Preliminary Questions 1. Which of the lines in Figure 10 are tangent to the curve? B C FIGURE 10

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1 3 DIFFERENTIATION 3. Definition of te Derivative Preliminar Questions. Wic of te lines in Figure 0 are tangent to te curve? A D B C FIGURE 0 Lines B an D are tangent to te curve.. Wat are te two was of writing te ifference quotient? or as Te ifference quotient ma be written eiter as f fa a fa+ fa. 3. Fin a an suc tat fa+ fa is equal to te slope of te secant line between 3,f3 an 5,f5. fa+ fa Wit a 3 an, is equal to te slope of te secant line between te points 3,f3 an 5,f5 on te grap of f. 4. Wic erivative is approimate b tan π ? tan π is a goo approimation to te erivative of te function f tan at π Wat o te following quantities represent in terms of te grap of f sin? sin.3 sin 0.9 a sin.3 sin 0.9 b c f Consier te grap of sin. a Te quantit sin.3 sin 0.9 represents te ifference in eigt between te points 0.9, sin 0.9 an.3, sin.3. sin.3 sin 0.9 b Te quantit represents te slope of te secant line between te points 0.9, sin 0.9 an.3, sin on te grap. c Te quantit f 0.9 represents te slope of te tangent line to te grap at 0.9. Eercises. Let f 5. Sow tat f Ten sow tat f3 + f3 an compute f 3 b taking te limit as April 4, 0

2 SECTION 3. Definition of te Derivative 9 Wit f 5, it follows tat f Using tis result, we fin f3 + f As 0, , so f Let f 3 5. Sow tat te secant line troug,f an +, f + as slope + 5. Ten use tis formula to compute te slope of: a Te secant line troug,f an 3,f3 b Te tangent line at b taking a limit Te formula for te slope of te secant line is f + f + [ ] a To fin te slope of te secant line troug,f an 3,f3, we take, so te slope is b As 0, te slope of te secant line approaces Hence, te slope of te tangent line at is5. In Eercises 3 6, compute f a in two was, using Eq. an Eq.. 3. f + 9, a 0 Let f + 9. Ten Alternatel, f f0 + f f f f f + 9, a Let f + 9. Ten f f + f Alternatel, f f f April 4, 0

3 9 CHAPTER 3 DIFFERENTIATION 5. f , a Let f Ten f f + f Alternatel, f 3, a Let f 3. Ten f f f f f + f Alternatel, In Eercises 7 0, refer to Figure. f f f f FIGURE 7. Fin te slope of te secant line troug,f an.5,f.5. Is it larger or smaller tan f? Eplain. From te grap, it appears tat f.5.5 an f. Tus, te slope of te secant line troug,f an.5,f.5 is f.5 f From te grap, it is also clear tat te secant line troug,f an.5,f.5 as a larger slope tan te tangent line at. In oter wors, te slope of te secant line troug,f an.5,f.5 is larger tan f. f + f 8. Estimate for 0.5. Wat oes tis quantit represent? Is it larger or smaller tan f? Eplain. Wit 0.5, +.5. Moreover, from te grap it appears tat f.5.7 an f. Tus, f + f Tis quantit represents te slope of te secant line troug te points,f an.5,f.5. It is clear from te grap tat te secant line troug te points,f an.5,f.5 as a smaller slope tan te tangent line at. f + f In oter wors, for 0.5 is smaller tan f. April 4, 0

4 SECTION 3. Definition of te Derivative Estimate f an f. From te grap, it appears tat te tangent line at woul be orizontal. Tus, f 0. Te tangent line at appears to pass troug te points 0.5, 0.8 an,. Tus f Fin a value of for wic f + f 0. In orer for f + f to be equal to zero, we must ave f + f. Now, f, an te onl oter point on te grap wit a -coorinate of is f0. Tus, + 0, or. In Eercises 4, refer to Figure FIGURE Grap of f.. Determine f a for a,, 4, 7. Remember tat te value of te erivative of f at a can be interprete as te slope of te line tangent to te grap of fat a. From Figure, we see tat te grap of fis a orizontal line tat is, a line wit zero slope on te interval 0 3. Accoringl, f f 0. On te interval 3 5, te grap of fis a line of slope ; tus, f 4. Finall, te line tangent to te grap of fat 7 is orizontal, so f For wic values of is f < 0? If f < 0, ten te slope of te tangent line at is negative. Grapicall, tis woul mean tat te value of te function was ecreasing for increasing. From te grap, it follows tat f < 0 for 7 <<9. 3. Wic is larger, f 5.5 or f 6.5? Te line tangent to te grap of fat 5.5 as a larger slope tan te line tangent to te grap of fat 6.5. Terefore, f 5.5 is larger tan f Sow tat f 3 oes not eist. Because f3 + f3 f3 + f3 lim 0 but lim 0 0+, it follows tat f f3 + f3 3 0 oes not eist. April 4, 0

5 94 CHAPTER 3 DIFFERENTIATION In Eercises 5 8, use te limit efinition to calculate te erivative of te linear function. 5. f f 7. gt 8 3t 8. kz 4z + fa+ fa 7a + 9 7a 9 lim fa+ fa lim ga + ga 8 3a + 8 3a 3 lim ka + ka 4a + + 4a + 4 lim Fin an equation of te tangent line at 3, assuming tat f3 5 an f 3? B efinition, te equation of te tangent line to te grap of fat 3is f3 + f Fin f3 an f 3, assuming tat te tangent line to fat a 3 as equation 5 +. Te slope of te tangent line to f at a 3isf 3 b efinition, terefore f 3 5. Also b efinition, te tangent line to fat a 3 goes troug 3,f3, sof3 7.. Describe te tangent line at an arbitrar point on te curve + 8. Since + 8 represents a straigt line, te tangent line at an point is te line itself, Suppose tat f + f Calculate: a Te slope of te secant line troug,f an 6,f6 b f Let f be a function suc tat f + f a We take 4 to compute te slope of te secant line troug,f an 6,f6: f4 + f 4 + b f f + f Let f. Does f + equal + or +? Compute te ifference quotient at a wit 0.5. Let f. Ten Wit a an 0.5, te ifference quotient is fa+ fa f + +. f.5 f April 4, 0

6 SECTION 3. Definition of te Derivative Let f. Does f5 + equal 5 + or 5 +? Compute te ifference quotient at a 5 wit. Let f. Ten f Wit a 5 an, te ifference quotient is fa+ fa f5 + f Let f /. Compute f 5 b sowing tat f5 + f Let f /. Ten f5 + f Tus, f f5 + f Fin an equation of te tangent line to te grap of f / at 9. Let f /. Ten f9 + f Tus, f f9 + f Because f9 3, it follows tat an equation of te tangent line to te grap of f / at 9is f f April 4, 0

7 96 CHAPTER 3 DIFFERENTIATION In Eercises 7 44, use te limit efinition to compute f a an fin an equation of te tangent line. 7. f + 0, a 3 Let f + 0. Ten At a 3, te tangent line is f f3 + f f f f 4, a Let f 4. Ten At a, te tangent line is f f + f f + + f ft t t, a 3 Let ft t t. Ten At a 3, te tangent line is f f3 + f f 3t 3 + f3 t 3 5 t f 8 3, a Let f 8 3. Ten f f + f At a, te tangent line is f + f April 4, 0

8 SECTION 3. Definition of te Derivative f 3 +, a 0 Let f 3 +. Ten f f f At a 0, te tangent line is f f0. 3. ft t 3 + 4t, a 4 Let ft t 3 + 4t. Ten At a 4, te tangent line is f f4 + f f 4t 4 + f4 00t t f, a 8 Let f. Ten f f8 + f Te tangent at a 8is f f f +, a 4 Let f +. Ten f f4 + f Te tangent at a 4is f f f + 3, a Let f +3. Ten f f + f 0 0 Te tangent line at a is f + + f April 4, 0

9 98 CHAPTER 3 DIFFERENTIATION 36. ft t, a Let ft t. Ten f f + f 0 0 At a, te tangent line is f + + f f + 4, a Let f + 4. Ten f f + f Te tangent line at a is f + f ft 3t + 5, a Let ft 3t + 5. Ten f f + f Te tangent line at a is 39. f, a 4 Let f. Ten f t + + f 3 t t + 7. f f4 + f At a 4 te tangent line is 40. f +, a f f Let f +. Ten f f4 + f April 4, 0

10 SECTION 3. Definition of te Derivative 99 At a 4 te tangent line is 4. ft t +, a 3 Let ft t +. Ten Te tangent line at a 3is f f f f3 + f f 3t 3 + f3 3 0 t t f, a Let f. Ten f f + f Te tangent line at a is f + + f f +, a 0 Let f +. Ten Te tangent line at a 0is f f0 + f f0 + f ft t 3, a Let ft t 3. Ten f f + f 0 0 Te tangent line at a is f t + f 3t + 3t Figure 3 isplas ata collecte b te biologist Julian Hule on te average antler weigt W of male re eer as a function of age t. Estimate te erivative at t 4. For wic values of t is te slope of te tangent line equal to zero? For wic values is it negative? Antler Weigt kg Age ears FIGURE 3 April 4, 0

11 00 CHAPTER 3 DIFFERENTIATION Let Wtenote te antler weigt as a function of age. Te tangent line sketce in te figure below passes troug te points, an 6, 5.5. Terefore W kg/ear. 6 If te slope of te tangent is zero, te tangent line is orizontal. Tis appears to appen at rougl t 0 an at t.6. Te slope of te tangent line is negative wen te eigt of te grap ecreases as we move to te rigt. For te grap in Figure 3, tis occurs for 0 <t< Figure 4A sows te grap of f. Te close-up in Figure 4B sows tat te grap is nearl a straigt line near 6. Estimate te slope of tis line an take it as an estimate for f 6. Ten compute f 6 an compare wit our estimate A Grap of B Zoom view near 6, 4 FIGURE 4 From te close-up in Figure 4B, te line appears to pass troug te points 5.9, 3.99 an 6.08, 4.0. Tus, f Wit f, f wic is consistent wit te approimation obtaine from te close-up grap , 47. Let f 4 +. a Plot fover [, ]. Ten zoom in near 0 until te grap appears straigt, an estimate te slope f 0. b Use a to fin an approimate equation to te tangent line at 0. Plot tis line an fon te same set of aes. a Te figure below at te left sows te grap of f + 4 over [, ]. Te figure below at te rigt is a close-up near 0. From te close-up, we see tat te grap is nearl straigt an passes troug te points 0.,.5 an 0.,.85. We terefore estimate f April 4, 0

12 SECTION 3. Definition of te Derivative 0 b Using te estimate for f 0 obtaine in part a, te approimate equation of te tangent line is f f Te figure below sows te grap of fan te approimate tangent line Let f cot. Estimate f π grapicall b zooming in on a plot of fnear π. Te figure below sows a close-up of te grap of f cot near π From te close-up, we see tat te grap is nearl straigt an passes troug te points.53, 0.04 an.6, We terefore estimate f π Determine te intervals along te -ais on wic te erivative in Figure 5 is positive FIGURE 5 Te erivative tat is, te slope of te tangent line is positive wen te eigt of te grap increases as we move to te rigt. From Figure 5, tis appears to be true for <<.5 an for > Sketc te grap of f sin on [0,π] an guess te value of f π. Ten calculate te ifference quotient at π for two small positive an negative values of. Are tese calculations consistent wit our guess? Here is te grap of sin on [0,π] At π, we re at te peak of te sine grap. Te tangent line appears to be orizontal, so te slope is 0; ence, f π appears to be sin π Tese numerical calculations are consistent wit our guess. April 4, 0

13 0 CHAPTER 3 DIFFERENTIATION In Eercises 5 56, eac limit represents a erivative f a. Fin fan a lim 0 Te ifference quotient lim 5 5 Te ifference quotient 3 5 as te form 5 sin π 53. lim Te ifference quotient sin π lim 4 4 Te ifference quotient as te form as te form lim 0 Te ifference quotient 5+ 5 as te form lim 0 Te ifference quotient 5 as te form fa+ fa f fa a were f 3 an a 5. were f 3 an a 5. fa+ fa 4 f fa as te form were f a an a 4. 4 fa+ fa were f sin an a π 6. were f 5 an a. fa+ fa were f 5 an a Appl te meto of Eample 6 to f sin to etermine f π 4 accuratel to four ecimal places. We know tat Creating a table of values of close to zero: f fπ/4 + fπ/4 π/4 0 0 sinπ/4 + / sin π 4 + / Accurate up to four ecimal places, f π Appl te meto of Eample 6 to f cos to etermine f π 5 accuratel to four ecimal places. Use a grap of fto eplain ow te meto works in tis case. We know tat f π fπ/5 + fπ/5 cos π 5 + cos π We make a cart using values of close to zero: f π cos π 5 + cos π cos π 5 + cos π April 4, 0

14 SECTION 3. Definition of te Derivative 03 Te figures sown below illustrate w tis proceure works. From te figure on te left, we see tat for <0, te slope of te secant line is greater less negative tan te slope of te tangent line. On te oter an, from te figure on te rigt, we see tat for >0, te slope of te secant line is less more negative tan te slope of te tangent line. Tus, te slope of te tangent line must fall between te slope of a secant line wit >0 an te slope of a secant line wit <0. Tangent line cos cos 0.8 Tangent line For eac grap in Figure 6, etermine weter f is larger or smaller tan te slope of te secant line between an + for >0. Eplain. f f A FIGURE 6 B On curve A,f is larger tan f + f ; te curve is bening ownwars, so tat te secant line to te rigt is at a lower angle tan te tangent line. We sa suc a curve is concave own, an tat its erivative is ecreasing. On curve B, f is smaller tan f + f ; te curve is bening upwars, so tat te secant line to te rigt is at a steeper angle tan te tangent line. We sa suc a curve is concave up, an tat its erivative is increasing. 60. Refer to te grap of f in Figure 7. a Eplain grapicall w, for >0, f f0 f 0 f f0 b Use a to sow tat f c Similarl, compute f to four ecimal places for,, 3, 4. Now compute te ratios f /f 0 for,, 3, 4. Can ou guess an approimate formula for f? 3 FIGURE 7 Grap of f. April 4, 0

15 04 CHAPTER 3 DIFFERENTIATION a In te grap, te inequalit f f f0 0 ols for positive values of, since te ifference quotient f f0 is an increasing function of. Te slopes of te secant lines between 0,f0 an a nearb point increase as te nearb point moves from left to rigt. Hence te slopes of te secant lines between 0,f0 an a nearb point to te rigt,, f were is positive ecee f 0. Similarl, for >0, is negative an 0 lies to te rigt of. Consequentl, te slope of te secant line between 0,f0 an a nearb point to te left,, f is less tan f 0. Terefore, te inequalit ols for >0. b For , we ave f 0 f f0 f f , an f f In ligt of a, f c We ll use te same values of ± an compute ifference quotients at,, 3, 4. Since f.38699, we conclue tat f.3863 to four ecimal places. Since f.77598, we conclue tat f.776 to four ecimal places. Since f , we conclue tat f to four ecimal places. Wit ± , f , so we conclue tat f to four ecimal places. 3 4 f /f Looking at tis table, we guess tat f /f 0. In oter wors, f f Sketc te grap of f 5/ on [0, 6]. a Use te sketc to justif te inequalities for >0: f4 f4 f 4 f4 + f4 b Use a to compute f 4 to four ecimal places. c Use a graping utilit to plot fan te tangent line at 4, using our estimate for f 4. a Te slope of te secant line between points 4,f4 an 4 +, f 4 + is f4 + f4. 5/ is a smoot curve increasing at a faster rate as. Terefore, if >0, ten te slope of te secant line is greater tan te slope of te tangent line at f4, wic appens to be f 4. Likewise, if <0, te slope of te secant line is less tan te slope of te tangent line at f4, wic appens to be f 4. b We know tat f f4 + f / 3. April 4, 0

16 SECTION 3. Definition of te Derivative 05 Creating a table wit values of close to zero: / Tus, f c Using te estimate for f 4 obtaine in part b, te equation of te line tangent to f 5/ at 4is f f Verif tat P, lies on te graps of bot f / + an L + m for ever slope m. Plot fan L on te same aes for several values of m until ou fin a value of m for wic L appears tangent to te grap of f. Wat is our estimate for f? Let f + an L + m. Because f + an L + m, it follows tat P, lies on te graps of bot functions. A plot of f an L on te same aes for several values of m is sown below. Te grap of L wit m appears to be tangent to te grap of fat. We terefore estimate f. 0.8 m m /4 m / Use a plot of f to estimate te value c suc tat f c 0. Fin c to sufficient accurac so tat fc+ fc for ±0.00 Here is a grap of f over te interval [0,.5] Te grap sows one location wit a orizontal tangent line. Te figure below at te left sows te grap of ftogeter wit te orizontal lines 0.6, 0.7 an 0.8. Te line 0.7 is ver close to being tangent to te grap of f. Te figure below at te rigt refines tis estimate b graping fan 0.69 on te same set of aes. Te point of tangenc as an -coorinate of rougl 0.37, so c April 4, 0

17 06 CHAPTER 3 DIFFERENTIATION We note tat an f f < f f < 0.006, so we ave etermine c to te esire accurac. 64. Plot f an + a on te same set of aes for several values of a until te line becomes tangent to te grap. Ten estimate te value c suc tat f c. Te figure below on te left sows te graps of te function f togeter wit te lines,, an ; te figure on te rigt sows te graps of f togeter wit te lines,., an.4. Te grap of. appears to be tangent to te grap of fat.4. We terefore estimate tat f In Eercises 65 7, estimate erivatives using te smmetric ifference quotient SDQ, efine as te average of te ifference quotients at an : fa+ fa + fa fa fa+ fa Te SDQ usuall gives a better approimation to te erivative tan te ifference quotient. 65. Te vapor pressure of water at temperature T in kelvins is te atmosperic pressure P at wic no net evaporation takes place. Use te following table to estimate P T for T 303, 33, 33, 333, 343 b computing te SDQ given b Eq. 4 wit 0. T K P atm Using equation 4, P 303 P 33 P 33 P 333 P 343 P33 P93 0 P33 P303 0 P333 P33 0 P343 P33 0 P353 P atm/k; atm/k; atm/k; atm/k; atm/k April 4, 0

18 SECTION 3. Definition of te Derivative Use te SDQ wit ear to estimate P T in te ears 000, 00, 004, 006, were PTis te U.S. etanol prouction Figure 8. Epress our answer in te correct units. P billions of gallons Using equation 4, P 000 P 00 P 004 P FIGURE 8 U.S. Etanol Prouction P00 P999 P003 P00 P005 P003 P007 P billions of gallons/r; 0.5 billions of gallons/r; billions of gallons/r;. billions of gallons/r In Eercises 67 an 68, traffic spee S along a certain roa in km/ varies as a function of traffic ensit q number of cars per km of roa. Use te following ata to answer te questions: q ensit S spee Estimate S 80. Let Sq be te function etermining S given q. Using equation 4 wit 0, wit 0, S 80 S 80 S90 S70 0 S00 S ; Te mean of tese two smmetric ifference quotients is kp km/car. 0.45; 68. Eplain w V qs, calle traffic volume, is equal to te number of cars passing a point per our. Use te ata to estimate V 80. Te traffic spee S as units of km/our, an te traffic ensit as units of cars/km. Terefore, te traffic volume V Sq as units of cars/our. A table giving te values of V follows. q V To estimate V /q, we take te mean of te smmetric ifference quotients. Wit 0, wit 0, V 80 V 80 V90 V70 0 V00 V ; 3.5; Te mean of te smmetric ifference quotients is 3.5. Hence V /q 3.5 cars per our wen q 80. April 4, 0

19 08 CHAPTER 3 DIFFERENTIATION Eercises 69 7: Te current in amperes at time t in secons flowing in te circuit in Figure 9 is given b Kircoff s Law: it Cv t + R vt were vt is te voltage in volts, C te capacitance in faras, an R te resistance in oms,. i + v R C FIGURE Calculate te current at t 3if vt 0.5t + 4V were C 0.0 F an R 00. Since vt is a line wit slope 0.5, v t 0.5 volts/s for all t. From te formula, i3 Cv 3 + /Rv / amperes. 70. Use te following ata to estimate v 0 b an SDQ. Ten estimate i0, assuming C 0.03 an R 000. t vt Tus, Taking 0., we fin v 0 v0. v volts/s. i amperes Assume tat R 00 but C is unknown. Use te following ata to estimate v 4 b an SDQ an euce an approimate value for te capacitance C. t vt it Solving i4 Cv 4 + /Rv4 for C iels i4 /Rv C v 00 4 v. 4 To compute C, we first approimate v 4. Taking 0., we fin v v4. v Plugging tis in to te equation above iels C faras. 60. Furter Insigts an Callenges 7. Te SDQ usuall approimates te erivative muc more closel tan oes te orinar ifference quotient. Let f an a 0. Compute te SDQ wit 0.00 an te orinar ifference quotients wit ±0.00. Compare wit te actual value, wic is f 0 ln. April 4, 0

20 SECTION 3. Definition of te Derivative 09 Let f an a 0. Te orinar ifference quotient for 0.00 is an for 0.00 is Te smmetric ifference quotient for 0.00 is Clearl te smmetric ifference quotient gives a better estimate of te erivative f 0 ln Eplain ow te smmetric ifference quotient efine b Eq. 4 can be interprete as te slope of a secant line. Te smmetric ifference quotient fa+ fa is te slope of te secant line connecting te points a, f a an a +, f a + on te grap of f ; te ifference in te function values is ivie b te ifference in te -values. 74. Wic of te two functions in Figure 0 satisfies te inequalit fa+ fa for >0? Eplain in terms of secant lines. fa+ fa a a A FIGURE 0 B Figure A satisfies te inequalit fa+ fa fa+ fa since in tis grap te smmetric ifference quotient as a larger negative slope tan te orinar rigt ifference quotient. [In figure B, te smmetric ifference quotient as a larger positive slope tan te orinar rigt ifference quotient an terefore oes not satisf te state inequalit.] 75. Sow tat if f is a quaratic polnomial, ten te SDQ at a for an 0 is equal to f a. Eplain te grapical meaning of tis result. Let f p + q + r be a quaratic polnomial. We compute te SDQ at a. fa+ fa pa + + qa + + r pa + qa + r pa + pa + p + qa + q + r pa + pa p qa + q r 4pa + q pa + q pa + q Since tis oesn t epen on, te limit, wic is equal to f a, is also pa + q. Grapicall, tis result tells us tat te secant line to a parabola passing troug points cosen smmetricall about a is alwas parallel to te tangent line at a. 76. Let f. Compute f b taking te limit of te SDQs wit a as 0. Let f. Wit a, te smmetric ifference quotient is f + f Terefore, f 0 +. April 4, 0

21 0 CHAPTER 3 DIFFERENTIATION 3. Te Derivative as a Function Preliminar Questions. Wat is te slope of te tangent line troug te point,f if f 3? Te slope of te tangent line troug te point,f is given b f. Since f 3, it follows tat f Evaluate f g an 3f + g assuming tat f 3 an g 5. f g f g 3 5 an 3f + g 3f + g To wic of te following oes te Power Rule appl? a f b f e c f e f e e f f f 4/5 a Yes. is a power function, so te Power Rule can be applie. b Yes. e is a constant function, so te Power Rule can be applie. c Yes. e is a power function, so te Power Rule can be applie. No. e is an eponential function te base is constant wile te eponent is a variable, so te Power Rule oes not appl. e No. is not a power function because bot te base an te eponent are variable, so te Power Rule oes not appl. f Yes. 4/5 is a power function, so te Power Rule can be applie. 4. Coose a or b. Te erivative oes not eist if te tangent line is: a orizontal b vertical. Te erivative oes not eist wen: b te tangent line is vertical. At a orizontal tangent, te erivative is zero. 5. Wic propert istinguises f e from all oter eponential functions g b? Te line tangent to f e at 0 as slope equal to. Eercises In Eercises 6, compute f using te limit efinition.. f 3 7 Let f 3 7. Ten, f f+ f f + 3 Let f + 3. Ten, f f+ f f 3 Let f 3. Ten, f f+ f April 4, 0

22 SECTION 3. Te Derivative as a Function 4. f Let f. Ten, f f+ f f Let f. Ten, f f+ f lim lim lim f / Let f /. Ten, f f+ f Multipling te numerator an enominator of te epression b + +, we obtain: f In Eercises 7 4, use te Power Rule to compute te erivative so t t 3 t4 t 3 3t 4 so t t t4 t t t/3 t8 t /3 t 3 t /3 so t t8 t/3 3 8 / t t /5 t t /5 t 5 t 7/5 so t t t /5 5 7/ April 4, 0

23 CHAPTER 3 DIFFERENTIATION. 4/3 3. t t 7 4. t t π 4/3 4 4/ /3. t 7 7t 7 t t t π π t π In Eercises 5 8, compute f an fin an equation of te tangent line to te grap at a. 5. f 4, a Let f 4. Ten, b te Power Rule, f 4 3. Te equation of te tangent line to te grap of fat is f + f f, a 5 Let f. Using te Power Rule, f 3. Te equation of te tangent line to te grap of fat 5is f f f 5 3, a 4 Let f 5 3 /. Ten f 5 6 /. In particular, f 4 3. Te tangent line at 4 is f f f 3, a 8 Let f 3 /3. Ten f 3 /3 3 /3. In particular, f f8, so te tangent line at 8is f f Calculate: a e b t 5t 8et Hint for c: Write e t 3 as e 3 e t. a e e e. b t 5t 8et 5 t t 8 t et 5 8e t. c t et 3 e 3 t et e 3 e t e t Fin an equation of te tangent line to 4e at. c t et 3 Let f 4e. Ten f 4e, f 4e, an f 4e. Te equation of te tangent line is f + f 4e + 4e. In Eercises 3, calculate te erivative.. f April 4, 0

24 SECTION 3. Te Derivative as a Function 3. f f 4 5/ / / f 5/ / + 5/ / /4 6 5/ gz 7z 5/4 + z z 5/4 + z z z 9/4 5z t 6 t + t 6t / + t / 3t / t t 3/. 7. fs 4 s + 3 s fs 4 s + 3 s s /4 + s /3. In tis form, we can appl te Sum an Power Rules. s /4 + s /3 s 4 s/4 + 3 s/3 4 s 3/4 + 3 s /3. 8. W /3 9. g e / /3. Because e is a constant, e f 3e 3 3e 3 3e t 5e t 3 t 5et 3 5e 3 t et 5e 3 e t 5e t f 9 /3 + 8e 9 /3 + 8e 4 /3 + 8e. In Eercises 33 36, calculate te erivative b epaning or simplifing te function. 33. Ps 4s 3 Ps 4s 3 6s 4s + 9. Tus, P s 3s 4. April 4, 0

25 4 CHAPTER 3 DIFFERENTIATION 34. Qr r3r + 5 Qr r3r + 5 6r 7r + 5. Tus, Q r r g + 4 / g + 4 / + 4 3/. Tus, g 6 5/. 36. st t t / st t t / t / t /. Tus, s t t 3/ t /. In Eercises 37 4, calculate te erivative inicate. T 37., T 3C /3 C C8 Wit TC 3C /3, we ave T C C /3. Terefore, T C 8 /3. C8 38. P, V V 39. P 7 V Wit P 7V, we ave P V 7V. Terefore, s, s 4z 6z z z 40. R W, W 4. r t, t4 P V 7 7 V 4. Wit s 4z 6z, we ave s 4 3z. Terefore, z s z z R W π Let RW W π. Ten R/W πw π. Terefore, R W π π π. W r t e t Wit r t e t, we ave r t et. Terefore, r t e 4. t4 April 4, 0

26 4. p, 4 p 7e Wit p 7e, we ave p 7e. Terefore, p 7e 4 7e. 4 SECTION 3. Te Derivative as a Function Matc te functions in graps A D wit teir erivatives I III in Figure 3. Note tat two of te functions ave te same erivative. Eplain w. A B C D I II III FIGURE 3 Consier te grap in A. On te left sie of te grap, te slope of te tangent line is positive but on te rigt sie te slope of te tangent line is negative. Tus te erivative soul transition from positive to negative wit increasing. Tis matces te grap in III. Consier te grap in B. Tis is a linear function, so its slope is constant. Tus te erivative is constant, wic matces te grap in I. Consier te grap in C. Moving from left to rigt, te slope of te tangent line transitions from positive to negative ten back to positive. Te erivative soul terefore be negative in te mile an positive to eiter sie. Tis matces te grap in II. Consier te grap in D. On te left sie of te grap, te slope of te tangent line is positive but on te rigt sie te slope of te tangent line is negative. Tus te erivative soul transition from positive to negative wit increasing. Tis matces te grap in III. Note tat te functions wose graps are sown in A an D ave te same erivative. Tis appens because te grap in D is just a vertical translation of te grap in A, wic means te two functions iffer b a constant. Te erivative of a constant is zero, so te two functions en up wit te same erivative. 44. Of te two functions f an g in Figure 4, wic is te erivative of te oter? Justif our answer. f g FIGURE 4 g is te erivative of f. For fte slope is negative for negative values of until 0, were tere is a orizontal tangent, an ten te slope is positive for positive values of. Notice tat g is negative for negative values of, goes troug te origin at 0, an ten is positive for positive values of. 45. Assign te labels f, g, an to te graps in Figure 5 in suc a wa tat f g an g. A B C FIGURE 5 April 4, 0

27 6 CHAPTER 3 DIFFERENTIATION Consier te grap in A. Moving from left to rigt, te slope of te tangent line is positive over te first quarter of te grap, negative in te mile alf an positive again over te final quarter. Te erivative of tis function must terefore be negative in te mile an positive on eiter sie. Tis matces te grap in C. Now focus on te grap in C. Te slope of te tangent line is negative over te left alf an positive on te rigt alf. Te erivative of tis function terefore nees to be negative on te left an positive on te rigt. Tis escription matces te grap in B. We soul terefore label te grap in A as f, te grap in B as, an te grap in C as g. Ten f g an g. 46. Accoring to te peak oil teor, first propose in 956 b geopsicist M. Hubbert, te total amount of crue oil Qt prouce worlwie up to time t as a grap like tat in Figure 6. a Sketc te erivative Q t for 900 t 50. Wat oes Q t represent? b In wic ear approimatel oes Q t take on its maimum value? c Wat is L Qt? An wat is its interpretation? t Wat is te value of lim t Q t?.3.0 Q trillions of barrels t ear FIGURE 6 Total oil prouction up to time t a One possible erivative sketc is sown below. Because te grap of Qt is rougl orizontal aroun t 900, te grap of Q t begins near zero. Until rougl t 000, te grap of Qt increases more an more rapil, so te grap of Q t increases. Tereafter, te grap of Qt increases more an more grauall, so te grap of Q t ecreases. Aroun t 50, te grap of Qt is again rougl orizontal, so te grap of Q t returns to zero. Note tat Q t represents te rate of cange in total worlwie oil prouction; tat is, te number of barrels prouce per ear b Te grap of Qt appears to be increasing most rapil aroun te ear 000, so Q t takes on its maimum value aroun te ear 000. c From Figure 6 L Qt.3 t trillion barrels of oil. Tis value represents te total number of barrels of oil tat can be prouce b te planet. Because te grap of Qt appears to approac a orizontal line as t, it appears tat lim t Q t Use te table of values of fto etermine wic of A or B in Figure 7 is te grap of f. Eplain f April 4, 0

28 SECTION 3. Te Derivative as a Function 7 A B FIGURE 7 Wic is te grap of f? Te increment between successive values in te table is a constant 0.5 but te increment between successive f values ecreases from 45 to 43 to 4 to 38 an so on. Tus te ifference quotients ecrease wit increasing, suggesting tat f ecreases as a function of. Because te grap in B epicts a ecreasing function, B migt be te grap of te erivative of f. 48. Let R be a variable an r a constant. Compute te erivatives: a R R b R r c R r R 3 a R, since R is a linear function of R wit slope. R b r 0, since r is a constant. R c We appl te Linearit an Power Rules: R r R 3 r R R3 r 3R 3r R. 49. Compute te erivatives, were c is a constant. a t ct3 b 9c 3 4c c z 5z + 4cz a b c t ct3 3ct. z 5z + 4cz 5 + 8cz. 9c 3 4c 7c. 50. Fin te points on te grap of f 3 were te tangent line is orizontal. Let f 3. Solve f 0 to obtain ± 6. Tus, te grap of f 3 as a orizontal tangent line at two points: 6, 6 6 an 6, Fin te points on te grap of at wic te slope of te tangent line is equal to 4. Let Solving / iels Fin te values of were 3 an + 5 ave parallel tangent lines. Let f 3 an g + 5. Te graps ave parallel tangent lines wen f g. Hence, we solve f g to obtain 3 5 an. 53. Determine a an b suc tat p + a + b satisfies p 0 an p 4. Let p + a + b satisf p 0 an p 4. Now, p + a. Terefore 0 p + a + b an 4 p + a; i.e., a an b 3. April 4, 0

29 8 CHAPTER 3 DIFFERENTIATION 54. Fin all values of suc tat te tangent line to is steeper tan te tangent line to 3. Let f an let g 3. We nee all suc tat f >g. f >g 8 + > 3 0 > >3 +. Te prouct wen an wen 3. We terefore eamine te intervals <, << 3 an > 3. For < an for > 3, 3 + >0, wereas for << 3, 3 + <0. Te set is terefore << Let f Sow tat f 3 for all an tat, for ever m> 3, tere are precisel two points were f m. Inicate te position of tese points an te corresponing tangent lines for one value of m in a sketc of te grap of f. Let P a, b be a point on te grap of f Te erivative satisfies f since 3 is nonnegative. Suppose te slope m of te tangent line is greater tan 3. Ten f a 3a 3 m, wence a m m + 3 > 0 an tus a ±. 3 Te two parallel tangent lines wit slope are sown wit te grap of fere Sow tat te tangent lines to 3 3 at a an at b are parallel if a b or a + b. Let P a, f a an Q b, f b be points on te grap of f 3 3. Equate te slopes of te tangent lines at te points P an Q: a a b b. Tus a a b + b 0. Now, a a b + b a ba + b a b a + ba b; terefore, eiter a b i.e., P an Q are te same point or a + b. 57. Compute te erivative of f 3/ using te limit efinition. Hint: Sow tat f+ f Once we ave te ifference of square roots, we multipl b te conjugate to solve te problem. f + 3/ 3/ Te first factor of te epression in te last line is clearl te limit efinition of te erivative of 3, wic is 3. Te secon factor can be evaluate, so 3/ /. April 4, 0

30 SECTION 3. Te Derivative as a Function Use te limit efinition of mb to approimate m4. Ten estimate te slope of te tangent line to 4 at 0 an. Recall m Using a table of values, we fin Tus m Knowing tat m4 4, it follows tat an Let f e. Use te limit efinition to compute f 0, an fin te equation of te tangent line at 0. Let f e. Ten f0 0, an f f0 + f0 e e. Te equation of te tangent line is f f Te average spee in meters per secon of a gas molecule is v avg 8RT πm were T is te temperature in kelvins, M is te molar mass in kilograms per mole, an R 8.3. Calculate v avg /T at T 300 K for ogen, wic as a molar mass of 0.03 kg/mol. Using te form v av 8RT /πm / 8R/πMT /, were M an R are constant, we use te Power Rule to compute te erivative v av /T. 8R/πMT / 8R/πM T T T / 8R/πM T /. In particular, if T 300 K, T v av 88.3/π / m/s K. 6. Biologists ave observe tat te pulse rate P in beats per minute in animals is relate to bo mass in kilograms b te approimate formula P 00m /4. Tis is one of man allometric scaling laws prevalent in biolog. Is P /m an increasing or ecreasing function of m? Fin an equation of te tangent line at te points on te grap in Figure 8 tat represent goat m 33 an man m 68. Pulse beats/min Guinea pig Goat Man FIGURE 8 Cattle Mass kg April 4, 0

31 0 CHAPTER 3 DIFFERENTIATION P /m 50m 5/4. For m>0, P /m 50m 5/4. P /m 0asmgets larger; P /m gets smaller as m gets bigger. For eac m c, te equation of te tangent line to te grap of P at m is P cm c + Pc. For a goat m 33 kg, P beats per minute bpm an P m / bpm/kg. Hence, 0.636m For a man m 68 kg, we ave P bpm an P m / bpm/kg. Hence, te tangent line as formula m Some stuies suggest tat kine mass K in mammals in kilograms is relate to bo mass m in kilograms b te approimate formula K 0.007m Calculate K/m at m 68. Ten calculate te erivative wit respect to m of te relative kine-to-mass ratio K/m at m 68. ence, K m m m 0.5 ; K m m68 Because we fin m K m m m 0.007m 0.5, K m m m m.5, an K m m m68 6 kg. 63. Te Clausius Claperon Law relates te vapor pressure of water P in atmosperes to te temperature T in kelvins: P T k P T were k is a constant. Estimate P /T for T 303, 33, 33, 333, 343 using te ata an te approimation P T PT + 0 PT 0 0 T K P atm Do our estimates seem to confirm te Clausius Claperon Law? Wat is te approimate value of k? Using te inicate approimation to te first erivative, we calculate P 303 P 33 P 33 P33 P93 0 P33 P303 0 P333 P atm/k; atm/k; atm/k; April 4, 0

32 P 333 P 343 P343 P33 0 P353 P SECTION 3. Te Derivative as a Function atm/k; atm/k If te Clausius Claperon law is vali, ten T P soul remain constant as T varies. Using te ata for vapor P T pressure an temperature an te approimate erivative values calculate above, we fin T K T P P T Tese values are rougl constant, suggesting tat te Clausius Claperon law is vali, an tat k Let L be te tangent line to te perbola at a, were a>0. Sow tat te area of te triangle boune b L an te coorinate aes oes not epen on a. Let f. Te tangent line to f at a is f a a + fa a a + a. Te -intercept of tis line were 0 is a. Its -intercept were 0 is a. Hence te area of te triangle boune b te tangent line an te coorinate aes is A b a a, wic is inepenent of a. P 0, a R a, a Q a, In te setting of Eercise 64, sow tat te point of tangenc is te mipoint of te segment of L ling in te first quarant. In te previous eercise, we saw tat te tangent line to te perbola or at a as -intercept P 0, a an -intercept Q a,0. Te mipoint of te line segment connecting P an Q is tus 0 + a a, + 0 a,, a wic is te point of tangenc. 66. Matc functions A C wit teir erivatives I III in Figure 9. A B C I II III FIGURE 9 Note tat te grap in A as tree locations wit a orizontal tangent line. Te erivative must terefore cross te -ais in tree locations, wic matces III. Te grap in B as onl one location wit a orizontal tangent line, so its erivative soul cross te -ais onl once. Tus, I is te grap corresponing to te erivative of B. Finall, te grap in B as two locations wit a orizontal tangent line, so its erivative soul cross te -ais twice. Tus, II is te grap corresponing to te erivative of C. April 4, 0

33 CHAPTER 3 DIFFERENTIATION 67. Make a roug sketc of te grap of te erivative of te function in Figure 0A A FIGURE 0 B Te grap as a tangent line wit negative slope approimatel on te interval, 3.6, an as a tangent line wit a positive slope elsewere. Tis implies tat te erivative must be negative on te interval, 3.6 an positive elsewere. Te grap ma terefore look like tis: Grap te erivative of te function in Figure 0B, omitting points were te erivative is not efine. On, 0, te grap is a line wit slope 3, so te erivative is equal to 3. Te erivative on 0, is. Finall, on, 4 te function is a line wit slope, so te erivative is equal to. Combining tis information leas to te grap: Sketc te grap of f. Ten sow tat f 0 eists. For <0, f, an f. For >0, f, an f.at 0, we fin f0 + f0 lim an f0 + f0 lim Because te two one-sie limits eist an are equal, it follows tat f 0 eists an is equal to zero. Here is te grap of f. 4 4 April 4, 0

34 SECTION 3. Te Derivative as a Function Determine te values of at wic te function in Figure is: a iscontinuous, an b nonifferentiable. 3 4 FIGURE Te function is iscontinuous at tose points were it is unefine or tere is a break in te grap. On te interval [0, 4], tere is onl one suc point, at. Te function is nonifferentiable at tose points were it is iscontinuous or were it as a corner or cusp. In aition to te point we alrea know about, te function is nonifferentiable at an 3. In Eercises 7 76, fin te points c if an suc tat f c oes not eist. 7. f Here is te grap of f. Its erivative oes not eist at. At tat value of tere is a sarp corner. 7. f[] Here is te grap of f []. Tis is te integer step function grap. Its erivative oes not eist at all values tat are integers. At tose values of te grap is iscontinuous. 73. f /3 Here is te grap of f /3. Its erivative oes not eist at 0. At tat value of, tere is a sarp corner or cusp f 3/ Te function is ifferentiable on its entire omain, { : 0}. Te formula is 3/ 3 /. April 4, 0

35 4 CHAPTER 3 DIFFERENTIATION 75. f Here is te grap of f. Its erivative oes not eist at orat. At tese values of, te grap as sarp corners f 3 3 Tis is te grap of f. Its erivative eists everwere. In Eercises 77 8, zoom in on a plot of fat te point a, f a an state weter or not fappears to be ifferentiable at a. If it is nonifferentiable, state weter te tangent line appears to be vertical or oes not eist. 77. f, a 0 Te grap of f for near 0 is sown below. Because te grap as a sarp corner at 0, it appears tat f is not ifferentiable at 0. Moreover, te tangent line oes not eist at tis point f 3 5/3, a 3 Te grap of f 3 5/3 for near 3 is sown below. From tis grap, it appears tat f is ifferentiable at 3, wit a orizontal tangent line f 3 /3, a 3 Te grap of f 3 /3 for near 3 is sown below. From tis grap, it appears tat f is not ifferentiable at 3. Moreover, te tangent line appears to be vertical April 4, 0

36 SECTION 3. Te Derivative as a Function f sin /3, a 0 Te grap of f sin /3 for near 0 is sown below. From tis grap, it appears tat f is not ifferentiable at 0. Moreover, te tangent line appears to be vertical f sin, a 0 Te grap of f sin for near 0 is sown below. Because te grap as a sarp corner at 0, it appears tat f is not ifferentiable at 0. Moreover, te tangent line oes not eist at tis point f sin, a 0 Te grap of f sin for near 0 is sown below. From tis grap, it appears tat f is ifferentiable at 0, wit a orizontal tangent line Plot te erivative f of f 3 0 for >0 set te bouns of te viewing bo appropriatel an observe tat f > 0. Wat oes te positivit of f tell us about te grap of fitself? Plot fan confirm tis conclusion. Let f 3 0. Ten f Te grap of f is sown in te figure below at te left an it is clear tat f > 0 for all >0. Te positivit of f tells us tat te grap of fis increasing for >0. Tis is confirme in te figure below at te rigt, wic sows te grap of f April 4, 0

37 6 CHAPTER 3 DIFFERENTIATION 84. Fin te coorinates of te point P in Figure at wic te tangent line passes troug 5, 0. f 9 9 P FIGURE Grap of f 9. Let f 9, an suppose P as coorinates a, 9 a. Because f, te slope of te line tangent to te grap of fat P is a, an te equation of te tangent line is f a a + fa a a + 9 a a a. In orer for tis line to pass troug te point 5, 0, we must ave 0 0a a a 9a. Tus, a ora 9. We eclue a 9 because from Figure we are looking for an -coorinate between 0 an 5. Tus, te point P as coorinates, 8. Eercises refer to Figure 3. Lengt QR is calle te subtangent at P, an lengt RT is calle te subnormal. f P, f Tangent line Q R T FIGURE Calculate te subtangent of f + 3 at Let f + 3. Ten f + 3, an te equation of te tangent line at is f + f Tis line intersects te -ais at 4 7. Tus Q as coorinates 4 7, 0, R as coorinates, 0 an te subtangent is Sow tat te subtangent of f e is everwere equal to. Let f e. Ten f e, an te equation of te tangent line at a is f a a + fa e a a + e a. Tis line intersects te -ais at a. Tus, Q as coorinates a, 0, R as coorinates a, 0 an te subtangent is a a. 87. Prove in general tat te subnormal at P is f f. Te slope of te tangent line at P is f. Te slope of te line normal to te grap at P is ten /f, an te normal line intersects te -ais at te point T wit coorinates + ff, 0. Te point R as coorinates, 0, so te subnormal is + ff ff. April 4, 0

38 SECTION 3. Te Derivative as a Function Sow tat PQas lengt f + f. Te coorinates of te point P are, f, te coorinates of te point R are, 0 an te coorinates of te point Q are f f, 0. Tus, PR f, QR f f, an b te Ptagorean Teorem f PQ f + f f + f. 89. Prove te following teorem of Apollonius of Perga te Greek matematician born in 6 bce wo gave te parabola, ellipse, an perbola teir names: Te subtangent of te parabola at a is equal to a/. Let f. Te tangent line to f at a is Te -intercept of tis line were 0 is a as claime. f a a + fa a a + a a a. a, a, a Sow tat te subtangent to 3 at a is equal to 3 a. Let f 3. Ten f 3, an te equation of te tangent line t a is f a a + fa 3a a + a 3 3a a 3. Tis line intersects te -ais at a/3. Tus, Q as coorinates a/3, 0, R as coorinates a, 0 an te subtangent is a 3 a 3 a. 9. Formulate an prove a generalization of Eercise 90 for n. Let f n. Ten f n n, an te equation of te tangent line t a is f a a + fa na n a + a n na n n a n. Tis line intersects te -ais at n a/n. Tus, Q as coorinates n a/n, 0, R as coorinates a, 0 an te subtangent is a n n a n a. Furter Insigts an Callenges 9. Two small arces ave te sape of parabolas. Te first is given b f for an te secon b g 4 4 for 6. A boar is place on top of tese arces so it rests on bot Figure 4. Wat is te slope of te boar? Hint: Fin te tangent line to ftat intersects g in eactl one point. FIGURE 4 April 4, 0

39 8 CHAPTER 3 DIFFERENTIATION At te points were te boar makes contact wit te arces te slope of te boar must be equal to te slope of te arces an ence te are equal to eac oter. Suppose t, f t is te point were te boar touces te left an arc. Te tangent line ere te line te boar efines is given b f t t + ft. Tis line must it te oter arc in eactl one point. In oter wors, if we plug in g to get g f t t + ft tere can onl be one for in terms of t. Computing f an plugging in we get wic simplifies to t + t + t t 8 + t Tis is a quaratic equation a + b + c 0 wit a, b t 8 an c t + 3. B te quaratic formula we know tere is a unique for iff b 4ac 0. In our case tis means t + 8 4t + 3. Solving tis gives t 3/8 an plugging into f sows te slope of te boar must be 3/ A vase is forme b rotating aroun te -ais. If we rop in a marble, it will eiter touc te bottom point of te vase or be suspene above te bottom b toucing te sies Figure 5. How small must te marble be to touc te bottom? FIGURE 5 Suppose a circle is tangent to te parabola at te point t, t. Te slope of te parabola at tis point is t, so te slope of te raius of te circle at tis point is t since it is perpenicular to te tangent line of te circle. Tus te center of te circle must be were te line given b t t + t crosses te -ais. We can fin te -coorinate b setting 0: we get + t. Tus, te raius etens from 0, + t to t, t an r + t t + t 4 + t. Tis raius is greater tan wenever t>0; so, if a marble as raius > / it sits on te ege of te vase, but if it as raius / it rolls all te wa to te bottom. 94. Let fbe a ifferentiable function, an set g f+ c, were c is a constant. Use te limit efinition to sow tat g f + c. Eplain tis result grapicall, recalling tat te grap of g is obtaine b sifting te grap of fcunits to te left if c>0 or rigt if c<0. Let g f+ c. Using te limit efinition, g g + g f + + c f+ c 0 0 f + c + f+ c f + c. 0 Te grap of g is obtaine b sifting fto te left b c units. Tis implies tat g is equal to f sifte to te left b c units, wic appens to be f + c. Terefore, g f + c. April 4, 0

40 SECTION 3. Te Derivative as a Function Negative Eponents Let n be a wole number. Use te Power Rule for n to calculate te erivative of f n b sowing tat f+ f Let f n were n is a positive integer. Te ifference quotient for f is + n n n + n f+ f + n n + n n + n n n + n. n + n n + n Terefore, f f+ f + n n 0 0 n + n 0 n + n lim + n n n n. 0 From above, we continue: f n n n n n n n. Since n is a positive integer, k nis a negative integer an we ave for negative integers k. k n n n k k ; i.e. k k k 96. Verif te Power Rule for te eponent /n, were n is a positive integer, using te following trick: Rewrite te ifference quotient for /n at b in terms of u b + /n an a b /n. Substituting b + /n an a b /n into te left-an sie of equation 3 iels n a n a b + b b + /n b /n b + /n b /n wereas substituting tese same epressions into te rigt-an sie of equation 3 prouces ence, n a n a n b + n + b + n n b /n + b + n 3 n b /n + +b n n ; b + /n b /n b + n n + b + n n b /n + b + n 3 n b /n + +b n n. If we take f /n, ten, using te previous epression, Replacing b b, we ave f n n. f b + /n b /n b 0 nb n n n b n. 97. Infinitel Rapi Oscillations Define sin 0 f 0 0 Sow tat fis continuous at 0 but f 0 oes not eist see Figure 4. { sin if 0 Let f.as 0, 0 if 0 f f0 sin 0 sin 0 April 4, 0

41 30 CHAPTER 3 DIFFERENTIATION since te values of te sine lie between an. Hence, b te Squeeze Teorem, lim f f0 an tus f is 0 continuous at 0. As 0, te ifference quotient at 0, sin 0 f f0 0 0 oes not converge to a limit since it oscillates infinitel troug ever value between an. Accoringl, f 0 oes not eist. 98. For wic value of λ oes te equation e λ ave a unique? For wic values of λ oes it ave at least one? For intuition, plot e an te line λ. First, note tat wen λ 0, te equation e 0 0 as no real. For λ 0, we observe tat s to te equation e λ correspon to points of intersection between te graps of e an λ. Wen λ<0, te two graps intersect at onl one location see te grap below at te left. On te oter an, wen λ>0, te graps ma ave zero, one or two points of intersection see te grap below at te rigt. Note tat te graps ave one point of intersection wen λ is te tangent line to e. Tus, not onl o we require e λ, but also e λ. It ten follows tat te point of intersection satisfies λ λ, so. Tis ten gives λ e. Terefore te equation e λ: a as at least one wen λ<0an wen λ e; b as a unique wen λ<0an wen λ e. sin Prouct an Quotient Rules Preliminar Questions. Are te following statements true or false? If false, state te correct version. a fg enotes te function wose value at is fg. b f/g enotes te function wose value at is f /g. c Te erivative of te prouct is te prouct of te erivatives. fg f4g 4 g4f 4 4 e fg f0g 0 + g0f 0 0 a False. Te notation fg enotes te function wose value at is fg. b True. c False. Te erivative of a prouct fg is f g + fg. False. fg f4g 4 + g4f 4. 4 e True.. Fin f/g if f f g an g 4. f/g [gf fg ]/g [ 4]/. 3. Fin g if f 0, f, an fg 0. fg fg + f g, so0 0 g + g an g 5. April 4, 0

42 SECTION 3.3 Prouct an Quotient Rules 3 Eercises In Eercises 6, use te Prouct Rule to calculate te erivative.. f 3 + Let f 3 +. Ten f f Let f Ten f f e Let f e. Ten 4. f 94e + Let f 94e +. Ten f e + e e + e e +. f 9 4e + + 4e e + 4e + 8e 8e s, s s / + s7 s s4 Let s s / + s7 s. Ten Terefore, s s / + s 7 s + 7 s s s / + ss + 7 s 6. t, t 8t e t + t t Let t t 8t e t + t. Ten s 3/ + s / + s s 7 s4 4 3/ / t 8t t t et + t + e t + t t t 8t t 8t e t + t + e t + t + 8t. 7 s 3/ + 3 s 5/ + 4. Terefore, t 4e e e + 4. t In Eercises 7, use te Quotient Rule to calculate te erivative. 7. f Let f. Ten f. April 4, 0

43 3 CHAPTER 3 DIFFERENTIATION 8. f Let f Ten 9. g t, gt t + t t Let gt t + t. Ten f Terefore, g t t t t + t + t t t t t t + t t g t 4 t t t. 0. w z, w z z9 z + z Let wz z z + z. Ten w z z + z z z z z z + z z + z z z + z z /z / + z + z 3/z3/ + z z + z. Terefore, w z 3/93/ + 9 z g + e Let g + e. Ten. f e + Let f e +. Ten g + e + e + e + e 0 e e + e + e. f + e e e e + e +. In Eercises 3 6, calculate te erivative in two was. First use te Prouct or Quotient Rule; ten rewrite te function algebraicall an appl te Power Rule irectl. 3. ft t + t Let ft t + t. Ten, using te Prouct Rule, f t t + t + t 6t + t 4. Multipling out first, we fin ft t 3 + t 4t. Terefore, f t 6t + t 4. April 4, 0

44 SECTION 3.3 Prouct an Quotient Rules f 3 + Let f 3 +. Ten, using te prouct rule, an ten power an sum rules, f Multipling out first, we fin f 3 +. Ten f t t t Let t t t. Using te quotient rule, f t t t t t t t + t for t. Simplifing first, we fin for t, t t + t t +. t Hence t for t. 6. g Let g Using te quotient rule an te sum an power rules, an simplifing g Simplifing first iels g + + 3, from wic we calculate g In Eercises 7 38, calculate te erivative. 7. f Let f Ten f f 4e 3 + Let f 4e 3 +. Ten f 4e e e , Let +0. Using te quotient rule: Terefore, z 0., z 3 + Let z 3. Using te quotient rule: + z Terefore, z April 4, 0

45 34 CHAPTER 3 DIFFERENTIATION. f + Let f +. Multipling troug first iels f for 0. Terefore, f for 0. If we carr out te prouct rule on f / + /, weget f / + / + / / + / + /.. f 95/ Let f 95/ 9 3/. Ten f 7 / +. 3., Let Ten Terefore, f 4 + e + Let f 4 + e +. Ten f e 4 + e e 4 e +. z 5., z 3 + Let z 3. Using te quotient rule: + z Terefore, 6. f 33 + z Let f /. Using te quotient rule, an ten simplifing b taking out te greatest negative factor: f / / 5 3/ / Alternatel, since tere is a single eponent of in te enominator, we coul also simplif f first, getting f 3 5/ 3/ + /. Ten f 5 3/ 3 / 3/. Te two answers are te same. April 4, 0

46 t 7. t t + t + t Let t t + t + t t 3 + t + t +. Ten t 3 + t + t + t t t 3 + t + t + 3t + t + SECTION 3.3 Prouct an Quotient Rules 35 t3 t + t 3 + t + t f 3/ / Let f 3/ /. We multipl troug te 3/ to get f / 3 5/ +. Ten f 9/ 5 3/ ft 3 / 5 / Let ft 3 5. Ten f t 0, since ftis a constant function! 30. π Let π. Ten π. 3. f Let f Using te Prouct Rule insie te Prouct Rule wit a first factor of + 3 an a secon factor of 5, wefin f Alternativel, f Terefore, f f e Let f e Using te Prouct Rule insie te Prouct Rule wit a first factor of e an a secon factor of + + 4, wefin f e e e. 33. f e + Let f 34. g e+ + e e + Let Ten g e. 35. gz z 4 z Let e e + +. Ten f e + + e e e e e e e + + e + +. gz for z an z. Ten, z z + z 4 z g e+ + e e + Hint: Simplif first. z z + e e + e + e. z + z z + z z z + g z z + + z z. z z + April 4, 0

47 36 CHAPTER 3 DIFFERENTIATION 36. a + bab + a,b constants Let f a + bab +. Ten f a + bab + ab + a 3a b + a + ab. 37. t t 4 t constant Let ft t 4 t. Using te quotient rule: 38. a + b c + f t t t 4t t t t + 8t t t + 8t t. Let f a, b, c, constants a + b. Using te quotient rule: c + f c + a a + bc c + a bc c +. In Eercises 39 4, calculate te erivative using te values: f4 f 4 g4 g fg 4 an f/g 4. Let fg an H f/g. Ten fg + gf an H gf fg. Finall, an 4 f4g 4 + g4f , H 4 g4f 4 f4g 4 g4 g F 4, were F f. Let F f. Ten F f + f, an F 4 6f 4 + 8f G 4, were G g. Let G g gg. Ten G gg + gg gg, an 4. H 4, were H gf. Let H gf. Ten an G 4 g4g H gf gf + fg gf, H April 4, 0

48 SECTION 3.3 Prouct an Quotient Rules Calculate F 0, were F Hint: Do not calculate F. Instea, write F f /g an epress F 0 irectl in terms of f0, f 0, g0, g 0. Taking te int, let f an let g Ten F f g. Now, f an g Moreover, f0 0, f 0 7, g0, an g 0. Using te quotient rule: F 0 g0f 0 f0g 0 g Procee as in Eercise 43 to calculate F 0, were F + + 4/3 + 5/ Write F f g/, were f + + 4/3 + 5/3 g an Now, f , g , an Moreover, f0, f 0, g0, g 0 5, 0, an 0 0. From te prouct an quotient rules, F 0 f0 0g 0 g f 5 0 0g0/0 + / Use te Prouct Rule to calculate e. Note tat e e e. Terefore e e e e e + e e e. 46. Plot te erivative of f / + over [ 4, 4]. Use te grap to etermine te intervals on wic f > 0 an f < 0. Ten plot fan escribe ow te sign of f is reflecte in te grap of f. Let f +. Ten f Te erivative is sown in te figure below at te left. From tis plot we see tat f > 0 for << an f < 0 for >. Te original function is plotte in te figure below at te rigt. Observe tat te grap of fis increasing wenever f > 0 an tat fis ecreasing wenever f < 0. April 4, 0

49 38 CHAPTER 3 DIFFERENTIATION Plot f / in a suitabl boune viewing bo. Use te plot to etermine weter f is positive or negative on its omain { : ±}. Ten compute f an confirm our conclusion algebraicall. Let f. Te grap of fis sown below. From tis plot, we see tat fis ecreasing on its omain { : ±}. Consequentl, f must be negative. Using te quotient rule, we fin wic is negative for all ±. f +, Let P V R/R + r as in Eample 7. Calculate P/r, assuming tat r is variable an R is constant. Note tat V is also constant. Let Using te quotient rule: fr V R R + r V R R + Rr + r. f r R + Rr + r 0 V RR + r R + r 4 V RR + r R + r 4 V R R + r Fin a>0suc tat te tangent line to te grap of f e at a passes troug te origin Figure 4. f e a FIGURE 4 Let f e. Ten fa a e a, f e + e e, f a a a e a, an te equation of te tangent line to f at a is For tis line to pass troug te origin, we must ave f a a + fa a a e a a + a e a. 0 a a e a a + a e a e a a a + a 3 a e a a. Tus, a 0ora. Te onl value a>0 suc tat te tangent line to f e passes troug te origin is terefore a. April 4, 0

50 SECTION 3.3 Prouct an Quotient Rules Current I amperes, voltage V volts, an resistance R oms in a circuit are relate b Om s Law, I V/R. a Calculate I R if V is constant wit value V 4. R6 b Calculate V R if I is constant wit value I 4. R6 a Accoring to Om s Law, I V/R VR. Tus, using te power rule, I R VR. Wit V 4 volts, it follows tat I R 46 R6 3 amps. b Solving Om s Law for V iels V RI. Tus V R I an V R 4 amps. I4 5. Te revenue per mont earne b te Couture cloting cain at time t is Rt NtSt, were Ntis te number of stores an St is average revenue per store per mont. Couture embarks on a two-part campaign: A to buil new stores at a rate of 5 stores per mont, an B to use avertising to increase average revenue per store at a rate of $0,000 per mont. Assume tat N0 50 an S0 $50,000. a Sow tat total revenue will increase at te rate R t 5St + 0,000Nt Note tat te two terms in te Prouct Rule correspon to te separate effects of increasing te number of stores on te one an, an te average revenue per store on te oter. b Calculate R t. t0 c If Couture can implement onl one leg A or B of its epansion at t 0, wic coice will grow revenue most rapil? a Given Rt NtSt, it follows tat R t NtS t + StN t. We are tol tat N t 5 stores per mont an S t 0,000 ollars per mont. Terefore, R t 5St + 0,000Nt. b Using part a an te given values of N0 an S0, wefin R t 550, ,00050,50,000. t0 c From part b, we see tat of te two terms contributing to total revenue growt, te term 5S0 is larger tan te term 0,000N0. Tus, if onl one leg of te campaign can be implemente, it soul be part A: increase te number of stores b 5 per mont. 5. Te tip spee ratio of a turbine Figure 5 is te ratio R T/W, were T is te spee of te tip of a blae an W is te spee of te win. Engineers ave foun empiricall tat a turbine wit n blaes etracts maimum power from te win wen R π/n. Calculate R/t t in minutes if W 35 km/ an W ecreases at a rate of 4 km/ per minute, an te tip spee as constant value T 50 km/. April 4, 0

51 40 CHAPTER 3 DIFFERENTIATION FIGURE 5 Turbines on a win farm Let R T/W. Ten R t WT TW W. Using te values T 50, T 0, W 35 an W 4, we fin R t Te curve / + is calle te witc of Agnesi Figure 6 after te Italian matematician Maria Agnesi , wo wrote one of te first books on calculus. Tis strange name is te result of a mistranslation of te Italian wor la versiera, meaning tat wic turns. Fin equations of te tangent lines at ±. 3 3 FIGURE 6 Te witc of Agnesi. Let f +. Ten f At, te tangent line is f + + f At, te tangent line is f + f Let f g. Sow tat f/g f /g. f/g /, so f/g 0. On te oter an, f /g / /. We see tat Use te Prouct Rule to sow tat f ff. Let g f ff. Ten g f ff ff + ff ff. 56. Sow tat f 3 3f f. Let g f 3 fff. Ten g f 3 [f ff ] f ff + ff + ff f 3f f. April 4, 0

52 SECTION 3.3 Prouct an Quotient Rules 4 Furter Insigts an Callenges 57. Let f, g, be ifferentiable functions. Sow tat fg is equal to Hint: Write fg as fg. Let p fg. Ten fg + fg + f g p fg f g + g + gf f g + fg + fg. 58. Prove te Quotient Rule using te limit efinition of te erivative. Let p f. Suppose tat f an g are ifferentiable at a an tat ga 0. Ten g p pa + pa a 0 0 fa+ ga + fa ga 0 fa+ ga faga+ ga + ga fa+ ga f aga + f aga faga+ 0 ga + ga fa+ fa ga + ga ga fa 0 ga + ga fa+ fa ga + ga ga lim fa lim 0 ga + ga 0 0 gaf a fag gaf a fag a a ga ga In oter wors, p f gf fg g g. 59. Derivative of te Reciprocal Use te limit efinition to prove f f f Hint: Sow tat te ifference quotient for /f is equal to f f+ f f + 7 Let g f. We ten compute te erivative of g using te ifference quotient: g g + g 0 0 f+ f f+ f 0 ff+ f+ f lim. 0 ff+ We can appl te rule of proucts for limits. Te first parentetical epression is te ifference quotient efinition of f. Te secon can be evaluate at 0 to give f. Hence g f f f. 60. Prove te Quotient Rule using Eq. 7 an te Prouct Rule. Let f g. We can write f g. Appling Eq. 7, f g + f f g g g + f g fg + f g g. 6. Use te limit efinition of te erivative to prove te following special case of te Prouct Rule: f f + f April 4, 0

53 4 CHAPTER 3 DIFFERENTIATION First note tat because fis ifferentiable, it is also continuous. It follows tat lim f+ f. 0 Now we tackle te erivative: + f + f f+ f f + f+ 0 0 f+ f lim + lim f+ 0 0 f + f. 6. Carr out Maria Agnesi s proof of te Quotient Rule from er book on calculus, publise in 748: Assume tat f, g, an f/g are ifferentiable. Compute te erivative of g f using te Prouct Rule, an solve for. Suppose tat f, g, an are ifferentiable functions wit f/g. Ten g f an via te prouct rule g + g f. Solving for iels f g g f f g g g gf fg g. 63. Te Power Rule Revisite If ou are familiar wit proof b inuction, use inuction to prove te Power Rule for all wole numbers n. Sow tat te Power Rule ols for n ; ten write n as n an use te Prouct Rule. Let k be a positive integer. If k, ten k. Note tat 0. Hence te Power Rule ols for k. Assume it ols for k n were n. Ten for k n +, we ave k n+ n n + n n n + n n + n k k Accoringl, te Power Rule ols for all positive integers b inuction. Eercises 64 an 65: A basic fact of algebra states tat c is a root of a polnomial fif an onl if f cg for some polnomial g. We sa tat c is a multiple root if f c, were is a polnomial. 64. Sow tat c is a multiple root of fif an onl if c is a root of bot fan f. Assume first tat fc f c 0 an let us sow tat c is a multiple root of f. We ave f cg for some polnomial g an so f cg + g. However, f c 0 + gc 0, so c is also a root of g an ence g c for some polnomial. We conclue tat f c, wic sows tat c is a multiple root of f. Conversel, assume tat c is a multiple root. Ten fc 0 an f c g for some polnomial g. Ten f c g + g c. Terefore, f c c c g c + gcc c Use Eercise 64 to etermine weter c is a multiple root: a b a To sow tat is a multiple root of f , it suffices to ceck tat f f 0. We ave f an f f April 4, 0

54 SECTION 3.4 Rates of Cange 43 b Let f Ten f Because but f f , it follows tat is a root of f, but not a multiple root. 66. Figure 7 is te grap of a polnomial wit roots at A, B, an C. Wic of tese is a multiple root? Eplain our reasoning using Eercise 64. A B C FIGURE 7 A on te figure is a multiple root. It is a multiple root because f 0atA an because te tangent line to te grap at A is orizontal, so tat f 0atA. For te same reasons, f also as a multiple root at C. 67. Accoring to Eq. 6 in Section 3., b mb b. Use te Prouct Rule to sow tat mab ma + mb. mabab ab a b a b + b a mba b + maa b ma + mbab. Tus, mab ma + mb. 3.4 Rates of Cange Preliminar Questions. Wic units migt be use for eac rate of cange? a Pressure in atmosperes in a water tank wit respect to ept b Te rate of a cemical reaction cange in concentration wit respect to time wit concentration in moles per liter a Te rate of cange of pressure wit respect to ept migt be measure in atmosperes/meter. b Te reaction rate of a cemical reaction migt be measure in moles/liter our.. Two trains travel from New Orleans to Mempis in 4 ours. Te first train travels at a constant velocit of 90 mp, but te velocit of te secon train varies. Wat was te secon train s average velocit uring te trip? 90 mp. Since bot trains travel te same istance in te same amount of time, te ave te same average velocit: 3. Estimate f6, assuming tat f5 43, f f f5 + f 5 5, sof Te population Ptof Freeonia in 009 was P009 5 million. a Wat is te meaning of P 009? b Estimate P00 if P a Because Ptmeasures te population of Freeonia as a function of time, te erivative P 009 measures te rate of cange of te population of Freeonia in te ear 009. b P00 P009 + P 00. Tus, if P , ten P million. April 4, 0

55 44 CHAPTER 3 DIFFERENTIATION Eercises In Eercises 8, fin te rate of cange.. Area of a square wit respect to its sie s wen s 5. Let te area be A fs s. Ten te rate of cange of A wit respect to s is /ss s. Wen s 5, te area canges at a rate of 0 square units per unit increase. Draw a 5 5 square on grap paper an trace te area ae b increasing eac sie lengt b, ecluing te corner, to see wat tis means.. Volume of a cube wit respect to its sie s wen s 5. Let te volume be V fs s 3. Ten te rate of cange of V wit respect to s is s s3 3s. Wen s 5, te volume canges at a rate of cubic units per unit increase. 3. Cube root 3 wit respect to wen, 8, 7. Let f 3. Writing f /3, we see te rate of cange of f wit respect to is given b f 3 /3. Te requeste rates of cange are given in te table tat follows: c ROC of fwit respect to at c. f f / f / Te reciprocal / wit respect to wen,, 3. Let f. Te rate of cange of fwit respect to is given b f. Te requeste rates of cange are ten wen, 4 wen an 9 wen Te iameter of a circle wit respect to raius. Te relationsip between te iameter of a circle an its raius r is r. Te rate of cange of te iameter wit respect to te raius is ten. 6. Surface area A of a spere wit respect to raius r A 4πr. Because A 4πr, te rate of cange of te surface area of a spere wit respect to te raius is A 8πr. 7. Volume V of a cliner wit respect to raius if te eigt is equal to te raius. Te volume of te cliner is V πr πr 3. Tus V /r 3πr. 8. Spee of soun v in m/s wit respect to air temperature T in kelvins, were v 0 T. Because, v 0 T 0T /, te rate of cange of te spee of soun wit respect to temperature is v 0T / 0. T In Eercises 9, refer to Figure 0, te grap of istance st from te origin as a function of time for a car trip. Distance km t FIGURE 0 Distance from te origin versus time for a car trip. 9. Fin te average velocit over eac interval. a [0, 0.5] b [0.5, ] c [,.5] [, ] a Te average velocit over te interval [0, 0.5] is km/our April 4, 0

56 SECTION 3.4 Rates of Cange 45 b Te average velocit over te interval [0.5, ] is c Te average velocit over te interval [,.5] is Te average velocit over te interval [, ] is km/our. 0km/our. 50 km/our. 0. At wat time is velocit at a maimum? Te velocit is maimum wen te slope of te istance versus time curve is most positive. Tis appears to appen wen t 0.5 ours.. Matc te escriptions i iii wit te intervals a c. i Velocit increasing ii Velocit ecreasing iii Velocit negative a [0, 0.5] b [.5, 3] c [.5, ] a i : Te istance curve is increasing, an is also bening upwar, so tat istance is increasing at an increasing rate. b ii : Over te interval [.5, 3], te istance curve is flattening, sowing tat te car is slowing own; tat is, te velocit is ecreasing. c iii : Te istance curve is ecreasing, so te tangent line as negative slope; tis means te velocit is negative.. Use te ata from Table in Eample to calculate te average rate of cange of Martian temperature T wit respect to time t over te interval from 8:36 am to 9:34 am. Te time interval from 8:36 am to 9:34 am as lengt 58 minutes, an te cange in temperature over tis time interval is Te average rate of cange is ten T C. T t C/min C/r. 3. Use Figure 3 from Eample to estimate te instantaneous rate of cange of Martian temperature wit respect to time in egrees Celsius per our at t 4 am. Te segment of te temperature grap aroun t 4 am appears to be a straigt line passing troug rougl :36, 70 an 4:48, 75. Te instantaneous rate of cange of Martian temperature wit respect to time at t 4 am is terefore approimatel T t C/our. 4. Te temperature in C of an object at time t in minutes is Tt 3 8 t 5t + 80 for 0 t 0. At wat rate is te object cooling at t 0? Give correct units. Given Tt 3 8 t 5t + 80, it follows tat T t 3 4 t 5 an T C/min. At t 0, te object is cooling at te rate of 7.5 C/min. 5. Te velocit in cm/s of bloo molecules flowing troug a capillar of raius cm is v r, were r is te istance from te molecule to te center of te capillar. Fin te rate of cange of velocit wit respect to r wen r cm. April 4, 0

57 46 CHAPTER 3 DIFFERENTIATION Te rate of cange of te velocit of te bloo molecules is v r 0.00r. Wen r cm, tis rate is /s. 6. Figure isplas te voltage V across a capacitor as a function of time wile te capacitor is being carge. Estimate te rate of cange of voltage at t 0 s. Inicate te values in our calculation an inclue proper units. Does voltage cange more quickl or more slowl as time goes on? Eplain in terms of tangent lines. V volts FIGURE 40 t s Te tangent line sketce in te figure below appears to pass troug te points 0, 3 an 30, 4. Tus, te rate of cange of voltage at t 0 secons is approimatel V/s As we move to te rigt of te grap, te tangent lines to it grow sallower, inicating tat te voltage canges more slowl as time goes on Use Figure to estimate T / at 30 an 70, were T is atmosperic temperature in egrees Celsius an is altitue in kilometers. Were is T / equal to zero? Temperature C Tropospere Stratospere Mesospere Termospere Altitue km FIGURE Atmosperic temperature versus altitue. At 30 km, te grap of atmosperic temperature appears to be linear passing troug te points 3, 50 an 40, 0. Te slope of tis segment of te grap is ten so ; T.94 C/km. 30 At 70 km, te grap of atmosperic temperature appears to be linear passing troug te points 58, 0 an 88, 00. Te slope of tis segment of te grap is ten ; 30 April 4, 0

58 SECTION 3.4 Rates of Cange 47 so T 3.33 C/km. 70 T 0 at tose points were te tangent line on te grap is orizontal. Tis appears to appen over te interval [3, 3], an near te points 50 an Te eart eerts a gravitational force of Fr /r newtons on an object wit a mass of 75 kg locate r meters from te center of te eart. Fin te rate of cange of force wit respect to istance r at te surface of te eart. Te rate of cange of force is F r /r 3. Terefore, F / N/m. 9. Calculate te rate of cange of escape velocit v esc r / m/s wit respect to istance r from te center of te eart. Te rate tat escape velocit canges is v esc r.4 07 r 3/. 0. Te power elivere b a batter to an apparatus of resistance R in oms is P.5R/R watts. Fin te rate of cange of power wit respect to resistance for R 3 an R 5. P R R RR + R Terefore, P W/ an P W/.. Te position of a particle moving in a straigt line uring a 5-s trip is st t t + 0 cm. Fin a time t at wic te instantaneous velocit is equal to te average velocit for te entire trip. Let st t t + 0, 0 t 5, wit s in centimeters cm an t in secons s. Te average velocit over te t-interval [0, 5] is s5 s cm/s. Te instantaneous velocit is vt s t t. Solving t 4 iels t 5 s, te time at wic te instantaneous velocit equals te calculate average velocit.. Te eigt in meters of a elicopter at time t in minutes is st 600t 3t 3 for 0 t. a Plot st an velocit vt. b Fin te velocit at t 8 an t 0. c Fin te maimum eigt of te elicopter. a Wit st 600t 3t 3, it follows tat vt 600 9t. Plots of te position an te velocit are sown below st vt b From part a, we ave vt 600 9t. Tus, v 8 4 meters/minute an v meters/minute. c From te grap in part a, we see tat te elicopter acieves its maimum eigt wen te velocit is zero. Solving 600 9t 0 for t iels 600 t minutes. 3 Te maimum eigt of te elicopter is ten 0 s meters. 3 3 April 4, 0

59 48 CHAPTER 3 DIFFERENTIATION 3. A particle moving along a line as position st t 4 8t m at time t secons. At wic times oes te particle pass troug te origin? At wic times is te particle instantaneousl motionless tat is, it as zero velocit? Te particle passes troug te origin wen st t 4 8t t t 8 0. Tis appens wen t 0 secons an wen t secons. Wit st t 4 8t, it follows tat vt s t 4t 3 36t 4tt 9. Te particle is terefore instantaneousl motionless wen t 0 secons an wen t 3 secons. 4. Plot te position of te particle in Eercise 3. Wat is te fartest istance to te left of te origin attaine b te particle? Te plot of te position of te particle in Eercise 3 is sown below. Positive values of position correspon to istance to te rigt of te origin an negative values correspon to istance to te left of te origin. Te most negative value of st occurs at t 3 an is equal to s Tus, te particle acieves a maimum istance to te left of te origin of 8 meters A bullet is fire in te air verticall from groun level wit an initial velocit 00 m/s. Fin te bullet s maimum velocit an maimum eigt. We emplo Galileo s formula, st s 0 + v 0 t gt 00t 4.9t, were te time t is in secons s an te eigt s is in meters m. Te velocit is vt t. Te maimum velocit of 00 m/s occurs at t 0. Tis is te initial velocit. Te bullet reaces its maimum eigt wen vt t 0; i.e., wen t 0.4 s. At tis point, te eigt is m. 6. Fin te velocit of an object roppe from a eigt of 300 m at te moment it its te groun. We emplo Galileo s formula, st s 0 + v 0 t gt t, were te time t is in secons s an te eigt s is in meters m. Wen te ball its te groun its eigt is 0. Solve st t 0 to obtain t s. We iscar te negative time, wic took place before te ball was roppe. Te velocit at impact is v m/s. Tis signifies tat te ball is falling at m/s. 7. A ball tosse in te air verticall from groun level returns to eart 4 s later. Fin te initial velocit an maimum eigt of te ball. Galileo s formula gives st s 0 + v 0 t gt v 0 t 4.9t, were te time t is in secons s an te eigt s is in meters m. Wen te ball its te groun after 4 secons its eigt is 0. Solve 0 s4 4v to obtain v m/s. Te ball reaces its maimum eigt wen s t 0, tat is, wen t 0, or t s. At tis time, t s, s m. 8. Olivia is gazing out a winow from te tent floor of a builing wen a bucket roppe b a winow waser passes b. Se notes tat it its te groun.5 s later. Determine te floor from wic te bucket was roppe if eac floor is 5 m ig an te winow is in te mile of te tent floor. Neglect air friction. Suppose H is te unknown eigt from wic te bucket fell starting at time t 0. Te eigt of te bucket at time t is st H 4.9t. Let T be te time wen te bucket its te groun tus ST 0. Olivia saw te bucket at time T.5. Te winow is locate 9.5 floors or 47.5 m above groun. So we ave te equations st.5 H 4.9T an st H 4.9T 0 Subtracting te secon equation from te first, we obtain 4.9 3T , so T 4 s. Te secon equation gives us H 4.9T m. Since tere are 5minafloor, te bucket was roppe 78.4/5 5.7 floors above te groun. Te bucket was roppe from te top of te 5t floor. 9. Sow tat for an object falling accoring to Galileo s formula, te average velocit over an time interval [t,t ] is equal to te average of te instantaneous velocities at t an t. Te simplest wa to procee is to compute bot values an sow tat te are equal. Te average velocit over [t,t ] is st st t t s 0 + v 0 t gt s 0 + v 0 t gt v 0t t + t t t t v 0t t t t g t + t v 0 g t + t g t t April 4, 0

60 SECTION 3.4 Rates of Cange 49 Wereas te average of te instantaneous velocities at te beginning an en of [t,t ] is s t + s t v 0 gt + v 0 gt v 0 g t + t v 0 g t + t. Te two quantities are te same. 30. An object falls uner te influence of gravit near te eart s surface. Wic of te following statements is true? Eplain. a Distance travele increases b equal amounts in equal time intervals. b Velocit increases b equal amounts in equal time intervals. c Te erivative of velocit increases wit time. For an object falling uner te influence of gravit, Galileo s formula gives st s 0 + v 0 t gt. a Since te eigt of te object varies quaraticall wit respect to time, it is not true tat te object covers equal istance in equal time intervals. b Te velocit is vt s t v 0 gt. Te velocit varies linearl wit respect to time. Accoringl, te velocit ecreases becomes more negative b equal amounts in equal time intervals. Moreover, its spee te magnitue of velocit increases b equal amounts in equal time intervals. c Acceleration, te erivative of velocit wit respect to time, is given b at v t g. Tis is a constant; it oes not cange wit time. Hence it is not true tat acceleration te erivative of velocit increases wit time. 3. B Faraa s Law, if a conucting wire of lengt l meters moves at velocit v m/s perpenicular to a magnetic fiel of strengt B in teslas, a voltage of size V Blv is inuce in te wire. Assume tat B an l 0.5. a Calculate V /v. b Fin te rate of cange of V wit respect to time t if v 4t + 9. a Assuming tat B an l 0.5, V.5v v. Terefore, V v. b If v 4t + 9, ten V.54t + 9 4t + 9. Terefore, V t Te voltage V, current I, an resistance R in a circuit are relate b Om s Law: V IR, were te units are volts, amperes, an oms. Assume tat voltage is constant wit V volts. Calculate specifing units: a Te average rate of cange of I wit respect to R for te interval from R 8toR 8. b Te rate of cange of I wit respect to R wen R 8 c Te rate of cange of R wit respect to I wen I.5 Let V IR or I V/R /R since we are assuming V volts. a Te average rate of cange is I I8. I8 R A/. b I/R /R / A/. c Wit R /I, we ave R/I /I / /A. 33. Etan fins tat wit ours of tutoring, e is able to answer correctl S percent of te problems on a mat eam. Wic woul ou epect to be larger: S 3 or S 30? Eplain. One possible grap of S is sown in te figure below on te left. Tis grap inicates tat in te earl ours of working wit te tutor, Etan makes rapi progress in learning te material but eventuall approaces eiter te limit of is abilit to learn te material or te maimum possible score on te eam. In tis scenario, S 3 woul be larger tan S 30. An alternative grap of S is sown below on te rigt. Here, in te earl ours of working wit te tutor little progress is mae peraps te tutor is assessing ow muc Etan alrea knows, is learning stle, is personalit, etc.. Tis is followe b a perio of rapi improvement an finall a leveling off as Etan reaces is maimum score. In tis scenario, S 3 an S 30 migt be rougl equal. Percentage correct Percentage correct Hours of tutoring Hours of tutoring April 4, 0

61 50 CHAPTER 3 DIFFERENTIATION 34. Suppose θt measures te angle between a clock s minute an our ans. Wat is θ t at 3 o clock? Te minute an makes one full revolution ever 60 minutes, so te minute an moves at a rate of π 60 π 30 ra/min. Te our an makes one-twelft of a revolution ever 60 minutes, so te our an moves wit a rate of π 360 ra/min. At 3 o clock, te movement of te minute an works to ecrease te angle between te minute an our ans wile te movement of te our an works to increase te angle. Terefore, at 3 o clock, θ t π 360 π 30 π 360 ra/min. 35. To etermine rug osages, octors estimate a person s bo surface area BSA in meters square using te formula BSA m/60, were is te eigt in centimeters an m te mass in kilograms. Calculate te rate of cange of BSA wit respect to mass for a person of constant eigt 80. Wat is tis rate at m 70 an m 80? Epress our result in te correct units. Does BSA increase more rapil wit respect to mass at lower or iger bo mass? Assuming constant eigt 80 cm, let f m m/ m be te formula for bo surface area in terms of weigt. Te rate of cange of BSA wit respect to mass is f 5 5 m 0 m / 0 m. If m 70 kg, tis is f m kg. If m 80 kg, f m 80 kg. Because te rate of cange of BSA epens on / m, it is clear tat BSA increases more rapil at lower bo mass. 36. Te atmosperic CO level At at Mauna Loa, Hawaii at time t in parts per million b volume is recore b te Scripps Institution of Oceanograp. Te values for te monts Januar December 007 were 38.45, , 384.3, 386.6, , , , 38.78, , 380.8, 38.33, a Assuming tat te measurements were mae on te first of eac mont, estimate A t on te 5t of te monts Januar November. b In wic monts i A t take on its largest an smallest values? c In wic mont was te CO level most nearl constant? a Te rate of cange in te atmosperic CO level on te 5t of eac mont can be estimate using te montl ifferences An An for n. Te estimates we obtain are:.3, 0.55,.03, 0.3, 0.5,.48,.6,.05, 0.08,.5,.36 t Jan Feb Mar Apr Ma Jun Jul Aug Sep Oct Nov P t b Accoring to te table in part a, te maimum rate of cange occurs in Marc an te minimum rate is in Jul. c Accoring to te table in part a, te CO level is most nearl constant in September. 37. Te tangent lines to te grap of f grow steeper as increases. At wat rate o te slopes of te tangent lines increase? Let f. Te slopes s of te tangent lines are given b s f. Te rate at wic tese slopes are increasing is s/. April 4, 0

62 SECTION 3.4 Rates of Cange Figure 3 sows te eigt of a mass oscillating at te en of a spring. troug one ccle of te oscillation. Sketc te grap of velocit as a function of time. Time FIGURE 3 Te position grap appears to break into four equal-size components. Over te first quarter of te time interval, te position grap is rising but bening ownwar, eventuall reacing a orizontal tangent. Tus, over te first quarter of te time interval, te velocit is positive but ecreasing, eventuall reacing 0. Continuing to eamine te structure of te position grap prouces te following grap of velocit: Velocit Time In Eercises 39 46, use Eq. 3 to estimate te unit cange. 39. Estimate an Compare our estimates wit te actual values. Let f /. Ten f /. We are using te erivative to estimate te average rate of cange. Tat is, so tat + f, + f. Tus, f. Te actual value, to si ecimal places, is Also, 0 00 f Te actual value, to si ecimal places, is Estimate f4 f3 if f. Ten estimate f4, assuming tat f3. Using te estimate tat f+ f so tat f+ f f wit 3,, we get f, f4 f If f3, ten f Let Fs.s s be te stopping istance as in Eample 3. Calculate F65 an estimate te increase in stopping istance if spee is increase from 65 to 66 mp. Compare our estimate wit te actual increase. Let Fs.s +.05s be as in Eample 3. F s. + 0.s. Ten F ft an F ft/mp. F 65 F66 F65 is approimatel equal to te cange in stopping istance per mp increase in spee wen traveling at 65 mp. Increasing spee from 65 to 66 terefore increases stopping istance b approimatel 7.6 ft. Te actual increase in stopping istance wen spee increases from 65 mp to 66 mp is F66 F feet, wic iffers b less tan one percent from te estimate foun using te erivative. 4. Accoring to Kleiber s Law, te metabolic rate P in kilocalories per a an bo mass m in kilograms of an animal are relate b a tree-quarter-power law P 73.3m 3/4. Estimate te increase in metabolic rate wen bo mass increases from 60 to 6 kg. April 4, 0

63 5 CHAPTER 3 DIFFERENTIATION Let P m 73.3m 3/4 be te function relating bo mass m to metabolic rate P. Ten, P m m / m /4 P6 P60 P / As bo mass is increase from 60 to 6 kg, metabolic rate is increase b approimatel kcal/a. 43. Te ollar cost of proucing bagels is C / Determine te cost of proucing 000 bagels an estimate te cost of te 00st bagel. Compare our estimate wit te actual cost of te 00st bagel. Epaning te power of 3 iels C Tis allows us to get te erivative C Te cost of proucing 000 bagels is C / ollars. Te cost of te 00st bagel is, b efinition, C00 C000. B te erivative estimate, C00 C000 C 000, so te cost of te 00st bagel is approimatel C $0.44. C , so te eact cost of te 00st bagel is inistinguisable from te estimate cost. Te function is ver nearl linear at tis point. 44. Suppose te ollar cost of proucing vieo cameras is C a Estimate te marginal cost at prouction level 5000 an compare it wit te actual cost C500 C5000. b Compare te marginal cost at 5000 wit te average cost per camera, efine as C/. Let C Ten C a Te cost ifference is approimatel C Te actual cost is C500 C , wic is quite close to te marginal cost compute using te erivative. b Te average cost per camera is C wic is sligtl iger tan te marginal cost , 45. Deman for a commoit generall ecreases as te price is raise. Suppose tat te eman for oil per capita per ear is Dp 900/p barrels, were p is te ollar price per barrel. Fin te eman wen p $40. Estimate te ecrease in eman if p rises to $4 an te increase if p eclines to $39. Dp 900p,soD p 900p. Wen te price is $40 a barrel, te per capita eman is D40.5 barrels per ear. Wit an increase in price from $40 to $4 a barrel, te cange in eman D4 D40 is approimatel D barrels a ear. Wit a ecrease in price from $40 to $39 a barrel, te cange in eman D39 D40 is approimatel D An increase in oil prices of a ollar leas to a ecrease in eman of barrels a ear, an a ecrease of a ollar leas to an increase in eman of barrels a ear. 46. Te reprouction rate f of te fruit fl Drosopila melanogaster, grown in bottles in a laborator, ecreases wit te number p of flies in te bottle. A researcer as foun te number of offspring per female per a to be approimatel fp pp a Calculate f5 an f 5. b Estimate te ecrease in ail offspring per female wen p is increase from 5 to 6. Is tis estimate larger or smaller tan te actual value f6 f5? c Plot fp for 5 p 5 an verif tat fp is a ecreasing function of p. Do ou epect f p to be positive or negative? Plot f p an confirm our epectation. Let Ten fp pp p p f p.37p p April 4, 0

64 SECTION 3.4 Rates of Cange 53 a f offspring per female per a an f offspring per female per a per fl. b f6 f5 f Te ecrease in ail offspring per female is estimate at f6 f Te actual ecrease in ail offspring per female is Te actual ecrease in ail offspring per female is less tan te estimate ecrease. Tis is because te grap of te function bens towars te ais. c Te function fpis plotte below at te left an is clearl a ecreasing function of p; we terefore epect tat f p will be negative. Te plot of te erivative sown below at te rigt confirms our epectation p p 47. Accoring to Stevens Law in pscolog, te perceive magnitue of a stimulus is proportional approimatel to a power of te actual intensit I of te stimulus. Eperiments sow tat te perceive brigtness B of a ligt satisfies B ki /3, were I is te ligt intensit, wereas te perceive eaviness H of a weigt W satisfies H kw 3/ k is a constant tat is ifferent in te two cases. Compute B/I an H /W an state weter te are increasing or ecreasing functions. Ten eplain te following statements: a A one-unit increase in ligt intensit is felt more strongl wen I is small tan wen I is large. b Aing anoter poun to a loa W is felt more strongl wen W is large tan wen W is small. a B/I k 3 I /3 k 3I /3. As I increases, B/I srinks, so tat te rate of cange of perceive intensit ecreases as te actual intensit increases. Increase ligt intensit as a iminise return in perceive intensit. A sketc of B against I is sown: See tat te eigt of te grap increases more slowl as ou move to te rigt. b H /W 3k W /.AsW increases, H /W increases as well, so tat te rate of cange of perceive weigt increases as weigt increases. A sketc of H against W is sown: See tat te grap becomes steeper as ou move to te rigt. 48. Let Mt be te mass in kilograms of a plant as a function of time in ears. Recent stuies b Niklas an Enquist ave suggeste tat a remarkabl wie range of plants from algae an grass to palm trees obe a tree-quarter-power growt law tat is, M/t CM 3/4 for some constant C. a If a tree as a growt rate of 6 kg/r wen M 00 kg, wat is its growt rate wen M 5 kg? b If M 0.5 kg, ow muc more mass must te plant acquire to ouble its growt rate? a Suppose a tree as a growt rate M/t of 6 kg/r wen M 00, ten 6 C00 3/4 0C 0, so tat C Wen M 5, M t C5 3/ / b Te growt rate wen M 0.5kg is M/t C0.5 3/4. To ouble te rate, we must fin M so tat M/t CM 3/4 C0.5 3/4. We solve for M. CM 3/4 C0.5 3/4 April 4, 0

65 54 CHAPTER 3 DIFFERENTIATION M 3/ /4 M 0.5 3/4 4/ Te plant must acquire te ifference kg in orer to ouble its growt rate. Note tat a oubling of growt rate requires more tan a oubling of mass. Furter Insigts an Callenges Eercises 49 5: Te Lorenz curve Fris use b economists to stu income istribution in a given countr see Figure 4. B efinition, Fris te fraction of te total income tat goes to te bottom rt part of te population, were 0 r. For eample, if F , ten te bottom 40% of ouseols receive 4.5% of te total income. Note tat F0 0 an F..0 Fr Fr L L P Q r A Lorenz curve for Sween in 004 B Two Lorenz curves: Te tangent lines at P an Q ave slope. FIGURE 4 r 49. Our goal is to fin an interpretation for F r. Te average income for a group of ouseols is te total income going to te group ivie b te number of ouseols in te group. Te national average income is A T/N, were N is te total number of ouseols an T is te total income earne b te entire population. a Sow tat te average income among ouseols in te bottom rt part is equal to F r/ra. b Sow more generall tat te average income of ouseols belonging to an interval [r, r + r] is equal to Fr + r Fr A r c Let 0 r. A ouseol belongs to te 00rt percentile if its income is greater tan or equal to te income of 00r % of all ouseols. Pass to te limit as r 0 in b to erive te following interpretation: A ouseol in te 00rt percentile as income F ra. In particular, a ouseol in te 00rt percentile receives more tan te national average if F r > an less if F r <. For te Lorenz curves L an L in Figure 4B, wat percentage of ouseols ave above-average income? a Te total income among ouseols in te bottom rt part is FrT an tere are rn ouseols in tis part of te population. Tus, te average income among ouseols in te bottom rt part is equal to FrT rn Fr r T N Fr A. r b Consier te interval [r, r + r]. Te total income among ouseols between te bottom rt part an te bottom r + r-t part is Fr + rt FrT. Moreover, te number of ouseols covere b tis interval is r + rn rn rn. Tus, te average income of ouseols belonging to an interval [r, r + r] is equal to Fr + rt FrT rn Fr + r Fr r T N c Take te result from part b an let r 0. Because Fr + r Fr lim F r, r 0 r Fr + r Fr A. r we fin tat a ouseol in te 00rt percentile as income F ra. Te point P in Figure 4B as an r-coorinate of 0.6, wile te point Q as an r-coorinate of rougl Tus, on curve L, 40% of ouseols ave F r > an terefore ave above-average income. On curve L, rougl 5% of ouseols ave above-average income. April 4, 0

66 SECTION 3.4 Rates of Cange Te following table provies values of Fr for Sween in 004. Assume tat te national average income was A 30,000 euros. r Fr a Wat was te average income in te lowest 40% of ouseols? b Sow tat te average income of te ouseols belonging to te interval [0.4, 0.6] was 6,700 euros. c Estimate F 0.5. Estimate te income of ouseols in te 50t percentile? Was it greater or less tan te national average? a Te average income in te lowest 40% of ouseols is F 0.4A , euros. b Te average income of te ouseols belonging to te interval [0.4, 0.6] is euros. c We estimate F0.6 F0.4 A , , F 0.5 F0.6 F Te income of ouseols in te 50t percentile is ten F 0.5A ,000 6,700 euros, wic is less tan te national average. 5. Use Eercise 49 c to prove: a F r is an increasing function of r. b Income is istribute equall all ouseols ave te same income if an onl if Fr r for 0 r. a Recall from Eercise 49 c tat F ra is te income of a ouseol in te 00r-t percentile. Suppose 0 r < r. Because r >r, a ouseol in te 00r -t percentile must ave income at least as large as a ouseol in te 00r -t percentile. Tus, F r A F r A, orf r F r. Tis implies F r is an increasing function of r. b If Fr r for 0 r, ten F r an ouseols in all percentiles ave income equal to te national average; tat is, income is istribute equall. Alternatel, if income is istribute equall all ouseols ave te same income, ten F r for 0 r. Tus, F must be a linear function in r wit slope. Moreover, te conition F0 0 requires te F intercept of te line to be 0. Hence, Fr r + 0 r. 5. Stuies of Internet usage sow tat website popularit is escribe quite well b Zipf s Law, accoring to wic te nt most popular website receives rougl te fraction /n of all visits. Suppose tat on a particular a, te nt most popular site a approimatel V n 0 6 /n visitors for n 5,000. a Verif tat te top 50 websites receive nearl 45% of te visits. Hint: Let TN enote te sum of V n for n N. Use a computer algebra sstem to compute T45 an T5,000. b Verif, b numerical eperimentation, tat wen Eq. 3 is use to estimate Vn+ V n, te error in te estimate ecreases as n grows larger. Fin again, b eperimentation an N suc tat te error is at most 0 for n N. c Using Eq. 3, sow tat for n 00, te nt website receive at most 00 more visitors tan te n + st website. a In Matematica, using te comman Sum[0 ˆ 6/n,{n,50}] iels an te comman Sum[0 ˆ 6/n,{n,5000}] iels We see tat te first 50 sites get aroun 4.4 million its, nearl alf te 0.9 million its of te first 5000 sites. b We use V[n_] :0ˆ6/n, an compute te error Vn+ V n V n for various values of n. Te table of values compute follows: n V n + V n V n Te error ecreases in ever entr. Furtermore, for n>50, te error appears to be less tan 0. c Since V n 0 6 n, V n 0 6 n. Te marginal erivative estimate Eq. 3 tells us tat V n Vn+ V n 0 6 n. If n 00, V n Terefore V n Vn+ <00 for n 00. April 4, 0

67 56 CHAPTER 3 DIFFERENTIATION In Eercises 53 an 54, te average cost per unit at prouction level is efine as C avg C/, were C is te cost function. Average cost is a measure of te efficienc of te prouction process. 53. Sow tat C avg is equal to te slope of te line troug te origin an te point, C on te grap of C. Using tis interpretation, etermine weter average cost or marginal cost is greater at points A, B, C, D in Figure 5. Cost D A B C Prouction level FIGURE 5 Grap of C. B efinition, te slope of te line troug te origin an, C, tat is, between 0, 0 an, C is C 0 0 C C av. At point A, average cost is greater tan marginal cost, as te line from te origin to A is steeper tan te curve at tis point we see tis because te line, tracing from te origin, crosses te curve from below. At point B, te average cost is still greater tan te marginal cost. At te point C, te average cost an te marginal cost are nearl te same, since te tangent line an te line from te origin are nearl te same. Te line from te origin to D crosses te cost curve from above, an so is less steep tan te tangent line to te curve at D; te average cost at tis point is less tan te marginal cost. 54. Te cost in ollars of proucing alarm clocks is C were is in units of 000. a Calculate te average cost at 4, 6, 8, an 0. b Use te grapical interpretation of average cost to fin te prouction level 0 at wic average cost is lowest. Wat is te relation between average cost an marginal cost at 0 see Figure 6? Cost ollars 0,000 5, Units tousans FIGURE 6 Cost function C Let C a Te slope of te line troug te origin an te point, C is te average cost. C 0 0 C C av, C C av b Te average cost is lowest at te point P 0 were te angle between te -ais an te line troug te origin an P 0 is lowest. Tis is at te point 8, 70, were te line troug te origin an te grap of C meet in te figure above. You can see tat te line is also tangent to te grap of C, so te average cost an te marginal cost are equal at tis point. April 4, 0

68 SECTION 3.5 Higer Derivatives Higer Derivatives Preliminar Questions. On September 4, 003, te Wall Street Journal printe te ealine Stocks Go Higer, Toug te Pace of Teir Gains Slows. Reprase tis ealine as a statement about te first an secon time erivatives of stock prices an sketc a possible grap. Because stocks are going iger, stock prices are increasing an te first erivative of stock prices must terefore be positive. On te oter an, because te pace of gains is slowing, te secon erivative of stock prices must be negative. Stock price Time. True or false? Te tir erivative of position wit respect to time is zero for an object falling to eart uner te influence of gravit. Eplain. Tis statement is true. Te acceleration of an object falling to eart uner te influence of gravit is constant; ence, te secon erivative of position wit respect to time is constant. Because te tir erivative is just te erivative of te secon erivative an te erivative of a constant is zero, it follows tat te tir erivative is zero. 3. Wic tpe of polnomial satisfies f 0 for all? Te tir erivative of all quaratic polnomials polnomials of te form a + b + c for some constants a, b an c is equal to 0 for all. 4. Wat is te milliont erivative of f e? Ever erivative of f e is e. Eercises In Eercises 6, calculate an.. 4 Let 4. Ten 8, 8, an Let 7. Ten, 0, an Let Ten , 50, an t 3 9t + 7 Let 4t 3 9t + 7. Ten t 8t, 4t 8, an πr3 Let 4 3 πr3. Ten 4πr, 8πr, an 8π. 6. Let /. Ten /, 4 3/, an 3 8 5/. 7. 0t 4/5 6t /3 Let 0t 4/5 6t /3. Ten 6t /5 4t /3, 6 5 t 6/ t 4/3, an 96 5 t /5 6 9 t 7/ /5 Let 9/5. Ten 9 5 4/5, 6 5 9/5, an /5. 9. z 4 z Let z 4z. Ten + 4z, 8z 3, an 4z 4. April 4, 0

69 58 CHAPTER 3 DIFFERENTIATION 0. 5t 3 + 7t 8/3 Let 5t 3 + 7t 8/3. Ten 5t t /3, 60t t 4/3, an 300t t 7/3.. θ θ + 7 Let θ θ + 7 θ 3 + 7θ. Ten 6θ + 4θ, θ + 4, an Since we on t want to appl te prouct rule to an ever growing list of proucts, we multipl troug first. Let Ten , , an Let 4 4. Ten 4, 8 3, an Let. Appling te quotient rule: 5. 5 e Let 5 e. Ten 5 e e e e e e e e e. 6. e Let e e. Ten e + e e e + e e + 3 e + e e. In Eercises 7 6, calculate te erivative inicate. 7. f 4, f 4 Let f 4. Ten f 4 3, f, f 4, an f 4 4. Tus f g, gt 4t 5 Let gt 4t 5. Ten g t 0t 6, g t 0t 7, an g t 840t 8. Hence g t, 4t 3 + 3t t Let 4t 3 + 3t. Ten t t 4 + 6t an t 48t Hence t t April 4, 0

70 SECTION 3.5 Higer Derivatives f t 4, ft 6t 9 t 5 t 4 f t 4 Let ft 6t 9 t 5. Ten f t 844t5 40t. Terefore, 54t 8 0t 4, f t 4 f t t 43t 7 40t 3, 3 f t 3 304t 6 0t, an. 4 t 4 Tus, t 3/4 t6 Let t t 3/4. Ten t 3 4 t 7/4, t 6 t /4, 3 t t 5/4, an 4 t t 9/4. 4 t /4 t f 4, ft t t Since ft t t, f t 4t, f t 4, an f t 0 for all t. In particular, f f 3, f 4e 3 Let f 4e 3. Ten f 4e 3, f 4e 6, f 4e 6, an f 3 4e f, ft t t + Let ft t t +. Ten an Tus, f 4. 5., w we w f t t + t t + t + t + t + f t t + t + 0 t + t + t + t + t + 4 t + 3. Let w we w w / e w. Ten w w / e w + e w w / w / + w / e w an w w / + w / e w + e w w / 4 w 3/ w / + w / 4 w 3/ e w. Tus, 7 4 e. 6. g 0, gs es s + Let gs es s +. Ten an Tus, g 0. g s s + es e s s + se s s + s + g s s + s + se s + e s se s s + s + s + s + e s s + 3. April 4, 0

71 60 CHAPTER 3 DIFFERENTIATION 7. Calculate k 0 for 0 k 5, were 4 + a 3 + b + c + wit a, b, c, te constants. Appling te power, constant multiple, an sum rules at eac stage, we get note 0 is b convention: k k a 3 + b + c a + b + c + 6a + b a from wic we get 0 0, 0 c, 0 b, 3 0 6a, 4 0 4, an Wic of te following satisf f k 0 for all k 6? a f b f 3 c f f 6 e f 9/5 f f Equations b an f go to zero after te sit erivative. We on t ave to take te erivatives to see tis. Look at a. f 8 3. Ever time we take iger erivatives of f, te negative eponent will keep ecreasing, an will never become zero. In te case of b, we see tat ever erivative ecreases te egree te igest eponent of te polnomial b one, so tat f 4 0. For c, f / /. Ever furter erivative of fis going to make te eponent more negative, so tat it will never go to zero. In te case of, like b, te igest eponent will ecrease wit ever erivative, but 6 erivatives will leave te eponent zero, f 6 will be 6!. Tis is eas to verif. e is like c. Since te eponent is not a wole number, successive erivatives will make te eponent pass over zero, an go to negative infinit. In te case of f, f 5 is constant, so tat f 6 0 for all. 9. Use te result in Eample 3 to fin 6 6. Te equation in Eample 3 inicates tat ! 6. 6 an 6! , so Calculate te first five erivatives of f. a Sow tat f n is a multiple of n+/. b Sow tat f n alternates in sign as n for n. c Fin a formula for f n for n. Hint: Verif tat te coefficient is ± 3 5 n 3 n. We use te Power Rule: f / 4 f / f 3/ 5 f / 3 f 3 3 5/ 6 f / April 4, 0

72 SECTION 3.5 Higer Derivatives 6 Te pattern we see ere is tat te nt erivative is a multiple of ± n+. Wic multiple? Te coefficient is te prouct of te o numbers up to n 3 ivie b n. Terefore we can write a general formula for te nt erivative as follows: f n n 3 5 n 3 n n+ In Eercises 3 36, fin a general formula for f n. 3. f f 3, f 6 4, f 4 5, f ,...From tis we can conclue tat te nt erivative can be written as f n n n +! n+. 3. f + Let f + +. Ten f +, f + 3,f 6 + 4, f ,... From tis we conclue tat te nt erivative can be written as 33. f / 34. f 3/ 35. f e f n n n! + n+. f 3/. We will avoi simplifing numerators an enominators to fin te pattern: f 3 5/ 3 5/ f 5 3 7/ /. f n n n n 3... n n+/. f 3 5/. We will avoi simplifing numerators an enominators to fin te pattern: f 5 3 7/ 5 3 7/ f / /. f n n n + n... 3 n n+3/. Let f e. Ten f e + e e e f e e e f e + e 3 e 3e From tis we conclue tat te nt erivative can be written as f n n ne. 36. f e Let f e. Ten f e + e + e f + e + e e f e + e e f e + e e From tis we conclue tat te nt erivative can be written as f n + n + nn e. April 4, 0

73 6 CHAPTER 3 DIFFERENTIATION 37. a Fin te acceleration at time t 5 min of a elicopter wose eigt is st 300t 4t 3 m. b Plot te acceleration t for 0 t 6. How oes tis grap sow tat te elicopter is slowing own uring tis time interval? a Let st 300t 4t 3, wit t in minutes an s in meters. Te velocit is vt s t 300 t an acceleration is at s t 4t. Tus a5 0 m/min. b Te acceleration of te elicopter for 0 t 6 is sown in te figure below. As te acceleration of te elicopter is negative, te velocit of te elicopter must be ecreasing. Because te velocit is positive for 0 t 6, te elicopter is slowing own Fin an equation of te tangent to te grap of f at 3, were f 4. Let f 4 an g f 4 3. Ten g. Te tangent line to g at 3 is given b g g Figure 5 sows f, f, an f. Determine wic is wic A B FIGURE 5 C a f b f c f. Te tangent line to c is orizontal at an 3, were b as roots. Te tangent line to b is orizontal at an 0, were a as roots. 40. Te secon erivative f is sown in Figure 6. Wic of A or B is te grap of f an wic is f? f'' A FIGURE 6 B f A an f B. 4. Figure 7 sows te grap of te position s of an object as a function of time t. Determine te intervals on wic te acceleration is positive. Position 0 0 FIGURE Time Rougl from time 0 to time 0 an from time 30 to time 40. Te acceleration is positive over te same intervals over wic te grap is bening upwar. April 4, 0

74 SECTION 3.5 Higer Derivatives Fin a polnomial ftat satisfies te equation f + f. Since f + f, an is a polnomial, it seems reasonable to assume tat fis a polnomial of some egree, call it n. Te egree of f is n, so te egree of f is n, an te egree of f + f is n. Hence, n, since te egree of is. Terefore, let f a + b + c. Ten f a + b an f a. Substituting into te equation f + f iels a + a + b + c, an ientit in. Equating coefficients, we ave a, a + b 0, c 0. Terefore, b an f. 43. Fin a value of n suc tat n e satisfies te equation 3. Let n e. Ten n e + n n e n e + n, an n e + n + n. Tus, n e satisfies te equation 3 for n Wic of te following escriptions coul not appl to Figure 8? Eplain. a Grap of acceleration wen velocit is constant b Grap of velocit wen acceleration is constant c Grap of position wen acceleration is zero Distance FIGURE 8 Time a Does NOT appl to te figure because if vt C were C is a constant, ten at v t 0, wic is te orizontal line going troug te origin. b Can appl because te grap as a constant slope. c Can appl because if we took tis as a position grap, te velocit grap woul be a orizontal line an tus, acceleration woul be zero. 45. Accoring to one moel tat takes into account air resistance, te acceleration at in m/s of a skiver of mass m in free fall satisfies at k m vt were vt is velocit negative since te object is falling an k is a constant. Suppose tat m 75 kg an k 4 kg/m. a Wat is te object s velocit wen at 4.9? b Wat is te object s velocit wen at 0? Tis velocit is te object s terminal velocit. Solving at m k vt for te velocit an taking into account tat te velocit is negative since te object is falling, we fin m 75 vt at at k 4 a Substituting at 4.9 into te above formula for te velocit, we fin b Wen at 0, 75 vt m/s. 75 vt m/s. April 4, 0

75 64 CHAPTER 3 DIFFERENTIATION 46. Accoring to one moel tat attempts to account for air resistance, te istance st in meters travele b a falling rainrop satisfies s t g D s t were D is te rainrop iameter an g 9.8 m/s. Terminal velocit v term is efine as te velocit at wic te rop as zero acceleration one can sow tat velocit approaces v term as time procees. a Sow tat v term 000gD. b Fin v term for rops of iameter 0 3 m an 0 4 m. c In tis moel, o rainrops accelerate more rapil at iger or lower velocities? a v term is foun b setting s t b If D ft, If D ft, 0, an solving for s t v. 0 g s D t g s D t s t D g gD v /. v term 000g m/s. v term 000g m/s. c Te greater te velocit, te more gets subtracte from g in te formula for acceleration. Terefore, assuming velocit is less tan v term, greater velocities correspon to lower acceleration. 47. Aservomotor controls te vertical movement of a rill bit tat will rill a pattern of oles in seet metal. Te maimum vertical spee of te rill bit is 4 in./s, an wile rilling te ole, it must move no more tan.6 in./s to avoi warping te metal. During a ccle, te bit begins an ens at rest, quickl approaces te seet metal, an quickl returns to its initial position after te ole is rille. Sketc possible graps of te rill bit s vertical velocit an acceleration. Label te point were te bit enters te seet metal. Tere will be multiple ccles, eac of wic will be more or less ientical. Let vt be te ownwar vertical velocit of te rill bit, an let at be te vertical acceleration. From te narrative, we see tat vt can be no greater tan 4 an no greater tan.6 wile rilling is taking place. During eac ccle, vt 0 initiall, vt goes to 4 quickl. Wen te bit its te seet metal, vt goes own to.6 quickl, at wic it stas until te seet metal is rille troug. As te rill pulls out, it reaces maimum non-rilling upwar spee vt 4 quickl, an maintains tis spee until it returns to rest. A possible plot follows: 4 Metal A grap of te acceleration is etracte from tis grap: Metal April 4, 0

76 SECTION 3.5 Higer Derivatives 65 In Eercises 48 an 49, refer to te following. In a 997 stu, Boarman an Lave relate te traffic spee S on a two-lane roa to traffic ensit Q number of cars per mile of roa b te formula for 60 Q 400 Figure 9. S 88Q 0.05Q Spee S mp Densit Q FIGURE 9 Spee as a function of traffic ensit. 48. Calculate S/Q an S/Q. S/Q 88Q 0.05 S/Q 5764Q a Eplain intuitivel w we soul epect tat S/Q < 0. b Sow tat S/Q > 0. Ten use te fact tat S/Q < 0 an S/Q > 0 to justif te following statement: A one-unit increase in traffic ensit slows own traffic more wen Q is small tan wen Q is large. c Plot S/Q. Wic propert of tis grap sows tat S/Q > 0? a Traffic spee must be reuce wen te roa gets more crowe so we epect S/Q to be negative. Tis is inee te case since S/Q /Q < 0. b Te ecrease in spee ue to a one-unit increase in ensit is approimatel S/Q a negative number. Since S/Q 5764Q 3 > 0 is positive, tis tells us tat S/Q gets larger as Q increases an a negative number wic gets larger is getting closer to zero. So te ecrease in spee is smaller wen Q is larger, tat is, a one-unit increase in traffic ensit as a smaller effect wen Q is large. c S/Q is plotte below. Te fact tat tis grap is increasing sows tat S/Q > Use a computer algebra sstem to compute f k for k,, 3 for te following functions. a f + 3 5/3 a Let f + 3 5/3. Using a computer algebra sstem, b f f /3 ; f / /3 ; an f / / /3. 4 b Let f. Using a computer algebra sstem, 5 6 f ; April 4, 0

77 66 CHAPTER 3 DIFFERENTIATION f ; an f Let f +. Use a computer algebra sstem to compute te f k for k 4. Can ou fin a general formula for f k? Let f +. Using a computer algebra sstem, f ; f ; f ! 3 ; an 3+ f 4 From te pattern observe above, we conjecture ! 4+. f k k 3 k! k+. Furter Insigts an Callenges 5. Fin te 00t erivative of p Tis is a polnomial of egree , so its 00t erivative is zero. 53. Wat is p 99 for p as in Eercise 5? First note tat for an integer n 98, n 0. Now, if we epan p, wefin p 99 + terms of egree at most 98; terefore, p terms of egree at most Using logic similar to tat use to compute te erivative in Eample 3, we compute: , so tat p 99!. 54. Use te Prouct Rule twice to fin a formula for fg in terms of f an g an teir first an secon erivatives. Let fg. Ten fg + gf f g + fg an f g + gf + fg + g f f g + f g + fg. April 4, 0

78 SECTION 3.6 Trigonometric Functions Use te Prouct Rule to fin a formula for fg an compare our result wit te epansion of a + b 3. Ten tr to guess te general formula for fg n. Continuing from Eercise 54, we ave Te binomial teorem gives f g + gf + f g + g f + fg + g f f g + 3f g + 3f g + fg a + b 3 a 3 + 3a b + 3ab + b 3 a 3 b 0 + 3a b + 3a b + a 0 b 3 an more generall a + b n n k0 n a n k b k, k were te binomial coefficients are given b n k Accoringl, te general formula for fg n is given b fg n kk k n +. n! n k0 n f n k g k, k were p k is te kt erivative of p or p itself wen k Compute f+ + f f f 0 for te following functions: a f b f c f 3 Base on tese eamples, wat o ou tink te limit f represents? For f, we ave Hence, 0. For f, f+ + f f f+ + f f , so. Working in a similar fasion, we fin 3 6. One can prove tat for twice ifferentiable functions, f f. It is an interesting fact of more avance matematics tat tere are functions f for wic f eists at all points, but te function is not ifferentiable. 3.6 Trigonometric Functions Preliminar Questions. Determine te sign + or tat iels te correct formula for te following: a sin + cos ±sin ± cos b sec ±sec tan c cot ±csc Te correct formulas are a sin + cos sin + cos April 4, 0

79 68 CHAPTER 3 DIFFERENTIATION b sec sec tan c cot csc. Wic of te following functions can be ifferentiate using te rules we ave covere so far? a 3 cos cot b cos c e sin a 3 cos cot is a prouct of functions wose erivatives are known. Tis function can terefore be ifferentiate using te Prouct Rule. b cos is a composition of te functions cos an. We ave not et iscusse ow to ifferentiate composite functions. c cos is a prouct of functions wose erivatives are known. Tis function can terefore be ifferentiate using te Prouct Rule. 3. Compute + cos witout using te erivative formulas for sin an cos. Recall tat sin + cos for all. Tus, sin + cos How is te aition formula use in eriving te formula sin cos? Te ifference quotient for te function sin involves te epression sin +. Te aition formula for te sine function is use to epan tis epression as sin + sin cos + sin cos. Eercises In Eercises 4, fin an equation of te tangent line at te point inicate.. sin, π 4 Let f sin. Ten f cos an te equation of te tangent line is f π π π + f π π. 4. cos, π 3 Let f cos. Ten f sin an te equation of te tangent line is f π π π 3 + f π π tan, π 4 Let f tan. Ten f sec an te equation of te tangent line is f π π π + f π + + π sec, π 6 Let f sec. Ten f sec tan an te equation of te tangent line is f π π π + f π π 3 9. In Eercises 5 4, compute te erivative. 5. f sin cos Let f sin cos. Ten f sin sin + cos cos sin + cos. April 4, 0

80 SECTION 3.6 Trigonometric Functions f cos Let f cos. Ten 7. f sin Let f sin sin sin. Ten f sin + cos cos sin. f sin cos + sin cos sin cos. 8. f 9 sec + cot Let f 9 sec + cot. Ten f 9 sec tan csc. 9. Ht sin t sec t Let Ht sin t sec t. Ten 0. t 9 csc t + t cot t Let t 9 csc t + t cot t. Ten. fθ tan θ sec θ Let fθ tan θ sec θ. Ten H t sin t t sec t sec t + sec tcos t sin tsec t sec t tan t + sec t sec t tan t + sec t sin t sec t tan t + sec t. t 9 csc t cot t + t csc t + cot t cot t 9 csc t cot t t csc t. f θ tan θ sec θ tan θ + sec θ sec θ sec θ tan θ + sec 3 θ tan θ + sec θ sec θ.. kθ θ sin θ Let kθ θ sin θ. Ten Here we use te result from Eercise f 4 4 sec Let f 4 4 sec. Ten k θ θ sin θ cos θ + θ sin θ θ sin θ cos θ + θ sin θ. f 4 4 sec tan + sec fz z tan z Let fz z tan z. Ten f z zsec z + tan z. 5. sec θ θ Let sec θ. Ten θ 6. Gz tan z cot z Let Gz tan z cot z. Ten θ sec θ tan θ sec θ θ. G z tan z cot z0 sec z + csc z tan z cot z sec z + csc z tan z cot z. April 4, 0

81 70 CHAPTER 3 DIFFERENTIATION 7. R 3 cos 4 sin Let R 3 cos 4. Ten sin 8. f R sin 3 sin 3 cos 4cos sin sin + Let f 9. f + tan tan 0. fθ θ tan θ sec θ + sin. Ten f Let f + tan tan. Ten f 4 cos 3sin + cos sin + sin cos + sin cos + sin + sin. tan sec + tan sec tan sec tan. 4 cos 3 sin. Let fθ θ tan θ sec θ. Ten f θ θ tan θ sec θ+ tan θ sec θ θ θtan θ sec θ tan θ + sec θ sec θ+ tan θ sec θ θ tan θ sec θ + θ sec 3 θ + tan θ sec θ.. f e sin Let f e sin. Ten f e cos + sin e e cos + sin.. t e t csc t Let t e t csc t. Ten t e t csc t cot t + csc te t e t csc t cot t. 3. fθ e θ 5 sin θ 4 tan θ Let fθ e θ 5 sin θ 4 tan θ. Ten f θ e θ 5 cos θ 4 sec θ+ e θ 5 sin θ 4 tan θ e θ 5 sin θ + 5 cos θ 4 tan θ 4 sec θ. 4. f e cos Let f e cos. Ten f e cos + e cos e sin + cos e + e cos e cos sin + cos. In Eercises 5 34, fin an equation of te tangent line at te point specifie cos, 0 Let f 3 + cos. Ten f 3 sin an f 0 0. Te tangent line at 0is f f April 4, 0

82 SECTION 3.6 Trigonometric Functions 7 6. tan θ, θ π 6 Let fθ tan θ. Ten f θ sec θ an f π Te tangent line at π 6 is 7. sin + 3 cos, 0 f π θ π π + f 4 θ π θ + 3 π 9. Let f sin + 3 cos. Ten f cos 3 sin an f 0. Te tangent line at 0is f f sin t + cos t, t π 3 Let ft sin t +cos t. Ten f t + cos tcos t sin t sin t + cos t + cos t + cos t + cos t, an f π 3 + / 3. Te tangent line at π 3 is f π π π + f π π sin θ + cos θ, θ π 3 π 3 is Let fθ sin θ + cos θ. Ten f θ cos θ sin θ an f π 3 3. Te tangent line at f π π π + f 3 π csc cot, π 4 Let f csc cot. Ten an Hence te tangent line is f π 4 f csc csc cot f π. 4 π π + f π π e cos, 0 Let f e cos. Ten f e sin + e cos e cos sin, an f 0 e 0 cos 0 sin 0. Tus, te equation of te tangent line is f f0 +. April 4, 0

83 7 CHAPTER 3 DIFFERENTIATION 3. e cos, π 4 Let f e cos. Ten f e cos cos + e cos e cos sin + cos sin + e cos e cos sin cos, an f π e π/4 4 eπ/4. Te tangent line at π 4 is f π π π + f eπ/4 π + 4 eπ/ e t cos t, t π Let ft e t cos t. Ten f t e t sin t + e t cos t e t + sin t cos t, an f π eπ/. Te tangent line at π is f π t π π + f e π/ t π + e π/. 34. e θ sec θ, θ π 4 Let fθ e θ sec θ. Ten f θ e θ sec θ tan θ + e θ sec θ e θ sec θtan θ +, an f π e π/4 sec π tan π e π/4. Tus, te equation of te tangent line is f π 4 π 4 In Eercises 35 37, use Teorem to verif te formula. 35. cot csc π + f e π/4 π + e π/ cot cos. Using te quotient rule an te erivative formulas, we compute: sin cot cos sin sin cos cos sin sin sin + cos sin sin csc. 36. sec sec tan Since sec, we can appl te quotient rule an te known erivatives to get: cos sec cos 0 sin cos cos sin cos sin cos tan sec. cos 37. csc csc cot Since csc, we can appl te quotient rule an te two known erivatives to get: sin csc sin 0 cos sin sin cos sin cos sin cot csc. sin April 4, 0

84 SECTION 3.6 Trigonometric Functions Sow tat bot sin an cos satisf. Let sin. Ten cos an sin. Similarl, if we let cos, ten sin an cos. In Eercises 39 4, calculate te iger erivative. 39. f θ, fθ θ sin θ Let fθ θ sin θ. Ten 40. t cos t f θ θ cos θ + sin θ f θ θ sin θ+ cos θ + cos θ θ sin θ + cos θ. t cos t cos t cos t cos t sin t + cos t sin t sin t cos t t t cos t t sin t cos t sin t sin t + cos tcos t cos t sin t. 4.,, tan Let tan. Ten sec an b te Cain Rule, 4.,, e t sin t Let e t sin t. Ten sec sec sec tan sec tan sec sec + sec tan tan sec 4 +4 sec 4 tan e t cos t + e t sin t e t cos t + sin t e t sin t + cos t + e t cos t + sin t e t cos t e t sin t + e t cos t e t cos t sin t. 43. Calculate te first five erivatives of f cos. Ten etermine f 8 an f 37. Let f cos. Ten f sin, f cos, f sin, f 4 cos, an f 5 sin. Accoringl, te successive erivatives of f ccle among { sin, cos,sin,cos } in tat orer. Since 8 is a multiple of 4, we ave f 8 cos. Since 36 is a multiple of 4, we ave f 36 cos. Terefore, f 37 sin. 44. Fin 57, were sin. Let f sin. Ten te successive erivatives of f ccle among {cos, sin, cos,sin } in tat orer. Since 56 is a multiple of 4, we ave f 56 sin. Terefore, f 57 cos. 45. Fin te values of between 0 an π were te tangent line to te grap of sin cos is orizontal. Let sin cos. Ten sin sin + cos cos cos sin. Wen 0, we ave sin ±cos. In te interval [0, π], tis occurs wen π 4, 3π 4, 5π 4, 7π 4. April 4, 0

85 74 CHAPTER 3 DIFFERENTIATION 46. Plot te grap fθ sec θ + csc θ over [0, π] an etermine te number of s to f θ 0 in tis interval grapicall. Ten compute f θ an fin te s. Te grap of fθ sec θ + csc θ over [0, π] is given below. From te grap, it appears tat tere are two locations were te tangent line woul be orizontal; tat is, tere appear to be two s to f θ 0. Now f θ sec θ tan θ csc θ cot θ. Setting sec θ tan θ csc θ cot θ 0 an ten multipling b sin θ tan θ an rearranging terms iels tan 3 θ. Tus, on te interval [0, π], tere are two of f θ 0: θ π 4 an θ 5π Let gt t sin t. a Plot te grap of g wit a graping utilit for 0 t 4π. b Sow tat te slope of te tangent line is nonnegative. Verif tis on our grap. c For wic values of t in te given range is te tangent line orizontal? Let gt t sin t. a Here is a grap of g over te interval [0, 4π] b Since g t cos t 0 for all t, te slope of te tangent line to g is alwas nonnegative. c In te interval [0, 4π], te tangent line is orizontal wen t 0, π, 4π. 48. Let f sin / for 0 an f0. a Plot fon [ 3π, 3π]. b Sow tat f c 0ifc tan c. Use te numerical root finer on a computer algebra sstem to fin a goo approimation to te smallest positive value c 0 suc tat f c 0 0. c Verif tat te orizontal line fc 0 is tangent to te grap of fat c 0 b plotting tem on te same set of aes. a Here is te grap of fover [ 3π, 3π] b Let f sin. Ten f cos sin. To ave f c 0, it follows tat c cos c sin c 0, or tan c c. Using a computer algebra sstem, we fin tat te smallest positive value c 0 suc tat f c 0 0isc April 4, 0

86 SECTION 3.6 Trigonometric Functions 75 c Te orizontal line fc an te function fare bot plotte below. Te orizontal line is clearl tangent to te grap of f Sow tat no tangent line to te grap of f tan as zero slope. Wat is te least slope of a tangent line? Justif b sketcing te grap of tan. Let f tan. Ten f sec cos. Note tat f cos as numerator ; te equation f 0 terefore as no. Because te maimum value of cos is, te minimum value of f cos is. Hence, te least slope for a tangent line to tan is. Here is a grap of f Te eigt at time t in secons of a mass, oscillating at te en of a spring, is st sin t cm. Fin te velocit an acceleration at t π 3 s. Let st sin t be te eigt. Ten te velocit is an te acceleration is vt s t 40 cos t at v t 40 sin t. At t π 3, te velocit is v π 3 0 cm/sec an te acceleration is a π3 0 3 cm/sec. 5. Te orizontal range R of a projectile launce from groun level at an angle θ an initial velocit v 0 m/s is R v0 /9.8 sin θ cos θ. Calculate R/θ. Ifθ 7π/4, will te range increase or ecrease if te angle is increase sligtl? Base our answer on te sign of te erivative. Let Rθ v0 /9.8 sin θ cos θ. R θ R θ v0 /9.8 sin θ + cos θ. If θ 7π/4, π 4 <θ< π,so sin θ > cos θ, an R/θ < 0 numericall, R/θ v 0. At tis point, increasing te angle will ecrease te range. 5. Sow tat if π <θ<π, ten te istance along te -ais between θ an te point were te tangent line intersects te -ais is equal to tan θ Figure 4. sin tan FIGURE 4 Let f sin. Since f cos, tis means tat te tangent line at θ, sin θ is cos θ θ + sin θ. Wen 0, θ tan θ. Te istance from to θ is ten θ θ tan θ tan θ. April 4, 0

87 76 CHAPTER 3 DIFFERENTIATION Furter Insigts an Callenges 53. Use te limit efinition of te erivative an te aition law for te cosine function to prove tat cos sin. Let f cos. Ten f cos + cos 0 0 sin sin cos + cos cos cos sin sin cos 0 sin + cos 0 sin. 54. Use te aition formula for te tangent tan + tan tan + + tan tan to compute tan tan irectl as a limit of te ifference quotients. You will also nee to sow tat lim. 0 First note tat tan sin lim 0 0 lim. 0 cos Now, using te aition formula for tangent, Terefore, tan + tan tan +tan +tan tan tan tan tan + tan tan tan sec + tan tan. tan tan 0 sec + tan tan tan 0 lim sec 0 + tan tan sec sec. 55. Verif te following ientit an use it to give anoter proof of te formula sin cos. sin + sin cos + sin Hint: Use te aition formula to prove tat sina + b sina b cos a sin b. Recall tat an Subtracting te secon ientit from te first iels sina + b sin a cos b + cos a sin b sina b sin a cos b cos a sin b. sina + b sina b cos a sin b. If we now set a + an b, ten te previous equation becomes sin + sin cos + Finall, we use te limit efinition of te erivative of sin to obtain In oter wors, sin + sin sin 0 0 sin cos. cos + lim 0 0 sin. cos + sin sin cos cos. April 4, 0

88 SECTION 3.6 Trigonometric Functions Sow tat a nonzero polnomial function fcannot satisf te equation. Use tis to prove tat neiter sin nor cos is a polnomial. Can ou tink of anoter wa to reac tis conclusion b consiering limits as? Let p be a nonzero polnomial of egree n an assume tat p satisfies te ifferential equation + 0. Ten p + p 0 for all. Tere are eactl tree cases. a If n 0, ten p is a constant polnomial an tus p 0. Hence 0 p + p p or p 0 i.e., p is equal to 0 for all or p is ienticall 0. Tis is a contraiction, since p is a nonzero polnomial. b If n, ten p is a linear polnomial an tus p 0. Once again, we ave 0 p + p p or p 0, a contraiction since p is a nonzero polnomial. c If n, ten p is at least a quaratic polnomial an tus p is a polnomial of egree n 0. Tus q p + p is a polnomial of egree n. B assumption, owever, p + p 0. Tus q 0, a polnomial of egree 0. Tis is a contraiction, since te egree of q is n. CONCLUSION: In all cases, we ave reace a contraiction. Terefore te assumption tat p satisfies te ifferential equation + 0isfalse. Accoringl, a nonzero polnomial cannot satisf te state ifferential equation. Let sin. Ten cos an sin. Terefore,. Now, let cos. Ten sin an cos. Terefore,. Because sin an cos are nonzero functions tat satisf, it follows tat neiter sin nor cos is a polnomial. 57. Let f sin an g cos. a Sow tat f g + sin an g f+ cos. b Verif tat f f+ cos an g g sin. c B furter eperimentation, tr to fin formulas for all iger erivatives of f an g. Hint: Te kt erivative epens on weter k 4n, 4n +, 4n +, or 4n + 3. Let f sin an g cos. a We eamine first erivatives: f cos + sin g + sin an g sin + cos f+ cos ; i.e., f g + sin an g f+ cos. b Now look at secon erivatives: f g + cos f + cos an g f sin g sin ; i.e., f f+ cos an g g sin. c Te tir erivatives are f f sin g 3 sin an g g cos f 3 cos ; i.e., f g 3 sin an g f 3 cos. Te fourt erivatives are f 4 g 3 cos f 4 cos an g 4 f + 3 sin g + 4 sin ; i.e., f 4 f 4 cos an g 4 g + 4 sin. We can now see te pattern for te erivatives, wic are summarize in te following table. Here n 0,,,... k 4n 4n + 4n + 4n + 3 f k f k cos g + k sin f+ k cos g k sin g k g + k sin f+ k cos g k sin f k cos 58. Figure 5 sows te geometr bein te erivative formula sin θ cos θ. Segments BA an BD are parallel to te - an -aes. Let sin θ sinθ + sin θ. Verif te following statements. a sin θ BC b BDA θ Hint: OA AD. c BD cos θad Now eplain te following intuitive argument: If is small, ten BC BD an AD, so sin θ cos θ an sin θ cos θ. D C B A O θ FIGURE 5 April 4, 0

89 78 CHAPTER 3 DIFFERENTIATION a We note tat sinθ + is te -coorinate of te point C an sin θ is te -coorinate of te point A, an terefore also of te point B. Now, sin θ sinθ + sin θ can be interprete as te ifference between te -coorinates of te points B an C; tat is, sin θ BC. b From te figure, we note tat OAB θ, so BAD π θ an BDA θ. c Using part b, it follows tat cos θ BD AD or BD cos θad. For small, BC BD an AD is rougl te lengt of te arc subtene from A to C; tat is, AD. Tus, using a an c, sin θ BC BD cos θad cos θ. In te limit as 0, so sin θ cos θ. sin θ sin θ, 3.7 Te Cain Rule Preliminar Questions. Ientif te outsie an insie functions for eac of tese composite functions. a b tan + c sec 5 + e 4 a Te outer function is, an te inner function is b Te outer function is tan, an te inner function is +. c Te outer function is 5, an te inner function is sec. Te outer function is 4, an te inner function is + e.. Wic of te following can be ifferentiate easil witout using te Cain Rule? a tan7 + b + c sec cos e e f e sin Te function + sec an e can be ifferentiate using te Prouct Rule. Te functions tan7 +, cos an e sin require te Cain Rule. 3. Wic is te erivative of f5? a 5f b 5f 5 c f 5 Te correct answer is b: 5f Suppose tat f 4 g4 g 4. Do we ave enoug information to compute F 4, were F fg? If not, wat is missing? If F fg, ten F f gg an F 4 f g4g 4. Tus, we o not ave enoug information to compute F 4. We are missing te value of f. Eercises In Eercises 4, fill in a table of te following tpe: fg f u f g g f g. f u u 3/, g 4 + fg f u f g g f g 4 + 3/ 3 u/ / / April 4, 0

90 SECTION 3.7 Te Cain Rule 79. f u u 3, g fg f u f g g f g u f u tan u, g 4 fg f u f g g f g tan 4 sec u sec sec 4 4. f u u 4 + u, g cos fg f u f g g f g cos 4 + cos 4u 3 + 4cos 3 + sin 4 sin cos 3 sin In Eercises 5 an 6, write te function as a composite fgan compute te erivative using te Cain Rule sin 4 Let f 4, g + sin, an fg + sin 4. Ten f gg 4 + sin 3 + cos. 6. cos 3 Let f cos, g 3, an fg cos 3. Ten f gg 3 sin Calculate cos u for te following coices of u: a u 9 b u c u tan a cosu cos9. cosu sinuu sin9 sin9. b cosu cos. cosu sinuu sin sin. c cosu costan. cosu sinuu sintan sec sec sintan. 8. Calculate f + for te following coices of f u: a f u sin u b f u 3u 3/ c f u u u a Let sinu sin +. Ten sin + cos + + cos + cos +. April 4, 0

91 80 CHAPTER 3 DIFFERENTIATION b Let 3u 3/ 3 + 3/. Ten 3 + 3/ / / 9 + /. c Let u u + +. Ten + + [ + ] + [ + ] Compute f f u if an u 6. Assuming f is a function of u, wic is in turn a function of, f f u u Compute f if f u u, u 5, an u 5. Because f u u, it follows tat f u u. Terefore, f f uu uu In Eercises, use te General Power Rule or te Sifting an Scaling Rule to compute te erivative Using te General Power Rule, Using te General Power Rule, Using te Sifting an Scaling Rule / / Using te General Power Rule, Using te General Power Rule, / Using te General Power Rule, / / / /3. April 4, 0

92 SECTION 3.7 Te Cain Rule 8 7. cos 4 θ Using te General Power Rule, θ cos4 θ 4 cos 3 θ θ cos θ 4 cos3 θ sin θ. 8. cos9θ + 4 Using te Sifting an Scaling Rule cos9θ + 4 9sin9θ + 4. θ 9. cos θ + 5 sin θ 9 Using te General Power Rule, θ cos θ + 5 sin θ9 9 cos θ + 5 sin θ 8 θ cos θ + 5 sin θ 9 cos θ + 5 sin θ8 5 cos θ sin θ sin Using te General Power Rule, sin sin / sin + cos sin.. e Using te Sifting an Scaling Rule, e e e.. e 8+9 Using te Sifting an Scaling Rule, e8+9 8e 8+9. In Eercises 3 6, compute te erivative of f g. 3. f u sin u, g + Let fg sin +. Ten, appling te sifting an scaling rule, cos +. Alternatel, 4. f u u +, g sin fg f gg cos + cos +. Let fg sin +. Ten cos. Alternatel, fg f gg cos. 5. f u e u, g + Let fg e +. Ten fg f gg e +. April 4, 0

93 8 CHAPTER 3 DIFFERENTIATION 6. f u u u, g csc Let f g. Ten, appling te quotient rule: Alternatel, csc csc cot csc csc cot csc cot csc csc. fg f gg csc cot csc cot csc csc, were we ave use te quotient rule to calculate f u u. In Eercises 7 an 8, fin te erivatives of fgan gf. 7. f u cos u, u g + 8. f u u 3, u g + fg f gg sin + sin +. gf g f f cos sin sin cos. Te erivative of + is taken using te sifting an scaling rule. fg f gg gf g f f In Eercises 9 4, use te Cain Rule to fin te erivative. 9. sin Let sin. Ten cos cos. 30. sin Let sin sin. Ten sin cos. 3. t + 9 Let t + 9 t + 9 /. Ten t + 9 / t t t t + 3t + 5/ Let t + 3t + 5/. Ten 5 t 7/ 5 t t + t + 3 t + 3t + 7/ /3 Let 4 3 /3. Ten / April 4, 0

94 SECTION 3.7 Te Cain Rule / 3/. Let + / Here, we note tat calculating te erivative of te insie function, +, requires te cain rule in te form of te scaling an sifting rule. After two applications of te cain rule, we ave Let Ten 4 + / / + / cos 3 θ After two applications of te cain rule, 3 cos θ sinθ 36 cos θsinθ. 37. sec Let f sec. Ten f sec tan sec / tan /. 38. tanθ 4θ Let tanθ 4θ. Ten sec θ 4θ θ 4 θ 4 sec θ 4θ. 39. tanθ + cos θ Let tan θ + cos θ. Ten sec θ + cos θ sin θ sin θ sec θ + cos θ. 40. e Let e. Ten e 4 4e. 4. e 9t Let e 9t. Ten e 9t 8t 8te 9t. 4. cos 3 e 4θ Let cos 3 e 4θ. After two applications of te cain rule, we ave 3 cos e 4θ sine 4θ 4e 4θ e 4θ cos e 4θ sine 4θ. April 4, 0

95 84 CHAPTER 3 DIFFERENTIATION In Eercises 43 7, fin te erivative using te appropriate rule or combination of rules. 43. tan + 4 Let tan + 4. B te cain rule, sec sec sin + 4 Let sin + 4. B te cain rule, + 4 cos cos 3 Let cos 3. Appling te prouct rule an ten te scaling an sifting rule, sin cos 3 3 sin 3 + cos sin cos We start b using a trig ientit to rewrite sin cos sin. Ten, b te cain rule, cos 4 cos t + 9 / Let 4t + 9 /. B te sifting an scaling rule, t 4 4t + 9 / 4t + 9 /. 48. z + 4 z 3 Let z + 4 z 3. Appling te prouct rule an te general power rule, z z + 4 3z + z 3 4z + 3 z + 3 z 6z + + 4z z + 3 z 4z cos 4 Let 3 + cos 4. B te general power rule, cos 5 3 sin 4sin cos sincossin Let sin cos sin. Appling te cain rule twice, cos cos sin sin sin cos cos sin sin cos cos sin. 5. sin cos We start b using a trig ientit to rewrite sin cos sin sin /. Ten, after two applications of te cain rule, sin cos / cos. sin April 4, 0

96 Let Appling te cain rule twice, we fin SECTION 3.7 Te Cain Rule cos 6 + sin / Let cos 6 + sin /. Appling te general power rule followe b bot te scaling an sifting rule an te cain rule, cos 6 + sin / sin cos cos 3 sin 6 cos 6 + sin / + Let +/ +. Appling te quotient rule an te sifting an scaling rule, we get + + / + / tan 3 + tan 3 Let tan 3 + tan 3 tan 3 + tan 3. Appling te general power rule to te first term an te cain rule to te secon term, 3tan sec + sec sec 3 + sec tan cos Let 4 3 cos /. B te general power rule, 57. z + z Let 4 3 cos / 3 sin 3 sin 4 3 cos. z + /. Appling te general power rule followe b te quotient rule, z z z + / z z + z z z + z 3/. 58. cos cos Let cos cos + 7 Appling te general power rule followe b te sum rule, wit te first term requiring te general power rule, 9 cos cos cos sin 3 sin 7 sin cos cos cos. cos cos Let cos + + cos. Ten, appling te quotient rule an te sifting an scaling rule, + cos sin + + cos + sin cos + sin cos sin + sin + + cos + cos sin sin + + cos. April 4, 0

97 86 CHAPTER 3 DIFFERENTIATION Te last line follows from te ientit sina B sin A cos B cos A sin B wit A an B sec t 9 Let sec t 9. Appling te cain rule followe b te general power rule, t sec t 9 tan t 9 t 9 / t sec t 9 t t 9 tan t cot 7 5 Let cot 7 5. Appling te general power rule followe b te cain rule, 7 cot6 5 csc cot 6 5 csc cos/ + Let cos/ + cos +. Ten, appling te quotient rule an te cain rule, we get: + sin cos + sin cos + sin cot Let + cot Appling te general power rule, te cain rule, an te general power rule in succession, 9 + cot cot csc cot csc cot e + 7e Let 4e + 7e. Using te cain rule twice, once for eac eponential function, we obtain 65. e 3 + 3e 4 4e 4e. Let e 3 + 3e 4. Appling te general power rule followe b two applications of te cain rule, one for eac eponential function, we fin 4e3 + 3e 3 6e 3 6e 4e 3 + 3e 3 e 3 e. 66. coste t Let coste t. Appling te cain rule an te prouct rule, we ave 67. e ++3 t sinte t te t + e t e t t sinte t. Let e ++3. B te cain rule an te general power rule, we obtain e e ++3. April 4, 0

98 SECTION 3.7 Te Cain Rule e e Let e e. Appling te cain rule, we ave ee e. Let + + / / /. Appling te general power rule twice, + + / / / + / / / Let + + / /. Appling te general power rule twice, + + / / + / k + b /3 ; k an b an constants Let k + b /3, were b an k are constants. B te scaling an sifting rule, 3 k + b 4/3 k k 3 k + b 4/3. 7. kt 4 + b ; k, b constants, not bot zero /, Let kt 4 + b were b an k are constants. B te cain rule, kt 4 3/ + b 4kt 3 kt 3 kt 4 + b 3/. In Eercises 73 76, compute te iger erivative. 73. sin rule, Let f sin. Ten, b te cain rule, f cos an, b te prouct rule an te cain f sin + cos cos 4 sin Let f Ten, b te general power rule, an, b te prouct rule an te general power rule, f f April 4, 0

99 88 CHAPTER 3 DIFFERENTIATION Let f 9 8. Ten, b repeate use of te scaling an sifting rule, f f , f sin Let f sin. Ten, b repeate use of te scaling an sifting rule, f cos f 4 sin f 8 cos. 77. Te average molecular velocit v of a gas in a certain container is given b v 9 T m/s, were T is te temperature in kelvins. Te temperature is relate to te pressure in atmosperes b T 00P. Fin v P. P.5 First note tat wen P.5 atmosperes, T K. Tus, v P v P.5 T T T 300 P 9 P m s atmosperes. Alternatel, substituting T 00P into te equation for v gives v 90 P. Terefore, so v P v P 90 P 90, P m P s atmosperes. 78. Te power P in a circuit is P Ri, were R is te resistance an i is te current. Fin P /t at t 3 if R 000 an i varies accoring to i sin4πt time in secons. Ri t/3 Ri i t t 0004π sin4πtcos4πt t/3 000π 3. t 79. An epaning spere as raius r 0.4t cm at time t in secons. Let V be te spere s volume. Fin V /t wen a r 3 an b t 3. Let r 0.4t, were t is in secons s an r is in centimeters cm. Wit V 4 3 πr3, we ave Tus V t V r V r 4πr. r t 4πr 0.4.6πr. a Wen r 3, V.6π cm/s. t b Wen t 3, we ave r.. Hence V.6π. 7.4 cm/s. t 80. A 005 stu b te Fiseries Researc Services in Abereen, Scotlan, suggests tat te average lengt of te species Clupea arengus Atlantic erring as a function of age t in ears can be moele b Lt 3 e 0.37t cm for 0 t 3. See Figure. a How fast is te lengt canging at age t 6 ears? b At wat age is te lengt canging at a rate of 5 cm/r? April 4, 0

100 SECTION 3.7 Te Cain Rule L cm t ear FIGURE Average lengt of te species Clupea arengus Let Lt 3 e 0.37t. Ten L t 30.37e 0.37t.84e 0.37t. a At age t 6, b Te lengt will be canging at a rate of 5 cm/r wen Solving for t iels L t.84e e..9 cm/r..84e 0.37t 5. t 0.37 ln 5.33 ears A 999 stu b Starke an Scarneccia evelope te following moel for te average weigt in kilograms at age t in ears of cannel catfis in te Lower Yellowstone River Figure 3: Fin te rate at wic weigt is canging at age t 0. Wt e t W kg Lower Yellowstone River t ear FIGURE 3 Average weigt of cannel catfis at age t Let Wt e t Ten W t e t e t e t.406 e t. At age t 0, W kg/r. 8. Te functions in Eercises 80 an 8 are eamples of te von Bertalanff growt function Mt a + b ae kmt /m m 0 introuce in te 930s b Austrian-born biologist Karl Luwig von Bertalanff. Calculate M 0 in terms of te constants a, b, k an m. April 4, 0