, f (4.02) = f (4) + Δy

.14 = 5.14.

37. We have Δx = 3.92 − 4 = −0.08. From the tangent line approximation we get: Δy

f ¨ (4)Δx = 7(−0.08) = −0.56.

Thus, f (3.92) = f (4) + Δy

.56 = 4.44.

38. We have Δx = 1.95 − 2 = −0.05. From the tangent line approximation we get: Δy

f ¨ (2)Δx = −3(−0.05) = 0.15.

Thus, f (1.95) = f (2) + Δy

−4 + 0.15 = −3.85.

39. We have Δx = −2.99 − (−3) = 0.01. From the tangent line approximation we get: Δy

f ¨ (−3)Δx = 2(0.01) = 0.02.

Thus, f (−2.99) = f (−3) + Δy

−4 + 0.02 = −3.98.

40. We have Δx = 2.99 − 3 = −0.01. From the tangent line approximation we get: Δy

f ¨ (3)Δx = −2(−0.01) = 0.02.

Thus, f (2.99) = f (3) + Δy

−4 + 0.02 = −3.98.

41. (a) We have ΔR 3ΔS. (b) We have ΔS = 10.2 − 10 = 0.2. Therefore ΔR 3(0.2) = 0.6. (c) Since R = f (10) = 13 and the change in ΔR = 0.6, we have R = f (10.2)

13 + 0.6 = 13.6.

2.3 SOLUTIONS

157

(i) The units of R = f (50) are miles. It means that at 50 degrees Fahrenheit the Leaf has an average range of 72 miles. (ii) The units of f ¨ (50) are miles per degree Fahrenheit. It means that, when it is 50 F outside, an increase in the temperature of 1 F increases the range by 0.6 miles. (b) At 55 F we can expect a range of 72 + (0.6) ⋅ 5 = 75 miles.

42. (a)

43. (a) kilograms per week (b) At week 24 the fetus is growing at a rate of 0.096 kg/week, or the fetus will weigh about 0.096 kilograms more in one week. 44. (a) The tangent line to the weight graph is steeper at 36 weeks then at 20 weeks, so g ¨ (36) is greater than g ¨ (20). (b) The fetus increases its weight more rapidly at week 36 than at week 20. 45. Compare the secant line to the graph from week 0 to week 40 to the tangent lines at week 16 and week 36. (a) At week 16 the secant line is steeper than the tangent line. The instantaneous weight growth rate is less than the average. (b) At week 36 the tangent line is steeper than the secant line. The instantaneous weight growth rate is greater than the average. 46. (a) We use the interval from t = 20 to t = 24 to estimate g ¨ (20), g ¨ (20)

g(24) − g(20) 0.60 − 0.25 = = 0.0875 kg/week. 24 − 20 4

(b) We use the interval from t = 36 to t = 40 to estimate g ¨ (36), g(40) − g(36) 3.1 − 2.3 = = 0.20 kg/week. 40 − 36 4

g¨

3.1 − 0 = 0.078 kg/week. 40 − 0

47. (a) The statement f (2) = 27.2 tells us that beef production was 27.2 billion pounds in 2019. The statement f ¨ (2) = 0.3 means that in 2019, US beef production was increasing by about 0.3 billion pounds per year. (b) We assume that the growth rate remains constant until t = 6. Since production in 2019 is 27.2 billion pounds and is increasing at 0.3 billion pounds a year, we expect production in 2023 to be approximately f (6)

27.2 + 4 ⋅ (0.3) = 27.2 + 1.2 = 28.4 billion pounds.

48. (a) The statement f (20) = 0.36 means that 20 minutes after smoking a cigarette, there will be 0.36 mg of nicotine in the body. The statement f ¨ (20) = −0.002 means that 20 minutes after smoking a cigarette, about 0.002 mg of nicotine leaves the body in the next minute. The units are 20 minutes, 0.36 mg, and −0.002 mg/minute. (b) f (21)

f (20) + change in f in one minute = 0.36 + (−0.002) = 0.358

f (30)

f (20) + change in f in 10 minutes = 0.36 + (−0.002)(10) = 0.36 − 0.02 = 0.34

49. (a) Since t represents the number of days from now, we are told f (0) = 80 and f ¨ (0) = 0.50. (b) f (10)

value now + change in value in 10 days = 80 + 0.50(10) = 80 + 5 = 85.

In 10 days, we expect that the mutual fund will be worth about $85 a share.

158

Chapter Two /SOLUTIONS

50. (a) The slope of the tangent line at 2 kg can be approximated by the slope of the secant line passing through the points (2, 6) and (3, 2). So v(3) − v(2) 2−6 Slope of tangent line ≈ = = −4 (cm∕sec)∕kg. 3−2 1 (b) Since 50 grams = 0.050 kg, the contraction velocity changes by about −4(cm∕sec)∕kg ⋅ 0.050kg = −0.20 cm∕sec. The velocity is reduced by about 0.20 cm∕sec or 2.0 mm∕sec. (c) Since v(x) is the contraction velocity in cm/sec with a load of x kg, we have v¨ (2) = −4. 51. (a) The slope of the tangent line at 2 hours can be approximated by the slope of the secant line passing through the points (2, 4.2) and (2.5, 4.75). So Slope of tangent line ≈

g(2.5) − g(2) 4.75 − 4.2 = = 1.1 (liters∕minute)∕hour. 2.5 − 2 0.5

(b) The rate of change of the pumping rate is the slope of the tangent line. One minute = 1∕60 hour, so in one minute the Pumping rate increases by about 1.1

(liter∕minute) 1 ⋅ hour = 0.018 liter∕minute. hour 60

(c) Since g(t) is the pumping rate in liters/minute at time t hours, we have g ¨ (2) = 1.1. 52. The consumption rates (kg/week) are the rates at which the quantities are decreasing, that is, −1 times the derivatives of the storage functions. To compare rates at a given time, compare the steepness of the tangent lines to the graphs at that time. (a) At 3 weeks, the tangent line to the fat storage graph is steeper than the tangent line to the protein storage graph. During the third week, fat is consumed at a greater rate than protein. (b) At 7 weeks, the protein storage graph is steeper than the fat storage graph. During the seventh week, protein is consumed at a greater rate than fat. 53. Where the graph is linear, the derivative of the fat storage function is constant. The derivative gives the rate of fat consumption (kg/week). Thus, for the frst four weeks the body burns fat at a constant rate. 54. The fat consumption rate (kg/week) is the rate at which the quantity of fat is decreasing, that is, −1 times the derivative of the fat storage function. We estimate the derivatives at 3, 6, and 8 weeks by calculating the slope of the secant line. (a) The tangent line at 3 weeks can be approximated by the secant line containing (3, 6) and (4, 4). So Slope of the tangent line ≈

4 − 12 = −2.0 kg∕week. 4−0

The consumption rate is 2.0 kg∕week. (b) Two points on the secant line are (6, 1.2) and (7, 0.8). So Slope of the tangent line =

0.8 − 1.2 = −0.4 kg∕week. 7−6

The consumption rate is 0.4 kg∕week. (c) Two points on the secant line are (8, 0.5) and (7, 0.8). So Slope of the tangent line =

0.8 − 0.5 = −0.3 kg∕week. 7−8

The consumption rate is 0.3 kg∕week. 55. The body changes from burning more fat than protein to burning more protein. This is done by reducing the rate at which it burns fat and simultaneously increasing the rate at which it burns protein. The physiological reason is that the body has begun to run out of fat. 56. The graph of fat storage is linear for four weeks, then becomes concave up. Thus, the derivative of fat storage is constant for four weeks, then increases. This matches graph I. The graph of protein storage is concave up for three weeks, then becomes concave down. Thus, the derivative of protein storage is increasing for three weeks and then becomes decreasing. This matches graph II. 57. The way N changes corresponds to the way the slope of P changes. When N increases, the slope of P increases. When N decreases, the slope of P decreases. When N is constant, P is straight. In (I) the daily cases N quickly increase to a maximum and then decrease. Since N is the slope of the total cases curve, P , we expect the slopes of P to initially increase and then to decrease (sigmoid-shaped). Thus, the total cases curve for (I)

2.3 SOLUTIONS

159

should be one of South Korea or Israel. Note that N stalls at a fxed positive value after the peak. This will correspond in the graph of P to linear growth with positive slope, so (I) corresponds to South Korea. In (II), the daily cases curve is similar to (I), but in this case the daily cases do continue decreasing all the way down to zero. This corresponds to a total cases curve that is sigmoid-shaped but that does fatten out at the end (slope zero). Thus (II) corresponds to Israel. In (IV), the daily cases rise rapidly and then fall, which again corresponds to an initial sigmoid-shaped total cases curve. However, in (IV) the daily cases rise again, beyond the frst peak, which corresponds to the slope of the total cases curve P increasing again, with steeper slope than before. Thus, (IV) corresponds to Azerbaijan. The remaining choice is Poland, which must correspond to (III). Note that, in Poland, N rises rapidly and then fuctuates around a fxed positive value. This corresponds to the approximately straight segment with positive slope in the plot of P for Poland. 58. The derivative f ¨ (T ) gives us the rate of change of the range with respect to the temperature. It tells us by approximately how much the range changes with a 1 F temperature increase. It is also the slope of the curve. (a) Small increases in temperature increase the range the most when the slope of the graph is positive and steepest. This occurs at about T = 45 F. At this point f ¨ (T ) is at its maximum on the graph. Small increases in temperature decrease the range the most when the slope of the graph is negative and steepest. This occurs at about T = 95 F. At this point f ¨ (T ) is at its minimum on the graph. (b) Small increases in temperature change the range the least when the slope of the graph is zero. This occurs at about T = 65 F. At this point f ¨ (T ) = 0. 59. (a) If f ¨ (80) = 2, it means that if the budget for materials is $80,000, another thousand dollars spent on materials will bring in about $2,000 more in revenue. If f ¨ (80) = 0.5, another thousand dollars spent on materials will bring in about $500 more in revenue. (b) If f ¨ (80) = 2, then as we saw in part (a), spending slightly more than $80,000 will increase revenue by an amount greater than the additional expense, and thus more should be spent on materials. If f ¨ (80) = 0.5, then the increase in revenue is less than the additional expense, and it does not make sense to spend more on materials. 60. (a) If dG∕db = 1.3, then producing 1 more gallon of biofuel requires about 1.3 more gallons of gasoline. This is not sustainable since we are using more gallons than we are producing. (b) If dG∕db = 0.2, then producing 1 more gallon of biofuel requires about 0.2 more gallons of gasoline. This might be sustainable since we are able to use the gasoline to produce many more gallons of biofuel. 61. (a) The statement f (5) = 5500 tells us that there were 5,500 thousand square kilometers of rain forest in the Amazon in 2015. The derivative f ¨ (5) = −10.9 tells us that in 2015, after one more year the Amazon’s rain forests will have shrunk by about 10.9 thousand square kilometers. (b) We have f ¨ (5) −10.9 Relative rate of change = = = −0.00198. f (5) 5500 The Amazon’s rain forests are shrinking at a continuous rate of 0.198% per year. 62. Since t = 12 is the end of December 2019, we have f (12) = 2.498 billion users. We use a di˙erence quotient with t = 6 for the end of June 2019 and t = 12 for the end of December 2019 to estimate the derivative: f ¨ (12)

f (12) − f (6) 2.498 − 2.414 = = 0.014 billion users per month. 12 − 6 12 − 6

To estimate the relative rate of change, we use Relative rate of change =

f ¨ (12) 0.014 = = 0.006 per month. f (12) 2.498

The number of active Facebook users at the end of December 2019 was 2.498 billion. The number was increasing at 0.014 billion, (14 million) users per month, which represents a relative growth rate of 0.6% per month. 63. Estimating the relative rate of change using Δt = 0.01, we have 1 Δf f Δt

1 f (4.01) − f (4) 1 4.012 − 42 = 2 = 0.50. 0.01 f (4) 0.01 4

64. Estimating the relative rate of change using Δt = 0.01, we have 1 Δf f Δt

1 f (10.01) − f (10) 1 10.012 − 102 = 2 = 0.20. f (10) 0.01 0.01 10

160

Chapter Two /SOLUTIONS

65. Let P = f (t). In 2022 we have t = 2. The relative rate of change of f in 2022 is f ¨ (2)∕f (2). We estimate f ¨ (2) using a di˙erence quotient. (a) Estimating the relative rate of change using Δt = 1 at t = 2, we have f ¨ (2) f (2)

1 f (3) − f (2) = 0.010353 = 1.035% per year. f (2) 1

f ¨ (2) f (2)

1 f (2.1) − f (2) = 0.0103053 = 1.031% per year. f (2) 0.1

dP ∕dt P (b) With Δt = 0.1 and t = 2, we have dP ∕dt P

f ¨ (2) f (2)

1 f (2.01) − f (2) = 0.0103005 = 1.03% per year. f (2) 0.01

The relative rate of change is approximately 1.03% per year. This is as we would expect since the function P = 7.68e0.0103t has a continuous rate of change of 1.03% per year for all t. 66. (a) At 2 1/2 months, the baby weighs 5.67 kilograms. (b) At 2 1/2 months, the baby’s weight is increasing at a relative rate of 13% per month. 67. March 2020 corresponds to t = 2. Let B = f (t). The relative rate of change of f at t = 2 is f ¨ (2)∕f (2). We estimate f ¨ (2) using a di˙erence quotient. (a) Estimating the relative rate of change using Δt = 1 at t = 2, we have dB∕dt f ¨ (2) = B f (2)

1 f (3) − f (2) = −0.03246 = 3.246% reduction per month. f (2) 1

(b) With Δt = 0.1 and t = 2, we have dB∕dt f ¨ (2) = B f (2)

1 f (2.1) − f (2) = −0.03295 = 3.295% reduction per month. f (2) 0.1

1 f (2.01) − f (2) = −0.03299 = 3.299% reduction per month. f (2) 0.01

The oil production is falling approximately 3.3% per month. 68. Since O¨ (2016) = −1.4, we know the ODGI is decreasing at 1.4 units per year. Assuming that it continues to decrease at the same rate, to reduce the ODGI from 81.6 to 0 will take 81.6∕1.4 = 58.3 years. Thus, the ozone hole is predicted to recover in 2016 + 58.3 = 2074.3—in other words, by 2075.

Solutions for Section 2.4 1. (a) Since g(x) is decreasing at x = 0, the value of g ¨ (0) is negative. (b) Since g(x) is concave down at x = 0, the value of g ¨¨ (0) is negative. 2. At E both dy∕dx and d 2 y∕dx2 could be negative because y is decreasing and the graph is concave down there. At all the other points one or both of the derivatives could not be negative. 3. (a)

f (x)

(b)

f (x)

2.4 SOLUTIONS (d)

(c)

161

f (x)

f (x) x

4. f ¨ (x) > 0 f ¨¨ (x) > 0 5. f ¨ (x) = 0 f ¨¨ (x) = 0 6. f ¨ (x) < 0 f ¨¨ (x) = 0 7. f ¨ (x) < 0 f ¨¨ (x) > 0 8. f ¨ (x) > 0 f ¨¨ (x) < 0 9. f ¨ (x) < 0 f ¨¨ (x) < 0 10. The slope of the graph is larger at x = 0 than at x = 4, so f ¨ (0) > f ¨ (4). 11. The downward slope of the graph at x = 2 is steeper than at x = 6. Thus, f ¨ (2) is farther below zero than f ¨ (6), so f ¨ (2) < f ¨ (6). 12. The graph is concave down at x = 1, so f ¨¨ (1) is negative. On the other hand the graph is concave up at x = 3, so f ¨¨ (3) is positive. Thus, f ¨¨ (1) < f ¨¨ (3). 13. Recall that f (x) gives the y-coordinate at a point, f ¨ (x) gives the slope at a point, and the sign of f ¨¨ (x) depends on whether the graph of f (x) is concave up or concave down at that point. For a = −2, we see that the y-coordinate is positive, so f (−2) is positive. The function is increasing at x = −2 so f ¨ (−2) is positive. The graph is concave down at x = −2 so f ¨¨ (−2) is negative. The signs, in order, are: +, +, −. 14. Recall that f (x) gives the y-coordinate at a point, f ¨ (x) gives the slope at a point, and the sign of f ¨¨ (x) depends on whether the graph of f (x) is concave up or concave down at that point. For a = −1, we see that the y-coordinate is positive, so f (−1) is positive. The function is decreasing at x = −1 so f ¨ (−1) is negative. The graph is concave down at x = −1 so f ¨¨ (−1) is negative. The signs, in order, are: +, −, −. 15. Recall that f (x) gives the y-coordinate at a point, f ¨ (x) gives the slope at a point, and the sign of f ¨¨ (x) depends on whether the graph of f (x) is concave up or concave down at that point. For a = 1, we see that the y-coordinate is negative, so f (1) is negative. The function is decreasing at x = 1 so f ¨ (1) is negative. The graph is concave up at x = 1 so f ¨¨ (1) is positive. The signs, in order, are: −, −, +. 16. Recall that f (x) gives the y-coordinate at a point, f ¨ (x) gives the slope at a point, and the sign of f ¨¨ (x) depends on whether the graph of f (x) is concave up or concave down at that point. For a = 2, we see that the y-coordinate is negative, so f (2) is negative. The function is increasing at x = 2 so f ¨ (2) is positive. The graph is concave up at x = 2 so f ¨¨ (2) is positive. The signs, in order, are: −, +, +. 17. The derivative is positive on those intervals where the function is increasing and negative on those intervals where the function is decreasing. Therefore, the derivative is positive on the intervals 0 < t < 0.4 and 1.7 < t < 3.4, and negative on the intervals 0.4 < t < 1.7 and 3.4 < t < 4. The second derivative is positive on those intervals where the graph of the function is concave up and negative on those intervals where the graph of the function is concave down. Therefore, the second derivative is positive on the interval 1 < t < 2.6 and negative on the intervals 0 < t < 1 and 2.6 < t < 4. 18. The derivative is positive on those intervals where the function is increasing and negative on those intervals where the function is decreasing. Therefore, the derivative is positive on the interval −2.3 < t < −0.5 and negative on the interval −0.5 < t < 4. The second derivative is positive on those intervals where the graph of the function is concave up and negative on those intervals where the graph of the function is concave down. Therefore, the second derivative is positive on the interval 0.5 < t < 4 and negative on the interval −2.3 < t < 0.5.

162

Chapter Two /SOLUTIONS

19. (a) Negative. We are told the graph lies below the x-axis for 2 ≤ x ≤ 6, so f (3) < 0. (b) Positive. We are told the graph rises from left to right on 2 ≤ x ≤ 6. This means that f is increasing at x = 3, so f ¨ (3) > 0. (c) Negative. We are told that the graph is concave down on 2 ≤ x ≤ 6. This means the second derivative is negative on this interval, so f ¨¨ (3) < 0. (d) Positive. We are told the graph rises from left to right on 2 ≤ x ≤ 6. This means that f is increasing on this interval, so that f (5) > f (3), making f (5) − f (3) > 0. (e) Negative. We are told that the graph is concave down on 2 ≤ x ≤ 6. This means the second derivative is negative on this interval, so f ¨ (x) is decreasing here. Thus, f ¨ (5) < f ¨ (3), so f ¨ (5) − f ¨ (3) < 0. (f) Not enough information to decide. We are told that the graph is concave down on 2 ≤ x ≤ 6 and so f ¨¨ (x) is negative on this interval. However, we don’t know how the value of f ¨¨ (x) changes over the interval. The value f ¨¨ (5) could be larger, smaller or the same as f ¨¨ (3), and so f ¨¨ (5) − f ¨¨ (3) could be positive, negative or zero. 20. The derivative of w(t) appears to be negative since the function is decreasing over the interval given. The second derivative, however, appears to be positive since the function is concave up, i.e., it is decreasing at a decreasing rate. 21. The derivative, s¨ (t), appears to be positive since s(t) is increasing over the interval given. The second derivative also appears to be positive or zero since the function is concave up or possibly linear between t = 1 and t = 3, i.e., it is increasing at a non-decreasing rate. 22. The graph must be everywhere decreasing and concave up on some intervals and concave down on other intervals. One possibility is shown in Figure 2.31.

Figure 2.31 23. Since all advertising campaigns are assumed to produce an increase in sales, a graph of sales against time would be expected to have a positive slope. A positive second derivative means the rate at which sales are increasing is increasing. If a positive second derivative is observed during a new campaign, it is reasonable to conclude that this increase in the rate sales are increasing is caused by the new campaign–which is therefore judged a success. A negative second derivative means a decrease in the rate at which sales are increasing, and therefore suggests the new campaign is a failure. 24. (a) The function appears to be decreasing and concave down, and so we conjecture that f ¨ is negative and that f ¨¨ is negative. (b) We use di˙erence quotients to the right: 137−145 f¨ = −4 4−2 f¨

56−98 = −21. 10−8 ¨

25. (a) The derivative, f (t), appears to be positive during the period 2011–2017 since the number of cars is increasing. The second derivative, f ¨¨ (t), appears to be positive during the period 2011–2017 because the rate of change is increasing. For example, between 2011 and 2013, the rate of change is (255.9 − 253.1)∕2 = 1.4 million cars per year. Between 2013 and 2015 the rate of change is (263.6 − 255.9)∕2 = 3.85 million cars per year, and between 2015 and 2017 the rate of change is (272.5 − 263.6)∕2 = 4.45 million cars per year . (b) The number of cars is increasing between 2005 and 2007, but decreasing between 2007 and 2009. Thus, the derivative, f ¨ (t), appears to be positive between 2005 and 2007 and negative from 2007 to 2009. The second derivative, f ¨¨ (t), appears to be negative during the period 2005–2009 because the rate of change is decreasing. For example, between 2005 and 2007, the rate of change is (254.4 − 247.4)∕2 = 3.5 million cars per year, while between 2007 and 2009, the rate of change is (254.2 − 254.4)∕2 = −0.1 million cars per year. (c) To estimate f ¨ (2017) we consider the interval 2015–2017 f (2017) − f (2015) 272.5 − 263.6 = 4.45. 2017 − 2015 2 We estimate that f ¨ (2017) 4.45 million cars per year. The number of passenger cars in the US was increasing at a rate of about 4,450,000 cars per year in 2017. f ¨ (2017)

2.4 SOLUTIONS

163

26. This graph is increasing for all x, and is concave down to the left of 2 and concave up to the right of 2. One possible answer is shown in Figure 2.32:

x 2

Figure 2.32

27. Since f (2) = 5, the graph goes through the point (2, 5). Since f ¨ (2) = 1∕2, the slope of the curve is 1∕2 when it passes through this point. Since f ¨¨ (2) > 0, the graph is concave up at this point. One possible graph is shown in Figure 2.33. Many other answers are also possible.

x 2

Figure 2.33

28. The two points at which f ¨ = 0 are A and B. Since f ¨ is nonzero at C and D and f ¨¨ is nonzero at all four points, we get the completed Table 2.3: Table 2.3 Point

f¨

f ¨¨

−

29. (b). The positive frst derivative tells us that the temperature is increasing; the negative second derivative tells us that the rate of increase of the temperature is slowing. 30. (e). Since the smallest value of f ¨ (t) was 2◦ C∕hour, we know that f ¨ (t) was always positive. Thus, the temperature rose all day. 31. To the right of x = 5, the function starts by increasing, since f ¨ (5) = 2 > 0 (though f may subsequently decrease) and is concave down, so its graph looks like the graph shown in Figure 2.34. Also, the tangent line to the curve at x = 5 has slope 2 and lies above the curve for x > 5. If we follow the tangent line until x = 7, we reach a height of 24. Therefore, f (7) must be smaller than 24, meaning 22 is the only possible value for f (7) from among the choices given.

164

Chapter Two /SOLUTIONS y

T 24

f (x)

x 5

Figure 2.34 32. (a) The EPA will say that the rate of discharge is still rising. The industry will say that the rate of discharge is increasing less quickly, and may soon level o˙ or even start to fall. (b) The EPA will say that the rate at which pollutants are being discharged is leveling o˙, but not to zero—so pollutants will continue to be dumped in the lake. The industry will say that the rate of discharge has decreased signifcantly. 33. (a) Let N(t) be the number of people below the poverty line. See Figure 2.35. N(t)

Figure 2.35 (b) dN∕dt is positive, since people are still slipping below the poverty line. d 2 N∕dt2 is negative, since the rate at which people are slipping below the poverty line, dN∕dt, is decreasing. 34. (a) The author is talking about the number of students enrolled in law schools in the United States over time. We could let f (t) be the total number of students enrolled in US law schools in year t. (b) enrollment

time

(c) Since law school enrollment is decreasing, f ¨ (t) < 0. Since the slide accelerates, f (t) is concave down, so f ¨¨ (t) < 0. 35. (a)

utility

quantity

(b) As a function of quantity, utility is increasing but at a decreasing rate; the graph is increasing but concave down. So the derivative of utility is positive, but the second derivative of utility is negative.

2.4 SOLUTIONS

165

36. (a) dP ∕dt > 0 and d 2 P ∕dt2 > 0. (b) dP ∕dt < 0 and d 2 P ∕dt2 > 0 (but dP ∕dt is close to zero). (i) N ¨ (t) is positive because nitrous oxide concentration is increasing with time. (ii) N ¨¨ (t) is positive because nitrous oxide concentration is increasing faster with time. (b) (i) N(20) is larger than N(10) since there is more nitrous oxide in the air at t = 20 than at t = 10. (ii) N ¨ (20) is larger than N ¨ (10) since the nitrous oxide concentration is increasing faster with time.

37. (a)

38. (a) IV, (b) III, (c) II, (d) I, (e) IV, (f) II 39. (a) Given that Arctic Sea ice extent grows from October to February, G(t) must be increasing over 0 < t < 4, which means G¨ (t) > 0 over this same interval. (b) The sign of the second derivative indicates G is concave down for 0 < t < 4. Under this assumption, the Arctic Sea ice extent grows quickly at the beginning of the winter period and slower toward the end of the period, as warmer months approach. (c) The graph of G(t) must be increasing and concave down over the interval 0 < t < 4. One possible graph1 appears in Figure 2.36. extent (millions of km2 ) 15 10 5 0 1

t (months)

Figure 2.36: Arctic Sea ice extent from Oct. 1, 2018 to Feb. 1, 2019 40. (a) Since the sea level is rising, we know that a¨ (t) > 0 and m¨ (t) > 0. Since the rate is accelerating, we know that a¨¨ (t) > 0 and m¨¨ (t) > 0. (b) The rate of change of sea level for the mid-Atlantic states is between 2 and 4, we know 2 < a¨ (t) < 4. (Possibly also a¨ (t) = 2 or a¨ (t) = 4.) Similarly, 2 < m¨ (t) < 10. (Possibly also m¨ (t) = 2 or m¨ (t) = 10.) (c) (i) If a¨ (t) = 2, then sea level rise = 2 ⋅ 100 = 200 mm. If a¨ (t) = 4, then sea level rise = 4 ⋅ 100 = 400 mm. So sea level rise is between 200 mm and 400 mm. (ii) The shortest amount of time for the sea level in the Gulf of Mexico to rise 1 meter occurs when the rate is largest, 10 mm per year. Since 1 meter = 1000 mm, shortest time to rise 1 meter = 1000∕10 = 100 years. 41. (a) For each time interval we can calculate the average rate of change of the number of Facebook users per month over this interval. For example, from the quarter ending in December 2018 to the quarter ending in March 2019, a period of 3 months, Average rate ΔN 2375 − 2320 55 = = = = 18.33. Δt 3 3 of change Thus, between December 2018 and March 2019, there were approximately 18,330,000 more active users each month. Values of ΔN∕Δt are listed in Table 2.4. Table 2.4 Time interval Average rate of change, ΔN∕Δt (millions∕month)

Dec 2018–Mar 2019

Mar–June 2019

June–Sep 2019

18.33

13.0

11.67

(b) We assume the data lies on a smooth curve. Since the values of ΔN∕Δt are decreasing between December 2018 and September 2019, this suggests that dN∕dt also decreases, so d 2 N∕dt2 is negative for this period. 1 An actual graph of the Arctic Sea ice extent over this period can be obtained from the National Snow and Ice Center, Boulder, CO, nsidc.org/arcticseaicenews/fles/2018/03/Figure2.png, accessed September 30, 2019.

166

Chapter Two /SOLUTIONS

42. (a) For each time interval we can calculate the average rate of change of the number of yeast population per hour over this interval. For example, between 0 and 2 hours Average rate ΔP 29.0 − 9.6 = = = 9.7. Δt 2−0 of change Values of ΔP ∕Δt are listed in Table 2.5. Table 2.5 Time interval Average rate of change, ΔP ∕Δt

0–2

2–4

4–6

6–8

8–10

10–12

12–14

14–16

16–18

9.7

21.1

51.8

88.1

81.3

40.8

23.0

7.6

3.0

(b) We assume the data lies on a smooth curve. Since the values of ΔP ∕Δt are increasing for 0 < t < 8, this suggests that dP ∕dt also increases, so d 2 P ∕dt2 is positive for this period. Since the values of ΔP ∕Δt are decreasing for 8 < t < 18, this suggests that dP ∕dt also decreases, so d 2 P ∕dt2 is negative for this period. 43. (a) If the second derivative is zero, that is P ¨¨ (t , then the graph of P (t) is approximately linear. This occurs between about t = 180 and about t = 280. (b) The graph of P (t) is concave up at t = 110 and concave down at about t = 170. In between, there must be a point where the concavity changes; for example, at about t = 140. Similarly, graph of P (t) is concave up at t = 290 and concave down at t = 350. In between, there is a point where the concavity changes; for example, at about t = 320. To confrm, we can look at the graph of N(t), the daily new cases is South Carolina. The graph of N(t) in Figure 2.37 shows the intervals where P (t) is concave up—N(t) is increasing there—and where P (t) is concave down—N(t) is decreasing there. South Carolina, new cases N 5000 3000 1000 0 100

200

300

t (days)

Figure 2.37

Solutions for Section 2.5 1. The marginal cost is approximated by the di˙erence quotient MC

ΔC Δq

4830 − 4800 = 3. 1305 − 1295

The marginal cost is approximately $3 per item. 2. (a) Marginal cost is the derivative C ¨ (q), so its units are dollars/barrel. (b) It costs about $3 more to produce 101 barrels of olive oil than to produce 100 barrels.

2.5 SOLUTIONS

167

3. Drawing in the tangent line at the point (600, R(600)), we get Figure 2.38. $ R(q)

20,000

10,000

q 600

Figure 2.38 We see that each vertical increase of 2500 in the tangent line gives a corresponding horizontal increase of roughly 150. The marginal revenue at the production level of 600 units is Slope of tangent line

R¨ (600) =

to R(q) at q = 600

2500 = 16.67. 150

This tells us that after producing 600 units, the revenue for producing the 601st product will be roughly $16.67. 4. Marginal cost = C ¨ (q). Therefore, marginal cost at q is the slope of the graph of C(q) at q. We can see that the slope at q = 5 is greater than the slope at q = 30. Therefore, marginal cost is greater at q = 5. At q = 20, the slope is small, whereas at q = 40 the slope is larger. Therefore, marginal cost at q = 40 is greater than marginal cost at q = 20. 5. Drawing in the tangent line at the point (10000, C(10000)) we get Figure 2.39. $ C(q) 20,000

10,000

q 0

10,000

Figure 2.39 We see that each vertical increase of 2500 in the tangent line gives a corresponding horizontal increase of roughly 6000. Thus the marginal cost at the production level of 10,000 units is C ¨ (10,000) =

Slope of tangent line to C(q) at q = 10,000

2500 = 0.42. 6000

This tells us that after producing 10,000 units, it will cost roughly $0.42 to produce one more unit. 6. (a) For q = 500 Proft = (500) = R(500) − C(500) = 9400 − 7200 = 2200 dollars. (b) As production increases from q = 500 to q = 501, ΔR ≈ R¨ (500)Δq = 20 ⋅ 1 = 20 dollars, ΔC ≈ C ¨ (500)Δq = 15 ⋅ 1 = 15 dollars, Thus Change in proft = Δ = ΔR − ΔC = 20 − 15 = 5 dollars.

Chapter Two /SOLUTIONS

168

7. We have

C(2500) − C(2000) 3825 − 3640 = = $0.37/ton. 2500 − 2000 500 This means that recycling the 2001st ton of paper will cost around $0.37. The marginal cost is smallest at the point where the derivative of the function is smallest. Thus the marginal cost appears to be smallest on the interval 2500 ≤ q ≤ 3000. C ¨ (2000) ≈

8. (a) The fxed cost is $8500. (b) The marginal cost per item is $4.65. (c) The price per item is $5.15. (d) The company makes a proft when revenues exceed costs, so we fnd the values of q for which R > C. R>C 5.15q > 8500 + 4.65q 0.50q > 8500 q > 17000. The company makes a proft when it produces more than 17,000 units. (e) Each additional unit costs the company $4.65 and the company can sell it for $5.15, so the company makes 5.15 − 4.65 = $0.50 on each additional unit sold. 9. Since marginal revenue is smaller than marginal cost around q = 4500, as you produce more of the product your revenue increases slower than your costs, so proft goes down. It does not pay to increase production, and maximal proft occurs at a production level below 4500. 10. The slope of the revenue curve is greater than the slope of the cost curve at both q1 and q2 , so the marginal revenue is greater at both production levels. 11. Since Price = MR, these graphs show price graphed against quantity. We see that Graph (I) corresponds to (b), price frst decreasing with quantity, then increasing. Graph (II) corresponds to (a), constant price because the graph is a horizontal line. Graph (III) corresponds to (d), price increases with quantity. Graph (IV) corresponds to (c), price is decreasing. 12. (a) We can approximate C(16) by adding C ¨ (15) to C(15), since C ¨ (15) is an estimate of the cost of the 16th item. C(16) ≈ C(15) + C ¨ (15) = $2300 + $108 = $2408. (b) We approximate C(14) by subtracting C ¨ (15) from C(15), where C ¨ (15) is an approximation of the cost of producing the 15th item. C(14) ≈ C(15) − C ¨ (15) = $2300 − $108 = $2192. 13. We know MC ≈ C(1,001) − C(1,000). Therefore, C(1,001) ≈ C(1,000) + MC or C(1,001) ≈ 5000 + 25 = 5025 dollars. Since we do not know MC(999), we will assume that MC(999) = MC(1,000). Therefore: MC(999) = C(1,000) − C(999). Then: C(999) ≈ C(1,000) − MC(999) = 5,000 − 25 = 4,975 dollars. Alternatively, we can reason that MC(1,000) ≈ C(1,000) − C(999), so C(999) ≈ C(1,000) − MC(1,000) = 4,975 dollars. Now for C(1,000), we have C(1,100) ≈ C(1,000) + MC ⋅ 100. Since 1,100 − 1,000 = 100, C(1,100) ≈ 5,000 + 25 × 100 = 5,000 + 2,500 = 7,500 dollars.

2.5 SOLUTIONS

169

14. (a) The cost to produce 50 units is $4300 and the marginal cost to produce additional items is about $24 per unit. Producing two more units (from 50 to 52) increases cost by $48. We have C(52)

4300 + 24(2) = $4348.

(b) When q = 50, the marginal cost is $24 per item and the marginal revenue is $35 per item. The proft on the 51st item is approximately 35 − 24 = $11. (c) When q = 100, the marginal cost is $38 per item and the marginal revenue is $35 per item, so the company loses about $3 by producing the 101st item. Since the company will lose money, it should not produce the 101st item. 15. The company makes money when revenue exceeds cost and loses money when cost exceeds revenue. Proft is maximized when revenue exceeds cost and the vertical distance between the graphs is as large as possible. This occurs when R(q) > C(q) and R¨ (q) = C ¨ (q). It appears that maximum proft in this case is at about q = 190. (a) At q = 75, we see that the cost function is larger than the revenue function, so the company is losing money. The company needs to increase production in order to make a proft. (b) At q = 150, the revenue function is larger than the cost function so the company is making money. Since marginal revenue is greater than marginal cost here, the company should increase production. (c) At q = 225, the revenue function is larger than the cost function so the company is making money. Since marginal revenue is less than marginal cost here, the company should decrease production. (d) At q = 300, the cost function is larger than the revenue function so the company is losing money. The company needs to decrease production in order to make a proft. 16. At q = 50, the slope of the revenue is larger than the slope of the cost. Thus, at q = 50, marginal revenue is greater than marginal cost and the 50th bike should be added. At q = 90 the slope of the revenue is less than the slope of the cost. Thus, at q = 90 the marginal revenue is less than the marginal cost and the 90th bike should not be added. 17. (a) At q = 2.1 million, Proft = (2.1) = R(2.1) − C(2.1) = 6.9 − 5.1 = 1.8 million dollars. (b) If Δq = 0.04, R¨ (2.1)Δq = 0.7(0.04) = 0.028 million dollars = $28,000.

Change in revenue, ΔR

Thus, revenues increase by about $28,000. (c) If Δq = −0.05, R¨ (2.1)Δq = 0.7(−0.05) = −0.035 million dollars = −$35,000.

Change in revenue, ΔR

Thus, revenues decrease by about $35,000. (d) We fnd the change in cost by a similar calculation. For Δq = 0.04, Change in cost, ΔC

C ¨ (2.1)Δq = 0.6(0.04) = 0.024 million dollars = $24,000

Change in proft, Δ

$28,000 − $24,000 = $4000.

Thus, increasing production 0.04 million units increases profts by about $4000. For Δq = −0.05, Change in cost, ΔC

C ¨ (2.1)Δq = 0.6(−0.05) = −0.03 million dollars = −$30,000

Change in proft, Δ

−$35,000 − (−$30,000) = −$5000.

Thus, decreasing production 0.05 million units decreases profts by about $5000. 18. (a) At q = 2000, we have Proft = R(2000) − C(2000) = 7780 − 5930 = 1850 dollars. (b) If q increases from 2000 to 2001, ΔR

R¨ (2000) ⋅ Δq = 2.5 ⋅ 1 = 2.5 dollars,

ΔC

C ¨ (2000) ⋅ Δq = 2.1 ⋅ 1 = 2.1 dollars,

Thus, Change in proft = ΔR − ΔC .5 − 2.1 = 0.4 dollars. Since increasing production increases proft, the company should increase production. (c) By a calculation similar to that in part (b), as q increases from 2000 to 2001, Change in proft

.32 − 4.77 = −0.45 dollars.

Since increasing production reduces the proft, the company should decrease production.

170

Chapter Two /SOLUTIONS

19. For each q, we calculate the average rate of change of the cost and the revenue over the interval to the right. For example, for q = 0 C(1) − C(0) Average rate 10 − 9 C¨ = = = 1, 1−0 1−0 of change of C(q) while R(1) − R(0) Average rate 5−0 R¨ = = = 5. 1−0 1−0 of change of R(q) Estimates for C ¨ (q) and R¨ (q) are in Table 2.6. Table 2.6 q

C ¨ (q)

6 7

R¨ (q)

20. For each value of q, we calculate the average rate of change of the cost and the revenue over the interval to the right. For example, for q = 1, C(2) − C(1) Average rate 60 − 20 MC = = = 40, 2−1 2−1 of change of C(q) while R(2) − R(1) Average rate 220 − 100 MR = = = 120. 2−1 2−1 of change of R(q) Estimates for MC(q) and MR(q) are in Table 2.7. Table 2.7 q

MC(q)

100

120

MR(q)

120

110

Solutions for Chapter 2 Review 1. (a) Let s = f (t). (i) We wish to fnd the average velocity between t = 1 and t = 1.1. We have Average velocity =

f (1.1) − f (1) 3.63 − 3 = = 6.3 m/sec. 1.1 − 1 0.1

(ii) We have Average velocity =

f (1.01) − f (1) 3.0603 − 3 = = 6.03 m/sec. 1.01 − 1 0.01

(iii) We have f (1.001) − f (1) 3.006003 − 3 = = 6.003 m/sec. 1.001 − 1 0.001 (b) We see in part (a) that as we choose a smaller and smaller interval around t = 1 the average velocity appears to be getting closer and closer to 6, so we estimate the instantaneous velocity at t = 1 to be 6 m/sec. Average velocity =

SOLUTIONS to Review Problems For Chapter Two

171

2. Since f ¨ (x) = 0 where the graph is horizontal, f ¨ (x) = 0 at x = d. The derivative is positive at points b and c, but the graph is steeper at x = c. Thus f ¨ (x) = 0.5 at x = b and f ¨ (x) = 2 at x = c. Finally, the derivative is negative at points a and e but the graph is steeper at x = e. Thus, f ¨ (x) = −0.5 at x = a and f ¨ (x) = −2 at x = e. See Table 2.8. Thus, we have f ¨ (d) = 0, f ¨ (b) = 0.5, f ¨ (c) = 2, f ¨ (a) = −0.5, f ¨ (e) = −2. Table 2.8 x

f ¨ (x)

0.5

−0.5

−2

3. (a) From Figure 2.40 we can see that for x = 1 the value of the function is decreasing. Therefore, the derivative of f (x) at x = 1 is negative. 4 3 2

f (x) = 2 − x3

1 x −4 −3 −2 −1 −1

−2 −3

Figure 2.40 (b) f ¨ (1) is the derivative of the function at x = 1. This is the rate of change of f (x) = 2 − x3 at x = 1. We estimate this by computing the average rate of change of f (x) over intervals near x = 1. Using the intervals 0.999 ≤ x ≤ 1 and 1 ≤ x ≤ 1.001, we see that 0 1 [2 − 13 ] − [2 − 0.9993 ] Average rate of change 1 − 1.002997 = = = −2.997, 1 − 0.999 0.001 on 0.999 ≤ x ≤ 1 0 1 [2 − 1.0013 ] − [2 − 13 ] Average rate of change 0.996997 − 1 = = = −3.003. 1.001 − 1 0.001 on 1 ≤ x ≤ 1.001 It appears that the rate of change of f (x) at x = 1 is approximately −3, so we estimate f ¨ (1) = −3. 4. We estimate f ¨ (2) using the average rate of change formula on a small interval around 2. We use the interval x = 2 to x = 2.001. (Any small interval around 2 gives a reasonable answer.) We have f¨

f (2.001) − f (2) 32.001 − 32 9.00989 − 9 = = = 9.89. 2.001 − 2 2.001 − 2 0.001

5. (a) f ¨ (x) is negative when the function is decreasing and positive when the function is increasing. Therefore, f ¨ (x) is positive at C and G. f ¨ (x) is negative at A and E. f ¨ (x) is zero at B, D, and F . (b) f ¨ (x) is the largest when the graph of the function is increasing the fastest (i.e. the point with the steepest positive slope). This occurs at point G. f ¨ (x) is the most negative when the graph of the function is decreasing the fastest (i.e. the point with the steepest negative slope). This occurs at point A. 6. (a) Since the point B = (2, 5) is on the graph of g, we have g(2) = 5. (b) The slope of the tangent line touching the graph at x = 2 is given by Slope = Thus, g ¨ (2) = −0.4.

Rise 5 − 5.02 −0.02 = = = −0.4. Run 2 − 1.95 0.05

172

Chapter Two /SOLUTIONS

7. (a) Let s = f (t). (i) We wish to fnd the average velocity between t = 1 and t = 1.1. We have Average velocity =

f (1.1) − f (1) 0.808496 − 0.909297 = = −1.00801 m/sec. 1.1 − 1 0.1

Average velocity =

f (1.01) − f (1) 0.900793 − 0.909297 = = −0.8504 m/sec. 1.01 − 1 0.01

(ii) We have (iii) We have f (1.001) − f (1) 0.908463 − 0.909297 = = −0.834 m/sec. 1.001 − 1 0.001 (b) We see in part (a) that as we choose a smaller and smaller interval around t = 1 the average velocity appears to be getting closer and closer to −0.83, so we estimate the instantaneous velocity at t = 1 to be −0.83 m/sec. In this case, more estimates with smaller values of ℎ would be very helpful in making a better estimate. Average velocity =

8. (a) The average velocity between t = 3 and t = 5 is s(5) − s(3) Distance 25 − 9 16 = = = = 8 ft/sec. Time 5−3 2 2 (b) Using an interval of size 0.1, we have H I s(3.1) − s(3) Instantaneous velocity 9.61 − 9 = = 6.1. 3.1 − 3 0.1 at t = 3 Using an interval of size 0.01, we have H I Instantaneous velocity at t = 3

s(3.01) − s(3) 9.0601 − 9 = = 6.01. 3.01 − 3 0.01

From this we guess that the instantaneous velocity at t = 3 is about 6 ft/sec. 9. We use the interval x = 2 to x = 2.01: f (2.01) − f (2) 52.01 − 52 25.4056 − 25 = = = 40.56. f¨ 2.01 − 2 0.01 0.01 For greater accuracy, we can use the smaller interval x = 2 to x = 2.001: f¨

f (2.001) − f (2) 52.001 − 52 25.040268 − 25 = = = 40.268. 2.001 − 2 0.001 0.001

10. P ¨ (0) is the derivative of the function P (t) = 200(1.05)t at t = 0. This is the same as the rate of change of P (t) at t = 0. We estimate this by computing the average rate of change over intervals near t = 0. If we use the intervals −0.001 ≤ t ≤ 0 and 0 ≤ t ≤ 0.001, we see that: 0 1 200(1.05)0 − 200(1.05)−0.001 Average rate of change 200 − 199.990242 = = = 9.758, 0 − (−0.001) 0.001 on − 0.001 ≤ t ≤ 0 0 1 200(1.05)0.001 − 200(1.05)0 Average rate of change 200.009758 − 200 = = = 9.758. 0.001 − 0 0.001 on 0 ≤ t ≤ 0.001 It appears that the rate of change of P (t) at t = 0 is 9.758, so we estimate P ¨ (0) = 9.758. 11. See Figure 2.41.

x −1.5 f ¨ (x)

Figure 2.41

SOLUTIONS to Review Problems For Chapter Two

173

12. This is a line with slope −2, so the derivative is the constant function f ¨ (x) = −2. The graph is a horizontal line at y = −2. See Figure 2.42. 1 x 1

−1 f ¨ (x)

−2

Figure 2.42 13. See Figure 2.43. f ¨ (x)

x 2

Figure 2.43 14. See Figure 2.44.

1 f ¨ (x)

Figure 2.44 15. See Figure 2.45. 20

f ¨ (x) x

−3

3 −20

Figure 2.45 16. See Figure 2.46. f ¨ (x)

Figure 2.46

174

Chapter Two /SOLUTIONS

17. See Figure 2.47. f ¨ (x)

Figure 2.47 18. See Figure 2.48.

f ¨ (x) x

Figure 2.48 19. One possible graph is shown in Figure 2.49. Notice that f (x) is increasing for x < 1 and decreasing for x > 1, so f }(x) is above the x-axis to the left of 1 and below the x-axis to the right of 1. Also as x gets large, the graph of f (x) gets more and more horizontal. Thus, as x gets large, f ¨ (x) gets closer and closer to 0, which means the graph gets closer and closer to the x-axis.

f ¨ (x) x 1

Figure 2.49 20. One possible graph is shown in Figure 2.50. Notice that f (x) is linear with negative slope for x < 0, so f }(x) is constant and negative to the left of 0. We see that f (x) is increasing and getting steeper to the right of 0, so f }(x) is positive and increasing there.

f ¨ (x) −1

x 1

Figure 2.50

SOLUTIONS to Review Problems For Chapter Two

175

21. (a) We use the interval to the right of x = 2 to estimate the derivative. (Alternately, we could use the interval to the left of 2, or we could use both and average the results.) We have f (4) − f (2) 24 − 18 6 = = = 3. 4−2 4−2 2

f¨

We estimate f ¨ (2) 3. (b) We know that f ¨ (x) is positive when f (x) is increasing and negative when f (x) is decreasing, so it appears that f ¨ (x) is positive for 0 < x < 4 and is negative for 4 < x < 12. 22. (a) The function f is increasing where f ¨ is positive, so for x1 < x < x3 . (b) The function f is decreasing where f ¨ is negative, so for 0 < x < x1 or x3 < x < x5 . 23. Since W , the weight, is a function of the height x, in meters, the units are kilograms∕meter 24. The statement f (33) = 73 means that when t = 33, we have P = 73. This tells us that in 2015, 73% of adults owned a personal computer. The statement f ¨ (33) = −3 tells us that in 2015, the percent of adults owning a personal computer is decreasing at a rate of about 3% a year. 25. (Note that we are considering the average temperature of the yam, since its temperature is di˙erent at di˙erent points inside it.) (a) It is positive, because the temperature of the yam increases the longer it sits in the oven. (b) The units of f ¨ (20) are F/min. The statement f ¨ (20) = 2 means that at time t = 20 minutes, the temperature T would increase by approximately 2 F if the yam is in the oven an additional minute. 26. (a) The statement f (15) = 200 tells us that when the price is $15, we sell about 200 units of the product. (b) The statement f ¨ (15) = −25 tells us that if we increase the price by $1 (from 15), we will sell about 25 fewer units of the product. 27. The derivative f ¨ (10) is the slope of the tangent line to the curve at t = 10. See Figure 2.51. Taking two points on the tangent line, we calculate its slope: 100 − 70 Slope = 6. 5 ¨ Since the slope is about 6, we have f (10) 6 cm/yr. At t = 10, the sturgeon was growing in length at a rate of about 6 centimeters a year. length (cm) 150 100 100 − 70 = 30 ✻ ❄ 50

✛5✲

t (years)

Figure 2.51 28. Using the approximation Δy

f ¨ (x)Δx with Δx = 2, we have Δy f (22)

f ¨ (20) ⋅ 2 = 6 ⋅ 2, so

f (20) + f ¨ (20) ⋅ 2 = 345 + 6 ⋅ 2 = 357.

29. Moving away slightly from the center of the hurricane from a point 15 kilometers from the center moves you to a point with stronger winds. For example, the wind is stronger at 15.1 kilometers from the center of the hurricane than it is at 15 kilometers from the center. 30. After falling 20 meters the speed of the rock is increasing at a rate of 0.5 meters∕second per meter. 31. Since f (t) = 1.434(1.0043)t , we have f (3) = 1.434(1.0043)3 = 1.453 ¨

To estimate f (3), we use a small interval around t = 3: f¨

f (3.001) − f (3) 1.434(1.0043)3.001 − 1.434(1.0043)3 = = 0.00623. 3.001 − 3 0.001

176

Chapter Two /SOLUTIONS We see that f (3) = 1.453 billion people and f ¨ (3) = 0.00623 billion (that is, 6.23 million) people per year. Since t = 3 in 2022, this model predicts that the population of China will be about 1,453,000,000 people in 2022 and growing at a rate of about 6,230,000 people per year at that time.

32. Since B is measured in dollars and t is measured in years, dB∕dt is measured in dollars per year. We can interpret dB as the extra money added to your balance in dt years. Therefore dB∕dt represents how fast your balance is growing, in units of dollars/year. 33. (a) If f ¨ (t) > 0, the depth of the water is increasing. If f ¨ (t) < 0, the depth of the water is decreasing. (b) The depth of the water is increasing at 20 cm/min when t = 30 minutes. (c) We use 1 meter = 100 cm, 1 hour = 60 min. At time t = 30 minutes Rate of change of depth = 20

cm cm 60 min 1m = 20 ⋅ ⋅ = 12 meters/hour. min min 1 hr 100 cm

34. (a) Since the graph is below the x-axis at x = 2, the value of f (2) is negative. (b) Since f (x) is decreasing at x = 2, the value of f ¨ (2) is negative. (c) Since f (x) is concave up at x = 2, the value of f ¨¨ (2) is positive. 35. At B both dy∕dx and d 2 y∕dx2 could be positive because y is increasing and the graph is concave up there. At all the other points one or both of the derivatives could not be positive. ℎ¨ (6000)Δx = (0.5)(3) = 1.5 meters. The elevation increases approximately 1.5 36. We have Δℎ = ℎ(6003) − ℎ(6000) meters as the climber moves from a position 6000 meters from the start of the trail to a position 6003 meters from the start. Thus the climber’s elevation increases from 8000 meters to about 8001.5 meters. The new elevation is about 8001.5 meters above sea level. 37. The fact that f (80) = 0.05 means that when the car is moving at 80 km/hr is it using 0.05 liter of gasoline for each kilometer traveled. The derivative f ¨ (v) is the rate of change of gasoline consumption with respect to speed. That is, f ¨ (v) tells us how the consumption of gasoline changes as speeds vary. We are told that f ¨ (80) = 0.0005. This means that a 1-kilometer increase in speed results in an increase in consumption of about 0.0005 liter per km. At higher speeds, the vehicle burns more gasoline per km traveled than at lower speeds. 38. (a) Since the traÿc fow is the number of cars per hour, it is the slope of the graph of C(t). It is greatest where the graph of the function C(t) is the steepest and increasing. This happens at approximately t = 3 hours, or 7am. (b) By reading the values of C(t) from the graph we see: C(2) − C(1) Average rate of change 1000 − 400 600 = = = = 600 cars/hour, 2−1 1 1 on 1 ≤ t ≤ 2 C(3) − C(2) Average rate of change 2000 − 1000 1000 = = = = 1000 cars/hour. 3−2 1 1 on 2 ≤ t ≤ 3 A good estimate of C ¨ (2) is the average of the last two results. Therefore: (600 + 1000) 1600 = = 800 cars/hour. 2 2

C¨ (c) Since t = 2 is 6 am, the fact that C ¨ (2)

800 cars/hour means that the traÿc fow at 6 am is about 800 cars/hour.

39. (a) As the cup of co˙ee cools, the temperature decreases, so f ¨ (t) is negative. (b) Since f ¨ (t) = dH∕dt, the units are degrees Celsius per minute. The quantity f ¨ (20) represents the rate at which the co˙ee is cooling, in degrees per minute, 20 minutes after the cup is put on the counter. 40. The statements f (100) = 35 and f ¨ (100) = 3 tell us that at x = 100, the value of the function is 35 and the function is increasing at a rate of 3 units for a unit increase in x. Since we increase x by 2 units in going from 100 to 102, the value of the function goes up by approximately 2 ⋅ 3 = 6 units, so f (102)

35 + 2 ⋅ 3 = 35 + 6 = 41.

41. Units of P ¨ (t) are dollars/year. The practical meaning of P ¨ (t) is the rate at which the monthly payments change as the duration of the mortgage increases. Approximately, P ¨ (t) represents the change in the monthly payment if the duration is increased by one year. P ¨ (t) is negative because increasing the duration of a mortgage decreases the monthly payments.

SOLUTIONS to Review Problems For Chapter Two 42. Let P = f (t). (a) Estimating the relative rate of change using Δt = 1, we have 1 ΔP 1 f (4) − f (3) = = 0.0356197 = 3.56% per year P Δt f (3) 1 (b) With Δt = 0.1 we have 1 ΔP 1 f (3.1) − f (3) = = 0.0350613 = 3.51% per year P Δt f (3) 0.1 (c) With Δt = 0.01 we have 1 ΔP 1 f (3.01) − f (3) = = 0.0350061 = 3.50% per year P Δt f (3) 0.01 In fact, the relative rate of change at t = 3 of the population for this city is exactly 0.035. 43. (a) This needs to be increasing, concave up everywhere.

(b) This needs to be increasing, concave down everywhere.

(d) This needs to be decreasing, concave down everywhere.

44. (a) This function is increasing, so f ¨ > 0. It appears to be increasing by increasing amounts, so f ¨¨ > 0. (b) This function is decreasing, so f ¨ < 0. It appears to be decreasing faster and faster, so f ¨¨ < 0. (c) This function is increasing, so f ¨ > 0. It appears to be increasing by decreasing amounts, so f ¨¨ < 0. 45. (a) minutes∕kilometer. (b) minutes∕kilometer2 .

177

178

Chapter Two /SOLUTIONS

46. The function is everywhere increasing and concave up. One possible graph is shown in Figure 2.52.

Figure 2.52

47. (a) At x4 and x5 , because the graph is below the x-axis there. (b) At x3 and x4 , because the graph is sloping down there. (c) At x3 and x4 , because the graph is sloping down there. This is the same condition as part (b). (d) At x2 and x3 , because the graph is bending downward there. (e) At x1 , x2 , and x5 , because the graph is sloping upward there. (f) At x1 , x4 , and x5 , because the graph is bending upward there. 48. Since velocity is positive and acceleration is negative, we have f ¨ > 0 and f ¨¨ < 0, and so the graph is increasing and concave down. See Figure 2.53. height

time

Figure 2.53

49. (a) At t3 , t4 , and t5 , because the graph is above the t-axis there. (b) At t2 and t3 , because the graph is sloping up there. (c) At t1 , t2 , and t5 , because the graph is concave up there (d) At t1 , t4 , and t5 , because the graph is sloping down there. (e) At t3 and t4 , because the graph is concave down there. 50. (a) For the two years ending in 2007 we have ΔP 68.1 − 72.4 −4.3 = = Δt 2007 − 2005 2

.15 %/year.

For the two years ending in 2009 we have ΔP 65.1 − 68.1 −3.0 = = Δt 2009 − 2007 2

.5 %/year.

For the two years ending in 2011 we have ΔP 63.5 − 65.1 −1.6 = = Δt 2011 − 2009 2

.8 %/year.

For the two years ending in 2013 we have ΔP 62.5 − 63.5 −1.0 = = Δt 2013 − 2011 2

.5 %/year.

(b) The negative numbers for the value of the derivative are increasing; that is, they are moving towards zero from the ΔP d2P left. The fact that is increasing from 2005 to 2013 suggests that 2 is positive. Δt dt

STRENGTHEN YOUR UNDERSTANDING

179

ΔP (c) The values of P and are troublesome because they indicate that the percent of students graduating is low, and Δt that the number is getting smaller each year. d2P (d) Since is positive, the percent of students graduating is not decreasing as fast as it once was. Also, in 2011 and dt2 2013 the magnitude of ΔP ∕Δt is less than 1% a year, so the percentage of students graduating does in fact seem to be hitting its minimum at around 60%. 51. (a) 6

Student B’s

✛ answer = slope

Student C’s answer =slope of this line

❘

of this line

✛ Student A’s answer =slope of this line

x 1

(b) The slope of f appears to be somewhere between student A’s answer and student B’s, so student C’s answer, halfway in between, is probably the most accurate. 52. (a) The rate of energy consumption required when v = 0 is the vertical intercept, about 1.8 joules/sec. (b) The graph shows f (v) frst decreases and then increases as v increases. This tells us that the bird expends more energy per second to remain still than to travel at slow speeds (say 0.5 to 1 meter/sec), but that the rate of energy consumption required increases again at speeds beyond 1 meter/sec. The upward concavity of the graph tells us that as the bird speeds up, it uses energy at a faster and faster rate. (c) Figure 2.54 shows a possible graph of the derivative f ¨ (v). Other answers are possible. (joules/sec)∕(m/sec)

f ¨ (v)

−1

v (m/sec)

Figure 2.54

STRENGTHEN YOUR UNDERSTANDING 1. True, this is the defnition of the derivative. 2. True, one way to see is that the graph of f (x) is increasing for all x > 0. Since the graph is increasing at x = 1, the instantaneous rate of change, f ¨ (1), is positive. 3. True, since g ¨ (1)

(g(1.001) − g(1))∕(1.001 − 1) = (21.001 − 2)∕(1.001 − 1).

4. False, since the function H(x) =

x is increasing, the slope is always positive.

5. False, the function f (x) = x is such a function, with a = 0. 6. False, the average rate of change between 0 and 1 is given by the slope of the line connecting the points (0, f (0)) and (1, f (1)). So, for example, the graph of f (x) = x2 has slope 0 at x = 0, but the average rate of change between 0 and 1 is (f (1) − f (0))∕(1 − 0) = 1.

180

Chapter Two /SOLUTIONS

7. False, the function r appears to be increasing, and therefore would have a positive derivative. 8. False, the derivative of f at 3 is approximated by a di˙erence quotient on a small interval containing 3, and is not the same as the average rate of change on the interval x = 0 to x = 3. 9. False, R(w) is increasing for all w so the derivative can not be negative at any point. 10. True, consider any constant function. 11. True. 12. False, the opposite is true: If the derivative of f is negative on an interval, then f is decreasing on that interval. 13. False. a function can be increasing while its slope is decreasing. 14. False, since g could be less than f at x = 2 but increasing faster than f near x = 2. For example, f (x) = 5 and g(x) = x2 has f (2) = 5 > 4 = g(2), but f ¨ (2) = 0 since f is constant, and g ¨ (2) > 0 since g is increasing near x = 2. 15. True, the derivative is the slope of the curve, which is always 0 for a horizontal line. 16. False, since a function f (x) can be decreasing and concave up, which means the function is decreasing less rapidly as x → ∞. Therefore, the derivative of the function is less and less negative, so the derivative is increasing. 17. False, the function ln(t) is an increasing function, and therefore has positive derivative. 18. False, if we let f (x) = 3 − x, then f (3) = 0, but f is linear, and so the slope is the derivative, which is −1. 19. True. 20. True. (x) 21. True, since the derivative is the limit of the di˙erence quotient f (x+ℎ)−f , the units of the derivative are the units of the ℎ numerator over the units of the denominator, which is dollars per student.

22. True, as long as it is understood that x is the independent variable, then the two quantities are equal. 23. True, this interpretation is as specifed in the text. 24. True, the local linear approximation is given by f (10) + f ¨ (10)(1) = 20 + 3(1) = 23. 25. True, dA∕dB represents the change in A with respect to a change in B, and so the units are the units of A divided by the units of B. 26. False, since f ¨ (B) = dA∕dB, the units of f ¨ (B) are the units of A divided by the units of B. 27. True, since the derivative is the approximate change in f when the independent variable is increased by one. 28. True, since f ¨ (D) = dt∕dD the units of f ¨ (D) are the units of t divided by the units of D, which is minutes per milligram. 29. True, since for the frst 10 years, height is an increasing function of age. 30. True, since if the derivative is negative, then W is a decreasing function of R. 31. False. If we let f (x) = x2 then f ¨¨ (x) > 0 because f is concave up, but f (x) is decreasing for x < 0. 32. True, if f ¨¨ < 0, then the derivative is decreasing, which means that the graph is concave down. 33. False, since f ¨¨ > 0 means only that the derivative is increasing, that is, the graph of f is concave up. However, f can still be decreasing, for example f (x) = 1∕x for x > 0. 34. True, if f ¨ is decreasing, then f ¨¨ < 0 which means that f is concave down. 35. True. If the car is slowing down, then the derivative is decreasing, which means the second derivative is negative. 36. True. The function f (x) = ex is positive, and since it is increasing and concave up, f ¨ and f ¨¨ are both positive. 37. True, any decaying exponential function has these properties, for example f (x) = e−x . 38. True, any linear function f has f ¨ = constant and so also has f ¨¨ = 0. 39. True, since ex is concave up everywhere. 40. False, since the graph of f (x) = ln(x) is always concave down, so f ¨¨ (x) < 0. 41. True. 42. True. 43. True, if marginal revenue is greater than marginal cost, then the amount of revenue earned on producing an additional unit will be more than the amount it costs to produce, and so it will increase the proft. 44. True, since the total cost function never decreases, so its derivative is never negative. 45. False, if the revenue function is linear, then the marginal revenue is constant slope of the line, or 5.

PROJECTS FOR CHAPTER TWO

181

46. False, the units of marginal cost are the units of cost divided by the units of quantity, which is dollars/unit. 47. True, both the marginal revenue and the marginal cost have dollars as the dependent variable and units as the dependent variable, so their units are the same. 48. False, most frms operate at a proft maximizing level. At this level, P = R − C where P is proft, R is revenue, and C is cost. If the proft is maximal at q ∗ , the the derivative of the proft function must be zero at q ∗ . This gives P ¨ (q ∗ ) = 0 = R(q ∗ ) − C(q ∗ ) which says that R(q ∗ ) = C(q ∗ ). 49. False, the cost and revenue functions can be equal at q ∗ but have di˙erent slopes. 50. True. If two graphs intersect, then the values of the functions are equal at the intersection point. 51. True. If P = f (t), then the relative rate of change is f ¨ (t) 1 dP = ⋅ . f (t) P dt 52. True, since the relative rate of change is f ¨ ∕f = 2∕10 = 0.20 = 20% per minute. 53. True, since 3% of 100 is 3 and the quantity is decreasing. 54. False. The rate of change is about 100 animals per month. To fnd the relative rate of change, we divide by the size of the population. The relative rate of change is −100∕500 = 0.20 = 20% per month. 55. True. This is one of the reasons that linear functions and exponential functions are so important.

PROJECTS FOR CHAPTER TWO 1. (a) A possible graph is shown in Figure 2.55. At frst, the yam heats up very quickly, since the di˙erence in temperature between it and its surroundings is so large. As time goes by, the yam gets hotter and hotter, its rate of temperature increase slows down, and its temperature approaches the temperature of the oven as an asymptote. The graph is thus concave down. (We are considering the average temperature of the yam, since the temperature in its center and on its surface will vary in di˙erent ways.) temperature

200 C

20 C

time

Figure 2.55

(b) If the rate of temperature increase were to remain 2 /min, in ten minutes the yam’s temperature would increase 20 , from 120 to 140 . Since we know the graph is not linear, but concave down, the actual temperature is between 120 and 140 . (c) In 30 minutes, we know the yam increases in temperature by 45 at an average rate of 45∕30 = 1.5 /min. Since the graph is concave down, the temperature at t = 40 is therefore between 120 + 1.5(10) = 135 and 140 . (d) If the temperature increases at 2 /minute, it reaches 150 after 15 minutes, at t = 45. If the temperature increases at 1.5 /minute, it reaches 150 after 20 minutes, at t = 50. So t is between 45 and 50 minutes. 2. (a) Since illumination is concave up and temperature is concave down, the graph on the left side corresponds to the graph of illumination as a function of distance and the graph on the right side corresponds to the graph of temperature as a function of distance. (b) The illumination drops from 75% at d = 2 to 56% at d = 5. Since T = 47 when d = 5 and T = 53.5 when d = 2, 47 − 53.5 −6.5 Average rate of change of temperature = = = −2.17 per foot. 5−2 3

182

Chapter Two /SOLUTIONS (c) A good estimate of the illumination when the distance is 3.5 feet is the average of the values of the illumination at 3 feet and at 4 feet. Therefore: Illumination at 3.5 feet =

67 + 60 = 63.5% < 65%. 2

Since illumination is concave up, the 63.5% is likely to be an overestimate, so you are not likely to be able to read the watch. (d) Let’s represent the illumination as a function of the distance by I(d) and the temperature as a function of the distance by T (d). Therefore: I(7) = I(6) + change in I(d) ≈ I(6) + Average rate of change of I(d) = 53% + (−3%) = 50%. And T (7) = T (6) + change in T (d) ≈ T (6) + Average rate of change of T (d) = 43.5◦ + (−4.5◦) = 39◦ F . (e) Let’s calculate the distance when T (d) = 40 (when we are cold): T (d) = T (6) + T ¨ (6) ⋅ (d − 6) 40 = 43.5 + (−4.5)(d − 6) 40 = 43.5 − 4.5d + 27 40 = 70.5 − 4.5d −30.5 = −4.5d −30.5 d= = 6.78 feet. −4.5 From part (d) we know that the illumination is 50% (darkness) when the distance is 7 feet. Therefore, as we walk away from the candle, we frst get cold and then we are in darkness. 3. (a) From the information given, C(2000) = 1908 ppt and C(2014) = 1638 ppt. (b) Since the change has been approximately linear, the rate of change is constant: C ¨ (2000) = C ¨ (2014) =

1638 − 1908 = −19.286 ppt per year . 2014 − 2000

(c) The slope is −19.286 and C(2000) = 1908. If t is the year, we have C(t) = 1908 − 19.286(t − 2000). (d) We solve 1500 = 1908 − 19.286(t − 2000) 19.286(t − 2000) = 1908 − 1500 1908 − 1500 t = 2000 + = 2021.16. 19.286 The CFC level in the atmosphere above the US is predicted to return to the original level in 2021. (e) Since C ¨¨ (t) > 0, the graph bends upward, so the answer to part (d) is too early. The CFCs are expected to reach their original level later than 2021.

SOLUTIONS TO PROBLEMS ON LIMITS AND THE DEFINITION OF THE DERIVATIVE

Solutions to Problems on Limits and the Definition of the Derivative 1. The answers to parts (a)–(f) are marked in Figure 2.56.

f (x)

✻ ✻

f (a + ℎ) − f (a)

❨

❄

(a) Slope = f (a+ℎ)−f ℎ

✻

✛

f (a)

f (a + ℎ)

a ❄

❄

✲ ✛ ℎ ✲

a+ℎ

Figure 2.56

2. The answers to parts (a)–(f) are marked in Figure 2.57.

(a) Slope = f (a+ℎ)−f ℎ

✻

✠

f (a + ℎ) − f (a) (which is negative) f (a)

❄

f (a + ℎ) ✻

❄ ✛a a + ℎ ✛ ℎ

✲ ✲

f (x)

❄

Figure 2.57

3. (a) When we substitute ℎ = 0,we get 0∕0. The expression is undefned at ℎ = 0. (b) We have Substituting ℎ = 0.1: Substituting ℎ = 0.01: Substituting ℎ = 0.001: Substituting ℎ = 0.0001:

e5(0.1) − 1 = 6.487. 0.1 e5(0.01) − 1 = 5.127. 0.01 5(0.001) e −1 = 5.013. 0.001 e5(0.0001) − 1 = 5.001. 0.0001

lim

183

184

Chapter Two /SOLUTIONS

4. (a) When we substitute ℎ = 0,we get 0∕0. The expression is undefned at ℎ = 0. (b) We have 2e3(0.1) − 2 = 6.997. 0.1 3(0.01) 2e −2 = 6.091. 0.01 2e3(0.001) − 2 = 6.009. 0.001 2e3(0.0001) − 2 = 6.001. 0.0001

Substituting ℎ = 0.1: Substituting ℎ = 0.01: Substituting ℎ = 0.001: Substituting ℎ = 0.0001: (c) It appears that

2e3ℎ − 2 ℎ

lim

ℎ→0

5. (a) When we substitute ℎ = 0,we get 0∕0. The expression is undefned at ℎ = 0. (b) We have ln(1.1) = 0.953. 0.1 ln(1.01) = 0.995. 0.01 ln(1.001) = 0.9995. 0.001 ln(1.0001) = 0.99995. 0.0001

Substituting ℎ = 0.1: Substituting ℎ = 0.01: Substituting ℎ = 0.001: Substituting ℎ = 0.0001: (c) It appears that lim

ℎ→0

ln(ℎ + 1) ℎ

6. (a) When we substitute ℎ = 0,we get 0∕0. The expression is undefned at ℎ = 0. (b) We have sin(3(0.1)) = 2.955. 0.1 sin(3(0.01)) = 2.9996. 0.01 sin(3(0.001)) = 2.999996. 0.001 sin(3(0.0001)) = 2.99999996. 0.0001

Substituting ℎ = 0.1: Substituting ℎ = 0.01: Substituting ℎ = 0.001: Substituting ℎ = 0.0001: (c) It appears that lim

ℎ→0

sin(3ℎ) ℎ

7. Figure 2.58 shows that as x approaches 0 from either side, the values of lim x→0

5x − 1 x

5x − 1 appear to approach 1.6, suggesting that x

.6.

Zooming in on the graph near x = 0 provides further support for this conclusion. Notice that

5x − 1 is undefned at x = 0. x

SOLUTIONS TO PROBLEMS ON LIMITS AND THE DEFINITION OF THE DERIVATIVE 3

f (x) =

185

5x − 1 x

2 1 x −1

Figure 2.58 sin x appears to approach 1, suggesting that x sin x sin x lim = 1. Zooming in on the graph near x = 0 provides further support for this conclusion. Notice that is x→0 x x undefned at x = 0.

8. Figure 2.59 shows that as x approaches 0 from either side, the value of

1 f (x) =

−2

−

sin x x

Figure 2.59 9. Using ℎ = 0.1, 0.01, 0.001, we see 70.1 − 1 = 2.148 0.1 70.01 − 1 = 1.965 0.01 70.001 − 1 = 1.948 0.001 0.0001 7 −1 = 1.946. 0.0001 Using ℎ = −0.1, −0.01, −0.001, we see 7−0.1 − 1 = 1.768 −0.1 −0.01 7 −1 = 1.927 −0.01 −0.001 −1 7 = 1.944 −0.001 7−0.0001 − 1 = 1.946. −0.0001 7ℎ − 1 ℎ→0 ℎ 10. Using ℎ = 0.1, 0.01, 0.001, we see This suggests that lim

.9.

(3 + 0.1)3 − 27 = 27.91 0.1 (3 + 0.01)3 − 27 = 27.09 0.01 (3 + 0.001)3 − 27 = 27.009. 0.001

186

Chapter Two /SOLUTIONS Using ℎ = −0.1, −0.01, −0.001, we see (3 − 0.1)3 − 27 = 26.11 −0.1 (3 − 0.01)3 − 27 = 26.91 −0.01 3 (3 − 0.001) − 27 = 26.99. −0.001 These calculations suggest that lim ℎ→0

(3 + ℎ)3 − 27 = 27. ℎ

11. Using radians,

These values suggest that lim ℎ→0

ℎ

(cos ℎ − 1)∕ℎ

0.01

−0.005

0.001

−0.0005

0.0001

−0.00005

-0.0001

0.00005

-0.001

0.0005

-0.01

0.005

cos ℎ − 1 = 0. ℎ

12. Using ℎ = 0.1, 0.01, 0.001, we see ℎ

(e1+ℎ − e)∕ℎ

0.01

2.7319

0.001

2.7196

0.0001

2.7184

-0.0001

2.7182

-0.001

2.7169

-0.01

2.7047

e1+ℎ − e = 2.7. In fact, this limit is e. ℎ→0 ℎ 13. No, f (x) is not continuous on 0 ≤ x ≤ 2, but it is continuous on the interval 0 ≤ x ≤ 0.5. These values suggest that lim

14. Yes, f (x) is continuous on 0 ≤ x ≤ 2. 15. No, f (x) is not continuous on 0 ≤ x ≤ 2, but it is continuous on 0 ≤ x ≤ 0.5. 16. No, f (x) is not continuous on 0 ≤ x ≤ 2, but it is continuous on 0 ≤ x ≤ 0.5. 17. (a) Yes, f (x) is continuous on 1 ≤ x ≤ 3. (b) Yes, f (x) is continuous on 0.5 ≤ x ≤ 1.5. (c) No, f (x) is not continuous on 3 ≤ x ≤ 5 because of the jump at x = 4. (d) No, f (x) is not continuous on 2 ≤ x ≤ 6 because of the jump at x = 4. 18. (a) No, f (x) is not continuous on 1 ≤ x ≤ 3 because of the jump at x = 2. (b) No, f (x) is not continuous on 0.5 ≤ x ≤ 1.5 because of the jump at x = 2. (c) Yes, f (x) is continuous on 3 ≤ x ≤ 5. (d) Yes, f (x) is continuous on 4 ≤ x ≤ 5. 19. We have lim

ℎ→0

(5 + 2ℎ) − 5 2ℎ = lim = lim 2 = 2. ℎ→0 ℎ ℎ→0 ℎ

SOLUTIONS TO PROBLEMS ON LIMITS AND THE DEFINITION OF THE DERIVATIVE 20. We have lim ℎ→0

21. We have lim ℎ→0

22. We have lim ℎ→0

187

(3ℎ − 7) + 7 3ℎ = lim = lim 3 = 3. ℎ→0 ℎ ℎ→0 ℎ

(ℎ2 + 2ℎ + 1) − 1 (ℎ + 1)2 − 1 ℎ2 + 2ℎ = lim = lim = lim(ℎ + 2) = 0 + 2 = 2. ℎ→0 ℎ→0 ℎ→0 ℎ ℎ ℎ

(ℎ − 2)2 − 4 (ℎ2 − 4ℎ + 4) − 4 ℎ2 − 4ℎ = lim = lim = lim(ℎ − 4) = 0 − 4 = −4. ℎ→0 ℎ→0 ℎ→0 ℎ ℎ ℎ

23. Yes: f (x) = x + 2 is continuous for all values of x. 24. Yes: f (x) = 2x is continuous function for all values of x. 25. Yes: f (x) = x2 + 2 is a continuous function for all values of x. 26. Yes: f (x) = 1∕(x − 1) is a continuous function on any interval that does not contain x = 1. 27. No: f (x) = 1∕(x − 1) is not continuous on any interval containing x = 1. 28. Yes: f (x) = 1∕(x2 + 1) is continuous for all values of x because the denominator is never 0. 29. This function is not continuous. Each time someone is born or dies, the number jumps by one. 30. Even though the car is stopping and starting, the distance traveled is a continuous function of time. 31. Continuous 32. Since we can’t make a fraction of a pair of pants, the number increases in jumps, so the function is not continuous. 33. The time is not a continuous function of position as distance from your starting point, because every time you cross from one time zone into the next, the time jumps by 1 hour. 34. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ 5(x + ℎ) − 5x = lim ℎ→0 ℎ 5x + 5ℎ − 5x = lim ℎ→0 ℎ 5ℎ = lim . ℎ→0 ℎ

f ¨ (x) = lim ℎ→0

As long as we let ℎ get close to zero without actually equaling zero, we can cancel the ℎ in the numerator and denominator, and we are left with f ¨ (x) = 5. 35. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ (3(x + ℎ) − 2) − (3x − 2) = lim ℎ→0 ℎ 3x + 3ℎ − 2 − 3x + 2 = lim ℎ→0 ℎ 3ℎ . = lim ℎ→0 ℎ

f ¨ (x) = lim ℎ→0

As ℎ gets very close to zero without actually equaling zero, we can cancel the ℎ in the numerator and denominator to obtain f ¨ (x) = lim(3) = 3. ℎ→0

36. Using the defnition of the derivative, we have f ¨ (x) = lim ℎ→0

f (x + ℎ) − f (x) ℎ

188

Chapter Two /SOLUTIONS ((x + ℎ)2 + 4) − (x2 + 4) ℎ x2 + 2xℎ + ℎ2 + 4 − x2 − 4 = lim ℎ→0 ℎ 2xℎ + ℎ2 = lim ℎ→0 ℎ ℎ(2x + ℎ) = lim . ℎ→0 ℎ = lim ℎ→0

As ℎ gets very close to zero without actually equaling zero, we can cancel the ℎ in the numerator and denominator to obtain f ¨ (x) = lim(2x + ℎ) = 2x. ℎ→0

37. Using the defnition of the derivative, we have f (x + ℎ) − f (x) 3(x + ℎ)2 − 3x2 = lim ℎ→0 ℎ→0 ℎ ℎ 3(x2 + 2xℎ + ℎ2 ) − 3x2 = lim ℎ→0 ℎ 3x2 + 6xℎ + 3ℎ2 − 3x2 = lim ℎ→0 ℎ ℎ(6x + 3ℎ) 6xℎ + 3ℎ2 = lim = lim . ℎ→0 ℎ→0 ℎ ℎ

f ¨ (x) = lim

As ℎ gets very close to zero (but not equal to zero), we can cancel the ℎ in the numerator and denominator to leave the following: f ¨ (x) = lim(6x + 3ℎ). ℎ→0

As ℎ → 0, we have f ¨ (x) = 6x. 38. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ ((x + ℎ) − (x + ℎ)2 ) − (x − x2 ) = lim ℎ→0 ℎ (x + ℎ − (x2 + 2xℎ + ℎ2 )) − x + x2 = lim ℎ→0 ℎ x + ℎ − x2 − 2xℎ − ℎ2 − x + x2 = lim ℎ→0 ℎ ℎ − 2xℎ − ℎ2 = lim . ℎ→0 ℎ

f ¨ (x) = lim ℎ→0

As long as we let ℎ get close to zero without actually equaling zero, we can cancel the ℎ in the numerator and denominator, and we are left with 1 − 2x − ℎ. Taking the limit as ℎ goes to zero, we get f ¨ (x) = 1 − 2x since the other term goes to zero. 39. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ (5(x + ℎ)2 + 1) − (5x2 + 1) = lim ℎ→0 ℎ 5(x2 + 2xℎ + ℎ2 ) + 1 − 5x2 − 1 = lim ℎ→0 ℎ 5x2 + 10xℎ + 5ℎ2 + 1 − 5x2 − 1 = lim ℎ→0 ℎ 10xℎ + 5ℎ2 = lim ℎ→0 ℎ ℎ(10x + 5ℎ) = lim . ℎ→0 ℎ

f ¨ (x) = lim ℎ→0

SOLUTIONS TO PROBLEMS ON LIMITS AND THE DEFINITION OF THE DERIVATIVE

189

As ℎ gets very close to zero without actually equaling zero, we can cancel the ℎ in the numerator and denominator to obtain f ¨ (x) = lim(10x + 5ℎ) = 10x. ℎ→0

40. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ (2(x + ℎ)2 + (x + ℎ)) − (2x2 + x) = lim ℎ→0 ℎ 2(x2 + 2xℎ + ℎ2 ) + (x + ℎ) − 2x2 − x = lim ℎ→0 ℎ 2x2 + 4xℎ + 2ℎ2 + x + ℎ − 2x2 − x = lim ℎ→0 ℎ 4xℎ + 2ℎ2 + ℎ = lim ℎ→0 ℎ ℎ(4x + 2ℎ + 1) = lim . ℎ→0 ℎ

f ¨ (x) = lim ℎ→0

As ℎ gets very close to zero without actually equaling zero, we can cancel the ℎ in the numerator and denominator to obtain f ¨ (x) = lim(4x + 2ℎ + 1) = 4x + 1. ℎ→0

41. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ −2(x + ℎ)3 − (−2x3 ) = lim ℎ→0 ℎ −2(x3 + 3x2 ℎ + 3xℎ2 + ℎ3 ) + 2x3 = lim ℎ→0 ℎ −2x3 − 6x2 ℎ − 6xℎ2 − 2ℎ3 + 2x3 = lim ℎ→0 ℎ −6x2 ℎ − 6xℎ2 − 2ℎ3 = lim . ℎ→0 ℎ

f ¨ (x) = lim ℎ→0

As long as we let ℎ get close to zero without actually equaling zero, we can cancel the ℎ in the numerator and denominator, and we are left with −6x2 − 6xℎ − 2ℎ2 . Taking the limit as ℎ goes to zero, we get f ¨ (x) = −6x2 since the other two terms go to zero. 42. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ (1 − (x + ℎ)3 ) − (1 − x3 ) = lim ℎ→0 ℎ (1 − (x3 + 3x2 ℎ + 3xℎ2 + ℎ3 )) − 1 + x3 = lim ℎ→0 ℎ 1 − x3 − 3x2 ℎ − 3xℎ2 − ℎ3 − 1 + x3 = lim ℎ→0 ℎ −3x2 ℎ − 3xℎ2 − ℎ3 = lim . ℎ→0 ℎ

f ¨ (x) = lim ℎ→0

As long as we let ℎ get close to zero without actually equaling zero, we can cancel the ℎ in the numerator and denominator, and we are left with −3x2 − 3xℎ − ℎ2 . Taking the limit as ℎ goes to zero, we get f ¨ (x) = −3x2 since the other two terms go to zero.

190

Chapter Two /SOLUTIONS

43. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ 1∕(x + ℎ) − 1∕x = lim . ℎ→0 ℎ

f ¨ (x) = lim ℎ→0

Writing the numerator over a common denominator and simplifying, we get (x − (x + ℎ))∕(x(x + ℎ)) ℎ −ℎ∕(x(x + ℎ)) = lim ℎ→0 ℎ −ℎ = lim . ℎ→0 ℎx(x + ℎ)

f ¨ (x) = lim ℎ→0

As long as we let ℎ get close to zero without actually equaling zero, we can cancel the ℎ in the numerator and denominator, and we are left with −1∕(x(x + ℎ)). Taking the limit as ℎ goes to zero, we get f ¨ (x) = −1∕x2 since ℎ goes to zero.

3.1 SOLUTIONS

CHAPTER THREE Solutions for Section 3.1 dy =3 dx dy =0 2. dx ¨ 3. y = −12x−13 . 1.

4. y¨ = 12x11 . 5. y¨ = 24t2 6. y¨ = 43 x1∕3 . dy =5 dx ¨ 8. y = 12t3 − 4t 7.

9. f ¨ (q) = 3q 2 10. f ¨ (x) = −4x−5 . 11. y¨ = 18x2 + 8x − 2. 12. y¨ = 2x + 5. dy 13. = 24t2 − 8t + 12. dt dy 14. = 6x + 7. dx 15. y¨ = −12x3 − 12x2 − 6. dy 16. = 8.4q − 0.5. dq 1 17. Since f (z) = − 6.1 = −z−6.1 , we have f ¨ (z) = −(−6.1)z−7.1 = 6.1z−7.1 . z 1 18. Since g(t) = 5 = t−5 , we have g ¨ (t) = −5t−6 . t √ dy 1 19. Since y = x = x1∕2 , we have = x−1∕2 . dx 2 dy 1 7 20. Since y = 7∕2 = r−7∕2 , we have = − r−9∕2 . dx 2 r u 1 1 3 21. Since f (x) = = 3∕2 = x−3∕2 , we have f ¨ (x) = − x−5∕2 . 2 x3 x 1 1 22. Since ℎ( ) = √ = −1∕3 , we have ℎ¨ ( ) = − −4∕3 . 3 3 23. Because z = (t − 1)(t + 1) = t2 − 1 we have dz = 2t. dt 24. Because R = (s2 + 1)2 = s4 + 2s2 + 1 we have dR = 4s3 + 4s. ds 25. y¨ = 2z − 2z12 . 26. y¨ = 15t4 − 25 t−1∕2 − t72 . 27. Since ℎ(t) =

3 4 + = 3t−1 + 4t−2 , we have ℎ¨ (t) = −3t−2 − 8t−3 . t t2

191

192

Chapter Three /SOLUTIONS

6 28. y¨ = 6t − t3∕2 + t23 .

1 −1∕2 + −2 . 2 √ dy 3 1 = x1∕2 + x−1∕2 . 30. Since y = x(x + 1) = x1∕2 x + x1∕2 ⋅ 1 = x3∕2 + x1∕2 , we have dx 2 2 31. Using the power rule and the fact that the derivative of the sum of terms equals the sum of the derivatives of the terms, we have y¨ = 2ax + b. 29. Since ℎ( ) = ( −1∕2 − −2 ) = −1∕2 − −2 = 1∕2 − −1 , we have ℎ¨ ( ) =

d (x2 ) = 2kx. 32. f ¨ (x) = k ⋅ dx

33. Since we have

b v = at2 + 2 = at2 + bt−2 t dv b = 2at + b(−2)t−3 = 2at − 2 3 . dt t

34. The derivatives of P 2 and P 3 are 2P and 3P 2 , respectively, so dQ = 2aP + 3bP 2 . dP 35. Since 4∕3, , and b are all constants, we have dV 4 8 = (2r)b = rb. dr 3 3 36. Since a and b are constants, we have

dP 1 b = 0 + b t−1∕2 = √ . dt 2 2 t

ax + b a b a = x + , we have ℎ¨ (x) = . c c c c 38. Since w is a constant times q, we have dw∕dq = 3ab2 .

37. Since ℎ(x) =

39. (a) 8 6 4

P (q) = 6q − q 2

2 q 1

Figure 3.1 Since P is increasing when q = 1, P ¨ (1) is positive. At (3, 9) we observe that the function has a horizontal tangent line, and horizontal lines have slope zero. Thus P ¨ (3) = 0. Finally, at q = 4 the function is decreasing; therefore the derivative P ¨ (4) is negative. (b) The derivative of P (q) = 6q − q 2 is P ¨ (q) = 6 − 2q. Therefore, P ¨ (1) = 6 − 2 = 4 P ¨ (3) = 6 − 2(3) = 6 − 6 = 0 P ¨ (4) = 6 − 2(4) = 6 − 8 = −2 40. f ¨ (x) = 3x2 − 8x + 7, so f ¨ (0) = 7, f ¨ (2) = 3, and f ¨ (−1) = 18.

3.1 SOLUTIONS

193

41. (a) f ¨ (t) = 2t − 4. (b) f ¨ (1) = 2(1) − 4 = −2 f ¨ (2) = 2(2) − 4 = 0 (c) We see from part (b) that f ¨ (2) = 0. This means that the slope of the line tangent to the curve at x = 2 is zero. From Figure 3.2, we see that indeed the tangent line is horizontal at the point (2, 1). The fact that f ¨ (1) = −2 means that the slope of the line tangent to the curve at x = 1 is −2. If we draw a line tangent to the graph at x = 1 (the point (1, 2)) we see that it does indeed have a slope of −2. 5 4 3 f (t) = t2 − 4t + 5

2 1

t 1

Figure 3.2 42. The rate of change of the population is given by the derivative. For P (t) = t3 + 4t + 1 the derivative is P ¨ (t) = 3t2 + 4. At t = 2, the rate of change of the population is 3(2)2 + 4 = 12 + 4 = 16, meaning the population is growing by 16 units per unit of time. 43. Since f (t) = 700 − 3t2 , we have f (5) = 700 − 3(25) = 625 cm. Since f ¨ (t) = −6t, we have f ¨ (5) = −30 cm/year. In the year 2021, the sand dune will be 625 cm high and eroding at a rate of 30 centimeters per year. 44. We have f ¨ (t) = 3, so the relative rate of change when t = 5 is f ¨ (5) 3 3 = = = 0.176 = 17.6% per year. f (5) 3(5) + 2 17 45. We have f ¨ (t) = 6t2 , so the relative rate of change at t = 4 is f ¨ (4) 6(42 ) 96 = = = 0.696 = 69.6% per year. f (4) 138 2(43 ) + 10 46. We have Δx = 1.04 − 1 = 0.04. Di˙erentiating,we get f ¨ (x) = 4x3 − 2x, so f ¨ (1) = 2. Δy ≈ f ¨ (1)Δx = 2(0.04) = 0.08. Thus, f (1.04) = f (1) + Δy ≈ 3 + 0.08 = 3.08. 47. We have Δx = 0.97 − 1 = −0.03. Di˙erentiating,we get f ¨ (x) = 3x2 + 2x, so f ¨ (1) = 5. Δy ≈ f ¨ (1)Δx = 5(−0.03) = −0.15. Thus, f (0.97) = f (1) + Δy ≈ −4 + (−0.15) = −4.15. √ √ √ 48. When t = 9, we have N = 120 9 = 360 acres. Since f ¨ (t) = 120(1∕2)t−1∕2 = 60∕ t, we have f ¨ (9) = 60∕ 9 = 60∕3 = 20 acres per year. Then Relative rate of change =

f ¨ (9) 20 = = 0.0555 = 5.55% per year. f (9) 360

Nine years after farming started in the region, there are 360 acres being harvested. The number of acres being harvested is increasing at a continuous rate of 20 acres per year, or 5.55% per year.

194

Chapter Three /SOLUTIONS

49. After four months, there are 300(4)2 = 4800 mussels in the bay. The population is growing at the rate Z ¨ (4) mussels per month, where Z ¨ (t) = 600t, so the rate of increase is 2400 mussels per month. 50. When t = 10, we have Q = 3(102 ) + 100 = 400 tons. Since f ¨ (t) = 6t, we have f ¨ (10) = 6(10) = 60 tons per year. Then Relative rate of change =

f ¨ (10) 60 = = 0.15 = 15% per year. f (10) 400

In 2010, there were 400 tons of waste at the site. The quantity was growing at a rate of 60 tons per year, which is 15% per year. 51. f ¨ (t) = 6t2 − 8t + 3

and

f ¨¨ (t) = 12t − 8.

52. The derivative of f (t) is f ¨ (t) = 4t3 − 6t + 5. The second derivative is the derivative of the derivative, and thus f ¨¨ (t) = 12t2 − 6. 53. (a) Di˙erentiating, we have k¨ (x) = 3x2 − 1, so k¨ (−1) = 3(−1)2 − 1 = 2. (b) Taking the second derivative, we have k¨¨ (x) = 6x, so k¨¨ (−1) = 6(−1) = −6. (c) Since k¨ (−1) = 2, we see that k(x) is increasing at x = −1. The only function that is increasing at x = −1 is (III), so (III) must be the graph of k(x). 54. (a) Di˙erentiating, we have f ¨ (x) = 6x2 + 6x, so f ¨ (−1) = 6(−1)2 + 6(−1) = 0. (b) Taking the second derivative, we have f ¨¨ (x) = 12x + 6, so f ¨¨ (−1) = 12(−1) + 6 = −6. (c) Since f ¨ (−1) = 0, we see that f (x) has a horizontal tangent line at x = −1. This means it could match (I) or (IV). Since f ¨¨ (−1) = −6, we see that f (x) is concave down at x = −1, so f (x) matches (IV). 55. (a) Di˙erentiating, we have g ¨ (x) = 4x3 − 2x − 2, so g ¨ (−1) = 4(−1)3 − 2(−1) − 2 = −4. (b) Taking the second derivative, we have g ¨¨ (x) = 12x2 − 2, so g ¨¨ (−1) = 12(−1)2 − 2 = 10. (c) Since g ¨ (−1) = −4, g(x) is decreasing at x = −1. The only function which is decreasing at x = −1 is (II), so (II) must be the graph of g(x). 56. (a) Di˙erentiating, we have ℎ¨ (x) = 8x3 + 24x2 + 30x + 14, so ℎ¨ (−1) = 8(−1)3 + 24(−1)2 + 30(−1) + 14 = 0. (b) Taking the second derivative, we have ℎ¨¨ (x) = 24x2 + 48x + 30, so ℎ¨¨ (−1) = 24(−1)2 + 48(−1) + 30 = 6. (c) Since ℎ¨ (−1) = 0, we see that ℎ(x) has a horizontal tangent line at x = −1. This means it could match (I) or (IV). Since ℎ¨¨ (−1) = 6, we see that ℎ(x) is concave up at x = −1, so ℎ(x) matches (I). 57. (a) The slope of f at x = 1 is negative, so f ¨ (1) < 0. On the other hand, the slopes of f at x = −1, x = 0, and x = 4 are all positive, with the least steep slope occurring at x = 0 and the steepest slope occurring at x = 4. Therefore, we must have f ¨ (1) < f ¨ (0) < f ¨ (−1) < f ¨ (4). (b) The derivative function of f is given by f ¨ (x) = 3x2 − 6x + 2. ¨

Evaluating at x = −1, 0, 1, 4, we have f (1) = −1, f ¨ (0) = 2, f ¨ (−1) = 11 and f ¨ (4) = 26 which confrms the ordering of the four quantities.

3.1 SOLUTIONS

195

58. Since f (1) = 2(13 ) − 5(12 ) + 3(1) − 5 = −5, the point (1, −5) is on the line. We use the derivative to fnd the slope. Di˙erentiating gives f ¨ (x) = 6x2 − 10x + 3, and so the slope at x = 1 is f ¨ (1) = 6(12 ) − 10(1) + 3 = −1. The equation of the tangent line is y − (−5) = −1(x − 1) y + 5 = −x + 1 y = −4 − x.

59. (a) The y coordinate where x = 1 is f (1) = 12 + 5(1) + 2 = 8, so a point on the line is (1, 8). We have f ¨ (x) = 2x + 5, so the slope of the line is f ¨ (1) = 2(1) + 5 = 7. Using the point-slope equation for a line, we have y − 8 = 7(x − 1) y = 7x + 1. The equation of the tangent line at x = 1 is y = 7x + 1. (b) The y coordinate where x = −1 is f (−1) = (−1)2 + 5(−1) + 2 = −2, so a point on the line is (−1, −2). We have f ¨ (x) = 2x + 5, so the slope of the line is f ¨ (−1) = 2(−1) + 5 = 3. Using the point-slope equation for a line, we have y + 2 = 3(x + 1) y = 3x + 1. The equation of the tangent line at x = −1 is y = 3x + 1. 60. To fnd the equation of a line we need to have a point on the line and its slope. We know that this line is tangent to the curve f (t) = 6t − t2 at t = 4. From this we know that both the curve and the line tangent to it will share the same point and the same slope. At t = 4, f (4) = 6(4) − (4)2 = 24 − 16 = 8. Thus we have the point (4, 8). To fnd the slope, we need to fnd the derivative. The derivative of f (t) is f ¨ (t) = 6 − 2t. The slope of the tangent line at t = 4 is f ¨ (4) = 6 − 2(4) = 6 − 8 = −2. Now that we have a point and the slope, we can fnd an equation for the tangent line: y = b + mt 8 = b + (−2)(4) b = 16. Thus, y = −2t + 16 is the equation for the line tangent to the curve at t = 4. See Figure 3.3.

8 6 4

f (t) = 6t − t2

2 t 1

Figure 3.3

61. (a) We have f (2) = 8, so a point on the tangent line is (2, 8). Since f ¨ (x) = 3x2 , the slope of the tangent is given by m = f ¨ (2) = 3(2)2 = 12. Thus, the equation is y − 8 = 12(x − 2)

y = 12x − 16.

196

Chapter Three /SOLUTIONS (b) See Figure 3.4. The tangent line lies below the function f (x) = x3 , so estimates made using the tangent line are underestimates. 30

y = x3 y = 12x − 16

20 10

x −2

−10

Figure 3.4 62. (a) We have f ¨ (x) = 0.009(3x2 ) = 0.027x2 . (b) Since y = f (x) with x in cm, we see that the units of A and C are both cm. Since y is in grams, we see that B is in grams. Finally, since f ¨ (x) = dy∕dx, we see that D is in grams/cm. 63. (a) Using the power rule, we have dQ = (7700)(0.67)v0.67−1 = 5159v−0.33 . dv (b) We have dQ || = (5159)(0.1)−0.33 = 11030 m3 ∕sec per km3 . dv ||v=0.1 For glacial lakes with volumes near 0.1 km3 , a small increase of Δv km3 in volume corresponds to a change in the maximum outfow rate of about 11, 030Δv m3 ∕sec. Larger lakes have a larger maximum outfow during a jökulhlaup. √ 64. (a) f (100) = 1.111 100 = 11.11 seconds. This tells us that it takes approximately 11.11 seconds for one complete oscillation of a 100 foot long pendulum. √ (b) To fnd f ¨ (100), we frst rewrite our equation with a fractional exponent. Thus, f (L) = 1.111 L = 1.111L1∕2 . 1 0.5555 0.5555 Di˙erentiating, we get f ¨ (L) = (1.111)L−1∕2 = = √ . 2 L1∕2 L 0.5555 Therefore, f ¨ (100) = √ = 0.05555 seconds per foot. 100 This tells us, when a pendulum is 100 feet long, an increase of one foot in the length of the pendulum results in an increase of about 0.05555 seconds in the time for one complete oscillation. 65. (a) We have C(w) = 42w0.75 . Using the derivative rules, we fnd C ¨ (w) = 42(0.75w0.75−1 ) = 31.5w−0.25 . (b)

(i) Substituting w = 10, we have C(10) = 42(100.75 ) = 236.183

and

C ¨ (10) = 31.5(10−0.25 ) = 17.714.

A mammal weighing 10 pounds needs about 236 calories a day. If weight increases by one pound, calorie consumption increases by about 17.7 calories a day. (ii) Substituting w = 100, we have C(100) = 42(1000.75 ) = 1328.157

and

C ¨ (100) = 31.5(100−0.25 ) = 9.961.

A mammal weighing 100 pounds needs about 1328 calories a day. If weight increases by one pound, calorie consumption increases by about 10 calories a day. (iii) Substituting w = 1000, we have C(1000) = 42(10000.75 ) = 7468.774

and

C ¨ (1000) = 31.5(1000−0.25 ) = 5.602.

A mammal weighing 1000 pounds needs about 7,469 calories a day. If weight increases by one pound, calorie consumption increases by about 5.6 calories a day. As expected, as weight increases, calorie requirements also increase. However, the derivative decreases as weight goes up, so an additional pound has a greater impact on calorie requirements for a small mammal than it does for a large mammal.

3.1 SOLUTIONS

197

66. (a) If the air temperature is 20◦ F, and the wind is blowing at 40 mph, we substitute v = 40 into the formula W (v) = 48.17 − 27.2v0.16 , giving W (40) = 48.17 − 27.2(40)0.16 = −0.909. The windchill temperature is approximately −1◦ F. (b) To fnd W ¨ (40), we frst determine W ¨ (v) = −0.16 ⋅ 27.2v−0.84 = −4.352v−0.84 . We substitute 40 for v to get W ¨ (40) = −4.352(40)−0.84 = −0.196◦ F/mph. This tells us that, when the temperature is 20◦ F and the wind speed is 40 mph, then an increase of 1 mph in wind speed results in a drop of about 0.196◦ F in the windchill temperature. 67. (a) We have f ¨ (m) = 17.4(0.25m−0.75 ) = 4.35 ⋅ m−0.75 . (b) We see that f (70) = 17.4 ⋅ 700.25 = 50.330 and f ¨ (70) = 4.35 ⋅ 70−0.75 = 0.180. This means that a human with body mass 70 kg will have a circulation time of about 50.330 seconds, and if the human’s body mass goes up by 1 kg, we expect the circulation time to go up by about 0.180 seconds. 68. (a) We have f ¨ (A) = 8.3(0.25A−0.75 ) = 2.075 ⋅ A−0.75 . (b) We see that f (40) = 8.3 ⋅ 400.25 = 20.873 and f ¨ (40) = 2.075 ⋅ 40−0.75 = 0.130. (c) The statement f (40) = 20.873 tells us that Key Largo, with a land mass of 40 km2 , has about 20.873 di˙erent species of beetles on it, that is, about 21 di˙erent species. (d) The statement f ¨ (40) = 0.130 tells us that the number of species of beetles will go down by about 0.130 if Key Largo shrinks by 1 square kilometer. 69. (a) A = r2 dA = 2 r. dr (b) This is the formula for the circumference of a circle. (c) A¨ (r) ≈ A(r+ℎ)−A(r) for small ℎ. When ℎ > 0, the numerator of the di˙erence quotient denotes the area of the region ℎ contained between the inner circle (radius r) and the outer circle (radius r + ℎ). See fgure below. As ℎ approaches 0, this area can be approximated by the product of the circumference of the inner circle and the “width” of the region, i.e., ℎ. Dividing this by the denominator, ℎ, we get A¨ = the circumference of the circle with radius r.

✛

ℎ

We can also think about the derivative of A as the rate of change of area for a small change in radius. If the radius increases by a tiny amount, the area will increase by a thin ring whose area is simply the circumference at that radius times the small amount. To get the rate of change, we divide by the small amount and obtain the circumference. 70. Since W is proportional to r3 , we have W = kr3 for some constant k. Thus, dW ∕dr = k(3r2 ) = 3kr2 . Thus, dW ∕dr is proportional to r2 . 71. If f (x) = xn , then f ¨ (x) = nxn−1 . This means f ¨ (1) = n ⋅ 1n−1 = n ⋅ 1 = n, because any power of 1 equals 1. 72. The marginal cost of producing the 25th item is C ¨ (25), where C ¨ (q) = 4q, so the marginal cost is $100. This means that the cost of production increases by about $100 when we add one unit to a production level of 25 units. 73. (a) We have R(p) = pq = p(300 − 3p) = 300p − 3p2 (b) Since R¨ (p) = 300 − 6p, we have R¨ (10) = 300 − 6 ⋅ 10 = 240. This means that revenues are increasing at a rate of $240 per dollar of price increase when the price is $10. (c) R¨ (p) = 300 − 6p is positive for p < 50 and negative for p > 50. d 74. (a) Velocity v(t) = dy = dt (1250 − 16t2 ) = −32t. dt Since t ≥ 0, the ball’s velocity is negative. This is reasonable, since its height y is decreasing. (b) The ball hits the ground when its height y = 0. This gives

1250 − 16t2 = 0 t ≈ ±8.84 seconds

198

Chapter Three /SOLUTIONS We discard t = −8.84 because time t is nonnegative. So the ball hits the ground about 8.84 seconds after its release, at which time its velocity is v(8.84) = −32(8.84) = −282.88 feet/sec = −192.87 mph.

75. (a) The yield is f (5) = 320 + 140(5) − 10(5)2 = 770 bushels per acre. (b) f ¨ (x) = 140 − 20x, so f ¨ (5) = 40 bushels per acre per pound of fertilizer. For each acre, yield will go up by about 40 bushels if an additional pound of fertilizer is used. (c) More should be used, because at this level of use, more fertilizer will result in a higher yield. Fertilizer’s use should be increased until an additional unit results in a decrease in yield. i.e. until the derivative at that point becomes negative. 76. (a) The p-intercept is the value of p when q = 0. p = f (0) = 50 − 0.03(0)2 = 50. The p-intercept occurs at p = 50. The q-intercept is the value of q such that p = f (q) = 0. p = f (q) = 50 − 0.03q 2 = 0 −0.03q 2 = −50 50 q2 = and q ≥ 0 0.03 u 50 q= ≈ 40.825 0.03 The q-intercept occurs at q ≈ 40.825. The p-intercept represents the price at which quantity demanded is zero. That is, when the price reaches 50 dollars, demand for the product will be zero. The q-intercept represents the demand for the product if the product were being given away free of charge. In this case, 40.825 units of the product would be consumed if the product were free (p = 0). (b) f (20) = 50 − 0.03(20)2 = 50 − 0.03(400) = 50 − 12 = 38 dollars. This tells us that if the price per unit is $38, then a total of 20 units are demanded. (c) To fnd f ¨ (20) we frst fnd f ¨ (q) = 2(−0.03)q = −0.06q. Therefore, f ¨ (20) = −0.06(20) = −1.2 dollars per unit demanded. This tells us given a demand of 20 units, which, according to our answer to part (b), occurs when the unit price is $38, an increase of $1.20 in the price will result in the reduction of consumption by approximately 1 unit while a decrease in price by the same amount will lead to an increase of approximately 1 unit in sales. 77. (a) The marginal cost function equals C ¨ (q) = 0.08(3q 2 ) + 75 = 0.24q 2 + 75. (b) C(50) = 0.08(50)3 + 75(50) + 1000 = $14,750. C(50) tells us how much it costs to produce 50 items. From above we can see that the company spends $14,750 to produce 50 items. The units for C(q) are dollars. C ¨ (50) = 0.24(50)2 + 75 = $675 per item. C ¨ (q) tells us the approximate change in cost to produce one additional item of product. Thus at q = 50 costs will increase by about $675 for one additional item of product produced. The units are dollars/item. 78. Since f (x) = x3 − 6x2 − 15x + 20, we have f ¨ (x) = 3x2 − 12x − 15. To fnd when f ¨ (x) = 0, we solve 3x2 − 12x − 15 = 0 3(x2 − 4x − 5) = 0 3(x + 1)(x − 5) = 0. We see that f ¨ (x) = 0 at x = −1 and at x = 5. We see that the graph of f (x) in Figure 3.5 is horizontal at x = −1 and at x = 5, which confrms what we found using the derivative.

3.2 SOLUTIONS 100

x −1

−100

Figure 3.5: f (x) = x3 − 6x2 − 15x + 20 79. (a) R(q) = q(b + mq) = bq + mq 2 . (b) R¨ (q) = b + 2mq.

Solutions for Section 3.2 dP = 9t2 + 2et . dt 2. f ¨ (x) = 2ex + 2x. 1.

3. f ¨ (x) = 3x2 + 3x ln 3 4. y¨ = 10t + 4et . dy 5. = 5 ⋅ 5t ln 5 + 6 ⋅ 6t ln 6 dx dy 2 6. Since y = 2x + 3 = 2x + 2x−3 , we have = (ln 2)2x − 6x−4 . dx x dy 7. = 4(ln 10)10x − 3x2 . dx 8. f ¨ (x) = (ln 2)2x + 2(ln 3)3x . dy 9. = 5(ln 2)(2x ) − 5. dx dy 10. = 3 − 2(ln 4)4x . dx dy 11. = e0.7t ⋅ (0.7) = 0.7e0.7t . dt 12. f ¨ (t) = e3t ⋅ 3 = 3e3t . 13. P ¨ = −0.2e−0.2t . 14. y¨ = −4e−4t . dP 15. = 200(0.12)e0.12t = 24e0.12t . dt 16. P ¨ = 50(−0.6)e−0.6t = −30e−0.6t . 17. P ¨ (t) = 12.41(ln 0.94)(0.94)t . 18. P ¨ (t) = 3000(ln 1.02)(1.02)t . 19. y¨ = Aet 20. P ¨ (t) = Cet . 21. Since y = 10x + 10x−1 , we have dy 10 = (ln 10)10x − 10x−2 = (ln 10)10x − 2 . dx x 22. f ¨ (x) = Aex − 2Bx. 23. D¨ = −1∕p.

199

200

Chapter Three /SOLUTIONS

24. R¨ = 3q . 25. R¨ (q) = 2q − 2∕q. 26. y¨ = 2t + 5∕t B 27. Aet + . t dy = 2x + 4 − 3∕x. 28. dx 29. Since f ¨ (t) = 15, we have

f ¨ (t) 15 = . f (t) 15t + 12

30. Since f ¨ (t) = 10, we have

f ¨ (t) 10 = . f (t) 10t + 5

31. Since f ¨ (t) = −7 ⋅ 30e−7t , we have

f ¨ (t) −7 ⋅ 30e−7t = = −7. f (t) 30e−7t

32. Since f ¨ (t) = 5 ⋅ 8e5t , we have

33. Since f ¨ (t) = −4 ⋅ 35t−5 , we have

f ¨ (t) 5 ⋅ 8e5t = = 5. f (t) 8e5t f ¨ (t) −4 ⋅ 35t−5 4 = = −4t−1 = − . t f (t) 35t−4

34. Since f ¨ (t) = 2 ⋅ 6t = 12t, we have

f ¨ (t) 12t 2 = 2 = . f (t) t 6t

35. f (t) = 4 − 2et , so f ¨ (t) = −2et and f ¨ (−1) = −2e−1 ≈ −0.736. f ¨ (0) = −2e0 = −2. f ¨ (1) = −2e1 ≈ −5.437. As expected, the slopes of the line segments do match the derivatives found. See Figure 3.6. y y = 4 − 2et 1

−1

Figure 3.6

36. (a) The slope of f appears to be positive at x = 3, so f ¨ (3) > 0. On the other hand, the slopes of f are negative at both x = 1 and x = 2, with a steeper negative slope at x = 2. Therefore, we must have f ¨ (2) < f ¨ (1) < f ¨ (3). (b) The derivative function of f is given by 15 3∕2 x . 2 ¨ ¨ Evaluating at x = 1, 2, 3, we have f (2) = −6.435, f (1) = −2.063 and f ¨ (3) = 1.200 which confrms the ordering of the three quantities. f ¨ (x) = 2ex −

3.2 SOLUTIONS

201

37. See Figure 3.7. The slope of the line tangent to the graph of the function at x = 1 will be the derivative of the function dy = (ln 3)3x . At x = 1, the derivative is 3 ln 3 ≈ 3.3; this is the slope of the line tangent evaluated at x = 1. Since y = 3x , dx to the curve at x = 1. The function evaluated at x = 1 yields the point (1, 3). Thus the equation of the line is y − 3 = 3.3(x − 1) y = 3.3x − 0.3 Thus the equation of the line is y = 3.3x − 0.3. y 9 8 y = 3x

7 6 5 4 3 2

y = 3.3x − 0.3

x −3

−2

−1

Figure 3.7 38. y = e−2t , y¨ = −2e−2t . At t = 0, y = 1 and y¨ = −2. Thus the tangent line at (0, 1) is y = −2t + 1. y

1 y = e−2t t y = −2t + 1

Figure 3.8 39. To fnd the equation of the line, we need a point and the slope. Since f (4) = 10e−0.2(4) = 4.493, the point on the line is (4, 4.493). We use the derivative to fnd the slope: f ¨ (x) = 10e−0.2x (−0.2) = −2e−0.2x . Substituting x = 4, we see that the slope is m = f ¨ (4) = −2e−0.2(4) = −0.899. Using y − y0 = m(x − x0 ), we fnd that the equation for the tangent line is: y − 4.493 = −0.899(x − 4) Simplifying, we have y = −0.899x + 8.089.

202

Chapter Three /SOLUTIONS

40. (a) P (12) = 10e0.6(12) = 10e7.2 ≈ 13,394 fsh. There are 13,394 fsh in the area after 12 months. (b) We di˙erentiate to fnd P ¨ (t), and then substitute in to fnd P ¨ (12): P ¨ (t) = 10(e0.6t )(0.6) = 6e0.6t P ¨ (12) = 6e0.6(12) ≈ 8037 fsh/month. The population is growing at a rate of approximately 8037 fsh per month. 41. (a) We have f ¨ (t) = 1000e0.03t ⋅ 0.03 = 30e0.03t . (b) Substituting t = 10, we have f (10) = 1000e0.03(10) = 1349.86 dollars, while f ¨ (10) = 30e0.03(10) = 40.50 dollars per year. Ten years after the money is deposited, the balance in the account is $1349.86, and the balance is increasing at a rate of 40.50 $/year. 42. (a) We have f ¨ (t) = 120e−0.17t ⋅ (−0.17) = −20.4e−0.17t . (b) Substituting t = 2, we have f (2) = 120e−0.17(2) = 85.412 mg, while f ¨ (2) = −20.4e−0.17(2) = −14.520 mg per hour. Two hours after drinking the co˙ee, the amount of ca˙eine in the body is 85.412 mg and the amount is decreasing at a rate of 14.520 mg/hour. 43. We have f (0) = 7.41 and f (10) = 7.41e0.011(10) = 8.27. The derivative of f (t) is f ¨ (t) = 7.41e0.011t ⋅ 0.011 = 0.08151e0.011t , so f ¨ (0) = 0.08151 and f ¨ (10) = 0.091. These values tell us that in July 2017 (at t = 0), the population of the world was 7.41 billion people and the population was growing at a rate of 0.08151 billion people per year. In July 2027 (at t = 10), this model predicts that the population of the world will be 8.27 billion people and growing at a rate of 0.091 billion people per year. 44. f (p) = 10, 000e−0.25p , f (2) = 10, 000e−0.5 = 6065. If the product sells for $2, then 6065 units can be sold. f ¨ (p) = 10, 000e−0.25p (−0.25) = −2500e−0.25p f ¨ (2) = −2500e−0.5 = −1516. Thus, at a price of $2, a $1 increase in price results in a decrease in quantity sold of about 1516 units . 45. (a) Substituting t = 4 gives V (4) = 30(0.85)4 = 30(0.522) = 15.660. Thus the value of the car after 4 years is $15,660. (b) We have a function of the form f (t) = Cat . We know that such functions have a derivative of the form (C ln a) ⋅ at . Thus, V ¨ (t) = (30 ln 0.85) ⋅ (0.85)t = −4.876(0.85)t . The units are the change in value (in thousands of dollars) with respect to time (in years), or thousands of dollars/year. (c) Substituting t = 4 gives V ¨ (4) = −4.876(0.85)4 = −4.876(0.522) = −2.545. This means that at the end of the fourth year, the value of the car is decreasing by $2545 per year. (d) The function V (t) is positive and decreasing, so that the value of the automobile is positive and decreasing. The function V ¨ (t) is negative, and its magnitude is decreasing, meaning the value of the automobile is always dropping, but the yearly loss of value decreases as time goes on. The graphs of V (t) and V ¨ (t) confrm that the value of the car decreases with time. What they do not take into account are the costs associated with owning the vehicle. At some time, t, it is likely that the yearly costs of owning the vehicle will outweigh its value. At that time, it may no longer be worthwhile to keep the car. 46. When t = 5, we have R = 350 ln 5 = $563.30. Since f ¨ (t) = 350∕t, we have f ¨ (5) = 350∕5 = 70 dollars per week. Then Relative rate of change =

f ¨ (5) 70 = = 0.124 = 12.4% per week. f (5) 563.30

Five weeks after the game was released, the revenue from downloads is $563.30, and is increasing at a continuous rate of $70 per week, or 12.4% per week. 47. (a) Since July 2014 corresponds to t = 0, July 2025 corresponds to t = 11. We have P = 9.919(0.998)11 = 9.703 million. (b) Di˙erentiating, we have

dP = 9.919(ln 0.998)(0.998)t dt dP || = 9.919(ln 0.998)(0.998)11 = −0.0194 million/year. | dt ||t=11

Thus in July 2025, Hungary’s population is predicted to be decreasing by about 19,400 people per year.

3.2 SOLUTIONS

203

48. Since P = 35,000(0.98)t , the rate of change of the population is given by dP = 35,000 ⋅ (ln 0.98)(0.98t ). dt On January 1, 2033, we have t = 13. At t = 13, the rate of change is 35,000(ln 0.98)(0.9813 ) = −544 people/year. The negative sign indicates that the population is decreasing. 49. Since P = 1 ⋅ (1.02)t , dP = ln(1.02)1.02t . When t = 10, dt dP = (ln 1.02)(1.02)10 ≈ $0.02414∕year ≈ 2.41c/∕year. dt 50. Since y = 2x , y¨ = (ln 2)2x . At (0, 1), the tangent line has slope ln 2 so its equation is y = (ln 2)x + 1. At c, y = 0, so 0 = (ln 2)c + 1, thus c = − ln12 . 51. The concentration of the drug in the body after 4 hours is f (4) = 27e−0.14(4) = 15.4 ng/ml. The rate of change of the concentration is the derivative f ¨ (t) = 27e−0.14t (−0.14) = −3.78e−0.14t . At t = 4, the concentration is changing at a rate of f ¨ (4) = −3.78e−0.14(4) = −2.16 ng/ml per hour.

52. C(q) = 1000 + 30e0.05q C(50) = 1000 + 30e2.5 ≈ 1365 so it costs about $1365 to produce 50 units. C ¨ (q) = 30(0.05)e0.05q = 1.5e0.05q C ¨ (50) = 1.5e2.5 ≈ 18.27 It costs about $18.27 to produce an additional unit when the production level is 50 units. 53. (a) We frst substitute A0 = 185, di˙erentiate A(t), and substitute t = 500. A(t) = 185e−0.000121t A¨ (t) = 185e−0.000121t (−0.000121) = −0.022385e−0.000121t A¨ (500) = −0.022385e−0.000121⋅500 = −0.021. This tell us that when t = 500, the quantity of carbon-14 in the tree is decaying by approximately 0.021 micrograms per year. (b) We are told that A(t) = 0.91A0 , so 0.91A0 = A0 e−0.000121t 0.91 = e−0.000121t ln 0.91 = −0.000121t ln 0.91 t=− = 779.4. 0.000121 The shroud was approximately 779.4 years old in 1988. 300 = 0.6. , C ¨ (500) = 500 54. C(500) = 1000 + 300 ln(500) ≈ 2864.38; it costs about $2864 to produce 500 units. C ¨ (q) = 300 q When the production level is 500, each additional unit costs about $0.60 to produce.

204

Chapter Three /SOLUTIONS

55. (a) For t in years since 2017, the population of Mexico is given by the formula M = 125(1 + 0.0112)t = 125(1.0112)t million. The rate of change of the Mexican population in 2017, in people/year, is given by | | dM || d | | | = 125 (1.0112)t | = 125(1.0112)t ln(1.0112)| = 1.392 million people per year. | | | dt |t=0 dt |t=0 |t=0 (b) The population of the US is given by U = 327(1 + 0.0081)t = 327(1.0081)t million. The rate of change of the US population in 2017, in people/year, is given by | | dU || d | | | = 327 (1.0081)t | = 327(1.0081)t ln(1.0081)| = 2.638 million people per year. | | dt ||t=0 dt |t=0 |t=0 Since

dU || dM || | > | , the population of the US was growing faster in 2017. dt ||t=0 dt ||t=0

56. (a) Since the initial population (at t = 0) is 1.339 and the growth rate is 1.04%, we have P = 1.339(1 + 0.0104)t = 1.339(1.0104)t billion. (b) Di˙erentiating gives dP d = 1.339 (1.0104)t = 1.339(1.0104)t (ln 1.0104). dt dt dP || | = 1.339(1.0104)0 ln 1.0104 = 0.014 billion people per year. dt ||t=0 dP || = 1.339(1.0104)25 ln 1.0104 = 0.018 billion people per year. | dt ||t=25 dP dP || dP || is the rate of growth of India’s population; are the rates of growth in the years | and | | dt dt |t=0 dt ||t=25 2021 and 2046, respectively.

The derivative

57. (a) We model P in the form P = P0 (1.062)t where t is the number of days after the cumulative number of cases reached P0 . We have dP = P0 (1.062)t ln(1.062) = 0.06P0 (1.062)t . dt Thus, the number of new cases per day is also an exponentially increasing function. (b) The number of new cases per day increases at the same rate as P , that is 6.2% per day. 58. (a) For y = ln x, we have y¨ = 1∕x, so the slope of the tangent line is f ¨ (1) = 1∕1 = 1. The equation of the tangent line is y − 0 = 1(x − 1), so, on the tangent line, y = g(x) = x − 1. (b) Using a value on the tangent line to approximate ln(1, 1), we have ln(1.1) ≈ g(1.1) = 1.1 − 1 = 0.1. Similarly, ln(2) is approximated by ln(2) ≈ g(2) = 2 − 1 = 1. (c) From Figure 3.9, we see that f (1.1) and f (2) are below g(x) = x − 1. Similarly, f (0.9) and f (0.5) are also below g(x). This is true for any approximation of this function by a tangent line since f is concave down (f ¨¨ (x) = − x12 < 0 for all x > 0). Thus, for a given x-value, the y-value given by the function is always below the value given by the tangent line.

3.2 SOLUTIONS

205

g(x) = x − 1 4 f (x) = ln x x 2

−4

Figure 3.9 59. (a) Let R = 21(1.05)t . Then the annual extraction is changing at a rate of � dR d � = 21 1.05t = 21 ⋅ ln 1.05 1.05t million tonnes per year. dt dt (b) At the start of 2025, we have t = 7, so � dR || | = 21 ⋅ ln 1.05 1.057 = 1.4417 million tonnes per year. dt ||t=7 (c) If the extraction rate continues to change at 1.4417 million tonnes per year, in fve years the extraction rate will have increased by 5(1.4417) = 7.209 million tonnes. (d) Since the graph of R = 21(1.05)t is concave up, 7.209 million tonnes is smaller than the predicted increase from the model. In fact, the model predicts the increase to be R(12) − R(7) = 21(1.05)12 − 21(1.05)7 = 8.16 million tonnes. 60. (a) Since N(t) = 931(1.145)t , the number is changing at a rate of N ¨ (t) = 931

� d � 1.145t = 931 ⋅ ln 1.145 1.145t = 126.06(1.145)t thousand parcels per year. dt

(b) Since the rate, N ¨ (t), is an exponential function with base 1.145 > 1, it is increasing with time. 61. g(x) = ax2 + bx + c

f (x) = ex

g ¨ (x) = 2ax + b

f ¨ (x) = ex

¨¨

f ¨¨ (x) = ex

g (x) = 2a

So, using g ¨¨ (0) = f ¨¨ (0), etc., we have 2a = 1, b = 1, and c = 1, and thus g(x) = 12 x2 +x+1, as shown in Figure 3.10. ex 1 2 x +x+1 2

Figure 3.10 The two functions do look very much alike near x = 0. They both increase for large values of x, but ex increases much more quickly. For very negative values of x, the quadratic goes to ∞ whereas the exponential goes to 0. By choosing a function whose frst few derivatives agreed with the exponential when x = 0, we got a function which looks like the exponential for x-values near 0.

206

Chapter Three /SOLUTIONS

62. (a) We use a negative sign for the rate 1.77 since the number of restaurant workers is decreasing. Thus, the number of open restaurants R, in thousands, is given by R = 647e−1.77t where t is in years since the start of 2020. (b) The number of restaurant workers W , in millions, is given by W = 15.6e−1.52t . (c) The derivative of R = 647e−1.77t is dR d � = 647e−1.77t = 647e−1.77t (−1.77) = −1145.19e−1.77t . dt dt The year 2021 corresponds to t = 1 and the value of the derivative at this point is dR = −1145.19e−1.77(1) = −195.0636 thousand restaurants per year. dt This tells us that restaurants were being closed at a rate of approximately 195 thousand restaurants per year at the start of 2021. (d) Similarly, dW d � = 15.6e−1.52t = 15.6e−1.52t (−1.52) = −23.712e−1.52t . dt dt When t = 1 we have dW = −23.712e−1.52(1) = −5.1861 million workers per year. dt This tells us that restaurant workers were being laid o˙ a rate of approximately 5 million workers per year at the start of 2021. (e) The average number of workers per restaurant at the start of 2020 is 15,600,000 workers 15.6 million workers = = 24.11 workers/restaurant. 647 thousand restaurants 647,000 restaurants Thus, at the start of 2020 there were approximately 24 workers per restaurant. Regarding the average number of workers per restaurant as time elapsed, note that 1.77 > 1.52 tells us that decay rate of open restaurants is higher than the decay rate of the number of workers (this is the same as noticing −1.77 < −1.52). In other words, we know workers are getting laid o˙ slower than restaurants are closing, so the average is expected to increase with time. To see this algebraically we write the average, given by W ∕R, in terms of t: W 15.6e−1.52t = R 647e−1.77t = 0.0241e−1.52t−(−1.77t) = 0.0241e0.25t million workers per thousand restaurants. Since W ∕R = 0.0241e0.25t is an increasing function of t, this confrms that the average is increasing with time. 63. (a) Using the chain rule and the fact that P0 is constant, we have P¨ =

d d d (P ekt ) = P0 (ekt ) = P0 ekt (kt) = P0 ekt ⋅ k = kP0 ekt . dt 0 dt dt

(b) Using the properties of the natural logarithm, we have ln P ¨ = ln(kP0 ekt ) = ln(kP0 ) + ln(ekt ) = ln(kP0 ) + kt. Thus, y = ln(P ¨ ) is of the form y = b + mt, so it is a line. (c) The slope of y = ln(P ¨ ) = ln(kP0 ) + kt is k. (d) The vertical intercept of y = ln(P ) = ln(kP0 ) + kt is ln(kP0 ). 64. (a) From Problem 63 we know the slope of the trend line is the continuous growth rate k. We use the points (t, ln(P ¨ (t))) = (20, 9.1) and (t, ln(P ¨ (t))) = (50, 9.5) to estimate the slope: k=

9.5 − 9.1 = 0.0133. 50 − 20

This tells us that daily new cases in Florida were growing approximately exponentially with a continuous growth rate k = 0.0133 = 1.33% per day.

3.3 SOLUTIONS

207

(b) We estimate the vertical intercept to be 8.8, so 8.8 = ln(P ¨ (0)) which gives P ¨ (0) = e8.8 = 6634.244 ≈ 6634 cases per day. (c) Using parts (a) and (b), we know the slope of the trend line of the natural log of P ¨ (t)—that is, ln(P ¨ (t))—is 0.0133 and the vertical intercept is 8.8, giving ln(P ¨ (t)) = 8.8 + 0.0133t. Taking e to the power of both sides gives ¨

eln(P (t)) = e8.8+0.0133t P ¨ (t) = e8.8 ⋅ e0.0133t P ¨ (t) = 6634e0.0133t . Thus, P ¨ (t) ≈ 6634e0.0133t .

Solutions for Section 3.3 1. f ¨ (x) = 99(x + 1)98 ⋅ 1 = 99(x + 1)98 . d � 2 d 2. g ¨ (x) = (4x + 1)7 = 7(4x2 + 1)6 (4x2 + 1) = 7(4x2 + 1)6 ⋅ 8x = 56x(4x2 + 1)6 . dx dx 3. w¨ = 100(t2 + 1)99 (2t) = 200t(t2 + 1)99 . dR 4. = 4(q 2 + 1)3 ⋅ 2q = 8q(q 2 + 1)3 . dq dw 5. = 3(5r − 6)2 ⋅ 5 = 15(5r − 6)2 . dr 6. If we let z = x3 + x2 then dz∕dx = 3x2 + 2x so that df dz d −90 (z ) = −90z−90−1 = = −90z−91 (3x2 + 2x) = −90(x3 + x2 )−91 (3x2 + 2x). dx dx dx � dy = −3(2x) + 2 3e3x = −6x + 6e3x . dx 3s2 . 8. y¨ = √ 2 s3 + 1 7.

9. f ¨ (x) = 6(e5x )(5) + (e−x )(−2x) = 30e5x − 2xe−x . dC 10. By the chain rule, = 12 ⋅ 3(3q 2 − 5)2 ⋅ 6q = 216q(3q 2 − 5)2 . dq 11. Using the chain rule, we have 2 dw = −6te−3t . dt dy = 5(5e5t+1 ) = 25e5t+1 . dt dy 5 13. = . dt 5t + 1 √ 1 14. w¨ = √ e s . 2 s 2t ¨ . 15. f (t) = 2 t +1 −1 1 16. f ¨ (x) = = . 1−x x−1 1 17. f ¨ (x) = x ⋅ ex . e +1 e−x 1 ⋅ (−e−x )(−1) = . 18. f ¨ (x) = 1 − e−x 1 − e−x 12.

208

Chapter Three /SOLUTIONS

19. If we let z = ln x then dz∕dx = 1∕x so that df d 1 dz 11 1 = (ln z) = = = . dx dx z dx zx x ln x 20. If we let z = ln x then dz∕dx = 1∕x so that 3(ln x)2 df dz 1 d 3 = (z ) = 3z2 = 3z2 = . dx dx dx x x dy 1 3 = ⋅3= . dx 3t + 2 3t + 2 dy = 2(5 + ex )ex . 22. dx dy 1 1 23. =5+ ⋅1 =5+ . dx x+2 x+2 dy d ex d √ x d x 1 = ( e + 1) = (e + 1)1∕2 = (ex + 1)−1∕2 (ex + 1) = √ . 24. dx dx dx 2 dx 2 ex + 1 dP 1 0.5 25. = 0.5(1 + ln x)−0.5 = . dx x x(1 + ln x)0.5 26. The power and chain rules give 21.

f ¨ ( ) = −1(e + e− )−2 ⋅ 27. If we let z = 2 +

d (e + e− ) = −(e + e− )−2 (e + e− (−1)) = − d

e − e− (e + e− )2

1 .

√ √ x then dz∕dx = 1∕(2 x) so that df d √ 1 dz 1 1 1 = ( z) = √ = √ √ = . t dx dx 2 z dx 2 z 2 x 4√x 2 + √x

28. We have f ¨ (t) = 2e0.3t (0.3), so the relative rate of change is f ¨ (t) 2e0.3t (0.3) = = 0.3 = 30% per year. f (t) 2e0.3t In particular, since the answer does not depend on the value of t, the relative rate of change at any value of t, including t = 7, is 30% per year. 29. We have f (2) = ln (22 + 1) = ln(5) = 1.609. We have f ¨ (t) = (2t)∕(t2 + 1), so f ¨ (2) = 4∕5 = 0.8. The relative rate of change at t = 2 is f ¨ (2) 0.8 = = 0.497 = 49.7% per year. f (2) 1.609 30. Since ln f (t) = ln(6.8e−0.5t ) = ln 6.8 − 0.5t we have d d ln f (t) = (ln 6.8 − 0.5t) = 0 − 0.5 = −0.5. dt dt 31. Since ln f (t) = ln(5e1.5t ) = ln 5 + 1.5t we have d d ln f (t) = (ln 5 + 1.5t) = 0 + 1.5 = 1.5. dt dt 32. Since ln f (t) = ln 4.5t−4 = ln 4.5 − 4 ln t we have d d 4 4 ln f (t) = (ln 4.5 − 4 ln t) = 0 − = − . dt dt t t

3.3 SOLUTIONS

209

33. Since ln f (t) = ln 3t2 = ln 3 + 2 ln t we have d d 2 2 ln f (t) = (ln 3 + 2 ln t) = 0 + = . dt dt t t 34. (a) The y-coordinate where x = 0 is f (0) = (2 ⋅ 0 − 3)3 = (−3)3 = −27, so a point on the line is (0, −27). Using the chain rule, we fnd f ¨ (x) = 3(2x − 3)2 (2) = 6(2x − 3)2 , so the slope of the line is f ¨ (0) = 6(−3)2 = 54. The equation of the tangent line at x = 0 is y = 54x − 27. (b) The y-coordinate where x = 2 is f (2) = (1)3 = 1, so a point on the line is (2, 1). We have f ¨ (x) = 6(2x − 3)2 , so the slope of the line is f ¨ (2) = 6(1)2 = 6. Using the point-slope equation for a line, we have y − 1 = 6(x − 2) y = 6x − 11. The equation of the tangent line at x = 2 is y = 6x − 11. 35. We have f (2) = (2 − 1)3 = 1, so (2, 1) is a point on the tangent line. Since f ¨ (x) = 3(x − 1)2 , the slope of the tangent line is m = f ¨ (2) = 3(2 − 1)2 = 3. The equation of the line is y − 1 = 3(x − 2)

y = 3x − 5.

dB dB r t r tells us how fast the amount of money in the bank is changing =P 1+ ln 1 + . The expression dt dt 100 100 with respect to time for fxed initial investment P and interest rate r. dB r t−1 1 dB (b) indicates how fast the amount of money changes with respect to = Pt 1 + . The expression dr 100 100 dr the interest rate r , assuming fxed initial investment P and time t.

36. (a)

37. (a) Since B = 13.525e0.052t , an exponential function with rate 0.052 > 0, oil consumption is increasing with time. (b) Oil consumption is changing at a rate given by dB d � 0.052t = 13.525 e = 13.525 ⋅ (0.052)e0.052t = 0.7033e0.052t million barrels per day per year. dt dt The rate is positive, as oil consumption is increasing. 38. (a) Since H = 5 + 95e−0.054t , the temperature is decreasing with time as the quantity e−0.054t = 1∕e0.054t decreases with time. (b) The temperature is changing at a rate given by dH d � −0.054t = 95 e = 95 ⋅ (−0.054)e−0.054t = −5.130e−0.054t degrees Celsius per min. dt dt The rate is negative, as the temperature is decreasing. 39. Using the Chain Rule, we have f ¨ (t) = 50

t∕30 t∕30 1 d t 1 1 ln(1∕2) ⋅ = 50 ln(1∕2) ⋅ 2 dt 30 2 30 t∕30 5 1 = ln(1∕2) . 3 2

Therefore, since

5∕30 5 1 ln(1∕2) ≈ −1.029, 3 2 the substance is decaying about 1.029 grams per year at t = 5 years. f ¨ (5) =

40. If the distance s(t) = 20et∕2 , then the velocity, v(t), is given by ¨ 1 � t∕2 � v(t) = s¨ (t) = 20et∕2 = 20e = 10et∕2 . 2 41. Estimates may vary. From the graphs, we estimate g(1) ≈ 2, g ¨ (1) ≈ 1, and f ¨ (2) ≈ 0.8. Thus, by the chain rule, ℎ¨ (1) = f ¨ (g(1)) ⋅ g ¨ (1) ≈ f ¨ (2) ⋅ g ¨ (1) ≈ 0.8 ⋅ 1 = 0.8.

210

Chapter Three /SOLUTIONS

42. Estimates may vary. From the graphs, we estimate f (1) ≈ −0.4, f ¨ (1) ≈ 0.5, and g ¨ (−0.4) ≈ 2. Thus, by the chain rule, k¨ (1) = g ¨ (f (1)) ⋅ f ¨ (1) ≈ g ¨ (−0.4) ⋅ 0.5 ≈ 2 ⋅ 0.5 = 1. 43. Estimates may vary. From the graphs, we estimate g(2) ≈ 1.6, g ¨ (2) ≈ −0.5, and f ¨ (1.6) ≈ 0.8. Thus, by the chain rule, ℎ¨ (2) = f ¨ (g(2)) ⋅ g ¨ (2) ≈ f ¨ (1.6) ⋅ g ¨ (2) ≈ 0.8(−0.5) = −0.4. 44. Estimates may vary. From the graphs, we estimate f (2) ≈ 0.3, f ¨ (2) ≈ 1.1, and g ¨ (0.3) ≈ 1.7. Thus, by the chain rule, k¨ (2) = g ¨ (f (2)) ⋅ f ¨ (2) ≈ g ¨ (0.3) ⋅ f ¨ (2) ≈ 1.7 ⋅ 1.1 ≈ 1.9. 45. The chain rule gives | d 1 1 | f (g(x))| = f ¨ (g(30))g ¨ (30) = f ¨ (55)g ¨ (30) = (1)( ) = . | 2 2 dx |x=30 46. The chain rule gives | d | f (g(x))| = f ¨ (g(70))g ¨ (70) = f ¨ (60)g ¨ (70) = (1)(0) = 0. | dx |x=70 47. The chain rule gives | d | g(f (x))| = g ¨ (f (30))f ¨ (30) = g ¨ (20)f ¨ (30) = (1∕2)(−2) = −1. | dx |x=30 48. The chain rule gives | d 1 1 | = g ¨ (f (70))f ¨ (70) = g ¨ (30)f ¨ (70) = (1)( ) = . g(f (x))| | 2 2 dx |x=70 49. The chain rule gives | d | = f ¨ (g(20))g ¨ (20) = f ¨ (50)g ¨ (20) = (1)(0.5) = 0.5. f (g(x))| | dx |x=20 50. The chain rule gives | d | g(f (x))| = g ¨ (f (60))f ¨ (60) = g ¨ (20)f ¨ (60) = (0.5)(1) = 0.5. | dx |x=60 51. (a) Di˙erentiating g(x) =

√ f (x) = (f (x))1∕2 , we have g ¨ (x) =

f ¨ (x) 1 (f (x))−1∕2 ⋅ f ¨ (x) = √ 2 2 f (x)

f ¨ (1) 3 3 g ¨ (1) = √ = √ = . 4 2 f (1) 2 4 √ (b) Di˙erentiating ℎ(x) = f ( x), we have √ 1 ℎ¨ (x) = f ¨ ( x) ⋅ √ 2 x √ f ¨ (1) 3 1 ℎ¨ (1) = f ¨ ( 1) ⋅ √ = = . 2 2 2 1 52. The marginal revenue, MR, is given by di˙erentiating the total revenue function, R. We use the chain rule so MR =

dR 1 d � 1 1000q 2 = = ⋅ ⋅ 2000q. dq 1 + 1000q 2 dq 1 + 1000q 2

When q = 10, 2000 ⋅ 10 = $0.20/item. 1 + 1000 ⋅ 102 When 10 items are produced, each additional item produced gives approximately $0.20 in additional revenue. Marginal revenue =

3.4 SOLUTIONS 53. (a) We have

f (x) = 32.7 ln

x 244.5

211

= 32.7 ln(x) − 32.7 ln(244.5)

so f ¨ (x) =

32.7 . x

(b) We have 32.7 = 0.01635 mm of leaf width per mm of rain. 2000 (c) We can approximate this di˙erence using the derivative f ¨ (2000) as follows: f ¨ (2000) =

Δw ≈ f ¨ (2000)Δx = (0.01635)(150) = 2.4525 mm. The average leaf width in the rainier forest is about 2.5 mm greater. To check, we can calculate this di˙erence exactly; it is Δw = f (2150) − f (2000) = 32.7 ln(2150∕244.5) − 32.7 ln(2000∕244.5) = 2.365 mm. 54. (a) With four additional year of education, your wages are 20(1.12)4 = 31.47 dollars per hour. (b) In 20 years time, Wages without additional education = 20e0.035(20) = 40.28 dollars/hour. Wages with additional education = 31.47e0.035(20) = 63.37 dollars/hour. Thus, the di˙erence is 63.37 − 40.28 = 23.09 dollars per hour. (c) The di˙erence has increased from $11.47 now (using the answer to part (a)) to $23.09 in 20 years time. Thus, the di˙erence is increasing. To calculate the rate, we frst fnd a formula for the di˙erence. Wages in year t without additional education = 20e0.035t dollars/hour. Wages in year t with additional education = 31.47e0.035t dollars/hour. Di˙erence = 31.47e0.035t − 20e0.035t = 11.47e0.035t dollars/hour. Di˙erentiating to fnd the rate of change, we have d (11.47e0.035t ) = 11.47(0.035)e0.035t = 0.40145e0.035t dollars per hour per year. dt When t = 20, Rate = 0.40145e0.035(20) = 0.808 dollars per hour per year. Thus, during your twentieth year of working, your wage increases about $0.80 per hour more because of the four additional years of education. 55. Since the graphs of f (t) and ℎ(t) are tangent at t = a, we have f (a) = ℎ(a) and f ¨ (a) = ℎ¨ (a). From ℎ(t) = Aekt we get Relative rate of change of f =

f ¨ (a) ℎ¨ (a) kAekt = = = k. f (a) ℎ(a) Aekt

Thus the relative rate of change of f at a point equals the continuous rate of change of the exponential function that is tangent to f at that point.

Solutions for Section 3.4 1. By the product rule, f ¨ (x) = 2x(x3 + 5) + x2 (3x2 ) = 2x4 + 3x4 + 10x = 5x4 + 10x. Alternatively, f ¨ (x) = (x5 + 5x2 )¨ = 5x4 + 10x. The two answers should, and do, match.

212

Chapter Three /SOLUTIONS

2. By the product rule, f ¨ (x) = 2(3x − 2) + (2x + 1) ⋅ 3 = 12x − 1. Alternatively, f ¨ (x) = (6x2 − x − 2)¨ = 12x − 1. The two answers match. 3. f ¨ (t) = (1)e−2t + t(−2e−2t ) = e−2t − 2te−2t . 4. f ¨ (x) = x ⋅ ex + ex ⋅ 1 = ex (x + 1). 5. Di˙erentiating with respect to t, we have dy d d d 2 = (t2 (3t + 1)3 ) = (t ) (3t + 1)3 + t2 ((3t + 1)3 ) dt dt dt dt = (2t)(3t + 1)3 + t2 (3(3t + 1)2 ⋅ 3) = 2t(3t + 1)3 + 9t2 (3t + 1)2

6. Di˙erentiating with respect to x, we have 2 dy 2 2 d d d = (5xex ) = (5x) ex + 5x (ex ) dx dx dx dx 2 2 = (5)ex + 5x(ex ⋅ 2x) 2

= 5ex + 10x2 ex .

7. Di˙erentiating with respect to x, we have dy d d d = (x ln x) = (x) ln x + x (ln x) dx dx dx dx 1 = 1 ⋅ ln x + x ⋅ x = ln x + 1. dy = 2tet + (t2 + 3)et = et (t2 + 2t + 3). dt 9. y¨ = (3t2 − 14t)et + (t3 − 7t2 + 1)et = (t3 − 4t2 − 14t + 1)et . dz 10. = (3t + 1)5 + 3(5t + 2) = 15t + 5 + 15t + 6 = 30t + 11. dt We could have started by multiplying the factors to obtain 15t2 + 11t + 2, and then taken the derivative of the result. dR 11. = (3q)(−e−q ) + (e−q )(3) = −3qe−q + 3e−q . dq dP 1 12. = (t2 )( ) + (2t)(ln t) = t + 2t ln t. dt t 13. Divide and then di˙erentiate 3 f (x) = x + x 3 f ¨ (x) = 1 − 2 . x 8.

√ 1 14. f ¨ (z) = √ e−z − ze−z . 2 z 2

15. y¨ = 1 ⋅ e−t + te−t (−2t) 16. f ¨ (t) = 1 ⋅ e5−2t + te5−2t (−2) = e5−2t (1 − 2t). 0 1 2p 2 17. g ¨ (p) = p + ln(2p + 1)(1) = + ln(2p + 1). 2p + 1 2p + 1

3.4 SOLUTIONS 18. y¨ = 2x + x(ln 2)2x = 2x (1 + x ln 2). 19. 2

f ¨ (w) = (ew )(10w) + (5w2 + 3)(ew )(2w) w2

= 2we (5 + 5w2 + 3) 2

= 2wew (5w2 + 8). 20. w¨ = (3t2 + 5)(t2 − 7t + 2) + (t3 + 5t)(2t − 7). 21. Using the product and chain rules, we have dz d = 9(te3t + e5t )8 ⋅ (te3t + e5t ) = 9(te3t + e5t )8 (1 ⋅ e3t + t ⋅ e3t ⋅ 3 + e5t ⋅ 5) dt dt = 9(te3t + e5t )8 (e3t + 3te3t + 5e5t ). ex (1 − x) 1−x ex ⋅ 1 − x ⋅ ex = = . ex (ex )2 (ex )2 23. Using the quotient rule gives

22. f ¨ (x) =

dz d = dt dt

1−t 1+t

−1 ⋅ (1 + t) − (1 − t) ⋅ 1 −1 − t − 1 + t −2 = = . (1 + t)2 (1 + t)2 (1 + t)2

24. Using the quotient rule, we have dw d = dz dz

3z 1 + 2z

3(1 + 2z) − 3z(2) 3 = . (1 + 2z)2 (1 + 2z)2

25. Using the quotient rule, dw d = dy dy

3y + y2 5+y

1 =

(3 + 2y)(5 + y) − (3y + y2 ) ⋅ 1 (5 + y)2

15 + 13y + 2y2 − 3y − y2 15 + 10y + y2 = . 2 (5 + y) (5 + y)2

26. Using the quotient rule, x ex (1 + ex ) − ex (ex ) dy ex + e2x − e2x ex d e = = = = . dx dx 1 + ex (1 + ex )2 (1 + ex )2 (1 + ex )2 27. We use the quotient rule. We have f ¨ (x) =

(cx + k)a − (ax + b)c acx + ak − acx − bc ak − bc = = . (cx + k)2 (cx + k)2 (cx + k)2

28. Using the quotient rule, we have 1 ⋅ ln z − (1 + z)(1∕z) dy d 1+z z ln z − 1 − z = = = . dz dz ln z (ln z)2 z(ln z)2 29. We use the product rule. We have f ¨ (x) = (ax)(e−bx (−b)) + (a)(e−bx ) = −abxe−bx + ae−bx . 30. Since a and b are constants, we have f ¨ (x) = 3(ax2 + b)2 (2ax) = 6ax(ax2 + b)2 .

213

214

Chapter Three /SOLUTIONS

31. Using the product and chain rules, we have −2 d ( e−2 ) = e e (1 ⋅ e−2 + e−2 (−2)) dx −2 = e e (e−2 − 2 e−2 ) −2

g¨( ) = e e

⋅

−2

= (1 − 2 )e−2 e e

32. Since a and b are constants, we have f ¨ (t) = aebt (b) = abebt . 33. We multiply out f (x) = (3x + 8)(2x − 5) = 6x2 + x − 40, and di˙erentiate term by term to get f ¨ (x) = 12x + 1

and

f ¨¨ (x) = 12.

34. Since f (0) = −5∕1 = −5, the tangent line passes through the point (0, −5), so its vertical intercept is −5. To fnd the slope of the tangent line, we fnd the derivative of f (x) using the quotient rule: f ¨ (x) =

(x + 1) ⋅ 2 − (2x − 5) ⋅ 1 7 = . (x + 1)2 (x + 1)2

At x = 0, the slope of the tangent line is m = f ¨ (0) = 7. The equation of the tangent line is y = 7x − 5. 35. (a) We have f (0) = 1 and

1 1 ⋅ (−1) = , (1 − x)2 (1 − x)2 so f ¨ (0) = 1. Thus, the tangent line approximation is f ¨ (x) = −

1 ≈1+x 1−x

for x near 0.

(b) For 1∕0.99, we take x = 0.01. Then f (0.01) =

1 1 = ≈ 1 + 0.01 = 1.01. 1 − 0.01 0.99

36. Since f (0) = 5(0)e0 = 0, the tangent line passes through the point (0, 0), so its vertical intercept is 0. To fnd the slope of the tangent line, we fnd the derivative of f (x) using the product rule: f ¨ (x) = (5x) ⋅ ex + 5 ⋅ ex . At x = 0, the slope of the tangent line is m = f ¨ (0) = 5(0)e0 + 5e0 = 5. The equation of the tangent line is y = 5x. 37. f (x) = x2 e−x , f (0) = 0 f ¨ (x) = 2xe−x +x2 e−x ⋅(−1) = e−x (2x−x2 ), so f ¨ (0) = 0. Thus the tangent line is y = 0 (the x-axis). See Figure 3.11. y y = x2 e−x

Figure 3.11

3.4 SOLUTIONS

215

38. Since f (2) = 23 ⋅ e2 = 59.112, the tangent line passes through the point (2, 59.112). To fnd the slope of the tangent line, we fnd the derivative of f (x) using the product rule: f ¨ (x) = (x3 ) ⋅ ex + (3x2 ) ⋅ ex . At x = 2, the slope of the tangent line is m = f ¨ (2) = 23 ⋅ e2 + 3 ⋅ 22 ⋅ e2 = 147.781. Using the point-slope form of a line, we have: y − 59.112 = 147.781(x − 2) y − 59.112 = 147.781x − 295.562 y = 147.781x − 236.450.

39. By the product rule,

d tf (t) = f (t) + tf ¨ (t). Thus, using the information given in the problem, we have dt f (t) + tf ¨ (t) = 1 + f (t).

Subtracting f (t) from both sides gives tf ¨ (t) = 1, so f ¨ (t) = 1∕t. 40. (a) We use the product rule to fnd f ¨ (t) and we have f ¨ (t) = 4 ⋅ e−0.08t + 4t ⋅ e−0.08t (−0.08) = 4e−0.08t − 0.32te−0.08t . (b) Substituting t = 15, we have f (15) = 4(15)e−0.08(15) = 18.072 ng/ml, while f ¨ (15) = −0.32(15)e−0.08(15) +4e−0.08(15) = −0.241 ng/ml per minute. Fifteen minutes after smoking the cigarette, the concentration of nicotine in the body is 18.072 ng/ml and the concentration is decreasing at a rate of 0.241 (ng/ml)/minute. 41. We have f (t) = 100te−0.5t so f (1) = 100(1)e−0.5 = 60.65 mg, and f (5) = 100(5)e−0.5(5) = 41.04 mg. We use the product rule to fnd f ¨ (t): f ¨ (t) = 100e−0.5t + 100t(e−0.5t (−0.5)) = 100e−0.5t − 50te−0.5t . Therefore, f ¨ (1) = 100e−0.5 − 50e−0.5 = 30.33 mg/hour, and f ¨ (5) = 100e−2.5 − 250e−2.5 = −12.31 mg/hour. One hour after the drug was administered, the quantity of drug in the body is 60.65 mg and the quantity is increasing at a rate of 30.33 mg per hour. Five hours after the drug was administered, the quantity in the body is 41.04 mg and the quantity is decreasing at a rate of 12.31 mg per hour. 42. (a) See Figure 3.12. C 210 180 150 f (t) = 20te−0.04t

120 90 60 30

t 15

Figure 3.12 Looking at the graph of C, we can see that the see that at t = 15, C is increasing. Thus, the slope of the curve at that point is positive, and so f ¨ (15) is also positive. At t = 45, the function is decreasing, i.e. the slope of the curve is negative, and thus f ¨ (45) is negative.

216

Chapter Three /SOLUTIONS (b) We begin by di˙erentiating the function: f ¨ (t) = (20t)(−0.04e−0.04t ) + (e−0.04t )(20) f ¨ (t) = e−0.04t (20 − 0.8t). At t = 30, f (30) = 20(30)e−0.04⋅(30) = 600e−1.2 ≈ 181 mg/ml f ¨ (30) = e−1.2 (20 − (0.8)(30)) = e−1.2 (−4) ≈ −1.2 mg/ml/min. These results mean the following: At t = 30, or after 30 minutes, the concentration of the drug in the body (f (30)) is about 181 mg/ml. The rate of change of the concentration (f ¨ (30)) is about −1.2 mg/ml/min, meaning that the concentration of the drug in the body is dropping by 1.2 mg/ml each minute at t = 30 minutes.

43. (a) q(10) = 5000e−0.8 ≈ 2247 units. (b) q ¨ = 5000(−0.08)e−0.08p = −400e−0.08p , q ¨ (10) = −400e−0.8 ≈ −180. This means that at a price of $10, a $1 increase in price will result in a decrease in quantity demanded by 180 units. 44. R(p) = p ⋅ q = 5000pe−0.08p R(10) = 50,000e−0.8 ≈ 22,466; revenues of about $22,466 can be expected when the selling price is $10. R¨ (p) = 5000e−0.08p + 5000p(−0.08)e−0.08p = (5000 − 400p)e−0.08p R¨ (10) = 1000e−0.8 ≈ 449; if price is increased by one dollar over $10, revenue will increase by about $449. 45. (a) R(p) = p ⋅ 1000e−0.02p = 1000pe−0.02p . (b) R¨ (p) = 1000e−0.02p + 1000pe−0.02p (−0.02) = e−0.02p (1000 − 20p) (c) R(10) = 10, 000e−0.2 ≈ 8187; you will have about 8187 dollars in revenue if you sell the product for $10. R¨ (10) = e−0.2 (1000 − 200) ≈ 655; a one dollar increase in price over $10 will generate about $655 in additional revenue. 46. (a) f (140) = 15,000 says that 15,000 skateboards are sold when the cost is $140 per board. f ¨ (140) = −100 means that if the price is increased from $140, every dollar of increase will decrease the total sales by about 100 boards. dR d d = (p ⋅ q) = (b) (p ⋅ f (p)) = f (p) + pf ¨ (p). dp dp dp So, dR || = f (140) + 140f ¨ (140) | dp ||p=140 = 15,000 + 140(−100) = 1000. (c) From (b) we see that raised by $1.

dR || = 1000 > 0. This means that the revenue will increase by about $1000 if the price is | dp ||p=140

47. (a) Even though the total cholesterol increases, it is possible for the cholesterol ratio to remain constant if the good cholesterol level increases correspondingly. (b) The diet starts on January 1 when t = 0. In this week, we know the cholesterol ratio remains unchanged, so R¨ (0) = 0, the total cholesterol is 120 mg/dl, so T (0) = 120, and the good cholesterol is 30 mg/dl and increasing at a rate of 1 mg/dl per week, so G(0) = 30 and G¨ (0) = 1. The rate of change of R(t) can be calculated using the quotient rule as R¨ (t) =

G¨ (t)T (t) − G(t)T ¨ (t) . T (t)2

Substituting in the known values for t = 0, we get 0= giving T ¨ (0) = 4 mg/dl per week.

1 ⋅ 120 − 30 ⋅ T ¨ (0) 1202

3.4 SOLUTIONS

217

48. (a) At C = 2 ⋅ 10−4 molar, the growth rate is R=

(1.35)(2) = 1.216 cell divisions per hour. 0.22 + 2

(b) Using the quotient rule we have (0.22 + C)1.35 − 1.35C(1) dR = dC (0.22 + C)2 0.297 = . (0.22 + C)2 (c) We have 0.297 dR || = 0.0603 cell divisions per hour per 10−4 molar. = dC ||C=2 (0.22 + 2)2 (d) At C = 2 we have R = 1.216 cell divisions per hour. Using a point-slope formula, we fnd the equation of the tangent line at C = 2: R ≈ 1.216 + 0.0603(C − 2) cell divisions per hour. (e) The tangent line approximation gives R = 1.216 + 0.0603(2.2 − 2) = 1.228 cell divisions per hour. The original model gives R=

(1.35)(2.2) = 1.227 cell divisions per hour. 0.22 + 2.2

The two values are very close. 49. This problem can be solved by using either the product rule or the fact that f¨ d = (ln f ) f dx

and

g¨ d = (ln g). g dx

We use the second method. The relative rate of change of f g is (f g)¨ ∕(f g), so f ¨ g¨ (f g)¨ d d d d = ln (f g) = (ln f + ln g) = (ln f ) + (ln g) = + . fg dx dx dx dx f g Thus, the relative rate of change of f g is the sum of the relative rates of change of f and g. 50. This problem can be solved by using either the quotient rule or the fact that f¨ d = (ln f ) f dx

and

g¨ d = (ln g). g dx

We use the second method. The relative rate of change of f ∕g is (f ∕g)¨ ∕(f ∕g), so 0 1 (f ∕g)¨ f f ¨ g¨ d d d d = ln = (ln f − ln g) = (ln f ) − (ln g) = − . f ∕g dx g dx dx dx f g Thus, the relative rate of change of f ∕g is the di˙erence between the relative rates of change of f and g. 51. The chain rule gives (f n )¨ = nf n−1 f ¨ . Dividing by f n yields nf n−1 f ¨ f¨ (f n )¨ = = n . fn fn f 52. (a) Since A(t) gives the total value in millions of dollars, the total value in dollars is 1,000,000A(t) and since N(t) is the number of shares in millions, the total number of shares is 1,000,000N(t). Therefore, the price per share on day t is P (t) =

1,000,000A(t) A(t) = . 1,000,000N(t) N(t)

(b) Even though the price per share decreases, more shares can be purchased to increase A(t).

218

Chapter Three /SOLUTIONS (c) On January 1, 2016, we know N(0) = 2 and A(0) = 32. Using the quotient rule, the rate of change in dollars per day of P (t) = A(t)∕N(t) when t = 0 is P ¨ (0) =

A¨ (0)N(0) − A(0)N ¨ (0) 2A¨ (0) − 32N ¨ (0) = . 4 (N(0))2

Since the value of each share is dropping at a rate of $0.23 per day, we know P ¨ (0) = −0.23 and since we want the total value A(t) not to change, we want A¨ (0) = 0. Therefore, we get −0.23 =

0 − 32N ¨ (0) 4

giving N ¨ (0) = 0.02875 million shares per day. This means 28,750 additional shares must be purchased per day in order for A(t) not to change. 53. We use the quotient rule: (k + r) ⋅ c − cr ⋅ 1 dP kc = = . dr (k + r)2 (k + r)2 The derivative dP ∕dr gives the approximate change in the size of the population given a one unit increase in the available quantity of the resource. 54. (a) With I = 0.2 and E = 0.8 we have 0.2 = 0.555 = 55.5%. 1 − (1 − 0.2)0.8 (b) The equation I = 1 means that initially there is 100% chance the lost person is not in the search area. This is unrealistic. (c) Writing O(E) = I∕(1 − (1 − I)E) as O(E) = I(1 − (1 − I)E)−1 and di˙erentiating with respect to E gives O(E) =

O¨ (E) =

I(1 − I) . (1 − (1 − I)E)2

This is positive, assuming I ≠ 1. It means that the probability of the subject being outside the search area continues to increase as more resources are poured into the feld.

Solutions for Section 3.5 dP = − sin t. dt dy 2. = 5 cos x. dx dy 3. = A cos t. dt 4. We have 1.

dy = 2t − 5 sin t. dt

dy = 5 cos x − 5. dx 6. R¨ (q) = 2q + 2 sin q. 5.

7. R¨ = 5 cos(5t). 8. f ¨ (x) = cos(3x) ⋅ 3 = 3 cos(3x). 9. Using the chain rule, we have dy = 2(− sin(5t))(5) = −10 sin(5t). dt 10.

dW = 4(− sin(t2 )) ⋅ 2t = −8t sin(t2 ). dt

3.5 SOLUTIONS

219

11. Using the chain rule, we have dy = A(cos(Bt)) ⋅ B = AB cos(Bt). dt 12. Using the chain rule, we have dy = 2x cos(x2 ). dx dy = 6 ⋅ 2 cos(2t) + (−4 sin(4t)) = 12 cos(2t) − 4 sin(4t). dt 14. z¨ = −4 sin(4 ).

13.

15. f ¨ (x) = 2 ⋅ [sin(3x)] + 2x[cos(3x)] ⋅ 3 = 2 sin(3x) + 6x cos(3x) 16. f ¨ (x) = (2x)(cos x) + x2 (− sin x) = 2x cos x − x2 sin x. 17. Using the quotient and chain rules 2

d (et + t) ⋅ sin(2t) − (et + t) ddt (sin(2t)) dz = dt dt (sin(2t))2 2 2 et ⋅ ddt (t2 ) + 1 sin(2t) − (et + t) cos(2t) ddt (2t) = sin2 (2t) 2

(2tet + 1) sin(2t) − (et + t)2 cos(2t) 2

sin (2t)

18. f ¨ ( ) = 3 2 cos − 3 sin . 19. Using the quotient rule, we have d d

sin

(cos )( ) − (sin )(1) cos − sin = . 2 2

This problem can also be done by writing (sin )∕ = (sin ) −1 and using the product rule. 20. Using the quotient rule, we get d dt

21. At x = , y = sin = 0, and the slope y = −(x − ) = −x + . See Figure 3.13.

t2 cos t

1 =

2t cos t − t2 (− sin t) (cos t)2

2t cos t + t2 sin t . (cos t)2

dy | | = cos x| = −1. Therefore the equation of the tangent line is | |x= dx |x=

y 3 2

y = −x +

1 2 0

−1

y = sin x

−2 −3

Figure 3.13

220

Chapter Three /SOLUTIONS

22. To calculate the equation of the tangent line, we need to fnd the y-coordinate and the slope at x = 0. The y-coordinate is y = f (0) = 3(0) + cos(0) = 0 + 1 = 1, so a point on the line is (0, 1) and the y-intercept is 1. The slope is found using the derivative: f ¨ (x) = 3 − 5 sin(5x). At the point x = 0, we have Slope = f ¨ (0) = 3 − 5 sin(0) = 3 − 0 = 3. The equation of the line is y = 3x + 1. 23. (a) Since sin 0 = 0 and

d (sin x) = cos x, with cos 0 = 1, we have dx sin x ≈ sin 0 + cos 0 ⋅ x sin x ≈ x

for x near 0

(b) From Figure 3.14, we see that the approximation is above the curve for x > 0 and the approximation is below the curve for x < 0. y=x sin x −

Figure 3.14 d (sin x) = cos x. The tangent line at x = 0 has slope f ¨ (0) = cos 0 = 1 and dx goes through the point (0, 0). Consequently, its equation is y = g(x) = x. The approximate value of sin( ∕6) given by this equation is g( ∕6) = ∕6 ≈ 0.524. Similarly, the tangent line at x = 3 has slope

24. The tangent lines to f (x) = sin x have slope

f¨ and goes through the point ( ∕3,

1 = cos = 3 3 2

√ 3∕2). Consequently, its equation is √ 3 3− 1 y = ℎ(x) = x + . 2 6

The approximate value of sin( ∕6) given by this equation is then √ 6 3− ℎ = ≈ 0.604. 6 12 The actual value of sin( ∕6) is 12 , so the approximation from 0 is better than that from ∕3. This is because the slope of the function changes less between x = 0 and x = ∕6 than it does between x = ∕6 and x = ∕3. This is illustrated by the following fgure. y y = g(x) y = sin x

y = ℎ(x) 6

3.5 SOLUTIONS

221

25. We begin by taking the derivative of y = sin(x4 ) and evaluating at x = 10: dy = cos(x4 ) ⋅ 4x3 . dx Evaluating cos(10,000) on a calculator (in radians) we see cos(10,000) < 0, so we know that dy∕dx < 0, and therefore the function is decreasing. Next, we take the second derivative and evaluate it at x = 10, giving sin(10,000) < 0: d2y = cos(x4 ) ⋅ (12x2 ) + 4x3 ⋅ (− sin(x4 ))(4x3 ) . dx2 «¯¬ «¯¬ negative

positive, but much larger in magnitude

From this we can see that d 2 y∕dx2 > 0, thus the graph is concave up. 26. (a) See Figure 3.15. The number of bird species is highest in June and lowest in December. We see that f ¨ (1) is negative since the function is decreasing there, and f ¨ (10) is positive since the function is increasing there. N (species) 30 20 10

t (months)

Figure 3.15: N = f (t) = 19 + 9 cos( t∕6) (b) We have

t = −4.712 sin t . f ¨ (t) = 0 + 9 − sin 6 6 6 (c) We have f (1) = 19 + 9 cos( ∕6) = 26.8 and f ¨ (1) = −4.712 sin( ∕6) = −2.36. In July, there are about 26.8 bird species, and the number of species is decreasing at a rate of 2.36 species per month. We have f (10) = 19 + 9 cos(10 ∕6) = 23.5 and f ¨ (10) = −4.712 sin(10 ∕6) = 4.08. In April, there are about 23.5 bird species, and the number of species is increasing at a rate of 4.08 species per month.

27. (a) The population is a maximum when the sin( t∕6) is 1. This occurs when t = 6 2 t = 3. Thus, the population is a maximum when t = 3, that is, at the end of March, or April 1. (b) The rate at which the population is changing is t p¨ (t) = 200 ⋅ cos . 6 6 So p¨ (t) is a maximum when t = 0, that is January 1. 28. (a) One cycle is completed in 60∕12 = 5 seconds. (b) We di˙erentiate to get 2 2 4 2 A¨ (t) = 2 sin t = sin t . 5 5 5 5 Substitute t = 1 to get A¨ (1) =

4 2 sin (1) = 2.390 hundred cubic centimeters/second. 5 5

This tell us that, one second after the cycle begins, the patient is inhaling at a rate of approximately 2.39 hundred cubic centimeters/second; that is 239 cubic centimeters/second.

222

Chapter Three /SOLUTIONS

29. We have: C ¨ (t) = 3.5

d 1 3.5 1 7 1 sin t +0+ = cos t + = cos t + . dt 6 5 6 6 5 12 6 5

Thus, 7 1 7 1 cos(8 ) + = + = 2.033 ppm/month. 12 5 12 5 This means that on December 1, 2017, the concentration of CO2 in the air was increasing by about 2.033 parts per million per month. C ¨ (48) =

30. We have: C ¨ (t) = 3.5

d 1 3.5 1 7 1 sin t +0+ = cos t + = cos t + . dt 6 5 6 6 5 12 6 5

Thus, 7 1 7 1 cos(12 ) + = + = 2.033 ppm/month. 12 5 12 5 This means that on December 1, 2019, the concentration of CO2 in the air was increasing by about 2.033 parts per million per month. C ¨ (72) =

31. We have: C ¨ (t) = 3.5

d 1 3.5 1 7 1 sin t +0+ = cos t + = cos t + . dt 6 5 6 6 5 12 6 5

Thus, 7 1 7 1 cos(7 ) + = − + = −1.633 ppm/month. 12 5 12 5 This means that on June 1, 2017, the concentration of CO2 in the air was decreasing by about 1.633 parts per million per month. C ¨ (42) =

32. We have:

(3.5 sin(12 ) + 398 + 72 ) − (3.5 sin(0) + 398) C(72) − C(0) ΔC 1 5 = = = ppm/month Δt 72 72 5 This means over the 72 month period between December 1, 2013 and December 1, 2019 the concentration of CO2 in the air increased, on average, by 0.2 parts per million per month.

33. (a) Using the chain rule we have: H ¨ (t) = 0 + 0.6

d sin t = 0.6 cos t = cos t ◦ C/hr. dt 12 12 12 20 12

(b) Using part (a), we have: cos = 0.079 ◦ C/hr 20 3 H ¨ (12) = cos( ) = −0.157 ◦ C/hr. 20 This means that at 1 pm, our body temperature is increasing by about 0.08 degrees Celsius per hour, whereas at 9 pm, our body temperature is decreasing nearly twice as fast, by about 0.16 degrees Celsius per hour. H ¨ (4) =

34. (a) Looking at the graph in Figure 3.16, we see that the maximum is $2600 per month and the minimum is $1400 per month. If t = 0 is January 1, then the sales are highest on April 1. S S(t) = 2000 + 600 sin( 6 t)

2600 2000 1400

t 3

Figure 3.16

3.5 SOLUTIONS

223

(b) S(2) is the monthly sales on March 1, S(2) = 2000 + 600 sin( ) 3 √ = 2000 + 600 3∕2 ≈ 2519.62

dollars/month

S ¨ (2) is the rate of change of monthly sales on March 1, and since S ¨ (t) = 600[cos( t)]( ) 6 6 = 100 cos( t), 6 We have,

35. (a) v(t) = (b)

S ¨ (2) = 100 cos( ) = 50 ≈ 157.08 3

dy d = (15 + sin(2 t)) = 2 cos(2 t). dt dt y

15 14

2 v = 2 cos 2 t

y = 15 + sin 2 t

t 1

36.

−2

dy = −7.5(0.507) sin(0.507t) = −3.80 sin(0.507t) dt

dy = −3.80 sin(0.507 ⋅ 6) = −0.38 meters/hour. So the tide is falling at 0.38 meters/hour. dt dy (b) When t = 9, we have = −3.80 sin(0.507 ⋅ 9) = 3.76 meters/hour. So the tide is rising at 3.76 meters/hour. dt dy (c) When t = 12, we have = −3.80 sin(0.507 ⋅ 12) = 0.75 meters/hour. So the tide is rising at 0.75 meters/hour. dt dy (d) When t = 18, we have = −3.80 sin(0.507 ⋅ 18) = −1.12 meters/hour. So the tide is falling at 1.12 meters/hour. dt 37. (a) We substitute t = 40: (a) When t = 6, we have

2 (t − 172) + 12 365 2 D(40) = 4 cos (40 − 172) + 12 = 9.4186 hours. 365 D(t) = 4 cos

This tells us that, on the 40th day of the year, February 9, 2021, Paris had approximately 9.4 hours of daylight. We di˙erentiate to get 2 2 D¨ (t) = −4 sin (t − 172) 365 365 8 2 =− sin (t − 172) . 365 365 Substitute t = 40 to get

8 2 sin (40 − 172) = 0.053 hours/day. 365 365 This tell us that, on February 9th, the number of hours of daylight in Paris was increasing at a rate of about 0.053 hours/day. (This is approximately 3.2 minutes per day.) D¨ (40) = −

224

Chapter Three /SOLUTIONS (b) We substitute t = 172 into D(t) to get D(172) = 4 cos

2 (172 − 172) + 12 = 4 cos(0) + 12 = 16 hours. 365

This tells us that, on the 172nd day of the year, June 21, 2021, Paris had approximately 16 hours of daylight. Substitute t = 172 into D¨ (t) to get 8 2 8 D¨ (172) = − sin (172 − 172) = − sin(0) = 0 hours/day. 365 365 365 This tell us that, on June 21st, the rate of change of the number of hours of daylight in Paris was zero. June 21st was the summer solstice (the longest day of the year), so the maximum number of hours of daylight in Paris in 2021 was about 16 hours. 38. (a) We di˙erentiate to get 10 H ¨ (t) = −50 sin t =− sin t . 15 15 3 15 The derivative gives the rate of change in the percent of the moon that is illuminated. The units are percent/day. (b) We set H ¨ (t) = 0 and solve for 0 ≤ t ≤ 30. 10 − sin t =0 3 15 sin t =0 15 We know that

sin

when

t =0 15

t = 0, , 2 , or any other multiple of . 15

This gives t = 0, 15, 30, or any other multiple of 15. The only solutions with 0 ≤ t ≤ 30 are at t = 0, 15, and 30. Thus H ¨ (t) = 0 at t = 0, 15, 30 days. When t = 0 and t = 30, we have H(0) = H(30) = 100%, so the percent of moon illuminated is a maximum. This means there is a full moon. When t = 15, we have H(15) = 0%, so the percent of moon illuminated is a minimum. This means there is a new moon. From Figure 3.17, we see there are no other full or new moons, so H ¨ (0) tells us when there is a full or new moon. % illumination 100 H(t) 50

10 15 20 25 30

t (days since Mar. 18)

Figure 3.17 (c) We can graph H ¨ (t) = −10 ∕3 sin( t∕15) to see that it is negative for 0 < t < 15. A negative derivative tells us that the percentage of the moon illuminated is decreasing. We see that H ¨ (t) = −10 ∕3 sin( t∕15) is positive for 15 < t < 30. A positive derivative tells us that the percentage of the moon illuminated is increasing. This makes sense because after the moon is full at t = 0, the percentage that is illuminated decreases until the new moon, approximately 15 days later. Then, the percentage of the moon that is illuminated increases until the next full moon, in approximately 15 more days.

SOLUTIONS to Review Problems For Chapter Three

225

Solutions for Chapter 3 Review 1. f ¨ (t) = 24t3 . 2. f ¨ (x) = 3x2 − 6x + 5. 3. P ¨ (t) = 2e2t . dW 4. = 3r2 + 5. dr dC 5. = 0.08e0.08q . dq dy = 5(−0.2)e−0.2t = −e−0.2t . 6. dt dy 7. = e3x + x ⋅ 3e3x = e3x (1 + 3x). dx 8. s¨ (t) = 2t(5t − 1) + (t2 + 4) ⋅ 5 = 15t2 − 2t + 20. 2 d 2 2 d (1+3t)2 e = e(1+3t) (1 + 3t)2 = e(1+3t) ⋅ 2(1 + 3t) ⋅ 3 = 6(1 + 3t)e(1+3t) . 9. dt dt 10. f ¨ (x) = 2x + x3 . 11. Q¨ (t) = 5 + 3.6e1.2t . d 2 (z + 5) = 3(z2 + 5)2 (2z) = 6z(z2 + 5)2 . dz d (5x − 1) = 18(5x − 1)2 (5) = 90(5x − 1)2 . 13. f ¨ (x) = 6(3(5x − 1)2 ) ⋅ dx 14. f ¨ (z) = z21+1 (2z) = z22z+1 . 12. g ¨ (z) = 3(z2 + 5)2 ⋅

15. We use the chain rule with z = g(x) = 1+ex as the inside function and f (z) = z10 as the outside function. Since g ¨ (x) = ex and f ¨ (z) = 10z9 we have d � ℎ¨ (x) = (1 + ex )10 = 10z9 ex = 10ex (1 + ex )9 . dx dq = 100(−0.05)e−0.05p = −5e−0.05p . dp dy 1 17. = 2x ln x + x2 ⋅ = x(2 ln x + 1). dx x 18. s¨ (t) = 2t + 2t .

16.

dP = 8t + 7 cos t. dt 20. y¨ = 5 sin4 cos . 1 � −t −e − 1 . 21. ℎ¨ (t) = −t e −t 22. f ¨ (x) = 2 cos(2x). 50xex − 25x2 ex 50x − 25x2 23. g ¨ (x) = = . ex e2x (1)(t − 4) − (1)(t + 4) t−4−t−4 −8 24. ℎ¨ (t) = = = . (t − 4)2 (t − 4)2 (t − 4)2 dy 25. = 2x cos x + x2 (− sin x) = 2x cos x − x2 sin x. dx 26. We use the chain rule with z = g(x) = 1+ex as the inside function and f (z) = ln z as the outside function. Since g ¨ (x) = ex and f ¨ (z) = 1∕z we have d 1 ex ℎ¨ (x) = . (ln(1 + ex )) = ex = dx z 1 + ex 19.

27. We use the chain rule with z = g(w) = ln w + 1 as the inside function and f (z) = ez as the outside function. Since

226

Chapter Three /SOLUTIONS g ¨ (w) = 1∕w and f ¨ (z) = ez we have ℎ¨ (w) =

d � ln w+1 1 eln w+1 eln w ⋅ e w⋅e e = (ez ) = = = = e. dw w w w w

Alternatively, note that eln w+1 = eln w e1 = we so that its derivative with respect to w is just e. 28. We use the chain rule with z = g(x) = x3 + x as the inside function and f (z) = ln(z)as the outside function. Since g ¨ (x) = 3x2 + 1 and f ¨ (z) = 1∕z we ℎ¨ (x) =

1 d � 3x2 + 1 ln(x3 + x) = ⋅ (3x2 + 1) = 3 . dx z x +x

29. This is a quotient where u(x) = 1 + ex and v(x) = 1 − e−x so that q(x) = u(x)∕v(x). Using the quotient rule the derivative is vu¨ − uv¨ q ¨ (x) = , v2 ¨ x ¨ −x where u = e and v = e . Therefore q ¨ (x) =

(1 − e−x )ex − e−x (1 + ex ) ex − 2 − e−x = . (1 − e−x )2 (1 − e−x )2

30. This is a quotient where u(x) = x and v(x) = 1 + x so that q(x) = u(x)∕v(x). Using the quotient rule the derivative is q ¨ (x) = where u¨ = 1 and v¨ = 1. Therefore q ¨ (x) =

vu¨ − uv¨ , v2

(1 + x) ⋅ 1 − x ⋅ 1 1 = . (1 + x)2 (1 + x)2

31. Using the product rule, the derivative is f ¨ (x) = x

d x d (x) ex = xex + ex . (e ) + dx dx

32. We use the chain rule with z = g(x) = ex as the inside function and f (z) = sin z as the outside function. Since g ¨ (x) = ex and f ¨ (z) = cos z, we have d ℎ¨ (x) = (sin(ex )) = (cos z)ex = ex cos(ex ). dx 33. We use the chain rule with z = g(x) = x3 as the inside function and f (z) = cos z as the outside function. Since g ¨ (x) = 3x2 and f ¨ (z) = − sin z, we have ℎ¨ (x) =

d � cos(x3 ) = − sin z ⋅ (3x2 ) = −3x2 sin(x3 ). dx

3(5t + 2) − (3t + 1)5 dz 15t + 6 − 15t − 5 1 = = = . dt (5t + 2)2 (5t + 2)2 (5t + 2)2 (2t + 5)(t + 3) − (t2 + 5t + 2) t2 + 6t + 13 35. z¨ = = . (t + 3)2 (t + 3)2 36. Using the quotient rule: 34.

ℎ¨ (p) =

37.

3 dy 33 1 = (ln 3)3x − (x− 2 ). dx 3 2

2p 6p + 4p3 − 4p − 4p3 2p(3 + 2p2 ) − 4p(1 + p2 ) = = . 2 2 (3 + 2p ) (3 + 2p2 )2 (3 + 2p2 )2

SOLUTIONS to Review Problems For Chapter Three

227

38. Using the chain rule twice: √ √ √ √ et + 1 d √ t 1 d t 1 t t cos t t t f (t) = cos e + 1 e + 1 = cos e + 1 √ ⋅ (e + 1) = cos e + 1 √ e =e √ . dt 2 et + 1 dt 2 et + 1 2 et + 1 ¨

39. Rewriting the function, we use the chain rule f (t) = so

7 = 7e−kt , ekt

f ¨ (t) = −7ke−kt . 40. Using the quotient rule gives g ¨ (t) = g ¨ (t) =

k +1 kt

(ln(kt) − t) − (ln(kt) + t) (ln(kt) − t)2

1 +1 t

(ln(kt) − t) − (ln(kt) + t)

k −1 kt

1 −1 t

(ln(kt) − t)2 ln(kt)∕t − 1 + ln(kt) − t − ln(kt)∕t − 1 + ln(kt) + t g ¨ (t) = (ln(kt) − t)2 2 ln(kt) − 2 . g ¨ (t) = (ln(kt) − t)2 41. Figure 3.18 shows the graph of f (x) = x2 + 1. d We have f ¨ (x) = dx (x2 + 1) = 2x, thus, f ¨ (0) = 2(0) = 0. We check this by seeing in that Figure 3.18 the tangent line at x = 0 has slope 0. We have f ¨ (1) = 2(1) = 2, f ¨ (2) = 2(2) = 4. and f ¨ (−1) = 2(−1) = −2. Thus, the slope is positive at x = 2 and x = 1, and negative at x = −1. Moreover, it is greater at x = 2 than at x = 1. This agrees with the graph in Figure 3.18.

✛ Slope = f ¨ (2) = 4

f (x)

6 5 4 Slope = f ¨ (−1) = −2

✲

✛ Slope = f ¨ (1) = 2

3 2 1

✛ Slope = f ¨ (0) = 0 x

−2

−1

Figure 3.18: Using slopes to check values for derivatives 42. f ¨ (x) = 2x + 3, so f ¨ (0) = 3, f ¨ (3) = 9, and f ¨ (−2) = −1. 43. Di˙erentiating gives f ¨ (x) = 6x2 − 4x

so f ¨ (1) = 6 − 4 = 2.

Thus the equation of the tangent line is (y − 1) = 2(x − 1) or y = 2x − 1. 44. Di˙erentiating gives Rate of change of price =

dV = 75(1.35)t ln 1.35 ≈ 22.5(1.35)t dollar/yr. dt

228

Chapter Three /SOLUTIONS

45. The rate of growth, in billions of people per year, was dP = 7.41(0.011)e0.011t . dt On July 1, 2017, we have t = 0, so dP = 7.41(0.011)e0 = 0.0815 billion∕year = 81.5 million people∕year. dt 46. (a) Figure 3.19 shows the height of the ball at time t. dℎ d (b) Velocity v(t) = = (32t − 16t2 ) = 32 − 32t. dt dt (c) We substitute 1 into v(t) = 32 − 32t to get v(1) = 32 − 32 ⋅ 1 = 0 ft/sec. Since the velocity at this time is 0 ft/sec, the ball is not going up or down. When we look at the graph of ℎ(t) from part (a), we see that the football has reached its maximum height when t = 1. Since ℎ(t) = 32t − 16t2 , when t = 1, the height of the ball is ℎ(1) = 32 ⋅ 1 − 16 ⋅ 12 = 16 feet. ℎ(t) 15 10 5 t 1

Figure 3.19

47. y¨ = 3x2 − 18x − 16 5 = 3x2 − 18x − 16 0 = 3x2 − 18x − 21 0 = x2 − 6x − 7 0 = (x + 1)(x − 7) x = −1 or x = 7. When x = −1, y = 7; when x = 7, y = −209. Thus, the two points are (−1, 7) and (7, −209). 48. (a) f (x) = 1 − ex crosses the x-axis where 0 = 1 − ex , which happens when ex = 1, so x = 0. Since f ¨ (x) = −ex , f ¨ (0) = −e0 = −1. (b) y = −x 49. To fnd the equation of the line tangent to the graph of P (t) = t ln t at t = 2 we must fnd the point (2, P (2)) as well as the slope of the tangent line at t = 2. P (2) = 2(ln 2) ≈ 1.386. Thus we have the point (2, 1.386). To fnd the slope, we must frst fnd P ¨ (t): 1 P ¨ (t) = t + ln t(1) = 1 + ln t. t At t = 2 we have P ¨ (2) = 1 + ln 2 ≈ 1.693 Since we now have the slope of the line and a point, we can solve for the equation of the line: Q(t) − 1.386 = 1.693(t − 2) Q(t) − 1.386 = 1.693t − 3.386 Q(t) = 1.693t − 2. The equation of the tangent line is Q(t) = 1.693t − 2. We see our results displayed graphically in Figure 3.20.

SOLUTIONS to Review Problems For Chapter Three

229

P 6 P (t) = t ln t 3

Q(t) = 1.693t − 2

1.39 1 t

0 2

−2

Figure 3.20

50. We can fnd the rate the balance changes by di˙erentiating B with respect to time: B ¨ (t) = 5000e0.08t ⋅ 0.08 = 400e0.08t . Calculating B ¨ at time t = 5, we have B ¨ (5) = $596.73/yr. After 5 years, the account generates about $597 interest in the next year. 51. (a) When the co˙ee was frst left on the counter, t = 0. Thus, C(0) = 74 + 103e0 = 74 + 103 ⋅ 1 = 177. The temperature was 177◦ F. 1 1 (b) Since e−0.033t = 0.033t , as t gets larger, 0.033t gets smaller and tends to zero. Thus, 103e−0.033t gets very small, so the e e temperature tends to 74 + 0 = 74◦ F. This is room temperature. (c) To fnd C(5), we substitute t = 5. C(5) = 74 + 103e−0.033⋅5 = 161.333. This tells us that, after the co˙ee sits on the counter for fve minutes, its temperature is approximately 161◦ F To fnd C ¨ (5), we di˙erentiate C(t) and substitute. C ¨ (t) = 103e−0.033t (−0.033) = −3.399e−0.033t . C ¨ (5) = −3.399e−0.033⋅5 = −2.882. This tell us that, after the co˙ee has sat on the counter for 5 minutes, it is cooling at a rate of 2.882◦ F/minute. (d) The magnitude of C ¨ (t) is the rate at which the co˙ee is cooling at time t. We expect the magnitude of C ¨ (50) to be less than the magnitude of C ¨ (5) because, when the co˙ee is frst put on the counter (at t = 5), it cools fast. When the co˙ee has been on the counter for some time (at t = 50), it is cooling more slowly. 52. (a) The rate of change of the period is given by dT 2 d √ 2 1 1 . = √ ( l) = √ ⋅ l−1∕2 = √ ⋅√ = √ dl dl 2 l 9.8 9.8 9.8 9.8l (b) The rate decreases since

√ l is in the denominator.

53. (a) Di˙erentiating using the chain rule gives dQ d = e−0.000121t = −0.000121e−0.000121t . dt dt (b) The following graph shows the rate, dQ∕dt, as a function of time.

230

Chapter Three /SOLUTIONS 10,000 0

60,000 t

−0.00012 dQ∕dt

54. (a) dH d = (40 + 30e−2t ) = 30(−2)e−2t = −60e−2t . dt dt dH < 0; this makes sense because the temperature of the soda is decreasing. (b) Since e−2t is always positive, dt dH (c) The magnitude of is dt | dH | | dH | | −2t −2t | | | | | | dt | = ||−60e || = 60e ≤ 60 = | dt t=0 | , | | | | since e−2t ≤ 1 for all t ≥ 0 and e0 = 1. This is just saying that at the moment that the can of soda is put in the refrigerator (at t = 0), the temperature di˙erence between the soda and the inside of the refrigerator is the greatest, so the temperature of the soda is dropping the quickest. d(ax ) 55. We are interested in when the derivative is positive and when it is negative. The quantity ax is always positive. dx However ln a > 0 for a > 1 and ln a < 0 for 0 < a < 1. Thus the function ax is increasing for a > 1 and decreasing for a < 1. 56. We have f (t) = 40(1.3)t , so f (0) = 40(1.3)0 = 40 gigawatts, and f (8) = 40(1.3)8 = 326 gigawatts. Since f ¨ (t) = 40 ln(1.3)(1.3)t , we have f ¨ (0) = 40 ln(1.3)(1.3)0 = 10 gigawatts/year, and f ¨ (8) = 40 ln(1.3)(1.3)8 = 86 gigawatts/year. In 2010, solar power production was 40 gigawatts and production was increasing at 10 gigawatts per year. In 2018, solar power production was 326 gigawatts and production was increasing at about 86 gigawatts per year. 57. (a) To fnd the temperature of the yam when it was placed in the oven, we need to evaluate the function at t = 0. In this case, the temperature of the yam to begin with equals 350(1 − 0.7e0 ) = 350(0.3) = 105◦ . (b) By looking at the function we see that the temperature which the yam is approaching is 350◦ . That is, if the yam were left in the oven for a long period of time (i.e. as t → ∞) the temperature would move closer and closer to 350◦ (because e−0.008t would approach zero, and thus 1 − 0.7e−0.008t would approach 1). Thus, the temperature of the oven is 350◦ . (c) The yam’s temperature will reach 175◦ when Y (t) = 175. Thus, we must solve for t: Y (t) = 175 175 = 350(1 − 0.7e−0.008t ) 175 = 1 − 0.7e−0.008t 350 0.7e−0.008t = 0.5 e−0.008t = 5∕7 ln e−0.008t = ln 5∕7 −0.008t = ln 5∕7 ln 5∕7 t= ≈ 42 minutes. −0.008

SOLUTIONS to Review Problems For Chapter Three

231

Thus the yam’s temperature will be 175◦ approximately 42 minutes after it is put into the oven. (d) The rate at which the temperature is increasing is given by the derivative of the function. Y (t) = 350(1 − 0.7e−0.008t ) = 350 − 245e−0.008t . Therefore, Y ¨ (t) = 0 − 245(−0.008e−0.008t ) = 1.96e−0.008t . At t = 20, the rate of change of the temperature of the yam is given by Y ¨ (20): Y ¨ (20) = 1.96e−0.008(20) = 1.96e−.16 = 1.96(0.8521) ≈ 1.67 degrees/minute. Thus, at t = 20 the yam’s temperature is increasing by about 1.67 degrees each minute. 58. (a) We substitute k = 3.98461 ⋅ 10−20 and take the square root of both sides to give us P (d) = 1.99615 ⋅ 10−10 d 3∕2 . Since P must be positive, we do not choose P = −1.99615 ⋅ 10−10 d 3∕2 . (b) We substitute 57.91 million for d. P (57,910,000) = 1.99615 ⋅ 10−10 (57,910,000)3∕2 = 87.968 days Thus, it takes Mercury approximately 88 days to orbit the sun. (c) We di˙erentiate to get P (d) = 1.99615 ⋅ 10−10 d 3∕2 3 P ¨ (d) = ⋅ 1.99615 ⋅ 10−10 d 1∕2 2 = 2.994225 ⋅ 10−10 d 1∕2 days per kilometer Since d 1∕2 is positive, P ¨ (d) is always positive. This tells us that as the planet gets further from the sun, the time it takes for the planet to orbit the sun increases. This makes sense, because as a planet’s distance from the sun increases, the length of its orbit also increases. Thus, it is not surprising that the length of time it takes to complete this orbit also increases. 59. All of the functions go through the origin. They will look the same if they have the same tangent line, or equivalently, the same slope at x = 0. Therefore for each function we fnd the derivative and evaluate it at x = 0 ∶ For y = x, √ For y = x,

y¨ = 1,

so y¨ (0) = 1.

y¨ = 2√1 x ,

so y¨ (0) is undefned.

For y = x2 ,

y¨ = 2x,

so y¨ (0) = 0.

For y = x

+ 21 x2 ,

For y = x3 ,

y = 3x + x, so y¨ (0) = 0. ¨

so y¨ (0) = 0.

y¨ = 3x2 ,

1 For y = ln(x + 1), y¨ = x+1 , 1 x 2 ¨ For y = 2 ln(x + 1), y = x2 +1 , √ For y = 2x − x2 , y¨ = √ 1−x 2 , 2x−x

so y¨ (0) = 1. so y¨ (0) = 0. so y¨ (0) is undefned.

So near the origin, functions with y¨ (0) = 1 will all be indistinguishable resembling the line y = x. These functions are: y=x ¨

and

y = ln(x + 1).

Functions with y (0) = 0 will be indistinguishable near the origin and resemble the line y = 0 (a horizontal line). These functions are: 1 1 y = x2 , y = x3 + x2 , y = x3 , and y = ln(x2 + 1). 2 2 Functions that have undefned derivatives at x = 0 look like vertical lines at the origin. These functions are √ √ y= x and y = 2x − x2 .

Chapter Three /SOLUTIONS

232

60. Since f (x) = axn , f ¨ (x) = anxn−1 . We know that f ¨ (2) = (an)2n−1 = 3, and f ¨ (4) = (an)4n−1 = 24. Therefore, f ¨ (4) 24 = f ¨ (2) 3 n−1 n−1 (an)4 4 = =8 2 (an)2n−1 2n−1 = 8, and thus n = 4. Substituting n = 4 into the expression for f ¨ (2), we get 3 = a(4)(8), or a = 3∕32. 61. (a) H ¨ (2) = r¨ (2) + s¨ (2) = −1 + 3 = 2. (b) H ¨ (2) = 5s¨ (2) = 5(3) = 15. (c) H ¨ (2) = r¨ (2)s(2) + r(2)s¨ (2) = −1 ⋅ 1 + 4 ⋅ 3 = 11. r¨ (2) −1 1 (d) H ¨ (2) = √ = √ =− . 4 2 r(2) 2 4 62. (a) H(x) = F (G(x)) H(4) = F (G(4)) = F (2) = 1 (b) H(x) = F (G(x)) H ¨ (x) = F ¨ (G(x)) ⋅ G¨ (x) H ¨ (4) = F ¨ (G(4)) ⋅ G¨ (4) = F ¨ (2) ⋅ 6 = 5 ⋅ 6 = 30 (c) H(x) = G(F (x)) H(4) = G(F (4)) = G(3) = 4 (d) H(x) = G(F (x)) H ¨ (x) = G¨ (F (x)) ⋅ F ¨ (x) H ¨ (4) = G¨ (F (4)) ⋅ F ¨ (4) = G¨ (3) ⋅ 7 = 8 ⋅ 7 = 56 63. (a) The rate of change of temperature change is 0 1 dT d CD2 D3 2CD 3D2 = − = − = CD − D2 . dD dD 2 3 2 3 (b) We want to know for what values of D the value of dT ∕dD is positive. This occurs when dT = (C − D) D > 0. dD We only consider positive values of D, since a zero dosage obviously has no e˙ect and a negative dosage does not make sense. If D > 0, then (C − D)D > 0 when C − D > 0, or D < C. So the rate of change of temperature change is positive for doses less than C. 64. Since we’re given that the instantaneous rate of change of T at t = 30 is 2, we want to choose a and b so that the derivative of T agrees with this value. Di˙erentiating, T ¨ (t) = ab ⋅ e−bt . Then we have 2 = T ¨ (30) = abe−30b or e−30b =

2 . ab

We also know that at t = 30, T = 120, so 120 = T (30) = 200 − ae−30b or e−30b = Thus

80 . a

80 2 = e−30b = , a ab

so b=

1 = 0.025 40

and

a = 169.36.

65. Estimates may vary. From the graphs, we estimate f (1) ≈ −0.4, f ¨ (1) ≈ 0.5, g(1) ≈ 2, and g ¨ (1) ≈ 1. By the product rule, ℎ¨ (1) = f ¨ (1) ⋅ g(1) + f (1) ⋅ g ¨ (1) ≈ (0.5)2 + (−0.4)1 = 0.6.

SOLUTIONS to Review Problems For Chapter Three

233

66. Estimates may vary. From the graphs, we estimate f (1) ≈ −0.4, f ¨ (1) ≈ 0.5, g(1) ≈ 2, and g ¨ (1) ≈ 1. By the quotient rule, to one decimal place k¨ (1) =

f ¨ (1) ⋅ g(1) − f (1) ⋅ g ¨ (1) (0.5)2 − (−0.4)1 ≈ = 0.4. 22 (g(1))2

67. Estimates may vary. From the graphs, we estimate f (2) ≈ 0.3, f ¨ (2) ≈ 1.1, g(2) ≈ 1.6, and g ¨ (2) ≈ −0.5. By the product rule, to one decimal place ℎ¨ (2) = f ¨ (2) ⋅ g(2) + f (2) ⋅ g ¨ (2) ≈ 1.1(1.6) + 0.3(−0.5) = 1.6. 68. Estimates may vary. From the graphs, we estimate f (2) ≈ 0.3, f ¨ (2) ≈ 1.1, g(2) ≈ 1.6, and g ¨ (2) ≈ −0.5. By the quotient rule, to one decimal place k¨ (2) =

f ¨ (2) ⋅ g(2) − f (2) ⋅ g ¨ (2) 1.1(1.6) − 0.3(−0.5) ≈ = 0.7. (g(2))2 (1.6)2

69. Estimates may vary. From the graphs, we estimate f (1) ≈ −0.4, f ¨ (1) ≈ 0.5, g(1) ≈ 2, and g ¨ (1) ≈ 1. By the quotient rule, to one decimal place l¨ (1) =

1(−0.4) − 2(0.5) g ¨ (1) ⋅ f (1) − g(1) ⋅ f ¨ (1) ≈ = −8.8. (f (1))2 (−0.4)2

70. Estimates may vary. From the graphs, we estimate f (2) ≈ 0.3, f ¨ (2) ≈ 1.1, g(2) ≈ 1.6, and g ¨ (2) ≈ −0.5. By the quotient rule, to one decimal place l¨ (2) =

g ¨ (2) ⋅ f (2) − g(2) ⋅ f ¨ (2) (−0.5)0.3 − 1.6(1.1) ≈ = −21.2. (0.3)2 (f (2))2

71. Decreasing means f ¨ (x) < 0: f ¨ (x) = 4x3 − 12x2 = 4x2 (x − 3), ¨

so f (x) < 0 when x < 3 and x ≠ 0. Concave up means f ¨¨ (x) > 0: f ¨¨ (x) = 12x2 − 24x = 12x(x − 2) so f ¨¨ (x) > 0 when 12x(x − 2) > 0 x < 0 or

x > 2.

So, both conditions hold for x < 0 or 2 < x < 3. 72. (a) We have p(x) = x2 − x. We see that p¨ (x) = 2x − 1 < 0 when x < 21 . So p is decreasing when x < 12 . (b) We have p(x) = x1∕2 − x, so p¨ (x) =

1 −1∕2 −1 < 0 x 2 1 −1∕2 x <1 2 x−1∕2 < 2 1 x1∕2 > 2 1 x> . 4

Thus p(x) is decreasing when x > 41 . (c) We have p(x) = x−1 − x, so p¨ (x) = −1x−2 − 1 < 0 −x−2 < 1 x−2 > −1, which is always true where x−2 is defned since x−2 = 1∕x2 is always positive. Thus p(x) is decreasing for x < 0 and for x > 0.

234

Chapter Three /SOLUTIONS

73. The frst and second derivatives of ex are ex . Thus, the graph of y = ex is concave up. The tangent line at x = 0 has slope e0 = 1 and equation y = x + 1. A graph that is always concave up is always above any of its tangent lines. Thus ex ≥ x + 1 for all x, as shown in Figure 3.21. y y = ex y =x+1

(0, 1) x

Figure 3.21

74. (a) Di˙erentiating gives

dy 6.3 =− sin t . dt 6 6 The derivative represents the rate of change of the depth of the water in feet/hour. (b) The derivative, dy∕dt, is zero where the tangent line to the curve y is horizontal. This occurs when dy∕dt = 0, so sin( t∕6) = 0, that is when t = 0, 6, 12, 18 and 24 (midnight, 6 am, noon, 6 pm, and midnight). When dy∕dt = 0, the depth of the water is no longer changing. Therefore, it has either just fnished rising or just fnished falling, and we know that the harbor’s level is at a maximum or a minimum.

75. The slopes of the tangent lines to y = x2 − 2x + 4 are given by y¨ = 2x − 2. A line through the origin has equation y = mx. So, at the tangent point, x2 − 2x + 4 = mx where m = y¨ = 2x − 2. x2 − 2x + 4 = (2x − 2)x x2 − 2x + 4 = 2x2 − 2x −x2 + 4 = 0 −(x + 2)(x − 2) = 0 x = 2, −2. Thus, the points of tangency are (2, 4) and (−2, 12). The lines through these points and the origin are y = 2x and y = −6x, respectively. Graphically, this can be seen in Figure 3.22. y

y = x2 − 2x + 4

(−2, 12)

y = 2x y = −6x (2, 4) x

Figure 3.22

76. (a) If the museum sells the painting and invests the proceeds P (t) at time t, then t years have elapsed since 2015, and the time span up to 2025 is 10 − t. This is how long the proceeds P (t) are earning interest in the bank. Each year the money is in the bank it earns 2% interest, which means the amount in the bank is multiplied by a factor of 1.02. So, at the end of (10 − t) years, the balance is given by B(t) = P (t)(1 + 0.02)10−t = P (t)(1.02)10−t .

SOLUTIONS to Review Problems For Chapter Three

235

(b) We have B(t) = P (t)(1.02)10 (1.02)−t = (1.02)10 (c) By the quotient rule,

4 B ¨ (t) = (1.02)10

P (t) . (1.02)t

5 P ¨ (t)(1.02)t − P (t)(1.02)t ln 1.02 . (1.02)2t

So, 4 B ¨ (5) = (1.02)10

2000(1.02)5 − 150,000(1.02)5 ln 1.02 (1.02)10

= (1.02)5 (2000 − 150,000 ln 1.02) = −1071.39 dollars∕year. 77. (a) On the interval 0 < M < 70, we have Slope =

ΔG 2.8 = = 0.04 gallons per mile. ΔM 70

On the interval 70 < M < 100, we have ΔG 4.6 − 2.8 1.8 = = = 0.06 gallons per mile. ΔM 100 − 70 30 (b) Gas consumption, in miles per gallon, is the reciprocal of the slope, in gallons per mile. On the interval 0 < M < 70, gas consumption is 1∕(0.04) = 25 miles per gallon. On the interval 70 < M < 100, gas consumption is 1∕(0.06) = 16.667 miles per gallon. (c) In the second fgure in the problem, we see that the velocity for the frst hour of this trip is 70 mph and the velocity for the second hour is 30 mph. The frst hour may have been spent driving on an interstate highway and the second hour may have been spent driving in a city. The answers to part (b) would then tell us that this car gets 25 miles to the gallon on the highway and about 17 miles to the gallon in the city. (d) Since M = ℎ(t), we have G = f (M) = f (ℎ(t)) = k(t). The function k gives the total number of gallons of gas used t hours into the trip. We have G = k(0.5) = f (ℎ(0.5)) = f (35) = 1.4 gallons. Slope =

The car consumes 1.4 gallons of gas during the frst half hour of the trip. (e) Since k(t) = f (ℎ(t)), by the chain rule, we have dG = k¨ (t) = f ¨ (ℎ(t)) ⋅ ℎ¨ (t). dt Therefore: dG || = k¨ (0.5) = f ¨ (ℎ(0.5)) ⋅ ℎ¨ (0.5) = f ¨ (35) ⋅ 70 = 0.04 ⋅ 70 = 2.8 gallons per hour, | dt ||t=0.5 and

dG || = k¨ (1.5) = f ¨ (ℎ(1.5)) ⋅ ℎ¨ (1.5) = f ¨ (85) ⋅ 30 = 0.06 ⋅ 30 = 1.8 gallons per hour. | dt ||t=1.5

Gas is being consumed at a rate of 2.8 gallons per hour at time t = 0.5 and is being consumed at a rate of 1.8 gallons per hour at time t = 1.5. Notice that gas is being consumed more quickly on the highway, even though the gas mileage is signifcantly better there. 78. Since the population is 308.75 million on April 1, 2010 and growing exponentially, P = 308.75ekt , where P is the population in millions and t is time in years since the census in 2010. Then dP = 308.75kekt , dt so, since the population is growing at 2.85 million∕year on April 1, 2010, dP || | = 308.75kek⋅0 = 308.75k = 2.85 dt ||t=0 k= so P = 308.75e0.00923t .

2.85 = 0.00923, 308.75

236

Chapter Three /SOLUTIONS

79. We have f (0) = 331.3 −1∕2 1 T 1 ⋅ 331.3 1 + 2 273.15 273.15 f ¨ (0) = 0.606.

f ¨ (T ) =

Thus, for temperatures, T , near zero, we have Speed of sound = f (T ) ≈ f (0) + f ¨ (0)T = 331.3 + 0.606T meters∕second.

80. (a) The function f (t) is linear; g(t) is quadratic (polynomial of degree 2); and ℎ(t) is exponential. (b) In 2010, we have t = 130. For f (t), the rate of change is f ¨ (t) = 0.006 f ¨ (130) = 0.006◦ C per year. For g(t) g ¨ (t) = 0.00012t − 0.0017 g ¨ (130) = 0.00012 ⋅ 130 − 0.0017 = 0.0139◦ C per year. For ℎ(t) ℎ¨ (t) = 13.63(0.0004)e0.0004t = 0.00545e0.0004t ℎ¨ (130) = 0.00545e0.0004(130) = 0.00574◦ C per year. (c) For f (t) ∶ Change = 0.006 ⋅ 130 = 0.78◦ C. For g(t) ∶ Change = 0.0139 ⋅ 130 = 1.807◦ C. For ℎ(t) ∶ Change = 0.00574 ⋅ 130 = 0.746◦ C. (d) For f (t) ∶ Predicted change = f (130) − f (0) = (13.625 + 0.006 ⋅ 130) − 13.625 = 0.78◦ C. For g(t) ∶ Predicted change = g(130) − g(0) = (0.00006(1302 ) − 0.0017(130) + 13.788) − 13.788 = 0.793◦ C For ℎ(t) ∶ Predicted change = ℎ(130) − ℎ(0) = 13.63e0.0004(130) − 13.63 = 0.728◦ C. (e) For the linear model, the answers in parts (c) and (d) are equal. (f) For the quadratic model, the discrepancy is largest.

STRENGTHEN YOUR UNDERSTANDING 1. True. If f (x) = 5x2 + 1 then f ¨ (x) = 10x so f ¨ (−1) = 10(−1) = −10. 2. True. We see that f ¨ (x) = g ¨ (x) = 15x4 . 3. False. Since ℎ(t) = (3t2 + 1)(2t) = 6t3 + 2t, we see that ℎ¨ (t) = 18t2 + 2. In a later section, we learn a product rule which gives another way to fnd ℎ¨ without multiplying the factors of ℎ together. √ √ 4. False. We write k(s) = s3 = s3∕2 . Then k¨ (s) = (3∕2)s1∕2 = 1.5 s. 5. False. The slope of the tangent line is given by the derivative. We have f ¨ (x) = 5x4 so f ¨ (1) = 5. The slope of the tangent line at x = 1 is 5 (not 9) so the statement is false. 6. False. We frst write f (r) as a power function: f (r) = r−5 . Then we see f ¨ (r) = −5r−4 = −5∕r4 . 7. True. Since g ¨ (w) = 3w2 − 3 = 3(w2 − 1) = 3(w + 1)(w − 1), when we set g ¨ (w) = 0, there are exactly two solutions, at w = 1 and w = −1. 8. False. Since f ¨ (x) = 9x2 − 2x + 2, we see that f ¨ (1) = 9. Since the derivative is positive at x = 1, the function is increasing at x = 1. 9. True. We see that f ¨ (x) = 9x2 − 2x + 2 and f ¨¨ (x) = 18x − 2. Then f ¨¨ (1) = 16. Since the second derivative is positive at x = 1, the graph of f is concave up at x = 1. 10. True. Since g ¨ (t) = t −1 , we have g ¨ (1) = (1 −1 ) = . 11. False. The derivative of ex is ex . 12. True. We have g ¨ (s) = 5(1∕s) so g ¨ (2) = 5(1∕2) = 5∕2.

STRENGTHEN YOUR UNDERSTANDING

237

13. True. We have f ¨ (x) = 3ex + 1 so the slope of the tangent line at x = 0 is f ¨ (0) = 3e0 + 1 = 3 + 1 = 4. Since f (0) = 3e0 + 0 = 3, the vertical intercept is 3. 14. False. The derivative is ℎ¨ (x) = 2∕x − 2x so ℎ¨ (1) = 2∕1 − 2(1) = 2 − 2 = 0. The graph of ℎ(x) is horizontal (since the slope is 0) at x = 1. 15. True. We fnd the second derivative at x = 1. We have f ¨ (x) = 2∕x − 2x so f ¨¨ (x) = −2∕x2 − 2. Substituting x = 1, we have f ¨¨ (1) = −2∕(12 ) − 2 = −2 − 2 = −4. Since the second derivative is negative at x = 1, the graph of the function is concave down at x = 1. 16. False. Since ln 2 is a constant, the derivative is zero. 17. False. We see that f ¨ (x) = 2e2x . 18. True. 19. False. Since the derivative of ep is ep , we have k¨ (p) = 5ep . 20. True. We have f ¨ (x) = 3(5e5x ) = 15e5x . 2

21. False. The derivative is f ¨ (t) = 2tet . 22. False. The derivative is dy∕dx = 5(x + x2 )4 (1 + 2x). 23. True.

√ 24. False. If y = (1 − 2t)1∕2 then y¨ = (1∕2)(1 − 2t)−1∕2 ⋅ (−2) = −1∕ 1 − 2t.

25. True. 26. False. The chain rule says d∕dt(f (g(t))) = f ¨ (g(t)) ⋅ g ¨ (t). 1 27. False. By the chain rule, we see that g ¨ (x) = 2 (2x + 3). x + 3x ¨ 1−x ¨ 28. True. The derivative is f (x) = −e so f (1) = −e0 = −1. Since the derivative is negative at x = 1, the function is decreasing at x = 1. 29. True. The derivative is f ¨ (x) = −e1−x so f ¨¨ (x) = e1−x and f ¨¨ (1) = e0 = 1. Since the second derivative is positive at x = 1, the function is concave up at x = 1. 30. False. By the chain rule, we need to multiply by the derivative of the inside function. The derivative is dB∕dr = 150(1 + 2r)4 ⋅ 2. 31. False. Since y is the product of two functions, we use the product rule to fnd the derivative: y¨ = ex ∕x + ex ln x. 32. False. Since y is the product of two functions, we need to use the product rule to fnd the derivative. 33. True. The derivative is found using the quotient rule. 34. True. 35. False. We need to use the product rule to fnd the derivative. 36. False. We need to use the chain rule when we di˙erentiate ln(q 2 + 1). We have P ¨ = q∕(q 2 + 1) ⋅ (2q) + ln(q 2 + 1). 37. True. We use the chain rule and then the product rule to fnd the derivative of the exponent. Notice that ex ln x = xx so this method shows us how to fnd the derivative of xx . 38. True. Since x2 ⋅ x2 = x4 , both ways of fnding the derivative give the same answer. Try it to see! 39. False. The product rule must be used to di˙erentiate the product of two functions. 40. False. The quotient rule must be used to fnd the derivative of the quotient of two functions. 41. True. 42. False, the answer is o˙ by a minus sign. We have f ¨ (t) = cos t so f ¨¨ (t) = − sin t. 43. True. We have g ¨ (t) = − sin t so g ¨¨ (t) = − cos t. 44. False. By the chain rule, we need to multiply by the derivative of the inside function. We have y¨ = 2 cos 2t. 45. False. We use the chain rule to get y¨ = −2t sin t2 . 46. False. We use the product rule to see that z¨ = (sin 2t)(−3 sin 3t) + (2 cos 2t) cos(3t). 47. False. We use the chain rule to fnd this derivative. We do not use the product rule since this is not the product of two functions. Using the chain rule, we have y¨ = cos(cos t) ⋅ (− sin t). 48. False. By the chain rule, the derivative of (sin q)−1 is −(sin q)−2 (cos q). 49. True. 50. True.

238

Chapter Three /SOLUTIONS

PROJECTS FOR CHAPTER THREE 1. (a) Assuming that T (1) = 98.6 − 2 = 96.6, we get 96.6 = 68 + 30.6e−k⋅1 28.6 = 30.6e−k 0.935 = e−k . So k = − ln(0.935) ≈ 0.067. (b) We’re looking for a value of t which gives T ¨ (t) = −1. First we fnd T ¨ (t): T (t) = 68 + 30.6e−0.067t T ¨ (t) = (30.6)(−0.067)e−0.067t ≈ −2e−0.067t . Setting this equal to −1 per hour gives −1 = −2e−0.067t ln(0.5) = −0.067t ln(0.5) t=− ≈ 10.3. 0.067 Thus, when t ≈ 10.3 hours, we have T ¨ (t) ≈ −1◦ F per hour. (c) The coroner’s rule of thumb predicts that in 24 hours the body temperature will decrease 25◦ F, to about 73.6◦F. The formula predicts a temperature of T (24) = 68 + 30.6e−0.067⋅24 ≈ 74.1◦F. 2. (a) See Figure 3.23. P 30 P = 30e−3.23×10

−5 ℎ

ℎ Figure 3.23

(b) Using the chain rule, we have −5 dP = 30e−3.23×10 ℎ (−3.23 × 10−5 ) dℎ

dP || = −30(3.23 × 10−5 ) = −9.69 × 10−4 dℎ ||ℎ=0

Hence, at ℎ = 0, the slope of the tangent line is −9.69 × 10−4 , so the equation of the tangent line is y − 30 = (−9.69 × 10−4 )(ℎ − 0) y = (−9.69 × 10−4 )ℎ + 30 = 30 − 0.000969ℎ.

PROJECTS FOR CHAPTER THREE

239

(c) The rule of thumb says Drop in pressure from ℎ = 1000 sea level to height ℎ But since the pressure at sea level is 30 inches of mercury, this drop in pressure is also (30 − P ), so 30 − P =

ℎ 1000

giving P = 30 − 0.001ℎ. (d) The equations in (b) and (c) are almost the same: both have P intercepts of 30, and the slopes are almost the same (9.69 × 10−4 ≈ 0.001). The rule of thumb calculates values of P which are very close to the tangent lines, and therefore yields values very close to the curve. (e) The tangent line is slightly below the curve, and the rule of thumb line, having a slightly more negative slope, is slightly below the tangent line (for ℎ > 0). Thus, the rule of thumb values are slightly smaller. 3. (a) (i) The GDP per capita is Y ∕P , where Y is GDP and P is population. Since Y ln = ln Y − ln P , P di˙erentiating this relationship tells us that Relative rate of change of Y ∕P = Relative rate of change of Y − Relative rate of change of P , so we have Relative rate of change of P = Relative rate of change of Y − Relative rate of change of Y ∕P . This equation expresses the relative rate of change of population as the vertical distance between the graphs of the relative rates of change of GDP and of GDP per capita. (ii) In 2000 the relative growth rate of GDP was 4.4% and the relative growth rate of GDP per capita was 3%. Thus the relative growth rate of population was 4.4% − 3% = 1.4% per year. In 2010 the relative growth rate of GDP was 4.3% and of GDP per capita was 3.1%. Thus the relative growth rate of population was 4.3% − 3.1% = 1.2% per year. (b) The per capita GDP is the quotient GDP∕P , where P is the world population. Since Relative rate of change of GDP∕P = Relative rate of change of GDP − Relative rate of change of P we have Relative rate of change of GDP∕P = 4.6% − 1.2% = 3.4%. In 2020 the per capita GDP in developing countries is forecast to grow at a relative rate of 3.4% per year. (c) The world’s per capita production is the quotient Y ∕P , where Y is the world’s annual production and P is the world population. Since Relative rate of change of Y ∕P = Relative rate of change of Y − Relative rate of change of P we have 1.4% = 2.5% − Relative rate of change of P . In 2019 the world population grew at a relative rate of 2.5% − 1.4% = 1.1% per year.

240

Chapter Three /SOLUTIONS

4. (a) The function f (t) is linear; g(t) is exponential; ℎ(t) is quadratic (polynomial of degree 2). (b) In 2020, we have t = 70. For f (t), the rate of change is f ¨ (t) = 1.3, f ¨ (70) = 1.3 ppm per year. For g(t), the rate of change is g ¨ (t) = 304(0.0038)e0.0038t = 1.1552e0.0038t, g ¨ (70) = 1.1552e0.0038(70) = 1.507 ppm per year . For ℎ(t), the rate of change is ℎ¨ (t) = 0.0135 ⋅ 2t + 0.5133 = 0.027t + 0.5133, ℎ¨ (70) = 0.027(70) + 0.5133 = 2.403 ppm per year . (c) Since 1.3 < 1.507 < 2.403, we see Linear prediction < Exponential < Quadratic. (d) The linear growth rate remains constant and will always be the smallest. The other two growth rates increase with time, and eventually the exponential growth rate will overtake the quadratic growth rate. For example, if t = 1000, we have f ¨ (1000) = 1.3 g ¨ (1000) = 1.1552e0.0038(1000) = 51.639 ℎ¨ (1000) = 0.027(1000) + 0.5133 = 27.513. Thus, at t = 1000, the exponential growth rate is largest.

Solutions to Problems on Establishing the Derivative Formulas 1. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ 2(x + ℎ) + 1 − (2x + 1) = lim ℎ→0 ℎ 2x + 2ℎ + 1 − 2x − 1 = lim ℎ→0 ℎ 2ℎ = lim . ℎ→0 ℎ

f ¨ (x) = lim ℎ→0

As long as ℎ is very close to, but not actually equal to, zero we can say that lim ℎ→0

2ℎ = 2, and thus conclude that f ¨ (x) = 2. ℎ

2. Using the defnition of the derivative, we have f (x + ℎ) − f (x) 5(x + ℎ)2 − 5x2 = lim ℎ→0 ℎ→0 ℎ ℎ 5(x2 + 2xℎ + ℎ2 ) − 5x2 5x2 + 10xℎ + 5ℎ2 − 5x2 = lim = lim ℎ→0 ℎ→0 ℎ ℎ 2 ℎ(10x + 5ℎ) 10xℎ + 5ℎ = lim = lim . ℎ→0 ℎ→0 ℎ ℎ

f ¨ (x) = lim

As ℎ gets close to zero, but not equal to zero, we can cancel the ℎ’s in the numerator and denominator to obtain the following limit which is equal to f ¨ (x): lim(10x + 5ℎ) = 10x. Thus, f ¨ (x) = 10x. ℎ→0

SOLUTIONS TO PROBLEMS ON ESTABLISHING THE DERIVATIVE FORMULAS

241

3. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ 2(x + ℎ)2 + 3 − (2x2 + 3) = lim ℎ→0 ℎ 2x2 + 4xℎ + 2ℎ2 + 3 − 2x2 − 3 = lim ℎ→0 ℎ 4xℎ + 2ℎ2 = lim ℎ→0 ℎ ℎ(4x + 2ℎ) = lim ℎ→0 ℎ As ℎ gets close to zero (but not equal to zero), we can cancel the ℎ in the numerator and denominator to obtain the following: f ¨ (x) = lim ℎ→0

f ¨ (x) = lim(4x + 2ℎ) = 4x ℎ→0

Thus, we get f (x) = 4x. 4. Using the defnition of the derivative, we have f (x + ℎ) − f (x) (x + ℎ)2 + (x + ℎ) − (x2 + x) = lim ℎ→0 ℎ→0 ℎ ℎ x2 + 2xℎ + ℎ2 + x + ℎ − x2 − x = lim ℎ→0 ℎ ℎ(2x + ℎ + 1) 2xℎ + ℎ2 + ℎ = lim = lim . ℎ→0 ℎ→0 ℎ ℎ

f ¨ (x) = lim

As ℎ approaches, but does not equal, zero we can cancel ℎ’s in the numerator and denominator to obtain the following limit equal to f ¨ (x): lim(2x + ℎ + 1) = 2x + 1. ℎ→0

Thus, f ¨ (x) = 2x + 1. 5. The defnition of the derivative states that f ¨ (x) = lim ℎ→0

f (x + ℎ) − f (x) . ℎ

Using this defnition, we have 4(x + ℎ)2 + 1 − (4x2 + 1) ℎ→0 ℎ 4x2 + 8xℎ + 4ℎ2 + 1 − 4x2 − 1 = lim ℎ→0 ℎ 8xℎ + 4ℎ2 = lim ℎ→0 ℎ ℎ(8x + 4ℎ) = lim . ℎ→0 ℎ As long as ℎ approaches, but does not equal, zero we can cancel ℎ in the numerator and denominator. The derivative now becomes lim(8x + 4ℎ) = 8x. f ¨ (x) = lim

ℎ→0

Thus, f ¨ (x) = 6x as we stated above. 6. Using the defnition of the derivative, we have f (x + ℎ) − f (x) (x + ℎ)4 − x4 = lim ℎ→0 ℎ→0 ℎ ℎ x4 + 4x3 ℎ + 6x2 ℎ2 + 4xℎ3 + ℎ4 − x4 = lim ℎ→0 ℎ 4x3 ℎ + 6x2 ℎ2 + 4xℎ3 + ℎ4 = lim ℎ→0 ℎ ℎ(4x3 + 6x2 ℎ + 4xℎ2 + ℎ3 ) = lim . ℎ→0 ℎ

f ¨ (x) = lim

242

Chapter Three /SOLUTIONS We can cancel a factor of ℎ as long as ℎ ≠ 0; thus we get lim(4x3 + 6x2 ℎ + 4xℎ2 + ℎ3 ) which goes to 4x3 as ℎ → 0. Thus, ℎ→0

f ¨ (x) = 4x3 . 7. Using the defnition of the derivative, we have f (x + ℎ) − f (x) (x + ℎ)5 − x5 = lim ℎ→0 ℎ→0 ℎ ℎ x5 + 5x4 ℎ + 10x3 ℎ2 + 10x2 ℎ3 + 5xℎ4 + ℎ5 − x5 = lim ℎ→0 ℎ 5x4 ℎ + 10x3 ℎ2 + 10x2 ℎ3 + 5xℎ4 + ℎ5 = lim ℎ→0 ℎ ℎ(5x4 + 10x3 ℎ + 10x2 ℎ2 + 5xℎ3 + ℎ4 ) = lim . ℎ→0 ℎ As ℎ → 0 but does not equal it, we can safely factor ℎ out of the numerator and denominator and cancel, leaving us with the following limit which equals f ¨ (x): f ¨ (x) = lim

lim(5x4 + 10x3 ℎ + 10x2 ℎ2 + 5xℎ3 + ℎ4 ) = 5x4 .

ℎ→0

Thus, f ¨ (x) = 5x4 . 8. (a)

2ℎ −1 ℎ

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 −1

−0.5

0.5

ℎ

Figure 3.24 2ℎ − 1 is given in Figure 3.24. We can see that as ℎ gets closer to zero, the value of ℎ the function approaches 0.6931. Thus, we can conclude that The graph of the function

2ℎ − 1 ≈ 0.6931. ℎ→0 ℎ You may also want to consider plugging some values of ℎ close to zero into your calculator so that you can observe that the function is indeed approaching 0.6931 as the values of ℎ get closer and closer to zero. (b) Using the defnition of the derivative and the results from part (a), we have lim

f (x + ℎ) − f (x) 2x+ℎ − 2x = lim ℎ→0 ℎ→0 ℎ ℎ 2x ⋅ 2ℎ − 2x = lim ℎ→0 ℎ 2x (2ℎ − 1) = lim ℎ→0 ℎ 0 1 2ℎ − 1 x . = 2 ⋅ lim ℎ→0 ℎ

f ¨ (x) = lim

From part (a) we know that lim ℎ→0

2ℎ − 1 ≈ 0.6931, and thus ℎ 0 1 2ℎ − 1 f ¨ (x) = 2x ⋅ lim ≈ (0.6931)2x . ℎ→0 ℎ

SOLUTIONS TO PROBLEMS ON ESTABLISHING THE DERIVATIVE FORMULAS

243

9. Since f (x) = C for all x, we have f (x + ℎ) = C. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ C −C = lim ℎ→0 ℎ 0 = lim . ℎ→0 ℎ

f ¨ (x) = lim ℎ→0

As ℎ gets very close to zero without actually equaling zero, we have 0∕ℎ = 0, so f ¨ (x) = lim(0) = 0. ℎ→0

10. Using the defnition of the derivative, we have f (x + ℎ) − f (x) ℎ (b + m(x + ℎ)) − (b + mx) = lim ℎ→0 ℎ b + mx + mℎ − b − mx = lim ℎ→0 ℎ mℎ = lim . ℎ→0 ℎ

f ¨ (x) = lim ℎ→0

Provided ℎ ≠ 0, we can cancel the ℎ in the numerator and denominator. Thus, as ℎ gets very close to zero without actually equaling zero, we obtain f ¨ (x) = lim(m) = m. ℎ→0

11. Using the defnition of the derivative and properties of limits, we have f (x + ℎ) − f (x) ℎ k ⋅ u(x + ℎ) − k ⋅ u(x) = lim ℎ→0 ℎ k(u(x + ℎ) − u(x)) = lim ℎ→0 ℎ u(x + ℎ) − u(x) = lim k ⋅ ℎ→0 ℎ u(x + ℎ) − u(x) = k ⋅ lim ℎ→0 ℎ = k ⋅ u¨ (x).

f ¨ (x) = lim ℎ→0

12. Using the defnition of the derivative and properties of limits, we have f (x + ℎ) − f (x) ℎ (u(x + ℎ) + v(x + ℎ)) − (u(x) + v(x)) = lim ℎ→0 ℎ (u(x + ℎ) − u(x)) + (v(x + ℎ) − v(x)) = lim ℎ→0 ℎ 0 1 u(x + ℎ) − u(x) v(x + ℎ) − v(x) = lim + ℎ→0 ℎ ℎ u(x + ℎ) − u(x) v(x + ℎ) − v(x) = lim + lim ℎ→0 ℎ→0 ℎ ℎ ¨ ¨ = u (x) + v (x).

f ¨ (x) = lim ℎ→0

244

Chapter Three /SOLUTIONS

Solutions to Practice Problems on Differentiation 1. f ¨ (t) = 2t + 4t3 2. g ¨ (x) = 20x3 3. y¨ = 15x2 + 14x − 3 4. s¨ (t) = −12t−3 + 9t2 − 2t−1∕2 −2 5 5. f ¨ (x) = −2x−3 + 5 12 x−1∕2 = 3 + √ x 2 x 6. P ¨ (t) = 100e0.05t (0.05) = 5e0.05t 7. f ¨ (x) = 10e2x − 2 ⋅ 3x (ln 3) 8. P ¨ (t) = 1000(1.07)t (ln 1.07) ≈ 68(1.07)t 2

9. D¨ (p) = 2pep + 10p � � 10. y¨ = t2 5e5t + 2t e5t = 5t2 e5t + 2te5t √ √ x3 11. y¨ = 2x x2 + 1 + x2 12 (x2 + 1)−1∕2 ⋅ 2x = 2x x2 + 1 + √ x2 + 1 2x 12. f ¨ (x) = 2 x +1 16 13. s¨ (t) = 2t + 1 1 14. g ¨ (w) = 2w ln w + w2 = 2w ln w + w w 15. f ¨ (x) = 2x (ln 2) + 2x t 16. P ¨ (t) = 12 (t2 + 4)−1∕2 (2t) = √ 2 t +4 17. C ¨ (q) = 3(2q + 1)2 ⋅ 2 = 6(2q + 1)2 18. g ¨ (x) = 5(x + 3)2 + 5x(2(x + 3)) = 5(x + 3)2 + 10x(x + 3) 19. P ¨ (t) = bkekt 20. Since a, b, and c are all constants, we have dy = a(2x) + b(1) + 0 = 2ax + b. dx 2x2 2x + 1 22. f ¨ (t) = 3(et + 4)2 (et ) = 3et (et + 4)2

21. y¨ = 2x ln (2x + 1) + 23. f ¨ (x) = 10 cos (2x)

24. W ¨ (r) = 2r cos r − r2 sin r 25. g ¨ (t) = 15 cos (5t) 26. y¨ = 3e3t sin (2t) + 2e3t cos (2t) 27. y¨ = 2ex + 3 cos x 28. f ¨ (t) = 6t − 4. 29. y¨ = 17 + 12x−1∕2 . 1 30. g ¨ (x) = − (5x4 + 2). 2 2 . x3 2x 2 2x 2x 2 dy 2e (x + 1) − e (2x) 2e (x + 1 − x) = = 32. dx (x2 + 1)2 (x2 + 1)2 (x + 2)(x + 1) x2 + 3x + 2 33. Either notice that f (x) = can be written as f (x) = which reduces to f (x) = x + 2, giving x+1 x+1 31. The power rule gives f ¨ (x) = 20x3 −

SOLUTIONS TO PRACTICE PROBLEMS ON DIFFERENTIATION f ¨ (x) = 1, or use the quotient rule which gives (x + 1)(2x + 3) − (x2 + 3x + 2) (x + 1)2 2 2x + 5x + 3 − x2 − 3x − 2 = (x + 1)2 2 x + 2x + 1 = (x + 1)2 (x + 1)2 = (x + 1)2 = 1.

f ¨ (x) =

0 2 1 2x 4 � x +2 = x x2 + 2 3 3 9 d d 35. sin(2 − 3x) = cos(2 − 3x) (2 − 3x) = −3 cos(2 − 3x). dx dx 0 1 1 z2 − 1 z 1 1� 1� 36. f (z) = + z−1 = z + z−1 , so f ¨ (z) = 1 − z−2 = . 3 3 3 3 3 z2 3(5r + 2) − 3r(5) 15r + 6 − 15r 6 = = 37. q ¨ (r) = (5r + 2)2 (5r + 2)2 (5r + 2)2 dy 1 38. = ln x + x − 1 = ln x dx x ax ae 39. j ¨ (x) = ax (e + b) (t + 4) − (t − 4) 8 40. g ¨ (t) = = . (t + 4)2 (t + 4)2 41. ℎ¨ (w) = 5(w4 − 2w)4 (4w3 − 2) 10 = 3w2 ln(10w) + w2 . 42. Using the product and chain rules gives ℎ¨ (w) = 3w2 ln(10w) + w3 10w cos x − sin x . 43. Using the chain rule gives f ¨ (x) = sin x + cos x 44. We can write w(r) = (r4 + 1)1∕2 , so 1 2r3 w¨ (r) = (r4 + 1)−1∕2 (4r3 ) = √ . 2 r4 + 1 34. y¨ = 2

3 45. ℎ¨ (w) = 6w−4 + w−1∕2 2 0 2 11∕2 x +9 , so 46. We can write ℎ(x) = x+3 ℎ¨ (x) =

1 2

u 0 2 1−1∕2 4 4 5 5 2x(x + 3) − (x2 + 9) 1 x + 3 x2 + 6x − 9 x +9 = . x+3 2 x2 + 9 (x + 3)2 (x + 3)2

47. Using the product rule gives v¨ (t) = 2te−ct − ce−ct t2 = (2t − ct2 )e−ct . 48. Using the quotient rule gives f ¨ (x) =

1 + ln x − x( x1 )

(1 + ln x)2 ln x = . (1 + ln x)2

49. Using the chain rule, g ¨ ( ) = (cos )esin . 50. p¨ (t) = 4e4t+2 .

245

246

Chapter Three /SOLUTIONS

3x2 2ax + −c 51. j ¨ (x) = a b H I √ z d z2 + 1 d 32 3 12 1 − 23 − 12 52. = (z + z ) = z − z = (3 − z−2 ). √ dz dz 2 2 2 z 0 2 1 2r(r + 1) (2r)(2r + 1) − 2r2 d r 53. ℎ¨ (r) = = = . dr 2r + 1 (2r + 1)2 (2r + 1)2 d 1 54. g ¨ (x) = + 3x ln 3. (2x − x−1∕3 + 3x − e) = 2 + 4 dx 3x 3 H I d 1 1 t ¨ 2te − √ 55. f (t) = = 2et + 2tet + 3∕2 . dt 2t t 56. (−3)(5 + 3z) − (5 − 3z)(3) dw = dz (5 + 3z)2 −15 − 9z − 15 + 9z −30 = = (5 + 3z)2 (5 + 3z)2 3x2 x3 3 x2 x2 + = x2 ln x (3 ln x − 1) + = x2 ln x − 9 9 x 3 3 1 3 d 1 1 3 5 x 2 + x−1 + x− 2 = x− 2 − x−2 − x− 2 . 58. g ¨ (x) = dx 2 2 59. Using the product and chain rules, we have

57. f ¨ (x) =

dy = 3(x2 + 5)2 (2x)(3x3 − 2)2 + (x2 + 5)3 [2(3x3 − 2)(9x2 )] dx = 3(2x)(x2 + 5)2 (3x3 − 2)[(3x3 − 2) + (x2 + 5)(3x)] = 6x(x2 + 5)2 (3x3 − 2)[6x3 + 15x − 2].

60. Using the quotient rule gives f ¨ (x) =

(−2x)(a2 + x2 ) − (2x)(a2 − x2 ) −4a2 x = 2 . (a2 + x2 )2 (a + x2 )2

61. Using the quotient rule gives 2ar(b + r3 ) − 3r2 (ar2 ) (b + r3 )2 2abr − ar4 = . (b + r3 )2

w¨ (r) =

62. Using the product rule gives H ¨ (t) = 2ate−ct − c(at2 + b)e−ct = (−cat2 + 2at − bc)e−ct .

63. Since g(w) = 5(a2 − w2 )−2 , g ¨ (w) = −10(a2 − w2 )−3 (−2w) =

20w (a2 − w2 )3

4.1 SOLUTIONS

247

CHAPTER FOUR Solutions for Section 4.1 1. We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0, or equivalently where the tangent line to the graph is horizontal. These points are shown in Figure 4.1.

f (x)

Figure 4.1 As we can see, there is one critical point. Since it is higher than nearby points, it is a local maximum. 2. We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve f (x) is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0, or equivalently where the tangent line to the graph is horizontal. These points are shown in Figure 4.2. As we can see, there are three critical points. The leftmost one is a local maximum, because points near it are all lower; similarly, the middle critical point is surrounded by higher points, and is a local minimum. The critical point to the right is a local maximum.

✻

local max

❥ local max

✻ local min

Figure 4.2

3. We fnd a critical point by noting where f ¨ (t) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (t) = 0, or equivalently where the tangent line to the graph is horizontal. These points are shown in Figure 4.3. f (t) C A D B

Figure 4.3 As we can see, there are four labeled critical points. Critical point A is a local maximum because points near it are all lower; similarly, point B is a local minimum, point C is a local maximum, and point D is a local minimum.

248

Chapter Four /SOLUTIONS

4. We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0, or equivalently where the tangent line to the graph is horizontal. These points are shown in Figure 4.4.

f (x)

Figure 4.4 As we can see, there is one critical point. Since some nearby points (those to the left) are lower, this point is not a local minimum; since nearby points to the right are higher, it is not a local maximum. So the one critical point is neither a local minimum nor a local maximum. 5. (a) One possible answer is shown in Figure 4.5. (b) One possible answer is shown in Figure 4.6 local maximum

❘ ✻

✻

✛

critical points

✛

local minima

Figure 4.5

Figure 4.6

(b)

6. (a)

g x

x C

E E 7. There was a critical point after the frst eighteen hours when temperature was at its highest point, a local maximum for the temperature function. 8. (a) We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0. We have: f ¨ (x) = 5 − 2x = 0 x = 2.5. There is only one critical point, at x = 2.5. (b) We know that the graph of this function is a parabola opening down, so the critical point at x = 2.5 is a local maximum. We verify this in Figure 4.7.

4.1 SOLUTIONS

249

15 f (x) 10

x 2.5

Figure 4.7 9. (a) We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0. We have: f ¨ (x) = 3x2 − 75 = 0 x2 = 25 x = ±5. There are two critical points, at x = 5 and x = −5. (b) We see in Figure 4.8 that f has a local maximum at x = −5 and a local minimum at x = 5. 300 f (x)

x −10

−5

−300

Figure 4.8 10. (a) We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0. We have: f ¨ (x) = 18x − 3x2 = 0 3x(6 − x) = 0 x = 0 or x = 6. There are two critical points, at x = 0 and x = 6. (b) We see in Figure 4.9 that f has a local minimum at x = 0 and a local maximum at x = 6. 200

f (x) x −6

6 −100

Figure 4.9

250

Chapter Four /SOLUTIONS

11. (a) We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0. We have: f ¨ (x) = 4x3 − 16x = 0 4x(x2 − 4) = 0 x = 0, x = 2 or x = −2. There are three critical points, at x = 0 and x = 2 and x = −2. (b) We see in Figure 4.10 that f has a local maximum at x = 0 and local minima at x = 2 and at x = −2. 25 f (x) x −4

−2

−25

Figure 4.10

12. (a) We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0. We have: f ¨ (x) = 4x3 − 12x2

= 0

4x (x − 3) = 0 x = 0 and x = 3. There are two critical points, at x = 0 and x = 3. (b) We see in Figure 4.11 that f has neither a local maximum nor a local minimum at x = 0 and that f has a local minimum at x = 3. 35 f (x) x −2

−35

Figure 4.11

13. (a) We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0. We have: f ¨ (x) = 5x4 − 45x2 = 0 5x2 (x2 − 9) = 0 x = 0 and x = 3 and x = −3. There are three critical points, at x = 0 and x = 3 and x = −3. (b) We see in Figure 4.12 that f has neither a local maximum nor a local minimum at x = 0, that f has a local maximum at x = −3, and that f has a local minimum at x = 3.

4.1 SOLUTIONS

251

200 f (x) x −5

−200

Figure 4.12 14. The graph of f in Figure 4.13 appears to be increasing for all x, with no critical points. Since f ¨ (x) = 3x2 + 6 and x2 ≥ 0 for all x, we have f ¨ (x) > 0 for all x. That explains why f is increasing for all x. f (x) = x3 + 6x + 1

x −1

1 −8

Figure 4.13 15. The graph of f in Figure 4.14 appears to be increasing for x < −1.4, decreasing for −1.4 < x < 1.4, and increasing for x > 1.4. There is a local maximum near x =√−1.4 and local minimum near x = 1.4. The derivative of f is f ¨ (x) = 3x2 − 6. Thus f ¨ (x) = 0 when x2 = 2,√ that is x = ± 2. This explains the critica√ l points near x = ±1.4. Since f ¨ (x) changes √ from is a local positive to negative at x there maximum at = − 2, x = 2, x = − 2 and a and from negative to positive at √ local minimum at x = 2. (−1.4, 6.6) f (x) = x3 − 6x + 1

6 1 −1 −6

(1.4, −4.6)

Figure 4.14 16. The graph of f in Figure 4.15 appears to be increasing for x < −1, decreasing for −1 < x < 1 although it is fat at x = 0, and increasing for x > 1. There are critical points at x = −1 and x = 1, and apparently also at x = 0. Since f ¨ (x) = 15x4 − 15x2 = 15x2 (x2 − 1), we have f ¨ (x) = 0 at x = 0, −1, 1. Notice that although f ¨ (0) = 0, making x = 0 a critical point, there is no change in sign of f ¨ (x) at x = 0; the only sign changes are at x = ±1. Thus the graph of f must alternate increasing/decreasing for x < −1, −1 < x < 1, x > 1, just as we described. (−1, 2) 2

f (x) = 3x5 − 5x3 x

−1

1 −2 (1, −2)

Figure 4.15

252

Chapter Four /SOLUTIONS

17. The graph of f in Figure 4.16 appears to be decreasing for x < 2.3 (almost like a straight line for x < 0), and increasing sharply for x > 2.3. Here f ¨ (x) = ex − 10, so f ¨ (x) = 0 when ex = 10, that is x = ln 10 = 2.302... This is the only place where f ¨ (x) changes sign, and it is a minimum of f . Notice that ex is small for x < 0 so f ¨ (x) ≈ −10 for x < 0, which means the graph looks like a straight line of slope −10 for x < 0. However, ex gets large quickly for x > 0, so f ¨ (x) gets large quickly for x > ln 10, meaning the graph increases sharply there. f (x) = ex − 10x x

(2.3, −13.0)

Figure 4.16

18. The graph of f below appears to be decreasing for 0 < x < 0.37, and then increasing for x > 0.37. We have f ¨ (x) = ln x + x(1∕x) = ln x + 1, so f ¨ (x) = 0 when ln x = −1, that is, x = e−1 ≈ 0.37. This is the only place where f ¨ changes sign and f ¨ (1) = 1 > 0, so the graph must decrease for 0 < x < e−1 and increase for x > e−1 . Thus, there is a local minimum at x = e−1 . f (x) = x ln x

x (0.37, −0.37)

19. The graph of f in Figure 4.17 looks like a climbing sine curve, alternately increasing and decreasing, with more time spent increasing than decreasing. Here f ¨ (x) = 1 + 2 cos x, so f ¨ (x) = 0 when cos x = −1∕2; this occurs when x=±

2 4 8 10 14 16 ,± ,± ,± ,± ,± ... 3 3 3 3 3 3

Since f ¨ (x) changes sign at each of these values, the graph of f must alternate increasing/decreasing. However, the distance between values of x for critical points alternates between (2 )∕3 and (4 )∕3, with f ¨ (x) > 0 on the intervals of length (4 )∕3. For example, f ¨ (x) > 0 on the interval (4 )∕3 < x < (8 )∕3. As a result, f is increasing on the intervals of length (4 ∕3) and decreasing on the intervals of length (2 ∕3). (8.4, 10.1) f (x) = x + 2 sin x (2.1, 3.8)

(10.5, 8.7)

✻

(4.2, 2.5)

Figure 4.17

20. We have g ¨ (x) = e−3x − 3xe−3x = (1 − 3x)e−3x . ¨

To fnd critical points, we set g (x) = 0. Then (1 − 3x)e−3x = 0.

4.1 SOLUTIONS

253

Therefore, the critical point of g is x = 1∕3. To the left of x = 1∕3, we have g ¨ (x) > 0. To the right of x = 1∕3, we have g ¨ (x) < 0. Thus g(1∕3) is a local maximum. See Figure 4.18. y g(x) = xe−3x

x 1∕3

Figure 4.18

21. We have ℎ¨ (x) = 1 − 1∕x2 . To fnd critical points, we set ℎ¨ (x) = 0. Then 1−

1 =0 x2 1 1= 2 x x = ±1.

Therefore, the critical points of ℎ are x = −1 and x = 1. For 0 < x < 1, we have ℎ¨ (x) < 0, and for x > 1, we have ℎ¨ (x) > 0. Thus we have a local minimum at x = 1. For x < −1, we have ℎ¨ (x) > 0 and for −1 < x < 0, we have ℎ¨ (x) < 0. Thus x = −1 is a local maximum. See Figure 4.19. y

x −1 ℎ(x) = x +

1 x

Figure 4.19

22. f ¨ (x) = 12x3 − 12x2 . To fnd critical points, we set f ¨ (x) = 0. This implies 12x2 (x − 1) = 0. So the critical points of f are x = 0 and x = 1. To the left of x = 0, f ¨ (x) < 0. Between x = 0 and x = 1, f ¨ (x) < 0. To the right of x = 1, f ¨ (x) > 0. Therefore, f (1) is a local minimum, but f (0) is not a local extremum. See Figure 4.20.

254

Chapter Four /SOLUTIONS y

f (x) = 3x4 − 4x3 + 6

x 1

Figure 4.20

23. f ¨ (x) = 7(x2 − 4)6 2x = 14x(x − 2)6 (x + 2)6 . The critical points of f are x = 0, x = ±2. To the left of x = −2, f ¨ (x) < 0. Between x = −2 and x = 0, f ¨ (x) < 0. Between x = 0 and x = 2, f ¨ (x) > 0. To the right of x = 2, f ¨ (x) > 0. Thus, f (0) is a local minimum, whereas f (−2) and f (2) are not local extrema. See Figure 4.21. y

x −2

f (x) = (x2 − 4)7

Figure 4.21

24. f ¨ (x) = 4(x3 − 8)3 3x2 = 12x2 (x − 2)3 (x2 + 2x + 4)3 . So the critical points are x = 0 and x = 2. To the left of x = 0, f ¨ (x) < 0. Between x = 0 and x = 2, f ¨ (x) < 0. To the right of x = 2, f ¨ (x) > 0. Thus, f (2) is a local minimum, whereas f (0) is not a local extremum. See Figure 4.22.

4.1 SOLUTIONS y

f (x) = (x3 − 8)4

x 2

Figure 4.22 25.

(1 − x)(1 + x) x2 + 1 − x ⋅ 2x 1 − x2 = 2 = . (x + 1)2 (x2 + 1)2 (x2 + 1)2 ¨ Critical points are x = ±1. To the left of x = −1, f (x) < 0. Between x = −1 and x = 1, f ¨ (x) > 0. To the right of x = 1, f ¨ (x) < 0. So, f (−1) is a local minimum, f (1) a local maximum. See Figure 4.23. f ¨ (x) =

x −1 1 f (x) =

x x2 + 1

Figure 4.23 26. (a) There is a critical point at (1, 5) since f ¨ (1) is zero. The point is a local maximum since f ¨¨ (1) is negative. (b) With x as the independent variable, the graph near the maximum is shown in Figure 4.24. (1, 5)

f (x)

x 2

Figure 4.24

255

256

Chapter Four /SOLUTIONS

27. (a) There is a critical point at (−5, 4) since g ¨ (−5) is zero. The point is a local minimum since g ¨¨ (−5) is positive. (b) With x as the independent variable, the graph near the minimum is shown in Figure 4.25.

g(x) (−5, 4) x −6

−4

Figure 4.25 28. (a) There is a critical point at (2, −5) since ℎ¨ (2) is zero. The point is a local maximum since ℎ¨¨ (2) is negative. (b) With x as the independent variable, the graph near the maximum is shown in Figure 4.26. x 1

3 (2, −5) ℎ(x)

Figure 4.26 29. (a) There is a critical point at (3, 5) since j ¨ (3) is undefned. The point is a local minimum since j ¨ (x) is negative to the left of x = 3 and positive to the right of x = 3. (b) The graph near the minimum is shown in Figure 4.27.

j(x)

(3, 5) x 2

Figure 4.27 30. Since f ¨ (x) = 4x3 − 12x2 + 8, we see that f ¨ (1) = 0, as we expected. We apply the second-derivative test to f ¨¨ (x) = 12x2 − 24x. Since f ¨¨ (1) = −12 < 0, the graph is concave down at the critical point x = 1, making it a local maximum. 31. Di˙erentiating using the product rule gives f ¨ (x) = 3x2 (1 − x)4 − 4x3 (1 − x)3 = x2 (1 − x)3 (3(1 − x) − 4x) = x2 (1 − x)3 (3 − 7x). The critical points are the solutions to f ¨ (x) = x2 (1 − x)3 (3 − 7x) = 0 3 x = 0, 1, . 7

4.1 SOLUTIONS

257

For x < 0, since 1 − x > 0 and 3 − 7x > 0, we have f ¨ (x) > 0. For 0 < x < 73 , since 1 − x > 0 and 3 − 7x > 0, we have f ¨ (x) > 0.

For 73 < x < 1, since 1 − x > 0 and 3 − 7x < 0, we have f ¨ (x) < 0. For 1 < x, since 1 − x < 0 and 3 − 7x < 0, we have f ¨ (x) > 0. Thus, x = 0 is neither a local maximum nor a local minimum; x = 3∕7 is a local maximum; x = 1 is a local minimum. 32. We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0. We have: f ¨ (x) = 2Ax + B = 0 −B x= . 2A There is one critical point, at x = −B∕2A. 33. We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0. We have: f ¨ (x) = 3Ax2 − B = 0 B x2 = 3A u x=± There are two critical points, at x =

B . 3A

√ √ B∕3A and x = − B∕3A.

34. We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0. We have: f ¨ (x) = 2Ax − 3Bx2

= 0

x(2A − 3Bx) = 0 2A . 3B

x = 0 and x = There are two critical points, at x = 0 and x = 2A∕3B.

35. We fnd a critical point by noting where f ¨ (x) = 0 or f ¨ is undefned. Since the curve is smooth throughout, f ¨ is always defned, so we look for where f ¨ (x) = 0. We have: f ¨ (x) = 4Ax3 − 2Bx = 0 2x(2Ax2 − B) = 0

x = 0 and x = ± There are three critical points, at x = 0 and x =

B . 2A

√ √ B∕2A and x = − B∕2A.

36. To fnd the critical points, we set the derivative equal to zero and solve for t. F ¨ (t) = U et + V e−t (−1) = 0 V U et − t = 0 e V U et = t e U e2t = V V e2t = U 2t = ln(V ∕U ) ln(V ∕U ) t= . 2 The derivative F ¨ (t) is never undefned, so the only critical point is t = 0.5 ln(V ∕U ).

258

Chapter Four /SOLUTIONS

37. The critical points of f are zeros of f ¨ . Just to the left of the frst critical point f ¨ > 0, so f is increasing. Immediately to the right of the frst critical point f ¨ < 0, so f is decreasing. Thus, the frst point must be a maximum. To the left of the second critical point, f ¨ < 0, and to its right, f ¨ > 0; hence it is a minimum. On either side of the last critical point, f ¨ > 0, so it is neither a maximum nor a minimum. See the fgure below. See Figure 4.28. f ¨ (x)

local min

❄

✻

local max

neither max nor min

Figure 4.28 38. (a) Increasing for x > 0, decreasing for x < 0. (b) f (0) is a local and global minimum, and f has no global maximum. 39. (a) Increasing for all x. (b) No maxima or minima. 40. (a) Decreasing for x < 0, increasing for 0 < x < 4, and decreasing for x > 4. (b) f (0) is a local minimum, and f (4) is a local maximum. 41. (a) Decreasing for x < −1, increasing for −1 < x < 0, decreasing for 0 < x < 1, and increasing for x > 1. (b) f (−1) and f (1) are local minima, f (0) is a local maximum. 42. The function f has critical points at x = 1, x = 3, x = 5. By the frst derivative test, since f ¨ is positive to the left of x = 1 and negative to the right, x = 1 is a local maximum. Since f ¨ is negative to the left of x = 3 and positive to the right, x = 3 is a local minimum. Since f ¨ does not change sign at x = 5, this point is neither a local maximum nor a local minimum. 43. Since P ¨ (t) ≈ N(t), the slope of the graph of P (t) approximates the values of N(t). (a) Local maxima of N(t) correspond to points on the graph of P (t) at which the slope is larger than all nearby values of t. The only numbers in the list where this happens are t = 130 and t = 320, so these are the only t-values among the given options where local maxima of N(t) occur. Further local maxima are visible in the graph of N(t) in Figure 4.29 (for example, at t = 205 and t = 290) but they are diÿcult to identify in the graph of P (t). (b) Similarly, local minima of N(t) correspond to points on the graph of P (t) at which the slope is smaller than all nearby values of t. The only numbers in the list where this happens are t = 0 and t = 190, so these are the only t-values among the given options where local minima of N(t) occur. Further local minima are visible in the graph of N(t) in Figure 4.29 (for example, at t = 210 and t = 300) but they are diÿcult to identify in the graph of P (t). Arizona, new cases N 10,000 7500 5000 2500 0 100

200

300

400

t (days)

Figure 4.29 44. (a) The demand for the product is increasing when f ¨ (t) is positive, and decreasing when f ¨ (t) is negative. Inspection of the table suggests that demand is increasing during weeks 0 to 2 and weeks 6 to 10, and decreasing during weeks 3 to 5. (b) Because the demand for the product is increasing during weeks 0 to 2, decreasing during weeks 3 to 5, and increasing during weeks 6 to 10, the demand has a local minimum during week 0, a local maximum sometime during week 2 or 3, a local minimum during week 5 or 6, and a local maximum during week 10.

4.1 SOLUTIONS

259

45. (a) A critical point occurs when f ¨ (x) = 0. Since f ¨ (x) changes sign between x = 2 and x = 3, between x = 6 and x = 7, and between x = 9 and x = 10, we expect critical points at around x = 2.5, x = 6.5, and x = 9.5. (b) Since f ¨ (x) goes from positive to negative at x ≈ 2.5, a local maximum should occur there. Similarly, x ≈ 6.5 is a local minimum and x ≈ 9.5 a local maximum. 46. Looking at the graph of the derivative function in Figure 4.30, we see that f (t) is increasing when 0 < t < 2 or when t > 4, because f ¨ (t) > 0 for these values of t. f (t) is decreasing when 2 < t < 4, because f ¨ (t) < 0 for these values of t. f (t) has a local maximum at t = 2, since f (t) is increasing to the left of t = 2 and decreasing to the right of t = 2. In addition, f (t) has a local maximum at t = 5 since f (t) is increasing to the left of t = 5 and t = 5 is an endpoint. Also, f (t) has a local minimum at t = 4, since it is decreasing to the left of t = 4 and increasing to the right of t = 4. In addition, f (t) has a local minimum at t = 0 since it is increasing to the right of t = 0 and t = 0 is an endpoint. f ¨ (t) 14 10 6 2 t −2

Figure 4.30

47. (a) To fnd the critical points, we set the derivative equal to zero and solve for x: f ¨ (x) = 3x2 − a = 0 a x2 = 3

u a . x=± 3

√ (b) We want the local extrema to occur at ±2, so we need the critical points ± a∕3 to occur at ±2. Thus, we need to have √ a∕3 = 2. We solve for a: u a =2 3 a =4 3 a = 12. 48. (a) To fnd the critical points, we set the derivative equal to zero and solve for x: f ¨ (x) = 5a − 4x = 0 5a x= . 4 (b) We want the local maximum to occur at x = 6, so we need the critical point 5a∕4 to occur at 6. We set these equal and solve for a: 5a =6 4 24 a= = 4.8. 5 Notice that we know this critical point gives us a local maximum since the graph of f (x) is a parabola with negative leading coeÿcient.

260

Chapter Four /SOLUTIONS

49. Since f (x) has its minimum at x = 3, then x = 3 must be a critical point. So f ¨ (3) = 0. f ¨ (x) = 2x + a

f ¨ (3) = 6 + a.

Since f ¨ (3) = 0, then 6 + a = 0 or a = −6. Since (3, 5) is a point on the graph of f (x) we must have f (3) = 5: f (3) = 32 + a(3) + b = 9 + 3a + b = 5. Since we know that a = −6, we have: f (3) = 9 + 3(−6) + b = b − 9 = 5

b = 14.

Thus, we have found that a = −6, b = 14, giving us f (x) = x − 6x + 14. 50. If the minimum of f (x) is at (−2, −3), then the derivative of f must be equal to 0 there. In other words, f ¨ (−2) = 0. If f (x) = x2 + ax + b,

then

f (x) = 2x + a f ¨ (−2) = 2(−2) + a = −4 + a = 0 so a = 4. Since (−2, −3) is on the graph of f (x) we know that f (−2) = −3. So f (−2) = (−2)2 + a(−2) + b = −3 (−2)2 + 4(−2) + b = −3

a = 4, so

4 − 8 + b = −3 −4 + b = −3 b=1 so a = 4 and b = 1, and f (x) = x2 + 4x + 1. 51. We wish to have f ¨ (3) = 0. Di˙erentiating to fnd f ¨ (x) and then solving f ¨ (3) = 0 for a gives: f ¨ (x) = x(aeax ) + 1(eax ) = eax (ax + 1) f ¨ (3) = e3a (3a + 1) = 0 3a + 1 = 0 1 a=− . 3 Thus, f (x) = xe−x∕3 . 52. Recall that the natural logarithm is undefned for x ≤ 0, so the domain of f is x > 0. We see from looking at the graph of f (x) = x − ln x in the text that this function has one local minimum. We want to assign values to a and b so that this local minimum occurs at x = 2. The function must therefore have a critical point at x = 2. We fnd the derivative of f (x) = a(x − b ln x) and the critical points in terms of a and b. 1 f ¨ (x) = a 1 − b =0 x 1 1−b =0 x b 1= x x=b We see that f (x) has only one critical point, at x = b. Since we want a critical point at x = 2, we choose b = 2. Since b = 2, we have f (x) = a(x − 2 ln x). We now use the condition that f (2) = 5 to fnd a: f (2) = 5 a(2 − 2 ln 2) = 5 a = 5∕(2 − 2 ln 2) a ≈ 8.147. We let a = 8.147 and b = 2, so the function is f (x) = 8.147(x − 2 ln x). If we sketch a graph of this function, we see that this function does indeed have a local minimum approximately at the point (2, 5).

4.1 SOLUTIONS 53. a = −2 a=0 a=2

261

✲ ✲ ✲ ✲ ✲

a=4 a=6

✛ y = − 1 x3 2 dy d � 3 = 0. Since dx x − ax2 = 3x2 − 2ax, we want 3x2 − 2ax = 0, so x = 0 or To solve for the critical points, we set dx

x = 32 a. At x = 0, we have y = 0. This frst critical point is independent of a and lies on the curve y = − 21 x3 . At x = 32 a, 3 4 3 we calculate y = − 27 a = − 12 23 a . Thus the second critical point also lies on the curve y = − 12 x3 . 54. (a) The function f (x) is defned for x ≥ 0. We set the derivative equal to zero and solve for x to fnd critical points: 1 f ¨ (x) = 1 − ax−1∕2 = 0 2 a 1− √ = 0 2 x √ 2 x=a x=

a2 . 4

Notice that f ¨ is undefned at x = 0 so there are two critical points: x = 0 and x = a2 ∕4. (b) We want the critical point x = a2 ∕4 to occur at x = 5, so we have: a2 4 20 = a2 √ a = ± 20. 5=

Since a is positive, we use the positive square root. The second derivative, f ¨¨ (x) =

1 −3∕2 1 √ ax = 20x−3∕2 4 4

is positive for all x > 0, so the function is concave up and x = 5 gives a local minimum. See Figure 4.31. y 5

x 5

Figure 4.31

262

Chapter Four /SOLUTIONS

55. Using the product rule on the function f (x) = axebx , we have f ¨ (x) = aebx + abxebx = aebx (1 + bx). We want f ( 31 ) = 1, and since this is to be a maximum, we require f ¨ ( 13 ) = 0. These conditions give f (1∕3) = a(1∕3)eb∕3 = 1, f ¨ (1∕3) = aeb∕3 (1 + b∕3) = 0. Since ae(1∕3)b is non-zero, we can divide both sides of the second equation by ae(1∕3)b to obtain 0 = 1 + 3b . This implies

b = −3. Plugging b = −3 into the frst equation gives us a( 31 )e−1 = 1, or a = 3e. How do we know we have a maximum at x = 31 and not a minimum? Since f ¨ (x) = aebx (1 + bx) = (3e)e−3x (1 − 3x), and (3e)e−3x is always positive, it follows that f ¨ (x) > 0 when x < 31 and f ¨ (x) < 0 when x > 31 . Since f ¨ is positive to the left of x = 13 and negative to the right

of x = 13 , f ( 13 ) is a local maximum. 56. (a) See Figure 4.32.

large a small a

❄

❘

Figure 4.32 (b) We see in Figure 4.32 that in each case the graph of f is a parabola with one critical point, its vertex, on the positive x-axis. The critical point moves to the right along the x-axis as a increases. (c) To fnd the critical points, we set the derivative equal to zero and solve for x. f ¨ (x) = 2(x − a) = 0 x = a. The only critical point is at x = a. As we saw in the graph, and as a increases, the critical point moves to the right. 57. (a) See Figure 4.33.

✻

■ small a

large a

Figure 4.33 (b) We see in Figure 4.33 that in each case f has two critical points placed symmetrically about the origin, one in each of quadrants II and IV. As a increases, they appear to move farther apart, the one in quadrant II up and to the left, the one in quadrant IV down and to the right. (c) To fnd the critical points, we set the derivative equal to zero and solve for x. f ¨ (x) = 3x2 − a = 0 a x2 = 3

u a x=± . 3

√ √ There are two critical points, at x = a∕3 and x = − a∕3. (Since the parameter a is positive, the critical points exist.) As we saw in the graph, and as a increases, the critical points both move away from the vertical axis.

4.2 SOLUTIONS

263

58. (a) See Figure 4.34. small a

❄

large a

❄

Figure 4.34 (b) We see in Figure 4.34 that in each case f appears to have two critical points. One critical point is a local minimum at the origin and the other is a local maximum in quadrant I. As the parameter a increases, the critical point in quadrant I appears to move down and to the left, closer to the origin. (c) To fnd the critical points, we set the derivative equal to zero and solve for x. Using the product rule, we have: f ¨ (x) = x2 ⋅ e−ax (−a) + 2x ⋅ e−ax = 0 xe−ax (−ax + 2) = 0 2 x = 0 and x = . a There are two critical points, at x = 0 and x = 2∕a. As we saw in the graph, as a increases the nonzero critical point moves to the left. 59. By the product rule f ¨ (x) =

d (xm (1 − x)n ) = mxm−1 (1 − x)n − nxm (1 − x)n−1 dx = xm−1 (1 − x)n−1 (m(1 − x) − nx) = xm−1 (1 − x)n−1 (m − (m + n)x).

We have f ¨ (x) = 0 at x = 0, x = 1, and x = m∕(m + n), so these are the three critical points of f . We can classify the critical points by determining the sign of f ¨ (x). If x < 0, then f ¨ (x) has the same sign as (−1)m−1 : negative if m is even, positive if m is odd. If 0 < x < m∕(m + n), then f ¨ (x) is positive. If m∕(m + n) < x < 1, then f ¨ (x) is negative. If 1 < x, then f ¨ (x) has the same sign as (−1)n : positive if n is even, negative if n is odd. If m is even, then f ¨ (x) changes from negative to positive at x = 0, so f has a local minimum at x = 0. If m is odd, then f ¨ (x) is positive to both the left and right of 0, so x = 0 is an infection point of f . At x = m∕(m + n), the derivative f ¨ (x) changes from positive to negative, so x = m∕(m + n) is a local maximum of f. If n is even, then f ¨ (x) changes from negative to positive at x = 1, so f has a local minimum at x = 1. If n is odd, then f ¨ (x) is negative to both the left and right of 1, so x = 1 is an infection point of f .

Solutions for Section 4.2 1. We fnd an infection point by noting where the concavity changes. Such points are shown in Figure 4.35. There are two infection points. f (x)

Figure 4.35

264

Chapter Four /SOLUTIONS

2. We fnd an infection point by noting where the concavity changes. Such points are shown in Figure 4.36: There are two infection points. f (x)

Figure 4.36

3. We fnd an infection point by noting where the concavity changes. Such points are shown in Figure 4.37. There are three infection points. f (t)

Figure 4.37

4. We fnd an infection point by noting where the concavity changes. Looking at Figure 4.38, we see that in fact the concavity changes only at the critical point. So there is one infection point, which is also a critical point. f (x)

Figure 4.38

5. One possible answer is shown in Figure 4.39.

✠

Critical point Inflection

✠ point

Figure 4.39

4.2 SOLUTIONS

265

6. (a) One possible answer is shown in Figure 4.40. local maxima

✮

❄

✻ ❦

✗

✒

✻ inflection points

local minima

Figure 4.40

Figure 4.41

(b) This function is concave down at each local maximum and concave up at each local minimum, so it changes concavity at least three times. This function has at least 3 infection points. See Figure 4.41 7. See Figure 4.42. y

Figure 4.42

8. (a) Critical point. (b) Infection point. 9. (a) There was a critical point at 6 pm when the temperature was at a local minimum. (b) The graph of temperature was decreasing but concave up in the morning. In the early afternoon the graph was decreasing but concave down. There was an infection point at noon when the northerly wind started blowing. By 6 pm when the temperature was at a local minimum, the graph must have been concave up again so there must have been a second infection point between noon and 6 pm. See Figure 4.43. temperature

9 am noon 3 pm 6 pm

time

Figure 4.43

10. We have f ¨ (x) = 3x2 − 36x − 10 and f ¨¨ (x) = 6x − 36. The infection point occurs where f ¨¨ (x) = 0, hence 6x − 36 = 0. The infection point is at x = 6. A graph is shown in Figure 4.44.

266

Chapter Four /SOLUTIONS f (x)

x 6

Figure 4.44

11. From the graph of f (x) in the fgure below, we see that the function must have two infection points. We calculate f ¨ (x) = 4x3 + 3x2 − 6x, and f ¨¨ (x) = 12x2 + 6x − 6. Solving f ¨¨ (x) = 0 we fnd that: x1 = −1 and

1 . 2

x2 =

Since f ¨¨ (x) > 0 for x < x1 , f ¨¨ (x) < 0 for x1 < x < x2 , and f ¨¨ (x) > 0 for x2 < x, it follows that both points are infection points.

f (x)

4 2

✛

Inflection point

−2 Inflection point

✲

−2

12. A critical point will occur whenever f ¨ (x) = 0 or f ¨ is undefned. Since f ¨ (x) is always defned, we set f ¨ (x) = 2x − 5 = 0 2x = 5 5 x= 2 To fnd the infection points of f (x), we fnd where f ¨¨ (x) goes from negative to positive or vice versa. For a point to satisfy this condition, it must have at least f ¨¨ (x) = 0 or f ¨¨ undefned. Since f ¨¨ (x) = 2, we know f ¨¨ (x) is always defned and never equal to zero, so f (x) cannot have infection points. So x = 52 is a critical point of f (x), and there are no infection points. To identify the nature of the critical point x = 52 that we have found, we can look at a graph of f (x) for values of x

near the critical point. Such a graph is shown in Figure 4.45. From the graph we see that f ( 25 ) is a local minimum of f .

x 2.5

local min

✲ (2.5, −3.25)

Figure 4.45

4.2 SOLUTIONS

267

13. A critical point will occur whenever f ¨ (x) = 0 or f ¨ is undefned. Since f ¨ (x) is always defned, we set f ¨ (x) = 3x2 − 3 = 3(x2 − 1) = 0. Factoring, we get f ¨ (x) = 3(x − 1)(x + 1) = 0. So, x = 1 or x = −1. To fnd the infection points of f (x), we fnd where f ¨¨ (x) goes from negative to positive or vice versa. For a point to satisfy this condition, it must have at least f ¨¨ (x) = 0 or f ¨¨ undefned. Since f ¨¨ (x) = 6x, we know f ¨¨ (x) is always defned. It is zero when 6x = 0, so x = 0. Since f ¨¨ (x) = 6x is negative for x < 0 and positive for x > 0, x = 0 must be an infection point for f (x). So x = 1 and x = −1 are critical points of f (x), and x = 0 is an infection point for f (x). To identify the nature of the critical points x = 1 and x = −1 that we have found, we can look at a graph of f (x) for values of x near the critical points. Such a graph is shown in Figure 4.46. From the graph we see that f (−1) is a local maximum of f and f (1) is a local minimum of f . y f (x) (−1, 12)

✲

Local max

✛ (0, 10)

Point of inflection

✛

Local min

(1, 8)

Figure 4.46

14. f ¨ (x) = 6x2 + 6x − 36. To fnd critical points, we set f ¨ (x) = 0. Then 6(x2 + x − 6) = 6(x + 3)(x − 2) = 0. Therefore, the critical points of f are x = −3 and x = 2. To fnd the infection points of f (x) we look for the points at which f ¨¨ (x) goes from negative to positive or vice-versa. Since f ¨¨ (x) = 12x + 6, x = −1∕2 is an infection point. From Figure 4.47, we see the critical point x = −3 is a local maximum and the critical point x = 2 is a local minimum. y

(−3, 86)

local max

✲

(−.5, 23.5)

✒ point of inflection

✛ (2, −39)

Figure 4.47

local min

268

Chapter Four /SOLUTIONS

15. A critical point will occur whenever f ¨ (x) = 0 or f ¨ is undefned. Since f ¨ (x) is always defned, we set f ¨ (x) =

x2 x 3x2 2x + −1 = + − 1 = 0. 6 4 2 2

Factoring, we get f ¨ (x) =

1 2 1 (x + x − 2) = (x − 1)(x + 2) = 0. 2 2

So x = 1 or x = −2. To fnd the infection points of f (x), we fnd where f ¨¨ (x) goes from negative to positive or vice versa. For a point to satisfy this condition, it must have at least f ¨¨ (x) = 0 or f ¨¨ undefned. Since f ¨¨ (x) = x + 21 , we know

f ¨¨ (x) is always defned and is zero when x = − 12 . Since f ¨¨ (x) = x + 12 is negative for x < − 12 and positive for x > − 12 , x = − 12 must be an infection point for f (x).

So x = 1 and x = −2 are critical points of f (x), and x = − 12 is an infection point for f (x). To identify the nature of the critical points x = 1 and x = −2 that we have found, we can look at a graph of f (x) for values of x near the critical points. Such a graph is shown in Figure 4.48. From the graph we see that f (−2) is a local maximum of f and f (1) is a local minimum of f . (−2, 11∕3)

✛

− 12 , 61 24

Local max ✛

Point of inflection

✛ 1, 17 12

Local min

Figure 4.48

16. A critical point will occur whenever f ¨ (x) = 0 or f ¨ is undefned. Since f ¨ (x) is always defned, we set f ¨ (x) = 4x3 − 4x = 4(x3 − x) = 0. Factoring, we get, f ¨ (x) = 4x(x2 − 1) = 4x(x − 1)(x + 1) = 0. So x = −1, x = 0 or x = 1. To fnd the infection points of f (x), we fnd where f ¨¨ (x) goes from negative to positive or vice versa. For a point to satisfy this condition, it must have at least f ¨¨ (x) = 0 or f ¨¨ undefned. Since f ¨¨ (x) = 12x2 − 4, we know f ¨¨ (x) is always defned. Factoring, we get H IH I 1 1 1 ¨¨ 2 2 = 0, x+ √ = 12 x − √ f (x) = 12x − 4 = 12 x − 3 3 3 √ √ √ in numbers close to −1∕ 3, we see that f ¨¨ (x) is positive for which is √ zero when x = −1∕ 3 or x√= 1∕ 3. Putting √ point for f (x).√ Putting in numbers close x < −1∕ √ 3 and negative for x > −1∕ 3, so x = −1∕ √ 3 must be an infection√ to 1∕ 3, we see that f ¨¨ (x) is negative for x < 1∕ 3 and positive for x > 1∕ 3, so x = 1∕ 3 must be an infection point for f (x). √ √ So x = −1, x = 0 and x = 1 are critical points of f (x), and x = −1∕ 3 and x = 1∕ 3 are infection points for f (x). To identify the nature of the critical points x = −1, x = 0 and x = 1 that we have found, we can look at a graph of f (x) for values of x near the critical points. Such a graph is shown in Figure 4.49. From the graph we see that f (−1) is a local minimum of f , f (0) is a local maximum of f and f (1) is a local minimum of f .

4.2 SOLUTIONS

269

y f (x) Local max 0

(0, 0) ✠

− √1 , − 59 3

Local min

1 5 1 √ ,− 9 3

✻

■

✲

Points of (1, −1) inflection ■

(−1, −1)

Local min

Figure 4.49

17. f ¨ (x) = 12x3 − 12x2 . To fnd critical points, we set f ¨ (x) = 0. This implies 12x2 (x − 1) = 0. So the critical points of f are x = 0 and x = 1. To fnd the infection points of f (x) we look for points at which f ¨¨ (x) goes from negative to positive or vice-versa. At any such point f ¨¨ (x) is either zero or undefned. Since f ¨¨ (x) = 36x2 − 24x = 12x(3x − 2), our candidate points are x = 0 and x = 2∕3. Both x = 0 and x = 2∕3 are infection points, since at both, f ¨¨ (x) changes sign. From Figure 4.50, we see that the critical point x = 1 is a local minimum and the critical point x = 0 is neither a local maximum nor a local minimum. y

point of inflection (0, 6)

❄

✒(2∕3, 5.4)

(1, 5)

✻

point of inflection

local min

x −1

Figure 4.50

18. The derivative of f (x) is f ¨ (x) = 4x3 − 16x. The critical points of f (x) are points at which f ¨ (x) = 0. Factoring, we get 4x3 − 16x = 0 4x(x2 − 4) = 0 4x(x − 2)(x + 2) = 0 So the critical points of f will be x = 0, x = 2, and x = −2. To fnd the infection points of f we look for the points at which f ¨¨ (x) cha√ nges sign. At any such point f ¨¨ (x) is either ¨¨ 2 zero or undefned. Since f (x) = 12x − 16 our candidate points are x = ±2∕ 3. At both of these points f ¨¨ (x) changes sign, so both of these points are infection points. From Figure 4.51, we see that the critical points x = −2 and x = 2 are local minima and the critical point x = 0 is a local maximum.

270

Chapter Four /SOLUTIONS

(0, ✛5)

local max x

−2

■ local min

✲

✒ ✛ points of inflection (2, −11)

(−2, −11)

local min

Figure 4.51

19. We frst fnd the critical points of f (x) = x4 − 4x3 + 10. Since f ¨ (x) = 4x3 − 12x2 , setting the derivative equal to 0 and factoring yields 4x3 − 12x2 = 0 4x2 (x − 3) = 0 So x = 0 and x = 3 are the critical points of f (x). We now fnd the infection points of f (x). Since f ¨¨ (x) = 12x2 − 24x, setting the second derivative equal to 0 and factoring yields 12x2 − 24x = 0 12x(x − 2) = 0. So x = 0 and x = 2 may be points of infection of f (x). Since f ¨¨ (x) changes sign at both x = 0 and x = 2, both are points of infection for f (x). From Figure 4.52, we see that the critical point x = 3 is a local minimum and the critical point x = 0 is neither a local minimum nor a local maximum. y

f (x)

✛ (0, 10)

points of inflection

❄ (2, −6)

local min

✲ (3, −17)

Figure 4.52

20. A critical point will occur whenever f ¨ (x) = 0 or f ¨ is undefned. Since f ¨ (x) is always defned, we set f ¨ (x) = 5x4 − 20x3 = 5(x4 − 4x3 ) = 0. We get f ¨ (x) = 5x3 (x − 4) = 0,

4.2 SOLUTIONS

271

so x = 0 or x = 4. To fnd the infection points of f (x), we fnd where f ¨¨ (x) goes from negative to positive or vice versa. For a point to satisfy this condition, it must have at least f ¨¨ (x) = 0 or f ¨¨ undefned. Since f ¨¨ (x) = 20x3 − 60x2 , we know f ¨¨ (x) is always defned. It is zero when 20x3 − 60x2 = 0. Factoring, we get f ¨¨ (x) = 20(x3 − 3x2 ) = 20x2 (x − 3) = 0. So x = 0 or x = 3. Putting in values of x very close to zero, we see that f ¨¨ (x) = 20x2 (x − 3) is negative for x < 0 and negative for x > 0, so x = 0 is not an infection point for f (x). Putting in values close to 3, we see that f ¨¨ (x) = 20x2 (x−3) is negative for x < 3 and positive for x > 3, so x = 3 is an infection point for f (x). So x = 0 and x = 4 are critical points of f (x), and x = 3 is an infection point for f (x). To identify the nature of the critical points x = 0 and x = 4 that we have found, we can look at a graph of f (x) for values of x near the critical points. Such a graph is shown in Figure 4.53. From the graph we see that f (0) is a local maximum of f and f (4) is a local minimum of f . y f (x)

(0, 35)

✒

Local max Point of inflection

❄ (3, −127) Local min

✠ (4, −221)

Figure 4.53 21. We see that f ¨ (x) = 15x4 − 15x2 and f ¨¨ (x) = 60x3 − 30x. Since f ¨ (x) = 15x2 (x2 − 1), the critical points are x = 0, ±1. ¨¨ ¨¨ 2 ¨¨ To fnd possible inf √ection points, we determine when f (x) = 0. Since f (x) = 30x(2x − 1), we have f (x) = 0 ¨¨ when x = 0 or x = ±1∕ 2. Since f (x) changes sign at each of these points, all are infection points. We see from Figure 4.54 that the critical point x = −1 is a local maximum, x = 1 is a local minimum, and x = 0 is neither. inflection point (−1, 2)

inflection point

❄

✠ ✻(1, −2) inflection point

Figure 4.54 2

22. (a) A graph of f (x) = e−x is shown in Figure 4.55. It appears to have one critical point, at x = 0, and two infection points, one between 0 and 1 and the other between 0 and −1. 2

Inflection point

Critical 1 point Inflection point x

−2

−1

Figure 4.55

272

Chapter Four /SOLUTIONS 2

(b) To fnd the critical points, we set f ¨ (x) = 0. Since f ¨ (x) = −2xe−x = 0, there is one solution, x = 0. The only critical point is at x = 0. To fnd the infection points, we frst use the product rule to fnd f ¨¨ (x). We have 2

f ¨¨ (x) = (−2x)(e−x (−2x)) + (−2)(e−x ) = 4x2 e−x − 2e−x . We set f ¨¨ (x) = 0 and solve for x by factoring: 2

4x2 e−x − 2e−x = 0 2

(4x2 − 2)e−x = 0. 2

Since e−x is never zero, we have 4x2 − 2 = 0 1 x2 = 2

√ x = ±1∕ 2.

√ √ There are exactly two infection points, at x = 1∕ 2 ≈ 0.707 and x = −1∕ 2 ≈ −0.707. 23. (a) We set the derivative equal to zero and solve for x to fnd critical points: f ¨ (x) = 4x3 − 4ax = 0 4x(x2 − a) = 0. We see that there are three critical points: Critical points:

x = 0,

√ a,

√ x = − a.

To fnd possible infection points, we set the second derivative equal to zero and solve for x: f ¨¨ (x) = 12x2 − 4a = 0. There are two possible infection points: u Possible infection points:

a , 3

u a x=− . 3

To see if these are infection points, we determine whether concavity changes by evaluating f ¨¨ at values on either side of each of the potential infection points. We see that u a a ) = 12(4 ) − 4a = 16a − 4a = 12a > 0, f ¨¨ (−2 3 3 √ so f is concave up to the left of x = − a∕3. Also, f ¨¨ (0) = −4a < 0, √ √ so f is concave down between x = − a∕3 and x = a∕3. Finally, we see that u a a f ¨¨ (2 ) = 12(4 ) − 4a = 16a − 4a = 12a > 0, 3 3 √ √ √ so f is concave up to the right of x = a∕3. Since f (x) changes concavity at x = a∕3 and x = − a∕3, both points are infection points. √ (b) The only positive critical point is at x = a, so to have a critical point at x = 2, we substitute: √ a √ 2= a

a = 4.

4.2 SOLUTIONS

273

Since the critical point is at the point (2, 5), we have f (2) = 5 24 − 2(4)22 + b = 5 16 − 32 + b = 5 b = 21. The function is f (x) = x4 − 8x2 + 21. √ √ (c) We have seen that a = 4, so the infection points are at x = 4∕3 and x = − 4∕3. 24. The infection points of f are the points where f ¨¨ changes sign. See Figure 4.56. inflection points of f

☛

❯

f ¨¨ (x)

Figure 4.56

25. (a) At a critical point, the derivative is zero or undefned. The derivative appears to be defned at all points for this function, so we look for places where the derivative is zero, which means a line tangent to the function is horizontal. This occurs at three x-values: B, D, and F . (b) At an infection point, the concavity of the function changes, from concave down to concave up or vice versa. This occurs at three x-values: B, C, and E. (c) There appears to be a local maximum at the point with x-value F , so f has one local maximum. There appears to be a local minimum at the point with x-value D, so f has one local minimum. These are the only local extrema for this graph. 26. (a) At a critical point, the derivative is zero or undefned. We see from the graph of the derivative that it is defned at all points so we look for places where the derivative is zero, which is where the graph hits the x-axis, which is at x-values D and G. (b) At an infection point, the function changes concavity. This can occur at a point where the second derivative is zero or undefned. Since the second derivative is the derivative of the derivative, we look for places on the graph of f ¨ (x) where the derivative (or slope) is zero or undefned. This appears to happen at three x-values: B, D, and F . Which of these might be infection points? At an infection point, the concavity must change, which means the second derivative must change signs. That means the derivative of the derivative must change signs, or that the derivative (slope) of the function shown in the graph must change from positive to negative or vice versa. This is the case at D and F , but not at B. Thus, the function f has two infection points, at D and F . (c) We see in the graph that f ¨ (x) is positive or zero for all x-values to the left of G, and is negative for all x-values to the right of G. This tells us that the function f is increasing or horizontal everywhere to the left of G and is decreasing everywhere to the right of G. Thus, f has a local maximum at the point with x-value G and it does not have a local minimum. The function has one local maximum and no local minima. 27. See Figure 4.57.

274

Chapter Four /SOLUTIONS y

y¨ = 0 y¨¨ = 0

y¨¨ = 0

y¨ = 0

y = f (x) y¨ > 0 y¨¨ < 0

y¨ > 0 y¨¨ > 0

y¨ < 0 y¨¨ < 0

y¨ > 0 y¨¨ < 0

Figure 4.57

28. See Figure 4.58. y¨ < 0 everywhere

y¨¨ = 0

y¨¨ > 0

y¨¨ = 0

y¨¨ < 0

y¨¨ = 0

y¨¨ > 0

y¨¨ < 0

y = f (x) x1

Figure 4.58

29. See Figure 4.59. y

y¨ = 0 y¨¨ = 0

y¨¨ = 0 y = f (x)

y¨¨ < 0

y¨¨ > 0

y¨¨ < 0

y¨ ≥ 0 everywhere

Figure 4.59

30. See Figure 4.60.

4.2 SOLUTIONS

275

y y = f (x)

y¨ = 2 y¨¨ = 0

y¨ > 0 y¨¨ > 0

Figure 4.60

31. Infection points are points where the concavity changes. There the slope changes from increasing to decreasing, or vice versa. (a) Of the points given, the only one that looks like the concavity changes is t = 260 (November 16, 2020). On this day, P (t) changes from being concave up to being concave down. Further t-values for infection points of P (t) are visible in the graph of N(t) in Figure 4.61 (for example, at t = 290 and t = 320 where N(t) has peaks) but these are diÿcult to see in the graph of P (t). (b) Since P ¨ (t) is approximately equal to the number of new daily cases, t = 260 corresponds to a peak in the number of new daily cases. After this day, the number of daily new cases, represented by the slope of P (t), fell sharply. Montana, new cases N 1000 500 0 100

200

300 365

t (days)

Figure 4.61

32. Infection points are points where the concavity changes. In terms of the slope, they correspond to points where slope changes from being increasing to being decreasing (or vice versa), which means the slope is maximum or minimum at these points. In the graph for P (t) in the problem, we see two points where concavity changes from concave up to concave down at days t = 120 (July 30, 2020) and t = 275 (January 1, 2021). These correspond to points where the slope, P ¨ (t), is maximum: Maxima of P ¨ (t) at t = 120 and t = 275. In terms of the outbreak, since P ¨ (t) is approximately equal to the number of new daily cases, t = 120 and t = 275 correspond to peaks in the number of new daily cases. There is a third infection point at t = 190 (October 8, 2020). At this point P ¨ (t) changes from decreasing to increasing, so it corresponds to a date in which the daily new cases stopped falling and started to increase again. 33. Note that P ¨ (t), with units cases/day, is telling us approximately how many new Covid-19 cases there were on day t. (a) The function is concave up on the interval from t = 50 to t = 120. In that interval the total number of cases is increasing at an increasing rate, which means the number of new cases on day t is growing. Thus, P ¨ (t) is increasing. (b) The function is approximately straight on the interval from t = 160 to t = 200. The total number of cases is increasing at a more or less constant rate, this means the number of new cases on day t is steady. In other words, P ¨ (t) is approximately constant. (c) The function is concave down on the interval from t = 280 to t = 330. The total number of cases is increasing at a decreasing rate, which means the number of new cases on day t is falling. P ¨ (t) is decreasing.

276

Chapter Four /SOLUTIONS

34. (a) An infection point occurs whenever the concavity of f (x) changes. If the graph shown is that of f (x), then an infection point will occur whenever its concavity changes, or equivalently when the tangent line moves from above the curve to below or vice-versa. Such points are shown in Figure 4.62. f ¨ (x)

f (x)

Figure 4.62

Figure 4.63

(b) To fnd infection points of the function f we must fnd points where f ¨¨ changes sign. However, because f ¨¨ is the derivative of f ¨ , any point where f ¨¨ changes sign will be a local maximum or minimum on the graph of f ¨ . Such points are shown in Figure 4.63. (c) The infection points of f are the points where f ¨¨ changes sign. If the graph shown is that of f ¨¨ (x), then we are looking for where the given graph passes from above the x-axis to below, or vice versa. Such points are shown in Figure 4.64: f ¨¨ (x)

Figure 4.64

35. (a) The infection points of P (t) are points where the concavity changes, so points where P ¨ (t) changes from increasing to decreasing, or vice versa. Since P ¨ (t) ≈ N(t), we can fnd the t-values of the infection points of P (t) by identifying points where N(t) changes from increasing to decreasing, or vice versa in Figure 4.32 in the problem statement. (b) From part (a), we see that P (t) has infection points around t = 50, t = 110, t = 290, t = 300 and t = 320. Since N(t) changes from being increasing to decreasing at t = 50, t = 290, t = 320, the graph of P (t) changes from being concave up to concave down at these infection points. For the other infection points at t = 110 and t = 300, since N(t) changes from being decreasing to increasing, the graph of P (t) changes from being concave down to concave up. These conclusions are refected in the graph of P (t) in Figure 4.65. Massachusetts, total cases P 600,000 400,000 200,000 0 100

200

Figure 4.65

36. (a) Centimeters per week. (b) At week 24 the fetus is growing at a rate of 1.6 cm/week.

300 365

t (days)

4.2 SOLUTIONS

277

37. (a) The tangent line to the graph is steeper at 20 weeks then at 36 weeks, so f ¨ (20) is greater than f ¨ (36). (b) The fetus increases its length more rapidly at week 20 than at week 36. 38. (a) The infection point occurs at about week 14 where the graph changes from concave up to concave down. (b) The fetus increases its length faster at week 14 than at any other time during its gestation. 39. We estimate the derivatives at 20 and 36 weeks by drawing tangent lines to the length graph, shown in Figure 4.66, and calculating their slopes. (a) Two points on the tangent line at 20 weeks are (6, 0) and (32, 49). Thus, 49 − 0 = 1.9 cm∕week. 32 − 6 (b) Two points on the tangent line at 36 weeks are (0, 20) and (40, 50). Thus f ¨ (20) =

50 − 20 = 0.75 cm∕week. 40 − 0 (c) The average rate of growth is the slope of the secant line from (0, 0) to (40, 50). Thus f ¨ (36) =

Average rate of change = length (centimeters)

✛

50 − 0 = 1.25 cm∕week. 40 − 0 Slope = f ¨ (20)

Slope = f ¨ (36)

50 40

❄

30 20 10 4

8 12 16 20 24 28 32 36 40

t, age of fetus (weeks after last menstruation)

Figure 4.66 40. (a) The graph has two infection points. One is at T = 10◦ F and the other at T = 45◦ F. (b) The maximum slope in the graph occurs at the second infection point T = 45◦ F. At this point, an increase in temperature adds the most to the range of the Leaf. 41. (a) The infection point on the graph occurs close to 30 kg of seed. (b) It corresponds to the point where the derivative is largest, meaning that at that point an extra kilogram of seed increases yield the most. 42. (a) Since the volume of water in the container is proportional to its depth, and the volume is increasing at a constant rate, d(t) = Depth at time t = Kt, where K is some positive constant. So the graph is linear, as shown in Figure 4.67. Since initially no water is in the container, we have d(0) = 0, and the graph starts from the origin. depth of water

depth of water d(t)

d(t)

slope = K

Figure 4.67

Figure 4.68

(b) As time increases, the additional volume needed to raise the water level by a fxed amount increases. Thus, although the depth, d(t), of water in the cone at time t, continues to increase, it does so more and more slowly. This means d ¨ (t) is positive but decreasing, i.e., d(t) is concave down. See Figure 4.68.

278

Chapter Four /SOLUTIONS

43. See Figure 4.69. depth of water

time at which water reaches widest part of urn

✠

time

Figure 4.69 44. (a) The concavity changes at t1 and t3 , as shown in Figure 4.70.

f (t)

y2 y1

Figure 4.70 (b) f (t) grows most quickly where the vase is skinniest (at y3 ) and most slowly where the vase is widest (at y1 ). The diameter of the widest part of the vase looks to be about 4 times as large as the diameter at the skinniest part. Since the area of a cross section is given by r2 , where r is the radius, the ratio between areas of cross sections at these two places is about 42 , so the growth rates are in a ratio of about 1 to 16 (the wide part being 16 times slower). 45. Since the x3 term has coeÿcient 1, the polynomial is of the form y = x3 + ax2 + bx + c. Di˙erentiating gives dy = 3x2 + 2ax + b. dx There is a critical point at x = 2, so dy∕dx = 0 at x = 2. Thus dy || | = 3(22 ) + 2a(2) + b = 12 + 4a + b = 0, so 4a + b = −12. dx ||x=2 We take the second derivative to look for the infection point. We fnd d2y = 6x + 2a, dx2 and for an infection point at x = 1, we have 6 + 2a = 0, so a = −3. We now use a = −3 and the relationship 4a + b = −12, which gives 4(−3) + b = −12, so b = 0. We now have y = x3 − 3x2 + c, and using the point (1,4) gives 1 − 3 + c = 4, c = 6. 3

Thus, y = x − 3x + 6.

4.2 SOLUTIONS

279

46. Since y(0) = a∕(1 + b) = 2, we have a = 2 + 2b. To fnd a point of infection, we calculate dy abe−t = , dt (1 + be−t )2 and using the quotient rule, d2y −abe−t (1 + be−t )2 − abe−t 2(1 + be−t )(−be−t ) = 2 dx (1 + be−t )4 −t −t abe (−1 + be ) = . (1 + be−t )3 The second derivative is equal to 0 when be−t = 1, or b = et . When t = 1, we have b = e. The second derivative changes sign at this point, so we have an infection point. Thus y=

2 + 2e . 1 + e1−t

47. The maximum of y = e−(x−a) ∕b occurs at x = a. (This is because the exponent −(x−a)2 ∕b is zero when x = a and negative for all other x-values. The same result can be obtained by taking derivatives.) Thus we know that a = 2. Points of infection occur where d 2 y∕dx2 changes sign, that is, where d 2 y∕dx2 = 0. Di˙erentiating gives dy 2(x − 2) −(x−2)2 ∕b =− e dx b 2 d y 4(x − 2)2 −(x−2)2 ∕b 2 −(x−2)2 ∕b 2 2 2 = − e−(x−2) ∕b + e = e −1 + (x − 2)2 . 2 2 b b b b dx 2

Since e−(x−2) ∕b is never zero, d 2 y∕dx2 = 0 where 2 −1 + (x − 2)2 = 0. b We know d 2 y∕dx2 = 0 at x = 1, so substituting x = 1 gives 2 −1 + (1 − 2)2 = 0. b Solving for b gives −1 +

2 =0 b b = 2.

Since a = 2, the function is

y = e−(x−2) ∕2 . You can check that at x = 2, we have d2y 2 = e−0 (−1 + 0) < 0 2 dx2 so the point x = 2 does indeed give a maximum. See Figure 4.71. y

Point of inflection at x = 1

✛ ✲

Max at x = 2

✛

Point of inflection at x = 3

x 1

3 2

Figure 4.71: Graph of y = e−(x−2) ∕2

280

Chapter Four /SOLUTIONS

Solutions for Section 4.3 1. See Figure 4.72. y

Global and local max

8 6 Local min

4 2

Global and local min x 1

Figure 4.72

2. The global maximum is achieved at the two local maxima, which are at the same height. See Figure 4.73.

Local and global max

4 Local min

3 2

Local min

Local and global min x 2

Figure 4.73

3. We consider the part of the graph between x = 0 and x = 4. A global maximum or minimum occurs at a critical point or an endpoint. On this interval, the global minimum is at the endpoint x = 4 and global maximum is at the critical point x = 2. 4. We consider the part of the graph between x = −3 and x = 0. A global maximum or minimum occurs at a critical point or an endpoint. On this interval, the global maximum is at the endpoint x = −3 and the global minimum is at the critical point x = −1. 5. We consider the part of the graph between x = −2 and x = 3. A global maximum or minimum occurs at a critical point or an endpoint. On this interval, the global minimum is at the critical point at x = −1 and the global maximum is at the critical point at x = 2. 6. We consider the part of the graph between x = −2 and x = 1. A global maximum or minimum occurs at a critical point or an endpoint. On this interval, the global minimum is at the critical point x = −1 and the global maximum is at the endpoint x = 1. 7. We consider the part of the graph between x = 0 and x = 3. A global maximum or minimum occurs at a critical point or an endpoint. On this interval, the global minimum is at the endpoint x = 0 and the global maximum is at the critical point x = 2. 8. We consider the part of the graph between x = −3 and x = 4. A global maximum or minimum occurs at a critical point or an endpoint. On this interval, the global minimum is at the right endpoint x = 4 and the global maximum the left endpoint x = −3. 9. (a) (IV) (b) (I) (c) (III) (d) (II)

4.3 SOLUTIONS 10. See Figure 4.74. y

x 3

Figure 4.74 11. See Figure 4.75. y

x 3

Figure 4.75 12. See Figure 4.76. y

Figure 4.76 13. See Figure 4.77. y

x 3

Figure 4.77

281

282

Chapter Four /SOLUTIONS

14. True. If the maximum is not at an endpoint, then it must be at critical point of f . But x = 0 is the only critical point of f (x) = x2 and it gives a minimum, not a maximum. 15. Using a computer to graph the function, f (x) = x3 − ex , and its derivative, f ¨ (x) = 3x2 − ex , we fnd that the derivative crosses the x-axis three times in the interval −1 ≤ x ≤ 4 and twice in the interval −3 ≤ x ≤ 2. See Figures 4.78 and 4.79. 10 5

f (x) x

−2

−5

f ¨ (x) x

−10

−2

−15 −10 −20

Figure 4.78: Function, f

Figure 4.79: Derivative, f ¨

Through trial and error, we obtain approximations: local maximum at x ≈ 3.73, local minimum at x ≈ 0.91 and local maximum at x ≈ −0.46. We can use the approximate values at these points, along with a picture as a guide, to fnd the global maximum and minimum on any interval. (a) On the interval −1 ≤ x ≤ 4, f has local minima at the endpoints x = −1 and x = 4, in addition to the local extrema at the critical points listed above. We fnd the global minimum and maximum on the interval by examining the critical points as well as the endpoints. Since f (−1) = −1.3679, f (−0.46) = −0.7286, f (0.91) = −1.7308, f (3.73) = 10.2160, f (4) = 9.4018, we see x ≈ 0.91 gives a global minimum on the interval and x ≈ 3.73 gives a global maximum. (b) On the interval −3 ≤ x ≤ 2, f has a local minimum at x = −3 and a local maximum at x = 2, in addition to the local extrema at the critical points listed above. We fnd the global minimum and maximum on the interval by examining the critical points as well as the endpoints. Since f (−3) = −27.0498, f (−0.46) = −0.7286, f (0.91) = −1.7308, f (2) = 0.6109, we see x = −3 gives a global minimum and x = 2 a global maximum. (Even though x ≈ −0.46 gives a local maximum, it does not give the greatest maximum on this interval; even though x ≈ 0.91 gives a local minimum, it is not the smallest minimum on this interval.) 16. See Figure 4.80. y

x 3

Figure 4.80

4.3 SOLUTIONS

283

17. See Figure 4.81. y

x 3

Figure 4.81 18. See Figure 4.82. y

x 10

Figure 4.82 19. See Figure 4.83. y

x 3

Figure 4.83 20. Since the function is positive, the graph lies above the x-axis. If there is a global maximum at x = 3, t¨ (x) must be positive, then negative. Since t¨ (x) and t¨¨ (x) have the same sign for x < 3, they must both be positive, and thus the graph must be increasing and concave up. Since t¨ (x) and t¨¨ (x) have opposite signs for x > 3 and t¨ (x) is negative, t¨¨ (x) must again be positive and the graph must be decreasing and concave up. A possible sketch of y = t(x) is shown in Figure 4.84. y (3, 3)

y = t(x) x

Figure 4.84

284

Chapter Four /SOLUTIONS

21. (a) The maximum photosynthesis rate occurs when t ≈ 50 days. (b) The leaf is always growing, since the photosynthesis rate is always positive. The leaf is growing fastest when the photosynthesis rate is greatest, that is t ≈ 50 days. 22. (a) Di˙erentiating f (x) = x3 − 3x2 produces f ¨ (x) = 3x2 − 6x. A second di˙erentiation produces f ¨¨ (x) = 6x − 6. (b) f ¨ (x) is defned for all x and f ¨ (x) = 0 when x = 0, 2. Thus 0 and 2 are the critical points of f . (c) f ¨¨ (x) is defned for all x and f ¨¨ (x) = 0 when x = 1. When x < 1, f ¨¨ (x) < 0 and when x > 1, f ¨¨ (x) > 0. Thus the concavity of the graph of f changes at x = 1. Hence x = 1 is an infection point. (d) We have f (−1) = −4, f (0) = 0, f (2) = −4, f (3) = 0. So f has a local maximum at x = 0 and at x = 3, a local minimum at x = −1 and at x = 2, global maxima at x = 0 and x = 3, and global minima at x = −1 and x = 2. (e) Plotting the function f (x) for −1 ≤ x ≤ 3 gives the graph shown in Figure 4.85.

x −1

−2 f (x) = x3 − 3x2 −4

Figure 4.85

23. (a) Di˙erentiating f (x) = 2x3 − 9x2 + 12x + 1 produces f ¨ (x) = 6x2 − 18x + 12. A second di˙erentiation produces f ¨¨ (x) = 12x − 18. (b) f ¨ (x) is defned for all x and f ¨ (x) = 0 when x = 1, 2. Thus x = 1, 2 are critical points. (c) f ¨¨ (x) is defned for all x and f ¨¨ (x) = 0 when x = 23 . Since the concavity of f changes at this point, it is an infection point. (d) We have f (−0.5) = −7.5, f (3) = 10, f (1) = 6, f (2) = 5. So f has a local maximum at x = 1 and at x = 3 and a local minimum at x = −0.5 and at x = 2, a global maximum at x = 3 and a global minimum at x = −0.5 (e) Plotting the function f (x) for −0.5 ≤ x ≤ 3 gives the graph shown in Figure 4.86. 10 5

f (x) = 2x3 − 9x2 + 12x + 1 x

−1

−5 −10

Figure 4.86

24. (a) Di˙erentiating f (x) = x3 − 3x2 − 9x + 15 produces f ¨ (x) = 3x2 − 6x − 9. A second di˙erentiation produces f ¨¨ (x) = 6x − 6. (b) f ¨ (x) is defned for all x and f ¨ (x) = 0 when x = −1, 3. Thus x = −1, 3 are critical points. (c) f ¨¨ (x) is defned for all x and f ¨¨ (x) = 0 when x = 1. Since the concavity of f changes at this point, it is an infection point. (d) We have f (−5) = −140, f (4) = −5, f (−1) = 2, f (3) = −12. So f has a global maximum at x = −1 and a global minimum at x = −5, and a local minimum at x = −5 and at x = 3 and a local maximum at x = −1 and at x = 4. (e) Plotting the function f (x) for −5 ≤ x ≤ 4 gives the graph shown in Figure 4.87:

4.3 SOLUTIONS

285

50 −4

x 4 −50 −100 f (x) =−150 x3 − 3x2 − 9x + 15

Figure 4.87

25. (a) Di˙erentiating f (x) = x + sin x produces f ¨ (x) = 1 + cos x. A second di˙erentiation produces f ¨¨ (x) = − sin x. (b) f ¨ (x) is defned for all x and f ¨ (x) = 0 when x = . Thus is the critical point of f . (c) f ¨¨ (x) is defned for all x and f ¨¨ (x) = 0 when x = 0, x = and x = 2 . Since the concavity of f changes at each of these points they are all infection points. (d) We have f (0) = 0, f (2 ) = 2 , f ( ) = . So f has a global and local minimum at x = 0 and a global and local maximum at x = 2 . (e) Plotting the function f (x) for 0 ≤ x ≤ 2 gives the graph shown in Figure 4.88:

6 5 4 3 2

f (x) = x + sin x

1 x

Figure 4.88

26. (a) We see in Figure 4.89 that the maximum occurs between x = 1 and x = 2 and the maximum value of y is about y = 30. (b) The maximum value occurs at a critical point, so we fnd all critical points of y: dy = 18 − 10x = 0 dx x = 1.8. Since y is a quadratic polynomial with negative leading coeÿcient, this critical point gives a local maximum, which is also a global maximum. We fnd the maximum value of y by substituting x = 1.8: Maximum value of y = 12 + 18(1.8) − 5(1.8)2 = 28.2. y 28.2

x 1.8

Figure 4.89

286

Chapter Four /SOLUTIONS

27. We fnd all critical points: dy = 2ax + b = 0 dx b . 2a Since y is a quadratic polynomial, its graph is a parabola which opens up if a > 0 and down if a < 0. The critical value is a maximum if a < 0 and a minimum if a > 0. x=−

28. Since the critical point at x = 1 is a local maximum (because f ¨¨ (1) < 0), and there are no other critical points, the global maximum is at x = 1 and the global minimum is at x = 3. See Figure 4.90. Global max

f (x)

x 1

Figure 4.90 29. Since the critical point at x = −5 is a local minimum (because f ¨¨ (−5) > 0), and there are no other critical points, the global minimum is at x = −5 and the global maximum is at x = −6. See Figure 4.91.

g(x) Global min x −6

−5

−4

Figure 4.91 30. Since the critical point at x = 2 is a local minimum (by the frst derivative test), and there are no other critical points, the global minimum is at x = 2 and the global maximum is at one of the endpoints, x = 1 or x = 4. Between x = 1 and x = 2, the value of the function changes by (2 − 1) ⋅ (−1) = −1. Between x = 2 and x = 4 the value of the function changes by (4 − 2) ⋅ 1 = 2. Thus, the global maximum is at x = 4. See Figure 4.92. Global max ℎ(x)

✻ +2

−1

❄ Global min x 1

Figure 4.92

4.3 SOLUTIONS

287

31. Since the critical point at x = 3 is a local maximum (by the frst derivative test), and there are no other critical points, the global maximum is at x = 3 and the global minimum is at one of the endpoints, x = 1 or x = 5. Between x = 1 and x = 3, the value of the function changes by (3 − 1) ⋅ 2 = 4. Between x = 3 and x = 5 the value of the function changes by (5 − 3) ⋅ (−2) = −4. Thus, the global minimum occurs at both x = 1 and x = 5. See Figure 4.93. Global max

✻

j(x)

−4

❄ Global min

Global min x

Figure 4.93

32. The description tells us that R is an increasing function of T for T < 68◦ F and a decreasing function for T > 68◦ F. In other words, the range has its maximum value at T = 68◦ F, and it is a global maximum. 33. Since the function goes up without bound on the left and goes down without bound on the right, there is no global maximum and there is no global minimum. 34. We see in the graph of f ¨ (x) that f ¨ is positive or zero for all x-values to the left of G, and is negative for all x-values to the right of G. This tells us that the function f is increasing or horizontal everywhere to the left of G and is decreasing everywhere to the right of G. Thus, the function f (x) has a global maximum at the point with x-value G. To see that the function does not have a global minimum, we see that, to the right of G, the derivative gets more and more negative. Therefore, to the right of G, the function f (x) is decreasing at a more and more rapid rate. The values of the function get more and more negative without bound, so there is no global minimum. 35. Since N(t) is approximately the derivative of P (t), the slope of the graph of P (t) approximates N(t). (a) The slope of the graph of P (t) is greater between about t = 20 and t = 60 (for the late March to the end of April, 2020) than at nearby points, so N(t) has a peak here. This was the frst wave. The maximum slope occurs at about t = 50, which corresponds to a local maximum of N(t). Similarly, P (t) climbs steeply between about t = 260 and t = 350 (from about mid November, 2020 until mid February, 2021), so N(t) also has a peak there. This was the second wave. The maximum slope in this second wave occurs at about t = 300, so this corresponds to a second local maximum of N(t). Further local maxima are visible in the graph of N(t) in Figure 4.94 but they are harder to identify in Figure 4.49 in the problem statement. For example, there is a third local maximum at around t = 290, which corresponds to a point in the graph of P (t) in which the slope is larger than at the surrounding points. (b) The slope is maximum at around day t = 300, so the global maximum of N(t) occurs there. (c) The number of new cases reaches its highest peak when the value of N(t) is maximum. That is, the highest peak of new cases occurs at the global maximum of N(t), so at around day t = 300. Massachusetts, new cases N 6000 4000 2000 0 100

200

Figure 4.94

300 365

t (days)

288

Chapter Four /SOLUTIONS

36. (a) If N(t) = 0 or small, the graph of P (t) is horizontal, or sloping very gently. Thus, N(t) has a local minimum at t = 0, and another around t = 150. (b) We know for sure that t = 0 is a global minimum because the frst confrmed case in Connecticut was on March 10, 2020, so N(0) = 0. It is hard to tell from Figure 4.50 in the problem statement if the slope is really zero at t = 150, so we cannot know for sure if this is a global minimum, but it could be. Figure 4.95 shows the actual data for N(t), daily new confrmed cases on day t in Connecticut in which these local minima are visible. We can see that N(150) is not zero, so in fact t = 150 does not correspond to a global minimum of N(t). Note that since N(t) = 0 for 0 ≤ t < 10, global minima occur at all these points. They correspond to points of time in which the outbreak in Connecticut had not started. (c) At the global minimum we know there were no new Covid-19 cases that day in Connecticut. The disease was not spreading on that day. Connecticut, new cases N 3000 2000 1000 0 100

200

300 365

t (days)

Figure 4.95 37. We want to maximize the height, y, of the grapefruit above the ground, as shown in the fgure below. Using the derivative we can fnd exactly when the grapefruit is at the highest point. We can think of this in two ways. By common sense, at the peak of the grapefruit’s fight, the velocity, dy∕dt, must be zero. Alternately, we are looking for a global maximum of y, so we look for critical points where dy∕dt = 0. We have dy −50 = −32t + 50 = 0 and so t= ≈ 1.56 sec. dt −32 Thus, we have the time at which the height is a maximum; the maximum value of y is then y ≈ −16(1.56)2 + 50(1.56) + 5 = 44.1 feet. y 60 40 20 t 1

38. Let the numbers be x and y. Then x + y = 100, so y = 100 − x. Since both numbers are nonnegative, we restrict to 0 ≤ x ≤ 100. The product is P = xy = x(100 − x) = 100x − x2 . Di˙erentiating to fnd the maximum, dP = 100 − 2x = 0 dx 100 x= = 50. 2 So there is a critical point at x = 50; the end points are at x = 0, 100. Evaluating gives At x = 0, we have P = 0. At x = 50, we have P = 50(100 − 50) = 2500. At x = 100, we have P = 100(100 − 100) = 0. Thus the maximum value is 2500.

4.3 SOLUTIONS

289

39. Let the numbers be x and y. Then xy = 784, so y = 784∕x. Since both numbers are positive, we restrict to x > 0. The sum is 784 S = x+y = x+ . x Di˙erentiating to fnd the minimum, dS 784 =1− 2 = 0 dx x x2 = 784

√ so x = ± 784 = ±28.

For x > 0, there is only one critical point at x = 28. We fnd 784 d2S =2 3 . dx2 x Since d 2 S∕dx2 > 0 for x > 0, there is a local minimum at x = 28. The derivative dS∕dx is negative for 0 < x < 28 and dS∕dx is positive for x > 28. Thus, x = 28 gives the global minimum for x > 0. The minimum value of the sum is 784 S = 28 + = 56. 28 40. Let the numbers be x, y, z and let y = 2x. Then x + y + z = 3x + z = 36,

so z = 36 − 3x.

Since all the numbers are nonnegative, we restrict to 0 ≤ x ≤ 12. The product is P = xyz = x ⋅ 2x ⋅ (36 − 3x) = 72x2 − 6x3 . Di˙erentiating to fnd the maximum, dP = 144x − 18x2 = 0 dx −18x(x − 8) = 0 x = 0, 8. So there are critical points at x = 0 and x = 8; the end points are at x = 0, 12. Evaluating gives: At x = 0, we have P = 0. At x = 8, we have P = 8 ⋅ 16 ⋅ (36 − 3 ⋅ 8) = 1536. At x = 12, we have P = 12 ⋅ 24 ⋅ (36 − 3 ⋅ 12) = 0. Thus, the maximum value of the product is 1536. 41. Let the sides be x and y cm. Then 2x + 2y = 64, so y = 32 − x. Since both sides are nonnegative, we restrict to 0 ≤ x ≤ 32. The area in cm2 is A = xy = x(32 − x) = 32x − x2 . Di˙erentiating to fnd the maximum, dA = 32 − 2x = 0 dx 32 x= = 16. 2 So there is a critical point at x = 16 cm; the end points are at x = 0 and x = 32 cm. Evaluating gives: At x = 0, we have A = 0 cm2 . At x = 16, we have A = 16(32 − 16) = 256 cm2 . At x = 32, we have A = 32(32 − 32) = 0 cm2 . Thus, the maximum area occurs in a square of side 16 cm.

290

Chapter Four /SOLUTIONS

42. This is a parabola opening downward. We fnd the critical points by setting g ¨ (x) = 0: g ¨ (x) = 4 − 2x = 0 x = 2. ¨

Since g (x) > 0 for x < 2 and g (x) < 0 for x > 2, the critical point at x = 2 is a local maximum. As x → ±∞, the value of g(x) → −∞. Thus, the local maximum at x = 2 is a global maximum of g(2) = 4⋅2−22 −5 = −1. There is no global minimum. See Figure 4.96. 2

−1

g(x) = 4x − x2 − 5

Figure 4.96 43. Di˙erentiating gives f ¨ (x) = 1 −

1 , x2

so the critical points satisfy 1−

1 =0 x2 x2 = 1 x=1

(We want x > 0).

Since f ¨ is negative for 0 < x < 1 and f ¨ is positive for x > 1, there is a local minimum at x = 1. Since f (x) → ∞ as x → 0+ and as x → ∞, the local minimum at x = 1 is a global minimum; there is no global maximum. See Figure 4.97. The the global minimum is f (1) = 2.

f (x) = x + 1∕x x 1

Figure 4.97 44. Di˙erentiating using the product rule gives g ¨ (t) = 1 ⋅ e−t − te−t = (1 − t)e−t , so the critical point is t = 1. Since g ¨ (t) > 0 for 0 < t < 1 and g ¨ (t) < 0 for t > 1, the critical point is a local maximum. As t → ∞, the value of g(t) → 0, and as t → 0+ , the value of g(t) → 0. Thus, the local maximum at x = 1 is a global maximum of g(1) = 1e−1 = 1∕e. In addition, the value of g(t) is positive for all t > 0; it tends to 0 but never reaches 0. Thus, there is no global minimum. See Figure 4.98.

1∕e

g(t) = te−t

t 1

Figure 4.98

4.3 SOLUTIONS

291

45. We have f ¨ (x) = 2ex − 3e−x . To fnd critical points, we set f ¨ (x) = 0. Then 2ex − 3e−x = 0 2ex = 3e−x e2x = 3∕2 2x = ln(3∕2) ln(3∕2) x= . 2 There is just one critical point of f , at x = (ln(3∕2))∕2. Since f ¨¨ (x) = 2ex + 3e−x > 0, the graph of f is concave up everywhere, and the critical point is a global minimum. The minimum value is u u 0 1 √ ln(3∕2) 3 2 f =2 +3 = 2 6. 2 2 3 46. We have f ¨ (x) = 3e3x − 2e2x . To fnd critical points, we set f ¨ (x) = 0. Then 3e3x − 2e2x = 0 3e3x = 2e2x ex = 2∕3 x = ln(2∕3). There is just one critical point of f , at x = ln(2∕3). We have f ¨¨ (x) = 9e3x − 4e2x . At the critical point the second derivative 3 2 2 2 2 8 =9 −4 = >0 f ¨¨ ln 3 3 3 9 is positive. Hence the critical point is a local minimum. Since it is the only critical point, it is also the global minimum. The minimum value is 3 2 2 2 2 4 f ln = − =− . 3 3 3 27 47. Di˙erentiating gives f ¨ (x) = 1 −

1 , x

so the critical points satisfy 1−

1 =0 x 1 =1 x x = 1.

292

Chapter Four /SOLUTIONS

f (x) = x − ln x

x 1

Figure 4.99 48. Di˙erentiating using the quotient rule gives f ¨ (t) =

1(1 + t2 ) − t(2t) 1 − t2 = . 2 2 (1 + t2 )2 (1 + t )

The critical points are the solutions to 1 − t2 =0 (1 + t2 )2 t2 = 1 t = ±1. ¨

Since f (t) > 0 for −1 < t < 1 and f (t) < 0 otherwise, there is a local minimum at t = −1 and a local maximum at t = 1. As t → ±∞, we have f (t) → 0. Thus, the local maximum at t = 1 is a global maximum of f (1) = 1∕2, and the local minimum at t = −1 is a global minimum of f (−1) = −1∕2. See Figure 4.100. 1∕2 −1

f (t) = 1+tt 2 t 1 −1∕2

Figure 4.100 49. Di˙erentiating using the product rule gives f ¨ (t) = 2 sin t cos t ⋅ cos t − (sin2 t + 2) sin t = 0 sin t(2 cos2 t − sin2 t − 2) = 0 sin t(2(1 − sin2 t) − sin2 t − 2) = 0 sin t(−3 sin2 t) = −3 sin3 t = 0. Thus, the critical points are where sin t = 0, so t = 0, ± , ±2 , ±3 , … . 3

Since f ¨ (t) = −3 sin t is positive for − < t < 0, negative for 0 < t < , positive for < t < 2 , and so on, we fnd that t = 0, ±2 , … give local maxima, while t = ± , ±3 , … give local minima. Evaluating gives f (0) = f (±2 ) = (0 + 2)1 = 2 f (± ) = f (±3 ) = (0 + 2)(−1) = −2. Thus, the global maximum of f (t) is 2, occurring at t = 0, ±2 , …, and the global minimum of f (t) is −2, occurring at t = ± , ±3 , … . See Figure 4.101. 2

−3 −2 −

2 3

−2

Figure 4.101

4.3 SOLUTIONS

293

50. To fnd the minimum value of E, we solve 2(1.7) dE = 0.25 − =0 dF F3 2(1.7) F3 = 0.25 0 11∕3 2(1.7) F = = 2.4 hours. 0.25 Since dE∕dF is negative for F < 2.4 and dE∕dF is positive for F > 2.4, a foraging time of F = 2.4 hours gives a local minimum. This is the global minimum for F > 0. See Figure 4.102. E 5

F (hours)

Figure 4.102 51. (a) Function f is increasing for all positive D, so it has no critical point. Function g has a critical point for some positive D. See Figures 4.103 and 4.104. √ (b) For D > 0, the function f is positive and increasing. The maximum value is at D = 300 where f (300) = 300 300 + 300 = 7348.469. For g, we fnd the derivative and set it equal to 0. We start by using the product and chain rules to get: H I √ 1 1 g ¨ (D) = 300 − D − D . √ 2 300 − D To fnd the critical point for g we set g ¨ (D) = 0: √ g (D) = 300 − D − D

Solving gives √

I 1 1 √ 2 300 − D

H 300 − D = D

= 0.

I 1 1 √ 2 300 − D

2(300 − D) = D D = 200. We can confrm that D = 200 gives the maximum impact from Figure 4.104. Alternatively, evaluate g at the endpoints and the critical point: g(0) = 0 and g(300) = 0 and g(200) = 2000. I, impact

I, impact

8000

3000 √ g(D) = D 300 − D √ f (D) = D 300 + D

100

200

Figure 4.103

300

D, dose

100

200

Figure 4.104

300

D, dose

294

Chapter Four /SOLUTIONS

52. Figure 4.105 shows the the pool has dimensions x by y and the deck extends 5 feet at either side and 10 feet at the ends of the pool. 10

y x 10

Figure 4.105 The dimensions of the plot of land containing the pool are then (x + 5 + 5) by (y + 10 + 10). The area of the land is then A = (x + 10)(y + 20), which is to be minimized. We also are told that the area of the pool is xy = 1800, so y = 1800∕x and

1800 + 20 x 18000 = 1800 + 20x + + 200. x

A = (x + 10)

We fnd dA∕dx and set it to zero to get dA 18000 = 20 − =0 dx x2 2 20x = 18000 x2 = 900 x = 30 feet. +

Since A → ∞ as x → 0 and as x → ∞, this critical point must be a global minimum. Also, y = 1800∕30 = 60 feet. The plot of land is therefore (30 + 10) = 40 by (60 + 20) = 80 feet. 53. (a) Suppose the farmer plants x trees per km2 . Then, for x ≤ 200, the yield per tree is 400 kg, so Total yield, y = 400x kg. For x > 200, the yield per tree is reduced by 1 kg for each tree over 200, so the yield per tree is 400 − (x − 200) kg = (600 − x) kg. Thus, Total yield, y = (600 − x)x = 600x − x2 kg. The graph of total yield is in Figure 4.106. It is a straight line for x ≤ 200 and a parabola for x > 200. y (kg) 100,000

50,000

200

400

600

Figure 4.106 (b) The maximum occurs where the derivative of the quadratic is zero, or dy = 600 − 2x = 0 dx x = 300 trees/km2 .

x (trees)

4.3 SOLUTIONS

295

54. (a) Since the crow makes n(x) trips to a height of x meters each, the total vertical distance traveled upward is ℎ = x ⋅ n(x) = x +

27 meters. x

(b) At the minimum, dℎ 27 =1− 2 = 0 dx x x2 = 27 x = 5.2 meters. Since dℎ∕dx is negative for x < 5.2 and dℎ∕dx is positive for x > 5.2, there is a local minimum at x = 5.2. This is the global minimum for x > 0. See Figure 4.107. total upward distance/whelk 30 20 10

5.2

✠ 5

x (meters)

Figure 4.107 The dropping height actually observed by biologists is about 5.2 meters. 55. Rewriting the expression for I using the properties of logs gives I = 192(ln S − ln 762) − S + 763. Di˙erentiating with respect to S gives dI 192 = − 1. dS S At a critical point 192 −1 = 0 S S = 192. Since

192 d2I =− 2 , dS 2 S we see that if S = 192, we have d 2 I∕dS 2 < 0, so S = 192 is a local maximum. From Figure 4.108, we see that it is a global maximum. The maximum possible number of infected children is therefore 192 I = 192 ln − 192 + 763 = 306 children. 762

S 192

Figure 4.108

296

Chapter Four /SOLUTIONS

56. (a) Di˙erentiating twice using the product rule yields dy∕dx = (1 − bx)ae−bx and d 2 y∕dx2 = −(2 − bx)abe−bx . Setting dy∕dx = 0, we fnd that there is a critical point where 1 − bx = 0 1 x= . b At x = 1b , we have y = ab e−1 . This is a local maximum since at this point d 2 y∕dx2 = −abe−1 < 0. (b) Since we are concerned only with positive values of x we consider x = 0 and limx→∞ y. When x = 0, y = 0, and limx→∞ y = 0 also. Since ab e−1 is positive, ( 1b , ab e−1 ) is in fact a global maximum. This implies that a large population may actually produce fewer o˙spring than a smaller one. 57. (a) The variables are S and H. At the maximum value of S, using the product rule, we have dS = ae−bH + aH(−b)e−bH = 0 dH ae−bH (1 − bH) = 0 bH = 1 1 H= . b Thus, the maximum value of S occurs when H = 1∕b. To fnd the maximum value of S, we substitute H = 1∕b, giving 1 a S = a e−b(1∕b) = e−1 . b b (b) Increasing a increases the maximum value of S. Increasing b decreases the maximum value of S. 58. (a) When t = 0, we have q(0) = 20(e−0 − e−2⋅0 ) = 20(1 − 1) = 0. This is as we would expect: initially there is none of the drug in the bloodstream. (b) The maximum value of q(t) occurs where q ¨ (t) = 20(−e−t + 2e−2t ) = 0 −e−t = −2e−2t e−t =2 e−2t et = 2 t = ln 2 = 0.69 hours. When t = 0.69 hours, we have q(0.69) = 5 mg. (c) To see what happens in the long run, we let t → ∞. Since e−t → 0 and e−2t → 0, we see that q(t) → 0 as t → ∞. This is as we would expect: in the long run, the drug will be metabolized or excreted and leave the body. 59. (a) The yield per kg of seed is given by Yield per kg of seed =

Crop yield (tons) Crop yield (tons) − 0 = , Seeds planted (kg) Seeds planted (kg) − 0

so graphically we can interpret the quantity (Crop yield (tons))∕(Seeds planted (kg)) as the slope of the line connecting the origin and the point on the graph of the crop yield corresponding to the amount of seed planted. Since the points on 20 kg and 75 kg lie on the same line through the origin, this means that both amounts of seed correspond to the same yield per kg of seed. (b) The yield per kg of seed is maximized when this slope is maximized. Figure 4.109 shows that this occurs approximately when 40 kg of seed are planted. crop yield (tons) 3 2 1 seeds (kg) 20

80 100 120

Figure 4.109

4.3 SOLUTIONS

297

60. (a) To maximize beneft (surviving young), we pick 10, because that’s the highest point of the beneft graph. (b) To optimize the vertical distance between the curves, we can either do it by inspection or note that the slopes of the two curves will be the same where the di˙erence is maximized. Either way, one gets approximately 9. 61. (a) At higher speeds, the bird uses more energy so the graph rises to the right. The initial drop is due to the fact that the energy it takes a bird to fy at very low speeds is greater than that needed to fy at a slightly higher speeds. (This is analogous to our swimming in a pool). (b) The value of f (v) measures energy per second; the value of a(v) measures energy per meter. In one second, a bird traveling at rate v travels v meters and consumes v ⋅ a(v) joules. Thus v ⋅ a(v) represents the energy consumption per second, so f (v) = v ⋅ a(v). (c) Since v ⋅ a(v) = f (v), we have f (v) a(v) = . v This ratio is the slope of a line passing from the origin through the point (v, f (v)) on the curve. (See Figure 4.110.) Thus a(v) is minimal when the slope of this line is minimal. This occurs where the line is tangent to the curve. To fnd the value of v minimizing a(v) symbolically, we solve a¨ (v) = 0. By the quotient rule, a¨ (v) =

vf ¨ (v) − f (v) . v2

Thus a¨ (v) = 0 when vf ¨ (v) = f (v), or when f ¨ (v) = f (v)∕v = a(v). Thus a(v) is minimized when a(v) = f ¨ (v). (d) Assuming the bird wants to go from one particular point to another, that is, when the distance is fxed, the bird should minimize a(v). Then minimizing a(v) minimizes the total energy used for the fight. energy

f (v) a(v)

Figure 4.110

62. (a) The maximum and minimum values of p can be found without taking derivatives, since the function 20 sin(2.5 t) has maximum and minimum values of 20 and −20, respectively. Thus, the maximum value of p is 120 mm Hg and the minimum value is 80 mm Hg. (b) The time between successive maxima is the period, which is 2 ∕(2.5 ) = 0.8 seconds. (c) See Figure 4.111. p (mm Hg) 120

✛

0.8 sec

✲

Maximum

100 Minimum

Figure 4.111

t (sec)

298

Chapter Four /SOLUTIONS

Solutions for Section 4.4 1. The proft function is positive when R(q) > C(q), and negative when C(q) > R(q). It’s positive for 5.5 < q < 12.5, and negative for 0 < q < 5.5 and 12.5 < q. Proft is maximized when R(q) > C(q) and R¨ (q) = C ¨ (q) which occurs at about q = 9.5. See Figure 4.112. $ (thousands) Profit function is positive

400

✛

C(q)

✲

R(q)

300 200 100

q (thousands)

Figure 4.112 2. The proft (q) is given by (q) = R(q) − C(q) = 500q − q 2 − (150 + 10q) = 490q − q 2 − 150. The maximum proft occurs when ¨ (q) = 490 − 2q = 0

q = 245 items.

¨¨

Since (q) = −2, this critical point is a maximum. Alternatively, we obtain the same result from the fact that the graph of is a parabola opening downward. 3. First fnd marginal revenue and marginal cost. MR = R¨ (q) = 450 MC = C ¨ (q) = 6q Setting MR = MC yields 6q = 450, so marginal cost is equal to marginal revenue when q=

450 = 75 units. 6

Is proft maximized at q = 75? Proft = R(q) − C(q); R(75) − C(75) = 450(75) − (10,000 + 3(75)2 ) = 33,750 − 26,875 = $6875. Testing q = 74 and q = 76: R(74) − C(74) = 450(74) − (10,000 + 3(74)2 ) = 33,300 − 26,428 = $6872. R(76) − C(76) = 450(76) − (10,000 + 3(76)2 ) = 34,200 − 27,328 = $6872. Since proft at q = 75 is more than proft at q = 74 and q = 76, we conclude that proft is maximized locally at q = 75. The only endpoint we need to check is q = 0. R(0) − C(0) = 450(0) − (10,000 + 3(0)2 ) = −$10,000. This is clearly not a maximum, so we conclude that the proft is maximized globally at q = 75, and the total proft at this production level is $6,875.

4.4 SOLUTIONS 4.

marginal cost

marginal revenue

profit

299

5. (a) See Figure 4.113. We fnd q1 and q2 by checking to see where the slope of the tangent line to C(q) is equal to the slope of R(q). Because the slopes of C(q) and R(q) represent marginal cost and marginal revenue, respectively, at q’s where the slopes are equal, the cost of producing an additional unit of q exactly equals the revenue gained from selling an additional unit. In other words, q1 and q2 are levels of production at which the additional or marginal proft from producing an additional unit of q is zero. $

C(q) R(q)

Figure 4.113 (b) We can think of the vertical distance between the cost and revenue curves as representing the frm’s total profts. The quantity q1 is the production level at which total profts are at a minimum. We see this by noticing that for points slightly to the left of q1 , the slope of C(q) is slightly greater than the slope of R(q). This means that the cost of producing an additional unit of q is greater than the revenue earned from selling it. So for points to the left of q1 , additional production decreases profts. For points slightly to the right of q1 , the slope of C(q) is less than the slope of R(q). Thus additional production results in an increase in profts. The production level q1 is the level at which the frm ceases to take a loss on each additional item and begins to make a proft on each additional item. Note that the total proft is still negative, and remains so until the graphs cross, i.e, where total cost equals total revenue. Similar reasoning applies for q2 , except it is the level of production at which profts are maximized. For points slightly to the left of q2 , the slope of C(q) is less than the slope of R(q). Thus the cost of producing an additional unit is less than the revenue gained from selling it. So by selling an additional unit, the frm can increase profts. For points to the right of q2 , the slope of C(q) is greater than the slope of R(q). This means that the proft from producing and selling an additional unit of q will be negative, decreasing total profts. The point q2 is the level of production at which the frm stops making a proft on each additional item sold and begins to take a loss. At both points q1 and q2 , note that the vertical distance between C(q) and R(q) is at a local maximum. This represents the fact that q1 and q2 are local proft minimum and local proft maximum points. 6. (a) The value of C(0) represents the fxed costs before production, that is, the cost of producing zero units, incurred for initial investments in equipment, and so on. (b) The marginal cost decreases slowly, and then increases as quantity produced increases. See Figure 4.114. (c) Concave down implies decreasing marginal cost, while concave up implies increasing marginal cost. (d) An infection point of the cost function is (locally) the point of maximum or minimum marginal cost. (e) One would think that the more of an item you produce, the less it would cost to produce extra items. In economic terms, one would expect the marginal cost of production to decrease, so we would expect the cost curve to be concave down. In practice, though, it eventually becomes more expensive to produce more items, because workers and resources may

300

Chapter Four /SOLUTIONS become scarce as you increase production. Hence after a certain point, the marginal cost may rise again. This happens in oil production, for example. marginal cost

Figure 4.114 7. (a) The proft earned by the 51st is the revenue earned by the 51st item minus the cost of producing the 51st item. This can be approximated by ¨ (50) = R¨ (50) − C ¨ (50) = 84 − 75 = $9. Thus the proft earned from the 51st item will be approximately $9. (b) The proft earned by the 91st item will be the revenue earned by the 91st item minus the cost of producing the 91st item. This can be approximated by ¨ (90) = R¨ (90) − C ¨ (90) = 68 − 71 = −$3. Thus, approximately three dollars are lost in the production of the 91st item. (c) If R¨ (78) > C ¨ (78), production of a 79th item would increase proft. If R¨ (78) < C ¨ (78), production of one less item would increase proft. Since proft is maximized at q = 78, we must have C ¨ (78) = R¨ (78). 8. (a) Proft is maximized when R(q) − C(q) is as large as possible. This occurs at q = 2500, where proft = 7500 − 5500 = $2000. (b) We see that R(q) = 3q and so the price is p = 3, or $3 per unit. (c) Since C(0) = 3000, the fxed costs are $3000. 9. (a) At q = 5000, MR > MC, so the marginal revenue to produce the next item is greater than the marginal cost. This means that the company will make money by producing additional units, and production should be increased. (b) Proft is maximized where MR = MC, and where the proft function is going from increasing (MR > MC) to decreasing (MR < MC). This occurs at q = 8000. 10. From the given graph, we can sketch the graph of marginal proft = M = MR − MC: see Figure 4.115. $/unit M = MR − MC

q 1000

3000 1000

q 3000

Figure 4.115

Figure 4.116

And, from the graph of marginal proft, we can make a sketch showing the shape of the shape of the graph of the proft function, since the marginal proft curve is the graph of the derivative of the proft function. See Figure 4.116. Because we don’t know the value of the proft when q = 0, the graph in Figure 4.116 may be shifted vertically.

4.4 SOLUTIONS

301

Since M < 0 for q < 1000 and for q > 3000, while M > 0 for 1000 < q < 3000 we see that profts appear to be at a minimum when q = 1000 and at a maximum for q = 3000. We see from Figure 4.116 that there is a local maximum at q = 3000. Therefore, the proft has a global maximum either at q = 3,000 or at the endpoint q = 0. Because the proft function could be shifted vertically downward, notice that the maximum proft could be zero or negative. 11. The company should increase production if MR > MC, since increasing production then adds more to revenue than to cost—a net gain for the company. (a) Since MC(25) = 17.75 and MR(25) = 30, the company should increase production. (b) Since MC(50) = 39 and MR(50) = 30, the company should decrease production. (c) Since MC(80) = 114 and MR(80) = 30, the company should decrease production. 12. We have MR = 30 for all quantities q. The proft is a maximum when MC = MR and MC < MR to the left of the point and MC > MR to the right of the point. There appear to be two points with MC = MR, one between q = 0 and q = 10 and another between q = 40 and q = 50. At q = 0 we have MC = 34 > MR, and at q = 10 we have MC = 23 < MR. Thus, the point between q = 0 and q = 10 does not give a maximum proft. At q = 40 we have MC = 26 < MR, and at q = 50 we have MC = 39 > MR. Thus, the point between q = 40 and q = 50 gives a maximum proft. 13. The proft is maximized at the point where the di˙erence between revenue and cost is greatest. Thus the proft is maximized at approximately q = 4000. 14. (a) We know that marginal cost = C ¨ (q) or the slope of the graph of C(q) at q. Similarly, marginal revenue = R¨ (q) or the slope of the graph of R(q) at q. From the graph, the slope of R(q) at q = 3000 is greater than the slope of C(q) at q = 3000. Therefore, marginal revenue is greater than marginal cost at q = 3000. Production should be increased. (b) The graph of C(q) is steeper than the graph of R(q) at q = 5000. Therefore, marginal cost is greater than marginal revenue at q = 5000. Production should be decreased. 15. Since marginal revenue is larger than marginal cost around q = 2000, as you produce more of the product your revenue increases faster than your costs, so proft goes up, and maximal proft will occur at a production level above 2000. 16. (a) The marginal cost at q = 400 is the slope of the tangent line to C(q) at q = 400. Looking at the graph, we can estimate a slope of about 1. Thus, the marginal cost is about $1. (b) At q = 500, we can see that slope of the cost function is greater than the slope of the revenue function. Thus, the marginal cost is greater than the marginal revenue and thus the 500th item will incur a loss. So, the company should not produce the 500th item. (c) The quantity which maximizes proft is at the point where marginal cost equals marginal revenue. This occurs when the slope of R(q) equals C(q), which occurs at approximately q = 400. Thus, the company should produce about 400 items. 17. At each of the endpoints, 0 and 100, the production function is Q = 0, so the maximum does not occur at an endpoint. To fnd the optimum value, we fnd the derivative dQ∕dx using the product rule, then set the derivative equal to zero and solve for x. dQ = 5(0.3x−0.7 )(100 − x)0.8 + 5x0.3 (0.8(100 − x)−0.2 (−1)) dx 1.5(100 − x)0.8 4x0.3 = − = 0. 0.7 (100 − x)0.2 x Multiplying through by the common denominator of x0.7 (100 − x)0.2 , we have: 1.5(100 − x)0.8 (100 − x)0.2 − 4x0.3 x0.7 = 0 1.5(100 − x) − 4x = 0 150 − 1.5x − 4x = 0 150 − 5.5x = 0 x = 27.273. The company should spend 27.273 units of investment capital on equipment and 100 − 27.273 = 72.727 units on labor. At this maximum, the production Q is Q = 5(27.273)0.3 (72.727)0.8 = 415.955 items.

302

Chapter Four /SOLUTIONS

18. (a) The total cost of x and y units of the two raw materials is 6x + 3y thousand dollars. If the budget is fully spent 6x + 3y = 12. Solving for y gives 12 − 6x = 4 − 2x. 3 Since both x and y are nonnegative, we have 0 ≤ x ≤ 2. (b) Substituting the expression for y into Q gives y=

Q = 3 ln(x + 1) + 2 ln(4 − 2x + 1)

for 0 ≤ x ≤ 2.

Di˙erentiating Q with respect to x gives a critical point: 2 ⋅ (−2) dQ 3 = + = 0. dx x+1 5 − 2x We solve for x: 2 ⋅ (−2) 3 + =0 x+1 5 − 2x 2 ⋅ (−2) 3 =− x+1 5 − 2x 3(5 − 2x) = 4(x + 1) 15 − 6x = 4x + 4 10x = 11 x = 1.1. Thus y = 4 − 2(1.1) = 1.8. To show this critical point is a local maximum for Q, we fnd the second derivative d2Q 3 −4 =− − ⋅ (−2). (x + 1)2 (5 − 2x)2 dx2 Substituting any positive x-value gives d2Q < 0. dx2 Thus the x = 1.1 gives a local maximum. Next we check for a global maximum. At x = 0, we have y = 4 and Q = 3 ln(0 + 1) + 2 ln(4 + 1) = 3.219. At x = 2 we have y = 0 and Q = 3 ln(2 + 1) + 2 ln(0 + 1) = 3.296. At x = 1.1 and y = 1.8 we have Q = 3 ln(1.1 + 1) + 2 ln(1.8 + 1) = 4.285. Thus, the global maximum occurs at x = 1.1 and y = 1.8. The maximum quantity that can be produced is 4.285 units and is made using 1.1 kg of x and 1.8 kg of y. 19. We frst need to fnd an expression for R(q), or revenue in terms of quantity sold. We know that R(q) = pq, where p is the price of one item. Here p = 45 − 0.01q, so we make the substitution R(q) = (45 − .01q)q = 45q − 0.01q 2 . This is the function we want to maximize. Finding the derivative and setting it equal to 0 yields R¨ (q) = 0 45 − 0.02q = 0 0.02q = 45 so q = 2250. Is this a maximum? R¨ (q) > 0 for q < 2250 and R¨ (q) < 0 for q > 2250.

4.4 SOLUTIONS

303

So we conclude that R(q) has a local maximum at q = 2250. Testing q = 0, the only endpoint, R(0) = 0, which is less than R(2250) = $50,625. So we conclude that revenue is maximized at q = 2250. The price of each item at this production level is p = 45 − .01(2250) = $22.50 and total revenue is pq = $22.50(2250) = $50,625, which agrees with the above answer. 20. (a) If q = 3000, the demand equation gives p = 70 − 0.02 ⋅ 3000 = 10. That is, at a price of $10, 3000 people attend. At this price, Revenue = 3000 people ⋅ 10 dollars/person = $30,000. To fnd total revenue at a price of $20, frst fnd the attendance at this price. Substituting p = 20 into the demand equation, p = 70 − 0.02q, gives 20 = 70 − 0.02q. Solving for q, we get −50 = −0.02q 2500 = q. That is, at a price of $20, attendance is 2500 people, and Revenue = 2500 ⋅ 20 = $50, 000. Notice that, although demand is reduced, the revenue is higher at a price of $20 than at $10. (b) Since Revenue = Price × Quantity = p ⋅ q and p = 70 − 0.002q, we have R(q) = (70 − 0.02q)q = 70q − 0.02q2 . (c) To maximize revenue, fnd the critical points of the revenue function R(q) = 70q − 0.02q 2 : R¨ (q) = 70 − 0.02 ⋅ 2q 0 = 70 − 0.04q 70 = 0.04q 1750 = q. The graph of revenue is a parabola opening downward, so an attendance of 1750 gives the maximum revenue. (d) Using the demand equation, we fnd the price corresponding to an attendance of 1750: p = 70 − 0.02 ⋅ 1750 = 70 − 35 = 35. The optimal price for a ticket at the amusement park is $35. (e) When the optimal price of $35 is charged, the attendance at the park is 1750 people. Thus, the maximum revenue is R = pq = 35 ⋅ 1750 = $61,250. The corresponding proft cannot be determined without knowing the costs. 21. To maximize revenue, we frst must fnd an expression for revenue in terms of price. We know that R(p) = pq, where p=price and q=quantity sold. We now need to fnd an expression for q in terms of p. Using the information given, we fnd that (4.00 − p) q = 4000 + (200) 0.25 Simplifcation of q yields q = 4000 + 800(4 − p) = 4000 + 3200 − 800p = 7200 − 800p We can now get an expression for revenue in terms of price. R(p) = qp = (7200 − 800p)p = 7200p − 800p2

304

Chapter Four /SOLUTIONS We want to maximize this function in terms of p. First fnd the critical points by fnding the derivative. R¨ (p) = 7200 − 1600p Setting R¨ (p) = 0 and solving for p yields 7200 − 1600p = 0 1600p = 7200 p = 4.5 ¨

Since R (p) > 0 for p < 4.5 and R (p) < 0 for p > 4.5, we conclude that revenue has a local maximum at p = 4.5. Since this is the only critical point, we conclude that it is the global maximum. So revenue is maximized at a price of $4.50. The quantity sold at this amount is given by q = 7200 − 800(4.50) = 3600 and the total revenue is R(4.5) = 7200(4.5) − 800(4.5)2 = $16,200. 22. We frst need to fnd an expression for revenue in terms of price. At a price of $8, 1500 tickets are sold. For each $1 above $8, 75 fewer tickets are sold. This suggests the following formula for q, the quantity sold for any price p. q = 1500 − 75(p − 8) = 1500 − 75p + 600 = 2100 − 75p. We know that R = pq, so substitution yields R(p) = p(2100 − 75p) = 2100p − 75p2 To maximize revenue, we fnd the derivative of R(p) and set it equal to 0. R¨ (p) = 2100 − 150p = 0 150p = 2100 so p = 2100 = 14. Does R(p) have a maximum at p = 14? Using the frst derivative test, 150 R¨ (p) > 0 if p < 14 and R¨ (p) < 0 if p > 14. So R(p) has a local maximum at p = 14. Since this is the only critical point for p ≥ 0, it must be a global maximum. So we conclude that revenue is maximized when the price is $14. 23. (a) The cost per acre is C = 2x dollars. The revenue per acre is R = 700 − 400e−x∕100 dollars. The proft per acre is P = R − C = 700 − 400e−x∕100 − 2x. The critical point of proft is given by dP =0 dx −x∕100 4e −2 = 0 1 −x∕100 e = 2 ex∕100 = 2 x = 100 ln 2 = 69.315 The second derivative

d2P −1 −x∕100 = e 25 dx2 is negative for all x, so the critical point x = 69.315 gives a maximum of proft. The farmer should apply about 69 pounds of fertilizer per acre. (b) The proft is P (69.315) = $361.37 per acre. Planting 200 acres produces Total proft = 361.37(200) = $72, 274.

4.4 SOLUTIONS

305

24. (a) In dollars, the revenue earned is R = pq = (−5q + 4000)q. Thus Proft, = R − C = (−5q + 4000)q − (6q + 5) = −5q 2 + 3994q − 5. (b) For maximum proft, d = −10q + 3994 = 0, dq so q = 399.4. (c) For q = 399.4, the proft is given in dollars by = −5(399.4)2 + 3994(399.4) − 5 = 797,596.80. 25. (a) Cost C = Fixed cost + Total variable cost = 10,000 + 2q dollars. (b) We express the demand equation in the form q = b + mp where q = 10,100 when p = 5 and q = 12,872 when p = 4.5. Thus Slope = m =

12,872 − 10,100 2772 = = −5544. 4.5 − 5 −0.5

To fnd b, we substitute q = 10,100 and p = 5. 10,100 = b − 5544 ⋅ 5. Thus, b = 37,820, so q = 37,820 − 5544p. (c) To fnd the proft as a function of q, we solve for p: p=

37,820 1 − q 5544 5544

so p = 6.822 − 0.00018q. Then we have Proft, = Revenue − Cost = pq − C = (6.822 − 0.00018q)q − (10,000 + 2q) = −0.00018q 2 + 4.822q − 10,000. (d) At the maximum proft, d = −0.00036q + 4.822 = 0, dq so q = 13,394.4. Thus the company should produce 13,394 items. At that production level, = −0.00018(13,394)2 + 4.822(13,394) − 10,000 = 22,294 dollars. 26. The answer depends on the fact that the maximum proft occurs when marginal cost equals marginal revenue, and where marginal cost is rising from below marginal revenue to above marginal revenue. (Then the marginal proft is at its maximum and about to decrease.) At q = 1, we know MR = MC, but MC is decreasing so the seller’s proft is increasing. Her marginal cost is at a minimum of $25 when the quantity sold is 100, which is still below the marginal revenue of $35. Thus she will continue to make a proft as the quantity increases, although her proft will start to decrease if the marginal cost goes above $35. So the quantity that maximizes proft is greater than 100.

306

Chapter Four /SOLUTIONS

27. Consider the rectangle of sides x and y shown in Figure 4.117.

y x

Figure 4.117

The total area is xy = 3000, so y = 3000∕x. Suppose the left and right edges and the lower edge have the shrubs and the top edge has the fencing. The total cost is C = 45(x + 2y) + 20(x) = 65x + 90y. Since y = 3000∕x, this reduces to C(x) = 65x + 90(3000∕x) = 65x + 270,000∕x. Therefore, C ¨ (x) = 65 − 270,000∕x2 . We set this to 0 to fnd the critical points: 65 −

270,000 =0 x2 270,000 = 65 x2 2 x = 4153.85 x = 64.450 ft

so that y = 3000∕x = 46.548 ft. +

Since C(x) → ∞ as x → 0 and x → ∞, we see that x = 64.450 is a minimum. The minimum total cost is then C(64.450) ≈ $8378.54.

28. Let x equal the number of chairs ordered in excess of 300, so 0 ≤ x ≤ 100. Revenue = R = (90 − 0.25x)(300 + x) = 27, 000 − 75x + 90x − 0.25x2 = 27, 000 + 15x − 0.25x2 At a critical point dR∕dx = 0. Since dR∕dx = 15 − 0.5x, we have x = 30, and the maximum revenue is $27, 225 since the graph of R is a parabola which opens downward. The minimum is $0 (when no chairs are sold). 29. (a) Since a∕q decreases with q, this term represents the ordering cost. Since bq increases with q, this term represents the storage cost. (b) At the minimum, dC −a = 2 +b=0 dq q giving u a a 2 q = so q = . b b Since d2C 2a = 3 > 0 for q > 0, dq 2 q √ √ we know that q = a∕b gives a local minimum. Since q = a∕b is the only critical point, this must be the global minimum.

4.4 SOLUTIONS

307

30. (a) The total revenue is given by R = pq so R = 400q − 2q 2 . (b) The marginal revenue, MR, is given by dR = 400 − 4q. dq When q = 10 the marginal revenue is 360 $/unit. (c) The total revenue when q = 10 is $3800, and for q = 11 it is $4158 so the increase in revenue is $358 compared with the marginal revenue of 360 $/unit calculated in part (b). 31. (a) The business must reorder often enough to keep pace with sales. If reordering is done every t months, then, Quantity sold in t months = Quantity reordered in each batch rt = q q t = months. r (b) The amount spent on each order is a + bq, which is spent every q∕r months. To fnd the monthly expenditures, divide by q∕r. Thus, on average, Amount spent on ordering per month =

a + bq ra = + rb dollars. q∕r q

(c) The monthly cost of storage is kq∕2 dollars, so C = Ordering costs + Storage costs kq ra C = + rb + dollars. q 2 (d) The optimal batch size minimizes C, so

dC −ra k = 2 + =0 dq 2 q ra k = 2 q2 2ra q2 = k u 2ra q= items per order. k

32. (a) Suppose n passengers sign up for the cruise. If n ≤ 100, then the cruise’s revenue is R = 2000n, so the maximum revenue is R = 2000 ⋅ 100 = 200,000. If n > 100, then the price is p = 2000 − 10(n − 100) and hence the revenue is R = n(2000 − 10(n − 100)) = 3000n − 10n2 . To fnd the maximum revenue, we set dR∕dn = 0, giving 20n = 3000 or n = 150. Then the revenue is R = (2000 − 10 ⋅ 50) ⋅ 150 = 225,000. Since this is more than the maximum revenue when n ≤ 100, the boat maximizes its revenue with 150 passengers, each paying $1500. (b) We approach this problem in a similar way to part (a), except now we are dealing with the proft function . If n ≤ 100, we have = 2000n − 80,000 − 400n, so is maximized with 100 passengers yielding a proft of = 1600 ⋅ 100 − 80,000 = $80,000. If n > 100, we have = n(2000 − 10(n − 100)) − (80,000 + 400n). We again set d ∕dn = 0, giving 2600 = 20n, so n = 130. The proft is then $89,000. So the boat maximizes proft by boarding 130 passengers, each paying $1700. This gives the boat $89,000 in proft.

308

Chapter Four /SOLUTIONS

33. (a) The government collects T dollars for each of q thousand items sold. Thus, the tax revenue R is R = qT = (10 − T )T = 10T − T 2 thousand dollars. Since

dR = 10 − 2T dT

the critical point occurs where 10 − 2T = 0 T = 5. The graph of R as a function of T is a parabola opening downward, so the critical point of R is a global maximum for tax revenue. To maximize tax revenue, the government should set the tax at $5 per item. (b) The number of items sold at this tax rate is q = 10 − T = 10 − 5 = 5 thousand. The government’s tax revenue is R = qT = 5 ⋅ 5 = 25 thousand dollars. (c) At this tax rate, the price of the taxed item is p = 50 + 0.2T = 50 + 0.2 ⋅ 5 = $51 so Producer’s revenue = Sales revenue − Taxes = pq − T q = (p − T )q = (51 − 5)5 = 230 thousand dollars. (d) If there is no tax, then the price is p = $50 per item and q = 10 thousand items are sold. Therefore Producer’s revenue = pq = 500 thousand dollars. 34. For each month, Proft = Revenue − Cost = pq − wL = pcK L − wL The variable on the right is L, so at the maximum d = pcK L −1 − w = 0 dL Now

− 1 is negative, since 0 <

< 1, so 1 −

is positive and we can write pcK =w L1−

giving

0 L=

pcK w

1 1

1−

Since − 1 is negative, when L is just above 0, the quantity L −1 is huge and positive, so d ∕dL > 0. When L is large, L −1 is small, so d ∕dL < 0. Thus the value of L we have found gives a global maximum, since it is the only critical point.

4.5 SOLUTIONS

309

35. (a) The units of f (L) are tons∕month. If f (1000) = 400, then with 1000 hours of labor in a month the company can produce 400 tons of product per month. (b) The units of L are hours∕month. For the derivative of f with respect to L we have Units of f ¨ (L) =

tons∕month Units of f tons = = . Units of L hours∕month hour

If f ¨ (1000) = 2, then increasing labor from 1000 to 1001 hours in a month increases production by about 2 tons per month. (c) At a price of p dollars per ton, Q tons of product costs pQ dollars. One hour’s labor earns w dollars, enough to purchase Q tons where w = pQ. Hence Real wage = Q =

w . p

(d) The company’s monthly cost is the price of the labor it employs. C = Hourly wage ⋅ Hours of labor = wL. The monthly revenue is R = Product price ⋅ Quantity = pf (L). Monthly proft is (L) = Revenue − Cost = pf (L) − wL. (e) At maximum proft, the marginal proft ¨ (L) is zero. We have ¨ (L) = pf ¨ (L) − w = 0. Hence pf ¨ (L) = w and

w , p which means that the marginal product of labor equals the real wage. The marginal product of labor and the real wage are also equal when the company operates at minimum proft. f ¨ (L) =

Solutions for Section 4.5 1. Since a(q) = (C(q) − 0)∕(q − 0) = C(q)∕q, we see that the average cost a(q) is the slope of the line from the origin to the point (q, C(q)). We use this fact for each cost function to visualize whether or not there is a minimum average cost. (a) From Figure 4.118 we see that the slope of the line from the origin to the point (q, C(q)) gets smaller as q gets larger, but it never reaches a minimum. Thus, there is no value of q for which the average cost is minimized. (b) From Figure 4.119 we see that the slope of the line from the origin to the point (q, C(q)) is minimized at the point q where the line is tangent to the cost curve. All the other lines joining the origin to points on the curve have larger slopes and therefore correspond to larger average values. (c) From Figure 4.120, we see that the slope of the line from the origin to the point (q, C(q)) gets smaller as q gets larger, but it never reaches a minimum. Thus, there is no value of q for which the average cost is minimized. (d) From Figure 4.121, we see that the slope of the line from the origin to the point (q, C(q)) gets smaller as q gets larger, but it never reaches a minimum. Thus, there is no value of q for which the average cost is minimized.

310

Chapter Four /SOLUTIONS C (cost)

C (cost) C(q)

C(q)

✛ Minimum slope q (quantity)

q (quantity)

Figure 4.118

Figure 4.119

C (cost)

C (cost) C(q) C(q)

q (quantity)

Figure 4.120

Figure 4.121

2. (a) The average cost of quantity q is given by the formula C(q)∕q. From the graph, we see that C(10,000) ≈ 16,000. So average cost at q = 10,000 is given by a(q) =

C(10,000) 16,000 ≈ = $1.60 per unit. 10,000 10,000

The economic interpretation of this is that $1.60 is each unit’s share of the total cost of producing 10,000 units. (b) Since a(q) = C(q)∕q = (C(q) − 0)∕(q − 0) we see that graphically, average cost a(q) is the slope of the line going through the origin and through the point (q, C(q)). Therefore, a(10,000) is the slope of a line connecting (0, 0) to (10,000, C(10,000)), see Figure 4.122. C (cost)

C (cost) C(q)

C(q)

20,000

10,000

Figure 4.122

q (quantity)

10,000

18,000

q (quantity)

Figure 4.123

(c) We know that a(q) is minimized where a(q) = C ¨ (q). Using the graphical interpretations from parts (b) and (c), this is equivalent to saying that the tangent line has the same slope as the line connecting the point on the curve to the origin. Since these two lines share a point, specifcally the point (q, C(q)) on the curve, and have the same slope, they

4.5 SOLUTIONS

311

are in fact the same line. So a(q) is minimized where the line passing from (q, C(q)) to the origin is also tangent to the curve. To fnd such points, a variety of lines passing through the origin and the curve are shown in Figure 4.123. From this plot, we see that the line with the desired properties intersects the curve at q ≈ 18,000. So q ≈ 18,000 units minimizes a(q). 3. (a)

(i) The average cost of quantity q is given by the formula C(q)∕q. So average cost at q = 25 is given by C(25)∕25. From the graph, we see that C(25) ≈ 200, so a(q) ≈

200 ≈ 8 dollars per unit. 25

To interpret this graphically, note that a(q) = C(q)∕q = (C(q) − 0)∕(q − 0). This is the formula for the slope of a line from the origin to a point (q, C(q)) on the curve. So a(25) is the slope of a line connecting (0, 0) to (25, C(25)). See Figure 4.124. C (cost)

C(q)

500

C(q)

500

400

✛ Slope=8

300

200

100

0 10

q (quantity)

Figure 4.124

Figure 4.125

(ii) The marginal cost is C ¨ (q). This derivative is the slope of the tangent line to C(q) at q = 25. To estimate this slope, we draw the tangent line, shown in Figure 4.125. From this plot, we see that the points (50, 300) and (0, 100) are approximately on this line, so its slope is approximately (300 − 100)∕(50 − 0) = 4. Thus, C ¨ (25) ≈ $4 per unit. (b) We know that a(q) is minimized where a(q) = C ¨ (q). Using the graphical interpretations from parts (i) and (ii), we see that a(q) is minimized where the line passing from (q, C(q)) to the origin is also tangent to the curve. To fnd such points, a variety of lines passing through the origin and the curve are shown in Figure 4.126. The line which is also a tangent touches the curve at q ≈ 30. So q ≈ 30 units minimizes a(q). C (cost) C(q)

500 400 300

✛ Minimum slope

200 100 0 10

q (quantity)

Figure 4.126

4. (a) The marginal cost of producing the q0th item is simply C ¨ (q0 ), or the slope of the graph at q0 . Since the slope of the cost function is always 12, the marginal cost of producing the 100th item and the marginal cost of producing the 1000th item is $12.

312

Chapter Four /SOLUTIONS (b) The average cost at q = 100 is given by a(100) =

C(100) 3700 = = $37. 100 100

The average cost at q = 1000 is given by a(1000) =

C(1000) 14,500 = = $14.50. 1000 1000

5. The cost function is C(q) = 1000 + 20q. The marginal cost function is the derivative C ¨ (q) = 20, so the marginal cost to produce the 200th unit is $20 per unit. The average cost of producing 200 units is given by a(200) =

C(200) 5000 = = $25∕unit 200 200

6. The graph of the average cost function is shown in Figure 4.127. $/unit

C(q)∕q

q (quantity)

Figure 4.127

7. For C(q) = 0.1q 2 + 1000 million dollars: (a) Fixed costs = 1000 million dollars, Marginal cost = MC(q) = C ¨ (q) = 0.2q million dollars per unit, and Average cost = AC(q) =

0.1q 2 + 1000 million dollars per unit. q

(b) At a critical point for average cost, MC(q) = AC(q), so setting them equal and solving for q gives: 0.1q 2 + 1000 q 2 2 0.2q = 0.1q + 1000 0.2q =

0.1q 2 = 1000 q 2 = 10,000

q = 100.

Thus, A(q) has a critical point at 100 units. (c) Di˙erentiating AC(q) = 0.1q + 1000∕q twice gives a positive second derivative at q = 100: AC ¨¨ (q) = 2000∕q 3 ,

AC ¨¨ (100) > 0.

Thus, at the critical point, AC(q) has a local minimum. See Figure 4.128. Alternatively, we compare MC(q) = 0.2q and AC(q) = 0.1q + 1000∕q on either side of q = 100. For q > 100, we know MC(q) > AC(q) (because AC(q) = 0.1q + 1000∕q is eventually smaller than MC(q) = 0.2q for large q). In addition, for q < 100, say as q → 0, we see MC(q) < AC(q) (because MC(q) → 0 while AC(q) → ∞).

4.5 SOLUTIONS

313

100 80 60

MC(q)

40 20

AC(q) q 100 200 300 400 500

Figure 4.128 8. (a) Since a(q) = C(q)∕q = (C(q) − 0)∕(q − 0), graphically, average cost a(q) is the slope of the line going through the origin and through the point (q, C(q)). From Figure 4.129, we see that the line connecting the origin and the graph of C(q) appears to have minimum slope at q = 6. Therefore average cost is minimized at about q = 6. C (cost)

C(q)

200 150 100 50 2

q (quantity)

Figure 4.129 (b) The average cost of the frst q items is given by a(q) =

C(q) q 3 − 12q 2 + 60q = = q 2 − 12q + 60 q q

We want to minimize a(q). Di˙erentiating gives a¨ (q) = 2q − 12. Setting this equal to 0 and solving yields q = 6. Since a¨ (q) < 0 if q < 6 and a¨ (q) > 0 if q > 6, we know q = 6 is a local minimum for a(q). From Figure 4.129, we see that q = 6 is in fact the global minimum for 0 ≤ q ≤ 8. 9. (a) Proft equals revenue minus cost. Your monthly revenue is 1200 slippers × $20∕slipper = $24000, your monthly cost equals 1200 slippers × $2∕slipper = $2400. Since you are earning a monthly proft of $24000 − $2400 = $21600, you are making money. (b) Since additional units produced cost about $3 each, which is above the average cost, producing them increases average cost. Since additional pairs of slippers cost about $3 to produce and can be sold for $20, you can increase your proft by making and selling them. This is a case where marginal revenue, which is $20 per slipper, is greater than marginal cost, which is $3 per pair of slippers. (c) You should recommend increase in production, since that increases proft. The fact that average cost of production increases is irrelevant to your decision. 10. (a) The average cost is a(q) = C(q)∕q, so the total cost is C(q) = 0.01q 3 − 0.6q 2 + 13q. (b) Taking the derivative of C(q) gives an expression for the marginal cost: C ¨ (q) = MC(q) = 0.03q 2 − 1.2q + 13. To fnd the smallest MC we take its derivative and fnd the value of q that makes it zero. So: MC ¨ (q) = 0.06q−1.2 = 0 when q = 1.2∕0.06 = 20. This value of q must give a minimum because the graph of MC(q) is a parabola opening upward. Therefore the minimum marginal cost is MC(20) = 1. So the marginal cost is at a minimum when the additional cost per item is $1.

314

Chapter Four /SOLUTIONS (c) Di˙erentiating gives a¨ (q) = 0.02q − 0.6. Setting a¨ (q) = 0 and solving for q gives q = 30 as the quantity at which the average is minimized, since the graph of a is a parabola which opens upward. The minimum average cost is a(30) = 4 dollars per item. (d) The marginal cost at q = 30 is MC(30) = 0.03(30)2 − 1.2(30) + 13 = 4. This is the same as the average cost at this quantity. Note that since a(q) = C(q)∕q, we have a¨ (q) = (qC ¨ (q) − C(q))∕q 2 . At a critical point, q0 , of a(q), we have 0 = a¨ (q0 ) =

q0 C ¨ (q0 ) − C(q0 ) q02

so C ¨ (q0 ) = C(q0 )∕q0 = a(q0 ). Therefore C ¨ (30) = a(30) = 4 dollars per item. Another way to see why the marginal cost at q = 30 must equal the minimum average cost a(30) = 4 is to view C ¨ (30) as the approximate cost of producing the 30th or 31st good. If C ¨ (30) < a(30), then producing the 31st good would lower the average cost, i.e. a(31) < a(30). If C ¨ (30) > a(30), then producing the 30th good would raise the average cost, i.e. a(30) > a(29). Since a(30) is the global minimum, we must have C ¨ (30) = a(30). 11. (a) The marginal cost tells us that additional units produced would cost about $10 each, which is below the average cost, so producing them would reduce average cost. (b) It is impossible to determine the e˙ect on proft from the information given. Proft depends on both cost and revenue, = R − C, but we have no information on revenue. 12. (a) We have N(x) = 100 + 20x, graphed in Figure 4.130. (b) (i) Di˙erentiating gives N ¨ (x) = 20 and the graph of N ¨ is a horizontal line. This means that rate of increase of the number of bees with acres of clover is constant—each acre of clover brings 20 more bees. (ii) On the other hand, N(x)∕x = 100∕x+20 means that the average number of bees per acre of clover approaches 20 as more acres are put under clover. See Figure 4.131. As x increases, 100∕x decreases to 0, so N(x)∕x approaches 20 (i.e. N(x)∕x → 20). Since the total number of bees is 20 per acre plus the original 100, the average number of bees per acre is 20 plus the 100 shared out over x acres. As x increases, the 100 are shared out over more acres, and so its contribution to the average becomes less. Thus the average number of bees per acre approaches 20 for large x. bees/acre number of bees

N(x) = 100 + 20x

2100

N(x)

20 100 50

100

= 100 + 20

x ✠ x ✐ N ¨ (x) = 20

x (acres)

Figure 4.130

100

x (acres)

Figure 4.131

13. (a) Since AC(q) = C(q)∕q, we fll in the another line of the table. For example, AC(5) = 89∕5 = 17.8.

C(q)

100

113

128

145

164

MC(q)

AC(q)

17.8

16.667

16.143

16.111

16.4

(b) At a critical point for average cost, MC(q) = AC(q), so q = 8 is a critical value. (Since MC(8) = 16 = AC(8).) (c) For q < 8 we see from the table that MC(q) < AC(q); for q > 8 we see that MC(q) > AC(q). Thus, by the frst derivative test, AC(q) has a local minimum at q = 8.

4.5 SOLUTIONS

315

14. (a) The costs and savings are in Table 4.1. Table 4.1 i

Marginal Cost (of ith flter)

Average Cost (for ith flter)

Marginal Savings

$64

$5.50

$32

$16

$6.50

$10

$7.50

$11

(b) She should install four flters. Up to the fourth flter, the marginal cost is less than the marginal savings. This tells us that for each of the frst four flters the developer buys, she will save more on the laundromat than she will have to pay for the flter. From the ffth flter onward, she pays more for each additional flter than she makes from the laundromat. (c) If the rack costs $100 she should not buy it. Instead she should let the laundromat protect itself. This is because, if she buys even one flter she has to spend $105 on the flter but her savings as a result of the purchase only amount to $64. If she buys two flters she has to spend $111 on the flters but her total savings as a result of the purchase only amount to $96. If she buys three flters she has to spend $118 on the flters but her total savings as a result of the purchase only amount to $112. If she buys four flters she has to spend $126 on the flters but her total savings as a result of the purchase only amount to $121 etc. (d) If the rack costs $50 she should buy it and install four flters since she covers the price of the rack with the frst purchase of flters i.e., having bought one flter she will have spent $50 + $5 = $55 but she will have saved $64 on the laundromat business. Her marginal cost and marginal savings do not change after this point and so she still saves more on laundromat than she spends on flters if she buys the second, third and fourth flter. 15. Since a(q) = C(q)∕q, we use the quotient rule to fnd a¨ (q) =

C ¨ (q) − C(q)∕q qC ¨ (q) − C(q) C ¨ (q) − a(q) = = . q q q2

Since marginal cost is C ¨ , if C ¨ (q) < a(q), then C ¨ (q) − a(q) < 0, so a¨ (q) < 0. 16. Since a(q) = C(q)∕q, we use the quotient rule to fnd a¨ (q) =

C ¨ (q) − C(q)∕q qC ¨ (q) − C(q) C ¨ (q) − a(q) = = . 2 q q q

Since marginal cost is C ¨ , if C ¨ (q) > a(q), then C ¨ (q) − a(q) > 0, so a¨ (q) > 0. 17. (a) Di˙erentiating C(q) gives K (1∕a)−1 q , a ¨¨ If a > 1, then C (q) < 0, so C is concave down. (b) We have C ¨ (q) =

C ¨¨ (q) =

K a

1 − 1 q (1∕a)−2 . a

C(q) Kq 1∕a + F = q q K (1∕a)−1 ¨ C (q) = q a a(q) =

so a(q) = C ¨ (q) means Kq 1∕a + F K = q (1∕a)−1 . q a Solving, K Kq 1∕a + F = q 1∕a a 1 1∕a K −1 q = F a 4 5a Fa q= . K(1 − a)

316

Chapter Four /SOLUTIONS

Solutions for Section 4.6 1. The e˙ect on the quantity demanded is approximately E times the change in price. A price increase causes a decrease in quantity demanded and a price decrease causes an increase in quantity demanded. (a) The quantity demanded decreases by about 0.5(3%) = 1.5%. (b) The quantity demanded increases by about 0.5(3%) = 1.5%. 2. The e˙ect on the quantity demanded is approximately E times the change in price. A price increase cause a decrease in quantity demanded and a price decrease cause an increase in quantity demanded. (a) The quantity demanded decreases by about 2(3%) = 6%. (b) The quantity demanded increases by about 2(3%) = 6%. 3. A price increase leads to a decrease in demand. The percent increase in price is 0.25∕1.75 = 0.14286 = 14.286%. The e˙ect on the quantity demanded is approximately E times the change in price. Thus, the quantity demanded decreases by about 0.35(14.286%) = 5%. 4. A price decrease leads to an increase in demand. The percent decrease in price is 0.50∕12.00 = 0.041667 = 4.1667%. The e˙ect on the quantity demanded is approximately E times the change in price. Thus, the quantity demanded increases by about 1.4(4.1667%) = 5.833%. 5. (a) We have E = |p∕q ⋅ dq∕dp| = |dollars∕tons ⋅ tons∕dollars|. All the units cancel, and so elasticity has no units. (b) We have E = |p∕q ⋅ dq∕dp| = |yen∕liters ⋅ liters∕yen|. All the units cancel, and so elasticity has no units. (c) Elasticity has no units. This is why it makes sense to compare elasticities of di˙erent products valued in di˙erent ways and measured in di˙erent units. Changing units of measurement will not change the value of elasticity. 6. Elasticity will be high. As soon as the price of a brand is raised, many people will switch to another brand causing a drop in sales. 7. Demand for smart TVs will be elastic, since it is not a necessary item. If the prices are too high, people will not choose to buy them, so price changes will cause relatively large demand changes. 8. Elasticity will be low. Nearly everyone regards internet service as a necessity. Since there is not another company they can turn to, they will keep their service even if the company raises the price. The number of sales will change very little. 9. Table 4.5 of Section 4.6 gives the elasticity of peaches as 1.49. Since E > 1, the demand is elastic, and a percentage change in price will cause a larger percentage change in demand. Specifcally, since E = 1.49, a 1% increase in price would cause an approximately 1.49% drop in demand: peaches are not essential, but a luxury item. 10. Table 4.5 of Section 4.6 gives the elasticity of potatoes as 0.27. Here 0 ≤ E ≤ 1, so the demand is inelastic, and a percentage change in price will cause a smaller percentage change in demand. Specifcally, since E = 0.27, a 1% increase in price would cause an approximately 0.27% drop in demand: potatoes are a staple and not a luxury food item. 11. The elasticity of demand for a product, E, is given by | p dq | |. E = || ⋅ | | q dp | We frst fnd dq∕dp = −4p. At a price of $5, the quantity demanded is q = 200 − 50 = 150 and dq∕dp = −20, so | 5 | 2 E = || ⋅ (−20)|| = . | 150 | 3 Since E < 1 demand is inelastic. 12. The elasticity of demand is given by

| p dq | |. E = || ⋅ | | q dp | To evaluate dq∕dp we solve p = 90 − 10q for q, so q = (90 − p)∕10 and hence dq 1 =− . dp 10

When p = 50, we fnd q = 4, dq∕dp = −1∕10 so | 50 1 | | = 1.25. E = || ⋅ − 10 || |4

4.6 SOLUTIONS

317

Since the elasticity is Percent change in demand Percent change in price when the price increases by 2% the percent change in demand is given by E=

Percent change in demand = E ⋅ Percent change in price = 2 ⋅ (1.25) = 2.5. Therefore, the percentage change in demand is 2.5%. 2.5% fall in demand. 13. (a) We use

| p dq | | | E=| ⋅ |. | q dp | | |

So we approximate dq∕dp at p = 1.00. dq 2440 − 2765 −325 ≈ = = −1300 dp 1.25 − 1.00 0.25 | p dq | | 1.00 | | | | | E=| ⋅ ⋅ (−1300)| = 0.470 |≈| | q dp | | 2765 | | | | | Since E = 0.470 < 1, demand for the candy is inelastic at p = 1.00. (b) At p = 1.25, dq 1980 − 2440 −460 ≈ = = −1840 dp 1.50 − 1.25 0.25 | | p dq | | 1.25 | | | | ⋅ (−1840)| = 0.943 so E=| ⋅ |≈| | | q dp | | 2440 | | | | At p = 1.5, dq 1660 − 1980 −320 ≈ = = −1280 dp 1.75 − 1.50 0.25 | | p dq | | 1.50 | | | | ⋅ (−1280)| = 0.970 so E=| ⋅ |≈| | | q dp | | 1980 | | | | At p = 1.75, dq 1175 − 1660 −485 ≈ = = −1940 dp 2.00 − 1.75 0.25 | | p dq | | 1.75 | | | | ⋅ (−1940)| = 2.05 so E=| ⋅ |≈| | | q dp | | 1660 | | | | At p = 2.00, dq 800 − 1175 −375 ≈ = = −1500 dp 2.25 − 2.00 0.25 | | p dq | | 2.00 | | | | ⋅ (−1500)| = 2.55 so E=| ⋅ |≈| | | q dp | | 1175 | | | | At p = 2.25, dq 430 − 800 −370 ≈ = = −1480 dp 2.50 − 2.25 0.25 | | p dq | | 2.25 | | | | ⋅ (−1480)| = 4.16 so E=| ⋅ |≈| | | q dp | | 800 | | | | Examination of the elasticities for each of the prices suggests that elasticity gets larger as price increases. In other words, at higher prices, an increase in price will cause a larger drop in demand than the same size price increase at a lower price level. This can be explained by the fact that people will not pay too much for candy, as it is somewhat of a “luxury” item. (c) Elasticity is approximately equal to 1 at p = $1.25 and p = $1.50. so

318

Chapter Four /SOLUTIONS (d)

Table 4.2 p($)

1.00

2765

Revenue = p ⋅ q ($) 2765

1.25

2440

3050

1.50

1980

2970

1.75

1660

2905

2.00

1175

2350

2.25

800

1800

2.50

430

1075

We can see that revenue is maximized at p = $1.25, with p = $1.50 a close second, which agrees with part (c). 14. (a) If the price of yams is $2/pound, the quantity sold will be q = 5000 − 10(2)2 = 5000 − 40 = 4960 so 4960 pounds will be sold. (b) Elasticity of demand is given by | | | | | | | p dq | | p d | |p | 20p2 E=| ⋅ (5000 − 10p2 )| = | ⋅ (−20p)| = |=| ⋅ | q dp | | q dp | |q | q | | | | | | Substituting p = 2 and q = 4960 yields E=

20(2)2 80 = = 0.016. 4960 4960

Since E < 1 the demand is inelastic, so it would be more accurate to say “People want yams and will buy them no matter what the price.” 15. (a) At a price of $2/pound, the quantity sold is q = 5000 − 10(2)2 = 5000 − 40 = 4960 so the total revenue is R = pq = 2 ⋅ 4960 = $9,920 (b) We know that R = pq, and that q = 5000 − 10p2 , so we can substitute for q to fnd R(p) R(p) = p(5000 − 10p2 ) = 5000p − 10p3 To fnd the price that maximizes revenue we take the derivative and set it equal to 0. R¨ (p) = 0 5000 − 30p2 = 0 30p2 = 5000 p2 = 166.67 p = ±12.91 We disregard the negative answer, so p = 12.91 is the only critical point. Is it the maximum? We use the frst derivative test. R¨ (p) > 0 if p < 12.91 and R¨ (p) < 0 if p > 12.91 So R(p) has a local maximum at p = 12.91. We also test the function at p = 0, which is the only endpoint. R(0) = 5000(0) − 10(0)3 = 0 R(12.91) = 5000(12.91) − 10(12.91)3 = 64,550 − 21,516.85 = $43,033.15 So we conclude that revenue is maximized at price of $12.91/pound.

4.6 SOLUTIONS

319

(c) At a price of $12.91/pound the quantity sold is q = 5000 − 10(12.91)2 = 5000 − 1666.68 = 3333.32 so the total revenue is R = pq = (3333.32)(12.91) = $43,033.16 which agrees with part (b). (d)

| | | | |p | 20p2 | p dq | | p d | | | (5000 − 10p2 )| = | ⋅ (−20p)| = E=| ⋅ |=| ⋅ | q dp | | q dp | |q | q | | | | | | Substituting p = 12.91 and q = 3333.32 yields E=

20(12.91)2 3333.36 = ≈1 3333.32 3333.32

which agrees with the result that maximum revenue occurs when E = 1. 16. We have

| | p | | p dq | | p d −0.6 | −1.6 | |=| | E = || ⋅ | | kp−0.6 ⋅ dp (kp )| = | kp−0.6 ⋅ (−0.6)kp )| = | − 0.6| = 0.6. q dp | | | | | |

17. The revenue is maximized by fnding the critical point of the revenue function: R = pq = p(1000 − 2p2 ) = 1000p − 2p3 . Di˙erentiate to fnd the critical points: dR = 1000 − 6p2 = 0 dp 1000 p2 = 6 p ≈ 12.91 To maximize revenues, the price of the product should be $12.91. 18. Demand is elastic at all prices. No matter what the price is, you can increase revenue by lowering the price. In the end, you would lower your prices all the way to zero. This is not a realistic example, but it is mathematically possible. It would correspond, for instance, to the demand equation q = 1∕p2 , which gives revenue R = pq = 1∕p which is decreasing for all prices p > 0. 19. Demand is inelastic at all prices. No matter what the price is, you can increase revenue by raising the price, so there is no actual price for which your revenue is maximized. This is not possible. It √ a realistic example, but it is mathematically √ would correspond, for instance, to the demand equation q = 1∕ p, which gives revenue R = pq = p which is increasing for all prices p > 0. 20. (a) High elasticity means that small changes in price lead to proportionally larger changes in quantity demanded and sold. As the the price increases, demand falls by a larger proportion so we expect revenue to fall. (b) Using the product rule on R = pq, we have dp dq dR = ⋅q+p⋅ . dp dp dp Since

| p dq | |, E = || | | q dp | and we know demand decreases as price increases, we have E=−

p dq . q dp

Thus, dR = q − qE = q(1 − E). dp (c) For large E (that is, for E > 1), we see dR∕dp < 0, so revenue decreases as price increases.

320

Chapter Four /SOLUTIONS

21. (a) Since q = k∕p, we have dq∕dp = −k∕p2 and p∕q = p∕(k∕p) = p2 ∕k. Therefore | p dq | | p2 −k | | | | | E=| ⋅ |=| ⋅ |=1 | q dp | | k p2 | | | | | Elasticity is a constant equal to 1, independent of the price. (b) Elasticity equal to 1 corresponds to critical points of the revenue function R. Since R = pq = k is constant, dR∕dp = 0, and so all prices are critical points of the revenue function. 22. Writing q = k∕pr as q = kp−r , we have | | p dq | | p | || pr+1 d | |=| E = || ⋅ ⋅ (−r)k∕pr+1 )| = r. ⋅ (kp−r )|| = | | | r | | q dp | | k∕p dp | || k | 23. Since marginal revenue equals dR∕dq and R = pq, we have, using the product rule, 0 1 ⎛ ⎞ d(pq) dp q dp dR 1 1 = =p⋅1+ ⋅q =p 1+ ⋅ = p ⎜1 − p dq ⎟ = p 1 − . ⎜ E dq dq dq p dq − q ⋅ dp ⎟⎠ ⎝ 24. Marginal cost equals C ¨ (q) = k. On the other hand, by the preceding exercise, marginal revenue equals p(1 − 1∕E). Maximum proft will occur when the two are equal k = p(1 − 1∕E) Thus k∕p = 1 − 1∕E 1∕E = 1 − k∕p On the other hand, kq Proft Revenue − Cost Cost k 1 = =1− =1− =1− = Revenue Revenue Revenue pq p E 25. (a) We see that d1 = p0 . In addition, we have dp Rise −d2 = Slope of demand curve = = dq Run q0 so d2 = −q0 ⋅ dp∕dq. Therefore

d1 p0 p dq = =− 0 ⋅ =E d2 q0 dp −q0 ⋅ dp dq

(b) For prices near the maximum possible, we have d1 > d2 and so elasticity E = d1 ∕d2 is greater than 1 for small quantities. For prices p near 0 (that is, near the q-axis), we have d1 < d2 and so E = d1 ∕d2 is less than 1 for large quantities. Elasticity equals 1 where d1 = d2 at exactly half the maximum possible quantity. See Figure 4.132. price

✛ 1

✛

E=1

✛

Figure 4.132

quantity

4.6 SOLUTIONS

321

26. Since R = pq, we have dR∕dp = p(dq∕dp) + q. We are assuming that | p dq | | | E=| ⋅ |>1 | q dp | | | so, removing the absolute values −E =

p dq < −1. q dp

Multiplication by q gives p and hence

dq < −q dp

dq dR =p +q <0 dp dp

27. Since R = pq, we have dR∕dp = p(dq∕dp) + q. We are assuming that | p dq | | | 0≤E=| ⋅ |<1 | q dp | | | so, removing the absolute values 0 ≥ −E =

p dq > −1 q dp

Multiplication by q gives p and hence

dq > −q dp

dq dR =p +q >0 dp dp

Δq∕q | shows that the cross-price elasticity measures the ratio of the fractional change in 28. The approximation Ecross ≈ | Δp∕p quantity of chicken demanded to the fractional change in the price of beef. Thus, for example, a 1% increase in the price of beef will stimulate a Ecross % increase in the demand for chicken, presumably because consumers will react to the price rise in beef by switching to chicken. The cross-price elasticity measures the strength of this switch.

29. If shoppers spend 1% more time in the store, then they spend about 1.3% more money. C∕C 30. (a) The approximation EC,q ≈ ΔΔq∕q shows that EC,q measures the ratio of the fractional change in the cost of production to the fractional change in the quantity produced. Thus, for example, a 1% increase in production will result in an EC,q % increase in the cost of production. (b) Since average cost equals C∕q and marginal cost equals dC∕dq, we have

Marginal cost dC∕dq q dC = = ⋅ = EC,q Average cost C∕q C dq | | | Δq∕q | 31. The approximation Eincome ≈ | ΔI∕I | shows that the income elasticity measures the ratio of the fractional change in | | | | quantity of the product demanded to the fractional change in the income of the consumer. Thus, for example, a 1% increase in income will translate into an Eincome % increase in the quantity purchased. After an increase in income, the consumer will tend to buy more. The income elasticity measures the strength of this tendency.

322

Chapter Four /SOLUTIONS

Solutions for Section 4.7 1.

t −15

Figure 4.133: C = 0.1

t −15

Figure 4.134: C = 1

t −15

Figure 4.135: C = 10

See Figures 4.133–4.135. The value of C appears to a˙ect where the curve cross the vertical axis. When C = 1, the graph crosses the vertical axis at t = 0 or at the point of diminishing returns. If C > 1, the graph crosses the the vertical axis before the point of diminishing returns; the higher C is, the sooner the graph crosses the vertical axis. If C < 1, the curve crosses the vertical axis after it changes concavity; the smaller C is, the closer to the limiting value the graph is before crossing the vertical axis. 2. The long-run total number of cases is given by L in the formula P =

L . 1 + Ce−kt

Thus, the country with the largest total number of cases is the Dominican Republic, with 539,227 cases. 3. The continuous growth rate for small P is given by k in the formula P =

L . 1 + Ce−kt

Thus, the country with the largest total number of cases is the Dominican Republic, with a continuous rate of 0.35 per week. 4. To fnd the initial number of cases from the formula P =

L , 1 + Ce−kt

we set t = 0 and evaluate

L L = . 1+C 1 + Ce−k⋅0 For the three countries given, rounding to the nearest integer, the initial number of cases were: P0 =

539,227 = 3033 1 + 176.8 3771 Dominica = =4 1 + 941.75 81,780 Guadeloupe = = 3. 1 + 27,259

Dominican Republic =

Thus, the country with the smallest number of cases initially was Guadeloupe, with 3 cases. 5. The logistic curve levels o˙ at the value L in the formula P =

L . 1 + ce−kt

Thus, the country with the largest total number of cases, the Dominican Republic, levels o˙ to 539,227 and is the top curve. Dominica levels o˙ to 3771, which can not be distinguished from the horizontal axis with the vertical scale shown. Guadeloupe levels o˙ to 81,780 and is the lower of the graphs visible. 6. (a) The three graphs rise to di˙erent limiting values. Since the parameter L gives the upper limit of the logistic curve, L is the varying parameter. (b) The curve with the highest limiting value, the top curve, corresponds to the largest value of L. (c) If L is larger, more people in the community catch the disease before the epidemic dies out. The population will try to fgure out ways to reduce the value of L.

4.7 SOLUTIONS

323

7. (a) The three curves rise to the same upper limit but they do so at di˙erent rates. Since the parameter k controls how fast the logistic curve climbs, k is the varying parameter. (b) The curve that rises the fastest, that has the steepest slope during its rise, corresponds to the largest value of k. This is the top curve in the fgure. (c) Once the graph of P (t) fattens out at its upper limit, the epidemic is over. If k is large, the epidemic passes quickly, with many people sick at the same time. If k is smaller, the same number of people catch the disease but there are fewer at once and it takes longer for the epidemic to pass. On the one hand, it may seem good to get the epidemic over with quickly. On the other hand, if it goes too quickly, the strain on the infrastructure of dealing with many cases at once might be too much to manage. 8. (a) The three graphs rise to the same upper limit, so they have the same L. They rise equally quickly, so they have the same k. The remaining parameter is C. Changing C causes a horizontal shift in the graph, which is what we see. So the varying parameter is C. (b) The largest C corresponds to the graph farthest to the right. (c) The graphs show similar epidemics occurring at di˙erent times. A community may want a larger C because that would give them more time to prepare. If C is large enough, a vaccine may be available soon enough to mitigate the impact of the epidemic. 9. (a) The data is plotted in Figure 4.136. On the intervals less than 10, the average rate of change increases, and greater than 10, this rate decreases, so the point of diminishing returns is about 10. total sales (thousands) Point of 60 diminishing returns 30 2

months since 14 game introduced

Figure 4.136 (b) The point of diminishing returns happens when total sales reach roughly 30,000. This predicts total sales of about 2(30) = 60,000. 10. (a) If we graph the data, we see that it looks approximately logistic. (See Figure 4.137.) But logistic growth is also reasonable from a common-sense point of view. As broadband internet became more available, the number of households which have it frst grew exponentially but then slowed down as more and more people had it. Eventually, nearly everyone who would subscribe already had and the number leveled o˙. broadband subscribers (millions) 30 20 10 0 5

years since 2000

Figure 4.137 (b) Since the number of households with broadband keeps increasing more quickly until about 2006, when t = 6, the point of diminishing returns happens around 13 million households. This predicts a carrying capacity of 26 million which is pretty close to the 26.59 million we see in 2018 and 26.7 million we see in 2019. (c) The numerator in the logistic function gives the limiting value: 25.475 million. (d) The limiting value predicts the number of households that will eventually have broadband internet. This model predicts that there will never be a time when more than 25.475 million households have broadband. We can see from the numbers in 2016 onward that this prediction is not accurate as the number of households with broadband is still growing, but at a lesser rate.

324

Chapter Four /SOLUTIONS

11. (a) At t = 0, which corresponds to 1935, we have P =

1 = 0.25 1 + 3e−0.0275(0)

showing that 25% of the land was in use in 1935. (b) This model predicts that as t gets very large, P approaches 1. That is, the model predicts that in the long run, all the land will be used for farming. (c) To solve this graphically, enter the function into a graphing calculator and trace the resulting curve until it reaches a height of 0.5, which occurs when t = 39.9 ≈ 40. Since t = 0 corresponds to 1935, t = 40 corresponds to 1935 + 40 = 1975. According to this model, the Tojolabal were using half their land in 1975. See Figure 4.138. P (proportion of land in use for farming) 1

0.5

80 120 160 200

t (years since 1935)

Figure 4.138 (d) The point of diminishing returns occurs when P = L∕2 or at one-half the carrying capacity. In this case, P = 1∕2 in 1975, as shown in part (c). 12. (a) Between day t = 0 and day t = 5, Average rate of change

ΔP 222 − 95 = = 25.4 new cases per day. Δt 5−0

Similarly, we fnd the values in Table 4.3. Table 4.3 Days

0–5

5–12

12–19

19–26

26–33

33–40

40–47

New cases per day

25.4

35.4

47.1

44.0

35.7

24.1

13.4

Days

47–54

54–61

61–68

68–75

75–81

81–87

New cases per day

7.6

5.1

2.0

2.1

1.8

0.8

(b) The frst computation showing a reduced rate of change was made with the data for t = 26, which corresponds to April 12, 2003. On this date, epidemiologists could report that the rate of new cases had begun to slow. (c) The rate of change increases from t = 0 to t = 19, so the graph is concave up there. After t = 19, the rate of change decreases and the graph is concave down. Exponential functions do not change concavity. (d) The infection point appears to occur near t = 19, when the average number of new cases per day stops increasing and begins decreasing. At this point, we have P = 800, so we estimate the limiting value of P to be about 2 ⋅ 800 = 1600 cases. (e) The predicted limiting value is 1760, which is close to the actual value of 1755. 13. Sales of a new product could very well follow a logistic curve. At frst, sales will grow exponentially as more and more people hear of the product and decide to buy it. Eventually, though, everyone will know about the product and while people may still buy it, not as many will (sales will slow down). Eventually, it’s possible that everyone who would buy the product already has, at which point sales will stop. It behooves the seller to notice the point of diminishing returns so they don’t make more of their product than people will want to buy. 14. (a), (b) The graph of P (t) with carrying capacity L and point of diminishing returns t0 is in Figure 4.139. The derivative P ¨ (t) is also shown.

4.7 SOLUTIONS

325

L P (t)

P (t0 )

P ¨ (t) t t0

Figure 4.139 (c) Keeping track of rate of sales is the same as keeping track of the derivative P ¨ (t). The point of diminishing returns happens when the concavity of P (t) changes, which is the when the derivative, P ¨ (t), switches from increasing to decreasing. This happens when P ¨ (t) reaches its maximum at t = t0 15. Substituting t = 0, 10, 20, … , 70 into the function P = 3.9(1.03)t gives the values in Table 4.4. Notice that the agreement is very close, refecting the fact that an exponential function models the growth well over the period 1790–1860. Table 4.4 Predicted versus actual US population 1790–1860, in millions. (exponential model) Year

Actual

Predicted

Year

Actual

Predicted

1790

3.9

1830

12.9

12.7

1800

5.3

5.2

1840

17.1

1810

7.2

7.0

1850

23.2

23.0

1820

9.6

9.5

1860

31.4

30.9

P =

187 1 + 47e−0.0318t

16. Substituting t = 0, 10, 20, 30, … into the function

gives the values in Table 4.5. Notice that the agreement is close between the predicted and actual values. Table 4.5 Year

Actual

Predicted

Year

Actual

Predicted

Year

Actual

Predicted

1790

3.9

1840

17.1

17.7

1890

62.9

63.3

1800

5.3

1850

23.1

23.5

1900

76.0

77.2

1810

7.2

1860

31.4

30.8

1910

92.0

91.9

1820

9.6

9.8

1870

38.6

39.9

1920

105.7

106.7

1830

12.9

13.2

1880

50.2

50.7

1930

122.8

120.8

1940

131.7

133.7

17. (a) We use k = 1.78 as a rough approximation. We let L = 5000 since the problem tells us that 5000 people eventually get the virus. This means the limiting value is 5000. (b) We know that 5000 P (t) = and P (0) = 10 1 + Ce−1.78t so 5000 5000 = 1+C 1 + Ce0 10(1 + C) = 5000 10 =

1 + C = 500 C = 499.

326

Chapter Four /SOLUTIONS (c) We have P (t) =

5000 . This function is graphed in Figure 4.140. 1 + 499e−1.78t number infected 5000 P (t) = 1+499e −1.78t

4000 3000 2000 1000 1

week

Figure 4.140

(d) The point of diminishing returns appears to be at the point (3.5, 2500); that is, after 3 and a half weeks and when 2500 people are infected. 18. (a) In a logistic epidemic model we have P (t) = L∕(1 + Ce−kt ) where k is the approximate exponential growth rate near the start of the epidemic. We see that k = 0.1363 = 13.63%, so the number of cases was initially growing 13.63% per day. (b) The number of daily cases is approximately P ¨ , the rate of change of P , so its maximum corresponds to the point of maximum slope on the graph of P . By plotting P we fnd that the maximum rate of change occurs at around t = 30. See Figure 4.141. Thus, according to the model, the maximum number of new confrmed cases in one day occurred around March 31. P (total cases) 7400

3700

t (days)

Figure 4.141

19. The slope of the curve, dy∕dt, is given by � � −2 � −2 dy = −50 1 + 6e−2t −12e−2t = 600e−2t 1 + 6e−2t . dt If the slope has a maximum, it occurs at a critical point of dy∕dt or at the endpoint t = 0. We have � � −2 � −3 � d2y −12e−2t = 600 −2e−2t 1 + 6e−2t − 1200e−2t 1 + 6e−2t dt2 1200e−2t � −2t = 6e − 1 . (1 + 6e−2t )3 At a critical point of dy∕dt, we have d 2 y∕dt2 = 0, so 6e−2t − 1 = 0 1 t = ln 6 = 0.896. 2 Since

dy || 600 = = 12.25 49 dt ||t=0

and

dy || = 25, dt ||t=(1∕2) ln 6

4.7 SOLUTIONS

327

the maximum slope occurs at t = 21 ln 6. We see in Figure 4.142 that the slope increases as t increases from 0 and tends to 0 as t → ∞, so the only critical point of slope, t = 12 ln 6, is a local and global maximum for the slope. At t = 12 ln 6, we have y = 25. The point 12 ln 6, 25 is the point where slope is maximum. y 50

50 1 + 6e−2t

t 1 ln 6 2

Figure 4.142

20. (a) The fact that f ¨ (15) = 11 means that the slope of the curve at the infection point, (15, 50) is 11. In terms of dose and response, this large slope tells us that the range of doses for which this drug is both safe and e˙ective is small. (b) As we can see from Figure 4.88 in the text, a dose-response curve starts out concave up (slope increasing) and switches to concave down (slope decreasing) at an infection point. Since the slope at the infection point (15, 50) is 11, and the slope is increasing before the infection point, f ¨ (10) is less than 11. Since the slope is decreasing after the infection point, f ¨ (20) is also less than 11. 21. (a) The dose-response curve for this drug is shown if Figure 4.143. 100 100 R = 1+100e −0.1x

x 150

Figure 4.143 (b) First solve for x: 100 1 + 100e−0.1x R + 100Re−0.1x = 100 R=

100Re−0.1x = 100 − R 100 − R e−0.1x = 100R 100 − R −0.1x = ln 100R 100 − R x = −10 ln . 100R Substituting the value of R = 50 we get x = −10 ln

100 − 50 100 ⋅ 50

≈ 46.0517019 ≈ 46.05

328

Chapter Four /SOLUTIONS (c) Evaluate the formula for x obtained in part (b) for R = 20 and R = 70. For R = 20, 100 − 20 x = −10 ln ≈ 32.18875825 ≈ 32.12 100 ⋅ 20 For R = 70, 100 − 70 x = −10 ln ≈ 54.5246805 ≈ 54.52 100 ⋅ 70 So the range of doses that is both safe and e˙ective is between 32.12 mg and 54.52 mg.

22. (a) The dose-response curve for product C crosses the minimum desired response line last, so it requires the largest dose to achieve the desired response. The dose-response curve for product B crosses the minimum desired response line frst, so it requires the smallest dose to achieve the desired response. (b) The dose-response curve for product A levels o˙ at the highest point, so it has the largest maximum response. The dose-response curve for product B levels o˙ at the lowest point, so it has the smallest maximum response. (c) Product C is the safest to administer because its slope in the safe and e˙ective region is the least, so there is a broad range of dosages for which the drug is both safe and e˙ective. 23. If the derivative of the dose-response curve is smaller, the slope is not as steep. Since the slope is not as steep, the response increases less at the same dosage. Therefore, there is a wider range of dosages that are both safe and e˙ective, and consequently the dosage given to the patient does not have to be as exact. 24. The range of safe and e˙ective doses begins at 10 mg where the drug is e˙ective for 100 percent of patients. It ends at 18 mg where the percent lethal curve begins to rise above zero. 25. When 50 mg of the drug is administered, it is e˙ective for 85 percent of the patients and lethal for 6 percent. 26. (a) We have L 1 + Ce−kt L(1 + Ce−kt ) − L = 1 + Ce−kt CLe−kt = 1 + Ce−kt = CP e−kt .

L−P = L−

Dividing by P gives L−P = Ce−kt . P (b) The existing population is P . Since the carrying capacity is L, the additional population the environment can support is L − P . 27. (a) We have f (V ) = =

1 1 + e−(V +25)∕2 1 1

1 + e−25∕2 e− 2 V

Comparison with L 1 + Ce−kV shows that L = 1, k = 1∕2 = 0.5, and C = e−25∕2 = 3.7 ⋅ 10−6 . (b) Ten percent of the channels are open at potential V where f (V ) =

f (V ) = 0.1 1

= 0.1 1 + e−(V +25)∕2 V = −29.4 mV. Half the channels are open when f (V ) = 0.5, or V = −25 mV. Ninety per cent of the channels are open when f (V ) = 0.9, or V = −20.6 mV.

4.8 SOLUTIONS

329

28. (a) Before di˙erentiating, we multiply out, giving dP P kP 2 = kP 1 − = kP − . dt L L Since k and L are constant, the chain rule gives 0 1 d2P d dP d kP 2 = = kP − dt dt dt L dt2 dP 2kP dP =k − . dt L dt (b) Setting d 2 P ∕dt2 = 0 gives 0=k

dP 2kP dP kdP − = dt L dt dt

So 1−

2P 1− . L

2P = 0, L

that is, when L , 2 we have d 2 P ∕dt2 = 0; that is, we are at the point of diminishing returns. P =

Solutions for Section 4.8 1. (a) See Figure 4.144. C (ng/ml)

20 C = 12.4te−0.2t 10

t (hours)

Figure 4.144 (b) The surge function y = atebt changes from increasing to decreasing at t = 1b . For this function b = 0.2 so the peak is 1 at 0.2 = 5 hours. We can now substitute this into the formula to compute the peak concentration:

C = 12.4(5)e−0.2(5) = 22.8085254 ng/ml ≈ 22.8 ng/ml. (c) Tracing along the graph of C = 12.4te−0.2t , we see it crosses the line C = 10 at t ≈ 1 hour and at t ≈ 14.4 hours. Thus, the drug is e˙ective for 1 ≤ t ≤ 14.4 hours. (d) The drug drops below C = 4 for t > 20.8 hours. Thus, it is safe to take the other drug after 20.8 hours. 2.

0.4

C 0.04

4 t

t −1

−0.4

Figure 4.145: a = 1

−1

−4

Figure 4.146: a = −10

t −1

−0.04

Figure 4.147: a = 0.1

The parameter a apparently a˙ects the height and direction of C = ate−bt . If a is positive, the “hump” is above the t-axis. If a is negative, it’s below the t-axis. If a > 0, the larger the value of a, the larger the maximum value of C. If a < 0, the more negative the value of a, the smaller the minimum value of C.

330

Chapter Four /SOLUTIONS

3. Newborn babies are generally not as e˙ective as adults in eliminating drugs from the body. See the fgure in the text. The level of the drug in the body begins to decline after two hours in an adult, but not until six hours have elapsed in a newborn. In addition, the concentration in the blood reaches a higher level in the newborn than it does in the adult. Graphs such as this explain why physicians modify the doses for drugs given to pregnant women and women who are breastfeeding. 4. Large quantities of water dramatically increase the value of the peak concentration, but do not change the amount of time it takes to reach the peak concentration. The e˙ect of the volume of water taken with the drug wears o˙ after approximately 6 hours. 5. Figure 4.148 has its maximum at t = 1.3 hours, C = 23.6 ng/ml. Concentration (ng/ml) 25 20 15 10 5

t (hours)

Figure 4.148 6. For cigarettes the peak concentration is approximately 17 ng/ml, the time until peak concentration is about 10 minutes, and the nicotine is at frst eliminated quickly, but after about 45 minutes it is eliminated more slowly. For chewing tobacco, the peak concentration is approximately 14 ng/ml and the time until peak concentration is about 30 minutes (when chewing stops). The nicotine is eliminated at a slow, somewhat erratic rate. For nicotine gum the peak concentration is about 10 ng/ml and the time until peak concentration is approximately 45 minutes. The nicotine is eliminated at a very slow but steady rate. 7. (a) Di˙erentiating using the product rule gives C ¨ (t) = 20e−0.03t + 20t(−0.03)e−0.03t = 20(1 − 0.03t)e−0.03t . At the peak concentration, C ¨ (t) = 0, so 20(1 − 0.03t)e−0.03t = 0 1 t= = 33.3 minutes. 0.03 When t = 33.3, the concentration is C = 20(33.3)e−0.03(33.3) ≈ 245 ng/ml. See Figure 4.149. The curve peaks after 33.3 minutes with a concentration of 244.9 ng/ml. concentration (ng/ml)

200

C(t) = 20te−0.03t

150 100 50

100

150

200

Figure 4.149

250

minutes

4.8 SOLUTIONS

331

(b) After 15 minutes, the drug concentration will be C(15) = 20(15)e(−0.03)(15) ≈ 191 ng/ml. After an hour, the concentration will be C(60) = 20(60)e−0.03(60) ≈ 198 ng/ml. (c) We want to know where C(t) = 10. We estimate from the graph. It looks like C(t) = 10 after 190 minutes or a little over 3 hours. 8. We fnd values of the parameters in the function C = ate−bt to create a local maximum at the point (1.3, 23.6). We frst set the derivative equal to zero and solve for t to fnd critical points. Using the product rule, we have: dC = at(e−bt (−b)) + a(e−bt ) = 0 dt ae−bt (−bt + 1) = 0 1 t= . b The only critical point is at t = 1∕b. Since we want a critical point at t = 1.3, we substitute and solve for b: 1 b 1 b= = 0.769. 1.3

1.3 =

To fnd the value of a, we use the fact that C = 23.6 when t = 1.3. We have: a(1.3)e−0.769(1.3) = 23.6 a ⋅ 1.3e−1 = 23.6 23.6 a= = 49.3. 1.3e−1 We have a = 49.3 and b = 0.769. 9. (a) The value f (t) = e−t − e−2t is the vertical distance between the two curves in Figure 4.150. That distance is 0 at t = 0. The distance frst increases as t increases from 0. As t gets larger and larger, the distance eventually begins to decrease toward 0 because both e−t and e−2t approach the same horizontal asymptote, the t-axis. Thus the graph of f (t) shows a rapid increase from zero followed by a slow decrease back toward zero; this is the shape of a surge. (b) At a critical point we have f ¨ (t) = −e−t + 2e−2t = 0. Hence e−t = 2e−2t et = 2 t = ln 2 = 0.693. At the critical point, f (ln 2) = 1∕4 = 0.250. The critical point is (ln 2, 1∕4) = (0.693, 0.250). At an infection point we have f ¨¨ (t) = e−t − 4e−2t = 0. Hence e−t = 4e−2t et = 4 t = ln 4 = 1.386 3 f (ln 4) = = 0.1875. 16 The infection point is (2 ln 2, 3∕16) = (1.386, 0.187). See Figure 4.151

332

Chapter Four /SOLUTIONS y

0.3

(log 2, 1∕4) (log 4, 3∕16)

0.2 0.5 0.1

e−t e−2t t 1

Figure 4.150

e−t − e−2t x 4 5

Figure 4.151

10. Food dramatically increases the value of the peak concentration but does not a˙ect the time it takes to reach the peak concentration. The e˙ect of food is stronger during the frst 8 hours. 11. (a) Products A and B have much higher peak concentrations than products C and D. Product A reaches its peak concentration slightly before products B, C, and D, which all take about the same time to reach peak concentration. (b) If the minimum e˙ective concentration were low, perhaps 0.2, and the maximum safe concentration were also low, perhaps 1.0, then products C and D would be the preferred drugs since they do not enter the unsafe range while being well within the e˙ective range. (c) If the minimum e˙ective concentration were high, perhaps 1.2, and the maximum safe concentration were also high, perhaps 2.0, then product A would be the preferred drug since it does not enter the unsafe range and it is the only drug that is in the e˙ective range for a substantial amount of time. 12. (a) The IV method reaches peak concentration the fastest, it in fact begins at its peak. The P-IM method reaches peak concentration the slowest. (b) The IV method has the largest peak concentration. The PO method has the smallest peak concentration. (c) The IV method wears o˙ the fastest. The P-IM method wears o˙ the slowest. (d) The P-IM method has the longest e˙ective duration. The IV method has the shortest e˙ective duration. (e) It is e˙ective for approximately 5 hours. 13. In general, a faster dissolution rate corresponds to a larger peak concentration. Dissolution rate does not a˙ect time to reach peak concentration, as the four products have di˙erent dissolution rates but virtually the same time to reach peak concentration.

Solutions for Chapter 4 Review 1. See Figure 4.152. Local and global max 50

Local max

f (x)

Local max

30 20

Local min

10 1

Local and global min x 4 5 6

Figure 4.152

SOLUTIONS to Review Problems For Chapter Four

333

2. See Figure 4.153. Local and global max 8

Local max

f (x) Critical point (not max or min)

6 4 2

Local and global min x 1

Figure 4.153 3. To fnd the critical points, we set f ¨ (x) = 0. Since f ¨ (x) = 3x2 − 18x + 24, we have 3x2 − 18x + 24 = 0 3(x2 − 6x + 8) = 0 3(x − 2)(x − 4) = 0 x = 2, 4. There are two critical points: at x = 2, x = 4. To fnd the infection points, we look for points where f ¨¨ is zero or undefned. Since f ¨¨ (x) = 6x − 18, it is defned everywhere, and setting f ¨¨ (x) = 0 we get 6x − 18 = 0 x = 3. Furthermore, 6x − 18 is positive when x > 3 and negative when x < 3, so f ¨¨ changes sign at x = 3. Thus, there is one infection point: x = 3. 4. To fnd the critical points, we set f ¨ (x) = 0. Since f ¨ (x) = 5x4 − 30x2 , we have 5x4 − 30x2 = 0 5x2 (x2 − 6) = 0

√ √ x = 0, − 6, 6. √ √ There are three critical points: x = 0, x = − 6, and x = 6. To fnd the infection points, we look for points where f ¨¨ is undefned or zero. Since f ¨¨ (x) = 20x3 − 60x, it is defned everywhere. Setting it equal to zero, we get 20x3 − 60x = 0 20x(x2 − 3) = 0

√ √ x = 0, − 3, 3. √ √ √ √ < − 3 or 0 √ < x < 3, so f ¨¨ Furthermore, 20x3 − 60x is positive when − 3 < x < 0 or x > 3 and negative when x√ changes sign at each of the solutions. Thus, there are three infection points: x = 0, x = − 3, and x = 3. 5. To fnd the critical points, we set f ¨ (x) = 0. Since f ¨ (x) = 5x4 + 60x3 , we have 5x4 + 60x3 = 0 5x3 (x + 12) = 0 x = 0, −12. There are two critical points: x = 0, x = −12. To fnd the infection points, look for points where f ¨¨ is undefned or zero. Since f ¨¨ (x) = 20x3 + 180x2 , it is defned everywhere. Setting it equal to zero, we get 20x3 + 180x2 = 0 20x2 (x + 9) = 0 x = 0, −9. 2

Furthermore, 20x (x + 9) is negative when x < −9, and positive or zero when x > −9. So f ¨¨ does change sign at x = −9, but not at x = 0. Thus, there is one infection point: x = −9.

334

Chapter Four /SOLUTIONS

6. To fnd the critical points, we set f ¨ (x) = 0. Since f ¨ (x) = 5 − 3(1∕x), we have 3 =0 x 3 5= x 5x = 3

5−

x = 3∕5. There is one critical point, at x = 3∕5. To fnd the infection points, we look for points where f ¨¨ (x) is undefned or zero. Since f ¨¨ (x) = 3x−2 , it is undefned at x = 0. However, f (x) is also undefned at x = 0, so this is not an infection point. Also, 3∕x2 is never equal to zero, so there are no infection points. 7. To fnd the critical points, we set f ¨ (x) = 0. Using the product rule, we have f ¨ (x) = 4x ⋅ (e3x ⋅ 3) + 4 ⋅ e3x . 12xe3x + 4e3x = 0 4e3x (3x + 1) = 0 x = −1∕3. There is one critical point, at x = −1∕3. To fnd the infection points, we look for points where f ¨¨ is zero or undefned. Since f ¨¨ (x) = 12x⋅(e3x ⋅3)+12⋅e3x +4e3x ⋅3 = 12e3x (3x + 2), it is defned everywhere. Setting it equal to zero, we get 12e3x (3x + 2) = 0 x = −2∕3. Since 12e3x is always positive, 12e3x (3x + 2) is negative when x < −2∕3 and positive when x > −2∕3, so f ¨¨ changes sign at x = −2∕3. Thus, there is one infection point, at x = −2∕3. 8. (a) We have f ¨ (x) = 10x9 − 10 = 10(x9 − 1). This is zero when x = 1, so x = 1 is a critical point of f . For values of x less than 1, x9 is less than 1, and thus f ¨ (x) is negative when x < 1. Similarly, f ¨ (x) is positive for x > 1. Thus f (1) = −9 is a local minimum. We also consider the endpoints f (0) = 0 and f (2) = 1004. Since f ¨ (0) < 0 and f ¨ (2) > 0, we see x = 0 and x = 2 are local maxima. (b) Comparing values of f shows that the global minimum is at x = 1, and the global maximum is at x = 2. 9. (a) f ¨ (x) = 1−1∕x. This is zero only when x = 1. Now f ¨ (x) is positive when 1 < x ≤ 2, and negative when 0.1 < x < 1. Thus f (1) = 1 is a local minimum. The endpoints f (0.1) ≈ 2.4026 and f (2) ≈ 1.3069 are local maxima. (b) Comparing values of f shows that x = 0.1 gives the global maximum and x = 1 gives the global minimum. 10. (a) Setting the derivative of p(1 − p)4 equal to 0 d (p(1 − p)4 ) = (1 − p)4 − 4p(1 − p)3 = 0 dp (1 − p)3 (1 − p − 4p) = 0 (1 − p)3 (1 − 5p) = 0 p = 1∕5, 1. Thus, the critical points are p = 1∕5 and p = 1. (b) Since d d2 (p(1 − p)4 ) = ((1 − p)4 − 4p(1 − p)3 ) dp dp2 = −4(1 − p)3 − 4(1 − p)3 + 12p(1 − p)2 = 4(1 − p)2 (−2(1 − p) + 3p) = 4(1 − p)2 (−2 + 5p), substituting p = 1∕5 and p = 1, we have | 2 | d2 4 4 | (p(1 − p) ) = 4 (−1) < 0 | 5 dp2 | |p=1∕5

and

| | d2 4 | ) (p(1 − p) | =0 dp2 | |p=1

SOLUTIONS to Review Problems For Chapter Four

335

Thus p = 1∕5 is a local maximum. The second derivative test does not enable us to classify p = 1. However, p(1 − p)4 is positive everywhere except at p = 0 and p = 1, where it is 0. Thus, p = 1 is a local minimum. (c) The global maximum occurs at the local maximum, at p = 1∕5, so Maximum =

1 4 44 1 256 1− = 5 = . 5 5 3125 5

The global minimum occurs at the end points, so Minimum = 0(1 − 0)4 = 1(1 − 1)4 = 0. 11. There are several possibilities. The revenue stream could have been increasing during the last few days of June, reaching a high point on July 1, then going back down during the frst few days of July. In this case there was a local maximum in the revenue stream on July 1. The revenue stream could have been decreasing during the last few days of June, reaching a low point on July 1, then going back up during the frst few days of July. In this case there was a local minimum in the revenue stream on July 1. It is also possible that there was neither a local maximum nor a local minimum in the revenue stream on July 1. This could have happened two ways. On the one hand, the revenue stream could have been rising in late June, then held steady with no change around July 1, after which the revenue stream increased some more. On the other hand, the revenue stream could have been falling in late June, then held steady with no change around July 1, after which the revenue stream fell some more. The key feature in these critical point scenarios is that there was no appreciable change in the revenue stream of the company around July 1. 12. (a) The function f (x) is defned for x ≥ 0. We set the derivative equal to zero to fnd critical points: 1 f ¨ (x) = 1 + ax−1∕2 = 0 2 a 1 + √ = 0. 2 x √ Since a > 0 and x > 0 for x > 0, we have f ¨ (x) > 0 for x > 0, so there is no critical point with x > 0. The only critical point is at x = 0 where f ¨ (x) is undefned. (b) We see in part (a) that the derivative is positive for x > 0 so the function is increasing for all x > 0. The second derivative is 1 f ¨¨ (x) = − ax−3∕2 . 4 Since a and x−3∕2 are positive for x > 0, the second derivative is negative for all x > 0. Thus the graph of f is concave down for all x > 0. 13. The critical points of f occur where f ¨ is zero. These two points are indicated in the fgure below.

f ¨ (x)

f has a local min.

f has crit. pt. Neither max or min

Note that the point labeled as a local minimum of f is not a critical point of f ¨ . 14. A critical point of f requires f ¨ (x) = 0 or f ¨ undefned. Since f ¨ is clearly defned over the relevant range, we fnd where f ¨ (x) = 0, that is, where the graph of f ¨ crosses the x-axis. These points are shown and labeled in Figure 4.154. To the left of critical point A, we see that f ¨ > 0 and f is increasing; to the right of the critical point, we see that ¨ f < 0 and f is decreasing. So there is a local maximum at A. To the left of critical point B, we see that f ¨ < 0 and f is decreasing; to the right of the critical point, we see that f ¨ > 0 and f is increasing. So there is a local minimum at B. The sketch of f (x) in Figure 4.155 shows A is a local maximum and B is a local minimum.

336

Chapter Four /SOLUTIONS

f (x) f ¨ (x) A

A x

Figure 4.154

Figure 4.155

15. A critical point of f would require f ¨ (x) = 0 or f ¨ undefned. Since f ¨ is clearly defned over the relevant range, we wish to fnd where f ¨ (x) = 0, or where the graph shown intersects the x-axis. These points are shown and labeled in Figure 4.156. f (x)

f ¨ (x)

x C

Figure 4.156: The critical points of f (x)

x C

Figure 4.157: A possible graph of f (x)

To the left of critical point A, f ¨ > 0 and f is increasing; to the right, f ¨ < 0 and f is decreasing. So there is a local maximum at A. To the left of critical point B, f ¨ < 0 and f is decreasing; to the right, f ¨ > 0 and f is increasing. So there is a local minimum at B. To both the left and right of critical point C, f ¨ > 0 and so f increases on both sides of point C. So, point C is neither a local maximum nor a local minimum. A sketch of f (x) is shown in Figure 4.157. 16. Local maximum for some , with 1.1 < < 1.2, since f ¨ (1.1) > 0 and f ¨ (1.2) < 0. Local minimum for some , with 1.5 < < 1.6, since f ¨ (1.5) < 0 and f ¨ (1.6) > 0. Local maximum for some , with 2.0 < < 2.1, since f ¨ (2.0) > 0 and f ¨ (2.1) < 0. 17. (a) The derivative of f (x) = x5 + x + 7 is f ¨ (x) = 5x4 + 1. The derivative is always positive. (b) Since f ¨ (x) ≠ 0 for any value of x, there are no critical points for the function. Since f ¨ (x) is positive for all x, the function is increasing for all x, it crosses the x-axis at most once. Since f (x) → +∞ as x → +∞ and f (x) → −∞ as x → −∞, the graph of f crosses the x-axis once. So we conclude that f (x) has one real root. 18. (a) This function is defned for x > 0. We set the derivative equal to zero and solve for x to fnd critical points: f ¨ (x) = 1 − b

1 =0 x b 1= x x = b.

The only critical point is at x = b. (b) Since f ¨ (x) = 1 − bx−1 , the second derivative is f ¨¨ (x) = bx−2 =

b . x2

Since b > 0, the second derivative is always positive. Thus, the function is concave up everywhere and f has a local minimum at x = b. 19. The domain is all real numbers except x = b. The function is undefned at x = b and has a vertical asymptote there. To fnd the critical points, we set the derivative equal to zero and solve for x. Using the quotient rule, we have: f ¨ (x) =

(x − b)2ax − (ax2 )1 =0 (x − b)2

SOLUTIONS to Review Problems For Chapter Four

337

2ax2 − 2abx − ax2 =0 (x − b)2 ax2 − 2abx = 0. (x − b)2 The frst derivative is equal to zero if ax2 − 2abx = 0 ax(x − 2b) = 0 x = 0 or

x = 2b.

There are two critical points: at x = 0 and x = 2b. 20. (a) In Figure 4.158, we see that f ( ) = − sin has a zero at = 0. To see if it has any other zeros near the origin, we use our calculator to zoom in. (See Figure 4.159.) No extra root seems to appear no matter how close to the origin we zoom. However, zooming can never tell you for sure that there is not a root that you have not found yet.

f ( ) = − sin

−0.1

−2

0.1

Figure 4.159: Graph of f ( ) Zoomed In

Figure 4.158: Graph of f ( )

(b) Using the derivative, f ¨ ( ) = 1 − cos , we can argue that there is no other zero. Since cos < 1 for 0 < ≤ 1, we know f ¨ ( ) > 0 for 0 < ≤ 1. Thus, f increases for 0 < ≤ 1. Consequently, we conclude that the only zero of f is the one at the origin. If f had another zero at x0 , with x0 > 0, then f would have to “turn around”, and recross the x-axis at x0 . But if this were the case, f ¨ would be nonpositive somewhere, which we know is not the case. y

f ¨ ( ) = 1 − cos

Figure 4.160: Graph of f ¨ ( ) 21. (a)

f ¨ (x)

x3 x2

(b) f ¨ (x) changes sign at x1 , x3 , and x5 . (c) f ¨ (x) has local extrema at x2 and x4 .

338

Chapter Four /SOLUTIONS

22. The local maxima and minima of f correspond to places where f ¨ is zero and changes sign or, possibly, to the endpoints of intervals in the domain of f . The points at which f changes concavity correspond to local maxima and minima of f ¨ . The change of sign of f ¨ , from positive to negative corresponds to a maximum of f and change of sign of f ¨ from negative to positive corresponds to a minimum of f . 23. We have f ¨ (x) = −ae−ax , so that f ¨ (0) = −a. We see that as a increases the slopes of the curves at the origin become more and more negative. Thus, A corresponds to a = 1, B to a = 2, and C to a = 5. 24. We have f ¨ (x) = (1 − ax)e−ax , so that the critical points occur when 1 − ax = 0, that is, when x = 1∕a. We see that as a increases the x-value of the extrema moves closer to the origin. Looking at the fgure we see that A’s local maximum is the farthest from the origin, while C’s is the closest. Thus, A corresponds to a = 1, B to a = 2, and C to a = 3. 25. See Figure 4.161.

depth of water

Time at which water reaches corner of vase

✠

time

Figure 4.161

26. See Figure 4.162. Suppose t1 is the time to fll the left side to the top of the middle ridge. Since the container gets wider as you go up, the rate dH∕dt decreases with time. Therefore, for 0 ≤ t ≤ t1 , graph is concave down. At t = t1 , water starts to spill over to right side and so depth of left side does not change. It takes as long for the right side to fll to the ridge as the left side, namely t1 . Thus the graph is horizontal for t1 ≤ t ≤ 2t1 . For t ≥ 2t1 , water level is above the central ridge. The graph is climbing because the depth is increasing, but at a slower rate than for t ≤ t1 because the container is wider. The graph is concave down because width is increasing with depth. Time t3 represents the time when container is full.

H (height)

2t1

Figure 4.162

27. (a) P (t) =

2000 1 + e(5.3−0.4t)

t (time)

SOLUTIONS to Review Problems For Chapter Four

339

population of rabbits 2000 1500 1000 500 years since 1774 10

From the graph we see that the population levels o˙ at about 2000 rabbits. (b) The population appears to have been growing fastest when there were about 1000 rabbits, approximately 13 years after Captain Cook left the original rabbits on the island. (c) The infection point coincides with the point of most rapid increase of the rabbit population, that is, the infection point occurs approximately 13 years after 1774 when the rabbit population is about 1000 rabbits. (d) The rabbits reproduce quickly, so their population initially grew very rapidly. Limited food and space availability and perhaps predators on the island probably account for the population being unable to grow past 2000. 28. (a) This is one of many possible graphs.

(b) Since f must have a bump between each pair of zeros, f could have at most four zeros. (c) f could well have no zeros at all. To see this, consider the graph of the above function shifted vertically downward. (d) f must have at least two infection points. Since f has 3 maxima or minima, it has 3 critical points. Consequently f ¨ will have 3 corresponding zeros. Between each consecutive pair of these zeroes f ¨ must have a local maximum or minimum. Thus f ¨ will have one local maximum and one local minimum, which implies that f ¨¨ will have two zeros. These values, where the second derivative is zero, correspond to points of infection on the graph of f . (e) The 3 critical points are zeros of f ¨ , so degree(f ¨ ) ≥ 3. Thus degree(f ) ≥ 4. (f) For example: f (x) = −(x + 1)(x − 1)(x − 3)(x − 5) will look something like the graph in part (a). Many other answers are possible. 29. To fnd the intercepts of f (x), we frst fnd the y-intercept, which occurs at f (0) = sin(02 ) = 0. To fnd the x-intercepts on the given interval, we use a calculator to graph f (x) as shown in Figure 4.163:

f (x) = sin(x2 ) x

f ¨¨ (x)

f ¨ (x) x

Figure 4.163

Figure 4.164

Figure 4.165

We can use a calculator’s root-fnding capability to fnd where f (x) = 0. We get: x = 0,

and

x = 1.77,

and

x = 2.51,

(0, 0),

(1.77, 0),

(2.51, 0).

so our intercepts are ¨

To fnd critical points, we look for where f (x) = 0 or f is undefned. Using a calculator’s di˙erentiation and graphing features provides a graph of f ¨ (x) shown in Figure 4.164.

340

Chapter Four /SOLUTIONS Since in Figure 4.164 all the critical points (x, f (x)) have f ¨ (x) = 0, we use the calculator’s root-fnding capability to fnd the critical points where f ¨ (x) = 0: x = 0,

x = 1.25,

x = 2.17,

and

x = 2.80.

Writing these out as coordinates, the critical points are at (0, 0),

(1.25, 1),

(2.80, 1).

(2.17, −1),

Similarly, infection points occur where f ¨¨ (x) changes from negative to positive or vice versa. We can look for such points on a graph of f ¨¨ (x), shown in Figure 4.165. We can use the calculator’s root-fnding capability on the f ¨¨ (x) to get these infection points: x = 0.81,

and

x = 1.81,

and

x = 2.52

The infection points have coordinates (0.81, f (0.81)), etcetera and so they are (0.81, 0.61),

(1.81, −0.13),

(2.52, 0.07).

2 2 Notice that the intercepts √ can also be computed algebraically, since sin(x ) = 0 when x = 0, , 2 . The solutions are x = 0, √ ¨ 2 x = = 1.77, x = 2 = 2.51. Similarly, if we set f (x) = 2x cos(x ) = 0, we get x = 0 or x2 = ∕2, 3 ∕2, 5 ∕2. √ √ √ Thus, the critical points are x = 0, x = ∕2 = 1.25, x = 3 ∕2 = 2.17, and x = 5 ∕2 = 2.80.

30. (a) See Figure 4.166. a = 0.5 10

a=3

x −10

−10

Figure 4.166 (b) The function f (x) = x + a sin x is increasing for all x if f ¨ (x) > 0 for all x. We have f ¨ (x) = 1 + a cos x. Because cos x varies between −1 and 1, we have 1 + a cos x > 0 for all x if −1 < a < 1 but not otherwise. When a = 1, the function f (x) = x + sin x is increasing for all x, as is f (x) = x − sin x, obtained when a = −1. Thus f (x) is increasing for all x if −1 ≤ a ≤ 1. 31. (a) See Figure 4.167.

100

a=1 a = 20

x −10

−100

Figure 4.167 (b) The function f (x) = x2 + a sin x is concave up for all x if f ¨¨ (x) > 0 for all x. We have f ¨¨ (x) = 2 − a sin x. Because sin x varies between −1 and 1, we have 2 − a sin x > 0 for all x if −2 < a < 2 but not otherwise. Thus f (x) is concave up for all x if −2 < a < 2.

SOLUTIONS to Review Problems For Chapter Four

341

32. One possible graph of g is in Figure 4.168. (−2, g(−2)) g(x) (0, g(0))

(2, g(2))

Figure 4.168 (a) From left to right, the graph of g(x) starts “fat”, decreases slowly at frst then more rapidly, most rapidly at x = 0. The graph then continues to decrease but less and less rapidly until fat again at x = 2. The graph should exhibit symmetry about the point (0, g(0)). (b) The graph has an infection point at (0, g(0)) where the slope changes from negative and decreasing to negative and increasing. (c) The function has a global maximum at x = −2 and a global minimum at x = 2. (d) Since the function is decreasing over the interval −2 ≤ x ≤ 2 g(−2) = 5 > g(0) > g(2). Since the function appears symmetric about (0, g(0)), we have g(−2) − g(0) = g(0) − g(2). 33. (a) We have

T (D) =

C D CD2 D3 − D2 = − , 2 3 2 3

and

dT = CD − D2 = D(C − D). dD Since, by this formula, dT ∕dD is zero when D = 0 or D = C, negative when D > C, and positive when D < C, we have (by the frst derivative test) that the temperature change is maximized when D = C. (b) The sensitivity is dT ∕dD = CD − D2 ; its derivative is d 2 T ∕dD2 = C − 2D, which is zero if D = C∕2, negative if D > C∕2, and positive if D < C∕2. Thus by the frst derivative test the sensitivity is maximized at D = C∕2. √ 34. The top half of the circle has equation y = 1 − x2 . The rectangle in Figure 4.169 has area, A, given by √ for 0 ≤ x ≤ 1. A = 2xy = 2x 1 − x2 , At a critical point, � √ −1∕2 dA 1 = 2 1 − x2 + 2x 1 − x2 (−2x) = 0 dt 2 √ 2x2 =0 2 1 − x2 − √ 1 − x2 √ 2 2 1 − x2 − 2x2 =0 √ 1 − x2 2(1 − x2 − x2 ) =0 √ 1 − x2 2(1 − 2x2 ) = 0 1 x = ±√ . 2

342

Chapter Four /SOLUTIONS √ Since A = 0 at the endpoints x = 0 and x = 1, and since A is positive at the only critical point, x = 1∕ 2, in the interval 0 ≤ √ x ≤ 1, the critical point is a local and global maximum. The vertices on the circle have y = √ 1 − (1∕2)2 = 1∕ 2. Thus the coordinates of the rectangle with maximum area are H I H I H I H I 1 1 1 1 1 1 , 0 ; , ; − , 0 ; − , √ √ √ √ √ √ 2 2 2 2 2 2 and the maximum area is

1 1 A = 2 √ ⋅ √ = 1. 2 2 y 1

✻y = √1 − x2 y

✛ x ✲

−1

❄

Figure 4.169 35. (a) Suppose the height of the box is ℎ. The box has six sides, four with area xℎ and two, the top and bottom, with area x2 . Thus, 4xℎ + 2x2 = A. So ℎ= Then, the volume, V , is given by

0 V = x2 ℎ = x2 =

A − 2x2 . 4x

A − 2x2 4x

x� A − 2x2 4

A 1 x − x3 . 4 2

(b) The graph is shown in Figure 4.170. We are assuming A is a positive constant. Also, we have drawn the whole graph, but we should only consider V > 0, x > 0 as V and x are lengths. V V = A4 x − 21 x3

x A 2

Figure 4.170 (c) To fnd the maximum, we di˙erentiate, regarding A as a constant: dV A 3 = − x2 . dx 4 2 So dV ∕dx = 0 if A 3 2 − x =0 4 2

x=±

A . 6

SOLUTIONS to Review Problems For Chapter Four For a real box, we must use x = √ at x = A∕6, we get

343

√ A∕6. Figure 4.170 makes it clear that this value of x gives the maximum. Evaluating

A V = 4

A 1 − 6 2

Hu I3 u u 3∕2 A A A 1 A A A = − ⋅ = . 6 4 6 2 6 6 6

36. (a) We want the maximum value of I. Using the properties of logarithms, we rewrite the expression for I as I = k(ln S − ln S0 ) − S + S0 + I0 . Since k and S0 are constant, di˙erentiating with respect to S gives dI k = − 1. dS S Thus, the critical point is at S = k. Since dI∕dS is positive for S < k and dI∕dS is negative for S > k, we see that S = k is a local maximum. We only consider positive values of S. Since S = k is the only critical point, it gives the global maximum value for I, which is I = k(ln k − ln S0 ) − k + S0 + I0 . (b) Since both k and S0 are in the expression for the maximum value of I, both the particular disease and how it starts infuence the maximum. 37. The triangle in Figure 4.171 has area, A, given by A=

1 1 x ⋅ y = x3 e−3x . 2 2

If the area has a maximum, it occurs where dA 3 3 = x2 e−3x − x3 e−3x = 0 dx 2 2 3 2 x (1 − x) e−3x = 0 2 x = 0, 1. The value x = 0 gives the minimum area, A = 0, for x ≥ 0. Since dA 3 = x2 (1 − x)e−3x , dx 2 we see that

dA > 0 for 0 < x < 1 dx Thus, x = 1 gives the local and global maximum of A=

and

dA < 0 for x > 1. dx

1 1 3 −3⋅1 1e = 3. 2 2e

y (x, x2 e−3x )

✛

✲

✻ y ❄

Figure 4.171

344

Chapter Four /SOLUTIONS

38. The area of the rectangle on the left is 2x. The entire rectangle has area 2, so the area of the rectangle on the right is 2 − 2x. We are to maximize the product f (x) = 2x(2 − 2x) = 4x − 4x2 of the two areas where 0 ≤ x ≤ 1. The critical points of f occur where f ¨ (x) = 4 − 8x = 0 at x = 1∕2. The maximum of f on the interval 0 ≤ x ≤ 1 is at the critical point x = 1∕2 or one of the endpoints x = 0 or x = 1. We have 1 f (0) = f (1) = 0 and f = 1. 2 The product of the areas is a maximum when x = 1∕2. 39. First fnd the length ℎ of the vertical segment at x. Since the top edge of the triangle has slope ℎ∕x = 2∕1, we have ℎ = 2x. Thus 1 1 Area of smaller triangle = × Base × Height = ⋅ x ⋅ 2x = x2 . 2 2 The entire triangle has area 1, so the area of the trapezoid on the right is 1 − x2 . We are to maximize the product f (x) = x2 (1 − x2 ) = x2 − x4 of the two areas where 0 ≤ x ≤ 1 The critical points of f occur where f ¨ (x) = 2x − 4x3 = 2x(1 − 2x2 ) = 0 √ √ √ at x = 0, x = 1∕ 2 or x = −1∕ 2. The only critical point in the interval 0 < x < 1 is x = 1∕ 2. Th√ e maximum of f on the interval 0 ≤ x ≤ 1 is at one of the endpoints x = 0 or x = 1 or at the critical point x = 1∕ 2. We have H I f (0) = f (1) = 0 and f

1 √ 2

1 . 4

√ The product of the areas is a maximum when x = 1∕ 2. √ 40. The distance d(x) from the point (x, 1 − x) on the curve to the origin is given by t √ √ d(x) = x2 + ( 1 − x)2 = x2 + (1 − x). √ Since x is in the domain of y = 1 − x, we have −∞ < x ≤ 1. Di˙erentiating gives 2x − 1 d ¨ (x) = √ , 2 x2 − x + 1 ¨ ¨ so x = 1∕2 is the only critical point. √ Since d (x) < 0 for x < 1∕2 and d (x) > 0 for x > 1∕2, the point x = 1∕2 is a minimum for x. The point (1∕2, 1∕ 2) is the closest point on the curve to the origin.

41. (a) For points (x, y) on the ellipse, we have y2 = 1 − x2 ∕9 and −3 ≤ x ≤ 3. We wish to minimize the distance u √ x2 2 2 D = (x − 2) + (y − 0) = (x − 2)2 + 1 − . 9 To do so, we fnd the value of x minimizing d = D2 for −3 ≤ x ≤ 3. This x also minimizes D. Since d = (x − 2)2 + 1 − x2 ∕9, we have 2x 16x d ¨ (x) = 2(x − 2) − = − 4, 9 9 ¨¨ which is 0 when x = 9∕4. Since d (9∕4) = 16∕9 > 0, we see d has a local minimum at x = 9∕4. Since the graph of d is a parabola, the local minimum is in fact a global minimum. Solving for y, we have 2 9 1 7 x2 =1− ⋅ = , 9 4 9 16 √ √ so y = ± 7∕4. Therefore, the points on the ellipse closest to (2, 0) are 9∕4, ± 7∕4 . y2 = 1 −

SOLUTIONS to Review Problems For Chapter Four (b) This time, we wish to minimize

√

345

x2 . 9 √ Again, let d = D2 and minimize d(x) for −3 ≤ x ≤ 3. Since d = (x − 8)2 + 1 − x2 ∕9, √ √ 2x 16x d ¨ (x) = 2(x − 2 2) − = − 4 2. 9 9 √ √ ¨ Therefore, d (x) = 0 when x = 9 2∕4. But 9 2∕4 > 3, so there are not any critical points on the interval −3 ≤ x ≤ 3. The minimum distance must be attained at an endpoint.√Since d ¨ (x) < 0 for all x between −3 and 3, the minimum is at x = 3. So (3, 0) is the point on the ellipse closest to ( 8, 0). D=

(x −

8)2 + 1 −

42. Let the radius of the can be r and let its height be ℎ. The surface area, S, and volume, V , of the can are given by S = Area of the sides of can + Area of top and bottom S = 2 rℎ + 2 r2 V = r2 ℎ. Since S = 280, we have 2 rℎ + 2 r2 = 280, we have ℎ=

140 − r2 r

and

140 − r2 = 140r − r3 . V = r2 r √ Since ℎ ≥ 0, we have r2 ≤ 140 and thus 0 < r ≤ 140∕ = 6.676. We have V ¨ (r) = 140 − 3 r2 , √ so the only critical point of V with r > 0 is r = 140∕3 = 3.854. The critical point is in the domain 0 < r ≤ 6.676. Since V ¨¨ (r) = −6 r is negative for all r > 0, the critical point r = 3.854 cm gives the maximum value of the volume. For this value of r, we have ℎ = 7.708 cm and V = 359.721 cm3 . 43. (a) Since Proft = Revenue − Cost, we can calculate (q) = R(q) − C(q) for each of the q values given: q

100

200

300

400

500

R(q)

500

1000

1500

2000

2500

C(q)

700

900

1000

1100

1300

1900

(q)

−700

−400

400

700

600

We see that maximum proft is $700 and it occurs when the production level q is 400. See Figure 4.172. R(q)

(500, 2500)

2500

❄ (400, 2000)

2000

(500, 1900)

✛

C(q)

(300, 1500)

1500

(400, 1300) 1000

(300, 1100)

(100, 900)

(200, 1000)

(0, 700) (100, 500)

500

q 0

100

200

300

Figure 4.172

400

500

346

Chapter Four /SOLUTIONS (b) Since revenue is $500 when q = 100, the selling price is $5 per unit. (c) Since C(0) = $700, the fxed costs are $700.

44. Since fxed costs are represented by the vertical intercept, they are $1.1 million. The quantity that maximizes proft is about q = 70, and the proft achieved is $(3.7 − 2.5) = $1.2 million 45. We know that the maximum (or minimum) proft can occur when Marginal cost = Marginal revenue

MC = MR.

From the table it appears that MC = MR at q ≈ 2500 and q ≈ 4500. To decide which one corresponds to the maximum proft, look at the marginal proft at these points. Since Marginal proft = Marginal revenue − Marginal cost (or M = MR − MC), we compute marginal proft at the di˙erent values of q in Table 4.6: Table 4.6 q

1000

2000

3000

4000

5000

6000

M = MR − MC

−22

−4

−5

−22

From the table, at q ≈ 2500, we see that proft changes from decreasing to increasing, so q ≈ 2500 gives a local minimum. At q ≈ 4500, proft changes from increasing to decreasing, so q ≈ 4500 is a local maximum. See Figure 4.173. Therefore, the global maximum occurs at q = 4500 or at the endpoint q = 1000. (3000, 4) 10

(4000, 7)

❄ 1000

2000

4000

5000

6000

0 (5000, −5)

(2000, −4) −10 P ¨ (q)

−20

(6000, −22) (1000, −22)

−30

Maximum Profit (q)

Minimum Profit 1000

2000

3000

4000

5000

6000

Figure 4.173

46. (a) Proft = = R − C; proft is maximized when the slopes of the two graphs are equal, at around q = 350. See Figure 4.174.

SOLUTIONS to Review Problems For Chapter Four $

347

$∕unit

C R

✛ 100 200 300 400 500

Max profit q

q (quantity)

200

Figure 4.174

400

Figure 4.175

(b) The graphs of MR and MC are the derivatives of the graphs of R and C. Both R and C are increasing everywhere, so MR and MC are everywhere positive. The cost function is concave down and then concave up, so MC is decreasing and then increasing. The revenue function is linear and then concave down, so MR is constant and then decreasing. See Figure 4.175. 47. (a) We know that Proft = Revenue − Cost, so di˙erentiating with respect to q gives: Marginal Proft = Marginal Revenue − Marginal Cost. We see from the fgure in the problem that just to the left of q = a, marginal revenue is less than marginal cost, so marginal proft is negative there. To the right of q = a marginal revenue is greater than marginal cost, so marginal proft is positive there. At q = a marginal proft changes from negative to positive. This means that proft is decreasing to the left of a and increasing to the right. The point q = a corresponds to a local minimum of proft, and does not maximize proft. It would be a terrible idea for the company to set its production level at q = a. (b) We see from the fgure in the problem that just to the left of q = b marginal revenue is greater than marginal cost, so marginal proft is positive there. Just to the right of q = b marginal revenue is less than marginal cost, so marginal proft is negative there. At q = b marginal proft changes from positive to negative. This means that proft is increasing to the left of b and decreasing to the right. The point q = b corresponds to a local maximum of proft. In fact, since the area between the MC and MR curves in the fgure in the text between q = a and q = b is bigger than the area between q = 0 and q = a, q = b is in fact a global maximum. 48. First fnd the marginal revenue and marginal cost. Note that each product sells for $788, so revenue is given by R(q) = 788q. MR = R¨ (q) = 788, MC = C ¨ (q) = 3q 2 − 120q + 1400. Setting MR = MC yields 3q 2 − 120q + 1400 = 788 3q 2 − 120q + 612 = 0. This factors to 3(q − 34)(q − 6) = 0, so MR = MC at q = 34 and q = 6. We now fnd the proft at these points: R(6) − C(6) = 788 ⋅ 6 − 63 − 60 ⋅ 62 + 1400 ⋅ 6 + 1000 = −$2728. 3 R(34) − C(34) = 788 ⋅ 34 − 34 − 60 ⋅ 342 + 1400 ⋅ 34 + 1000 = $8248 We must also try the endpoints R(0) − C(0) = 788 ⋅ 0 − 03 − 60 ⋅ 02 + 1400 ⋅ 0 + 1000 = −$1000 R(50) − C(50) = 788 ⋅ 50 − 503 − 60 ⋅ 502 + 1400 ⋅ 50 + 1000 = −$6600. From this we see that proft is maximized at q = 34 units. The total cost at q = 34 is C(34) = $18,544. The total revenue at q = 34 is R(34) = $26,792 and Total proft = 26,792 − 18,544 = $8248.

348

Chapter Four /SOLUTIONS C(q)

$ 40,000

R(q)

Profit ✻✻

20,000

Cost

❄ ✻ Revenue ❄❄

Figure 4.176 49. (a) The fxed cost is 0 because C(0) = 0. (b) Proft, (q), is equal to money from sales, 7q, minus total cost to produce those items, C(q). = 7q − 0.01q 3 + 0.6q 2 − 13q ¨ = −0.03q 2 + 1.2q − 6 √ −1.2 ± (1.2)2 − 4(0.03)(6) ¨ = 0 if q = ≈ 5.9 or 34.1. −0.06 ¨¨ ¨¨ ¨¨ Now = −0.06q + 1.2, so (5.9) > 0 and (34.1) < 0. This means q = 5.9 is a local min and q = 34.1 a local max. We now evaluate the endpoint, (0) = 0, and the points nearest q = 34.1 with integer q-values: (35) = 7(35) − 0.01(35)3 + 0.6(35)2 − 13(35) = 245 − 148.75 = 96.25, (34) = 7(34) − 0.01(34)3 + 0.6(34)2 − 13(34) = 238 − 141.44 = 96.56. So the (global) maximum proft is (34) = 96.56. The money from sales is $238, the cost to produce the items is $141.44, resulting in a proft of $96.56. (c) The money from sales is equal to price×quantity sold. If the price is raised from $7 by $x to $(7 + x), the result is a reduction in sales from 34 items to (34 − 2x) items. So the result of raising the price by $x is to change the money from sales from (7)(34) to (7 + x)(34 − 2x) dollars. If the production level is fxed at 34, then the production costs are fxed at $141.44, as found in part (b), and the proft is given by: (x) = (7 + x)(34 − 2x) − 141.44 This expression gives the proft as a function of change in price x, rather than as a function of quantity as in part (b). We set the derivative of with respect to x equal to zero to fnd the change in price that maximizes the proft: d = (1)(34 − 2x) + (7 + x)(−2) = 20 − 4x = 0 dx So x = 5, and this must give a maximum for (x) since the graph of is a parabola which opens downward. The proft when the price is $12 (= 7 + x = 7 + 5) is thus (5) = (7 + 5)(34 − 2(5)) − 141.44 = $146.56. This is indeed higher than the proft when the price is $7, so the smart thing to do is to raise the price by $5. 50. Note that proft = (q) = R(q) − C(q), so to fnd proft, we must fnd an expression for R(q). If p is the price of a single item, R(q) = p ⋅ q. Substituting p = b1 − a1 q gives R(q) = (b1 − a1 q) ⋅ q = b1 q − a1 q 2

and

(q) = R(q) − C(q) = b1 q − a1 q 2 − b2 − a2 q. Finding the derivative and setting it equal to 0 yields ¨ (q) = b1 − 2a1 q − a2 = 0 −a2 + b1 so q = 2a1

SOLUTIONS to Review Problems For Chapter Four

349

is a critical point for (q). Using the second derivative test, we see is concave down, ¨¨ (q) = −2a1 Since a1 is positive, −2a1 is negative. Thus ¨¨ (q) < 0 for all q, so q = (−a2 + b1 )∕(2a1 ) is a local maximum. This q is the only critical point, so it is the global maximum for (q). 51. (a) (q) is maximized when R(q) > C(q) and they are as far apart as possible. See Figure 4.177. (b) ¨ (q0 ) = R¨ (q0 ) − C ¨ (q0 ) = 0 implies that C ¨ (q0 ) = R¨ (q0 ) = p. Graphically, the slopes of the two curves at q0 are equal. This is plausible because if C ¨ (q0 ) were greater than p or less than p, the maximum of (q) would be to the left or right of q0 , respectively. In economic terms, if the cost were rising more quickly than revenues, the proft would be maximized at a lower quantity (and if the cost were rising more slowly, at a higher quantity). (c) See Figure 4.178. $

C(q)

C ¨ (q)

R(q)

■

p maximum (q)

Figure 4.177

Figure 4.178

52. (a) We know that the average cost is given by a(q) =

C(q) . q

Thus the average cost is a(q) = 0.04q 2 − 3q + 75 +

96 . q

(b) Average cost is graphed in Figure 4.179. a a(q)

150 100 50

q 20

100

Figure 4.179

(c) Looking at the graph of the average value we see that it is decreasing for q < 38. Likewise we see that a(q) is increasing for q > 38. (d) Thus the average cost hits its minimum at about q = 38 with the value of about $21.

350

Chapter Four /SOLUTIONS

53. If the minimum average cost occurs at a production level of 15,000 units, the line from the origin to the curve is tangent to the curve at that point. The slope of this line is 25, so the cost of producing 15,000 units is 25(15,000) = 375,000. See Figure 4.180. Cost

375,000

15,000

Quantity

Figure 4.180 54. The point F representing the fxed costs is the vertical intercept of the cost function, C(q). The break-even level of production is where R(q) = C(q), so the point B is where the two graphs intersect. There are two such points in Figure 4.181. We have marked the one at which production frst becomes proftable. The marginal cost is C ¨ (q), so the point M is where the slope of C(q) is minimum. The average cost is minimized when a(q) = C ¨ (q). Since a(q) equals the slope of a line joining the origin to a point (q, C(q)) on the graph of C(q), the point A is where such a line is tangent to the graph of C(q). The proft is P (q) = R(q) − C(q), so the point P occurs where the two graphs are furthest apart and R(q) > C(q). See Figure 4.181. C(q)

R(q)

q B M

A P

Figure 4.181 55. C ¨ (q) decreases with q because C(q) is concave down. Therefore, C ¨ (2) is larger then C ¨ (3). 56. Note that (C(5) − C(3))∕(5 − 3) is the slope of the secant line between q = 3 and q = 5. The value of (C(5) − C(3))∕(5 − 3) is larger because the slope of the secant line is larger than the slope of the tangent line at its right endpoint. 57. Both quantities are slopes of secant lines with left endpoint at q = 50. The value of (C(75) − C(50))∕25 is larger as the slope of the secant line decreases as the right endpoint moves to the right. 58. Both quantities are slopes of secant lines. Since the secant line between q = 25 and q = 75 is to the left of the secant line between q = 50 and q = 100, and C(q) is concave down, (C(75) − C(25))∕50 is larger. 59. Since C ¨ (3) is the slope at q = 3, and C(3)∕3 is the slope of the line from the origin to the point q = 3 on the curve C(q), the value of C(3)∕3 is larger. 60. Since C(q)∕q is the slope of the line from the origin to the point q on the curve C(q), the concavity of C(q) tells us that the value of C(10)∕10 is larger. | p dq | | p d | | | 5p | | | | |p | 61. E = | (2000 − 5p)| = | ⋅ (−5)|, so E = . At a price of $20.00, the number of items produced is |= | ⋅ | q dp | | q dp | |q | q | | | | | | q = 2000 − 5(20) = 1900, so at p = 20, we have E=

5(20) ≈ 0.05. (1900)

SOLUTIONS to Review Problems For Chapter Four

351

Since 0 ≤ E < 1, this product has inelastic demand–a 1% change in price will only decrease demand by 0.05%. 62. (a) Quadratic polynomial (degree 2) with negative leading coeÿcient. (b) Exponential. (c) Logistic. (d) Logarithmic. (e) This is a quadratic polynomial (degree 2) and positive leading coeÿcient. (f) Exponential. (g) Surge 63. Let l and w be the length and width of the rectangle. Since the perimeter is 200 meters, we have 2w + 2l = 200 l = 100 − w. The area is A(w) = w ⋅ l = w(100 − w) = 100w − w2 . At the critical points A¨ (w) = 100 − 2w = 0, so w = 50, l = 50. The rectangle of perimeter 200 meters with maximum area is the 50 meter by 50 meter square. Its area is w ⋅ l = 50 ⋅ 50 = 2500 meters2 . 64. The volume is given by V = x2 y. The surface area is given by S = 2x2 + 4xy = 2x2 + 4xV ∕x2 = 2x2 + 4V ∕x. To fnd the dimensions which minimize the area, fnd x such that dS∕dx = 0: 4V dS = 4x − 2 = 0 dx x x3 = V .

√ 3 Solving for x gives x = V = y. To see that this gives a minimum, note that for small x, S ≈ 4V ∕x is decreasing. For 2 large x, S ≈ 2x is increasing. Since there is only one critical point, it must give a global minimum. Therefore, when the width equals the height, the surface area is minimized. 65. The triangle in Figure 4.182 has area, A, given by A=

1 x xy = e−x∕3 . 2 2

At a critical point, dA 1 x = e−x∕3 − e−x∕3 = 0 dx 2 6 1 −x∕3 e (3 − x) = 0 6 x = 3. Substituting the critical point and the endpoints into the formula for the area gives: For x = 1, we have A = 21 e−1∕3 = 0.358 For x = 3, we have A = 32 e−1 = 0.552

For x = 5, we have A = 52 e−5∕3 = 0.472 Thus, the maximum area is 0.552 and the minimum area is 0.358. y (x, e−x∕3 ) y✻

✛

✲

❄

Figure 4.182

352

Chapter Four /SOLUTIONS

66. For x > 0, the line in Figure 4.183 has Slope =

y x2 e−3x = = xe−3x . x x

If the slope has a maximum, it occurs where d (Slope) = 1 ⋅ e−3x − 3xe−3x = 0 dx e−3x (1 − 3x) = 0 1 x= . 3 For this x-value, 1 1 −3(1∕3) 1 −1 e = e = . 3 3 3e Figure 4.183 shows that the slope tends toward 0 as x → ∞; the formula for the slope shows that the slope tends toward 0 as x → 0. Thus the only critical point, x = 1∕3, must give a local and global maximum. Slope =

y (x, x2 e−3x ) y✻

✛

✲

❄

Figure 4.183

67. (a) The rectangle in Figure 4.184 has area, A, given by A = xy = xe−2x . At a critical point, we have dA = 1 ⋅ e−2x − 2xe−2x = 0 dx e−2x (1 − 2x) = 0 1 x= . 2 Since A = 0 when x = 0 and A → 0 as x → ∞, the critical point x = 1∕2 is a local and global maximum. Thus the maximum area is 1 1 A = e−2(1∕2) = . 2 2e (b) The rectangle in Figure 4.184 has perimeter, P , given by P = 2x + 2y = 2x + 2e−2x . At a critical point, we have dP = 2 − 4e−2x = 0 dx 1 e−2x = 2 1 2 1 1 1 x = − ln = ln 2. 2 2 2

−2x = ln

To see if this critical point gives a maximum or minimum, we fnd d2P = 8e−2x . dx2

SOLUTIONS to Review Problems For Chapter Four

353

Since d 2 P ∕dx2 > 0 for all x, including x = 21 ln 2, the critical point is a local and global minimum. Thus, the minimum perimeter is 1 1 −2 1 ln 2 = ln 2 + 2e− ln 2 = ln 2 + 2 ⋅ = ln 2 + 1. P =2 ln 2 + 2e 2 2 2 y

✻

y = e−x∕2

✛ x ✲

❄

Figure 4.184

68. (a) The distance the pigeon fies over water is BP =

AB 500 = , sin sin

and over land is P L = AL − AP = 2000 −

500 500 cos = 2000 − . tan sin

Therefore the energy required is

500 500 cos + e 2000 − sin sin 2 − cos 500 = 500e + 2000e, for arctan ≤ ≤ . sin 2000 2

E = 2e

2 − cos must have the same critical points since the graph of E is just a sin 2 − cos 500 stretch and a vertical shift of the graph of f . The graph of for arctan( 2000 ) ≤ ≤ 2 in Figure 4.185 shows sin that E has precisely one critical point, and that a minimum for E occurs at this point.

(b) Notice that E and the function f ( ) =

500 arctan 2000

500 Figure 4.185: Graph of f ( ) = 2−cos for arctan( 2000 ) ≤ ≤ 2 sin

To fnd the critical point , we solve f ¨ ( ) = 0 or 0 1 sin ⋅ sin − (2 − cos ) ⋅ cos E ¨ = 0 = 500e sin2 0 1 1 − 2 cos = 500e . sin2 Therefore 1 − 2 cos = 0 and so = ∕3.

354

Chapter Four /SOLUTIONS (c) Letting a = AB and b = AL, our formula for E becomes a a cos E = 2e +e b− sin sin 2 − cos a = ea + eb, for arctan ≤ ≤ . sin b 2 2 − cos . Thus, the critical point = ∕3 sin is independent of e, a, and b. But the maximum of E on the domain arctan(a∕b) ≤ ≤ 2 is dependent on the ratio

Again, the graph of E is just a stretch and a vertical shift of the graph of AB

. In other words, the optimal angle is = ∕3 provided arctan(a∕b) ≤ 3 ; otherwise, the optimal angle is AL arctan(a∕b), which means the pigeon should fy over the lake for the entire trip—this occurs when a∕b > 1.733.

a∕b =

69. (a) Di˙erentiating using the chain rule gives p¨ (x) =

0 1 (x − ) (x − ) 2 2 1 −(x− )2 ∕(2 2 ) e ⋅ −2 = − √ e−(x− ) ∕(2 ) . √ 2 2 3 2 2

So p¨ (x) = 0 where x − = 0, so x = . Di˙erentiating again using the product rule gives 0 0 11 −2(x − ) −1 −(x− )2 ∕(2 2 ) −(x− )2 ∕(2 2 ) ¨¨ p (x) = √ 1⋅e + (x − )e ⋅ 2 2 3 2 2

=−

e−(x− ) ∕(2 ) 2 ( − (x − )2 ). √ 5 2

Substituting x = gives e0 1 2 √ =− √ . 5 3 2 2 Since p¨¨ ( ) < 0, there is a local maximum at x = . Since this local maximum is the only critical point, it is a global maximum. See Figure 4.186. (b) Using the formula for p¨¨ (x), we see that p¨¨ (x) = 0 where p¨¨ ( ) = −

2 − (x − )2 = 0 x − = ± x = ± . Since p¨¨ (x) changes sign at x = + and x = − , these are both points of infection. See Figure 4.186. Maximum 1 Inflection point

p(x)

✲

−

✛

Inflection point

x +

Figure 4.186 70. (a) Varying the a parameter changes the peak concentration proportionally, but does not change the time to reach peak concentration . (b) The value of the peak concentration decreases as b increases and the time until peak concentration decreases as b increases. (c) As a gets larger, the time until peak concentration decreases because it is only a˙ected by the constant in the exponent. The value of the peak concentration is not a˙ected as a changes because the peak always occurs at t = 1∕b and if a = b, the value of the peak concentration is c = a ⋅ a1 e−a⋅1∕a = e−1 , independent of a.

SOLUTIONS to Review Problems For Chapter Four

355

71. (a) Figure 4.187 shows the rate of sales as a function of time. The point of diminishing returns is reached when the rate of sales is at a maximum; this happens during the 4th month. rate of sales 800 700 600 500 400 300 200 100 0 1

month

Figure 4.187

(b) Total sales for the frst four months are 140 + 520 + 680 + 750 = 2090 sales. (c) Assuming logistic growth, the limiting value should be twice the value at the point of diminishing returns, so total sales should be 2(2090) = 4180 sales. 72. When bemetizide is taken with triamterene, rather than by itself, the peak concentration is lower, the time it takes to reach peak concentration is about the same, the time until onset of e˙ectiveness is slightly less, and the duration of e˙ectiveness is less. If bemetizide became unsafe in a concentration greater than 70 ng/ml, then it would be wise to take it with triamterene. 73. (a) For a point (t, s), the line from the origin has rise = s and run = t; See Figure 4.188. Thus, the slope of the line OP is s∕t. (b) Sketching several lines from the origin to points on the curve (see Figure 4.189), we see that the maximum slope occurs at the point P , where the line to the origin is tangent to the graph. Reading from the graph, we see t ≈ 2 hours at this point. s (km)

s (km) (t, s) P✻ P s

O✛

✲

❄

Figure 4.188

t (hours)

Figure 4.189

(c) The instantaneous speed of the cyclist at any time is given by the slope of the corresponding point on the curve. At the point P , the line from the origin is tangent to the curve, so the quantity s∕t equals the cyclist’s speed at the point P. 74. (a) The line in Figure 4.190 has slope equal to the rate worms arrive. To understand why, see line (1) in Figure 4.191. (This is the same line.) For any point Q on the loading curve, the line P Q has slope QT QT load = = . PT P O + OT traveling time + searching time (b) The slope of the line P Q is maximized when the line is tangent to the loading curve, which happens with line (2). The load is then approximately 7 worms.

356

Chapter Four /SOLUTIONS (c) If the traveling time is increased, the point P moves to the left, to point P ¨ , say. If line (3) is tangent to the curve, it will be tangent to the curve further to the right than line (2), so the optimal load is larger. This makes sense: if the bird has to fy further, you’d expect it to bring back more worms each time. load (number of worms) 8 (3)

load (number of worms) 8

(2) Number of worms

Number of worms Q

(1) time

P¨

time P

O T

Traveling time

Figure 4.190

Searching time

Figure 4.191

75. (a) See Figure 4.192. The capacity appears to be 200 cars. The parking lot is full just before 8:30 am. total number of cars in parking lot 200 150 100 50

5 ∶ 00

6 ∶ 00

7 ∶ 00

8 ∶ 00

9 ∶ 00

time (am)

Figure 4.192 (b) The rate of arrival between 5:00 and 5:30 is (5 − 4)∕0.5 = 2 cars/hours. We assign this the time 5:15 (rather than either 5:00 or 5:30). The other values in Table 4.7 are calculated in a similar way. The data is plotted in Figure 4.193. Table 4.7 Time (am)

5:15

5:45

6:15

6:45

7:15

7:45

8:15

8:45

Rate of arrival (cars/hour)

120

rate of arrival of cars 140 100 60 20 5 ∶ 15

6 ∶ 15

7 ∶ 15

8 ∶ 15

time (am)

Figure 4.193 (c) Rush hour occurs around the time when the rate of arrival of cars is maximum, namely, about 7:30 am. (d) The maximum of the rate of arrival occurs at the infection point of the total number of cars in the lot.

SOLUTIONS to Review Problems For Chapter Four

357

dA . d H√

76. (a) To minimize A with ℎ and s fxed we have to fnd

I 3 − cos dA 3 2 d = s d 2 d sin H I √ 2 3 2 sin − cos ( 3 − cos ) = s 2 sin2 H I √ sin2 − 3 cos + cos2 3 = s2 2 sin2 H I √ 3 2 1 − 3 cos = s 2 sin2 H I √ √ dA 3 2 1 − 3 cos Set = s = 0. Then 1 − 3 cos = 0 and cos = √1 , so ≈ 54.7◦ is a critical point of 2 3 d 2 sin A. Since H√ I √ 3 sin3 − 2 sin cos (1 − 3 cos ) || 3 2 d 2 A || = s | | | 2 d 2 || =54.7 sin4 | =54.7 3 ≈ s2 (2.122) > 0, 2 = 54.7 is indeed a minimum. (b) Since 55◦ is very close to 54.7◦ , we conclude that bees attempt to minimize the surface areas of their honey combs. b−c⋅0

77. (a) The vertical intercept is W = Ae−e = Ae−e . There is no horizontal intercept since the exponential function is always positive. There is a horizontal asymptote. As t → ∞, we see that eb−ct = eb ∕ect → 0, since t is positive. Therefore W → Ae0 = A, so there is a horizontal asymptote at W = A. (b) The derivative is b−ct b−ct dW = Ae−e (−eb−ct )(−c) = Ace−e eb−ct . dt Thus, dW ∕dt is always positive, so W is always increasing and has no critical points. The second derivative is b−ct b−ct d d d2W = (Ace−e )eb−ct + Ace−e (eb−ct ) dt dt dt2 b−ct b−ct = Ac 2 e−e eb−c t eb−ct + Ace−e (−c)eb−ct b−ct

= Ac 2 e−e

eb−ct (eb−ct − 1).

Now eb−ct decreases from eb > 1 when t = 0 toward 0 as t → ∞. The second derivative changes sign from positive to negative when eb−ct = 1, i.e., when b − ct = 0, or t = b∕c. Thus the curve has an infection point at t = b∕c, where b−(b∕c)c W = Ae−e = Ae−1 . (c) See Figure 4.194. W

A = 50, b = 2, c = 5

A = 50, b = 2, c = 1

A = 20, b = 2, c = 1 t

Figure 4.194 (d) The fnal size of the organism is given by the horizontal asymptote W = A. The curve is steepest at its infection point, which occurs at t = b∕c, W = Ae−1 . Since e = 2.71828 … ≈ 3, the size the organism when it is growing fastest is about A∕3, one third its fnal size. So yes, the Gompertz growth function is useful in modeling such growth.

358

Chapter Four /SOLUTIONS

STRENGTHEN YOUR UNDERSTANDING 1. False. The function f has a local minimum at p if f (p) ≤ f (x) for points x near p. 2. True—according to the defnition, critical points of f occur where f ¨ (x) = 0 or f ¨ (x) is undefned. 3. False. The function f can also have critical points where f ¨ is undefned. 4. True, since f has lower values to the left of p and to the right of p than it does at p. 5. False. If f ¨ (p) = 0 and f ¨¨ (p) > 0 then f has a local minimum at p. 6. True. If f ¨ (p) = 0 and f ¨¨ (p) > 0 then by the second derivative test f has a local minimum at p. 7. False. If f ¨¨ (p) > 0, the function is concave up, but p must be a critical point before we can conclude that f has a local minimum at p. 8. False. Consider f (x) = x3 . The function has a critical point at x = 0, but x = 0 is neither a local maximum nor a local minimum. 9. False. Consider f (x) = ex . The function and its derivative are defned for all x, and f ¨ (x) is never equal to zero. Thus, f has no critical points. 10. False. Consider f (x) = |x|. The function has a local minimum at x = 0, but f ¨ (0) is undefned. 11. True, by the defnition of infection point. 12. False. Consider f (x) = x4 . The second derivative is zero at x = 0, but the function is concave up for all x. 13. True. Consider f (x) = x3 . The point x = 0 is both a critical point and an infection point of f . 14. True. The graph of f (x) = x3 is concave down to the left of x = 0 and concave up to the right of x = 0, so there is an infection point at x = 0. 15. False. The function H(x) = x4 is concave up for all x, so H has no infection points. 16. True. For example, the function f (x) = x3 + x has no critical points and has one infection point, at x = 0. 17. True. For example, you can sketch the graph of such a function with a local maximum at (−1, 1), a local minimum at (1, −1), and horizontal asymptote y = 0 as x → ±∞. For example, f (x) = −2x∕(1 + x2 ). 18. True. The second derivative is equal to zero at x = −1 and is positive to the left of x = −1 and negative to the right of x = −1. Thus, f has an infection point at x = −1. 19. False. The second derivative is equal to zero at x = −1 but the sign of f ¨¨ is negative both to the left and right of x = −1. Thus, there is no change of concavity at x = −1. 20. True. Note that ex is positive for all x and that the factor (x + 1) changes sign from negative to positive at x = −1. Thus the function changes concavity at x = −1 and x = −1 is an infection point. 21. True. Consider f (x) = −x2 . The function has a local and global maximum at x = 0. 22. False. The global maximum may also occur at an endpoint x = 1 or x = 2. 23. False. Consider f (x) = −x2 over the entire real line. There is no global minimum. 24. True. If f is increasing on the interval,then the largest value occurs at the right endpoint. 25. False. If f (x) < 0 then f is decreasing on the interval, and the largest value occurs at the left endpoint. The global maximum will be at x = a. 26. True. For example, let S(x) = x. On the interval 1 ≤ x ≤ 2, the global maximum is 2 and occurs at the right-hand endpoint x = 2. On the interval 2 ≤ x ≤ 3, the global maximum of 3 occurs at x = 3. 27. False. The function k has no critical points. As x → 0 from the right, the function increases without bound. Thus, k does not have a global maximum when x > 0. 28. False. According to the defnition, the function f has a global minimum at p on the interval given that f (p) ≤ f (x) for all −5 ≤ x ≤ 5. 29. True, as defned in the text. 30. False. By the second derivative test, we have that p is a local minimum. However, a global minimum could occur at another critical point or at an endpoint if the function is defned on a closed interval. 31. True. Cost is increasing at a slower rate than revenue–thus the cost of the next item will be less than the revenue for that item. 32. True, since for the last item made and sold, we have received less in revenue than it cost to make, so by not making and selling this item, proft would increase.

STRENGTHEN YOUR UNDERSTANDING

359

33. False, since a point where marginal revenue equals marginal cost may also indicate minimum proft. 34. True. 35. False. Maximum proft occurs at a critical point of the proft function—or at an endpoint of the proft function if there are constraints on the number of items that can be made or sold. 36. True. When marginal proft is zero we have a critical point. The point must be tested to see if it gives a maximum of the proft function. 37. False. The cost and revenue functions cross when a proft is turning into a loss, in which case proft is decreasing, or when a loss is turning into a proft, in which case proft is increasing. In neither case is proft a maximum. 38. True. Since Proft = Revenue - Cost, when the cost and revenue curves are equal (and therefore the graphs cross), the proft is zero. 39. True. Revenue is price times quantity. If the price is constant at price p, the graph of marginal revenue is the horizontal line y = p. 40. False. If prices are constant at price p, the graph of revenue is a line with slope p. 41. True, as specifed in the text. 42. True. The units of marginal cost and average cost have the same units—for example, if the units of cost are dollars, the units of marginal cost and average cost are $/item. 43. False. Marginal cost gives the rate at which cost is changing—or, once q items have been produced, the approximate cost of the next item. Average cost gives the average cost of producing q items. In general, these are not the same. 44. False. Average cost of q items can be visualized as the slope of a line from the origin to the point (q, C(q)) on the graph of the cost function C. 45. False. If marginal cost is less than average cost, increasing production decreases average cost. 46. False. Marginal cost equals average cost at critical points of average cost. 47. True. If marginal cost is greater than average cost, increasing production increases average cost. 48. True, since the example in the section shows that the average cost function has critical points exactly where average cost (which is the slope of the line from the origin to the cost function) is equal to the marginal cost (which is the slope of the tangent to the cost function). 49. False. Not necessarily. Average cost may be increasing or decreasing as a functions of quantity, depending on the concavity of the cost function. 50. False. Average cost is minimized when equal to marginal cost, but this is not the same quantity at which marginal cost is minimized. 51. False. The elasticity of demand is given by E = |p∕q ⋅ dq∕dp| . 52. False. If E > 1 then we say that demand is elastic. 53. False. If 0 ≤ E < 1 then we say that demand is inelastic. 54. True. This is what elasticity tells us. 55. False. If a product is considered a necessity, the demand is generally inelastic. 56. False. An increase in price may cause an increase or a decrease in revenue, depending on elasticity. 57. False. If elasticity E > 1, demand is elastic and revenue increases when price decreases. 58. True. At a critical point of the revenue function E = 1, and the demand is neither elastic nor inelastic. 59. True, since by increasing the price, demand drops, so costs are lower, but revenue is higher, so proft increases. 60. False. If E > 1 then increasing price causes a decrease in revenue, demand, and costs. We have seen examples where profts decrease because revenue decreases more than costs. 61. True. The function P fts the model f (t) = L∕(1 + Ce−kt ) with L = 1000, C = 2, and k = 3. 62. False. Note that the end behavior of P does not approach the limiting value L. Instead, as t → ∞, the function values approach zero. The fact that the exponent of e is not negative causes this behavior. 63. False. The function P has an infection point when P = 1000∕2 = 500. This occurs at approximately t = 0.2310. 64. True, since the logistic function P = L∕(1 + Ce−kt ) has an infection point at L∕2. 65. True. Note that as t increases, the denominator (1 + Ce−kt ) approaches 1, so P (t) approaches L. 66. True, since the slope of the dose/response curve is the rate of change of the response as a function of the dose.

360

Chapter Four /SOLUTIONS

67. True. The logistic model approaches the carrying capacity as time increases. 68. True. If k is large the function approaches L more rapidly than when k is small. 69. True. At the infection point, P = L∕2, the function changes from concave up to concave down. 70. True. We can show this analytically by noting that P ¨ (t) = LkCe−kt (1 + Ce−kt )−2 > 0 for all t. 71. False. The formula for a surge function is y = ate−bt while the formula for an exponential decay function is y = ae−bt . 72. True. 73. False. The function is concave down and then concave up as it approaches the t-axis asymptotically. 74. True, the surge function will cross the line when it is increasing and again when it is decreasing. 75. False. The drug will never be e˙ective since the concentration will always be below the minimum e˙ective concentration. 76. True. There is exactly one infection point, and it is always to the right of the one critical point. 77. True, as specifed in the text. 78. True. Note that y can be written as y = (1∕12)te−3t . This is a surge function with a = 1∕12, b = 3. 79. False. The function y can be rewritten as y = (1∕12)te3t . This function does not ft the model of a surge function because the exponent of e is positive. Note that y → ∞ as t increases. 80. True, since the surge function y = ate−bt has maximum at t = 1∕b.

PROJECTS FOR CHAPTER FOUR 1. Since a(q) = C(q)∕q, we have C(q) = a(q) ⋅ q. Thus C ¨ (q) = a¨ (q)q + a(q), and so C ¨ (q0 ) = a¨ (q0 )q0 + a(q0 ). Since t1 is the line tangent to a(q) at q = q0 , the slope of t1 is a¨ (q0 ), and the equation of t1 is � y = a(q0 ) + a¨ (q0 ) ⋅ (q − q0 ) = a¨ (q0 ) ⋅ q + a(q0 ) − a¨ (q0 ) ⋅ q0 . Thus the y-intercept of t1 is given by a(q0 ) − a¨ (q0 )q0 , and the equation of the line t2 is � y = 2 ⋅ a¨ (q0 ) ⋅ q + a(q0 ) − a¨ (q0 ) ⋅ q0 since t2 has twice the slope of t1 . Let’s compute the y-value on t2 when q = q0 : � y = 2 ⋅ a¨ (q0 ) ⋅ q0 + a(q0 ) − a¨ (q0 ) ⋅ q0 = a¨ (q0 )q0 + a(q0 ) = C ¨ (q0 ). Hence C ¨ (q0 ) is given by the point on t2 where q = q0 . 2. (a) (i) We want to minimize A, the total area lost to the forest, which is made up of n frebreaks and 1 stand of trees lying between frebreaks. The area of each frebreak is (50 km)(0.01 km) = 0.5 km2 , so the total area lost to the frebreaks is 0.5n km2 . There are n total stands of trees between frebreaks. The area of a single stand of trees can be found by subtracting the frebreak area from the forest and dividing by n, so 2500 − 0.5n Area of one stand of trees = . n Thus, the total area lost is A = Area of one stand + Area lost to frebreaks 2500 − 0.5n 2500 = + 0.5n = − 0.5 + 0.5n. n n We assume that A is a di˙erentiable function of a continuous variable, n. Di˙erentiating this function yields dA 2500 =− + 0.5. dn n2

PROJECTS FOR CHAPTER FOUR

361

√ At critical points, dA∕dn = 0, so 0.5 = 2500∕n2 or n = 2500∕0.5 ≈ 70.7. Since n must be an integer, we check that when n = 71, A = 70.211 and when n = 70, A = 70.214. Thus, n = 71 gives a smaller area lost. We can check that this is a local minimum since the second derivative is positive everywhere d 2 A 5000 = > 0. dn2 n3 Finally, we check the endpoints: n = 1 means that the entire forest lost after a fre, since there is only one stand of trees in this case and it all burns. The largest n is 5000, and in this case the frebreaks remove the entire forest. Both of these cases maximize the area of forest lost. Thus, n = 71 is a global minimum. So 71 frebreaks minimizes the area of forest lost. (ii) Repeating the calculation using b for the width gives A=

2500 − 50b + 50bn, n

and

dA 2500 =− + 50b, dn n2 √ with a critical point when b = 50∕n2 so n = 50∕b. So, for example, if we make the width b four times as large we need half as many frebreaks. (b) We want to minimize A, the total area lost to the forest, which is made up of n frebreaks in one direction, n frebreaks in the other, and one square of trees surrounded by frebreaks. The area of each frebreak is 0.5 km2 , and there are 2n of them, giving a total of 0.5 ⋅ 2n. But this is larger than the total area covered by the frebreaks, since it counts the small intersection squares, of size (0.01)2, twice. Since there are n2 intersections, we must subtract (0.01)2n2 from the total area of the 2n frebreaks. Thus, Area covered by the frebreaks = 0.5 ⋅ 2n − (0.01)2n2 . To this we must add the area of one square patch of trees lost in a fre. These are squares of side (50 − 0.01n)∕n = 50∕n − 0.01. Thus the total area lost is 2 50 A = n − 0.0001n2 + − 0.01 n Treating n as a continuous variable and di˙erentiating this function yields 1 0 dA 50 50 = 1 − 0.0002n + 2 − 0.01 − . dn n n2 Using a computer algebra system to fnd critical points we fnd that dA∕dn = 0 when n ≈ 17 and n = 5000. Thus n = 17 gives a minimum lost area, since the endpoints of n = 1 and n = 5000 both yield A = 2500 or the entire forest lost. So we use 17 frebreaks in each direction. 3. (a) Since the company can produce more goods if it has more raw materials to use, the function f (x) is increasing. Thus, we expect the derivative f ¨ (x) to be positive. (b) The cost to the company of acquiring x units of raw material is wx, and the revenue from the sale of f (x) units of the product is pf (x). The company’s proft is (x) = Revenue − Cost = pf (x) − wx. (c) Since proft (x) is maximized at x = x∗ , we have ¨ (x∗ ) = 0. From ¨ (x) = pf ¨ (x) − w, we have pf ¨ (x∗ ) − w = 0. Thus f ¨ (x∗ ) = w∕p. (d) Computing the second derivative of (x) gives ¨¨ (x) = pf ¨¨ (x). Since (x) has a maximum at x = x∗ , the second derivative ¨¨ (x∗ ) = pf ¨¨ (x∗ ) is negative. Thus f ¨¨ (x∗ ) is negative. (e) Di˙erentiate both sides of pf ¨ (x∗ ) − w = 0 with respect to w. The chain rule gives d ¨ ∗ f (x ) − 1 = 0 dw dx∗ pf ¨¨ (x∗ ) −1=0 dw 1 dx∗ = . dw pf ¨¨ (x∗ ) p

362

Chapter Four /SOLUTIONS

Since f ¨¨ (x∗ ) < 0, we see dx∗ ∕dw is negative. (f) Since dx∗ ∕dw < 0, the quantity x∗ is a decreasing function of w. If the price w of the raw material goes up, the company should buy less. 4. (a) (i) The slope, dV ∕dt, is the rate of change of the volume of air that has been exhaled over time. Thus, it is the rate of fow of the air out of the lungs, measured in liters per second. The greater the slope, the faster air is being expelled from the lungs. Notice that dV ∕dt is a volume fow; that is, the rate at which the volume of air exhaled is changing. It is not the speed at which air particles are leaving the lungs, which would be measured in meters per second. (ii) As the horizontal asymptote on the graph, the vital capacity VC is the volume of air expelled at the end of the exhalation. It is the maximum volume of air the patient can exhale in one breath. (b) (i) The slope of the fow-volume curve tells you how much the fow rate of air out of the lungs changes in response to exhalation of an additional volume of air. (ii) For V near 0, when little air has been exhaled, the slope of the fow-volume curve is steep and positive. This means that the fow rate dV ∕dt is rapidly increasing as air is exhaled. The patient is expelling air faster and faster. The fow volume curve has zero slope when approximately 1 liter has been exhaled. This corresponds to the maximum fow rate dV ∕dt; air is expelled most rapidly at this stage. Thereafter, the slope of the fow-volume curve is negative, so the fow rate decreases and air is expelled slower and slower until approximately 5.5 liters of air have been exhaled. At that point, the patient has no more air to exhale and must take a breath. (iii) Suppose the fow rate, dV ∕dt, is given as function of V by the function f , so dV ∕dt = f (V ); the fow-volume curve we were given is a graph of f . Now we are asked to plot f ¨ (V ), whose graph is in Figure 4.195. flow rate/liter

20 15 10 5

f ¨ (V ) 1

V , liters

Figure 4.195

(iv) The maximal rate is identifed as the highest point on the fow-volume curve, which occurs at about V = 1 liter. Thus, the patient’s peak expiratory fow occurs when 1 liter of air has been exhaled. The peak expiratory fow is about 11.5 liters per second. On the slope graph in Figure 4.195, a critical point occurs where the slope is 0; that is, where the slope graph intersects the horizontal axis. This occurs at V = 1. The frst derivative test tells us that this critical point gives a maximum fow rate since the slope is positive to the left of V = 1 and negative to the right of V = 1. (c) The increased resistance to airfow means that the patient exhales more slowly. This makes the volumetime curve rise more slowly. In fact, sometimes the patient exhales so slowly that they “run out of time” and cannot exhale all the volume they want to, leading to air trapping. This corresponds to a lowered vital capacity, VC. See Figure 4.196. The peak expiratory fow will be lower. This means that the fow-volume curve does not rise as high. See Figure 4.197.

PROJECTS FOR CHAPTER FOUR V , liters 6 5 4 3 VC 2 1 t, sec 1 2 3 4 5 6 7 8 9 10 Figure 4.196: Volume-time curve for an asthmatic

dV ∕dt, liters/sec 12 10 8 6 4 2 V , liters 1 2 3 4 5 6 Figure 4.197: Flow-volume curve for an asthmatic

363

5.1 SOLUTIONS

365

CHAPTER FIVE Solutions for Section 5.1 1. (a) We compute the distances traveled for each of the three legs of the trip and add them to fnd the total distance traveled: Distance = (30 miles/hour)(2 hours) + (40 miles/hour)(1∕2 hour) + (20 miles/hour)(4 hours) = 60 miles + 20 miles + 80 miles = 160 miles. You travel 160 miles on this trip. (b) The velocity is 30 miles/hour for the frst 2 hours, 40 miles/hour for the next 1∕2 hour, and 20 miles/hour for the last 4 hours. The entire trip lasts 2 + 1∕2 + 4 = 6.5 hours, so we need a scale on our horizontal (time) axis running from 0 to 6.5. Between t = 0 and t = 2, the velocity is constant at 30 miles/hour, so the velocity graph is a horizontal line at 30. Likewise, between t = 2 and t = 2.5, the velocity graph is a horizontal line at 40, and between t = 2.5 and t = 6.5, the velocity graph is a horizontal line at 20. The graph is shown in Figure 5.1. (c) How can we visualize distance traveled on the velocity graph given in Figure 5.1? The velocity graph looks like the top edges of three rectangles. The distance traveled on the frst leg of the journey is (30 miles/hour)(2 hours), which is the height times the width of the frst rectangle in the velocity graph. The distance traveled on the frst leg of the trip is equal to the area of the frst rectangle. Likewise, the distances traveled during the second and third legs of the trip are equal to the areas of the second and third rectangles in the velocity graph. It appears that distance traveled is equal to the area under the velocity graph. In Figure 5.2, the area under the velocity graph in Figure 5.1 is shaded. Since this area is three rectangles and the area of each rectangle is given by Height × Width, we have Total area = (30)(2) + (40)(1∕2) + (20)(4) = 60 + 20 + 80 = 160. The area under the velocity graph is equal to distance traveled. velocity (miles/hour)

velocity (miles/hour)

10 1

time (hours)

Figure 5.1: Velocity graph

Area = Distance traveled

time (hours)

Figure 5.2: The area under the velocity graph gives distance traveled

2. Using n = 2 subdivisions on the interval 0 ≤ t ≤ 6, we have Δt = 3. (a) The left sum is v(0) ⋅ Δt + v(3) ⋅ Δt = 10 ⋅ 3 + 8 ⋅ 3 = 54 meters. This estimate gives an upper estimate of the actual distance traveled. (b) The right sum is v(3) ⋅ Δt + v(6) ⋅ Δt = 8 ⋅ 3 + 2 ⋅ 3 = 30 meters. This estimate gives a lower estimate of the actual distance traveled.

366

Chapter Five /SOLUTIONS

3. Using n = 2 subdivisions on the interval 0 ≤ t ≤ 8, we have Δt = 4. (a) The left sum is v(0) ⋅ Δt + v(4) ⋅ Δt = 6 ⋅ 4 + 8 ⋅ 4 = 56 km. This estimate gives a lower estimate of the actual distance traveled. (b) The right sum is v(4) ⋅ Δt + v(8) ⋅ Δt = 8 ⋅ 4 + 14 ⋅ 4 = 88 km. This estimate gives an upper estimate of the actual distance traveled. 4. The distance traveled is represented by area under the velocity curve. We can approximate the area using left- and righthand sums. Alternatively, counting the squares (each of which has area 10), and allowing for the broken squares, we can see that the area under the curve from 0 to 6 is between 140 and 150. Hence the distance traveled is between 140 and 150 meters. 5. (a) Left sum (b) Upper estimate (c) 6 (d) Δt = 2 (e) Upper estimate is approximately 4 ⋅ 2 + 2.9 ⋅ 2 + 2 ⋅ 2 + 1.5 ⋅ 2 + 1 ⋅ 2 + 0.8 ⋅ 2 = 24.4. 6. (a) Right sum (b) Lower estimate (c) n = 5 (d) Δt = 3 seconds (e) Lower estimate is approximately 15.5 ⋅ 3 + 14 ⋅ 3 + 11.7 ⋅ 3 + 8.3 ⋅ 3 + 4 ⋅ 3 = 160.5 feet. 7. (a) Right sum (b) Upper estimate (c) n = 8 (d) Δt = 3 (e) Upper estimate is approximately 4.5 ⋅ 3 + 6 ⋅ 3 + 7.5 ⋅ 3 + 8.5 ⋅ 3 + 9.5 ⋅ 3 + 10.5 ⋅ 3 + 11.5 ⋅ 3 + 12 ⋅ 3 = 210. 8. (a) Lower estimate = (45)(2) + (16)(2) + (0)(2) = 122 feet. Upper estimate = (88)(2) + (45)(2) + (16)(2) = 298 feet. v (b) 80 60 40 20 t 2

9. (a) We estimate how many new cases there were between days t = 0 and t = 7 by assuming the rate was constant during that interval. Since we are computing the left sum, we assume the rate was constant at the left endpoint of 325 cases per day during those 7 days, which gives Total new cases during the frst seven days ≈ 325 ⋅ 7 = 2275 cases. We do the same for the other four 7-day periods which gives Left sum = 325 ⋅ 7 + 367 ⋅ 7 + 475 ⋅ 7 + 717 ⋅ 7 + 919 ⋅ 7 = 19,621. Thus, there were a total of around 20,000 new Covid-19 cases between October 15 and November 19, 2020. (b) Since the rate r(t) seems to be increasing and we are fnding the left sum, we are always taking the lower estimate for the constant rate in each interval. This means we expect to obtain an underestimate for the total number of new cases between the two dates. Note, however, that we cannot be sure that we are getting an underestimate because r(t) may have increased in between the values of t we are seeing in the table.

5.1 SOLUTIONS

367

10. We estimate the number of new cases in the frst interval, between days t = 0 and t = 14, by assuming the rate was constant during that interval. Since we want an underestimate and the rate is decreasing, we use the rate at the right endpoint, 1614 cases per day. Thus Total new cases during the frst fourteen days ≈ 1614 ⋅ 14 = 22,596 cases. We do the same for the other three 14-day periods which gives Understimate = 1614 ⋅ 14 + 934 ⋅ 14 + 550 ⋅ 14 + 323 ⋅ 14 = 47,894 cases. Thus, there were a total of around 47,800 new Covid-19 cases between January 4 and March 1, 2021. This is an underestimate. 11. We use Distance = Rate × Time on each subinterval with Δt = 3. Underestimate = 0 ⋅ 3 + 10 ⋅ 3 + 25 ⋅ 3 + 45 ⋅ 3 = 240, Overestimate = 10 ⋅ 3 + 25 ⋅ 3 + 45 ⋅ 3 + 75 ⋅ 3 = 465. 240 ≤ Distance traveled ≤ 465.

We know that A better estimate is the average. We have

Distance traveled ≈

240 + 465 = 352.5. 2

The car travels about 352.5 feet during these 12 seconds. 12. Using the data in Table 5.2 of Example 2, we construct Figure 5.3. rate of sales (games/week) 2000 1500

Underestimate

1000 500 time (weeks) 10

Figure 5.3

13. (a) Since the bike takes 3 seconds to stop, a = 3. Since the bike is initially going 20 ft/sec, b = 20. See Figure 5.4. (b) The distance traveled is the area under the velocity graph, so Distance traveled =

1 1 ⋅ Base ⋅ Height = ⋅ 3 ⋅ 20 = 30 feet. 2 2

velocity (ft/sec) 20

Figure 5.4

t (secs)

368

Chapter Five /SOLUTIONS

14. (a) See Figure 5.5. (b) The distance traveled is the area under the graph of the velocity in Figure 5.5. The region is a triangle of base 2.2 seconds and height 88 ft/sec, so Distance traveled = Area of triangle =

1 ⋅ 2.2 ⋅ 88 = 96.8 ft. 2

v (ft/sec) 88

2.2

t (sec)

Figure 5.5 15. (a) Note that 15 minutes equals 0.25 hours. Lower estimate = 11(0.25) + 10(0.25) = 5.25 miles. Upper estimate = 12(0.25) + 11(0.25) = 5.75 miles. (b) Lower estimate = 11(0.25) + 10(0.25) + 10(0.25) + 8(0.25) + 7(0.25) + 0(0.25) = 11.5 miles. Upper estimate = 12(0.25) + 11(0.25) + 10(0.25) + 10(0.25) + 8(0.25) + 7(0.25) = 14.5 miles. 16. The distance traveled is represented by area under the velocity curve. We can approximate the area using left- and rightthe graph appears to hand sums. Alternatively, by counting squares and fractions of squares, we fnd that the area under 5280 ≈ 455 feet, within about be around 310 (miles/hour) sec, within about 10. So the distance traveled was about 310 3600 5280 ≈ 15 feet. (Note that 455 feet is about 0.086 miles.) 10 3600 17. Figure 5.6 shows the graph of f (t). The region under the graph of f (t) from t = 0 to t = 10 is a triangle of base 10 seconds and height 50 meter/sec. Then Distance traveled = Area of triangle =

1 ⋅ 10 ⋅ 50 = 250 meters. 2

Thus the distance traveled is 250 meters. 50 f (t) = 5t

t 10

Figure 5.6 18. (a) Since the bike takes 10 second to reach its maximum velocity of 15 ft/sec, b = 15. See Figure 5.7 (b) The distance traveled is the area under the velocity graph, so Distance traveled =

1 1 ⋅ Base ⋅ Height = ⋅ 10 ⋅ 15 = 75 feet. 2 2

velocity (ft/sec) 15

Figure 5.7

t (secs)

5.1 SOLUTIONS

369

19. (a) Since the car starts at a velocity of 44 ft/sec, a = 44. Since the car takes 5 seconds to reach its maximum velocity of 88 ft/sec, b = 88 and c = 5. See Figure 5.8 (b) The distance traveled is the area under the velocity graph, which can be viewed as a rectangle with base 5 and height 44 and a triangle with base 5 and height 44. So Distance traveled = 5 ⋅ 44 +

1 ⋅ 5 ⋅ 44 = 330 feet. 2

velocity (ft/sec) 88 44

t (secs)

Figure 5.8 20. (a) Let’s begin by graphing the data given in the table; see Figure 5.9. The total amount of pollution entering the lake during the 30-day period is equal to the shaded area. The shaded area is roughly 40% of the rectangle measuring 30 units by 35 units. Therefore, the shaded area measures about (0.40)(30)(35) = 420 units. Since the units are kilograms, we estimate that 420 kg of pollution have entered the lake. Alternatively, we can average the left- and right-hand sums to arrive at an answer. pollution rate (kg/day) 35 28 21 14 7 0

time (days)

Figure 5.9 (b) Using left and right sums, we have Underestimate = (7)(6) + (8)(6) + (10)(6) + (13)(6) + (18)(6) = 336 kg. Overestimate = (8)(6) + (10)(6) + (13)(6) + (18)(6) + (35)(6) = 504 kg. 21. (a) See Figure 5.10. (b) The distance traveled is the area under the graph of the velocity in Figure 5.10. The region is a triangle of base 5 seconds and altitude 50 ft/sec, so the distance traveled is (1∕2)5 ⋅ 50 = 125 feet. (c) The slope of the graph of the velocity function is the same, so the triangular region under it has twice the altitude and twice the base (it takes twice as long to stop). See Figure 5.11. Thus, the area is 4 times as large and the car travels 4 times as far, or 500 ft. 100

50 v(t) v(t) t 5

Figure 5.10

t 5

Figure 5.11

370

Chapter Five /SOLUTIONS

22. The table gives the rate of oil consumption in billions of barrels per year. To fnd the total consumption, we use left-hand and right-hand sums. We have Left-hand sum = (24.2)(5) + (25.6)(5) + (27.9)(5) + (30.6)(5) + (31.7)(5) = 700 bn barrels. Right-hand sum = (25.6)(5) + (27.9)(5) + (30.6)(5) + (31.7)(5) + (33.8)(5) = 748 bn barrels. 700 + 748 Average of left- and right-hand sums = = 724 bn barrels. 2 The consumption of oil between 1990 and 2015 is about 724 billion barrels. 23. (a) We have (i) The left end-points give Left sum = 23.5 ⋅ 2 + 25.6 ⋅ 2 + 26.8 ⋅ 2 + 29.5 ⋅ 2 + 31.4 ⋅ 2 + 32.6 ⋅ 2 = 338.8 trillion cubic feet. (ii) The right end-points give Right sum = 25.6 ⋅ 2 + 26.8 ⋅ 2 + 29.5 ⋅ 2 + 31.4 ⋅ 2 + 32.6 ⋅ 2 + 37.3 ⋅ 2 = 366.4 trillion cubic feet. (b) The annual natural gas production in the US is increasing between 2006 and 2018, so the right sum gives an overestimate for the total natural gas production between 2006 and 2018. 24. The rate at which the fsh population grows varies between 10 fsh per month and about 22 fsh per month. If the rate of change were constant at the lower bound of 10 during the entire 12-month period, we obtain an underestimate for the total change in the fsh population of (10)(12) = 120 fsh. An overestimate is (22)(12) = 264 fsh. The actual increase in the fsh population is equal to the area under the curve. (Notice that the units of the area are height units times width units, or fsh per month times months, as we want.) We can estimate the area using left- and right-hand sums. Alternatively, there are approximately 11 grid squares below the graph. The area of each grid square represents an increase of (5 fsh per month) ⋅ (4 months) = 20 fsh. We have Total change in fsh population = Area under curve ≈ 11 ⋅ 20 = 220. We estimate that the fsh population grew by 220 fsh during this 12-month period. 25. (a) Note that the rate r(t) sometimes increases and sometimes decreases in the interval. We can calculate an upper estimate of the volume by choosing Δt = 5 and then choosing the highest value of r(t) on each interval, and similarly a lower estimate by choosing the lowest value of r(t) on each interval: Upper estimate = 5[20 + 24 + 24] = 340 liters. Lower estimate = 5[12 + 20 + 16] = 240 liters. (b) A graph of r(t) along with the areas represented by the choices of r(t) in calculating the lower estimate is shown in Figure 5.12. 24 20 16 12

r(t)

t 5

Figure 5.12 26. (a) Since car B starts at t = 2, the tick marks on the horizontal axis (which we assume are equally spaced) are 2 hours apart. Thus car B stops at t = 6 and travels for 4 hours. Car A starts at t = 0 and stops at t = 8, so it travels for 8 hours. (b) Car A’s maximum velocity is approximately twice that of car B, that is 100 km/hr. (c) The distance traveled is given by the area under the velocity graph. Using the formula for the area of a triangle, the distances are given approximately by 1 1 ⋅ Base ⋅ Height = ⋅ 8 ⋅ 100 = 400 km 2 2 1 1 Car B travels = ⋅ Base ⋅ Height = ⋅ 4 ⋅ 50 = 100 km. 2 2 Car A travels =

5.1 SOLUTIONS

371

27. (a) Car A has the largest maximum velocity because the peak of car A’s velocity curve is higher than the peak of B’s. (b) Car A stops frst because the curve representing its velocity hits zero (on the t-axis) frst. (c) Car B travels farther because the area under car B’s velocity curve is the larger. 28. (a) See Figure 5.13. Since v(t) is increasing, we use a left sum to get a lower estimate with Δt = 2. Evaluating v(t) at t = 0, 2, 4: Lower estimate = v(0) ⋅ 2 + v(2) ⋅ 2 + v(4) ⋅ 2 = 0 ⋅ 2 + 20 ⋅ 2 + 40 ⋅ 2 = 120. To get an upper estimate, we use a right sum. Evaluating v(t) at t = 2, 4, 6: Upper estimate = v(2) ⋅ 2 + v(4) ⋅ 2 + v(6) ⋅ 2 = 20 ⋅ 2 + 40 ⋅ 2 + 60 ⋅ 2 = 240 Thus, the lower estimate for the distance traveled is 120 ft, the upper estimate is 240 ft, and the average of the two is 180 ft. v(t) 60

t 2

Figure 5.13

(b) See Figure 5.14. The distance traveled is the area between the t-axis and the graph of v(t) = 10t for 0 ≤ t ≤ 6. This is a triangle with base 6 seconds and height 60 feet/sec, so the distance traveled is 1 ⋅ 6 ⋅ 60 = 180 ft. 2 This is the same as the average of the upper and lower estimates in part (a). Area of the triangle =

v(t)

t 6

Figure 5.14 29. We begin by graphing v(t) (see Figure 5.15). Adding up the areas using an overestimate with data every 1 second, we get s ≈ 2 + 5 + 10 + 17 + 26 = 60 m. The actual distance traveled is less than 60 m. (m/sec) 30

Figure 5.15

t (sec)

372

Chapter Five /SOLUTIONS

30. (a) Since the stopping distance, s, is proportional to v2 , we have s = kv2 for some constant k. We fnd k using the fact that s = 50 when v = 20, so s = kv2 50 = k ⋅ 202 50 = 400k 1 k= . 8 Thus, s = v2 ∕8. When v = 40, we have s = (1∕8)402 = 200, so the stopping distance is 200 ft when the velocity is 40 mi/hr. Likewise, when v = 60, we have s = (1∕8)602 = 450, so the stopping distance is 450 ft when the velocity is 60 mi/hr. There is another way to fnd the stopping distances without fnding k. Since we know that s = kv2 , we know that if the initial speed v doubles, we have s = k(2v)2 = 4kv2 , so the stopping distance at 40 mi/hr is 4 ⋅ 50 = 200 ft. If the speed triples, s increases by a factor of 9, so the stopping distance at 60 mi/hr is 9 ⋅ 50 = 450 ft. (b) Since v represents the velocity of the van and v ≥ 0, the distance traveled over a time period is the area between the t-axis and the graph of v. In this case, this area is the stopping distance. The time period is from 0 to 5, so the stopping distance is the area under the graph of v from t = 0 to t = 5. To make the graph of v, we fnd the intercepts. When t = 0, we have the v-intercept of 30; when v = 0, we have 0 = 30 − 6t, which gives t = 5, as we would expect. Looking at the graph of v = 30 − 6t in Figure 5.16, we see that the stopping distance is equal to the area of the triangle under the graph. (c) We know that the graph of v = 30 − 6t is a line with v-intercept 30 and t-intercept 5. If the initial speed increases to 60, we know that the v-intercept is 60, not 30. Because the deceleration is the same, we know that the slope of the line is the same, −6. So we have v = 60 − 6t, which has a t-intercept of 10. Figure 5.16 shows the graphs of v = 30 − 6t for initial velocity 30 ft/sec and v = 60 − 6t for initial velocity 60 ft/sec. The stopping distances are equal to the triangular areas under the graphs. For the initial velocity of 30 ft/sec, the area under the graph of v is a triangle with base 5 and height 30, so Stopping distance = Area =

1 ⋅ 5 ⋅ 30 = 75 ft. 2

Similarly, for the initial velocity of 60 ft/sec, the area under the graph of v is a triangle with base 10 and height 60, so Stopping distance = Area = Thus, we have:

1 ⋅ 10 ⋅ 60 = 300 ft. 2

Stopping distance at 60 ft/sec 300 = = 4. Stopping distance at 30 ft/sec 75 v 60 v = 60 − 6t

v = 30 − 6t t 5

Figure 5.16 There is an alternative method of explaining why one stopping distance is four times the other without fnding either one. The base and height for the triangle with initial velocity 60 ft/sec are each twice that of the triangle with initial velocity 30 ft/sec. Since the area of a triangle is half of the base times the height, if the frst triangle has base b and height ℎ, its area, A, is 1 A = bℎ. 2

5.1 SOLUTIONS

373

The bigger triangle has base 2b and height 2ℎ, so its area is 1 1 ⋅ 2b ⋅ 2ℎ = 4 ⋅ bℎ = 4A. 2 2 Thus, its area is 4 times as big, so the stopping distance is 4 times as far. 31. (a) See Table 5.1. Table 5.1 t

500

541.64

586.76

635.62

688.56

745.91

(b) The total change is the rate of change, in dollars per year, times the number of years in the time interval, summed up over all time intervals. We fnd an underestimate and an overestimate and average the two: Underestimate = 500 ⋅ 2 + 541.64 ⋅ 2 + 586.76 ⋅ 2 + 635.62 ⋅ 2 + 688.56 ⋅ 2 = $5905.16. Overestimate = 541.64 ⋅ 2 + 586.76 ⋅ 2 + 635.62 ⋅ 2 + 688.56 ⋅ 2 + 745.91 ⋅ 2 = $6396.98. A better estimate of the total change in the value of the fund is the average of the two: Total change in the value of the fund ≈

5905.16 + 6396.98 = $6151. 2

32. Suppose f (t) is the fow rate in m3 /hr at time t. We are only given two values of the fow rate, so in making our estimates of the fow, we use one subinterval, with Δt = 3∕1 = 3: Left estimate = 3[f (6 am)] = 3 ⋅ 100 = 300 m3

(an underestimate)

Right estimate = 3[f (9 am)] = 3 ⋅ 280 = 840 m3

(an overestimate).

The best estimate is the average of these two estimates, Best estimate =

Left + Right 2

300 + 840 = 570 m3 . 2

33. (a) Based on the data, we will calculate the underestimate and the overestimate of the total change. A good estimate will be the average of both results. Underestimate of total change = 47 ⋅ 10 + 57 ⋅ 10 + 76 ⋅ 10 + 79 ⋅ 10 + 76 ⋅ 10 + 76 ⋅ 10 + 82 ⋅ 10 = 4930. 76 is considered three times since we needed to calculate the area under the graph. Overestimate of total change = 57 ⋅ 10 + 76 ⋅ 10 + 79 ⋅ 10 + 88 ⋅ 10 + 88 ⋅ 10 + 83 ⋅ 10 + 83 ⋅ 10 = 5540. 88 and 83 were considered twice since we needed to calculate the area over the graph. The average is: (4930 + 5540)∕2 = 5235 million people. (b) The total change in the world’s population between 1950 and 2020 is given by the di˙erence between the populations in those two years. That is, the change in population equals 7684 (population in 2020) − 2526 (population in 1950) = 5158 million people. Our estimate of 5235 million people and the actual di˙erence of 5158 million people are close to each other, suggesting our estimate was a good one.

374

Chapter Five /SOLUTIONS

34. To make this estimate, we frst observe that, for constant speed, Fuel used =

Miles traveled Speed ⋅ Time = . Miles/gallon Miles/gallon

We make estimates at the start and end of each of the six intervals given. The 60 mph increase in speed (from 10 mph to 70 mph) takes half an hour, or 30 minutes. Since the speed increases at a constant rate, each 10 mph increase takes 5 minutes. At the start of the frst 5 minutes, the speed is 10 mph and at the end, the speed is 20 mph. Since 5min = 5∕60 hour, during the frst 5 minutes the distance traveled is between 10 ⋅

5 mile 60

and

20 ⋅

5 mile. 60

During this frst fve minute period, the speed is between 10 and 20 mph, so the fuel eÿciency is between 15 mpg and 18 mpg. So the fuel used during this period is between 1 5 ⋅ 10 ⋅ gallons 18 60

and

1 5 ⋅ 20 ⋅ gallons. 15 60

Thus, an underestimate of the fuel used is 1 1 1 1 1 1 5 Fuel = ⋅ 10 + ⋅ 20 + ⋅ 30 + ⋅ 40 + ⋅ 50 + ⋅ 60 = 0.732 gallons. 18 21 23 24 25 26 60 An overestimate of the fuel used is 1 1 1 1 1 1 5 Fuel = ⋅ 20 + ⋅ 30 + ⋅ 40 + ⋅ 50 + ⋅ 60 + ⋅ 70 = 1.032 gallons. 15 18 21 23 24 25 60 We can get better bounds by using the actual distance traveled during each fve minute period. For example, in the frst fve minutes the average speed is 15 mph (halfway between 10 and 20 mph because the speed is increasing at a constant rate). Since 5 minutes is 5/60 of an hour, in the frst fve minutes the car travels 15(5/60)=5/4 miles. Thus the fuel used during this period was between 5 1 5 1 ⋅ and ⋅ . 4 18 4 15 Using this method for each fve minute period gives a lower estimate of 0.843 gallons and an upper estimate of 0.909 gallons. 35. No, the 2010 Prius Plug-in Prototype cannot travel half a mile in EV-only mode in under 25 seconds, because the upper estimate for the distance traveled in 25 seconds is only about 0.3 miles. To see this, we convert seconds to hours (1 sec = 1∕3600 hours) so that the upper distance estimate is in miles: Total distance traveled between t = 0 and t = 25 (Upper estimate)

= 20 ⋅

5 5 5 5 5 + 33 ⋅ + 45 ⋅ + 50 ⋅ + 59 ⋅ < 0.3 miles 3600 3600 3600 3600 3600

36. To estimate the distance traveled during the frst 15 seconds, we calculate upper and lower bound estimates for this distance, and then average these estimates. Note that because the speed of the Prius is in miles per hour, we need to convert seconds to hours (1 sec = 1∕3600 hours) to calculate the distance estimates in miles. We then convert miles to feet(1 mile = 5280 feet).

Since:

Total distance traveled between t = 0 and t = 15 (Lower Estimate)

= 0⋅

Total distance traveled between t = 0 and t = 15 (Upper Estimate)

= 20 ⋅

5 5 5 + 20 ⋅ + 33 ⋅ = 0.0736 miles or 389 feet 3600 3600 3600

5 5 5 + 33 ⋅ + 45 ⋅ = 0.1361 miles or 719 feet 3600 3600 3600

389 + 719 Average of Lower and Upper = = 554 feet, Estimates 2 the 2010 Prius Prototype travels about 554 feet, roughly a tenth of a mile, during the frst 15 seconds of movement in EV-only mode.

5.1 SOLUTIONS

375

37. To estimate the distance between the two cars at t = 5, we calculate upper and lower bound estimates for this distance and then average these estimates. Note that because the speed of the Prius is in miles per hour, we need to convert seconds to hours (1 sec = 1∕3600 hours) to calculate the distance estimates in miles. We may then convert miles to feet (1 mile = 5280 feet). Distance between cars at t = 5 (Upper Estimate)

≈ (Di˙erence in speeds at t = 5) ⋅ (Travel time) = (33 − 20) ⋅

Distance between cars at t = 5 (Lower Estimate)

5 = 0.018 miles 3600

≈ (Di˙erence in speeds at t = 0) ⋅ (Travel time) = (0 − 0) ⋅

5 = 0 miles 3600

Since:

0.018 + 0 Average of Lower and Upper Estimates of = = 0.009 miles = 48 feet, distance between cars at t = 5 2 the distance between the two cars is about 48 feet, fve seconds after leaving the stoplight. 38. To estimate the distance between the two cars at t = 15, we calculate upper and lower bound estimates for this distance and then average these estimates. Note that because the speed of the Prius is in miles per hour, we need to convert seconds to hours (1 sec = 1∕3600 hours) to calculate the distance estimates in miles. We may then convert miles to feet (1 mile = 5280 feet). Distance between cars at t = 15 (Upper Estimate)

≈ (Di˙erence in speeds at t = 5) ⋅ (Travel time) + (Di˙erence in speeds at t = 10) ⋅ (Travel time) + (Di˙erence in speeds at t = 15) ⋅ (Travel time) 5 5 5 = (33 − 20) ⋅ + (53 − 33) ⋅ + (70 − 45) ⋅ = 0.081 miles 3600 3600 3600

Distance between cars at t = 15 (Lower Estimate)

≈ (Di˙erence in speeds at t = 0) ⋅ (Travel time) + (Di˙erence in speeds at t = 5) ⋅ (Travel time) + (Di˙erence in speeds at t = 10) ⋅ (Travel time) 5 5 5 = (0 − 0) ⋅ + (33 − 20) ⋅ + (53 − 33) ⋅ = 0.046 miles 3600 3600 3600

Since:

0.081 + 0.046 Average of Lower and Upper Estimates of = = 0.0635 miles = 335 feet, distance between cars at t = 15 2 the distance between the two cars is about 335 feet, 15 seconds after leaving the stoplight. 39. (a) From the table we have t0 = 0, t1 = 3, t2 = 6, t3 = 9, t4 = 12 and

f (t0 ) = 80, f (t1 ) = 62, f (t2 ) = 47, f (t3 ) = 32, f (t4 ) = 22.

Note that we have Δt = 3 and n = 4. Choosing the values on the left end of each interval, we have Left sum = (80)(3) + (62)(3) + (47)(3) + (32)(3) = 663 ft (b) Since the velocity is decreasing, choosing the velocity on the left end of each interval will give us an upper estimate for the total distance traveled. (c) Choosing the values on the right end of each interval, we have Right sum = (62)(3) + (47)(3) + (32)(3) + (22)(3) = 489 ft (d) Since the velocity is decreasing, choosing the velocity on the right end of each interval will give us a lower estimate for the total distance traveled.

376

Chapter Five /SOLUTIONS (e) We know the actual distance the car has traveled over the 12-second interval is greater than 489 feet but less than 663 feet. A better estimate would be the average of the upper and lower estimates, 663 ft + 489 ft = 576 ft. 2

40. Lower estimate: 2.7 ⋅ 2 + 2.7 ⋅ 2 + 4 ⋅ 2 + 6.3 ⋅ 2 + 8.5 ⋅ 2 + 11.6 ⋅ 2 + 13.4 ⋅ 2 + 17.4 ⋅ 2 + 21.9 ⋅ 2 + 29.1 ⋅ 2 = 235.2 meters. Upper estimate: 2.7 ⋅ 2 + 4 ⋅ 2 + 6.3 ⋅ 2 + 8.5 ⋅ 2 + 11.6 ⋅ 2 + 13.4 ⋅ 2 + 17.4 ⋅ 2 + 21.9 ⋅ 2 + 29.1 ⋅ 2 + 32.6 ⋅ 2 = 295.0 meters. 41. (a) With n = 5, we have Δt = 2. Then t0 = 2008, t1 = 2010, t2 = 2012, t3 = 2014, t4 = 2016,

and

f (t0 ) = 32.0, f (t1 ) = 33.1, f (t2 ) = 35.0, f (t3 ) = 35.6, f (t4 ) = 35.7. Choosing the values on the left end of each interval, we have Left sum = (32.0)(2) + (33.1)(2) + (35.0)(2) + (35.6)(2) + (35.7)(2) = 342.8. Thus, the left sum leads to an estimate of 342.8 billion tons of CO2 emissions. (b) Since the rate at which CO2 is being emitted is increasing, choosing the values on the left of each interval will give a lower estimate for the world’s total CO2 emissions between 2008 and 2018. (c) With n = 5, we have Δt = 2. Then t1 = 2010, t2 = 2012, t3 = 2014, t4 = 2016, t5 = 2018,

and

f (t1 ) = 33.1, f (t2 ) = 35.0, f (t3 ) = 35.6, f (t4 ) = 35.7, f (t5 ) = 37.1. Choosing the values on the right end of each interval, we have Right sum = (33.1)(2) + (35.0)(2) + (35.6)(2) + (35.7)(2) + (37.1)(2) = 353.0 Thus, the right sum leads to an estimate of 353 billion tons of CO2 emissions. (d) Since the rate at which CO2 is being emitted is increasing, choosing the values on the right of each interval will give an upper estimate for the world’s total CO2 emissions between 2008 and 2018. 42. (a) If Δt = 4, then n = 2. We have: Underestimate of total change = f (0)Δt + f (4)Δt = 1 ⋅ 4 + 17 ⋅ 4 = 72. Overestimate of total change = f (4)Δt + f (8)Δt = 17 ⋅ 4 + 65 ⋅ 4 = 328. See Figure 5.17. f (t) = t2 + 1

f (t) = t2 + 1

30 10

10 t 4

Figure 5.17

Figure 5.18

(b) If Δt = 2, then n = 4. We have: Underestimate of total change = f (0)Δt + f (2)Δt + f (4)Δt + f (6)Δt = 1 ⋅ 2 + 5 ⋅ 2 + 17 ⋅ 2 + 37 ⋅ 2 = 120. Overestimate of total change = f (2)Δt + f (4)Δt + f (6)Δt + f (8)Δt = 5 ⋅ 2 + 17 ⋅ 2 + 37 ⋅ 2 + 65 ⋅ 2 = 248. See Figure 5.18.

5.2 SOLUTIONS (c) If Δt = 1, then n = 8. Underestimate of total change = f (0)Δt + f (1)Δt + f (2)Δt + f (3)Δt + f (4)Δt + f (5)Δt + f (6)Δt + f (7)Δt = 1 ⋅ 1 + 2 ⋅ 1 + 5 ⋅ 1 + 10 ⋅ 1 + 17 ⋅ 1 + 26 ⋅ 1 + 37 ⋅ 1 + 50 ⋅ 1 = 148. Overestimate of total change = f (1)Δt + f (2)Δt + f (3)Δt + f (4)Δt + f (5)Δt + f (6)Δt + f (7)Δt + f (8)Δt = 2 ⋅ 1 + 5 ⋅ 1 + 10 ⋅ 1 + 17 ⋅ 1 + 26 ⋅ 1 + 37 ⋅ 1 + 50 ⋅ 1 + 65 ⋅ 1 = 212. See Figure 5.19. f (t) = t2 + 1 70 50 30 10 t 1

Figure 5.19

43. (a) With n = 4, we have Δt = 2. Then t0 = 15, t1 = 17, t2 = 19, t3 = 21, t4 = 23 and

f (t0 ) = 10, f (t1 ) = 13, f (t2 ) = 18, f (t3 ) = 20, f (t4 ) = 30

(b) Left sum = (10)(2) + (13)(2) + (18)(2) + (20)(2) = 122 Right sum = (13)(2) + (18)(2) + (20)(2) + (30)(2) = 162. (c) With n = 2,we have Δt = 4. Then t0 = 15, t1 = 19, t2 = 23 and

f (t0 ) = 10, f (t1 ) = 18, f (t2 ) = 30

(d) Left sum = (10)(4) + (18)(4) = 112 Right sum = (18)(4) + (30)(4) = 192.

Solutions for Section 5.2 1. (a) Right sum, because the top edge of each shaded rectangle touches the graph of g on its right end. (b) Upper estimate, because the total area of the 3 shaded rectangles is greater than the area under the graph of g. (c) n = 3, because there are 3 rectangles. (d) Δx = 2, because the interval of length 6 is divided into 3 subintervals, each of width Δx. 2. (a) Left sum, because the top edge of each shaded rectangle touches the graph of g on its left end. (b) Lower estimate, because the total area of the 3 shaded rectangles is less than the area under the graph of g. (c) n = 3, because there are 3 rectangles. (d) Δx = 2, because the interval of length 6 is divided into 3 subintervals, each of width Δx.

377

378

Chapter Five /SOLUTIONS

3. (a) Left-hand sum. Right-hand sum would be larger. 6−2 1 = . (b) We have a = 2, b = 6, n = 8, Δx = 8 2 4. Dividing the interval from 0 to 6 into 2 equal subintervals gives Δx = 3. Using f (x) = 2x , we have Left-hand sum = f (0) ⋅ Δx + f (3) ⋅ Δx = 20 ⋅ 3 + 23 ⋅ 3 = 27.

5. Dividing the interval from 0 to 12 into 3 equal subintervals gives Δx = 4. Using f (x) = 1∕(x + 1), we have Left-hand sum = f (0) ⋅ Δx + f (4) ⋅ Δx + f (8) ⋅ Δx 1 1 1 = ⋅4+ ⋅4+ ⋅4 0+1 4+1 8+1 = 5.244. 2

6. Since e−x is decreasing between x = 0 and x = 1, the left sum is an overestimate and the right sum is an underestimate 2 of the integral. Letting f (x) = e−x , we divide the interval 0 ≤ x ≤ 1 into n = 5 sub-intervals to create Table 5.2. Table 5.2 x

0.0

0.2

0.4

0.6

0.8

1.0

f (x)

1.000

0.961

0.852

0.698

0.527

0.368

(a) Letting Δx = (1 − 0)∕5 = 0.2, we have: Left-hand sum = f (0)Δx + f (0.2)Δx + f (0.4)Δx + f (0.6)Δx + f (0.8)Δx = 1(0.2) + 0.961(0.2) + 0.852(0.2) + 0.698(0.2) + 0.527(0.2) = 0.808. (b) Again letting Δx = (1 − 0)∕5 = 0.2, we have: Right-hand sum = f (0.2)Δx + f (0.4)Δx + f (0.6)Δx + f (0.8)Δx + f (1)Δx = 0.961(0.2) + 0.852(0.2) + 0.698(0.2) + 0.527(0.2) + 0.368(0.2) = 0.681. 7. Using n = 4, for the left-hand sum we have Left-hand sum = f (0) ⋅ 3 + f (3) ⋅ 3 + f (6) ⋅ 3 + f (9) ⋅ 3 = 0 ⋅ 3 + 9 ⋅ 3 + 36 ⋅ 3 + 81 ⋅ 3 = 378. Using n = 4, for the right-hand sum we have Right-hand sum = f (3) ⋅ 3 + f (6) ⋅ 3 + f (9) ⋅ 3 + f (12) ⋅ 3 = 9 ⋅ 3 + 36 ⋅ 3 + 81 ⋅ 3 + 144 ⋅ 3 = 810. 8. Using n = 5, for the left-hand sum we have Left-hand sum = f (−2) ⋅ 2 + f (0) ⋅ 2 + f (2) ⋅ 2 + f (4) ⋅ 2 + f (6) ⋅ 2 = 4 ⋅ 2 + 0 ⋅ 2 + 4 ⋅ 2 + 64 ⋅ 2 + 324 ⋅ 2 = 792. Using n = 5, for the right-hand sum we have Right-hand sum = f (0) ⋅ 2 + f (2) ⋅ 2 + f (4) ⋅ 2 + f (6) ⋅ 2 + f (8) ⋅ 2 = 0 ⋅ 2 + 4 ⋅ 2 + 64 ⋅ 2 + 324 ⋅ 2 + 1024 ⋅ 2 = 2832. 9. Using n = 3, for the left-hand sum we have Left-hand sum = f (−1) ⋅ 3 + f (2) ⋅ 3 + f (5) ⋅ 3 =

1 ⋅ 3 + 4 ⋅ 3 + 32 ⋅ 3 = 109.5. 2

Using n = 3, for the right-hand sum we have Right-hand sum = f (2) ⋅ 3 + f (5) ⋅ 3 + f (8) ⋅ 3 = 4 ⋅ 3 + 32 ⋅ 3 + 256 ⋅ 3 = 876.

5.2 SOLUTIONS

379

10. Using n = 4, for the left-hand sum we have Left-hand sum = f (−4) ⋅ 2 + f (−2) ⋅ 2 + f (0) ⋅ 2 + f (2) ⋅ 2 = 17 ⋅ 2 + 5 ⋅ 2 + 2 ⋅ 2 + 1.25 ⋅ 2 = 50.5. Using n = 4, for the right-hand sum we have Right-hand sum = f (−2) ⋅ 2 + f (0) ⋅ 2 + f (2) ⋅ 2 + f (4) ⋅ 2 = 5 ⋅ 2 + 2 ⋅ 2 + 1.25 ⋅ 2 + 1.0625 ⋅ 2 = 18.625. 11. Using n = 4, for the left-hand sum we have Left-hand sum = f (1) ⋅

1 1 1 1 1 1 1 1 77 + f (1.5) ⋅ + f (2) ⋅ + f (2.5) ⋅ = 3 ⋅ + 2 ⋅ + 1.5 ⋅ + 1.2 ⋅ = . 2 2 2 2 2 2 2 2 20

Using n = 4, for the right-hand sum we have Right-hand sum = f (1.5) ⋅

1 1 1 1 1 1 1 1 57 + f (2) ⋅ + f (2.5) ⋅ + f (3) ⋅ = 2 ⋅ + 1.5 ⋅ + 1.2 ⋅ + 1 ⋅ = . 2 2 2 2 2 2 2 2 20

12. Using n = 3, for the left-hand sum we have Left-hand sum = f (5.5) ⋅ 0.5 + f (6) ⋅ 0.5 + f (6.5) ⋅ 0.5 = 46.75 ⋅ 0.5 + 54 ⋅ 0.5 + 61.75 ⋅ 0.5 = 81.25. Using n = 3, for the right-hand sum we have Right-hand sum = f (6) ⋅ 0.5 + f (6.5) ⋅ 0.5 + f (7) ⋅ 0.5 = 54 ⋅ 0.5 + 61.75 ⋅ 0.5 + 70 ⋅ 0.5 = 92.875. 13. Using n = 3, for the left-hand sum we have Left-hand sum = f (1) ⋅ 1 + f (2) ⋅ 1 + f (3) ⋅ 1 = 1 ⋅ 1 +

√

2⋅1+

√ √ √ 3 ⋅ 1 = 1 + 2 + 3.

Using n = 3, for the right-hand sum we have Right-hand sum = f (2) ⋅ 1 + f (3) ⋅ 1 + f (4) ⋅ 1 =

√ √ √ √ 2 ⋅ 1 + 3 ⋅ 1 + 2 ⋅ 1 = 2 + 3 + 2.

14. Using n = 4, for the left-hand sum we have √ √ 2 2 3 √ Left-hand sum = f (0) ⋅ + f ⋅ +f ⋅ +f ⋅ =0⋅ + ⋅ +1⋅ + ⋅ = ( 2 + 1). 4 4 4 2 4 4 4 4 2 4 4 2 4 4 Using n = 4, for the right-hand sum we have √ √ 2 2 3 √ Right-hand sum = f ⋅ +f ⋅ +f ⋅ + f ( ) ⋅ = ⋅ +1⋅ + ⋅ + 0 ⋅ = ( 2 + 1). 4 4 2 4 4 4 4 2 4 4 2 4 4 4 15. We estimate the integral by fnding left- and right-hand sums and averaging them: Left-hand sum = 100 ⋅ 4 + 88 ⋅ 4 + 72 ⋅ 4 + 50 ⋅ 4 = 1240 Right-hand sum = 88 ⋅ 4 + 72 ⋅ 4 + 50 ⋅ 4 + 28 ⋅ 4 = 952. We have

Ê10

f (x) dx ≈

1240 + 952 = 1096. 2

16. We estimate ∫0 f (x)dx using left- and right-hand sums: 40

Left sum = 350 ⋅ 10 + 410 ⋅ 10 + 435 ⋅ 10 + 450 ⋅ 10 = 16,450. Right sum = 410 ⋅ 10 + 435 ⋅ 10 + 450 ⋅ 10 + 460 ⋅ 10 = 17,550. We estimate that Ê0

f (x)dx ≈

In this estimate, we used n = 4 and Δx = 10.

16450 + 17550 = 17,000. 2

380

Chapter Five /SOLUTIONS

17. Left-hand sum = 50 ⋅ 3 + 48 ⋅ 3 + 44 ⋅ 3 + 36 ⋅ 3 + 24 ⋅ 3 = 606 Right-hand sum = 48 ⋅ 3 + 44 ⋅ 3 + 36 ⋅ 3 + 24 ⋅ 3 + 8 ⋅ 3 = 480 606 + 480 Average = = 543 2 15

f (x) dx ≈ 543. Ê0 18. Since we are given a table of values, we must use Riemann sums to approximate the integral. Values are given every 0.2 units, so Δt = 0.2 and n = 5. Our best estimate is obtained by calculating the left-hand and right-hand sums, and then averaging the two. So we have

Left-hand sum = 25(0.2) + 23(0.2) + 20(0.2) + 15(0.2) + 9(0.2) = 18.4 Right-hand sum = 23(0.2) + 20(0.2) + 15(0.2) + 9(0.2) + 2(0.2) = 13.8. We average the two sums to obtain our best estimate of the integral: Ê3 19.

W (t)dt ≈

18.4 + 13.8 = 16.1. 2

24 f (t)

f (t)

8 t 2

Figure 5.20: Left Sum, Δt = 4

Figure 5.21: Right Sum, Δt = 4

(a) Left-hand sum = 32 ⋅ 4 + 24 ⋅ 4 = 224. (b) Right-hand sum = 24 ⋅ 4 + 0 ⋅ 4 = 96. 32

f (t)

8 t 2

Figure 5.22: Left Sum, Δt = 2

t 2

Figure 5.23: Right Sum, Δt = 2

(c) Left-hand sum = 32 ⋅ 2 + 30 ⋅ 2 + 24 ⋅ 2 + 14 ⋅ 2 = 200. (d) Right-hand sum = 30 ⋅ 2 + 24 ⋅ 2 + 14 ⋅ 2 + 0 ⋅ 2 = 136. 20

20.

f (x) dx is equal to the area shaded. We can use Riemann sums to estimate the area, or we can count grid squares in Ê0 Figure 5.24. There are about 15 grid squares and each grid square represents 4 square units, so the area shaded is about 60. We have Ê0

f (x) dx ≈ 60.

5.2 SOLUTIONS

381

5 f (x)

4 3 2 1

x 4

Figure 5.24

21. We know that

Ê−10

f (x)dx = Area under f (x) between x = −10 and x = 15.

The area under the curve consists of approximately 14 boxes, and each box has area (5)(5) = 25. Thus, the area under the curve is about (14)(25) = 350, so 15

Ê−10

f (x)dx ≈ 350.

22.

f (x) dx is equal to the area shaded. We can use Riemann sums to estimate this area, or we can count grid squares. Ê0 There are 3 whole grid squares and about 4 half-grid squares, for a total of 5 grid squares. Since each grid square represents 4 square units, our estimated area is 5(4) = 20. We have

Ê0

f (x) dx ≈ 20. See Figure 5.25.

16 12 f (x) 8 4 x 1

Figure 5.25

23. To estimate the integral, we count the rectangles under the curve and above the x-axis for the interval [0, 3]. There are approximately 17 of these rectangles, and each has area 1, so Ê0

f (x)dx ≈ 17.

24. The integral represents the area below the graph of f (x) but above the x-axis. Since each square has area 1, by counting squares and half-squares we fnd Ê1

f (x) dx = 8.5.

25. The values of the two integrals are equal because the Riemann sums formed to evaluate each of the integrals have the same value for each n.

382

Chapter Five /SOLUTIONS

26. We have 3

Ê2

1 f (x) dx + 1 = 3(4) + 1 = 13.

27. We use a calculator or computer to see that

1 dx = 1.2. Ê1 √1 + x2 1

28. We use a calculator or computer to see that

1 dt = 2.350. Ê−1 et

29. We use a calculator or computer to see that ∫−1 xx2 +1 dx = −0.75. −4 2

30. We use a calculator or computer to see that ∫1.1 et ln t dt = 0.865. 1.7

31. We use a calculator or computer to see that

Ê0

ln(y2 + 1) dy = 3.406. 4√

ez + z dz = 6.111. Ê3 33. The values of the two integrals are equal, so evaluating with a calculator or computer, we have

32. We use a calculator or computer to see that

Ê0

1 1 dx + dy = 1.107 + 1.107 = 2.214. Ê0 1 + y2 1 + x2

34. The values of the two integrals are equal, so their di˙erence is 0. 35. (a) See Figure 5.26. y

y = x3

0 1

Figure 5.26

Ê0

x3 dx = area shaded, which is less than 0.5. Rough estimate is about 0.3.

x3 dx = 0.25 Ê0 36. (a) See Figure 5.27. (b)

f (x) =

2 1.73

√

x 0

Figure 5.27 A rough estimate of the shaded area is 70% of the area of the rectangle 3 by 1.7. Thus, Ê0 (b)

Ê0

3√

x dx = 3.4641

3√

x dx ≈ 70% of 3 ⋅ 1.7 = 0.70 ⋅ 3 ⋅ 1.7 = 3.6.

5.2 SOLUTIONS

383

37. (a) See Figure 5.28. f (t) 4 3 2 1 0

f (t) = 3t

t 1

Figure 5.28 The shaded area is approximately 60% of the area of the rectangle 1 unit by 3 units. Therefore, Ê0

3t dt ≈ 0.60 ⋅ 1 ⋅ 3 = 1.8.

3t dt = 1.8205. Ê0 38. (a) See Figure 5.29. (b)

f (x) = xx 4 3 2 1 x 1

Figure 5.29

The shaded area appears to be approximately 2 units, and so 2

Ê1

xx dx ≈ 2.

xx dx = 2.05045 Ê1 39. The graph given shows that f is positive for 0 ≤ t ≤ 1. Since the graph is contained within a rectangle of height 100 and 1 length 1, the answers −98.35 and 100.12 are both either too small or too large to represent ∫0 f (t)dt. Since the graph of f is above the horizontal line y = 80 for 0 ≤ t ≤ 0.95, the best estimate is IV, 93.47 and not 71.84. (b)

40. (a) The integral gives the shaded area in Figure 5.30. We fnd Ê0

(x2 + 1) dx = 78.

y 40 y = x2 + 1 20

x 3

Figure 5.30

384

Chapter Five /SOLUTIONS (b) Using n = 3, we have Left-hand sum = f (0) ⋅ 2 + f (2) ⋅ 2 + f (4) ⋅ 2 = 1 ⋅ 2 + 5 ⋅ 2 + 17 ⋅ 2 = 46. This sum is an underestimate. See Figure 5.31. y

y = x2 + 1

y = x2 + 1 20

x 3

Figure 5.31

Figure 5.32

(c) Using n = 3 gives Right-hand sum = f (2) ⋅ 2 + f (4) ⋅ 2 + f (6) ⋅ 2 = 5 ⋅ 2 + 17 ⋅ 2 + 37 ⋅ 2 = 118. This sum is an overestimate. See Figure 5.32. tn − t0 15 − 3 = = 3. Since we are taking steps of Δt = 3, n 4 we know that t0 = 3, t1 = 6, t2 = 9, t3 = 12, t4 = 15. This gives

41. (a) This is a left-hand Riemann sum of 4 terms. We have Δt = n−1 É i=0

� � � � � f ti Δt = f t0 Δt + f t1 Δt + f t2 Δt + f t3 Δt = f (3) ⋅ 3 + f (6) ⋅ 3 + f (9) ⋅ 3 + f (12) ⋅ 3 = (−40 + 23 + 68 + 95) ⋅ 3 = 438.

tn − t0 15 − 3 = = 4. Since we are taking steps of n 3 Δt = 4, we know that t0 = 3, t1 = 7, t2 = 11, t3 = 15. This gives

(b) This is a right-hand Riemann sum of 3 terms. We have Δt = n É i=1

� � � � f ti Δt = f t1 Δt + f t2 Δt + f t3 Δt = f (7) ⋅ 4 + f (11) ⋅ 4 + f (15) ⋅ 4 = (40 + 88 + 104) ⋅ 4 = 928.

tn − t0 13 − 5 = = 2. Since we are taking steps of n 4 Δt = 2, we know that t0 = 5, t1 = 7, t2 = 9, t3 = 11, t4 = 13. This gives

� � � � � f ti Δt = f t1 Δt + f t2 Δt + f t3 Δt + f t4 Δt = f (7) ⋅ 2 + f (9) ⋅ 2 + f (11) ⋅ 2 + f (13) ⋅ 2 = (40 + 68 + 88 + 100) ⋅ 2 = 592.

42. (a) Since Δx = 3 and x0 = 2 we have x0 = 2, x1 = 5, x2 = 8, x3 = 11. Thus, 3 É � g xi Δx = g(2)Δx + g(5)Δx + g(8)Δx + g(11)Δx i=0

= 7 ⋅ 3 + 19 ⋅ 3 + 31 ⋅ 3 + 43 ⋅ 3 = 300

5.2 SOLUTIONS (b) Since Δx = 2 and x0 = 4 we have x0 = 4, x1 = 6, x2 = 8, x3 = 10, x4 = 12. Thus, 4 É � g xi Δx = g(6)Δx + g(8) + g(10)Δx + g(12)Δx i=1

= 23 ⋅ 2 + 31 ⋅ 2 + 39 ⋅ 2 + 47 ⋅ 2 = 280 (c) Since Δx = 3 and x0 = 1 we have x0 = 1, x1 = 4, x2 = 7, x3 = 10, x4 = 13, x5 = 16. Thus, 5 É � g xi Δx = g(7)Δx + g(10)Δx + g(13)Δx + g(16)Δx i=2

= 27 ⋅ 3 + 39 ⋅ 3 + 51 ⋅ 3 + 63 ⋅ 3 = 540 43. (a) Since Δx = 2 and x0 = 2 we have x0 = 2, x1 = 4, x2 = 6, x3 = 8, x4 = 10. Thus, 4 É � ℎ xi Δx = ℎ(2)Δx + ℎ(4)Δx + ℎ(6)Δx + ℎ(8)Δx + ℎ(10)Δx i=0

= 6 ⋅ 2 + 7 ⋅ 2 + 8 ⋅ 2 + 9 ⋅ 2 + 10 ⋅ 2 = 80 (b) Since Δx = 3 and x0 = 0 we have x0 = 0, x1 = 3, x2 = 6, x3 = 9, x4 = 12, x5 = 15. Thus, 5 É � ℎ xi Δx = ℎ(6)Δx + ℎ(9) + ℎ(12)Δx + ℎ(15)Δx i=2

= 8 ⋅ 3 + 9.5 ⋅ 3 + 11 ⋅ 3 + 12.5 ⋅ 3 = 123 (c) Since Δx = 2 and x0 = 1 we have x0 = 1, x1 = 3, x2 = 5, x3 = 7, x4 = 9, x5 = 11, x6 = 13, x7 = 15 Thus, 7 É � ℎ xi Δx = ℎ(9)Δx + ℎ(11)Δx + ℎ(13)Δx + ℎ(15)Δx i=4

= 9.5 ⋅ 2 + 10.5 ⋅ 2 + 11.5 ⋅ 2 + 12.5 ⋅ 2 = 88

385

386

Chapter Five /SOLUTIONS

Solutions for Section 5.3 1. Since f (x) is positive along the interval from 0 to 6 the area is simply

Ê0

(x2 + 2) dx = 84.

2. See Figure 5.33. Area =

Ê0

100(0.6)t dt ≈ 192.47

P 100

P = 100(0.6)t

t 2

Figure 5.33

3. The graph of y = 4 − x2 crosses the x-axis at x = 2 since solving y = 4 − x2 = 0 gives x = ±2. See Figure 5.34. To fnd the total area, we fnd the area above the axis and the area below the axis separately. We have Ê0

(4 − x2 )dx = 5.3333

and

Ê2

(4 − x2 )dx = −2.3333.

As expected, the integral from 2 to 3 is negative. The area above the axis is 5.3333 and the area below the axis is 2.3333, so Total area = 5.3333 + 2.3333 = 7.6666. Thus the area is 7.667. y 4 x 2 y = 4 − x2

Figure 5.34 4. On the interval 0 ≤ x ≤ 2, we see that x + 5 is greater than 2x + 1. We have Area =

Ê0

((x + 5) − (2x + 1)) dx = 6.

5. The two functions intersect at x = 0 and x = 3. Between these values, 3x is greater than x2 . We have Area =

Ê0

(3x − x2 ) dx = 4.5.

5.3 SOLUTIONS

387

6. (a) The total area between f (x) and the x-axis is the sum of the two given areas, so Area = 7 + 6 = 13. (b) To fnd the integral, we note that from x = 3 to x = 5, the function lies below the x-axis, and hence makes a negative contribution to the integral. So Ê0

f (x) dx =

Ê0

f (x)dx +

Ê3

f (x)dx = 7 − 6 = 1.

7. (a) Negative, since f ≤ 0 everywhere on the interval −5 ≤ x ≤ −4. (b) Positive, since f ≥ 0 everywhere on the interval −4 ≤ x ≤ 1. (c) Negative, since the graph of f has more area under the x-axis than above on the interval 1 ≤ x ≤ 3. (d) Positive, since the graph of f has more area above the x-axis than underneath on the interval −5 ≤ x ≤ 3.

8. The integral ∫−5 f dx is the smallest, since it has a negative value on the interval −5 ≤ x ≤ −4 plus a small positive value −3

on −4 ≤ x ≤ −3. The integral ∫−5 f dx is larger than ∫−5 f dx, since f > 0 on the interval −3 ≤ x ≤ −1. The integral −1

−3

1 −1 ∫−5 f dx is larger than ∫−5 f dx, since f > 0 on the interval −2 ≤ x ≤ 1. However, since f < 0 during part of the interval 3 1 3 −1 1 ≤ x ≤ 3, the integral ∫−5 f dx is less than ∫−5 f dx. But ∫−5 f dx > ∫−5 f dx, since there is more area above the x-axis

on the interval −1 ≤ x ≤ 1 than there is below the x-axis on the interval 1 ≤ x ≤ 3. Thus in ascending order the integrals −3 −1 3 1 are ∫−5 f dx < ∫−5 f dx < ∫−5 f dx < ∫−5 f dx. 9. (a) Counting the squares yields an estimate of 16.5, each with area = 1, so the total shaded area is approximately 16.5. (b) Ê0

f (x)dx = (shaded area above x-axis) − (shaded area below x-axis)

≈ 6.5 − 10 = −3.5 (c) The answers in (a) and (b) are di˙erent because the shaded area below the x-axis is subtracted in order to fnd the value of the integral in (b). 10. We know that

Ê−3

f (x)dx = Area above the axis − Area below the axis.

The area above the axis is about 3 boxes. Since each box has area (1)(5) = 5, the area above the axis is about (3)(5) = 15. The area below the axis is about 11 boxes, giving an area of about (11)(5) = 55. We have 5

Ê−3

f (x)dx ≈ 15 − 55 = −40.

11. (a) The area shaded between 0 and 1 is about the same as the area shaded between −1 and 0, so the area between 0 and 1 is about 0.25. Since this area lies below the x-axis, we estimate that Ê0

f (x) dx = −0.25.

(b) Between −1 and 1, the area above the x-axis approximately equals the area below the x-axis, and so 1

Ê−1

f (x) dx = (0.25) + (−0.25) = 0.

(c) We estimate Total area shaded = 0.25 + 0.25 = 0.5. 12. In Figure 5.35 the area A1 is largest, A2 is next, and A3 is smallest. We have I= IV =

Êa

Êb

f (x) dx = A1 ,

II =

f (x) dx = −A2 + A3 ,

Êa

f (x) dx = A1 − A2 , V=

Êb

III =

f (x) dx = −A2 .

Êa

f (x) dx = A1 − A2 + A3 ,

388

Chapter Five /SOLUTIONS The relative sizes of A1 , A2 , and A3 mean that I is positive and largest, III is next largest (since −A2 + A3 is negative, but less negative than −A2 ), II is next largest, but still positive (since A1 is larger than A2 ). The integrals IV and V are both negative, but V is more negative. Thus V < IV < 0 < II < III < I. f (x) A1 A3

b A2 c

Figure 5.35 13. (a) The area between the graph of f and the x-axis between x = a and x = b is 13, so Êa

f (x) dx = 13.

(b) Since the graph of f (x) is below the x-axis for b < x < c, Êb

f (x) dx = −2.

f (x) dx = 13 − 2 = 11.

(d) The graph of |f (x)| is the same as the graph of f (x) except that the part below the x-axis is refected to be above it. See Figure 5.36. Thus Êa

|f (x)| dx = 13 + 2 = 15. |f (x)| Area = 13

✠

Area = 2

✠ a

Figure 5.36 14. The area above the x-axis appears to be about half the area of the rectangle with area 10 ⋅ 3 = 30, so we estimate the area above the x-axis to be approximately 15. The area below the x-axis appears to be about half the area of the rectangle with area 5 ⋅ 2 = 10, so we estimate the area below the x-axis to be approximately 5. See Figure 5.37. The value of the integral is the area above the x-axis minus the area below the x-axis, so we estimate Ê0

f (x)dx ≈ 15 − 5 = 10.

The correct match for this function is IV. 10 5 3 −5

Figure 5.37

5.3 SOLUTIONS

389

15. The area below the x-axis is bigger than the area above the x-axis, so the integral is negative. The area above the x-axis appears to be about half the area of the rectangle with area 20 ⋅ 2 = 40, so we estimate the area above the x-axis to be approximately 20. The area below the x-axis appears to be about half the area of the rectangle with area 15 ⋅ 3 = 45, so we estimate the area below the x-axis to be approximately 22.5. See Figure 5.38. The value of the integral is the area above the x-axis minus the area below the x-axis, so we estimate Ê0

f (x)dx ≈ 20 − 22.5 = −2.5.

The correct match for this function is II. 20 10 x 2

−10 −15

Figure 5.38

16. The area below the x-axis is bigger than the area above the x-axis, so the integral is negative. The area above the x-axis appears to be about half the area of the rectangle with area 5 ⋅ 2 = 10, so we estimate the area above the x-axis to be approximately 5. The area below the x-axis appears to be about half the area of the rectangle with area 10 ⋅ 3 = 30, so we estimate the area below the x-axis to be approximately 15. See Figure 5.39. The value of the integral is the area above the x-axis minus the area below the x-axis, so we estimate Ê0

f (x)dx ≈ 5 − 15 = −10.

The correct match for this function is I. 5 x 2

−5 −10

Figure 5.39 17. On the interval 0 ≤ x ≤ 5, the entire graph lies above the x-axis, so the value of the integral is positive and equal to the area between the graph of the function and the x-axis. This area appears to be about half the area of the rectangle with area 2 ⋅ 5 = 10, so we estimate the area to be approximately 5. See Figure 5.40. Since f (x) is positive on this interval, the value of the integral is equal to this area, so we have Ê0

f (x)dx = Area ≈ 5.

The correct match for this function is III. 2 1 x 5 −1

Figure 5.40

390

Chapter Five /SOLUTIONS y

18. (a) y = f (x) 2 A1

−2

(b) A1 =

Ê−2

f (x) dx = 2.667. 1

f (x) dx = 0.417. Ê0 So total area = A1 + A2 ≈ 3.084. Note that while A1 and A2 are accurate to 3 decimal places, the quoted value for A1 + A2 is accurate only to 2 decimal places.

A2 = −

(c)

Ê−2

f (x) dx = A1 − A2 = 2.250.

19. (a)

Ê−3

f (x) dx = −2.

A A f (x) dx = −2 + 2 − = − . Ê3 Ê−3 Ê0 Ê−3 2 2 20. It appears that f (x) is positive on the whole interval 0 ≤ x ≤ 20, so we have (b)

f (x) dx =

f (x) dx +

Area =

Ê0

f (x)dx.

We estimate the value of the integral using left and right sums: Left-hand sum = 15 ⋅ 5 + 18 ⋅ 5 + 20 ⋅ 5 + 16 ⋅ 5 = 345. Right-hand sum = 18 ⋅ 5 + 20 ⋅ 5 + 16 ⋅ 5 + 12 ⋅ 5 = 330. A better estimate of the area is the average of the two: Area ≈

345 + 330 = 337.5. 2

21. We can compute each integral in this problem by fnding the di˙erence between the area that lies above the x-axis and the area that lies below the x-axis on the given interval. (a) For ∫0 f (x) dx, on 0 ≤ x ≤ 1 the area under the graph is 1; on 1 ≤ x ≤ 2 the areas above and below the x-axis are 2

equal and cancel each other out. Therefore, ∫0 f (x) dx = 1. (b) On 3 ≤ x ≤ 7 the graph of f (x) is the upper half circle of radius 2 centered at (5, 0). The integral is equal to the area between the graph of f (x) and the x-axis, which is the area of a semicircle of radius 2. This area is 2 , and so 2

Ê3

Ê2

f (x) dx =

22 = 2 . 2

(c) On 2 ≤ x ≤ 7 we are looking at two areas: We already know that the area of the semicircle on 3 ≤ x ≤ 7 is 2 . On 2 ≤ x ≤ 3, the graph lies below the x-axis and the area of the triangle is 21 . Therefore, 1 f (x) dx = − + 2 . 2

(d) For this portion of the problem, we can split the region between the graph and the x-axis into a quarter circle on 5 ≤ x ≤ 7 and a trapezoid on 7 ≤ x ≤ 8 below the x-axis. The semicircle has area , the trapezoid has area 3∕2. Therefore, 8 3 f (x) dx = − . Ê5 2

5.3 SOLUTIONS 22. A graph of y = 6x3 − 2 shows that this function is nonnegative on the interval x = 5 to x = 10. Thus, Area =

Ê5

(6x3 − 2) dx = 14,052.5.

The integral was evaluated on a calculator.

23. The graph of y = 5 ln(2x) is above the line y = 3 for 3 ≤ x ≤ 5. See Figure 5.41. Therefore Ê3

Area =

(5 ln(2x) − 3) dx = 14.688.

The integral was evaluated on a calculator. y

y = 5 ln(2x)

y=3 x 3

Figure 5.41 24. The graph of y = sin x + 2 is above the line y = 0.5 for 6 ≤ x ≤ 10. See Figure 5.42. Therefore Area =

Ê6

sin x + 2 − 0.5 dx = 7.799.

The integral was evaluated on a calculator. y

y = sin x + 2

y = 0.5 x 6

Figure 5.42 25. The graph of y = cos x + 7 is above y = ln(x − 3) for 5 ≤ x ≤ 7. See Figure 5.43. Therefore Area = The integral was evaluated on a calculator.

Ê5

cos x + 7 − ln(x − 3) dx = 13.457.

391

392

Chapter Five /SOLUTIONS y y = cos x + 7

y = ln(x − 3) x 7

Figure 5.43 √ 4 26. The graph of y = x4 − 8 has intercepts x = ± 8. See Figure 5.44. Since the region is below the x-axis, the integral is negative, so √ 4

Area = −

The integral was evaluated on a calculator.

8 (x4 − 8) dx = 21.527. √ 4 Ê− 8

y y = x4 − 8

√ 4 − 8

√ 4

Figure 5.44

27. The graph of y = −ex + e2(x−1) has intercepts where ex = e2(x−1) , or where x = 2(x − 1), so x = 2. See Figure 5.45. Since the region is below the x-axis, the integral is negative, so Area = −

Ê0

−ex + e2(x−1) dx = 2.762.

The integral was evaluated on a calculator. y y = −ex + e2(x−1)

Figure 5.45

5.3 SOLUTIONS 28.

Area= 3.99

❄ 1

✛

393

f (x) = x x−3 x

Area= 0.65

−2

Figure 5.46 Using a calculator or computer, we see that Ê1

x2 − 3 dx ≈ 3.34. x

√ x2 − 3 is shown in Figure 5.46. The function is negative to the left of x = 3 ≈ 1.73 and positive x to the right of x = 1.73. We compute

The graph of f (x) =

Ê1

1.73

and

x2 − 3 dx = −Area below axis ≈ −0.65 x

x2 − 3 dx = Area above axis ≈ 3.99. Ê1.73 x See Figure 5.46. Then Ê1

x2 − 3 dx = Area above axis−Area below axis = 3.99 − 0.65 = 3.34. x

29. Using a calculator or computer, we fnd Ê1

(x − 3 ln x) dx ≈ −0.136.

The function f (x) = x − 3 ln x crosses the x-axis at x ≈ 1.86. See Figure 5.47.

f (x) = x − 3 ln x 1 2

✻

−1

1 A1

✻

x 5

Figure 5.47 We fnd Ê1

1.86

(x − 3 ln x) dx = Area above axis ≈ 0.347 4

Ê1.86

(x − 3 ln x) dx = −Area below axis ≈ −0.483.

Thus, A1 ≈ 0.347 and A2 ≈ 0.483, so Ê1

(x − 3 ln x) dx = Area above axis−Area below axis ≈ 0.347 − 0.483 = −0.136.

394

Chapter Five /SOLUTIONS

30. We frst fnd where the graph of f (x) = x − 5 crosses the x-axis by solving the equation x − 5 = 0, so at x = 5. Making a sketch. In Figure 5.48 we see that the graph is composed of two right triangles. The area of the triangle that lies above the x-axis is 12.5 and the area of the triangle that lies below the x-axis is also 12.5. The integral is equal to the area of the triangle above the x-axis minus the area of the triangle below the x-axis. Thus, Ê0

(x − 5) dx = 12.5 − 12.5 = 0.

5 f (x) +12.5 x 5

−12.5

−5

Figure 5.48 31. We frst fnd where the graph of f (x) = 6 − 2x crosses the x-axis by solving the equation 6 − 2x = 0, so at x = 3. Making a sketch. In Figure 5.49 we see that the graph is composed of two right triangles. The area of the triangle that lies above the x-axis is 9 and the area of the triangle that lies below the x-axis is 25. The integral is equal to the area of the triangle above the x-axis minus the area of the triangle below the x-axis. Thus, Ê0

(6 − 2x) dx = 9 − 25 = −16.

−25 f (x) −10

Figure 5.49 32. We frst fnd where the graph of f (x) = 12 x + 3 crosses the x-axis by solving the equation 12 x + 3 = 0, so at x = −6. Making a sketch. In Figure 5.50 we see that the graph is composed of two right triangles. The area of the triangle that lies above the x-axis is 36 and the area of the triangle that lies below the x-axis is 1. The integral is equal to the area of the triangle above the x-axis minus the area of the triangle below the x-axis. Thus, 6 1 x + 3 dx = 36 − 1 = 35. Ê−8 2 6 f (x)

−1 −8

+36

❄

−6

x −1

Figure 5.50

5.3 SOLUTIONS

395

−4x − 16 −4x − 16 crosses the x-axis by solving the equation 33. We frst fnd where the graph of f (x) = = 0, so at x = −4. 3 3 Making a sketch. In Figure 5.51 we see that the graph is composed of two right triangles. The area of the triangle that 50 lies above the x-axis is 24 and the area of the triangle that lies below the x-axis is . 3 The integral is equal to the area of the triangle above the x-axis minus the area of the triangle below the x-axis. Thus, 1

Ê−10

−4x − 16 50 22 dx = 24 − = . 3 3 3 8 f (x) +24

−10

−4

− 50 3 − 20 3

Figure 5.51 34. The graph of y = x2 − 2 is shown in Figure 5.52, and the relevant area is shaded. If you compute the integral ∫0 (x2 − 2)dx, you fnd that 3

(x2 − 2)dx = 3.0. Ê0 However, since part of the area lies below the x-axis and part of it lies above the x-axis, this computation does not help us at all. (In fact, it is clear from the graph that the shaded area is more than 3.) We have to fnd the area above the x-axis and the area below the x-axis separately. We fnd that the graph crosses the x-axis at x = 1.414, and we compute the two areas separately: 1.414

(x2 − 2)dx = −1.886 and (x2 − 2)dx = 4.886. Ê0 Ê1.414 As we expect, we see that the integral between 0 and 1.414 is negative and the integral between 1.414 and 3 is positive. The total area shaded is the sum of the absolute values of the two integrals: Area shaded = 1.886 + 4.886 = 6.772

square units.

y 7 6 5 4 3 2 1 −1 −2

y = x2 − 2

x 2

Figure 5.52 35. (a) The graph of f (x) = x3 − x crosses the x-axis at x = 1 since solving f (x) = x3 − x = 0 gives x = 0 and x = ±1. See Figure 5.53. To fnd the total area, we fnd the area above the axis and the area below the axis separately. We have Ê0

(x3 − x)dx = −0.25

and

Ê1

(x3 − x)dx = 16.

As expected, the integral from 0 to 1 is negative. The area above the axis is 16 and the area below the axis is 0.25 so Total area = 16.25.

Chapter Five /SOLUTIONS

396

(b) We have Ê0

(x3 − x)dx = 15.75.

Notice that the integral is equal to the area above the axis minus the area below the axis, as expected. (c) No, they are not the same, since the integral counts area below the axis negatively while total area counts all area as positive. The two answers are not expected to be the same unless all the area lies above the horizontal axis. y

f (x) = x3 − x

x 1

Figure 5.53

36.

Ê0

cos

√ x dx = 0.80 = Area A1 – Area A2 . See Figure 5.54. 1

cos A1

√

x A2 4 ( 2 )2

❄

−1

Figure 5.54

37. We know Ê0

f (x) dx = Area above x-axis − Area below x-axis

between x = 0 and x = b. Therefore, to make the area below the x-axis is largest.

Ê0

f (x) dx largest, we need to choose b so the area above the x-axis minus b

f (x) dx as we move the upper limit b to the right starting from b = 0. Ê0 If we move b to the right over an interval where f (x) > 0 then more area above the x-axis is included between the We consider what happens to the value of b

f (x) dx gets larger. Ê0 If we move b to the right over an interval where f (x) < 0 then more area below the x-axis is included between the

limits of integration, so

Ê0

f (x) dx gets smaller. b

Ê0 ure 5.55, this gives either b = 1 or b = 4 We know Therefore, to make

f (x) dx largest, we choose b at a point where

Ê0

f (x) dx = Area between 0 and 1

f (x) dx stops getting larger. From Fig-

5.3 SOLUTIONS and

397

f (x) dx = Area between 0 and 1 + Area between 3 and 4 − Area between 1 and 3. Ê0 From Figure 5.55 we see the area between x = 1 and x = 3 is larger than the area between x = 3 and x = 4, so the area under the x-axis is larger in

Ê0

f (x) dx than in

Ê0

f (x) dx. Therefore,

Ê0

f (x) dx <

Ê0

f (x) dx and so b = 1.

x 1

f (x)

Figure 5.55 38. We know Êa

f (x) dx = Area above x-axis − Area below x-axis

between x = a and x = 4. Therefore, to make minus the area below the x-axis is smallest.

Êa

f (x) dx smallest, we need to choose a so the area above the x-axis

f (x) dx as we move the lower limit a to the left starting from a = 4. Êa If we move a to the left over an interval where f (x) > 0 then more area above the x-axis is included between the We consider what happens to the value of 4

f (x) dx gets larger. Êa If we move a to the left over an interval where f (x) < 0 then more area below the x-axis is included between the

limits of integration, so

Êa

f (x) dx gets smaller. 4

Êa Figure 5.56, this gives either a = 1 or a = 4. We know Therefore, to make

f (x) dx smallest, we choose a at a point where

Ê4 and

Êa

f (x) dx stops getting smaller. From

f (x) dx = 0

f (x) dx = Area between 3 and 4 − Area between 1 and 3. Ê1 From Figure 5.56 we see the area under the x-axis between x = 1 and x = 3 is larger than the area above the x-axis between x = 3 and x = 4, so

Ê1

f (x) dx is negative. Therefore, the integral

x 1

f (x)

Figure 5.56

Ê1

f (x) dx is smallest, so a = 1.

Chapter Five /SOLUTIONS

398

39. Since Êa

f (x) dx = Area above x-axis − Area below x-axis

between x = a and x = b, we need to choose a and b so that the total area above the x-axis minus the total area below the x-axis is largest. b

f (x) dx as we move the upper limit b to the right starting Êa from b = a. If we move b to the right over an interval where f (x) > 0 then more area above the x-axis is included between the For a fxed a, we consider what happens to the value of

f (x) dx gets larger. Êa If we move b to the right over an interval where f (x) < 0 then more area below the x-axis is included between the

limits of integration, so

Êa

f (x) dx gets smaller. b

f (x) dx stops getting larger. From FigÊa Êa ure 5.57, this gives either b = 1 or b = 4. By similar reasoning, for a fxed b, if we move a to the left from a = b over an interval where f (x) > 0, then Therefore, to make

Êa

f (x) dx largest, we choose b at a point where

f (x) dx gets larger and if we move a to the left over an interval where f (x) < 0, then b

Êa ure 5.57, this gives either a = 0 or a = 3. Since a < b, this gives three possible integrals: Therefore, to make

f (x) dx largest, we choose a at a point where

Ê0

f (x) dx,

Ê0

f (x) dx, or

Ê3

Êa

f (x) dx gets smaller.

f (x) dx stops getting larger. From Fig-

f (x) dx.

From Figure 5.57, the area between x = 3 and x = 4 is less than that between x = 0 and x = 1, so largest. For the remaining two integrals, We know Ê0 and Ê0

Ê3

f (x) dx cannot be

f (x) dx = Area between 0 and 1

f (x) dx = Area between 0 and 1 + Area between 3 and 4 − Area between 1 and 3.

From Figure 5.57 we see the area between x = 1 and x = 3 is larger than the area between x = 3 and x = 4, so the area under the x-axis is larger in b = 1.

Ê0

f (x) dx than in

Ê0

f (x) dx. Therefore,

x 1

f (x)

Figure 5.57

Ê0

f (x) dx <

Ê0

f (x) dx and so a = 0 and

5.3 SOLUTIONS 40. Since Êa

399

f (x) dx = Area above x-axis − Area below x-axis

between x = a and x = b, we need to choose a and b so that the total area above the x-axis minus the total area below the x-axis is smallest. For a fxed a, we consider what happens to the value of

Êa

f (x) dx as we move the upper limit b to the right starting

from b = a. If we move b to the right over an interval where f (x) > 0 then more area above the x-axis is included between the b

f (x) dx gets larger. Êa If we move b to the right over an interval where f (x) < 0 then more area below the x-axis is included between the

limits of integration, so

Êa

f (x) dx gets smaller. b

f (x) dx smallest, we choose b at a point where f (x) dx stops getting smaller. From Êa Êa Figure 5.58, this gives either b = 3 or b = 5. By similar reasoning, for a fxed b, if we move a to the left from a = b over an interval where f (x) > 0, then Therefore, to make

Êa

f (x) dx gets larger and if we move a to the left over an interval where f (x) < 0, then b

Êa

f (x) dx gets smaller.

f (x) dx smallest, we choose a at a point where f (x) dx stops getting smaller. From Êa Êa Figure 5.58, this gives either a = 1 or a = 4. Since a < b, this gives three possible integrals: Therefore, to make

Ê1

f (x) dx,

Ê1

f (x) dx, or

Ê4

f (x) dx.

From Figure 5.58, the area below the x-axis between x = 4 and x = 5 is less than that between x = 1 and x = 3, so Ê4

f (x) dx cannot be smallest. We know Ê1

and Ê1

f (x) dx = − Area between 1 and 3

f (x) dx = Area between 3 and 4 − Area between 1 and 3 − Area between 4 and 5.

From Figure 5.58 we see the area between x = 4 and x = 5 is larger than the area between x = 3 and x = 4, so the area under the x-axis is larger in

Ê1

f (x) dx than in

Ê1

f (x) dx. Therefore,

x 1

f (x)

Figure 5.58

Ê1

f (x) dx is smallest, so a = 1 and b = 5.

400

Chapter Five /SOLUTIONS

Solutions for Section 5.4 1. The integral ∫1 v(t) dt represents the change in position between time t = 1 and t = 3 seconds; it is measured in meters. 3

2. The integral ∫0 a(t) dt represents the change in velocity between times t = 0 and t = 6 hours; it is measured in km/hr. 6

3. The integral ∫2005 f (t) dt represents the change in the world’s population between the years 2005 and 2011. It is measured in billions of people. 2011

4. The integral ∫0 s(x) dx represents the change in salinity (salt concentration) in the frst 5 cm of water; it is measured in gm/liter. 5

5. Since v(t) ≥ 0 for 0 ≤ t ≤ 3, we can fnd the total distance traveled by integrating the velocity from t = 0 to t = 3. Using a calculator to compute the integral gives Distance =

Ê0

ln(t2 + 1) dt = 3.406 ft.

6. For any t, consider the interval [t, t + Δt]. During this interval, oil is leaking out at an approximately constant rate of f (t) gallons/minute. Thus, the amount of oil which has leaked out during this interval can be expressed as Amount of oil leaked = Rate × Time = f (t) Δt and the units of f (t) Δt are gallons/minute × minutes = gallons. The total amount of oil leaked is obtained by adding all these amounts between t = 0 and t = 60. (An hour is 60 minutes.) The sum of all these infnitesimal amounts is the integral Total amount of oil leaked, in gallons

Ê0

f (t) dt.

7. (a) On day 12 pollution is removed from the lake at a rate of 500 kg∕day. (b) The limits of the integral are t = 5 and t = 15. Since t is time in days, the units of the 5 and 15 are days. The units of the integral are obtained by multiplying the units of f (t), kg∕day, by the units of dt, day. Thus the units of the integral are kg × day = kg. day The 4000 has units of kilograms. (c) The integral of a rate gives the total change. Here f (t) is the rate of change of the quantity of pollution that has been removed from the lake. The integral gives the change in the quantity of pollution that has been removed during the time interval; in other words, the total quantity removed during that time period. During the 10 days from day 5 to day 15, a total of 4000 kg were removed from the lake. 8. (a) On April 4, the total number of Covid-19 cases in California is increasing at a rate of 821 cases per day. (b) The units of the 2 and 22 are days, the units of 23,044 are cases. (c) Between April 2 and April 22, California had a total of 23,044 new Covid-19 cases. 9. (a) We integrate the rate of change to fnd the total change, so Total number of new Covid-19 cases between t = 0 and t = 35 =

Ê0

r(t) dt.

(b) The table is giving intervals with Δt = 7, and there are n = 5 intervals in total. The right-Riemann sum is Right sum = 7179 ⋅ 7 + 5901 ⋅ 7 + 5139 ⋅ 7 + 4188 ⋅ 7 + 3200 ⋅ 7 = 179,249, which tells us that there were approximately 179,249 cases in Texas between t = 0 and t = 35 (August 5 and September 9, 2020). (c) Since the rate r(t) seems to be decreasing, and we are computing a right-hand sum, we expect to the estimate in part (b) to be an underestimate. 10. We use left- and right-hand sums to estimate the total amount of coal produced during this period: Left sum = (1.132)2 + (1.147)2 + (1.075)2 + (1.096)2 + (0.985)2 + (0.897)2 = 12.664. Right sum = (1.147)2 + (1.075)2 + (1.096)2 + (0.985)2 + (0.897)2 + (0.775)2 = 11.950.

5.4 SOLUTIONS

401

We see that

12.664 + 11.950 = 12.307 billion tons. 2 The total amount of coal produced is the defnite integral of the rate of coal production r = f (t) given in the table. Since t is in years since 2005, the limits of integration are t = 0 and t = 12. We have Total amount of coal produced ≈

Total amount of coal produced =

Ê0

f (t) dt billion tons.

11. (a) The integral ∫2 H(t) dt represents the total US emissions of super greenhouse gases, in millions of metric tons of carbon-dioxide equivalent, during the period 2002 to 2014. (b) We estimate the integral using left and right sums: 14

Left sum = (142.3)2 + (139.6)2 + (144.1)2 + (157.5)2 + (164.0)2 + (170.1)2 = 1835.2 million metric tons. Right sum = (139.6)2 + (144.1)2 + (157.5)2 + (164.0)2 + (170.1)2 + (180.1)2 = 1910.8 million metric tons. We can also average the left and right estimates to get another estimate of the integral: Ê2

H(t) dt ≈

1835.2 + 1910.8 = 1873 millions of metric tons. 2

12. (a) The plane is going up when the velocity is positive and down when the velocity is negative. Thus, it is going up between 0 and 90 seconds and down between 100 and 190 seconds. (b) The velocity is given every ten seconds. A left sum estimate for the maximum altitude of the plane above the airport is Altitude above airport = (10+20+60+490+890+980+830+970+300+10) ft/min ⋅(1 min ∕60 sec )⋅(10 sec ) = 760 ft. (The velocity is neither increasing nor decreasing over the trip, so we do not know if the left estimate is an over- or underestimate.) Thus, the maximum altitude, including the altitude of the airport, is 760 + 110 = 870 ft. 13. (a) In 2008, when t = 0, gas production was 2965.96 billion cubic meters. In 2022, when t = 14, gas production was N = 2965.96 + 81.05 ⋅ 14 = 4628.7 billion cubic meters. (b) We use an integral to approximate the total amount produced over the 10-year period from 2018 to 2028: 20

Total amount of gas produced =

Ê10

(2965.96 + 81.05t) dt = 41,817 billion cubic meters.

14. The quantity S gives the rate at which PV cells were manufactured at time t in years since 2018. The integral of S with respect to t gives the total energy-generating capacity for the PV cells manufactured during the period from 2019 to 2024: Total PV energy-generating capacity =

Ê1

87.55e0.17t dt = 817.76 gigawatts.

15. The particle is moving to the right at a constant speed of 10 feet per second. In 4 seconds it moves 40 feet. This is also given by the area under the curve. 16. The particle moves right from t = 0 to t = 4 because the velocity graph is above the t-axis. The distance moved is the area under the velocity graph. 1 Distance = ⋅ 4 ⋅ 10 = 20 feet. 2 The particle moves 20 feet right. 17. The particle moves right for 2 seconds at 10 ft/sec, then to the left for 1 second at 10 ft/sec, and then sits still for 1 second. The particle ends up at Position = 10 ⋅ 2 − 10 ⋅ 1 = 10 feet. Thus the particle ends up 10 feet to the right. 18. The particle moves right by 10 ⋅ 2 = 20 feet and then left by Distance = 10 ⋅ 2 − Thus the particle ends up 10 feet to the right.

1 ⋅ 2 ⋅ 10 = 10 feet. Thus 2 1 ⋅ 2 ⋅ 10 = 10 feet. 2

402

Chapter Five /SOLUTIONS

19. The velocity is constant and negative, so the change in position is −3 ⋅ 5 cm, that is, 15 cm to the left. Since the particle does not change direction, total distance is 15 cm. 20. From t = 0 to t = 3 the velocity is constant and positive, so the change in position is 2 ⋅ 3 cm, that is, 6 cm to the right. From t = 3 to t = 5, the velocity is negative and constant, so the change in position is −3 ⋅ 2 cm, that is, 6 cm to the left. Thus, the total change in position is 0. The particle moves 6 cm to the right, followed by 6 cm to the left, and returns to where it started. For total distance, we add the distance traveled in each direction, giving 6 + 6 = 12 cm. 21. From t = 0 to t = 5 the velocity is positive so the change in position is to the right. The area under the velocity graph gives the distance traveled. The region is a triangle, and so has area (1∕2)bℎ = (1∕2)5 ⋅ 10 = 25. Thus, the change in position is 25 cm to the right. Since the particle does not change direction, total distance is also 25 cm. 22. From t = 0 to t = 4 the velocity is positive so the change in position is to the right. The distance traveled to the right is the area of the triangle above the axis, (1∕2)bℎ = (1∕2)4 ⋅ 8 = 16. Thus, the change in position is 16 cm to the right for t = 0 to t = 4. From t = 4 to t = 5, the velocity is negative so the change in position is to the left. The distance traveled to the left is given by the area of the triangle below the axis, (1∕2)bℎ = (1∕2)1 ⋅ 2 = 1. Thus, the total change in position is 16 − 1 = 15 cm to the right. For total distance, we add the distance traveled in each direction giving 16 + 1 = 17 cm. 23. The change in the number of acres is the integral of the rate of change. We have Change in number of acres =

Ê0

√ (8 t)dt = 627.

The number of acres the fre covers after 24 hours is the original number of acres plus the change, so we have Acres covered after 24 hours = 2000 + 627 = 2627 acres. 24. Since W is in tons per week and t is in weeks since January 1, 2016, the integral ∫0 W dt gives the amount of waste, in tons, produced during the year 2016. 52

Total waste during the year =

Ê0

3.75e−0.008t dt = 159.5249 tons.

Since waste removal costs $150∕ton, the cost of waste removal for the company was 159.5249 ⋅ 150 = $23,928.74. 25. (a) The amount when t = 0 is four times the maximum acceptable limit, or 4 ⋅ 0.6 = 2.4. Therefore the level of radiation is given by R(t) = 2.4(0.996)t . We want to fnd the value of t making R(t) = 0.6. The graph of the function in Figure 5.59 shows that R(t) = 0.6 at about t = 346, so the level of radiation reaches an acceptable level after about 346 hours (or about 2 weeks). Alternatively, using logarithms gives 0.6 = (0.996)t 2.4 ln(0.25) = ln(0.996)t = t ln(0.996) ln(0.25) t= = 346. ln(0.996)

5.4 SOLUTIONS

403

2.4

R(t) = 2.4(0.996)t

0.6 t 346

Figure 5.59 (b) Total radiation emitted between t = 0 and t = 346 is Ê0

346

2.4(0.996)t dt ≈ 449 millirems.

26. Notice that the area of a square on the graph represents 10∕6 miles. At t = 1∕3 hours, v = 0. The area between the curve 1∕3 v and the t-axis over the interval 0 ≤ t ≤ 1∕3 is − ∫0 v dt ≈ 5∕3. Since v is negative here, she is moving toward the lake. At t = 1∕3, she is about 5 − 5∕3 = 10∕3 miles from the lake. Then, as she moves away from the lake, v is positive for 1∕3 ≤ t ≤ 1. At t = 1, Ê0

v dt =

Ê0

1∕3

v dt +

Ê1∕3

5 10 35 = = 11.667 miles, v dt ≈ − + 8 ⋅ 3 6 3

and the cyclist is about 5 + 35∕3 = 50∕3 = 16.667 miles from the lake. Since, starting from the moment t = 1∕3, she moves away from the lake, the cyclist will be farthest from the lake at t = 1. The maximal distance equals 16.667 miles. 27. From t = 0 to t = 3, you are moving away from home (v > 0); thereafter you move back toward home. So you are the farthest from home at t = 3. To fnd how far you are then, we can measure the area under the v curve as about 9 squares, or 9 ⋅ 10 km/hr ⋅ 1 hr = 90 km. To fnd how far away from home you are at t = 5, we measure the area from t = 3 to t = 5 as about 25 km, except that this distance is directed toward home, giving a total distance from home during the trip of 90 − 25 = 65 km. 28. The total number of “worker-hours” is equal to the area under the curve. The total area is about 14.5 boxes. Since each box represents (10 workers)(8 hours) = 80 worker-hours, the total area is 1160 worker-hours. At $10 per hour, the total cost is $11,600. 29. The time period 9am to 5pm is represented by the time t = 0 to t = 8 and t = 24 to t = 32. The area under the curve, or total number of worker-hours for these times, is about 9 boxes or 9(80) = 720 worker-hours. The total cost for 9am to 5pm is (720)(10) = $7200. The area under the rest of the curve is about 5.5 boxes, or 5.5(80) = 440 worker-hours. The total cost for this time period is (440)(15) = $6600. The total cost is about 7200 + 6600 = $13,800. 30. The area under A’s curve is greater than the area under B’s curve on the interval from 0 to 6, so A had the most total sales in the frst 6 months. On the interval from 0 to 12, the area under B’s curve is greater than the area under A, so B had the most total sales in the frst year. At approximately nine months, A and B appear to have sold equal amounts. Each square corresponds to 2 × 10 = 20 sales. There are about 8.5 boxes under the graph of A’s rate, so A’s sales for the year are about 8.5 × 20 = 170. There are about 12.5 boxes under the graph of B’s rate, so B’s sales for the year are about 12.5 × 20 = 250. 31. The change in height of each tree from t = 0 is the area under the tree’s rate of growth graph. Since they start at the same height, looking at the areas, we see that after 5 years Tree B is taller while Tree A is taller after 10 years. 32. We use the graph to fnd the energy produced by a 1-square-meter solar panel, in watt-hours, by estimating the area under the curve. This area is approximately 24 grid rectangles, and each grid rectangle has width 2 hours and height 50 w/m2 , so an area of 100 watt-hours/m2 . Thus, Energy from solar radiation in one day ≈ 2400 watt-hours/m2 . Since 18% is converted to energy, over the entire day the energy generated is Energy produced in one day ≈ 2400(0.18) = 432 watt-hours/m2 . Therefore, a 20 square-meter-sized solar array generates approximately 20 ⋅ 432 = 8640 watt-hours or 8.6 kwh.

404

Chapter Five /SOLUTIONS

33. The change in the amount of water is the integral of rate of change, so we have Number of liters pumped out =

Ê0

(5 − 5e−0.12t )dt = 258.4 liters.

Since the tank contained 1000 liters of water initially, we see that Amount in tank after one hour = 1000 − 258.4 = 741.6 liters.

34. (a) Since velocity is the rate of change of distance, we have Distance traveled =

Ê0

(10 + 8t − t2 ) dt.

This distance is the shaded area in Figure 5.60. (b) A graph of this velocity function is given in Figure 5.60. Finding the distance traveled is equivalent to fnding the area under this curve between t = 0 and t = 5. We estimate that this area is about 100 since the average height appears to be about 20 and the width is 5. (c) We use a calculator or computer to calculate the defnite integral Distance traveled =

Ê0

(10 + 8t − t2 ) dt ≈ 108.33

velocity (m/sec)

meters.

v(t) = 10 + 8t − t2

25 20 15

✛

Distance traveled = Area shaded

10 5 t

0 1

Figure 5.60: A velocity function 35. (a) Solving v(t) = −4t2 + 16t = 0 gives t = 0 and t = 4. At t = 0 the jump has just started; at t = 4 the jumper is momentarily stopped before starting back up. In Figure 5.61, we see the velocity is positive from t = 0 to t = 4 and negative from t = 4 to t = 5. The total number of meters traveled is given by the total area between the velocity curve and the t-axis. We fnd the total area by fnding the areas above and below the axis separately. We have Ê0

(−4t2 + 16t) dt = 42.6667

and

Ê4

(−4t2 + 16t) dt = −9.3333.

The jumper traveled 42.6667 meters downward and then 9.3333 meters upward, so Total number of meters traveled = 42.6667 + 9.3333 = 52 meters. (b) Since the jumper fell 42.6667 meters downward and came back up 9.3333 meters, Ending position = 42.6667 − 9.3333 = 33.333 meters below the starting point. (c) We fnd that 5

(−4t2 + 16t) dt = 33.333 meters. Ê0 Thus, the net change in position is a positive 33.333 meters, or 33.333 meters below the starting point. The net change is given by the integral, while the total number of meters traveled is given by the total area.

5.4 SOLUTIONS

405

v(t) = −4t2 + 16t 15 10 5 x 4

−5 −10 −15

Figure 5.61

36. The total amount of antibodies produced is Total antibodies =

Ê0

r(t)dt ≈ 1.417 thousand antibodies

The integral can be evaluated on a calculator or computer. 37. (a) April has 30 days and May has 31 days. The integral representing the total number of new cases in April and May is 61 ∫0 r(t) dt. (b) See Figure 5.62. (c) A lower estimate is 61,000 cases since each of the 61 days had more than 1000 cases per day. An upper estimate is 183,000 cases, since each day had less than 3000 cases per day. r (cases per day) 7000 5000 3000 1000 20

t (days)

Figure 5.62

38. (a) The integral representing the total number of new cases between November 5, 2020 (t = 35) and January 20, 2021 (t = 111) is 111

Total new Covid cases between day 35 and day 111 =

Ê35

r(t) dt.

(b) See Figure 5.63. (c) The lower estimate for the total number of new cases is given by the area of the rectangle with base from t = 35 to t = 111 and height given by the lowest number of cases per day between t = 35 and t = 111. This gives (see Figure 5.64) Lower estimate ≈ (111 − 35)( 2700 ) = 205,200 cases «¯¬ «¯¬ Base

Height

which tells us that a lower estimate for the total number of new Covid cases in Missouri between days t = 35 and t = 111 is 205,200 cases, so about 200 thousand cases. Similarly, to fnd the upper estimate we take the height of the rectangle to be the highest number cases per day in the given period. Thus (see Figure 5.64) Upper estimate ≈ (111 − 35)(5200) = 395,200 cases which tells us that an upper estimate for the total number of new Covid cases in Missouri between days t = 35 and t = 111 is 395,200 cases, so about 400 thousand cases.

Chapter Five /SOLUTIONS

406

r (cases per day) 6000 4000 2000

111

150

t (days)

Figure 5.63 r (cases per day) 6000 5200 2700 1000 35

111

150

t (days)

Figure 5.64

39. (a) The length growth rate is the derivative of the length function. A graph of the length function has an infection point when its derivative has a maximum. (b) The total growth in length during the forty week gestation equals the integral of the growth rate over the time period. Thus, the birth length equals the area under the graph. The region consists of a triangle (between t = 2 and t = 14) of base 12 weeks and height 2.3 cm/week, another triangle (between t = 14 and t = 40) of base 26 weeks and height 1.8 cm/week, and a rectangle (between t = 14 and t = 40) of length 26 weeks and height 0.5 cm/week. Thus, approximately Total length =

1 1 12(2.3) + 26(1.8) + 26(0.5) = 50.2 cm. 2 2

40. The quantities consumed equal the areas under the graphs of the rates of consumption. There is more area under the graph of fat consumption rate, so more fat is consumed than protein. 41. (a)

(i) The income curve shows the rate of change of the value of the fund due to infow of money. The area under the curve, 2020

I(t) dt, Ê2016 represents the total change in the value of the fund that is due to income. It is the quantity of money, in billions of dollars, that is projected to fow into the fund between 2016 and 2020. (ii) The expenditure curve shows the rate of change of the value of the fund due to outfow of money. The area under the curve, 2020

E(t) dt, Ê2016 represents the magnitude of the change in the value of the fund that is due to expenses. It is the quantity of money, in billions of dollars, that is projected to fow out of the fund between 2016 and 2020. (iii) The area between the income and expenditure curves, 2020

Ê2016

I(t) − E(t) dt,

represents the di˙erence between total income and total expenses between 2016 and 2020. It is the projected change in value of the fund between 2016 and 2020. (b) In the fgure, we see that the value of the fund was about 2790 billion dollars in 2016 and is projected to be about 2840 billion dollars in 2020. The fund is projected to increase in value by about 50 billion dollars, and that is the area between the income and expenditure curves on the graph.

5.4 SOLUTIONS

407

42. (a) When income is greater than expenses, before 2021, the value of the fund increases. When the expenses are greater than income, after 2021, the value of the fund decreases. The value increases from 2016 through 2021, and then declines. The value is a maximum in 2021, where the income and expenditure curves cross. (b) The value of the fund is represented by the height of the curve. Its highest point is in 2021, when the value is a maximum. 43. The rate of change of the value of the fund is the di˙erence I(t) − E(t) between the rate of increase due to income and the rate of decrease due to expenses. To fnd the total change in the value, integrate the rate of change: 2025

Change in value of fund =

Ê2016

I(t) − E(t) dt.

44. (a) The dam broke at 3 minutes before midnight and there was no water fowing about 70 minutes past midnight, so it took about 73 minutes to empty the reservoir. (b) The maximum discharge rate was slightly less than 1800 acre-feet per minute and occurred about 2 minutes after midnight, or about 5 minutes after the dam broke. (c) The volume of water discharged is the area under the discharge rate curve. Each square represents 200 ⋅ 10 = 2000 acre-feet, and there are approximately 22 squares under the graph. Thus Total discharge = 22 ⋅ 2000 = 44,000 acre-feet. (d) The reservoir was half empty when about 22,000 acre-feet of water had fowed out. This is represented by 11 squares and occurred sometime between 10 and 20 minutes, so at approximately 15 minutes past midnight. So it took approximately 15+3=18 minutes for the reservoir to be half empty. 45. (a) The black curve is for boys, the colored one for girls. The area under each curve represents the change in growth in centimeters. Since men are generally taller than women, the curve with the larger area under it is the height velocity of the boys. (b) Each square below the height velocity curve has area 1 cm∕yr ⋅ 1 yr = 1 cm. Counting squares lying below the black curve gives about 43 cm. Thus, on average, boys grow about 43 cm between ages 3 and 10. (c) Counting squares lying below the black curve gives about 23 cm growth for boys during their growth spurt. Counting squares lying below the colored curve gives about 18 cm for girls during their growth spurt. (d) We can measure the di˙erence in growth by counting squares that lie between the two curves. Between ages 2 and 12.5, the average girl grows faster than the average boy. Counting squares yields about 5 cm between the colored and black curves for 2 ≤ x ≤ 12.5. Counting squares between the curves for 12.5 ≤ x ≤ 18 gives about 18 squares. Thus, there is a net increase of boys over girls by about 18 − 5 = 13 cm. 46. (a) Figure 5.65 is the graph of a rate of blood fow versus time. The total quantity of blood pumped during the three hours is given by the area under the rate graph for the three-hour time interval. The area can be estimated by counting grid boxes under the graph. Each grid rectangle has area 30 minutes × 1∕2 liter/minute = 15 liters, representing 15 liters of blood pumped. The grid boxes in the graph are stacked in six columns. Estimating the number of boxes in each column under the graph gives Number of boxes = 10 + 7 + 3.75 + 3.5 + 3 + 1.5 = 28.75 boxes. Approximately Amount of blood pumped = (28.75)(15) = 431.25 liters. Thus, about 431 liters of blood are pumped during the three hours leading to death. (b) Since f (t) is the pumping rate in liters∕minute at time t hours, 60f (t) is the pumping rate in liters∕hour. Thus 3 ∫0 60f (t) dt gives the total quantity of blood pumped in liters during the three hours. (c) During three hours with no bleeding, the heart pumps 5 liters∕minute for 3 ⋅ 60 = 180 minutes. Thus Total blood pumped = 5 ⋅ 180 = 900 liters. This is 900 − 431.25 = 468.75 liters more than pumped with 2 liter bleeding. Thus, about 470 liters are pumped. This corresponds to the area on the graph between the 5 liters∕minute line and the pumping rate for the 2 liter bleed. See Figure 5.65.

408

Chapter Five /SOLUTIONS Return to normal pumping rate of heart (liters per minute)

❄

5 4 3

Bled 1 liter

2 1

Death

Bled 2 liters

✠ 1

hours

Figure 5.65

47. (a) Figure 5.66 is the graph of a rate of blood fow versus time. The total quantity of blood pumped during the three hours is given by the area under the rate graph for the three-hour time interval. The area can be estimated by counting grid boxes under the graph. Each grid rectangle has area 30 minutes × 1∕2 liter/minute = 15 liters, corresponding to 15 liters of blood pumped. The grid boxes in the graph are stacked in six columns. Estimating the number of boxes in each column under the graph gives Number of boxes = 10 + 7.75 + 6 + 7.5 + 9 + 9.75 = 50 boxes. Approximately Amount of blood pumped = 50 ⋅ 15 = 750 liters. Thus, about 750 liters of blood are pumped during the three hours leading to full recovery. 3 (b) Since g(t) is the pumping rate in liters/minute at time t hours, 60g(t) is the pumping rate in liters/hour. Thus ∫0 60g(t) dt gives the total quantity of blood pumped in liters during the three hours. (c) During three hours with no bleeding, the heart pumps 5 liters/minute for 3 ⋅ 60 = 180 minutes. Thus, Total blood pumped = 5 ⋅ 180 = 900 liters. This is 900 − 750 = 150 liters more than pumped with 1 liter bleeding. This corresponds to the area on the graph between the 5 liters/minute line and the pumping rate for the 1 liter bleed. See Figure 5.66. Return to normal pumping rate of heart (liters per minute)

❄

5 4 3

Bled 1 liter

2 1

Death

Bled 2 liters

✠ 1

hours

Figure 5.66

48. In the fgure, we see that Product B gives a greater peak concentration than Product A; Product A peaks sooner than Product B; Product B provides a greater overall exposure than Product A. Since we are looking for the product providing the faster response, Product A should be used as it peaks sooner. 49. Looking at the fgure in the problem, we note that Product A has a greater peak concentration than Product B; Product A peaks much faster that Product B; Product B provides greater overall exposure than Product A. On the other hand, Product A provides faster relief.

5.5 SOLUTIONS

409

50. See Figure 5.67: C

Product A

Product B

t (time)

Figure 5.67

51. (a) The distance traveled is the integral of the velocity, so in T seconds you fall Ê0

49(1 − 0.8187t ) dt.

(b) We want the number T for which Ê0

49(1 − 0.8187t ) dt = 5000.

We can use a calculator or computer to experiment with di˙erent values for T , and we fnd T ≈ 107 seconds.

52. (a) The acceleration is positive for 0 ≤ t < 40 and for a tiny period before t = 60, since the slope is positive over these intervals. Just to the left of t = 40, it looks like the acceleration is approaching 0. Between t = 40 and a moment just before t = 60, the acceleration is negative. (b) The maximum altitude was about 500 feet, when t was a little greater than 40 (here we are estimating the area under the graph for 0 ≤ t ≤ 42). (c) The total change in altitude for the Montgolfers and their balloon is the defnite integral of their velocity, or the total area under the given graph (counting the part after t = 42 as negative, of course). As mentioned before, the total area of the graph for 0 ≤ t ≤ 42 is about 500. The area for t > 42 is about 220. So subtracting, we see that the balloon fnished 280 feet or so higher than where it began.

Solutions for Section 5.5 1. If H(t) is the temperature of the co˙ee at time t, by the Fundamental Theorem of Calculus Change in temperature = H(10) − H(0) = Ê

H ¨ (t) dt = 0

Ê0

−7(0.9t ) dt.

Therefore, H(10) = H(0) + 900

2. The units for the integral 900

Ê800

Ê0

−7(0.9t ) dt ≈ 90 − 43.3 = 46.7◦ C.

C ¨ (q)dq are dollars tons ⋅ (tons) = dollars.

C ¨ (q)dq represents the cost of increasing production from 800 tons to 900 tons.

Chapter Five /SOLUTIONS

410

3. If we integrate the rate of change we fnd fnd the total change, so 17

Ê10

P ¨ (t) dt = P (17) − P (10).

Note that P (17) = Total number of Covid-19 cases confrmed by t = 17 P (10) = Total number of Covid-19 cases confrmed by t = 10, so Total number of Covid-19 cases confrmed

P (17) − P (10) =

from t = 10 to t = 17

17 Thus, ∫10 P ¨ (t) dt = P (17) − P (10) is the total number of cases in Colorado that were confrmed from t = 10 to t = 17. 17 The statement ∫10 P ¨ (t) dt = 6790 tells us that 6790 Covid-19 cases were confrmed in Colorado from t = 10 to

t = 17 (March 11 to March 18, 2021).

4. If we integrate the rate of change we fnd fnd the total change, so Ê7

P ¨ (t) dt = P (15) − P (7).

Note that P (15) = Total number of Covid-19 cases confrmed by t = 15 P (7) = Total number of Covid-19 cases confrmed by t = 7, so P (15) − P (7) =

Total number of Covid-19 cases confrmed from t = 7 to t = 15

15 Thus, ∫7 P ¨ (t) dt = P (15) − P (7) is the total number of cases in South Carolina that were confrmed from t = 7 to t = 15. 15 The statement ∫7 P ¨ (t) dt = 7388 tells us that 7388 Covid-19 cases were confrmed in South Carolina from t = 7 to

t = 15 (April 12 to April 20, 2021).

5. The fxed cost is C(0) = 1,000,000. Total variable cost =

Ê0

500

C ¨ (x) dx =

Ê0

500

(4000 + 10x) dx = 3,250,000.

Therefore, Total cost = Fixed Cost + Total variable cost = 4,250,000 riyals.

6. The area between 1970 and 1990 is about 15.3 grid squares, each of which has area 0.1(5) = 0.5 million people. So 1990

Change in population =

Ê1970

P ¨ (t) dt ≈ 15.3(0.5) ≈ 7.65 million people.

The population of Tokyo increased by about 8 million people between 1970 and 1990. 7. Each grid square has height 0.1 million people/year and length 4 years, so it has area 0.1(4) = 0.4 million people. The area under the curve between 2002 and 2018 is about 28.3 grid squares. Therefore, 2018

Change in population =

Ê2002

P ¨ (t) dt ≈ 28.3(0.4) ≈ 11.32 million people.

The population of Delhi increased by about 11.32 million people between 2002 and 2018.

5.5 SOLUTIONS

411

8. (a) There are approximately 5.5 squares under the curve of C ¨ (q) from 0 to 30. Each square represents $100, so the total variable cost to produce 30 units is around $550. To fnd the total cost, we add the fxed cost Total cost = fxed cost + total variable cost = 10, 000 + 550 = $10, 550. (b) There are approximately 1.5 squares under the curve of C ¨ (q) from 30 to 40. Each square represents $100, so the additional cost of producing items 31 through 40 is around $150. (c) Examination of the graph tells us that C ¨ (25) = 10. This means that the cost of producing the 26th item is approximately $10. 9. Suppose F (t) represents the total quantity of water in the water tower at time t, where t is in days since April 1. Then the graph shown in the problem is a graph of F ¨ (t). By the Fundamental Theorem, F (30) − F (0) =

Ê0

F ¨ (t)dt.

We can calculate the change in the quantity of water by calculating the area under the curve. If each box represents about 300 liters, there is about one box, or −300 liters, from t = 0 to t = 12, and 6 boxes, or about +1800 liters, from t = 12 to t = 30. Thus Ê0

F ¨ (t)dt = 1800 − 300 = 1500,

so the fnal amount of water is given by F (30) = F (0) +

Ê0

F ¨ (t)dt = 12,000 + 1500 = 13,500 liters.

10. (a) The statement means that 1 hour after the drug is administered the concentration of the medicine in the plasma is increasing at a rate of 50 mg/ml per hour. 3 (b) The quantity ∫0 ℎ(t) dt = 480 represents the increase in the concentration of the drug in the plasma during the frst three hours after it is administered. Since there is an initial concentration of 250 mg/ml when the drug is administered, at the end of three hours we have Concentration = 250 +

Ê0

ℎ(t) dt = 250 + 480 = 730 mg/ml in the plasma.

11. (a)

C ¨ (q)

150

Figure 5.68 The total variable cost of producing 150 units is represented by the area under the graph of C ¨ (q) between 0 and 150, or 150

(0.005q 2 − q + 56)dq. Ê0 (b) An estimate of the total cost of producing 150 units is given by 20,000 +

Ê0

150

(0.005q 2 − q + 56)dq.

This represents the fxed cost ($20,000) plus the variable cost of producing 150 units, which is represented by the integral. Using a calculator, we see Ê0

150

(0.005q 2 − q + 56)dq ≈ 2,775.

So the total cost is approximately $20,000 + $2,775 = $22,775.

412

Chapter Five /SOLUTIONS (c) C ¨ (150) = 0.005(150)2 − 150 + 56 = 18.5. This means that the marginal cost of the 150th item is 18.5. In other words, the 151st item will cost approximately $18.50. (d) C(151) is the total cost of producing 151 items. This can be found by adding the total cost of producing 150 items (found in part (b)) and the additional cost of producing the 151st item (C ¨ (150), found in (c)). So we have C(151) ≈ 22,775 + 18.50 = $22,793.50.

12. (a) Total variable cost in producing 400 units is Ê0

400

C ¨ (q) dq.

We estimate this integral: Left–hand sum = 25(100) + 20(100) + 18(100) + 22(100) = 8500; Right–hand sum = 20(100) + 18(100) + 22(100) + 28(100) = 8800; and so

Ê0

400

C ¨ (q) dq ≈

8500 + 8800 = $8650. 2 Total cost = Fixed cost + Variable cost = $10,000 + $8650 = $18,650.

(b) C (400) = 28, so we would expect that the 401st unit would cost an extra $28. 13. (a) If we integrate the rate of change we fnd the total change, so Ê0

P ¨ (t) dt = P (28) − P (0).

More specifcally, since P (28) = Total number of Covid-19 cases confrmed by t = 28 P (0) = Total number of Covid-19 cases confrmed by t = 0 we have P (28) − P (0) =

Total number of Covid-19 cases confrmed between t = 0 and t = 28.

28 In other words, ∫0 P ¨ (t) dt = P (28) − P (0) is the total number of cases in Illinois that were confrmed during the

frst 28 days of 2021. (b) We can use the values in the table to compute Riemann sums with n = 4 intervals and Δt = 7. The right Riemann sum is Right sum = 6380 ⋅ 7 + 6376 ⋅ 7 + 4807 ⋅ 7 + 4292 ⋅ 7 = 152,985 and the left Riemann sum is Left sum = 5469 ⋅ 7 + 6380 ⋅ 7 + 6376 ⋅ 7 + 4807 ⋅ 7 = 161,224. We average the left and right sums to obtain an estimate for the integral Ê0

P ¨ (t) dt ≈

152,985 + 161,224 = 157,104.5. 2

This tells us that there were approximately 160,000 Covid-19 cases in Illinois during the frst 28 days of 2021. 14. Since C(0) = 500, the fxed cost must be $500. The total variable cost to produce 20 units is Ê0

C ¨ (q) dq =

Ê0

(q 2 − 16q + 70) dq = $866.67 (using a calculator).

The total cost to produce 20 units is the fxed cost plus the variable cost of producing 20 units. Thus, Total cost = $500 + $866.67 = $1,366.67.

5.5 SOLUTIONS

413

15. (a) The total cost of producing 30 bicycles is the fxed cost plus the variable cost of producing 30 bicycles. Thus C(30) = C(0) + or $2000 +

Ê0

C ¨ (q)dq

15q + 600 dq ≈ $4701.22. 0.3q + 5

(b) If the bikes are sold for $200 each the total revenue from 30 bicycles is 30 ⋅ 200 = $6000, and so the total proft is $6000 − $4701.22 = $1298.78. st

(c) The marginal proft on the 31 bicycle is the di˙erence between the marginal cost of producing the 31st bicycle and the marginal revenue, which is the price. Thus the marginal proft is 200 − C ¨ (30) = 200 − 16. (a)

1050 = $125. 14

200 150

R¨ (q)

100 50 q 50

100

150

Figure 5.69 (b) By the Fundamental Theorem, Ê0

100

R¨ (q)dq = R(100) − R(0).

R(0) = 0 because no revenue is produced if no units are sold. Thus we get R(100) =

Ê0

100

R¨ (q)dq ≈ $12,000.

(c) The marginal revenue in selling the 101st unit is given by R¨ (100) = $80/unit. The total revenue in selling 101 units is: R(100) + R¨ (100) = $12,080. 17. (a) The area under the curve of P ¨ (t) from 0 to t gives the change in the value of the stock. Examination of the graph suggests that this area is greatest at t = 5, so we conclude that the stock is at its highest value at the end of the 5th week. (Some may also conclude that the area is greatest at t = 1.5, making the stock most valuable in the middle of the second week. Both are valid answers.) Since P ¨ (t) < 0 from t = 1.5 to about t = 3.8, we know that the value of the stock decreases in this interval. This is the only interval in which the stock’s value is decreasing, so the stock will reach its lowest value at the end of this interval, which is near the end of the fourth week. (b) We know that P (t) − P (0) is the area under the curve of P ¨ (t) from 0 to t, so examination of the graph leads us to conclude that P (4) < P (3) ≈ P (0) < P (1) ≈ P (2) < P (5).

Chapter Five /SOLUTIONS

414

18. The total change in the net worth of the company from 2020 (t = 0) to 2030 (t = 10) is found using the Fundamental Theorem: Ê0

Change in net worth = f (10) − f (0) =

f ¨ (t) dt =

Ê0

(2000 − 12t2 ) dt = 16,000 dollars.

The worth of the company in 2030 is the worth of the company in 2020 plus the change in worth between 2020 and 2030. Thus, in 2030, Net worth = f (10) = f (0) + Change in worth = Worth in 2020 + Change in worth between 2020 and 2030 = 40,000 + 16,000 = $56,000.

19. We fnd the changes in f (x) between any two values of x by counting the area between the curve of f ¨ (x) and the x-axis. Since f ¨ (x) is linear throughout, this is quite easy to do. From x = 0 to x = 1, we see that f ¨ (x) outlines a triangle of area 1∕2 below the x-axis (the base is 1 and the height is 1). By the Fundamental Theorem, Ê0

f ¨ (x) dx = f (1) − f (0),

so f (0) +

Ê0

f ¨ (x) dx = f (1) 1 3 = 2 2

f (1) = 2 −

Similarly, between x = 1 and x = 2 we can see that f ¨ (x) outlines a rectangle below the x-axis with area 1, so f (2) = 3∕2 − 1 = 1∕2. Continuing with this procedure (note that at x = 4, f ¨ (x) becomes positive), we get the table below.

f (x)

3∕2

1∕2

−1∕2

−1

−1∕2

1∕2

20. We fnd the changes in f (x) between any two values of x by counting the area between the curve of f ¨ (x) and the x-axis. Since f ¨ (x) is linear throughout, this is quite easy to do. From x = 0 to x = 1, we see that f ¨ (x) outlines a rectangle of area 8 above the x-axis (the width is 1 and the height is 8). By the Fundamental Theorem, Ê0

f ¨ (x) dx = f (1) − f (0),

so 1

f ¨ (x) dx = f (1) Ê0 f (1) = −10 + 8 = −2

f (0) +

Similarly, between x = 1 and x = 2 we can see that f ¨ (x) outlines a rectangle above the x-axis with area 8, so f (2) = −2 + 8 = 6. Continuing with this procedure (note that at x = 4, f ¨ (x) becomes negative, and we need to subtract area below the x-axis), we get the table below.

f (x)

−10

−2

5.5 SOLUTIONS

415

21. (a) Graph A achieves its maximum almost immediately after the dose is taken, suggesting that the maximum rate of change in concentration happens almost as soon as the drug has been administered. Graph B, however, increases gradually from 0 to its maximum value about an hour after the medicine is administered. So we choose Graph A as the rate of change graph for the immediate-release capsule and Graph B for the delayedrelease capsule. (b) Since the graph shows the rate of change of concentration, concentration is given by area under the curve. The maximum concentration occurs at the point at which the rate curve goes beneath the t-axis. The maximum concentration of the drug from the immediate-release capsule is represented by the area under Graph A and above the horizontal axis for t between 0 and approximately 0.5. We approximate this area as the area of a triangle of base 0.5 and height 3200, so for A: 1 ⋅ 3200 ng/ml per hour ⋅ 0.5 hour = 800 ng/ml. 2 Similarly, the maximum concentration of the drug from the delayed-release capsule is represented by the area under Graph B and above the horizontal axis for t between 0.5 and 2. We approximate this area as the area of a triangle of base 1.5 and height 500, so for B: Concentration =

1 ⋅ 500 ng/ml per hour ⋅ 1.5 hours = 375 ng/ml. 2 Thus, the immediate-release capsule gives a larger maximum concentration. Concentration =

22. The value of this integral tells us how much oil is pumped from the well between day t = 0 and day t = t0 . 23. The time period from t0 to 2t0 is shorter (and contained within) the time period from 0 to 2t0 . Thus, the amount of oil pumped out during the shorter time period, period,

Ê0

2t0

Êt0

2t0

r(t) dt, is less than the amount of oil pumped out in the longer timer

r(t) dt. This means Êt 0

2t0

r(t) dt <

Ê0

2t0

r(t) dt.

The length of the time period from 2t0 to 3t0 is the same as the length from t0 to 2t0 : both are t0 days. But the rate at which oil is pumped is going down, since r¨ (t) < 0. Thus, less oil is pumped out during the later time period, so 3t0

Ê2t0 3t0

We conclude that

Ê2t0

r(t) dt <

Êt0

2t0

r(t) dt <

r(t) dt < Ê0

Êt 0

2t0

r(t) dt.

2t0

r(t) dt.

24. Since C is an antiderivative of c, we know by the Fundamental Theorem that 24

Ê15

c(n) dn = C(24) − C(15) = 13 − 8 = 5.

Since C(24) is the cost to plow a 24-inch snowfall, and C(15) is the cost to plow a 15-inch snowfall, this tells us that it costs $5 million more to plow a 24-inch snowfall than a 15-inch snowfall. 25. Since C is an antiderivative of c, we know from the Fundamental Theorem that 15

c(n) dn = C(15) −C(0) Ê0 «¯¬ «¯¬ 8 7.5

so C(0) = 8 − 7.5 = 0.5. This tells us the cost of preparing for a storm, even if no snow falls, is $0.5 million, or $500,000. 26. By the Fundamental Theorem, we have 24

c(15) +

Ê15

c ¨ (n) dn = c(15) + c(24) − c(15) = c(24) = 0.4.

Since c(n) = C (n), we have C (24) = 0.4, which tells us the cost of plowing increases at the instantaneous rate of $0.4 million/inch, or by $400,000 per each additional inch, after 24 inches have already fallen.

416

Chapter Five /SOLUTIONS

27. The expression ∫0 r(t) dt represents the amount of water that leaked from the ruptured pipe during the frst two hours. 2

Likewise, the expression ∫2 r(t) dt represents the amount of water that leaked from the ruptured pipe during the next two hours. Since the leak “worsened throughout the afternoon,” we know that r is an increasing function, which means that more water leaked out during the second two hours than during the frst two hours. Therefore, 4

Ê0

r(t) dt <

Ê2

r(t) dt.

28. The expression ∫0 r(t) dt represents the amount of water that leaked from the ruptured pipe during the frst four hours. The expression r(4) represents the rate water was leaking from the pipe at time t = 4, or four hours after the leak began. Had water leaked at this rate during the whole four-hour time period, 4

Total amount of water leaked

= Rate × Time elapsed = 4r(4) gallons. «¯¬ «¯¬

at a constant rate of r(4)

r(4)

According to the article, the leak “worsened throughout the afternoon,” so the rate that water leaked out initially was less than r(4). Thus, 4r(4) overestimates the total amount of water leaked during the frst four hours. We conclude that Ê0

r(t) dt < 4r(4).

29. According to the article, the leak “worsened throughout the afternoon,” eventually reaching (before being shut o˙) 8 million gallons/hour. Thus, 0 ≤ r(t) ≤ 8 million gal/hr. The leak began around 10 am and ended after 6 pm. The expression 8 ∫0 r(t) dt represents the total amount of water leaked during the frst 8 hours (or almost the total amount of water leaked altogether). Since r(t) < 8 million gal/hr, Total amount of water leaked during frst 8 hours

< Rate (million gal/hr) × Time elapsed = 64 million gallons. «¯¬ «¯¬ 8

Thus, Ê0

r(t) dt < 64 million gallons.

30. Expression (I) gives the total number of calories burned up to time t = 10, which is the area under the graph of r(t) from t = 0 to t = 10. Expression (II) gives the total number of calories burned up to time t = 12, which is the area under the graph of r(t) from t = 0 to t = 12. Since r(t) > 0, we have that expression (II) is larger than expression (I). Expression (III) gives the area from t = 0 to t = 10, namely R(10), and then estimates the area between t = 10 to t = 12 as a rectangle with height r(10) and width 2. Since r¨ (t) < 0, the rate r(t) is decreasing, so this rectangle gives an overestimate to the actual area between t = 10 and t = 12. Thus, expression (III) is an overestimate of the actual area R(12). Thus, the order from smallest to largest is (I)<(II)<(III). 31. Expression (I) and expression (II) represent the area under the graph of r(t) over di˙erent three-hour intervals. Since R¨ (T ) = r(t), expression (III) also represents area under the graph of r(t) over a three-hour interval. We have that r(t) > 0 and r¨ (t) < 0, so r(t) is a positive, decreasing function. Thus, the areas over any three-hour interval get smaller as t increases. We have that (I) is the earliest interval, (II) is the next earliest, and (III) is the latest. Thus, the order from smallest to greatest is (III)<(II)<(I).

Solutions for Section 5.6 1. (a) Since f (x) is positive on the interval from 0 to 6, the integral is equal to the area under the curve. By examining the graph, we can measure and see that the area under the curve is 20 square units, so Ê0

f (x)dx = 20.

5.6 SOLUTIONS

417

(b) The average value of f (x) on the interval from 0 to 6 equals the defnite integral we calculated in part (a) divided by the size of the interval. Thus 6 1 1 Average Value = f (x)dx = 3 . 6 Ê0 3 2. (a) Counting the squares yields an estimate of 25 squares, each with area = 1, so we conclude that Ê0

f (x) dx ≈ 25.

(b) The average height appears to be around 5. (c) Using the formula, we get

∫0 f (x) dx 5

Average value =

5−0

≈

25 ≈ 5, 5

which is consistent with (b). 2

3. Average value =

1 1 (1 + t) dt = (4) = 2. 2 − 0 Ê0 2 10

1 1 et dt = (22025) = 2202.5 10 10 − 0 Ê0 1√ 5. (a) Average value = 1 − x2 dx = 0.79. Ê0

4. Average value =

(b) The area between the graph of y = 1 − x and the x-axis is 0.5. Because the graph of y = it lies above the line y = 1 − x, so its average value is above 0.5. See Figure 5.70. 1

0.79

√

√ 1 − x2 is concave down,

1 − x2

✻

0.5 Average value Area = 0.5

❄x 1

Figure 5.70 6. By a visual estimate, the average value is ≈ −3. 7. By a visual estimate, the average value is ≈ 8. 8. It appears that the area under a line at about y = 2.5 is approximately the same as the area under f (x) on the interval x = a to x = b, so we estimate that the average value is about 2.5. See Figure 5.71. f (x) 3

Figure 5.71

418

Chapter Five /SOLUTIONS

9. It appears that the area under a line at about y = 17 is approximately the same as the area under f (x) on the interval x = a to x = b, so we estimate that the average value is about 17. See Figure 5.72. f (x)

20 15 10 5

Figure 5.72 1 10. The expression 25−10 ∫10 r(t) dt is the average value of r(t) from t = 10 to t = 25. The statement tells us that there were on average 1417 Covid-19 cases per day from t = 10 to t = 25 (April 11 to April 26, 2021). 25

1 ∫7 r(t) dt is the average value of r(t) from t = 7 to t = 21. The statement tells us that there were on 11. The expression 21−7 average 374 Covid-19 cases per day from t = 7 to t = 21 (February 12 to February 26, 2021). 21

12. Since t = 0 in 1975 and t = 35 in 2010, we want: 35

1 225(1.15)t dt 35 − 0 Ê0 1 = (212,787) = $6080. 35

Average Value =

13. (a) The average inventory is given by the formula 1 90 − 0 Ê0

5000(0.9)t dt.

Using a calculator yields Ê0

5000(0.9)t dt ≈ 47452.5

so the average inventory is 47452.5 ≈ 527.25. 90 (b) The function is graphed in Figure 5.73. The area of a rectangle of height 527.25 is equal to the area under the curve. 5000

I(t)

527.25 t 90

Figure 5.73

5.6 SOLUTIONS

419

14. (a) When t = 10 we have P = 6.77e0.012(10) = 7.63. The predicted population of the world in the year 2020 is 7.63 billion people. (b) We fnd the average value of P on the interval t = 0 to t = 10: Average value =

1 10 − 0 Ê0

6.77e0.012t dt =

1 (71.9295) = 7.2. 10

The average population of the world over this decade is about 7.2 billion people. 15. (a) We want to fnd the integral 1 12 Ê0

Average =

10 + 5 sin

t dt. 12

Evaluating numerically gives 13.183 feet. (b) We take the limits of the integration to be from 0 to 24 hours, so again evaluating numerically we get Average =

1 24 Ê0

10 + 5 sin

t dt = 10. 12

The average depth is 10 feet. 16. (a) Figure 5.74 shows P (t), daily oil production in thousands of barrels per day as a function of the year, t. Since P (t) is linear, 1970

1 P (t) dt 1970 − 1920 Ê1920 1 = (Area triangle shaded + Area rectangle shaded) 50 1 1 ⋅ 50 ⋅ 8427 + 50 ⋅ 1210 = 5423.5 thousand barrels per day. = 50 2

Average daily production =

daily oil production (thousand barrels per day) (50, 9637)

10000

✻ P (t)

5423.5 8427 50

1210 ✛

✲ ❄ t, year

1920

1970

Figure 5.74 (b) Using the 1920 and 1970 values, we have Average daily oil production =

1 (1210 + 9637) = 5423.5 thousand barrels per day. 2

This answer is the same as the answer to part (a) because P (t) is linear. Thus, the trapezoid-shaped area under the sloped line representing P (t) equals the area of the rectangle with height 5423.5. In other words: 1970

Ê1920

P (t) dt = 50 ⋅ 5423.5.

420

Chapter Five /SOLUTIONS

17. (a) Let f (t) be the annual income at age t. Between ages 25 and 85, 85

Average annual income =

1 f (t) dt. 85 − 25 Ê25

The integral equals the area of the region under the graph of f (t). The region is a rectangle of height 40,000 dollars∕year and width 40 years, so we have 85

Ê25

f (t) dt = Height × Width = 1,600,000 dollars.

Therefore

1,600,000 = $26,667 per year. 85 − 25 (b) This person spends less than their income for their entire working life, age 25 to 65. In retirement, ages 65 to 85, they have no income but can continue spending at the same rate because they saved in their working years. Average annual income =

18. (a) Let f (t) be the annual income at age t. Between ages 25 and 85, 85

Average annual income =

1 f (t) dt. 85 − 25 Ê25

The integral equals the area of the region under the graph of f (t). The region consists of 10 full grid rectangles and 4 half grid rectangles, or 12 full rectangles in all. Each grid rectangle has an area of 10 years ⋅ 20,000 dollars∕year = 200,000 dollars. Thus, we have 85

Ê25

f (t) dt = 12(200,000) = 2,400,000 dollars,

2,400,000 = $40,000 per year. 85 − 25 (b) This person’s income increases steadily over forty years from $20,000∕year to $100,000∕year. Every working year the income increases by $2,000∕year. Their income reaches $40,000, their spending rate, when they turn 35. They spend less than their income every year from age 35 to 65. They spend more than their income every year from age 25 to 35 and from age 65 to 85. Average annual income =

19. (a) The average value is closer to 65. On the graph we see that if we draw a horizontal line at the various options, the one that gives a rectangle with area closest to to the area under the graph is R = 65. See Figure 5.75. (b) Various answers are possible. One reasonable answer is to say that the graph is more useful because it enables the consumer to select information relevant to their particular circumstances. For example, if the consumer lives in an area where the temperature rarely dips below 30◦ F, then the graph can be used to estimate the average over the temperature interval from 30◦ F to 95◦ F which is more relevant than the average computed in part (a). In fact, from the fgure we see that average value of the function restricted to the temperature interval from 30◦ F to 95◦ F is higher than of the average of the function over the full domain from −10◦ F to 95◦ F. In general, the graph contains more information than a single average, which is just a very concise summary. R (miles) Area under curve = Area in rectange 80

✻ R ≈ 65

−10 5

20 35 50 65 80 95

T (◦ F)

Figure 5.75 20. (a) Systolic, or maximum, blood pressure = 120 mm Hg. (b) Diastolic, or minimum, blood pressure = 80 mm Hg. (c) Average = (120 + 80)∕2 = 100 mm Hg (d) The pressure is at the diastolic level for a longer time than it is at or near the systolic level. Thus the average arterial pressure over the cardiac cycle is closer to the diastolic pressure than to the systolic pressure. The average pressure over the cycle is less than the average that was computed in part (c).

5.6 SOLUTIONS

421

21. The shaded region under the pressure curve in Figure 5.76 is well approximated by a rectangle, of width 1 and height 80, surmounted by a triangle of base 0.45 and height 40 (See Figure 5.77). Thus 1

Average value =

1 1 f (x) dx = Area of the shaded region ≈ 1 ⋅ 80 + (0.45)(40) = 89 mm Hg. 1 − 0 Ê0 2

Thus, the average blood pressure is about 90 mm Hg. To check the answer graphically, draw the horizontal line at average pressure 89 mm Hg as in Figure 5.78. Observe that the area of the region above the average line but below the pressure curve appears equal to the area of the region below the average line but above the pressure curve. arterial pressure (mm Hg)

arterial pressure (mm Hg) 120

Systolic pressure

✠

100

✻ 40 80 ✛ ❄ 0.45

Diastolic pressure

✲

Area of rectangle = 80

one cardiac cycle

Figure 5.77: Calculating approximate area under pressure curve

Figure 5.76: Total area under pressure curve arterial pressure (mm Hg)

Systolic pressure

✠

Average arterial pressure Diastolic pressure

one cardiac cycle

Figure 5.78: Average arterial pressure = 89 mm Hg 22. (a) Over the interval [−1, 3], we estimate that the total change of the population is about 1.5, by counting boxes between the curve and the x-axis; we count about 1.5 boxes below the x-axis from x = −1 to x = 1 and about 3 above from x = 1 to x = 3. So the average rate of change is just the total change divided by the length of the interval, that is 1.5∕4 = 0.375 thousand/hour. (b) We can estimate the total change of the algae population by counting boxes between the curve and the x-axis. Here, there is about 1 box above the x-axis from x = −3 to x = −2, about 0.75 of a box below the x-axis from x = −2 to x = −1, and a total change of about 1.5 boxes thereafter (as discussed in part (a)). So the total change is about 1 − 0.75 + 1.5 = 1.75 thousands of algae. 23. (a) Since t = 0 to t = 31 covers January: Average number of 1 = 31 Ê0 daylight hours in January

(12 + 2.4 sin(0.0172(t − 80))) dt.

Evaluating the integral numerically gives Average ≈

306 ≈ 9.9 hours. 31

(b) Assuming it is not a leap year, the last day of May is t = 151(= 31 + 28 + 31 + 30 + 31) and the last day of June is t = 181(= 151 + 30). Again fnding the integral numerically: 181

Average number of 1 = (12 + 2.4 sin(0.0172(t − 80))) dt 30 Ê151 daylight hours in June 431 ≈ ≈ 14.4 hours. 30

422

Chapter Five /SOLUTIONS (c) Assuming it is not a leap year, 365

1 (12 + 2.4 sin(0.0172(t − 80))) dt 365 Ê0 4381 ≈ ≈ 12.0 hours. 365

Average for whole year =

(d) The average over the whole year should be 12 hours, as computed in part (c). Since Madrid is in the northern hemisphere, the average for a winter month, such as January, should be less than 12 hours (it is 9.9 hours) and the average for a summer month, such as June, should be more than 12 hours (it is 14.4 hours). 24. (a) At the end of one hour t = 60, and H = 22◦ C. (b) 60

1 (20 + 980e−0.1t )dt 60 Ê0 1 = (10976) = 183◦ C. 60

Average temperature =

(c) Average temperature at beginning and end of hour = (1000 + 22)∕2 = 511◦ C. The average found in part (b) is smaller than the average of these two temperatures because the bar cools quickly at frst and so spends less time at high temperatures. Alternatively, the graph of H against t is concave up. H(◦ C) 1000◦ H = 20 + 980e−0.1t 511◦

22◦

t (mins)

25. (a) See Figure 5.79. More sales were made in the second half of the year, because the area under the second half of the curve is greater. sales/month 1200 800 400 t 3

Figure 5.79 (b) The total sales for the frst six months amount to Ê0

r(t)dt = $1531.20

The total sales for the last 6 months of the year amount to Ê6

r(t)dt = $1963.20

5.6 SOLUTIONS

423

r(t)dt =

Ê0

r(t)dt +

Ê6

r(t)dt = $1531.20 + $1963.20 = $3494.40

(d) The average sales per month is the quotient of the total sales with 12 months giving Total sales 3494.4 = = $291.20/month. 12 months 12 26. (a) E(t) = 1.4e0.07t (b) Average Yearly Consumption =

Total Consumption for the Century 100 years 100

1 1.4e0.07t dt 100 Ê0 ≈ 219 million megawatt-hours. =

(c) We are looking for t such that E(t) ≈ 219: 1.4e0.07t ≈ 219 e0.07t = 156.4. Taking natural logs, 0.07t = ln 156.4 5.05 t≈ ≈ 72.18. 0.07 Thus, consumption was closest to the average during 1972. (d) Between the years 1900 and 2000 the graph of E(t) looks like E(t) 1.4e7

E(t) = 1.4e0.07t

219 1900 (t = 0)

1950 (t = 50)

t 2000 (t = 100)

From the graph, we can see the t value such that E(t) = 219. It lies to the right of t = 50, and is thus in the second half of the century. 27. Looking at the graph of f (x), we see that the curve has a horizontal tangent at x = 1. Thus f ¨ (1) = 0. Next we must determine if the average value of f (x) is positive or negative on 0 ≤ x ≤ 4. The average value of the function from x = 0 to x = 4 will have the same sign as the integral of the function over the same interval. Because the area between the curve and the x-axis that is above the x-axis is greater than the area that is below the x-axis, the integral is positive. Thus the average value is positive. The area between the graph of f (x) and the x-axis between x = 0 and x = 1 lies entirely below the x-axis. Hence 1 ∫0 f (x) dx is negative. We can now arrange the quantities in order from least to greatest: (c) < (a) < (b).

424

Chapter Five /SOLUTIONS

28. In (a), f ¨ (1) is the slope of a tangent line at x = 1, which is negative. As for (c), the rate of change in f (x) is given by f ¨ (x), and the average value of this over 0 ≤ x ≤ a is a

f (a) − f (0) 1 f ¨ (x) dx = . a−0 a − 0 Ê0 This is the slope of the line through the points�(0, 1) and (a, 0), which is lessa negative that the tangent line at x = 1. a Therefore, (a) < (c) < 0. The quantity (b) is ∫0 f (x) dx ∕a and (d) is ∫0 f (x) dx, which is the net area under the a graph of f (counting the area as negative for f below the x-axis). Since a > 1 and ∫0 f (x) dx > 0, we have 0 <(b)<(d). Therefore (a) < (c) < (b) < (d). 29. (a) The expression is the average value of P ¨ (t) from t = 2 to t = 6. Specifcally, this expression gives the average value of the rate of change, in cases per day, of the total number of cases in Tennessee from t = 2 to t = 6 (July 10 to July 14, 2020). (b) From the table we have P (6) − P (2) 61,164 − 54,305 = = 1714.75. 6−2 4 (c) Since P (6) = Total number of Covid-19 cases confrmed by t = 6 P (2) = Total number of Covid-19 cases confrmed by t = 2, we have P (6) − P (2) =

Total number of Covid-19 cases confrmed from t = 2 to t = 6

Using this, we see that Average number of cases per day from t = 2 to t = 6.

P (6) − P (2) P (6) − P (2) = . 4 6−2

From part (b) we conclude that there were on average 1715 cases per day in Tennessee from t = 2 to t = 6 (July 10 to July 14, 2020). (d) By the Fundamental Theorem of Calculus, if we integrate the rate of change we fnd fnd the total change, so Ê2

P ¨ (t) dt = P (6) − P (2).

Thus, 6

P (6) − P (2) 1 . P ¨ (t) dt = 6 − 2 Ê2 6−2

Solutions for Chapter 5 Review 1. (a) Since the velocity is decreasing, for an upper estimate, we use a left sum. With n = 5,we have Δt = 2. Then Upper estimate = (44)(2) + (42)(2) + (41)(2) + (40)(2) + (37)(2) = 408. (b) For a lower estimate, we use a right sum, so Lower estimate = (42)(2) + (41)(2) + (40)(2) + (37)(2) + (35)(2) = 390.

SOLUTIONS to Review Problems For Chapter Five 2. (a)

425

(i) Since the velocity is increasing, for an upper estimate we use a right sum. Using n = 4, we have Δt = 3, so Upper estimate = (37)(3) + (38)(3) + (40)(3) + (45)(3) = 480. (ii) Using n = 2, we have Δt = 6, so Upper estimate = (38)(6) + (45)(6) = 498.

(b) The answer using n = 4 is more accurate as it uses the values of v(t) when t = 3 and t = 9. (c) Since the velocity is increasing, for a lower estimate we use a left sum. Using n = 4, we have Δt = 3, so Lower estimate = (34)(3) + (37)(3) + (38)(3) + (40)(3) = 447. 3. Calculating both the LHS and RHS and averaging the two, we get 1 (5(100 + 82 + 69 + 60 + 53) + 5(82 + 69 + 60 + 53 + 49)) = 1692.5 2 4. ∫a f (t)dt is measured in b

miles ⋅ (hours) = miles. hours

5. (a) Right sum (b) Upper estimate (c) 4 (d) Δt = 2 (e) Upper estimate is approximately 12 ⋅ 2 + 14.8 ⋅ 2 + 16.5 ⋅ 2 + 18 ⋅ 2 = 122.6. 6. The units of measurement are meters per second (which are units of velocity). 7. The units of measurement are dollars. 8. The units of measurement are foot-pounds (which are units of work). 9. (a) Using rectangles under the curve, we get Acres defaced ≈ (1)(0.2 + 0.4 + 1 + 2) = 3.6 acres. (b) Using rectangles above the curve, we get Acres defaced ≈ (1)(0.4 + 1 + 2 + 3.5) = 6.9 acres. (c) The number of acres defaced is between 3.6 and 6.9, so we estimate the average, 5.25 acres. 10. The area = Figure 5.80.

Ê0

(x3 + 2)dx. Using technology to evaluate the integral, we see that Area =

y 10

y = x3 + 2

x 1

Figure 5.80

Ê0

(x3 + 2)dx = 8. See

426

Chapter Five /SOLUTIONS

11. The area we want is shaded in Figure 5.81. A rough estimate of this area is 6, since it has about the same area as a rectangle of width 3 and height 2. y 3 y = 10x(3−x ) 2 1 x 1

Figure 5.81 To fnd the area more accurately, we say Area shaded =

Ê0

10x(3−x )dx.

Using a calculator or computer to evaluate the integral, we obtain Area shaded =

Ê0

10x(3−x ) dx = 6.967 ≈ 7 square units.

Ê1

13. We use a calculator or computer to see that

Ê1

14. We use a calculator or computer to see that

Ê0

1√

12. We use a calculator or computer to see that

15. We use a calculator or computer to see that 16. We use a calculator or computer to see that

Ê1

2x dx = 2.9. (x2 + x) dx = 28.5. (3x + 1)2 dx = 448.0. 1 + t2 dt = 1.15.

(1.03)t dt = 1.0.

−1 17. We use a calculator or computer to see that ∫2 (r+1) dr = −0.083. 2 3

18. We use a calculator or computer to see that ∫1 z z+1 dz = 5.10. 3

19. We use a calculator or computer to see that ∫−3 e−t dt = 2 ∫0 e−t dt = 2(0.886) = 1.772. 3

20. (a) Since the car starts at a velocity of 90 ft/sec, b = 90. Since the car takes 12 seconds to reach its minimum velocity of 20 ft/sec, a = 20 and c = 12. See Figure 5.82. (b) The distance traveled is the area under the velocity graph, which can be viewed as a rectangle with base 12 and height 20 and a triangle with base 12 and height 70. So Distance traveled = 12 ⋅ 20 +

1 ⋅ 12 ⋅ 70 = 660 feet. 2

velocity (ft/sec) 90

20 12

Figure 5.82

t (sec)

SOLUTIONS to Review Problems For Chapter Five

427

21. (a) With n = 4, we have Δt = 4. Then t0 = 0, t1 = 4, t2 = 8, t3 = 12, t4 = 16 and

f (t0 ) = 25, f (t1 ) = 23, f (t2 ) = 22, f (t3 ) = 20, f (t4 ) = 17

(b) Left sum = (25)(4) + (23)(4) + (22)(4) + (20)(4) = 360 Right sum = (23)(4) + (22)(4) + (20)(4) + (17)(4) = 328. (c) With n = 2,we have Δt = 8. Then t0 = 0, t1 = 8, t2 = 16 and

f (t0 ) = 25, f (t1 ) = 22, f (t2 ) = 17

(d) Left sum = (25)(8) + (22)(8) = 376 Right sum = (22)(8) + (17)(8) = 312. 22. Since x3 ≤ x2 for 0 ≤ x ≤ 1, we have Ê0

Area =

(x2 − x3 ) dx = 0.083.

The integral was evaluated on a calculator.

23. Since x1∕2 ≤ x1∕3 for 0 ≤ x ≤ 1, we have

Ê0

Area =

(x1∕3 − x1∕2 ) dx = 0.0833.

The integral was evaluated on a calculator. 24.

y = x2 9 6 3

x −3 −2

−1 −3

−6 −9

y = 3x

Figure 5.83 Inspection of the graph tells us that the curves intersect at (0, 0) and (3, 9), with 3x ≥ x2 for 0 ≤ x ≤ 3, so we can fnd the area by evaluating the integral Ê0

(3x − x2 )dx.

Using technology to evaluate the integral, we see Ê0 So the area between the graphs is 4.5.

(3x − x2 )dx = 4.5.

428

Chapter Five /SOLUTIONS

25. See Figure 5.84. y y=x √ y= x

2 1

x −2

−1

−1 −2

Figure 5.84 Inspection of the graph tell us that the curves intersect at (0, 0) and (1, 1), with the area by evaluating the integral 1 √ ( x − x)dx. Ê0

√ x ≥ x for 0 ≤ x ≤ 1, so we can fnd

Using technology to evaluate the integral, we see

Ê0

1 √

( x − x)dx ≈ 0.1667.

So the area between the graphs is about 0.1667. 26. (a) An overestimate is 7 tons. An underestimate is 5 tons. (b) An overestimate is 7 + 8 + 10 + 13 + 16 + 20 = 74 tons. An underestimate is 5 + 7 + 8 + 10 + 13 + 16 = 59 tons. 27. To fnd the distance the car moved before stopping, we estimate the distance traveled for each two-second interval. Since speed decreases throughout, we know that the left-hand sum will be an overestimate to the distance traveled and the righthand sum an underestimate. Applying the formulas for these sums with Δt = 2 gives: LEFT = 2(100 + 80 + 50 + 25 + 10) = 530 ft. RIGHT = 2(80 + 50 + 25 + 10 + 0) = 330 ft. (a) The best estimate of the distance traveled will be the average of these two estimates, or Best estimate =

530 + 330 = 430 ft. 2

(b) All we can be sure of is that the distance traveled lies between the upper and lower estimates calculated above. In other words, all the GPS data tells us for sure is that the car traveled between 330 and 530 feet before stopping. So we don’t know whether it hit the skunk or not; the answer is (ii). 28. The change in position is calculated from the area between the velocity graph and the t-axis, with the region below the axis corresponding to negatives velocities and counting negatively. Figure 5.85 shows the graph of f (t). From t = 0 to t = 3 the velocity is positive. The region under the graph of f (t) is a triangle with height 6 cm/sec and base 3 seconds. Thus, from t = 0 to t = 3, the particle moves Distance moved to right =

1 ⋅ 3 ⋅ 6 = 9 centimeters. 2

From t = 3 to t = 4, the velocity is negative. The region between the graph of f (t) and the t-axis is a triangle with height 2 cm/sec and base 1 second, so in this interval the particle moves Distance moved to left =

1 ⋅ 1 ⋅ 2 = 1 centimeter. 2

Thus, the total change in position is 9 − 1 = 8 centimeters to the right.

SOLUTIONS to Review Problems For Chapter Five

429

y 6

Motion to right: 9 cm

✠ 4 3 −2

✒

f (t)

Motion to left: 1 cm

Figure 5.85

29. (a) See Figure 5.86. (b) The peak of the fight is when the velocity is 0, namely t = 3. The height at t = 3 is given by the area under the graph of the velocity from t = 0 to t = 3; see Figure 5.86. The region is a triangle of base 3 seconds and altitude 96 ft/sec, so the height is (1∕2)3 ⋅ 96 = 144 feet. (c) The velocity is negative from t = 3 to t = 5, so the motion is downward then. The distance traveled downward can be calculated by the area of the triangular region which has base of 2 seconds and altitude of −64 ft/sec. Thus, the baseball travels (1∕2)2 ⋅ 64 = 64 feet downward from its peak height of 144 feet at t = 3. Thus, the height at time t = 5 is the total change in position, 144 − 64 = 80 feet. 96

t 3

6 v(t)

−96

Figure 5.86 «««< .mine ||||||| .r594

30. Change in income = ∫0 r(t) dt = ∫0 40(1.002)t dt = $485.80 ======= »»»> .r595 12

31. (a) The area under the curve is greater for species B for the frst 5 years. Thus species B has a larger population after 5 years. After 10 years, the area under the graph for species B is still greater so species B has a greater population after 10 years as well. (b) Unless something happens that we cannot predict now, species A will have a larger population after 20 years. It looks like species A will continue to quickly increase, while species B will add only a few new plants each year. 32. You start traveling toward home. After 1 hour, you have traveled about 5 miles. (The distance traveled is the area under the frst bump of the curve. The area is about 1∕2 grid square; each grid square represents 10 miles.) You are now about 10 − 5 = 5 miles from home. You then decide to turn around. You ride for an hour and a half away from home, in which time you travel around 13 miles. (The area under the next part of the curve is about 1.3 grid squares below the axis.) You are now 5+13 = 18 miles from home and you decide to turn around again at t = 2.5. The area above the curve from t = 2.5 to t = 3.5 has area 18 miles. This means that you ride for another hour, at which time you reach your house (t = 3.5.) You still want to ride, so you keep going past your house for a half hour. You make about 8 miles in this time and you decide to stop. So, at the end of the ride, you end up 8 miles away from home. 33. (a) The weight growth rate is the derivative of the weight function. The fact that the weight growth rate is increasing means that the weight function has an increasing derivative. Thus the graph of the weight function is concave up. (b) The total weight growth during the forty week gestation equals the integral of the growth rate over the time period. Thus the birth weight equals the area under the graph. The region is triangular, with base 28 weeks and height 0.22 kg/week. Thus, 1 Total weight = 28(0.22) = 3.08 kg ≈ 3.1 kg. 2 34. (a) At t = 20 minutes, she stops moving toward the lake (with v > 0) and starts to move away from the lake (with v < 0). So at t = 20 minutes the cyclist turns around.

430

Chapter Five /SOLUTIONS (b) The cyclist is going the fastest when v has the greatest magnitude, either positive or negative. Looking at the graph, we can see that this occurs at t = 40 minutes, when v = −25 and the cyclist is pedaling at 25 km/hr away from the lake. (c) From t = 0 to t = 20 minutes, the cyclist comes closer to the lake, since v > 0; thereafter, v < 0 so the cyclist moves away from the lake. So at t = 20 minutes, the cyclist comes the closest to the lake. To fnd out how close she is, note that between t = 0 and t = 20 minutes the distance she has come closer is equal to the area under the graph of v. Each box represents 5/6 of a kilometer, and there are about 2.5 boxes under the graph, giving a distance of about 2 km. Since she was originally 5 km away, she then is about 5 − 2 = 3 km from the lake. (d) At t = 20 minutes she turns around, since v changes sign then. Since the area below the t-axis is greater than the area above, the farthest she is from the lake is at t = 60 minutes. Between t = 20 and t = 60 minutes, the area under the graph is about 10.8 km. (Since 13 boxes ⋅ 5∕6 = 10.8.) So at t = 60 she will be about 3 + 10.8 = 13.8 km from the lake. (e) For time t = 0 to t = 20, since v > 0, the area between the graph and t-axis represents the distance traveled toward the lake, about 2 km. For time t = 20 to t = 60, since v < 0, the area between the graph and t-axis represents the distance traveled away from the lake, about 10.8 km. Thus, the total distance she traveled is about 2 + 10.8 = 12.8 km.

35. The area below the x-axis is greater than the area above the x-axis, so the integral is negative. 36. The entire graph of the function lies above the x-axis, so the integral is positive. 37. The areas above and below the x-axis are approximately equal in size, so the integral is approximately zero. 38. The area above the x-axis is larger than the area below the x-axis, so the integral is positive. 39. The region shaded between x = 0 and x = 2 appears to have approximately the same area as the region shaded between 0 x = −2 and x = 0, but it lies below the axis. Since ∫−2 f (x) dx = 4, we have the following results: (a) ∫0 f (x) dx ≈ − ∫−2 f (x) dx = −4. 2 (b) ∫−2 f (x) dx ≈ 4 − 4 = 0. (c) The total area shaded is approximately 4 + 4 = 8. 2

1 y

40.

f (x) = 2−x x −1

Figure 5.87 Since the shaded area is positive and less than the area of the rectangle measuring two units by one unit, we can say that

Ê−1

2−x dx < 2.

√ 41. Notice that f (x) = 1 + x3 is increasing for 0 ≤ x ≤ 2, since x3 gets bigger as x increases. This means that f (0) ≤ f (x) ≤ f (2). For this function, f (0) = 1 and f (2) = 3. Thus, the area under f (x) lies between the area under the line y = 1 and the area under the line y = 3 on the interval 0 ≤ x ≤ 2. See Figure 5.88. That is, 1(2 − 0) ≤

Ê0

2√

1 + x3 dx ≤ 3(2 − 0).

SOLUTIONS to Review Problems For Chapter Five y

431

y=3

3 f (x) =

√

1 + x3 y=1

x 2

Figure 5.88 42. (a) The distance traveled in the frst 3 hours (from t = 0 to t = 3) is given by Ê0

(40t − 10t2 )dt.

(b) The shaded area in Figure 5.89 represents the distance traveled. v(t) 40

t 1

Figure 5.89 (c) Using a calculator, we get Ê0

(40t − 10t2 )dt = 90.

So the total distance traveled is 90 miles.

43. (a) Quantity used = ∫0 f (t) dt. (b) Using a left hand-sum, since Δt = 2, our approximation is 10

35.53(1.014)0 ⋅ 2 + 35.53(1.014)2 ⋅ 2 + 35.53(1.014)4 ⋅ 2 + 35.53(1.014)6 ⋅ 2 + 35.53(1.014)8 ⋅ 2 ≈ 375.91. Since f is an increasing function, this represents an underestimate. (c) Each term is a lower estimate of 2 years’ consumption of oil.

44. (a) Since v = 6 − 2t, we see that v > 0 (the car is moving forward) if 0 ≤ t < 3 and that v < 0 (the car is moving backward) if t > 3. When t = 3, we have v = 0, so the car is not moving at the instant t = 3. (b) The distance the car moves forward is given by the area above the t-axis in Figure 5.90. The distance moved backward is given by the area below the t-axis. Since the area above between t = 0 and t = 3 is equal to the area below between t = 3 and t = 6, the car is back at its starting point at t = 6. (c) The car moves forward on the interval 0 ≤ t < 3, so it is farthest forward when t = 3. For all t > 3, the car is moving backward. There is no upper bound on the car’s distance behind its starting point since it is moving backward for all t > 3. 6 v = 6 − 2t 3

6 t

−6

Figure 5.90

432

Chapter Five /SOLUTIONS

45. We are told that C ¨ (q) = q 2 − 50q + 700 for 0 ≤ q ≤ 50. We also know that 50

Ê0

C ¨ (q)dq = C(50) − C(0).

Since we are told that fxed costs are $500 we know that C(0) = $500. Thus C(50) =

Ê0

C ¨ (q)dq + $500 ≈ $14,667.

46. The area under the curve represents the number of cubic feet of storage times the number of days the storage was used. This area is given by Area under graph = Area of rectangle + Area of triangle 1 = 30 ⋅ 10,000 + ⋅ 30(30,000 − 10,000) 2 = 600,000. Since the warehouse charges $5 for every 10 cubic feet of storage used for a day, the company will have to pay (5)(60,000) = $300,000. 47. (a)

CCl4 dumped 16

t 7

Figure 5.91 (b) 7 years, because t2 − 14t + 49 = (t − 7)2 indicates that the rate of fow was zero after 7 years. (c) Area under the curve = 3(16) +

Ê3

(t2 − 14t + 49) dt

= 69.3. So 69 31 cubic yards of CCl4 entered the waters from the time the EPA frst learned of the situation until the fow of CCl4 was stopped. y

48. (a)

2 1

x 4

Figure 5.92: Graph of y = x3 − 5x2 + 4x (b) I1 = ∫0 (x3 − 5x2 + 4x)dx is positive, since 1

x3 − 5x2 + 4x > 0 for 0 < x < 1.

I2 = ∫0 (x3 − 5x2 + 4x)dx seems to be negative, since the area under the x-axis is bigger than the area above the x-axis for 0 ≤ x ≤ 2. I3 and I4 will also be negative, since the area under the x-axis will also be larger than the area above the x-axis for 0 ≤ x ≤ 3 and 0 ≤ x ≤ 4. I5 will also be negative, since the area above the x-axis will still be less than the area below the x-axis for 0 ≤ x ≤ 5. From this we conclude that I1 is the largest and I4 is the smallest. 2

SOLUTIONS to Review Problems For Chapter Five

433

(i) The mouse changes direction (when its velocity is zero) at about times 17, 23, and 27. (ii) The mouse is moving most rapidly to the right at time 10 and most rapidly to the left at time 40. x (iii) The mouse is farthest to the right when the integral of the velocity, ∫0 v(t) dt, is most positive. Since the integral is the sum of the areas above the t-axis minus the areas below the t-axis, the integral is largest when the velocity is zero at about 17 seconds. The mouse is farthest to the left of center when the integral is most negative at 40 seconds. (iv) The mouse’s speed decreases during seconds 10 to 17, from 20 to 23 seconds, and from 24 seconds to 27 seconds. (v) The mouse is at the center of the tunnel at any time t for which the integral from 0 to t of the velocity is zero. This is true at time 0 and again somewhere around 35 seconds. (b) To fnd total distance traveled we estimate area between the velocity graph and the t-axis. Each box represents 10 cm/sec ⋅ 5 sec = 50 cm traveled. Area above the t-axis gives distance moved to the right and area below gives distance moved to the left. From time t = 0 to t = 17 the mouse travels about 225 cm to the right. From t = 17 to t = 23 it travels about 85 cm to the left. From t = 23 to t = 27 it travels about 15 cm to the right and from t = 27 to t = 40 it travels about 290 cm to the left. Thus, the total distance the mouse traveled is about 225+85+15+290 = 615 cm.

49. (a)

50. The graph of rate against time is the straight line shown in Figure 5.93. Since the shaded area is 270, we have 1 (10 + 50) ⋅ t = 270 2 270 t= ⋅ 2 = 9 years 60 rate (kg/yr) 50

✛

Area = 270

10 time (yr)

Figure 5.93 51. Using the Fundamental Theorem of Calculus with f (x) = 4 − x2 and a = 0, we see that F (b) − F (0) =

Ê0

F ¨ (x)dx =

Ê0

(4 − x2 )dx.

We know that F (0) = 0, so b

(4 − x2 )dx. Ê0 Using a calculator or computer to estimate the integral for values of b, we get F (b) =

Table 5.3 b

0.0

0.5

1.0

1.5

2.0

2.5

F (b)

1.958

3.667

4.875

5.333

4.792

52. From the graph we see that F ¨ (x) < 0 for x < −2 and x > 2, and that F ¨ (x) > 0 for −2 < x < 2, so we conclude that F (x) is decreasing for x < −2 and x > 2 and F (x) is increasing for −2 < x < 2.

x −2

F ¨ (x) = 4 − x2

Figure 5.94

434

Chapter Five /SOLUTIONS

53. Since F ¨ (x) is positive for 0 < x < 2 and F ¨ (x) is negative for 2 < x < 2.5, F (x) increases on 0 < x < 2 and decreases on 2 < x < 2.5. From this we conclude that F (x) has a maximum at x = 2. From the process used in Problem 51 we see that the chart agrees with this assumption and that F (2) = 5.333. 54. By counting grid squares, we fnd 1 10 − 0 Ê0 56. The average value equals 55. Average value =

Ê1

f (x) dx = 8.5, so the average value of f is

8.5 8.5 = = 1.7. 6−1 5

et dt ≈ 2202.55 3

1 24 f (x) dx = = 8. 3 3 Ê0 57. (a) When t = 0, f (0) = 100e0 = 100. So there are 100 cases at the start of the six months. When t = 6, f (6) = 100e−3 = 4.98, so there are almost 5 cases left at the end of the half year. (b) Average number of cases in the inventory can be given by

6 1 190 f (t) dt ≈ ≈ 32 cases. 6 6 − 0 Ê0

58. (a) Q(10) = 4(0.96)10 ≈ 2.7. Q(20) = 4(0.96)20 ≈ 1.8 (b) Q(10) + Q(20) ≈ 2.21 2 (c) The average value of Q over the interval is about 2.18. (d) Because the graph of Q is concave up between t = 10 and t = 20, the area under the curve is less than what is obtained by connecting the endpoints with a straight line. 59. It appears that the area under a line at about y = 8.5 is approximately the same as the area under f (x) on the interval x = a to x = b, so we estimate that the average value is about 8.5. See Figure 5.95. 10 f (x) 6

2 x

Figure 5.95 60. It appears that the area under a line at about y = 45 is approximately the same as the area under f (x) on the interval x = a to x = b, so we estimate that the average value is about 45. See Figure 5.96. 100 f (x)

75 50 25

Figure 5.96

STRENGTHEN YOUR UNDERSTANDING

435

61. We’ll show that in terms of the average value of f , I > II = IV > III Using the defnition of average value and the fact that f is even, we have 1 2 ∫−2 f (x)dx Average value ∫0 f (x)dx = = 2 of f on II 2 2 2

∫−2 f (x)dx 2

4 = Average value of f on IV.

Since f is decreasing on [0,5], the average value of f on the interval [0, c], where 0 ≤ c ≤ 5, is decreasing as a function of c. The larger the interval the more low values of f are included. Hence Average value of f Average value of f Average value of f > > on [0, 2] on [0, 5] on [0, 1] 62. (a) Average value of f = 15 ∫0 f (x) dx. 5

(b) Average value of |f | = 51 ∫0 |f (x)| dx = 51 (∫0 f (x) dx − ∫2 f (x) dx). 5

STRENGTHEN YOUR UNDERSTANDING 1. True, as specifed in the text. 2. True, as specifed in the text. 3. False, since using the smaller velocity at the beginning of each subinterval gives an underestimate of the distance traveled. 4. False, since the width of the interval is 2 seconds and an underestimate for the distance traveled is given by 10 ⋅ 2 = 20 ft and an overestimate for the distance traveled is given by 20 ⋅ 2 = 40 ft. The distance traveled is between 20 and 40 feet during this 2-second interval. 5. True, since the width of the interval is 2 seconds and an underestimate for the distance traveled is given by 10 ⋅ 2 = 20 ft and an overestimate for the distance traveled is given by 20 ⋅ 2 = 40 ft. The distance traveled is between 20 and 40 feet during this 2-second interval. 6. True, since 20 gallons per minute times 10 minutes is 200 gallons and that is an underestimate. 7. False. The estimate of 50 gallons per minute times 10 minutes is 500 gallons but it is an overestimate not an underestimate. 8. False. Since the width of the interval is 2 days, an overestimate is 2⋅1000 = $2000 and an underestimate is 2⋅800 = $1600. 9. True, since the triangular area under the graph of v(t) = 3t when 0 ≤ t ≤ 10 is (1∕2)(30)(10) = 150 feet. 10. False, since the rate of change has units quantity per time, while total change has units of quantity. 11. False, since the left-hand value on each subinterval is the least value f has on that subinterval, so the left-hand sum gives an underestimate of the integral. 12. False, since the left-hand sum of a decreasing function will be an overestimate whether the function is concave up or concave down. 13. False. Since the width of the interval is 2, the left-hand sum estimate is 2 ⋅ 1000 = $2000. 14. True. Since the width of the interval is 2, the right-hand sum estimate is 2 ⋅ 800 = 1600. 15. True. If the number of rectangles increases, the width of each rectangle decreases. 16. False. If n = 4 on this interval, then Δt = 0.5. 17. False, since right-hand rectangles could be above, below, or neither. Only when f is increasing are the right-hand rectangles guaranteed to be above the graph. 18. True, as specifed in the text.

436

Chapter Five /SOLUTIONS

19. True, since f is decreasing over the interval −2 ≤ t ≤ −1. 20. False, since 25 − 5 = 20 which divided into fve pieces gives Δt = 4.

21. False, since we don’t know whether f (x) ≥ 0 between x = 0 and x = 2. 22. False, since the graph of f could just have equal areas above and below the x-axis between x = 0 and x = 2. 23. False, since the graph of f could just have more area above than below the x-axis between x = 0 and x = 2. 24. True, since all of the terms in a left- or right-hand sum defning ∫0 f (x) dx are positive. 2

25. True, since the graph of f (x) = x2 − 1 is entirely on or below the x-axis between x = −1 and x = 1.

26. False, since ∫0 f (x) dx is the sum of areas where f (x) > 0 and (−1) times the areas where f (x) < 0. 2

27. False, since the graph of f (x) = 1 − x is not always above the x-axis between x = 0 and x = 3. 28. True, since f (x) = ex is always positive. 29. True, as specifed in the text.

30. True, since ∫1 f (x) dx is the sum of areas above the x-axis and (−1) times the sum of areas below the x-axis. 10

31. True, since the units of the integral are the product of units of r(t) and units of t. 32. False, the integral has units dollars per day times days, or dollars. 33. True, since the units of the integral are the product of the units of a(t) and the units of t. 34. False, since the units of the integral are the product of the units of w(t) and the units of t, giving worker-days. 35. True, since the units of the integral are the product of the units of Ct) and the units of t, giving dollars. 36. True, since the integral of the rate of change, dollars per day, gives total change in the cost. 37. False. The integral gives the total change in the volume of water. We have to add the change to the starting volume at t = 0 to fnd the volume at t = 30. 38. True. The integral gives the total change in the volume of water. We can add the change to the starting volume at t = 0 to fnd the volume at t = 30. 39. True, as specifed in the text. 40. False, since bioavailability depends on the area under the drug concentration curve, not its peak. So drug B might last longer in the bloodstream over time, while drug A might grow quickly to a high concentration, but then go quickly back to zero.

41. False, since the correct statement is if F ¨ (t) is continuous, then ∫a F ¨ (t) dt = F (b) − F (a). b

42. True, as specifed in the text.

43. False, since the correct statement is if F ¨ (t) is continuous, then ∫a F ¨ (t) dt = F (b) − F (a). b

44. True, since by the Fundamental Theorem, ∫100 C ¨ (q) dq = C(200) − C(100). 200

45. False, since by the Fundamental Theorem, ∫0 subtracted the fxed costs of C(0).

1000

C ¨ (q) dq = C(1000) − C(0) which is the total variable costs, as we have

0.03q +0.1 dq = 15,100 gives only the variable costs. The total cost is the sum of the variable 46. False, since the integral ∫0 cost and the fxed cost, or $16,600. 1000

47. False, since the integral of the marginal cost function only gives the change in cost, or variable costs.

48. False, since ∫0 F ¨ (t) dt = F (5) − F (0) = 0 only says that F (0) = F (5), not that F has the same value between t = 0 and t = 5. 5

49. True, since if the total change of F over the interval 0 ≤ t ≤ 5 is negative, then F (0) must be greater than F (5). 50. False, since if F ¨ (t) = k, then ∫0 k dt = 5k. 5

1 51. False. The average value of f (x) on the interval 0 ≤ x ≤ 10 is 10 ∫0 f (x) dx. 10

52. False, since f could have values much larger than 50 or much less than 0 between 0 and 50.

53. True, since 0 < ∫0 f (x) dx ≤ ∫0 50 dx = 500, so the average value of f is between 0 and 50. 10

54. True, since the units of ∫a f (x) dx are the units of f (x) multiplied by the units of x, and the average value is 1∕(b − b a) ∫a f (x) dx so we divide the integral units by the units of x, giving the same units as f . b

55. False. A counterexample is f (x) = x2 , since (1∕10) ∫0 x2 dx = 100∕3, while (1∕20) ∫0 x2 dx = 400∕3. 10

PROJECTS FOR CHAPTER FIVE

437

56. True, since all of the function values are doubled for 2f (x). Also, we see that (1∕10) ∫0 2f (x) dx = 2⋅(1∕10) ∫0 f (x) dx. 10

57. False, since, for example, f (x) could be +1 over half of the interval 0 ≤ x ≤ 10 and −1 over the other half. 58. True. If f (x) > g(x) ≥ 0, then ∫0 f (x) dx > ∫0 g(x) ≥ 0, so (1∕10) ∫0 f (x) dx > (1∕10) ∫0 g(x). 10

59. True. This follows from the defnition. 50 = 5 dollars per 60. False. For example, consider the cost function C(q) = 5q. The average cost of q = 10 items is Cq(q) = 10 1 = 25 dollars. item. The average value of the cost function from q = 0 to q = 10 is 10 ∫0 C(q) dq = 101 500 2 10

PROJECTS FOR CHAPTER FIVE 1. (a) CO2 is being taken out of the water during the day and returned at night. The pond must therefore contain some plants. (The data is in fact from pond water containing both plants and animals.) (b) Suppose t is the number of hours past dawn. The graph in Figure 5.94 of the text shows that the CO2 content changes at a greater rate for the frst 6 hours of daylight, 0 < t < 6, than it does for the fnal 6 hours of daylight, 6 < t < 12. It turns out that plants photosynthesize more vigorously in the morning than in the afternoon. Similarly, CO2 content changes more rapidly in the frst half of the night, 12 < t < 18, than in the 6 hours just before dawn, 18 < t < 24. The reason seems to be that at night plants quickly use up most of the sugar that they synthesized during the day, and then their respiration rate is inhibited. So the constant rate hypothesis is false, if we assume plants are the main cause of CO2 changes in the pond. (c) The question asks about the total quantity of CO2 in the pond, rather than the rate at which it is changing. We will let f (t) denote the CO2 content of the pond water (in mmol∕l) at t hours past dawn. Then Figure 5.94 of the text is a graph of the derivative f ¨ (t). There are 2.600 mmol∕l of CO2 in the water at dawn, so f (0) = 2.600. The CO2 content f (t) decreases during the 12 hours of daylight, 0 < t < 12, when f ¨ (t) < 0, and then f (t) increases for the next 12 hours. Thus, f (t) is at a minimum when t = 12, at dusk. By the Fundamental Theorem, f (12) = f (0) +

Ê0

f ¨ (t) dt = 2.600 +

Ê0

f ¨ (t) dt.

We must approximate the defnite integral by a Riemann sum. From the graph in Figure 5.94 of the text, we estimate the values of the function f ¨ (t) in Table 5.4. Table 5.4

Rate, f ¨ (t), at which CO2 is entering or leaving water

f ¨ (t)

0.000

−0.039

−0.026

0.000

0.035

0.020

−0.042

−0.038

−0.023

0.054

0.030

0.015

−0.044

−0.035

−0.020

0.045

0.027

0.012

−0.041

−0.030

−0.008

0.040

0.023

0.005

The left Riemann sum with n = 12 terms, corresponding to Δt = 1, gives Ê0

f ¨ (t) dt ≈ (0.000)(1) + (−0.042)(1) + (−0.044)(1) + ⋯ + (−0.008)(1) = −0.346.

At 12 hours past dawn, the CO2 content of the pond water reaches its lowest level, which is approximately 2.600 − 0.346 = 2.254 mmol∕l. (d) The increase in CO2 during the 12 hours of darkness equals 24

f (24) − f (12) =

Ê12

f ¨ (t) dt.

438

Chapter Five /SOLUTIONS Using Riemann sums to estimate this integral, we fnd that about 0.306 mmol∕l of CO2 was released into the pond during the night. In part (c) we calculated that about 0.346 mmol∕l of CO2 was absorbed from the pond during the day. If the pond is in equilibrium, we would expect the daytime absorption to equal the nighttime release. These quantities are suÿciently close (0.346 and 0.306) that the di˙erence could be due to measurement error, or to errors from the Riemann sum approximation. If the pond is in equilibrium, the area between the rate curve in Figure 5.94 of the text and the t-axis for 0 ≤ t ≤ 12 will equal the area between the rate curve and the t-axis for 12 ≤ t ≤ 24. In this experiment the areas do look approximately equal. (e) We must evaluate b

f ¨ (t) dt Ê0 Ê0 for the values b = 0, 3, 6, 9, 12, 15, 18, 21, 24. Left Riemann sums with Δt = 1 give the values for the CO2 content in Table 5.5. The graph is shown in Figure 5.97. f (b) = f (0) +

f ¨ (t) dt = 2.600 +

CO2 content (mmol/l)

2.6

f (t)

2.4 2.2

Time (hours past dawn)

Figure 5.97: CO2 content in pond water throughout the day Table 5.5

CO2 content throughout the day

b (hours after dawn)

f (b) (CO2 content)

2.600

2.514

2.396

2.305

2.254

2.353

2.458

2.528

2.560

2. (a) About 300 meter3 /sec. (b) About 250 meter3 /sec. (c) Looking at the graph, we can see that the 1996 food reached its maximum just between March and April, for a high of about 1250 meter3 /sec. Similarly, the 1957 food reached its maximum in mid-June, for a maximum fow rate of 3500 meter3 /sec. (d) The 1996 food lasted about 1∕3 of a month, or about 10 days. The 1957 food lasted about 4 months. (e) The area under the controlled food graph is about 2∕3 box. Each box represents 500 meter3 /sec for one month. Since days hours minutes seconds ⋅ 24 ⋅ 60 ⋅ 60 1 month = 30 month day hour minute = 2.592 ⋅ 106 ≈ 2.6 ⋅ 106 seconds, each box represents Flow ≈ (500 meter3 ∕sec) ⋅ (2.6 ⋅ 106 sec) = 13 ⋅ 108 meter3 of water. So, for the artifcial food, 2 ⋅ 13 ⋅ 108 = 8.7 ⋅ 108 meter3 ≈ 109 meter3 . 3 (f) The 1957 food released a volume of water represented by about 12 boxes above the 250 meter/sec baseline. Thus, for the natural food, Additional fow ≈

Additional fow ≈ 12 ⋅ 13 ⋅ 108 = 1.95 ⋅ 1010 ≈ 2 ⋅ 1010 meter3 . So, the natural food was nearly 20 times larger than the controlled food and lasted much longer.

PROJECTS FOR CHAPTER FIVE

439

3. (a) The hydrostatic pressure pℎ is a linear function of distance x, so pℎ = b + mx. At the artery end, x = 0, and pℎ = 35. At the vein end, x = L = 0.1 and pℎ = 15. This means the points (0, 35) and (0.1, 15) are on the graph of pℎ , so the slope is given by m=

mm Hg 15 − 35 = −200 . 0.1 − 0 cm

We know that b = 35, so pℎ = 35 − 200x mm Hg. (b) The net pressure, p, is the di˙erence between the hydrostatic pressure and the oncotic pressure. Using part (a), we have p = pℎ − po = (35 − 200x) − 23 = 12 − 200x mm Hg. (c) The average net pressure is given by an integral: L

Average pressure =

1 1 p dx = L Ê0 0.1 − 0 Ê0

0.1

12 − 200x dx = 2.0 mm Hg.

(d) Since j = k ⋅ p, we have j units = (k units) × (p units) =

cm cm × mm Hg = . sec ⋅ mm Hg sec

In addition Volume/time/area units =

1 cm Volume units 1 cm3 × = × = . 2 sec Time units Area units sec cm

Thus, j has units of volume per time per area. (e) The rate of movement of fuid across the capillary wall at a point per unit area of the wall is given by j = k⋅p. To estimate the net fow rate through the entire capillary, we use the average pressure and multiply by the area 2 rL of the entire capillary wall. We have Net Flow = k ⋅ pavg ⋅ 2 rL = 10−7 ⋅ 2 ⋅ 2 (0.0004)(0.1) = 5.03 ⋅ 10−11

cm3 . sec

4. (a) The volume of a cylinder of radius r and height ℎ is V = r2 ℎ. Since the total volume of the urine sample is 2000 ml and the diameter is 10 cm (so the radius is 5 cm), we have: r2 ℎ = 52 ℎ = 2000 2000 ℎ= = 25.46 cm. 25 Thus, the depth of urine in the cylinder is 25.46 cm. The protein concentration c is a linear function of height y, so c = b + my. At the bottom of the sample, y = 0, and c = 0.96. At the top, y = 25.46 and c = 0.14. This means the points (0, 0.96) and (25.46, 0.14) are on the graph of c, so the slope is given by m=

mg 0.14 − 0.96 = −0.0322 . 25.46 − 0 ml ⋅ cm

We know that b = 0.96, so c = 0.96 − 0.0322y mg∕ml. (b) Because the urine is collected in a cylinder with straight sides, with the same cross-sectional area at every height, the average concentration of protein in the container equals the average value of the concentration function. Thus Average concentration =

1 25.46 − 0 Ê0

25.46

0.96 − 0.0322y dy = 0.55

mg . ml

440

Chapter Five /SOLUTIONS (c) To fnd the total quantity of protein, we multiply: mg ⋅ 2000 ml ml = 1100 mg.

Total mass = 0.55

Thus, the patient excreted approximately 1100 mg, or 1.1 gm, of protein during the 24-hour period in which urine was collected. This is above the threshold of 1 gm per 24-hour period at which active treatment is recommended.

Solutions to Problems on the Second Fundamental Theorem of Calculus 1. By the Second Fundamental Theorem, G¨ (x) = x3 . 2. By the Second Fundamental Theorem, G¨ (x) = 3x . 3. By the Second Fundamental Theorem, G¨ (x) = xex . 4. The variable of integration does not a˙ect the value of the integral, so by the Fundamental Theorem of Calculus, G¨ (x) = ln x. 5. (a) If F (b) = ∫0 2x dx then F (0) = ∫0 2x dx = 0 since we are calculating the area under the graph of f (x) = 2x on the interval 0 ≤ x ≤ 0, or on no interval at all. (b) Since f (x) = 2x is always positive, the value of F will increase as b increases. That is, as b grows larger and larger, the area under f (x) on the interval from 0 to b will also grow larger. (c) Using a calculator or a computer, we get 0

F (1) = F (2) = F (3) = 6.

Ê0

2x dx ≈ 1.4, 2x dx ≈ 4.3, 2x dx ≈ 10.1.

Table 5.6 x

0.5

1.5

I(x)

0.50

1.09

2.03

3.65

7. Using the Fundamental Theorem, we know that the change in F between x = 0 and x = 0.5 is given by Ê0

F (0.5) − F (0) =

0.5

sin t cos t dt ≈ 0.115.

Since F (0) = 1.0, we have F (0.5) ≈ 1.115. The other values are found similarly, and are given in Table 5.7. Table 5.7 b

0.5

1.5

2.5

F (b)

1.11492

1.35404

1.4975

1.41341

1.17908

1.00996

8. Ê1

(f (x) + g(x)) dx =

Ê1

f (x) dx +

Ê1

g(x) dx = 5 + 2 = 7.

SOLUTIONS TO PROBLEMS ON THE SECOND FUNDAMENTAL THEOREM OF CALCULUS 9. 3

Ê1

3f (x) − 7g(x) dx = 3

Ê1

f (x) dx − 7

Ê1

Êa

g(x) dx = 3 ⋅ 5 − 7 ⋅ 2 = 1.

10. Note that ∫a g(x) dx = ∫a g(t) dt. Thus, we have b

Êa

(f (x) + g(x)) dx =

Êa

f (x) dx +

g(x) dx = 8 + 2 = 10.

11. Note that ∫a (g(x))2 dx = ∫a (g(t))2 dt. Thus, we have b

Êa

�

(f (x))2 − (g(x))2 dx =

12. We have Êa

(f (x))2 dx −

Êa

(f (x))2 dx −

Êa

(g(x))2 dx = 12 − 3 = 9.

12 f (x) dx = 12 − 82 = −52.

13. Note that ∫a f (z) dz = ∫a f (x) dx. Thus, we have b

Êa

cf (z) dz = c

Êa

f (z) dz = 8c.

441

6.1 SOLUTIONS

443

CHAPTER SIX Solutions for Section 6.1 1. Apply the Fundamental Theorem with F ¨ (x) = 2x2 + 5 and a = 0 to get values for F (b). Since F (b) − F (0) =

Ê0

F ¨ (x) dx =

Ê0

2x2 + 5 dx

and F (0) = 3, we have F (b) = 3 +

Ê0

2x2 + 5 dx.

We use a calculator or computer to estimate the defnite integral ∫0 2x2 + 5 dx for each value of b. For example, when b b = 0.1, we fnd that ∫0 2x2 + 5 dx = 0.501. Thus F (0.1) = 3.501. Continuing in this way gives the values in Table 6.1. b

Table 6.1 b

0.1

0.2

0.5

1.0

F (b)

3.501

4.005

5.583

6.667

2. Apply the Fundamental Theorem with G¨ (t) = (1.12)t and a = 5 to get values for G(b). Since G(b) − G(5) =

Ê5

G¨ (t) dt =

Ê5

(1.12)t dt

and G(5) = 1, we have G(b) = 1 +

Ê5

(1.12)t dt.

We use a calculator or computer to estimate the defnite integral ∫5 (1.12)t dt for each value of b. For example, when b b = 5.1, we fnd that ∫5 (1.12)t dt = 0.177. Thus G(5.1) = 1.177. Continuing in this way gives the values in Table 6.2. b

Table 6.2 b

5.1

5.2

5.5

6.0

G(b)

1.177

1.356

1.907

2.866

3. Apply the Fundamental Theorem with f ¨ (t) = (0.82)t and a = 2 to get values for f (b). Since f (b) − f (2) =

Ê2

f ¨ (t) dt =

Ê2

(0.82)t dt

and f (2) = 9, we have f (b) = 9 +

Ê2

(0.82)t dt.

We use a calculator or computer to estimate the defnite integral ∫2 (0.82)t dt for each value of b. For example, when b = 4, b we fnd that ∫2 (0.82)t dt = 1.110. Thus f (4) = 10.110. Continuing in this way gives the values in Table 6.3. b

Table 6.3 b

f (b)

10.110

10.856

11.696

12.293

444

Chapter Six /SOLUTIONS

4. By the Fundamental Theorem, the change in F (x) on an interval is the integral of the rate of change on that interval. We have Ê0

Change in F (x) = F (2) − F (0) =

3e−x dx = 2.646,

where the value of the integral is found using a calculator or computer. Since F (0) = 5, we see that F (2) = 5 + 2.646 = 7.646.

5. By the Fundamental Theorem, the change in G(x) on an interval is the integral of the rate of change on that interval. We have Change in G(x) = G(4) − G(1) =

Ê1

ln x dx = 2.545,

where the value of the integral is found using a calculator or computer. Since G(1) = 50, we see that G(4) = 50 + 2.545 = 52.545.

6. The Fundamental Theorem tells us that Change in F = F (5) − F (0) =

Ê0

f (x) dx = 12.

Since F (0) = 50, we have F (5) = 50 + 12 = 62. 7. The Fundamental Theorem tells us that Change in F = F (4) − F (1) =

Ê1

f (x) dx = −7.

Since F (1) = 20, we have F (4) = 20 + (−7) = 13. 8. (a) The value of the integral is negative since the area below the x-axis is greater than the area above the x-axis. We count boxes: The area below the x-axis includes approximately 11.5 boxes and each box has area (2)(1) = 2, so Ê0

f (x)dx ≈ −23.

The area above the x-axis includes approximately 2 boxes, each of area 2, so Ê5 So we have Ê0

f (x)dx =

Ê0

f (x)dx ≈ 4.

f (x)dx +

Ê5

f (x)dx ≈ −23 + 4 = −19.

(b) By the Fundamental Theorem of Calculus, we have F (7) − F (0) = so, F (7) = F (0) +

Ê0

f (x)dx

f (x)dx = 25 + (−19) = 6.

6.1 SOLUTIONS 9. Since F (0) = 0, F (b) = ∫0 f (t) dt. For each b we determine F (b) graphically as follows: F (0) = 0 F (1) = F (0) + Area of 1 × 1 rectangle = 0 + 1 = 1 F (2) = F (1) + Area of triangle ( 21 ⋅ 1 ⋅ 1) = 1 + 0.5 = 1.5 F (3) = F (2) + Negative of area of triangle = 1.5 − 0.5 = 1 F (4) = F (3) + Negative of area of rectangle = 1 − 1 = 0 F (5) = F (4) + Negative of area of rectangle = 0 − 1 = −1 F (6) = F (5) + Negative of area of triangle = −1 − 0.5 = −1.5 The graph of F (t), for 0 ≤ t ≤ 6, is shown in Figure 6.1. b

1.5 1

F (t) t 1

−1 −1.5

Figure 6.1 10. Finding the areas of the rectangular region under the curve between t = 0 and t = 5: G(5) = G(0) +

Ê0

g(t) dt = 4 + 2 ⋅ 5 = 14.

Similarly, fnding the area of trapezoidal or triangular regions G(10) = G(5) +

Ê5

g(t) dt = 14 +

1 ⋅ 5(2 + 4) = 29, 2

1 g(t) dt = 29 + ⋅ 10 ⋅ 4 = 49. Ê10 2 The integral between t = 20 and t = 25 is negative because g is negative, so G(20) = G(10) +

G(25) = G(20) +

Ê20

g(t) dt = 49 −

1 ⋅ 5 ⋅ 2 = 44. 2

11. By the Fundamental Theorem, 1

F ¨ (t) dt Ê0 = 5 − 1.5 = 3.5

F (1) = F (0) +

F ¨ (t) dt Ê1 = 3.5 − 1.5 = 2

F (2) = F (1) +

F ¨ (t) dt Ê2 = 2 − 0.5 = 1.5

F (3) = F (2) +

F ¨ (t) dt Ê3 = 1.5 + 0.5 = 2

F (4) = F (3) +

F ¨ (t) dt Ê4 = 2 + 0.5 = 2.5

F (5) = F (4) +

Thus, our table is as follows: Table 6.4 t

F (t)

3.5

1.5

2.5

445

446

Chapter Six /SOLUTIONS

12. (a) The area under the curve is a triangle with base 4 and height 3, so Ê0

1 ⋅ 4 ⋅ 3 = 6. 2

f (x) dx =

(b) Since f (x) is positive on this interval, the antiderivative F (x) is increasing on the interval. (c) The Fundamental Theorem tells us that Change in F = F (4) − F (0) =

Ê0

f (x) dx = 6.

Since F (0) = 20, we have F (4) = 20 + 6 = 26. 13. First, we observe that g is increasing when g ¨ is positive, which is when 0 < x < 4. g is decreasing when g ¨ is negative, which is when 4 < x < 6. Since a function has a local maximum at a point where its derivative is zero and when it decreases immediately before and decreases immediately after that point, we see that x = 4 is a local maximum. Likewise, since a function has a local minimum at a point where its derivative is zero and when it decreases immediately before and increases immediately after that point, we see that g has no local minima. Table 6.5 shows the area between the curve and the x-axis for the intervals 0–1, 1–2, etc. It also shows the corresponding change in the value of g. These changes are used to compute the values of g using the Fundamental Theorem of Calculus: g(1) − g(0) =

Ê0

g ¨ (x) dx =

1 . 2

Since g(0) = 0, g(1) =

1 . 2

Similarly, g(2) − g(1) =

Ê1

g ¨ (x) dx = 1

g(2) = g(1) + 1 =

3 . 2

Continuing in this way gives the values of g in Table 6.6. Table 6.5 Interval

Table 6.6 Area

Total change in g =

Êa

g ¨ (x)dx

g(x)

0–1

1/2

1∕2

1/2

1–2

3/2

2–3

5/2

3–4

1/2

1∕2

4–5

1/2

−1∕2

5/2

5–6

1/2

−1∕2

Notice: the graph of g will be a straight line from 1 to 3 because g ¨ is horizontal there. Furthermore, the tangent line will be horizontal at x = 4, x = 0 and x = 6. The maximum is at (4, 3). See Figure 6.2. y 3 2 1 x 1

Figure 6.2

6.1 SOLUTIONS

447

14. The rate of change is negative for t < 5 and positive for t > 5, so the concentration of adrenaline decreases until t = 5 and then increases. Since the area under the t-axis is greater than the area over the t-axis, the concentration of adrenaline goes down more than it goes up. Thus, the concentration at t = 8 is less than the concentration at t = 0. See Figure 6.3. adrenaline concentration ( g/ml)

t (minutes)

Figure 6.3

15. (a) f (x) is increasing when f ¨ (x) is positive. f ¨ (x) is positive when 2 < x < 5. So f (x) is increasing when 2 < x < 5. f (x) is decreasing when f ¨ (x) is negative. f ¨ (x) is negative when x < 2 or x > 5. So f (x) is decreasing when x < 2 or x > 5. A function has a local minimum at a point x when its derivative is zero at that point, and when it decreases immediately before x and increases immediately after x. f ¨ (2) = 0, f decreases to the left of 2, and f increases immediately after 2, therefore f (x) has a local minimum at x = 2. A function has a local maximum at a point x when its derivative is zero at that point, and when it increases immediately before x and decreases immediately after x. f ¨ (5) = 0, f increases before 5, and f decreases after 5. Therefore f (x) has a local maximum at x = 5. (b) Since we do not know any areas or vertical values, we can only sketch a rough graph. We start with the minimum and the maximum, then connect the graph between them. The graph could be more or less steep and further above or below the x-axis. See Figure 6.4. f (x)

x 1

Figure 6.4

16. (a) The function f is increasing where f ¨ is positive, so f is increasing for x < −1 or x > 1. The function f is decreasing where f ¨ is negative, so f is decreasing for −1 < x < 1. The function f has critical points at x = −1, 0, 1. The point x = −1 is a local maximum (because f is increasing to the left of x = −1 and decreasing to the right of x = −1). The point x = 1 is a local minimum (because f decreases to the left of x = 1 and increases to the right). The point x = 0 is neither a local maximum nor a local minimum, since f (x) is decreasing on both sides. (b) See Figure 6.5.

448

Chapter Six /SOLUTIONS f (x)

x −2

−1

Figure 6.5 17. See Figure 6.6. F (0) = 1

F (0) = 0

x 1

Figure 6.6 18. See Figure 6.7. 1

x 1 F (0) = 1

F (0) = 0

Figure 6.7 19. See Figure 6.8. 1 F (0) = 1

x 1

Figure 6.8

F (0) = 0

6.1 SOLUTIONS

449

20. See Figure 6.9.

F (0) = 1 1

F (0) = 0 x 1

Figure 6.9

21. See Figure 6.10.

F (0) = 1 1 F (0) = 0 x 1

Figure 6.10

22. See Figure 6.11. 1 F (0) = 1 x 1 F (0) = 0

Figure 6.11

23. We know that F (x) increases for x < 50 because the derivative of F is positive for x < 50. Similarly, F (x) decreases for x > 50 because F ¨ (x) is negative for x > 50. Therefore, the graph of F rises until the point at which x = 50, and then it begins to fall. So the highest point on the graph of F is at x = 50 and the maximum value attained by F is F (50). By the Fundamental Theorem: 50

F (50) − F (20) =

Ê20

F ¨ (x) dx.

Since F (20) = 150, we have 50

F (50) = F (20) +

Ê20

F ¨ (x) dx = 150 +

Ê20

F ¨ (x) dx.

The defnite integral is the area of the shaded region under the graph of F ¨ , which is roughly a triangle of base 30 and height 20. Therefore, the shaded area is about 300 and the maximum value attained by F is F (50) ≈ 150 + 300 = 450.

450

Chapter Six /SOLUTIONS

24. (a) The water stored in Lake Sonoma on March 1, 2014, is S(0) = 182,566 acre-feet. Since t = 3 is June 1, we fnd the water stored in Lake Sonoma then by starting with the water stored in March and adding the change per month for the next three months: S(3) = 182,566 + 3003 − 5631 − 8168 = 171,770 acre-feet. (b) We estimate the maxima and minima of S(t) by examining where the changes go from positive to negative values or vice versa. We can think of the change in water stored as S ¨ (t). When the data are positive more water is fowing into the reservoir than fowing out; when the data are negative more water is fowing out than in. Therefore, we have a local maximum at the beginning of April at t = 1 and a local minimum at the beginning of November at t = 8. (c) To locate an infection point, we compute the rate of change of the change in water stored per month, that is S ¨¨ (t). See Table 6.7. To estimate S ¨¨ (t) we fnd the di˙erence between successive months of S ¨ (t). Since S ¨¨ (t) changes sign only once, from negative in June to positive in July 2014, an infection point is located somewhere in that time. Until June, the water stored was decreasing more and more each month. Starting in July it was still decreasing but more slowly. Table 6.7 Month

Mar. 2014

Apr. 2014

May 2014

June 2014

July 2014

S ¨ (t)

3003

−5631

−8168

−8620

−8270

−8634

−2537

−452

350

S ¨¨ (t) Month

Aug. 2014

Sept. 2014

Oct. 2014

Nov. 2014

S ¨ (t)

−7489

−6245

−4593

54,743

S ¨¨ (t)

781

1244

1652

59,336

25. (a) The total volume emptied must increase with time and cannot decrease. The smooth graph (I) that is always increasing is therefore the volume emptied from the bladder. The jagged graph (II) that increases then decreases to zero is the fow rate. (b) The total change in volume is the integral of the fow rate. Thus, the graph giving total change (I) shows an antiderivative of the rate of change in graph (II). Alternatively, since (I) is always positive, any antiderivative of (I) will always be increasing. Since (II) both increases and decreases, it is clearly not an antiderivative of (I), so (I) must be an antiderivative of (II). 26. For every number b, the Fundamental Theorem tells us that Ê0

F ¨ (x) dx = F (b) − F (0) = F (b) − 0 = F (b).

Therefore, the values of F (1), F (2), F (3), and F (4) are values of defnite integrals. The defnite integral is equal to the area of the regions under the graph above the x-axis minus the area of the regions below the x-axis above the graph. Let A1 , A2 , A3 , A4 be the areas shown in Figure 6.12. The region between x = 0 and x = 1 lies above the x-axis, so F (1) is positive, and we have 1

F ¨ (x) dx = A1 . Ê0 The region between x = 0 and x = 2 also lies entirely above the x-axis, so F (2) is positive, and we have F (1) =

F (2) =

Ê0

F ¨ (x) dx = A1 + A2 .

We see that F (2) > F (1). The region between x = 0 and x = 3 includes parts above and below the x-axis. We have F (3) =

Ê0

F ¨ (x) dx = (A1 + A2 ) − A3 .

Since the area A3 is approximately the same as the area A2 , we have F (3) ≈ F (1). Finally, we see that F (4) =

Ê0

F ¨ (x) dx = (A1 + A2 ) − (A3 + A4 ).

6.1 SOLUTIONS

451

Since the area A1 + A2 appears to be larger than the area A3 + A4 , we see that F (4) is positive, but smaller than the others. The largest value is F (2) and the smallest value is F (4). None of the numbers are negative. Area = A1

❄F ¨ (x) ✛ Area = A2 3 2 1 Area = A3

✲

✛

x Area = A4

Figure 6.12 27. (a) The amount of oxygen, E(t), released by the leaf in an antiderivative of p(t). If we assume there is no oxygen released by the leaf at zero days of age, that is E(0) = 0, we have: amount of oxygen released E(t)

t (days)

Figure 6.13 (b) The infection point occurs at t = 4 days. This point tells us that the rate at which oxygen is released grows more and more each day up to the 4th day of life, then slows more and more each day after the 4th day of life. During the 4th day of life this leaf releases the most amount of oxygen per day. (c) The total amount of oxygen released during the frst ten days of life is given by the defnite integral Ê0

p(t) dt.

Alternatively, we can use the Fundamental theorem to describe this defnite integral as: E(10) − E(0) = E(10), since E(0) = 0. (d) The leaf releases more oxygen during the frst 10 days of its life. This is because the area under the graph of p(t) is much larger over the interval 0 ≤ t ≤ 10, than over the interval 10 ≤ t ≤ 20: Ê0

p(t) dt >

Ê10

p(t) dt

28. (a) In the beginning, both birth and death rates are small; this is consistent with a very small population. Both rates begin climbing, the birth rate faster than the death rate, which is consistent with a growing population. The birth rate is then high, but it begins to decrease after the population reaches a certain level. (b) bacteria/hour bacteria/hour B B−D

≈6

time (hours)

≈6

Figure 6.14: Di˙erence between B and D is greatest at t ≈ 6

time (hours)

452

Chapter Six /SOLUTIONS The bacteria population is growing most quickly when B − D, the rate of change of population, is maximal; that happens when B is farthest above D, which is at a point where the slopes of both graphs are equal. That point is t ≈ 6 hours. (c) Total number born by time t is the area under the B graph from t = 0 up to time t. See Figure 6.15. Total number alive at time t is the number born minus the number that have died, which is the area under the B graph minus the area under the D graph, up to time t. See Figure 6.16. bacteria bacteria

D N 5

time (hours)

Figure 6.15: Number born by time t is t ∫0 B(x) dx

≈ 11 15

time (hours)

Figure 6.16: Number alive at time t is t ∫0 (B(x) − D(x)) dx

From Figure 6.16, we see that the population is at a maximum when B = D, that is, after about 11 hours. This stands to reason, because B − D is the rate of change of population, so population is maximized when B − D = 0, that is, when B = D. 29. (a) Since P (t) is the total number of cases on or before day t, the graph of P (t) is always increasing or constant, never decreasing. Thus graphs (II) and (IV) must be P (t). Alternatively, note that since the number of new Covid-19 cases on a given day can never be negative, we know that N(t) ≥ 0. Since P ¨ (t) ≈ N(t), this tells us that the slopes of P (t) can never be negative, so the graph of P (t) never decreases. (b) Since P ¨ (t) ≈ N(t) and P (t) is an antiderivative of N(t), the peaks of N(t) correspond to the sharp upward sloping parts of P (t). Notice that the N(t) in (III) has peaks at about t = 50 and t = 150. Graph (IV) has two steeper sloping segments at the same t-values. Thus, (III) and (IV) represent the same location. Similarly, the N(t) in (I) has two high values at around t = 140 and t = 320 and (II) slopes steeply upward there. So (I) and (II) represent the same location. Side note: Graphs (I) and (II) are for Panama, and Graphs (III) and (IV) are for Singapore, where t = 0 is March 1, 2020. 30. (a) Since P (t) is the total number of cases on or before day t, the graph of P (t) is always increasing or constant, never decreasing. Thus graphs (II) and (III) must be P (t). Alternatively, note that since the number of new Covid-19 cases on a given day can never be negative, we know that N(t) ≥ 0. Since P ¨ (t) ≈ N(t), this tells us that the slopes of P (t) can never be negative, so the graph of P (t) never decreases. (b) Since P ¨ (t) ≈ N(t) and P (t) is an antiderivative of N(t), the peaks of N(t) correspond to the sharp upward sloping parts of P (t). Notice that the N(t) in (IV) has three peaks at about t = 50, t = 240 and t = 320. Graph (II) has three steeper sloping segments at the same t-values. Thus, (II) and (IV) represent the same location. Similarly, the N(t) in (I) has two high values at around t = 280 and t = 400 and (III) slopes steeply upward there. So (III) and (I) represent the same location. Side note: Graphs (I) and (III) are for Moldova, and Graphs (II) and (IV) are for Ireland, where t = 0 is March 1, 2020. 31. The graph of P (t) starts at P = 0 as New Zealand has no cases when t = 0. Since P (t) is the total number of cases, the function is always increasing (or constant if there are no new cases). P (t) is the antiderivative of N(t). Right at the start, N(t) was small, so P (t) climbed very slowly. Then came a big spike in N(t) which led to a steep increase in P (t). Later on, after the reduction in N(t), the slope of P (t) was greatly reduced, frst to almost zero (between about t = 50 and t = 150) and then to small positive values. See Figure 6.17. Estimating the number of cases on the vertical axis requires the Fundamental Theorem of Calculus: the size of a jump is the area under a peak. Using more detailed data, we can get the graph in Figure 6.18.

6.1 SOLUTIONS P (total cases)

453

P (total cases) 2500 1500 500

100

200

300

t (days)

100

Figure 6.17

200

300

t (days)

Figure 6.18

32. The graph of P (t) starts at P = 0 as Hawaii had no cases when t = 0. Since P (t) is the total number of cases, the function is always increasing (or constant if there are no new cases). P (t) is the antiderivative of N(t). Early in the year, at about t = 40, we see N(t) has a small peak. This causes P (t) to climb. But since this peak is much smaller than the later peaks, the increase in P is tiny in comparison to later increases. Around t = 100, the value of N is almost 0, so the graph of P is approximately horizontal there. At about t = 175, there is a large peak in N which corresponds to a large jump in P . Between t = 200 and t = 300, the value of N is lower but positive, so P continues to climb, albeit more slowly. At about t = 320, another peak in N leads to another climb in P ; this climb is smaller than the one at t = 175. Then, as N drops, P starts to level o˙. See Figure 6.19. Estimating the number of cases on the vertical axis requires the Fundamental Theorem of Calculus. The size of a jump in P is determined by the area under the peak. With more detailed data, we can get the graph in Figure 6.20. P (total cases)

P (total cases) 30,000 20,000 10,000

100

200

300

t (days)

100

Figure 6.19

200

300

t (days)

Figure 6.20

33. The critical points are at (0, 5), (2, 21), (4, 13), and (5, 15). A graph is given in Figure 6.21. y

(2, 21)

G(t) (5, 15)

15 (4, 13) 10 5

(0, 5) t 1

Figure 6.21

34. We can start by fnding four points on the graph of F (x). The frst one is given: F (2) = 3. By the Fundamental Theorem of 6 Calculus, F (6) = F (2) + ∫2 F ¨ (x)dx. The value of this integral is −7 (the area is 7, but the graph lies below the x-axis), so

454

Chapter Six /SOLUTIONS F (6) = 3 − 7 = −4. Similarly, F (0) = F (2) − 2 = 1, and F (8) = F (6) + 4 = 0. We sketch a graph of F (x) by connecting these points, as shown in Figure 6.22. (2, 3) F (x)

(8, 0)

(0, 1)

x 4

8 (6, −4)

Figure 6.22

35. We see that F decreases when x < 1.5 or x > 4.67, because F ¨ is negative there. F increases when 1.5 < x < 4.67, because F ¨ is positive there. So F has a local minimum at x = 1.5. F has a local maximum at x = 4.67. We have F (0) = 14. Since F ¨ is negative between 0 and 1.5, the Fundamental Theorem of Calculus gives us 1.5

F ¨ (x) dx = −34 Ê0 F (1.5) = 14 − 34 = −20.

F (1.5) − F (0) =

Similarly 4.67

F (4.67) = F (1.5) +

Ê1.5

F ¨ (x) dx = −20 + 25 = 5. 6

F (6) = F (4.67) +

Ê4.67

F ¨ (x) dx = 5 − 5 = 0.

A graph of F is in Figure 6.23. The local maximum is (4.67, 5) and the local minimum is (1.5, −20). 14 10

F (x) x 1

−10 −20

Figure 6.23

36. The areas given enable us to calculate the changes in the function F as we move along the t-axis. Areas above the axis count positively and areas below the axis count negatively. We know that F (0) = 3, so F (2) − F (0) =

Ê0

F ¨ (t) dt =

Area under F ¨ =5 0≤t≤2

Thus, F (2) = F (0) + 5 = 3 + 5 = 8. Similarly, F (5) − F (2) =

Ê2

F ¨ (t) dt = −16

6.1 SOLUTIONS

455

F (5) = F (2) − 16 = 8 − 16 = −8 and F (6) = F (5) +

Ê5

F ¨ (t) dt = −8 + 10 = 2.

A graph is shown in Figure 6.24. (2, 8) F (t) (6, 2) (0, 3) t

(5, −8)

Figure 6.24

37. (a) The integral is the area above the x-axis counted positively plus the area below the x-axis counted negatively, so the value of the integral is negative. The area below the x-axis is a triangle with base 3 and height 12; the area above the x-axis is a triangle with base 1 and height 4. So Ê0

1 1 f (x) dx = − ⋅ 3 ⋅ 12 + ⋅ 1 ⋅ 4 = −18 + 2 = −16. 2 2

(b) The Fundamental Theorem tells us that Change in F = F (4) − F (0) =

Ê0

f (x) dx = −16.

Since F (0) = 100, we have F (4) = 100 + (−16) = 84. 38. (a) Critical points of F (x) are the zeros of f : x = 1 and x = 3. (b) F (x) has a local minimum at x = 1 and a local maximum at x = 3. (c) See Figure 6.25. x 1

F (x)

Figure 6.25 Notice that the graph could also be above or below the x-axis at x = 3. 39. (a) Critical points of F (x) are x = −1, x = 1 and x = 3. (b) F (x) has a local minimum at x = −1, a local maximum at x = 1, and a local minimum at x = 3. (c) See Figure 6.26.

456

Chapter Six /SOLUTIONS

x −2

4 F (x)

Figure 6.26 40. Since F (x) is an antiderivative of f , we know F ¨ (x) = f (x). (a) We need to know where F ¨ (x) is positive. We have that F ¨ (x) = f (x) = 1 − x2 . Thus, we want 0 < 1 − x2 so x2 < 1. Thus, F ¨ (x) is positive for −1 < x < 1, so F (x) is increasing on that interval. (b) We need to know where F ¨¨ (x) is positive. We know F ¨ (x) = 1 − x2 so F ¨¨ (x) = −2x. Thus, F ¨¨ (x) > 0 when x < 0. The graph of F (x) is concave up for x < 0. 41. See Figure 6.27. Point of inflection f (x)

❄ x1

Figure 6.27 42. See Figure 6.28. Inflection point Local min

❄

❄ f (x) ✛ Inflection point

❄ x1

Local max

Figure 6.28 43. By the Fundamental Theorem, f (1) − f (0) =

Ê0

f ¨ (x) dx,

Since f ¨ (x) is negative for 0 ≤ x ≤ 1, this integral must be negative and so f (1) < f (0).

6.1 SOLUTIONS

457

44. First rewrite each of the quantities in terms of f ¨ , since we have the graph of f ¨ . If A1 and A2 are the positive areas shown in Figure 6.29: f (3) − f (2) = f (4) − f (3) =

Ê2

Ê3

f ¨ (t) dt = −A1 f ¨ (t) dt = −A2

4 A + A2 f (4) − f (2) 1 = f ¨ (t) dt = − 1 2 2 2 Ê2

Since Area A1 > Area A2 , A2 < so

A1 + A2 < A1 2

−A1 < − and therefore f (3) − f (2) <

A1 + A2 < −A2 2

f (4) − f (2) < f (4) − f (3). 2

y 3

2 1

A2 y = f ¨ (x)

Figure 6.29 45. See Figure 6.30. f (x)

✻

f (b) − f (a)

❄ x

Figure 6.30 46. See Figure 6.31. (a) Slope= f (b)−f b−a

f (x)

✻ ❘

f (b) − f (a)

❄ x

Figure 6.31

458

Chapter Six /SOLUTIONS

47. See Figure 6.32. f (x)

Figure 6.32

48. See Figure 6.33. Note that we are using the interpretation of the defnite integral as the length of the interval times the average value of the function on that interval, which we developed in Section 5.6. f (x)

✻ F (b)−F (a) b−a

❄x

Figure 6.33

Solutions for Section 6.2 1. Since, using the chain rule, F ¨ (x) = 2e2x = f (x), F (x) is an antiderivative of f (x). 2. Since, using the chain rule, F ¨ (x) = 2e2x = f (x), F (x) is an antiderivative of f (x). � F ¨ (x) = 5 2e2x = 10e2x ≠ f (x),

3. Since, using the chain rule, F (x) is not an antiderivative of f (x).

� F ¨ (x) = e2x + x 2e2x = (1 + 2x)e2x ≠ f (x),

4. Since F (x) is not an antiderivative of f (x).

� F ¨ (x) = 2 2e2x = 4e2x ≠ f (x),

5. Since F (x) is not an antiderivative of f (x).

6. Since the defnite integral in the expression for F (x) is a constant, its derivative is zero and we have F ¨ (x) = Thus, F (x) is an antiderivative of f (x).

d � 2x e + constant = 2e2x + 0 = f (x), dx

6.2 SOLUTIONS

459

7. Recall that F (x) is an antiderivative of f (x) if F ¨ (x) = f (x), so to determine which are antiderivatives, we di˙erentiate each and check whether the derivative is equal to f (x). We have d x∕2 1 e = ex∕2 ≠ f (x) dx 2 d 1 (2ex∕2 ) = 2 ⋅ ex∕2 = ex∕2 = f (x) dx 2 d d 1∕2 x∕2 1 (1+x)∕2 (2e ) = 2⋅ (e e ) = 2e1∕2 ex∕2 ⋅ = e1∕2 ex∕2 ≠ f (x) dx dx 2 d 1 (2ex∕2 + 1) = 2 ⋅ ex∕2 = ex∕2 = f (x) dx 2 0 21 2 2 d x2 ∕4 d x x x ∕4 e =e ⋅ = ex ∕4 ⋅ ≠ f (x). dx dx 4 2 Therefore, (II) and (IV) are antiderivatives. 8. Recall that F (x) is an antiderivative of f (x) if F ¨ (x) = f (x), so to determine which are antiderivatives, we di˙erentiate each and check whether the derivative is equal to f (x). We have d 1 ln x = = f (x) dx x d 1 d − 2 = (−x−2 ) = 2x−3 ≠ f (x) dx dx x d 1 (ln x + ln 3) = + 0 = f (x) dx x d 1 1 ln(2x) = ⋅ 2 = = f (x) dx 2x x d 1 ln(x + 1) = ≠ f (x). dx x+1 Therefore, (I), (III) and (IV) are antiderivatives. 9. Recall that F (x) is an antiderivative of f (x) if F ¨ (x) = f (x), so to determine which are antiderivatives, we di˙erentiate each and check whether the derivative is equal to f (x). We have d (−2 sin x cos x) = −2 sin x(− sin x) + (−2 cos x)(cos x) = 2 sin2 x − 2 cos2 x ≠ f (x) dx

d (2 cos2 x − 2 sin2 x) = −4 cos x sin x − 4 sin x cos x = −8 sin x cos x ≠ f (x) dx d (sin2 x) = 2 sin x cos x = f (x) dx d (− cos2 x) = −2 cos x(− sin x) = 2 sin x cos x = f (x) dx d (2 sin2 x + cos2 x) = 4 sin x cos x + 2 cos x(− sin x) = 2 sin x cos x = f (x). dx Therefore, (III), (IV) and (V) are antiderivatives. 10. Since 5+

2f (x) dx = 5 + 2

f (x) dx,

and ∫ f (x) dx is a family of functions, the expression is a family of functions. 11. Since this is a defnite integral, the expression is a number, not a family of functions. 12. Since 5 is a constant, an antiderivative of 5 is 5x + C, so the expression is a family of functions. 13. Since f (5) is a constant, an antiderivative of f (5) is f (5)x + C, so the expression is a family of functions. 14. Since the expression is an indefnite integral, it is a family of functions.

460

Chapter Six /SOLUTIONS

15. Since the expression is a defnite integral, it is a number. 16. The frst term of the expression is a defnite integral, so it is a number. The second term is an indefnite integral, hence a family of functions. Since adding a number to a family of functions gives us a new family of functions, the expression is a family of functions. 17. 18. 19.

5 2 t 2 1 3 x 3 1 3 t + 12 t2 3

20. 5x x5 21. . 5 t8 t4 22. + . 8 4 5q 3 23. . 3 x4 3x4 + 4x. 24. 6( ) + 4x = 4 2 25. We break the antiderivative into two terms. Since y3 is an antiderivative of 3y2 and −y4 ∕4 is an antiderivative of −y3 , an antiderivative of 3y2 − y3 is y4 y3 − . 4 x4 ) = 10x + 2x4 . 4 27. P (r) = r2 + C x2 x6 x−4 28. Antiderivative F (x) = + − +C 2 6 4 26. 10x + 8(

29. 32 z 2 x3 x2 x3 − 6( ) + 17x = − 3x2 + 17x. 3 2 3 5 2 3 31. x2 − x 2 2 3 t4 t3 t2 − − 32. 4 6 2 1 33. − t 34. Since g(z) = z−3 , an antiderivative of g(z) is 30.

z−3+1 z−2 1 = = − 2. −3 + 1 −2 2z 35. F (z) = ez + 3z + C x7 1 x−5 x7 1 36. F (x) = − ( )+C = + x−5 + C 7 7 −5 7 35 1 1 37. ln |x| − − 2 + C x 2x √ √ 38. Since ( z)3 = z3∕2 , an antiderivative of ( z)3 is z(3∕2)+1 2 = z5∕2 . (3∕2) + 1 5 e−3t −e−3t = . −3 3 40. sin t 39.

41. G(t) = 5t + sin t + C 42. G( ) = − cos − 2 sin + C

6.2 SOLUTIONS 43. Since g ¨ (x) =

461

d � 1∕2 1 −1∕2 1 2x =2 x = √ , dx 2 x

we have g ¨ (x) = f (x). So g(x) is an antiderivative of f (x). 44. Since

d (cos x − sin x) = − sin x − cos x, dx we have g ¨ (x) = f (x). So g(x) is an antiderivative of f (x). g ¨ (x) =

45. Since

d 2 3x 2 e = (3)e3x = 2e3x , dx 3 3 we have f ¨ (x) = g(x). So f (x) is an antiderivative of g(x). f ¨ (x) =

46. Since

1 d � −1 x + x = (−1)x−2 + 1 = 1 − 2 , dx x we have g ¨ (x) = f (x). So g(x) is an antiderivative of f (x). g ¨ (x) =

47. f (x) = 3, so F (x) = 3x + C. F (0) = 0 implies that 3 ⋅ 0 + C = 0, so C = 0. Thus F (x) = 3x is the only possibility. 48. f (x) = 2 + 4x + 5x2 , so F (x) = 2x + 2x2 + 53 x3 + C . F (0) = 0 implies that C = 0. Thus F (x) = 2x + 2x2 + 35 x3 is the only possibility. 2

49. f (x) = 14 x, so F (x) = x8 + C . F (0) = 0 implies that 81 ⋅ 02 + C = 0, so C = 0. Thus F (x) = x2 ∕8 is the only possibility. 50. f (x) = x1∕2 , so F (x) = 23 x3∕2 + C. F (0) = 0 implies that 23 ⋅ 03∕2 + C = 0, so C = 0. Thus F (x) = 32 x3∕2 is the only possibility. x3 x3 03 + C. F (0) = 0 implies that + C = 0, so C = 0. Thus F (x) = is the only possibility. 51. f (x) = x2 , so F (x) = 3 3 3 d x (e ) = ex , we take F (x) = ex + C. Now 52. Since dx F (0) = e0 + C = 1 + C = 0, so C = −1 and F (x) = ex − 1. 53. Since P and Q are antiderivatives of p, we know Q(t) = P (t) + C. Therefore, since Q(2) = P (2) + C and since Q(2) = 25, we have 25 = 24 + C C = 25 − 16 = 9. Thus Q(t) = P (t) + C = t4 + 9. 54. 52 x2 + 7x + C 55. 3x3 + C. −1 −0.05t e + C = −20e−0.05t + C. 56. 0.05 57. p + ln |p| + C t13 58. + C. 13 x3 x−1 x3 1 + +C = − +C 59. (x2 + x−2 ) dx = Ê 3 −1 3 x t3 t2 2 60. (t + 5t + 1) dt = + 5 ⋅ + t + C Ê 3 2 61. 5ez + C 2 62. 3 ln |t| + + C t

Chapter Six /SOLUTIONS

462

t4 t3 t4 + 6 ⋅ + C = + 2t3 + C 4 3 4 3∕2 w 3w1∕2 dw = 3 ⋅ + C = 2w3∕2 + C 64. Ê 3∕2 x3 + 2x2 − 5x + C 65. 3 e2t + C. 66. 2 2 x + 2x1∕2 + C 67. 2 x4 5x3 (x3 + 5x2 + 6) dx = + + 6x + C 68. Ê 4 3 69. ex + 5x + C x3 70. + ln |x| + C. 3 6 x − 3x4 + C 71. 6 1 72. e3r dr = e3r + C Ê 3 d 73. Since cos t = − sin t, we have dt 63.

(t3 + 6t2 ) dt =

Ê 74.

25e−0.04q dq = 25

sin t dt = − cos t + C, where C is a constant.

1 e−0.04q + C = −625e−0.04q + C −0.04

75. 25e4x + C 76. sin + C 77. 2 ln |x| − cos x + C 1 78. − cos(3x) + C 3 79. 3 sin x + 7 cos x + C 80. 2 sin(3x) + C 81. 10x − 4 cos(2x) + C 82. 2ex − 8 sin x + C 83. −6 cos(2x) + 3 sin(5x) + C x2 84. 2 ln |x| + 4 √ √ 85. Since x x = x(x1∕2 ) = x3∕2 , an antiderivative of x x is: x(3∕2)+1 2 = x5∕2 . (3∕2) + 1 5 √ √ √ √ 86. Since x∕ x = x(x−1∕2 ) = x1∕2 , and similarly x∕x = x1∕2 (x−1 ) = x−1∕2 , an antiderivative of x∕ x + x∕x is: √ x(1∕2)+1 x(−1∕2)+1 2 + = x3∕2 + 2 x. (1∕2) + 1 (−1∕2) + 1 3 e2x e−2x e2x + e−2x − = . 2 −2 2 0 1 − cos(7x) cos(7x) = − 7 cos x. 88. −7 cos x − 7 7 87.

89. An antiderivative is F (x) = 3x2 −5x+C. Since F (0) = 5, we have 5 = 0+C, so C = 5. The answer is F (x) = 3x2 −5x+5.

6.2 SOLUTIONS

463

x3 + x + C. Since F (0) = 5, we have 5 = 0 + C, so C = 5. The answer is F (x) = x3 ∕3 + x + 5. 3 91. An antiderivative is F (x) = −4 cos(2x) + C. Since F (0) = 5, we have 5 = −4 cos 0 + C = −4 + C, so C = 9. The answer is F (x) = −4 cos(2x) + 9. 90. An antiderivative is F (x) =

92. An antiderivative is F (x) = 2e3x + C. Since F (0) = 5, we have 5 = 2e0 + C = 2 + C, so C = 3. The answer is F (x) = 2e3x + 3. 93. (a) The marginal revenue, MR, is given by di˙erentiating the total revenue function, R, with respect to q so dR = MR. dq Therefore, R=

MR dq

(20 − 4q) dq

= 20q − 2q 2 + C. We can check this by noting

dR d � = 20q − 2q 2 + C = 20 − 4q = MR. dq dq

When no goods are produced the total revenue is zero so C = 0 and the total revenue is R = 20q − 2q 2 . (b) The total revenue, R, is given by pq where p is the price, so the demand curve is p=

R = 20 − 2q. q

94. The marginal cost, MC, is given by di˙erentiating the total cost function, C, with respect to q so dC = MC. dq Therefore, C=

MC dq �

3q 2 + 4q + 6 dq

= q 3 + 2q 2 + 6q + D,

where D is a constant. We can check this by noting dC d � 3 = q + 2q 2 + 6q + D = 3q 2 + 4q + 6 = MC. dq dq The fxed costs are given to be 200 so C = 200 when q = 0, thus D = 200. The total cost function is C = q 3 + 2q 2 + 6q + 200. 95. Since C ¨ (x) = 4000 + 10x we want to evaluate the indefnite integral Ê

(4000 + 10x) dx = 4000x + 5x2 + K

where K is a constant. Thus C(x) = 5x2 +4000x+K, and the fxed cost of 1,000,000 riyal means that C(0) = 1,000,000 = K. Therefore, the total cost is C(x) = 5x2 + 4000x + 1,000,000 riyals. Since C(x) depends on x2 , the square of the depth drilled, costs will increase dramatically when x grows large.

464

Chapter Six /SOLUTIONS

96. (a) I: Integrating we have C(t) = 1.73t + C0 . Substituting t = 0 gives C0 = 339, so C(t) = 1.73t + 339. II: Integrating we have C(t) = 1.24t + 0.03 Substituting t = 0 gives C0 = 339, so

t2 + C0 . 2

C(t) = 1.24t + 0.015t2 + 339.

III: Integrating we have C(t) =

1.4 0.005t e + C0 0.005

Substituting t = 0 and C(0) = 339, we have 339 = 280e0.005(0) + C0 339 = 280 + C0 C0 = 59. Thus C(t) = 280e0.005t + 59. (b) In 2030, we have t = 50, so I C(50) = 1.73 ⋅ 50 + 339 = 425.5 ppm. II C(50) = 1.24 ⋅ 50 + (0.015)502 + 339 = 438.5 ppm. III C(50) = 280e0.005(50) + 59 = 418.527 ppm.

Solutions for Section 6.3 1. Since F ¨ (x) = 6x, we use F (x) = 3x2 . By the Fundamental Theorem, we have Ê0

|4 | 6xdx = 3x2 | = 3 ⋅ 42 − 3 ⋅ 02 = 48 − 0 = 48. | |0

2. Since F ¨ (x) = 5, we use F (x) = 5x. By the Fundamental Theorem, we have Ê1 3. If F ¨ (t) = t3 , then F (t) =

|3 | 5dx = 5x| = 5(3) − 5(1) = 15 − 5 = 10. | |1

t4 . By the Fundamental Theorem, we have 4 Ê0

t3 dt = F (3) − F (0) =

34 0 t4 || 81 − = . | = 4 ||0 4 4 4

4. Since F ¨ (x) = 12x2 + 1, we use F (x) = 4x3 + x. By the Fundamental Theorem, we have |2 | (12x2 + 1)dx = (4x3 + x)| = (4 ⋅ 23 + 2) − (4 ⋅ 03 + 0) = 34 − 0 = 34. | Ê0 |0 2

6.3 SOLUTIONS 5. If f (t) = 3t2 + 4t + 3, then F (t) = t3 + 2t2 + 3t. By the Fundamental Theorem, we have |2 | (3t2 + 4t + 3) dt = (t3 + 2t2 + 3t)| = 23 + 2(22 ) + 3(2) − 0 = 22. | Ê0 |0 2

6. If f (x) = 1∕x, then F (x) = ln |x| (since

d 1 ln |x| = ). By the Fundamental Theorem, we have dx x 2 |2 1 | dx = ln |x|| = ln 2 − ln 1 = ln 2. | Ê1 x |1

√ 1 7. Since F ¨ (x) = √ = x−1∕2 , we use F (x) = 2x1∕2 = 2 x. By the Fundamental Theorem, we have x 4

Ê1

√ √ √ ||4 1 √ dx = 2 x|| = 2 4 − 2 1 = 4 − 2 = 2. x |1

8. If f (q) = 6q 2 + 4, then F (q) = 2q 3 + 4q. By the Fundamental Theorem, we have |1 | (6q + 4) dq = (2q + 4q)| = 2(1) + 4(1) − (0 + 0) = 6. | Ê0 |0 1

9. Since F ¨ (x) = 3x2 , we take F (x) = x3 . Then Ê0

3x2 dx = F (5) − F (0) = 53 − 03 = 125.

|1 | 2ex dx = 2ex | = 2e − 2 ≈ 3.437. | Ê0 |0 11. Since F ¨ (t) = 5t3 , we take F (t) = 45 t4 . Then 1

10.

Ê1

5t3 dt = F (2) − F (1) 5 5 4 (2 ) − (14 ) 4 4 5 5 = ⋅ 16 − 4 4 75 = 4

12. If F ¨ (x) = 6x2 , then F (x) = 2x3 . By the Fundamental Theorem, we have Ê1

13. Since F ¨ (y) = y2 + y4 , we take F (y) = Ê0

|3 | 6x2 dx = 2x3 | = 2(27) − 2(1) = 54 − 2 = 52. | |1 y3 y5 + . Then 3 5 1

(y2 + y4 ) dy = F (3) − F (0) 0 3 1 0 3 1 15 0 05 1 = + − + 3 5 3 5 1 1 8 = + = . 3 5 15

465

Chapter Six /SOLUTIONS

466

14. Since F ¨ (x) =

1 x−1 1 = x−2 , we use F (x) = = − . By the Fundamental Theorem, we have −1 x x2 Ê1

15.

Ê2

(x3 − x2 ) dx =

2 1 || 1 1 1 1 1 dx = − = − − − = − +1 = . | x ||1 2 1 2 2 x2

0 4 1 5 x3 || x 609 − − 39 ≈ 29.728. | = | 4 4 3 |2

d 16. If f (t) = e−0.2t , then F (t) = −5e−0.2t . (This can be verifed by observing that (−5e−0.2t ) = e−0.2t .) By the Fundamental dt Theorem, we have 1 |1 | e−0.2t dt = (−5e−0.2t )| = −5(e−0.2 ) − (−5)(1) = 5 − 5e−0.2 ≈ 0.906. | Ê0 |0 17. If F ¨ (t) = cos t, we can take F (t) = sin t, so |1 | cos t dt = sin t| = sin 1 − sin(−1). | Ê−1 |−1 1

Since sin(−1) = − sin 1, we can simplify the answer and write 1

Ê−1

18.

Ê0

∕4

| ∕4 | (sin t + cos t) dt = (− cos t + sin t)| = | |0

cos t dt = 2 sin 1

H √ √ I 2 2 − + − (−1 + 0) = 1. 2 2

d 19. If f (t) = e0.05t , then F (t) = 20e0.05t (you can check this by observing that (20e0.05t ) = e0.05t ). By the Fundamental dt Theorem, we have |3 | e0.05t dt = 20e0.05t | = 20e0.15 − 20e0 = 20(e0.15 − 1). | Ê |0 20. Since F ¨ (x) =

√ x3∕2 2 x, we take F (x) = = x3∕2 . Then 3∕2 3 Ê4

9√

x dx = F (9) − F (4) 2 3∕2 2 3∕2 ⋅9 − ⋅4 3 3 2 2 = ⋅ 27 − ⋅ 8 3 3 38 = . 3 =

21. To apply the Fundamental Theorem of Calculus, we need a function F (t) such that F ¨ (t) = f (t) = 6t. We try F (t) = 3t2 and check that F ¨ (t) = 6t. Then the area under the graph is given by Ê0

6t dt = F (5) − F (0) = 3 ⋅ 52 − 3 ⋅ 02 = 75.

In Figure 6.34, the shaded region under the graph is a triangle with base 5 and height 6⋅5 = 30, so the area is 21 ⋅5⋅30 = 75.

6.3 SOLUTIONS

467

f (t)

t 5

Figure 6.34 22. To apply the Fundamental Theorem of Calculus, we need a function F (t) such that F ¨ (t) = f (t) = 10 − 2t. We try F (t) = 10t − t2 and check that F ¨ (t) = 10 − 2t. Then the area under the graph is given by Ê0

(10 − 2t) dt = F (5) − F (0) = (10 ⋅ 5 − 52 ) − (10 ⋅ 0 − 02 ) = 25.

In Figure 6.35, the shaded region under the graph is a triangle with base 5 and height 10 so the area is 12 ⋅ 5 ⋅ 10 = 25. 10

f (t)

t 5

Figure 6.35 23. To apply the Fundamental Theorem of Calculus, we need a function F (t) such that F ¨ (t) = f (t) = t. We try F (t) = 12 t2 and check that F ¨ (t) = t. Then the area under the graph is given by Ê0

t dt = F (5) − F (0) =

1 2 1 25 ⋅5 − ⋅0= . 2 2 2

In Figure 6.36, the shaded region under the graph is a triangle with base 5 and height 5. The area of the triangle is 1 ⋅ 5 ⋅ 5 = 25 . 2 2 5

f (t)

t 5

Figure 6.36

468

Chapter Six /SOLUTIONS

24. To apply the Fundamental Theorem of Calculus, we need a function F (t) such that F ¨ (t) = f (t) = −2t. We try F (t) = −t2 and check that F ¨ (t) = −2t. Then the integral is given by Ê0

−2t dt = F (5) − F (0) = −52 − (−02 ) = −25.

In Figure 6.37, the shaded region under the graph is a triangle below the t-axis, so the area is the negative of the integral, that is, 25. Since the base of the triangle is 5 and the height is 10, the area is 21 ⋅ 5 ⋅ 10 = 25. 5 t

f (t)

−10

Figure 6.37

25. Solving x2 = x shows that the curves intersect at x = 0 and x = 1. See Figure 6.38. Note that x is greater than x2 on the interval 0 < x < 1. To fnd the area between the curves for 0 ≤ x ≤ 1 we can compute the area under the the top curve y = x and the area under the the bottom curve y = x2 separately. Then we subtract the area under the bottom curve to compute the di˙erence in areas. This is the same as integrating the top curve y = x minus the bottom curve y = x2 . Thus Area between curves =

Ê0

(x − x2 ) dx =

1 1 2 1 3 || 1 1 1 2 1 3 1 x − x | = ⋅ 12 − ⋅ 13 − ⋅0 − ⋅0 = . 3 2 3 6 2 3 |0 2

y = x2

y=x

Area between curves 1

❄

x 1

Figure 6.38

26. We begin by fnding the point of intersection where f (x) = g(x): 2x = 6 − x 3x = 6 x = 2. See Figure 6.39.

6.3 SOLUTIONS

469

Note that g(x) is greater than f (x) on the interval 0 < x < 2. To fnd the area between the curves for 0 ≤ x ≤ 2 we can compute the area under the the top curve y = g(x) and the area under the the bottom curve y = f (x) separately. Then we subtract the area under the bottom curve to compute the di˙erence in areas. This is the same as integrating the top curve y = g(x) minus the bottom curve y = f (x): Area =

Ê0

6 − x − 2x dx =

Ê0

2 3 | 3 3 6 − 3x dx = 6x − x2 || = 6 ⋅ 2 − ⋅ 22 − 6 ⋅ 0 − ⋅ 02 = 6. 2 |0 2 2

Area between curves

f (x) = 2x

❄ g(x) = 6 − x

x 2

Figure 6.39

27. Since y = x3 − x = x(x − 1)(x + 1), the graph crosses the axis at the three points shown in Figure 6.40. The two regions have the same area (by symmetry). Since the graph is below the axis for 0 < x < 1, we have 0 1 1 � 3 Area = 2 − x − x dx Ê0 4 4 51 x x2 1 1 1 − = . = −2 − = −2 4 2 0 4 2 2 y y = x3 − x

x −1

Figure 6.40

470

Chapter Six /SOLUTIONS

28. One antiderivative of f (x) = e0.5x is F (x) = 2e0.5x . Thus, the defnite integral of f (x) on the interval 0 ≤ x ≤ 3 is Ê0

|3 | e0.5x dx = F (3) − F (0) = 2e0.5x | . | |0

The average value of a function on a given interval is the defnite integral over that interval divided by the length of the interval: H I 1 0 3 |3 1 1 1 0.5x 0.5x | Average value = ⋅ e dx = 2e | = (2e1.5 − 2e0 ) ≈ 2.32. | Ê0 3−0 3 3 |0 From the graph of y = e0.5x in Figure 6.41 we see that an average value of 2.32 on the interval 0 ≤ x ≤ 3 does make sense. y 5 y = e0.5x

2.32

x 3

Figure 6.41

29. We have Area =

Ê0

x2 dx =

b3 x3 || | = . | 3 |0 3

We fnd the value of b making the area equal to 100: b3 3 300 = b3

100 =

b = (300)1∕3 = 6.694.

30. We have Area =

Ê1

|b 2|

4x dx = 2x | = 2b2 − 2. | |1

We fnd the value of b making the area equal to 240: 240 = 2b2 − 2 242 = 2b2 121 = b2 b = 11.

31. (a) Since r gives the rate of energy use, between 2020 and 2025 (where t = 0 and t = 5), we have Total energy used =

Ê0

583.9e0.013t dt exajoules.

(b) The Fundamental Theorem of Calculus states that Êa

f (t) dt = F (b) − F (a)

6.3 SOLUTIONS

471

provided that F ¨ (t) = f (t). To apply this theorem, we need to fnd F (t) such that F ¨ (t) = 583.9e0.013t ; we take 583.9 0.013t e = 44,915.385e0.013t . 0.013

F (t) = Thus, Total energy used =

Ê0

583.9e0.013t dt = F (5) − F (0) |5 | = 44,915.385e0.013t | | |0 0.065 = 44,915.385(e − e0 ) = 3016.473 exajoules.

Approximately 3016 exajoules of energy will be consumed between 2020 and 2025. 32. (a) Since t = 0 in 2019 and t = 99 in 2118, by the Fundamental Theorem of Calculus, we have Change in population =

Ê0

2.303e0.007t dt

|99 2.303 0.007t || = e | 0.007 | |0 � 2.303 0.007⋅99 − e0 = e 0.007 = 329 million. (b) Note that 329 million is both the 2019 population and the increase between 2019 and 2118. This is as we expect, since the doubling time is 99 years and we have 2118 population = 2019 population + Change in population = 329 + 329 = 658 million = 2 × 2019 population. 33. (a) At time t = 0, the rate of oil leakage = r(0) = 50 thousand liters/minute. At t = 60, rate = r(60) = 15.06 thousand liters/minute. (b) To fnd the amount of oil leaked during the frst hour, we integrate the rate from t = 0 to t = 60: |60 | 50e Oil leaked = | | Ê0 |0 −1.2 0 + 2500e = 1747 thousand liters. = −2500e 60

−0.02t

50 −0.02t dt = − e 0.02

34. (a) In the frst case, we are given that R0 = 1000 widgets/year. So we have R = 1000e0.125t . To determine the total number sold, we need to integrate this rate over the time period from 0 to 10. Therefore Total number of widgets sold =

Ê0

1000e0.125t dt = 19,923 widgets.

In the second case, Total number of widgets sold =

Ê0

1,000,000e0.125t dt = 19.9 million widgets.

(b) We want to determine T such that Ê0

1000e0.125t dt =

19,923 . 2

Trying a few values of T , we get T ≈ 6.47 years.

472

Chapter Six /SOLUTIONS Similarly, in the second case, we want T so that Ê0

1,000,000e0.125t dt =

19,900,000 2

we get T ≈ 6.47 years. So the half way mark is reached at the same time regardless of the initial rate. (c) Since half the widgets are sold in the last 3 12 years of the decade, if each widget is expected to last at least 3.5 years, their claim could easily be true. 35. (a) Using a calculator or computer, we get 3

Ê0 Ê1

e−2t dt = 0.4988

e−2t dt = 0.4999996

Ê0

e−2t dt = 0.49998 e−2t dt = 0.499999999.

The values of these integrals are getting closer to 0.5. A reasonable guess is that the improper integral converges to 0.5. (b) Since − 12 e−2t is an antiderivative of e−2t , we have Ê0

|b 1 1 1 1 1 | e−2t dt = − e−2t | = − e−2b − − e0 = − e−2b + . | 2 2 2 2 2 |0

b → ∞,

e−2b =

1 → 0. e2b

Therefore,

1 1 1 1 =0+ = . e−2t dt = lim − e−2b + b→∞ b→∞ Ê0 2 2 2 2 So the improper integral converges to 1∕2 = 0.5: b

lim

Ê0 36. (a) An antiderivative of F ¨ (x) =

∞

e−2t dt =

1 . 2

1 1 d −1 1 is F (x) = − since = . So by the Fundamental Theorem we have: x dx x x2 x2 Ê1

. (b) Taking a limit, we have

b 1 1 1 || = − + 1. dx = − | x ||1 b x2

1 lim − + 1 = 0 + 1 = 1. b→∞ b

Since the limit is 1, we know that b→∞ Ê1

lim

So the improper integral converges to 1: Ê1

∞

1 dx = 1. x2

6.3 SOLUTIONS ∞

37. (a) The graph of y = e−x is in Figure 6.42. The integral shaded.

Ê−∞

473

e−x dx represents the entire area under the curve, which is

f (x) = e−x

x −5

−3

Figure 6.42 (b) Using a calculator or computer, we see that 1

Ê−1

Ê−2

e−x dx = 1.494,

e−x dx = 1.764,

Ê−3

e−x dx = 1.772,

Ê−5

e−x dx = 1.772

(c) From part (b), we see that as we extend the limits of integration, the area appears to get closer and closer to about 1.772. We estimate that ∞ Ê−∞

e−x dx = 1.772

38. Figure 6.43 shows the graphs of y = 1∕x2 and y = 1∕x3 . We see that ∫1 larger than the area under 1∕x3 .

∞ 1 dx is larger, since the area under 1∕x2 is x2

1∕x2 1∕x3

✠ ✠x 1

Figure 6.43

39. The original dose equals the quantity of drug eliminated. The quantity of drug eliminated is the defnite integral of the rate. Thus, letting t → ∞, we have the improper integral Total quantity of drug eliminated = Using the fact that ∫ ekt dt = k1 ekt + C, we have

Ê0

∞

50(e−0.1t − e−0.2t ) dt.

b 1 −0.1t 1 −0.2t || Total quantity = lim 50 − e + e | b→∞ 0.1 0.2 |0 = lim 50(−10e−0.1b + 5e−0.2b − (−10e0 + 5e0 )) b→∞

−0.1b

Since e

−0.2b

→ 0 and e

→ 0 as b → ∞, we have Total quantity = 50(10 − 5) = 250 mg.

474

Chapter Six /SOLUTIONS

40. (a) The total number of people that get sick is the integral of the rate. The epidemic starts at t = 0. Since the rate is positive for all t, we use ∞ for the upper limit of integration. Total number getting sick = −0.5t

(b) The graph of r = 1000te sick.

Ê0

∞

�

1000te−0.5t dt

is shown in Figure 6.44. The shaded area represents the total number of people who get r

Figure 6.44 41. (a) No, it is not reached. We have Total number of rabbits =

Ê1

∞

1 dt t2

b→∞ Ê1

= lim

1 dt t2

b 1 || = lim − | b→∞ t ||1 1 1 = lim − − − b→∞ b 1 =1

since −1∕b → 0 as b → ∞. Thus, the total number of rabbits reached is 1000. (b) Yes, it is reached. We have Total number of rabbits =

Ê1

∞

t dt

b→∞ Ê1

= lim

t dt b

t2 || | b→∞ 2 | |1 0 2 1 1 b = lim − b→∞ 2 2 = lim

∞

t dt does not converge to a fnite value, which means that infnitely many Ê1 rabbits could be produced, and therefore 1 million is certainly reached. (c) Yes, it is reached. We have Since b2 ∕2 → ∞ as b → ∞, the integral

Total number of rabbits =

Ê1

∞

1 √ dt t

b→∞ Ê1

= lim

t− 2 dt

|b 1| = lim 2t 2 | | b→∞ |1 √ = lim 2 b − 2 b→∞

6.4 SOLUTIONS √ Since 2 b → ∞ as b → ∞, the integral

475

∞

1 dt does not converge to a fnite value, which means that infnitely Ê1 √t many rabbits could be produced, and therefore 1 million is certainly reached. 42. (a) A calculator or computer gives Ê1

100

Ê1

1 √ dx = 18 x

1000

1 √ dx = 61.2 x

Ê1

10000

1 √ dx = 198. x

These values do not seem to be converging.

H I √ 1 d √ 1 (b) An antiderivative of F (x) = √ is F (x) = 2 x since (2 x) = √ . So, by the Fundamental Theorem, we dx x x have b √ √ √ √ ||b 1 dx = 2 x| = 2 b − 2 1 = 2 b − 2. √ | Ê1 x |1 √ √ (c) The limit of 2 b − 2 as b → ∞ does not exist, as b grows without bound. Therefore b √ 1 lim dx = lim (2 b − 2) does not exist. √ b→∞ Ê1 b→∞ x ¨

So the improper integral

Ê1

∞

1 √ dx does not converge. x

Solutions for Section 6.4 1. (a) The equilibrium price is $30 per unit, and the equilibrium quantity is 6000. (b) The region representing the consumer surplus is the shaded triangle in Figure 6.45 with area 12 ⋅ 6000 ⋅ 70 = 210,000. The consumer surplus is $210,000. The area representing the producer surplus, shaded in Figure 6.46, is about 7 grid squares, each of area 10,000. The producer surplus is about $70,000. p ($/unit) 100

p ($/unit) 100 S

50 D 5000

D 10000

q (quantity)

5000

Figure 6.45

10000

q (quantity)

Figure 6.46

2. Looking at the graph we see that the supply and demand curves intersect at roughly the point (345, 8). Thus the equilibrium price is $8 per unit and the equilibrium quantity is 345 units. Figures 6.47 and 6.48 show the shaded areas corresponding to the consumer surplus and the producer surplus. Counting grid squares we see that the consumer surplus is roughly $2000 while the producer surplus is roughly $1400. p ($/unit)

p ($/unit)

Supply

500

Demand q (quantity) 1000

Figure 6.47: Consumer surplus

Supply

500

Demand q (quantity) 1000

Figure 6.48: Producer surplus

476

Chapter Six /SOLUTIONS

3. When q ∗ = 5, the equilibrium price is p∗ = 100 − 3 ⋅ 52 = 25. Then Ê0

Consumer Surplus =

(100 − 3q 2 ) dq − 5 ⋅ 25.

Using a calculator or computer to evaluate the integral we get Consumer Surplus = 250. Alternatively, using the Fundamental Theorem of Calculus, we get Consumer Surplus =

Ê0

(100 − 3q 2 ) dq − 5 ⋅ 25

5 � || = 100q − q 3 | − 125 | |0 = 500 − 125 − 125

= 250. 4. To calculate the producer surplus we need to fnd the market equilibrium price, p∗ , and quantity q ∗ . In equilibrium, when supply equals demand, 35−q 2 = 3+q 2 so q 2 = 16. We need only consider the positive solution q ∗ = 4. The corresponding equilibrium price is p∗ = 35 − 42 = 19. From Figure 6.30 on page 320 of the text, Producer surplus = q ∗ p∗ −

Ê0

q∗

(3 + q 2 ) dq = 4 ⋅ 19 −

Ê0

(3 + q 2 ) dq.

Using a calculator or computer to evaluate the integral we get Producer surplus = 42.667. Using the Fundamental Theorem of Calculus, we get: Producer surplus = 4(19) −

Ê0

(3 + q 2 ) dq

0 1 4 q 3 || = 76 − 3q + | 3 ||0 = 76 − (3(4) + =

64 − 0) 3

128 . 3

5. When q ∗ = 10, the equilibrium price is p∗ = 100 − 4 ⋅ 10 = 60. Then Consumer surplus =

Ê0

(100 − 4q) dq − 60 ⋅ 10.

Using a calculator or computer to evaluate the integral we get Consumer surplus = 200. Using the Fundamental Theorem of Calculus, we get: Consumer surplus =

Ê0

(100 − 4q) dq − 60 ⋅ 10 10

|| � = 100q − 2q 2 | − 600 | |0 = 1000 − 200 − 600 = 200.

6.4 SOLUTIONS

477

6. (a) Solving this system of linear equations we fnd that the equilibrium price is p = $10 and equilibrium quantity is q = 15. The consumer surplus is the area under the demand curve and above the line p = 10. We have Consumer surplus =

1 ⋅ 7.5 ⋅ 15 = 56.25 dollars. 2

(b) The producer surplus is the area above the supply curve and below the line p = 10. We have Producer surplus =

1 ⋅ 2 ⋅ 15 = 15 dollars. 2

See Figure 6.49. p ($/unit) 17.5 p∗ = 10 8

Consumer surplus

✠ ✛

S Producer surplus D q ∗ = 15

q (quantity in units)

Figure 6.49

7. (a) We fnd the equilibrium price by setting the supply equal to the demand. We have 100 − 2p = 3p − 50 5p = 150 p = 30. Hence the equilibrium price is $30. We can fnd the equilibrium quantity by substituting p = 30 into either of the equations. So the equilibrium quantity is q = 100 − 2 ⋅ 30 = 40. The consumer surplus is the area under the demand curve and above the line p = 30. We have Consumer surplus =

1 ⋅ 20 ⋅ 40 = 400 dollars. 2

(b) The producer surplus is the area above the supply curve and below the line p = 30. We have Producer surplus =

1 40 ⋅ ⋅ 40 = 266.7 dollars. 2 3

See Figure 6.50. p ($/unit) 50 p∗ = 30

Consumer surplus S

✠

✛

Producer surplus

50∕3 D q ∗ = 40

Figure 6.50

100

q (quantity in units)

478

Chapter Six /SOLUTIONS

8. (a) The quantity demanded at a price of $50 is calculated by substituting p = 50 into the demand equation p = 100e−0.008q . Solving 50 = 100e−0.008q for q gives q ≈ 86.6. In other words, at a price of $50, consumer demand is about √ 87 units. The quantity supplied at a price of $50 is calculated by substituting by p = 50 into the supply equation p = 4 q + 10. √ Solving 50 = 4 q + 10 for q gives q = 100. So at a price of $50, producers supply about 100 units. At a price of $50, the supply is larger than the demand, so some goods remain unsold. We can expect prices to be pushed down. (b) The supply and demand curves are shown in Figure 6.51. The equilibrium price is about p∗ = $48 and the equilibrium quantity is about q ∗ = 91 units. The market will push prices downward from $50 toward the equilibrium price of $48. This agrees with the conclusion to part (a) that prices will drop. p ($/unit) S

p∗ = 48 D q ∗ = 91

q (quantity in units)

Figure 6.51: Demand and supply curves for a product (c) See Figure 6.52. We have Consumer surplus = Area between demand curve and horizontal line p = p∗ . The demand curve has equation p = 100e0.008q . Using a calculator or computer to evaluate the defnite integral, we have 91

100e−0.008q dq − p∗ q ∗ = 6464 − 48 ⋅ 91 = 2096. Ê0 Using instead the Fundamental Theorem of Calculus to evaluate the defnite integral, we get: Consumer surplus =

Ê0

100e−0.008q dq = 100

|91 1 | e−0.008q | = 6464. | −0.008 |0

Consumers gain $2096 by buying goods at the equilibrium price instead of the price they would have been willing to pay. For producer surplus, see Figure 6.53. We have Producer surplus = Area between supply curve and horizontal line p = p∗ √ The supply curve has equation p = 4 q + 10. Using a calculator or computer to evaluate the defnite integral, we get Producer surplus = p∗ q ∗ −

Ê0

√ (4 q + 10) dq = 48 ⋅ 91 − 3225 = 1143.

Using instead the Fundamental Theorem of Calculus to evaluate the integral, we get 0 0 1 1 |91 91 √ 1 | (4 q + 10) dq = 4 q 3∕2 + q | = 3225. | Ê0 3∕2 |0 Producers gain $1143 by supplying goods at the equilibrium price instead of the price at which they would have been willing to provide the goods. p ($/unit)

p ($/unit) Consumer surplus

p∗

✠

p∗ D

q∗

q (quantity)

Figure 6.52: Consumer surplus

✛

Producer surplus D

q∗

q (quantity)

Figure 6.53: Producer surplus

6.4 SOLUTIONS

479

9. (a) Consumer surplus is greater than producer surplus in Figure 6.54. price

price Supply

✲

Consumer surplus

✲

Consumer surplus

Producer surplus

p∗

Supply

✲ q∗

Producer surplus

Demand

✲

Demand quantity

q∗

Figure 6.54

quantity

Figure 6.55

(b) Producer surplus is greater than consumer surplus in Figure 6.55. 10. (a) Looking at the fgure in the problem we see that the equilibrium price is roughly $30 giving an equilibrium quantity of 125 units. (b) Consumer surplus is the area above p∗ and below the demand curve. Graphically this is represented by the shaded area in Figure 6.56. From the graph we can estimate the shaded area to be roughly 14 squares where each square represents ($25/unit)⋅(10 units). Thus the consumer surplus is approximately 14 ⋅ $250 = $3500. p ($/unit)

p ($/unit)

100

Consumer surplus

✲ p∗

p∗ Producer 20✲ surplus

100

200

q (quantity)

100

Figure 6.56

200

q (quantity)

Figure 6.57

Producer surplus is the area under p∗ and above the supply curve. Graphically this is represented by the shaded area in Figure 6.57. From the graph we can estimate the shaded area to be roughly 8 squares where each square represents ($25/unit)⋅(10 units). Thus the producer surplus is approximately 8 ⋅ $250 = $2000 (c) We have Total gains from trade = Consumer surplus + producer surplus = $3500 + $2000 = $5500. 11. (a) The consumer surplus is the area the between demand curve and the price $40—roughly 9 squares. See Figure 6.58. Since each square represents ($25/unit)⋅(10 units), the total area is 9 ⋅ $250 = $2250.

480

Chapter Six /SOLUTIONS At a price of $40, about 90 units are sold. The producer surplus is the area under $40, above the supply curve, and to the left of q = 90. See Figure 6.58. The area is 10.5 squares or 10.5 ⋅ $250 = $2625. The total gains from the trade is Total gain = Consumer surplus + Producer surplus = $4875. p (price/unit) 100 80 Consumer ✲ surplus 60 40 Producer surplus

✲

100

150

200

250

q (quantity)

Figure 6.58

(b) The consumer surplus is less with the price control. The producer surplus is greater with the price control. The total gains from trade are less with the price control. 12. (a) To fnd the equilibrium price and quantity, we fnd the price at which quantity supplied is equal to quantity demanded: 10p − 30 = −2p + 30 12p = 60 p = 5. At a price of 5, the quantity supplied is equal to the quantity demanded, −2 ⋅ 5 + 30 = 20. So the equilibrium price is 5 dollars, and the equilibrium quantity is 20 million. (b) To fnd the consumer surplus, we fnd the area between the demand curve and the line p = 5 from q = 0 to q = 20. This area is equal to the integral of the demand curve minus the area of the rectangle under the line p = 5. We frst write the demand curve in terms of price: q = −2p + 30 q − 30 = −2p q p = 15 − . 2 Now we integrate: 0 Consumer surplus =

Ê0

15 −

q dq 2

1 − Area rectangle

= 15q −

q 2 || | − 5 ⋅ 20 4 ||0

= 15 ⋅ 20 −

202 − 100 = 100 million dollars. 4

See Figure 6.59. Similarly, for producer surplus, we fnd the area between the supply curve and the line p = 5 from q = 0 and q = 20. This area is equal to the area of the rectangle under the line p = 5 minus the integral of the supply curve. We

6.4 SOLUTIONS

481

frst write the supply curve in terms of price: q = 10p − 30 q + 30 = 10p q . 10

p = 3+ Now we integrate:

0 Producer surplus = Area rectangle −

Ê0

q dq 10

|20

= 5 ⋅ 20 − 3q −

q2 | | 20 ||0

= 100 − 3 ⋅ 20 −

202 = 20 million dollars. 20

See Figure 6.59. p (price/unit)

Consumer surplus = 100

✠

Producer surplus = 20 Supply: p = 3 + q∕10

✠

Demand: p = 15 − q∕2 q (quantity)

Figure 6.59

13. (a) It appears that the equilibrium price is p∗ = 6 dollars per unit and the equilibrium quantity is q ∗ = 400 units. See Figure 6.60. p ($/unit) 12

4 D 200

400

600

800

1000

q (quantity)

Figure 6.60

(b) To fnd the consumer surplus, we estimate the area shaded in Figure 6.61 by counting grid squares. There appear to be about 5.5 grid squares in this shaded area, and each grid square has area 200, so the total area is about 1100. The consumer surplus is about $1100. Similarly, to fnd the producer surplus, we estimate the area shaded in Figure 6.62 to be about 4 grid squares, for a total area of 800. The producer surplus is about $800.

482

Chapter Six /SOLUTIONS Consumer surplus

p ($/unit) 12

✠

Producer surplus

p ($/unit) 12 8

✠

200

600

400

D q (quantity) 800 1000

200

400

Figure 6.61

600

D q (quantity) 800 1000

Figure 6.62

(c) The consumer surplus is given by the area shaded in Figure 6.63. This area is about 6 grid squares, for a total area of about 1200. The consumer surplus is about $1200, which is larger than the consumer surplus of $1100 at the equilibrium price. The producer surplus is the area shaded in Figure 6.64. This area is about 200, so the producer surplus is about $200. This is less than the producer surplus of $800 at the equilibrium price. Notice that the sum of the consumer surplus and the producer surplus is $1900 at the equilibrium price and is $1400 at the artifcial price. The total gains from trade are always lower at an artifcial price. p ($/unit) 12

p ($/unit) 12

4 D 200

400

600

800 1000

q (quantity)

Figure 6.63

200

400

600

D q (quantity) 800 1000

Figure 6.64

14. Figure 6.65 shows the consumer and producer surplus for the price, p− . For comparison, Figure 6.66 shows the consumer and producer surplus at the equilibrium price. price

price

Supply

✲

Consumer surplus p∗ p− Producer surplus

✲

Consumer surplus p∗

Demand

✲ q−

q∗

Figure 6.65

quantity

Producer surplus

✲

Demand quantity

Figure 6.66

(a) The producer surplus is the area on the graph between p− and the supply curve. Lowering the price also lowers the producer surplus. (b) The consumer surplus — the area between the supply curve and the line p− — may increase or decrease depends on the functions describing the supply and demand, and the lowered price. (For example, the consumer surplus seems to be increased in Figure 6.65 but if the price were brought down to $0 then the consumer surplus would be zero, and hence clearly less than the consumer surplus at equilibrium.) (c) Figure 6.65 shows that the total gains from the trade are decreased.

6.4 SOLUTIONS

483

15. (a) In Table 6.3, the quantity q increases as the price p decreases, while in Table 6.4, q increases as p increases. Therefore, the demand data is in Table 6.3 and the supply data is in Table 6.4. (b) It appears that the equilibrium price is p∗ = 25 dollars per unit and the equilibrium quantity is q ∗ = 400 units sold at this price. (c) To estimate the consumer surplus, we use the demand data in Table 6.3. We use a Riemann sum using the price from the demand data minus the equilibrium price of 25. Left sum = (60 − 25) ⋅ 100 + (50 − 25) ⋅ 100 + (41 − 25) ⋅ 100 + (32 − 25) ⋅ 100 = 8300. Right sum = (50 − 25) ⋅ 100 + (41 − 25) ⋅ 100 + (32 − 25) ⋅ 100 + (25 − 25) ⋅ 100 = 4800. We average the two to estimate that Consumer surplus ≈

8300 + 4800 = 6550. 2

To estimate the producer surplus, we use the supply data in Table 6.4. We use a Riemann sum using the equilibrium price of 25 minus the price from the supply data. Left sum = (25 − 10) ⋅ 100 + (25 − 14) ⋅ 100 + (25 − 18) ⋅ 100 + (25 − 22) ⋅ 100 = 3600. Right sum = (25 − 14) ⋅ 100 + (25 − 18) ⋅ 100 + (25 − 22) ⋅ 100 + (25 − 25) ⋅ 100 = 2100. We average the two to estimate that Producer surplus ≈

3600 + 2100 = 2850. 2

16. (a) If the price is artifcially high, the consumer surplus at the artifcial price is always less than the consumer surplus at the equilibrium price, but the producer surplus may be larger or smaller. See Figure 6.67. (b) If the price is artifcially low, the producer surplus at the artifcial price is always less than the producer surplus at the equilibrium price, but the consumer surplus may be larger or smaller. See Figure 6.68.

Consumer surplus

p ($/unit)

Large producer surplus

p ($/unit)

✠

p ($/unit)

✠

q (quantity)

Small producer surplus

q (quantity)

Figure 6.67

p ($/unit)

Small consumer p ($/unit) surplus

Large consumer surplus

p ($/unit)

✠

✠ ✛

Producer surplus q (quantity)

q (quantity)

Figure 6.68 17. The supply curve, S(q), represents the minimum price p per unit that the suppliers will be willing to supply some quantity q of the good for. See Figure 6.69. If the suppliers have q ∗ of the good and q ∗ is divided into subintervals of size Δq, then if the consumers could o˙er the suppliers for each Δq a price increase just suÿcient to induce the suppliers to sell an additional Δq of the good, the consumers’ total expenditure on q ∗ goods would be É p1 Δq + p2 Δq + ⋯ = pi Δq.

q (quantity)

484

Chapter Six /SOLUTIONS q∗

q∗

S(q) dq is the amount the consumers would Ê0 Ê0 pay if suppliers could be forced to sell at the lowest price they would be willing to accept. S(q) dq. Thus

As Δq → 0 the Riemann sum becomes the integral

Price S(q)

P2 P1

Δq

Quantity

q∗

Δq

Figure 6.69

18. Ê0

q∗

(p∗ − S(q)) dq =

Ê0

q∗

p∗ dq −

= p∗ q ∗ −

Ê0

q∗

S(q) dq

q∗

S(q) dq.

Using Problem 17, this integral is the extra amount consumers pay (i.e., suppliers earn over and above the minimum they would be willing to accept for supplying the good). It results from charging the equilibrium price. 19. (a) p∗ q ∗ = the total amount paid for q ∗ of the good at equilibrium. See Figure 6.70. q∗ (b) ∫0 D(q) dq = the maximum consumers would be willing to pay if they had to pay the highest price acceptable to them for each additional unit of the good. See Figure 6.71. price

price

Supply ∶ S(q) p∗

q∗

Figure 6.70

Demand ∶ D(q) quantity

q∗

Demand ∶ D(q) quantity

Figure 6.71

(c) ∫0 S(q) dq = the minimum suppliers would be willing to accept if they were paid the minimum price acceptable to them for each additional unit of the good. See Figure 6.72. q∗ (d) ∫0 D(q) dq − p∗ q ∗ = consumer surplus. See Figure 6.73. q∗

6.5 SOLUTIONS price

485

price

Supply ∶ S(q)

p∗

q∗

Demand ∶ D(q) quantity

q∗

Figure 6.72

Demand ∶ D(q) quantity

Figure 6.73

(e) p∗ q ∗ − ∫0 S(q) dq = producer surplus. See Figure 6.74. q∗

(f) ∫0 (D(q) − S(q)) dq = producer surplus and consumer surplus. See Figure 6.75. q∗

price

Supply ∶ S(q)

Supply ∶ S(q) p∗

p∗

q∗

Demand ∶ D(q) quantity

q∗

Figure 6.74

Demand ∶ D(q) quantity

Figure 6.75

Solutions for Section 6.5 1.

$/year

t (years from present)

The graph reaches a peak each summer, and a trough each winter. The graph shows sunscreen sales increasing from cycle to cycle. This gradual increase may be due in part to infation and to population growth. 2. Using a calculator or computer, we get: Present value =

Ê0

12000e−0.06t dt = $139,761.16.

Using instead the Fundamental Theorem of Calculus, we get: P =

Ê0

12000e−0.06t dt

|20 1 | e−0.06t | | −0.06 |0 � −1.2 = −200000 e − e0 = $139,761.16

= 12000

486

Chapter Six /SOLUTIONS We use the present value to fnd the future value: Future value = P erM = 139761.16e0.06(20) = $464,023.39.

3. Using a calculator of computer to evaluate the defnite integral, we fnd the present value: Present value =

Ê0

2000e−0.05t dt = $21,105.

Using instead the Fundamental Theorem of Calculus, we get: Present value =

Ê0

2000e−0.05t dt

|15 1 | e−0.05t | | −0.05 |0 � −0.75 = −40000 e − e0 = $21,105.

= 2000

We use the present value to fnd the future value: Future value = P erM = 21105e0.05(15) = $44,680. 4. (a) We frst fnd the present value, P , of the income stream. Using a calculator or computer, we get P =

Ê0

5000e−0.02t dt = $36,964.053.

Using instead the Fundamental Theorem of Calculus, we get: P =

Ê0

5000e−0.02t dt

|8 1 | e−0.02t | | −0.02 |0 � −0.16 0 = −250, 000 e − e = $36,964.053

= 5000

We use the present value to fnd the future value: Future value = P erM = 36964.053e0.02(8) = $43,377.718. (b) A stream of income of $5000 per year invested in a bank account paying 2% annual interest rate will grow to $43,377.72 in 8 years. Alternatively, one lump sum deposit of $36,964.05 today in a bank account with an interest rate of 2% compounded continuously will grow to $43,377.72 in 8 years. 5. (a) The future value is Future value = 10,000e0.03(10) = $13,498.59. (b) We fnd the present value of the income stream frst, using a calculator or computer to evaluate the defnite integral: Present value =

Ê0

1000e−0.03t dt = $8,639.39.

Using instead the Fundamental Theorem of Calculus, we get: Present value =

Ê0

1000e−0.03t dt

|10 1 −0.03t | = 1000 e | | −0.03 |0 � −0.3 0 = −33333 e − e = $8,639.39

We use the present value to fnd the future value. Future value = 8639.39e0.03(10) = $11,661.96. (c) Although we deposit the exact same amount in the two situations, the future value is larger for the lump sum. It is always fnancially preferable to receive the money earlier rather than later, since it has more time to earn interest.

6.5 SOLUTIONS

487

6. (a) The future value is Future value = 12,000e0.05(6) = $16,198.31. (b) We fnd the present value of the income stream frst. Using a calculator or computer to evaluate the defnite integral, we get 6

2000e−0.05t dt = $10,367.27. Ê0 Using instead the Fundamental Theorem of Calculus, we get: Present value =

Ê0

Present value =

2000e−0.05t dt

|6 1 | e−0.05t | | −0.05 |0 � −0.3 0 = −40000 e − e = $10,367.27.

= 2000

We use the present value to fnd the future value. Future value = 10,367.27e0.05(6) = $13,994.35. (c) Although we deposit the exact same amount in the two situations, the future value is larger for the lump sum. It is always fnancially preferable to receive the money earlier rather than later, since it has more time to earn interest. 7. (a) We frst compute the present value of this income stream, using a calculator or computer to evaluate the defnite integral: 15

1000e−0.05t dt = $10,552.67. Ê0 Using instead the Fundamental Theorem of Calculus, we get: Present value =

Present value =

Ê0

1000e−0.05t dt

|15 1 | e−0.05t | | −0.05 |0 � −0.75 = −20000 e − e0 = $10,552.67.

= 1000

We use the present value to fnd the future value: Future value = 10,552.67e0.05(15) = $22,340.00. After 15 years, the account will contain $22,340. (b) The person has deposited $1000 every year for 15 years, for a total of $15,000. (c) The total interest earned is $22,340 − $15,000 = $7340.00. 8. (a) We frst compute the present value of this income stream, using a calculator or computer to evaluate the defnite integral: 30

1000e−0.06t dt = $13,911.69. Ê0 Using instead the Fundamental Theorem of Calculus, we get: Present value =

Present value =

Ê0

1000e−0.06t dt

|30 1 | e−0.06t | | −0.06 |0 � −1.8 = −16666.67 e − e0 = $13,911.69.

= 1000

We use the present value to fnd the future value: Future value = 13,911.69e0.06(30) = $84,160.82. After 30 years, the account will contain $84,160.82. (b) The person has deposited $1000 every year for 30 years, for a total of $30,000. (c) The total interest earned is $84,160.82 − $30,000 = $54,160.82.

488

Chapter Six /SOLUTIONS

9. (a) We frst fnd the present value, P , of the income stream. Using a calculator or computer, we get P =

Ê0

6000e−0.05t dt = $47,216.32.

Using instead the Fundamental Theorem of Calculus, we get: P =

Ê0

6000e−0.05t dt

|10 1 | e−0.05t | | −0.05 |0 � −0.5 = −120, 000 e − e0 = $47,216.32

= 6000

We use the present value to fnd the future value, F : F = P erM = 47126.32e0.05(10) = $77,846.55. (b) The income stream contributed $6000 per year for 10 years, or $60,000. The interest earned was 77,846.55−60,000 = $17,846.55. 10. (a)

(i) Using a calculator or computer with an interest rate of 3%, we have Present value =

Ê0

5000e−0.03t dt = $18,846.59.

Using instead the Fundamental Theorem of Calculus, we get: P =

Ê0

5000e−0.03t dt

|4 1 | e−0.03t | | −0.03 |0 � −0.12 = −166666.67 e − e0 = $18,846.59

= 5000

(ii) If the interest rate is 10%, using a calculator or computer we get Present value =

Ê0

5000e−0.10t dt = $16,484.00.

Using instead the Fundamental Theorem of Calculus, we get: P =

Ê0

5000e−0.10t dt

|4 1 −0.10t | = 5000 e | | −0.10 |0 � −0.4 0 = −50000 e − e = $16,484.00

(b) At the end of the four-year period, if the interest rate is 3%, Value = 18,846.59e0.03(4) = $21,249.47. At 10%, Value = 16,484.00e0.10(4) = $24,591.24. 11. The present value of an income stream of $120,000 per year at 2% per year for 20 years is given by the Fundamental Theorem of Calculus: Ê0

120000e−0.02t dt =

20 � 120000 −0.02t || e | = 6,000,000 1 − e−0.02⋅20 = 1,978,080 dollars. | −0.02 |0

Thus, an o˙er of a $2 million is a reasonable deal for Company A, since it is greater than the present value of its expected income stream. We do not have enough information to determine whether this is a reasonable deal for Company B. Either it is making a mistake, or its directors believe that the owners of Company A are underestimating the income stream, or there are synergies making Company A more valuable as part of Company B than as an independent company.

6.5 SOLUTIONS

489

12. (a) The present value of the net sales over this fve-year time period is given by Present value =

Ê0

(7.4 + 0.19t)e−0.02t dt = 37.43 billion dollars.

(b) The future value, on January 1, 2021 is given by Value 5 years later = 37.43e0.02(5) = 41.37 billion dollars. The value, on January 1, 2021, of Hershey’s net sales during this time period is 41.37 billion dollars. 13. We begin by fnding the present value. Since the fee was charged at a continuous rate of 200 euros per year for time, t, in years between t = 0 in 1805 and t = 203 in 2008, with a continuous interest rate of 3% per year, the present value, P , is given by P =

Ê0

203

200e−0.03t dt.

Using a calculator or computer, we get P =

Ê0

203

200e−0.03t dt = 6651.56394 euros.

Using instead the Fundamental Theorem of Calculus, we get: P =

Ê0

203

200e−0.03t dt

|203 1 | e−0.03t | | −0.03 |0 � 200 −0.03⋅203 = e − e−0.03⋅0 = 6651.56394 euros. −0.03

= 200

To convert to future value, 203 years later, we multiply the P by e0.03(203) : 6651.56394e0.03(203) = 2,936,142.74 euros. Thus, Schiller would have owed nearly 3 million euros. 14. (a) The present value of the revenue during the frst year is the sum of the present value during the frst six months (from t = 0 to t = 1∕2) and the present value during the second six months (from t = 1∕2 to t = 1). Present value of revenue = during the frst year Ê0

1∕2

(45,000 + 60,000t)e−0.07t dt +

Ê1∕2

75,000e−0.07t dt

= 29,438.08 + 35,583.85 = 65,021.93. The present value of the revenue earned by the machine during the frst year of operation is about 65,022 dollars. (b) The present value over a time interval 0 ≤ t ≤ T with T > 1∕2 is Present value during = frst T years Ê0

1∕2

(45,000 + 60,000t)e−0.07t dt +

Ê1∕2

75,000e−0.07t dt

= 29,438.08 +

Ê1∕2

75,000e−0.07t dt.

If the present value of revenue equals the cost, then T

150,000 = 29,438.08 +

Ê1∕2

75,000e−0.07t dt.

Solving for the integral, we get T

120,561.92 =

Ê1∕2

75,000e−0.07t dt.

Trying a few values for T gives T ≈ 2.27. It will take approximately 2.27 years for the present value of the revenue to equal the cost of the machine.

490

Chapter Six /SOLUTIONS

15. (a) The future value in 10 years is $100,000. We frst fnd the present value, P : 100000 = P e0.10(10) P = $36,787.94 We solve for the income stream S, using a calculator or computer to evaluate the defnite integral: 36,787.94 =

Ê0

Se−0.10t dt 10

e−0.10t dt Ê0 36,787.94 = S(6.321)

36,787.94 = S

Using instead the Fundamental Theorem of Calculus to evaluate the defnite integral, we get Ê0

e−0.10t dt =

|10 1 | e−0.10t | = 6.321. | −0.10 |0

Solving for S we get 36,787.94 = $5820.00 per year. 6.321 The income stream required is about $5820 per year (or about $112 per week). (b) The present value is $36,787.94. This is the amount that would have to be deposited now. S=

16. (a) If P is the present value, then the value in two years at 9% interest is P e0.09(2) : 500,000 = P e0.09(2) 500,000 P = 0.09(2) = 417, 635.11 e The present value of the renovations is $417,635.11. (b) If money is invested at a constant rate of $S per year, then Present value of deposits =

Ê0

Se−0.09t dt

Since S is constant, we can take it out in front of the integral sign: Present value of deposits = S

Ê0

e−0.09t dt

We want the rate S so that the present value is 417,635.11. Evaluating the integral with a calculator, we get 417,635.11 = S(1.830330984). Using instead the Fundamental Theorem of Calculus, we get: Ê0

e−0.09t dt =

|2 1 | e−0.09t | = 1.830330984 | −0.09 |0

Solving for S, we have 417,635.11 = 228, 174.64. 1.830330984 Money deposited at a continuous rate of 228,174.64 dollars per year and earning 9% interest per year has a value of $500,000 after two years. S=

6.5 SOLUTIONS

491

17. We compute the present value of the company’s earnings over the next 8 years. Using a calculator or computer to evaluate the integral, we get 8

Ê0 Using instead the Fundamental Theorem of Calculus, we get: Present value of earnings =

Ê0

Present value of earnings =

50,000e−0.07t dt = $306,279.24.

50,000e−0.07t dt

|7 1 | e−0.07t | | −0.07 |0 � −0.49 = −714,285.71 e − e0 = $306,279.24.

= 50,000

If you buy the rights to the earnings of the company now for $350,000, you want the present value to be worth more than $350,000. Since the present value of the earnings is less than this amount, you should not buy. 18. At any time t, the company receives income of s(t) = 50,000e−t dollars per year. Thus the present value is Present value =

Ê0

s(t)e−0.06t dt 2

(50,000e−t )e−0.06t dt Ê0 = $41,508,

where we used a calculator or computer to evaluate the integral. If we simplify the integrand and use instead the Fundamental Theorem of Calculus, we get Ê0

�

50,000e−t e−0.06t dt =

Ê0

50,000e−1.06t dt

2 � || 1 e−1.06t | | 1.06 |0 � −2.12 0 = −47, 170 e − e = $41,508.

= −50,000

19. (a) The income stream is $42.1 billion per year and the interest rate is 2%. Using a calculator or computer to evaluate the integral, we get 1

42.1e−0.02t dt Ê0 = 41.68 billion dollars.

Present value =

Using instead the Fundamental Theorem of Calculus, we get: Present value =

Ê0

42.1e−0.02t dt

|1 1 | e−0.02t | | −0.02 |0 � −0.02 0 = −(2105) e − e = $41.68 billion dollars.

= 42, 1

The present value of Intel’s profts over the one-year time period is about 41.68 billion dollars. (b) The value at the end of the year is 42.1e0.02(1) = 42.95, or about 42.95 billion dollars. 20. (a) Gross income in 2018, when t = 0, is 1.97 billion dollars. Income in 2022, when t = 4, is projected to be 1.97 − 0.13(4) = 1.45 billion dollars. (b) January 1, 2018 through January 1, 2022 is a four-year time period, and t = 0 corresponds to January 1, 2018, so the value on January 1, 2022 of the income over this four-year time period is Value on Jan. 1, 2018 =

Ê0

(1.97 − 0.13t)e−0.02t dt = 6.59 billion dollars.

The value, on January 1, 2018, of Harley-Davidson income over the four-year time period is about 6.59 billion dollars. (c) We have Value on Jan. 1, 2022 = 6.59e0.02(4) = 7.14 billion dollars.

492

Chapter Six /SOLUTIONS

21. (a) Since the rate at which revenue is generated is at least 21.1 and at most 24.6 billion dollars per year, the present value of the revenue over a fve-year time period is at least 5

Ê0 and at most

21.1e−0.045t dt = 94.465 billion dollars

24.6e−0.045t dt = 110.144 billion dollars. Ê0 Using instead the Fundamental Theorem of Calculus to evaluate the integral, we have Ê0

e−0.045t dt = −

1 � −0.225 e − e0 = 4.4774. 0.045

We get the same numbers, with revenue between 21.1(4.4774) = 94.465 and 24.6(4.4774) = 110.144 billion dollars. (b) The present value of the revenue over a twenty-fve year time period is at least Ê0 and at most

21.1e−0.045t dt = 316.663 billion dollars 25

24.6e−0.045t dt369.190 billion dollars. Ê0 Using instead the Fundamental Theorem of Calculus to evaluate the integral, we have Ê0

e−0.045t dt = −

1 � −1.125 e − e0 = 15.00772. 0.045

We get the same numbers, with revenue between 21.1(15.00772) = 316.663 and 24.6(15.00772) = 369.190 billion dollars. 22. We want to fnd the value of T making the present value of income stream ($80,000∕year) equal to $130,000. Thus we want to fnd the time T at which T

80,000e−0.085t dt. Ê0 Trying a few values of T and evaluating the integral with a calculator or computer, we get T ≈ 1.75. It takes approximately one year and nine months for the present value of the proft generated by the new machinery to equal the cost of the machinery. Using instead the Fundamental Theorem of Calculus, to evaluate the integral, we get 130,000 =

130,000 = −80,000 Thus 0.085

1 � −0.085T e − e0 . 0.085

130,000 = 0.138 = 1 − e−0.085T . 80,000

We get e−0.085T = 0.862, so ln 0.862 = 1.747 years. 0.085 23. We fnd the present value of the income stream, S(t) = 400,000, from t = 0 to t = 20 years, at a continuous interest rate of r = 5% = 0.05: T =−

Present value of income stream =

Ê0

S(t)e−rt dt 20

400,000e−0.05t dt Ê0 = 5,056,964.47.

This means that the Present value of the company = $5,056,964.47. Since there are 200 shares, we divide the present value of the company by 200 to fnd the value of each share: Value of a share =

$5,056,964.47 = $25,284.822. 200

6.5 SOLUTIONS

493

24. We fnd the present value of the income stream, S(t) = 600,000 − 15,000t, from t = 0 to t = 40 years, at a continuous interest rate of r = 5% = 0.05: Present value of income stream =

Ê0

S(t)e−rt dt 40

(600,000 − 15,000t)e−0.05t dt Ê0 = 6,812,011.70. =

This means that the Present value of the company = $6,812,011.70. Since there are 400 shares, we divide the present value of the company by 400 to fnd the value of each share: Value of a share =

$6,812,011.70 = $17,030.029. 400

25. We fnd the present value of the income stream, S(t) = 500,000e0.01t , from t = 0 to t = ∞, at a continuous interest rate of r = 5% = 0.05 by setting up an improper integral: Ê0

Ê0

∞

Ê0

∞

Present value of income stream =

S(t)e−rt dt 500,000e0.01t e−0.05t dt 500,000e−0.04t dt

= 500,000

Ê0

∞

e−0.04t dt.

We can evaluate the improper integral numerically: Ê0

∞

500,000e−0.04t dt = 12,500,000.

This means that the Present value of the company = $12,500,000. Since there are 1000 shares, we divide the present value of the company by 1000 to fnd the value of each share: Value of a share =

$12,500,000 = $12,500. 1000

26. (a) We split the income stream into the part happening in 2021 and the part from 2022–2040, fnd the present value of each part, and add them together. Let t = 0 be the beginning of January, 2021. The income stream in 2021 is S(t) = 3.59, so we fnd its present value from t = 0 to t = 1, at a continuous interest rate of r = 5% = 0.05: Present value of 2021 income stream =

Ê0

S(t)e−rt dt 1

3.59e−0.05t dt Ê0 = 3.50173,

so the present value of the frst year of expected earnings is $3.50 per share. The income stream from 2022–2040 is S(t) = 4.36, so we fnd its present value from t = 1 to t = 20, at a continuous interest rate of r = 5% = 0.05: Present value of 2022–2040 income stream =

Ê1

S(t)e−rt dt 20

4.36e−0.05t dt Ê1 = 50.8681,

494

Chapter Six /SOLUTIONS so the present value of the 19 years of expected earnings from 2022 to 2040 is $50.87 per share. Adding the two presesent values together, we get Total present value = $3.50 + $50.87 = $54.37 per share. (b) The present value we found is much higher than the share price: Present value − price = 54.37 − 41.22 = $13.15, which means our estimated value is $13.15—over 30%!—higher than the share price. One reason for that might be that people expect Exxon’s earnings to go down, perhaps because they expect oil to run out or fossil fuel to be replaced by renewable energy. Another reason might be that we chose too low an interest rate, because people perceive Exxon to be a risky investment.

27. The interest rate is 1.5% = 0.015. The present value of the payments is $73.3 million. With t in years since 2020, we write the present value of the payments as an integral and factor out S since it is constant: Present value = 73.3 =

Ê0

Se−0.015t dt = S

Ê0

e−0.015t dt = 43.337S.

73.3 = 1.691. 43.337 Thus, the city will need to pay at a rate of $1.691 million per year. S=

28. The interest rate is 2.5% = 0.025. Let damages be accrued at a rate of $S million per year. With t in years since the wall is built, we write the present value of the damages as an integral and factor out S, since it is constant: Present value, in millions =

Ê0

Se−0.025t dt = S

Ê0

e−0.025t dt = 28.540S.

The present value of the cost of the seawall is $20 million, so 20 = 28.540S

20 = 0.701 million dollars per year. 28.540

If the city sustains more than $701,000 per year in damages, then the wall is cheaper—as well as avoiding considerable disruption. 29. (a) To fnd the annual income stream, we solve for S(t) in the formula for the future value of a cost stream: Future value =

Ê0

S(t)er(M−t) dt.

Since S(t) is a constant annual income stream, we have S(t) = S for some constant S. We know the future value is $60 billion, the time over which we are integrating is M = 30 years and the interest rate is r = 0.02. Thus, we solve: 60 =

Ê0

Se0.02(30−t) dt.

Since S is constant, we can factor it out of the integral, and using a calculator or computer, we get 60 = S

Ê0

e0.02(30−t) dt = S ⋅ 41.1059.

Thus S = 60∕41.1059 = 1.4596 billion dollars per year. (b) An income stream of S = $1.4596 billion per year gives a future value of Future value =

Ê0

1.4596e0.02(70−t) dt = 222.968 billion dollars.

This is suÿcient to cover the costs in 2090. 30. With t in years from the present, the present value of the health expenses over the next 50 years, in millions of dollars, is Present value =

Ê0

100e0.01t e−0.03t dt =

Ê0

100e−0.02t = −

50 100 −0.02t || e | = 3160.603. | 0.02 |0

Just on the basis of cost, it is the same to pay $3160.603 million now to prevent emissions and consequent health care costs as to pay the health care costs as they arise.

6.5 SOLUTIONS

495

31. We use time t in years since 2020, so 2030 is t = 10. The interest rate is r = 0.03 per year. (a) We solve for S making the future value $700 in 2030, when t = 10: Future value =

Ê0

Se0.03(10−t) dt = 700.

Since S is constant, we can factor it out of the integral, and using substitution, we get 700 = S

Ê0

|10 1 � 0.03(10) 1 | e e0.03(10−t) | = S − e0 = S ⋅ 11.662. | −0.03 0.03 |0

e0.03(10−t) dt = S ⋅

Thus S = 700∕11.662 = 60.024 bn dollars per year. (b) Similarly, for a future value $1000 bn in 2030, when t = 10, we solve for S: 1000 =

Ê0

Se0.03(10−t) dt = S ⋅ 11.662.

We get S = 1000∕11.662 = 85.749 bn dollars per year. (c) The present value in 2020 of $700 bn in 2030 is Present value = 700e−0.03⋅10 = 518.573 bn dollars. Similarly, the present value in 2020 of $1000 bn in 2030 is Present value = 1000e−0.03⋅10 = 740.818 bn dollars. 32. In order to compare damages, we need to compare their present or future values at the same time. We choose the future value in 2100. For the coastal cities, we are measuring time from 2050, so t = 50 in 2100, and we have Future value =

Ê0

(60 + t)e0.01(50−t) dt = 5379.540 billion dollars.

Since the total projected damage from sea-level rise through 2100 is $14,000 billion, the fraction that occurs in the costal cities is 5379.540∕14,000 = 0.384, so 38.4% of the damages are projected to be in these coastal cities. Alternatively, we could also compute the present value in 2050 of the stream of payments: Present value =

Ê0

(60 + t)e−0.01t dt = 3262.856 billion dollars,

and the present value in 2050 of the $14,000 payment in 2100 is Present value = 14,000e−0.01(50) = 8491.429 billion dollars. Then we compare present values: 3262.856∕8491.429 = 0.384 = 38.4%, the same as we found with the future values. 33. (a) For 2021, we have Present value =

Ê0

100e0.02t e−0.035t dt = 3007.922 thousand dollars.

Thus, the present value is $3007.922 thousand, or about $3 million. (b) For 2021, with 2000 homes, Present value of total savings over forty years = 2000 ⋅ 3007.922 thousand dollars = 6,015,844 thousand dollars = 6,015.844 million dollars. This tells us homeowners would save over $6000 million if a seawall were built.

496

Chapter Six /SOLUTIONS

34. (a) Suppose the oil extracted over the time period 0 ≤ t ≤ M is E. (See Figure 6.76.) Since q(t) is the rate of oil extraction, we have: E=

Ê0

q(t)dt =

(a − bt)dt =

Ê0

(10 − 0.1t) dt.

To calculate the time at which the oil is exhausted, set E = 100 and try di˙erent values of M. We fnd M = 10.6 gives Ê0

10.6

(10 − 0.1t) dt = 100,

so the oil is exhausted in 10.6 years. Extraction Curve

✠

q(t) Area below the extraction curve is the total oil extracted

✲ t 0

Figure 6.76 (b) Suppose p is the oil price, C is the extraction cost per barrel, and r is the interest rate. We have the present value of the proft as Present value of proft =

Ê0

(p − C)q(t)e−rt dt 10.6

(20 − 10)(10 − 0.1t)e−0.1t dt Ê0 = 624.9 million dollars. =

35. We know that the

√ Price in the future = P 1 + 20 t .

The present value V of price P satisfes √ Present value V = P 1 + 20 t e−0.05t . We want to maximize V . To do so, we fnd the critical points of V (t) for t ≥ 0. Note that H I √ � dV 20 −0.05t −0.05t =P + 1 + 20 t −0.05e √ e dt 2 t H I √ 10 −0.05t = Pe √ − 0.05 − t . t Setting

√ t is not di˙erentiable at t = 0.

dV = 0 gives dt H

I √ 10 0 = Pe √ − 0.05 − t t √ 10 0 = √ − 0.05 − t. t −0.05t

Using a calculator, we fnd t ≈ 10 years. Since V ¨ (t) > 0 for 0 < t < 10 and V ¨ (t) < 0 for t > 10, we confrm that this is a maximum. Thus, the best time to sell the wine is in about 10 years.

6.6 SOLUTIONS

497

Solutions for Section 6.6 d sin(x2 + 1) = 2x cos(x2 + 1); 1. (a) dx

(b) (i) 12 sin(x2 + 1) + C (c) (i)

− 12 cos(x2 + 1) + C 2

d sin(x3 + 1) = 3x2 cos(x3 + 1) dx 3

(ii) 13 sin(x + 1) + C

(ii) − 13 cos(x3 + 1) + C

2. Setting w = 1 − 5x , we see that dw = −10x dx. Since the integrand has a factor of x that di˙ers from dw by a constant factor, substitution is appropriate and allows us to replace the integral by one that can be found directly using the power rule: 1 1 x(1 − 5x2 )5 dx = w5 − dw = − w5 dw. Ê Ê 10 10 Ê 3. Setting w = ln x, we see that dw = (1∕x) dx. Since the integrand has a factor of 1∕x and no additional factors, substitution is appropriate and allows us to replace the integrand by one we can integrate directly using the power rule: √ √ ln x 1 dx = ln x dx = w1∕2 dw. Ê Ê Ê x x 4. Setting w = ln x, we see that dw = (1∕x) dx. Since the integrand does not have a factor of 1∕x, substitution is not appropriate in this case. The integrand has a factor of x, but x and 1∕x do not di˙er by a constant. 5. Setting w = sin t, we see that dw = cos t dt. Since the integrand has a factor of cos t and no additional factors, substitution is appropriate and allows us to replace the integrand by one we can integrate directly using the power rule: Ê

sin9 t (cos t dt) =

w9 dw.

6. We use the substitution w = x2 + 1, dw = 2xdx. Ê Ê d 1 2 (x + 1)6 + C = 2x(x2 + 1)5 . Check: dx 6 7. We use the substitution w = x2 + 4, dw = 2xdx. 2x(x2 + 1)5 dx =

w5 dw =

w6 1 + C = (x2 + 1)6 + C. 6 6

√ x 1 1 w1∕2 dx = w−1∕2 dw = + C = x2 + 4 + C. √ Ê 2Ê 2 1∕2 x2 + 4 Check:

d √ 2 x ( x + 4 + C) = √ . dx x2 + 4

8. We use the substitution w = 5x − 7, dw = 5dx. Ê

(5x − 7)10 dx =

1 1 w11 1 w10 dw = +C = (5x − 7)11 + C. Ê 5 5 11 55

d 1 (5x − 7)11 + C = (5x − 7)10 . dx 55 9. We use the substitution w = x2 + 1, dw = 2x dx. √ 1 1 1 x x2 + 1 dx = w1∕2 dw = w3∕2 + C = (x2 + 1)3∕2 + C. Ê 2Ê 3 3 √ d 1 2 1 3 (x + 1)3∕2 + C = ⋅ (x2 + 1)1∕2 ⋅ 2x = x x2 + 1. Check: dx 3 3 2 10. We use the substitution w = q 2 + 1, dw = 2qdq. Check:

Ê Check:

2 d q2 +1 (e + C) = 2qeq +1 . dq

2qeq +1 dq =

ew dw = ew + C = eq +1 + C.

498

Chapter Six /SOLUTIONS

11. We use the substitution w = 5t + 2, dw = 5dt. Ê

5e5t+2 dt =

ew dw = ew + C = e5t+2 + C.

d 5t+2 (e + C) = 5e5t+2 . dt 12. We use the substitution w = −x2 , dw = −2x dx. Check:

2 1 1 e−x (−2x dx) = − ew dw 2Ê 2Ê 1 2 1 = − ew + C = − e−x + C. 2 2

xe−x dx = −

d Check: dx (− 21 e−x + C) = (−2x)(− 21 e−x ) = xe−x .

13. We use the substitution w = −0.1t + 4, dw = −0.1 dt: Ê

e−0.1t+4 dt = −

Check:

1 1 w ew dw = − e + C = −10e−0.1t+4 + C. 0.1 Ê 0.1

d � −10e−0.1t+4 + C = −10e−0.1t+4 (−0.1) = e−0.1t+4 . dt

14. We use the substitution w = −0.2t, dw = −0.2dt. Ê

100e−0.2t dt =

100 ew dw = −500ew + C = −500e−0.2t + C. −0.2 Ê

d (−500e−0.2t + C) = 100e−0.2t . dt 15. We use the substitution w = t3 − 3, dw = 3t2 dt. Check:

1 1 (t3 − 3)10 (3t2 dt) = w10 dw Ê 3Ê 3 11 1w 1 3 = +C = (t − 3)11 + C. 3 11 33

t2 (t3 − 3)10 dt =

d 1 3 1 [ (t − 3)11 + C] = (t3 − 3)10 (3t2 ) = t2 (t3 − 3)10 . dt 33 3 16. We use the substitution w = x2 , dw = 2xdx. Check:

x sin(x2 )dx =

1 1 1 sin w dw = − cos w + C = − cos(x2 ) + C. 2Ê 2 2

d 1 1 − cos(x2 ) + C = − (− sin(x2 )) ⋅ 2x = x sin(x2 ). dx 2 2 17. We use the substitution w = x2 − 4, dw = 2x dx. Check:

1 1 (x2 − 4)7∕2 (2xdx) = w7∕2 dw 2Ê 2Ê 1 2 9∕2 1 = w + C = (x2 − 4)9∕2 + C. 2 9 9 d 1 2 1 9 Check: (x − 4)9∕2 + C = (x2 − 4)7∕2 2x = x(x2 − 4)7∕2 . dx 9 9 2 18. We use the substitution w = x2 + 3, dw = 2x dx. x(x2 − 4)7∕2 dx =

1 1 w3 1 x(x2 + 3)2 dx = w2 ( dw) = + C = (x2 + 3)3 + C. Ê Ê 2 2 3 6 d 1 2 1 2 3 2 2 2 (x + 3) + C = 3(x + 3) (2x) = x(x + 3) . Check: dx 6 6

6.6 SOLUTIONS 19. We use the substitution w = 3x + 1, dw = 3 dx. 1 w−1 1 1 1 1 1 dw = w−2 dw = +C =− + C. dx = Ê (3x + 1)2 3 Ê w2 3Ê 3 −1 3(3x + 1) 0 1 d 1 1 − +C = . Check: dx 3(3x + 1) (3x + 1)2 20. We use the substitution w = x4 + 1, dw = 4x3 dx. Ê

4x3 x4 + 1

1 dw = ln |w| + C = ln(x4 + 1) + C. Ê w

dx =

4x3 d . (ln(x4 + 1) + C) = 4 dx x +1 3 21. We use the substitution w = x , dw = 3x2 dx. Check:

x2 cos(x3 )dx =

12 cos(w)dw = 4 sin(w) + C = 4 sin(x3 ) + C. 3 Ê

d (4 sin(x3 ) + C) = 12x2 cos(x3 ). dx 22. We use the substitution w = 2t − 7, dw = 2 dt. Check:

(2t − 7)73 dt =

1 1 1 w73 dw = w74 + C = (2t − 7)74 + C. 2Ê (2)(74) 148

d 1 74 (2t − 7)74 + C = (2t − 7)73 (2) = (2t − 7)73 . dt 148 148 23. In this case, it seems easier not to substitute. Check:

x5 (x2 + 3)2 dx = (x4 + 6x2 + 9) dx = + 2x3 + 9x + C. Ê 5 4 5 d x5 + 2x3 + 9x + C = x4 + 6x2 + 9 = (x2 + 3)2 . Check: dx 5 24. In this case, it seems easier not to substitute. Ê

y2 (1 + y)2 dy = =

y2 (y2 + 2y + 1) dy =

(y4 + 2y3 + y2 ) dy

y5 y4 y3 + + + C. 5 2 3

0 1 d y5 y4 y3 + + + C = y4 + 2y3 + y2 = y2 (y + 1)2 . dy 5 2 3 25. We use the substitution w = cos + 5, dw = − sin d . Check:

1 w7 dw = − w8 + C Ê 8 1 = − (cos + 5)8 + C. 8

sin (cos + 5)7 d = −

Check: d 1 1 − (cos + 5)8 + C = − ⋅ 8(cos + 5)7 ⋅ (− sin ) d 8 8 = sin (cos + 5)7

499

500

Chapter Six /SOLUTIONS

26. We use the substitution w = sin , dw = cos d . Ê Ê 4 5 d sin7 + C = sin6 cos . Check: d 7 27. We use the substitution w = cos 3t, dw = −3 sin 3t dt. sin6 cos d =

w6 dw =

w7 sin7 +C = + C. 7 7

√ √ 1 cos 3t sin 3t dt = − w dw Ê 3Ê 3 2 1 2 3 = − ⋅ w 2 + C = − (cos 3t) 2 + C. 3 3 9 Check: 3 1 2 3 d 2 − (cos 3t) 2 + C = − ⋅ (cos 3t) 2 ⋅ (− sin 3t) ⋅ 3 dt 9 9 2 √ = cos 3t sin 3t.

28. We use the substitution w = 3 − t, dw = − dt. Ê

sin(3 − t)dt = −

sin(w)dw = −(− cos(w)) + C = cos(3 − t) + C.

d (cos(3 − t) + C) = − sin(3 − t)(−1) = sin(3 − t). Check: dt

29. We use the substitution w = 1 + 3t2 , dw = 6t dt. t 1 1 1 1 dt = ( dw) = ln |w| + C = ln(1 + 3t2 ) + C. Ê w 6 Ê 1 + 3t2 6 6 2 (We can drop the absolute value signs since 1 + 3t > 0 for all t). d 1 1 1 t 2 Check: ln(1 + 3t ) + C = (6t) = . dt 6 6 1 + 3t2 1 + 3t2 30. We use the substitution w = x3 + 1, dw = 3x2 dx, to get Ê

x2 ex +1 dx =

1 1 1 3 ew dw = ew + C = ex +1 + C. 3Ê 3 3

3 d 1 x3 +1 1 3 e + C = ex +1 ⋅ 3x2 = x2 ex +1 . dx 3 3 31. We use the substitution w = sin , dw = cos d . Check:

Check:

d d

sin4 4

1 +C

sin3 cos d =

1 ⋅ 4 sin3 4

⋅ cos

w3 dw =

sin4 w4 +C = 4 4

+ C.

= sin3 cos .

32. We use the substitution w = 4x2 , dw = 8xdx. Ê

x sin(4x2 )dx =

1 1 1 sin(w)dw = − cos(w) + C = − cos(4x2 ) + C. 8Ê 8 8

d 1 − cos(4x2 ) + C = x sin(4x2 ). dx 8 33. We use the substitution w = sin x, dw = cos xdx: Check:

Ê Check:

d dx

(sin x)3 +C 3

sin2 x cos xdx =

1 = sin2 x cos x.

w2 dw =

(sin x)3 w3 +C = + C. 3 3

6.6 SOLUTIONS 34. We use the substitution w = 3x − 4, dw = 3dx. Ê

e3x−4 dx =

1 1 1 ew dw = ew + C = e3x−4 + C. 3Ê 3 3

1 d 1 3x−4 e + C = e3x−4 ⋅ 3 = e3x−4 . dx 3 3 35. We use the substitution w = x + ex , dw = (1 + ex ) dx. Check:

√ √ 1 + ex dw dx = √ √ = 2 w + C = 2 x + ex + C. Ê x w x+e 1 1 + ex d √ 1 (2 x + ex + C) = 2 ⋅ (x + ex )− 2 ⋅ (1 + ex ) = √ . Check: dx 2 x + ex Ê

36. We use the substitution w = 3x2 , dw = 6xdx. Ê

xe3x dx =

1 1 1 2 ew dw = ew + C = e3x + C. 6Ê 6 6

2 d 1 3x2 1 2 e + C = e3x ⋅ 6x = xe3x . dx 6 6 37. We use the substitution w = 3x2 + 4, dw = 6x dx. √ 1 1 w3∕2 1 w1∕2 dw = + C = (3x2 + 4)3∕2 + C. x 3x2 + 4 dx = Ê 6Ê 6 3∕2 9 √ d 1 (3x2 + 4)3∕2 + C = x 3x2 + 4. Check: dx 9 38. We use the substitution w = ex + e−x , dw = (ex − e−x ) dx. Check:

ex − e−x dw dx = = ln |w| + C = ln(ex + e−x ) + C. Ê ex + e−x Ê w (We can drop the absolute value signs since ex + e−x > 0 for all x). d 1 [ln(ex + e−x ) + C] = x (ex − e−x ). Check: dx e + e−x 39. We use the substitution w = ln z, dw = z1 dz. (ln z)2 (ln z)3 w3 +C = + C. dz = w2 dw = Ê Ê 3 z 3 4 5 (ln z)2 1 1 d (ln z)3 Check: + C = 3 ⋅ (ln z)2 ⋅ = . dz 3 3 z z 40. We use the substitution w = y2 + 4, dw = 2y dy. y 1 dw 1 1 = ln |w| + C = ln(y2 + 4) + C. dy = Ê y2 + 4 2Ê w 2 2 2 (We can drop the absolute value signs since y + 4 ≥ 0yfor all y.) d 1 1 1 Check: ln(y2 + 4) + C = ⋅ 2 ⋅ 2y = 2 . dy 2 2 y +4 y +4 41. We use the substitution w = 5q 2 + 8, dw = 10qdq. q 1 1 1 1 dq = dw = ln |w| + C = ln(5q 2 + 8) + C. Ê 5q 2 + 8 10 Ê w 10 10 q d 1 1 1 ⋅ ⋅ 10q = 2 . ln(5q 2 + 8) + C = Check: dq 10 10 5q 2 + 8 5q + 8 42. We use the substitution w = et + t, dw = (et + 1) dt. et + 1 1 dt = dw = ln |w| + C = ln |et + t| + C. Ê et + t Ê w d et + 1 Check: (ln |et + t| + C) = t . dt e +t

501

502

Chapter Six /SOLUTIONS

43. We use the substitution w =

√ 1 y, dw = √ dy. 2 y √

√ e y w w y √ dy = 2 e dw = 2e + C = 2e + C. Ê Ê y √

√ √ d 1 e y Check: (2e y + C) = 2e y ⋅ √ = √ . dy 2 y y √ 1 44. We use the substitution w = x, dw = √ dx. 2 x √ √ cos x dx = cos w(2 dw) = 2 sin w + C = 2 sin x + C. √ Ê Ê x H I √ √ √ cos x d 1 Check: (2 sin x + C) = 2 cos x = √ . √ dx 2 x x

45. We use the substitution w = x2 + 2x + 19, dw = 2(x + 1)dx. (x + 1)dx 1 dw 1 1 = ln |w| + C = ln(x2 + 2x + 19) + C. = Ê x2 + 2x + 19 2Ê w 2 2 (We can drop the absolute value signs, since x2 + 2x + 19 = (x + 1)2 + 18 > 0 for all x.) 1 1 1 1 x+1 Check: [ ln(x2 + 2x + 19)] = (2x + 2) = 2 . dx 2 2 x2 + 2x + 19 x + 2x + 19 46. We use the substitution w = et + 1, dw = et dt: Ê

et et + 1

dt =

1 dw = ln |w| + C = ln(et + 1) + C. Ê w

Note: The absolute value can be dropped because et + 1 is positive for all t. d et Check: (ln(et + 1) + C) = t . dt e +1 47. We use the substitution w = sin(x2 ), dw = 2x cos(x2 ) dx. √ 1 1 x cos(x2 ) 1 1 w− 2 dw = (2w 2 ) + C = sin(x2 ) + C. dx = Ê √sin(x2 ) Ê 2 2 Check:

x cos(x2 ) d √ 1 ( sin(x2 ) + C) = √ [cos(x2 )]2x = √ . dx 2 sin(x2 ) sin(x2 )

48. We substitute w = x2 + 1, so dw = 2xdx. Ê

x(x2 + 1)2 dx =

1 1 w3 1 w2 dw = + C = (x2 + 1)3 + C. 2Ê 2 3 6

Using the Fundamental Theorem, we have Ê0

x(x2 + 1)2 dx =

|2 1 1 125 1 124 62 1 2 | (x + 1)3 | = ⋅ 53 − ⋅ 13 = − = = . | 6 6 6 6 6 6 3 |0

49. Let w = x2 + 1, then dw = 2xdx. When x = 0, w = 1 and when x = 3, w = 10. Thus we have Ê0

2x dx = Ê1 x2 + 1

|10 dw | = ln |w|| = ln 10 − ln 1 = ln 10. | w |1

50. Since d(− cos )∕d = sin , we have Ê0

∕2

| ∕2 1 | e− cos sin d = e− cos | = e− cos( ∕2) − e− cos(0) = 1 − . | e |0

6.6 SOLUTIONS 51. Let

√ 1 dx = 2 dw. If x = 1 then w = 1, and if x = 4 so w = 2. So we have x = w, 12 x− 2 dx = dw, √ x Ê1

√

2 |2 e x ew ⋅ 2 dw = 2ew || = 2(e2 − e). √ dx = Ê1 |1 x

52. We substitute w = t + 1, so dw = dt. √ 1 1 dt = dw = w−1∕2 dw = 2w1∕2 + C = 2 t + 1 + C. Ê √t + 1 Ê √w Ê Using the Fundamental Theorem, we have Ê0

|3 √ √ √ 1 | dt = 2 t + 1| = 2 4 − 2 1 = 4 − 2 = 2. √ | t+1 |0

53. We substitute w = t + 2, so dw = dt. t=e−2

Êt=−1

w=e |e 1 dw | dt = = ln |w|| = ln e − ln 1 = 1. | Êw=1 w t+2 |1

54. We substitute w = 1 + x2 . Then dw = 2x dx. x=2

Êx=0

w=5

x 1 dx = Êw=1 w2 (1 + x2 )2

5 1 1 1 || 2 dw = − | = . 2 2 w ||1 5

55. Let w = −t2 , then dw = −2tdt so tdt = − 12 dw. When t = 0, w = 0 and when t = 1, w = −1. Thus we have Ê0

2te−t dt =

Ê0

−1

−1 1 2ew − dw = − ew dw Ê0 2

|−1 | = −ew | = −e−1 − (−e0 ) = 1 − e−1 . | |0 56.

57.

2√

Ê−1 Ê1

x + 2 dx =

|2 2 14 2 3∕2 2 | (x + 2)3∕2 | = (4) − (1)3∕2 = (7) = | 3 3 3 3 |−1

3 1 1 1 dt −1 || = = = − − − | t + 7 ||1 10 8 40 (t + 7)2 2

58. To fnd the area under the graph of f (x) = xex , we need to evaluate the defnite integral Ê0

xex dx.

This is done in Example 8, Section 6.6, using the substitution w = x2 , the result being Ê0

xex dx =

1 4 (e − 1). 2

59. Since f (x) = 1∕(x + 1) is positive on the interval x = 0 to x = 2, we have Area = The area is ln 3 ≈ 1.0986.

Ê0

|2 1 | dx = ln(x + 1)| = ln 3 − ln 1 = ln 3. | x+1 |0

503

504

Chapter Six /SOLUTIONS

60. If f (x) =

1 , the average value of f on the interval 0 ≤ x ≤ 2 is defned to be x+1 2

x=2

Êx=0

Thus, the average value of f (x) on 0 ≤ x ≤ 2 is 21 ln 3 ≈ 0.5493. See Figure 6.77.

1 f (x) = 1+x

0.54931 x 2

Figure 6.77

61. (a) If w = t∕2, then dw = (1∕2)dt. When t = 0, w = 0; when t = 4, w = 2. Thus, Ê0

g(t∕2) dt =

Ê0

g(w) 2dw = 2

Ê0

g(w) dw = 2 ⋅ 5 = 10.

(b) If w = 2 − t, then dw = −dt. When t = 0, w = 2; when t = 2, w = 0. Thus, Ê0 62. (a)

g(2 − t) dt =

Ê2

g(w) (−dw) = +

Ê0

g(w) dw = 5.

(i) Multiplying out gives Ê

(x2 + 10x + 25) dx =

x3 + 5x2 + 25x + C. 3

(ii) Substituting w = x + 5, so dw = dx, gives Ê

(x + 5)2 dx =

w2 dw =

(x + 5)3 w3 +C = + C. 3 3

(b) The results of the two calculations are not the same since (x + 5)3 x3 15x2 75x 125 +C = + + + + C. 3 3 3 3 3 However they di˙er only by a constant, 125∕3, as guaranteed by the Fundamental Theorem of Calculus. 4x(x2 + 1) dx = (4x3 + 4x) dx = x4 + 2x2 + C. Ê Ê (b) If w = x2 + 1, then dw = 2x dx.

63. (a)

4x(x2 + 1) dx =

2w dw = w2 + C = (x2 + 1)2 + C.

(c) The expressions from parts (a) and (b) look di˙erent, but they are both correct. Note that (x2 +1)2 +C = x4 +2x2 +1+C. In other words, the expressions from parts (a) and (b) di˙er only by a constant, so they are both correct antiderivatives.

6.7 SOLUTIONS

505

64. (a) We frst try the substitution w = sin , dw = cos d . Then w2 sin2 sin cos d = w dw = +C = + C. Ê Ê 2 2 (b) If we instead try the substitution w = cos , dw = − sin d , we get w2 cos2 sin cos d = − w dw = − +C =− + C. Ê Ê 2 2 (c) Once we note that sin 2 = 2 sin cos , we can also say 1 sin cos d = sin 2 d . Ê 2Ê Substituting w = 2 , dw = 2 d , the above equals 1 cos w cos 2 sin w dw = − +C =− + C. 4Ê 4 4 (d) All these answers are correct. Although they have di˙erent forms, they di˙er from each other only in terms of a 2 2 constant, and thus they are all acceptable antiderivatives. For example, 1 − cos2 = sin2 , so sin2 = − cos2 + 12 . Thus the frst two expressions di˙er only by a constant C. 2 Similarly, cos 2 = cos2 − sin2 = 2 cos2 − 1, so − cos42 = − cos2 + 41 , and thus the second and third expressions di˙er only by a constant. Of course, if the frst two expressions and the last two expressions di˙er only in the constant C, then the frst and last only di˙er in the constant as well. 65. (a) The rate at which copper is mined is 21e0.05t million tons per year, so in T years Total quantity extracted =

Ê0

21e0.05t dt million tons.

We evaluate by substituting w = 0.05t so dw = 0.05 dt, or by guess-and-check, giving Total quantity extracted =

Ê0

21e0.05t dt =

T 21 0.05t || e | = 420(e0.05T − 1) million tons. | 0.05 |0

(b) The copper runs out when 830 million tons have been extracted: 420(e0.05T − 1) = 830. Solving this equation gives 830 +1 420 1 830 T = ln + 1 = 21.813 ≈ 22 years. 0.05 420 Thus, copper is predicted to run out toward the end of 2039. e0.05T =

Solutions for Section 6.7 1. Let u = t and v¨ = e5t , so u¨ = 1 and v = 15 e5t .

1 5t Then ∫ te5t dt = 51 te5t − ∫ 51 e5t dt = 51 te5t − 25 e + C.

2. Let u = p and v¨ = e(−0.1)p , u¨ = 1. Thus, v = ∫ e(−0.1)p dp = −10e(−0.1)p . With this choice of u and v, integration by parts gives: Ê

pe(−0.1)p dp = p(−10e(−0.1)p ) −

(−10e(−0.1)p ) dp

e(−0.1)p dp

= −10pe(−0.1)p + 10

= −10pe(−0.1)p − 100e(−0.1)p + C.

506

Chapter Six /SOLUTIONS

3. Let u = ln y, v¨ = y. Then, v = 21 y2 and u¨ = Ê

1 . Integrating by parts, we get: y 1 2 1 2 1 y ln y − y ⋅ dy Ê 2 2 y 1 1 y dy = y2 ln y − 2 2Ê 1 1 = y2 ln y − y2 + C. 2 4

y ln y dy =

4. Let u = z + 1, v¨ = e2z . Thus, v = 21 e2z and u¨ = 1. Integrating by parts, we get: Ê

1 1 2z e dz (z + 1)e2z dz = (z + 1) ⋅ e2z − Ê 2 2 1 1 = (z + 1)e2z − e2z + C 2 4 1 = (2z + 1)e2z + C. 4

5. Let u = ln 5q, v¨ = q 5 . Then v = 16 q 6 and u¨ = Ê

1 . Integrating by parts, we get: q 1 6 1 1 q ln 5q − (5 ⋅ ) ⋅ q 6 dq Ê 6 5q 6 1 1 = q 6 ln 5q − q 6 + C. 6 36

q 5 ln 5q dq =

6. Let u = y and v¨ = (y + 3)1∕2 , so u¨ = 1 and v = 32 (y + 3)3∕2 : Ê

√ 2 2 2 4 y y + 3 dy = y(y + 3)3∕2 − (y + 3)3∕2 dy = y(y + 3)3∕2 − (y + 3)5∕2 + C. Ê 3 3 3 15 4

7. Let u = ln x and v¨ = x3 , so u¨ = x1 and v = x4 . Then Ê 8. Let u = t + 2 and v¨ =

x3 ln x dx =

x4 x3 x4 x4 ln x − dx = ln x − + C. Ê 4 4 4 16

√ 2 + 3t, so u¨ = 1 and v = 29 (2 + 3t)3∕2 . Then √ 2 2 (t + 2) 2 + 3t dt = (t + 2)(2 + 3t)3∕2 − (2 + 3t)3∕2 dt Ê 9 9Ê 2 4 = (t + 2)(2 + 3t)3∕2 − (2 + 3t)5∕2 + C. 9 135

9. Let u = y and v¨ = √ 1 , so u¨ = 1 and v = −2(5 − y)1∕2 . 5−y

4 dy = −2y(5 − y)1∕2 + 2 (5 − y)1∕2 dy = −2y(5 − y)1∕2 − (5 − y)3∕2 + C. √ Ê 3 5−y 10. Let u = z, v¨ = e−z . Thus v = −e−z and u¨ = 1. Integration by parts gives: y

(−e−z ) dz Ê = −ze−z − e−z + C

ze−z dz = −ze−z −

= −(z + 1)e−z + C.

6.7 SOLUTIONS

507

11. We can split the integral into two Ê

t+7 t dt = dt + 7 (5 − t)−1∕2 dt. √ √ Ê Ê 5−t 5−t

To calculate the frst integral, we use integration by parts. Let u = t and v¨ = √1 , so u¨ = 1 and v = −2(5 − t)1∕2 . Then 5−t

t 4 dt = −2t(5 − t)1∕2 + 2 (5 − t)1∕2 dt = −2t(5 − t)1∕2 − (5 − t)3∕2 + C. Ê √5 − t Ê 3 We can calculate the second integral directly: 7 Ê

(5 − t)−1∕2 = −14(5 − t)1∕2 + C1 . Thus

t+7 4 dt = −2t(5 − t)1∕2 − (5 − t)3∕2 − 14(5 − t)1∕2 + C2 . √ 3 5−t

12. Let u = ln x, v¨ = x−2 . Then v = −x−1 and u¨ = x−1 . Integrating by parts, we get: Ê

x−2 ln x dx = −x−1 ln x −

(−x−1 ) ⋅ x−1 dx

= −x−1 ln x − x−1 + C.

13. Let u = t, v¨ = sin t. Thus, v = − cos t and u¨ = 1. With this choice of u and v, integration by parts gives: Ê

(− cos t) dt Ê = −t cos t + sin t + C.

t sin t dt = −t cos t −

14. Let u = + 1 and v¨ = sin( + 1), so u¨ = 1 and v = − cos( + 1). Ê

cos( + 1) d Ê = −( + 1) cos( + 1) + sin( + 1) + C.

( + 1) sin( + 1) d = −( + 1) cos( + 1) +

15. Let u = ln x and v¨ = x1∕2 , so u¨ = 1∕x and v = 32 x3∕2 . Then √ 2 2 2 2 3∕2 1 2 4 x ln x dx = x3∕2 ln x − x ⋅ dx = x3∕2 ln x − x1∕2 dx = x3∕2 ln x − x3∕2 + C. Ê Ê 3 3 x 3 3Ê 3 9 16. Let u = y and v¨ =

√ 1 − y, so u¨ = 1 and v = − 32 (1 − y)3∕2 . Then

√ 2 2 2 4 y 1 − y dy = − y(1 − y)3∕2 + (1 − y)3∕2 dy = − y(1 − y)3∕2 − (1 − y)5∕2 + C. Ê Ê 3 3 3 15 17. A calculator gives ∫1 ln x dx = 0.386. An antiderivative of ln x is x ln x − 1, so the Fundamental Theorem of Calculus gives 2 |2 | ln x dx = (x ln x − x)| = 2 ln 2 − 1. | Ê1 |1 Since 2 ln 2 − 1 = 0.386, the value from the Fundamental Theorem agrees with the numerical answer. 2

508

Chapter Six /SOLUTIONS

18. We use integration by parts. Let u = z and v¨ = e−z , so u¨ = 1 and v = −e−z . Then Ê0

10 −z

10 |10 dz = −ze | + e−z dz | Ê0 |0 |10 | = −10e−10 + (−e−z )| | |0 −10 = −11e + 1 −z |

≈ 0.9995. |5 | ln t dt = (t ln t − t)| = 5 ln 5 − 4 ≈ 4.047 | Ê1 |1 5 |5 | 20. ln(1 + t) dt = ((1 + t) ln(1 + t) − (1 + t)) | = 6 ln 6 − 5 ≈ 5.751. | Ê0 |0 3 3 | 1 2 1 9 | 21. t ln t dt = t ln t − t2 | = ln 3 − 2 ≈ 2.944. | Ê1 2 4 2 |1 5 |5 | 22. x cos x dx = (cos x + x sin x)| = cos 5 + 5 sin 5 − cos 3 − 3 sin 3 ≈ −3.944. | Ê3 |3 2 ln t 23. Let u = (ln t)2 and v¨ = 1, so u¨ = and v = t. Then t 5

19.

Ê (We use the fact that

(ln t)2 dt = t(ln t)2 − 2

ln t dt = t(ln t)2 − 2t ln t + 2t + C.

ln x dx = x ln x − x + C, a result which can be derived using integration by parts.)

24. Let u = t2 and v¨ = e5t , so u¨ = 2t and v = 51 e5t . Then ∫ t2 e5t dt = 51 t2 e5t − 25 ∫ te5t dt.

1 5t Using Problem 1, we have ∫ t2 e5t dt = 15 t2 e5t − 25 ( 15 te5t − 25 e )+C

2 2 5t = 15 t2 e5t − 25 te5t + 125 e + C.

25. (a) We evaluate this integral using the substitution w = 1 + x3 . (b) We evaluate this integral using the substitution w = x2 . (c) We evaluate this integral using the substitution w = x3 + 1. (d) We evaluate this integral using the substitution w = 3x + 1. (e) This integral can be evaluated using integration by parts with u = ln x, v¨ = x2 . (f) This integral can be evaluated using integration by parts with u = ln x, v¨ = 1. 26. Using integration by parts with u¨ = e−t , v = t, so u = −e−t and v¨ = 1, we have |2 2 | Area = te dt = −te | − −1 ⋅ e−t dt Ê0 Ê0 | |0 |2 | = (−te−t − e−t )|| = −2e−2 − e−2 + 1 = 1 − 3e−2 . | |0 2

−t |

−t

27. Since ln(x2 ) = 2 ln x and ∫ ln x dx = x ln x − x + C, we have Area =

Ê1

(ln(x2 ) − ln x) dx =

Ê1

(2 ln x − ln x) dx

|2 | ln x dx = (x ln x − x)|| = 2 ln 2 − 2 − (1 ln 1 − 1) = 2 ln 2 − 1. | |1

SOLUTIONS to Review Problems For Chapter Six 28. We have Exposure =

Ê0

509

15te−0.2t dt.

We frst use integration by parts to evaluate the indefnite integral of this function. Let u = 15t and v¨ = e−0.2t dt, so u¨ = 15dt and v = −5e−0.2t . Then, Ê

15te−0.2t dt = (15t)(−5e−0.2t ) − = −75te−0.2t + 75

(−5e−0.2t )(15dt) e−0.2t dt = −75te−0.2 − 375e−0.2t + C.

Thus, |3 | 15te−0.2t dt = (−75te−0.2t − 375e−0.2t )| = −329.29 + 375 = 45.71. | Ê0 |0 The exposure provided by the drug over this time interval is 45.71 (ng/ml)-hours. 3

dE = r, so the total energy E used in the frst T hours is given by E = te−at dt. We use integration Ê0 dt by parts. Let u = t, v¨ = e−at . Then u¨ = 1, v = − a1 e−at .

29. (a) We know that

Ê0

te−at dt

T |T t 1 − e−at dt = − e−at || − a a |0 Ê 0 T

1 1 = − T e−aT + e−at dt a a Ê0 1 1 = − T e−aT + 2 (1 − e−aT ). a a (b)

1 T 1 1 lim + 2 1 − lim aT . aT T →∞ e T →∞ a T →∞ e a Since a > 0, the second limit on the right hand side in the above expression is 0. In the frst limit, although both the numerator and the denominator go to infnity, the denominator eaT goes to infnity more quickly than T does. So in the T end the denominator eaT is much greater than the numerator T . Hence lim aT = 0. (You can check this by graphing T →∞ e T 1 y = aT on a calculator or computer for some values of a.) Thus lim E = 2 . T →∞ e a lim E = −

Solutions for Chapter 6 Review 1. Since dP ∕dt is negative for t < 3 and positive for t > 3, we know that P is decreasing for t < 3 and increasing for t > 3. Between each two integer values, the magnitude of the change is equal to the area between the graph dP ∕dt and the t-axis. For example, between t = 0 and t = 1, we see that the change in P is −1. Since P = 2 at t = 0, we must have P = 1 at t = 1. The other values are found similarly, and are shown in Table 6.8. Table 6.8 t

−1∕2

510

Chapter Six /SOLUTIONS

2. See Figure 6.78. F (0) = 1

F (0) = 0

x 1

Figure 6.78

3. See Figure 6.79. F (0) = 1

x 1

F (0) = 0

Figure 6.79

4. First notice that F will be decreasing on the interval 0 < x < 1 and on the interval 3 < x < 4 and will be increasing on the interval 1 < x < 3. The areas tell us how much the function increases or decreases. By the Fundamental Theorem, we have F (1) = F (0) + F (3) = F (1) + F (4) = F (3) +

Ê0

Ê1

Ê3

F ¨ (x)dx = 5 + (−6) = −1. F ¨ (x)dx = −1 + 8 = 7. F ¨ (x)dx = 7 + (−2) = 5.

5. (a) The function f (x) is increasing when f ¨ (x) is positive, so f (x) is increasing for x < −2 or x > 2. The function f (x) is decreasing when f ¨ (x) is negative, so f (x) is decreasing for −2 < x < 2. Since f (x) is increasing to the left of x = −2, decreasing between x = −2 and x = 2, and increasing to the right of x = 2, the function f (x) has a local maximum at x = −2 and a local minimum at x = 2. (b) See Figure 6.80.

f (x) x −4

−2

Figure 6.80

SOLUTIONS to Review Problems For Chapter Six

511

6. (a) The function f (x) is increasing when f ¨ (x) is positive, so f (x) is increasing for −1 < x < 3 or x > 3. The function f (x) is decreasing when f ¨ (x) is negative, so f (x) is decreasing for x < −1. Since f (x) is decreasing to the left of x = −1 and increasing to the right of x = −1, the function has a local minimum at x = −1. Since f (x) is increasing on both sides of x = 3, it has neither a local maximum nor a local minimum at that point. (b) See Figure 6.81.

f (x)

x −4

−2

Figure 6.81 7t2 + t. 2 2 3 3 4 8. 3 t + 4 t + 54 t5 7. t3 +

9. 2x3 − 4x2 + 3x. y5 + ln |y| 10. 5 11. x3 + 5x. 12. ln |z| 13. P (y) = ln |y| + y2 ∕2 + y + C 14. − cos t 15. We use substitution with w = 2x + 1 and dw = 2 dx. Then Ê 16. Antiderivative G(x) =

f (x) dx =

(2x + 1)3 dx =

(2x + 1)4 1 w4 +C = + C. w3 dw = Ê 2⋅4 8 2

(x + 1)4 x4 3x2 + x3 + +x+C = +C 4 2 4

17. 2t2 + 7t + C 3x2 18. +C 2 x4 x2 19. − + C. 4 2 20. 4t2 + 3t + C. 5 3 21. − − 2 + C t t 22. 2x4 + ln |x| + C. 23. 2x2 + 2ex + C 24. 5 sin x + 3 cos x + C 25. 2x3 + C. 26.

(x + 1)3 + C. Ê 3 Another way to work the problem is to expand (x + 1)2 to x2 + 2x + 1 as follows: (x + 1)2 dx =

(x + 1)2 dx =

These two answers are the same, since constant.

(x2 + 2x + 1) dx =

x3 + x2 + x + C. 3

(x + 1)3 x3 + 3x2 + 3x + 1 x3 1 x3 = = + x2 + x + , which is + x2 + x, plus a 3 3 3 3 3

512

Chapter Six /SOLUTIONS

x+1 1 = 1 + , the indefnite integral is x + ln |x| + C x x 28. Since F ¨ (x) = 2x + 3, we use F (x) = x2 + 3x. By the Fundamental Theorem, we have 27. Since f (x) =

|2 | (2x + 3) dx = (x2 + 3x)| = (22 + 3 ⋅ 2) − (12 + 3 ⋅ 1) = 10 − 4 = 6. | Ê1 |1 2

29. Since F ¨ (t) = 1∕t2 = t−2 , we take F (t) =

t−1 = −1∕t. Then −1 2

1 dt = F (2) − F (1) Ê1 t2 1 1 =− − − 2 1 1 = . 2 −1

30.

Ê−3

|−1 2 1 −2 | dr = −r | = −1 + = −8∕9 ≈ −0.889. | 9 r3 |−3

|1 | sin d = − cos | = 1 − cos 1 ≈ 0.460. | Ê0 |0 1 0 4 1 |2 20 3 4 x x | + 2x dx = + x2 | = + 4 = 16∕3 ≈ 5.333. 32. | Ê0 3 12 |0 3 1

31.

33. Since F ¨ (t) = 1∕(2t), we take F (t) = 21 ln |t|. Then 2

1 dt = F (2) − F (1) Ê1 2t 1 1 = ln |2| − ln |1| 2 2 1 = ln 2. 2 34. We have Area =

Ê1

x2 dx =

x3 || 64 − 1 43 13 − = = 21. | = 3 ||1 3 3 3

35. The integral which represents the area under this curve is Area = Since

Ê0

(6x2 + 1) dx.

d (2x3 + x) = 6x2 + 1, we can evaluate the defnite integral: dx |2 | (6x2 + 1) dx = (2x3 + x)| = 2(23 ) + 2 − (2(0) + 0) = 16 + 2 = 18. | Ê0 |0 2

36. We have Average value =

1 10 − 0 Ê0

(x2 + 1)dx =

1 10

0 3 1 |10 0 1 103 x 1 103 | +x | = + 10 − 0 = . | 3 10 3 3 |0

We see in Figure 6.82 that the average value of 103∕3 ≈ 34.33 for f (x) looks right.

SOLUTIONS to Review Problems For Chapter Six

513

f (x) = x2 + 1

100

50 34.33 x 5

4b2 − 4 = 192 4b2 = 196 b2 = 49 b = ±7. Since b is larger than 1, we have b = 7.

38. Since y = x3 (1 − x) is positive for 0 ≤ x ≤ 1 and y = 0, when x = 0, 1, the area is given by Area =

Ê0

x3 (1 − x) dx =

Ê0

(x3 − x4 ) dx =

1 x4 x5 || 1 − | = . 4 5 |0 20

39. Since y = 0 only when x = 0 and x = 1, the area lies between these limits and is given by Area = =

Ê0

x2 (1 − x)2 dx =

Ê0

x2 (1 − 2x + x2 ) dx =

Ê0

(x2 − 2x3 + x4 ) dx

1 x3 2 4 x5 || 1 − x + | = . 3 4 5 |0 30

40. (a) We sketch f (x) = xe−x ; see Figure 6.83. The shaded area to the right of the y-axis represents the integral

f (x) = xe−x x

Figure 6.83 (b) Using a calculator or computer, we obtain Ê0

xe−x dx = 0.9596

Ê0

xe−x dx = 0.9995

Ê0

xe−x dx = 0.99999996.

Ê0

∞

xe−x dx.

514

Chapter Six /SOLUTIONS

41. (a) Evaluating the integrals with a calculator gives Ê0

Ê0

xe−x∕10 dx = 26.42 xe−x∕10 dx = 95.96

Ê0

100

Ê0

200

xe−x∕10 dx = 99.95 xe−x∕10 dx = 100.00

(b) The results of part (a) suggest that Ê0

∞

xe−x∕10 dx ≈ 100

42. (a) Since v(t) = 60∕50t is never 0, the car never stops. (b) For time t ≥ 0, Distance traveled =

∞

Ê0

60 dt. 50t

60 dt for b = 1, 5, 10 gives Ê0 50t Ê0

60 dt = 15.0306 50t

Ê0

60 dt = 15.3373 50t

Ê0

60 dt = 15.3373, 50t

so the integral appears to converge to 15.3373; so we estimate the distance traveled to be 15.34 miles. 43. (a) In July 2025, we have P = 7.41e0.011⋅8 = 8.092 billion people. In July 2030, we have P = 7.41e0.011⋅13 = 8.549 billion people. (b) Since the year 2020 corresponds to t = 3, and the year 2030 to t = 13, we have 13 13 1 1 7.41 0.011t || 7.41e0.011t dt = ⋅ e | | 10 0.011 13 − 3 Ê3 |3 1 7.41 � 0.011⋅13 0.011⋅3 = e −e = 8.096. 10 0.011

Average population between 2020 and 2030 =

The average population of the world between 2020 and 2030 is predicted to be 8.096 billion people. 44. (a) The supply and demand curves are shown in Figure 6.84. Tracing along the graphs, we fnd that they intersect approximately at the point (322, 11.43). Thus, the equilibrium price is about $11.43 per unit, and the equilibrium quantity is about 322 units. p ($/unit) 30 20 S 10

D 100

200

300

400

500

q (quantity)

Figure 6.84 (b) The consumer surplus is shown in Figure 6.85. This is an area between two curves. Using a calculator or computer to evaluate the defnite integral, we have Consumer surplus =

Ê0

322

(30e−0.003q − 11.43) dq = $2513.52.

SOLUTIONS to Review Problems For Chapter Six

515

Using instead the Fundamental Theorem of Calculus to evaluate the defnite integral, we get Ê0

322

� −0.003q − 11.43 dq = 30 30e

|322 1 | e−0.003q − 11.43q | = $2513.52. | −0003 |0

The producer surplus is shown in Figure 6.86. This is an area between two curves. Using a calculator or computer to evaluate the defnite integral, we have Producer surplus =

322

Ê0

(11.43 − (5 + 0.02q))dq = $1033.62.

Using instead the Fundamental Theorem of Calculus to evaluate the defnite integral, we get Ê0

322

322 � || (11.43 − (5 + 0.02q)) dq = 11.43q − (5q + 0.01q 2 ) | = $1033.62 | |0

p ($/unit) 30

20 S

S 10

D 100

200

300

400

q (quantity)

500

D 100

200

Figure 6.85

300

400

500

q (quantity)

Figure 6.86

45. The total gains from trade at the equilibrium price is shaded in Figure 6.87. We see in Figures 6.88 and 6.89 that if the price is artifcially high or low, the quantity sold is less than q ∗ . Thus, the total gains from trade are reduced. p

p+ p∗ p− D

q∗

Figure 6.87: Shade area: Total gains from trade at equilibrium price

Figure 6.88: Shade area: Gains when price is artifcially high

D q−

Figure 6.89: Shade area: Gains when price is artifcially low

46. Measuring money in thousands of dollars, the equation of the line representing the demand curve passes through (50, 980) and (350, 560). Its slope is (560−980)∕(350−50) = −420∕300. See Figure 6.90. So the equation is y−560 = − 420 (x−350), 300 i.e. y − 560 = − 75 x + 490. Thus

Consumer surplus =

Ê0

350

7 − x + 1050 dx − 350 ⋅ 560 = 85,750. 5

(Note that 85,750 = 12 ⋅ 490 ⋅ 350, the area of the triangle in Figure 6.90. We could have used this instead of the integral to fnd the consumer surplus.) Recalling that our unit measure for the price axis is $1000/car, the consumer surplus is $85,750,000.

516

Chapter Six /SOLUTIONS price (1000s of dollars/car) 1050

(50, 980) Demand (350, 560)

quantity (number of cars)

Figure 6.90

47. (a) The price at which 300 units are demanded = 20e−0.002⋅300 ≈ $11. The price at which 300 units are supplied = 0.02 ⋅ 300 + 1 = $7. Therefore, the price at which 300 units are demanded is higher. See Figure 6.91. (b) Looking at Figure 6.91 we see that p∗ ≈ $9 = equilibrium price

q ∗ ≈ 400 = equilibrium quantity.

and

price/unit Supply

20 15 10 5

Demand 200

400

600

800

quantity

Figure 6.91 (c) Consumer surplus is the area under the demand curve and above $9. Thus, using a calculator or computer, we have Consumer surplus =

Ê0

400

20e−0.002q dq − 400 ⋅ 9 ≈ 1907 dollars.

Using the Fundamental Theorem of Calculus, we get Consumer surplus =

Ê0

400

20e−0.002q dq − 400 ⋅ 9

|400 1 | e−0.002q | − 3600 | −.002 |0 � −0.8 − 3600 ≈ 1907 dollars = 10, 000 1 − e

= 20

See Figure 6.92. Consumers gain $1907 in buying goods at the equilibrium price instead of at the price they would be willing to pay. Producer surplus is the area under $9 and above the supply curve. Thus, using a calculator or computer, we get Producer surplus = 400 ⋅ 9 −

Ê0

400

(0.02q + 1) dq = 1600 dollars.

SOLUTIONS to Review Problems For Chapter Six

517

Using the Fundamental Theorem of Calculus, we get Producer surplus = 400 ⋅ 9 −

Ê0

400

(0.02q + 1) dq

400 � || = 3600 − 0.01q 2 + q | | |0 � 2 = 3600 − 0.01(400) + 400 = 1600 dollars .

See Figure 6.93. Producers gained $1600 in supplying goods at the equilibrium price instead of the price at which they would have been willing to supply the goods. price/unit

price/unit Supply

20 Consumer surplus

Supply

20 15

✲

10 Producer surplus

Demand 200

400

600

800

✲

quantity

Demand 200

Figure 6.92

400

600

800

quantity

Figure 6.93

48. The present value is given by M

Se−rt dt, Ê0 with S = 1000, r = 0.09 and M = 5. Hence, using a calculator or computer to evaluate the defnite integral: Present value =

Present value =

Ê0

1000e−0.09t dt = $4026.35.

Using instead the Fundamental Theorem of Calculus, we get: Present value =

Ê0

1000e−0.09t dt

|5 1 | e−0.09t | | −0.09 |0 � −0.45 = −11, 111.11 e − e0 = $4026.35.

= 1000

49. The future value is $20,000 in 5 years. We frst fnd the present value. Since 20,000 = P e0.06(5) = P e0.3 . Solving for P gives 20,000 = $14,816.36. e0.3 Now we solve for the income stream S which gives a present value of $14,816.36. P =

14,816.36 =

Ê0

Se−0.06t dt.

Since S is a constant, we bring it outside the integral sign. Using a calculator or computer to evaluate the integral, we get: 14,816.36 = S

Ê0

e−0.06t dt = S(4.320)

518

Chapter Six /SOLUTIONS Using the Fundamental Theorem of Calculus instead, we get: Ê0

e−0.06t dt =

|5 1 | e−0.06t | = 4.320. | −0.06 |0

Solving for S gives 14816.36 ≈ 3430. 4.320 Money must be deposited at a rate of about $3430 per year, or about $66 per week. S=

50. 10

(100 + 10t)e−.05t dt Ê0 = $1,147.75.

Present Value =

51. The present value, P , is given by P =

Ê0

Se−rt dt,

where S is the income stream, r = 0.05 and M = 10. Using a calculator or computer to evaluate the defnite integral, we have P =

Ê0

Se−0.05t dt = S

Ê0

e−0.05t dt = 7.8694S.

Using instead the Fundamental Theorem of Calculus to evaluate the defnite integral, we get: Ê0

e−0.05t dt =

|10 1 | e−0.05t | = 7.8694. | −0.05 |0

Solving 5000 = 7.8694S for S gives an income stream of S = $635.37 per year. 52. One good way to approach the problem is in terms of present values. In 1980, the present value of Germany’s loan was 20 billion DM. Now let’s fgure out the rate that the Soviet Union would have to give money to Germany to pay o˙ 10% interest on the loan by using the formula for the present value of a continuous stream. Since the Soviet Union sends gas at a constant rate, the rate of deposit, P (t), is a constant c. Since they don’t start sending the gas until after 5 years have passed, the present value of the loan is given by: Present Value =

Ê5

∞

P (t)e−rt dt.

We want to fnd c so that 20,000,000,000 =

Ê5

∞

ce−rt dt = c

Ê5

∞

e−rt dt

∞

e−0.10t dt Ê5 ≈ 6.065c.

Dividing, we see that c should be about 3.3 billion DM per year. At 0.10 DM per m3 of natural gas, the Soviet Union must deliver gas at the constant, continuous rate of about 33 billion m3 per year. 53. f (x) = 2x, so F (x) = x2 + C. F (0) = 0 implies that 02 + C = 0, so C = 0. Thus F (x) = x2 is the only possibility. 2

+ C. F (0) = 0 implies that − 72 ⋅ 02 + C = 0, so C = 0. Thus F (x) = −7x2 ∕2 is the only 54. f (x) = −7x, so F (x) = −7x 2 possibility. 55. We use the substitution w = x2 + 1, dw = 2xdx. √ 2x dx = w−1∕2 dw = 2w1∕2 + C = 2 x2 + 1 + C. √ Ê Ê x2 + 1 Check:

d √ 2 2x (2 x + 1 + C) = √ . dx x2 + 1

SOLUTIONS to Review Problems For Chapter Six 56. We use the substitution w = t2 , dw = 2t dt. Ê

t cos(t2 )dt =

1 1 1 cos(w)dw = sin(w) + C = sin(t2 ) + C. 2Ê 2 2

d 1 1 ( sin(t2 ) + C) = cos(t2 )(2t) = t cos(t2 ). dt 2 2 57. We use the substitution w = x3 + 1, dw = 3x2 dx. Check:

3x2 (x3 + 1)4 dx =

w4 dw =

w5 1 + C = (x3 + 1)5 + C. 5 5

58. We use the substitution w = x2 + 1, dw = 2x dx. 1 2x dx = dw = ln |w| + C = ln(x2 + 1) + C. Ê w Ê x2 + 1 59. We use the substitution w = x + 10, dw = dx. Ê

(x + 10)3 dx =

w3 dw =

1 w4 + C = (x + 10)4 + C. 4 4

d 1 ( (x + 10)4 + C) = (x + 10)3 . dx 4 60. We use the substitution w = x2 + 9, dw = 2x dx: Check:

x(x2 + 9)6 dx =

1 2 1 1 w7 w6 dw = +C = (x + 9)7 + C. 2Ê 2 7 14

61. We use the substitution w = t2 , dw = 2t dt. Ê 2

tet dt =

1 1 2 1 ew dw = ew + C = et + C. 2Ê 2 2

d 1 t Check: dt ( 2 e + C) = 12 et (2t) = tet .

62. We use the substitution w = −x, dw = − dx. Ê

e−x dx = −

ew dw = −ew + C = −e−x + C.

d Check: dx (−e−x + C) = −(−e−x ) = e−x .

63. We use the substitution w = 1 + 2x3 , dw = 6x2 dx. Ê

x2 (1 + 2x3 )2 dx =

1 w3 1 1 (1 + 2x3 )3 + C. w2 ( dw) = ( ) + C = 6 6 3 18

1 d 1 (1 + 2x2 )3 + C = [3(1 + 2x3 )2 (6x2 )] = x2 (1 + 2x3 )2 . 18 dx 18 64. We use the substitution w = 4 − x, dw = −dx. Check:

√ √ 1 1 dx = − dw = −2 w + C = −2 4 − x + C. √ √ Ê Ê w 4−x √ 1 1 d 1 ⋅ −1 = √ . Check: (−2 4 − x + C) = −2 ⋅ ⋅ √ 2 dx 4−x 4−x 65. We use the substitution w = y + 5, dw = dy, to get dy dw = = ln |w| + C = ln |y + 5| + C. Ê y+5 Ê w Check:

d 1 (ln |y + 5| + C) = . dy y+5

519

520

Chapter Six /SOLUTIONS

66. Since f ¨ (x) = 2x, integration by parts tells us that Ê0

10 |10 f (x)g ¨ (x) dx = f (x)g(x)|| − f ¨ (x)g(x) dx Ê0 |0

= f (10)g(10) − f (0)g(0) − 2

Ê0

xg(x) dx.

We can use left and right Riemann Sums with Δx = 2 to approximate ∫0 xg(x) dx: 10

Left sum ≈ 0 ⋅ g(0)Δx + 2 ⋅ g(2)Δx + 4 ⋅ g(4)Δx + 6 ⋅ g(6)Δx + 8 ⋅ g(8)Δx = (0(2.3) + 2(3.1) + 4(4.1) + 6(5.5) + 8(5.9)) 2 = 205.6. Right sum ≈ 2 ⋅ g(2)Δx + 4 ⋅ g(4)Δx + 6 ⋅ g(6)Δx + 8 ⋅ g(8)Δx + 10 ⋅ g(10)Δx = (2(3.1) + 4(4.1) + 6(5.5) + 8(5.9) + 10(6.1)) 2 = 327.6. A good estimate for the integral is the average of the left and right sums, so Ê0

xg(x) dx ≈

205.6 + 327.6 = 266.6. 2

Substituting values for f and g, we have Ê0

f (x)g ¨ (x) dx = f (10)g(10) − f (0)g(0) − 2

Ê0

xg(x) dx

≈ 102 (6.1) − 02 (2.3) − 2(266.6) = 76.8 ≈ 77.

67. We integrate by parts. Since we know what the answer is supposed to be, it’s easier to choose u and v¨ . Let u = xn and v¨ = ex , so u¨ = nxn−1 and v = ex . Then Ê

xn ex dx = xn ex − n

xn−1 ex dx.

68. We integrate by parts, with u = y, v¨ = sin y. We have u¨ = 1, v = − cos y, and Ê Check:

y sin y dy = −y cos y −

(− cos y) dy = −y cos y + sin y + C.

d (−y cos y + sin y + C) = − cos y + y sin y + cos y = y sin y. dy

69. Integration by parts with u = x, v¨ = cos x gives Ê

x cos x dx = x sin x −

sin x dx + C = x sin x + cos x + C.

ln x dx = 2x ln x − 2x + C.

70. Remember that ln(x2 ) = 2 ln x. Therefore, Ê Check:

ln(x2 ) dx = 2

d 2x (2x ln x − 2x + C) = 2 ln x + − 2 = 2 ln x = ln(x2 ). dx x

STRENGTHEN YOUR UNDERSTANDING 1. True.

STRENGTHEN YOUR UNDERSTANDING

521

2. True. If a function is concave up, its second derivative is positive which implies that its derivative is increasing. 3. True. This is the Fundamental Theorem of Calculus. 4. False. The limits of integration on the integral need to be from 0 to 3 to make this a true statement. 5. True. Since f ¨ is positive on the interval 3 to 4, the function is increasing on that interval. 6. False. Since f ¨ is negative on the interval 1 to 2, the function is decreasing on that interval. 7. True. Since f ¨ is positive on the interval 2 to 3, the function is increasing on that interval. 8. True. Since f ¨ is negative on the interval 5 to 6, the function is decreasing on that interval. 9. False. Since f ¨ is negative on the interval 0 to 1, the function is decreasing on that interval. 10. True. The area below the curve of f ¨ between x = 1 and x = 2 is similar in size to the area above the curve between x = 2 and x = 3. Between x = 1 and x = 3, the function f increases approximately the same amount that it decreases, so f (1) ≈ f (3). 11. True. We see that the derivative of t3 ∕3 + 5 is t2 . 12. False. When we add one to the exponent −2, we get −1. The function −x−1 is an antiderivative of x−2 . 13. False. Antiderivatives of e3x are of the form (1∕3)e3x + C. 14. True. This is a correct integral statement. 15. True. We know

z−1∕2 dz =

√ z1∕2 + C = 2 z + C. 1∕2

16. False. We know that ∫ ex dx = ex + C.

17. False. The derivative of ln |t| is 1∕t so the correct integral statement is ∫ (1∕t) dt = ln |t| + C.

18. True, since the derivative of 2x is (ln 2)2x . 19. True. 20. True. We know that all antiderivatives di˙er only by a constant. 21. False. We need to substitute the endpoints into an antiderivative of 1∕x. 22. True, since ln x is an antiderivative of 1∕x. 23. True, since x2 is an antiderivative of 2x. 24. False. We need to frst fnd an antiderivative of 3x2 . 25. False. For a defnite integral, we need to substitute the endpoints into the antiderivative. 26. True. An antiderivative is et and we substitute the limits of integration and subtract. 27. False. When we make the substitution w = x2 , we must also substitute for the limits of integration. Since w = 52 = 25 25 when x = 5 and w = 0 when x = 0, the result of the substitution is ∫0 ew dw. 28. True, since dw = (1∕x) dx and w = 1 when x = e and w = 0 when x = 1. 29. False. The two defnite integrals represent two di˙erent quantities. 30. True. The function y = e−kx is positive, so the integral represents the area under the curve between x = 1 and x = 2 and so is positive. 31. True, as specifed in the text. 32. True, as specifed in the text. 33. False. The equilibrium price is the price where the supply curve crosses the demand curve. 34. False. The consumer surplus is the total amount gained by consumers by purchasing items at prices lower than they are willing to pay. 35. True, they both have units of dollars. 36. False. The units of producer surplus are dollars∕quantity ⋅ quantity = dollars. 37. False. Total gains from trade is the sum of consumer and producer surplus. 38. False. The integral is the amount the consumers who bought would have spent if they had paid as much as they were willing to pay. The consumers actually spend p∗ q ∗ . 39. True, as specifed in the text. 40. False. Producer surplus at price p∗ is the area between the supply curve and the horizontal line at p∗ .

522

Chapter Six /SOLUTIONS

41. True. In M years, future value = present value ⋅ erM , and since the interest rate r is positive, the future value is greater. 42. True. The amount deposited is $5000 and the future value also includes the accumulated interest. 43. False, since if there were no interest, a deposit of $5000 today would pay for an income stream of $1000 per year for 5 years. With an interest rate of 2% less than $5000 would have to be deposited today to generate the same income stream. 44. False, since the single deposit grows to 1000e0.03 = 1030.45 dollars in one year, while the income stream has a future value 1 in one year of e0.03 ∫0 1000e−0.03t dt = 1015.15. 45. False. An income stream has units of dollars/year. 46. False. The future value is 3000e5r . 47. True, since present value = future value∕e5r = future value ⋅e−5r . 48. True. Over the fve-year period, the amount deposited is $10,000 so the rest is the amount of interest earned. 49. False. The present value of an income stream of 2000 dollars per year that starts now and pays out over 6 years with a 6 continuous interest rate of 2% is ∫0 2000e−.02t dt. 50. True, since the amount deposited is $18,000 and the future value also includes the compounded interest. 51. False, since dw = (3q 2 + 6q − 1) dq cannot be substituted. 52. True, since dw = (1∕x) dx.

53. False. We have dw = 2x dx. Since the integral ∫ ex dx does not include an x dx to be substituted for dw, this integral cannot be evaluated using this substitution. 2

54. True, since dw = 2x dx. 55. False. This is almost true, but is o˙ by a minus sign since dw = −1 ds. 56. True. Since dw = 2t dt, we have t 1 1 1 dt = √ dw = Ê √ dw. Ê √t2 + 1 2Ê w 2 w 57. True, since dw = (ex − e−x ) dx.

58. False. The substitution w = q 3 + 5 would give the integral ∫ (1∕3)w10 dw.

59. True, since dw = cos d . 60. False. This is almost true, but is o˙ by a minus sign, since dw = − sin x dx.

61. False. The integral has to be ∫ uv¨ dx when u and v¨ are substituted. In this case, we should have u = x2 and v¨ = ex . 62. True.

63. False. We integrate v¨ to fnd v. We see v = ∫ e3x dx = (1∕3)e3x . 64. False. We integrate v¨ to fnd v, and 1∕x is not the antiderivative of ln x. It is the derivative of ln x. In this case, the assignment of parts is wrong. We should try u = ln x and v¨ = x2 . 65. False. This integral is more appropriately evaluated using the method of substitution. 66. True. 67. True. 68. False. This integral is more appropriately evaluated using the method of substitution. 69. False. This integral is more appropriately evaluated using the method of substitution. 70. True. We use u = ln x and dv = x3 dx.

PROJECTS FOR CHAPTER SIX 1. (a) Suppose Q(t) is the amount of water in the reservoir at time t. Then Q¨ (t) =

Rate at which water in reservoir is changing

Infow rate

−

Outfow rate

Thus the amount of water in the reservoir is increasing when the infow curve is above the outfow, and decreasing when it is below. This means that Q(t) is a maximum where the curves cross in July 2016 (as

PROJECTS FOR CHAPTER SIX

523

shown in Figure 6.94), and Q(t) is decreasing fastest when the outfow is farthest above the infow curve, which occurs about October 2016 (see Figure 6.94). To estimate values of Q(t), we use the Fundamental Theorem which says that the change in the total quantity of water in the reservoir is given by t

Q(t) − Q(Jan 2016) =

ÊJan 16

(Infow rate − Outfow rate) dt

Q(t) = Q(Jan 2016) +

ÊJan 16

(Infow rate − Outfow rate) dt.

rate of flow (millions of gallons/day)

Q(t) is max Q(t) is min

❄

Outflow

Inflow

April

Jan (16)

July

Oct

Jan(17)

Q(t)

millions of gallons

Q(t) is increasing most rapidly

❘

Q(t) is decreasing most rapidly

✠

April

Jan (16)

July

Oct

Jan(17)

Figure 6.94

(b) See Figure 6.94. Maximum in July 2016. Minimum in Jan 2017. (c) See Figure 6.94. Increasing fastest in May 2016. Decreasing fastest in Oct 2016. (d) In order for the water to be the same as Jan 2016 the total amount of water which has fowed into the reservoir minus the total amount of water which has fowed out of the reservoir must be 0. Referring to Figure 6.95, we have July 17

ÊJan 16 giving A1 + A3 = A2 + A4

(Infow − Outfow) dt = −A1 + A2 − A3 + A4 = 0

524

Chapter Six /SOLUTIONS rate of flow (millions of gallons/day)

Outflow

Jan (‘16)

Inflow

April

July

Oct

Jan (‘17)

April

July

Figure 6.95

2. (a) If the poorest p% of the population has exactly p% of the goods, then F (x) = x. (b) Any such F is increasing. For example, the poorest 50% of the population includes the poorest 40%, and so the poorest 50% must own more than the poorest 40%. Thus F (0.4) ≤ F (0.5), and so, in general, F is increasing. In addition, it is clear that F (0) = 0 and F (1) = 1. The graph of F is concave up by the following argument. Consider F (0.05) − F (0.04). This is the fraction of resources the ffth poorest percent of the population has. Similarly, F (0.20) − F (0.19) is the fraction of resources that the twentieth poorest percent of the population has. Since the twentieth poorest percent owns more than the ffth poorest percent, we have F (0.05) − F (0.04) ≤ F (0.20) − F (0.19).

More generally, we can see that

F (x1 + Δx) − F (x1 ) ≤ F (x2 + Δx) − F (x2 )

for any x1 smaller than x2 and for any increment Δx. Dividing this inequality by Δx and taking the limit as Δx → 0, we get F ¨ (x1 ) ≤ F ¨ (x2 ). So, the derivative of F is an increasing function, i.e. F is concave up. (c) G is twice the shaded area below in the following fgure. y=x

F (x)

x 1

(d) The most equitable distribution is the one modeled by the equation F (x) = x, as this tells us that the wealth is evenly distributed. The less equitable the distribution the further the function is from resembling F (x) = x. Thus in our case, Country A has more equitable distribution. (e) The function F goes through the points (0, 0) and (1, 1) and is increasing and concave up. Graphical representations of the two extremes for G are shown in Figures 6.96 and 6.97. Since Gini’s index is twice the area shown, the maximum possible value is 1 and the minimum possible value is 0. If Gini’s index is 1, the

PROJECTS FOR CHAPTER SIX

525

distribution of resources is as inequitable as it can get, with one person holding all resources and everyone else having none. (See Figure 6.96.) If Gini’s index is 0, the distribution of resources is totally equitable, with F (x) = x as in part (a). (See Figure 6.97.) The closer Gini’s index is to 0, the more equitable the distribution. fraction of resources

fraction of resources

x (poorest 1 fraction) Figure 6.96: Maximum value of Gini’s index: Function F with G = 1

x (poorest 1 fraction) Figure 6.97: Minimum value of Gini’s index: Function F with G = 0

3. (a) In Figure 6.98, the area of the shaded region is F (M). Thus, F (M) = ∫0 y(t) dt and, by the Fundamental Theorem, F ¨ (M) = y(M). M

y (annual yield)

F (M)

t (time in years)

M Figure 6.98

(b) Figure 6.99 is a graph of F (M). Note that the graph of y looks like the graph of a quadratic function. Thus, the graph of F looks like a cubic. F (total yield) F (M)

20000 15000 10000 5000 10

M (time in years)

Figure 6.99

1 1 F (M) = M M Ê0

y(t) dt.

526

Chapter Six /SOLUTIONS (d) If the function a(M) takes on its maximum at some point M, then a¨ (M) = 0. Since a(M) =

1 F (M), M

di˙erentiating using the quotient rule gives a¨ (M) =

MF ¨ (M) − F (M) = 0, M2

so MF ¨ (M) = F (M). Since F ¨ (M) = y(M), the condition for a maximum may be written as My(M) = F (M) or as y(M) = a(M). To estimate the value of M which satisfes My(M) = F (M), use the graph of y(t). Notice that F (M) is the area under the curve from 0 to M, and that My(M) is the area of a rectangle of base M and height y(M). Thus, we want the area under the curve to be equal to the area of the rectangle, or A = B in Figure 6.100. This happens when M ≈ 50 years. In other words, the orchard should be cut down after about 50 years. y (annual yield)

Area B

❄

Area A

✲

t (time in years)

Figure 6.100

Solutions to Practice Problems on Integration 1. We use the substitution w = y2 + 5, dw = 2y dy. Ê

1 (y2 + 5)8 (2y dy) 2Ê 1 w9 1 = w8 dw = +C 2Ê 2 9 1 2 = (y + 5)9 + C. 18

y(y2 + 5)8 dy =

d 1 2 1 ( (y + 5)9 + C) = [9(y2 + 5)8 (2y)] = y(y2 + 5)8 . dy 18 18 q2 q4 (q 3 + 8q + 15) dq = +8⋅ + 15q + C 2. Ê 4 2 q4 = + 4q 2 + 15q + C 4 Check:

SOLUTIONS TO PRACTICE PROBLEMS ON INTEGRATION 3.

(u4 + 5) du =

u5 + 5u + C 5

x3 + x + C. 3 1 d −3t e−3t dt = − e−3t + C. (e ) = −3e−3t , we have 5. Since Ê dx 3 6. 4x3∕2 + C x3 7. (ax2 + b) dx = a ⋅ + bx + C Ê 3 q 1∕2 8. q −1∕2 dq = + C = 2q 1∕2 + C Ê 1∕2 x4 9. + 2x2 + 8x + C. 4 1 −0.5t 100e−0.5t dt = 100 10. e + C = −200e−0.5t + C Ê −0.5 q 2 q −2 q2 1 11. (q + q −3 ) dq = + +C = − 2 +C Ê 2 −2 2 2q w4 w3 w5 12. (w4 − 12w3 + 6w2 − 10) dw = − 12 ⋅ +6⋅ − 10 ⋅ w + C Ê 5 4 3 5 w = − 3w4 + 2w3 − 10w + C 5 q 3 5q 2 13. + + 2q + C 3 2 Ax4 Bx2 (Ax3 + Bx) dx = + +C 14. Ê 4 2 4 5x−1 5 15. + 5x−2 dx = 4 ln |x| + + C = 4 ln |x| − + C Ê x −1 x √ x1∕2 x3∕2 16 16. (6x−1∕2 + 8x1∕2 )dx = 6 +8 = 12 x + x3∕2 + C Ê 1∕2 3∕2 3 4.

17.

3 sin d = −3 cos + C

18. −150e−0.2t + C p3 5 19. + 5 ln |p| + C (p2 + ) dp = Ê p 3 1 0.075t 20. 1000e0.075t dt = 1000 e + C = 13333e0.075t + C Ê 0.075 21.

(5 sin x + 3 cos x) dx = −5 cos x + 3 sin x + C

22.

(10 + 5 sin x) dx = 10x − 5 cos x + C

23.

(Aq + B)dq =

Aq 2 + Bq + C 2

5 dw = 5 ln |w| + C Ê w 0 31 r r2 ℎ dr = ℎ 25. + C = ℎr3 + C Ê 3 3 26. 24.

where C is a constant.

ex dx Ê = xex − ex + C,

xex dx = xex −

(let x = u, ex = v¨ , ex = v)

527

Chapter Six /SOLUTIONS

528

0 31 p q 4 + C = 5p3 q 4 + C Ê 3 0 51 q 15p2 q 4 dq = 15p2 + C = 3p2 q 5 + C 28. Ê 5 15p2 q 4 dp = 15

27.

29.

(3x2 + 6e2x ) dx = 3 ⋅

e2x x3 +6⋅ +C 3 2 3 2x = x + 3e + C

1 5e2q dq = 5 ⋅ e2q + C = 2.5e2q + C 2 0 1 p4 1 31. dp = + ln |p| + C p3 + Ê p 4 30.

12 cos(4x)dx = 3 sin(4x) + C Ê 33. We use the substitution w = y + 2, dw = dy:

32.

1 1 dy = dw = ln |w| + C = ln |y + 2| + C. Ê y+2 Ê w 34.

(6x1∕2 + 15) dx = 6 ⋅

35.

(x2 + 8 + ex ) dx =

x3∕2 + 15x + C = 4x3∕2 + 15x + C 3∕2

x3 + 8x + ex + C 3

t3 − 3t2 + 5t + C. 3 1 x−1 b + bx−2 dx = a ln |x| + b 37. a + C = a ln |x| − + C Ê x −1 x 38. 36.

Ê0

10 |10 −e−z dz ze−z dz = [−ze−z ]|| − Ê0 |0 |10 = −10e−10 − [e−z ]|| |0 = −10e−10 − e−10 + 1

(let z = u, e−z = v¨ , −e−z = v)

= −11e−10 + 1. 1 (e2t + 5)dt = e2t + 5t + C Ê 2 1 40. sin(4x) + C 4 P 1 kt 41. e + C = 0 ekt + C P0 ekt dt = P0 Ê k k 1 42. sin(3x)dx = − cos(3x) + C Ê 3 A 43. A sin(Bt)dt = − cos(Bt) + C Ê B 44. We use the substitution w = 3x + 1, dw = 3dx: 39.

√ 1 1 w3∕2 2 3x + 1dx = w1∕2 dw = + C = (3x + 1)3∕2 + C. Ê 3 3 3∕2 9

45. We use the substitution w = 2 + ex , dw = ex dx. ex dw dx = = ln |w| + C = ln(2 + ex ) + C. Ê 2 + ex Ê w

SOLUTIONS TO PRACTICE PROBLEMS ON INTEGRATION (We can drop the absolute value signs since 2 + ex ≥ 0 for all x.) ex d 1 x [ln(2 + ex ) + C] = Check: ⋅ e = . 2 + ex dx 2 + ex 46. We use the substitution w = 1 + sin x, dw = cos xdx: √ cos x w1∕2 + C = 2 1 + sin x + C. dx = w−1∕2 dw = √ Ê Ê 1∕2 1 + sin x 47. Integration by parts with u = ln x, v¨ = x gives Ê

x ln x dx =

x2 1 1 1 ln x − x dx = x2 ln x − x2 + C. Ê 2 2 2 4

529

7.1 SOLUTIONS

531

CHAPTER SEVEN Solutions for Section 7.1 1. We use the fact that the area of a rectangle is Base × Height. (a) The fraction less than 5 meters high is the area to the left of 5, so Fraction = 5 ⋅ 0.05 = 0.25. (b) The fraction above 6 meters high is the area to the right of 6, so Fraction = (20 − 6)0.05 = 0.7. (c) The fraction between 2 and 5 meters high is the area between 2 and 5, so Fraction = (5 − 2)0.05 = 0.15. 1 2. We use the fact that the area of a triangle is ⋅ Base ⋅ Height. Since p(x) is a line with slope 0.1∕20 = 0.005, its equation 2 is p(x) = 0.005x. (a) The fraction less than 5 meters high is the area to the left of 5. Since p(5) = 0.005(5) = 0.025, Fraction =

1 ⋅ 5(0.025) = 0.0625. 2

(b) The fraction more than 6 meters high is the area to the right of 6. Since p(6) = 0.005(6) = 0.03, Fraction = 1 − (Area to left of 6) 1 = 1 − ⋅ 6(0.03) = 0.91. 2 (c) Fraction between 2 and 5 meters high is area between 2 and 5. Since p(2) = 0.005(2) = 0.01, Fraction = (Area to left of 5) − (Area to left of 2) 1 = 0.0625 − ⋅ 2 ⋅ (0.01) = 0.0525. 2 3. We use the fact that the area of a triangle is intercept 0.1, its equation is

1 ⋅ Base ⋅ Height. Since p(x) is a line with slope −0.1∕20 = −0.005 and vertical 2 p(x) = 0.1 − 0.005x.

(a) The fraction less than 5 meters high is the area to the left of 5. Since p(5) = 0.1 − 0.005(5) = 0.075, Fraction = 1 − (Area to the right of 5) 1 = 1 − ⋅ (20 − 5)0.075 = 0.4375. 2 (b) The fraction more than 6 meters high is the area to the right of 6. Since p(6) = 0.1 − 0.005(6) = 0.07, Fraction =

1 (20 − 6)0.07 = 0.49. 2

(c) The fraction between 2 meters and 5 meters high is the area between 2 and 5. Since p(2) = 0.1 − 0.005(2) = 0.09, Fraction = (Area to the right of 2) − (Area to the right of 5) 1 1 = (20 − 2)(0.09) − (20 − 5)(0.075) 2 2 = 0.81 − 0.5625 = 0.2475

532

Chapter Seven /SOLUTIONS

4. We use the fact that the area of a rectangle is Base × Height. (a) The fraction less than 5 meters high is the area to the left of 5, so Fraction = 5 ⋅ 0.1 = 0.5. (b) The fraction above 6 meters high is the area to the right of 6, so Fraction = (10 − 6) ⋅ 0.1 = 0.4. (c) The fraction between 2 and 5 meters high is the area between 2 and 5, so Fraction = (5 − 2) ⋅ 0.1 = 0.3. 5. We can determine the fractions by estimating the area under the curve. Counting the squares for insects in the larval stage between 10 and 12 days we get about 4.5 squares, with each square representing (2)⋅(3%) giving a total of 27% of the insects in the larval stage between 10 and 12 days. Likewise we get about 2 squares for the insects in the larval stage for less than 8 days, giving 12% of the insects in the larval stage for less than 8 days. Likewise we get about 7.5 squares for the insects in the larval stage for more than 12 days, giving 45% of the insects in the larval stage for more than 12 days. Since the peak of the graph occurs between 12 and 13 days, the length of the larval stage is most likely to fall in this interval. 6. (a) Most of the earth’s surface is below sea level. Much of the earth’s surface is either around 3 miles below sea level or exactly at sea level. It appears that essentially all of the surface is between 4 miles below sea level and 2 miles above sea level. Very little of the surface is around 1 mile below sea level. (b) The fraction below sea level corresponds to the area under the curve from −4 to 0 divided by the total area under the curve. This appears to be about 34 . 7. The statement p(70) = 0.05 means that for some small interval Δx around 70, the fraction of families with incomes in that interval around $70,000 is about 0.05Δx. 8. For a small interval Δx around 68, the fraction of the population of American men with heights in this interval is about (0.2)Δx. For example, taking Δx = 0.1, we can say that approximately (0.2)(0.1) = 0.02 = 2% of American men have heights between 68 and 68.1 inches. 9. Since p(x) is a density function, Area under graph =

1 ⋅ 50c = 25c = 1, 2

so c = 1∕25 = 0.04. 10. Since p(t) is a density function, Area under graph =

1 ⋅ c ⋅ 0.01 = 0.005c = 1, 2

so c = 1∕0.005 = 200. 11. Since p(t) is a density function, Area under graph = 50 ⋅ 2c + 25 ⋅ c = 125c = 1, so c = 1∕125 = 0.008. 12. Since p(x) = cx, we know p(2) = 2c. Since p(x) is a density function, Area under graph =

1 ⋅ 2 ⋅ 2c = 2c = 1, 2

so c = 1∕2 = 0.5. 13. (a) The total area under the graph must be 1, so Area = 5(0.01) + 5C = 1 So 5C = 1 − 0.05 = 0.95 C = 0.19

7.1 SOLUTIONS

533

(b) The machine is more likely to break in its tenth year than frst year. It is equally likely to break in its frst year and second year. (c) Since p(t) is a density function, Fraction of machines lasting up to 2 years Fraction of machines lasting between 5 and 7 years Fraction of machines lasting between 3 and 6 years

= Area from 0 to 2 = 2(0.01) = 0.02.

= Area from 5 to 7 = 2(0.19) = 0.38.

= Area from 3 to 6 = Area from 3 to 5 + Area from 5 to 6 = 2(0.01) + 1(0.19) = 0.21.

14. We need to fnd a nonnegative function p(x) for which the area between y = p(x) and the x axis is 1. One example is a decreasing linear function p(x) = b + mx with m < 0. We choose b and m so that p(5) = 0 and the triangular area under the graph of p(x) for 0 ≤ x ≤ 5 is 1. Since b is the height of the triangle, we have Area =

1 b⋅5=1 2

or b = 2∕5. Then p(5) = 0 gives 2 + 5m = 0 5 or m = −2∕25. Thus p(x) =

2 2 − x 5 25

when

0≤x≤5

and

p(x) = 0 otherwise.

15. If x is the yield in kg, the density function is a horizontal line at p(x) = 1∕100 for 0 ≤ x ≤ 100. See Figure 7.1. p(x)

1∕100

100

x (kg)

Figure 7.1 16. See Figure 7.2. Many other answers are possible. p(x)

200

x (kg)

Figure 7.2 17. See Figure 7.3. Many other answers are possible.

p(x)

Figure 7.3

x (kg)

534

Chapter Seven /SOLUTIONS

18. It is not (a) since a probability density must be a non-negative function; not (c) since the total integral of a probability density must be 1; (b) and (d) are probability density functions, but (d) is not a good model. According to (d), the probability that the next customer comes after 4 minutes is 0. In real life there should be a positive probability of not having a customer in the next 4 minutes. So (b) is the best answer.

Solutions for Section 7.2 1. (a) The cumulative distribution function P (t) is defned to be the fraction of patients who get in to see the doctor within t hours. No one gets in to see the doctor in less than 0 minutes, so P (0) = 0. We saw in Example 2 part (c) that 60% of patients wait less than 1 hour, so P (1) = 0.60. We saw in part (b) of Example 2 that an additional 37.5% of patients get in to see the doctor within the second hour, so 97.5% of patients will see the doctor within 2 hours; P (2) = 0.975. Finally, all patients are admitted within 3 hours, so P (3) = 1. Notice also that P (t) = 1 for all values of t greater than 3. A table of values for P (t) is given in Table 7.1. Table 7.1 Cumulative distribution function for the density function in Example 2 t

⋯

P (t)

0.60

0.975

⋯

(b) The graph is in Figure 7.4. fraction of patients 1 P (t)

0.8 0.6 0.4 0.2 1

t (hours)

Figure 7.4 2. Splitting the fgure into four pieces, as in Figure 7.5, we see that Area under the curve = A1 + A2 + A3 + A4 1 1 = (0.16)4 + 4(0.08) + (0.12)2 + 2(0.12) 2 2 = 1. ∞

We expect the area to be 1, since

Ê−∞

p(x) dx = 1 for any probability density function, and p(x) is 0 except when 2 ≤ x ≤ 8. p(x)

0.24

0.12 0.08

A2 2

Figure 7.5

x (tons of fish)

7.2 SOLUTIONS

535

3. (a) The two functions are shown below. The choice is based on the fact that the cumulative distribution does not decrease. (b) The cumulative distribution levels o˙ to 1, so the top mark on the vertical scale must be 1. 1 0.8 0.6 0.4 0.2

Cumulative

✛

Density

✠ 2

The total area under the density function must be 1. Since the area under the density function is about 2.5 boxes, each box must have area 1∕2.5 = 0.4. Since each box has a height of 0.2, the base must be 2. 4. Since the function takes on the value of 4, it cannot be a cdf (whose maximum value is 1). In addition, the function decreases for x > c, which means that it is not a cdf. Thus, this function is a pdf. The area under a pdf is 1, so 4c = 1 giving c = 41 . The pdf is p(x) = 4 for 0 ≤ x ≤ 14 , so the cdf is given in Figure 7.6 by ⎧ ⎪ 0 ⎪ P (x) = ⎨ 4x ⎪ ⎪ 1 ⎩

for x < 0 1 for 0 ≤ x ≤ 4 1 for x > 4 P (x)

x 1 4

Figure 7.6 5. Since the function is decreasing, it cannot be a cdf (whose values never decrease). Thus, the function is a pdf. The area under a pdf is 1, so, using the formula for the area of a triangle, we have 1 4c = 1, 2

giving

The pdf is p(x) =

1 1 − x 2 8

1 . 2

for 0 ≤ x ≤ 4,

so the cdf is given in Figure 7.7 by ⎧ ⎪ 0 2 ⎪ P (x) = ⎨ x − x ⎪ 2 16 ⎪ 1 ⎩

for x < 0

for 0 ≤ x ≤ 4 for x > 4. P (x)

x 4

Figure 7.7

536

Chapter Seven /SOLUTIONS

6. Since the function levels o˙ at the value of c, the area under the graph is not fnite, so it is not 1. Thus, this function cannot be a pdf. It is a cdf and c = 1. The cdf is given by ⎧ 0 ⎪ x P (x) = ⎨ ⎪ 5 ⎩ 1

for

x<0

for

0≤x≤5

for

x > 5.

The pdf in Figure 7.8 is given by ⎧ 0 for x < 0 ⎪ p(x) = ⎨ 1∕5 for 0 ≤ x ≤ 5 ⎪ for x > 5. ⎩ 0 P (x) 1∕5

x 5

Figure 7.8 7. This function decreases, so it cannot be a cdf. Since the graph must represent a pdf, the area under it is 1. The region consists of two rectangles, each of base 0.5, and one of height 2c and one of height c, so Area = 2c(0.5) + c(0.5) = 1 c=

1 2 = 1.5 3

The pdf is therefore ⎧ 0 for x<0 ⎪ ⎪ 4∕3 for 0 ≤ x ≤ 0.5 p(x) = ⎨ ⎪ 2∕3 for 0.5 < x ≤ 1 ⎪ for x > 1. ⎩ 0 The cdf P (x) is the antiderivative of this function with P (0) = 0. See Figure 7.9. The formula for P (x) is ⎧ 0 ⎪ ⎪ 4x∕3 P (x) = ⎨ ⎪ 2∕3 + (2∕3)(x − 0.5) ⎪ 1 ⎩

for

x<0

for for

0.5 < x ≤ 1

for

x > 1.

0 ≤ x ≤ 0.5

P (x) 1 2 3

x 0.5

Figure 7.9

7.2 SOLUTIONS

537

8. This function increases and levels o˙ to c. The area under the curve is not fnite, so it is not 1. Thus, the function must be a cdf, not a pdf, and 3c = 1, so c = 1∕3. The pdf, p(x) is the derivative, or slope, of the function shown, so, using c = 1∕3, ⎧ 0 ⎪ ⎪ (1∕3 − 0)∕(2 − 0) = 1∕6 p(x) = ⎨ ⎪ (1 − 1∕3)∕(4 − 2) = 1∕3 ⎪ 0 ⎩

for x < 0

for 0 ≤ x ≤ 2

for 2 < x ≤ 4 for x > 4.

See Figure 7.10. 1 3

p(x) 1 6

x 2

Figure 7.10 9. This function does not level o˙ to 1, and it is not always increasing. Thus, the function is a pdf. Since the area under the curve must be 1, using the formula for the area of a triangle, 1 ⋅c⋅1=1 2

c = 2.

Thus, the pdf is given by ⎧ 0 for x<0 ⎪ ⎪ 4x for 0 ≤ x ≤ 0.5 p(x) = ⎨ ⎪ 2 − 4(x − 0.5) = 4 − 4x for 0.5 < x ≤ 1 ⎪ 0 for x > 0. ⎩ To fnd the cdf, we integrate each part of the function separately, making sure that the constants of integration are arranged so that the cdf is continuous. Since ∫ 4xdx = 2x2 + C and P (0) = 0, we have 2(0)2 + C = 0 so C = 0. Thus P (x) = 2x2 on 0 ≤ x ≤ 0.5. At x = 0.5, the cdf has value P (0.5) = 2(0.5)2 = 0.5. Thus, we arrange that the integral of 4 − 4x goes through the point (0.5, 0.5). Since ∫ (4 − 4x) dx = 4x − 2x2 + C, we have 4(0.5) − 2(0.5)2 + C = 0.5 giving Thus

⎧ 0 ⎪ ⎪ 2x2 P (x) = ⎨ 2 ⎪ 4x − 2x − 1 ⎪ 1 ⎩

for for

C = −1.

x<0

0 ≤ x ≤ 0.5

for

0.5 < x ≤ 1

for

x > 1.

See Figure 7.11. P (x) 1

0.5

x 0.5

Figure 7.11

538

Chapter Seven /SOLUTIONS

10. (a) (i) The probability density function is (III). (ii) The cumulative distribution function is (VI). (b) (i) The probability density function is (I). (ii) The cumulative distribution function is (V). (c) (i) The probability density function is (IV). (ii) The cumulative distribution function is (II). 11. (a) The shaded region in Figure 7.12 represents the probability that the bus will be from 2 to 4 minutes late. 0.25

1 P (t)

p(t)

0.125

−2

−1

time late (minutes)

−2

−1

Figure 7.12

time late (minutes)

Figure 7.13

(b) The probability that the bus will be 2 to 4 minutes late (the area shaded in Figure 7.12) is P (4) − P (2). The infection point on the graph of P (t) in Figure 7.13 corresponds to where p(t) is a maximum. To the left of the infection point, P is increasing at an increasing rate, while to the right of the infection point P is increasing at a decreasing rate. Thus, the infection point marks where the rate at which P is increasing is a maximum (i.e., where the derivative of P , which is p, is a maximum). 12. (a) F (7) = 0.6 tells us that 60% of the trees in the forest have height 7 meters or less. (b) F (7) > F (6). There are more trees of height less than 7 meters than trees of height less than 6 meters because every tree of height ≤ 6 meters also has height ≤ 7 meters. 13. (a) The probability of waiting between 10 and 20 minutes is given by 20

Ê10

p(t) dt = P (20) − P (10) = 65% − 50% = 15%.

(b) The probability of waiting between 5 and 10 minutes is given by Ê5

p(t) dt = P (10) − P (5) = 50% − 35% = 15%.

The probability of waiting longer than 30 minutes is: ∞

Ê30

p(t) dt = 100% − 80% = 20%.

Therefore you are more likely to wait longer than 30 minutes. 14. (a) The fraction of students passing is given by the area under the curve from 2 to 4 divided by the total area under the curve. This appears to be about 23 . (b) The fraction with honor grades corresponds to the area under the curve from 3 to 4 divided by the total area. This is about 13 . (c) The peak around 2 probably exists because many students work to get just a passing grade. (d) fraction of students 1

GPA

7.2 SOLUTIONS

539

15. (a) The area under the graph of the height density function p(x) is concentrated in two humps centered at 0.5 m and 1.1 m. The plants can therefore be separated into two groups, those with heights in the range 0.3 m to 0.7 m, corresponding to the frst hump, and those with heights in the range 0.9 m to 1.3 m, corresponding to the second hump. This grouping of the grasses according to height is probably close to the species grouping. Since the second hump contains more area than the frst, there are more plants of the tall grass species in the meadow. (b) As do all cumulative distribution functions, the cumulative distribution function P (x) of grass heights rises from 0 to 1 as x increases. Most of this rise is achieved in two spurts, the frst as x goes from 0.3 m to 0.7 m, and the second as x goes from 0.9 m to 1.3 m. The plants can therefore be separated into two groups, those with heights in the range 0.3 m to 0.7 m, corresponding to the frst spurt, and those with heights in the range 0.9 m to 1.3 m, corresponding to the second spurt. This grouping of the grasses according to height is the same as the grouping we made in part (a), and is probably close to the species grouping. (c) The fraction of grasses with height less than 0.7 m equals P (0.7) ≈ 0.25 = 25%. The remaining 75% are the tall grasses. 16. (a) See Figure 7.14. This is a cumulative distribution function. fraction of cost overruns 1

probability density

0.5

−20

C (in %)

−20

Figure 7.14

C (in %)

Figure 7.15

(b) The density function is the derivative of the cumulative distribution function. See Figure 7.15. (c) Let’s call the cumulative distribution function F (C). The probability that there will be a cost overrun of more than 50% is 1 − F (50) = 0.01, a 1% chance. The probability that it will be between 20% and 50% is F (50) − F (20) = 0.99 − 0.50 = 0.49, or 49%. The most likely amount of cost overrun occurs when the slope of the tangent line to the cumulative distribution function is a maximum. This occurs at the infection point of the cumulative distribution graph, at about C = 28%. 17. (a) The probability that a banana lasts between 1 and 2 weeks is given by Ê1

p(t)dt = 0.25

Thus there is a 25% probability that the banana will last between one and two weeks. (b) The formula given for p(t) is valid for up to four weeks; for t > 4 we have p(t) = 0. So a banana lasting more than 3 weeks must last between 3 and 4 weeks. Thus the probability is Ê3

p(t)dt = 0.325

32.5% of the bananas last more than 3 weeks. (c) Since p(t) = 0 for t > 4, the probability that a banana lasts more than 4 weeks is 0. 18. (a) The cumulative distribution, P (t), is the function whose slope is the density function p(t). So P ¨ (t) = p(t). The graph of P (t) starts out with a small slope at t = 0; its slope increases as t increases to 3. The graph of P (t) levels o˙ at 1 for t ≥ 4. See Figure 7.16.

540

Chapter Seven /SOLUTIONS fraction of bananas 1

P (t)

0.75 0.5 0.25

−1

t (weeks)

1 2 3 4 5

Figure 7.16 (b) The probability that a banana will last between 1 and 2 weeks is given by the di˙erence P (2)−P (1) where P ¨ (t) = p(t) and p(t) is the density function. Looking at Figure 7.16 we see that the di˙erence is roughly 0.25 = 25%. 19. (a) Let P (x) be the cumulative distribution function of the heights of the unfertilized plants. As do all cumulative distribution functions, P (x) rises from 0 to 1 as x increases. The greatest number of plants will have heights in the range where P (x) rises the most. The steepest rise appears to occur at about x = 1 m. Reading from the graph we see that P (0.9) ≈ 0.2 and P (1.1) ≈ 0.8, so that approximately P (1.1) − P (0.9) = 0.8 − 0.2 = 0.6 = 60% of the unfertilized plants grow to heights between 0.9 m and 1.1 m. Most of the plants grow to heights in the range 0.9 m to 1.1 m. (b) Let PA (x) be the cumulative distribution function of the plants that were fertilized with A. Since PA (x) rises the most in the range 0.7 m ≤ x ≤ 0.9 m, many of the plants fertilized with A will have heights in the range 0.7 m to 0.9 m. Reading from the graph of PA , we fnd that PA (0.7) ≈ 0.2 and PA (0.9) ≈ 0.8, so PA (0.9) − PA (0.7) ≈ 0.8 − 0.2 = 0.6 = 60% of the plants fertilized with A have heights between 0.7 m and 0.9 m. Fertilizer A had the e˙ect of stunting the growth of the plants. On the other hand, the cumulative distribution function PB (x) of the heights of the plants fertilized with B rises the most in the range 1.1 m ≤ x ≤ 1.3 m, so most of these plants have heights in the range 1.1 m to 1.3 m. Fertilizer B caused the plants to grow about 0.2 m taller than they would have with no fertilizer. 20. The cumulative distribution function P (t) =

Ê0

p(x)dx = Area under graph of density function p(x) for 0 ≤ x ≤ t = Fraction of population who die t years or less after treatment = Fraction of population who are dead at time t.

21. (a) Since d(e−ct )∕dt = ce−ct , we have c

Ê0

e−ct dt = −e−ct ||0 = 1 − e−6c = 0.1, 6

1 c = − ln 0.9 ≈ 0.0176. 6

(b) Similarly, with c = 0.0176, we have c

Ê6

e−ct dt = −e−ct ||6

= e−6c − e−12c = 0.9 − 0.81 = 0.09, so the probability is 9%. 22. (a) The probability you dropped the glove within a kilometer of home is given by Ê0

|1 | 2e−2x dx = −e−2x | = −e−2 + 1 ≈ 0.865. | |0

7.3 SOLUTIONS

541

(b) Since the probability that the glove was dropped within y km = ∫0 p(x)dx = 1 − e−2y , we solve y

1 − e−2y = 0.95 e−2y = 0.05 ln 0.05 y= ≈ 1.5 km. −2

Solutions for Section 7.3 1. The median daily catch is the amount of fsh such that half the time a boat will bring back more fsh and half the time a boat will bring back less fsh. Thus the area under the curve and to the left of the median must be 0.5. There are 25 squares under the curve so the median occurs at 12.5 squares of area. Now 5

Ê2

p(x)dx = 10.5 squares

and

p(x)dx = 5.5 squares, Ê5 so the median occurs at a little over 5 tons. We must fnd the value a for which Ê5

p(t)dt = 2 squares,

and we note that this occurs at about a = 5.4. Hence Ê2

5.4

p(t) dt ≈ 12.5 squares ≈ 0.5.

The median is about 5.4 tons. 2. (a) The median corresponds to the value of t such that P (t) = 0.5. Since P (36) ≈ 0.5, the median ≈ 36. (b) The density function is positive wherever the derivative of P (t) is positive, namely from about t = 5 until roughly t = 65. The derivative function is increasing everywhere P (t) is concave up. So that the density function is increasing until about t = 35 and is decreasing after that. The local maximum is where the function is changing from increasing to decreasing so that t = 35 is a local maximum. 3. (a) The percentage of fsh having a mass in the range from 9 to 15 grams is equal to the area under the graph of the density function on the interval 9 ≤ x ≤ 15. This area can be approximated by the area under the trapezoid with vertices (9, 0.06), (15, 0.03), (9, 0) and (15, 0). See Figure 7.17. The area under the trapezoid is 0.5(0.06+0.03)(15−9) = 0.27, so approximately 27% of the fsh have a mass between 9 and 15 grams. (b) From the fgure, we see that the area from 0 ≤ x ≤ 5 is clearly too small, and the area from 0 ≤ x ≤ 12 (or any larger value of x) is clearly too large. About half the area under the curve appears to lie on the interval 0 ≤ x ≤ 8, so about half the fsh weigh 8 gm or less, and half weigh 8 gm or more. Thus, 8 gm is closest to the median fsh mass. (c) Visually, the area under the curve on the interval 0 ≤ x ≤ 5 is greater (about 25%) than the area on the interval 15 ≤ x ≤ 25 (about 16%). Thus, there are more fsh with mass less than 5 gm. fraction of fish per gram of mass 0.09 0.06 0.03

Figure 7.17

mass (grams)

542

Chapter Seven /SOLUTIONS

4. The mean is the value of the integral Ê0

x ⋅ 0.5(2 − x) dx =

The median is the value of T such that Ê0

2 . 3

0.5(2 − x) dx = 0.5.

Integrating gives the equation

x2 || T2 =T − = 0.5 | 4 2 |0 T

x − 0.5 which is √ a quadratic with solutions T = 2 ± T = 2 − 2 = 0.586.

√

2. Since the median must be between 0 and 2, the solution we want is

5. The median is the value of T such that P (T ) = 0.5, so we solve T−

1 T2 = 4 2

√ √ to get T = 2 ± 2. Since the median is between 0 and 2 we discard the larger solution, so the median is T = 2 − 2. To fnd the mean, we frst calculate the probability density Density = p(x) = P ¨ (x) = 1 −

x 2

and then evaluate the integral Mean =

Ê0

xp(x) dx =

Ê0

x−

x2 2 dx = . 2 3

So the mean is 2∕3. 6. We know that the median is given by T such that T

Ê−∞

p(t) dt = 0.5.

That is, the median splits the area into two equal parts. Looking at the graph suggests that the median is between 2 and 3 weeks. We can use a calculator or computer to evaluate the integral for various T -values between these two points as in Table 7.2. From Table 7.2, we see that the median is between 2.4 and 2.6 weeks. With more evaluations we fnd that the median is approximately T = 2.48 weeks. Figure 7.18 supports the conclusion that t = 2.48 is in fact the median. fraction of bananas per week of age 0.4 0.3 0.2 0.1

Table 7.2 T

∫−∞ p(t) dt T

2.0

2.2

2.4

2.6

2.8

3.0

0.35

0.4114

0.4752

0.5408

0.6076

0.675

7. We know that the mean is given by

∞

Ê−∞

tp(t)dt.

Figure 7.18

weeks

7.3 SOLUTIONS Thus we get Mean =

Ê0

tp(t)dt 4

Ê0 ≈ 2.4 =

(−0.0375t3 + 0.225t2 )dt

Thus the mean is 2.4 weeks. Figure 7.19 supports the conclusion that t = 2.4 is in fact the mean. fraction of bananas per week of age 0.4 0.3 0.2 0.1

t (weeks)

Figure 7.19

8. (a) We can fnd the proportion of students by integrating the density p(x) between x = 1.5 and x = 2: 2

x3 dx Ê1.5 4

P (2) − P (1.5) = =

x4 || | 16 ||1.5

(2)4 (1.5)4 − = 0.684, 16 16

so that the proportion is 0.684 ∶ 1 or 68.4%. (b) We fnd the mean by integrating x times the density over the relevant range: 2 0 31 x dx Mean = x Ê0 4 2

x4 dx 4

Ê0

x5 || | 20 ||0

25 = 1.6 hours. 20

(c) The median will be the time T such that exactly half of the students are fnished by time T , or in other words 1 = 2 Ê0

x3 dx 4

T 1 x4 || = | 2 16 ||0

1 T4 = 2 16 √ 4 T = 8 = 1.682 hours.

543

544

Chapter Seven /SOLUTIONS

9. (a) See Figure 7.20. The mean is larger than the median for this distribution; both are less than 15. fraction of people waiting

fraction of people waiting

0.1

0.05

✛ Median = 6.9 minutes ✛ Mean = 10 minutes 15

minutes

Figure 7.20

minutes

Figure 7.21

(b) We know that the median is the value T such that T

Ê−∞

p(t)dt = 0.5

In our case this gives 0.5 = =

Ê0

p(t)dt 0.1e−0.1t dt

Substituting di˙erent values of T we get Median = T ≈ 6.9. See Figure 7.21. We know that the mean is given by ∞

Mean =

Ê−∞

tp(t)dt

(0.1te−0.1t )dt Ê0 ≈ 9.83. =

(c) The median tells us that exactly half of the people waiting at the stop wait less than 6.9 minutes. The fact that the mean is 9.83 minutes can be interpreted in the following way: If all the people waiting at the stop were to wait exactly 9.83 minutes, the total time waited would be the same. 10. (a) The normal distribution of car speeds with = 58 and = 4 is shown in Figure 7.22.

x 52

60 = 58

Figure 7.22

7.3 SOLUTIONS

545

The probability that a randomly selected car is going between 60 and 65 is equal to the area under the curve from x = 60 to x = 65, 65 2 2 1 Probability = √ e−(x−58) ∕(2⋅4 ) dx ≈ 0.2685. Ê 4 2 60 We obtain the value 0.2685 using a calculator or computer. (b) To fnd the fraction of cars going under 52 km/hr, we evaluate the integral 1 Fraction = √ 4 2 Ê0

e−(x−58) ∕32 dx ≈ 0.067.

Thus, approximately 6.7% of the cars are going less than 52 km/hr. 11. (a) Since = 100 and = 15:

1 − 1 x−100 √ e 2 15 . 15 2 (b) The fraction of the population with IQ scores between 115 and 120 is (integrating numerically) p(x) =

120

Ê115

120

p(x) dx =

Ê115

(x−100)2 1 − √ e 450 dx 15 2

120

(x−100)2 1 e− 450 dx √ 15 2 Ê115 ≈ 0.067 = 6.7% of the population.

12. (a) Since P is the cumulative distribution function, the percentage of households that made between $50,000 and $75,000 is P (75) − P (50) = 57.1% − 39.9% = 17.2%. Therefore 17.2% of the households made between $50,000 and $75,000. The percentage of households making over $150,000 is 100% − 84.5% = 15.5%. (b) The median income is the income such that half the households make less than this amount. Looking at the table, we see that the 50% mark occurs between $50,000 and $75,000. Since P (50) = 39.9%, we know 39.9% of the households made less than $50,000. Assuming local linearity we calculate the slope of the line connecting (50, 39.9) with (75, 57.1) as (57.1−39.9)∕25 = 0.688. If x is the additional income it takes to achieve the median, then 0.688x = 50.0 − 39.9, so x = 14.68. Since 50 + 14.68 = 64.68, the median income is approximately $65,000. (c) The percentage of households that made between $50,000 and $100,000 is 69.6 − 39.9 = 29.7. Since this percentage is less than 1∕3, the statement is false. 13. Since p(t) = 0.04 − 0.0008t, the cumulative distribution which satisfes P ¨ (t) = p(t) is given by 0.0008t2 +C 2 = 0.04t − 0.0004t2 + C.

P (t) = 0.04t −

Since P (0) = 0, we have C = 0, so P (t) = 0.04t − 0.0004t2 . For the median T , P (T ) = 0.04T − 0.0004T 2 = 0.5. Solving the quadratic equation 0.0004T 2 − 0.04T + 0.5 = 0 gives

√ (0.04)2 − 4(0.0004)(0.5) T = . 2(0.0004) Evaluating gives T = 85.4 and 14.6. Since p(t) is not defned for t > 50, the median is T = 14.6 days. 0.04 ±

546

Chapter Seven /SOLUTIONS

Solutions for Chapter 7 Review 1. Since p(x) is a density function, the area under the graph of p(x) is 1, so Area = Base ⋅ Height = 15a = 1 1 a= . 15 2. The area under the graph of p(x) is the sum of two rectangles, each of base 5. Since ∫0 p(x) dx = 1, we have 10

Ê0

p(x) dx = 5 ⋅ a + 5 ⋅ 2a = 15a = 1 a=

1 . 15

3. Since p(x) is a density function, the area under the graph of p(x) is 1, so Area =

1 1 Base ⋅ Height = ⋅ 10 ⋅ a = 5a = 1 2 2 a=

1 . 5

4. Since p(x) is a density function, the area under the graph of p(x) is 1, so Area =

1 1 ⋅ Base ⋅ Height = ⋅ 100 ⋅ a = 50a = 1 2 2 a=

1 . 50

5. Suppose x is the age of death; a possible density function is graphed in Figure 7.23. fraction dying per year

x (years)

Figure 7.23

6. If x is the height in feet of each person, a possible density function is graphed in Figure 7.24. The teachers are included. fraction of population per foot

Figure 7.24

x (ft)

SOLUTIONS to Review Problems For Chapter Seven

547

7. The two humps of probability in density (a) correspond to two intervals on which its cumulative distribution function is increasing. Thus (a) and (II) correspond. A density function increases where its cumulative distribution function is concave up, and it decreases where its cumulative distribution function is concave down. Density (b) matches the distribution with both concave up and concave down sections, which is (I). Density (c) matches (III) which has a concave down section but no interval over which it is concave up. 8.

% of population per dollar of income

% of population having at least this income

income

Figure 7.25: Density function 9.

Figure 7.26: Cumulative distribution function

% of population per dollar of income

% of population having less than this income

income

Figure 7.27: Density function 10.

Figure 7.28: Cumulative distribution function

% of population per dollar of income

% of population having at least this income

income

Figure 7.29: Density function

income

Figure 7.30: Cumulative distribution function

11. (a) More insects die in the twelfth month than the frst month because P (t) is larger in the twelfth month. This means that 12 1 the area ∫11 p(t) dt is larger than the area ∫0 p(t) dt. (b) We want to fnd the fraction dying within the frst 6 months. Since p(12) = 1∕6, we have p(6) = 1∕12, so 6

p(t) dt Ê0 = Area from t = 0 to t = 6 1 1 1 = . = ⋅6⋅ 2 12 4

Fraction living up to 6 months =

So 1/4 of population lives no more than 6 months. (c) We can frst fnd the fraction of insects who die within the frst 9 months. Using p(9) = 61 ⋅ 129 = 81 , we have Fraction living up to 9 months =

Ê0

p(t) dt = Area from t = 0 to t = 9 =

1 1 9 ⋅9⋅ = . 2 8 16

So the quantity we want is Fraction living more than 9 months = 1 −

9 7 = . 16 16

12. No. Though the density function has its maximum value at 50, this does not mean that a large fraction of the population receives scores near 50. The value p(50) can not be interpreted as a probability. Probability corresponds to area under the graph of a density function. Most of the area in this case is in the broad hump covering the range 0 ≤ x ≤ 40, very little in the peak around x = 50. Most people score in the range 0 ≤ x ≤ 40.

548

Chapter Seven /SOLUTIONS

13. The fact that most of the area under the graph of the density function is concentrated in two humps, centered at 8 and 12 years, indicates that most of the population belong to one of two groups, those who leave school after fnishing approximately 8 years and those who fnish about 12 years. There is a smaller group of people who fnish approximately 16 years of school. The percentage of adults who have completed less than ten years of school is equal to the area under the density function to the left of the vertical line at t = 10. (See Figure 7.31.) We know that the total area is 1, so we are estimating the percentage of the total area that lies in this shaded part shown in Figure 7.31. A rough estimate of this area is about 30%. fraction of population per year of education

8 10 12

t (years of education)

Figure 7.31: What percent has less than 10 years of education? 14. (a) The frst item is sold at the point at which the graph is frst greater than zero. Thus the frst item is sold at t = 30 or January 30. The last item is sold at the t value at which the function is frst equal to 100%. Thus the last item is sold at t = 240 or August 28, unless its a leap year. (b) Looking at the graph at t = 121 we see that roughly 65% of the inventory has been sold by May 1. (c) The percent of the inventory sold during May and June is the di˙erence between the percent of the inventory sold on the last day of June and the percent of the inventory sold on the frst day of May. Thus, the percent of the inventory sold during May and June is roughly 25%. (d) The percent of the inventory left after half a year is 100 − (percent inventory sold after half year). Thus, roughly 10% of the inventory is left after half a year. (e) The items probably went on sale on day 100 and were on sale until day 120. Roughly from April 10 until April 30. 15. (a) The density function f (r) is zero outside the range 0 < r < 5 and equal to a nonzero constant k inside this range. The area of the region under the density curve equals 5k, which must equal 1, so k = 0.2. We have T 0 if r ≤ 0 f (r) = 0.2 if 0 < r < 5 0 if 5 ≤ r. The graph of f (r) is given in Figure 7.32. f (r)

0.2

r 0

Figure 7.32 (b) The cumulative distribution function F (r) equals the area of the region under the density function up to r. From the graph in Figure 7.32, we see that the area is zero if r ≤ 0; for 0 < r < 5 the region is rectangular of height 0.2, width r, and area 0.2r; and for r ≥ 5 the area is 1. Thus T 0 if r ≤ 0 F (r) = 0.2r if 0 < r < 5 1 if 5 ≤ r.

SOLUTIONS to Review Problems For Chapter Seven

549

16. Since Product D is absorbed most quickly, this is the solution. We see that 80% of the theophylline solution is absorbed within an hour, and all of it within 5 or 6 hours. As we would expect, the timed-release capsules are absorbed more slowly. Product A is absorbed slightly faster than Product B, although both are close to fully absorbed after 24 hours. However, we see very slow absorption with Product C. Even after 28 hours have passed, only about 60% of the drug has been absorbed. 17. Figure 7.33 is a graph of the density function; Figure 7.34 is a graph of the cumulative distribution. 1

Mar

Jun

Sep

Dec

Figure 7.33: Density function Mar

Jun

Sep

Dec

Figure 7.34: Cumulative distribution function 18. (a) The percentage of calls lasting from 1 to 2 minutes is given by the integral Ê1

p(x) dx ≈ 0.221 = 22.1%.

(b) A similar calculation (changing the limits of integration) gives the percentage of calls lasting 1 minute or less as Ê0

p(x) dx ≈ 0.33 = 33.0%.

∞

p(x) dx ≈ 0.301 = 30.1%.

(d) The cumulative distribution function is the integral of the probability density; thus, C(ℎ) =

ℎ

Ê0

p(x) dx =

19. The median is the value T such that

Ê0

ℎ

0.4e−0.4x dx = 1 − e−0.4ℎ .

Ê−∞

p(x)dx = 0.5

Thus we get T

0.5 = =

Ê−∞ Ê0

p(x)dx

0.122e−0.122x dx

Substituting di˙erent values for T we get T ≈ 5.68. Thus the median occurs at 5.68 seconds We know that the mean is ∞ Ê−∞

xp(x)dx.

Thus we get ∞

Mean = =

Ê−∞ Ê0

xp(x)dx

x(0.122e−0.122x )dx ≈ 7.83 seconds.

The median tells us that ffty percent of the time gaps between cars are less than 5.68 seconds, and ffty percent of the time gaps between cars are more than 5.68 seconds. The mean tells us that over all time gaps, the average time gap between cars is 7.83 seconds.

550

Chapter Seven /SOLUTIONS

20. False. Note that p is the density function for the population, not the cumulative density function. Thus p(10) = 1∕2 means that the probability of x lying in a small interval of length Δx around x = 10 is about (1∕2)Δx. 21. True. This follows directly from the defnition of the cumulative density function. 22. True. The interval from x = 9.98 to x = 10.04 has length 0.06. Assuming that the value of p(x) is near 1∕2 for 9.98 < 10.04 x < 10.04, the fraction of the population in that interval is ∫9.98 p(x) dx ≈ (1∕2)(0.06) = 0.03. 23. False. Note that p is the density function for the population, not the cumulative density function. Thus p(10) = p(20) means that x values near 10 are as likely as x values near 20. 24. True. By the defnition of the cumulative distribution function, P (20) − P (10) = 0 is the fraction of the population having x values between 10 and 20.

STRENGTHEN YOUR UNDERSTANDING 1. False, since we also need p(x) ≥ 0.

2. True, since both ∫−∞ p(x) dx = 1 and p(x) ≥ 0. ∞

3. True, since this is the interpretation of density functions given in the text. 4. True, since this is the interpretation of density functions given in the text.

5. False. The integral ∫0 p(x) dx represents the fraction of the population between the ages of 0 and 70 years old. The fraction will be signifcantly more than 50% of the population. 70

6. False, since we have to multiply p(50) by an age interval to obtain a fraction of population. We can interpret p(50) ⋅ 1 = p(50)(50.5 − 49.5) as approximately the fraction of the population between 49.5 and 50.5 years old. 7. True, since p(10)(10.5 − 9.5) = 0.014, so about 1.4 percent has ages between 9.5 and 10.5 years.

8. False. For example, if p(t) is an age density distribution, then ∫0 p(x) dx = 1. ∞

9. True, since the population with values between 0 and 1 is included in the population with values between 0 and 2.

10. False. It is possible for p(x) = 0 for 100 ≤ x ≤ 200.

11. True, since as t increases, the fraction of the population that is less than t must stay the same or increase. 12. False. It should be P ¨ = p. 13. False. P (30) is the fraction of the population with values at or below 30. 14. True, since P (20) is the fraction of the population with values less than 20, and P (10) is the fraction of the population with values less than 10. 15. False. We have P (30) − P (10) is the fraction of the population with values between 10 and 30, while P (20) is the fraction of the population with values at or below 20. Clearly, the two can be di˙erent (for example, for age distribution). 16. True, since this is the defnition of the cumulative distribution function. 17. False. The units of p(x) are fraction of population per unit of x, while the units for P (x) are fraction of population. 18. True, since the fraction having value at most t cannot be less than zero or more than one. 19. False. We know that P (25) is the fraction less than 25, which we cannot determine from the information about the fraction less than 10 or less than 15. 20. True. This follows from the Fundamental Theorem of Calculus, since P ¨ = p. Both expressions represent the fraction of values between 25 and 50. 21. True, this is the defnition of mean value as given in the text. 22. False. The integral for the mean value uses the distribution function p(t), not the cumulative distribution function. 23. False. The median T satisfes P (T ) = 0.5. 24. True. The given integral is just P (T ), which is 0.5. 25. True, since the median T is the value for which one-half of the population is greater than or equal to T . 26. False. For example, if p(x) is a density function with a normal distribution having mean 0, then 0 is both the median and the mean. 27. True, this is the defnition of the normal distribution as specifed in the text. 28. False, the given distribution has mean = 5. 29. True, since the integrand is a normal distribution with mean = 7 and standard deviation = 1.

PROJECTS FOR CHAPTER SEVEN

551

30. False, T is the median, not the mean.

PROJECTS FOR CHAPTER SEVEN 1. (a) The area under the graph of p(x) from x = a to x = b should be 1. 2 . Therefore 1 = 21 (base) ⋅ (height) = 21 (b − a) ⋅ p(c), and so p(c) = b−a

2 2 , we have p(9) = 10−6 = 0.5. (b) We have a = 6, b = 10, and c = 9. Using p(c) = b−a (c) To fnd the equation for the frst line, we note that (6, 0) and (9, 0.5) are points on the line, so m1 = Δy∕Δx = (0.5 − 0)∕(9 − 6) = 1∕6. And if y = m1 x + b1 , substituting in x = 6, y = 0, yields 0 = (1∕6)6 + b1 , so b1 = −1. To fnd the equation for the second line, we note that (10, 0) and (9, 0.5) are points on the line, so m2 = Δy∕Δx = (0.5 − 0)∕(9 − 10) = −1∕2. And if y = m2 x + b2 , substituting in x = 10, y = 0, yields 0 = (−1∕2)10 + b2 , so b2 = 5. Thus <x −1 6≤x≤9 p(x) = 6 x − 2 + 5 9 ≤ x ≤ 10.

(d) The probability that production cost is under $8 is given by the area under p(x) from x = 6 to x = 8. The area is 21 (8 − 6) ⋅ p(8) = p(8) = 86 − 1 = 13 .

(e) The median cost is the value x = m such that the area under p(x) from x = 6 to x = m is 12 . To fnd m, we

solve 12 (m − 6) ⋅ p(m) = 12 , or (m − 6) ⋅ p(m) = 1. We know that 6 ≤ m ≤ 9 since the area under p(x) from

x = 6 to x = 9 is greater than 12 . Therefore p(m) = m6 − 1 and so (m − 6) ⋅ (

m − 1) = 1 6

m2 − 2m + 6 = 1 6 m2 − 12m + 36 = 6 m2 − 12m + 30 = 0 By the quadratic formula, m=

12 ±

√ √ 144 − 120 = 6 ± 6. 2

√ Since 6 ≤ m ≤ 9, we must have m = 6 + 6. (f) The cumulative probability distribution function P (x) gives the area under p(x) from x = a to x. (i) For 6 ≤ x ≤ 9, the area is given by 1 1 x 1 x2 (x − 6) ⋅ p(x) = (x − 6) ⋅ ( − 1) = ( − 2x + 6) 2 2 6 2 6 x2 = −x+3 12

(ii) For 9 ≤ x ≤ 10, the area is given by the area under p(x) from x = 6 to x = 10 , which is equal to 1, minus the area under p(x) from x to x = 10, which is given by 21 ⋅ (10 − x) ⋅ p(x). Since 9 ≤ x ≤ 10, p(x) = − x2 + 5. Therefore 1 x x2 P (x) = 1 − (10 − x) ⋅ (− + 5) = − + 5x − 24. 2 2 4 Therefore T x2 −x+3 6≤x≤9 12 P (x) = x2 − 4 + 5x − 24 9 ≤ x ≤ 10.

552

Chapter Seven /SOLUTIONS and its graph is given in Figure 7.35. P (x) 1

x 6 Figure 7.35

9 10

8.1 SOLUTIONS

553

CHAPTER EIGHT Solutions for Section 8.1 1. We make a table by calculating values for C = f (m, d) for each value of m and d. Such a table is shown in Table 8.1. Table 8.1 m

1000

130

170

210

2000

140

180

220

260

3000

190

230

270

310

4000

240

280

320

360

2. (a) We have f (3, 2000) = 40(3) + 0.05(2000) = 220. Insuring a car for three months and driving it 2000 miles costs $220. (b) Since f (3, d) has m fxed at 3, it is the cost, C, of insuring a car for three months in terms of miles driven, d, over that time period. A plot of C = f (3, d) against d is shown in Figure 8.1. C ($) 600

C ($) 400 300

400

200 200

100 1000

3000

5000

d (miles)

Figure 8.1

m (months)

Figure 8.2

(c) Since f (m, 1000) has d fxed at 1000, it is the cost, C, of driving a car 1000 miles in terms of months, m, taken to drive that distance. A plot of C = f (m, 1000) against m is shown in Figure 8.2. 3. (a) Since each cable subscription costs $100, the monthly revenue earned from s subscriptions is 100s dollars, and since each special feature movie viewed costs $5, the monthly revenue earned from m special feature movies viewed is 5m dollars. The total monthly revenue is therefore given in dollars by R = 100s + 5m. (b) We have f (1000, 5000) = 100(1000) + 5(5000) = 125,000. This means if 1000 customers subscribe to the cable company, and these customers watch a total of 5000 special feature movies, the monthly revenue will be $125,000. 4. (a) Decreasing, because, other things being equal, we expect the sales of cars to drop if gas prices increase. (b) Decreasing, because, other things being equal, we expect car sales to drop if car prices increase. 5. The BMI of a person 60 inches tall is given by Row 1 of Table 8.2. Table 8.2 w

120

140

160

180

200

f (w, 60)

23.4

27.3

31.2

35.2

39.1

554

Chapter Eight /SOLUTIONS

6. From row 5 of Table 8.2 a woman of height 6 feet, that is 72 inches, and weight 120 lbs has a BMI of 16.3 which would be considered slightly underweight. A woman with weight 200 lbs has a BMI of 27.1 which is outside the given range. Using Table 8.2 the range of healthy weights is between 140 and 180 lbs. 7. There are 2.2 lb in each kilogram and 39.4 inches in each meter. We convert 90 kilograms to 90 ⋅ 2.2 = 198 lbs and 1.9 meters to 1.9 ⋅ 39.4 = 74.9 inches. The closest values in Table 8.2 are w = 200 and ℎ = 75, giving an estimated BMI of 25. 8. Beef consumption by households making $20,000/year is given by Row 1 of Table 8.3 on page 369 of the text. Table 8.3 p

3.00

3.50

4.00

4.50

f (20, p)

2.65

2.59

2.51

2.43

For households making $20,000/year, beef consumption decreases as price goes up. Beef consumption by households making $100, 000/year is given by Row 5 of Table 8.3 on page 369 of the text. Table 8.4 p

3.00

3.50

4.00

4.50

f (100, p)

5.79

5.77

5.60

5.53

For households making $100,000/year, beef consumption also decreases as price goes up. Beef consumption by households when the price of beef is $3.00/lb is given by Column 1 of Table 8.3 on page 369 of the text. Table 8.5 I

100

f (I, 3.00)

2.65

4.14

5.11

5.35

5.79

When the price of beef is $3.00/lb, beef consumption increases as income increases. Beef consumption by households when the price of beef is $4.00/lb is given by Column 3 of Table 8.3 on page 369 of the text. Table 8.6 I

100

f (I, 4.00)

2.51

3.94

4.97

5.19

5.60

When the price of beef is $4.00/lb, beef consumption increases as income increases. 9. If the price of beef is held constant, beef consumption for households with various incomes can be read from a fxed column in Table 8.3 on page 369 of the text. For example, the column corresponding to p = 3.00 gives the function ℎ(I) = f (I, 3.00); it tells you how much beef a household with income I will buy at $3.00/lb. Looking at the column from the top down, you can see that it is an increasing function of I. This is true in every column. This says that at any fxed price for beef, consumption goes up as household income goes up—which makes sense. Thus, f is an increasing function of I for each value of p. 10. The amount of money spent on beef equals the product of the unit price p and the quantity C of beef consumed: M = pC = pf (I, p). Thus, we multiply each entry in Table 8.3 on page 369 of the text by the price at the top of the column. This yields Table 8.7. Table 8.7

Amount of money spent on beef ($/household/week) Price 3.00

Income

3.50

4.00

4.50

7.95

9.07

10.04

10.94

12.42

14.18

15.76

17.46

15.33

17.50

19.88

21.78

16.05

18.52

20.76

22.82

100

17.37

20.20

22.40

24.89

8.1 SOLUTIONS

555

11. Table 8.8 gives the amount M spent on beef per household per week. Thus, the amount the household spent on beef in a year is 52M. Since the household’s annual income is I thousand dollars, the proportion of income spent on beef is 52M M = 0.052 . 1000I I Thus, we need to take each entry in Table 8.8, divide it by the income at the left, and multiply by 0.052. Table 8.9 shows the results. P =

Table 8.8 Money spent on beef ($/household/week)

Table 8.9

Proportion of annual income spent on beef Price of Beef ($)

Price of Beef ($)

Income ($1,000)

3.00

3.50

4.00

4.50

3.00

3.50

4.00

4.50

0.021

0.024

0.026

0.028

7.95

9.07

10.04

10.94

0.016

0.018

0.020

0.023

12.42

14.18

15.76

17.46

0.013

0.015

0.017

0.019

15.33

17.50

19.88

21.78

0.010

0.012

0.013

0.015

16.05

18.52

20.76

22.82

100

0.009

0.011

0.012

0.013

100

17.37

20.20

22.40

24.89

Income ($1,000)

12. In the answer to Problem 11 we saw that P = 0.052

M , I

and in the answer to Problem 10 we saw that M = pf (I, p). Putting the expression for M into the expression for P , gives: P = 0.052

pf (I, p) . I

13. (a) It feels like 81◦ F. (b) At 30% relative humidity, 90◦ F feels like 90◦ F. (c) By fnding the temperature which has heat index 105◦ F for each humidity level, we get Table 8.10: Table 8.10

Estimates of danger temperatures

Relative humidity(%)

Temperature(◦ F)

117

110

105

101

(d) With a high humidity your body cannot cool itself as well by sweating. With a low humidity your body is capable of cooling itself to below the actual temperature. Therefore a high humidity feels hotter and a low humidity feels cooler. 14.

heat index (degrees Fahrenheit) T = 100

130 120 110 100 90 80

T = 70

70 60 0 10 20 30 40 50 60

Figure 8.3

relative humidity (%)

556

Chapter Eight /SOLUTIONS Both graphs are increasing because at any fxed temperature the air feels hotter as the humidity increases. The fact that the graph for T = 100 increases more rapidly with humidity than the graph for T = 70 tells us that when it is hot (100◦ F), high humidity has more e˙ect on how we feel than at lower temperatures (70◦ F).

15. Asking if f is an increasing or decreasing function of p is the same as asking how does f vary as we vary p, when we hold a fxed. Intuitively, we know that as we increase the price p, total sales of the product will go down. Thus, f is a decreasing function of p. Similarly, if we increase a, the amount spent on advertising, we can expect f to increase and therefore f is an increasing function of a. 16. We have M = f (B, t) = B(1.05)t . Figure 8.4 gives the graphs of f as a function of t for B fxed at 10, 20, and 30. For each fxed initial investment B, the function f (B, t) is an increasing function of t, which means that the amount of money in the bank account grows for any initial investment. The larger the fxed value of B, the larger the value of f (B, t), which means that larger initial investments lead to larger account balances at any given time. Figure 8.5 gives the graphs of f as a function of B for t fxed at 0, 5, and 10. For each fxed time t after the initial investment, f (B, t) is an increasing (and in fact linear) function of B. This means that at a fxed time, the account with the larger initial investment will have more money in it. The larger t is, the larger the slope of the line. This means that at later times, the account balance depends more strongly on the size of the initial investment than at earlier times. M ($1000s)

M ($1000s) B = 30 f (30, t)

50 40

B = 20 f (20, t)

B = 10 f (10, t)

20 10 0

t (years)

f (B, 5) t=5

f (B, 10) t = 10

f (B, 0) t=0

40 30 20 10 0

Figure 8.4

B ($1000s)

Figure 8.5

17. (a) We have f (59, 180, 28) = 14 ⋅ 59 + 5 ⋅ 180 − 7 ⋅ 28 + 66 = 1596 kcal/day. (b) We have g(52, 162, 43) = 10 ⋅ 52 + 2 ⋅ 162 − 5 ⋅ 43 + 655 = 1284 kcal/day. (c) The BMR of a 40-year-old man 175 cm tall weighing 77 kg is f (77, 175, 40) = 14 ⋅ 77 + 5 ⋅ 175 − 7 ⋅ 40 + 66 = 1739 kcal/day. Since the caloric intake is less than his BMR, he should expect to lose weight. 18. (a) f (250, 200, 12) = 0.25 ⋅ 250 + 0.2 ⋅ 200 + 2 ⋅ 12 = 126.5. Talking for 250 minutes, sending 200 texts and using 12 megabytes of data results in a monthly bill of $126.5. (b) If data is disabled, then the data usage d will be zero. This means the cost P will be P = 0.25t + 0.2m where t is the number of minutes talked and m is the number of messages sent. (c) Your average monthly cost is f (120, 100, 20) = 0.25 ⋅ 120 + 0.2 ⋅ 100 + 2 ⋅ 20 = 90, or $90. Therefore, it makes sense to switch as it is cheaper. 19. We expect P to be an increasing function of A and r. (If you borrow more, your payments go up; if the interest rates go up, your payments go up.) However, P is a decreasing function of t. (If you spread out your payments over more years, you pay less each month.) 20. (a) We wish to fnd H(20, 0.3). The plot of H(T , 0.3) is shown on the given graph. To fnd H(20, 0.3), we can just fnd the value of H(T , 0.3) where T = 20. Looking at the graph, we see that H(20, 0.3) ≈ 260 cal/m3 . It takes about 260 calories to clear one cubic meter of air at 20◦ C with 0.3 grams of water in it. (b) Proceeding in an analogous manner we can estimate H for the other combinations of T and w; these are shown in Table 8.11:

8.1 SOLUTIONS

557

Table 8.11 w (gm/m3 )

T (◦ C)

0.1

0.2

0.3

0.4

150

290

425

590

110

240

330

450

100

180

260

350

150

220

300

140

200

270

21. (a) The quantity E(15, 12) is the annual energy production of a wind turbine with diameter 15 feet with a wind at 12 mph. To fnd E(15, 12) from the graph, we locate the point that is on both (1) the vertical line corresponding to a diameter of 15 feet and (2) the middle curve corresponding to 12 mph wind speed. Energy production can be read from the vertical axis. It appears to be just above the midpoint between 4000 kWh and 6000 kWh, so we estimate E(15, 12) = 5200 kilowatt-hours of energy produced per year. (b) Reading the graph we estimate the energy produced by a 10-foot turbine to be E(10, 12) = 2300 kWh with 12 mph wind E(10, 15) = 4500 kWh with 15 mph wind. Thus, the 10-foot turbine produces 4500 − 2300 = 2200 kWh more energy per year when the wind is blowing 15 mph instead of 12 mph. (c) Reading the graph we estimate the energy produced when the wind is 12 mph to be E(10, 12) = 2300 kWh with 10 ft diameter E(15, 12) = 5200 kWh with 15 ft diameter. Thus when the wind is at 12 mph, the 15-foot turbine produces 5200 − 2300 = 2900 kWh more energy per year than the 10-ft turbine. (d) We have seen that the higher wind increases production by 2200 kWh per year, whereas the larger turbine would increase production by 2900 kWh per year. You will increase energy production more by replacing the 10-ft turbine with the larger 15-ft turbine. You can discover the same result without evaluating the energy production function. Starting on the graph at the point corresponding to 10-ft diameter and 12 mph wind, you can move straight up to the 15 mph curve or move along the 12 mph curve to the 15-ft-diameter line. Moving to the 15-ft-diameter line causes a larger vertical motion in the graph, which indicates a greater increase in energy production. 22. What is the BMI of a 1.72 m tall man weighing 72 kg? We use the formula to fnd BMI: B(72, 1.72) = 72∕1.722 = 24.34. 23. A 1.58 m tall woman has a BMI of 23.2. What is her weight? We know the height and the BMI, so using the formula, we know B(W , 1.58) =

W = 23.2. 1.582

Solving for W , we get W = 57.9 kg. 24. With a BMI less than 18.5, a person is considered underweight. What is the possible range of weights for an underweight person 1.58 m tall? A person is considered underweight with BMI less than 18.5, so when B(W , H) = W ∕H 2 < 18.5. We know H = 1.58, so we are looking for all W that satisfy the inequality W ∕1.582 < 18.5, or W < 46.18 kg. 25. Converting w lbs to W kilograms gives W = 0.4536w and ℎ inches to H meters gives H = 0.0254ℎ so B(W , H) =

W 0.4536w 0.4536 w w w = = = 703.08 2 ≈ 703 2 . H2 (0.0254ℎ)2 0.02542 ℎ2 ℎ ℎ

558

Chapter Eight /SOLUTIONS

26. At any fxed time t, as we get farther from the volcano, the fallout decreases exponentially. At a fxed distance, d, at the start of the eruption, there is no fallout; as time passes, the fallout increases. This models the behavior of an actual volcano, as we expect the fallout to increase with time but diminish with distance. See Figure 8.6. V

t=2

0.5

t=1

d 0.5

1.5

2.5

Figure 8.6

27. At any fxed distance, d, as time passes, the fallout increases, although increasing more slowly as time passes. Also, the fallout decreases with distance. This answer is as we would expect: the fallout increases with time but decreases with distance from the volcano. See Figure 8.7. V 2.5

d=0

2 1.5 1

d=1

0.5

d=2 t 1

Figure 8.7

28. We have f (2, 12) = (950 + 24)e−2∕7 = 731.9, which means that, at an elevation of 2 km above sea level after 12 hours, the atmospheric pressure is 731.9 millibars. 29. (a) Holding y fxed at 3 and letting t vary means that, at an elevation of 3 km above sea level, we are considering the e˙ect of the passage of time on barometric pressure. Figure 8.8 shows the graph of P = f (3, t) = (950 + 2t)e−3∕7 . The linear shape of the graph refects the fact that the barometric pressure at a constant elevation of 3 km is increasing at a constant rate as time goes by. P (millibars) 700 600 500 400 300 200 100 12

P (millibars) 1000 800 600 400 200 24

t (hours)

Figure 8.8: f (3, t) = (950 + 2t)e−3∕7

y (km)

Figure 8.9: f (y, 24) = 998e−y∕7

8.2 SOLUTIONS

559

(b) Holding t fxed at 24 and letting y vary means that, at a fxed time of t = 24 hours, we are considering the e˙ects of elevation above sea level on barometric pressure. Figure 8.9 shows the graph of P = f (y, 24) = (950 + 48)e−y∕7 = 998e−y∕7 . The decreasing and concave upward shape of the graph indicates that, at t = 24 hours, the barometric pressure decreases as elevation increases, but the rate of decrease slows down as the elevation increases. 30. (a) For a cross-section with T fxed, we let T = k for some constant k, and get the equation P = 10k∕V . This equation shows that for a fxed temperature, the pressure of the gas in the container decreases as the volume of the container increases, so it matches the cross-sections in Figure 8.7. The concave upward shape of the contours shows that, at a fxed temperature, the pressure of the gas will decrease quickly at frst, and then more slowly, as the volume of the cylinder increases. (b) For a cross-section with V fxed, we let V = k for some constant k, and get the equation P = 10T ∕k = (10∕k)T . This equation shows that for a fxed volume, the pressure of the gas is a linear function of its temperature, so it matches the cross-sections in Figure 8.6. The linear shape of the contours shows that, at a fxed volume, the pressure of the gas is directly proportional to the temperature of the gas in the cylinder.

Solutions for Section 8.2 1. The value of the function f (x, y) at a point is the value of the contour through the point. We are looking for a point that lies on the 0.5 contour. One such point is approximately (1.4, 2). Other answers are possible. 2. The point A lies on the 0 contour, therefore f (A) = 0. 3. Since the point B lies on the same contour as the point A, f (A) = f (B). 4. We can see that as we move horizontally to the right, we are increasing x but not changing y. As we take such a path at y = 2, we cross decreasing contour lines, starting at the contour line 6 at x = 1 to the contour line 1 at around x = 5.7. This trend holds true for all of horizontal paths. Thus, z is a decreasing function of x. Similarly, as we move up along a vertical line, we cross increasing contour lines and thus z is an increasing function of y. 5. Using our economic intuition, we know that the total sales of a product should be an increasing function of the amount spent on advertising. From the graph, Q is a decreasing function of x and an increasing function of y. Thus, the y-axis corresponds to the amount spent on advertising and the x-axis corresponds to the price of the product. 6. (a) The value of the function f (x, y) at a point is the value of the contour through that point. Therefore, since the 0 contour passes through (1, −1), we have f (1, −1) = 0. (b) The value of the function f (x, y) at a point is the value of the contour through that point. Therefore, since the 4 contour passes through (2, −1), we have f (2, −1) = 4. (c) The value of the function f (x, y) at a point is the value of the contour through that point. Therefore, since the 2 contour passes through (−1, 0), we have f (−1, 0) = 2. 7. (a) The value of the function f (x, y) at a point is the value of the contour through that point. Therefore, since y = −1, we are looking for a point with that y-value which passes through the 2 contour. One such point is (0, −1). Other answers are possible. (b) The value of the function f (x, y) at a point is the value of the contour through that point. Therefore, since y = 2, we are looking for a point with that y-value which passes through the 0 contour. One such point is (−2, 2). Other answers are possible. 8. (a) The value of the function f (x, y) at a point is the value of the contour through that point. Therefore, since x = 1, we are looking for a point with that x-value which passes through the −4 contour. The only such point is (1, −2). (b) The value of the function f (x, y) at a point is the value of the contour through that point. Therefore, since x = 0, we are looking for a point with that x-value which passes through the 0 contour. One such point is (0, 0). Other answers are possible. 9. Many di˙erent answers are possible. Answers are in degrees Celsius. (a) Minnesota in winter. See Figure 8.10.

560

Chapter Eight /SOLUTIONS 10

−10

Figure 8.10

Figure 8.11

(b) San Francisco in winter. See Figure 8.11. (c) Houston in summer. See Figure 8.12. 30

Figure 8.12

Figure 8.13

(d) Oregon in summer. See Figure 8.13. 10. (a) If t = 0, then the amount of money in the bank will be the same as the initial investment, so we have f (B, 0) = B for any initial investment of B thousand dollars. This means the point (B, 0) must lie on the contour labeled B. Thus, we use the B-intercept of the contour lines to determine the contour labels as shown in Figure 8.14. (b) We need to fnd a value for B so that f (B, 15) = 4. This means the point (B, 15) has to lie on the contour labeled 4. We use the contour map to fnd the point on that contour that has t-coordinate 15. This is the point (3, 4). So we have to invest $3, 000 to accrue $4, 000 in 15 t (years) 30 1 2 3 4 5 6

B (thousands of dollars)

Figure 8.14

11. We see that if we fx a value on the horizontal axis and increase the values of the vertical axis, the contour values decrease. This is consistent with the medication leaving the blood stream over time after administration. On the other hand, if we fx a value on the vertical axis and increase the values of the horizontal axis, the contour values increase. This is consistent with the fact that a higher initial dosage would result in a higher amount of the medication remaining in the blood stream after the same amount of time. So we conclude that the horizontal axis is the initial dosage administered, D. 12. To fnd the hills, we look for areas of the map with concentric contour arcs that increase as we move closer to the center of the area. We see three hills: a height 9 hill centered near (0, 2), a height 7 hill with a peak at (3, 7), and a height 7 hill near (7, 1.5). The highest hill is just the one with the highest contour line, in this case the hill near (0, 2). We can expect the river to look like a valley, which is represented on a contour diagram as roughly parallel lines. We see such a pattern running from the bottom left at around (2, 0) to the upper right at (8, 8). To see where the river is fowing, we have to see which end of the river is at a higher altitude. At the bottom left of the map, the river has altitude between 3 and 4 and at the upper right, it reaches an altitude of 0. Thus, since rivers fow from areas of high altitude to areas of low altitude, the river fows toward the upper right—or northeast—section of the map.

8.2 SOLUTIONS

561

13. The contour where f (x, y) = x + y = c, or y = −x + c, is the graph of the straight line with slope −1 as shown in Figure 8.15. Note that we have plotted the contours for c = −3, −2, −1, 0, 1, 2, 3. The contours are evenly spaced. y 2

c = 3

= 2 c = 1

x −2

−1

c c

= 0

−2

−1

= c

−3

= −1

−2

Figure 8.15 14. The contour where f (x, y) = 3x + 3y = c or y = −x + c∕3 is the graph of the straight line of slope −1 as shown in Figure 8.16. Note that we have plotted the contours for c = −9, −6, −3, 0, 3, 6, 9. The contours are evenly spaced. y 2 c = 9

= 6 c = 3

x −2

−1

−6

c c

−1

= c

−9

= −3

−2

Figure 8.16 15. The contour where f (x, y) = x + y + 1 = c or y = −x + c − 1 is the graph of the straight line of slope −1 as shown in Figure 8.17. Note that we have plotted the contours for c = −2, −1, 0, 1, 2, 3, 4. The contours are evenly spaced. y 2

c = 4

= 3 c = 2

x −2

−1

c c

−1

= c

−2

= 0

−2

Figure 8.17

562

Chapter Eight /SOLUTIONS

16. The contour where f (x, y) = 2x − y = c is the graph of the straight line y = 2x − c of slope 2 as shown in Figure 8.18. Note that we have plotted the contours for c = −3, −2, −1, 0, 1, 2, 3. The contours are evenly spaced. y

2 c=

−1 c= 0 c= 1

c= −2

−2

−1

−2

Figure 8.18 17. The contour where f (x, y) = −x − y = c or y = −x − c is the graph of the straight line of slope −1 as shown in Figure 8.19. Note that we have plotted contours for c = −3, −2, −1, 0, 1, 2, 3. The contours are evenly spaced. y 2 c = =

−3

−2

c = −1

−2

−1

x 2

−1

= c

= 1

−2

Figure 8.19 18. The contour where f (x, y) = y − x2 = c is the graph of the parabola y = x2 + c, with vertex (0, c) and symmetric about the y-axis, shown in Figure 8.20. Note that we have plotted the contours for c = −2, −1, 0, 1. The contours become more closely packed as we move farther from the y-axis. y

−2

−1

−2

Figure 8.20

2 x

8.2 SOLUTIONS

563

√ 19. The contour where f (x, y) = x2 +y2 = c, where c ≥ 0, is the graph of the circle centered at (0, 0), with radius c as shown in Figure 8.21. Note that we have plotted the contours for c = 0, 1, 2, 3, 4. The contours become more closely packed as we move further from the origin. y 2 c = =

c 3

c =

c = 1 c=0

−2

−1

−2

Figure 8.21

20. The contour where f (x, y) = xy = c, is the graph of the hyperbola y = c∕x if c ≠ 0 and the coordinate axes if c = 0, as shown in Figure 8.22. Note that we have plotted contours for c = −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5. The contours become more closely packed as we move further from the origin.

= c

4 3

−1

−2

−3

−4

0 =

= c

−4

−3

= c 4

−2

−1

Figure 8.22

21. The contour for C = 50 is given by 40d + 0.15m = 50. This is the equation of a line with intercepts d = 50∕40 = 1.25 and m = 50∕0.15 ≈ 333. (See Figure 8.23.) The contour for C = 100 is given by 40d + 0.15m = 100. This is the equation of a parallel line with intercepts d = 100∕40 = 2.5 and m = 100∕0.15 ≈ 667. The contours for C = 150 and C = 200 are parallel lines drawn similarly. (See Figure 8.23.)

564

Chapter Eight /SOLUTIONS m 1400 1200 1000 800 200

600

150

400 100

200

d 1

Figure 8.23: A contour diagram for C = 40d + 0.15m 22. If we let x represent the high temperature and y represent the low temperature, then the contours should be highest for large x values and small y values. The contour values are greatest in the lower right corner of Quadrant I. One possible contour diagram is shown in Figure 8.24. Other answers are also possible. low temperature 10 20 30 40

high temperature

Figure 8.24 23. The vertical spacing between the contours just north and just south of the trail increases as you move eastward along the trail. A possible contour diagram is in Figure 8.25.

10 20

Elevation in meters

0 101

Trail 1000

0 98

990

Figure 8.25 24. (a) 80-90◦ F (b) 60-72◦ F (c) 60-100◦ F

8.2 SOLUTIONS 25.

565

predicted high temperature 100 90 80 70

✛

Topeka south

✲

north

distance from Topeka

Figure 8.26 26. 80

Boise

100

Boise

100

✠

South

North

West

Figure 8.27

East

Figure 8.28

27. (a) The proft is given by the following: = (Revenue from q1 ) + ( Revenue from q2 ) − Cost. Measuring in thousands, we obtain: = 4q1 + 13q2 − (4 + q1 + q2 ) = 3q1 + 12q2 − 4. (b) A contour diagram of is in Figure 8.29. Note that the units of are in thousands. q2 3

2 = 30

20 10 0

Figure 8.29

566

Chapter Eight /SOLUTIONS (i) As money increases, with love fxed, your happiness goes up, reaches a maximum and then goes back down. Evidently, there is such a thing as too much money. (ii) On the other hand, as love increases, with money fxed, your happiness keeps going up. (b) A cross-section with love fxed will show your happiness as money increases; the curve goes up to a maximum then back down, as in Figure 8.30. The higher cross-section, showing more overall happiness, corresponds to a larger amount of love, because as love increases so does happiness. Figure 8.31 shows two cross-sections with money fxed. Happiness increases as love increases. We cannot say, however, which cross-section corresponds to a larger fxed amount of money, because as money increases happiness can either increase or decrease.

28. (a)

happiness

money

love

Figure 8.30: Cross-sections with love fxed

Figure 8.31: Cross-sections with money fxed

29. The contours of z = y − sin x are of the form y = sin x + c for a constant c. They are sinusoidal graphs shifted vertically by the value of z on the contour line. The contours are equally spaced vertically for equally spaced z values. See Figure 8.32. y 4 z=3

z=1

−2

−

x 2 z = −1

z = −3 −4

Figure 8.32

30. One possible answer follows. (a) If there is a city at the center of the diagram, then the population is very dense at the center, but progressively less dense as you move into the suburbs, further from the city center. This scenario corresponds to diagram (I) or (II). We pick (I) because it has the highest density, as we would expect in a city. (b) If the center of the diagram is a lake, and is a very busy and thriving center, where lake front property is considered the most desirable, then the most dense area will be at lakeside, and decrease as you move further from the lake in the center, as in diagram (I) or (II). We pick (II) because we expect the population density at lake front to be less than that in the center of a city. In an alternative solution, if the lake were in the middle of nowhere, the entire area would be very sparsely populated, and there would be slightly fewer people living on the actual lake shore, as in diagram (III). (c) If the center of the diagram is a power plant and if the plant is not in a densely populated area, where people can and will choose not to live anywhere near it, the population density will then be very low nearby, increasing slightly further from the plant, as in diagram (III).

8.2 SOLUTIONS

567

31. (A) In graph I, L = 1, K = 1 gives us F = 1, and L = 3, K = 3 gives us F = 3. So tripling all inputs in graph I triples output; graph I corresponds to statement (A). (B) In graph II, L = 1, K = 1 gives us F = 1, and L = 2.2, K = 2.2 gives us F = 1.5. Extrapolating from this ratio, L = 4, K = 4 should gives us F = 2. So, quadrupling all inputs in graph II doubles output; graph II corresponds to statement (B). (C) In graph III, L = 1, K = 1 gives us F = 1, and L = 2, K = 2 gives us F = 2.8. So, doubling the inputs in graph III almost tripled output; graph III corresponds to statement (C). 32. (a) False. The values on the level curves are decreasing as you go northward in Canada. (b) True. In general, the contour levels are increasing from peninsulas to mainland. This is true for all three examples. For instance, the density on the Baja peninsula is mostly below 180 whereas on the mainland nearby it is 200 and up to 280. (c) False. It is below 100, since the values on the level curves are decreasing as you go southward in Florida and Miami is south of the 100 level curve. 33. We frst investigate the behavior of f with t fxed. We choose a value for t and move horizontally across the diagram looking at how the values on the contours change. For example, for t = 4 hours, as we move from the left at x = 0 to the right at x = 5, we cross contours of 0.1, 0.2, 0.3, 0.4, and 0.5. The concentration of the drug is increasing as the initial dose, x, increases. (This is what we saw in Figure 8.2 in Section 8.1.) For each choice of f with t fxed (each horizontal line), we see that the contour values are increasing as we move from left to right, showing that, at any given time, the concentration of the drug increases as the size of the dose increases. The values of f with x fxed are read vertically. For x = 1, as we move up from t = 0 to t = 5, we see that the contours go up from 0.1 to 0.2 to 0.3, and then back down from 0.3 to 0.2 to 0.1. The maximum value of C is very close to 0.4 and is reached at about t = 1 hour. The concentration increases quickly at frst (the contour lines are closer together), reaches its maximum, and then decreases slowly (the contour lines are farther apart.) All of the contours of f with x fxed are similar to this, although the maximum value varies. 34. (a) Find the point where the horizontal line for 15 mph meets the contour for −20◦ F wind chill. The actual temperature is about 0◦ F. (b) The horizontal line for 10 mph meets the vertical line for 0◦ F about 1∕5 of the way from the contour for −20◦ F to the contour for 0◦ F wind chill. We estimate the wind chill to be about −16◦ F. (c) We look for the point on the vertical line for −20◦ F where the wind chill is −50◦ F, the danger point for humans. This is a point on the line that is about half way between the contours for −60◦ F and −40◦ F. The point can not be determined exactly, but we estimate that it occurs where the wind speed is about 23 mph. (d) A temperature drop of 20◦ F corresponds to moving left from one vertical grid line to the next on the horizontal line for 15 mph. This horizontal movement appears to correspond to about 1 1∕4 the horizontal distance between contours crossing the line. Since contours are spaced at 20◦ F wind chill, we estimate that the wind chill drops about 25◦ F when the air temperature goes down 20◦ F during a 15 mph wind. 35. (a) We have N = 300 and V = 200 so production P = 2(300)0.6 (200)0.4 = 510.17 thousand pages per day. (b) If the labor force, N, is doubled to 600 workers and the value of the equipment, V , remains at 200 thousand, we have P = 2(600)0.6 (200)0.4 = 773.27. The production P has increased from 510.17 to 773.27, an increase of about 263 thousand pages per day. (c) If the value of the equipment, V , is doubled to 400 thousand and the labor force, N, remains at 300 workers, we have P = 2(300)0.6 (400)0.4 = 673.17. The production P has increased from 510.17 to 673.17, an increase of about 163 thousand pages per day. We see that doubling the work force has a greater e˙ect on production than doubling the value of the equipment. We might have expected this, since the exponent for N is larger than the exponent for V . (d) What happens if we double both N and V ? Then P = 2(600)0.6 (400)0.4 = 1020.34 thousand pages per day. This is exactly double our original production level of 510.17. In this case, doubling both N and V doubled P . 36. (a) The contour lines are much closer together on path A, so path A is steeper. (b) If you are on path A and turn around to look at the countryside, you fnd hills to your left and right, obscuring the view. But the ground falls away on either side of path B, so you are likely to get a a wider view of the horizon from path B. (c) There is more likely to be a stream alongside path A, because water follows the direction of steepest descent.

568

Chapter Eight /SOLUTIONS

37. (a) To fnd cardiac output o when pressure p = 4 mm Hg and time t = 0 hours (i.e. when the patient goes into shock) we go to the coordinate (0, 4) on the graph; we see that this point falls at where o = 12 L/min, so initial cardiac output for p = 4 mm Hg is 12 L/min. Three hours later, t = 3 and p still = 4. Going to (3, 4) on the graph, we fnd that we are somewhat to the right of where o = 8. Since output is decreasing as we move to the right, output at (3, 4) will be somewhat less than 8. Estimating, 3 hours after shock with pressure of 4 mm Hg, o ≈ 7.5 L/min. Since o = 12 at t = 0, we wish to fnd the time corresponding to o = 12∕2 = 6 and p still = 4. Going along the line of o = 6, we fnd that this intersects p = 4 at t ≈ 4.2. So about 4.2 hours have elapsed when cardiac output halves. (b) Looking at the graph, we see that output increases as we move up a vertical line. (c) Looking at the graph, we see that output decreases as we move to the right along a horizontal line. (d) If p = 3, the cardiac output decreases rather slowly between t = 0 and t = 2, the frst two hours. We can see this because there is a large gap between initial output, which is approximately 9 L/min, and o = 8. In these two hours, output only falls by 1 L/min, a fairly gradual change. Between t = 2 and t = 4, output falls from 8 L/min to 6 L/min. In the last hour, output falls all the way from 6 L/min to 0 L/min, corresponding to death. So the decrease in output accelerates rapidly. This information might be very useful to physicians treating shock. It tells them that in the frst two hours the patient’s condition will not worsen very much; in the next two hours, it will deteriorate at a faster rate. After four hours, the patient may well die unless something is done to maintain the patient’s cardiac output. 38. (a) The TMS map of an eye of constant curvature will have only one color, with no contour lines dividing the map. (b) The contour lines are circles, because the cross-section is the same in every direction. The largest curvature is in the center. See picture below.

✛ 54

✛ 50 48

39. (a) Since P is proportional to d 2 and to v3 , a formula for P is P (d, v) = kd 2 v3 , where k is the constant of proportionality. (b) Let d be the diameter of the original windmill, and let v1 be the wind speed at which the windmill produces 100 kW. Then 100 . kd 2 v31 = 100, and thus k = d 2 v31 The second windmill has diameter 2d and we want to fnd a speed v2 such that k(2d)2 v32 = 100. We solve for v2 : v3 d 2 v31 100 100 = = = 1 2 2 3 2 2 4 k(2d) 4d 100∕(d v1 ) ⋅ 4d v1 v2 = √ 3 4 v32 =

√ 3 So v2 needs to be 1∕ 4 of v1 , or about 63% of v1 . √ (c) Contours of P are curves of the form kd 2 v3 = c, or v = 3 c∕(kd 2 ). Thus, a contour diagram for P looks like the diagram in Figure 8.33. y 40

10, 000 ,00 0

100, 000 1000

x 50

Figure 8.33

8.2 SOLUTIONS

569

40. (a) The P = 30 contour crosses the horizontal line GFR = 50 at about t = 4. In such a patient, it takes about 4 hours to excrete 30% of the dose. (b) When GFR = 60 and t = 5, we are approximately on the 40 contour. Forty percent of the dose has been excreted after 5 hours. (c) The contour lines of P are approximately horizontal lines for t ≥ 12. If we look across the horizontal line for any fxed GFR, we see that the value of P changes very little for t ≥ 12. This means that the percent excreted changes very little then. (d) If we fx GFR and increase t, we see that the values of P increase. The percent excreted is an increasing function of time. This makes sense because as time goes by, more of the drug will pass through the patient’s system. (e) If we fx time and increase GFR along a vertical line, the values of P increase. The percent excreted is an increasing function of GFR. We see that in sicker patients with lower GFRs, less of the dose is excreted. This is why physicians are careful not to administer frequent doses of antibiotics to patients with kidney disease. 41. (a) is (II). According to the diagram, increasing X without changing Y leads to greater satisfaction, but increasing Y without changing X leads to reduced satisfaction. More X is desirable, and more Y is undesirable. This fts the option where X is income and Y is hours worked. (b) is (I). According to the diagram, increasing X without changing Y and increasing Y without changing X both lead to greater satisfaction. This fts the option where X is income and Y is leisure time. (c) is (III). According to the diagram, increasing X or increasing Y each lead to lower satisfaction. More X is undesirable; more Y is also undesirable. This fts the option where X is hours worked and Y is time spent commuting. 42. (a) R: The contour through R is nearly horizontal. Moving to the right on the contour, which corresponds to increasing the wages, with little change in the safety, does not change the job satisfaction. This job has a poor safety level, which can not be compensated for by an increase in wages. (b) Q: For jobs with satisfaction near Q on the contour through Q, safety and wages are equally desirable. Moving to the right along the contour represents accepting a little less safety in exchange for a little higher wage. (c) P : The contour through P is nearly vertical. Moving up on the contour, which corresponds to increasing the safety with little change in the wages, does not change the job satisfaction. This job has low wages, which can not be compensated for by increased safety. 43. We can get a sense of the shape of the contours by thinking about what it would take to keep the total productivity of the oÿce constant. If the oÿce has a certain productivity at 75◦ F, then increasing or decreasing the temperature will increase the number of workers needed to maintain the same level of productivity. As the temperature becomes more extreme, it may be necessary to increase the number of workers greatly to maintain the same productivity. Therefore, the contours of the function may be shaped somewhat like parabolas in Figure 8.34. Other answers are possible. T ◦F 100 90 80 75

70 60 50 20

100

n, workers

Figure 8.34: 44. To fnd the value of BMI on a contour, we evaluate f (w, ℎ) at any point (w, ℎ) on that contour. The contour bounding the underweight region crosses the vertical line w = 100, at ℎ ≈ 61.6, so its contour value is approximately BMI ≈ f (100, 61.6) = 703 ⋅ 100∕61.62 = 18.5. The contour bounding the normal region to the right crosses the horizontal line ℎ = 60, when w ≈ 128, so its contour value is BMI ≈ f (128, 60) = 703 ⋅ 128∕602 = 25.

570

Chapter Eight /SOLUTIONS Therefore, the underweight region for BMI is below 18.5 and the normal region is between 18.5 and 25. ℎ (in) 78

Underweight Normal

Overweight 18 .5

69 Obese 30

66 63

120

140

160

180

200

220

w (lbs)

Figure 8.35 45. (a) Since s(x, y) = xy = c, substituting x = 8, y = 2 gives s(8, 2) = 8 ⋅ 2 = 16, so the indi˙erence curve through the point (8, 2) has equation s(x, y) = xy = 16. See Figure 8.36. (b) If the shopper buys 4 units of A, so x = 4, we have 4y = 16, giving y = 4. Thus, buying 4 units of item B gives the same satisfaction. (c) From the points shown on the indi˙erence curve in Figure 8.36, we see that if the purchase of item A is reduced by a fxed amount, the increase in item B required to maintain satisfaction is larger for a smaller initial purchase of A. Thus, an initial purchase of 6 units of A leads to a larger increase in B. y

4 s(x, y) = xy = 16

x 4

Figure 8.36

Solutions for Section 8.3 1. (a) The point A is on the 10 contour, so f (A) = 10. (b) Moving right from point A in the direction of increasing x increases f , so fx (A) is positive. (c) Moving up from A in the direction of increasing y stays on the same contour and does not change the value of f , so fy (A) is zero. 2. (a) The point A is on the 15 contour, so f (A) = 15. (b) Moving right from point A in the direction of increasing x stays on the same contour and does not change the value of f , so fx (A) is zero. (c) Moving up from A in the direction of increasing y decreases the value of f , so fy (A) is negative.

8.3 SOLUTIONS

571

3. (a) The point A is on the 58 contour, so f (A) = 58. (b) Moving right from point A in the direction of increasing x increases the value of f , so fx (A) is positive. (c) Moving up from A in the direction of increasing y increases the value of f , so fy (A) is positive. 4. (a) The point A is on the 88 contour, so f (A) = 88. (b) Moving right from point A in the direction of increasing x decreases the value of f , so fx (A) is negative. (c) Moving up from A in the direction of increasing y decreases the value of f , so fy (A) is negative. 5. (a) The point A is on the 25 contour, so f (A) = 25. (b) Moving right from point A in the direction of increasing x increases the value of f , so fx (A) is positive. (c) Moving up from A in the direction of increasing y decreases the value of f , so fy (A) is negative. 6. (a) The point A is on the 40 contour, so f (A) = 40. (b) Moving right from point A in the direction of increasing x decreases the value of f , so fx (A) is negative. (c) Moving up from A in the direction of increasing y increases the value of f , so fy (A) is positive. 7. (a) Positive. (b) Negative. (c) Positive. (d) Zero. 8. fx (5, 2), because the function is increasing in the x-direction and the contour lines in the positive x direction are closer together at (5, 2) than at (3, 1). 9. We estimate )I∕)H and )I∕)T by using di˙erence quotients. We have f (H + ΔH, T ) − f (H, T ) )I ≈ )H ΔH

and

f (H, T + ΔT ) − f (H, T ) )I ≈ )T ΔT

Choosing ΔH = 10 and reading the values from Table 8.5 on page 376 in the text we get f (10 + 10, 100) − f (10, 100) )I || 99 − 95 ≈ = = 0.4. | )H ||(10,100) 10 10

Similarly, choosing ΔT = 5 we get

f (10, 100 + 5) − f (10, 100) )I || 100 − 95 ≈ = = 1. )T ||(10,100) 5 5

)I || ≈ 0.4 means that the rate of change of the heat index per unit increase in humidity is about | )H ||(10,100) 0.4. This means that the heat index increases by approximately 0.4◦ F for every percentage point increase in humidity. This rate is positive, because as the humidity increases, the heat index increases. The partial derivative )I∕)T gives the rate of increase of heat index with respect to temperature. It is positive because )I || ≈ 1 tells us that as the temperature increases the heat index increases as temperature increases. Knowing that | )T ||(10,100) by 1◦ F, the temperature you feel increases by 1◦ F also. In other words, at this humidity and temperature, the changes in temperature that you feel are approximately equal to the actual changes in temperature. )I )I > The fact that at (10, 100) tells us that a unit increase in temperature has a greater e˙ect on the heat index )T )H in Tucson than a unit increase in humidity. The fact that

10. We estimate )I∕)H and )I∕)T by using di˙erence quotients. We have f (H + ΔH, T ) − f (H, T ) )I ≈ )H ΔH

and

f (H, T + ΔT ) − f (H, T ) )I ≈ )T ΔT

Choosing ΔH = 10 and reading the values from Table 8.5 on page 376 in the text we have f (50 + 10, 80) − f (50, 80) )I || 82 − 81 = = 0.1. ≈ )H ||(50,80) 10 10

Similarly, choosing ΔT = 5 we get

f (50, 80 + 5) − f (50, 80) )I || 88 − 81 ≈ = = 1.4. )T ||(50,80) 5 5

572

Chapter Eight /SOLUTIONS )I || ≈ 0.1 means that the average rate of change of the heat index per unit increase in humidity is about | )H ||(50,80) 0.1◦ F. This means that the heat index increases by approximately 0.1 for every percentage point increase in humidity. This rate is positive, because as the humidity increases, the heat index increases. The partial derivative )I∕)T gives the rate of increase of heat index with respect to temperature. It is positive because )I || the heat index increases as temperature increases. Knowing that ≈ 1.4 tells us that as the temperature increases | )T ||(50,80) by 1◦ F, the temperature you feel increases by 1.4◦ F. Thus, at this humidity and temperature , the changes in temperature you feel are even larger then the actual change. The fact that )I∕)T > )I∕)H at (50,80) tells the residents of Boston that a unit increase in temperature has a greater e˙ect on the heat index than a unit increase in humidity. The fact that

11. (a) We expect the demand for co˙ee to decrease as the price of co˙ee increases (assuming the price of tea is fxed.) Thus we expect fc to be negative. We expect people to switch to co˙ee as the price of tea increases (assuming the price of co˙ee is fxed), so that the demand for co˙ee will increase. We expect ft to be positive. (b) The statement f (3, 2) = 780 tells us that if co˙ee costs $3 per pound and tea costs $2 per pound, we can expect 780 pounds of co˙ee to sell each week. The statement fc (3, 2) = −60 tells us that, if the price of co˙ee then goes up $1 and the price of tea stays the same, the demand for co˙ee will go down by about 60 pounds. The statement 20 = ft (3, 2) tells us that if the price of tea goes up $1 and the price of co˙ee stays the same, the demand for co˙ee will go up by about 20 pounds. )q 12. (a) An increase in the price of a new car will decrease the number of cars bought annually. Thus 1 < 0. Similarly, an )x )q2 < 0. increase in the price of gasoline will decrease the amount of gas sold, implying )y (b) Since the demands for a car and gas complement each other, an increase in the price of gasoline will decrease the total )q )q number of cars bought. Thus 1 < 0. Similarly, we may expect 2 < 0. )y )x )q > 0 because, other things being constant, as people get richer, more beer will be bought. 13. )I )q < 0 because, other things being constant, if the price of beer rises, less beer will be bought. )p1 )q > 0 because, other things being constant, if the price of other goods rises, but the price of beer does not, more beer )p2 will be bought. 14. (a) If you borrow $8000 at an interest rate of 1% per month and pay it o˙ in 24 months, your monthly payments are $376.59. (b) The increase in your monthly payments for borrowing an extra dollar under the same terms as in (a) is about 4.7 cents. (c) If you borrow the same amount of money for the same time period as in (a), but if the interest rate increases by 1%, the increase in your monthly payments is about $44.83. 15. (a) The units of )c∕)x are units of concentration/distance. (For example, (gm/cm3 )/cm.) The practical interpretation of )c∕)x is the rate of change of concentration with distance as you move down the blood vessel at a fxed time. We expect )c∕)x < 0 because the further away you get from the point of injection, the less of the drug you would expect to fnd (at a fxed time). (b) The units of )c∕)t are units of concentration/time. (For example, (gm/cm3 )/sec.) The practical interpretation of )c∕)t is the rate of change of concentration with time, as you look at a particular point in the blood vessel. We would expect the concentration to frst increase (as the drug reaches the point) and then decrease as the drug dies away. Thus, we expect )c∕)t > 0 for small t and )c∕)t < 0 for large t. 16. The partial derivative, )Q∕)b is the rate of change of the quantity of beef purchased with respect to the price of beef, when the price of chicken stays constant. If the price of beef increases and the price of chicken stays the same, we expect consumers to buy less beef and more chicken. Thus when b increases, we expect Q to decrease, so )Q∕)b < 0. On the other hand, )Q∕)c is the rate of change of the quantity of beef purchased with respect to the price of chicken, when the price of beef stays constant. An increase in the price of chicken is likely to cause consumers to buy less chicken and more beef. Thus when c increases, we expect Q to increase, so )Q∕)c > 0. 17. (a) We see that fw is positive since B increases as w increases, when s is held constant. We see that fs is also positive since B increases as s increases, when w is held constant. This means that a heavier person will burn more calories per minute while roller-blading than a lighter person, and the number of calories burned per minute goes up as the speed goes up.

8.3 SOLUTIONS

573

(b) We have f (180, 11) − f (160, 11) ΔB 14.0 − 12.4 = = = 0.08 (cal/min)/lb. Δw 180 − 160 180 − 160 For a 160 lb person roller-blading at a speed of 11 mph, the number of calories burned per minute increases by about 0.08 if the person’s weight is 1 pound more. To fnd fs , we have fw (160, 11) ≈

fs (160, 11) ≈

f (160, 13) − f (160, 11) ΔB 15.6 − 12.4 = = = 1.6 (cal/min)/mph. Δs 13 − 11 13 − 11

For a 160 lb person roller-blading at a speed of 11 mph, the number of calories burned per minute increases by about 1.6 as the speed of the roller-blader increases by 1 mph. 18. (a) At the point (−2, −1), if we hold y fxed and increase x from −2 to 0, the value of f (x, y) changes from 7 to 8, so f (x, y) seems to increase, meaning fx (−2, −1) > 0. (b) At the point (2, 1), if we hold x fxed and increase y from 1 to 3, the value of f (x, y) changes from 7 to 5, so f (x, y) seems to decrease, meaning fy (2, 1) < 0. (c) At the point (2, 1), if we hold y fxed and increase x from 2 to 4, the value of f (x, y) changes from 7 to 10, so f (x, y) seems to increase, meaning fx (2, 1) > 0. (d) At the point (0, 3), if we hold x fxed and increase y from 3 to 5, the value of f (x, y) changes from 3 to 2, so f (x, y) seems to decrease, meaning fy (0, 3) < 0. 19. We know zx (1, 0) is the rate of change of z in the x-direction at (1, 0).Therefore Δz 1 ≈ = 2, so zx (1, 0) ≈ 2. Δx 0.5 We know zx (0, 1) is the rate of change of z in the x-direction at the point (0, 1). Since we move along the contour, the change in z Δz 0 zx (0, 1) ≈ ≈ = 0. Δx Δx We know zy (0, 1) is the rate of change of z in the y-direction at the point (0, 1) so zx (1, 0) ≈

zy (0, 1) ≈

Δz 1 ≈ = 10. Δy 0.1

20. (a) This means you must pay a mortgage payment of $1090.08/month if you have borrowed a total of $92,000 at an interest rate of 14%, on a 30-year mortgage. (b) This means that the rate of change of the monthly payment with respect to the interest rate is $72.82; i.e., your monthly payment will go up by approximately $72.82 for one percentage point increase in the interest rate for the $92,000 borrowed under a 30-year mortgage. (c) It should be positive, because the monthly payments will increase if the total amount borrowed is increased. (d) It should be negative, because as you increase the number of years in which to pay the mortgage, you should have to pay less each month. 21. (a) We expect fp to be negative because if the price of the product increases, the sales usually decrease. (b) If the price of the product is $8 per unit and if $12000 has been spent on advertising, sales increase by approximately 150 units if an additional $1000 is spent on advertising. )−f (R,T ) . Approximating with ΔR = 3 22. (a) We know that we can approximate fR at a point (R, T ) by fR (R, T ) ≈ f (R+ΔR,T ΔR at the point (15, 76) gives:

f (15 + 3, 76) − f (15, 76) 3 f (18, 76) − f (15, 76) = 3

fR (15, 76) ≈

Looking at the graph, we see that f (15, 76) ≈ 100 and f (18, 76) ≈ 110, so fR (15, 76) ≈

110 − 100 ≈ 3.3% per in. 3

The next additional inch of rainfall increases corn yield by about 3.3% from the present production level.

574

Chapter Eight /SOLUTIONS (b) We estimate fT similarly; we approximate with ΔT = 2. f (15, 76 + 2) − f (15, 76) 2 f (15, 78) − f (15, 76) = 2

fT (15, 76) ≈

Looking at the graph, we see that f (15, 76) ≈ 100 and f (15, 78) ≈ 90, so fT (15, 76) ≈

90 − 100 = −5% per ◦ F. 2

When the temperature increases by one degree, corn yield falls by about 5% from the current production level. 23. Estimate )P ∕)r and )P ∕)L by using di˙erence quotients and reading values of P from the graph: P (15, 5000) − P (8, 5000) )P || ≈ | )r ||(8,5000) 15 − 8 120 − 100 = 2.9, 7

and P (8, 4000) − P (8, 5000) )P || ≈ | )L ||(8,5000) 4000 − 5000 =

80 − 100 = 0.02. −1000

Pr (8, 5000) ≈ 2.9 means that at an interest rate of 8% and a loan amount of $5000 the monthly payment increases by approximately $2.90 for an additional one percent increase of the interest rate. PL (8, 5000) ≈ 0.02 means the monthly payment increases by approximately $0.02 for an additional $1 increase in the loan amount at an 8% rate and a loan amount of $5000. 24. Reading values of H from the graph gives Table 8.12. In order to compute HT (T , w) at T = 30, it is useful to have values of H(T , w) for T = 40◦ C. The column corresponding to w = 0.4 is not used in this problem. Table 8.12 Estimated values of H(T , w) (in calories/meter3 )

Table 8.13 Estimated values of HT (T , w) (in calories/meter3 )

w (gm/m3 )

T (◦ C)

0.1

0.2

0.3

0.4

110

240

330

450

100

180

260

350

150

220

300

140

200

270

w (gm/m3 ) 0.1 T (◦ C)

0.2

0.3

−1.0

−6.0

−7.0

−3.0

−4.0

−0.5

−1.0

−2.0

The estimates for HT (T , w) in Table 8.13 are now computed using the formula HT (T , w) ≈

H(T + 10, w) − H(T , w) . 10

Practically, the values HT (T , W ) show how much less heat (in calories per cubic meter of fog) is needed to clear the fog if the temperature is increased by 10◦ C, with w fxed. 25. The sign of )f ∕)P1 tells you whether f (the number of people who ride the bus) increases or decreases when P1 is increased. Since P1 is the price of taking the bus, as it increases, f should decrease. This is because fewer people will be )f willing to pay the higher price, and more people will choose to ride the train. On the other hand, the sign of tells you )P2 the change in f as P2 increases. Since P2 is the cost of riding the train, as it increases, f should increase. This is because fewer people will be willing to pay the higher fares for the train, and more people will choose to ride the bus. )f )f Therefore, < 0 and > 0. )P1 )P2

8.3 SOLUTIONS

575

)q1 is the rate of change of the quantity of the frst brand sold as its price increases. Since this brand competes with )x another brand, an increase in the price of the frst brand should result in a decrease in the quantity sold of this same )q )q brand. Thus 1 < 0. Similarly, 2 < 0. )x )y (b) Again we take into consideration the competition between the two brands. If the second brand were to increase its )q )q price, then more of the frst brand should sell. Thus 1 > 0. Similarly, 2 > 0. )y )x 27. (a) The table gives f (200, 400) = 150,000. This means that sales of 200 full-price tickets and 400 discount tickets generate $150,000 in revenue. (b) The notation fx (200, 400) represents the rate of change of f as we fx y at 400 and increase x from 200. What happens to the revenue as we look along the row y = 400 in the table? Revenue increases, so fx (200, 400) is positive. The notation fy (200, 400) represents the rate of change of f as we fx x at 200 and increase y from 400. What happens to the revenue as we look down the column x = 200 in the table? Revenue increases, so fy (200, 400) is positive. Both partial derivatives are positive. This makes sense since revenue increases if more of either type of ticket is sold. (c) To estimate fx (200, 400), we calculate ΔR∕Δx as x increases from 200 to 300 while y is fxed at 400. We have

26. (a)

fx (200, 400) ≈

ΔR 185,000 − 150,000 = = 350 dollars/ticket. Δx 300 − 200

The partial derivative of f with respect to x is 350 dollars per full-price ticket. This means that the price of a full-price ticket is $350. To estimate fy (200, 400), we calculate ΔR∕Δy as y increases from 400 and 600 while x is fxed at 200. We have fy (200, 400) ≈

190,000 − 150,000 ΔR = = 200 dollars/ticket. Δy 600 − 400

The partial derivative of f with respect to y is 200 dollars per discount ticket. This means that the price of a discount ticket is $200. 28. (a) The partial derivative fx = 350 tells us that R increases by $350 as x increases by 1. Thus, f (201, 400) = f (200, 400)+ 350 = 150,000 + 350 = 150,350. (b) The partial derivative fy = 200 tells us that R increases by $200 as y increases by 1. Since y is increasing by 5, we have f (200, 405) = f (200, 400) + 5(200) = 150,000 + 1,000 = 151,000. (c) Here x is increasing by 3 and y is increasing by 6. We have f (203, 406) = f (200, 400) + 3(350) + 6(200) = 150,000 + 1,050 + 1,200 = 152,250. In this problem, the partial derivatives gave exact results, but in general they only give an estimate of the changes in the function. 29. We can use the formula Δf ≈ Δxfx + Δyfy . Applying this formula in order to estimate f (105, 21) from the known value of f (100, 20) gives f (105, 21) ≈ f (100, 20) + (105 − 100)fx (100, 20) + (21 − 20)fy (100, 20) = 2750 + 5 ⋅ 4 + 1 ⋅ 7 = 2777.

30. We can use the formula Δf ≈ Δrfr + Δsfs . Applying this formula in order to estimate f (52, 108) from the known value of f (50, 100) gives f (52, 108) ≈ f (50, 100) + (52 − 50)fr (50, 100) + (108 − 100)fs (50, 100) = 5.67 + 2(0.60) + 8(−0.15) = 5.67.

31. We have g(Q) ≈ g(P ) + gx (P )Δx + gy (P )Δy = 100 + (2)(0.5) + (−3)(2) = 95.

576

Chapter Eight /SOLUTIONS

32. We have g(Q) ≈ g(P ) + gx (P )Δx + gy (P )Δy = 2500 + (10)(−2) + (20)(3) = 2540.

33. We have g(Q) ≈ g(P ) + gx (P )Δx + gy (P )Δy = 12 + (−0.1)(−0.03) + (−0.2)(−0.01) = 12.005.

34. We have g(Q) ≈ g(P ) + gx (P )Δx + gy (P )Δy = 840 + (4)(−5) + (1.5)(10) = 835.

35. We are asked to fnd f (26, 15). The closest to this point in the data given in the table is f (24, 15), so we will approximate based on this value. First we need to calculate the partial derivative ft . Using Δt = −2, we have f (24, 15) ≈

f (22, 15) − f (24, 15) 58 − 36 = = −11. −2 −2

Given this, we can use Δf ≈ Δtft + Δcfc to obtain f (26, 15) ≈ f (24, 15) + 2 ⋅ ft (24, 15) + 0 ⋅ fc (26, 15) ≈ 36 − 2 ⋅ 11 + 0 = 14%. )p = fc (c, s) = rate of change in blood pressure as cardiac output increases while systemic vascular resistance )c remains constant. (b) Suppose that p = kcs. Note that c (cardiac output), a volume, s (SVR), a resistance, and p, a pressure, must all be positive. Thus k must be positive, and our level curves should be confned to the frst quadrant. Several level curves are shown in Figure 8.37. Each level curve represents a di˙erent blood pressure level. Each point on a given curve is a combination of cardiac output and SVR that results in the blood pressure associated with that curve.

36. (a)

s (SVR) s (SVR)

After Nitroglycerine, SVR drops

✠

3 p = 9k

❄

p = 1k

✲B ✻

p = 4k

c (vol) c (vol) 1

Figure 8.37

After Dopamine, cardiac output increases

Figure 8.38

(c) Point B in Figure 8.38 shows that if the two doses are correct, the changes in pressure will cancel. The patient’s cardiac output will have increased and his SVR will have decreased, but his blood pressure won’t have changed.

8.3 SOLUTIONS

577

(d) At point F in Figure 8.39, the patient’s blood pressure is normalized, but his/her cardiac output has dropped and his SVR is up. s (SVR) F resident gives drug increasing SVR

✲

✻

E ✛

during heart attack cardiac output drops ✠ D

c (vol)

Figure 8.39 Note: c1 and c2 are the cardiac outputs before and after the heart attack, respectively. 37. (a) Since fx > 0, the values on the contours increase as you move to the right. Since fy > 0, the values on the contours increase as you move upward. See Figure 8.40. y

−2

−3

−4

−5

−1

x 1

−1

−2

−3

−4

−5

Figure 8.40: fx > 0 and fy > 0

Figure 8.41: fx > 0 and fy < 0

(b) Since fx > 0, the values on the contours increase as you move to the right. Since fy < 0, the values on the contours decrease as you move upward. See Figure 8.41. (c) Since fx < 0, the values on the contours decrease as you move to the right. Since fy > 0, the values on the contours increase as you move upward. See Figure 8.42. y

−5 −4

−3

0 1

−4

−3

−2

−2 −1

−1

−5

4 5

Figure 8.42: fx < 0 and fy > 0

Figure 8.43: fx < 0 and fy < 0

(d) Since fx < 0, the values on the contours decrease as you move to the right. Since fy < 0, the values on the contours decrease as you move upward. See Figure 8.43.

578

Chapter Eight /SOLUTIONS

38. Since fx = 0, as we move in the x-direction the value of f (x, y) does not change and since fy ≠ 0, as we move in the y-direction, the value of f (x, y) does change. Therefore, the contours must be horizontal lines. Figure 8.44 shows one possibility. There are many others. y 8

−2 −4 −6

−8

Figure 8.44

39. Since fy = 0, as we move in the y-direction the value of f (x, y) does not change, and since fx ≠ 0, as we move in the x-direction, the value of f (x, y) does change.Therefore, the contours must be vertical lines. Figure 8.45 shows one possibility. There are many others. y

4 2

−8

−2

−6

−4

Figure 8.45

40. Since fx = 1, the slope of f (x, y) in the x-direction is 1. This means for each unit we move in the x-direction, the value of f (x, y) goes up by 1, so the contours are equally spaced a distance 1 apart in the x-direction for each unit increase of f (x, y). Figure 8.46 shows one possibility. There are many others.

8.4 SOLUTIONS

579

y 2 0

−1

−3

−2

x −1

2 2

−2

−1

−2

Figure 8.46 41. Since fy = −2, the slope of f (x, y) in the y-direction is −2. This means for each unit we move in the y-direction, the value of f (x, y) goes down by 2, so the contours are equally spaced a distance 1 apart in the y-direction for each decrease of 2 of f (x, y). Figure 8.47 shows one possibility. There are many others. y 2 −4

1 −2

x −2

−1

−1 2

−2

Figure 8.47 42. The fact that fx (P ) > 0 tells us that the values of the function on the contours increase as we move to the right in Figure 8.8 past the point P . Thus, the values of the function on the contours (a) Decrease as we move upward past P . Thus fy (P ) < 0. (b) Decrease as we move upward past Q (since Q and P are on the same contour line.) Thus fy (Q) < 0. (c) Decrease as we move to the right past Q. Thus fx (Q) < 0 43. (a) Since the contours are parallel straight lines, they all have the same slope. Two points on the contour for happiness level 10 are (0 cherries, 10 grapes) and (5 cherries, 0 grapes). Thus Slope =

Grapes 10 − 0 = −2 . 0−5 Cherry

(b) Dan’s happiness with his snack stays the same if he replaces two grapes with one cherry.

Solutions for Section 8.4 1. fx (x, y) = 2x + 2y, fy (x, y) = 2x + 3y2 .

580

Chapter Eight /SOLUTIONS

2. fx (x, y) = 4x + 0 = 4x fy (x, y) = 0 + 6y = 6y 3. fx (x, y) = 2 ⋅ 100xy = 200xy fy (x, y) = 100x2 ⋅ 1 = 100x2 4. fu (u, v) = 2u + 5v + 0 = 2u + 5v fv (u, v) = 0 + 5u + 2v = 5u + 2v )z = 2xey 5. )x )Q = 5a2 − 9ap2 6. )p 7. ft (t, a) = 3 ⋅ 5a2 t2 = 15a2 t2 . 8. fx (x, y) = 10xy3 + 8y2 − 6x and fy (x, y) = 15x2 y2 + 16xy. 9. fx (x, y) = 20xe3y , fy (x, y) = 30x2 e3y . 10. zx = 2xy + 10x4 y ) 1 2 1 mv = v2 11. )m 2 2 )P = 100tert 12. )r )A 1 = (a + b) 13. )ℎ 2 14. f (1, 2) = (1)3 + 3(2)2 = 13 fx (x, y) = 3x2 + 0 ⇒ fx (1, 2) = 3(1)2 = 3 fy (x, y) = 0 + 6y ⇒ fy (1, 2) = 6(2) = 12

15. f (3, 1) = 5(3)(1)2 = 15 fu (u, v) = 5v2 ⇒ fu (3, 1) = 5(1)2 = 5 fv (u, v) = 10uv ⇒ fv (3, 1) = 10(3)(1) = 30

16. We have fx = 2x and fy = 2y, so fx = 0 when x = 0 and fy = 0 when y = 0. Therefore, they are both 0 only at the point (0, 0). 17. We have fx = ey and fy = xey , so fx is never 0. Therefore, there are no points where both fx and fy are 0. 18. We have fx = 2x + 2 and fy = 2y, so fx = 0 when x = −1 and fy = 0 when y = 0. Therefore, they are both 0 only at the point (−1, 0). 19. We have fx = 3x2 + 6x = 3x(x + 2) and fy = 3y2 − 3 = 3(y − 1)(y + 1) so fx = 0 when x = 0 or −2 and fy = 0 when y = ±1. Therefore, they are both 0 at the points (0, 1), (0, −1), (−2, 1) and (−2, −1). 20. Di˙erentiating with respect to m gives )C = 40 + 0 = $40 per month, )m so the cost of renting a car an additional month is $40. Di˙erentiating with respect to d gives )C = 0 + 0.05 = $0.05 per mile, )d so the cost of driving the car an additional mile is $0.05. 21. To calculate )B∕)t, we hold P and r constant and di˙erentiate B with respect to t: )B ) = (P ert ) = P rert . )t )t

8.4 SOLUTIONS

581

In fnancial terms, )B∕)t represents the change in the amount of money in the bank as one unit of time passes by. To calculate )B∕)r, we hold t and P constant and di˙erentiate with respect to r: )B ) = (P ert ) = P tert . )r )r In fnancial terms, )B∕)r represents the rate of change in the amount of money in the bank as the interest rate changes. To calculate )B∕)P , we hold t and r constant and di˙erentiate B with respect to P : )B ) = (P ert ) = ert . )P )P In fnancial terms, )B∕)P represents the rate of change in the amount of money in the bank at time t as you change the amount of money that was initially deposited. 22. N is the number of workers, so N = 80. V is the value of equipment in units of $25,000, so V = (value of equipment)∕$25,000 = 30. f (80, 30) = 5(80)0.75 (30)0.25 = 313 tons. Using 80 workers and $750,000 worth of equipment produces about 313 tons of output. fN (N, V ) = 0.75 ⋅ 5N −0.25 V 0.25 ,

so fN (80, 30) = 0.75 ⋅ 5(80)−0.25 (30)0.25 = 2.9 tons/worker.

When the company is using 80 workers and $750,000 worth of equipment, each additional worker would add about 2.9 tons to total output. fV (N, V ) = 0.25 ⋅ 5N 0.75 V −0.75 ,

so fV (80, 30) = 0.25 ⋅ 5(80)0.75 (30)−0.75 = 2.6 tons/$25,000.

When the company is using 80 workers and $750,000 worth of equipment, each additional $25,000 of equipment would add about 2.6 tons to total output. 23. (a) QK = 18.75K −0.25 L0.25 , QL = 6.25K 0.75 L−0.75 . (b) When K = 60 and L = 100, Q = 25 ⋅ 600.75 ⋅ 1000.25 = 1704.33 QK = 18.75 ⋅ 60−0.25 1000.25 = 21.3 QL = 6.25 ⋅ 600.75 100−0.75 = 4.26 (c) Q is actual quantity being produced. QK is how much more could be produced if you increased K by one unit. QL is how much more could be produced if you increased L by 1. 24. (a) We have fx (x, y) = 2x and fy (x, y) = 2y, so fx (1, 1) = 2 and fy (1, 1) = 2. (b) We estimate fx and fy from a contour diagram by using di˙erence quotients. Since fx = fy at (1, 1), these di˙erence quotients must be approximately the same, so the horizontal and vertical spacing of the contours must be the same. The only contour diagram to exhibit this property is (II). 25. (a) We have fx (x, y) = 4x3 and fy (x, y) = 2y, so fx (1, 1) = 4 and fy (1, 1) = 2. (b) We estimate fx and fy from a contour diagram by using di˙erence quotients. Since fx > fy > 0 at (1, 1) the values of f increase more rapidly in the x-direction than in the y-direction. This means that the contours in the x-direction must be much closer together than those in the y-direction. The only contour diagram to exhibit this property is (III). 26. (a) We have fx (x, y) = 2x and fy (x, y) = 4y3 , so fx (1, 1) = 2 and fy (1, 1) = 4. (b) We estimate fx and fy from a contour diagram by using di˙erence quotients. Since fy > fx > 0 at (1, 1) the values of f increase more rapidly in the y-direction than in the x-direction. This means that the contours in the y-direction must be much closer together than those in the x-direction. The only contour diagram to exhibit this property is (I). 27. (a) From contour diagram, f (2.3, 1) − f (2, 1) 2.3 − 2 6−5 = = 3.3, 0.3 f (2, 1.4) − f (2, 1) fy (2, 1) ≈ 1.4 − 1 6−5 = = 2.5. 0.4

fx (2, 1) ≈

582

Chapter Eight /SOLUTIONS (b) A table of values for f is given in Table 8.14. Table 8.14 y

0.9

1.0

1.1

1.9

4.42

4.61

4.82

2.0

4.81

5.00

5.21

2.1

5.22

5.41

5.62

From Table 8.14 we estimate fx (2, 1) and fy (2, 1) using di˙erence quotients: 5.41 − 5.00 = 4.1 2.1 − 2 5.21 − 5.00 = 2.1. fy (2, 1) ≈ 1.1 − 1

fx (2, 1) ≈

(c) fx (x, y) = 2x, fy (x, y) = 2y. So the exact values are fx (2, 1) = 4, fy (2, 1) = 2. 28. f (500, 1000) = 16 + 1.2(500) + 1.5(1000) + 0.2(500)(1000) = $102,116 The cost of producing 500 units of item 1 and 1000 units of item 2 is $102,116. fq1 (q1 , q2 ) = 0 + 1.2 + 0 + 0.2q2 So fq1 (500, 1000) = 1.2 + 0.2(1000) = $201.20 per unit. When the company is producing at q1 = 500, q2 = 1000, the cost of producing one more unit of item 1 is $201.20. fq2 (q1 , q2 ) = 0 + 0 + 1.5 + 0.2q1 So fq2 (500, 1000) = 1.5 + 0.2(500) = $101.50 per unit. When the company is producing at q1 = 500, q2 = 1000, the cost of producing one more unit of item 2 is $101.50. 29. (a) At Q, R, we have fx < 0 because f decreases as we move in the x-direction. (b) At Q, P , we have fy > 0 because f increases as we move in the y-direction. (c) At all four points, P , Q, R, S, we have fxx > 0, because fx is increasing as we move in the x-direction. (At P , S, we see that fx is positive and getting larger; at Q, R, we see that fx is negative and getting less negative.) (d) At all four points, P , Q, R, S, we have fyy > 0, so there are none with fyy < 0. The reasoning is similar to part (c). 30. fx = 2xy and fy = x2 , so fxx = 2y, fxy = 2x, fyy = 0 and fyx = 2x. 31. Since f (x, y) = xey , the partial derivatives are fx = ey ,

fy = xey

fxx = 0,

fxy = ey = fyx ,

fyy = xey .

32. fx = 2x + 2y and fy = 2x + 2y, so fxx = 2, fxy = 2, fyy = 2 and fyx = 2. 33. fx = 2∕y and fy = −2x∕y2 , so fxx = 0, fxy = −2∕y2 , fyy = 4x∕y3 and fyx = −2∕y2 . 34. fx = 2xy2 and fy = 2x2 y, so fxx = 2y2 , fxy = 4xy, fyy = 2x2 and fyx = 4xy. 35. fx = yexy and fy = xexy , so fxx = y2 exy , fxy = xyexy +exy = (xy+1)exy , fyy = x2 exy and fyx = xyexy +exy = (xy+1)exy . 36. Bx = 5e−2t and Bt = −10xe−2t , so Bxx = 0, Btt = 20xe−2t and Bxt = Btx = −10e−2t . 37. fx = −8xt and ft = 3t2 − 4x2 , so fxx = −8t, ftt = 6t and fxt = ftx = −8x. 38. fr = 100tert and ft = 100rert , so frr = 100t2 ert , frt = ftr = 100trert + 100ert = 100(rt + 1)ert and ftt = 100r2 ert . 39. Qp1 = 10p1 p2−1 and Qp2 = −5p12 p2−2 , so Qp1 p1 = 10p−1 , Qp2 p2 = 10p21 p2−3 and Qp1 p2 = Qp2 p1 = −10p1 p2−2 . 2 40. Vr = 2 rℎ and Vℎ = r2 , so Vrr = 2 ℎ, Vℎℎ = 0 and Vrℎ = Vℎr = 2 r.

8.5 SOLUTIONS

583

41. PK = 2L2 and PL = 4KL, so PKK = 0, PLL = 4K and PKL = PLK = 4L. 42. Since fx (x, y) = 4x3 y2 − 3y4 , we could have f (x, y) = x4 y2 − 3xy4 . In that case, ) 4 2 (x y − 3xy4 ) = 2x4 y − 12xy3 )y as expected. More generally, we could have f (x, y) = x4 y2 − 3xy4 + C, where C is any constant. fy (x, y) =

43. We compute the partial derivatives: )Q )Q = b K −1 L1− so K = b K L1− )K )K )Q )Q = b(1 − )K L− so L = b(1 − )K L1− )L )L Adding these two results, we have: K

)Q )Q +L = b( + 1 − )K L1− = Q. )K )L

44. (a) We have )M c +1 = . )B c+r (b) Positive, because c > 0 and r > 0. (c) If people and banks hold constant the ratio, c, of cash to checking deposits and the fraction, r, of checking account deposits held as cash, then increasing the amount of cash in an economy increases its money supply. 45. (a) Since M = (c + 1)(c + r)−1 B we have )M c +1 = −(c + 1)(c + r)−2 B = − B. )r (c + r)2 (b) Negative, because c > 0, r > 0 and B > 0. (c) Suppose that people hold a constant ratio of cash to checking account balances, and that the amount of cash in an economy is constant. If banks start to keep a greater fraction of their deposits in cash, then the money supply decreases. 46. (a) The quotient rule gives (c + r)1 − (c + 1)1 )M 1−r = B=− B. )c (c + r)2 (c + r)2 (b) Negative sign, because c > 0, 0 < r < 1, and B > 0. (c) Suppose that banks hold a constant ratio of cash to checking account balances, and that the amount of cash in an economy is constant. If people start to keep a greater fraction of their money in cash and less in checking accounts, then the money supply decreases.

Solutions for Section 8.5 1. We can identify local extreme points on a contour diagram because these points will be at the center of a series of concentric circles that close around them. Looking at the graph, we see that (2, 10) and (6, 4) appear to be such points. Since the points near (2, 10) decrease in functional value as they close around (2, 10), f (2, 10) will be somewhat less than its nearest contour. So f (2, 10) ≈ 0.5. Similarly, since the contours near (2, 10) are greater in functional value than f (2, 10), f (2, 10) is a local minimum. Applying analogous arguments to the point (6, 4), we see that f (6, 4) ≈ 9.5 and is a local maximum. The point (2, 10) is a global minimum on the region shown since f (x, y) takes no smaller value in this region. Neither (6, 4) nor (2, 10) are global maximums since f (x, y) takes a larger value than both at the point (6.5, 16). 2. We can identify local extreme points on a contour diagram because these points will be at the center of a series of concentric circles that close around them. Looking at the graph, we see that (13, 30), (37, 18), and (32, 34) appear to be such points. Since the points near (13, 30) decrease in functional value as they close around (13, 30), f (13, 30) will be somewhat less than its nearest contour. So f (13, 30) ≈ 4.5. Similarly, since the contours near (13, 30) are greater in functional value than f (13, 30), f (13, 30) is a local minimum. Applying analogous arguments to the other extreme points, we see that f (37, 18) ≈ 2.5 and is a local minimum; and f (32, 34) ≈ 10.5 and is a local maximum. The point (37, 18) is a global minimum on the region shown since f (x, y) takes no smaller value in this region. None of the points are global maximums since f (x, y) takes larger values close to (0, 0).

584

Chapter Eight /SOLUTIONS

3. The maximum value, which is about 11, occurs at (5.1, 4.9). The minimum value, which is about −1, occurs at (1, 3.9). 4. The maximum value, which is slightly above 30, say 30.5, occurs approximately at the origin. The minimum value, which is about 20.5, occurs at (2.5, 5). 5. The maxima occur at about ( ∕2, 0) and ( ∕2, 2 ). The minimum occurs at ( ∕2, ). The maximum value is about 1, the minimum value is about −1. 6. (a) Critical points of f occur at points where each of the partial derivatives are either zero or undefned. Reading from the table, this occurs at the points (0, 0), (1, 0) and (−1, 0). (b) We apply the second derivative test at each point. At the point (0, 0), we have D = (−2)(−3) − (2)2 = 6 − 4 = 2 > 0, and fxx < 0, so f has a local maximum.At the point (1, 0), , we have D = (1)(−1) − (2)2 = −1 − 4 = −5 < 0, so f has a saddle point. At the point (−1, 0), we have D = (−1)(1) − (2)2 = −1 − 4 = −5 > 0, so f has a saddle point. 7. We have fx = y and fy = x so fxx = 0, fyy = 0, fxy = 1. Evaluating these at (0, 0) we get D = (0)(0) − (1)2 = −1 < 0. Therefore (0, 0) is a saddle point. 8. We have fx = 2x and fy = 2y so fxx = 2, fyy = 2, fxy = 0. Evaluating these at (0, 0) we get D = (2)(2) − (0)2 = 4 > 0. Therefore (0, 0) is either a local minimum or maximum. Since fxx > 0, it is a local minimum. 9. At the origin g(0, 0) = 0. Since y3 ≥ 0 for y > 0 and y3 < 0 for y < 0, the function g takes on both positive and negative values near the origin, which must therefore be a saddle point. The second derivative test does not tell you anything since D = 0. 10. At the origin f (0, 0) = 0. Since x6 ≥ 0 and y6 ≥ 0, the point (0, 0) is a local (and global) minimum. The second derivative test does not tell you anything since D = 0. 11. We have fx = ey − 1 and fy = xey so

fxx = 0, fyy = xey , fxy = ey .

Evaluating these at (0, 0) we get D = (0)(0) − (e0 )2 = −1 < 0. Therefore (0, 0) is a saddle point. 12. We have fx = 1 − ex and fy = −2y so

fxx = −ex , fyy = −2, fxy = 0.

Evaluating these at (0, 0) we get D = (−e0 )(−2) − (0)2 = 2 > 0. Therefore (0, 0) is either a minimum or maximum. Since fxx < 0, it is a local maximum. 13. We have fx = 2x and fy = sin y so fxx = 2, fyy = cos y, fxy = 0. Evaluating these at (0, 0) we get D = (2)(1) − (0)2 = 2 > 0. Therefore (0, 0) is a local maximum or minimum. Since fxx (0, 0) = 2, we see that (0, 0) is a local minimum.

8.5 SOLUTIONS

585

14. We have fx = sin y and fy = x cos y so fxx = 0, fyy = −x sin y, fxy = cos y. Evaluating these at (0, 0) we get D = (0)(0) − (1)2 = −1 < 0. Therefore (0, 0) is a saddle point. 15. At a critical point fx = 2x + 4 = 0 and fy = 2y = 0, so (−2, 0) is the only critical point. Since fxx = 2 > 0 and 2 = 4 > 0, the point (−2, 0) is a local minimum. fxx fyy − fxy 2 16. At a critical point fx = 2x + y = 0 and fy = x + 3 = 0, so (−3, 6) is the only critical point. Since fxx fyy − fxy = −1 < 0, the point (−3, 6) is a saddle point.

17. At a critical point fx = 2x + 6 = 0 and fy = 2y − 10 = 0, so (−3, 5) is the only critical point. Since fxx = 2 > 0 and 2 fxx fyy − fxy = 4 > 0, the point (−3, 5) is a local minimum. 2 18. At a critical point fx = −3y+6 = 0 and fy = 3y2 −3x = 0, so (4, 2) is the only critical point. Since fxx fyy −fxy = −9 < 0, the point (4, 2) is a saddle point.

19. To fnd the critical points, we solve fx = 0 and fy = 0 for x and y. Solving fx = 3x2 − 6x = 0 fy = 2y + 10 = 0 shows that x = 0 or x = 2 and y = −5. There are two critical points: (0, −5) and (2, −5). We have D = (fxx )(fyy ) − (fxy )2 = (6x − 6)(2) − (0)2 = 12x − 12. When x = 0, we have D = −12 < 0, so f has a saddle point at (0, −5). When x = 2, we have D = 12 > 0 and fxx = 6 > 0, so f has a local minimum at (2, −5). 20. To fnd the critical points, we solve fx = 0 and fy = 0 for x and y. Solving fx = 3x2 − 3 = 0 fy = 3y2 − 12y = 0 shows that x = −1 or x = 1 and y = 0 or y = 4. There are four critical points: (−1, 0), (1, 0), (−1, 4), and (1, 4). We have D = (fxx )(fyy ) − (fxy )2 = (6x)(6y − 12) − (0)2 = (6x)(6y − 12). At critical point (−1, 0), we have D > 0 and fxx < 0, so f has a local maximum at (−1, 0). At critical point (1, 0), we have D < 0, so f has a saddle point at (1, 0). At critical point (−1, 4), we have D < 0, so f has a saddle point at (−1, 4). At critical point (1, 4), we have D > 0 and fxx > 0, so f has a local minimum at (1, 4). 21. To fnd the critical points, we solve fx = 0 and fy = 0 for x and y. Solving fx = 3x2 − 6x = 0 fy = 3y2 − 3 = 0 shows that x = 0 or x = 2 and y = −1 or y = 1. There are four critical points: (0, −1), (0, 1), (2, −1), and (2, 1). We have D = (fxx )(fyy ) − (fxy )2 = (6x − 6)(6y) − (0)2 = (6x − 6)(6y). At the point (0, −1), we have D > 0 and fxx < 0, so f has a local maximum. At the point (0, 1), we have D < 0, so f has a saddle point. At the point (2, −1), we have D < 0, so f has a saddle point. At the point (2, 1), we have D > 0 and fxx > 0, so f has a local minimum. 22. To fnd the critical points, we solve fx = 0 and fy = 0 for x and y. Solving fx = 2x − 2y = 0, fy = −2x + 6y − 8 = 0. We see from the frst equation that x = y. Substituting this into the second equation shows that y = 2. The only critical point is (2, 2). We have D = (fxx )(fyy ) − (fxy )2 = (2)(6) − (−2)2 = 8. Since D > 0 and fxx = 2 > 0, the function f has a local minimum at the point (2, 2).

586

Chapter Eight /SOLUTIONS

23. To fnd the critical points, we solve fx = 0 and fy = 0 for x and y. Solving fx = 3x2 − 3 = 0 fy = 3y2 − 3 = 0 shows that x = ±1 and y = ±1. There are four critical points: (1, 1), (−1, 1), (1, −1) and (−1, −1). We have D = (fxx )(fyy ) − (fxy )2 = (6x)(6y) − (0)2 = 36xy. At the points (1, −1) and (−1, 1), we have D = −36 < 0, so f has saddle points at (1, −1) and (−1, 1). At (1, 1), we have D = 36 > 0 and fxx = 6 > 0, so f has a local minimum at (1, 1). At (−1, −1), we have D = 36 > 0 and fxx = −6 < 0, so f has a local maximum at (−1, −1). 24. Setting fx = 0 and fy = 0 to fnd the critical point, we have fx = −6x − 4 + 2y = 0

and

fy = 2x − 10y + 48 = 0, or

2y − 6x = 4

and

10y − 2x = 48.

Solving these equations simultaneously gives x = 1 and y = 5. Since fxx = −6, fyy = −10 and fxy = 2 for all (x, y), at (1, 5) the discriminant D = (−6)(−10) − (2)2 = 56 > 0, and fxx < 0. Thus, f (x, y) has a local maximum value at (1, 5). 25. Mississippi lies entirely within a region designated as 80s so we expect both the maximum and minimum daily high temperatures within the state to be in the 80s. The southwestern-most corner of the state is close to a region designated as 90s, so we would expect the temperature here to be in the high 80s, say 87-88. The northern-most portion of the state is located near the center of the 80s region. We might expect the high temperature there to be between 83-87. Alabama also lies completely within a region designated as 80s so both the high and low daily high temperatures within the state are in the 80s. The southeastern tip of the state is close to a 90s region so we would expect the temperature here to be about 88-89 degrees. The northern-most part of the state is near the center of the 80s region so the temperature there is 83-87 degrees. Pennsylvania is also in the 80s region, but it is touched by the boundary line between the 80s and a 70s region. Thus we expect the low daily high temperature to occur there and be about 80 degrees. The state is also touched by a boundary line of a 90s region so the high will occur there and be 89-90 degrees. New York is split by a boundary between an 80s and a 70s region, so the northern portion of the state is likely to be about 74-76 while the southern portion is likely to be in the low 80s, maybe 81-84 or so. California contains many di˙erent zones. The northern coastal areas will probably have the daily high as low as 65-68, although without another contour on that side, it is diÿcult to judge how quickly the temperature is dropping o˙ to the west. The tip of Southern California is in a 100s region, so there we expect the daily high to be 100-101. Arizona will have a low daily high around 85-87 in the northwest corner and a high in the 100s, perhaps 102-107 in its southern regions. Massachusetts will probably have a high daily high around 81-84 and a low daily high of 70. 26. The function f can not have a minimum at (1, 2) because (1, 2) is not a critical point of f . At a local minimum, when the derivatives exist, we must have fx = fy = 0. 27. At a local maximum value of f , )f = −2x − B = 0. )x We are told that this is satisfed by x = −2. So −2(−2) − B = 0 and B = 4. In addition, )f = −2y − C = 0 )y and we know this holds for y = 1, so −2(1) − C = 0, giving C = −2. We are also told that the value of f is 15 at the point (−2, 1), so 15 = f (−2, 1) = A − ((−2)2 + 4(−2) + 12 − 2(1)) = A − (−5), so A = 10. Now we check that these values of A, B, and C give f (x, y) a local maximum at the point (−2, 1). Since fxx (−2, 1) = −2,

8.5 SOLUTIONS

587

fyy (−2, 1) = −2 and fxy (−2, 1) = 0, 2 we have that fxx (−2, 1)fyy (−2, 1) − fxy (−2, 1) = (−2)(−2) − 0 > 0 and fxx (−2, 1) < 0. Thus, f has a local maximum

value 15 at (−2, 1). 28. We are given that (0, 0) is a critical point. We have fx = 6x + 9y and fy = 2ky + 9x, so fxx = 6, fxy = 9, and fyy = 2k. The discriminant is D = (fxx )(fyy ) − (fxy )2 = (6)(2k) − (9)2 = 12k − 81. Since D = 12k − 81, we see that D < 0 when k < 6.75. The function has a saddle point at the point (0, 0) when k < 6.75. When k > 6.75, we have D > 0 and fxx > 0, so the function has a local minimum at the point (0, 0). When k = 6.75, the discriminant is zero, and we get no information about this critical point from the second derivative test. By looking at the values of the function in Table 8.15, it appears that f has a local minimum at the point (0, 0) when k = 6.75. We can confrm this by rewriting f (x, y) = 3x2 + 6.75y2 + 9xy as f (x, y) = 3(x + 1.5y)2 , which has the value 0 at (0, 0) but is never negative. Table 8.15 x −0.1 y

−0.1

0.1

0.1875

0.0675

0.0075

0.03

0.1

0.0075

0.0675

0.1875

(a) The function f (x, y) has a saddle point at (0, 0) if k < 6.75. (b) There are no values of k for which this function has a local maximum at the point (0, 0). (c) The function f (x, y) has a local minimum at (0, 0) if k ≥ 6.75. 29. We are given that (0, 0) is critical point. We have fx = 3x2 − 5y and fy = 2ky − 5x, so fxx = 6x, fxy = −5, and fyy = 2k. The discriminant is D = (fxx )(fyy ) − (fxy )2 = (6x)(2k) − (−5)2 = 12kx − 25. Since D = 12kx − 25, we see that D = −25 at the critical point (0, 0). Since D < 0 for all values of k, the function has a saddle point for all values of k. (a) The function f (x, y) has a saddle point at (0, 0) for all values of k. (b) There are no values of k for which this function has a local maximum at the point (0, 0). (c) There are no values of k for which this function has a local minimum at the point (0, 0). 30. (a) We frst express the revenue R in terms of the prices p1 and p2 : R(p1 , p2 ) = p1 q1 + p2 q2 = p1 (517 − 3.5p1 + 0.8p2 ) + p2 (770 − 4.4p2 + 1.4p1 ) = 517p1 − 3.5p21 + 770p2 − 4.4p22 + 2.2p1 p2 . (b) We compute the partial derivatives and set them to zero: )R = 517 − 7p1 + 2.2p2 = 0, )p1 )R = 770 − 8.8p2 + 2.2p1 = 0. )p2 Solving these equations, we fnd that p1 = 110 and

p2 = 115.

To see whether or not we have a found a local maximum, we compute the second-order partial derivatives: )2 R = −7, )p21

)2 R = −8.8, )p22

)2 R = 2.2. )p1 )p2

588

Chapter Eight /SOLUTIONS Therefore,

0 2 12 )2 R )2 R ) R − = (−7)(−8.8) − (2.2)2 = 56.76, )p1 )p2 )p21 )p22 and so we have found a local maximum point. The graph of R(p1 , p2 ) has the shape of an upside down bowl. Therefore, (110, 115) is a global maximum point. D=

31. (a) This tells us that an increase in the price of either product causes a decrease in the quantity demanded of both products. An example of products with this relationship is tennis rackets and tennis balls (supposing that people tend to use balls and rackets from the same company). An increase in the price of either product is likely to lead to a decrease in the quantity demanded of both products as they are used together. In economics, it is rare for the quantity demanded of a product to increase if its price increases, so for q1 , the coeÿcient of p1 is negative as expected. The coeÿcient of p2 in the expression could be either negative or positive. In this case, it is negative showing that the two products are complementary in use. If it were positive, however, it would indicate that the two products are competitive in use, for example Coke and Pepsi. (b) The revenue from the frst product would be q1 p1 = 150p1 − 2p21 − p1 p2 , and the revenue from the second product would be q2 p2 = 200p2 − p1 p2 − 3p22 . The total sales revenue of both products, R, would be R(p1 , p2 ) = 150p1 + 200p2 − 2p1 p2 − 2p21 − 3p22 . Note that R is a function of p1 and p2 . To fnd the critical points of R, we solve )R )R = = 0. )p1 )p2 This gives )R = 150 − 2p2 − 4p1 = 0 )p1 and

)R = 200 − 2p1 − 6p2 = 0 )p2 Solving simultaneously, we have p1 = 25 and p2 = 25. Therefore the point (25, 25) is a critical point for R. Further, )2 R )2 R )2 R = −4, 2 = −6, = −2, 2 )p )p1 )p2 1 )p2 so )2 R )2 R − ) 2 p12 ) 2 p22

)2 R )p1 )p2

= (−4)(−6) − (−2)2 = 20.

Since D > 0 and ) 2 R∕)p21 < 0, this critical point is a local maximum. Since R is quadratic in p1 and p2 , this is a global maximum. Therefore, the maximum possible revenue is R = 150(25) + 200(25) − 2(25)(25) − 2(25)2 − 3(25)2 = 6(25)2 + 8(25)2 − 7(25)2 = 4375. This is obtained when p1 = p2 = 25. Note that at these prices, q1 = 75 units, and q2 = 100 units. 32. (a) The revenue R = p1 q1 + p2 q2 . Proft = P = R − C = p1 q1 + p2 q2 − 2q12 − 2q22 − 10. p )P = p1 − 4q1 = 0 gives q1 = 1 4 )q1 p2 )P = p2 − 4q2 = 0 gives q2 = )q2 4 2

P Since ))qP2 = −4, ))qP2 = −4 and )q) )q = 0, at (p1 ∕4, p2 ∕4) we have that D = (−4)(−4) > 0 and ))qP2 < 0, thus P has 1

a local maximum value at (q1 , q2 ) = (p1 ∕4, p2 ∕4). Since P is quadratic in q1 and q2 , this is a global maximum. So the maximum proft is p2 p2 p2 p2 p2 p2 P = 1 + 2 − 2 1 − 2 2 − 10 = 1 + 2 − 10. 4 4 16 16 8 8 (b) The rate of change of the maximum proft as p1 increases is 2p p ) (max P ) = 1 = 1 . 8 4 )p1

8.6 SOLUTIONS

589

33. The total revenue is R = pq = (60 − 0.04q)q = 60q − 0.04q 2 , and as q = q1 + q2 , this gives

R = 60q1 + 60q2 − 0.04q12 − 0.08q1 q2 − 0.04q22 .

Therefore, the proft is P (q1 , q2 ) = R − C1 − C2 = −13.7 + 60q1 + 60q2 − 0.07q12 − 0.08q22 − 0.08q1 q2 . At a local maximum point, we would have: )P = 60 − 0.14q1 − 0.08q2 = 0, )q1 )P = 60 − 0.16q2 − 0.08q1 = 0. )q2 Solving these equations, we fnd that q1 = 300

and q2 = 225.

To see whether or not we have found a local maximum, we compute the second-order partial derivatives: )2 P = −0.14, )q12

)2 P = −0.16, )q22

)2 P = −0.08. )q1 )q2

Therefore, D=

)2 P )2 P )2 P − = (−0.14)(−0.16) − (−0.08)2 = 0.016, 2 2 )q1 )q2 )q1 )q2

and so we have found a local maximum point. The graph of P (q1 , q2 ) has the shape of an upside down paraboloid since P is quadratic in q1 and q2 , hence (300, 225) is a global maximum point.

Solutions for Section 8.6 1. Our objective function is f (x, y) = x + y and our equation of constraint is g(x, y) = x2 + y2 = 1. To optimize f (x, y) with Lagrange multipliers, we solve the following system of equations fx (x, y) = gx (x, y),

so 1 = 2 x

fy (x, y) = gy (x, y),

so 1 = 2 y

so x2 + y2 = 1

g(x, y) = 1, Solving for gives

1 1 = , 2x 2y which tells us that x = y. Going back to our equation of constraint, we use the substitution x = y to solve for y: =

g(y, y) = y2 + y2 = 1 2y2 = 1 1 y2 = 2

y=± √

√

√ 2 1 =± . 2 2

Since x = y, our critical points are ( 22 , 22 ) and (− 22 , − 22 ). Since the constraint is closed and bounded, maximum and minimum values of f subject to the constraint exist. Evaluating f at the critical points we fnd that the maximum value is √ √ √ √ √ √ 2 2 2 2 f ( 2 , 2 ) = 2 and the minimum value is f (− 2 , − 2 ) = − 2.

590

Chapter Eight /SOLUTIONS

2. We wish to optimize f (x, y) = x2 + 4xy subject to the constraint g(x, y) = x + y = 100. To do this we must solve the following system of equations: fx (x, y) = gx (x, y),

so 2x + 4y =

fy (x, y) = gy (x, y), g(x, y) = 100,

so 4x = so x + y = 100

Solving these equations produces: x ≈ 66.7

y ≈ 33.3

≈ 266.8

corresponding to optimal f (x, y) ≈ (66.7) + 4(66.7)(33.3) ≈ 13,333. 3. We wish to optimize f (x, y) = xy subject to the constraint g(x, y) = 5x+2y = 100. To do this we must solve the following system of equations: fx (x, y) = gx (x, y),

so y = 5

fy (x, y) = gy (x, y), g(x, y) = 100,

so x = 2 so 5x + 2y = 100

We substitute in the third equation to obtain 5(2 ) + 2(5 ) = 100, so = 5. Thus, x = 10

y = 25 = 5

corresponding to optimal f (x, y) = (10)(25) = 250. 4. We wish to optimize f (x, y) = x2 + 3y2 + 100 subject to the constraint g(x, y) = 8x + 6y = 88. To do this we must solve the following system of equations: fx (x, y) = gx (x, y),

so x = 4

fy (x, y) = gy (x, y),

so y =

g(x, y) = 88,

so 8x + 6y = 88

Solving these equations produces: x = 9.26 y = 2.32 = 2.32 2

corresponding to optimal f (x, y) = (9.26) + 3(2.32)2 + 100 = 201.9. 5. We wish to optimize f (x, y) = 5xy subject to the constraint g(x, y) = x + 3y = 24. To do this we must solve the following system of equations: fx (x, y) = gx (x, y), so 5y = fy (x, y) = gy (x, y), so 5x = 3 g(x, y) = 24, so x + 3y = 24 Solving these equations produces: x = 12

y = 4 = 20

corresponding to optimal f (x, y) = 5(12)(4) = 240. 6. Our objective function is f (x, y) = 3x − 2y and our equation of constraint is g(x, y) = x2 + 2y2 = 44. To optimize f (x, y) with Lagrange multipliers we solve the following system of equations: fx (x, y) = gx (x, y),

so 3 = 2 x

fy (x, y) = gy (x, y),

so − 2 = 4 y

g(x, y) = 44,

so x2 + y2 = 44

Solving for gives us =

3 −2 = , 2x 4y

which we can use to fnd x in terms of y: 3 −2 = 2x 4y −4x = 12y x = −3y.

8.6 SOLUTIONS

591

Using this relation in our equation of constraint, we can solve for y: x2 + 2y2 = 44 (−3y)2 + 2y2 = 44 9y2 + 2y2 = 44 11y2 = 44 y2 = 4 y = ±2. Thus, the critical points are (−6, 2) and (6, −2). Since the constraint is closed and bounded, maximum and minimum values of f subject to the constraint exist. Evaluating f at the critical points, we fnd that the maximum is f (6, −2) = 18 + 4 = 22 and the minimum value is f (−6, 2) = −18 − 4 = −22. 7. Our objective function is f (x, y) = x2 + y and our equation of constraint is g(x, y) = x2 − y2 = 1. We have fx = 2x,

fy = 1

gx = 2x,

gy = −2y.

The Lagrange multiplier equations are 2x = 2x 1 = − 2y But x cannot be zero, since the constraint equation, −y2 = 1, would then have no real solution for y. Thus, the frst equation gives = 1. Then the second equation gives −2y = 1 1 y=− . 2 Substituting this into our equation of constraint we fnd 2 1 1 =1 g(x, − ) = x2 − − 2 2 5 x2 = 4

√ 5 x=± . 2

√

So the critical points are ( 25 , − 12 ) and (− 25 , − 12 ). Evaluating f at these points we fnd f ( 25 , − 12 ) = f (− 25 , − 12 ) = 5 − 21 4

= 43 . This is the minimum value for f (x, y) constrained to g(x, y) = 1. To see this, note that for x2 = y2 + 1, f (x, y) = y2 + 1 + y = (y + 1∕2)2 + 3∕4 ≥ 3∕4. Alternatively, see Figure 8.48. To see that f has no maximum on g(x, y) = 1, note that f → ∞ as x → ∞ and y → ∞ on the part of the graph of g(x, y) = 1 in quadrant I. y 2 x 2 − y2 = 1 2.25 1.5 0.75 0

√ −2 (− 5∕2, −1∕2)

x 2 √ ( 5∕2, 1∕2)

−2

Figure 8.48: Graph of x2 − y2 = 1

592

Chapter Eight /SOLUTIONS

8. Our objective function is f (x, y) = xy and our equation of constraint is g(x, y) = 4x2 + y2 = 8. Their partial derivatives are fx = y,

fy = x

gx = 8x,

gy = 2y.

This gives 8x = y

and

2y = x.

Multiplying, we get 8x2 = 2y2 . If = 0, then x = y = 0, which does not satisfy the constraint equation. So ≠ 0 and we get 2y2 = 8x2 y2 = 4x2 y = ±2x. To fnd x, we substitute for y in our equation of constraint. 4x2 + y2 = 8 4x2 + 4x2 = 8 x2 = 1 x = ±1 So our critical points are (1, 2), (1, −2), (−1, 2) and (−1, −2). Evaluating f (x, y) at the critical points, we have f (1, 2) = f (−1, −2) = 2 f (1, −2) = f (−1, 2) = −2. Thus, the maximum value of f on g(x, y) = 8 is 2, and the minimum value is −2. 9. The objective function is f (x, y) = x2 + y2 and the constraint equation is g(x, y) = 4x − 2y = 15, so fx = 2x, fy = 2y and gx = 4, gy = −2. We have: 2x = 4 , 2y = −2 . From the frst equation we have = x∕2, and from the second equation we have = −y. Setting these equal gives y = −0.5x. Substituting this into the constraint equation 4x − 2y = 15 gives x = 3. The only critical point is (3, −1.5). We have f (3, −1.5) = (3)2 + (1.5)2 = 11.25. One way to determine if this point gives a maximum or minimum value or neither for the given constraint is to examine the contour diagram of f with the constraint sketched in, Figure 8.49. It appears that moving away from the point P = (3, −1.5) in either direction along the constraint increases the value of f , so (3, −1.5) is a point of minimum value. y Constraint: 4x − 2y = 15 1 5

10 15

x 1

−1 P = (3, −1.5) −2 −3

Figure 8.49

8.6 SOLUTIONS

593

10. The objective function is f (x, y) = x2 + y2 and the equation of constraint is g(x, y) = x4 + y4 = 2. Their partial derivatives are fx = 2x,

fy = 2y

gx = 4x3 ,

gy = 4y3 .

We have 2x = 4 x3 , 2y = 4 y3 . Now if x = 0, the frst equation is true for any value of . In particular, we can choose which satisfes the second equation. Similarly, y = 0 is solution. Assuming both x ≠ 0 and y ≠ 0, we can divide to solve for and fnd =

2y 2x = 3 4x3 4y 1 1 = 2 2x2 2y y2 = x2 y = ±x.

Going back to our equation of constraint, we fnd √ 4 so y = ± 2 √ 4 so x = ± 2

g(0, y) = 04 + y4 = 2, g(x, 0) = x4 + 04 = 2,

g(x, ±x) = x4 + (±x)4 = 2, so x = ±1. √ √ 4 4 Thus, the critical points are (0, ± 2), (± 2, 0), (1, ±1) and (−1, ±1). Evaluating f at the critical points, we fnd f (1, 1) = f (1, −1) = f (−1, 1) = f (−1, −1) = 2, √ √ √ √ √ 4 4 4 4 f (0, 2) = f (0, − 2) = f ( 2, 0) = f (− 2, 0) = 2. √ Thus, the minimum value of f (x, y) on g(x, y) = 2 is 2 and the maximum value is 2.

11. (a) Point F since the value of f is greatest at this point. (b) At points A, B, C, D, E, the level curve of f and the constraint curve are parallel. Point D has the greatest f value of these points.

g=c

❘ A

E F 13

B 11 10

594

Chapter Eight /SOLUTIONS

12. To maximize f (x, y) subject to g(x, y) = c, we wish to fnd the highest possible value of f (x, y) where (x, y) is on the line corresponding to our constraint. From our constraint, we see that f = 400 is the highest curve that intersects the constraint line. We can verify this either by noting that all higher contours lie above the constraint g(x, y) = c or by noting that the contour f = 400 is tangent to the constraint. The point of intersection, from looking at the graph, occurs at approximately x = 6, y = 6, giving us maximal f (x, y) = 400. 13. (a) The objective function gives the quantity that is to be maximized, in this case the production. Therefore, the objective function is Q = x10.3 x20.7 . (b) Constraint: 10x1 + 25x2 = 50,000. 14. We want to minimize cost C = 100L + 200K subject to Q = 900L1∕2 K 2∕3 = 36000. We solve the system of equations: CL = QL ,

so 100 = 450L−1∕2 K 2∕3

CK = QK ,

so 200 = 600L1∕2 K −1∕3

Q = 36000,

so 900L1∕2 K 2∕3 = 36000

Since ≠ 0 this gives 450L−1∕2 K 2∕3 = 300L1∕2 K −1∕3 . Solving, we get L = (3∕2)K. Substituting into Q = 36,000 gives 900

1∕2 3 K K 2∕3 = 36,000. 2

4 1∕2 56∕7 ≈ 19.85, so L ≈ 23 (19.85) = 29.78. We can thus calculate cost using K = 20 and Solving yields K = 40 ⋅ 23

L = 30 which gives C = $7, 000.

15. (a) Because of budget constraints, we are limited in the size of the labor force and the amount of total equipment. This constraint is described in the formula 300L + 100K = 15,000 We can let L range from 0 to about 50. Each choice of L determines a choice of K. We have, as a result, the Table 8.16. Table 8.16 L

135

120

105

50 0

111

135

156

172

184

189

181

(b) We wish to maximize P subject to that cost C satisfes C = 300L+100K = $15,000. This is accomplished by solving the following system of equations: PL = CL ,

so 4L−0.2 K 0.2 = 300

PK = CK ,

so L0.8 K −0.8 = 100

C = 15,000,

so 300L + 100K = 15,000

We can solve these equations by dividing the frst by the second and then substituting into the budget constraint. Doing so produces the solution K = 30, L = 40. So the optimal choices of labor and capital are L = 40, K = 30. 16. We want to minimize C = f (q1 , q2 ) = 2q12 + q1 q2 + q22 + 500 subject to the constraint q1 + q2 = 200 or g(q1 , q2 ) = q1 + q2 = 200. We solve the system of equations: Cq1 = gq1 ,

so 4q1 + q2 =

Cq2 = gq2 ,

so 2q2 + q1 =

g = 200,

so q1 + q2 = 200.

Solving we get 4q1 + q2 = 2q2 + q1

8.6 SOLUTIONS

595

so 3q1 = q2 . We want q1 + q2 = 200 q1 + 3q1 = 4q1 = 200. Therefore q1 = 50 units,

q2 = 150 units.

17. (a) To be producing the maximum quantity Q under the cost constraint given, the frm should be using K and L values given by )Q = 0.6aK −0.4 L0.4 = 20 )K )Q = 0.4aK 0.6 L−0.6 = 10 )L 20K + 10L = 150. L 20 4 0.6aK −0.4 L0.4 = 1.5 = = 2, so L = K. Substituting in 20K + 10L = 150, we obtain 20K + Hence 0.6 L−0.6 K 10 3 0.4aK 1 4 9 K = 150. Then K = and L = 6, so capital should be reduced by unit, and labor should be increased by 10 3 2 2 1 unit. 0.6 0.4 New production a4.5 6 = ≈ 1.01, so tell the board of directors, “Reducing the quantity of capital by 1∕2 unit (b) Old production a50.6 50.4 and increasing the quantity of labor by 1 unit will increase production by 1% while holding costs to $150.” 18. We use the equation y = x∕2 to fnd how much to change y: (a) We have Δy = Δx∕2 = 3000∕2 = 1500 (b) Since y = x∕2, we have 2y = (2x)∕2, so we should also double y. (c) Increasing x by 5% means multiplying x by 1.05. Since y = x∕2, we have 1.05y = (1.05x)∕2, so we should also increase y by 5% 19. Let g(x, y) = x + y. Then the constraint is g(x, y) = x + y = 100. At the maximum Qx = 5(0.3)x−0.7 y0.8 = ⋅ 1 Qy = 5(0.8)x0.3 y−0.2 = ⋅ 1. Dividing to eliminate we have 1.5 −1 1 x y =1 4 8 y = x. 3 Substituting into the constraint, we have 8 x + x = 100 3 so x = 27.273 and thus y = 72.727. The end points are x = 0, y = 100 and x = 100, y = 0. Evaluating Q at each end point gives Q = 0. Evaluating at x = 27.273, y = 72.727 gives the maximum of Q = 5(27.273)0.3 (72.727)0.8 = 415.955. 20. Since the budget is fully spent, the constraint is 6x + 3y = 12. At the maximum 3 = ⋅6 x+1 2 Qy = = ⋅ 3. y+1

Qx =

Dividing to eliminate we have 3∕(x + 1) =2 2∕(y + 1) 3y = 4x + 1.

596

Chapter Eight /SOLUTIONS Substituting into the constraint, we have 6x + 4x + 1 = 12 so x = 1.1 and y = (4(1.1) + 1)∕3 = 1.8. The end points are x = 0, y = 4 and x = 2, y = 0. Evaluating: At x = 0, we have y = 4 and Q = 3 ln(0 + 1) + 2 ln(1 + 4) = 3.219. At x = 2 we have y = 0 and Q = 3 ln(1 + 2) + 2 ln(0 + 1) = 3.296. At x = 1.1 and y = 1.8 we have Q = 3 ln(1.1 + 1) + 2 ln(1.8 + 1) = 4.285. Thus the global maximum occurs at x = 1.1 and y = 1.8.

21. The value of tells us how the optimum value of f changes when we change the constraint. In particular, if the constraint increases by 1, the optimum value of f increases by . (a) So, if we raise the quota by 1 product, cost rises by = 15. So C = 1200 + 15 = 1215. (b) Similarly, if we lower the quota by 1, cost falls by = 15. So C = 1200 − 15 = 1185. 22. (a) The objective function is the one we are trying to optimize, so P (x, y) is the objective function. The constraint equations tells us what condition must be satisfed; here, we must have costs equal to $50,000, so C(x, y) = 50,000 is the constraint equation. The meaning of is how much the objective would increase if the budget available increased by 1 and we optimized again with the new budget. Here, this means how much optimal production would increase if the budget increased to $50,001. (b) Here, we are trying to minimize cost, so C(x, y) is the objective function. Since we must satisfy P (x, y) = 2000, our constraint equation becomes P (x, y) = 2000. In this situation, represents the change in the minimal cost when production is increased by one unit to 2001. 23. (a) The objective function is the function that you want to maximize, that is, your grade as a function of the time spent on each project. Its units are points. (b) The constraint is the condition that limits your options. The limitation is the amount of time set aside, 20 hours. (c) The Lagrange multiplier tells you the rate at which the grade changes when the total time to work on the projects is increased. Its units are points per hour. (d) Working 21 hours instead of 20 hours will improve your grade by approximately 5 points. 24. (a) The objective function is the function that is optimized. Since the problem refers to maximum production, the objective function is the production function P (K, L). (b) Production is maximized subject to a budget restriction, which is the constraint. The constraint equation is C(K, L) = 900,000. (c) The Lagrange multiplier tells you the rate at which maximum production changes when the budget is increased. Its units are tons of steel per dollar of budget, or simply tons∕dollar. (d) Increasing the budget from $900,000 to $(900,000+a) increases the maximum possible production from 10,000 tons to approximately (10,000 + 0.0125a) tons. Every extra dollar of budget increases maximal production by approximately = 0.0125 tons. 25. (a) We wish to maximize q subject to the constraint that cost C = 10W + 20K = 3000. To optimize q according to this, we must solve the following system of equations: 9 − 14 41 W K = 10 2 3 3 3 so W 4 K − 4 = 20 2 so 10W + 20K = 3000

qW = CW ,

qK = CK , C = 3000,

Dividing yields K = 16 W , so substituting into C gives 1 40 W = W = 3000. 6 3 Thus W = 225 and K = 37.5. Substituting both answers to fnd gives 10W + 20

1 1 9 (225)− 4 (37.5) 4 2

10 3

= 0.2875. 1

We also fnd the optimum quantity produced, q = 6(225) 4 (37.5) 4 = 862.57. (b) When the budget is increased by one dollar, we substitute the relation K1 = 16 W1 into 10W1 + 20K1 = 3001 which gives 10W1 + 20( 16 W1 ) = 40 W1 = 3001. Solving yields W1 = 225.075 and K1 = 37.513, so q1 = 862.86 = q + 0.29. 3 Thus production has increased by 0.29 ≈ , the Lagrange Multiplier.

8.6 SOLUTIONS

597

26. (a) The curves are shown in Figure 8.50. s 1500

III II I

1000

s = 1000 − 10l

500

(50, 500)

l 20

100

Figure 8.50 (b) The income equals $10/hour times the number of hours of work: s = 10(100 − l) = 1000 − 10l. (c) The graph of this constraint is the straight line in Figure 8.50. (d) For any given salary, curve III allows for the most leisure time, curve I the least. Similarly, for any amount of leisure time, curve III also has the greatest salary, and curve I the least. Thus, any point on curve III is preferable to any point on curve II, which is preferable to any point on curve I. We prefer to be on the outermost curve that our constraint allows. We want to choose the point on s = 1000 − 10l which is on the most preferable curve. Since all the curves are concave up, this occurs at the point where s = 1000 − 10l is tangent to curve II. So we choose l = 50, s = 500, and work 50 hours a week. 27. The constraint is g(x1 , x2 ) = x1 + 3x2 = 100. We need to solve the equations )g )U = , )x1 )x1

)g )U = , )x2 )x2

x1 + 3x2 = 100.

These equations are 2x2 + 3 = 2x1 = 3 x1 + 3x2 = 100.

The frst equation gives x2 =

−3 2

and the second that x1 =

3 . 2

Substituting these into the third equation gives 3 −3 +3 = 100, 2 2 therefore, = 209∕6 so x1 = 209∕4 and x2 = 191∕12. The maximum utility is 209 191 209 191 209 U , =2 +3 = 1820.04. 4 12 4 12 4 The approximate change in maximum utility due to a one unit increase in the consumer’s disposable income is . For a $6 increase in disposable income, the maximum utility increases by about 6 which is 209.

598

Chapter Eight /SOLUTIONS

Solutions for Chapter 8 Review 1. To see whether f is an increasing or decreasing function of x, we need to see how f varies as we increase x and hold y fxed. We note that each column of the table corresponds to a fxed value of y. Scanning down the y = 2 column, we can see that as x increases, the value of the function decreases from 114 when x = 0 down to 93 when x = 80. Thus, f may be decreasing. In order for f to actually be decreasing however, we have to make sure that f decreases for every column. In this case, we see that f indeed does decrease for every column. Thus, f is a decreasing function of x. Similarly, to see whether f is a decreasing function of y we need to look at the rows of the table. As we can see, f increases for every row as we increase y. Thus, f is an increasing function of y. 2. (a) We expect B to be an increasing function of all three variables. (b) A deposit of $1250 at a 1% annual interest rate leads to a balance of $1276 after 25 months. 3. (a) This states that after 5 minutes, a point 2 cm from the center has a temperature of 24◦ C. (b) If d is held constant, H is an increasing function of t because, at a fxed distance from the center of the rod, the temperature will increase as time increases. (c) If t is held constant, H is a decreasing function of d because, at a fxed time, the temperature of the rod will be cooler the farther we move away from the heating element. 4. (a) This states that a $300,000 fxed rate 30-year mortgage with an annual interest rate of 5% has a monthly payment of $1610.46. (b) If P is held constant, then m is an increasing function of r because the higher the interest rate, the more interest paid each month. (c) If r is held constant, then m is an increasing function of P because the more borrowed, the higher the monthly payment. 5. (a) The coeÿcient of m is 30 dollars per month. It represents the monthly charge to use this service. The coeÿcient of t is 10 dollars per gigabyte. Each gigabyte the customer uses costs 10 dollars. (b) The intercept represents the base charge. It costs $99 just to get hooked up to this service. (c) We have f (3, 8) = 269. A customer who uses this service for three months and uses a total of 8 gigabytes is charged $269. 6. Contours are lines of the form 3x − 5y + 1 = c as shown in Figure 8.51. Note that for the regions of x and y given, the c values range from −12 < c < 12 and are evenly spaced. y 2 2 −1

−8 −4 0

x 4 8

−1

−2 −2

−1

Figure 8.51

7. Contours are ellipses of the form 2x2 + y2 = c as shown in Figure 8.52. Note that for the ranges of x and y given, the range of c value is 1 ≤ c < 9 and are closer together farther from the origin.

SOLUTIONS to Review Problems For Chapter Eight

599

y 2 1

−1 −2 −2

−1

Figure 8.52 8. (a) According to Table 8.2 of the problem, it feels like −19◦ F. (b) A wind of 20 mph, according to Table 8.2. (c) About 17.5 mph. Since at a temperature of 25◦ F, when the wind increases from 15 mph to 20 mph, the temperature adjusted for wind chill decreases from 13◦ F to 11◦ F, we can say that a 5 mph increase in wind speed causes a 2◦ F decrease in the temperature adjusted for wind chill. Thus, each 2.5 mph increase in wind speed brings about a 1◦ F drop in the temperature adjusted for wind chill. If the wind speed at 25◦ F increases from 15 mph to 17.5 mph, then the temperature you feel will be 13 − 1 = 12◦ F. (d) Table 8.2 shows that with wind speed 20 mph the temperature will feel like 0◦ F when the air temperature is somewhere between 15◦ F and 20◦ F. When the air temperature drops 5◦ F from 20◦ F to 15◦ F, the temperature adjusted for wind-chill drops 6◦ F from 4◦ F to −2◦ F. We can say that for every 1◦ F decrease in air temperature there is about a 6∕5 = 1.2◦ F drop in the temperature you feel. To drop the temperature you feel from 4◦ F to 0◦ F will take an air temperature drop of about 4∕1.2 = 3.3◦ F from 20◦ F. With a wind of 20 mph, approximately 20 − 3.3 = 16.7◦ F would feel like 0◦ F. 9. To draw a contour for a wind-chill of W = 20, we need a few combinations of temperature and wind velocity (T , v) such that W (T , v) = 20. Estimating from the table, some such points are (24, 5) and (33, 10). We can connect these points to get a contour for W = 20. Similarly, some points that have wind-chill of about 0◦ F are (5, 5), (17.5, 10), (23.5, 15), (27, 20), and (29, 25). By connecting these points we get the contour for W = 0. If we carry out this procedure for more values of W , we get a full contour diagram such as is shown in Figure 8.53:

−30

windspeed, v, (mph) 25

−2 0

−1 0

15 10

0 0

10 15 20 25 30 35

T , (◦ F)

Figure 8.53 10.

Table 8.17 20◦ F

Temperature adjusted for wind chill at

Wind speed (mph)

Adjusted temperature (◦ F)

Table 8.18 Temperature adjusted for wind chill at 0◦ F Wind speed (mph)

Adjusted temperature (◦ F)

−11

−16

−19

−22

−24

600

Chapter Eight /SOLUTIONS

11.

Table 8.19

Temperature adjusted for wind chill at 5 mph

Temperature (◦ F)

Adjusted temperature (◦ F)

−5

−11

Table 8.20

Temperature adjusted for wind chill at 20 mph

Temperature (◦ F)

Adjusted temperature (◦ F)

−2

−9

−15

−22

12. We have

1 (2)(10)2 = 100, 2 which means that an object moving at a speed of 10 m/sec with a mass of 2 kg has a kinetic energy of 100 joules. f (2, 10) =

13. (a) Holding m fxed at 6 and letting v vary means that, for an object with a mass of 6 kg, we are considering the e˙ects of various speeds on the kinetic energy of the object. Figure 8.54 shows the graph of E = f (6, v) = (1∕2)(6)v2 = 3v2 . The concave upward shape of the graph refects the fact that the kinetic energy of an object having a mass of 6 kg increases at an increasing rate as its speed increases. E (Joules) 300 250 200 150 100 50 0 2

Figure 8.54: f (6, v) = 3v2

v (m/s)

E (Joules) 1000 800 600 400 200 0 1

m (kg)

Figure 8.55: f (m, 20) = 200m

(b) Holding v fxed at 20 and letting m vary means that, for an object moving at a constant speed of 20 m/sec, we are considering the e˙ects that changing its mass has on its kinetic energy. Figure 8.55 shows the graph of E = f (m, 20) = (1∕2)m(20)2 = 200m. The linear shape of the graph refects the fact that the kinetic energy of an object moving at a speed of 20 m/sec increases at a constant rate as its mass increases. 14. Looking at the contour diagram, we can see that Q(x, y) is an increasing function of x and a decreasing function of y. It stands to reason that as the price of orange juice goes up, the demand for orange juice will go down. Thus, the demand is a decreasing function of the price of orange juice and thus the y-axis corresponds to the price of orange juice. Also, as the price of apple juice goes up, so will the demand for orange juice and therefore the demand for orange juice is an increasing function of the price of apple juice. Thus, the price of apple juice corresponds to the x-axis. 15. (a) In this company success only increases when money increases, so success will remain constant along the work axis. However, as money increases so does success, which is shown in Graph (III). (b) As both work and money increase, success never increases, which corresponds to Graph (II). (c) If the money does not matter, then regardless of how much the money increases success will be constant along the money axis. However, success increases as work increases. This is best represented in Graph (I). (d) This company’s success increases as both money and work increase, which is demonstrated in Graph (IV). 16. One possible answer follows. (a) If the middle line is a highway in the city, there may be many people living around it since it provides mass transit nearby that nearly everyone needs and uses. Thus, the population density would be greatest near the highway, and less dense farther from the highway, as is shown in diagram (I). In an alternative solution, highways may be considered inherently very noisy and dirty. In this scenario, the people in a certain community may purposely live away from the highway. Then the population density would follow the pattern in diagram (III), or in an extreme case, even diagram (II). (b) If the middle line is a sewage canal, there will be no one living within it, and very few people living close to it. This is represented in diagram (II).

SOLUTIONS to Review Problems For Chapter Eight

601

(c) If the middle line is a railroad line in the city, then one scenario is that very few people would enjoy living nearby a railroad yard, with all of the noise and diÿculty in crossing it. Then the population density would be smallest near the middle line, and greatest farther from the railroad, as in diagram (III). In an alternative solution, a community might well depend upon the railroad for transit, news or supplies, in which case there would be a denser population nearer the railroad than farther from it, as in diagram (I). 17. The temperature is decreasing away from the window, suggesting that heat is fowing in from the window. As time goes by the temperature at each point in the room increases. This could be caused by opening the window of an air conditioned room at t = 0 thus letting heat from the hot summer day outside raise the temperature inside. 18. (a) The point representing 8% and $6000 on the graph lies between the 120 and 140 contours. We estimate the monthly payment to be about $122. (b) Since the interest rate has dropped, we will be able to borrow more money and still make a monthly payment of $122. To fnd out how much we can a˙ord to borrow, we fnd where the interest rate of 6% intersects the $122 contour and read o˙ the loan amount to which these values correspond. Since the $122 contour is not shown, we estimate its position from the $120 and $140 contours. We fnd that we can borrow an amount of money that is more than $6000 but less than $6500. So we can borrow about $350 more without increasing the monthly payment. (c) The entries in the table will be the amount of loan at which each interest rate intersects the 122 contour. Using the $122 contour from (b) we make table 8.21. Table 8.21

Amount borrowed at a monthly payment of $122.

Interest Rate (%)

Loan Amount ($)

7400

7200

7000

6800

6650

6500

6350

6200

Interest rate (%)

Loan Amount ($)

6000

5850

5700

5600

5500

5400

5300

5200

19. The values in Table (a) are not constant along rows or columns and therefore cannot be the lines shown in (I) or (IV). Also observe that as you move away from the origin, whose contour value is 0, the z-values on the contours increase. Thus, this table corresponds to diagram (II). The values in Table (b) are also not constant along rows or columns. Since the contour values are decreasing as you move away from the origin, this table corresponds to diagram (III). Table (c) shows that for each fxed value of x, we have constant contour value, suggesting a straight vertical line at each x-value, as in diagram (IV). Table (d) also shows lines, however these are horizontal since for each fxed value of y we have constant contour values. Thus, this table matches diagram (I). 20. The values of z increase as we move in the direction of increasing x-values, so fx is positive. The values of z decrease as we move in the direction of increasing y-values, so fy is negative. We see in the contour diagram that f (2, 1) = 10. We estimate the partial derivatives: Δz 14 − 10 fx (2, 1) ≈ = = 2, Δx 4−2 Δz 6 − 10 fy (2, 1) ≈ = = −4. Δy 2−1 21. For fw (10, 25) we get fw (10, 25) ≈

f (10 + ℎ, 25) − f (10, 25) . ℎ

Choosing ℎ = 5 and reading values from Table 8.2 on page 2 of the text, we get fw (10, 25) ≈

f (15, 25) − f (10, 25) 13 − 15 = = −0.4◦ F∕mph 5 5

This means that when the wind speed is 10 mph and the true temperature is 25◦ F, as the wind speed increases from 10 mph by 1 mph we feel an approximately 0.4◦ F drop in temperature. This rate is negative because the temperature you feel drops as the wind speed increases.

602

Chapter Eight /SOLUTIONS

22. Using a di˙erence quotient with ℎ = 5, we get fT (5, 20) ≈

f (5, 20 + 5) − f (5, 20) 19 − 13 = = 1.2◦ F∕◦ F. 5 5

This means that when the wind speed is 5 mph and the true temperature is 20◦ F, the apparent temperature increases by approximately 1.2◦ F for an increase of 1◦ F in the true temperature. This rate is positive because the true temperature you feel increases as true temperature increases. 23. Since the average rate of change of the temperature adjusted for wind-chill is about −0.8 (drops by 0.8◦ F), with every 1 mph increase in wind speed from 5 mph to 10 mph, when the true temperature stays constant at 20◦ F, we know that fw (5, 20) ≈ −0.8.

24. A graph with kilometers north fxed at 50 is in Figure 8.56. Density of the fox population P 2.5 2 Density of the fox population P 1.5

1.5 1

1 0.5

0.5 60

120

180

East

Figure 8.56

120

150

180

East

Figure 8.57

A graph with kilometers north fxed at 100 is in Figure 8.57. A graph with kilometers east fxed at 60 is in Figure 8.58. Density of the fox population P

Density of the fox population P

2.5 2

1.5 1

0.5

0.5 35

North

Figure 8.58

100

150

North

Figure 8.59

A graph with kilometers east fxed at 120 is in Figure 8.59. 25. Estimating from the contour diagram, using positive increments for Δx and Δy, we have, for point A, 1∕2 )n || 1.5 − 1 1 = = ≈ 0.06 | ≈ | )x |(A) 67 − 59 8 16

foxes/km2 km

1∕2 1 )n || 0.5 − 1 =− =− ≈ −0.06 | ≈ )y ||(A) 60 − 51 9 18

foxes/km2 . km

SOLUTIONS to Review Problems For Chapter Eight

603

So, from point A the fox population density increases as we move eastward. The population density decreases as we move north from A. At point B, 1∕4 0.75 − 1 1 )n || =− =− ≈ −0.01 | ≈ 20 80 )x ||(B) 135 − 115 1∕2 0.5 − 1 1 )n || =− =− ≈ −0.05 | ≈ 10 20 )y ||(B) 120 − 110

foxes/km2 km foxes/km2 . km

So, fox population density decreases as we move both east and north of B. However, notice that the partial derivative )n∕)x at B is smaller in magnitude than the others. Indeed if we had taken a negative Δx we would have obtained an estimate of the opposite sign. This suggests that better estimates forB are )n || | ≈0 )x ||(B)

foxes/km2 km

)n || | ≈ −0.05 )y ||(B)

At point C,

foxes/km2 . km

1∕2 )n || 2 − 1.5 1 = = ≈ 0.02 | ≈ 20 40 )x ||(C) 135 − 115 1∕2 2 − 1.5 1 )n || = = ≈ 0.02 | ≈ 25 50 )y ||(C) 80 − 55

foxes/km2 km foxes/km2 . km

So, the fox population density increases as we move east and north of C. Again, if these estimates were made using negative values for Δx and Δy we would have had estimates of the opposite sign. Thus, better estimates are )n || | ≈0 )x ||(C) )n || | ≈0 )y ||(C)

26. fx = 2x + y, fy = 2y + x.

foxes/km2 km foxes/km2 . km

27. Pa = 2a − 2b2 , Pb = −4ab. )Q )Q = 50p2 , 28. = 50p1 − 2p2 . )p2 )p1 )f )f 29. = 5e−2t , = −10xe−2t . )x )t )P )P = 7K −0.3 L0.3 , = 3K 0.7 L−0.7 . 30. )K )L 31. fx = √ 2x 2 , fy = √ 2y 2 . x +y

x +y

32. The function in ℎ(x, t) tells us the height of the head of the spectator in seat x at time t seconds. Thus, ℎx (2, 5) is in feet per seat and ℎt (2, 5) is in feet per second. So ℎx (x, t) = −0.5 sin(0.5x − t) ℎx (2, 5) = −0.5 sin(0.5(2) − 5) ≈ −0.38 ft/seat. and ℎt (x, t) = sin(0.5x − t) ℎt (2, 5) = sin(0.5(2) − 5) = 0.76 ft/sec. The value of ℎx (2, 5) is the rate of change of height of heads as you move along the row of seats. The value of ℎt (2, 5) is the vertical velocity of the head of the person at seat 2 at time t = 5.

604

Chapter Eight /SOLUTIONS

33. Since )f and )f are defned everywhere, a critical point will occur where )f = 0 and )f = 0. So: )x )y )x )y )f = 3x2 − 3 = 0, so x2 = 1 and x = ±1 )x )f = 2y = 0, so y = 0 )y So (1, 0) and (−1, 0) are the critical points. To determine whether these are local extrema, we can examine values of f for (x, y) near the critical points. Functional values for points near (1, 0) are shown in Table 8.22: Table 8.22 x 0.99 y

1.00

1.01

−0.01

−1.9996

−1.9999

−1.9996

0.00

−1.9997

−2.0000

−1.9997

0.01

−1.9996

−1.9999

−1.9996

As we can see from the table, all the points close to (1, 0) have greater functional value, so f (1, 0) is a local minimum. A similar display of points near (−1, 0) is shown in Table 8.23: Table 8.23 x

−1.01

−1.00

−0.99

−0.01

1.9998

2.0001

1.9998

0.00

1.9997

2.0000

1.9997

0.01

1.9998

2.0001

1.9998

As we can see, some points near (−1, 0) have greater functional value than f (−1, 0) and others have less. So f (−1, 0) is neither a local maximum nor a local minimum. By computing D = fxx (−1, 0)fyy (−1, 0) − fxy (−1, 0)2 = −12, we see that (−1, 0) is a saddle point. )f )f )f )f are defned everywhere, a critical point will occur where 34. Since and = 0 and = 0. So: )y )y )x )y )f = 2x − 4 = 0 ⇒ x = 2 )x )f = 6y + 6 = 0 ⇒ y = −1 )y (2, −1) is the critical point of f (x, y). 35. The smallest value of f (x, y) on the line y = 100 shown in Figure 8.60 occurs at the right endpoint, at the point (300, 100). The minimum value is f (300, 100) = 35. y

300

200 y = 100

✠

100

x 100

200

Figure 8.60

300

SOLUTIONS to Review Problems For Chapter Eight

605

36. The largest value of f (x, y) on the line y = 100 shown in Figure 8.61 occurs at the left endpoint, at the point (0, 100). The maximum value is f (0, 100) ≈ 43. y

300

200 y = 100

✠

100

x 100

200

300

Figure 8.61 37. The constraint is the line y = x. The largest value of f (x, y) on the line y = x shown in Figure 8.62 occurs at the point where that line is tangent to a contour. Since the contour is not shown in the fgure, we make a rough approximation by imagining intermediate contours. The point (150, 150) seems about right. The maximum value is f (150, 150) ≈ 42. y

✛

y−x = 0

300

200

100

x 100

200

300

Figure 8.62 38. (a) Objective function: C = 127x1 + 92x2 . (b) Constraint: x10.6 x20.4 = 500. (c) The value of tells us that if the constraint is relaxed by one unit, the objective function changes by about units. In this context, this means that if the production quota of 500 is relaxed to 499, the cost decreases by about $219. 39. The largest value of f (x, y) on the constraint line 15x + 20y = 300 shown in Figure 8.63 occurs at the point where that line is tangent to a contour. Since the contour is not shown in the fgure, we make a rough approximation by imagining intermediate contours. The point (12, 6) seems about right. The maximum value is f (12, 6) ≈ 9000, achieved with inputs x = 12 and y = 6. y 20 0 00 20

15 0 00 10

10 0 200

x 5

Figure 8.63

606

Chapter Eight /SOLUTIONS

40. (a) We have 1500 workers and $15,000,000 per month of capital, so x = 1500, y = 15,000,000∕4000 = 3750. Substituting into the equation for Q gives us Q = (1500)0.4 (3750)0.6 = 2599 cars per month. (b) Now we are only producing 2000 cars per month. We wish to minimize cost subject to the constraint that the monthly production is 2000 cars. So, our objective function is cost C = 5000x+4000y and our constraint is that Q = x0.4 y0.6 = 2000. To minimize C according to this, we solve the following system of equations: Cx = Qx ,

so 0.4x−0.6 y0.6 = 5000

Cy = Qx ,

so 0.6x0.4 y−0.4 = 4000

Q = 2000,

so x0.4 y0.6 = 2000

Dividing the frst two equations gives 0.4x−0.6 y0.6 0.4 y 5000 = = 0.6 x 4000 0.6x0.4 y−0.4

giving y = 1.875x.

Substituting this into the constraint gives us x ≈ 1371.606 and y ≈ 2571.760. So our new level of production uses 905 workers and $6,786,916 of equipment. So 1500 − 1371 = 129 workers will be laid o˙, and monthly investment in capital will fall by $15,000,000 − $10,287,042 = $4,712,958. (c) Solving for the Lagrange multiplier from the above equations gives us =

5000 5000 = ≈ $8572.54 per car 0.4x−0.6 y0.6 0.4(1371.606)−0.6 (2571.760)0.6

This means that to produce on additional car per month would cost about $8572.54 with the lowest-cost use of capital and labor. 41. (a) We want to minimize C subject to g = x + y = 39. We solve the system of equations Cx = gx ,

so 10x + 2y =

Cy = gy ,

so 2x + 6y =

g = 39,

so x + y = 39.

The frst two equations give y = 2x. Solving with x + y = 39 gives x = 13, y = 26, = 182. Therefore C = $4349. (b) Since = 182, increasing production by 1 will cause costs to increase by approximately $182 . Similarly, decreasing production by 1 will save approximately $182. 42. (a) The proft is given by Proft = (q1 , q2 ) = Total Revenue − Total Cost = p1 q1 + p2 q2 − (10q1 + q1 q2 + 10q2 ) = (50 − q1 + q2 )q1 + (30 + 2q1 − q2 )q2 − (10q1 + q1 q2 + 10q2 ) = 40q1 − q12 + 2q1 q2 + 20q2 − q22 , subject to the constraint q1 + q2 = 15. Then we need to solve the equations ) = 0, )q1

) = 0, subject to q1 + q2 = 15. )q2

These equations are 40 − 2q1 + 2q2 = 2q1 + 20 − 2q2 = q1 + q2 = 15.

Adding the frst two equations gives 60 = 2

STRENGTHEN YOUR UNDERSTANDING

607

so = 30. Substituting this into the frst equation gives q1 − q2 = 5, therefore, q1 and q2 satisfy q1 + q2 = 15 q1 − q2 = 5.

Adding these equations gives 2q1 = 20 so q1 = 10 and q2 = 5. Substituting these values into the expression for the total proft gives (10, 5) = 40 ⋅ 10 − 102 + 2 ⋅ 10 ⋅ 5 + 20 ⋅ 5 − 52 = 475. The endpoints of the constraint are (15, 0) and (0, 15) giving (15, 0) = 40 ⋅ 15 − 152 = 375 (0, 15) = 20 ⋅ 15 − 152 = 75. Thus the maximum proft is 475. (b) The approximate change in the maximum proft due to a one unit increase in the production constraint is = 30. Thus a one unit increase in the production quota increases production by 30 units, to 475 + 30 = 505 units.

STRENGTHEN YOUR UNDERSTANDING 1. False, the units of 1.5 are kilometers. 2. True, since t = 6 hours. 3. False, the units of 4.1 are ppm, parts per million. 4. False. We expect the quantity of pollutants to go down as we get farther from the incinerator, so Q is a decreasing function of x. 5. True. 6. False. To be an increasing function of x, we must have that f increases whenever x increases and y is fxed. 7. True. For example, let f (x, y) = x − y. Then if y is fxed, f increases as x increases, and if x is fxed, f decreases as y increases. 8. True. If x is fxed, then f decreases as y increases. 9. False. The cross-section with x = 1 is f (1, y) = ey − y2 . 10. False. The cross section with y = 0 is f (x, 0) = e0 − 0 = 1. 11. True. When we set f (x, y) = c, we get y + 3x − 1 = c, which is the equation of a line. 12. True. When we set f (x, y) = c, we get x2 − y = c, which can be rewritten y = x2 − c, a parabola. 13. True. If they intersected at some point (a, b), we would have simultaneously f (a, b) = 1 and f (a, b) = 2. This is impossible since a function can have only one output for a given input. 14. True. If we set f (x, y) = c, we get 3x + 2y = c, which can be rewritten as y = −3∕2x + c∕2. Thus all the contours are lines with the same slope. 15. True. A Cobb-Douglas production function for variable N and V has the form P = cN a V b . 16. False, since the exponents of N and V must be between 0 and 1. 17. True, since a contour is defned as the set of (x, y) such that f (x, y) is constant. 18. True. For example, if f (x, y) = x2 + y2 , the contour f (x, y) = 0 consists of the single point (0, 0). 19. True, as illustrated in the text. 20. True. If (a, b) is any point in the domain of g and f (a, b) = c, then (a, b) lies on the contour f (x, y) = c. 21. True, as specifed in the text. 22. False. It has units of dollars per mile. 23. True. As mileage on a used car increases, the price generally decreases, so the partial derivative will be negative.

608

Chapter Eight /SOLUTIONS

24. True, since the units of the partial derivative are units of the output (pounds) per units of the changing input (calories). 25. False. In general, as calorie consumption goes up, the weight goes up, so the partial derivative would be positive. 26. True. For a fxed amount of exercise t, the weight should increase as the daily food intake increases. Thus the sign of )W ∕)c is positive. Also, for a fxed amount of daily food intake c, the weight should decrease as the amount of daily exercise t increases. Thus the sign of )W ∕)t is negative. 27. True, since when y is held constant at 2, we have fx (1, 2) ≈ Δz∕Δx = 0.5∕0.1 = 5. 28. False, since when x is held constant at 1, we have fy (1, 2) ≈ Δz∕Δy = 0.8∕0.1 = 8. 29. True, since when y is held constant at 3, we have fx (5, 3) ≈ Δz∕Δx = 0.9∕0.1 = 9. 30. False. The correct value is −6, since when x is held constant at 5, we have fy (5, 3) ≈ Δz∕Δy = −0.6∕0.1 = −6. 31. True, since we fnd the x-partial by treating y as a constant. 32. False, since fy = x2 , so fy (1, 2) = 12 = 1. 33. True, since gu = ev so gu (0, 0) = e0 = 1. 34. True. For example, let f (x, y) = 17, so fx = fy = 0. 35. False. We have Qy = 2x3 y, so Qy (1, 2) = 4 > 0. Thus Q is increasing in the y direction near (1, 2), not decreasing. 36. True, since we take the V -partial by treating N as a constant. 37. False. For example, let f (x, y) = x2 + y. Then fxx = 2 but fyy = 0. It is true that the mixed second order partial derivatives are equal: fxy = fyx if fxx , fxy , fyx , and fyy are all continuous. 38. False. We have fx = 6xe2y and fy = 6x2 e2y , so fx (1, 0) = fy (1, 0) = 6. 39. True, since we take the A-partial by treating B as a constant. 40. False, since )z∕)r = 2 rℎ, so taking the ℎ-partial of this result gives 2 r not 2 . 41. False. We need also fy (1, 2) = 0. 42. False. We have fx (1, 1) = 2(1) ≠ 0, so (1, 1) is not a critical point for f . 43. True. We have gu = 2(u − 3) and gv = 2(v − 2). Thus gu (3, 2) = gv (3, 2) = 0. 44. True. We have fx = ey , which is never 0. 45. False. For example, let f (x, y) = x2 − y2 . Then critical point (0, 0) is neither a local maximum nor a local minimum. 46. True. Because D > 0, we have either a local maximum or a local minimum at (a, b), and because fxx > 0 we have a local minimum. 47. True. The second derivative test applies only to critical points. 48. False. It has neither a local maximum not a local minimum at (0, 0) by the second derivative test since D = −4 < 0. 49. True. We have D = (0)(0) − 32 < 0, so the critical point (0, 0) is a saddle point. 50. True. For example, let f (x, y) = x2 + y2 . Then (0, 0) is both a local minimum and a global minimum. 51. False. For example, let f (x, y) = x and let the constraint be x2 + y2 = 1. Then since the constraint forces −1 ≤ x ≤ 1, the maximum occurs at P0 = (0, 1). Since fx = 1, fy = 0, we have that f has no critical points so P0 is not a critical point. 52. True. Solutions to a constrained optimization problem must satisfy the constraint equation. 53. False. The budget equation is the constraint equation. 54. True, since the production equation is the equation that must be met. 55. True. Since (0, 0) satisfes the constraint and f (x, y) ≥ 0 for all x, y, the point (0, 0) must be the minimum value of f subject to the constraint. 56. False. The point (0, 0) cannot be a solution to this constrained optimization problem since it does not satisfy the constraint 2x + 3y = 12. 57. True, since in this situation, represents the approximate change in cost given a one unit increase in production. 58. False, since in this situation, represents the approximate change in production given a one unit increase in the budget. We expect a budget of $80,001 to generate production of about 50,120 tons. 59. False. The second derivative test does not apply to constrained optimization problems. 60. False, since we also need to know the unit cost of each of the two raw materials in order to set up a budget constraint.

PROJECTS FOR CHAPTER EIGHT

609

PROJECTS FOR CHAPTER EIGHT 1. (a) About 15 feet along the wall, because that’s where there are regions of cold air (55◦F and 65◦F). (b) Roughly between 10 am and 12 noon, and between 4 pm and 6 pm. (c) Roughly between midnight and 2 am, between 10 am and 1 pm, and between 4 pm and 9 pm, since that is when the temperature near the heater is greater than 80◦ F. (d) ◦

◦

6 am

11 am

55 10

◦

3 pm (15 hours)

5 pm (17 hours)

55 10

Figure 8.64

(e) ◦

◦

55 8 16 24 Figure 8.65: Temp. vs. Time at heater

8 16 24 Figure 8.66: Temp. vs. Time at window

55 hr

8 16 24 Figure 8.67: Temp. vs. Time midway between heater and window

(f) The temperature at the window is colder at 5 pm than at 11 am because the outside temperature is colder at 5 pm than at 11 am. (g) The thermostat is set to roughly 70◦ F. We know this because the temperature in the room stays close to 70◦ F until we get close (a couple of feet) to the window. (h) We are told that the thermostat is about 2 feet from the window. Thus, the thermostat is either about 13 feet or about 17 feet from the wall. If the thermostat is set to 70◦F, every time the temperature at the thermostat goes over or under 70◦ F, the heater turns o˙ or on. Look at the point at which the vertical lines at 13 feet or about 17 feet cross the 70◦ F contours. We need to decide which of these crossings correspond best with the times that the heater turns on and o˙. (These times can be seen along the wall.) Notice that the 17 foot line does not cross the 70◦ F contour after 16 hours (4 pm). Thus, if the thermostat were 17 feet from the wall, the heater would not turn o˙ after 4 pm. However, the heater does turn o˙ at about 21 hours (9 pm). Since this is the time that the 13 foot line crosses the 70◦F contour, we estimate that the thermostat is about 13 feet away from the wall.

610

Chapter Eight /SOLUTIONS

2. We want to maximize the theater’s proft, , as a function of the two variables (prices) pc and pa . As always, = R − C,where R is the revenue, R = qc pc + qa pa , and C is the cost, which is of the form C = k(qc + qa ) for some constant k. Thus, (pc , pa ) = qc pc + qa pa − k(qc + qa ) −2 = rpc−3 + spa−1 − krp−4 c − kspa

To fnd the critical points, solve ) −5 = −3rp−4 c + 4krpc = 0 )pc ) −3 = −sp−2 a + 2kspa = 0. )pa We get pc = 4k∕3 and pa = 2k. This critical point is a global maximum by the following useful, general argument. Suppose that F (x, y) = f (x) + g(y), where f has a global maximum at x = b and g has a global maximum at y = d. Then for all x, y: F (x, y) = f (x) + g(y) ≤ f (b) + g(d) = F (b, d), so F has global maximum at x = b, y = d. The proft function in this problem has the form (pc , pa ) = f (pc ) + g(pa ), and the usual single-variable calculus argument using f ¨ and g ¨ shows that pc = 4k∕3 and pa = 2k are global maxima for f and g, respectively. Thus the maximum proft occurs when pc = 4k∕3 and pa = 2k. Thus, pc 4k∕3 2 = = . pa 2k 3 1∕3 2∕3

3. (a) The budget constraint is 4x1 + 27x2 = 324. Maximizing P = 270x1 x2 solve

subject to this constraint, we

−2∕3 2∕3 x2 = 4 1∕3 −1∕3 180x1 x2 = 27

90x1

4x1 + 27x2 = 324. Dividing the frst two equations gives −2∕3 2∕3 x2 4 = 1∕3 −1∕3 27 180x1 x2

90x1

x2 4 = . 27 2x1 Substituting 27x2 = 8x1 into 4x1 + 27x2 = 324 gives 4x1 + 8x1 = 324, so x1 = 27 and x2 = 8. This critical point gives a maximum because the endpoint values on the budget constraint, where either x1 = 0 or x2 = 0, give a production of 0. With x1 = 27 and x2 = 8, the maximum production is P0 = 270 ⋅ 271∕3 ⋅ 82∕3 = 270 ⋅ 3 ⋅ 4 = 3240.

SOLUTIONS TO PROBLEMS ON DERIVING THE FORMULA FOR REGRESSION LINES 1∕3 2∕3

(b) To minimize cost, C = 4x1 + 27x2 subject to the production constraint 270x1 x2 12, we solve

611

1∕3 2∕3

= 3240, or x1 x2

−2∕3 2∕3 x2 ) 1∕3 −1∕3 27 = (180x1 x2 ) 1∕3 2∕3 x1 x2 = 12.

4 = (90x1

Dividing the frst two equations gives −2∕3 2∕3

90x1 x2 4 = 27 180x1∕3x−1∕3 1

x 4 = 2 . 27 2x1 1∕3 2∕3

Substituting 8x1 = 27x2 into x1 x2

= 12 gives 0 1 27x2 1∕3 2∕3 x2 = 12 8 3 1∕3 2∕3 3 x x = x2 = 12, 2 2 2 2 so x2 = 8 and x1 = 27. Since there are points on this production constraint with arbitrarily large values of either x1 or x2 , as we move away from the critical point on the constraint, the cost increases without bound. Thus, the critical point gives a minimum. The minimum cost, in dollars, is C = 4 ⋅ 27 + 27 ⋅ 8 = 324. (c) The maximum production and the minimum cost occur for the same input values of x1 and x2 . The critical points in parts (a) and (b) are both x1 = 27, x2 = 8. Then production is P = 3240 and cost is C = 324 dollars. These input values give a maximum production of 3240 for a cost of $324, and a minimum cost of $324 for a production of 3240. The fact that the solution is the same when the objective function and the constraint are interchanged in this way is described as duality by economists.

Solutions to Problems on Deriving the Formula for Regression Lines 1. Let the line be in the form y = b + mx. When x equals −1, 0 and 1, then y equals b − m, b, and b + m, respectively. The sum of the squares of the vertical distances, which is what we want to minimize, is f (m, b) = (2 − (b − m))2 + (−1 − b)2 + (1 − (b + m))2 . To fnd the critical points, we compute the partial derivatives with respect to m and b, fm = 2(2 − b + m) + 0 + 2(1 − b − m)(−1) = 4 − 2b + 2m − 2 + 2b + 2m = 2 + 4m, fb = 2(2 − b + m)(−1) + 2(−1 − b)(−1) + 2(1 − b − m)(−1) = −4 + 2b − 2m + 2 + 2b − 2 + 2b + 2m = −4 + 6b. Setting both partial derivatives equal to zero, we get a system of equations: 2 + 4m = 0, −4 + 6b = 0. The solution is m = −1∕2 and b = 2∕3. You can check that it is a minimum. Hence, the regression line is y =

2 1 − x. 3 2

612

Chapter Eight /SOLUTIONS

2. Let the line be in the form y = b + mx. When x equals 0, 1, and 2, then y equals b, b + m, and b + 2m, respectively. The sum of the squares of the vertical distances, which is what we want to minimize, is f (m, b) = (2 − b)2 + (4 − (b + m))2 + (5 − (b + 2m))2 . To fnd the critical points, we compute the partial derivatives with respect to m and b, fm = 0 + 2(4 − b − m)(−1) + 2(5 − b − 2m)(−2) = −8 + 2b + 2m − 20 + 4b + 8m = −28 + 6b + 10m fb = 2(2 − b)(−1) + 2(4 − b − m)(−1) + 2(5 − b − 2m)(−1) = −4 + 2b − 8 + 2b + 2m − 10 + 2b + 4m = −22 + 6b + 6m Setting both partial derivatives equal to zero, we get a system of equations: −28 + 6b + 10m = 0, −22 + 6b + 6m = 0. The solution is m = 1.5 and b = 13∕6 ≈ 2.17. You can check that it is a minimum. Hence, the regression line is y = 2.17 + 1.5x. ∑ ∑ ∑ ∑ 3. We have xi = 0, yi = 2, x2i = 2, and yi xi = −1. Thus � b = (2 ⋅ 2 − 0 ⋅ (−1)) ∕ 3 ⋅ 2 − 02 = 2∕3 � m = (3 ⋅ (−1) − 0 ⋅ 2) ∕ 3 ⋅ 2 − 02 = −1∕2. 2 1 The line is y = − x, which agrees with the answer to Problem 1. 3 2 ∑ ∑ ∑ ∑ 4. We have xi = 3, yi = 11, x2i = 5, and yi xi = 14. Thus � b = (5 ⋅ 11 − 3 ⋅ 14) ∕ 3 ⋅ 5 − 32 = 13∕6 = 2.17. � m = (3 ⋅ 14 − 3 ⋅ 11) ∕ 3 ⋅ 5 − 32 = 9∕6 = 1.5. The line is y = 2.17 + 1.5x, which agrees with the answer to Problem 2. ∑ ∑ ∑ ∑ 5. We have xi = 6, yi = 5, x2i = 14, and yi xi = 12. Thus � b = (14 ⋅ 5 − 6 ⋅ 12) ∕ 3 ⋅ 14 − 62 = −1∕3. � m = (3 ⋅ 12 − 6 ⋅ 5) ∕ 3 ⋅ 14 − 62 = 1. 1 The line is y = x − , which agrees with the answer to Example 1. 3 6. (a) Let t be the number of years since 1960 and let P (t) be the population in millions in the year 1960 + t. We assume that P = Ceat , and therefore ln P = at + ln C. So, we plot ln P against t and fnd the line of best ft. Our data points are (0, ln 180), (10, ln 206), and (20, ln 226). Applying the method of least squares to fnd the best-ftting line, we fnd that ln 226 − ln 180 ≈ 0.0114, 20 ln 206 ln 226 5 ln 180 ln C = − + ≈ 5.20 3 6 6 a=

Then, C = e5.20 = 181.3 and so P (t) = 181.3e0.0114t . In 1990, we have t = 30 and the predicted population in millions is P (30) = 181.3e0.01141(30) = 255.3.

SOLUTIONS TO PROBLEMS ON DERIVING THE FORMULA FOR REGRESSION LINES

613

(b) The di˙erence between the actual and the predicted population is about 6 million or 2 12 %. Given that only three data points were used to calculate a and c, this discrepancy is not surprising. Thus, the 1990 census data does not mean that the assumption of exponential growth is unjustifed. (c) In 2010, we have t = 50 and P (50) = 320.7. 7. (a)

(i) Suppose N = kAp . Then the rule of thumb tells us that if A is multiplied by 10, the value of N doubles. Thus 2N = k(10A)p = k10p Ap . Thus, dividing by N = kAp , we have 2 = 10p so taking logs to base 10 we have p = log 2 = 0.3010. (where log 2 means log10 2). Thus,

N = kA0.3010 .

(ii) Taking natural logs gives ln N = ln(kAp ) ln N = ln k + p ln A ln N ≈ ln k + 0.301 ln A Thus, ln N is a linear function of ln A. (b) Table 8.24 contains the natural logarithms of the data: Table 8.24

ln N and ln A

Island

ln A

ln N

Redonda

1.1

1.6

Saba

3.0

2.2

Montserrat

2.3

2.7

Puerto Rico

9.1

4.3

Jamaica

9.3

4.2

Hispaniola

11.2

4.8

Cuba

11.6

4.8

Using a least squares ft we fnd the line: ln N = 1.20 + 0.32 ln A This yields the power function: N = e1.20 A0.32 = 3.32A0.32 Since 0.32 is pretty close to log 2 ≈ 0.301, the answer does agree with the biological rule.

9.1 SOLUTIONS

615

CHAPTER NINE Solutions for Section 9.1 1. (a) = (III), (b) = (IV), (c) = (I), (d) = (II). 2. (a) = (I), (b) = (IV), (c) = (III). Graph (II) represents an egg originally at 0 C which is moved to the kitchen table (20 C) two minutes after the egg in part (a) is moved. 3. The rate of change of P is proportional to P so we have dP = kP , dt for some constant k. Since the population P is increasing, the derivative dP ∕dt must be positive. Therefore, k is positive. 4. The rate at which the balance is changing is 5% times the current balance, so we have Rate of change of B = 0.05 ⋅ Current balance so we have

dB = 0.05B. dt

5. The rate of change of Q is proportional to Q so we have dQ = kQ, dt for some constant k. Since the radioactive substance is decaying, the quantity present, Q, is decreasing. The derivative dQ∕dt must be negative, so the constant of proportionality k is negative. 6. The balance in the account, B, is increasing at a rate of 4% times B and is decreasing at a rate of 2000 dollars per year. We have Rate of change of B = Rate in − Rate out. dB = 0.04B − 2000. dt Notice that the initial amount of $25,000 in the account is not used in the di˙erential equation. The di˙erential equation tells us only how things are changing. 7. The amount of the pollutant, P , is decreasing at a rate of 0.08 times P and is also decreasing at a constant rate of 30 gallons per day. Notice that both changes cause P to decrease, so both will have a negative e˙ect on dP ∕dt. We have dP = −0.08P − 30. dt 8. The amount of morphine, M, is increasing at a rate of 2.5 mg/hour and is decreasing at a rate of 0.347 times M. We have Rate of change of M = Rate in − Rate out. dM = 2.5 − 0.347M. dt 9. The amount of alcohol, A, is decreasing at a constant rate of 1 ounce per hour, so we have dA = −1. dt The negative sign indicates that the amount of alcohol is decreasing.

616

Chapter Nine /SOLUTIONS

10. The amount of toxin, A, is increasing at a rate of 10 micrograms per day and is decreasing at a rate of 0.03 times A. We have Rate of change of A = Rate in − Rate out. dA = 10 − 0.03A. dt 11. (a) The amount of ca˙eine, A, is decreasing at a rate of 17% times A, so we have dA = −0.17A. dt The negative sign indicates that the amount of ca˙eine is decreasing at a rate of 17% times A. Notice that the initial amount of ca˙eine, 100 mg, is not used in the di˙erential equation. The di˙erential equation tells us only how things are changing. (b) At the start of the frst hour, we have A = 100. Substituting this into the di˙erential equation, we have dA = −0.17A = −0.17(100) = −17 mg/hour. dt We estimate that the amount of ca˙eine decreases by about (17 mg/hr) ⋅ (1 hr) = 17 mg during the frst hour. This is only an estimate, however, since the derivative dA∕dt will not stay constant at −17 throughout the entire frst hour. 12. (a) The balance in the account, B, is increasing at a rate of $6000 per year and is also increasing at a rate of 0.07 times the balance B. Notice that both changes cause B to increase, rather than decrease, so both will have a positive e˙ect on dB∕dt. We have dB = 6000 + 0.07B. dt (b) If B = 10,000, we have dB = 6000 + 0.07B = 6000 + 0.07(10,000) = 6700. dt If the balance is $10,000, we expect the balance to increase at a rate of about $6700 per year. If B = 100,000, we have dB = 6000 + 0.07B = 6000 + 0.07(100,000) = 13,000. dt If the balance is $100,000, we expect the balance to increase at a rate of about $13,000 per year. 13. The di˙erential equations are: dP (a) = 500. dt dP (b) = 0.009P . dt 14. (a) To see if W is increasing or decreasing, we determine whether the derivative dW ∕dt is positive or negative. When W = 10, we have dW = 5W − 20 = 5(10) − 20 = 30 > 0. dt Since dW ∕dt is positive when W = 10, the quantity W is increasing. When W = 2, we have dW = 5W − 20 = 5(2) − 20 = −10 < 0. dt Since dW ∕dt is negative when W = 2, the quantity W is decreasing. (b) We set dW ∕dt = 0 and solve: dW =0 dt 5W − 20 = 0 W = 4. The rate of change of W is zero when W = 4. 15. The quantity y is increasing when dy∕dt is positive. Using the di˙erential equation, we see that dy∕dt is positive when −0.5y is positive, which means y is negative. The quantity y is increasing when y is negative. Similarly, the quantity y is decreasing when y is positive.

9.2 SOLUTIONS

617

16. The derivative dN∕dt represents the number of new Wikipedia articles per day. The model expresses this number as a sum of two terms. The frst term is B, the number of articles added every day by dedicated Wikipedians. The second term, representing the number of articles per day added by the general public, is proportional to the number of articles in the Wikipedia and can be written kN, where k is constant. Putting the terms together, we have: dN = B + kN. dt 17. The rate of change of the value of infrastructure, dK∕dt, is the sum of two terms. One term represents increase due to investment, and can be written k1 Y , where Y is national income and k1 is a constant of proportionality. The constant k1 is positive because investment increases the value of infrastructure. The other term represents the decrease due to depreciation, and can be written k2 K, where k2 is a second proportionality constant. The constant k2 is positive because depreciation decreases the value of infrastructure, which means that the −k2 K is negative. The di˙erential equation for K is dK = k1 Y − k2 K. dt

Solutions for Section 9.2 1. (a) Since y = x2 , we have y¨ = 2x. Substituting these functions into our di˙erential equation, we have xy¨ − 2y = x(2x) − 2(x2 ) = 2x2 − 2x2 = 0. Therefore, y = x2 is a solution to the di˙erential equation xy¨ − 2y = 0. (b) For y = x3 , we have y¨ = 3x2 . Substituting gives: xy¨ − 2y = x(3x2 ) − 2(x3 ) = 3x3 − 2x3 = x3 . Since x3 does not equal 0 for all x, we see that y = x3 is not a solution to the di˙erential equation. 2. Since y = t4 , the derivative is dy∕dt = 4t3 . We have dy = t(4t3 ) = 4t4 . dt Right-side = 4y = 4t4 . Left-side = t

Since the substitution y = t4 makes the di˙erential equation true, y = t4 is in fact a solution. 3. We are told that y is a function of t (since the derivative is dy∕dt) with derivative 2t. We need to think of a function with derivative 2t. Since y = t2 has derivative 2t, we see that y = t2 is a solution to this di˙erential equation. Since the function y = t2 + 1 also has derivative 2t, we see that y = t2 + 1 is also a solution. In fact, y = t2 + C is a solution for any constant C. The general solution is y = t2 + C. 4. We know that at time t = 0, the value of y is 8. Since we are told that dy∕dt = 4 − y, we know that at time t = 0 dy = 4 − 8 = −4. dt As t goes from 0 to 1, y will decrease by 4, so at t = 1, y=8−4=4 Likewise, we get that at t = 1, dy = 4−4 = 0 dt so that at t = 2, y = 4 + 0(1) = 4.

618

Chapter Nine /SOLUTIONS dy = 4 − 4 = 0 so that at t = 3, y = 4 + 0 = 4. dt dy At t = 3, = 4 − 4 = 0 so that at t = 4, y = 4 + 0 = 4. dt Thus we get the following table At t = 2,

5. We know that at time t = 0, the value of y is 8. Since we are told that dy∕dt = 0.5t, we know that at time t = 0 dy = 0.5(0) = 0. dt As t goes from 0 to 1, y will increase by 0, so at t = 1, y = 8 + 0(1) = 8. Likewise, we get that at t = 1, dy = 0.5(1) = 0.5 dt and so at t = 2 y = 8 + 0.5(1) = 8.5. At t = 2,

dy = 0.5(2) = 1 dt

then at t = 3 y = 8.5 + 1(1) = 9.5. dy At t = 3, = 0.5(3) = 1.5 so that at t = 4, y = 9.5 + 1.5(1) = 11. dt Thus we get the following table t

8.5

9.5

6. We know that at time t = 0, the value of y is 8. Since we are told that dy∕dt = 0.5y, we know that at time t = 0 dy = 0.5(8) = 4. dt As t goes from 0 to 1, y will increase by 4, so at t = 1, y = 8 + 4 = 12. Likewise, we get that at t = 1, dy = .5(12) = 6 dt so that at t = 2, y = 12 + 6 = 18. dy At t = 2, = .5(18) = 9 so that at t = 3, y = 18 + 9 = 27. dt dy At t = 3, = .5(27) = 13.5 so that at t = 4, y = 27 + 13.5 = 40.5. dt Thus we get the Table 9.1 Table 9.1 t

40.5

9.2 SOLUTIONS

619

7. When y = 100, the rate of change of y is

√ dy √ = y = 100 = 10 dt The value of y goes up by 10 units as t goes up 1 unit. When t = 1, we have y = Old value of y + Change in y = 100 + 10 = 110. Continuing in this way, we obtain the table: t

100

110

120.5

131.5

143.0

8. Since dy∕dx = −1, the slope of the curve must be −1 at all points. Since the slope is constant, the solution curve must be a line with slope −1. Graph C is a possible solution curve for this di˙erential equation. 9. Since dy∕dx = 0.1, the slope of the curve is 0.1 at all points. Thus the curve is a line with positive slope, such as Graph F. 10. Since −y2 is always less than or equal to zero, the derivative dy∕dx is always less than or equal to zero. A possible solution curve must have slope less than or equal to zero at all points, so possible answers are B or C. The slope of −y2 is steeper for large y-values and less steep for y-values close to zero, so the only possible solution curve for this di˙erential equation is Graph B. 11. Since dy∕dx = 2x, the slope of the solution curve will be negative when x is negative and positive when x is positive. A solution curve will be decreasing for negative x and increasing for positive x. The only graph with these features is Graph E. 12. Since dy∕dx = 2, the slope of the solution curve will be 2 at all points. Any possible solution curve for this di˙erential equation will be a line with slope 2. A possible solution curve for this di˙erential equation is Graph F. 13. Since dy∕dx = y, the slope of the solution curve will be positive for positive y-values and negative for negative y-values. In addition, the slope will be bigger for large y and closer to zero when the y-value is closer to zero. A possible solution curve for this di˙erential equation is Graph A. 14. Since dy∕dx is negative for all x, the slope of the solution curve is negative everywhere and becomes closer to horizontal as x increases, as in Graph B. 15. Since dy∕dx = 1 − x is positive for x < 1 and negative for x > 1, the slope of the solution curve is positive for x < 1 and negative for x > 1. The answer is Graph D. 16. Since dy∕dx is positive if y is positive, the slope of the solution curve is positive everywhere and increases as y increases, as in Graph A. 17. At t = 0, we know P = 70 and we can compute the value of dP ∕dt: At t = 0,

dP = 0.2P − 10 = 0.2(70) − 10 = 4. dt

we have

The population is increasing at a rate of 4 million fsh per year. At the end of the frst year, the fsh population will have grown by about 4 million fsh, and so we have: At t = 1,

we estimate

P = 70 + 4 = 74.

We can now use this new value of P to calculate dP ∕dt at t = 1: At t = 1,

we have

dP = 0.2P − 10 = 0.2(74) − 10 = 4.8, dt

and so: At t = 2,

we estimate

P = 74 + 4.8 = 78.8.

Continuing in this way, we obtain the values in Table 9.2. Table 9.2 t

78.8

84.56

620

Chapter Nine /SOLUTIONS

18. If P = P0 et , then

dP d = (P0 et ) = P0 et = P . dt dt

19. If Q = Cekt , then

dQ = Ckekt = k(Cekt ) = kQ. dt = −0.03Q, so we know that kQ = −0.03Q. Thus we either have Q = 0 (in which case C = 0 and k We are given that dQ dt is anything) or k = −0.03. Notice that if k = −0.03, then C can be any number.

20. Substitute y = ekx into the di˙erential equation. Since dy∕dx = kekx , we have 7ekx − 3kekx = 0 Since ekx ≠ 0, we need

(7 − 3k) ekx = 0.

7 − 3k = 0 7 k= . 3 dy dy d n 21. Yes. To see why, we substitute y = xn into the equation 13x = (x ) = nxn−1 . The = y. We frst calculate dx dx dx di˙erential equation becomes 13x(nxn−1 ) = xn But 13x(nxn−1 ) = 13n(x ⋅ xn−1 ) = 13nxn , so we have 13n(xn ) = xn This equality must hold for all x, so we get 13n = 1, so n = 1∕13. Thus, y = x1∕13 is a solution. 22. Since y = x2 + k we know that y¨ = 2x. Substituting y = x2 + k and y¨ = 2x into the di˙erential equation we get 10 = 2y − xy¨ = 2(x2 + k) − x(2x) = 2x2 + 2k − 2x2 = 2k. Thus, k = 5 is the only solution. 23. We frst compute dy∕dx for each of the functions on the right. If y = x3 then dy = 3x2 dx y =3 . x If y = 3x then dy =3 dx y = . x If y = e3x then dy = 3e3x dx = 3y.

9.3 SOLUTIONS

621

If y = 3ex then dy = 3ex dx = y. Finally, if y = x then dy =1 dx y = . x Comparing our calculated derivatives with the right-hand sides of the di˙erential equations we see that (a) is solved by (II) and (V), (b) is solved by (I), (c) is not solved by any of our functions, (d) is solved by (IV) and (e) is solved by (III).

Solutions for Section 9.3 1. See Figure 9.1. Other choices of solution curves are, of course, possible. y

Figure 9.1

2. We substitute x- and y-coordinates into the di˙erential equation to determine the slopes. At the point (−1, 2), we have dy∕dx = 22 = 4 so the slope is 4. At the point (0, 2), we have dy∕dx = 22 = 4, so the slope is again 4. Continuing in this way, we create Figure 9.2. y 2

x −1

Figure 9.2

622

Chapter Nine /SOLUTIONS

3. We substitute x- and y-coordinates into the di˙erential equation to determine the slopes. At the point (1, 1), we have dy∕dx = 1∕1 = 1 so the slope is 1. At the point (1, 0), we have dy∕dx = 1∕0, which is undefned. An undefned slope corresponds to a vertical line segment. Continuing in this way, we create Figure 9.3. y 1

x −1

−1

Figure 9.3

4. (a) (i), (ii), (iii) y

✛

x = 0, y = 12

16 12

✛

x = 0, y = 10

8 4

✛ 1

x = 0, y = 8 x 2

Figure 9.4 (b) When y = 10, we have dy∕dx = 0 and so the solution curve is horizontal. This is why the solution curve through y = 10 is a horizontal line. 5. (a) The slope at any point is given by the derivative, so we fnd the slope by substituting the x- and y-coordinates into the di˙erential equation dy∕dx = x2 − y2 to fnd dy∕dx. dy = 12 − 02 = 1. dx dy = 02 − 12 = −1. The slope at (0, 1) is dx dy = 12 − 12 = 0. The slope at (1, 1) is dx dy = 22 − 12 = 3. The slope at (2, 1) is dx dy = 12 − 22 = −3. The slope at (1, 2) is dx dy = 22 − 22 = 0. The slope at (2, 2) is dx The slope at (1, 0) is

(b) See Figure 9.5.

9.3 SOLUTIONS

623

y 2

x 1

Figure 9.5 6. (a) See Figure 9.6. y 4

(i)

x −4

(ii)

−4

(iii)

Figure 9.6 (b) The solution through (−1, 0) appears to be linear with equation y = −x − 1. (c) If y = −x − 1, then y¨ = −1 and x + y = x + (−x − 1) = −1, so this checks as a solution. 7. (a) Since y¨ = 1 + y2 , the slope is everywhere positive. This is true for slope feld (III). (b) Since y¨ = x, the slopes are positive to the right of the y-axis and negative to the left of the y-axis. Solution curves are parabolas, as in slope feld (VI). (c) Since y¨ = sin x, the slopes oscillate between −1 and 1. This corresponds to slope feld (V). (d) Since y¨ = y, the slopes are positive above the x-axis and negative below the x-axis. This is true for slope feld (I). (e) Since y¨ = x − y, the slopes are zero when x = y. This corresponds to slope feld (IV). (f) Since y¨ = 4 − y, the slopes are positive when y < 4 and negative when y > 4. This corresponds to slope feld (II). 8. (a) Since y¨ = −y, the slope is negative above the x-axis (when y is positive) and positive below the x-axis (when y is negative). The only slope feld for which this is true is II. (b) Since y¨ = y, the slope is positive for positive y and negative for negative y. This is true of both I and III. As y gets larger, the slope should get larger, so the correct slope feld is I. (c) Since y¨ = x, the slope is positive for positive x and negative for negative x. This corresponds to slope feld V. 1 (d) Since y¨ = , the slope is positive for positive y and negative for negative y. As y approaches 0, the slope becomes y larger in magnitude, which corresponds to solution curves close to vertical. The correct slope feld is III. (e) Since y¨ = y2 , the slope is always positive, so this must correspond to slope feld IV. 9. III. The slope feld appears to be near zero at P = 1 and P = 0, so this rules out dP ∕dt = P − 1, which has a slope of −1 at P = 0. Between P = 0 and P = 1, the slopes in the fgure are positive, so this rules out dP ∕dt = P (P − 1), which has negative values for 0 < P < 1. To decide between the remaining two possibilities, note that when P = 1∕2, the slopes in the fgure appear to be about 1. The di˙erential equation dP ∕dt = 3P (1 − P ) gives a slope of 3 ⋅ 1∕2 ⋅ (1 − 1∕2) = 3∕4, while the di˙erential equation dP ∕dt = 1∕3P (1 − P ) gives a slope of 1∕12 which is clearly too small. Thus the best answer is dP ∕dt = 3P (1 − P ). 10. (a) The slope at any point (x, y) is equal to dy∕dx, which is xy.Then: At point (2, 1), slope = 2 ⋅ 1 = 2,

624

Chapter Nine /SOLUTIONS At point (0, 2), slope = 0 ⋅ 2 = 0, At point (−1, 1), slope = −1 ⋅ 1 = −1, At point (2, −2), slope = 2(−2) = −4. (b) See Figure 9.7. y Slope = 0 ✲

Slope = 2

❄

Slope = −1 ✲

1 x

−2

−1

−1 −2

✻ Slope = −4

Figure 9.7

11. If the starting point has y > 0, then y → ∞ as x → ∞. If the starting point has y = 0, then the solution is constant; y = 0. If the starting point has y < 0, then y → −∞ as x → ∞. 12. As x increases, y → ∞. 13. As x → ∞, y → ∞, no matter what the starting point is. 14. As x → ∞, y seems to oscillate within a certain range. The range will depend on the starting point, but the size of the range appears independent of the starting point.

15. If y = 4 for the starting point, then y = 4 always, so y → 4 as x → ∞. If y ≠ 4 for the starting point, then y → 4 as x → ∞. 16. From the slope feld, the function looks like a parabola of the form y = x2 + C, where C depends on the starting point. In any case, y → ∞ as x → ∞. 17. When a = 1 and b = 2, the Gompertz equation is y¨ = −y ln(y∕2) = y ln(2∕y) = y(ln 2 − ln y). This di˙erential equation is similar to the di˙erential equation y¨ = y(2 − y) in certain ways. For example, in both equations y¨ is positive for 0 < y < 2 and negative for y > 2. Also, for y-values close to 2, the quantities (ln 2−ln y) and (2−y) are both close to 0, so y(ln 2−ln y) and y(2 − y) are approximately equal to zero. Thus around y = 2 the slope felds look almost the same. This happens again around y = 0, since around y = 0 both y(2 − y) and y(ln 2 − ln y) go to 0. Finally, for y > 2, ln y grows much slower than y, so the slope feld for y¨ = y(ln 2 − ln y) is less steep, negatively, than for y¨ = y(2 − y).

Solutions for Section 9.4 1. The equation given is in the form dP = kP . dt Thus we know that the general solution to this equation will be P = Cekt . And in our case, with k = 0.02 and C = 20 we get P = 20e0.02t .

9.4 SOLUTIONS 2. The equation is in the form dw∕dr = kw, so the general solution is the exponential function w = Ce3r . We fnd C using the initial condition that w = 30 when r = 0. w = Ce3r 30 = Ce0 C = 30. The solution is w = 30e3r . 3. The equation is in the form dy∕dx = ky, so the general solution is the exponential function y = Ce−0.14x . We fnd C using the initial condition that y = 5.6 when x = 0. y = Ce−0.14x 5.6 = Ce0 C = 5.6. The solution is y = 5.6e−0.14x . 4. The equation given is in the form dQ = kQ. dt Thus we know that the general solution to this equation will be Q = Cekt . And in our case, with k =

1 , we get 5

Q = Ce 5 t . We know that Q = 50 when t = 0. Thus we get 1

Q(t) = Ce 5 t Q(0) = 50 = Ce0 50 = C Thus we get 1

Q = 50e 5 t . 5. The equation is in the form dp∕dq = kp, so the general solution is the exponential function p = Ce−0.1q . We fnd C using the condition that p = 164.87 when q = 0. p = Ce−0.1q 164.87 = Ce−0.1(0) C = 164.87. The solution is p = 164.87e−0.1q .

625

Chapter Nine /SOLUTIONS

626

6. Rewriting we get dy 1 = − y. dx 3 We know that the general solution to an equation in the form dy = ky dx is y = Cekx . Thus in our case we get 1

y = Ce− 3 x . We are told that y(0) = 10 so we get 1

y(x) = Ce− 3 x y(0) = 10 = Ce0 C = 10 Thus we get 1

y = 10e− 3 x . 7. (a) Since interest is earned continuously, Rate of change of balance = 1.5%(Balance) so

dB = 0.015B. dt (b) B = Ce0.015t is the general solution. Since B = 5000 when t = 0, we have C = 5000. The solution is B = 5000e0.015t (c) When t = 10, B = 5000e0.015(10) = $5809.17. 8. (a) The rate of growth of the investment is proportional to the value of the investment. Thus, dM = rM. dt (b) We know that the equation dM = rM dt has the general solution M = Aert . We know that in 2010 (i.e. t = 0) we have M = 2000. Thus we get M = Aert M(0) = 2000 = Ae0r 2000 = Ae0 A = 2000. Thus we get M = 2000ert . (c) See Figure 9.8.

9.4 SOLUTIONS

627

M 40000

M = 2000e0.10t

10000

M = 2000e0.05t

2000

t t=0 2010

t = 30 2040

Figure 9.8

9. (a) This statement is described by equation (II). For instance, if there are Q = 100 grams of the radioactive substance at time t = 10, then the rate of change dQ∕dt is −0.05(100) = −5 grams/year. Also note that the rate of change is negative. (b) This statement is described by equation (III). For instance, suppose there are N = 1000 people in total and that Q = 100 have signed up for the service at time t = 10. Then the number who have yet to sign up is N − Q = 900, so the rate at which people are signing up is given by 5% of this, or 0.05(900) = 45 people/year. (c) This statement is described by equation (I). For instance, if there are Q = 1000 members at time t = 10, then the rate of change dQ∕dt is 0.05(1000) = 50 members/year. (d) This statement is described by equation (IV). For instance, suppose the natural salt level of the groundwater is N = 0.1 g/liter and that the pond’s salt level is Q = 0.5 grams/liter. Then the amount that the pond’s salt level falls above the natural level is Q − N = 0.4 grams/liter, so the rate at which this changes, dQ∕dt, is given by −0.05(0.4) = −0.02 grams/liter per year. 10. Since the rate of change is proportional to the amount present, we have dQ = kQ. We know the constant of proportionality dt is k = −0.0025, so a di˙erential equation for Q as a function of t is dQ = −0.0025Q. dt The solution to this di˙erential equation is Q = Ce−0.0025t , for some constant C. When t = 20, we have Q = Ce−0.0025(20) = C(0.951), so approximately 95% of the current ozone will still be here in 20 years. Approximately 5% will decay during this time. 11. Michigan: dQ r 158 =− Q=− Q dt V 4.9 × 103

−0.032Q

Q = Q0 e−0.032t . We want to fnd t such that 0.1Q0 = Q0 e−0.032t so t= Ontario: so

− ln(0.1) 0.032

72 years.

dQ r −209 =− Q= Q = −0.131Q dt V 1.6 × 103 Q = Q0 e−0.131t .

We want to fnd t such that 0.1Q0 = Q0 e−0.131t so

− ln(0.1) 18 years. 0.131 Lake Michigan will take longer because it is larger (4900 km3 compared to 1600 km3 ) and water is fowing through it at a slower rate (158 km3 /year compared to 209 km3 /year). t=

Chapter Nine /SOLUTIONS

628

12. Lake Superior will take the longest, because the lake is largest (V is largest) and water is moving through it most slowly (r is smallest). Lake Erie looks as though it will take the least time because V is smallest and r is close to the largest. For Erie, k = r∕V = 175∕460 0.38. The lake with the largest value of r is Ontario, where k = r∕V = 209∕1600 0.13. Since e−kt decreases faster for larger k, Lake Erie will take the shortest time for any fxed fraction of the pollution to be removed. For Lake Superior, dQ r 65.2 =− Q=− Q −0.0053Q dt V 12,200 so Q = Q0 e−0.0053t . When 80% of the pollution has been removed, 20% remains so Q = 0.2Q0 . Substituting gives us 0.2Q0 = Q0 e−0.0053t so

ln(0.2) years. 0.0053 65.2 (Note: The 301 is obtained by using the exact value of Vr = 12,200 , rather than 0.0053. Using 0.0053 gives 304 years.) For Lake Erie, as in the text dQ r 175 =− Q=− Q −0.38Q dt V 460 so Q = Q0 e−0.38t . t=−

When 80% of the pollution has been removed 0.2Q0 = Q0 e−0.38t ln(0.2) t=− 0.38

years.

So the ratio is

Time for Lake Superior 301 . Time for Lake Erie 4 In other words it will take about 75 times as long to clean Lake Superior as Lake Erie. 13. (a) Since the amount leaving the blood is proportional to the quantity in the blood, dQ = −kQ for some positive constant k. dt Thus Q = Q0 e−kt , where Q0 is the initial quantity in the bloodstream. Only 20% is left in the blood after 3 hours. Thus 0.20 .5365. Therefore Q = Q0 e−0.5365t . 0.20 = e−3k , so k = ln−3 (b) Since 20% is left after 3 hours, after 6 hours only 20% of that 20% will be left. Thus after 6 hours only 4% will be left, so if the patient is given 100 mg, only 4 mg will be left 6 hours later. 14. (a) Since the rate of change is proportional to the amount present, dy∕dt = ky for some constant k. (b) Solving the di˙erential equation, we have y = Aekt , where A is the initial amount. Since 100 grams become 54.9 grams in one hour, 54.9 = 100ek , so k = ln(54.9∕100) −0.5997. Thus, after 10 hours, there remains 100e(−0.5997)10 0.2486 grams. 15. (a) Suppose Y (t) is the quantity of oil in the well at time t. We know that the oil in the well decreases at a rate proportional to Y (t), so dY = −kY . dt Integrating, and using the fact that initially Y = Y0 = 106 , we have Y = Y0 e−kt = 106 e−kt . In six years, Y = 500, 000 = 5 ⋅ 105 , so

5 ⋅ 105 = 106 e−k⋅6

so 0.5 = e−6k ln 0.5 k=− = 0.1155. 6

9.4 SOLUTIONS

629

When Y = 600, 000 = 6 ⋅ 105 , dY = kY = 0.1155(6 ⋅ 105 ) = 69,300 barrels/year. dt

Rate at which oil decreasing = (b) We solve the equation

5 ⋅ 104 = 106 e−0.1155t 0.05 = e−0.1155t ln 0.05 t= = 25.9 years. −0.1155 16. (a) Since we are told that the rate at which the quantity of the drug decreases is proportional to the amount of the drug left in the body, we know the di˙erential equation modeling this situation is dQ = kQ. dt Since we are told that the quantity of the drug is decreasing, we know that k < 0. (b) We know that the general solution to the di˙erential equation dQ = kQ dt is Q = Cekt . (c) We are told that the half life of the drug is 3.8 hours. This means that at t = 3.8, the amount of the drug in the body is half the amount that was in the body at t = 0, or, in other words, 0.5Q(0) = Q(3.8). Solving this equation gives 0.5Q(0) = Q(3.8) 0.5Cek(0) = Cek(3.8) 0.5C = Cek(3.8) 0.5 = ek(3.8) ln(0.5) = k(3.8) ln(0.5) =k 3.8 k −0.182. (d) From part (c) we know that the formula for Q is Q = Ce−0.182t . We are told that initially there are 10 mg of the drug in the body. Thus, at t = 0, we get 10 = Ce−0.182(0) so C = 10. Thus, our equation becomes Q(t) = 10e−0.182t . Substituting t = 12, we get Q(t) = 10e−0.182t Q(12) = 10e−0.182(12) = 10e−2.184 Q(12)

.126 mg.

Chapter Nine /SOLUTIONS

630

17. (a) Assuming that the world’s population grows exponentially, satisfying dP ∕dt = cP , and that the land in use for crops is proportional to the population, we expect A to satisfy dA∕dt = kA. (b) We have A(t) = A0 ekt = 1.49 ⋅ 109 ekt , where t is the number of years after 1988. Since 30 years later the amount of land in use is 1.57 billion hectares, we have 1.57 ⋅ 109 = (1.49 ⋅ 109 )ek(30) , so e30k =

1.57 . 1.49

Solving for k gives k=

ln (1.57∕1.49) = 0.00174. 30

Thus, A = (1.49 ⋅ 109 )e0.00174t . We want to fnd t such that 4.2 ⋅ 109 = A(t) = (1.49 ⋅ 109 )e0.00174t . Taking logarithms gives ln (4.2∕1.49) = 595.58 years. 0.00174 This model predicts land will have run out 596 years after 1988, that is by the year 2584. t=

Solutions for Section 9.5 1. We know that the general solution to a di˙erential equation of the form dy = k(y − A) dt is y = A + Cekt . Thus in our case we get y = 200 + Ce0.5t . We know that at t = 0 we have y = 50, so solving for C we get y = 200 + Ce0.5t 50 = 200 + Ce0.5(0) −150 = Ce0 C = −150. Thus we get y = 200 − 150e0.5t . 2. We know that the general solution to a di˙erential equation of the form dP = k(P − A) dt is P = Cekt + A. Thus in our case we have k = 1, so we get P = Cet − 4.

9.5 SOLUTIONS We know that at t = 0 we have P = 100 so solving for C we get P = Cet − 4 100 = Ce0 − 4 104 = Ce0 C = 104. Thus we get P = 104et − 4. 3. We know that the general solution to a di˙erential equation of the form dH = k(H − A) dt is H = A + Cekt . Thus in our case we get H = 75 + Ce3t . We know that at t = 0 we have H = 0, so solving for C we get H = 75 + Ce3t 0 = 75 + Ce3(0) −75 = Ce0 C = −75. Thus we get H = 75 − 75e3t . 4. We know that the general solution to a di˙erential equation of the form dm = k(m − A) dt is m = Cekt + A. Factoring out a 0.1 on the left side we get dm −200 = 0.1 m − = 0.1(m − (−2000)). dt 0.1 Thus in our case we get m = Ce0.1t − 2000. We know that at t = 0 we have m = 1000 so solving for C we get m = Ce0.1t − 2000 1000 = Ce0 − 2000 3000 = Ce0 C = 3000. Thus we get m = 3000e0.1t − 2000.

631

Chapter Nine /SOLUTIONS

632

5. We know that the general solution to a di˙erential equation of the form dB = k(B − A) dt is B = A + Cekt . To get our equation in this form we factor out a 4 to get dB 100 =4 B− = 4(B − 25). dt 4 Thus in our case we get B = Ce4t + 25. We know that at t = 0 we have B = 20, so solving for C we get B = 25 + Ce4t 20 = 25 + Ce4(0) −5 = Ce0 C = −5. Thus we get B = 25 − 5e4t . 6. We know that the general solution to a di˙erential equation of the form dQ = k(Q − A) dt is Q = A + Cekt . To get our equation in this form we factor out a 0.3 to get dQ 120 = 0.3 Q − = 0.3(Q − 400). dt 0.3 Thus in our case we get Q = 400 + Ce0.3t . We know that at t = 0 we have Q = 50, so solving for C we get Q = 400 + Ce0.3t 50 = 400 + Ce0.3(0) −350 = Ce0 C = −350. Thus we get Q = 400 − 350e0.3t . 7. We know that the general solution to a di˙erential equation of the form dB = k(B − A) dt is B = Cekt + A. Rewriting we get dB = −2B + 50. dt

9.5 SOLUTIONS Factoring out a −2 on the right side we get dB −50 = −2 B − = −2(B − 25). dt −2 Thus in our case we get B = Ce−2t + 25. We know that at t = 1 we have B = 100 so solving for C we get B = Ce−2t + 25 100 = Ce−2 + 25 75 = Ce−2 C = 75e2 . Thus we get B = 75e2 e−2t + 25 = 75e2−2t + 25. 8. Rewrite the di˙erential equation as dB = −0.1B + 10 dt Factoring out −0.1 gives dB = −0.1(B − 100), dt which has solution B = 100 + Ce−0.1t . Substituting B = 3 and t = 2 gives 3 = 100 + Ce−0.1(2) . Solving for C we get 97 C = − −0.2 ≈ −118 e So the solution is B = 100 − 118e−0.1t . 9. In order to check that y = A + Cekt is a solution to the di˙erential equation dy = k(y − A), dt we must show that the derivative of y with respect to t is equal to k(y − A): y = A + Cekt dy = 0 + (Cekt )(k) dt = kCekt = k(Cekt + A − A) � = k (Cekt + A) − A = k(y − A)

10. (a) If B = f (t) (where t is in years) dB = Rate at which interest is earned + Rate at which money is deposited dt = 0.10B + 1000. (b) dB = 0.1(B + 10,000) dt

633

634

Chapter Nine /SOLUTIONS We know that a di˙erential equation of the form dB = k(B − A) dt has general solution: B = Cekt + A. Thus, in our case B = Ce0.1t − 10,000. For t = 0, B = 0, hence C = 10,000. Therefore, B = 10,000e0.1t − 10,000.

11. (a) For this situation, H

I Rate money added to account

H =

I Rate money added via interest

H +

I Rate money deposited

Translating this into an equation yields dB = 0.02B + 1200. dt (b) We know that the general solution to the di˙erential equation dB = k(B + A) dt is B = Cekt − A. We factor out 0.02 to put our equation in the form dB 1200 = 0.02 B + = 0.02(B + 60,000). dt 0.02 This equation has the solution B = Ce0.02t − 60,000. Solving for C with B(0) = 0 we get C = 60,000 and so B = f (t) = 60,000(e0.02t − 1). The solution is B = 60,000e0.02t − 60,000. (c) After 5 years, the balance is B = f (5) = 60,000(e0.02(5) − 1) = 6310.26 dollars. 12. (a) The value of the company satisfes Rate of change of value = Rate interest earned − Rate expenses paid so

dV = 0.02V − 80,000. dt

(b) We fnd V when dV =0 dt 0.02V − 80,000 = 0 0.02V = 80,000 V = 4,000,000 There is an equilibrium solution at V = $4,000,000. If the company has $4,000,000 in assets, its earnings will exactly equal its expenses. (c) The general solution is V = 4,000,000 + Ce0.02t .

9.5 SOLUTIONS

635

(d) If V = 3,000,000 when t = 0, we have C = −1,000,000. The solution is V = 4,000,000 − 1,000,000e0.02t . When t = 12, we have V = 4,000,000 − 1,000,000e0.02(12) = 4,000,000 − 1,271,249 = $2,728,751. The company is losing money. 13. The bank account is earning money at a rate of 8% times the current balance, and it is losing money at a constant rate of $5000 a year. We have Rate of change of B = Rate in − Rate out dB = 0.08B − 5000 = 0.08(B − 62,500). dt The solution to this di˙erential equation is B = 62,500 + Ce0.08t , for some constant C. To fnd C, we use the fact that B = 50,000 when t = 0: B = 62,500 + Ce0.08t 50,000 = 62,500 + Ce0 C = −12,500. The solution is B = 62,500 − 12,500e0.08t . This solution is shown in Figure 9.9. We see that the account loses money, and runs out of money in about 20 years. Algebraically, B = 0 where 12,500e0.08t = 62,500 62,500 e0.08t = =5 12,500 ln 5 t= ≈ 20.1. 0.08 So the account runs out of money in 20.1 years. B $50,000

t 20.1

Figure 9.9

14. (a) We know that the rate by which the account changes every year is Rate of change of balance = Rate of increase − Rate of decrease. Since $1000 will be withdrawn every year, we know that the account decreases by $1000 every year. We also know that the account accumulates interest at 7% compounded continuously. Thus the amount by which the account increases each year is Rate balance increases per year = 7%(Account balance) = 0.07(Account balance).

636

Chapter Nine /SOLUTIONS Denoting the account balance by B we get Rate balance increases per year = 0.07B. Thus we get Rate of change of balance = 0.07B − 1000. or

dB = 0.07B − 1000, dt

with t measured in years. (b) The equilibrium solution makes the derivative 0, so dB =0 dt 0.07B − 1000 = 0 1000 B= $14,285.71. 0.07 (c) We know that the general solution to a di˙erential equation of the form dB = k(B − A) dt is B = Cekt + A. To get our equation in this form we factor out a 0.07 to get dB 1000 .07(B − 14,285.71). = 0.07 B − dt 0.07 Thus in our case we get B = Ce0.07t + 14,285.71. We know that at t = 0 we have B = $10,000 so solving for C we get B = Ce0.07t + 14,285.71 10,000 = Ce4(0) + 14,285.71 −4285.71 = Ce0 C = −4285.71. Thus we get B = 14,285.71 − (4285.71)e0.07t . (d) Substituting the value t = 5 into our function for B we get B(t) = 14,285.71 − (4285.71)e0.07t B(5) = 14,285.71 − (4285.71)e0.07(5) = 14,285.71 − (4285.71)e0.35 B(5)

$8204

(e) From Figure 9.10 we see that in the long run there is no money left in the account. B ($) 10,000 8000 6000 4000 2000 4

Figure 9.10

t (years)

9.5 SOLUTIONS

637

15. (a) We know that the general solution to a di˙erential equation of the form dy = k(y − A) dt is y = Cekt + A. Factoring out a −1 on the left side we get dy = −(y − 100) dt Thus in our case we get y = Ce−t + 100.

Since one cannot know than 100% of a task, we take C ≤ 0. Since one cannot know less than 0% of a task, we take C ≥ −100. (b) See Figure 9.11. y

✛ C = −50

100

■

■ C = −100 C = −150

Figure 9.11 (c) Substituting y = 0 when t = 0 gives 0 = 100 − Ce−0 so C = 100. The solution is y = 100 − 100e−t 16. (a) The quantity increases with time. As the quantity increases, the rate at which the drug is excreted also increases, and so the rate at which the drug builds up in the blood decreases; thus the graph of quantity against time is concave down. The quantity rises until the rate of excretion exactly balances the rate at which the drug is entering; at this quantity there is a horizontal asymptote. (b) Theophylline enters at a constant rate of 43.2mg/hour and leaves at a rate of 0.082Q, so we have dQ = 43.2 − 0.082Q dt (c) We know that the general solution to a di˙erential equation of the form dy = k(y − A) dt is y = Cekt + A. Thus in our case, since dQ = 43.2 − 0.082Q ≈ −0.082(Q − 526.8), dt we have Q = 526.8 + Ce−0.082t . Since Q = 0 when t = 0, we can solve for C: Q = 526.8 + Ce−0.082t 0 = 526.8 + Ce0 C = −526.8

638

Chapter Nine /SOLUTIONS The solution is Q = 526.8 − 526.8e−0.082t . In the long run, the quantity in the body approaches 526.8 mg. See Figure 9.12. Q (mg) 526.8 Q = 526.8(1 − e−0.082t )

t 20

Figure 9.12

17. (a) The smoker smokes 5 cigarettes per hour, and each cigarette contributes 0.4 mg of nicotine, so, every hour, the amount of nicotine is increasing by 5(0.4) = 2.0 mg. At the same time, the nicotine is being eliminated at a rate of −0.346 times the amount of nicotine. Thus, we have dN = Rate in − Rate out = 2.0 − 0.346N. dt (b) We have dN = 2.0 − 0.346N = −0.346(N − 5.78), dt so the solution is N = 5.78 − 5.78e−0.346t . (c) At t = 16, we have N = 5.78 − 5.78e−0.346(16) = 5.76 mg. dy 18. (a) We have = −k(y − a), where k > 0 and a are constants. dt (b) We know that the general solution to a di˙erential equation of the form dy = −k(y − a) dt is y = Ce−kt + a. We can assume that right after the course is over (at t = 0) 100% of the material is remembered. Thus we get y = Ce−kt + a 1 = Ce0 + a C = 1 − a. Thus y = (1 − a)e−kt + a. (c) As t → ∞, e−kt → 0, so y → a. Thus, a represents the fraction of material which is remembered in the long run. The constant k tells us about the rate at which material is forgotten. 19. If a curve lies just above the line y = 20, so that y > 20, the value of dy∕dx is positive, so the curve climbs farther above the line y = 20. On the other hand, if a curve lies just below the line y = 20, so that y < 20, the value of dy∕dx is negative, so the curve falls farther below the line y = 20. In both cases, solutions near the line y = 20 pull away from it as time passes, so the equilibrium is unstable. 20. If a curve lies just above the line y = 20, so that y > 20, the value of dy∕dx is negative, so the curve falls closer to the line y = 20. On the other hand, if a curve lies just below the line y = 20, so that y < 20, the value of dy∕dx is positive, so the curve climbs closer to the line y = 20. In both cases, solutions near the line y = 20 draw closer to it as time passes, so the equilibrium is stable.

9.5 SOLUTIONS

639

21. (a) We know that the equilibrium solution is the solution satisfying the di˙erential equation whose derivative is everywhere 0. Thus we must solve dy = 0. dt Solving this gives dy =0 dt 0.5y − 250 = 0 y = 500 (b) We know that the general solution to a di˙erential equation of the form dy = k(y − A) dt is y = A + Cekt . To get our equation in this form we factor out a 0.5 to get dy 250 = 0.5 y − = 0.5(y − 500). dt 0.5 Thus in our case we get y = 500 + Ce0.5t . (c) The graphs of several solutions is shown in Figure 9.13. y C=1

520 515 510 505

C = .2 y = 500

495 C = −.4 t −4

Figure 9.13 (d) Looking at Figure 9.13 we see that as t → ∞, the value of y gets further and further away from the line y = 500. The equilibrium solution y = 500 is unstable. 22. (a) We know that the equilibrium solution are the functions satisfying the di˙erential equation whose derivative everywhere is 0. Thus we must solve the equation dy = 0. dt Solving we get dy =0 dt 0.2(y − 3)(y + 2) = 0 (y − 3)(y + 2) = 0 Thus the solutions are y = 3 and y = −2.

640

Chapter Nine /SOLUTIONS (b) Looking at Figure 9.14 we see that the line y = 3 is an unstable solution while the line y = −2 is a stable solution. y 3

t −2

Figure 9.14

23. Graphically, a function is an equilibrium solution if its graph is a horizontal line. From the slope feld we see that the equilibrium solutions are y = 1 and y = 3. An equilibrium solution is stable if a small change in the initial value conditions gives a solution which tends toward the equilibrium as t tends to positive infnity. Thus, by looking at the given slope felds, we see that y = 3 is a stable solution while y = 1 is an unstable solution. 24. (a) We know that the general solution to a di˙erential equation of the form dH = −k(H − 200) dt is H = Ce−kt + 200. We know that at t = 0 we have H = 20 so solving for C we get H = Ce−kt + 200 20 = Ce0 + 200 C = −180. Thus we get H = −180e−kt + 200. −80 , giving (b) Using part (a), we have 120 = 200 − 180e−k(30) . Solving for k, we have e−30k = −180

ln 49 −30

.027.

Note that this k is correct if t is given in minutes. (If t is given in hours, k =

ln 49 − 12

.622.)

25. (a) If the object starts out hotter than 68 (that is, H − 68 > 0), then the temperature of the object decreases (i.e., dH∕dt < 0) and so the constant a must be negative. (b) Similarly, if the object starts out colder than 68 (that is, H − 68 < 0), then the temperature of the object increases (i.e., dH∕dt > 0) and so the constant a must be negative too. 26. (a) Newton’s Law of Cooling says that Rate of change of temperature = Constant ⋅ Temperature di˙erence. Since the temperature di˙erence is H − 20, we have for some constant k dH = k(H − 20). dt The general solution is H = 20 + Cekt .

9.5 SOLUTIONS

641

To determine C, we use the fact that H = 35 at t = 0: 35 = 20 + Ce0 35 = 20 + C. So C = 15 and we have H = 20 + 15ekt . To fnd k, we use the fact that H = 33 when t = 2: 33 = 20 + 15ek(2) . We isolate the exponential and solve for k: 13 = 15e2k 13 = e2k 15 13 ln = 2k 15 ln(13∕15) k= = −0.072. 2 Therefore, the temperature, H, of the body as a function of time, t, is given by H = 20 + 15e−0.072t . (b) The graph of H = 20 + 15e−0.072t has a vertical intercept of H = 35, the initial temperature. The temperature decays exponentially with a horizontal asymptote of H = 20. (See Figure 9.15.) “In the long run” means as t → ∞. The graph shows that H → 20 as t → ∞. Algebraically, since e−0.072t → 0 as t → ∞, H = 20 + 15e−0.072t ⟶ 20 as t → ∞. «¯¬ Goes to 0 as t → ∞

(c) We want to know when the temperature was 37 C. We substitute H = 37 and solve for t: 37 = 20 + 15e−0.072t 17 = e−0.072t . 15 Taking natural logs on both sides gives

17 15

= −0.072t

ln(17∕15) = −1.74 hours. 0.072 The murder occurred about 1.74 hours before noon, that is, about 10:15 am. t=−

H( C) 35

H = 20 + 15e−0.072t

Figure 9.15

t (hours)

642

Chapter Nine /SOLUTIONS dT = −k(T − A), where A = 68 F is the temperature of the room, and t is time since 9 am. dt (b) We know that the general solution to a di˙erential equation of the form

27. (a)

dT = −k(T − 68) dt is T = Ce−kt + 68. We know that the temperature of the body is 90.3 F at 9 am. Thus, letting t = 0 correspond to 9 am, we get T = Ce−kt + 68 T (0) = 90.3 = Ce−k(0) + 68 90.3 = Ce0 + 68 C = 90.3 − 68 = 22.3 Thus T = 68 + 22.3e−kt . At t = 1, we have 89.0 = 68 + 22.3e−k 21 = 22.3e−k 21 k = − ln 22.3

.06.

Thus T = 68 + 22.3e−0.06t . We want to know when T was equal to 98.6 F, the temperature of a live body, so 98.6 = 68 + 22.3e−0.06t 30.6 ln = −0.06t 22.3 0 1 1 30.6 t= − ln 0.06 22.3 t −5.27. The victim was killed approximately 5 14 hours prior to 9 am, at 3:45 am. 28. (a) The di˙erential equation is dT = −k(T − A), dt where A = 10 F is the outside temperature. (b) We know that the general solution to a di˙erential equation of the form dT = −k(T − 10) dt is T = Ce−kt + 10. We know that initially T = 68 F. Thus, letting t = 0 correspond to 1 pm, we get T = Ce−kt + 10 68 = Ce0 + 10 C = 58. Thus T = 10 + 58e−kt . Since 10:00 pm corresponds to t = 9, 57 = 10 + 58e−9k 47 = e−9k 58 47 ln = −9k 58 1 47 .0234. k = − ln 9 58

9.5 SOLUTIONS

643

At 7:00 the next morning (t = 18) we have T

10 + 58e18(−0.0234) = 10 + 58(0.66) 48 F,

so the pipes won’t freeze. (c) We assumed that the temperature outside the house stayed constant at 10 F. This is probably incorrect because the temperature was most likely warmer during the day (between 1 pm and 10 pm) and colder after (between 10 pm and 7 am). Thus, when the temperature in the house dropped from 68 F to 57 F between 1 pm and 10 pm, the outside temperature was probably higher than 10 F, which changes our calculation of the value of the constant k. The house temperature will most certainly be lower than 48 F at 7 am, but not by much—not enough to freeze. 29. (a) We have dQ = r − Q. dt We know that the general solution to a di˙erential equation of the form dQ = k(Q − A) dt is Q = Cekt + A. Factoring out a − on the left side we get

dQ r =− Q− . dt

Thus in our case we get Q = Ce− t +

We know that at t = 0 we have Q = 0 so solving for C we get Q = Ce− t + 0 = Ce0 +

r C=− . Thus we get

r r Q = − e− t + .

So, Q∞ = lim Q = t→∞

Q = r (1 − e− t )

(b) Doubling r doubles Q∞ . Since Q∞ = r∕ , the time to reach 21 Q∞ is obtained by solving r r = (1 − e− t ) 2 1 = 1 − e− t 2 1 e− t = 2 ln(1∕2) ln 2 t=− = .

644

Chapter Nine /SOLUTIONS So altering r does not alter the time it takes to reach 21 Q∞ . See Figure 9.16. Q 2r

Q = 2r (1 − e− t ) r

Q = r (1 − e− t )

t ln 2

Figure 9.16 (c) Q∞ is halved by doubling , and so is the time, t = ln 2 , to reach 12 Q∞ . 30. We would like the equation to be in the form

dP P = kP 1 − . dt L Then is k the baseline growth rate and L is the carrying capacity at which the number of cases level o˙. We rewrite the equation by frst multiplying by P dP = 8000P − 0.2P 2 , dt

and then factoring the 8000P :

dP 0.2 = 8000P 1 − P dt 8000

Now use the fact that 8000∕0.2 = 40,000: dP = 8000P dt

0 1−

P 40,000

1 .

This tells us that the carrying capacity is 40,000. 31. This is a logistic di˙erential equation. (a) We would like the equation to be in the form P P ¨ = kP 1 − L Then k is the growth rate and L is the carrying capacity at which the number of cases level o˙. We rewrite the equation by factoring the 0.2091P : 0.0005 P . P ¨ = 0.2091P 1 − 0.2091 Now we use the fact that 0.2091∕0.0005 = 418.2: P ¨ = 0.2091P 1 −

P . 418.2

(b) The initial exponential growth rate was 20.91%. The di˙erential equation is given with time in days. Thus, the growth rate in confrmed cases is 20.91% per day. (c) The logistic equation predicts that cases will level o˙ at L; here L = 418.2. Since the number of cases must be an integer, the di˙erential equation predicts that the number of cases will level o˙ somewhere around 418. In fact the outbreak actually leveled o˙ at 443 total cases. (This is considered close!)

9.6 SOLUTIONS

645

32. This is a logistic di˙erential equation. (a) We would like the equation to be in the form P P ¨ = kP 1 − L Then k is the initial exponential growth rate and L is the carrying capacity at which the number of cases level o˙. We rewrite the equation by factoring the 0.1711P : 0.000095 P ¨ = 0.1711P 1 − P . 0.1711 Now we use the fact that 0.1711∕0.000095 = 1801.05: P ¨ = 0.1711P 1 −

P . 1801.05

(b) The initial exponential growth rate was 17.11%. The di˙erential equation is given with time in days. Thus, the growth rate in confrmed cases is 17.11% per day. (c) The logistic equation predicts that cases will level o˙ at L; here L = 1801.05. Since the number of cases must be an integer, the di˙erential equation predicts that the number of cases would level o˙ somewhere around 1800. In fact the outbreak actually leveled o˙ at 1804 total cases. 33. (a) Since P (t) is the total number of cases on or before day t, the graph of P (t) is always increasing or constant, never decreasing. Thus graphs (I) and (III) must be P (t). Alternatively, note that since the number of new Covid-19 cases on a given day can never be negative, we know that N(t) ≥ 0. Since P ¨ (t N(t), this tells us that the slopes of P (t) can never be negative, so the graph of P (t) never decreases. (b) Since P ¨ (t N(t), the peaks of N(t) correspond to the sharp upward sloping parts of P (t) and approximate constant values of N(t) correspond to segments of P (t) with approximately constant slope. Notice that the N(t) in (II) levels o˙ to approximately 200 cases per day which should correspond to a positive approximately constant slope in the graph of P (t). The graph of P (t) in (III) ends with this corresponding linear piece, so (II) and (III) represent the same location. Also, note that since N(t) in (IV) drops to near zero after reaching a peak, the corresponding P (t) must level o˙ to nearly constant after a rapid increase, as in (I). Thus, (I) and (IV) represent the same location. Side note: Graphs (I) and (IV) are for Iceland, and Graphs (II) and (III) are for Philippines. (c) The only P (t) which looks sigmoid shaped is (I). We expect this P (t) to be approximately the solution of a logistic di˙erential equation. (d) The P (t) in (I) seems to level o˙ at 1800 cases, so this is the carrying capacity. This tells us that the total number of cases in this country leveled o˙ at around 1800 in the frst wave of the pandemic.

Solutions for Section 9.6 1. (a) The x population is una˙ected by the y population—it grows exponentially no matter what the y population is, even if y = 0. If alone, the y population decreases to zero exponentially, because its equation becomes dy∕dt = −0.1y. (b) Here, interaction between the two populations helps the y population but does not e˙ect the x population. This is not a predator-prey relationship; instead, this is a one-way relationship, where the y population is helped by the existence of x’s. These equations could, for instance, model the interaction of rhinoceroses (x) and dung beetles (y). 2. (a) If alone, the x population grows exponentially, since if y = 0 we have dx∕dt = 0.01x. If alone, the y population decreases to 0 exponentially, since if x = 0 we have dy∕dt = −0.2y. (b) This is a predator-prey relationship: interaction between populations x and y decreases the x population and increases the y population. The interaction of lions and gazelles might be modeled by these equations. 3. (a) If alone, the x and y populations each grow exponentially, because the equations become dx∕dt = 0.01x and dy∕dt = 0.2y. (b) For each population, the presence of the other decreases their growth rate. The two populations are therefore competitors— they may be eating each other’s food, for instance. The interaction of gazelles and zebras might be modeled by these equations.

646

Chapter Nine /SOLUTIONS

4. (a) The species need each other to survive. Both would die out without the other, and they help each other. (b) If x = 2 and y = 1, dx = −3x + 2xy = −3(2) + 2(2)(1) < 0, dt and so population x decreases. If x = 2 and y = 1, dy = −y + 5xy = −1 + 5(2)(1) > 0, dx and so population y increases. dy −y + 5xy (c) = . dx −3x + 2xy (d) See Figure 9.17. (e) See Figure 9.17. y 4

x −4

−4

Figure 9.17 dy dx = −x − xy, = −y − xy dt dt dy dx = x − xy, 6. = y − xy dt dt dy dx 7. = −x + xy, =y dt dt 8. (a) Letting q = 400, we have 5.

1 dp = 0.03 − 0.0001(400) = −0.01. p dt Therefore, dp∕dt = −0.01p, and so letting p = 500, we get dp = −0.01(500) = −5. dt Thus, the growth rate is −5 members per year. 1 dq (b) If p = 0, we see that = −0.01, which means that population q will decrease exponentially by 1%/year in the q dt absence of population p. For dq∕dt > 0, that is, for population q to grow, population p must be at least 25. Thus, the presence of population p is benefcial to population q. 9. (I) Both companies start with about 4 million dollars, and both initially lose money. In the long run, however, Company A makes money and Company B looks like it goes out of business. (II) Initially, Company A has 2 million dollars and Company B has 4 million dollars. Both companies lose money in the beginning. Company A continues to lose money and probably goes out of business, but Company B eventually makes money and does well. (III) Company A starts with well under 1 million dollars and Company B starts with 1 million dollars. Company B makes money the whole time and does well. Company A shows a small proft for a while but then loses money and probably goes out of business. (IV) Both companies start with well under 1 million dollars. Company A makes money and does well, Company B holds steady for a time but then loses money and probably goes out of business.

9.6 SOLUTIONS

647

10. This is an example of a predator-prey relationship. Normally, we would expect the worm population, in the absence of predators, to increase without bound. As the number of worms w increases, so would the rate of increase dw∕dt; in other words, the relation dw∕dt = w might be a reasonable model for the worm population in the absence of predators. However, since there are predators (robins), dw∕dt won’t be that big. We must lessen dw∕dt. It makes sense that the more interaction there is between robins and worms, the more slowly the worms are able to increase their numbers. Hence we lessen dw∕dt by the amount wr to get dw∕dt = w − wr. The term −wr refects the fact that more interactions between the species means slower reproduction for the worms. Similarly, we would expect the robin population to decrease in the absence of worms. We’d expect the population decrease at a rate related to the current population, making dr∕dt = −r a reasonable model for the robin population in absence of worms. The negative term refects the fact that the greater the population of robins, the more quickly they are dying o˙. The wr term in dr∕dt = −r + wr refects the fact that the more interactions between robins and worms, the greater the tendency for the robins to increase in population. 11. If there are no worms, then w = 0, and dr = −r giving r = r0 e−t , where r0 is the initial robin population. If there are no dt = w giving w = w0 et , where w0 is the initial worm population. robins, then r = 0, and dw dt

dr w(r−1) 12. There is symmetry across the line r = w. Indeed, since dw = r(w−1) , if we switch w and r we get dw = r(1−w) , so w(1−r) dr dr r(1−w) = w(r−1) . Since switching w and r changes nothing, the slope feld must be symmetric across the line r = w. The slope dw

feld shows that the solution curves are either spirals or closed curves. Since there is symmetry about the line r = w, the solutions must in fact be closed curves. = −2 and dr = 2, so initially the number of worms decreases and the number of robins 13. If w = 2 and r = 2, then dw dt dt increases. In the long run, however, the populations will oscillate; they will even go back to w = 2 and r = 2. See Figure 9.18. r (robins in thousands) 3 (2500 robins) 2

w (worms in millions)

Figure 9.18 14. Sketching the trajectory through the point (2, 2) on the slope feld given shows that the maximum robin population is about 2500, and the minimum robin population is about 500. When the robin population is at its maximum, the worm population is about 1,000,000. 15.

population

Worms

❄

1.5

Robins

✠

t P0

Figure 9.19 16. It will work somewhat; the maximum number the robins reach will increase. However, the minimum number the robins reach will decrease as well. (See graph of slope feld.) In the long term, the robin-worm populations will again fall into a cycle. Notice, however, if the extra robins are added during the part of the cycle where there are the fewest robins, the new cycle will have smaller variation. See Figure 9.20.

648

Chapter Nine /SOLUTIONS Note that if too many robins are added, the minimum number may get so small the model may fail, since a small number of robins are more susceptible to disaster. r (robins in thousands) 3

✛ ✛

New trajectory Old trajectory

w (worms in millions)

Figure 9.20 17. The numbers of robins begins to increase while the number of worms remains approximately constant. See Figure 9.21. The numbers of robins and worms oscillate periodically between 0.2 and 3, with the robin population lagging behind the worm population. r

(3, 1)

w 1

Figure 9.21 18. (a) Substituting w = 2.2 and r = 1 into the di˙erential equations gives dw = 2.2 − (2.2)(1) = 0 dt dr = −1 + 1(2.2) = 1.2. dt (b) Since the rate of change of w with time is 0, At t = 0.1, we estimate w = 2.2 Since the rate of change of r is 1.2 thousand robins per unit time, At t = 0.1, we estimate r = 1 + 1.2(0.1) = 1.12 ≈ 1.1. (c) We must recompute the derivatives. At t = 0.1, we have dw = 2.2 − 2.2(1.12) = −0.264 dt dr = −1.12 + 1.12(2.2) = 1.344. dt

9.6 SOLUTIONS

649

Then at t = 0.2, we estimate w = 2.2 − 0.264(0.1) = 2.1736 ≈ 2.2 r = 1.12 + 1.344(0.1) = 1.2544 ≈ 1.3 Recomputing the derivatives at t = 0.2 gives dw = 2.1736 − 2.1736(1.2544) = −0.553 dt dr = −1.2544 + 1.2544(2.1736) = 1.472 dt Then at t = 0.3, we estimate w = 2.1736 − 0.553(0.1) = 2.1183 ≈ 2.1 r = 1.2544 + 1.472(0.1) = 1.4016 ≈ 1.4. 19. (a) See Figure 9.22. r (predator) 4 3 2 1

w (prey)

Figure 9.22 (b) The point moves in a counterclockwise direction. If w = 3 and r = 2, we have dw = w − wr = 3 − (3)(2) = 3 − 6 < 0 dt dr = −r + wr = −2 + (3)(2) = −2 + 6 > 0. dt So w is decreasing and r is increasing. The point moves up and to the left (counterclockwise). (c) We see in the trajectory that r achieves a maximum value of about 3.5, so the population of robins goes as high as 3.5 thousand robins. At this time, the worm population is at about 1 million. (d) The worm population goes as high as 3.5 million worms. At this time, the robin population is about 1 thousand. 20. (a) See Figure 9.23.

r (predator) 4 3 2 1

Figure 9.23

w (prey)

650

Chapter Nine /SOLUTIONS (b) If w = 0.5 and r = 3, we have dw = w − wr = 0.5 − (0.5)(3) < 0 dt dr = −r + wr = −3 + (0.5)(3) < 0 dt Both w and r are decreasing, so the point is moving down and to the left (counterclockwise). (c) The robin population goes up to about 3.3 thousand robins. At this time, the worm population is about 1 million. (d) The worm population goes up to about 3.3 million worm. At this point, the robin population is about 1 thousand.

21. (a) Both x and y decrease, since dx = 0.2x − 0.5xy = 0.2(2) − 0.5(2)(2) < 0, dt dy = 0.6y − 0.8xy = 0.6(2) − 0.8(2)(2) < 0. dt (b) Population x increases and population y decreases, since dx = −2x + 5xy = −2(2) + 5(2)(2) > 0, dt dy = −y + 0.2xy = −2 + 0.2(2)(2) < 0. dt (c) Both x and y increase, since dx = 0.5x = 0.5(2) > 0, dt dy = −1.6y + 2xy = −1.6(2) + 2(2)(2) > 0 dt (d) Population x decreases and population y increases, since dx = 0.3x − 1.2xy = 0.3(2) − 1.2(2)(2) < 0 dt dy = −0.7x + 2.5xy = 0.7(2) + 2.5(2)(2) > 0 dt 22. Here x and y both increase at about the same rate. 23. Initially x = 0, so we start with only y. Then y decreases while x increases. Then x continues to increase while y starts to increase as well. Finally y continues to increase while x decreases. 24. x decreases quickly while y increases more slowly. 25. The closed trajectory represents populations which oscillate repeatedly. 26. (a) The rate dQ1 ∕dt is the sum of three terms that represent the three changes in Q1 : dQ1 = A − k1 Q1 + k2 Q2 . dt The term A is the rate at which Q1 increases due to creation of new toxins. The term −k1 Q1 is the rate at which Q1 decreases due to fow of toxins into the blood. The constant k1 is a positive constant of proportionality. The term k2 Q2 is the rate at which Q1 increases due to fow of toxins out of the blood. The constant k2 is a second positive constant. (b) The rate dQ2 ∕dt is the sum of three terms that represent the three changes in Q2 : dQ2 = −k3 Q2 + k1 Q1 − k2 Q2 . dt The term −k3 Q2 is the rate at which Q2 decreases due to removal of toxins by dialysis. The constant k3 is a positive constant of proportionality. The term k1 Q1 is the rate at which Q2 increases due to fow of toxins into the blood. The constant k1 is the same positive constant as in part (a). The term −k2 Q2 is the rate at which Q2 decreases due to fow of toxins out of the blood. The constant k2 is the same positive constant as in part (a).

9.7 SOLUTIONS

651

dy 0.6y − 0.8xy = ; See Figure 9.24. dt 0.2x − 0.5xy dy −y + 0.2xy (b) = ; See Figure 9.25. dx −2x + 5xy dy −1.6y + 2xy (c) = ; See Figure 9.26. dx 0.5x dy −0.7y + 2.5xy (d) = ; See Figure 9.27. dx 0.3x − 1.2xy

27. (a)

1 x 1

dy 0.6y−0.8xy Figure 9.24: dx = 0.2x−0.5xy

dy Figure 9.25: dx = −y+0.2xy −2x+5xy

1 x 1

dy Figure 9.26: dx = −1.6y+2xy 0.5x

dy Figure 9.27: dx = −0.7y+2.5xy 0.3x−1.2xy

Solutions for Section 9.7 1. Susceptible people are infected at a rate proportional to the product of S and I. As susceptible people become infected, S decreases at a rate of aSI and (since these same people are now infected) I increases at the same rate. At the same time, infected people are recovering at a rate proportional to the number infected, so I is decreasing at a rate of bI. 2. Since dS = −aSI, dt dI = aSI − bI, dt dR = bI dt we have

dS dI dR + + = −aSI + aSI − bI + bI = 0. dt dt dt d Thus dt (S + I + R) = 0, so S + I + R = constant.

Chapter Nine /SOLUTIONS

652

3. We know that

dS = −aSI, dt

−40 = −a ⋅ 1000 ⋅ 200, which gives −40 = 0.0002. −1000 ⋅ 200 This means that for the new values of I and S, we have a=

dS = −aSI dt = −0.0002 ⋅ 900 ⋅ 250 = −45. Thus, dS∕dt = −45. 4. (a) Setting dI∕dt = 0, dI = aSI − bI = (aS − b)I = 0 dt aS − b = 0 b S= . a (b) For S > b∕a, we know aS − b > 0, so dI = (aS − b)I > 0. dt Thus, for S > b∕a, we know I is increasing. Similarly, for S < b∕a, we can show I is decreasing. (c) Since I increases initially if S0 > b∕a, we have an epidemic if S0 > b∕a. If S0 < b∕a, there is no epidemic as I decreases initially. Thus, the threshold value is b∕a. 5. The epidemic is over when the number of infected people is zero, that is, at the horizontal intercept of the trajectory. This intercept is a very small S value. When the epidemic is over, there are almost no susceptibles left— that is, almost everyone has become infected. 6. (a) The initial values are I0 = 1, S0 = 149. (b) Substituting for I0 and S0 , initially we have dI = 0.0026SI − 0.5I = 0.0026(149)(1) − 0.5(1) < 0. dt So, I is decreasing. The number of infected people goes down from 1 to 0. The disease does not spread. 7. (a) I0 = 1, S0 = 349 dI = 0.0026SI − 0.5I = 0.0026(349)(1) − 0.5(1) > 0, so I is increasing. The number of infected people will (b) Since dt increase, and the disease will spread. This is an epidemic. 8. (a) See Figure 9.28. I (infecteds) 400 200

192

400

800

S (susceptibles)

Figure 9.28 (b) I is at a maximum when S = 192. 9. The maximum value of I is approximately 300 boys. This represents the maximum number of infected boys who have not days. (yet) been removed from circulation. It occurs at about t

9.7 SOLUTIONS

653

10. In the system dS = −aSI dt dI = aSI − bI dt

the constant a represents how infectious the disease is; the larger a, the more infectious. The constant b represents 1∕(number of days before remova Thus, the larger b is, the quicker the infecteds are removed. For the fu example, a = 0.0026 and b = 0.5. (I) Since 0.0026 < 0.04, this is more infectious. Since 0.2 < 0.5, infecteds are being removed more slowly. So system (I) corresponds to (a). (II) Since 0.002 < 0.0026, this is less infectious. Since 0.3 < 0.5, infecteds are being removed more slowly. This corresponds to (c). (III) Since the second equation has no −bI term, we have b = 0. The infecteds are never removed. This corresponds to (e). A system of equations corresponding to (b) is dS = −0.04SI dt dI = 0.04SI − 0.7I. dt A system of equations corresponding to (d) is dS = −0.002SI dt dI = 0.002SI − 0.7I. dt 11. The threshold value of S is the value at which I is a maximum. When I is a maximum, dI = 0.04SI − 0.2I = 0, dt so S = 0.2∕0.04 = 5. 12. Since the threshold value of S is given by dI = 0.002SI − 0.3I = 0, dt we have

0.3 = 150. 0.002 So, if S0 = 100, the disease does not spread initially. If S0 = 200, the disease does spread initially. S=

13. (a) The SIR variable S represents the number of people who have not got the disease and could still get it. Therefore S = (80% of total Population) − Confrmed S = 0.8 ⋅ (38,048,738) − 1,234,912 = 29,204,078.4 ≈ 29,204,078 (b) The SIR variable R represents the number of people who can not spread the disease now or in the future. These people have gained immunity, by having the disease or by vaccination, or they have died. We assume that all those who recover actually have gained immunity. Therefore R = Recovered + Deaths R = 1,131,494 + 24,245 = 1,155,739.

654

Chapter Nine /SOLUTIONS (c) The SIR variable I represents the number of people who are currently infected and are spreading the disease. They are the confrmed cases who have not gained immunity or died. Therefore I = Confrmed − Recovered − Deaths I = 1,234,912 − 1,131,494 − 24,245 = 79,173.

14. (a) The terms representing the spread of the disease are −aSI in the equation for S ¨ and aSI in the equation for I ¨ . The −aSI in the equation for S ¨ is negative because as people get sick they leave the susceptible group, making S ¨ negative. The aSI in the equation for I ¨ is positive because the same people move into the infected group, contributing positively to I ¨ . (b) The terms representing the infected people recovering or being removed are −bI in the equation for I ¨ and bI in the equation for R¨ . The −bI in the equation for I ¨ is negative because people are leaving the infected group. The bI in the equation for R¨ is positive because the same people move into the recovered group. (c) Reduces a because the disease is transmitted less easily. (d) Increases b because the disease lasts for a shorter time, so people recover faster. 15. (a) I0 = 1, so S0 = 45,000 − 1 = 44,999. (b) If S0 is greater than the threshold value, we expect an epidemic. b 9.865 = = 36,948 people. a 0.000267 Since the camp contained 45,000 soldiers, more than the threshold, an epidemic is predicted. (c) Using the chain rule, and the values of a and b, we have Threshold =

dI∕dt 36,948 dI aSI − bI b = = = −1 + = −1 + . dS dS∕dt −aSI aS S Figure 9.29 shows the slope feld for this di˙erential equation and the solution curve for these initial conditions. The total number of soldiers who did not get the disease is the S-intercept, or about 30,000. Thus, approximately 15,000 soldiers were infected. (d) To solve the di˙erential equation, we integrate 36,948 dI = −1 + dS S giving, since S > 0, I = −S + 36,948 ln S + C. To fnd C, substitute I0 = 1, S0 = 44,999, so 1 = −44,999 + 36,948 ln(44,999) + C C = 45,000 − 36,948 ln(44,999), thus I = −S + 36,948 ln S + 45,000 − 36,948 ln(44,999) 0 1 S I = −S + 36,948 ln + 45,000. 44,999 We fnd the S-intercept, giving the number of people una˙ected, by setting I = 0. Then 0 1 S 0 = −S + 36,948 ln + 45,000. 44, 999 This equation cannot be solved algebraically. However, numerical methods, or tracing along a graph, gives S ≈ 30, 000. Thus, about 15,000 soldiers were infected.

SOLUTIONS to Review Problems For Chapter Nine

655

500

−

∨

20,000

40,000

(S0 , I0 ) ↙ ∙ S

Figure 9.29

Solutions for Chapter 9 Review 1. (a) (III) An island can only sustain the population up to a certain size. The population will grow until it reaches this limiting value. (b) (V) The ingot will get hot and then cool o˙, so the temperature will increase and then decrease. (c) (I) The speed of the car is constant, and then decreases linearly when the breaks are applied uniformly. (d) (II) Carbon-14 decays exponentially. (e) (IV) Tree pollen is seasonal, and therefore cyclical. dy 2. (a) (i) If y = Cx2 , then = C(2x) = 2Cx. We have dx x

dy = x(2Cx) = 2Cx2 dx

and 3y = 3(Cx2 ) = 3Cx2

dy ≠ 3y, this is not a solution. dx dy = C(3x2 ) = 3Cx2 . We have (ii) If y = Cx3 , then dx Since x

dy = x(3Cx2 ) = 3Cx3 , dx

and 3y = 3Cx3 . dy = 3y, and y = Cx3 is a solution. dx dy = 3x2 . We have (iii) If y = x3 + C, then dx Thus x

dy = x(3x2 ) = 3x3 dx

and 3y = 3(x3 + C) = 3x3 + 3C.

dy ≠ 3y, this is not a solution. dx (b) The solution is y = Cx3 . If y = 40 when x = 2, we have Since x

40 = C(23 ) 40 = C ⋅ 8 C = 5.

656

Chapter Nine /SOLUTIONS

3. Since y = x3 , we know that y¨ = 3x2 . Substituting y = x3 and y¨ = 3x2 into the di˙erential equation we get Left-side = xy¨ − 3y = x(3x2 ) − 3(x3 ) = 3x3 − 3x3 = 0. Since the left and right sides are equal for all x, we see that y = x3 is a solution. 4. When y = 125, the rate of change of y is dy = −0.20y = −0.20(125) = −25. dt The value of y goes down by 25 as t goes up by 1, so when t = 1, we have y = Old value of y + Change in y = 125 + (−25) = 100. Continuing in this way , we fll in the table as shown:

5. (a) Slope feld I corresponds to (b) See Figures 9.30 and 9.31.

125

100

51.2

dy dy = 1 + y and slope feld II corresponds to = 1 + x. dx dx

x −2

−1

−2

−1

−2

dy Figure 9.30: dx = 1+y

dy Figure 9.31: dx = 1+x

dy (c) Slope feld I has an equilibrium solution at y = −1, since = 0 at y = −1. We see in the slope feld that this dx equilibrium solution is unstable. Slope feld II does not have any equilibrium solutions. 6. (a) See Figure 9.32. y 2 1 x −2

−1

−1 −2

Figure 9.32 (b) The point (1, 0) satisfes the equation y = x − 1. If y = x − 1, then y¨ = 1 and x − y = x − (x − 1) = 1, so y = x − 1 is the solution to the di˙erential equation through (1, 0).

SOLUTIONS to Review Problems For Chapter Nine

657

7. III. The slope feld appears to be constant for a fxed value of y, regardless of the value of x. This feature says that y¨ does not depend on x, ruling out the formulas y¨ = 1 + x and y¨ = xy. The di˙erential equation y¨ = 1 + y would have a slope feld with zero slope at y = −1 and nowhere else, but the given slope feld has two areas of zero slope, so y¨ = 1 + y is ruled out and so is y¨ = 2 − y for the same reason. This leaves y¨ = (1 + y)(2 − y) as the correct answer, which fts the slope feld as it has zero slopes at y = 2 and y = −1, positive slopes for −1 < y < 2 and negative slopes for y < −1 and y > 2. 8. Figure (I) shows a line segment at (4, 0) with positive slope. The only possible di˙erential equation is (b), since at (4, 0) we have y¨ = cos 0 = 1. Note that (a) is not possible as y¨ (4, 0) = e−16 = 0.0000001, a much smaller positive slope than that shown. Figure (II) shows a line segment at (0, 4) with zero slope. The possible di˙erential equations are (d), since at (0, 4) we have y¨ = 4(4 − 4) = 0, and (f), since at (0, 4) we have y¨ = 0(3 − 0) = 0. Figure (III) shows a line segment at (4, 0) with negative slope of large magnitude. The only possible di˙erential equation is (f), since at (4, 0) we have y¨ = 4(3 − 4) = −4. Note that (c) is not possible as y¨ (4, 0) = cos(4 − 0) = −0.65, a negative slope of smaller magnitude than that shown. Figure (IV) shows a line segment at (4, 0) with a negative slope of small magnitude. The only possible di˙erential equation is (c), since at (4, 0) we have y¨ = cos(4 − 0) = −0.65. Note that (f) is not possible as y¨ (4, 0) = 4(3 − 4) = −4, a negative slope of larger magnitude than that shown. Figure (V) shows a line segment at (0, 4) with positive slope. Possible di˙erential equations are (a), since at (0, 4) we 2 have y¨ = e0 = 1, and (c), since at (0, 4) we have y¨ = cos(4 − 4) = 1. Figure (VI) shows a line segment at (0, 4) with a negative slope of large magnitude. The only possible di˙erential equation is (e), since at (0, 4) we have y¨ = 4(3 − 4) = −4. Note that (b) is not possible as y¨ (0, 4) = cos 4 = −0.65, a negative slope of smaller magnitude than that shown. 9. (a) We know that the balance, B, increases at a rate proportional to the current balance. Since interest is being earned at a rate of 2% compounded continuously we have Rate at which interest is earned = 2% (Current balance) or in other words, if t is time in years, dB = 2%(B) = 0.02B. dt (b) The equation is in the form dB = kB dt so we know that the general solution will be B = B0 ekt where B0 is the value of B when t = 0, i.e., the initial balance. In our case we have k = 0.02 so we get B = B0 e0.02t . (c) We are told that the initial balance, B0 , is $15,000 so we get B = 15,000e0.02t . (d) Substituting the value t = 10 into our formula for B we get B = 15,000e0.02t B(10) = 15,000e0.02(10) = 15,000e0.2 B(10) ≈ $18,321.04 10. If we let Q represent the amount of radioactive iodine present at time t, with t measured in days, then we have dQ = −0.09Q. dt The −0.09 is negative because the quantity of iodine is decreasing. The solution to this di˙erential equation is Q = Ce−0.09t , for some constant C.

Chapter Nine /SOLUTIONS

658

11. Integrating both sides gives P =

1 2 t + C, 2

where C is some constant. 12. The general solution is y = Ce5t . 13. Integrating both sides we get y=

5 2 t + C, 2

where C is a constant. 14. We know that the general solution to an equation of the form dP = kP dt is P = Cekt . Thus in our case the solution is P = Ce0.03t , where C is some constant. 15. For some constant C, the general solution is A = Ce−0.07t . 16. Multiplying both sides by Q gives

dQ = 2Q, where Q ≠ 0. dt We know that the general solution to an equation of the form dQ = kQ dt is Q = Cekt . Thus in our case the solution is Q = Ce2t ,

where C is some constant, C ≠ 0.

17. We know that the general solution to the di˙erential equation dP = k(P − A) dt is P = Cekt + A. Thus in our case we factor out −2 to get 10 dP = −2(P − 5). = −2 P + −2 dt Thus the general solution to our di˙erential equation is P = Ce−2t + 5, where C is some constant. dy 18. Since = −(y − 100), the general solution is y = 100 + Ce−t . dt

SOLUTIONS to Review Problems For Chapter Nine

659

19. We know that the general solution to the di˙erential equation dy = k(y − A) dx is y = Cekx + A. Thus in our case we factor out 0.2 to get dy 8 = 0.2 y − = 0.2(y − 40). dx 0.2 Thus the general solution to our di˙erential equation is y = Ce0.2x + 40, where C is some constant. 20. We know that the general solution to the di˙erential equation dH = k(H − A) dt is H = Cekt + A. Thus in our case we factor out 0.5 to get dH 10 = 0.5 H + = 0.5(H − (−20)). dt 0.5 Thus the general solution to our di˙erential equation is H = Ce0.5t − 20, where C is some constant. 21. We fnd the temperature of the orange juice as a function of time. Newton’s Law of Heating says that the rate of change of the temperature is proportional to the temperature di˙erence. If S is the temperature of the juice, this gives us the equation dS = −k(S − 65) dt

for some constant k.

Notice that the temperature of the juice is increasing, so the quantity dS∕dt is positive. In addition, S = 40 initially, making the quantity (S − 65) negative. We know that the general solution to a di˙erential equation of the form dS = −k(S − 65) dt is S = Ce−kt + 65. Since at t = 0, S = 40, we have 40 = 65 + C, so C = −25. Thus, S = 65 − 25e−kt for some positive constant k. See Figure 9.33 for the graph. S ( F) 65 40 t −kt

Figure 9.33: Graph of S = 65 − 25e for k > 0

660

Chapter Nine /SOLUTIONS

22. (a) Since the growth rate of the tumor is proportional to its size, we should have dS = kS. dt (b) We can solve this di˙erential equation by separating variables and then integrating: dS = k dt Ê S Ê ln |S| = kt + B S = Cekt . (c) This information is enough to allow us to solve for C: 5 = Ce0k C = 5. (d) Knowing that C = 5, this second piece of information allows us to solve for k: 8 = 5e3k 1 8 k = ln 3 5

.1567.

So the tumor’s size is given by S = 5e0.1567t . 23. (a) If C ¨ = −kC, then C = C0 e−kt . Since the half-life is 5730 years, 1 C = C0 e−5730k . 2 0 Solving for k, we have −5730k = ln

1 2

ln(1∕2) .000121. 5730 −kt (b) From the given information, we have 0.91 = e , where t is the age of the shroud in 1988. Solving for t, we have k=−

t=−

ln 0.91 = 779.4 years. k

24. (a) Use the fact that Rate balance changing

Rate interest accrued

−

Rate payments made

Thus, dB = 0.05B − 12,000. dt (b) We know that the general solution to a di˙erential equation of the form dB = k(B − A) dt is B = Cekt + A. Factoring out a 0.05 on the left side we get 0 1 12,000 dB = 0.05 B − = 0.05(B − 240,000). dt 0.05 Thus, in our case we get B = Ce0.05t + 240,000.

SOLUTIONS to Review Problems For Chapter Nine

661

We know that the initial balance is B0 , thus we get B0 = Ce0 + 240,000 C = B0 − 240,000. Thus, we get B = (B0 − 240,000)e0.05t + 240,000. (c) To fnd the initial investment such that the account has a 0 balance after 20 years, we solve 0 = (B0 − 240,000)e(0.05)20 + 240,000 = (B0 − 240,000)e1 + 240,000, B0 = 240,000 − 25. (a) Q0

240,000 e

$151,708.93.

1 Q 2 0

Q = Q0 e−0.0187t t 37

dQ = −kQ dt (c) Since 25% = 1∕4, it takes two half-lives = 74 hours for the drug level to be reduced to 25%. Alternatively, Q = Q0 e−kt and 12 = e−k(37) , we have ln(1∕2) k=− .0187. 37 Therefore Q = Q0 e−0.0187t . We know that when the drug level is 25% of the original level that Q = 0.25Q0 . Setting these equal, we get 0.25 = e−0.0187t .

(b)

giving t=−

ln(0.25) 0.0187

74 hours

3 days.

26. Let D(t) be the quantity of dead leaves, in grams per square centimeter. Then dD = 3 − 0.75D, where t is in years. We dt know that the general solution to a di˙erential equation of the form dD = k(D − B) dt is D = Aekt + B. Factoring out a −0.75 on the left side we get dD −3 = −0.75 D − = −0.75(D − 4). dt −0.75 Thus in our case we get D = Ae−0.75t + 4. If initially the ground is clear, the solution looks like the following graph: D 4

The equilibrium level is 4 grams per square centimeter, regardless of the initial condition.

662

Chapter Nine /SOLUTIONS

27. (a) We know that the rate at which morphine leaves the body is proportional to the amount of morphine in the body at that particular instant. If we let Q be the amount of morphine in the body, we get that Rate of morphine leaving the body = kQ, where k is the rate of proportionality. The solution is Q = Q0 ekt (neglecting the continuously incoming morphine). Since the half-life is 2 hours, we have 1 Q = Q0 ek⋅2 , 2 0 so ln(1∕2) k= = −0.347. 2 (b) Since Rate of change of quantity = Rate in − Rate out, we have

dQ = −0.347Q + 2.5. dt (c) Equilibrium occurs when dQ∕dt = 0, that is, when 0.347Q = 2.5 or Q = 7.2 mg. 28. (a) We know that the equilibrium solutions are those functions which satisfy the di˙erential equation and whose derivative is everywhere 0. Thus we must solve dy dx = 0.5y(y − 4)(2 + y)

Thus the equilibrium solutions are y = 0, y = 4, and y = −2. (b) The slope feld of the di˙erential equation is shown in Figure 9.34. An equilibrium solution is stable if a small change in the initial conditions gives a solution which tends toward the equilibrium as the independent variable tends to positive infnity. Looking at Figure 9.34 we see that the only stable solution is y = 0. y 7 5 3 1 x −5

−3

−1

−3 −5

Figure 9.34

29. (a) Since the rate of change of the weight is given by dW 1 = (Intake − Amount to maintain weight) dt 3500 we have

dW 1 = (I − 20W ). dt 3500 (b) To fnd the equilibrium, we solve dW ∕dt = 0, or 1 (I − 20W ) = 0. 3500

SOLUTIONS to Review Problems For Chapter Nine

663

Solving for W , we get W = I∕20. This means that if an athletic adult male weighing I∕20 pounds has a constant caloric intake of I Calories per day, his weight remains constant. We expect the equilibrium to be stable because an athletic adult male slightly over the equilibrium weight loses weight because his caloric intake is too low to maintain the higher weight. Similarly, an adult male slightly under the equilibrium weight gains weight because his caloric intake is higher than required to maintain his weight. (c) We know that the general solution to a di˙erential equation of the form dW = k(W − A) dt is W = Cekt + A. Factoring out a −20 on the left side we get −20 dW −I 2 I = W − =− W − . dt 3500 −20 350 20 Thus in our case we get I . 20 Let us call the person’s initial weight W0 at t = 0. Then W0 = I∕20 + Ce0 , so C = W0 − I∕20. Thus I I W = + W0 − e−(1∕175)t . 20 20 2

W = Ce− 350 t +

(d) Using part (c), we have W = 150 + 10e−(1∕175)t . This means that W → 150 as t → ∞. See Figure 9.35. W 160 150

t 100 days

Figure 9.35 30. (a) The graph of f (y) = y − y2 is shown Figure 9.36. (b) The slope feld dy∕dx = y − y2 is shown in Figure 9.37. y

1 x f (y)

Figure 9.36

−1

Figure 9.37

(c) The slopes in Figure 9.37 are positive for 0 < y < 1, where f (y) is above the horizontal axis in Figure 9.36. The slopes in Figure 9.37 are negative for y < 0 and y > 1, where f (y) is negative. The equilibrium solutions are y = 0 and y = 1, where the graph of f (y) crosses the horizontal axis. The equilibrium y = 1 is stable, which can be seen in Figure 9.36 from the fact that f (y) is positive for y < 1 and negative for y > 1. Thus, the solution curves increase toward y = 1 from below and decrease toward y = 1 from above. The equilibrium at y = 0 is unstable. Figure 9.36 shows that f (y) is negative for y < 0 and positive for y > 0. Thus, solution curves below y = 0 decrease away from y = 0; solution curves above y = 0 increase away from y = 0.

664

Chapter Nine /SOLUTIONS

31. (a) The equilibrium solutions are y = 1, y = 8, y = 16. For these values, f (y) = 0, so y¨ = 0. (b) See Figure 9.38. Since f (y) > 0 for 0 < y < 1, the slopes are upward for these y values; similarly for 8 < y < 16 and y > 16. For 1 < y < 8, the slopes are downward, since f (y) < 0 for these y values. (c) See Figure 9.38. The equilibrium solutions for the initial conditions y(0) = 1, y(0) = 8, y(0) = 16 are horizontal lines. The other solution curves follow the slope feld. (d) The equilibrium y = 1 is stable; y = 8 and y = 16 are unstable. y

y(0) = 17

y(0) = 16 (equilibrium) y(0) = 10

y(0) = 8 (equilibrium) y(0) = 6

y(0) = 1 (equilibrium) x

✻

y(0) = 0

Figure 9.38

STRENGTHEN YOUR UNDERSTANDING 1. True, since dQ∕dt represents the rate of change of Q. 2. False. Since the population is decreasing by 5% (rather than 5 units of P ), the correct di˙erential equation is dP ∕dt = −0.05P . 3. False. Since the population is decreasing, the rate of change must be negative. The correct di˙erential equation is dP ∕dt = −0.05P . 4. True. The balance increases at a rate of 3% of B, or 0.03B, each year so the rate of change of B is 0.03B. 5. True, since when two quantities are proportional, one is a constant times the other. 6. False. The 200 is not a percent, but in units of P over units of t, so the correct di˙erential equation is dP ∕dt = −200. 7. True. The balance is increasing due to the interest at a rate of 0.05B and is decreasing due to the payments at a rate of 8000. 8. False, the correct di˙erential equation is dQ∕dt = 200 − kQ (where k > 0). 9. True. The rate the drug is entering the body is 12 mg per hour and the rate the drug is leaving the body is 0.063Q mg per hour. 10. False. The deposit of $10,000 was a one-time deposit, and not a rate of change. This di˙erential equation would be correct if the deposits were being made at a continuous rate of $10,000 per year. 11. True, since the derivative of P is zero, so dP ∕dt = 0 and also substituting P = 10 in the expression 3P (10 − P ) gives zero, so substituting P = 10 on both sides of the di˙erential equation gives 0, and we have 0 = 0, a true equation. 12. True, since substituting P = 0 on both sides of the di˙erential equation gives 0. 13. False, since substituting P = 5 on the left-hand side of the di˙erential equation gives 0, but on the right-hand side gives 75. 14. True. When we substitute t = 0 and y = 40 in the general solution, we have 40 = Ce0.05(0) . Since e0 = 1, this gives C = 40. 15. False. When we substitute t = 0 and y = 40 in the general solution, we have 40 = 25 + Ce0.05(0) . Since e0 = 1, this gives 40 = 25 + C, which implies C = 15.

STRENGTHEN YOUR UNDERSTANDING

665

16. True. When we substitute t = 0 and y = 40 in the general solution, we have 40 = 25 + Ce0.05(0) . Since e0 = 1, this gives 40 = 25 + C, which implies C = 15. 17. False. Since y¨ = 0.2y, when y = 100 we have y¨ = 0.2 ⋅ 100 = 20. The variable y is changing at a rate of 20 units per unit of time. This tells us that y increases approximately 20 units between t = 0 and t = 1, so we expect y(1) ≈ 100 + 20 = 120. 18. True. Since y¨ = 0.2y, when y = 100 we have y¨ = 0.2 ⋅ 100 = 20. The variable y is changing at a rate of 20 units per unit of time. This tells us that y increases approximately 20 units between t = 0 and t = 1, so we expect y(1) ≈ 100 + 20 = 120. 19. True, since when Q = 10 we have dQ∕dt = 5 ⋅ 10 − 200 = −150 < 0. 20. True, since when Q = 40 both sides of the di˙erential equation give 0. 21. False, since when x = 3, we have dy∕dx = 2 ⋅ 3 = 6. 22. True, since when x = 3, we have dy∕dx = 2 ⋅ 3 = 6. 23. True, since when x = 1 and y = −2, we have dy∕dx = 3 ⋅ 1 ⋅ (−2) = −6. 24. False, since when x = 2 and y = 2, we have dy∕dx = 3 ⋅ 2 ⋅ 2 = 12. 25. True, since when x = 3 and y = 2, we have dy∕dx = 3 ⋅ 3 ⋅ 2 = 18. 26. False, since whenever x < 0 we have dy∕dx < 0 regardless of the sign of y. 27. True, since when y = 1, dy∕dx = 2 ⋅ 1 = 2, regardless of the value of x. 28. True, since when x = 3, dy∕dx = 5 ⋅ 0 ⋅ (y − 2) = 0. 29. False, since when P > 3 we have 12 − 4P < 0 so dP ∕dt < 0. 30. True, since the slope feld lines have slope dy∕dx which is the derivative of a solution y = f (x). 31. False; the general solution is y = Cekt . 32. True. 33. False. That solution would be the solution to the di˙erential equation dw∕dr = 0.3w. 34. True. The general solution is H = Ce0.5t and the particular solution H = 57e0.5t satisfes the initial condition H(0) = 57. 35. True, since the general solution is y = Ce3t and setting t = 0, y = 5 gives C = 5. 36. False, since y = Ce−2t is a general solution. The particular solution is y = 3e−2t . 37. False, the correct di˙erential equation is dB∕dt = 0.03B. 38. False, since if k were negative, −k would be positive, and Q(t) would grow exponentially, rather than decay. 39. False. The function Q = Cekt is the solution not the di˙erential equation. The di˙erential equation is dQ∕dt = kQ. 40. True. 41. True, as explained in the text. 42. False. We need to factor out a 2 to write the di˙erential equation in the form dP ∕dt = 2(P − 50). Then the general solution is P = 50 + Ce2t . 43. False. The function Q = 20 + Ce0.5t is a solution to the di˙erential equation dQ∕dt = 0.5(Q − 20). 00 + W ) = −3(W − 200), 44. True. We factor out the coeÿcient of W to rewrite the di˙erential equation as dW ∕dt = −3( 6−3 which has the solution shown.

45. False. The initial condition gives C = 10 so the correct solution is A = 40 + 10e0.25t . 46. True. 47. False; the correct di˙erential equation is dB∕dt = 0.04B − 12,000. The equation given has no derivatives so it cannot be a di˙erential equation. 48. True, since the drug is entering the body at a rate of 12 and leaving the body at a rate of 0.18A. 49. True, since setting dH∕dt = 0 gives H = 225. 50. True. 51. False. If X would do fne if Y did not exist, then the parameter a must be positive. 52. False. If species X eats species Y , then the parameter d must be negative. 53. True. Since X has a negative impact on Y , the coeÿcient of the interaction term must be negative.

666

Chapter Nine /SOLUTIONS

54. True. Since X would die out alone, the parameter a must be negative. Since Y helps X, the parameter b of the interaction term must be positive. 55. False. If Y is absent then the population of X grows at a continuous rate of 2%, and if X is absent the population of Y grows at a continuous rate of 5% 56. True, since the coeÿcients 0.02 and 0.05 are positive. 57. True, since the coeÿcients −0.15 and −0.18 of the interaction terms are negative. 58. True. Since the coeÿcients −0.12 and −0.10 are negative, both populations would die out in the absence of the other. Since the coeÿcients 0.07 and 0.25 of the interaction terms are positive, both populations are helped by the other. 59. True. Substituting the values for x and y into the di˙erential equations gives dx∕dt = −3(1) + (1)(2) = −3 + 2 = −1 < 0 and dy∕dt = 2(2) − 5(1)(2) = 4 − 10 = −6 < 0. Since the derivatives of x and y are both negative, both populations are decreasing. 60. False. Substituting the values for x and y into the di˙erential equations gives dx∕dt = −3(2) + (2)(5) = −6 + 10 = 4 > 0 and dy∕dt = 2(5) − 5(2)(5) = 10 − 50 = −40 < 0. Since the derivative of y is negative, the y population is decreasing. However, the derivative of x is positive, so the x population is increasing. 61. False. It is negative because people in the susceptible group are becoming sick and moving to the group of infected people. 62. True. People in the susceptible group are becoming sick and moving to the group of infected people. 63. False, since the negative of this quantity does not appear in the expression for dS∕dt. 64. True. 65. False. The parameter a will be larger for Type I fu. 66. True. The parameter a measures how infectious or contagious the disease is. 67. False. The parameter b will be smaller for Type I fu, since a smaller percentage recover per unit time. 68. True. The parameter b will be smaller for Type I fu, since a smaller percentage recover per unit time. 69. True. To see if I will increase, we see if dI∕dt is positive. We have dI∕dt = 0.001SI − 0.3I = 0.001(500)(100) − 0.3(100) = 50 − 30 = 20 > 0. 70. False. To see if I will increase, we see if dI∕dt is positive. We have dI∕dt = 0.001SI − 0.3I = 0.001(100)(500) − 0.3(500) = 50 − 150 = −100 < 0.

PROJECTS FOR CHAPTER NINE 1. (a) Equilibrium values are N = 0 (unstable) and N = 200 (stable). The graphs are shown in Figures 9.39 and 9.40. dN dt

N 220 200

100

200

300

40 −300

t Figure 9.39: dN∕dt = 2N − 0.01N 2

Figure 9.40: Solutions to dN∕dt = 2N − 0.01N 2

(b) When there is no fshing the rate of population change is given by dN = 2N −0.01N 2. If fshermen remove dt , by 75 fsh/year. This is fsh at a rate of 75 fsh/year, then this results in a decrease in the growth rate, dP dt refected in the di˙erential equation by including the −75.

PROJECTS FOR CHAPTER NINE (c)

dP dt

(d)

25 50 100 150 200

300

667

170 150

60 50 40 −375

Figure 9.41: dP ∕dt = 2P − 0.01P 2 − 75

Figure 9.42

(e) In Figure 9.41, we see that dP ∕dt = 0 when P = 50 and when P = 150, that dP ∕dt is positive when P is between 50 and 150, and that dP ∕dt is negative when P is less than 50 or greater than 150. (i) Since dP ∕dt = 0 at P = 50 and at P = 150, these are the two equilibrium values. (ii) Since dP ∕dt is positive when P is between 50 and 150, we know that P increases for initial values in this interval. It increases toward the equilibrium value of P = 150. (iii) Since dP ∕dt is negative for P less than 50 or P greater than 150, we know P decreases for starting values in these intervals. If the initial value of P is less than 50, then P decreases to zero and the fsh all die out. If the initial value of P is greater than 150, then the fsh population decreases toward the equilibrium value of 150. The solutions look like those shown in Figure 9.43. P

(f)

160 150 140

170 150

60 50 40

60 50 40 t

Figure 9.43: Solutions to dP ∕dt = 2P − 0.01P 2 − 75

Figure 9.44

(g) The two equilibrium populations are P = 50, 150. The stable equilibrium is P = 150, while P = 50 is unstable. Notice that P = 50 and P = 150 are solutions of dP ∕dt = 0: dP = 2P − 0.01P 2 − 75 = −0.01(P 2 − 200P + 7500) = −0.01(P − 50)(P − 150). dt (h) (i)

dP dt

25 −75 −100 −200

150 200

H = 200 −300

100

H = 100

300

✛ H = 75 ✲ ✲

668

Chapter Nine /SOLUTIONS

(ii) For H = 75, the equilibrium populations (where dP ∕dt = 0) are P = 50 and P = 150. If the population is between 50 and 150, dP ∕dt is positive. This means that when the initial population is between 50 and 150, the population will increase until it reaches 150, when dP ∕dt = 0 and the population no longer increases or decreases. If the initial population is greater than 150, then dP ∕dt is negative, and the population decreases until it reaches 150. Thus 150 is a stable equilibrium. However, 50 is unstable. For H = 100, the equilibrium population (where dP ∕dt = 0) is P = 100. For all other populations, dP ∕dt is negative and so the population decreases. If the initial population is greater than 100, it will decrease to the equilibrium value, P = 100. However, for populations less than 100, the population decreases until the species dies out. For H = 200, there are no equilibrium points where dP ∕dt = 0, and dP ∕dt is always negative. Thus, no matter what the initial population, the population always dies out eventually. (iii) If the population is not to die out, looking at the three cases above, there must be an equilibrium value where dP ∕dt = 0, i.e. where the graph of dP ∕dt crosses the P axis. This happens if H ≤ 100. Thus provided fshing is not more than 100 fsh/year, there are initial values of the population for which the population will not be depleted. (iv) Fishing should be kept below the level of 100 per year. 2. (a) In each generation, mutation causes the fraction of b genes to decrease k1 times the fraction of b genes (as b genes mutate to B genes). Likewise, in every generation, mutation causes the fraction of b genes to increase by k2 times the fraction of B genes (as B genes mutate to b genes). Therefore, q decreases by k1 q and increases by k2 (1 − q), and we have: dq = −k1 q + k2 (1 − q). dt (b) We have dq = −0.0001q + 0.0004(1 − q) dt = −0.0001q + 0.0004 − 0.0004q = −0.0005q + 0.0004 = −0.0005(q − 0.8). The solution to this di˙erential equation is q = 0.8 + Ce−0.0005t . If q0 = 0.1, then C = −0.7 and the solution is q = 0.8 − 0.7e−0.0005t . If q0 = 0.9, then C = 0.1 and the solution is q = 0.8 + 0.1e−0.0005t . These solution are in Figure 9.45. The equilibrium value is q = 0.8. From Figure 9.45, we see that as generations pass, the fraction of genes responsible for the recessive trait gets closer to 0.8. The equilibrium is given by the solution to the equation dq = −0.0005q + 0.0004 = 0.0005(q − 0.8) = 0. dt Therefore the equilibrium is given by 0.0004∕0.0005 = 0.8 and so is completely determined by the values of k1 and k2 . (c) We have dq = −0.0003q + 0.0001(1 − q) dt = −0.0003q + 0.0001 − 0.0001q = −0.0004q + 0.0001 = −0.0004(q − 0.25).

PROJECTS FOR CHAPTER NINE

669

The solution to this di˙erential equation is q = 0.25 + Ce−0.0004t . If q0 = 0.1, then C = −0.15 and the solution is q = 0.25 − 0.15e−0.0004t . If q0 = 0.9, then C = 0.65 and the solution is q = 0.25 + 0.65e−0.0004t . These solutions are shown in Figure 9.46. The equilibrium value is q = 0.25. As more generations pass, the fraction of genes responsible for the recessive trait gets and closer to 0.25. q

0.9 0.8

0.9

0.1

0.25 0.1

Figure 9.45

Figure 9.46

3. (a) Since I0 is the number of infecteds on day t = 0, March 17, we have I0 = 95. Since S0 is the initial number of susceptibles, which is the whole population of Hong Kong, S0 .8 million. (b) For a = 1.25 ⋅ 10−8 and b = 0.06, the system of equations is dS = −1.25 ⋅ 10−8 SI dt dI = 1.25 ⋅ 10−8 SI − 0.06I. dt So, by the chain rule, dI∕dt 4.8 ⋅ 106 dI 1.25 ⋅ 10−8 SI − 0.06I = = = −1 + . −8 S dS dS∕dt −1.25 ⋅ 10 SI The slope feld and trajectory are in Figure 9.47. I (infecteds) 0.4 ⋅ 106 0.3 ⋅ 106 0.2 ⋅ 106 (6.8 ⋅ 106 , 95)

0.1 ⋅ 106

✠ 5 ⋅ 106

S (susceptibles)

Figure 9.47

(c) The maximum value of I is about 300,000; this gives us the maximum number of infecteds at any one time—the total number of people infected during the course of the disease is much greater than this. The trajectory meets the S-axis at about 3.3 million; this tells us that when the disease dies out, there are still 3.3 million susceptibles who have never had the disease. Therefore 6.8 − 3.3 = 3.5 million people are predicted to have had the disease. The threshold value of S occurs where dI∕dt = 0 and I ≠ 0, so, for a = 1.25 ⋅ 10−8 and b = 0.06, dI = 1.25 ⋅ 10−8 SI − 0.06I = 0, dt

670

Chapter Nine /SOLUTIONS giving 0.06 = 4.8 ⋅ 106 people. 1.25 ⋅ 10−8 The threshold value tells us that if the initial susceptible population, S0 is more than 4.8 million, there will be an epidemic. If S0 is less than 4.8 million, there will not be an epidemic. Since the population of Hong Kong is over 4.8 million, an epidemic is predicted. (d) The value of b represents the rate at which infecteds are removed from circulation. Quarantine increases the rate people are removed and thus increases b. (e) For a = 1.25 ⋅ 10−8 and b = 0.24, the system of di˙erential equations is Threshold value = S =

dS = −1.25 ⋅ 10−8 SI dt dI = 1.25 ⋅ 10−8 SI − 0.24I. dt So, by the chain rule, dI∕dt 19.2 ⋅ 106 dI 1.25 ⋅ 10−8 SI − 0.24I = = = −1 + . S dS dS∕dt −1.25 ⋅ 10−8 SI The slope feld is in Figure 9.48. The solution trajectory does not show as the disease dies out right away. I (infecteds) 0.4 ⋅ 106 0.3 ⋅ 106 0.2 ⋅ 106 0.1 ⋅ 106 5 ⋅ 106

S (susceptibles)

Figure 9.48

(f) The threshold value of S occurs where dI∕dt = 0 and I ≠ 0, so, for b = 0.24 and the same value of a, dI = 1.25 ⋅ 10−8 SI − 0.24I = 0, dt giving 0.24 = 19.2 ⋅ 106 people. 1.25 ⋅ 10−8 The threshold value tells us that if S0 is less than 19.2 million, there will be no epidemic. The population of Hong Kong is 6.8 million, so S0 is below this value. Thus no epidemic is predicted. Policies, such as quarantine, which raise the value of b can be e˙ective in preventing an epidemic. In this case, the value of b increased suÿciently that the population of Hong Kong fell below the threshold value, and a potential epidemic was averted. However, we do not have evidence that the quarantine policy was responsible for the increase in b. (g) Policy I: Closing o˙ the city changes the initial values of S0 and I0 but not the values of a and b. If not one infected person enters the city, then I0 = 0 and the solution trajectory is an equilibrium point on the S-axis. However, in practice it is almost impossible to cut o˙ a city completely, so usually I0 > 0. Also, by the time a policy to close o˙ a city is put into e˙ect, there may already be infected people inside the city, so again I0 > 0. Thus, whether or not there is an epidemic depends on whether S0 is greater than the threshold value, not on the value of I0 (provided I0 > 0). Threshold value = S =

SOLUTIONS TO PROBLEMS ON SEPARATION OF VARIABLES

671

For example, in the case of Hong Kong with the March values of a and b, changing the value of I0 to 1 leaves the solution trajectory much as before; see Figure 9.49. The main di˙erence is that the epidemic occurs slightly later. So a policy of isolating a city only works if it keeps the disease out of the city of the city entirely. Thus, Policy I does not help the city except in the exceptional case that every infected person is kept out. I (infecteds) 0.4 ⋅ 106 0.3 ⋅ 106 0.2 ⋅ 106 (6.8 ⋅ 106 , 1)

0.1 ⋅ 106

✠ 5 ⋅ 106

S (susceptibles)

Figure 9.49

Policy II: From the analysis of the Hong Kong data, we see that a quarantine policy can help prevent an epidemic if the value of b is increased enough to bring S0 below the threshold value. Thus, Policy II can be very e˙ective.

Solutions to Problems on Separation of Variables 1. (a) This matches equation (III). Separating variables, we have dy = ex+y dx dy = ex ey dx dy = ex dx ey e−y dy = ex dx. Ê Ê (b) This matches equation (IV). Separating variables, we have dy = ex−y dx dy ex = y dx e ey dy = ex dx. Ê Ê (c) This matches equation (II). Separating variables, we have dy y = dx x dy dx = y x Ê

y−1 dy =

x−1 dx.

(d) This matches equation (I). Separating variables, we have

dy x = dx y y dy =

x dx.

Chapter Nine /SOLUTIONS

672

2. Separating variables gives 1 dP = − 2dt, Ê P Ê so ln |P | = −2t + C. Therefore P = ±e−2t+C = Ae−2t . The initial value P (0) = 1 gives 1 = A, so P = e−2t . 3. Separating variables and integrating both sides gives 1 1 dp dL = Ê L 2Ê or ln |L| =

1 p + C. 2

This can be written L = ±e(1∕2)p+C = Aep∕2 . The initial condition L(0) = 100 gives 100 = A, so L = 100ep∕2 . 4. Separating variables gives Ê so that or

P dP =

P2 =t+C 2 √ P = ± 2t + D

(where D = 2C). The initial condition P (0) = 1 implies we must take the positive root and that 1 = D, so √ P = 2t + 1. 5. Separating variables gives 1 dm = ds. Ê m Ê Hence ln |m| = s + C which gives m = ±es+C = Aes . The initial condition m(1) = 2 gives 2 = Ae1 or A = 2∕e, so m=

2 s e = 2es−1 . e

6. Separating variables gives 1 1 du = dt Ê u2 Ê 2 or −

1 1 = t + C. u 2

The initial condition gives C = −1 and so u=

1 . 1 − (1∕2)t

SOLUTIONS TO PROBLEMS ON SEPARATION OF VARIABLES 7. Separating variables and integrating gives 1 dz = ydy Ê z Ê which gives 1 2 y +C 2

ln |z| = or

z = ±e(1∕2)y +C = Aey ∕2 . The initial condition y = 0, z = 1 gives A = 1. Therefore 2

z = ey ∕2 .

8. Rearrange and write 1 dR = dy Ê 1−R Ê or − ln |1 − R| = y + C which can be written as 1 − R = ±e−C−y = Ae−y or R = 1 − Ae−y . The initial condition R(1) = 0.1 gives 0.1 = 1 − Ae−1 and so A = 0.9e. Therefore R = 1 − 0.9e1−y . 9. Write

1 1 dy = dt Ê y Ê 3+t

and so ln |y| = ln |3 + t| + C or ln |y| = ln D|3 + t| where ln D = C. Therefore y = D(3 + t). The initial condition y(0) = 1 gives D = 13 , so y=

1 (3 + t). 3

10. Separating variables gives dz = tez dt e−z dz = tdt Ê

e−z dz =

t dt,

t2 + C. 2 Since the solution passes through the origin, z = 0 when t = 0, we must have −e−z =

−e−0 =

0 + C, so C = −1. 2

673

Chapter Nine /SOLUTIONS

674

Thus

t2 − 1, 2 0 1 t2 z = − ln 1 − . 2 −e−z =

11. Separating variables gives dy 5y = dx x dy 5 = dx Ê y Ê x ln |y| = 5 ln |x| + C. Thus

|y| = e5 ln |x| eC = eC eln |x| = eC |x|5 , giving y = Ax5 ,

A = ±eC .

where

Since y = 3 when x = 1, so A = 3. Thus y = 3x5 . 12. Separating variables gives dy = y2 (1 + t) dt dy = (1 + t) dt, Ê y2 Ê so −

1 t2 = t + + C, y 2

giving y=−

1 . t + t2 ∕2 + C

Since y = 2 when t = 1, we have 2=−

1 , 1 + 1∕2 + C

Thus

2C + 3 = −1,

and

C = −2.

1 2 y=− 2 =− 2 . t + 2t − 4 t ∕2 + t − 2

13. Separating variables gives dz = z + zt2 = z(1 + t2 ) dt dz = (1 + t2 )dt, Ê z Ê so ln |z| = t +

t3 + C, 3

giving 3

z = Aet+t ∕3 . We have z = 5 when t = 0, so A = 5 and

z = 5et+t ∕3 . 14. (a) (d) (g) (j)

Yes No No Yes

(b) (e) (h) (k)

No Yes Yes Yes

SOLUTIONS TO PROBLEMS ON SEPARATION OF VARIABLES 15. Separating variables gives dP dt. = Ê P −a Ê Integrating yields ln |P − a| = t + C, so |P − a| = et+C = et eC P = a + Aet ,

where A = ±eC

or A = 0.

16. Separating variables gives dQ = dt. Ê b−Q Ê Integrating yields − ln |b − Q| = t + C, so |b − Q| = e−(t+C) = e−t e−C Q = b − Ae−t ,

where A = ±e−C

or A = 0.

17. Separating variables gives dP = k dt. Ê P −a Ê Integrating yields ln |P − a| = kt + C, so P = a + Aekt

where A = ±eC

or A = 0.

18. Factoring and separating variables gives dR b = a R+ dt a dR a dt = Ê R + b∕a Ê ln R +

= at + C

b R = − + Aeat , a

where A can be any constant.

19. Separating variables and integrating gives 1 dP = dt. Ê aP + b Ê This gives 1 ln |aP + b| = t + C a ln |aP + b| = at + aC aP + b = ±eat+aC = Aeat , or P =

where A = ±eaC

1 (Aeat − b). a

A = 0,

675

Chapter Nine /SOLUTIONS

676

20. Separating variables and integrating gives 1 dy = k(1 + t2 )dt Ê y2 Ê 1 1 − = k t + t3 + C. y 3

or Hence,

−1 k(t + 13 t3 ) + C

21. (a) Separating variables and integrating gives 1 dy = dt Ê 100 − y Ê so that − ln |100 − y| = t + C or y(t) = 100 − Ae−t . (b) See Figure 9.50. y 110

25 t

Figure 9.50 (c) The initial condition y(0) = 25 gives A = 75, so the solution is y(t) = 100 − 75e−t . The initial condition y(0) = 110 gives A = −10 so the solution is y(t) = 100 + 10e−t . (d) The increasing function, y(t) = 100 − 75e−t . 22. (a) The slope feld for dy∕dx = xy is in Figure 9.51. y

Figure 9.51

Figure 9.52

SOLUTIONS TO PROBLEMS ON SEPARATION OF VARIABLES (b) Some solution curves are shown in Figure 9.52. (c) Separating variables gives 1 dy = xdx Ê y Ê or ln |y| =

1 2 x + C. 2

Solving for y gives 2

y(x) = Aex ∕2 2

where A = ±eC . In addition, y(x) = 0 is a solution. So y(x) = Aex ∕2 is a solution for any A.

677

10.1 SOLUTIONS

679

CHAPTER TEN Digital only

Solutions for Section 10.1 1. Adding the terms, we see that 3 + 3 ⋅ 2 + 3 ⋅ 22 = 3 + 6 + 12 = 21. We can also fnd the sum using the formula for a fnite geometric series with a = 3, r = 2, and n = 3: 3 + 3 ⋅ 2 + 3 ⋅ 22 =

3(1 − 23 ) = 3(8 − 1) = 21. 1−2

2. Adding the terms, we see that 50 + 50(0.9) + 50(0.9)2 + 50(0.9)3 = 50 + 45 + 40.5 + 36.45 = 171.95. We can also fnd the sum using the formula for a fnite geometric series with a = 50, r = 0.9, and n = 4: 50 + 50(0.9) + 50(0.9)2 + 50(0.9)3 =

50(1 − (0.9)4 ) = 171.95. 1 − 0.9

3. Yes, a = 5, ratio = −2. 4. No. Ratio between successive terms is not constant:

1∕3 1∕4 = 0.66 …, while = 0.75. 1∕2 1∕3

5. Yes, a = 2, ratio = 1∕2. 6. Yes, a = 1, ratio = −1∕2. 7. Yes, a = 1, ratio = 2z. 8. No. Ratio between successive terms is not constant:

3x3 3 2x2 = 2x, while 2 = x. 2 x 2x

9. Yes, a = y2 , ratio = y. 10. The formula for the sum of a fnite geometric series with a = 5, r = 3, and n = 13 gives Sum = 5 + 5 ⋅ 3 + 5 ⋅ 32 + ⋯ + 5 ⋅ 312 =

5(1 − 313 ) = 3,985,805. 1−3

11. The formula for the sum of a fnite geometric series with a = 100, r = 0.85, and n = 11 gives Sum = 100 + 100(0.85) + ⋯ + 100(0.85)10 =

100(1 − (0.85)11 ) = 555.10. 1 − 0.85

12. This is an infnite geometric series with a = 1000 and r = 1.05. Since r > 1, the series diverges and the sum does not exist. 13. This is an infnite geometric series with a = 75 and r = 0.22. Since −1 < r < 1, the series converges and the sum is given by: 75 = 96.154. Sum = 75 + 75(0.22) + 75(0.22)2 + ⋯ = 1 − 0.22 14. The formula for the sum of a fnite geometric series with a = 20, r = 1.45, and n = 15 gives Sum = 20 + 20(1.45) + ⋯ + 20(1.45)14 =

20(1 − (1.45)15 ) = 11,659.64. 1 − 1.45

680

Chapter Ten /SOLUTIONS

15. Notice that the frst term in this series is 500(0.4) = 200 and each term is 0.4 times the preceding term. This is an infnite geometric series with a = 500(0.4) = 200 and r = 0.4. Since −1 < r < 1, the series converges and the sum is given by: Sum =

200 = 333.33. 1 − 0.4

16. Since 6300∕31500 = 0.2 and 1260∕6300 = 0.2 and 252∕1260 = 0.2, this is an infnite geometric series with a = 31500 and r = 0.2: Sum = 31500 + 6300 + 1260 + 252 + ⋯ = 31500 + 31500(0.2) + 31500(0.2)2 + 31500(0.2)3 + ⋯ . Since −1 < r < 1, the series converges and the sum is given by: Sum =

31500 = 39,375. 1 − 0.2

17. This is a fnite geometric series with a = 3, r = 1∕2, and n − 1 = 10, so n = 11. Thus Sum =

3(1 − (1∕2)11 ) (211 − 1) 3(211 − 1) =3⋅2 = . 1 − 1∕2 211 210

18. Each term in this series is 1.5 times the preceding term, so this is an infnite geometric series with a = 1000 and r = 1.5. Since r > 1, the series diverges and the sum does not exist. 19. Each term in this series is half the preceding term, so this is an infnite geometric series with a = 200 and r = 0.5. Since −1 < r < 1, the series converges and we have Sum =

200 = 400. 1 − 0.5

20. The formula for the sum of a fnite geometric series with a = 65, r = 1∕(1.02), and n = 19 gives Sum = 65 +

65(1 − 1∕(1.02)19 ) 65 65 65 + +⋯+ = = 1039.482. 2 18 1 − 1∕1.02 1.02 (1.02) (1.02)

21. This is an infnite geometric series with a = −2 and r = −1∕2. Since −1 < r < 1, the series converges and Sum =

−2 4 =− . 1 − (−1∕2) 3

22. Since a = 10 and r = 0.75, we fnd the partial sums using the formula Sn = For n = 5, we have

10(1 − (0.75)n ) . 1 − 0.75

S5 =

10(1 − (0.75)5 ) = 30.51. 1 − 0.75

S10 =

10(1 − (0.75)10 ) = 37.75. 1 − 0.75

S15 =

10(1 − (0.75)15 ) = 39.47. 1 − 0.75

For n = 10, we have

For n = 15, we have

For n = 20, we have

10(1 − (0.75)20 ) = 39.87. 1 − 0.75 As n gets larger, the partial sums appear to be approaching 40, as we expect. S20 =

10.1 SOLUTIONS 23. We fnd the partial sums using the formula Sn = For n = 5, we have

250(1 − (1.2)n ) . 1 − 1.2

S5 =

250(1 − (1.2)5 ) = 1,860.40. 1 − 1.2

S10 =

250(1 − (1.2)10 ) = 6,489.67. 1 − 1.2

S15 =

250(1 − (1.2)15 ) = 18,008.78. 1 − 1.2

For n = 10, we have

For n = 15, we have

681

For n = 20, we have

250(1 − (1.2)20 ) = 46,672.00. 1 − 1.2 As n gets larger, the partial sums appear to be growing without bound, as we expect, since r > 1. S20 =

24. (a) 0.232323 … = 0.23 + 0.23(0.01) + 0.23(0.01)2 + ⋯, which is a geometric series with a = 0.23 and r = 0.01. 0.23 0.23 23 (b) The sum is = = . 1 − 0.01 0.99 99 25. (a) Notice that the 6th deposit is made 5 months after the frst deposit, so the frst deposit has grown to 500(1.001)5 at that time. The balance in the account right after the 6th deposit is the sum Balance = 500 + 500(1.001) + 500(1.001)2 + ⋯ + 500(1.001)5 . We fnd the sum using the formula for a fnite geometric series with a = 500, r = 1.001, and n = 6: Balance right after 6th deposit =

500(1 − (1.001)6 ) = $3007.51. 1 − 1.001

Since each deposit is $500, the balance in the account right before the 6th deposit is 3007.51 − 500 = $2507.51. (b) Similarly, the 12th deposit is made 11 months after the frst deposit, so the frst deposit has grown to 500(1.001)11 at that time. The balance in the account right after the 12th deposit is the sum Balance = 500 + 500(1.001) + 500(1.001)2 + ⋯ + 500(1.001)11 . We fnd the sum using the formula for a fnite geometric series with a = 500, r = 1.001, and n = 12: Balance right after 12th deposit =

500(1 − (1.001)12 ) = $6033.11. 1 − 1.001

Since each deposit is $500, the balance in the account right before the 12th deposit is 6033.11 − 500 = $5533.11. 26. Notice that the 25th deposit is made 24 years after the frst deposit, so the frst deposit has grown to 5000(1.0125)24 at that time. The balance in the account from all 25 deposits is Balance = 5000 + 5000(1.0125) + 5000(1.0125)2 + ⋯ + 5000(1.0125)24 . We fnd the sum using the formula for a fnite geometric series with a = 5000, r = 1.0125, and n = 25: Balance =

5000(1 − (1.0125)25 ) = $145,677.18. 1 − 1.0125

27. If the person smokes a cigarette at 7 am, at 8 am, at 9 am, and every hour until 11 pm, the person has smoked 17 cigarettes. Sixteen hours have passed since the person smoked the cigarette at 7 am, and we have Quantity of nicotine at 11 pm = 0.4 + 0.4(0.71) + 0.4(0.71)2 + ⋯ + 0.4(0.71)16 . This is a fnite geometric series with a = 0.4, r = 0.71, and n = 17. We use the formula for the sum of a fnite geometric series: 0.4(1 − (0.71)17 ) Quantity of nicotine at 11 pm = = 1.375 mg. 1 − 0.71

682

Chapter Ten /SOLUTIONS

28. The quantity of drug in the body after the nth injection is given by Quantity after nth injection = 25 + 25(0.4) + 25(0.4)2 + ⋯ + 25(0.4)n−1 . This is a fnite geometric series with a = 25 and r = 0.4. Using the formula for the sum, we have Sum =

25(1 − (0.4)n ) . 1 − 0.4

(a) We have Quantity after 3rd injection = 25 + 25(0.4) + 25(0.4)2 =

25(1 − (0.4)3 ) = 39 mg. 1 − 0.4

Notice that we could also have found the sum by adding the three terms. (b) We have Quantity after 6th injection = 25 + 25(0.4) + 25(0.4)2 + ⋯ + 25(0.4)5 25(1 − (0.4)6 ) = 1 − 0.4 = 41.496 mg. Notice that we could also have found the sum by adding the six terms. (c) We fnd the quantity of drug in the long run by letting n → ∞. Right after an injection, we have Quantity of drug in the long term = 25 + 25(0.4) + 25(0.4)2 + ⋯ . This is an infnite geometric series with a = 25 and r = 0.4. Since −1 < r < 1, the series converges and we have Sum =

25 = 41.667 mg. 1 − 0.4

29. (a) The average quantity in the body is (65 + 15)∕2 = 40 mg. (b) The average concentration for this patient (in milligrams of quinine per kilogram of body weight) is (40 mg)/(70 kg) = 0.57 mg/kg. This average concentration falls within the range that is both safe and e˙ective. (c) (i) Since this treatment produces an average of 40 mg of quinine in the body, a body weight W kg produces an average concentration below 0.4 mg/kg if 40 < 0.4 W so W > 100. The treatment is not e˙ective for anyone weighing more than 100 kg (or about 220 pounds.) (ii) A body weight W kg produces an average concentration above 3.0 mg/kg if 40 > 3.0 W so W < 13.3. The treatment is unsafe for anyone weighing less than 13.3 kg (or about 30 pounds.) 30. (a) Let ℎn be the height of the nt h bounce after the ball hits the foor for the nth time. Then from Figure 10.1, ℎ0 = height before frst bounce = 10 feet, 3 feet, ℎ1 = height after frst bounce = 10 4 2 3 feet. ℎ2 = height after second bounce = 10 4 Generalizing gives ℎn = 10

n 3 . 4

10.2 SOLUTIONS

683

✻ ✻ 10 10( 3 ) 4

✻ 10( 3 )2 4

⋯

❄

✻

ℎn

⋯

❄

Figure 10.1 (b) When the ball hits the foor for the frst time, the total distance it has traveled is just D1 = 10 feet. (Notice that this is 3 the same as ℎ0 = 10.) Then the ball bounces back to a height of ℎ1 = 10 , comes down and hits the foor for the 4 second time. See Figure 10.1. The total distance it has traveled is 3 = 25 feet. D2 = ℎ0 + 2ℎ1 = 10 + 2 ⋅ 10 4 2 3 Then the ball bounces back to a height of ℎ2 = 10 , comes down and hits the foor for the third time. It has 4 traveled 2 2 3 3 3 + 2 ⋅ 10 = 25 + 2 ⋅ 10 = 36.25 feet. D3 = ℎ0 + 2ℎ1 + 2ℎ2 = 10 + 2 ⋅ 10 4 4 4 Similarly, D4 = ℎ0 + 2ℎ1 + 2ℎ2 + 2ℎ3 2 3 3 3 3 = 10 + 2 ⋅ 10 + 2 ⋅ 10 + 2 ⋅ 10 4 4 4 3 3 = 36.25 + 2 ⋅ 10 4 ≈ 44.69 feet. (c) When the ball hits the foor for the nth time, its last bounce was of height ℎn−1 . Thus, by the method used in part (b), we get Dn = ℎ0 + 2ℎ1 + 2ℎ2 + 2ℎ3 + ⋯ + 2ℎn−1 2 3 n−1 3 3 3 3 = 10 + 2 ⋅ 10 + 2 ⋅ 10 + 2 ⋅ 10 + ⋯ + 2 ⋅ 10 4 4 4 4 «¯¬ fnite geometric series

0 2 n−2 1 3 3 3 3 = 10 + 2 ⋅ 10 ⋅ 1+ + +⋯+ 4 4 4 4 n−1 ⎛ ⎞ 3 ⎟ ⎜1 − 4 = 10 + 15 ⎜ ⎟ ⎜ 1− 3 ⎟ 4 ⎠ ⎝ 0 n−1 1 3 = 10 + 60 1 − . 4

Solutions for Section 10.2 1. The 20th deposit has just been made and has not yet earned any interest, the 19th deposit has earned interest for one year and is worth 1000(1.02), the 18th deposit has earned interest for two years and is worth 1000(1.02)2 , and so on. The frst

684

Chapter Ten /SOLUTIONS deposit has earned interest for 19 years and is worth 1000(1.02)19 . The total balance in the account right after the 20th deposit is Balance = 1000 + 1000(1.02) + 1000(1.02)2 + ⋯ + 1000(1.02)19 . This is a fnite geometric series with a = 1000, r = 1.02, and n = 20. Using the formula for the sum of a fnite geometric series, we have 1000(1 − (1.02)20 ) Balance after the 20th deposit = = $24,297.37. 1 − 1.02 The twenty annual deposits of $1000 have contributed a total of $20,000 to this balance, and the remaining $4297.37 in the account comes from the interest earned.

2. Right after the 5th deposit has been made, the 5th deposit has not yet earned any interest, the 4th deposit has earned interest for one year and is worth 2000(e0.025 ), the 3rd deposit has earned interest for two years and is worth 2000(e0.025 )2 , and so on. The frst deposit has earned interest for 4 years and is worth 2000(e0.025 )4 . The total balance in the account right after the 5th deposit is Balance = 2000 + 2000(e0.025 ) + 2000(e0.025 )2 + 2000(e0.025 )3 + 2000(e0.025 )4 . This is a fnite geometric series with a = 2000, r = e0.025 , and n = 5. Using the formula for the sum of a fnite geometric series, we have 2000(1 − (e0.025 )5 ) Balance after the 5th deposit = = $10,519.28. 1 − e0.025 The balance right before the 5th deposit of $2000 is 10, 519.28 − 2000 = $8519.28. 3. (a) The present value of the payment made one year from now is 50,000(1.0365−1 ), and the present value of the payment made n years from now is 50,000(1.0365−1 )n . Since the frst payment is made now, the 10th payment is made 9 years from now. Summing, we have Present value = 50000 + 50000(1.0365−1 ) + 50000(1.0365−1 )2 + ⋯ + 50000(1.0365−1 )9 . This is a fnite geometric series with a = 50000, r = 1.0365−1 , and n = 10. Using the formula for the sum, we have Present value of the annuity =

50000(1 − (1.0365−1 )10 ) = $427,767.74. 1 − 1.0365−1

(b) If the payments are to continue indefnitely, the present value is the sum: Present value = 50000 + 50000(1.0365−1 ) + 50000(1.0365−1 )2 + 50000(1.0365−1 )3 + ⋯ . This is an infnite geometric series with a = 50000 and r = 1.0365−1 = 0.964785. Since −1 < r < 1, this series converges and we use the formula for the sum of an infnite geometric series: Present value of the annuity =

50000 = $1,419,863.01. 1 − 1.0365−1

4. The amount that must be deposited now is the present value of this annuity. The present value of the frst payment (made one year from now) is 5000(1.02−1 ), and the present value of the 20th payment (made 20 years from now) is 5000(1.02−1 )20 . The total is Present value = 5000(1.02−1 ) + 5000(1.02−1 )2 + 5000(1.02−1 )3 + ⋯ + 5000(1.02−1 )20 . Notice that the frst payment is not made until a year from now so the frst term of the series is 5000(1.02−1 ) rather than 5000. Since 5000(1.02−1 )n = 5000(1.02−1 )(1.02−1 )n−1 , we can rewrite the series as Present value = 5000(1.02−1 ) + 5000(1.02−1 )(1.02−1 ) + 5000(1.02−1 )(1.02−1 )2 + ⋯ + 5000(1.02−1 )(1.02−1 )19 . This is a fnite geometric series with a = 5000(1.02−1 ), r = 1.02−1 , and n = 20. Using the formula for the sum of a fnite geometric series, we have Present value of the annuity =

5000(1.02−1 )(1 − (1.02−1 )20 ) = $81,757.17 1 − 1.02−1

10.2 SOLUTIONS

685

5. We sum the geometric series Present value = 20,000 + =

20,000 20,000 + +⋯ 1.01 1.012

20,000 = 2,020,000. 1 − 1∕1.01

Thus, the present value is 2.02 million dollars. 6. The frst payment of $1000 is made now, so the present value of the frst payment is 1000. The second payment is made one month from now, so the present value of the second payment is 1000(1.0025−1 ), and so on. The present value of the 36th payment, made 35 months from now, is 1000(1.0025−1 )35 . Summing, we have Present value of the annuity = 1000 + 1000(1.0025−1 ) + 1000(1.0025−1 )2 + ⋯ + 1000(1.0025−1 )35 . This is a fnite geometric series with a = 1000, r = 1.0025−1 , and n = 36. Using the formula for the sum, we have Present value of the annuity =

1000(1 − (1.0025−1 )36 ) = $34,472.43. 1 − 1.0025−1

7. (a) The amount that must be deposited now is the present value of the annuity to fund this scholarship. The present value of the award made one year from now is 10,000(1.03−1 ), and the present value of the award made n years from now is 10,000(1.036−1 )n . Since the awards start immediately, the 20th award is made 19 years from now. We have Present value = 10000 + 10000(1.03−1 ) + 10000(1.03−1 )2 + ⋯ + 10000(1.03−1 )19 . This is a fnite geometric series with a = 10000, r = 1.03−1 , and n = 20. Using the formula for the sum, we have 10000(1 − (1.03−1 )20 ) = $153,237.99. 1 − 1.03−1 The donor must deposit $153,237.99 now in order to fund this scholarship for twenty years. (b) If the awards are to continue indefnitely, the present value is the sum of infnitely many terms: Present value of the annuity =

Present value = 10000 + 10000(1.03−1 ) + 10000(1.03−1 )2 + 10000(1.03−1 )3 + ⋯ . This is an infnite geometric series with a = 10000 and r = 1.03−1 = 0.97087378. Since −1 < r < 1, this series converges to the sum: 10000 Present value of the annuity = = $343,333.33. 1 − 1.03−1 The donor must deposit $343,333.33 now in order to fund this scholarship forever. We can check that this present value makes sense. After the frst payment, the amount in the account is $333,333.33. This amount will fund the awards forever because the interest earned on the account in one year exactly matches the money paid out of the account to fund the scholarship award: $333,333.33(0.03) = $10,000. 8. Her salary is $30,000 at the start of the frst year, $30,000(1.02) at the start of the second year, $30,000(1.02)2 at the start of the third year, and so on. At the start of the 11th year, her salary is $30,000(1.02)10 = $36,569.83. The employee’s total earnings for the frst ten years is the sum Total earnings = 30000 + 30000(1.02) + 30000(1.02)2 + ⋯ + 30000(1.02)9 . This is a fnite geometric series with a = 30000, r = 1.02, and n = 10. We use the formula for the sum: Total earnings =

30000(1 − (1.02)10 ) = $328,491.63. 1 − 1.02

9. You earn 1 cent the frst day, 2 cents the second day, 22 = 4 the third day, 23 = 8 the fourth day, and so on. On the nth day, you earn 2n−1 cents. We have Total earnings for n days = 1 + 2 + 22 + 23 + ⋯ + 2n−1 . This is a fnite geometric series with a = 1 and r = 2. We use the formula for the sum: Total earnings for n days =

1 − 2n = 2n − 1. 1−2

(a) Using n = 7, we see that Total earnings for 7 days = 27 − 1 = 127 = $1.27.

686

Chapter Ten /SOLUTIONS (b) Using n = 14, we see that Total earnings for 14 days = 214 − 1 = 16383 = $163.83. (c) Using n = 21, we see that Total earnings for 21 days = 221 − 1 = 2097151 = $20,971.51. (d) Using n = 28, we see that Total earnings for 28 days = 228 − 1 = 268435455 = $2,684,354.55.

10. In any given year, there are 10,000 units in use that were manufactured that year, and 10,000(0.75) units in use that were manufactured the previous year (since 25% of them failed), and 10,000(0.75)2 units in use that were manufactured two years ago, and so on. We have Total number of units in use = 10,000 + 10,000(0.75) + 10,000(0.75)2 + 10,000(0.75)3 + ⋯ . This is an infnite geometric series with a = 10,000 and r = 0.75. Since −1 < r < 1, the series converges and we have Total number of units in use =

10,000 = 40,000. 1 − 0.75

The market stabilization point for this product is 40,000 units. 11. (a) In any given year, the number of units manufactured is 1000, the number of units in use that were manufactured the previous year is 1000(0.80) (since 20% of them failed), the number of units that were manufactured two years ago is 1000(0.80)2 , and so on. We have Total number of units in use = 1000 + 1000(0.80) + 1000(0.80)2 + 1000(0.80)3 + ⋯ . This is an infnite geometric series with a = 1000 and r = 0.80. Since −1 < r < 1, this series converges to a fnite sum. We have a 1000 Total number of units in use = = = 5000 units. 1−r 1 − 0.80 This sum, 5000 units, is the market stabilization point for this product. (b) The total number in use after n = 5 production cycles is S5 = 1000 + 1000(0.80) + 1000(0.80)2 + 1000(0.80)3 + 1000(0.80)4 1000(1 − (0.80)5 ) 1 − 0.80 = 3362 units.

Similarly, we fnd the other values in Table 10.1 Table 10.1 Market’s approach to the stabilization point n

3362

4463

4824

4942

12. Let k be the fraction of the $1 bills in circulation that are removed each day. Then r = 1−k is the fraction of bills remaining in circulation the next day. On any day, there are 4.3 million new bills, and 4.3r million remaining from the day before, and 4.3r2 million remaining from the day before, and so on. In millions, the total number of $1 bills in circulation is given by N = 4.3 + 4.3r + 4.3r2 + ⋯ . This is an infnite geometric series with a = 4.3 and −1 < r < 1 (since the fraction of bills remaining is positive and less than 1). The series converges, and its sum is given by N=

a 4.3 = . 1−r 1−r

10.2 SOLUTIONS

687

Since we know the total number of $1 bills in circulation is 12.7 billion, or 12,700 million, we have 12,700 =

4.3 . 1−r

Solving for r gives 4.3 12,700 4.3 r = 1− = 0.999661. 12,700

1−r =

Thus 99.97% of the notes remain in circulation the next day, so 0.03% are removed each day. Notice that if 4.3 million new bills are introduced each day, then at steady state, the same number are removed each day. Thus, we can calculate the fraction of the $1 bills in circulation that are removed each day, k, without using a series: k=

4.3 million = 0.03%. 12.7 billion

13. The present value is the sum of the present values of the coupons plus the present value of the principal. For the coupons, we have 50 Present value of frst coupon = 1.04 50 Present value of second coupon = , etc. (1.04)2 Adding the coupons to the principal gives Total present value of the bond =

50 50 50 1000 + +⋯+ + 1.04 (1.04)2 (1.04)10 (1.04)10 «¯¬ «¯¬ coupons

principal

0 1 50 1 1 1000 = 1+ +⋯+ + 1.04 1.04 (1.04)9 (1.04)10 ⎛ 1 − 1 10 ⎞ ⎟ ⎜ 1.04 1000 50 = ⎟+ ⎜ 1 1.04 ⎜ 1 − (1.04)10 ⎟ 1.04 ⎠ ⎝ = 405.545 + 675.564 = $1081.11

14. The present value is the sum of the present values of the coupons plus the present value of the principal. For the coupons, we have 50 Present value of frst coupon = 1.06 50 Present value of second coupon = , etc. (1.06)2 Adding the coupons to the principal gives Total present value of the bond =

50 50 50 1000 + +⋯+ + 1.06 (1.06)2 (1.06)10 (1.06)10 «¯¬ «¯¬ coupons

principal

0 1 50 1 1 1000 = 1+ +⋯+ + 1.06 1.06 (1.06)9 (1.06)10 ⎛ 1 − 1 10 ⎞ ⎟ 1.06 1000 50 ⎜ = ⎟+ 10 1.06 ⎜⎜ 1 − 1 (1.06) ⎟ 1.06 ⎠ ⎝ = 368.004 + 558.395 = $926.40

688

Chapter Ten /SOLUTIONS

15. (a) 50 1.05 50 Present value of second coupon = , etc. (1.05)2 Present value of frst coupon =

Total present value =

50 50 50 1000 + +⋯+ + 1.05 (1.05)2 (1.05)10 (1.05)10 «¯¬ «¯¬ coupons

principal

0 1 50 1 1 1000 = 1+ +⋯+ + 1.05 1.05 (1.05)9 (1.05)10 ⎛ 1 − 1 10 ⎞ ⎟ 1.05 1000 50 ⎜ = ⎟+ 10 1.05 ⎜⎜ 1 − 1 (1.05) ⎟ 1.05 ⎠ ⎝ = 386.087 + 613.913 = $1000

(b) When the interest rate is 5%, the present value equals the principal. (c) When the interest rate is more than 5%, the present value is smaller than it is when interest is 5% and must therefore be less than the principal. Since the bond will sell for around its present value, it will sell for less than the principal; hence the description trading at discount. (d) When the interest rate is less than 5%, the present value is more than the principal. Hence the bond will be selling for more than the principal, and is described as trading at a premium. 16. (a) The people who receive the 100 billion dollars in tax rebates spend 80% of it, for an initial amount spent of 100(0.80) = 80 billion dollars. The people who receive this 80 billion dollars spend 80% of it, for an additional amount spent of 80(0.80), and so on. We have Total amount spent = 80 + 80(0.80) + 80(0.80)2 + 80(0.80)3 + ⋯ . Notice that the initial amount spent is not the original tax rebate of 100 billion dollars, but 80% of 100 billion dollars, or 80 billion dollars. The total amount spent is an infnite geometric series with a = 80 and r = 0.80. Since −1 < r < 1, the series converges and we have Total amount spent =

80 = 400 billion dollars. 1 − 0.80

(b) If everyone who receives money spends 90% of it, then the initial amount spent is 100(0.90) = 90 billion dollars. The people who receive this 90 billion dollars spend 90% of it, and so on. We have Total amount spent = 90 + 90(0.90) + 90(0.90)2 + 90(0.90)3 + ⋯ . This is an infnite geometric series with a = 90 and r = 0.90. Since −1 < r < 1, the series converges and we have Total amount spent =

90 = 900 billion dollars. 1 − 0.90

Notice that an increase in the spending rate from 80% to 90% causes a dramatic increase in the total e˙ect on spending. 17. (a) The people who receive the 300 billion dollars in tax rebates spend 80% of it, for an initial amount spent of 300⋅0.80 = 240 billion dollars. The people who receive this 240 billion dollars spend 80% of it, for an additional amount spent of 240 ⋅ 0.80, and so on. We have Total amount spent = 240 + 240 ⋅ 0.80 + 240 ⋅ 0.802 + 240 ⋅ 0.803 + ⋯ . Notice that the initial amount spent is not the original tax rebate of 300 billion dollars, but 80% of 300 billion dollars, or 240 billion dollars. The total amount spent is an infnite geometric series with a = 240 and r = 0.80. Since −1 < r < 1, the series converges and we have Total amount spent =

240 = 1200 billion dollars. 1 − 0.80

10.2 SOLUTIONS

689

(b) If everyone who receives money spends 90% of it, then the initial amount spent is 300 ⋅ 0.90 = 270 billion dollars. The people who receive this 270 billion dollars spend 90% of it, and so on. We have Total amount spent = 270 + 270 ⋅ 0.90 + 270 ⋅ 0.902 + 270 ⋅ 0.903 + ⋯ . This is an infnite geometric series with a = 270 and r = 0.90. Since −1 < r < 1, the series converges and we have 270 = 2700 billion dollars. 1 − 0.90 Notice that an increase in the spending rate from 80% to 90% causes a dramatic increase in the total e˙ect on spending. Total amount spent =

18. If we assume that everyone spends a proportion r of what they get, then the person who gets the dollar spends r dollars and the person who gets that money spends r ⋅ r = r2 dollars, and so on. Thus Total additional spending = r + r2 + r3 + ⋯ . This is an infnite geometric series with frst term a = r and constant multiplier r. Using the formula for the sum of the series and the fact that the total additional spending from the $1 tax cut is $3, we have: a =3 1−r r =3 1−r r = 3(1 − r) 4r = 3 r = 0.75. The economists are assuming that, on average, people spend 75% of what they receive. 19. (a) To meet the $1.4 billion payment, $P should grow to $1.4 in 50 years, so, in billions, P = 1.4e−0.01(50) = 0.849 That is, about $849 million. (b) Payments at di˙erent times have di˙erent present values. Over ten years, Present Value =

59 É

1.4e−0.01t

t=50

� = 1.4e−0.01(50) 1 + e−0.01 + e−0.01(2) + e−0.01(3) + ⋯ + e−0.01(9)

1 − e−0.01(10) 1 − e−0.01 = 8.12 billion dollars. = 1.4e−0.01(50)

69 É

1.4e−0.01t

t=50

� = 1.4e−0.01(50) 1 + e−0.01 + e−0.01(2) + e−0.01(3) + ⋯ + e−0.01(19)

1 − e−0.01(20) 1 − e−0.01 = 15.47 billion dollars.

= 1.4e−0.01(50)

(d) The payments in the 10–20 year interval are farther in the future than the payments in the 0–10 year interval. So the 10–20 year payments have a smaller present value than the 0–10 year payments. 20. (a) To meet the $1.4 billion payment, $P should grow to $1.4 in 50 years, so, in billions, P = 1.4e−0.01(50) = 0.849. (b) Since the present value of one of the $1.4 billion costs is $0.849 billion, the fraction is 0.56∕0.849 = 0.66, about two-thirds. (c) Assuming each year balances a year exactly 50 years later, the fraction does not depend on which year. (d) If is the discount rate, we are looking at 0.56∕(1.4e− ⋅50 ). Higher discount rates make the fraction larger, which is good for the Mayor. For example, if = 5%, the fraction is 0.56∕(1.4e−0.05⋅50 ) = 4.873! Costs far in the future have a lower present value (they seem less ominous) with high discount rates. Given the choice, the Mayor would like a high discount rate.

690

Chapter Ten /SOLUTIONS

21. (a) We fnd the total amount deposited by adding the deposits: Total amount of money deposited = 100 + 97 + 94.09 + ⋯ = 100 + 100 ⋅ 0.97 + 100 ⋅ 0.972 + ⋯ 100 = = 3333.333 dollars. 1 − 0.97 (b) Credit multiplier = 3333.333∕100 = 33.333. The 33.333 is the factor by which the bank has increased its deposits, from $100 to $3333.333. 22. (a) The frst person deposits N dollars, and the second person deposits (1 − r)N dollars. After the second deposit, the total value of the two deposits is N + (1 − r)N dollars. (b) The third person deposits 1 − r times the amount of the second deposit, that is, (1 − r)2 N dollars. The total value of the frst three deposits is N + (1 − r)N + (1 − r)2 N dollars. (c) Each person deposits 1 − r times the amount of the preceding deposit. If the process continues forever, the total value of all deposits is the geometric series Total value = N + (1 − r)N + (1 − r)2 N + (1 − r)3 N + ⋯ . Since 0 < r < 1, we have 0 < 1 − r < 1, so the series converges. We have Total value of all deposits =

N N = dollars. 1 − (1 − r) r

Solutions for Section 10.3 1. After receiving the nth injection, the quantity in the body is 50 mg from the nth injection, 50(0.60) from the injection the previous day, 50(0.60)2 from the injection two days before, and so on. The quantity remaining from the frst injection (which has been in the body for n − 1 days) is 50(0.60)n−1 . We have Quantity after nth injection = 50 + 50(0.60) + 50(0.60)2 + ⋯ + 50(0.60)n−1 . (a) The quantity of drug in the body after the 3rd injection is Quantity after 3rd injection = 50 + 50(0.60) + 50(0.60)2 = 98 mg. Alternately, we could fnd the sum using the formula for a fnite geometric series with a = 50, r = 0.60, and n = 3: Quantity after 3rd injection =

50(1 − (0.60)3 ) = 98 mg. 1 − 0.60

(b) Similarly, we have Quantity after 7th injection = 50 + 50(0.60) + 50(0.60)2 + ⋯ + 50(0.60)6 . We use the formula for the sum of a fnite geometric series with a = 50, r = 0.60, and n = 7: Quantity after 7th injection =

50(1 − (0.60)7 ) = 121.5 mg. 1 − 0.60

(c) The steady state quantity is the long-run quantity of drug in the body if injections are continued indefnitely. In the long run, Quantity right after injection = 50 + 50(0.60) + 50(0.60)2 + ⋯ . This is an infnite geometric series with a = 50 and r = 0.60. Since −1 < r < 1, the series converges. Its sum is Quantity right after injection =

50 = 125 mg. 1 − 0.60

10.3 SOLUTIONS

691

2. In 2019, the total quantity of oil consumed was 98.3 million barrels per day, or 98.3(365) = 35,879.5 or almost 36 billion barrels per year. Each subsequent year, the quantity is multiplied by 1.00945. Thus, in 2020, we have 35.8795(1.00945) = 36.219 billion barrels per year; in 2021 we have 35.8795(1.00945)2 = 36.561 billion barrels per year; and so on. Year

2019

2020

2021

2022

2023

2024

2025

Oil (bbls/year)

35.880

36.219

36.561

36.906

37.255

37.607

37.963

In 2025, the total quantity of oil consumed is expected to be about 35.8795(1.00945)6 = 37.963 or about 38 billion barrels per year. To fnd the total consumption, we sum the geometric series: Total consumption = 35.8795 + 35.8795(1.00945) + 35.8795(1.00945)2 + ⋯ + 35.8795(1.00945)6 during 2019–2025 =

35.8795(1 − (1.00945)7 ) = 258.390 billion barrels. 1 − 1.00945

3. If consumption changes by a factor of r each year (so we use r = 0.987 in part (a) and r = 1.035 in part (b)), then consumption in 2020 is predicted to be 35.9r billion barrels, in 2021 it is predicted to be 35.9r2 billion barrels, in 2022 it is predicted to be 35.9r3 billion barrels, and so on. Thus, if Qn is total consumption in n years, in billions of barrels, Qn = 35.9r + 35.9r2 + 35.9r3 + ⋯ + 35.9rn . Using the formula for the sum of a fnite geometric series with a = 35.9r, we have Qn = 35.9r

(1 − rn ) . 1−r

Since the total reserves are 1733.9 billion barrels, in each case we try to fnd n making Qn = 1733.9. (a) Using r = 0.987, we try to solve (1 − (0.987)n ) = 1733.9 1 − 0.987 n 2725.64(1 − (0.987) ) = 1733.9 1733.9 = 0.6361 1 − (0.987)n = 2725.64 n 0.987 = 0.3639.

35.9(0.987)

Taking logarithms and using ln(Ap ) = p ln A, we have n ln(0.987) = ln(0.3639) ln(0.3639) n= = 77.25 years. ln(0.987) Thus, the oil runs out about 77.25 years after the end of 2019. (b) Using r = 1.035, we have 1 − (1.035)n = 1733.9 1 − 1.035 n −1061.61(1 − (1.035) ) = 1733.9

35.9(1.035)

1061.61((1.035)n − 1) = 1733.9 1733.9 (1.035)n − 1 = = 1.6333 1061.61 1.035n = 2.6333 Taking logarithms and using ln(Ap ) = p ln A, we have n ln(1.035) = ln(2.6333) ln(2.6333) n= = 28.1453 years. ln(1.035) Thus, the oil runs out about 28.15 years after the end of 2019.

692

Chapter Ten /SOLUTIONS

4. Since the quantity of ampicillin excreted during the time interval between tablets is 200 mg, we have Quantity of ampicillin excreted = Original quantity − Final quantity 200 = Q − (0.12)Q. Solving for Q gives, as before, Q=

200 ≈ 227.27 mg. 1 − 0.12

5. (a) The steady state quantity is the quantity of drug in the body if tablets are taken daily for the long run. Right after a tablet is taken, in the long run we have Quantity after tablet = 120 + 120(0.70) + 120(0.70)2 + ⋯ . This is an infnite geometric series with a = 120 and r = 0.70. Since −1 < r < 1, the series converges and we use the formula for the sum of an infnite geometric series: In the long run, 120 = 400 mg. 1 − 0.70 (b) Right after a tablet is taken, at the steady state there are 400 mg of the drug in the body. In one day, 30% of this quantity, or 400(0.30) = 120 mg, is excreted. This is equal to the quantity that is ingested in one tablet. Quantity right after tablet =

6. (a) The quantity of morphine in the body right after taking the 6th tablet is Quantity after 6th tablet = 30 + 30(0.25) + 30(0.25)2 + 30(0.25)3 + 30(0.25)4 + 30(0.25)5 . This is a fnite geometric series with a = 30, r = 0.25, and n = 6. We have Quantity after 6th tablet =

30(1 − (0.25)6 ) = 39.99 mg. 1 − 0.25

Since the 6th tablet contributes 30 mg toward this total, the quantity of morphine in the body right before the 6th tablet is 39.99 − 30 = 9.99 mg. (b) The steady state quantity is the long-run quantity of drug in the body if the tablets are taken every 4 hours indefnitely. In the long run, we have Quantity after tablet = 30 + 30(0.25) + 30(0.25)2 + ⋯ . This is an infnite geometric series with a = 30 and r = 0.25. Since −1 < r < 1, the series converges and we use the formula for the sum of an infnite geometric series: 30 = 40 mg. 1 − 0.25 Right before a tablet is taken, the quantity in the body is 30 mg less, so Quantity after tablet =

Quantity before tablet = 40 − 30 = 10 mg. 7. (a) After a single dose of 50 mg of fuoxetine, the quantity, Q, in the body decays exponentially, so Q = 50bt . After 3 days, half the original dose remains. We use this information to solve for b: 25 = 50b3 0.5 = b3 b = (0.5)1∕3 = 0.7937. The fraction of a dose remaining after one day is 0.7937. (b) The quantity of fuoxetine remaining right after taking the 7th dose is the sum of a fnite geometric series with a = 50, r = 0.7937, and n = 7. Quantity after 7th dose = 50 + 50(0.7937) + 50(0.7937)2 + ⋯ + 50(0.7937)6 50(1 − (0.7937)7 ) = = 194.27 mg. 1 − 0.7937 (c) In the long run, the quantity of fuoxetine in the body right after taking a dose is the sum of an infnite geometric series with a = 50 and r = 0.7937. Since −1 < r < 1, the series converges. Its sum is Quantity after dose = 50 + 50(0.7937) + 50(0.7937)2 + ⋯ . 50 = = 242.37 mg. 1 − 0.7937

10.3 SOLUTIONS

693

8. (a) We use the half-life to fnd the fraction of the drug remaining after one week. After a single dose of 100 mg of the drug, the quantity, Q, in the body decays exponentially so Q = 100bt . We use the fact that the half-life is 18 weeks to solve for b: 50 = 100b18 0.5 = b18 b = (0.5)1∕18 . The fraction remaining after one week is (0.5)1∕18 , or 0.9622238. At the steady state, the quantity right after a dose is the sum of an infnite geometric series with a = 100 and r = (0.5)1∕18 . Since −1 < r = (0.5)1∕18 = 0.9622238 < 1, the series converges and we use the formula for the sum of an infnite geometric series: Long-run quantity = 100 + 100((0.5)1∕18 ) + 100((0.5)1∕18 )2 + ⋯ 100 = 1 − (0.5)1∕18 = 2647.17 mg. (b) Since at the steady state, the quantity is 2647.17 mg, the quantity of the drug in the body eventually passes 2000 mg. How many weeks does this take? The quantity of drug in the body after the nth dose (at the start of the nth week) is a fnite geometric series with a = 100 and r = (0.5)1∕18 . We have Quantity after dose in nth week = 100 + 100((0.5)1∕18 ) + 100((0.5)1∕18 )2 + ⋯ + 100((0.5)1∕18 )n−1 100(1 − ((0.5)1∕18 )n ) = . 1 − (0.5)1∕18 We want to fnd the value of n for which this quantity is 2000. We simplify and use logarithms to solve for n. 100(1 − ((0.5)1∕18 )n ) 1 − (0.5)1∕18 1 − ((0.5)1∕18 )n 20 = 1 − (0.5)1∕18 0.755523 = 1 − ((0.5)1∕18 )n 2000 =

0.244477 = (0.5)n∕18 n ln(0.244477) = ln(0.5) 18 18 ln(0.244477) n= = 36.58 weeks. ln(0.5) The drug frst becomes e˙ective in the 37th week. 9. Since the toxin is metabolized at a continuous rate of 0.5% per day, the quantity remaining one day after consuming a single quantity of 8 micrograms is 8e−0.005 . The person is consuming 8 micrograms every day, so the total accumulated toxin the person has in the body right after consuming the toxin is the sum of 8 (from the quantity just consumed) + 8(e−0.005 ) (from the quantity consumed the previous day) + 8(e−0.005 )2 (from the quantity consumed two days ago), and so on. The total accumulation in the body is the sum of an infnite geometric series with a = 8 and r = e−0.005 . Since −1 < r = 0.9950124 < 1, the series converges to the sum: Total accumulation right after lunch = 8 + 8(e−0.005 ) + 8(e−0.005 )2 + ⋯ 8 = = 1604.0 micrograms. 1 − e−0.005 Since the person consumes 8 micrograms each day, the total accumulation of the toxin right before lunch is 1604−8 = 1596 micrograms. 10. Since nicotine leaves the body at a continuous rate of 34.65%, the fraction remaining after T hours is e−0.3465T . (a) The fraction of nicotine remaining after one hour is e−0.3465 = 0.707. The steady state is the sum of the infnite geometric series with a = 1.2 and r = 0.707. Since −1 < r < 1, the series converges and we have Long-run quantity after cigarette = 1.2 + 1.2(e−0.3465 ) + 1.2(e−0.3465 )2 + ⋯ = Thus, the nicotine quantity does not reach the lethal level.

1.2 = 4.10 mg. 1 − e−0.3465

694

Chapter Ten /SOLUTIONS (b) The fraction of nicotine remaining after 0.5 hour is e−0.3465(0.5) = e−0.17325 = 0.841. The steady state is the sum of the infnite geometric series with a = 1.2 and r = 0.841. Since −1 < r < 1, the series converges and we have Long-run quantity after cigarette = 1.2 + 1.2(e−0.17325 ) + 1.2(e−0.17325 )2 + ⋯ =

1.2 = 7.54 mg. 1 − e−0.17325

Thus, the nicotine quantity does not reach the lethal level. (c) Since 15 minutes = 0.25 hour, the fraction of nicotine remaining after 15 minutes is e−0.3465(0.25) = e−0.086625 = 0.917. The steady state is the sum of the infnite geometric series with a = 1.2 and r = 0.917. Since −1 < r < 1, the series converges and we have Long-run quantity after cigarette = 1.2 + 1.2(e−0.086625 ) + 1.2(e−0.086625 )2 + ⋯ =

1.2 = 14.46 mg. 1 − e−0.086625

Thus, the nicotine quantity does not reach the lethal level. (d) Since 6 minutes = 0.1 hour, the fraction of nicotine remaining after 6 minutes is e−0.3465(0.1) = e−0.03465 = 0.966. The steady state is the sum of the infnite geometric series with a = 1.2 and r = 0.966. Since −1 < r < 1, the series converges and we have Long-run quantity after cigarette = 1.2 + 1.2(e−0.03465 ) + 1.2(e−0.03465 )2 + ⋯ =

1.2 = 35.24 mg. 1 − e−0.03465

Thus, the nicotine quantity does not reach the lethal level. (e) Since 3 minutes = 0.05 hour, the fraction of nicotine remaining after 3 minutes is e−0.3465(0.05) = e−0.017325 = 0.983. The steady state is the sum of the infnite geometric series with a = 1.2 and r = 0.983. Since −1 < r < 1, the series converges and we have Long-run quantity after cigarette = 1.2 + 1.2(e−0.017325 ) + 1.2(e−0.017325 )2 + ⋯ =

1.2 = 69.87 mg. 1 − e−0.017325

Notice that under these circumstances (smoking a high-nicotine cigarette every 3 minutes), the quantity of nicotine in the body does reach the lethal level. In fact, the lethal level is reached in less than 6 hours. 11. If Qn denotes the total amount of the mineral mined and used in the frst n years after 2015 (that is, 2016, 2017, . . . ), we have Qn = 5000 + 5000(1.05) + 5000(1.05)2 + ⋯ + 5000(1.05)n−1 . This is a fnite geometric series with a = 5000 and r = 1.05, so we have Qn =

a(1 − rn ) 5000(1 − (1.05)n ) = = −100, 000(1 − (1.05)n ) = 100, 000((1.05)n − 1). 1−r 1 − 1.05

We want the value of n for which the total consumption, Qn , reaches 250,000 m3 , the total amount available. We can estimate n numerically (using trial and error) or graphically, or we can fnd n analytically: 100,000((1.05)n − 1) = 250,000 (1.05)n − 1 = 2.5 (1.05)n = 3.5. Taking logarithms and using ln(Ap ) = p ln A, we have n ln(1.05) = ln(3.5) ln(3.5) = 25.7 years. ln(1.05) Thus, if consumption of this mineral continues to increase at 5% a year, all reserves will be exhausted after 25 years. If, however, the relative rate of increase changes, the length of time until the reserve runs out may be very di˙erent. n=

12. Assuming consumption increases by 3% per year, consumption in 2018 was expected to be 3.67(1.03) trillion m3 . In 2019, we expect 3.67(1.03)2 to be consumed, in 2020, we expect 3.67(1.03)3 , and so on. In trillion m3 , Qn , the total consumption in n years after the end of 2017, is given by Qn = 3.67(1.03) + 3.67(1.03)2 + 3.67(1.03)3 + ⋯ + 3.67(1.03)n . If we factor out 1.03, we get � Qn = (1.03) 3.67 + 3.67(1.03) + 3.67(1.03)2 + 3.67(1.03)3 + ⋯ + 3.67(1.03)n−1 .

10.3 SOLUTIONS

695

Using the formula for the sum of a fnite geometric series, we have 1 − (1.03)n = −126.0(1 − (1.03)n ) = 126.0((1.03)n − 1). 1 − 1.03 To fnd when reserves will be exhausted, we set Qn = 193.5 trillion. Qn = 3.67(1.03)

126.0((1.03)n − 1) = 193.5 193.5 = 1.5357 (1.03)n − 1 = 126.0 n (1.03) = 2.5357 n ln(1.03) = ln(2.5357) ln(2.5357) n= = 31.5 years. ln(1.03) Thus, natural gas reserves are predicted to be exhausted 31.5 years after the end of 2017, which is the end of 2017 + 31 = 2048 plus half a year, so during the year 2049. 13. (a) In this case, we assume consumption remains 3.67 trillion m3 per year. Since reserves are 193.5 trillion m3 , the reserves are exhausted in n years, where 3.67n = 193.5 193.5 n= = 52.725 years. 3.67 (b) We assume consumption increases each year by a factor of 1.05, and we know consumption in 2017 was 3.67 trillion m3 ; in 2018, it is predicted to be 3.67(1.05)1 ; in 2019, it is predicted to be 3.67(1.05)2 , and so on. Representing total usage in n years by Qn trillion m3 , we have Qn = 3.67(1.05) + 3.67(1.05)2 + 3.67(1.05)3 + ⋯ + 3.67(1.05)n . Using the formula for the sum of a fnite geometric series, we have 1 − (1.05)n = 77.07((1.05)n − 1). 1 − 1.05 To fnd how long reserves will last, we set Qn = 193.5 and solve for n: Qn = 3.67(1.05)

77.07((1.05)n − 1) = 193.5 193.5 = 2.5107 (1.05)n − 1 = 77.07 n (1.05) = 3.5107 n ln(1.05) = ln(3.5107) ln(3.5107) n= = 25.739 years. ln(1.05) 14. Consumption of the mineral this year is 1500 kg, consumption next year is predicted to be 1500(1.04), consumption the following year is predicted to be 1500(1.04)2 and so on. Total consumption during the next n years is given by Total consumption for n years = 1500 + 1500(1.04) + 1500(1.04)2 + ⋯ + 1500(1.04)n−1 . This is a fnite geometric series with a = 1500 and r = 1.04. We have 1500(1 − (1.04)n ) . 1 − 1.04 We wish to fnd the value of n making total consumption equal to 120,000. Total consumption for n years =

1500(1 − (1.04)n ) 1 − 1.04 1 − (1.04)n 80 = 1 − 1.04 −3.2 = 1 − (1.04)n

120,000 =

4.2 = (1.04)n ln(4.2) = n ln(1.04) ln(4.2) n= = 36.59 years. ln(1.04) Assuming the percent rate of increase remains constant, the reserves of this mineral will run out in 36 or 37 years.

696

Chapter Ten /SOLUTIONS

15. Consumption of the mineral this year is 5000 m3 , consumption next year is predicted to be 5000(1.04), consumption the following year is predicted to be 5000(1.04)2 and so on. Total consumption during the next n years is given by Total consumption for n years = 5000 + 5000(1.04) + 5000(1.04)2 + ⋯ + 5000(1.04)n−1 . This is a fnite geometric series with a = 5000 and r = 1.04. We have Total consumption for n years =

5000(1 − (1.04)n ) . 1 − 1.04

We wish to fnd the value of n making total consumption equal to 350,000. 5000(1 − (1.04)n ) 1 − 1.04 1 − (1.04)n 70 = 1 − 1.04 −2.8 = 1 − (1.04)n

350,000 =

3.8 = (1.04)n ln(3.8) = n ln(1.04) ln(3.8) n= = 34.04 years. ln(1.04) If the percent rate of increase stays constant at 4% per year, the total reserve of this mineral will run out in about 34 years. 16. If the mineral is used at a constant rate of 5000 m3 per year, the total reserves of 350,000 m3 will be used up in 350,000 = 70 years. 5000 17. If usage decreases by 4% each year, then consumption of the mineral this year is 5000 m3 , consumption next year is predicted to be 5000(0.96), consumption the following year is predicted to be 5000(0.96)2 and so on. Total consumption during the next n years is given by Total consumption for n years = 5000 + 5000(0.96) + 5000(0.96)2 + ⋯ + 5000(0.96)n−1 . This is a fnite geometric series with a = 5000 and r = 0.96. We have Total consumption for n years =

5000(1 − (0.96)n ) . 1 − 0.96

If we try to fnd the value of n making total consumption equal to 350,000, we see that there are no such values of n. Why? If we consider consumption of this mineral forever under these circumstances, we have the infnite geometric series: Total consumption forever = 5000 + 5000(0.96) + 5000(0.96)2 + ⋯ . Since −1 < r = 0.96 < 1, this infnite series converges and we have Total consumption =

5000 = 125,000 m3 . 1 − 0.96

If usage of this mineral decreases by 4% per year, we can use the mineral forever and the total reserve never runs out. 18. (a) Since doses are given at time intervals equal to the half-life, the fraction of one dose remaining when the next dose is given is 0.5. At the steady state, the quantity right after a dose is given is the sum of an infnite geometric series with a = D (the size of one dose) and r = 0.5. Since −1 < r < 1, the series converges to the sum: Steady state quantity after dose = D + D(0.5) + D(0.5)2 + ⋯ D = = 2D. 1 − 0.5 Under these conditions, at the steady state, the quantity is twice the quantity of a single dose. (b) Since at the steady state, the quantity is twice the quantity of a single dose, if we want the steady state quantity to be 300 mg, each dose should be 150 mg.

SOLUTIONS to Review Problems For Chapter Ten

697

19. Right after a dose, D . 1−r At the steady state, right after a dose is given, S units of the drug are in the body. Over one time interval (the dosage interval), a fraction r of the drug remains, so the fraction (1 − r) of the drug is excreted. Thus, Steady state quantity = S = D + Dr + Dr2 + Dr3 + ⋯ =

Quantity excreted over one time interval = S ⋅ (1 − r) =

D ⋅ (1 − r) = D = Quantity of one dose. 1−r

20. The quantity of cephalexin in the body is given by Q(t) = Q0 e−kt , where Q0 = Q(0) and k is a constant. Since the half-life is 0.9 hours, 1 1 1 = e−0.9k , k = − ln .8. 2 0.9 2 (a) After 6 hours Q = Q0 e−k(6) Q0 e−0.8(6) = Q0 (0.01). Thus, the percentage of the cephalexin that remains after 6 hours

1%.

(b) Q1 = 250 Q2 = 250 + 250(0.01) Q3 = 250 + 250(0.01) + 250(0.01)2 Q4 = 250 + 250(0.01) + 250(0.01)2 + 250(0.01)3 (c) 250(1 − (0.01)3 ) 1 − 0.01 252.5 250(1 − (0.01)4 ) Q4 = 1 − 0.01 252.5 Q3 =

Thus, by the time a patient has taken three cephalexin tablets, the quantity of drug in the body has leveled o˙ to 252.5 mg. (d) Looking at the answers to part (b) shows that Qn = 250 + 250(0.01) + 250(0.01)2 + ⋯ + 250(0.01)n−1 250(1 − (0.01)n ) = . 1 − 0.01 (e) In the long run, n → ∞. So, Q = lim Qn = n→∞

250 = 252.5. 1 − 0.01

Solutions for Chapter 10 Review 1. The sum can be rewritten as 2 + 2(2) + 2(22 ) + ⋯ + 2(29 ). This is a fnite geometric series with a = 2, r = 2, and n = 10. We have Sum =

2(1 − (2)10 ) = 2046. 1−2

698

Chapter Ten /SOLUTIONS

2. We use the formula for the sum of a fnite geometric series with a = 20, r = 1.4, and n = 9. We have Sum =

20(1 − (1.4)9 ) = 983.05. 1 − 1.4

3. This is an infnite geometric series with a = 1000 and r = 1.08. Since r > 1, the series diverges and the sum does not exist. 4. We use the formula for the sum of a fnite geometric series with a = 500, r = 0.6, and n = 16. We have Sum =

500(1 − (0.6)16 ) = 1249.65. 1 − 0.6

5. This is an infnite geometric series with a = 30 and r = 0.85. Since −1 < r < 1, the series converges and we have Sum =

30 = 200. 1 − 0.85

6. This is an infnite geometric series with a = 25 and r = 0.2. Since −1 < r < 1, the series converges and we have Sum =

25 = 31.25. 1 − 0.2

7. We use the formula for the sum of a fnite geometric series with a = 1, r = 1∕2, and n = 9. We have Sum =

1 − (0.5)9 = 1.9961. 1 − 0.5

8. This is an infnite geometric series with a = 1 and r = 1∕3. Since −1 < r < 1, the series converges and we have Sum =

9. (a)

1 = 1.5. 1 − 1∕3

(i) On the night of December 31, 1999: First deposit will have grown to 2(1.04)7 million dollars. Second deposit will have grown to 2(1.04)6 million dollars. ⋯ Most recent deposit (Jan.1, 1999) will have grown to 2(1.04) million dollars. Thus Total amount = 2(1.04)7 + 2(1.04)6 + ⋯ + 2(1.04) = 2(1.04)(1 + 1.04 + ⋯ + (1.04)6 ) «¯¬ fnite geometric series

1 1 − (1.04)7 1 − 1.04 = 16.43 million dollars. = 2(1.04)

(ii) Notice that if 10 payments were made, there are 9 years between the frst and the last. On the day of the last payment: First deposit will have grown to 2(1.04)9 million dollars. Second deposit will have grown to 2(1.04)8 million dollars. ⋯ Last deposit will be 2 million dollars.

SOLUTIONS to Review Problems For Chapter Ten

699

Therefore Total amount = 2(1.04)9 + 2(1.04)8 + ⋯ + 2 = 2(1 + 1.04 + (1.04)2 + ⋯ + (1.04)9 ) «¯¬ 0

fnite geometric series

1 1 − (1.04)10 =2 1 − 1.04 = 24.01 million dollars.

(b) In part (a) (ii) we found the future value of the contract 9 years in the future. Thus Present Value =

24.01 = 16.87 million dollars. (1.04)9

Alternatively, we can calculate the present value of each of the payments separately: 2 2 2 Present Value = 2 + + +⋯+ 1.04 (1.04)2 (1.04)9 0 1 1 − (1∕1.04)10 =2 = 16.87 million dollars. 1 − 1∕1.04 Notice that the present value of the contract ($16.87 million) is considerably less than the face value of the contract, $20 million. 10. Right after a dose is given, the quantity of the drug in the body is the quantity from that dose (100 mg) plus the amounts remaining from all previous doses. In the long run Quantity of drug in the body after a dose = 100 + 100(0.82) + 100(0.82)2 + ⋯ . This is an infnite geometric series with a = 100 and r = 0.82. Since −1 < r < 1, the series converges. Using the formula for the sum: Long-run quantity, right after a dose = 100 + 100(0.82) + 100(0.82)2 + ⋯ 100 = = 555.556 mg. 1 − 0.82 Since each dose is 100 mg, the long-run quantity right before a dose is 100 mg less than the long-run quantity right after a dose. We have Long-run quantity, right before a dose = 555.556 − 100 = 455.556 mg. 11. The cost of owning the car is the sum of the original cost, $15,000, plus the repairs, in dollars: Repairs = 500 + 500(1.2) + 500(1.2)2 + ⋯ + 500(1.2)9 500(1 − (1.2)10 ) = = 12,979.34. 1 − 1.2 Thus, the total cost of the car is $15,000 + $12,979.34 = $27,979.34. 12. Since the account is paying out more than it is earning in interest, the balance will decrease and the account will eventually run out of money. When will this happen? If n payments are made, the present value of the payments is given by Present value = 10000 + 10000(1.08−1 ) + 10000(1.08−1 )2 + ⋯ + 10000(1.08−1 )n−1 . We use the fact that the present value is $100,000 and the formula for the sum of a fnite geometric series with a = 10000 and r = 1.08−1 : 10000(1 − (1.08−1 )n ) Present value = 100,000 = . 1 − 1.08−1

700

Chapter Ten /SOLUTIONS We simplify and then use logarithms to solve for n: 10000(1 − (1.08−1 )n ) 1 − 1.08−1 1 − (1.08−1 )n 10 = 1 − 1.08−1 1 − (1.08−1 )n 10 = 0.074074 0.74074 = 1 − (1.08−1 )n 100000 =

0.25926 = (1.08−1 )n 0.25926 = (1.08)−n ln(0.25926) = −n ln(1.08) ln(0.25926) n=− = 17.54. ln(1.08) The account will be able to make about seventeen and a half payments before it runs out of money. 13. The amount of additional income generated directly by people spe nding their extra money is $100(0.8) = $80 million. This additional money in turn is spent, generating another $100(0.8) (0.8) = $100(0.8)2 million. This continues indefnitely, resulting in Total additional income = 100(0.8) + 100(0.8)2 + 100(0.8)3 + ⋯ =

100(0.8) = $400 million 1 − 0.8

14. (a) The people who receive the 5 billion dollars in tax rebates spend 80% of it, for an initial amount spent of 5(0.80) = 4 billion dollars. The people who receive this 4 billion dollars spend 80% of it, for an additional amount spent of 4(0.80), and so on. We have Total amount spent = 4 + 4(0.80) + 4(0.80)2 + 4(0.80)3 + ⋯ . Notice that the initial amount spent is not the original tax rebate of 5 billion dollars, but 80% of 5 billion dollars, or 4 billion dollars. The total amount spent is an infnite geometric series with a = 4 and r = 0.80. Since −1 < r < 1, the series converges and we have Total amount spent =

4 = 20 billion dollars. 1 − 0.80

(b) If everyone who receives money spends 90% of it, then the initial amount spent is 5(0.90) = 4.5 billion dollars. The people who receive this 4.5 billion dollars spend 90% of it, and so on. We have Total amount spent = 4.5 + 4.5(0.90) + 4.5(0.90)2 + 4.5(0.90)3 + ⋯ . This is an infnite geometric series with a = 4.5 and r = 0.90. Since −1 < r < 1, the series converges and we have Total amount spent =

4.5 = 45 billion dollars. 1 − 0.90

Notice that an increase in the spending rate from 80% to 90% causes a dramatic increase in the total e˙ect on spending. 15. (a) The people who receive the N dollars in tax rebates spend N ⋅ k of it. The people who receive this money spend N ⋅ k2 of it, and so on. We have Total amount spent = Nk + Nk2 + Nk3 + ⋯ . The total amount spent is an infnite geometric series with a = Nk and r = k. Since 0 < k < 1, the series converges. We use the formula for the sum to see Nk k Total amount spent = =N . 1−k 1−k (b) We substitute k = 0.85 into the formula from part (a): k 0.85 Total amount spent = N =N = 5.667N. 1−k 1 − 0.85 The total additional spending is more than 5 times the size of the original tax rebate.

STRENGTHEN YOUR UNDERSTANDING

701

16. A person should expect to pay the present value of the bond on the day it is bought. 10 1.04 10 Present value of second payment = , etc. (1.04)2 Present value of frst payment =

Therefore, Total present value = This is a geometric series with a =

10 10 10 + + + ⋯. 1.04 (1.04)2 (1.04)3

10 1 and x = , so 1.04 1.04 Total present value =

10 1.04 1 1 − 1.04

= £250.

17. (a) The quantity of atenolol in the blood is given by Q(t) = Q0 e−kt , where Q0 = Q(0) and k is a constant. Since the half-life is 6.3 hours, 1 1 1 = e−6.3k , k = − ln .11. 2 6.3 2 After 24 hours Q = Q0 e−k(24) Q0 e−0.11(24) Q0 (0.07). Thus, the percentage of the atenolol that remains after 24 hours

7%.

(b) Q0 = 50 Q1 = 50 + 50(0.07) Q2 = 50 + 50(0.07) + 50(0.07)2 Q3 = 50 + 50(0.07) + 50(0.07)2 + 50(0.07)3 ⋮ Qn = 50 + 50(0.07) + 50(0.07)2 + ⋯ + 50(0.07)n =

50(1 − (0.07)n+1 ) 1 − 0.07

(c) P1 = 50(0.07) P2 = 50(0.07) + 50(0.07)2 P3 = 50(0.07) + 50(0.07)2 + 50(0.07)3 P4 = 50(0.07) + 50(0.07)2 + 50(0.07)3 + 50(0.07)4 ⋮ Pn = 50(0.07) + 50(0.07)2 + 50(0.07)3 + ⋯ + 50(0.07)n � 0.07(50)(1 − (0.07)n ) = 50(0.07) 1 + (0.07) + (0.07)2 + ⋯ + (0.07)n−1 = 1 − 0.07 18. Total present value, in dollars = 1000 + 1000e−0.01 + 1000e−0.01(2) + 1000e−0.01(3) + ⋯ = 1000 + 1000(e−0.01 ) + 1000(e−0.01 )2 + 1000(e−0.01 )3 + ⋯ This is an infnite geometric series with a = 1000 and x = e(−0.01) , and sum Total present value, in dollars =

1000 = 100,500.833. 1 − e−0.01

STRENGTHEN YOUR UNDERSTANDING 1. True, since there is a constant ratio of 2 between successive terms.

702

Chapter Ten /SOLUTIONS

2. False. It has 11 terms. 3. False. It is not a geometric series since the ratio of consecutive terms is not constant. 4. True. There are 6 terms and consecutive terms all have the same ratio of −2. 5. True, since this is a geometric series with 11 terms, with frst term 1 and constant ratio 1∕3. 6. False. The sum is 3(1 − 221 )∕(1 − 2). 7. False. The series diverges because the constant ratio is 3 ≥ 1. 8. False. The frst term of the series is 1∕2 so the sum is (1∕2)∕(1 − (1∕2)). 9. True. The frst term is 1∕3, so the sum is (1∕3)∕(1 − (1∕3)) = (1∕3)∕(2∕3) = 1∕2. 10. True. The frst term is 5(1∕2) so the sum is (5∕2)∕(1 − 1∕2)) = 5. 11. True, as specifed in the text. 12. True, as specifed in the text. 13. True, since without any interest, we would need to deposit 6000 ⋅ 10 = 60,000 dollars in the annuity today to pay 6000 dollars for 10 years. With the addition of interest, we can deposit less than $60,000. 14. False. For example, an initial deposit of $1000 could pay $30 a year forever at an annual interest rate of 3%. 15. False. The fourth deposit would be at the beginning of the fourth year; the value given is for the end of the fourth year. 10

.02) 16. True. The amount after the tenth deposit would be 3000(1.02)9 + 3000(1.02)8 + ⋯ + 3000 = 3000 1−(1 . 1−1.02

17. True. The account will earn 735,000(0.05) = 36, 750 dollars a year so it can generate $35,000 annual payments in perpetuity. 18. False. The present value will be less than $2000, not more. It should be 2000(1.03)−5 dollars. 19. True, since the frst payment has present value 600, the second payment one year from now has present value 600(1.04)−1 and the third payment two years from now has present value 600(1.04)−2 . 20. True, since the present value of the series of ten payments is geometric, with frst term 600 and constant ratio (1.04)−1 . 21. False. Since the person is metabolizing the drug throughout the day, the person will have less than the full 100 mg in the body. 22. False. Steady state does not mean that the level stays constant, only that the level oscillates between a maximum level right after a dose is taken and a minimum level right before a dose is taken. 1 = 60. Alternatively, if the steady level is 23. False. The total amount in the long run is 3 + 3(0.95) + 3(0.95)2 + ⋯ = 3 1−0.95 L, then the amount eliminated in one day would be 0.05L and should equal the amount consumed 3. Thus 0.05L = 3 so L = 60.

24. True. The total oil consumption for the next 5 years is given by the geometric series 10(1.02) + ⋯ 10(1.02)5 , which has frst term 10(1.02) and constant ratio 1.02. 25. False. The level goes up right after each new dose. 26. True. If the amount eliminated were less than D, then the level would increase each day. Similarly, if the amount eliminated were greater than D, then the level would decrease each day. 27. False. If the drug is administered intravenously, the amount consumed is continuous, so a di˙erential equation is a better model than geometric series. 28. True, since nine hours is 3 half-lives, so the amount drops by a factor of (1∕2)3 = 1∕8. 29. False; the correct value for long-term quantity is 50∕(1 − e−0.05 ). 30. True, since we are looking for an equilibrium solution to the di˙erential equation dQ∕dt = 50 − 0.05Q, which can be found by solving 0 = 50 − 0.05Q, yielding Q = 1000.

PROJECTS FOR CHAPTER TEN 1. (a) Each person has 2 parents plus 22 grandparents plus 23 great-grandparents, and so on. The number of ancestors going back 3 generations is 2 + 22 + 23 . We have Number of ancestors going back n generations = 2 + 22 + 23 + ⋯ + 2n . (b) We estimate that a generation is about 25 years. (Many di˙erent answers are also possible.) Using 25 years

PROJECTS FOR CHAPTER TEN

703

for a generation, there are 4 generations in 100 years, 20 generations in 500 years, 40 generations in 1000 years, and 80 generations in 2000 years. (Others answers are also possible.) (c) The number of ancestors going back n generations is a fnite geometric series with a = 2 and r = 2 and fnal term 2(2n−1). We use the formula for the sum of a fnite geometric series. Number of ancestors in 100 years = 2 + 22 + 23 + 24 2(1 − 24 ) = 1−2 = 30. Number of ancestors in 500 years = 2 + 22 + 23 + ⋯ + 220 2(1 − 220 ) = 1−2 = 2, 097, 150. Number of ancestors in 1000 years = 2 + 22 + 23 + ⋯ + 240 2(1 − 240 ) = 1−2 = about 2.2 × 1012 . Number of ancestors in 2000 years = 2 + 22 + 23 + ⋯ + 280 2(1 − 280 ) = 1−2 = about 2.4 × 1024 . There are many other possible answers. (d) If there were no common ancestors, the number of ancestors 80 generations back (about 2000 years ago) would be 280 = 1.2×1024 which is far greater than the 2×108 people actually living at that time. In fact, the number of ancestors 40 generations back is 240 = 1.1 × 1012 , which is bigger than the current population of the world. It is clear that every person on earth has had many common ancestors. 2. (a) We have f (1) = kf (0) − ℎ = kC − ℎ, f (2) = kf (1) − ℎ = k(kC − ℎ) − ℎ = k2 C − kℎ − ℎ, f (3) = kf (2) − ℎ = k(k2 C − kℎ − ℎ) − ℎ = k3 C − k2 ℎ − kℎ − ℎ. (b) We have f (1) = kC − ℎ, and f (2) = k2 C − kℎ − ℎ = k2 C − (1 + k)ℎ, and f (3) = k3 C − k2 ℎ − kℎ − ℎ = k3 C − (1 + k + k2 )ℎ. Looking at the pattern, we guess that f (n) = kn C − (1 + k + k2 + ⋯ + kn−1 )ℎ. (c) Using the formula for the sum of a fnite geometric sequence with a = 1 and r = k, we have f (n) = kn C − (1 + k + k2 + ⋯ + kn−1 )ℎ 1 − kn = kn C − ℎ. 1−k

704

Chapter Ten /SOLUTIONS

3. (a) (i) p2 (ii) There are two ways to do this. One way is to compute your opponent’s probability of winning two in a row, which is (1 − p)2 . Then the probability that neither of you win the next points is: 1 − (Probability you win next two + Probability opponent wins next two) = 1 − (p2 + (1 − p)2 ) = 1 − (p2 + 1 − 2p + p2 ) = 2p2 − 2p = 2p(1 − p). The other way to compute this is to observe either you win the frst point and lose the second or vice versa. Both have probability p(1 − p), so the probability you split the points is 2p(1 − p). (iii) Probability = (Probability of splitting next two) ⋅ (Probability of winning two after that) = 2p(1 − p)p2 (iv) Probability = (Probability of winning next two) + (Probability of splitting next two, winning two after that) = p2 + 2p(1 − p)p2 (v) The probability is: w = (Probability of winning frst two) + (Probability of splitting frst two)⋅(Probability of winning next two) + (Prob. of split. frst two)⋅(Prob. of split. next two)⋅(Prob. of winning next two) +⋯ = p2 + 2p(1 − p)p2 + (2p(1 − p))2 p2 + ⋯ . This is an infnite geometric series with a frst term of p2 and a ratio of 2p(1 − p). Therefore the probability of winning is p2 w= . 1 − 2p(1 − p) 2

(0.5) (vi) For p = 0.5, w = 1−2(0.5)(1−(0.5)) = 0.5. This is what we would expect. If you and your opponent are equally likely to score the next point, you and your opponent are equally likely to win the next game. (0.6)2 For p = 0.6, w = 1−2(0.6)(0.4) = 0.69. Here your probability of winning the next point has been magnifed to a probability 0.69 of winning the game. Thus it gives the better player an advantage to have to win by two points, rather than the “sudden death” of winning by just one point. This makes sense: when you have to win by two, the stronger player always gets a second chance to overcome the weaker player’s winning the frst point on a “fuke.” (0.7)2 For p = 0.7, w = 1−2(0.7)(0.3) = 0.84. Again, the stronger player’s probability of winning is magnifed. (0.4)2 For p = 0.4, w = 1−2(0.4)(0.6) = 0.31. We already computed that for p = 0.6, w = 0.69. Thus the value for w when p = 0.4, should be the same as the probability of your opponent winning for p = 0.6, namely 1 − 0.69 = 0.31.

PROJECTS FOR CHAPTER TEN

705

(b) (i) S = (Prob. you score frst point) +(Prob. you lose frst point, your opponent loses the next, you win the next) +(Prob. you lose a point, opponent loses, you lose, opponent loses, you win) +⋯ = (Prob. you score frst point) +(Prob. you lose)⋅(Prob. opponent loses)⋅(Prob. you win) +(Prob. you lose)⋅(Prob. opponent loses)⋅(Prob. you lose) ⋅(Prob. opponent loses)⋅(Prob. you win)+ ⋯ = p + (1 − p)(1 − q)p + ((1 − p)(1 − q))2 p + ⋯ p = 1 − (1 − p)(1 − q) (ii) Since S is your probability of winning the next point, we can use the formula computed in part (v) of (a) for winning two points in a row, thereby winning the game: w= • When p = 0.5 and q = 0.5, S=

S2 . 1 − 2S(1 − S)

0.5 = 0.67. 1 − (0.5)(0.5)

Therefore

(0.67)2 S2 = = 0.80. 1 − 2S(1 − S) 1 − 2(0.67)(1 − 0.67) • When p = 0.6 and q = 0.5, w=

(0.75)2 0.6 = 0.75 and w = = 0.9. 1 − (0.4)(0.5) 1 − 2(0.75)(1 − 0.75)

4. (a) The undiluted strength is 2 mg/ml, so the concentration of a 10-fold dilution is 0.2 mg/ml and the concentration of a 100-fold dilution is 0.02 mg/ml. Since each of the frst 11 steps lasts 15 minutes = 0.25 hour, the volume infused at each step is given, in ml, by Volume = Rate × Time = (Rate, in ml/hr) × (0.25 hr). The dose administered is given, in mg, by Dose = Concentration × Volume = (Concentration, in mg/ml) × (Rate, in ml/hr) × (0.25 hr). Thus in the frst step, the concentration is 0.02 mg/ml and Dose administered = 0.02 × 2.5 × 0.25 = 0.0125 mg. At the second step, Dose administered = 0.02 × 5.0 × 0.25 = 0.0250 mg, and so on. See the frst 11 rows in Table 10.2. (b) Using the spreadsheet to fnd the cumulative dose given in the frst 11 steps, we fnd that 25.5875 mg have been administered. To reach the target dose of 500 mg, Dose at 12t h step = 500 − 25.5875 = 474.4125 mg.

Chapter Ten /SOLUTIONS

706

The time required to deliver the 12th dose is Time =

474.4125 mg Dose = = 4.633 hours = 278 minutes. Concentration × Rate 2 × 51.2 mg∕hr

The volume infused at the last step is 4.633 ⋅ 51.2 = 237.21 ml. See the last row in Table 10.2. Table 10.2 Ratio of dose administered Concentration

Rate

Time

Volume infused

Dose infused

Cumulative

in this step to dose

(ml/hr)

(min)

per step (ml)

per step (mg)

dose (mg)

administered in previous step

2.5

0.63

0.0125

Step

Solution

(mg/ml)

100-fold dilution

0.02

100-fold dilution

0.02

5.0

1.25

0.0250

0.0375

100-fold dilution

0.02

10.0

2.50

0.0500

0.0875

100-fold dilution

0.02

20.0

5.00

0.1000

0.1875

10-fold dilution

0.2

4.0

1.00

0.2000

0.3875

10-fold dilution

0.2

8.0

2.00

0.4000

0.7875

10-fold dilution

0.2

16.0

4.00

0.8000

1.5875

10-fold dilution

0.2

32.0

8.00

1.6000

3.1875

undiluted

6.4

1.60

3.2000

6.3875

undiluted

12.8

3.20

6.4000

12.7875

undiluted

25.6

6.40

12.8000

25.5875

undiluted

51.2

278

237.21

474.4125

500

(c) The dose starts at 0.0125 mg and increases by a factor of 2 at each step. Hence if Dn is the dose at step n, we have a geometric series with terms D1 = 0.0125, D2 = 0.0125 ⋅ 2, …, and Dn = 0.0125 ⋅ 2n−1 . Thus, Total dose in 11 steps = 0.0125 + 0.0125 ⋅ 2 + ⋯ + 0.0125 ⋅ 210 (211 − 1) = 0.0125 = 25.5875 mg. 2−1 which agrees with the value computed in the spreadsheet in part (a). (d) The frst 11 ffteen minute doses take 2 hours 45 minutes. Adding the 4.63 hours, or 4 hours and 38 minutes, for the 12th dose, we see that to administer the full target dose of 500 mg requires 7 hours 23 minutes, about 7 and a half hours.

PSA A / SOLUTIONS

Prerequisite Support Appendix A Solutions for PSA A.1 EXERCISES 1. No. Reordering the expression on the left gives us 3� ⋅ �, or 3�2 , but not 4�. 2. Yes. The expression �(5�) can be reordered to 5� ⋅ �, which is equivalent to 5�2 . 3. No. Reordering the two terms gives 2� + � = � + 2�, but � + 2� is not equivalent to these. 4. Yes. The expression 5 − � is equivalent to 5 + (−�), which can be reordered to −� + 5. 5. No. The expression (2�)(2�) is equivalent to 4��, which is not equivalent to 2��. 6. No. The expression (2�)(5�) can be reordered as 2 ⋅ 5 ⋅ � ⋅ �, which is equivalent to 10��, not 7��. 7. Yes. The expression (3�)(4�)(2�) can be reordered as 3 ⋅ 4 ⋅ 2 ⋅ � ⋅ � ⋅ �, which is equivalent to 24��. 8. Yes. The expression ( + 3) + (� + 2) can be reordered as + � + 2 + 3, which is equivalent to ( + �) + 5. 9. Yes. Since 4 + � is equivalent to � + 4, the two expressions are equivalent. 10. Incorrect. For � = 2, the left side of the equation equals 32, but the right side of the equation equals 160. 11. Incorrect. For � = 2 and � = 1, the left side of the equation equals 16, but the right side equals 40. 12. Correct. Think of ℎ2 as an object. If you have 3 objects and 2 of the same object, you have 5 objects altogether. 13. Incorrect. For � = 2, the left side of the equation equals 14, but the right side equals 40. 14. We have (2� + 1) + (5� + 8) = (2� + 5�) + (1 + 8) = 7� + 9. 15. We have (4 − 2�) + (5� − 9) = (−2� + 5�) + (4 − 9) = 3� − 5. 16. We can combine the �2 terms in 3�2 − 2� 2 + 6�� − �2 to get 2�2 − 2� 2 + 6��. 17. We have (� + 1) + (� + 2) + (� + 3) = (� + � + �) + (1 + 2 + 3) = 3� + 6. 18. We can combine the �3 terms and the �� terms in �3 + 2�� − 4�3 + � − 2�� to get −3�3 + �. 19. We can combine the �4 terms in 5�4 + 5�3 − 3�4 to get 2�4 + 5�3 . 20. We have (7� + 1) + (5 − 3�) + (2� − 4) = (7� − 3� + 2�) + (1 + 5 − 4) = 6� + 2. 21. We have 5�2 + 5� + 3�2 = (5�2 + 3�2 ) + 5� = 8�2 + 5�. 22. There are no like terms to combine. 23. We have (2�)(3�) + 4� + 5� + (6�)(3�) = 6�� + 4� + 5� + 18�� = (6�� + 18��) + 4� + 5� = 24�� + 4� + 5�. 24. We have (2�)(5�) + (3�)(2�) + 5(3�) + �(3�) = 10�2 + 6�2 + 15� + 3�2 = (10�2 + 6�2 + 3�2 ) + 15� = 19�2 + 15�.

PROBLEMS 25. We regroup in the expression ( + 5) + (� − 3) + (� + 8) so that we can use the information that + � + � = 12. We have ( + 5) + (� − 3) + (� + 8) =

+ 5 + � + (−3) + � + 8

+ � + � + 5 + (−3) + 8

= ( + � + �) + (5 − 3 + 8) = ( + � + �) + 10 = 12 + 10 = 22.

PSA A / SOLUTIONS 26. We regroup and reorder in the expression (� − 10) + (� + 8) + (� − 5) so that we can use the information that � + � + � = 25. We have (� − 10) + (� + 8) + (� − 5) = � + (−10) + � + 8 + � + (−5) = � + � + � + (−10) + 8 + (−5) = (� + � + �) + (−10 + 8 − 5) = (� + � + �) − 7 = 25 − 7 = 18. 27. We regroup in the expression (3�)(2�)(5�) so that we can use the information that �� = 100. We have (3�)(2�)(5�) = (3 ⋅ 2 ⋅ 5)(��) = 30(��) = 30(100) = 3000. � 28. We regroup in the expression (2�)( )(6�) so that we can use the information that �� = 20. We have 4 � 1 (2�)( )(6�) = 2 ⋅ � ⋅ ⋅ � ⋅ 6 ⋅ � 4 4 1 = (2 ⋅ ⋅ 6) ⋅ (��) 4 2⋅6 = ⋅ (��) 4 = 3(��) = 3(20) = 60. 29. We have + 2(� − ) − 3(� + �) =

+ (� − ) + (� − ) − (� + �) − (� + �) − (� + �) ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ 2 groups

3 groups

− − +�+�−�−�−�−�−�−�

= − − � − 3�

regrouping terms collecting like terms.

30. When the speed is � and the time is �, the distance traveled is � ⋅ �. We know the car travels 200 miles, so � ⋅ � = 200. If the car travels half as fast, its new rate is (1∕2)�, and if it travels three times as long, its new time is 3�. We have Distance = Speed × Time ( ) � = ⋅ (3�) 2 ( ) 3 = ⋅ (��) 2 ( ) 3 = ⋅ (200) 2 = 300. The car travels 300 miles.

PSA A / SOLUTIONS

31. The area of a rectangle is length times width. If the original length is � and the original width is �, the area is � ⋅ �, so we have � ⋅ � = 50. Since the length is increased by 25%, the new length is � + 0.25� = 1.25�. Since the width is increased by 10%, the new width is � + 0.10� = 1.10�. We have Area = Length × Width = (1.25�)(1.10�) = (1.25 ⋅ 1.10) ⋅ (��) = (1.375) ⋅ (50) = 68.75. The area of the new rectangle is 68.75 square meters. 32. (a) The amount at the end of the month is the starting amount plus the flow in minus the flow out. We have Amount at end of January = Amount at start of January + Flow in − Flow out = 412 + � − �. The amount at the end of January is 412 + � − � billion gallons. (b) The amount at the end of February is the amount at the end of January plus the flow in minus the flow out. The flow in in February is � − 20 and the flow out in February is � + 12. We have Amount at end of February = Amount at start of February + Flow in − Flow out = (412 + � − �) + (� − 20) − (� + 12) The amount at the end of February is (412 + � − �) + (� − 20) − (� + 12) billion gallons. (c) Combining like terms, we have (412 + � − �) + (� − 20) − (� + 12) = 412 + � − � + � − 20 − � − 12 = 412 − 20 − 12 + � + � − � − � = 380 + 2� − 2�.

Solutions for PSA A.2 EXERCISES 1. Distributing the 2, we have

2. Distributing the 3�, we have

3. Distributing the −5, we have

2(� + 3�) = 2 ⋅ � + 2 ⋅ 3� = 2� + 6�.

3�(� + 4) = 3� ⋅ � + 3� ⋅ 4 = 3�2 + 12�.

−5(2� − 3) = −5 ⋅ 2� − (−5) ⋅ 3 = −10� + 15.

4. Distributing the 3 �, we have 3 �(2 − 5�) = 3 � ⋅ 2 − 3 � ⋅ 5� = 6 2 � − 15 �2 .

PSA A / SOLUTIONS 5. Distributing the 2�, we have 2�(�2 − 3� + 4) = 2� ⋅ �2 − 2� ⋅ 3� + 2� ⋅ 4 = 2�3 − 6�2 + 8�. 6. Distributing the −3�, we have −3�(5 − 3� − 2�2 ) = −3� ⋅ 5 − (−3�) ⋅ 3� − (−3�) ⋅ 2�2 = −15� + 9�2 + 6�3 . 7. Distributing the 2 through the expression 5� − 3�, we have 2(5� − 3�) + 5 = 2 ⋅ 5� − 2 ⋅ 3� + 5 = 10� − 6� + 5. 8. We distribute the 3 through the expression � − �, and we distribute the � through the resulting expression. We have 3(� − �)� = (3� − 3�)� = 3�� − 3�2 . 9. We have 2(� + 5) + 3(� − 4) = 2� + 10 + 3� − 12 = 5� − 2. 10. We have 7(� − 2) − 5(2� − 5) = 7� − 14 − 10� + 25 = −3� + 11. 11. We have 3(2� − 5) − 2(5� + 4) + 5(10 − 3�) = 6� − 15 − 10� − 8 + 50 − 15� = −19� + 27. 12. We have 6(2� + 1) − 5(3� − 4) + 6� − 10 = 12� + 6 − 15� + 20 + 6� − 10 = 3� + 16. 13. We have 2�(3� + 4) + 3(�2 − 5� + 6) = 6�2 + 8� + 3�2 − 15� + 18 = 9�2 − 7� + 18. 14. We have 3�(� + 5) − 4�(3� − 1) + 5(6� − 3) = 3�2 + 15� − 12�2 + 4� + 30� − 15 = −9�2 + 49� − 15.

PSA A / SOLUTIONS

15. We have 2 ( + �) − 5�( + �) = 2 2 + 2 � − 5 � − 5�2 = 2 2 − 3 � − 5�2 . 16. We have 5 ( + 2�) − 3�(2 − 5�) + �(2 − 1) = 5 2 + 10 � − 6 � + 15�2 + 2 2 � − � = 5 2 + 3 � + 15�2 + 2 2 �. 17. We have ��(� + 2�) + 3��(2� + �) + 5�2 � = �2 � + 2��2 + 6�2 � + 3��2 + 5�2 � = 12�2 � + 5��2 . 18. We have 5(�2 − 3� − 5) − 2�(�2 − 5� − 7) = 5�2 − 15� − 25 − 2�3 + 10�2 + 14� = −2�3 + 15�2 − � − 25. 19. First multiply � by (� + 2), then multiply 5 by (� + 2). This gives �(� + 2) = �� + 2� and 5(� + 2) = 5� + 10. Thus, (� + 5)(� + 2) = �(� + 2) + 5(� + 2) = �� + 2� + 5� + 10. 20. First multiply � by (� − 1), then multiply 3 by (� − 1). This gives �(� − 1) = �� − � and 3(� − 1) = 3� − 3. Thus, (� + 3)(� − 1) = �(� − 1) + 3(� − 1) = �� − � + 3� − 3. 21. First multiply

by ( + � + �), then multiply −� by ( + � + �), and then multiply −� by ( + � + �). This gives us ( + � + �)( − � − �) = ( + � + �) − �( + � + �) − �( + � + �) =

+ � + � − � − �2 − �� − � − �� − � 2

− �2 − 2�� − � 2 .

22. We begin by distributing the � through the (4� − 7), giving �(4� − 7)(2� + 2) = (� ⋅ 4� − � ⋅ 7)(2� + 2) = (4� 2 − 7�)(2� + 2). Now multiply 4� 2 by (2� + 2), then multiply 7� by (2� + 2). This gives us �(4� − 7)(2� + 2) = (4� 2 − 7�)(2� + 2) = 4� 2 (2� + 2) − 7�(2� + 2) = 8� 3 + 8� 2 − 14� 2 − 14� = 8� 3 − 6� 2 − 14�. 23. First multiply � by (3� + �), then multiply 2� by (3� + �). This gives us (3� + �)(� + 2�) = �(3� + �) + 2�(3� + �) = 3�2 + �� + 6�� + 2� 2 = 3�2 + 7�� + 2� 2 . 24. Taking out the common factor of �, we have 2 � − 3�� = �(2 − 3�).

PSA A / SOLUTIONS 25. Since 100 = 5 ⋅ 20, we take out the common factor of 5. We have 5� + 100 = 5 ⋅ � + 5 ⋅ 20 = 5(� + 20). 26. Taking out the common factor of 1∕5, we have � � 1 1 1 + = ⋅ � + ⋅ � = (� + �). 5 5 5 5 5 27. Taking out the common factors of 2 and �, we have 2�2 − 6� = 2�(� − 3). 28. Taking out the common factors of −1 and � and �, we have −�2 � − 3��2 = −(�2 � + 3��2 ) = −��(� + 3�). 29. Taking out the common factor of 3, we have 9�2 + 18� + 3 = 3(3�2 + 6� + 1). 30. Taking out the common factors of −2 and

and �, we have

−4 2 � − 6 �2 − 2 � = −2(2 2 � + 3 �2 + �) = −2 �(2 + 3� + 1). 31. Taking out the common factor of (� + 1), we have 5�(� + 1) + 7(� + 1) = (� + 1)(5� + 7). 32. Taking out the common factor of (� + 3), we have �(� + 3) − 6(� + 3) = (� + 3)(� − 6). 33. Taking out the common factor of 6(� − 2), we have 6�(� − 2) − 12(� − 2) = 6(� − 2)(� − 2). 34. Taking out the common factor of 2�(� + 4), we have 4 �(� + 4) − 2�(� + 4) = 2�(� + 4)(2 − 1). 35. No. Distributing the −2 through � − 4, we have −2� + 8. 36. No. Taking out the common factor 3 in the expression on the left, we have 3(�2 + 2� + 1). 37. Yes. Distributing the � through ( + � + 1) gives us the expression on the right. 38. No. Factoring out the 5 in the expression on the left gives 5(� + 20). 39. Yes. In the expression on the left, we distribute the � − 3 through � + 2 to get the expression on the right. 40. No. You cannot distribute a factor through a product. The expression on the left is equivalent to 24�� while the expression on the right is equivalent to 48��.

PSA A / SOLUTIONS

42. No. You can see that the two expressions are not equivalent by substituting values for , �, and �. For example, if � = 1, and � = 1, then the expression on the left is equal to 12 , while the expression on the right is equal to 2.

= 1,

41. No. You cannot distribute an exponent through a sum.

PROBLEMS 43. To write the expression in the form �(� + �), we factor out the coefficient of �. Factoring out 2, we have: 2� + 50 = 2(� + 25). We see that � = 2 and � = 25. 44. To write the expression in the form �(� + �), we factor out the coefficient of �. Factoring out 3, we have: 3� − 18 = 3(� − 6) = 3(� + (−6)). We see that � = 3 and � = −6. 45. To write the expression in the form �(� + �), we factor out the coefficient of �. Factoring out −5, we have: 15 − 5� = −5(−3 + �) = −5(� + (−3)). We see that � = −5 and � = −3. 46. To write the expression in the form �(� + �), we factor out the coefficient of �. Factoring out 0.05, we have: ( ) 100 0.05� + 100 = 0.05 � + = 0.05(� + 2000). 0.05 We see that � = 0.05 and � = 2000. 47. To write the expression in the form �(� + �), we factor out the coefficient of �. Factoring out 0.2, we have: ( ) 60 0.2� − 60 = 0.2 � − = 0.2(� − 300) = 0.2(� + (−300)). 0.2 We see that � = 0.2 and � = −300. 48. To write the expression in the form �(� + �), we factor out the coefficient of �. Factoring out −0.1, we have: ( ) 50 50 − 0.1� = −0.1 + � = −0.1(−500 + �) = −0.1(� + (−500)). −0.1 We see that � = −0.1 and � = −500. 49. We have

3(�2 + 2) − 3�(1 − �) = 3�2 + 6 − 3� + 3�2 = 6�2 − 3� + 6,

which is equivalent to (ii). 50. We can rewrite the fraction (� + 3)∕� by writing the division in this expression as multiplication and using the distributive law: �+3 1 1 1 3 = ⋅ (� + 3) = ⋅ � + ⋅ 3 = 1 + . � � � � � Therefore, the two expressions are equivalent. 51. The two expressions are not equivalent. To see this, observe that if � = 0, then the ﬁrst expression is equal to 0, while the second expression is equal to 1. Note that if we rewrite the second expression as a single fraction, we get 1+

� 1 1 1 3+� = ⋅ 3 + ⋅ � = (3 + �) = , 3 3 3 3 3

which is not the same as the ﬁrst expression.

PSA A / SOLUTIONS

52.

−2

−1

−(1∕2)( + 1) + 1

3∕2

1∕2

−1∕2

−(1∕2) + (1∕2)

3∕2

1∕2

−1∕2

Since the values of the two expressions are the same for each value of , the two expressions could be equivalent. Rewriting the ﬁrst expression using the distributive law and collecting like terms, we get −(1∕2)( + 1) + 1 = (−1∕2) − (1∕2) + 1 = (−1∕2) + (1∕2), so the ﬁrst expression is equivalent to the second. 53. Writing the left-hand side as (2� + 3)3 = (2� + 3)2 ( 2� + 3 ) ⏟⏞⏟⏞⏟ ⏟⏟⏟ ⏟⏟⏟ �

�

= (2� + 3)2 ⋅ 2� + (2� + 3)2 ⋅ 3, ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ �

we see that

�

= (2� + 3) , � = 2�, � = 3.

54. Writing the left-hand side as �2 (� + � + 3) =

�2 ((� + �) + 3 ) ⏟⏟⏟ ⏟⏟⏟ ⏟⏟⏟ �

�

= �2 (� + �) + 3�2 , ⏟⏞⏟⏞⏟ ⏟⏟⏟ �

we see that

�

= �2 , � = � + �, � = 3.

55. Rewriting this expression gives �(2�� + 3�) + 3�(�� − �) = 2�� + 3�� + 3�� − 3�� = 5�� = 5 ⋅ 17 = 85. 56. We have 2( + 1) − (� + 3) + (2� − �) = 2 + 2 − � − 3 + 2� − � = 2 − 2� + 2� − 1 = 2( − � + �) − 1 = 2 ⋅ 17 − 1 = 33. 57. We know that: • There are � items costing � dollars, which amounts to a cost of � ⋅ � dollars. • There are twice as many items, amounting to 2� items, each costing $1 more, or � + 1 dollars. The total cost for these items is 2� ⋅ (� + 1) dollars. We conclude that the total cost (in dollars) is given by Total cost = �� + 2�(� + 1) = �� + 2� ⋅ � + 2� ⋅ 1 = 3�� + 2�.

PSA A / SOLUTIONS

58. (a) The total output in January is � + � + �, so Output in February = 2(� + � + �). (b) The output in February from the ﬁrst factory is 2�, from the second factory is 2�, and from the third factory is 2�. Adding them up, we have Output in February = 2� + 2� + 2�. (c) Yes. Both expressions double the total output from January. Using the distributive law, we see that 2(� + � + �) = 2� + 2� + 2�. 59. (a) We have Total cost = (Cost at site 1) + (Cost at site 2) + (Cost at site 3) = 12� + 2� + 4� +14� + 5� + 3� +17� + � + 5� = 43� + 8� + 12�. (b) We have Diﬀerence = (Cost at site 1) − (Cost at site 3) = 12� + 2� + 4� − (17� + � + 5�) = −5� + � − �. (c) We have Amount = (Original budget) − (Total cost) remaining = 50� + 10� + 20� − (43� + 8� + 12�) = 7� + 2� + 8�.

PSA B / SOLUTIONS

Prerequisite Support Appendix B Solutions for PSA B.1 EXERCISES 1. We see that � = 5 is a solution. 2. We see that � = 5 is a solution. 3. We see that � = 19 is a solution. 4.

0.5� − 5 = 9 0.5� = 14 � = 28.

5. We have −7 + 9� = −4� −7 = −13� 7 = 0.538. �= 13 6. Since 3∕3 = 1, we see that � = 3 is a solution. 7. We see that � = 19 is a solution. 8. We see that � = 4 is a solution. 9. We see that � = 30 is a solution. 10. We see that � = 7 is a solution. 11. Five times a number is 20. The number must be 4. 12. Five more than a number is 20. The number is 15. 13. A number divided by 5 is 20. The number is 100. 14. Five less than a number is 20. The number is 25. 15. A number is subtracted from 5 and the result is 20. The number is −15. 16.

0.5� − 3 = 11 0.5� = 14 � = 28.

17. We have 7 + 5� = 10 − 3� 8� = 3 3 � = = 0.375. 8

PSA B / SOLUTIONS 18. We have 2(� + 3) = 10 2� + 6 = 10 2� = 4 � = 2. 19. We have −9 + 10 = −3 −9 = −13 9 = = 0.692. 13 20. We have 4� − 1.3 = −6� − 16.7 10� − 1.3 = −16.7 10� = −15.4 � = −1.54. 21. We have 100 − 24� = 5� − 30 130 = 29� 130 = 4.483. �= 29 22. We have 6� − 3 = −2� + 37 8� − 3 = 37 8� = 40 � = 5. 23. We have 1 −2 = 3 +5 2 1 = 3 +7 2 5 − =7 2 14 =− = −2.8. 5 24. We have −6(2� − 1) = 5(3 − 2�) −12� + 6 = 15 − 10� −2� = 9 �=−

9 = −4.5. 2

PSA B / SOLUTIONS 25. We have 4 − ( − 3) = 6(1 − ) 4− +3 = 6−6 7− = 6−6 7+5 = 6 5 = −1 1 = − = −0.2. 5 26. We ﬁrst distribute 53 (� + 4) to obtain: 1 5 (� + 4) = − � 3 2 5 20 1 �+ = −� 3 3 2 5 1 20 �+� = − 3 2 3 3� 3 40 5 �+ = − 3 3 6 6 8� 37 =− 3 6 ) ( ) 8� ( ) ( 3 37 3 = − 8 3 8 6 37 �=− = −2.313. 16 27. We have 0.2(� − 6) = 0.6(� − 4) 0.2� − 1.2 = 0.6� − 2.4 0.2� = 0.6� − 1.2 −0.4� = −1.2 � = 3. 28. We have −4(2� − 5) = 5 −8� + 20 = 5 −8� = −15 15 �= = 1.875. 8 29. We have 1 (� − 6) 3 1 5 = �−2 3 1 7= � 3 � = 21. 5=

PSA B / SOLUTIONS 30. We have 2 3 (3� − 12) = (4� − 3) 3 4 9 2� − 8 = 3� − 4 9 −� − 8 = − 4 23 −� = 4 23 �=− = −5.75. 4 31. We have 1 5 3� − (2� − 4) = − (� + 4) 2 4 5 3� − � + 2 = − � − 5 4 5 2� + 2 = − � − 5 4 5 2� + 7 = − � 4 13 7=− � 4 28 �=− = −2.154. 13 32. Solving for �, � − 4(� − 3(1 − �)) = 57 � − 4(� − 3 + 3�) = 57 � − 4(4� − 3) = 57 � − 16� + 12 = 57 −15� + 12 = 57 −15� = 45 � = −3. 33. Expanding yields 1.06� − 0.01(240 − �) = 22.67� 1.06� − 2.40 + 0.01� = 22.67� −21.6� = 2.40 � = −0.111. 34. We have 5(2� − 6) + 10 = 6(� + 3) + 4(2� − 1) 10� − 30 + 10 = 6� + 18 + 8� − 4 10� − 20 = 14� + 14 −4� = 34 17 34 =− = −8.5. �=− 4 2

PSA B / SOLUTIONS 35. We have 2.3(2� + 5.9) = 0.1(24.7 + 54.2�) + 2.4� 4.6� + 13.57 = 2.47 + 5.42� + 2.4� 4.6� + 13.57 = 2.47 + 7.82� 11.1 = 3.22� 11.1 = 3.447. �= 3.22

PROBLEMS 36. Dividing by � gives � = �∕�. 37. Dividing by 3� gives �=

� . 3�

38. To solve for �, we subtract � from both sides of the equation. �+� = � � = � − �. 39. To solve for �, we divide both sides of the equation by �, which is permitted since � ≠ 0. �� = � � �= . � 40. To solve for �, we subtract � from both sides of the equation, and then divide both sides of the new equation by 2. 2� + � = � 2� = � − � �−� �= . 2 41. Solving for �, � = �0 + � � 2 2(� − �0 ) = �� 2 (� − �0 ) = �. �

� � 2

� − �0 =

42. We collect all terms involving � and then factor out the �. �� − � = �� + � �� − �� = � + � �(� − �) = � + � �+� . �= �−�

PSA B / SOLUTIONS 43. Putting �0 � on the other side of the equation: � − �0 � =

1 2 �� 2

2 (� − �0 �) = �. �2 44. We collect all terms involving � and then divide by 2�:

�� + �� = � − �� 2�� = � − �� − �� = . 2� 45. We have 3�� + 1 = 2� − 5� 3�� − 2� = −5� − 1 �(3� − 2) = −5� − 1 −5� − 1 �= . 3� − 2 We can divide by 3� − 2 since it cannot be zero if � ≠ 2∕3. 46. We collect all terms involving � and then factor out the �. �(� + 2) + �(� − 3) = �(� − 1) �� + 2� + �� − 3� = �� − � �� + �� − �� = 3� − 2� − � �(� + � − �) = 3� − 2� − � 3� − 2� − � �= . �+�−� We can divide by � + � − � since it cannot be zero. 47. We must ﬁrst get all of the ’s on one side of the equation, and all of the �’s on the other side. 2 − � = + 2� − � = 2� = 3�. 48. We ﬁrst get all of the �’s on one side of the equation and all of the �’s on the other side. Then solve for �. 6� − 4� = 3� + 5� 3� − 4� = 5� 3� = 9� � = 3�.

PSA B / SOLUTIONS

49. We ﬁrst distribute, then get all of the �’s on one side of the equation and all of ℎ’s on the other. 3(3� − ℎ) = 6(� − 2ℎ) 9� − 3ℎ = 6� − 12ℎ 3� − 3ℎ = −12ℎ 3� = −9ℎ � = −3ℎ. 50. Since we are solving for , we put the terms with on one side of the equation, and the terms without on the other side. We have �� − 0.2 � + 0.1�� = 5 − 1.8� �� − 0.2 � − 5 = −0.1�� − 1.8� (�� − 0.2� − 5) = −(0.1�� + 1.8�) 0.1�� + 1.8� . =− �� − 0.2� − 5 51. We begin by using the distributive law. Then, since we are solving for �, we put terms with � on one side of the equation, and terms without � on the other side. We have (�� − �� + � ) = 25� + �(�� + � ) � � − � � + � 2 = 25� + �� + �� − �� = 25� + �� + � � − � 2 (� − ��)� = 25� + �� + � � − � 2 25� + �� + � � − � 2 �= . � − �� 52. Since we are solving for �′ , we put terms with �′ on one side of the equation, and terms without �′ on the other side. We have �2 �2 + 2��′ + �2 �′ − 5� + 2�′ + 10 = 0 2��′ + �2 �′ + 2�′ = 5� − �2 �2 − 10 (2�� + �2 + 2)�′ = 5� − �2 �2 − 10 5� − �2 �2 − 10 . �′ = 2�� + �2 + 2 53. We ﬁrst use the distributive law. Then, since we are solving for [� ], we put terms with [� ] on one side of the equation, and terms without [� ] on the other side. We have 25�0 � 2 [� ] + 10(� 2 + �0 ) = �0 (3� + �0 ) 25�0 � 2 [� ] + 10� 2 + 10�0 = 3�0 � + �0 �0 25�0 � 2 [� ] = 3�0 � + �0 �0 − 10� 2 − 10�0 [� ] =

3�0 � + �0 �0 − 10� 2 − 10�0 . 25�0 � 2

PSA B / SOLUTIONS

54. The second equation can be transformed into the ﬁrst: 4�� + 5 = 2� + 3 4�� + 2 = 2�

Subtract 3 from both sizes

0.5(4�� + 2) = 0.5(2�) Multiply both sides by 0.5 2�� + 1 = �

Original equation.

We conclude that the two equations are equivalent, which tells us they have the same solutions. Thus, � = 3 is a solution to the second equation, too.

PSA C / SOLUTIONS

Prerequisite Support Appendix C Solutions for PSA C.1 IDENTIFYING ALGEBRAIC STRUCTURE 1. The equations can be found in slope-intercept form. The line with negative slope has slope � = 7−5 = −1. The line with 0−2 = 2. By inspection, the �-intercepts are 1 and 7, so our equations are � = 2� + 1 and positive slope has slope � = 1−5 0−2 � = −� + 7. Checking with � = 2, in each equation, we verify that � = 5. � = 2 ⋅ 2 + 1 = 5 and � = −2 + 7 = 5. = −1. The line 2. The equations can be found in slope-intercept form. The line with negative slope has slope � = −5−(−3) 0−(−2)

3−(−3) = 3. By inspection, the �-intercepts are 3 and −5, so our equations are � = 3� + 3 with positive slope has slope � = 0−(−2) and � = −� − 5. Checking with � = −2, in each equation, we verify that � = −3.

� = 3 ⋅ (−2) + 3 = −3 and � = −(−2) − 5 = −3. 3. The equations can be found by inspection. The horizontal line has zero slope and the equation � = 5. The vertical line has undeﬁned slope and the equation � = 3. Clearly (3, 5) is a solution to both equations. = 4 and �-intercept 0, so its equation is � = 4�. The line with negative 4. The line with positive slope has slope � = 4−0 1−0 = −2 and using the point-slope form we ﬁnd � = 4 − 2(� − 1) = −2� + 6. slope has slope � = 4−0 1−3 Checking with � = 1, in each equation, we verify that � = 4. � = 4 ⋅ 1 = 4 and � = −2 ⋅ 1 + 6 = 4.

5. (a) The first equation says that the sum of two numbers is four, and the second equation says that the difference of the two numbers is two. (b) The numbers � = 3 and � = 1 satisfy both equations. We can see this by solving the second equation for �, which gives � = � + 2, and substituting this into the first equation, which gives (� + 2) + � = 4 2� + 2 = 4 2� = 2 � = 1. Putting � = 1 in � = � + 2 gives � = 3. 6. A solution to this system would be a set of values for � and � such that both equations are true. However, it is not possible for 2� − 3� to be equal to 5 and equal to 7 at the same time. Therefore, the system cannot have any solutions. 7. Remember that if you multiply both sides of an equation by the same number, it does not change the solution(s) or graph of the equation. Notice that by multiplying both sides of the equation � + 13� = 5 by 2, we get the equation 2� + 26� = 10. Thus these two equations are equivalent and have exactly the same solution set. The solution set of � + 13� = 5 is a line in the ��-plane; this line has infinitely many points. All of these points are solutions to the system; thus there are infinitely many. 8. The two equations are represented by lines that have different slopes. Since the lines have different slopes, they are not parallel; thus they intersect at a single point. This point is the solution to the system.

PSA C / SOLUTIONS 9. For there to be many solutions, the graphs of the two equations must be overlapping lines. This will happen if � = 10, for then the two equations describe the same line. Otherwise, the system describes two non-overlapping parallel lines, so there are no solutions. 10. The left side of the equations are not multiples of each other, so there is one solution. 11. The second equation is three times the ﬁrst one, so there are inﬁnitely many solutions. 12. The equations can be written as

{

� + 3� = 2 � − 3� = 2.

Since the left sides are not multiples of each other, there is one solution. 13. The left side of the second equation is three times the left side of the first equation; however the right sides of the two equations are equal. Thus this system of equations has no solution. 14. Multiplying both sides of the second equation by �� gives 4 1 3 ⋅ �� − ⋅ �� = ⋅ �� so the equation becomes 4� − � = 3. Thus the two equations are the same, and there are infinitely many solutions. However, neither � nor � can equal zero, so neither (0, −3) nor (3∕4, 0) is a solution. 15. The equations � + � = 23 and 2� + 2� = 46 are equivalent and have the same solution set. Thus the solution of the first system, which satisfies the equations � + � = 23 and 5� − 3� = 1, also must satisfy the equation 2� + 2� = 46. Thus, it is the solution of the second system as well. 16. The solution of the first system must be a set of values of � and � such that � + � = 23. It is therefore impossible that this solution satisfies the equation � + � = 25, so it cannot be a solution of the second system. 17. System II does not have the same solutions as system I since the signs in the equations have been switched. System III has the same solutions as system I, since the first equation in III is 2 times the second equation in I. System IV has the same solution as system I, since the first equation in IV, when multiplied by 2, is the second equation in I. The second equation is the first equation in I if the � is put on the right. System V does not have the same solutions as either systems I or II. In system V, only the first term on each side of I has been multiplied by 2. Confirmation: The solutions to the equations are: System

III

Solution

(17.5, 8.5)

(17.5, −8.5)

(17.5, 8.5)

(17.5, 17)

PROBLEMS 18. Adding the two equations to eliminate �, we have 2� = 12 � = 6. Using � = 6 in the ﬁrst equation gives

6 + � = 5,

so � = −1.

PSA C / SOLUTIONS 19. Substituting the value of � from the first equation into the second equation, we obtain � + 2(10 − 3�) = 15 � + 20 − 6� = 15 −5� = −5 � = 1. Now we substitute � = 1 into the first equation, obtaining 3(1) + � = 10, hence � = 7. 20. Substituting the value of � from the second equation into the first, we obtain 3� − 4(4� − 5) = 7 3� − 16� + 20 = 7 −13� = −13 � = 1. From the second equation, we have � = 4(1) − 5 = −1 so � = −1. 21. Substituting the value of � from the first equation into the second equation, we obtain 4(� − 9) − � = 0 4� − 36 − � = 0 3� = 36 � = 12. From the first equation, we have � = 12 − 9 = 3. 22. Since the coefficients of � are the same in each of the two equations, we can add the two equations to eliminate �. 3 = 15 = 5. To solve for �, substitute the value for

into one of the equations: 2 + 3� = 4 2(5) + 3� = 4 10 + 3� = 4 3� = −6 � = −2.

Thus,

= 5 and � = −2 is the solution.

PSA C / SOLUTIONS 23. We need to make the coefficients of one of the variables the same or negatives of each other. We can multiply the bottom equation by 3 to make the coefficients of � the same. Or, we can multiply the top equation by 2 to make the coefficients of � negatives of each other. We’ll use the latter approach. We eliminate the � and solve for �. { 2(3� − �) = 2 ⋅ 4 � + 2� = 6 { 6� − 2� = 8 � + 2� = 6. We now add the second equation to the first: 7� = 14 � = 2. To solve for �, substitute the value for � into one of the equations: � + 2� = 6 2 + 2� = 6 2� = 4 � = 2. Thus, � = 2 and � = 2 is the solution. 24. We need to make the coefficients of one of the variables the same or negatives of each other. We can multiply the top equation by 5 and the bottom equation by 2 to make the coefficients of � negatives of each other. Or, we can multiply the top equation by 2 and the bottom equation by 3 to make the coefficients of � the same. We’ll use the first approach. We eliminate the � and solve for �. { 5(2� + 3�) = 5 ⋅ 10 2(−5� + 2�) = 2 ⋅ 13 { 10� + 15� = 50 −10� + 4� = 26 19� = 76 � = 4. To solve for �, substitute the value for � into one of the equations: 2� + 3� = 10 2� + 3(4) = 10 2� + 12 = 10 2� = −2 � = −1. Thus, � = −1 and � = 4 is the solution.

PSA C / SOLUTIONS

25. We can multiply the top equation by 4 and the bottom equation by 5 to make the coefficients of � the same. Or, we can multiply the top equation by 5 and the bottom equation by 4 to make the coefficients of � the same. We’ll use the first approach. We eliminate the � and solve for �. { 4(5� + 4�) = 4 ⋅ 2 5(4� + 5�) = 5 ⋅ 7 { 20� + 16� = 8 20� + 25� = 35 −9� = −27 � = 3. To solve for �, substitute the value for � into one of the equations: 5� + 4� = 2 5� + 4(3) = 2 5� + 12 = 2 5� = −10 � = −2. Thus, � = −2 and � = 3 is the solution. 26. Solving the second equation for � gives

� = 11 − 4�.

Substituting this into the ﬁrst equation gives 8� − 3 (11 − 4�) = 7 ⏟⏞⏞⏟⏞⏞⏟ �

8� − 3 ⋅ 11 + 3 ⋅ 4� = 7 8� + 12� = 7 + 33 20� = 40 � = 2. This means � = 11 − 4 ⋅ 2 = 11 − 8 = 3, so (�, �) = (2, 3). 27. Solving the second equation for � gives

� = 5 + 2�.

Substituting this into the ﬁrst equation gives 4� + 5 (5 + 2�) = 11 ⏟⏞⏟⏞⏟ �

4� + 5 ⋅ 5 + 5 ⋅ 2� = 11 4� + 10� = 11 − 25 14� = −14 � = −1. This gives � = 5 + 2(−1) = 3, so (�, �) = (−1, 3).

PSA C / SOLUTIONS 28. Multiply the first equation by 3, 3(20� + 50�) = 3 ⋅ 15 60� + 150� = 45, and the second by −2: −2(70� + 30�) = −2 ⋅ 22 −140� − 60� = −44. Add these two equations gives 60� + 150� − 140� − 60� = 45 − 44 10� = 1 � = 0.1. This means 20� + 50(0.1) = 15 20� + 5 = 15 20� = 10 � = 0.5, so (�, �) = (0.1, 0.5). 29. Subtracting the second equation from the first gives � + � − (� − 2�) = −3 − 6 � + � − � + 2� = −9 3� = −9 � = −3. From the second equation we have � − 2(−3) = 6 �+6 = 6 � = 0, so (�, �) = (−3, 0). 30. Since � equals both 20 − 4� and 30 − 5�, we have 20 − 4� = 30 − 5� 5� − 4� = 30 − 20 � = 10. Substituting � = 10 into the first equation gives � = 20 − 4 ⋅ 10 = 20 − 40 = −20, so (�, �) = (10, −20). 31. (a) 18, 22, 29, 30 (b) 20, 21, 30 (c) 18, 19, 22, 23, 26, 27, 29.

PSA C / SOLUTIONS

32. We see from Figure C.1 that the lines cross at approximately (3, 11). To see whether � = 3, � = 11 is the solution we need to check that they satisfy { � = 6� − 7

� = 3� + 2. We see that 6� − 7 = 6(3) − 7 = 11, which is �, so the first equation is satisfied. Because 3� + 2 = 3(3) + 2 = 11, which is �, the second equation is also satisfied, so the solution is � = 3, � = 11. �

� = 6� − 7 � = 3� + 2

(3, 11)

� 2

Figure C.1 33. We see from Figure C.2 that the lines cross at approximately (1, 5). To see whether � = 1, � = 5 is the solution we need to check that they satisfy { � = −2� + 7 � = 4� + 1. We see that −2� + 7 = −2(1) + 7 = 5, which is �, so the first equation is satisfied. Because 4� + 1 = 4(1) + 1 = 5, which is �, the second equation is also satisfied, so the solution is � = 1, � = 5. � 10

� = 4� + 1

(1, 5) � = −2� + 7

� 1

Figure C.2 34. We see from Figure C.3 that the lines cross at approximately (0.47, 1.21). To see whether � = 0.47, � = 1.21 is the solution we need to check that they satisfy { 2� + 5� = 7 −3� + 2� = 1. We see that 2� + 5� = 2(0.47) + 5(1.21) = 6.99, which is approximately 7, so the first equation is essentially satisfied. Because −3� + 2� = −3(0.47) + 2(1.21) = 1.01, which is approximately 1, the second equation is also essentially satisfied.

PSA C / SOLUTIONS Thus, � = 0.47, � = 1.21 is an approximate solution. To ﬁnd the exact solution, we need solve the system algebraically, getting � = 9∕19, � = 23∕19. � −3� + 2� = 1

(0.47, 1.21) 2� + 5� = 7

� 1

Figure C.3 35. We see from Figure C.4 that the lines cross at approximately (1.5, −4). To see whether � = 1.5, � = −4 is the solution we need to check that they satisfy { � = 22 + 4(� − 8)

� = 11 − 2(� + 6). We see that 22 + 4(� − 8) = 22 + 4(1.5 − 8) = −4, which is �, so the first equation is satisfied. Because 11 − 2(� + 6) = 11 − 2(1.5 + 6) = −4, which is �, the second equation is also satisfied, so the solution is � = 1.5, � = −4. �

� = 22 + 4(� − 8)

2 � 1

(1.5, −4) −5 � = 11 − 2(� + 6) −10

Figure C.4 36. Multiply the ﬁrst equation by 5: 5(2� + 5�) = 5 ⋅ 14 10� + 25� = 70. Multiply the second equation by 2: 2(5� − 3�) = 2 ⋅ 4 10� − 6� = 8. Subtract the second of these equations from the ﬁrst: 10� + 25� − (10� − 6�) = 70 − 8 10� + 25� − 10� + 6� = 62 31� = 62 � = 2.

PSA C / SOLUTIONS

Substituting for � in the ﬁrst equation gives 2� + 5 ⋅ 2 = 14 2� = 4 � = 2, so (�, �) = (2, 2). 37. Solving by elimination, we multiply the ﬁrst equation by 7 and the second equation by 9: 7(9� + 10�) = 7 ⋅ 21

multiply ﬁrst equation by 7

63� + 70� = 147 9(7� + 11�) = 9 ⋅ 26

multiply second equation by 9

63� + 99� = 234. Subtracting the first equation from the second gives 63� + 99� − (63� + 70�) = 234 − 147 29� = 87 87 �= = 3. 29 Substituting � = 3 into the original first equation gives 9� + 10 ⋅ 3 = 21 9� + 30 = 21 9� = −9 � = −1. We can check our answer by verifying that the point (�, �) = (−1, 3) satisfies the original second equation, 7� + 11� = 26: 7(−1) + 11 ⋅ 3 = −7 + 33 = 26. 38. Using the process of elimination, we will eliminate the variable �. We multiply the first equation by 8: 8(7� + 5�) = −1 ⋅ 8 56� + 40� = −8. We multiply the second equation by −5: −5(11� + 8�) = −1(−5) −55� − 40� = 5. Adding these two equations gives 56� + 40� + (−55� − 40�) = −8 + 5 � = −3. Substituting this value of � into the original first equation gives: 7(−3) + 5� = −1 −21 + 5� = −1 5� = 20 � = 4. We can verify the solution (�, �) = (−3, 4) by substituting these values into the original second equation: 11(−3) + 8(4) = −33 + 32 = −1, as required. So the solution is (�, �) = (−3, 4).

PSA C / SOLUTIONS

39. Solving by elimination, we multiply the ﬁrst equation by 3 and the second equation by 2: 3(5� + 2�) = 3 ⋅ 1

multiply ﬁrst equation by 3

15� + 6� = 3 2(2� − 3�) = 2 ⋅ 27

multiply second equation by 2

4� − 6� = 54. Adding these two equations gives 15� + 6� + 4� − 6� = 3 + 54 19� = 57 � = 3. Substituting � = 3 into the original ﬁrst equation gives 5 ⋅ 3 + 2� = 1 15 + 2� = 1 2� = −14 � = −7. We can check our answer by verifying that the point (�, �) = (3, −7) satisﬁes the original second equation, 2� − 3� = 27: 2 ⋅ 3 − 3(−7) = 6 + 21 = 27.

40. Solving by elimination, we multiply the ﬁrst equation by 8 and the second equation by 7: 8(11� + 7�) = 8 ⋅ 2

multiply ﬁrst equation by 8

88� + 56� = 16 7(13� + 8�) = 7 ⋅ 1 multiply second equation by 7 91� + 56� = 7. Subtracting the first equation from the second gives 91� + 56� − (88� + 56�) = 7 − 16 3� = −9 � = −3. Substituting � = −3 into the original first equation gives 11(−3) + 7� = 2 −33 + 7� = 2 7� = 35 � = 5. We can check our answer by verifying that (�, �) = (−3, 5) satisfies the original second equation, 13� + 8� = 1: 13(−3) + 8 ⋅ 5 = −39 + 40 = 1.

PSA C / SOLUTIONS 41. Solving by elimination, we multiply the ﬁrst equation by 4 and the second equation by −3: 4(3� + 2� ) = 4 ⋅ 4

Multiply ﬁrst equation by 4

12� + 8� = 16 −3(4� + 5� ) = −3(−11)

Multiply second equation by −3

−12� − 15� = 33. Adding these two equations gives 12� + 8� − 12� − 15� = 16 + 33 −7� = 49 � = −7. Substituting � = −7 into the first equation gives 3� + 2(−7) = 4 3� − 14 = 4 3� = 18 � = 6. We can check our answer by verifying that (�, � ) = (6, −7) satisfies the original second equation, 4� + 5� = −11: 4 ⋅ 6 + 5(−7) = 24 − 35 = −11. 42. We will eliminate the variable �. We multiply the first equation by 3: 3(4� − 7�) = 3 ⋅ 2 12� − 21� = 6. We multiply the second equation by 4: 4(5� − 3�) = 4(−1) 20� − 12� = −4. Adding these two equations gives 12� − 21� + 20� − 12� = 6 + (−4) −� = 2, so � = −2. Substituting this value of � into the original second equation gives 5(−2) − 3� = −1 −10 − 3� = −1 −3� = 9 � = −3. We can check our answer by substituting these values of � and � into the first equation: 4� − 7� = 4(−3) − 7(−2) = 2.

PSA C / SOLUTIONS

43. One approach is to multiply the first equation by 3 and the second by 2: { 6� + 15� = 3 4� − 6� = 16. Adding these two equations, we eliminate the variable �: 6� + 15� + 4� − 6� = 3 + 16 19� = 19 � = 1. Substitution � = 1 into the first equation of the original system gives 2� + 5 ⋅ 1 = 1 2� = −4 � = −2, so the solution to this system is (�, �) = (−2, 1). 44. One approach is to multiply the first equation by 4 and the second by 3: 4(7� − 3�) = 4 ⋅ 24 28� − 12� = 96 and 3(4� + 5�) = 3 ⋅ 11 12� + 15� = 33. Adding these two equations gives: 28� − 12� + 12� + 15� = 96 + 33 43� = 129 129 �= = 3. 43 Using the second equation we can solve for �: 4� + 5 ⋅ 3 = 11 4� + 15 = 11 4� = −4 � = −1, so the solution is (�, �) = (3, −1). 45. Multiplying the first equation by 3 and the second by 7 gives { 15� − 21� = 93 14� + 21� = −35. Adding these two equations gives 15� + 14� − 21� + 21�� = 93 − 35 29� = 58 � = 2.

PSA C / SOLUTIONS Substituting � = 2 into the original first equation gives 5 ⋅ 2 − 7� = 31 −7� = 31 − 10 21 �= = −3. −7 We have (�, �) = (2, −3). 46. One approach is to multiply the first equation by 4 and the second by 7: 44� − 28� = 124 28� − 21� = 14. Adding these two equations gives 44� − 28� + 28� − 21� = 124 + 14 23� = 138 138 �= = 6. 23 Substituting � = 6 into the second equation gives 4� − 3 ⋅ 6 = 2 4� = 20 � = 5, so (�, �) = (6, 5). 47. Adding the two equations gives 3� = 33 � = 11. Substituting for � in the first equation gives 3� + 11 = 32 3� = 21 � = 7. Thus, (�, �) = (7, 11) Alternatively, we could multiply the first equation by 2: 2(3� + �) = 2 ⋅ 32 6� + 2� = 64. Subtract the second equation from this: 6� + 2� − (2� − 3�) = 64 − 1 6� + 2� − 2� + 3� = 63 9� = 63 � = 7. Substituting for � in the first equation gives 3 ⋅ 7 + � = 32 � = 32 − 31 = 11, so (�, �) = (7, 11).

PSA C / SOLUTIONS

48. One approach is to multiply the first equation by 3 and the second equation by 2: { 9� − 6� = 12 6� − 10� = −10. Adding these equations, we eliminate the variable �: 9� − 6� + 6� − 10� = 12 − 10 9� − 10� = 2 −� = 2 � = −2. Substituting � = −2 into the original second equation gives 3� − 5(−2) = −5 3� + 10 = −5 3� = −15 � = −5, so the solution to the system is (�, �) = (−2, −5). 49. Solving the first equation for � gives 3(� + � ) = 5� + � + 2 3� + 3� = 5� + � + 2 3� − � = 5� − 3� + 2 2� = 2� + 2 � = � + 1. Substituting for � in the second equation gives 4(� − �) = � + 2� − 4 4((� + 1) − �) = � + 2(� + 1) − 4 4 = � + 2� + 2 − 4 3� = 6 � = 2. This means � = � + 1 = 2 + 1 = 3, so (�, � ) = (2, 3). 50. Multiplying the second equation by 3 gives 3(2� + 3�) = 3 ⋅ 1 6� + 9� = 3. Adding this to the first equation gives 6� + 9� + 7� − 9� = 3 + 23 13� = 26 � = 2. Substituting � = 2 into the second equation gives 2 ⋅ 2 + 3� = 1 3� = 1 − 4 � = −1. So (�, �) = (2, −1).

PSA C / SOLUTIONS

51. We set the equations � = � and � = 2 − � equal to one another. � = 2−� 2� = 2 �=1 � = 1. So the point of intersection is � = 1, � = 1. 52. Substituting � = � + 1 into 2� + 3� = 13 gives 2� + 3(� + 1) = 13 5� + 3 = 13 5� = 10 � = 2. If � = 2, then � = 2 + 1 = 3. Thus, the point of intersection is � = 2, � = 3. 53. Substituting � = 2� into 2� + � = 16 gives 2� + 2� = 16 4� = 16 � = 4. Thus, substituting � = 4 into � = 2� gives � = 8, so the point of intersection is � = 4, � = 8. 54. The point of intersection lies on the two lines 1 � = − � + 3.5. 2 To ﬁnd the point, we solve this system of equations simultaneously. Setting these two equations equal to each other and solving for �, we have � = 2� − 3.5

and

1 2� − 3.5 = − � + 3.5 2 1 2� + � = 3.5 + 3.5 = 7 2 2� + .5� = 7 2.5� = 7 � = 2.8 Since � = 2.8, we have

� = 2� − 3.5 = 2(2.8) − 3.5 = 5.6 − 3.5 = 2.1. Thus, the point of intersection is (2.8, 2.1). 55. Builder � charges � = 3� + 500 and Builder � charges � = 2.5� + 750. When � = 200, Builder � charges 3 ⋅ 200 + 500 = 1100 dollars, and Builder � charges 2.5 ⋅ 200 + 750 = 1250 dollars, so Builder � is cheaper. For a 1000 square foot patio, Builder � charges 3 ⋅ 1000 + 500 = 3500 dollars and Builder � charges 2.5 ⋅ 1000 + 750 = 2250 dollars, so Builder � is cheaper. The costs are the same when we solve the system { � = 3� + 500 � = 2.5� + 750, so 3� + 500 = 2.5� + 750 0.5� = 250 250 �= = 500. 0.5 So a 500 square foot patio costs the same with either builder.

PSA C / SOLUTIONS

56. We let � and � be the two numbers, and set up two equations in two unknowns: { � + � = 17 � − � = 12, and solve for � and �. Adding the equations we get 2� = 29 29 �= , 2 and subtracting the second from the first we get 2� = 5 5 �= . 2 57. We can write a system of equations to solve this problem. If we let � be the amount of quinoa, in grams, in one serving of pasta and � be the amount of corn, in grams, in one serving of pasta we have � + � = 57. We also know that 12% of quinoa is protein and 6% of corn is protein and one serving of pasta contains 4 grams of protein. We can summarize this information in the equation 0.12� + 0.06� = 4. Therefore, we have the following system of equations: { � + � = 57 0.12� + 0.06� = 6. We can solve this system using the method of substitution or the method of elimination. Using the method of substitution, we solve the first equation for �: � = 57 − �. We substitute this for � in the second equation and solve for � ∶ 0.12� + 0.06� = 4 0.12(57 − �) + 0.06� = 4 6.84 − 0.12� + 0.06� = 4 −0.06� = 4 − 6.84 −0.06� = −2.84 � = 47.33. Since � = 47.33 and � + � = 57, we have � = 9.67. Out of the 57 grams of pasta in one serving, 47.33 grams are corn and 9.67 grams are quinoa. In other words, 83% of the pasta blend is corn and 17% is quinoa. 58. We let � represent the number of chickens and � represent the number of pigs. Each animal has one head, so we know � + � = 95. Since chickens have 2 legs and pigs have 4 legs, we know 2� + 4� = 310. We solve the system of equations: { � + � = 95 2� + 4� = 310. We can solve this system using the method of substitution or the method of elimination. Using the method of substitution, we solve the first equation for �: � = 95 − �. We substitute this for � in the second equation and solve for �: 2� + 4� = 310 2� + 4(95 − �) = 310 2� + 380 − 4� = 310 −2� = 310 − 380 −2� = −70 � = 35. Since � = 35 and � + � = 95, we have � = 60. The farmer has 35 chickens and 60 pigs.

PSA C / SOLUTIONS

59. (a) Suppose there are � small rooms and � large rooms. Each small room uses 250 ft2 , and so � small rooms use a total of 250� ft2 . Similarly, each large room uses 500 ft2 , and so � large rooms use a total of 500� ft2 . Since the total square footage for rooms is 16,000 ft2 , we have 250� + 500� = 16,000. Also, each of the � small rooms can legally hold 2 occupants for a total of 2� occupants. Similarly, each of the � large rooms can legally hold 5 occupants for a total of 5� occupants. We have Maximum number

Maximum number +

of occupants in

small rooms ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

large rooms ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

2�

5�

Maximum total occupancy ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ 150

2� + 5� = 150. (b) We must solve the system of equations given by { 250� + 500� = 16,000 2� + 5� = 150.

(equation 1) (equation 2)

Solving for � in equation 1 gives � + 2� = 64

dividing by 250

� = 64 − 2�.

(equation 3)

Substituting for � in equation 2 gives 2(64 − 2�) + 5� = 150

using equation 3 to substitute for � in equation 2

128 − 4� + 5� = 150 � = 22. From equation 3, we have � = 64 − 2� = 64 − 2(22) = 20. Thus, the solution to this system is (�, �) = (20, 22). This tells us that in order to satisfy both floor space and fire code constraints, the motel should build 20 small rooms and 22 large rooms. 60. Taking the hint, we rewrite the first two equations as 3� + 2� + 5(2�) = 11 13� + 2� = 11 and 2� − 3� + 2� = 7 4� − 3� = 7. This gives us a new system of two equations in two variables (instead of 3 equations in 3 variables): { 13� + 2� = 11 4� − 3� = 7. We now multiply the first equation by 3 and the second by 2: { 39� + 6� = 33 8� − 6� = 14.

PSA C / SOLUTIONS We add these two equations to eliminate the variable �: 47� = 47 � = 1. Given that � = 1, from the third equation of original system we have � = 2� = 2. Given that � = 1, � = 2, from the second equation of the original system we have 3� + 2� + 5� = 11 3 ⋅ 1 + 2� + 5 ⋅ 2 = 11 2� = −2 � = −1, so the solution is � = 1, � = −1, � = 2. This can also be written (�, �, �) = (1, −1, 2).

61. (a) If we let � , �, and ℎ be the price in dollars of one fish, one order of chips, and one pair of hush-puppies, respectively, then we have � + � + ℎ = 2.27 2� + � + ℎ = 3.26 � + � + 2ℎ = 2.76. If we subtract the first equation from the second, we find (2� + � + ℎ) − (� + � + ℎ) = 3.26 − 2.27, or � = 0.99. So fish cost $0.99 a piece. If we subtract the first equation from the third, we find (� + � + 2ℎ) − (� + � + ℎ) = 2.76 − 2.27, or ℎ = 0.49. So one pair of hush-puppies costs $0.49. If we rewrite the first equation as � = 2.27−� −ℎ and use � = 0.99 and ℎ = 0.49, we find � = 2.27−0.99−0.49 = 0.79. So chips cost $0.79 an order. Thus, the cost of two fish, two orders of chips, and one pair of hush-puppies is 2� + 2� + ℎ = 2 ⋅ 0.99 + 2 ⋅ 0.79 + 0.49 = 4.05, that is, $4.05. (b) By distributing the right-hand side of 2� + 2� + � = 3(� + � + �) − (� + � + 2�) we have 3(� + � + �) − (� + � + 2�) = 3� + 3� + 3� − � − � − 2� = 2� + 2� + �. (c) If we let �, �, and � be the price in dollars of one fish, one order of chips, and on pair of hush-puppies, respectively, then 2� + 2� + � is the cost of two fish, two orders of chips, and one pair of hush-puppies (what we are looking for), � + � + � is the cost of one fish, one order of chips, and one pair of hush-puppies, so � + � + � = 2.27, and � + � + 2� is the cost of one fish, one order of chips, and two pairs of hush-puppies, so � + � + 2� = 2.76. Thus, 2� + 2� + � = 3(� + � + �) − (� + � + 2�) = 3 ⋅ 2.27 − 2.76 = 4.05. (d) In the solution in part (c), we did not use the information that “Two fish, one order of chips, and one pair of hushpuppies costs $3.26” to solve part (a).

SOLUTIONS to Review Problems for PSA C

Solutions for PSA C Review EXERCISES 1. When � = 0, we have � = 4.29, which is the initial value. This tells us that the company charges $4.29 to rent the video for up two days. The rate of growth (slope) is 3.99 which represents Δdollars = 3.99 . Thus, for each day beyond the first 1 Δdays two, an additional $3.99 is charged for each additional day the video is rented. 2. The initial value of 30 tells us that the student was 30 miles from home when he noticed the formula. The rate of change (slope) is 55. Since this represents Δmiles = 551 we see that the student’s distance from home is increasing at a rate of 55 Δhours miles per hour. 3. The initial value is 250, which represents the cost of $250 to start up the business. The rate of change (slope) is 1∕36. Since this represents Δdollars = 361 we know that it costs $1 to produce 36 donuts. Δdonuts 4. The initial value is 300, which is the number of people in the lecture hall when the lecture began. The rate of change (slope) Δpeople is −19∕3. Since this represents we see that people are leaving the lecture hall at the rate of 19 people = −19 3 Δminutes every 3 minutes. 5. Writing function IV in slope-intercept form, we have � = 10,000 + 200� − 2000 = 8000 + 200�. We see that investments III and IV both begin with $8000. 6. The value of investment III is $8000, no matter what the value of �. Notice that rewriting function III as � = 8000 + 0�, we have � = 0, which means the rate of change is $0/year. 7. Investment VI begins with $8500, which is more than the other investments. 8. We start with the fixed fee, $50 and add $45 times the number of hours, ℎ. The total cost is 50 + 45ℎ dollars. 9. We take the initial population, 23,400 and subtract 200 times the number of years, �. The population is 23,400 − 200�. 10. We start with the fixed cost, $2400 and subtract $500 times the number of years, �. The total value is 2400 − 500� dollars. 11. We take the initial cost, $7600 and add $3500 times the number of months, �. The total cost is 7600 + 3500� dollars. 12. The student’s score begins with an 80 and increases 2 points times � problems, giving us 80 + 2�. 13. The initial distance from home is 200. The distance increases by 50 miles per hour for ℎ hours. The total distance from home is 200 + 50ℎ miles. 14. We start with the membership fee, $350 and add $30 times the number of months, �. The total cost is 350 + 30� dollars. 15. The starting value is � = 12,000, and the growth rate is � = 225, so ℎ(�) = 12,000 + 225�. 16. The starting value is � = 77 and the rate of change is � = −3.2 cm/year, so � = 77 − 3.2�. 17. There are initially � = 8 million viewers. The viewership drops by 4 million in 5 episodes, or by 4∕5 = 0.8 million people per episode. This gives �(�) = 8 − 0.8�. 18. The slope is � = −12, and the �-intercept is � = 220. 19. The slope is � = 1∕3, and the �-intercept is � = −11. 20. (a) Since the slopes are −2 and −4, we see that � = 5 − 2� has the greater slope. (b) Since the �-intercepts are 5 and 8, we see that � = 8 − 4� has the greater �-intercept. 21. (a) Since the slopes are 3 and −10, we see that � = 7 + 3� has the greater slope. (b) Since the �-intercepts are 7 and 8, we see that � = 8 − 10� has the greater �-intercept.

PSA C / SOLUTIONS 2 � is in slope-intercept form. Put a point on the �-axis of −4. From there, for every 1 unit you 3 move to the right, also move up 2∕3 units, which is the same as saying for every 3 units you move to the right, move up 2 units. See Figure C.5.

22. The equation � = −4 +

� � −3

3 −2

−6

Figure C.5 23. We put this in slope-intercept form, � = 2∕5 − (4∕5)�. So the �-intercept is at 2∕5. We plot the point (0, 2∕5). Then we draw a line of slope −4∕5 through this point, so we move down 4∕5 for every unit increase in �, which is the same as moving down 4 for every increase of 5 in �. See Figure C.6. � 14∕5

� −3

3 −2

Figure C.6

24. To put in standard form, we must gather all the variables, with their coeﬃcients, on one side, and the constant(s) on the other: � = 3� − 2 � − 3� = −2. Writing the variables in the more standard order, we get −3� + � = −2. 25. To put in standard form, we must gather all the variables, with their coeﬃcients, on one side, and the constant(s) on the other: 3� = 2� − 1 3� − 2� = −1.

26. Positive, because 7� must be positive. 27. Positive, because 3� must be positive. 28. Positive, because 5� must be positive. 29. Negative, because −3� must be positive. 30. Negative, because −5� must be positive. 31. Positive, because, collecting like terms, we see that 5� must be positive. 32. Positive, because, collecting like terms, we see that 13� must be positive.

SOLUTIONS to Review Problems for PSA C 33. Positive, because, collecting like terms, we see that −10� must be negative. 34. Negative, because, collecting like terms, we see that 6� must be negative. 35. Negative, because, collecting like terms, 2� must be negative. 36. No solution, because the value of 8� added to 3 can’t equal the same value added to 11. 37. Zero, because, collecting like, we see that −2� must equal 0. 38. Linear, because it can be written as −7 + 7�. 39. Not linear, because of the term in �2 . 40. Linear, because �� + � + � 3 = (� + � 3 ) + ��. 41. Not linear, because �� + � + � 3 �2 = � 3 �2 + �� + �, so there is a term in �2 . 42. Not linear. We have � (� − �)(� − � ) = � (−� 2 + (� + �)� − ��) = −� 3 + (� + �)� 2 − �� , so the expression has terms in � 2 and � 3 . 43. Linear. We have

� (� − �)(� − � ) = −� 3 + (� + �)� 2 − �� = (−� 3 + �� 2 ) + (� 2 − �� )�,

which is linear in �. 44. Linear, because � (2 + � ) − � 2 = 2� + � 2 − � 2 = 2� . 45. Not linear, because � (2 + � ) − 2� 2 = 2� + � 2 − 2� 2 = 2� − � 2 has a term in � 2 . 46. Linear, because �� + � + �� + � = (�� + �) + (� + )�. 47. Linear, because �� + � + �� + � = ( � + �) + (� + �)�. 48. Linear, because �� + � + �� + � = (�� + ��) + (� + �) . 49. Linear, because �� + � + �� + � = (�� + �) + (� + )�. 50. Using point-slope form, we have � = 2 and (�0 , �0 ) = (4, 6), so � = �0 + �(� − �0 ) � = 6 + 2(� − 4). We need to convert this to slope-intercept form. � = 6 + 2(� − 4) � = 6 + 2� − 8 � = −2 + 2�. 51. Using point-slope form, we have � = −3 and (�0 , �0 ) = (−6, 2), so � = �0 + �(� − �0 ) � = 2 − 3(� − (−6)) � = 2 − 3(� + 6). We need to convert this to slope-intercept form. � = 2 − 3(� + 6) � = 2 − 3� − 18 � = −16 − 3�.

PSA C / SOLUTIONS

52. Using point-slope form, we have � = 1∕2 and (�0 , �0 ) = (−4, −8), so � = �0 + �(� − �0 ) 1 � = −8 + (� − (−4)) 2 1 � = −8 + (� + 4). 2 We need to convert this to slope-intercept form. 1 � = −8 + (� + 4) 2 1 � = −8 + � + 2 2 1 � = −6 + �. 2 53. Using the point-slope form, we have � = −2∕3 and (�0 , �0 ) = (9, 7), so � = �0 + �(� − �0 ) 2 � = 7 − (� − 9). 3 We need to convert this to slope-intercept form. 2 � = 7 − (� − 9) 3 2 � = 7− �+6 3 2 � = 13 − �. 3 54. We ﬁrst ﬁnd the slope:

Δ� 8 − 12 −4 = = = 2. Δ� 6−8 −2 Using the point-slope form, we have � = 2 and (�0 , �0 ) = (6, 8), so �=

� = �0 + �(� − �0 ) � = 8 + 2(� − 6). We need to convert this to slope-intercept form. � = 8 + 2(� − 6) � = 8 + 2� − 12 � = −4 + 2�. Note that if we use the point (8, 12), we obtain the same answer. 55. We ﬁrst ﬁnd the slope:

Δ� 1 5−4 = = . Δ� −3 − (−6) 3 Using the point-slope form, we have � = 1∕3 and (�0 , �0 ) = (−3, 5), so �=

� = �0 + �(� − �0 ) 1 � = 5 + (� − (−3)) 3 1 � = 5 + (� + 3). 3

SOLUTIONS to Review Problems for PSA C We need to convert this to slope-intercept form. 1 � = 5 + (� + 3) 3 1 � = 5+ �+1 3 1 � = 6 + �. 3 Note that if we use the point (−6, 4), we obtain the same answer. 56. We ﬁrst ﬁnd the slope:

Δ� −2 − 1 −3 1 = = =− . Δ� 4 − (−8) 12 4 Using the point-slope form, we have � = −1∕4 and (�0 , �0 ) = (4, −2), so �=

� = �0 + �(� − �0 ) 1 � = −2 − (� − 4). 4 We need to convert this to slope-intercept form. 1 � = −2 − (� − 4) 4 1 � = −2 − � + 1 4 1 � = −1 − �. 4 Note that if we use the point (−8, 1), we obtain the same answer. 57. We ﬁrst ﬁnd the slope:

Δ� 6−3 3 3 = = =− . Δ� 5 − 10 −5 5 Using the point-slope form, we have � = −3∕5 and (�0 , �0 ) = (5, 6), so �=

� = �0 + �(� − �0 ) 3 � = 6 − (� − 5). 5 We need to convert this to slope-intercept form. 3 � = 6 − (� − 5) 5 3 � = 6− �+3 5 3 � = 9 − �. 5 Note that if we use the point (10, 3), we obtain the same answer. 58. If two lines are parallel, their slopes are equal. Thus, the slope of our line is � = 1∕2. Using the point-slope form, we have � = 1∕2 and (�0 , �0 ) = (4, 7), so � = �0 + �(� − �0 ) 1 � = 7 + (� − 4). 2

PSA C / SOLUTIONS We need to convert this to slope-intercept form. 1 � = 7 + (� − 4) 2 1 � = 7+ �−2 2 1 � = 5 + �. 2

59. If two lines are parallel, their slopes are equal. Thus, the slope of our line is � = −4∕5. Using the point-slope form, we have � = −4∕5 and (�0 , �0 ) = (−10, −5), so � = �0 + �(� − �0 ) 4 � = −5 − (� − (−10)) 5 4 � = −5 − (� + 10). 5 We need to convert this to slope-intercept form. 4 � = −5 − (� + 10) 5 4 � = −5 − � − 8 5 4 � = −13 − �. 5 60. We can write the equation in slope-intercept form 3� + 5� = 6 5� = 6 − 3� 6 3 � = − �. 5 5 . Lines parallel to this line all have slope −3 . Since the line passes through (0, 5), its �-intercept is equal to The slope is −3 5 5

5. So � = 5 − 35 �.

61. Writing 12� − 4� = 8 in slope-intercept form, we have 12� − 4� = 8 12� = 4� + 8 1 �= (4� + 8) 12 1 2 = �+ , 3 3 so the slope of this line is � = 1∕3. Since the new line is perpendicular to this line, its slope is −1∕� = −1∕(1∕3) = −3. Using point-slope form with (�0 , �0 ) = (−10, −30), we have � = −30 + (−3)(� − (−10)) = −30 − 3(� + 10) = −30 − 3� − 30 = −60 − 3�.

SOLUTIONS to Review Problems for PSA C

62. We have � = � + �� where � = −80. The line contains that points (30, 0) and (0, −80) and so �=

0 − (−80) 8 = , 30 − 0 3

so � = 83 � − 80.

63. We have �(�) = � + �� where � = 8200. The growth rate is given by �=

12,700 − 8200 = 150, 30 − 0

so �(�) = 8200 + 150�. 64. The cost is found by adding the connection fee of $2.95 to the number of minutes times the cost per minute, so � = 2.95 + 0.35�. 65. We have �(4) = 20 and �(12) = −4. This gives � = �� + � where � = (�(12) − �(4))∕(12 − 4) = (−4 − 20)∕8 = −3. Solving for �, we have �(4) = � − 3(4) � = �(4) + 3(4) = 20 + 3(4) = 32, so �(�) = −3� + 32. 66. We have �(−30) = 20 and �(70) = 140. This gives � = �� + � where � = (�(70) − �(−30))∕(70 − (−30)) = (140 − 20)∕100 = 1.2. Solving for �, we have �(−30) = � + 1.2(−30) � = �(−30) − 1.2(−30) = 20 − 1.2(−30) = 56, so �(�) = 1.2� + 56. 67. We have �(�) = � + �� where �(0.7) − �(0.3) 0.7 − 0.3 0.01 − 0.07 = 0.4 = −0.15.

�=

We can use the point (0.3, 0.07) to solve for �: �(0.3) = � + �(0.3) 0.07 = � + (−0.15)(0.3) � = 0.115. Thus, �(�) = 0.115 − 0.15�. 68. We have �(5) = 50 and �(30) = 25. This gives � = �� + � where � = (�(30) − �(5))∕(30 − 5) = (25 − 50)∕25 = −1. Solving for �, we have �(5) = � − 1(5) � = �(5) + 1(5) = 50 + 1(5) = 55, so �(�) = −� + 55. 69. Since the graph is parallel to the line � = 20 − 4�, the slope is the same, so � = −4. Using the point-slope form with (�0 , �0 ) = (3, 12) we have � = 12 − 4(� − 3) = 12 − 4� + 12 = 24 − 4�.

PSA C / SOLUTIONS

70. We have � (−1) = 4 and � (2) = −11. This gives � = �� + � where � = (� (2) − � (−1))∕(2 − (−1)) = (−11 − 4)∕3 = −5. Solving for �, we have � (−1) = � − 5(−1) � = � (−1) + 5(−1) = 4 + 5(−1) = −1, so � (�) = −5� − 1. 71. Since the function is linear, we can use any two points to find its formula. We use the form � = � + �� to get temperature in ◦ C, �, as a function of temperature in ◦ F, �. We use the two points, (32, 0) and (41, 5). We begin by finding the slope, Δ�∕Δ� = (5 − 0)∕(41 − 32) = 5∕9. Next, we substitute a point into our equation using our slope of 5∕9◦ C per ◦ F and solve to find �, the �-intercept. We use the point (32, 0): 0 = �+ −

5 ⋅ 32 9

160 = �. 9

Therefore, 160 5 + �. 9 9 Traditionally, we give this formula as � = (5∕9)(� − 32), which is often easier to manipulate. You might want to check to see if the two are the same. �=−

72. Since the function is linear, we can choose any two points (from the graph) to find its formula. We use the form � = � + �ℎ to get the price of an apartment as a function of its height. We use the two points (10, 175,000) and (20, 225,000). We begin by finding the slope, Δ�∕Δℎ = (225,000 − 175,000)∕(20 − 10) = 5000. Next, we substitute a point into our equation using our slope of 5000 dollars per meter of height and solve to find �, the �-intercept. We use the point (10, 175,000): 175,000 = � + 5000 ⋅ 10 125,000 = �. Therefore, � = 125,000 + 5000ℎ. 73. Since the function is linear, we can use any two points (from the graph) to find its formula. We use the form � = � + �� to get the meters of shelf space used as a function of the number of different medicines stocked. We use the two points (60, 5) and (120, 10). We begin by finding the slope, Δ�∕Δ� = (10 − 5)∕(120 − 60) = 1∕12. Next, we substitute a point into our equation using our slope of 1∕12 meters of shelf space per medicine and solve to find �, the �-intercept. We use the point (60, 5): 5 = � + (1∕12) ⋅ 60 0 = �. Therefore, � = (1∕12)�. The fact that � = 0 is not surprising, since we would expect that, if no medicines are stocked, they should take up no shelf space.

SOLUTIONS to Review Problems for PSA C

74. Since the function is linear, we can use any two points (from the graph) to find its formula. We use the form � = � + �� to get the number of hours of sleep obtained as a function of the quantity of tea drunk. We use the two points (4, 7) and (12, 3). We begin by finding the slope, Δ�∕Δ� = (3 − 7)∕(12 − 4) = −0.5. Next, we substitute a point into our equation using our slope of −0.5 hours of sleep per cup of tea and solve to find �, the �-intercept. We use the point (4, 7): 7 = � − 0.5 ⋅ 4 9 = �. Therefore, � = 9 − 0.5�. 75. Since we are told that the function is linear, any two points will define the line for us. We will use the form � = � + �� to get temperature in ◦ Rankine, �, as a function of temperature in ◦ Fahrenheit, �. (Rankine is a rarely used absolute temperature scale.) We choose the two points, (0, 459.7) and (10, 469.7). We begin by finding the slope, Δ�∕Δ� = (469.7 − 459.7)∕(10 − 0) = 10∕10 = 1. Next, we substitute a point into our equation using our slope of 1 ◦ R per ◦ F and solve to find �, the �-intercept. We use the point (0, 459.7): 459.7 = � + 1 ⋅ 0 459.7 = �. Therefore, � = 459.7 + 1�. ◦

◦

Note that each R is the same as each F, but the two systems choose different starting points (zero ◦ R is absolute zero, while zero ◦ F is an arbitrary point). 76. Since the function is linear, we can choose any two points to find its formula. We use the form � = � + �� to get the number of bottles sold as a function of the price per bottle. We use the two points (0.50, 1500) and (1.00, 500). We begin by finding the slope, Δ�∕Δ� = (500 − 1500)∕(1.00 − 0.50) = −2000. Next, we substitute a point into our equation using our slope of −2000 bottles sold per dollar increase in price and solve to find �, the � -intercept. We use the point (1.00, 500): 500 = � − 2000 ⋅ 1.00 2500 = �. Therefore, � = 2500 − 2000�. 77. As the � values increase by 1, the � values decrease by 1. Since there is a constant rate of change, this could represent a linear function. 78. As the � values increase by 2, the � values increase by 0.5. Since there is a constant rate of change, this could represent a linear function. 79. As the � values increase by 1, the � values increase by different amounts. Since there is not a constant rate of change, this could not represent a linear function.

PSA C / SOLUTIONS

80. As the � values increase by 3, the � values ﬁrst decrease by 1, and then increase by 1. Since there is not a constant rate of change, this could not represent a linear function. 81. Although neither the �-values nor the �-values change by a constant amount, the �-values decrease by half the amount the �-values increase. We can see this by calculating the slope between successive data points. We ﬁnd 8 − 10 1 5−8 1 1−5 1 =− , = − , and =− . 5−1 2 11 − 5 2 19 − 11 2 Thus, the data are linear with slope � = − 12 , and could represent a linear function.

82. The �-values increase by 2, 6, and then 18. The �-values decrease in steps of 5. Since there is not a constant rate of change, this is not a linear equation. 83. We eliminate the variable �. We multiply the ﬁrst equation by 4: 4(2 − 3�) = 4 ⋅ 22 8 − 12� = 88. We multiply the second equation by 3: 3(3 + 4�) = 3 ⋅ 1 9 + 12� = −3. Adding these two equations gives 8 − 12� + 9 + 12� = 88 − 3 17 = 85 = 5. Substituting this value into the original ﬁrst equation gives 2 ⋅ 5 − 3� = 22 −3� = 12 � = −4. We check our answer by substituting these values of

and � into the second equation:

3 + 4� = 3 ⋅ 5 + 4(−4) = −1. 84. We eliminate the variable �. We multiply the ﬁrst equation by 3: 3(2� + 7�) = 3 ⋅ 1 6� + 21� = 3. We multiply the second equation by −2: −2(3� + 10�) = −2 ⋅ 3 −6� − 20� = −6. Adding these two equations gives 6� + 21� + (−6� − 20�) = 3 + (−6) � = −3.

SOLUTIONS to Review Problems for PSA C Substituting this value into the original ﬁrst equation gives 2� + 7(−3) = 1 2� − 21 = 1 2� = 22 � = 11. We check our answer by substituting these values of � and � into the second equation: 3� + 10� = 3 ⋅ 11 + 10(−3) = 3. 85. We have multiplying 1st equation by 3

3(10� + 4�) = 3(−3) 30� + 12� = −9 −5(6� − 5�) = −5 ⋅ 13

multiplying 2nd equation by −5

−30� + 25� = −65 30� + 12� + (−30� + 25�) = −9 + (−65)

adding these two equations

37� = −74 � = −2. Substituting this into the first equation gives 10� + 4(−2) = −3 10� − 8 = −3 10� = 5 1 �= . 2 Thus, (�, �) = (0.5, −2). 86. We eliminate the variable �. We multiply the first equation by 3: 3(2� + 3�) = 3 ⋅ 11 6� + 9� = 33. We multiply the second equation by 2: 2(2� − 3�) = 2 ⋅ 29 4� − 6� = 58. Adding these two equations gives 6� + 9� + (4� − 6�) = 33 + 58 13� = 91 � = 7. Substituting this value of � into the original first equation gives 2� + 3 ⋅ 7 = 11 2� + 21 = 11 2� = −10 � = −5. We check our answer by substituting these values of � and � into the second equation: 2� − 3� = 2(7) − 3(−5) = 29.

PSA C / SOLUTIONS

87. We have,

�(−4) = 9 + 4(−4) = 9 − 16 = −7.

88. We have, �(� − 4) = 9 + 4(� − 4) = 9 + 4� − 16 = 4� − 7.

89. We have, �(� + ℎ) − �(�) = 9 + 4(� + ℎ) − (9 + 4�) ⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ ⏟⏞⏟⏞⏟ �(�+ℎ)

�(�)

= 9 + 4� + 4ℎ − 9 − 4� = 4ℎ.

90. We have, �(9 + 4�) = 9 + 4(9 + 4�) = 9 + 4 ⋅ 9 + 4 ⋅ 4� = 9 + 36 + 16� = 45 + 16�. 91. Dividing by � � gives �=

� . ��

92. We have 9 � + 32 5 9 � − 32 = � 5 5 � = (� − 32). 9 � =

93. Solving for �: − �� + =0 � + �� − �� =− � + �� − �� = − (� + ��) = − � − �� − �� = − � − ( � − �)� = − (� + 1) (� + 1) �=− . �−� (Since we are given � ≠ �, we know that this is ok and not a division by zero.)

SOLUTIONS to Review Problems for PSA C

94. Multiplying on both sides by � − �(1 − 2�) gives �� − � = 3(� − � + 2��) �� − � = 3� − 3� + 6�� − 6�� = 3� − 3� + � = 3� − 2� �(� − 6�) = 3� − 2� 3� − 2� �= . � − 6� (Since we are given � ≠ 6�, we know that this is ok and not a division by zero.)

PROBLEMS 95. At � = 0 minutes, there are 100 + 4 ⋅ 0 = 100 gallons in the tank, so 100 represents the number of gallons initially in the tank. After 1 minute there are 100 + 4 ⋅ 1 = 104 gallons in the tank; after 2 minutes there are 100 + 4 ⋅ 2 = 108 gallons in the tank; and so on. Thus, 4 is the number of gallons per minute that are ﬂowing into the tank. The tank is full when � = 25, at which time it will hold 100 + 4 ⋅ 25 = 200 gallons. 96. Since the total cost involves multiplication of � by constants and the addition of constants, we expect the expression to be linear in � . The facility will cost 500 + 20� + 0.20(500 + 20� ) dollars. This is linear in � and can be written as 500 + 20� + 100 + 4� = 600 + 24� . 97. Since the total cost involves multiplication of � by constants and the addition of a constant, we expect the expression to be linear in �. The class will cost 30 + 12� + 3� + 0.07(12�) dollars. This is linear in � and can be written as 30 + 15� + 0.84� = 30 + 15.84�. 98. The constant 200 is the starting cost, the cost of buying the printer. The constant 1.00 tells us the rate at which the total cost goes up with each photo printed: Cost for 1 photo = 200 + 1.00 ⋅ 1 = $201 Cost for 2 photos = 200 + 1.00 ⋅ 2 = $202 Cost for 3 photos = 200 + 1.00 ⋅ 3 = $203 ⋮ 99. Since � is the daily rental fee, 9� must represent the accumulated daily rental fees, which means that the skier had the skis for 9 days. Thus 9 represents the length of the vacation in days. 100. Since the graphs of � and � intersect at � = −2, 3, we have �(−2) = �(−2) = 1 + (−2)3 = −7. �(3) = �(3) = 1 + 33 = 28. We have �(−2) = −7 and �(3) = 28. This gives � = �� + � where � = (�(3) − �(−2))∕(3 − (−2)) = (28 − (−7))∕5 = 7. Solving for �, we have �(−2) = � + 7(−2) � = �(−2) − 7(−2) = −7 − 7(−2) = 7, so �(�) = 7� + 7.

PSA C / SOLUTIONS

101. We have � (−2) = 0.5(−2)3 − 4 = −8 � (4) = 0.5(4)3 − 4 = 28. This gives � = �� + � where � = (� (4) − � (−2))∕(4 − (−2)) = (28 + 8)∕6 = 6. Solving for �, we have � (−2) = � + 6(−2) � = � (−2) − 6(−2) = −8 − 6(−2) = 4, so � (�) = 6� + 4. 102. The towns’ populations have the same, constant rate of change. The slopes of the lines describing them are each 1000 people per decade. Since the graphs are lines with the same slope, the graphs are parallel lines. 103. The bacteria populations do not have a constant rate of change, so the graphs that describe them are not lines. For example, if one of the populations started at 100, it would have 130 bacteria after one hour and 169 bacteria after 2 hours, which is not a constant rate of change. Since the graphs are not lines, they are not parallel lines either. 104. The countries’ populations have the same, constant rate of change. The slopes of the lines describing them are each 1 million people per decade or 100 thousand people per year (since there are 10 years in a decade, and 100 thousand multiplied by 10 is 1 million). Since the graphs are lines with the same slope, the graphs are parallel lines. 105. The villages’ populations do not have the same constant rate of change, so the graphs that describe them are not parallel. The graph of Village �’s population has a slope of 2 people per year, and the graph of Village �’s population has a slope of 1 person per year. 106. The � dishes cost 23� dollars and the tip is 15% of 23� which is 0.15(23�) = 3.45� dollars. The total cost is 23� +3.45� = 26.45� dollars which is linear in �. 107. The tip you pay is (�∕100) ⋅ 15 dollars. Your total cost is 15 + (15∕100)� dollars which is linear in �. 108. If your income is � dollars then your adjusted income is � − 5(1000) dollars. Your tax is 0.25(� − 5(1000)) = 0.25� − 1250 dollars which is linear in �. Your tax is linear in your income. 109. The cost is: 9 ⋅ 2 dollars for adult tickets; 7� dollars for child tickets; 3� dollars for popcorn; and 5 dollars for parking. The total is 9 ⋅ 2 + 7� + 3� + 5 = 23 + 10� dollars which is linear in �. 110. (a) After � hours the ﬁrst person has traveled 65� miles and the second person has traveled 75(� − 1) miles. The distance between them is 65� − 75(� − 1) miles. (b) This is valid between when the second person departs and when the cars meet. The second person departs at � = 1 hour, and the cars meet when 65� − 75(� − 1) = 0 −10� + 75 = 0 � = 75∕10 = 7.5. Thus the expression is valid for 1 ≤ � ≤ 7.5. 111. At time � = 0 the pond is full and contains 2 ⋅ 10 = 20 ft3 water. After 1 day (� = 1) the depth lost is 0.3 inches = 0.3∕12 = 0.025 feet, so the volume lost is 10 ⋅ 0.025 ft3 , so (20 − 10 ⋅ 0.025) ft3 remains. After 2 days (� = 2), the depth lost is 2 ⋅ 0.3 inches = 2 ⋅ 0.025 feet, so the volume lost is 10 ⋅ 2 ⋅ 0.025 ft3 , so (20 − 10 ⋅ 2 ⋅ 0.025) ft3 remains. Thus, after � days the depth lost is � ⋅ 0.3 inches = � ⋅ 0.025 feet, so the volume lost is 10 ⋅ � ⋅ 0.025 ft3 , so 20 − 10 ⋅ � ⋅ 0.025 = 20 − 0.25� ft3 remains. This is valid from � = 0 until the pond is empty, which will occur when 20 − 0.25� = 0, that is, � = 20∕0.25 = 80 days. 112. There is a net change of water of 1 − 5 = −4 gallons per minute, which is a net loss. At � = 0 there is 100 gallons, so after 1 minute there is 100 − 4 = 96 gallons, after 2 minutes there is 100 − 4 ⋅ 2 = 92 gallons. After � minutes the container has 100 − 4� gallons. The amount of water is going down, so the container is emptying. When there are 0 gallons in the container, it is empty. This will occur when 100 − 4� = 0, so that � = 25 minutes. Thus, there are 100 − 4� gallons in the container for 0 ≤ � ≤ 25. The container is empty when � = 25. After that the statement of the problem no longer makes sense, since the container is empty and cannot be losing more water.

SOLUTIONS to Review Problems for PSA C

113. (a) We ﬁrst ﬁnd the slope using the points (20, 68) and (25, 77). �=

Δ� 77 − 68 9 = = . Δ� 25 − 20 5

Using the point-slope form, we have � = 9∕5 and (�0 , �0 ) = (20, 68), so � = �0 + �(� − �0 ) 9 � = 68 + (� − 20). 5 We can rewrite the function in slope-intercept form. 9 � = 68 + (� − 20) 5 9 � = 68 + � − 36 5 9 � = 32 + �. 5 (b)

(i) Using the function from part (a), we can convert the temperature of 10◦ C to Fahrenheit. 9 � = 32 + � 5 9 � = 32 + (10) 5 � = 32 + 18 � = 50. Thus, a temperature of 10◦ C is equal to 50◦ F. (ii) Using the function from part (a), we can convert the temperature of 86◦ F to Celsius. 9 � = 32 + � 5 9 86 = 32 + � 5 9 54 = � 5 � = 30. Thus, a temperature of 86◦ F is equal to 30◦ C.

114. (a) Since the temperature drops about 3◦ F per 1000 foot rise, we know that the slope can be represented by: �=

Δ� 3 =− . Δ� 1000

Using the point-slope form, we have � = −3∕1000 and (�0 , �0 ) = (7000, 53), so � = �0 + �(� − �0 ) 3 � = 53 − (� − 7000) 1000 3 � = 53 − (� − 7000). 1000 (b)

(i) Using the function from part (a), we can ﬁnd the average temperature at Sun Visitor Center, which is located at 9745 feet. � = 53 −

3 (� − 7000) 1000

PSA C / SOLUTIONS 3 (9745 − 7000) 1000 3 � = 53 − (2745) 1000 � = 53 − 8.235 � = 53 −

� = 44.765. Thus, we can expect that the average temperature at Sun Visitor Center is approximately 45◦ F. (ii) Using the function from part (a), we can approximate the average temperature at sea level (elevation = 0). 3 (� − 7000) 1000 3 � = 53 − (0 − 7000) 1000 3 � = 53 − (−7000) 1000 � = 53 + 21 � = 53 −

� = 74. Thus, we can approximate the average temperature at sea level to be 74◦ F. 115. (a) We ﬁrst ﬁnd the slope using the points (3, 12.5) and (6, 17). �=

Δ� 17 − 12.5 4.5 = = = 1.5. Δ� 6−3 3

Using the point-slope form, we have � = 1.5 and (�0 , �0 ) = (3, 12.5), so � = �0 + �(� − �0 ) � = 12.5 + 1.5(� − 3). (b) � = 12.5 + 1.5(� − 3) � = 12.5 + 1.5� − 4.5 � = 8 + 1.5�. (c) We can use the function in part (b) to determine the slope and the intercept. In the function � = 8 + 1.5�, the slope is 1.5 and the intercept is 8. Δ�(cost) 1.5 . Thus, we can see that each additional ride costs $1.50. The slope of represents 1 Δ�(number of rides) The intercept of 8 tell us that the cost is $8 for no rides. Thus, the cost of admission to the carnival is $8. 116. (a) We ﬁrst ﬁnd the slope using the points (100, 320) and (500, 400). Δ� 400 − 320 80 = = = 0.2. Δ� 500 − 100 400 Using the point-slope form, we have � = 0.2 and (�0 , �0 ) = (100, 320), so �=

� = �0 + �(� − �0 ) � = 320 + 0.2(� − 100). (b) � = 320 + 0.2(� − 100) � = 320 + 0.2� − 20 � = 300 + 0.2�.

SOLUTIONS to Review Problems for PSA C

The slope is 0.2 and the intercept is 300. Δ�(cost) 0.2 . Thus, we can see that it costs $0.20 to produce each CD. The slope of represents 1 Δ�(number of CDs) The intercept of 300 tell us that the cost is $300 to produce 0 CDs. This $300 is the start-up cost of producing CDs. (c) To determine the cost of producing 750 CDs, we substitute 750 for � in our equation. � = 300 + 0.2� � = 300 + 0.2(750) � = 300 + 150 � = 450. Thus, it costs $450 to produce 750 CDs. (d) To determine the number of CDs that can be produced for $500, we substitute 500 for � in our equation. � = 300 + 0.2� 500 = 300 + 0.2� 200 = 0.2� � = 1000. If the band has $500, it can produce 1000 CDs. 117. The timeshare was bought at time 0, so the purchase price, including the ﬁrst maintenance fee, was $3000. After 1 year the total cost was about $3,500, so the maintenance fee was $3500 − $3000 = $500. Thus, the timeshare cost $3000 − $500 = $2500. 118. If is the number of pounds of apples purchased and � is the number of pounds of pears, then the total cost of the purchase is 0.99 + 1.25� dollars. If I spend $4 then 4 = 0.99 + 1.25�. If the weight of the apples I buy is twice the weight of the pears then = 2�. Thus, 4 = 0.99 × 2� + 1.25�, so � = 4∕(1.98 + 1.25) = 1.238 pounds. 119. Since the sum of the two distances represented by 2� and 15 equals the distance represented by 35, we have the equation 2� + 15 = 35. Then 2� must be 20. Therefore � = 10. 120. (iv), because 7� must be positive but less than 7. 121. (v), because 4� must be greater than 4. 122. (ii), because 2� must be negative but greater than −2. 123. (ii), because 11� must be negative but greater than −11. 124. (v), because 11� is greater than 11. 125. (iv), because 2� must be positive but less than 2. 126. (i), because, collecting terms, −2� equals 2. 127. (ii), because, collecting terms, 3� is negative but greater than −3. 128. (vi), because the quantity −4� added to 3 can’t equal the same quantity added to −3. 129. (iii), because, collecting terms, we ﬁnd that 3� = 0. 130. (iv), because �∕2 is less than 1∕2. 131. (v), because 2∕� is less than 2. 132. (iii), because (� + 8)∕2 = 4 = (0 + 8)∕2. 133. (i), because (� + 8)∕3 is less than (−1 + 8)∕3 = 7∕3. 134. (v), because the denominator must be 10. 135. (iii), because the denominator must be 5. 136. (vi), because the numerator must equal the denominator, but 2� plus 5 can’t equal the same quantity (2�) plus 6. 137. (i), because 2� + 2 = 3� + 5.

PSA C / SOLUTIONS

1 = 4. � 1 139. (vi), because can’t equal 0. � 140. (a) We need to solve � = 9�∕5+32 for �. We subtract 32 from both sides and multiply by 5∕9 to obtain 5(� −32)∕9 = �, so � = 5(� − 32)∕9. (b) When � = 350 we have � = 5(350 − 32)∕9 = 176.7◦ C. When � = 450 we have � = 5(450 − 32)∕9 = 232.2◦ C. 138. (iv), because

141. (a) Since number of degrees Fahrenheit is twice the number of degrees Celsius, we have � = 2�. (b) We need to solve the system of linear equations 9 � + 32 5 � = 2�,

� =

for � and �. Substituting the second equation into the ﬁrst gives 2� = 9�∕5 + 32, so � = 160. This is equivalent to � = 2 ⋅ 160 = 320. Thus, the two formulas agree when � = 320 and � = 160, that is, 320◦ F = 160◦ C. (c) The equations 9 � + 32 5 � = 2�,

� =

represent two lines which intersect at � = 160, which is below the lowest temperature, 176.7◦ C, used in the oven. Thus, the vertical distance between the two lines increases as the temperature increases. So this distance is largest (maximum error) at the upper end of this temperature range. � Formulas agree when � = 160

500 400

300 200 100

150

200

250

�

Figure C.7

142. (a) When � = 21 we have � = 9 ⋅ 21∕5 + 32 = 69.8, which is about 70◦ F. (b) No. When � = 42 we have � = 9 ⋅ 42∕5 + 32 = 107.6◦ F. 143. The rate of discharge is constant, so we assume � is a linear function with slope � = −0.25 mAh/s. Notice that � is negative because the battery is discharging, so the amount of charge is going down. The initial charge is � = 4500 mAh, so � (�) = 4500 − 0.25�. 144. The maximum charge is given by �(0) = 3200 mAh. This is 1300 mAh less, or about 29% less, than the new battery. The battery discharges at a rate of 0.4 mAh/s. This is 0.15 mAh/s faster, or 60% faster, than the new battery. 145. We have � (�1 ) = 0 4500 − 0.25�1 = 0 4500 �1 = = 18,000. 0.25 �(�2 ) = 0 3200 − 0.4�2 = 0

SOLUTIONS to Review Problems for PSA C

3200 = 8000. 0.4 This means that the new battery fully discharges after 18,000 seconds, and the old battery fully discharges after 8000 seconds. The expression �1 18,000 = = 2.5 8, 000 �2 tells us that the new batter lasts 2.5 times as long as the old battery. �2 =

146. The bird uses 5 calories for each hour spent singing for a total of 5� calories, and 10 calories for each hour spent defending its territory for a total of 10� calories. We have Calories spent singing

Calories spent defending territory

= 60

5� + 10� = 60. Setting � = 0, we see that the �-intercept is � = 6, which means the bird could spend a maximum of 6 hours defending its territory if it spends no time singing. Setting � = 0, we see that the �-intercept is � = 12, which means the bird could spend a maximum of 12 hours singing if it spends no time defending its territory. See Figure C.8. � (hours) 6 5 4 3 2 1 0

� (hours)

Figure C.8 147. The bird uses 4 calories for each hour spent singing for a total of 4� calories, and 12 calories for each hour spent defending its territory for a total of 12� calories. We have Calories spent singing

Calories spent defending territory

= 60

4� + 12� = 60. Setting � = 0, we see that the �-intercept is � = 5, which means the bird could spend a maximum of 5 hours defending its territory if it spends no time singing. Setting � = 0, we see that the �-intercept is � = 15, which means the bird could spend a maximum of 15 hours singing if it spends no time defending its territory. See Figure C.9. � (hours) 5 4 3 2 1 0

Figure C.9

� (hours)

PSA C / SOLUTIONS

148. We have a simultaneous system of linear equations: { 5� + 10� = 60 4� + 12� = 60. Multiplying the ﬁrst equation by 4 and the second by 5 gives { 20� + 40� = 240 20� + 60� = 300. Subtracting 20� + 40� = 240 from 20� + 60� = 300 gives 20� = 60 � = 3. Solving for � using the ﬁrst equation gives 5� + 10 ⋅ 3 = 60 5� = 30 � = 6. This means that both birds can spend 6 hours singing and 3 hours defending their territories with a budget of 60 calories. 149. (a) This matches equation IV: Total amount spent each week = 60 Amount spent on Italian ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟

Amount spent on Kenyan ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟

10�

= 60

15�

10� + 15� = 60. (b) This matches equation II: Total − Amount spent on Italian Price for Kenyan 60 − 10� = . 15

Amount of Kenyan =

(c) The total amount of Kenyan that can be purchased is 4 lbs, because 60∕15 = 4. Thus, this statement matches equation III: ( ) Amt. of Tot. possible Amt. of Kenyan = 1.5 − Italian amt. of Kenyan purchased = 1.5(4 − �). (d) The total amount of Italian that can be purchased is 60∕10 = 6 lbs, so this matches equation I: Amt. of Italian

Tot. possible amt.

of Italian � = 6−3⋅ 2 3 = 6 − �. 2

− 3 lbs. ×

Amt. Kenyan 2 lbs.

SOLUTIONS to Review Problems for PSA C 150. We substitute the expression

3 � + 6 for � in the ﬁrst equation. 4 2� + 4� = 44 ( ) 3 2� + 4 � + 6 = 44 4 12 2� + � + 24 = 44 4 5� = 20

� = 4. From the second equation, we have 3 ⋅4+6 4 � = 9. �=

151. One way to solve this system is by substitution. Solve the ﬁrst equation for �: 3� − � = 20 −� = 20 − 3� � = 3� − 20. In the second equation, substitute the expression 3� − 17 for �: −2� − 3� = 5 −2� − 3(3� − 20) = 5 −2� − 9� + 60 = 5 −11� = 5 − 60 = −55 −55 �= = 5. −11 Since � = 5 and � = 3� − 20, we have

� = 3(5) − 20 = 15 − 20 = −5. Thus, the solution to the system is � = 5 and � = −5. Check your results by substituting the values into the second equation: −2� − 3� = 5 Substituting, we get − 2(5) − 3(−5) = 5 −10 + 15 = 5 5 = 5. 152. From the first equation, we get 2� + 2� = 5. From the second equation we get −2� − � = −9. So, { 2� + 2� = 5 −2� − � = −9. Adding the two equations gives � = −4, and solving for � in either equation gives � = 6.5. 153. Because we are solving for � and �, we regard � as a constant. Multiplying the first equation by � and subtracting the second gives �2 � + �� = 2�2 � + �� = 1 + �2 so, subtracting (�2 − 1)� = �2 − 1. Since � ≠ ±1, we have �2 − 1 ≠ 0, so � = 1. Solving for � in the first equation gives � = 2� − �(1), so � = �.

PSA C / SOLUTIONS

154. Subtracting the second equation from the ﬁrst gives 3 2 − = 2� − � = �. 4 3 Thus, 1 . 12 So a UK tablespoon is 1∕12 a US cup. If we substitute this into �=

2 1 = �+� 3 2 we ﬁnd

2 1 1 = �+ , 3 2 12

2 1 1 − = �, 3 12 2

giving �=

7 . 6

So one UK cup is 7∕6 a US cup. 155. Multiplying the second equation by 2 and adding it to the ﬁrst gives 1+

1 3 1 = � + �, 2 4 2

or, 3 5 = �. 2 4 Thus, 6 . 5 This says that a UK cup is 6∕5 a US cup. If we substitute this into �=

1 1 = �−� 4 4 we ﬁnd

1 1 6 = ⋅ − �, 4 4 5

giving �=

1 . 20

This says that a UK dessertspoon is 1∕20 a UK cup.

Solutions for Solving Drill 1. We have 5� + 2 = 12 5� = 10 � = 2.

SOLUTIONS to Review Problems for PSA C 2. We have 7 − 3� = 25 −3� = 18 � = −6.

3. We have 5 + 9� = 72 9� = 67 67 �= . 9 4. We have 10 = 25 − 3� −15 = −3� � = 5.

5. We have 5� + 2 = 3� − 7 2� = −9 9 � = − = −4.5. 2 6. We have 1.25 + 0.07� = 3.92 0.07� = 2.67 2.67 �= = 38.143. 0.07 7. We have 0.5� − 13.4 = 25.8 0.5� = 39.2 39.2 �= = 78.4. 0.5 8. We have 12.53 + 5.67� = 45.1� − 125 137.53 = 39.43� 137.53 �= = 3.488. 39.43

PSA C / SOLUTIONS 9. We have 3(� − 2) + 15 = 5(� + 4) 3� − 6 + 15 = 5� + 20 3� + 9 = 5� + 20 −2� = 11 11 �=− = −5.5. 2

10. We have 3� − 5(� + 2) = 3(2 − 4�) + 8 3� − 5� − 10 = 6 − 12� + 8 −2� − 10 = 14 − 12� 10� = 24 24 12 �= = = 2.4. 10 5 11. We have 10.8 − 3.5(40 − 5.1�) = 3.2� + 4.5(25.4 − 5.6�) 10.8 − 140 + 17.85� = 3.2� + 114.3 − 25.2� −129.2 + 17.85� = 114.3 − 22� 39.85� = 243.5 243.5 �= = 6.110. 39.85 12. We have �+� = � � = �−� �−� �= . 13. We have �� + � = �� + � �� − �� = � − � (� − �)� = � − � �−� �= . �−� 14. Since we are solving for �, we put the terms with � on one side of the equation, and the terms without � on the other side. We have �� − 0.2�� + 0.1�� = 5� − 1.8� �� + 0.1�� + 1.8� = 5� + 0.2�� (�� + 0.1� + 1.8)� = 5� + 0.2�� 5� + 0.2�� = . �� + 0.1� + 1.8

SOLUTIONS to Review Problems for PSA C

15. We begin by using the distributive law. Then, since we are solving for �, we put terms with � on one side of the equation, and terms without � on the other side. We have �(�� − �� + ��) = 25� + �(�� + ��) �� − �� + ��2 = 25� + �� + �� −�� − 25� − �� = �� − �� − ��2 −(�� + �� + 25)� = �� − �� − ��2 �� − �� − ��2 �=− . �� + �� + 25 16. We begin by using the distributive law. Then, since we are solving for � , we put terms with � on one side of the equation, and terms without � on the other side. We have 0.2� �2 + ��(� − 1.5� + 2�) = 2.5 0.2� �2 + �� − 1.5��2 + 2�2 � = 2.5 0.2� �2 + �� = 2.5 + 1.5��2 − 2�2 � (0.2�2 + ��)� = 2.5 + 1.5��2 − 2�2 � 2.5 + 1.5��2 − 2�2 � � = . 0.2�2 + �� 17. We ﬁrst use the distributive law. Then, since we are solving for �0 , we put terms with �0 on one side of the equation, and terms without �0 on the other side. We have 25�0 � 2 [� ] + 10(� 2 + �0 ) = �0 (3� + �0 ) 25�0 � 2 [� ] + 10� 2 + 10�0 = 3�0 � + �0 �0 25�0 � 2 [� ] + 10�0 − �0 �0 = 3�0 � − 10� 2 (25� 2 [� ] + 10 − �0 )�0 = 3�0 � − 10� 2 �0 =

3�0 � − 10� 2 . 25� 2 [� ] + 10 − �0

PSA D / SOLUTIONS

Prerequisite Support Appendix D Solutions for PSA D.1 EXERCISES 1. We have 3 ⋅ 23 = 3 ⋅ (2 ⋅ 2 ⋅ 2) = 3 ⋅ 8 = 24. 2. We have −32 = −(3 ⋅ 3) = −9. 3. We have (−2)3 = (−2) ⋅ (−2) ⋅ (−2) = −8. 4. We have −23 ⋅ −32 = −(2 ⋅ 2 ⋅ 2) ⋅ −(3 ⋅ 3) = −8 ⋅ −9 = 72. 5. We have (−2)3 ⋅ (−3)2 = (−2 ⋅ −2 ⋅ −2) ⋅ (−3 ⋅ −3) = −8 ⋅ 9 = −72. 6. We have −43 ⋅ −15 = −(4 ⋅ 4 ⋅ 4) ⋅ −(1 ⋅ 1 ⋅ 1 ⋅ 1 ⋅ 1) = −64 ⋅ −1 = 64. 7. We have −2(−3)4 = −2 ⋅ (−3)(−3)(−3)(−3) = −2 ⋅ 81 = −162. 8. We have 5−2 = 1∕52 = 1∕25. 9. We have 9−1 = 1∕9. 10. We have 51 ⋅ 14 ⋅ 32 = 5 ⋅ 1 ⋅ 9 = 45. 11. We have 52 ⋅ 22 = (5 ⋅ 5) ⋅ (2 ⋅ 2) = 25 ⋅ 4 = 100. 12. We have

−13 ⋅ (−3)4 −1 ⋅ 81 −81 = = = −1. 81 81 92

13. We have (−5)3 ⋅ (−2)2 = (−5 ⋅ −5 ⋅ −5) ⋅ (−2 ⋅ −2) = −125 ⋅ 4 = −500. 14. We have −53 ⋅ −22 = −(5 ⋅ 5 ⋅ 5) ⋅ −(2 ⋅ 2) = −125 ⋅ −4 = 500. ( ) 15. We have −14 ⋅ (−3)2 −23 = −(1 ⋅ 1 ⋅ 1 ⋅ 1) ⋅ (−3)(−3) ⋅ (−(2 ⋅ 2 ⋅ 2)) = −1 ⋅ 9(−8) = 72. 16. We have

)2 ( −64 2 −43 = = 82 = 64. 3 −8 −2 Alternatively, we could have ﬁrst simpliﬁed the fraction as shown below: ( 3 )2 ( )2 ( )2 ( −1 ⋅ 43 −1 43 −4 64 2 = = ⋅ = = 82 = 64. −1 23 8 −1 ⋅ 23 −23 (

17. Any non-zero number to the zero power is one, so we have 30 = 1. 18. We have 03 = 0 ⋅ 0 ⋅ 0 = 0. 19. We have −�� = −(2)(−3)(−5) = −30. 20. We have �� = (−3)2 = (−3)(−3) = 9. 21. We have −�� = −(−3)2 = −(−3)(−3) = −9. (� � ( (−3)(−3) 9 −3 2 (−3)2 22. We have = = = = . � −5 (−5)(−5) 25 (−5)2 ( −� ( (2)(2)(2) (2)3 8 8 2 3 � = = = = =− 23. We have . � −5 (−5)(−5)(−5) −125 125 (−5)3 24. We have �−� = 25 = (2)(2)(2)(2)(2) = 32. 25. We have −�−� = −25 = −(2)(2)(2)(2)(2) = −32.

PSA D / SOLUTIONS )2 ( )2 ( ( 3 )2 ( 3 ( �� (2) (−3) (8)(−3) 144 −24 2 12 2 (12)2 . = = = = = = 25 2� (2)(−5) −10 −10 5 (5)2 ( )3 ( )3 ( ( 3� (3)(−3) 93 729 −9 3 9 3 27. We have . = = = = 3 = 1000 2� (2)(−5) −10 10 10 ( � + � � ( 2 + (−3) )2 ( 1 −1 2 (−1)2 . 28. We have = = = = 49 �−� 2 − (−5) 7 72 26. We have

29. We have �3 ⋅ �5 = �3+5 = �8 . 30. We have

�3 ⋅ �4 �7 = 2 = �7−2 = �5 . 2 � �

31. We have (�4 ⋅ �)2 = (�4+1 )2 = (�5 )2 = �10 . 32. We have

( 5 )4 � = (�5−2 )4 = (�3 )4 = �12 . �2

33. We have �5 ⋅ �3 �5+3 �8 = 4+2 = 6 = �8−6 = �2 . 4 2 � ⋅� � � 34. We have �7 �5 ⋅ = �7−4 ⋅ �5−1 = �3 ⋅ �4 = �3+4 = �7 . �4 � 35. We have (�3 )5 = �3⋅5 = �15 . 36. We have �7 � ⋅ �6 �1+6 7−6 1 ( )2 = �3⋅2 = �6 = � = � . 3 � Note that �1 = �. 37. We have ( 10 )2 ( 4 6 )2 ( 4+6 )2 � � ⋅� � �10⋅2 �20 20−15 = �5 . )3 = ( )3 = �5⋅3 = �15 = � ( )3 = ( 2 3 2+3 5 � ⋅� � � 38. We have 42 ⋅ 4� = 42+� . 39. We have 2� 22 = 2�+2 . 40. We have �5 5 = (� )5 . ( � �� . 41. We have � = 2� 3� 6� = 6�− . = 6 6 (22 )� 4� 22� 43. We have � = � = � = 22�−� . 2 2 2 44. We have ��+3 � � � 3 = ��+3 � �+3 = (��)�+3 . 42. We have

45. We have � � � �+1 = � �+(�+1) = � 2�+1 . 46. We have (�2 + �)3 (� + �2 )3 = ((�2 + �)(� + �2 ))3 . ( )5 47. We have (� + �)4 = (� + �)4⋅5 = (� + �)20 .

PSA D / SOLUTIONS

( )2 48. We have 162 �8 = 24 �8 = 28 �8 = (2�)8 . (� + ℎ)6 (� + ℎ)5 (� + ℎ)6+5 (� + ℎ)11 11−8 = (� + ℎ)3 . ( )4 = (� + ℎ)2⋅4 = (� + ℎ)8 = (� + ℎ) (� + ℎ)2 )5 ( 50. We have (� + )2 = (� + )2⋅5 = (� + )10 . 49. We have

(� + )5 = (� + )5−2 = (� + )3 . (� + )2 52. The quantity (−4)3 = (−4)(−4)(−4) = −64 is negative. When we multiply an odd number of negative numbers, we get a negative number.

51. We have

53. The quantity −43 = −(4)(4)(4) = −64 is negative. We multiply 4 by itself 3 times and then negate the answer. 54. The quantity (−3)4 = (−3)(−3)(−3)(−3) = 81 is positive. When we multiply an even number of negative numbers, we obtain a positive number. 55. The quantity −34 = −(3)(3)(3)(3) = −81 is negative. We multiply 3 by itself four times and then negate the answer. 56. The quantity (−23)42 is positive. We multiply −23 by itself 42 times. When we raise a negative number to an even power, we get a positive number. 57. The quantity −3166 is negative. We multiply 31 by itself 66 times and then negate the answer. 1 1 58. The quantity 17−1 = 1 = is positive. 17 17 1 1 59. The quantity (−5)−2 = = is positive. 25 (−5)2 1 1 60. The quantity −5−2 = − 2 = − is negative. 25 5 1 1 1 = =− 61. The quantity (−4)−3 = is negative. −64 64 (−4)3 62. The quantity (−73)0 = 1 is positive. 63. The quantity −480 = −(480 ) = −1 is negative. 1 64. The quantity (−47)−15 = is negative since we are raising a negative number to an odd power. (−47)15 1 65. The quantity (−61)−42 = is positive since we are raising a negative number to an even power. (−61)42 ( 5 ( � � �� 5 66. Since = � , we have = 5. ( 3 )4 ( 3 )4 � � 3⋅4 � 12 � 67. We have = = 4 = 4. 4 � � � � ( )5 5 5 5 2� 2� 32� 32�5 68. We have = ( )5 = 3⋅5 = 15 . 3 � � � �3 ( ) 3 ( 2 )3 43 �2 64�2⋅3 64�6 4� 69. We have . = = = ( ) 3 5�4 125�4⋅3 125�12 53 �4 (

34 81 81 = ( )4 = 4⋅4 = 16 . � � �4 ( )2 ( 5 )2 62 � 5 36� 5⋅2 36� 10 6� 71. We have . = = = 7 7ℎ 49ℎ7⋅2 49ℎ14 72 (ℎ7 )2 72. (�� )9 = � 9 � 9 . 70. We have

3 �4

73. (2�)5 = 25 �5 = 32�5 .

PSA D / SOLUTIONS

(

24 16 . = 81 34 ( 2 )2� ( ) 2� 75. 4 = 42� 2 = 16� 4� . ( )2 ( ) 76. 3 10(�3� ) = 3 102 �6� = 300�6� . ( ) 77. 3 (2� �� )4 = 3 24� �4� = 3 ⋅ 16� �4� . ( )2� ( )2� = 32� �2 78. 3�2 = 9� �4� . 74.

2 3

PROBLEMS 79. Choices (a), (c), and (d) are equivalent to 1∕9. Choices (b) and (e) are equivalent to 9. 3� 80. Choices (a), (b), (c), and (e) are equivalent to � . 2 1 � 81. Choices (a), (c), and (e) are equivalent to � . Choices (b), and (d) are equivalent to � . � �� 82. Choices (c), and (d) are equivalent to � . Choices (a), (b), and (e) are equivalent to � . � � �� 83. Choices (a) and (b) are equivalent to � . Choices (c), (d), and (e) are equivalent to � . � � 2+3 2 3 84. 3 = 3 ⋅ 3 . 85. �4+1 = �4 ⋅ �. 86. �2+� = �2 ⋅ �� . 104 87. 104−� = � . 10 �� 88. ��− = . � 89. 4�+3 = 4� ⋅ 43 . 6� 90. 6�−1 = . 6 91. (−�)�+ = (−�)� (−�) . 92. ��+ +1 = �� ⋅ � ⋅ �. � �1 93. �1−(�+ ) = �+ = � . �� 94. (� − �)�+� = (� − �)� (� − �)� . (� + �)� 95. (� + �)�− = . (� + �) �� (� + 1) �� 96. ��−1 (� + 1) = (� + 1) = . � � � +� � � = (� + 1) (� + 1) . 97. (� + 1) ( 2 )4 ( 4 )2 2⋅4 98. 4 = 4 or 4 Both expressions are equal to 65,536. ( )� 99. 23� = 23 = 8� . ( )� 100. 52� = 52 = 25� . ( )� 101. 34� = 34 = 81� . 2

102. 5� = (5� )� . √ ( )2 (√ 2 3�� . 103. 3�2� = ( 3)2 �� = )� ( 104. (� + 3)2� = (� + 3)2 .

PSA D / SOLUTIONS

105. Since 33� = (3� )3 and 3� = � then 33� = �3 . 106. Since 3�+2 = 3� 32 and 3� = �, we have 3�+2 = 9�. 4 � 107. Since 4 −3 = 3 and 4 = � we have 4 −3 = . 64 4 � � � 108. Substituting the values for � and � in �� = we have �� = = � = � −� . Therefore, � = − �. � � �

Solutions for PSA D.2 EXERCISES 1. We have 41∕2 =

√ 4 = 2.

√ 2. We have 25−1∕2 = 1∕(251∕2 ) = 1∕( 25) = 1∕5. √ ( −1∕2 ( 1∕2 √ 9 9 3 4 9 3. We have = = = √ = . 9 4 4 2 4 √ √ 3 ( ( 27 3 27 1∕3 64 −1∕3 3 27 4. We have = = = √ = . 3 27 64 64 4 64 ( 4� 4� )1∕2 ( 4� )1∕2 ( 4� )1∕2 5. 2 5 5 = 2 = 22� 52� = 102� = 100� . ( ) 1∕3 ( 6� )1∕3 106� 10 102� 2� � 6. = ( )1∕3 = 52� = 2 = 4 56� 56� √ ( 10 )1∕5 � �10 7. First we write 5 5 as . Then, using the laws of exponents, �5 � √ ( 10 )1∕5 10 � �2 5 � = = . 5 5 � � � √ √ √ √ √ √ √ ( )1∕4 4 4 4 4 � � 8. Using � � ⋅ = � , we have 4�3 4�5 = 16�8 . Then, 16�8 = 16�8 = 2�2 . √ √ √ √ √ � � and 9. Write 48�3 7 as (16�2 6 )(3� ). Notice that the ﬁrst factor is a perfect square. Using the fact that � = � � ⋅ the laws of exponents, we have √ √ √ ( )1∕2 = (16�2 6 )1∕2 (3� )1∕2 = 4� 3 3� . 48�3 7 = (16�2 6 )(3� ) = (16�2 6 )(3� ) √ 96�7 �8 96�7 �8 . Now using the exponent laws and then writing the result as product of two factors, 10. First, write √ as 3 3 3�4 � 3�4 � one of which is a perfect cube, we have √ 7 8 √ ( )1∕3 3 96� � = (32�3 �7 )1∕3 = (8�3 �6 )(4�) = 2��2 3 4�. 3�4 � √ 3

1 11. We have √ = �−1∕2 . The value of the exponent is � = −1∕2. � 1 12. We have 5 = �−5 . The value of the exponent is � = −5. �

PSA D / SOLUTIONS √ �3 = (�3 )1∕2 = �3∕2 . The value of the exponent is � = 3∕2. √ 14. We have ( 3 �)5 = (�1∕3 )5 = �5∕3 . The value of the exponent is � = 5∕3. 13. We have

15. We have

1 1 �−2

1 �2

�−2 .

So � = −2. 16. We have (�3 )2 �3⋅2 �6 = 2⋅3 = 6 = �0 . � � (�2 )3 Note that �0 = 1 provided � ≠ 0.

√ √ √ √ 17. Since the radicands are all the same, we can combine like terms. Therefore, 12 11 − 3 11 + 11 = 10 11. √ √ 18. Both 9 and 144 are perfect squares, so we have 5 9 − 2 144 = (5)(3) − (2)(12) = −9. √ √ √ √ 3 √ 5 3 = 3(2 + 1∕2) = . 19. The radicands are the same. Factoring out 3 from both terms gives 2 3 + 2 2 √ √ √ √ 3 3 3 3 20. The radicands are the same so we combine terms. 4� + 6 4� − 2 4� = 5 4�. √ √ √ 3 3 3 21. The ﬁrst two radicands are the same and have the same index, so we combine terms.(8 3 − 2 3 = 6 3. We cannot √ √ √ √ 3 3 combine terms whose indexes are diﬀerent. Therefore the solution is 6 3 − 2 3 = 2 3 3 − 3 . √ √ √ √ 22. Noticing that 98 = 49 ⋅ 2 and 8 = 4 ⋅ 2, we have √ √ √ √ √ √ −6 98 + 4 8 = −6 (49)(2) + 4 (4)(2) = −6(7) 2 + 4(2) 2. Since the radicands are now the same, we combine terms to get √ √ √ −6(7) 2 + 4(2) 2 = −34 2. 23. Since the radicands are not the same, we try to rewrite them. We have √ √ √ √ √ √ 6 48�4 + 2� 27�2 − 3�2 75 = 6 (16�4 )(3) + 2� (9�2 )(3) − 3�2 (25)(3). Using the product rule for radicals, we obtain √ √ √ √ √ √ 6 (16�4 )(3) + 2� (9�2 )(3) − 3�2 (25)(3) = 24�2 3 + 6�2 3 − 15�2 3. Combining like terms, we get

√ √ √ √ 24�2 3 + 6�2 3 − 15�2 3 = 15�2 3.

24. Since the radicands are not the same, we try to rewrite them. We have √ √ √ √ √ √ 5 12�3 + 2� 128� − 3� 48� = 5 (4�2 )(3�) + 2� (64)(2�) − 3� (16)(3�). Using the product rule for radicals, we obtain √ √ √ √ √ √ 5 (4�2 )(3�) + 2� (64)(2�) − 3� (16)(3�) = 10� 3� + 16� 2� − 12� 3�. Combining the radicals whose radicands are equal, we get √ √ √ √ √ 10� 3� + 16� 2� − 12� 3� = −2� 3� + 16� 2�.

PSA D / SOLUTIONS 25. Since the radicands are not the same, we try to rewrite them. We have √ √ √ √ √ √ (9)(5) 2 (4)(5) (16)(5) 45 2 20 80 − +√ = − + √ . 5 5 5 5 25 25 Using the product rule for radicals, we obtain √ √ √ √ √ √ √ (9)(5) 2 (4)(5) (16)(5) 3 5 4 5 4 5 3 5 − + √ = − + = . 5 5 5 5 5 5 25 26. Since the radicands are not the same, we try to rewrite them. We have √ √ √ √ √ √ 4�� 90�� + 2 40�3 �3 − 3�� 50�� = 4�� (9)(10��) + 2 (4�2 �2 )(10��) − 3�� (25)(2��). Using the product rule for radicals, we obtain √ √ √ √ √ √ 4�� (9)(10��) + 2 (4�2 �2 )(10��) − 3�� (25)(2��) = 12�� 10�� + 4�� 10�� − 15�� 2��. Combining terms with the same radicand, we have √ √ √ √ √ 12�� 10�� + 4�� 10�� − 15�� 2�� = 16�� 10�� − 15�� 2��. ( √ ( √ √ √ 27. The conjugate of 4 + 6 is 4 − 6. The product is 4 + 6 4 − 6 = 16 − 6 = 10. (√ ( √ √ √ 28. The conjugate of 13 − 10 is − 13 − 10. The product is 13 − 10 − 13 − 10 = −13 + 100 = 87. ( √ √ ( √ √ √ √ √ √ 29. A conjugate of − 5 − 6 is − 5 + 6. The product is − 5 − 6 − 5 + 6 = 5 − 6 = −1. ( √ √ √ √ √ √ ( √ √ 30. A conjugate of 7 2 − 2 7 is 7 2 + 2 7. The product is 7 2 − 2 7 7 2 + 2 7 = 98 − 28 = 70. ( √ √ √ √ ( √ √ √ √ 31. A conjugate of �+� �−� = � 2 − �2 . is �+� � − � . The product is ( ( √ √ √ √ 32. The conjugate of 1 − � + 1 is 1 + � + 1. The product is 1 − � + 1 1 + � + 1 = 1 − (� + 1) = −�. 33. Rationalizing the denominator gives

√ √ 3−1 2( 3 − 1) √ = √ ⋅√ = = 3 − 1. √ 2 3+1 3+1 3−1 2

34. Rationalizing the denominator gives

√

√ √ 5 5 5+ 5 √ = √ ⋅ √ 5− 5 5− 5 5+ 5 √ √ ( 5 5+ 5 = 25 − 5 √ 5 5+5 = 20 (√ 5 5+1 = 20 √ 5+1 = . 4

PSA D / SOLUTIONS

35. Rationalizing the numerator gives √

6+1 = √ ⋅√ √ 6−1 6−1 6+1 (√ 10 6+1 = 6−1 (√ =2 6+1 . 10

36. Rationalizing the denominator gives √ √ 3 2− 3 √ √ = √ √ ⋅ √ √ 3 2+ 3 3 2+ 3 3 2− 3 √ ( √ √ 3 3 2− 3 = 18 − 3 √ 3 6−3 = 15 (√ 3 6−1 = 15 √ 6−1 = . 5 √

√

37. Rationalizing the denominator gives √ √ √ √ √ 6+ 2 6+ 2 6+ 2 = ⋅ √ √ √ √ √ √ 6− 2 6− 2 6+ 2 √ 6 + 2 12 + 2 = 6−2 √ 8 + 2 12 = 4 √ 8+4 3 = 4 ( √ 4 2+ 3 = 4 √ = 2 + 3.

√

PROBLEMS

√ √ 2 + ℎ2 . Since ℎ2 > 0, we have �2 + ℎ2 > through and 38. (a) Thinking the calculation, we multiply � by �, then by � √ 2 � = �, √ �� 2 + ℎ2 > �� ⋅ � = ��2 . (b) ��2 is the area of the base, so it is smaller than the area of the cone sitting above it.

PSA D / SOLUTIONS 39. We can write ( 2 4 )3�+3 ( 2 4 )3(�+1) 2� � = 2� � = � �, where � = 2�2 �4 , = 3, and � = � + 1. Applying the given exponent rule, we have ( )� ��= � (( )3 �+1 = 2�2 �4 (( )3 ( 4 )3 �+1 = 2�2 � ( 6 12 )�+1 = 8� � , which equals the right-hand side, as required. 40. The left-hand side can be written

√ ( ) √ (� + 1) �2 + 2� + 1 = (� + 1)(� + 1)2 √ = (� + 1)3 ( )1∕2 = (� + 1)3 ( )� = � ,

where � = � + 1, = 3, and � = 1∕2. Applying the given exponent rule, we have ( )� =�� = (� + 1)3(1∕2) = (� + 1)3∕2 , which equals the right-hand side, as required.

PSA E / SOLUTIONS

Prerequisite Support Appendix E Solutions for PSA E.1 PROBLEMS 1. The dependent variable is �, the number of napkins used, the output of the function. The independent variable is �, the number of customers, the input of the function. 2. The dependent variable is � , the output of the function. The independent variable is �, the input of the function. 3. � = (�). 4. � = (�). 5. (a) The independent variable is �, and the dependent variable is � . (b) (i) This tells us that the cost of taking 3 credits is $3000. (ii) This tells us that the cost of taking 12 credits is the same as the cost of taking 16 credits. 6. The point � marks the input value for �, $20, and the point � marks the output value for � , $4. However, neither of these points tells us that the function connects the input with the output. The single point � tells us that (20) = 4. (i) The point on the graph with � = 0 is (0, 5), so (0) = 5. (ii) It is not as easy to see the point on the graph where � = 10. It is close to (10, 2), so we estimate that (10) is about 2. (iii) Using the graph of , we estimate that (16) is approximately 1. (b) (i) Evaluating (�) at � = 0, we get √ (0) = 5 − 0 = 5.

7. (a)

This is consistent with our answer in part (a). (ii) Evaluating (�) at � = 10, we get √ (10) = 5 − 10 = 1.838. This is very close to our estimate for (10) in part (a). (iii) Evaluating (�) at � = 16, we get √ (16) = 5 − 16 = 5 − 4 = 1. This is consistent with our answer in part (a). 7 9 8. To evaluate when � = −7, we substitute −7 for � in the function, giving (−7) = − − 1 = − . 2 2 9. To evaluate when � = −7, we substitute −7 for � in the function, giving (−7) = (−7)2 − 3 = 49 − 3 = 46. 10. We have (a) �(2) = (12 − 2)2 − (2 − 1)3 = 102 − 13 = 100 − 1 = 99. (b) �(5) = (12 − 5)2 − (5 − 1)3 = 72 − 43 = 49 − 64 = −15.

PSA E / SOLUTIONS (c) �(0) = (12 − 0)2 − (0 − 1)3 = 122 − (−1)3 = 144 − (−1) = 145. (d) �(−1) = (12 − (−1))2 − (−1 − 1)3 = 132 − (−2)3 = 169 − (−8) = 177. 11. To evaluate, we substitute the input value of the function for � in (�) = 2�2 + 7� + 5. (a) We substitute 3 for �: (b) We substitute � for �:

(3) = 2(3)2 + 7(3) + 5 = 2 ⋅ 9 + 21 + 5 = 18 + 26 = 44. (�) = 2(�)2 + 7(�) + 5 = 2�2 + 7� + 5.

(c) We substitute 2� for �: (2�) = 2(2�)2 + 7(2�) + 5 = 2 ⋅ (4�2 ) + 14� + 5 = 8�2 + 14� + 5. (d) We substitute −2 for �: (−2) = 2(−2)2 + 7(−2) + 5 = 2 ⋅ 4 − 14 + 5 = 8 − 9 = −1. 12. We have (0) = 13. We have

14. We have

2⋅0+1 1 = . 3−5⋅0 3

1 1 �(0) = √ = √ = 1. 2 0 +1 1 1 1 1 �(−1) = √ = √ = √ . (−1)2 + 1 1+1 2

This can be rewritten as

√ √ 2 2 1 . √ ⋅√ = 2 2 2

15. We have (10) = 16. We have (1∕2) =

2 ⋅ 10 + 1 21 21 = =− . 3 − 5 ⋅ 10 −47 47

2 ⋅ 12 + 1 3 − 5 ⋅ 12

1+1 3 − 52

= 6

− 25 2

2 = 1 = 4. 2

PSA E / SOLUTIONS 17. We have �

(√ ) 8 = √

1 1 1 1 = √ = √ = . (√ )2 3 8+1 9 8 +1

18. We have ℎ(�) = 10 − 3�. 19. We have ℎ(2�) = 10 − 3 ⋅ 2� = 10 − 6�. 20. We have ℎ(� − 3) = 10 − 3(� − 3) = 10 − 3� + 9 = 19 − 3�.

21. We have ℎ(4 − �) = 10 − 3(4 − �) = 10 − 12 + 3� = −2 + 3�. 22. We have ℎ(5�2 ) = 10 − 3 ⋅ 5�2 = 10 − 15�2 . 23. We have ℎ(4 − �3 ) = 10 − 3(4 − �3 ) = 10 − 12 + 3�3 = −2 + 3�3 . 24. We have (1 − �) = 1 − (1 − �)2 − (1 − �) 2

Applying formula for

= 1 − (1 − 2� + � ) − (1 − �)

Expanding

= 1 − 1 + 2� − �2 − 1 + �

Clearing parentheses

= −� + 3� − 1. 25. (a) Since (� ) gives the volume at temperature � , the 40 is the temperature, so its units are degrees Celsius (◦ C). Similarly, the 3 is the volume, so its units are liters. (b) Since (40) = 3, we know that when � = 40 that (� ) = 3. Since � = 40 at 40◦ C, the volume is 3 liters at 40◦ C. 26. Since the tax is 0.06� , the total cost would be the price of the item plus the tax,or � = � + 0.06� = 1.06� .

PSA E / SOLUTIONS 27. (a) Since �(�) is 5 when � = 0, we have �(0) = 5. (b) Since � = −5 when �(�) = 0, we have �(−5) = 0. (c) �(−5) = 0 (d) Since � = −10 when �(�) = −5, we have �(−10) = −5. 28. (a) Since ℎ(�) is −2 when � = 0, we have ℎ(0) = −2. (b) There are two � values leading to ℎ(�) = 0, namely � = −2 and � = 3. So ℎ(−2) = 0 and ℎ(3) = 0. (c) ℎ(−2) = 0 (d) There are two � values leading to ℎ(�) = −2, namely � = 0 and � = 2. So ℎ(0) = −2 and ℎ(2) = −2. 29. (a) Since ℎ(�) is −2 when � = 0, we have 2ℎ(0) = −4. Thus, we need to find the value(s) of � for which ℎ(�) = −4. There is only one � value leads to this, namely � = −1. (b) Since ℎ(�) is −1 when � = −3 and ℎ(�) is −2 when � = 2, we have 2ℎ(−3) + ℎ(2) = −4. Thus, we need to find the value(s) of � for which ℎ(�) = −4. There is only one � value that leads to this, namely � = −1. (c) Since ℎ(�) is 0 when � = −2, we have ℎ(−2) = 0. Thus, we need to find the value(s) of � for which ℎ(�) = 0. There are two � values leading to ℎ(�) = 0, namely � = −2 and � = 3. (d) Since ℎ(�) is −1 when � = 1 and ℎ(�) is −2 when � = 2, we have ℎ(1) + ℎ(2) = −3. Thus, we need to find the value(s) of � for which ℎ(�) = −3. There are no � values that lead to this. 30. (a) Since the point on the graph with � = 0 is (0, 2), we have (0) = 2. (b) Since (�) = 0 when � = −2 and � = 1, we have (−2) = 0 and (1) = 0. 31. (a) Since the point on the graph with � = 0 is (0, 1), we have ℎ(0) = 1. (b) Since ℎ(�) never equals zero, there are no values of � for which ℎ(�) = 0. 32. (a) Since the point on the graph with � = 0 is (0, 4), we have �(0) = 4. (b) Since �(�) = 0 when � = −2 and � = 2, we have �(−2) = 0 and �(2) = 0. 33. (a) When the company charges 15 dollars for one of its products, its total sales are 112,500 dollars. (b) When the company charges � dollars for one of its products, its total sales are 0 dollars. (c) When the company charges 1 dollar for one of its products, its total sales are � dollars. (d) The number � is the total sales, in dollars, that the company receives when it charges a price of � dollars for one of its products. 34. (a) The car takes 111 feet to stop when traveling at 30 mph. (b) The car takes 10 feet to stop when traveling at � mph. (c) The car takes � feet to stop when traveling at 10 mph. (d) The number � is the distance it takes the car to stop when traveling at � mph. 35. (a) In Figure 1.5, we see when � = 60, the point on the graph of the function has coordinates (60, 37). This means when the car’s speed is 60 miles per hour, its highway gas mileage is about 37 miles per gallon. (b) In Figure 1.5, we see the point on the graph that has the greatest value for � is approximately (50, 40). This means the car’s maximum highway gas mileage is 40 miles per gallons. This occurs when the car’s speed is 50 miles per hour. 36. (a) Because 5% = 0.05, the interest each year is 0.05 ⋅ (Face value), so the total interest after � years is given by � = (Number of years) ⋅ 0.05 ⋅ (Face value) = �(0.05)� = 0.05��. (b) The payout, � , is the sum of the face value and the total interest. That is, � = � + ��. Thus, since � = 0.05��, we have � = � + 0.05��. 37. The variable is � and the constant is �. 38. The variable is � and the constants are � and �. 39. The variable is � and the constants are � and �. 40. The variable is � and the constants are � and �. 41. We are given that � is a function of �, so � is the dependent variable and � is the independent variable. The symbol � is a constant (which stands for the speed of light).

PSA E / SOLUTIONS

42. (a) In this situation, we regard the tip as a function of the variable � and regard � as a constant. If we call this function , then we can write � Tip = (�) = �. 100 (b) Here we regard the tip as a function of the variable � and regard � as a constant. If we call this function �, then Tip = �(�) =

� �. 100

43. (a) Since your possibilities are all apartments of 1000 square feet, � is the constant (always 1000), and � is the variable (since you can have the same-sized apartment at diﬀerent distances from the station). (b) Since your possibilities are all apartments 1 mile from the station, � is the constant (always 1), and � is the variable (since you can have apartments of diﬀerent sizes that are the same distance from the station). (c) Here, both � and � are variables. In choosing the apartment, you must decide whether you want to be farther from the station or to have a smaller area. 44. �(99) gives the average number of times the song is downloaded in a day at a price of $0.99 per song. 45. The current price of the song is �0 , so �(�0 ) is the average number of times the song is downloaded in a day at the current price. 46. �0 − 10 means 10 cents less than the current price of the song, so �(�0 − 10) means the average number of times the song is downloaded per day at a price 10 cents lower than the current price. 47. �0 − 10 means 10 cents less than the current price of the song, so �(�0 − 10) means the average number of times the song is downloaded per day at a price 10 cents lower than the current price. So �(�0 − 10) − �(�0 ) means Number of daily downloads at lower price

−

Current number of daily downloads

or the change in the average number of daily downloads if the price drops 10 cents from the current price. 48. �(�0 ) is the average number of daily downloads of the song at the current price. Since there are 365 days in a year, 365�(�0 ) is the number times the song is downloaded in one year at current price. 49. 0.80�0 is 80% of the current price, or 20% less than the current price. Thus, �(0.80�0 ) is the average number times the song is downloaded per day after the price drops 20%. 50. �(�0 ) is the average number of times the song is downloaded per day at the current price. Since there are 24 hours in one day, �(�0 )∕24 is the average number of downloads every hour. 51. �(�0 ) is the average number of times the song is downloaded per day at its current price of �0 cents, so �0 ⋅ �(�0 ) = Price of the song ⋅ Number of times downloaded in a day = Average daily sales for the music store. We see that multiplying the average number of times the song is downloaded in a day by the price of the song gives the average daily sales of the music store—that is, the average amount of money it brings in each day. 52. We have 1 ⋅ 100(100 + 1) 2 = 50(101)

(100) =

= 5050.

PSA E / SOLUTIONS

53. Working this in steps, we ﬁrst ﬁnd (� + 1): 1 (� + 1)((� + 1) + 1) 2 1 = (� + 1)(� + 2). 2

(� + 1) =

Now we subtract: (� + 1) − (�) =

1 1 (� + 1)(� + 2) − �(� + 1) 2 2 ⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏟⏞⏞⏟ (�+1)

1 (� + 1) ((� + 2) − �) 2 1 = (� + 1) ⋅ 2 2 = � + 1.

(�)

factor out 12 (� + 1)

54. This is the number of people (in 1000s) infected by a strain whose open-air survival time is 3 minutes longer than the most common strain. 55. ℎ(�0 ) is the number eventually infected by the most common strain, and ℎ(2�0 ) is the number eventually infected by a strain that survives twice as long in open air. The ratio of these numbers gives how many times more people are infected by the longer-living strain. 56. This is the gas mileage of the car if its tire pressure is 90% of the recommended pressure, that is, if the tires are 10% underinﬂated. 57.

(�0 ) is the mileage at the recommended pressure, and (�0 − 5) is the mileage if the tires are underinflated by 5 lbs/in2 . So this is the difference in gas mileage between driving on properly inflated tires and tires underinflated by 5 lbs/in2 .

58. We have �(� − 5) = 4 − 7(� − 5) because �(�) = 4 − 7� = 4 − 7� + 35 = 39 − 7�. Thus, 3�(� − 5) = 3 (39 − 7�) ⏟⏞⏟⏞⏟ �(�−5)

= 117 − 21�. 59. We have ( ( � �2 − 2 = 4 − 7 �2 − 2

because �(�) = 4 − 7�

= 4 − 7�2 + 14 = 18 − 7�2 . Thus, ( � �2 − 2 + 2 = 18 − 7�2 + 2 2

= 20 − 7� .

because �(�2 − 2) = 18 − 7�2

PSA E / SOLUTIONS 60. We have �(7 − 4�) = 4 − 7(7 − 4�)

because �(�) = 4 − 7�

= 4 − (49 − 28�) = 4 − 49 + 28� = −45 + 28�. Thus, 4�(7 − 4�) − 7 = 4 (−45 + 28�) − 7 because �(7 − 4�) = −45 + 28� = −180 + 112� − 7 = −187 + 112�.