CHAPTER TWO. f(x) Slope = f (3) = Rate of change of f at 3. x 3. f(1.001) f(1) Average velocity = s(0.8) s(0) 0.8 0

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1 CHAPTER TWO 2.1 SOLUTIONS 99 Solutions for Section (a) Te average rate of cange is te slope of te secant line in Figure 2.1, wic sows tat tis slope is positive. (b) Te instantaneous rate of cange is te slope of te grap at 3, wic we see from Figure 2.2 is negative. Slope Average rate of cange of f f() Slope f (3) Rate of cange of f at 3 f() Figure 2.1 Figure (a) Te function N f(t) is decreasing wen t Terefore, f (1950) is negative. Tat means tat te number of farms in te US was decreasing in (b) Te function N f(t) is decreasing in 1960 as well as in 1980 but it is decreasing faster in 1960 tan in Terefore, f (1960) is more negative tan f (1980). 3. (a) Let s f(t). (i) We wis to find te average velocity between t 1 and t 1.1. We ave (ii) We ave Average velocity Average velocity f(1.1) f(1) f(1.01) f(1) m/sec m/sec. (iii) We ave f(1.001) f(1) Average velocity m/sec (b) We see in part (a) tat as we coose a smaller and smaller interval around t 1 te average velocity appears to be getting closer and closer to 6, so we estimate te instantaneous velocity at t 1 to be 6 m/sec. 4. For te interval 0 t 0.8, we ave ( Average velocity ( ( 0 t 0.8 Average velocity 0 t 0.2 Average velocity 0.2 t 0.4 ) ) ) s(0.8) s(0) s(0.2) s(0) s(0.4) s(0.2) ft/sec ft/sec ft/sec. 0.2 To find te velocity at t 0.2, we find te average velocity to te rigt of t 0.2 and to te left of t 0.2 and average tem. So a reasonable estimate of te velocity at t 0.2 is te average of 1 ( ) 4.5 ft/sec. 2

2 100 Capter Two /SOLUTIONS 5. (a) Te average velocity between t 3 and t 5 is Distance Time s(5) s(3) 5 3 (b) Using an interval of size 0.1, we ave ( Instantaneous velocity at t 3 Using an interval of size 0.01, we ave ( Instantaneous velocity at t 3 ) ) s(3.1) s(3) s(3.01) s(3) From tis we guess tat te instantaneous velocity at t 3 is about 6 ft/sec ft/sec (a) Te average rate of cange of a function over an interval is represented grapically as te slope of te secant line to its grap over te interval. See Figure 2.3. Segment AB is te secant line to te grap in te interval from 0 to 3 and segment BC is te secant line to te grap in te interval from 3 to 5. We can easily see tat slope of AB > slope of BC. Terefore, te average rate of cange between 0 and 3 is greater tan te average rate of cange between 3 and 5. y 4 4 y B C 3 Tis slope is larger A Figure 2.3 Figure 2.4 (b) We can see from te grap in Figure 2.4 tat te function is increasing faster at 1 tan at 4. Terefore, te instantaneous rate of cange at 1 is greater tan te instantaneous rate of cange at 4. (c) Te units of rate of cange are obtained by dividing units of cost by units of product: tousands of dollars/kilogram. 7. (a) Te size of te tumor wen t 0 monts is S(0) cubic millimeter. Te size of te tumor wen t 6 monts is S(6) cubic millimeters. Te total cange in te size of te tumor is S(6) S(0) mm 3. (b) Te average rate of cange in te size of te tumor during te first si monts is: Average rate of cange S(6) S(0) 6 0 (c) We will consider intervals to te rigt of t 6: cubic millimeters/mont. t (monts) S (cubic millimeters)

3 2.1 SOLUTIONS Average rate of cange Average rate of cange Average rate of cange We can continue taking smaller intervals but te value of te average rate will not cange muc. Terefore, we can say tat a good estimate of te growing rate of te tumor at t 6 monts is 44.4 cubic millimeters/mont. 8. (a) We ave Rate of cange of population billion people per year (b) To estimate f (2005), we use te rate of cange formula on an interval containing 2005: f (2005) 9. We use te interval 2 to 2.01: f (2) f(2.01) f(2) billion people per year For greater accuracy, we can use te smaller interval 2 to 2.001: f (2) f(2.001) f(2) (a) Figure 2.5 sows tat for t 2 te function g(t) (0.8) t is decreasing. Terefore, g (2) is negative. g(t) 2 1 g(t) (0.8) t 1 2 t Figure 2.5 (b) To estimate g (2) we take te small interval between t 2 and t to te rigt of t 2. g (2) g(2.001) g(2) (a) f () is negative wen te function is decreasing and positive wen te function is increasing. Terefore, f () is positive at C and G. f () is negative at A and E. f () is zero at B, D, and F. (b) f () is te largest wen te grap of te function is increasing te fastest (i.e. te point wit te steepest positive slope). Tis occurs at point G. f () is te most negative wen te grap of te function is decreasing te fastest (i.e. te point wit te steepest negative slope). Tis occurs at point A. 12. We estimate f (2) using te average rate of cange formula on a small interval around 2. We use te interval 2 to (Any small interval around 2 gives a reasonable answer.) We ave f (2) f(2.001) f(2)

4 102 Capter Two /SOLUTIONS 13. Since f () 0 were te grap is orizontal, f () 0 at d. Te derivative is positive at points b and c, but te grap is steeper at c. Tus f () 0.5 at b and f () 2 at c. Finally, te derivative is negative at points a and e but te grap is steeper at e. Tus, f () 0.5 at a and f () 2 at e. See Table 2.1. Tus, we ave f (d) 0, f (b) 0.5, f (c) 2, f (a) 0.5, f (e) 2. Table 2.1 f () d 0 b 0.5 c 2 a 0.5 e (a) Since 75.2 percent live in te city in 1990 and 35.1 percent in 1890, we ave Average rate of cange Tus te average rate of cange is percent/year. (b) By looking at te population in 1990 and 2000 we see tat Average rate of cange wic gives an average rate of cange of 0.38 percent/year between 1990 and Alternatively, we look at te population in 1980 and 1990 and see tat Average rate of cange giving an average rate of cange of 0.15 percent/year between 1980 and We see tat te rate of cange in te year 1990 is somewere between 0.38 and 0.15 percent/year. A good estimate is ( )/ percent per year. In fact, te definition of urban area was canged for te 2000 data, so tis estimate sould be used wit care. (c) By looking at te population in 1830 and 1860 we see tat Average rate of cange , giving an average rate of cange of 0.36 percent/year between 1830 and Alternatively we can look at te population in 1800 and 1830 and see tat Average rate of cange giving an average rate of cange of 0.10 percent/year between 1800 and We see tat te rate of cange at te year 1830 is somewere between 0.10 and Tis tells us tat in te year 1830 te percent of te population in urban areas is canging by a rate somewere between 0.10 percent/year and 0.36 percent/year. (d) Looking at te data we see tat te percent gets larger as time goes by. Tus, te function appears to be always increasing.

5 2.1 SOLUTIONS g() f() Figure 2.6 (a) Te tangent line to te grap of f() 2 at 0 coincides wit te -ais and terefore is orizontal (slope 0). Te tangent line to te grap of g() at 0 is te dased line indicated in te figure and it also as a slope equal to zero. Terefore bot tangent lines at 0 are parallel. We see in Figure 2.6 tat te tangent lines at 1 appear parallel, and te tangent lines at 2 appear parallel. Te slopes of te tangent lines at any value a will be equal. (b) Adding a constant sifts te grap vertically, but does not cange te slope of te curve. 16. (a) Since te values of P go up as t goes from 4 to 6 to 8, we see tat f (6) appears to be positive. Te percent of ouseolds wit cable television is increasing at t 6. (b) We estimate f (2) using te difference quotient for te interval to te rigt of t 2, as follows: f (2) P t Te fact tat f (2) 0.95 tells us tat te percent of ouseolds wit cable television in te United States was increasing at a rate of 0.95 percentage points per year wen t 2 (tat means 1992). Similarly: f (10) P t Te fact tat f (10) 0.55 tells us tat te percent of ouseolds in te United States wit cable television was increasing at a rate of 0.55 percentage points per year wen t 10 (tat means 2000). 17. P (0) is te derivative of te function P(t) 200(1.05) t at t 0. Tis is te same as te rate of cange of P(t) at t 0. We estimate tis by computing te average rate of cange over intervals near t 0. If we use te intervals t 0 and 0 t 0.001, we see tat: ( Average rate of cange on t 0 ) 200(1.05)0 200(1.05) ( 0.001) , ( Average rate of cange ) 200(1.05) (1.05) 0 on 0 t It appears tat te rate of cange of P(t) at t 0 is 10, so we estimate P (0) Te coordinates of A are (4,25). See Figure 2.7. Te coordinates of B and C are obtained using te slope of te tangent line. Since f (4) 1.5, te slope is 1.5

6 104 Capter Two /SOLUTIONS Tangent line B 1.5(0.2) 0.3 A (4, 25) 0.2 C Figure 2.7 From A to B, 0.2, so y 1.5(0.2) 0.3. Tus, at C we ave y Te coordinates of B are (4.2, 25.3). From A to C, 0.1, so y 1.5( 0.1) Tus, at C we ave y Te coordinates of C are (3.9, 24.85). 19. (a) Since te point B (2,5) is on te grap of g, we ave g(2) 5. (b) Te slope of te tangent line toucing te grap at 2 is given by Tus, g (2) Te answers to parts (a) (d) are sown in Figure 2.8. Slope Rise Run Slope f (3) f() Slope f(5) f(2) 5 2 f(4) f(2) f(4) Figure (a) Since f is increasing, f(4) > f(3). (b) From Figure 2.9, it appears tat f(2) f(1) > f(3) f(2). f(2) f(1) (c) Te quantity represents te slope of te secant line connecting te points on te grap at and 2. Tis is greater tan te slope of te secant line connecting te points at 1 and 3 wic is f(3) f(1). 3 1 (d) Te function is steeper at 1 tan at 4 so f (1) > f (4).

7 2.1 SOLUTIONS 105 f() f(3) f(2) f(2) f(1) slope f(3) f(1) 3 1 slope f(2) f(1) Figure Using a difference quotient wit 0.001, say, we find f ln(1.001) 1ln(1) (1) f ln(2.001) 2ln(2) (2) Te fact tat f is larger at 2 tan at 1 suggests tat f is concave up between 1 and Te quantity f(0) represents te population on October 17, 2006, so f(0) 300 million. Te quantity f (0) represents te rate of cange of te population (in millions per year). Since so we ave f (0) person 11 seconds 1/10 6 million people million people/year, 11/( ) years 24. (a) f (t) is negative or zero, because te average number of ours worked a week as been decreasing or constant over time. g (t) is positive, because ourly wage as been increasing. (t) is positive, because average weekly earnings as been increasing. (b) We use a difference quotient to te rigt for our estimates. (i) f (1970) 0.2 ours/year f (1995) ours/year. In 1970, te average number of ours worked by a production worker in a week was decreasing at te rate of 0.2 ours per year. In 1995, te number of ours was not canging. (ii) g (1970) $0.27 per year g (1995) $0.47 per year In 1970, te ourly wage was increasing at a rate of $0.27 per year. In 1995, te ourly wage was increasing at a rate of $0.47 per year. (iii) (1970) $8.90 per year (1995) $16.18 per year 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.

8 106 Capter Two /SOLUTIONS Solutions for Section Estimating te slope of te lines in Figure 2.10, we find tat f ( 2) 1.0, f ( 1) 0.3, f (0) 0.5, and f (2) Figure Te grap is tat of te line y Te slope, and ence te derivative, is 2. See Figure Figure See Figure Figure See Figure Figure 2.13

9 2.2 SOLUTIONS Te slope of tis curve is approimately 1 at 4 and at 4, approimately 0 at 2.5 and 1.5, and approimately 1 at 0. See Figure Figure See Figure Figure See Figure Figure We know tat f () f( + ) f(). For tis problem, we ll take te average of te values obtained for 1 and 1; tat s te average of f( + 1) f() and f() f( 1) wic equals f (0) f(1) f(0) f (1) (f(2) f(0))/2 (10 18)/2 4. f (2) (f(3) f(1))/2 (9 13)/2 2. f (3) (f(4) f(2))/2 (9 10)/ f( + 1) f( 1). Tus, 2

10 108 Capter Two /SOLUTIONS f (4) (f(5) f(3))/2 (11 9)/2 1. f (5) (f(6) f(4))/2 (15 9)/2 3. f (6) (f(7) f(5))/2 (21 11)/2 5. f (7) (f(8) f(6))/2 (30 15)/ f (8) f(8) f(7) Te rate of cange of f() is positive for 4 8, negative for 0 3. Te rate of cange is greatest at about For 0, 5, 10, and 15, we use te interval to te rigt to estimate te derivative. For 20, we use te interval to te left. For 0, we ave f f(5) f(0) (0) Similarly, we find te oter estimates in Table 2.2. Table f () (a) 3 (b) 4 (c) 5 (d) See Figure f () Figure Tis is a line wit slope 2, so te derivative is te constant function f () 2. Te grap is a orizontal line at y 2. See Figure Figure 2.18 f () 13. See Figure f () 2 4 Figure 2.19

11 2.2 SOLUTIONS See Figure f () Figure See Figure f () Figure See Figure f () Figure See Figure f () Figure See Figure f () Figure 2.24

12 110 Capter Two /SOLUTIONS 19. Te grap is increasing for 0 < t < 10 and is decreasing for 10 < t < 20. One possible grap is sown in Figure Te units on te orizontal ais are years and te units on te vertical ais are people. people f(t) years Figure 2.25 Te derivative is positive for 0 < t < 10 and negative for 10 < t < 20. Two possible graps are sown in Figure Te units on te orizontal aes are years and te units on te vertical aes are people per year. people/year people/year f (t) f (t) years years Figure Te value of g() is increasing at a decreasing rate for 2.7 < < 4.2 and increasing at an increasing rate for > 4.2. Tus g () sould be close to 3 near 5.2. y between 4.7 and 5.2 y between 5.2 and Te function is decreasing for < 2 and > 2, and increasing for 2 < < 2. Te matcing derivative must be negative (below te -ais) for < 2 and > 2, positive (above te -ais) for 2 < < 2, and zero (on te -ais) for 2 and 2. Te matcing derivative is in grap VIII. 22. Te function is a line wit negative slope, so f () is a negative constant, and te grap of f () is a orizontal line below te -ais. Te matcing derivative is in grap IV. 23. Te function is increasing for < 2 and decreasing for > 2. Te corresponding derivative is positive (above te -ais) for < 2, negative (below te -ais) for > 2, and zero at 2. Te matcing derivative is in grap II. 24. Te function is increasing for < 2 and decreasing for > 2. Te corresponding derivative is positive (above te -ais) for < 2, negative (below te -ais) for > 2, and zero at 2. Te derivatives in graps VI and VII bot satisfy tese requirements. To decide wic is correct, consider wat appens as gets large. Te grap of f() approaces an asymptote, gets more and more orizontal, and te slope gets closer and closer to zero. Te derivative in grap VI meets tis requirement and is te correct answer. 25. Since f () > 0 for < 1, f() is increasing on tis interval. Since f () < 0 for > 1, f() is decreasing on tis interval. Since f () 0 at 1, te tangent to f() is orizontal at 1. One possible sape for y f() is sown in Figure 2.27.

13 2.2 SOLUTIONS Figure Since f () > 0 for 1 < < 3, we see tat f() is increasing on tis interval. Since f () < 0 for < 1 and for > 3, we see tat f() is decreasing on tese intervals. Since f () 0 for 1 and 3, te tangent to f() will be orizontal at tese s. One of many possible sapes of y f() is sown in Figure y Figure (a) Grap II (b) Grap I (c) Grap III 28. (a) f f(1.1) f(1) ln(1.1) ln(1) (1) f f(2.1) f(2) ln(2.1) ln(2) (2) f f(3.1) f(3) ln(3.1) ln(3) (3) f f(4.1) f(4) ln(4.1) ln(4) (4) f f(5.1) f(5) ln(5.1) ln(5) (5) (b) It looks like te derivative of ln() is 1/. 29. Table 2.3 f() f() f() Near 2, te values of f() seem to be increasing by for eac increase of in, so te derivative appears to be Near 3, te values of f() are increasing by for eac step of 0.001, so te derivative appears to be 9. Near 4, f() increases by for eac step of 0.001, so te derivative appears to be 16. Te pattern seems to be, ten, tat at a point, te derivative of f() is f () 2.

14 112 Capter Two /SOLUTIONS 30. (I) y 3 (II) y 3 (III) y 3 (IV) y (a) III f (1) > 0 means tat f is increasing near 1. f decreasing means f is getting less steep, tat is, f is concave down. (b) II f (1) > 0 means tat f is increasing near 1. f increasing means f is getting steeper, tat is, f is concave up. (c) IV f (1) < 0 means tat f is decreasing near 1. f decreasing means f is sloping down more, tat is, f is concave down. (d) I f (1) < 0 means tat f is decreasing near 1. f increasing means f is sloping down less, tat is, f is concave up. 31. (a) t 3 (b) t 9 (c) t 14 (d) 1 V (t) t 2 Solutions for Section In Leibniz notation te derivative is dc/dw and te units are dollars per pound. 2. In Leibniz notation te derivative is dd/dt and te units are feet per minute. 3. In Leibniz notation te derivative is dp/dh and te units are dollars per our. 4. In Leibniz notation te derivative is dn/dd and te units are gallons per mile. 5. (a) Te 12 represents te weigt of te cemical; terefore, its units are pounds. Te 5 represents te cost of te cemical; terefore, its units are dollars. Te statement f(12) 5 means tat wen te weigt of te cemical is 12 pounds, te cost is 5 dollars. (b) We epect te derivative to be positive since we epect te cost of te cemical to increase wen te weigt bougt increases. (c) Again, 12 is te weigt of te cemical in pounds. Te units of te 0.4 are dollars/pound since it is te rate of cange of te cost as a function of te weigt of te cemical bougt. Te statement f (12) 0.4 means tat te cost is increasing at a rate of 0.4 dollars per pound wen te weigt is 12 pounds, or tat an additional pound will cost about an etra 40 cents. 6. (a) Te statement f(5) 18 means tat wen 5 milliliters of catalyst are present, te reaction will take 18 minutes. Tus, te units for 5 are ml wile te units for 18 are minutes. (b) As in part (a), 5 is measured in ml. Since f tells ow fast T canges per unit a, we ave f measured in minutes/ml. If te amount of catalyst increases by 1 ml (from 5 to 6 ml), te reaction time decreases by about 3 minutes.

15 2.3 SOLUTIONS (a) If te price is $150, ten 2000 items will be sold. (b) If te price goes up from $150 by $1 per item, about 25 fewer items will be sold. Equivalently, if te price is decreased from $150 by $1 per item, about 25 more items will be sold. 8. Te derivative f (10) is te slope of te tangent line to te curve at t 10. See Figure Taking two points on te tangent line, we calculate its slope: Slope 6. 5 Since te slope is about 6, we ave f (10) 6 cm/yr. At t 10, te sturgeon was growing in lengt at a rate of about 6 centimeters a year. lengt (cm) Figure 2.29 t (years) 9. (Note tat we are considering te average temperature of te yam, since its temperature is different at different points inside it.) (a) It is positive, because te temperature of te yam increases te longer it sits in te oven. (b) Te units of f (20) are F/min. Te statement f (20) 2 means tat at time t 20 minutes, te temperature T would increase by approimately 2 F if te yam is in te oven an additional minute. 10. Let p be te rating points earned by te CBS Evening News, let R be te revenue earned in millions of dollars, and let R f(p). Wen p 4.3, Tus Rate of cange of revenue $5.5 million 0.1 point f (4.3) million dollars/point. 11. (a) Te units of compliance are units of volume per units of pressure, or liters per centimeter of water. (b) Te increase in volume for a 5 cm reduction in pressure is largest between 10 and 15 cm. Tus, te compliance appears maimum between 10 and 15 cm of pressure reduction. Te derivative is given by te slope, so Compliance liters per centimeter. (c) Wen te lung is nearly full, it cannot epand muc more to accommodate more air. 12. (a) Te units of lapse rate are te same as for te derivative dt/dz, namely (units of T)/(units of z) C/km. (b) Since te lapse rate is 6.5, te derivative of T wit respect to z is dt/dz 6.5 C/km. Te air temperature drops about 6.5 for every kilometer you go up. 13. (a) Tis means tat investing te $1000 at 5% would yield $1649 after 10 years. (b) Writing g (r) as db/dr, we see tat te units of db/dr are dollars per percent (interest). We can interpret db as te etra money earned if interest rate is increased by dr percent. Terefore g (5) db r5 165 means tat dr te balance, at 5% interest, would increase by about $165 if te interest rate were increased by 1%. In oter words, g(6) g(5) Te units of f () are feet/mile. Te derivative, f (), represents te rate of cange of elevation wit distance from te source, so if te river is flowing downill everywere, te elevation is always decreasing and f () is always negative. (In fact, tere may be some stretces were te elevation is more or less constant, so f () 0.)

16 114 Capter Two /SOLUTIONS 15. (a) Since W f(c) were W is weigt in pounds and c is te number of Calories consumed per day: f(1800) 155 f (2000) 0 means tat means tat consuming 1800 Calories per day results in a weigt of 155 pounds. consuming 2000 Calories per day causes neiter weigt gain nor loss. (b) Te units of dw/dc are pounds/(calories/day). 16. (a) Te statement f(200) 1300 means tat it costs $1300 to produce 200 gallons of te cemical. (b) Te statement f (200) 6 means tat wen te number of gallons produced is 200, costs are increasing at a rate of $6 per gallon. In oter words, it costs about $6 to produce te net (te 201 st ) gallon of te cemical. 17. (a) Positive, since weigt increases as te cild gets older. (b) f(8) 45 tells us tat wen te cild is 8 years old, te cild weigs 45 pounds. (c) Te units of f (a) are lbs/year. f (a) tells te rate of growt in lbs/years at age a. (d) f (8) 4 tells us tat te 8-year-old cild is growing at a rate of 4 lbs/year. (e) As a increases, f (a) will decrease since te rate of growt slows down as te cild grows up. 18. (a) An additional dollar per year of government purcases increases national output for te year by about $0.60. Te derivative is called a fiscal policy multiplier because if government purcases increase by dollars per year, ten national output increases by about 0.60 dollars per year. (b) An additional ta dollar collected per year decreases national output for te year by about $0.26. Te derivative is called a fiscal policy multiplier because if government ta revenues increase by dollars per year, ten national output decreases by about 0.26 dollars per year. 19. Let f(t) be te age of onset of Alzeimer s, in weeks, of a typical man wo retires at age t years. Te derivative f (t) as units weeks/year. Te study reports tat eac additional year of employment is associated wit about a si week later age of onset, tat is, f (t) (a) Since f (c) is negative, te function P f(c) is decreasing: pelican eggsells are getting tinner as te concentration, c, of PCBs in te environment is increasing. (b) Te statement f(200) 0.28 means tat te tickness of pelican eggsells is 0.28 mm wen te concentration of PCBs in te environment is 200 parts per million (ppm). Te statement f (200) means tat te tickness of pelican eggsells is decreasing (eggsells are becoming tinner) at a rate of mm per ppm of concentration of PCBs in te environment wen te concentration of PCBs in te environment is 200 ppm. 21. (a) kilograms per week (b) At week 24 te fetus is growing at a rate of kg/week, or 96 grams per week. 22. (a) Te tangent line to te weigt grap is steeper at 36 weeks ten at 20 weeks, so g (36) is greater tan g (20). (b) Te fetus increases its weigt more rapidly at week 36 tan at week Compare te secant line to te grap from week 0 to week 40 to te tangent lines at week 20 and week 36. (a) At week 20 te secant line is steeper tan te tangent line. Te instantaneous weigt growt rate is less tan te average. (b) At week 36 te tangent line is steeper tan te secant line. Te instantaneous weigt growt rate is greater tan te average. 24. We estimate te derivatives at 20 and 36 weeks by drawing tangent lines to te weigt grap, sown in Figure 2.30, and calculating teir slopes. (a) Two points on te tangent line at 20 weeks are (16, 0) and (40, 1.5). Tus, g (20) kg/week (b) Two points on te tangent line at 36 weeks are (24, 0) and (40, 3.0). Tus, g (36) kg/week (c) Te average rate of growt is te slope of te secant line from (0,0) to (40, 3.1). Tus, Average rate of cange kg/week.

17 2.3 SOLUTIONS 115 weigt (kilograms) Slope g (20) Slope g (36) Figure 2.30 age of fetus (weeks after last menstruation) 25. Since we do not ave information beyond t 25, we will assume tat te function will continue to cange at te same rate. Terefore, Since , ten f(26) f(25) + f (25) ( 0.2) 3.4. f(30) f(25) + f (25)(5) ( 0.2)(5) Using te approimation y f () wit 2, we ave y f (20) 2 6 2, so f(22) f(20) + f (20) (a) Te statement f(8) 5.1 means tat annual net sales for te Hersey Company were 5.1 billion dollars in Te statement f (8) 0.22 tells us tat in 2008, annual net sales were increasing at a rate of 0.22 billion dollars per year, or 220 million dollars per year. (b) Since sales were 5.1 billion in 2008 and increasing at 0.22 billion dollars per year, we estimate tat, four years later in 2012, Sales in (0.22) 5.98 billion dollars. Tus f(12) 5.98, so annual net sales for te Hersey Company are projected to be 5.98 billion dollars in Tis prediction assumes tat te growt rate remains constant. 28. (a) Te statement f(5) 249 tells us tat M 249 wen t 5, and te statement f (5) 6.5 tells us tat te rate of cange of M is 6.5 wen t 5. In terms of meat production, tis means tat in 2005, world meat production was 249 million metric tons and increasing at a rate of 6.5 million metric tons per year. (b) We assume tat te growt rate remains constant until t 10. Since production in 2005 is 249 million metric tons and is increasing at 6.5 million metric tons a year, we epect production in 2010 to be approimately f(10) million metric tons. 29. (a) Te statement f(140) 120 means tat a patient weiging 140 pounds sould receive a dose of 120 mg of te painkiller. Te statement f (140) 3 tells us tat if te weigt of a patient increases by one pound (from 140 pounds), te dose sould be increased by about 3 mg. (b) Since te dose for a weigt of 140 lbs is 120 mg and at tis weigt te dose goes up by about 3 mg for one pound, a 145 lb patient sould get about an additional 3(5) 15 mg. Tus, for a 145 lb patient, te correct dose is approimately f(145) (5) 135 mg. 30. (a) Te statement f(20) 0.36 means tat 20 minutes after smoking a cigarette, tere will be 0.36 mg of nicotine in te body. Te statement f (20) means tat 20 minutes after smoking a cigarette, nicotine is leaving te body at a rate mg per minute. Te units are 20 minutes, 0.36 mg, and mg/minute.

18 116 Capter Two /SOLUTIONS (b) f(21) f(20) + cange in f in one minute ( 0.002) f(30) f(20) + cange in f in 10 minutes ( 0.002)(10) (a) Since t represents te number of days from now, we are told f(0) 80 and f (0) (b) f(10) value now + cange in value in 10 days (10) In 10 days, we epect tat te mutual fund will be wort about $85 a sare. 32. (a) Te tangent line is sown in Figure Two points on te line are (0,16) and (3.2, 0). Tus Slope (cm/sec)/kg (b) Since 50 grams kg, te contraction velocity canges by about 5(cm/sec)/kg 0.050kg 0.25 cm/sec. Te velocity is reduced by about 0.25 cm/sec or 2.5 mm/sec. (c) Since v() is te contraction velocity in cm/sec wit a load of kg, we ave v (2) 5. contraction velocity (cm/sec) v() , load opposing contraction (kg) Figure (a) Te tangent line is sown in Figure Two points on te line are (0,0.75) and (2.5, 5). Te Slope (liters/minute)/our. (b) Te rate of cange of te pumping rate is te slope of te tangent line. One minute 1/60 our, so in one minute te Pumping rate increases by about 1.7 (liter/minute) 1 our liter/minute. our 60 (c) Since g(t) is te pumping rate in liters/minute at time t ours, we ave g (2) 1.7.

19 2.3 SOLUTIONS 117 pumping rate of eart (liters pumped per minute) 5 4 g(t) Figure 2.32 t (ours) 34. Units of C (r) are dollars/percent. Approimately, C (r) means te additional amount needed to pay off te loan wen te interest rate is increased by 1%. Te sign of C (r) is positive, because increasing te interest rate will increase te amount it costs to pay off a loan. 35. (a) Te company opes tat increased advertising always brings in more customers instead of turning tem away. Terefore, it opes f (a) is always positive. (b).mine If f (100) 2, it means tat if te advertising budget is $100,000, eac etra dollar spent on advertising will bring in about $2 If f (100) 2, it means tat if te advertising budget is $100,000, eac etra dollar spent on advertising will bring in about $2.r3088 wort of sales. If f (100) 0.5, eac dollar above $100 tousand spent on advertising will bring in about $0.50 wort of sales. (c) If f (100) 2, ten as we saw in part (b), spending sligtly more tan $100,000 will increase revenue by an amount greater tan te additional epense, and tus more sould be spent on advertising. If f (100) 0.5, ten te increase in revenue is less tan te additional epense, ence too muc is being spent on advertising. Te optimum amount to spend, a, is an amount tat makes f (a) 1. At tis point, te increases in advertising ependitures just pay for temselves. If f (a) < 1, too muc is being spent; if f (a) > 1, more sould be spent. 36. (a) See part (b). (b) See Figure concentration of enzymes Normal level time f > 0 f < 0 Figure 2.33 (c) Te derivative, f, is te rate at wic te concentration is increasing or decreasing. We see tat f is positive at te start of te disease and negative toward te end. In practice, of course, f cannot be measured directly. Cecking te value of C in blood samples taken on consecutive days allows us to estimate f (t): f (t) f(t + 1) f(t) f(t + 1) f(t). (t + 1) t 37. Te consumption rates (kg/week) are te rates at wic te quantities are decreasing, tat is, 1 times te derivatives of te storage functions. To compare rates at a given time, compare te steepness of te tangent lines to te graps at tat time. (a) At 3 weeks, te tangent line to te fat storage grap is steeper tan te tangent line to te protein storage grap. During te tird week, fat is consumed at a greater rate tan protein. (b) At 7 weeks, te protein storage grap is steeper tan te fat storage grap. During te sevent week, protein is consumed at a greater rate tan fat.

20 118 Capter Two /SOLUTIONS 38. Were te grap is linear, te derivative of te fat storage function is constant. Te derivative gives te rate of fat consumption (kg/week). Tus, for te first four weeks te body burns fat at a constant rate. 39. Te fat consumption rate (kg/week) is te rate at wic te quantity of fat is decreasing, tat is, 1 times te derivative of te fat storage function. We estimate te derivatives at 3, 6, and 8 weeks by drawing tangent lines to te storage grap, sown in Figure 2.34, and calculating teir slopes. (a) Te tangent line at 3 weeks is te storage grap itself since tat part of te grap is straigt. Two points on te tangent line are (0,12) and (4, 4). Ten Slope Te consumption rate is 2.0 kg/week. (b) Two points on te tangent line are (0, 4.7) and (8, 0.1). Ten, Slope Te consumption rate is 0.6 kg/week. (c) Two points on te tangent line are (0, 2.7) and (8, 0.6). Ten Te consumption rate is 0.3 kg/week Slope quantities of stored fat (kilograms) kg/week. 0.6 kg/week. 0.3 kg/week weeks of starvation Figure 2.34: Rates of consumption are 1 times slopes of tangent lines 40. Te body canges from burning more fat tan protein to burning more protein. Tis is done by reducing te rate at wic it burns fat and simultaneously increasing te rate at wic it burns protein. Te pysiological reason is tat te body as begun to run out of fat. 41. Te grap of fat storage is linear for four weeks, ten becomes concave up. Tus, te derivative of fat storage is constant for four weeks, ten increases. Tis matces grap I. Te grap of protein storage is concave up for tree weeks, ten becomes concave down. Tus, te derivative of protein storage is increasing for tree weeks and ten becomes decreasing. Tis matces grap II. 42. (a) Te statement f(9) 740 tells us tat tere were 740 million acres of rain forest in Brazil in Te derivative f (9) 2.7 tells us tat in 2009 Brazil s rain forests were srinking at a rate of 2.7 million acres per year. (b) We ave Relative rate of cange f (9) f(9) Brazil s rain forests are srinking at a rate of 0.365% per year. 43. Since t 4 is te end of April 2009, we ave f(4) 200 million users. We use a difference quotient wit t 2 for te end of February 2009 and t 4 for te end of April 2009 to estimate te derivative: f f(4) f(2) (4) 4 2 To estimate te relative rate of cange, we use million users per mont. Relative rate of cange f (4) f(4) per mont. 200 Te number of active Facebook users at te end of April 2009 was 200 million. Te number was increasing at 12.5 million users per mont, wic represents a relative growt rate of 6.25% per mont.

21 2.4 SOLUTIONS (a) At 2 1/2 monts, te baby weigs 5.67 kilograms. (b) At 2 1/2 monts, te baby s weigt is increasing at a relative rate of 13% per mont. 45. Estimating te relative rate of cange using t 0.01, we ave 46. Let P f(t). 1 f f t 1 f(4.01) f(4) f(4) (a) Estimating te relative rate of cange using t 1, we ave (b) Wit t 0.1 we ave (c) Wit t 0.01 we ave 1 P P t 1 f(4) f(3) % per year f(3) 1 1 P P t 1 f(3.1) f(3) % per year f(3) P P t 1 f(3.01) f(3) % per year f(3) 0.01 In fact, te relative rate of cange at t 3 of te population for tis city is eactly Solutions for Section (a) Since te grap is below te -ais at 2, te value of f(2) is negative. (b) Since f() is decreasing at 2, te value of f (2) is negative. (c) Since f() is concave up at 2, te value of f (2) is positive. 2. At B bot dy/d and d 2 y/d 2 could be positive because y is increasing and te grap is concave up tere. At all te oter points one or bot of te derivatives could not be positive. 3. f () > 0 f () > 0 4. f () 0 f () 0 5. f () < 0 f () 0 6. f () < 0 f () > 0 7. f () > 0 f () < 0 8. f () < 0 f () < 0 9. (a) f() (b) f()

22 120 Capter Two /SOLUTIONS (c) f() (d) f() 10. Te derivative of w(t) appears to be negative since te function is decreasing over te interval given. Te second derivative, owever, appears to be positive since te function is concave up, i.e., it is decreasing at a decreasing rate. 11. Te derivative, s (t), appears to be positive since s(t) is increasing over te interval given. Te second derivative also appears to be positive or zero since te function is concave up or possibly linear between t 1 and t 3, i.e., it is increasing at a non-decreasing rate. 12. Te grap must be everywere decreasing and concave up on some intervals and concave down on oter intervals. One possibility is sown in Figure Figure Since all advertising campaigns are assumed to produce an increase in sales, a grap of sales against time would be epected to ave a positive slope. A positive second derivative means te rate at wic sales are increasing is increasing. If a positive second derivative is observed during a new campaign, it is reasonable to conclude tat tis increase in te rate sales are increasing is caused by te new campaign wic is terefore judged a success. A negative second derivative means a decrease in te rate at wic sales are increasing, and terefore suggests te new campaign is a failure. 14. (a) Te function appears to be decreasing and concave down, and so we conjecture tat f is negative and tat f is negative. (b) We use difference quotients to te rigt: f (2) f (8) (a) Te derivative, f (t), appears to be positive since te number of cars is increasing. Te second derivative, f (t), appears to be positive during te period because te rate of cange is increasing. For eample, between 1940 and 1950, te rate of cange is ( )/ million cars per year, wile between 1950 and 1960, te rate of cange is 2.14 million cars per year. (b) We use te average rate of cange formula on te interval 1970 to 1980 to estimate f (1975): f (1975) We estimate tat f (1975) 3.24 million cars per year. Te number of passenger cars in te US was increasing at a rate of about 3.24 million cars per year in Te derivative is positive on tose intervals were te function is increasing and negative on tose intervals were te function is decreasing. Terefore, te derivative is positive on te intervals 0 < t < 0.4 and 1.7 < t < 3.4, and negative on te intervals 0.4 < t < 1.7 and 3.4 < t < 4. Te second derivative is positive on tose intervals were te grap of te function is concave up and negative on tose intervals were te grap of te function is concave down. Terefore, te second derivative is positive on te interval 1 < t < 2.6 and negative on te intervals 0 < t < 1 and 2.6 < t < 4.

23 2.4 SOLUTIONS Te derivative is positive on tose intervals were te function is increasing and negative on tose intervals were te function is decreasing. Terefore, te derivative is positive on te interval 2.3 < t < 0.5 and negative on te interval 0.5 < t < 4. Te second derivative is positive on tose intervals were te grap of te function is concave up and negative on tose intervals were te grap of te function is concave down. Terefore, te second derivative is positive on te interval 0.5 < t < 4 and negative on te interval 2.3 < t < Tis grap is increasing for all, and is concave down to te left of 2 and concave up to te rigt of 2. One possible answer is sown in Figure 2.36: 2 Figure Te two points at wic f 0 are A and B. Since f is nonzero at C and D and f is nonzero at all four points, we get te completed Table 2.4: Table 2.4 Point f f f A 0 + B + 0 C + D (b). Te positive first derivative tells us tat te temperature is increasing; te negative second derivative tells us tat te rate of increase of te temperature is slowing. 21. (e). Since te smallest value of f (t) was 2 C/our, we know tat f (t) was always positive. Tus, te temperature rose all day. 22. Since f(2) 5, te grap goes troug te point (2, 5). Since f (2) 1/2, te slope of te curve is 1/2 wen it passes troug tis point. Since f (2) > 0, te grap is concave up at tis point. One possible grap is sown in Figure Many oter answers are also possible. 5 2 Figure 2.37

24 122 Capter Two /SOLUTIONS 23. To te rigt of 5, te function starts by increasing, since f (5) 2 > 0 (toug f may subsequently decrease) and is concave down, so its grap looks like te grap sown in Figure Also, te tangent line to te curve at 5 as slope 2 and lies above te curve for > 5. If we follow te tangent line until 7, we reac a eigt of 24. Terefore, f(7) must be smaller tan 24, meaning 22 is te only possible value for f(7) from among te coices given. y T f() 5 7 Figure (a) Te EPA will say tat te rate of discarge is still rising. Te industry will say tat te rate of discarge is increasing less quickly, and may soon level off or even start to fall. (b) Te EPA will say tat te rate at wic pollutants are being discarged is leveling off, but not to zero so pollutants will continue to be dumped in te lake. Te industry will say tat te rate of discarge as decreased significantly. 25. (a) Let N(t) be te number of people below te poverty line. See Figure N(t) Figure 2.39 t (b) dn/dt is positive, since people are still slipping below te poverty line. d 2 N/dt 2 is negative, since te rate at wic people are slipping below te poverty line, dn/dt, is decreasing. 26. (a) dp/dt > 0 and d 2 P/dt 2 > (a) (b) dp/dt < 0 and d 2 P/dt 2 > 0 (but dp/dt is close to zero). utility quantity (b) As a function of quantity, utility is increasing but at a decreasing rate; te grap is increasing but concave down. So te derivative of utility is positive, but te second derivative of utility is negative. 28. (a) IV, (b) III, (c) II, (d) I, (e) IV, (f) II

25 2.5 SOLUTIONS 123 Solutions for Section (a) Marginal cost is te derivative C (q), so its units are dollars/barrel. (b) It costs about $3 more to produce 101 barrels of olive oil tan to produce 100 barrels. 2. Te marginal cost is approimated by te difference quotient Te marginal cost is approimately $3 per item. MC C q Marginal cost C (q). Terefore, marginal cost at q is te slope of te grap of C(q) at q. We can see tat te slope at q 5 is greater tan te slope at q 30. Terefore, marginal cost is greater at q 5. At q 20, te slope is small, wereas at q 40 te slope is larger. Terefore, marginal cost at q 40 is greater tan marginal cost at q Drawing in te tangent line at te point (10000, C(10000)) we get Figure $ C(q) 20,000 10, ,000 Figure 2.40 q We see tat eac vertical increase of 2500 in te tangent line gives a corresponding orizontal increase of rougly Tus te marginal cost at te production level of 10,000 units is C (10,000) Slope of tangent line to C(q) at q 10, Tis tells us tat after producing 10,000 units, it will cost rougly $0.42 to produce one more unit. 5. Drawing in te tangent line at te point (600, R(600)), we get Figure ,000 $ R(q) 10, Figure 2.41 q We see tat eac vertical increase of 2500 in te tangent line gives a corresponding orizontal increase of rougly 150. Te marginal revenue at te production level of 600 units is R (600) Slope of tangent line to R(q) at q Tis tells us tat after producing 600 units, te revenue for producing te 601 st product will be rougly $16.67.

26 124 Capter Two /SOLUTIONS 6. (a) For q 500 Profit π(500) R(500) C(500) dollars. (b) As production increases from q 500 to q 501, R R (500) q dollars, C C (500) q dollars, Tus Cange in profit π R C dollars. 7. We ave C C(2500) C(2000) (2000) $0.37/ton Tis means tat recycling te 2001st ton of paper will cost around $0.37. Te marginal cost is smallest at te point were te derivative of te function is smallest. Tus te marginal cost appears to be smallest on te interval 2500 q Te slope of te revenue curve is greater tan te slope of te cost curve at bot q 1 and q 2, so te marginal revenue is greater at bot production levels. 9. (a) We can approimate C(16) by adding C (15) to C(15), since C (15) is an estimate of te cost of te 16 t item. C(16) C(15) + C (15) $ $108 $2408. (b) We approimate C(14) by subtracting C (15) from C(15), were C (15) is an approimation of te cost of producing te 15 t item. C(14) C(15) C (15) $2300 $108 $ We know MC C(1,001) C(1,000). Terefore, C(1,001) C(1,000) + MC or C(1,001) dollars. Since we do not know MC(999), we will assume tat MC(999) MC(1,000). Terefore: Ten: Alternatively, we can reason tat so Now for C(1,000), we ave Since 1,100 1, , MC(999) C(1,000) C(999). C(999) C(1,000) MC(999) 5, ,975 dollars. MC(1,000) C(1,000) C(999), C(999) C(1,000) MC(1,000) 4,975 dollars. C(1,100) C(1,000) + MC 100. C(1,100) 5, , ,500 7,500 dollars. 11. (a) Te cost to produce 50 units is $4300 and te marginal cost to produce additional items is about $24 per unit. Producing two more units (from 50 to 52) increases cost by $48. We ave C(52) (2) $4348. (b) Wen q 50, te marginal cost is $24 per item and te marginal revenue is $35 per item. Te profit on te 51 st item is approimately $11. (c) Wen q 100, te marginal cost is $38 per item and te marginal revenue is $35 per item, so te company loses about $3 by producing te 101 st item. Since te company will lose money, it sould not produce te 101 st item. 12. At q 50, te slope of te revenue is larger tan te slope of te cost. Tus, at q 50, marginal revenue is greater tan marginal cost and te 50 t bus sould be added. At q 90 te slope of revenue is less tan te slope of cost. Tus, at q 90 te marginal revenue is less tan marginal cost and te 90 t bus sould not be added.

27 2.5 SOLUTIONS (a) At q 2000, we ave (b) If q increases from 2000 to 2001, Tus, Profit R(2000) C(2000) dollars. R R (2000) q dollars, C C (2000) q dollars, Cange in profit R C dollars. Since increasing production increases profit, te company sould increase production. (c) By a calculation similar to tat in part (b), as q increases from 2000 to 2001, Cange in profit dollars. Since increasing production reduces te profit, te company sould decrease production. 14. (a) At q 2.1 million, (b) If q 0.04, Profit π(2.1) R(2.1) C(2.1) million dollars. Cange in revenue, R R (2.1) q 0.7(0.04) million dollars $28,000. Tus, revenues increase by about $28,000. (c) If q 0.05, Cange in revenue, R R (2.1) q 0.7( 0.05) million dollars $35,000. Tus, revenues decrease by about $35,000. (d) We find te cange in cost by a similar calculation. For q 0.04, Cange in cost, C C (2.1) q 0.6(0.04) million dollars $24,000 Cange in profit, π $28,000 $24,000 $4000. Tus, increasing production 0.04 million units increases profits by about $4000. For q 0.05, Cange in cost, C C (2.1) q 0.6( 0.05) 0.03 million dollars $30,000 Cange in profit, π $35,000 ( $30,000) $5000. Tus, decreasing production 0.05 million units decreases profits by about $ (a) Te value of C(0) represents te fied costs before production, tat is, te cost of producing zero units, incurred for initial investments in equipment, and so on. (b) Te marginal cost decreases slowly, and ten increases as quantity produced increases. marginal cost q 1 q 2 q (c) Concave down implies decreasing marginal cost, wile concave up implies increasing marginal cost. (d) An inflection point of te cost function is (locally) te point of maimum or minimum marginal cost. (e) One would tink tat te more of an item you produce, te less it would cost to produce etra items. In economic terms, one would epect te marginal cost of production to decrease, so we would epect te cost curve to be concave down. In practice, toug, it eventually becomes more epensive to produce more items, because workers and resources may become scarce as you increase production. Hence after a certain point, te marginal cost may rise again. Tis appens in oil production, for eample.

28 126 Capter Two /SOLUTIONS Solutions for Capter 2 Review 1. (a) Let s f(t). (i) We wis to find te average velocity between t 1 and t 1.1. We ave 2. (ii) We ave Average velocity Average velocity f(1.1) f(1) f(1.01) f(1) m/sec m/sec. (iii) We ave f(1.001) f(1) Average velocity m/sec (b) We see in part (a) tat as we coose a smaller and smaller interval around t 1 te average velocity appears to be getting closer and closer to 8, so we estimate te instantaneous velocity at t 1 to be 8 m/sec. Slope /2 1 2 Point F C E A B D 3. (a) From Figure 2.42 we can see tat for 1 te value of te function is decreasing. Terefore, te derivative of f() at 1 is negative Figure 2.42 f() 2 3 (b) f (1) is te derivative of te function at 1. Tis is te rate of cange of f() 2 3 at 1. We estimate tis by computing te average rate of cange of f() over intervals near 1. Using te intervals and , we see tat ( ) Average rate of cange [2 13 ] [ ] , on ( Average rate of cange ) [ ] [2 1 3 ] on It appears tat te rate of cange of f() at 1 is approimately 3, so we estimate f (1) Using te interval , we estimate f f(1.001) f(1) (1) Te grap of f() 3 is concave up so we epect our estimate to be greater tan f (1). 5. Te slope is positive at A and D; negative at C and F. Te slope is most positive at A; most negative at F. 6. (a) Since te point A (7, 3) is on te grap of f, we ave f(7) 3. (b) Te slope of te tangent line toucing te curve at 7 is given by Tus, f (7) 4. Slope Rise Run