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 4 + 0.000? 0.000 tan π 4 + 0.000 is a goo approimation to te erivative of te function f tan at π 0.000 4. 5. 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 0.9 0.4 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.3 0.4 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 f3 + 5 + 30 + 45. Ten sow tat f3 + f3 an compute f 3 b taking te limit as 0. 5 + 30 0 Ma 5, 0
0 CHAPTER 3 DIFFERENTIATION Wit f 5, it follows tat f3 + 53 + 59 + 6 + 45 + 30 + 5. Using tis result, we fin f3 + f3 45 + 30 + 5 5 9 45 + 30 + 5 45 30 + 5 30 + 5. As 0, 30 + 5 30, so f 3 30. In Eercises 3 6, compute Let f f 3 a in two was, using Eq. an Eq.. 5. Sow tat te secant line troug,f an +, f + as slope + 5. 3. Ten f use tis + formula 9, a to0compute te slope of: a Te secant Let f line troug + 9.,f Ten an 3,f3 b Te tangent line at b taking a limit f f0 + f0 0 + + 90 + 0 9 + 0 lim lim lim lim 9 + 9. 0 0 0 0 Alternatel, 5. f 3 + f 4 +, a + 9, a Let f 3 + 4 +. Ten f f f0 + 9 0 0 lim lim 0 0 0 lim + 9 9. 0 f f + f 3 + + 4 + + lim lim 0 0 Alternatel, 3 lim 0 lim 3. 0 f lim In Eercises 7 0, f 3 refer to Figure., a f f 3 + 4 + lim + 3 + + lim lim 3 +. + 3.0.5.0.5.0 0.5 f 0.5.0.5.0.5 3.0 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 Ma 5, 0 f.5 f.5.5.5. 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. 9. Estimate f an f. f + f Estimate From te grap, it appears for tat 0.5. te tangent Wat line oes at tis quantit woul be represent? orizontal. Is ittus, largerf or smaller tan f? 0. Te tangent line Eplain. at appears to pass troug te points 0.5, 0.8 an,. Tus f 0.8 0.5 0.8. Fin a value of for wic f + f 0.
SECTION 3. Definition of te Derivative 03 In Eercises 4, refer to Figure. 5 4 3 3 4 5 6 7 8 9 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 7 0. 3. Wic is larger, f 5.5 or f 6.5? For wic values of is f < 0? 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 6.5. In Eercises Sow 5 8, tat f use 3 te oes limit not efinition eist. to calculate te erivative of te linear function. 5. f 7 9 fa+ fa 7a + 9 7a 9 lim lim lim 0 0 7 7. 0 7. gt 8 3t f ga + ga 8 3a + 8 3a 3 lim lim lim 0 0 0 lim 3 3. 0 9. Fin an equation of te tangent line at 3, assuming tat f3 5 an f 3? kz 4z + B efinition, te equation of te tangent line to te grap of fat 3is f3 + f 3 3 5 + 3.. Describe te tangent Fin f3 an f line at an arbitrar point on te curve + 8. 3, assuming tat te tangent line to fat a 3 as equation 5 +. Since + 8 represents a straigt line, te tangent line at an point is te line itself, + 8. 3. LetSuppose f tat. Does f + f + f equal 3 + 5. + Calculate: or +? Compute te ifference quotient at a wit a 0.5. Te slope of te secant line troug,f an 6,f6 b f Let f. Ten Wit a an 0.5, te ifference quotient is fa+ fa f + +. f.5 f 0.5.5 0.5 3. 5. Let f / Let f. Compute f 5 b sowing. Does f5 + equal tat 5 + or 5 +? Compute te ifference quotient at a 5 wit. f5 + f5 5 5 + 5 + + 5 Ma 5, 0
04 CHAPTER 3 DIFFERENTIATION Tus, Let f /. Ten f5 + f5 5+ 5 5 5 + 5 5 + 5 5 + 5 + 5 + 5 5 + 5 + 5 + 5 5 + 5 5 + 5 + + 5. 5 5 + 5 + + 5 f f5 + f5 5 lim lim 0 0 5 5 + 5 + + 5 5 5 5 + 5 0 5. In Eercises Fin 7 44, an equation use te of te limit tangent efinition line to to compute te grap f of a f an fin / an equation at 9. of te tangent line. 7. f + 0, a 3 Let f + 0. Ten f f3 + f3 3 + + 03 + 48 3 lim lim 0 0 At a 3, te tangent line is 8 + + + 30 + 0 48 lim 0 lim +. 0 9. ft t t, f 4 a 3, a Let ft t t. Ten f 3 3 + f3 3 + 48 8. f f3 + f3 3 + 3 + 5 3 lim lim 0 0 At a 3, te tangent line is 3 + 8 + 5 lim 0 lim. 0 3. f 3 +, f 8 3 a 0, Let f 3 +. Ten f 3t 3 + f3 t 3 5 t + 8. f f f0 3 + 0 0 lim lim 0 0 At a 0, te tangent line is lim 0 +. 33. f, ft t 3 a 8 + 4t, a 4 Let f. Ten f f8 + f8 8 lim lim 0 0 f 0 0 + f0. 8+ 8 lim 0 8 8 88+ lim 0 64 + 8 64 Ma 5, 0
SECTION 3. Definition of te Derivative 05 Te tangent at a 8is 35. f f + 3 +, a, a 4 Let f +3. Ten f f + f lim lim 0 0 Te tangent line at a is f 8 8 + f8 64 8 + 8 64 + 4. ++3 lim 0 + lim 0 f + + f + +. + lim 0 +. 37. f + ft 4, a Let f t, a + 4. Ten f f + f + 5 5 + 5 5 + 5 + 5 lim lim lim 0 0 0 + 5 + 5 lim 0 Te tangent line at a is 39. f ft, 3t a + 5, 4 a Let f. Ten + 5 + 5 lim 0 + 5 + 5 5. f + f 5 + 5 5 + 9 5. f f4 + f4 4+ 4 lim lim 0 0 lim 0 4 4 + + 4 + 6. lim 0 4+ 4+ + 4+ + 4+ 4 4 4 4++4+ lim 0 At a 4 te tangent line is f 4 4 + f4 6 4 + 6 + 3 4. 4. ft t +, a 3 f, a 4 Let ft + t +. Ten Te tangent line at a 3is f f3 + f3 3 lim lim 0 0 0 + 6 + lim 0 0 lim 0 6 + 0 + 6 + + 0 lim 0 0 + 6 + 0 0 + 6 + + 0 0 + 6 + + 0 6 + 0 + 6 + + 0 3. 0 f 3t 3 + f3 3 0 t 3 + 0 3 0 t + 0. f, a Ma 5, 0
06 CHAPTER 3 DIFFERENTIATION 43. f +, a 0 Let f +. Ten Te tangent line at a 0is f f0 + f0 0 lim lim 0 0 0+ + lim 0 + f0 + f 0 0 + 0. lim 0 + 0. 45. Figure 3 isplas ft t 3 ata collecte b te biologist Julian Hule 887 975 on te average antler weigt W of male re eer as a function, of a 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 8 7 6 5 4 3 0 0 4 6 8 0 4 Age ears FIGURE 3 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 4 5.5 0.9 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<.6. 8 7 6 5 4 3 0 0 4 6 8 0 4 47. Figure Let4A f sows 4 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 a Plot fover [, ]. Ten zoom in near 0 until te grap appears straigt, an estimate te slope f 0. an compare wit our estimate. 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 0.85.5 0. 0. 0.3 0.44 0.68 3.5.5 0.5.4..0.8 0. 0. 0. 0. Ma 5, 0
SECTION 3. Definition of te Derivative 07 b Using te estimate for f 0 obtaine in part a, te approimate equation of te tangent line is f 0 0 + f0 0.68 +. Te figure below sows te grap of fan te approimate tangent line. 3.5.5 0.5 49. Determine te intervals along te -ais Let f cot. Estimate f π on wic te erivative in Figure 5 is positive. grapicall b zooming in on a plot of fnear π. 4.0 3.5 3.0.5.0.5.0 0.5 0.5.0.5.0.5 3.0 3.5 4.0 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 >3.5. In Eercises 5 56, eac limit represents a erivative f Sketc te grap of f sin on [0,π] an guess a. Fin fan te value of f a. π. Ten calculate te ifference quotient at π for two small positive an negative values of. Are tese calculations consistent wit our guess? 5 + 3 5 5. lim 0 Te ifference quotient 5 + 3 5 as te form sin π 53. lim 3 6 + 0.5 5 0lim 5 5 Te ifference quotient sin π 6 +.5 as te form 5 + 5 55. lim 4 0lim 4 4 Te ifference quotient 5+ 5 as te form fa+ fa fa+ fa fa+ fa were f 3 an a 5. were f sin an a π 6. were f 5 an a. 57. Appl te 5 meto of Eample 6 to f sin to etermine f π 4 accuratel to four ecimal places. lim 0 We know tat Creating a table of values of close to zero: f fπ/4 + fπ/4 π/4 lim lim 0 0 sinπ/4 + /. 0.00 0.000 0.0000 0.0000 0.000 0.00 sin π 4 + / 0.707460 0.7074 0.70703 0.707033 0.707074 0.706753 Accurate up to four ecimal places, f π 4 0.707. 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. Ma 5, 0
08 CHAPTER 3 DIFFERENTIATION 59. 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. 6. Sketc te grap of f Refer to te grap of f 5/ on [0, 6]. in Figure 7. a Use te sketc to justif te inequalities for >0: a Eplain grapicall w, for >0, f4 f f4 f0 f f4 f f + f0 f4 4 0 b Use b Use a to a compute to sow f tat 4 0.6934 to four ecimal f 0 places. 0.6935. c Use c Similarl, a grapingcompute utilit tof plot fan to four ecimal te tangent places line forat, 4,, using 3, 4. our estimate for f 4. Now compute te ratios f /f 0 for,, 3, 4. Can ou guess an approimate formula for f? 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 Creating a table wit values of close to zero: f f4 + f4 4 lim lim 0 0 4 + 5/ 3. 0.000 0.0000 0.0000 0.000 4 + 5/ 3 9.99965 9.99999 0.0000 0.0000375 Tus, f 4 0.0000. c Using te estimate for f 4 obtaine in part b, te equation of te line tangent to f 5/ at 4is f 4 4 + f4 0 4 + 3 0 48. Ma 5, 0
SECTION 3. Definition of te Derivative 09 80 60 40 0 0 40 60 3 4 5 6 63. Use a plot of f Verif tat P, to estimate te value c suc tat f c 0. Fin c to sufficient accurac so tat lies on te graps of bot f / + an L + m for ever slope m. Plot fan L on fc+ te same aes fc 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 0.006 for estimate for f ±0.00? Here is a grap of f over te interval [0,.5]..5 0.5 0. 0.4 0.6 0.8..4 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 0.37..5.5 0.5 0.5 0. 0.4 0.6 0.8..4 0. 0.4 0.6 0.8..4 We note tat f0.37 + 0.00 f0.37 0.00 0.0049 < 0.006 an f0.37 0.00 f0.37 0.00 0.00304 < 0.006, so we ave etermine c to te esire accurac. In Eercises 65 7, estimate Plot f erivatives using te smmetric ifference quotient SDQ, efine as te average of te an + a on te same set of aes for several values of a until te line becomes tangent ifference quotients at an : to te grap. Ten estimate te value c suc tat f c. fa+ fa fa fa fa+ fa + 4 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 93 303 33 33 333 343 353 P atm 0.078 0.048 0.0808 0.3 0.067 0.373 0.4754 Using equation 4, P 303 P33 P93 0 0.0808 0.078 0 0.0065 atm/k; Ma 5, 0
0 CHAPTER 3 DIFFERENTIATION P 33 P 33 P 333 P 343 P33 P303 0 P333 P33 0 P343 P33 0 P353 P333 0 0.3 0.048 0 0.067 0.0808 0 0.373 0.3 0 0.4754 0.067 0 0.00445 atm/k; 0.00695 atm/k; 0.0093 atm/k; 0.03435 atm/k In Eercises Use te 67 an SDQ 68, wit traffic spee ear S along to estimate a certain P T roa in in te km/ ears varies 000, as 00, a function 004, 006, of traffic were ensit PT q number is te U.S. of cars etanol per km prouction of roa. Use Figure te following 8. Epress ata our to answer answer te in questions: te correct units. q ensit 60 70 80 90 00 S spee 7.5 67.5 63.5 60 56 67. Estimate S 80. Let Sq be te function etermining S given q. Using equation 4 wit 0, wit 0, S 80 S90 S70 0 60 67.5 0 0.375; S 80 S00 S60 40 56 7.5 40 Te mean of tese two smmetric ifference quotients is 0.39375 kp km/car. 0.45; Eercises 69 7: Te current in amperes at time t in secons flowing in te circuit in Figure 9 is given b Kircoff s Eplain w V qs, calle traffic volume, is equal to te number of cars passing a point per our. Use te Law: ata to estimate V 80. 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 9 69. 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 + /Rv3 0.00.5 + /005.5 0.005 + 0.055 0.06 amperes. 7. Assume tat R 00 but C is unknown. Use te following ata to estimate v Use te following ata to estimate v 4 b an SDQ an euce an approimate value for te capacitance C. 0 b an SDQ. Ten estimate i0, assuming C 0.03 an R 000. t 9.8 9.9 0 0. 0. t 3.8 3.9 4 4. 4. vt 56.5 57.3 58. 58.9 59.69 vt 388.8 404. 40 436. 45.8 it 3.34 33. 34. 34.98 35.86 Solving i4 Cv 4 + /Rv4 for C iels C i4 /Rv4 v 4 34. 40 00 v. 4 Ma 5, 0
SECTION 3. Te Derivative as a Function To compute C, we first approimate v 4. Taking 0., we fin v v4. v3.9 4 0. Plugging tis in to te equation above iels Furter Insigts an Callenges C 34.. 60 436. 404. 0. 0. faras. 60. 73. Eplain ow te smmetric ifference quotient efine b Eq. 4 can be interprete as te slope of a secant line. Te SDQ usuall approimates te erivative muc more closel tan oes te orinar ifference quotient. Let f Te an smmetric a 0. Compute ifference te quotient SDQ wit 0.00 an te orinar ifference quotients wit ±0.00. Compare wit te actual value, wic is f 0 fa+ ln. 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. 75. Wic Sow of te tat two if functions f is ainquaratic Figure 0 polnomial, satisfies teten inequalit te SDQ at a for an 0 is equal to f a. Eplain te grapical meaning of tis result. fa+ fa fa+ fa Let f p + q + r be a quaratic polnomial. We compute te SDQ at a. for >0? fa+ Eplain fa in terms of pa secant + lines. + 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. Let f. Compute f b taking te limit of te SDQs wit a as 0. 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 3 8.. Evaluate f g an 3f + g assuming tat f 3 an g 5. f g f g 3 5 an 3f + g 3f + g 33 + 5 9. 3. 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. Ma 5, 0
CHAPTER 3 DIFFERENTIATION Eercises In Eercises 6, compute f using te limit efinition.. f 3 7 Let f 3 7. Ten, f f+ f 3 + 7 3 7 3 lim lim lim 0 0 0 3. 3. f 3 f + 3 Let f 3. Ten, f f+ f + 3 3 3 + 3 + 3 + 3 3 lim lim lim 0 0 0 3 + 3 + 3 lim lim 0 0 3 + 3 + 3. 5. f f Let f. Ten, f f+ f + + + + + lim lim lim 0 0 0 + + lim 0 + + + lim 0 + +. In Eercises 7 4, f / use te Power Rule to compute te erivative. 7. 4 4 4 3 so 4 4 3 3. 9. t t/3 t t 3 t8 t4 t /3 t 3 t /3 so t t8 t/3 3 8 /3 3.. 0.35 t t /5 t 0.35 0.35 0.35 0.35 0.65. 3. t t 7 4/3 t 7 7t 7 t In Eercises 5 8, t t π 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 3 + 6 3 48. 7. f 5 f 3, a 4, a 5 Let f 5 3 /. Ten f 5 6 /. In particular, f 4 3. Te tangent line at 4 is f 4 4 + f4 3 4 44 3 3. f 3, a 8 Ma 5, 0
SECTION 3. Te Derivative as a Function 3 9. 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 3. c t et 3 In Eercises Fin 3, an equation calculate of te te tangent erivative. line to 4e at.. f 3 3 + 5 3 3 + 5 6 6. 3. f 4 5/3 f 3 3 3 + 4 5/3 3 0 3 /3 + 6 3. 5. gz 7z 5/4 f 5/4 + z 5 + 4 3/ + 9 + 7z 5/4 + z 5 + 9 5 z z 9/4 5z 6. 7. fs 4 s + t 6 3 s t + fs 4 s t + 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. 9. g e W 6 4 + 7 /3 Because e is a constant, e 0. 3. t 5e t 3 f 3e 3 t 5et 3 5e 3 t et 5e 3 e t 5e t 3. In Eercises 33 36, calculate f 9 /3 te + 8e erivative b epaning or simplifing te function. 33. Ps 4s 3 Ps 4s 3 6s 4s + 9. Tus, 35. g Qr + 4 / r3r + 5 P s 3s 4. g + 4 / + 4 3/. Tus, g 6 5/. In Eercises 37 4, st t calculate te erivative inicate. T t 37., T / 3C /3 C C8 Wit TC 3C /3, we ave T C C /3. Terefore, T C 8 /3. C8 P, V V P 7 V Ma 5, 0
4 CHAPTER 3 DIFFERENTIATION s 39., s 4z 6z z z Wit s 4z 6z, we ave s 4 3z. Terefore, z s z 4 3 60. z r 4. R t, r t e t W, R W π t4 W Wit r t e t, we ave r t et. Terefore, r t e 4. t4 43. Matc te functions in graps A D wit teir erivatives I III in Figure 3. Note tat two of te functions p ave te same erivative., p 7eEplain w. 4 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. 45. Assign te labels f, g, an to te graps in Figure 5 in suc a wa tat f g an g. Of te two functions f an g in Figure 4, wic is te erivative of te oter? Justif our answer. A B C FIGURE 5 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. Ma 5, 0
SECTION 3. Te Derivative as a Function 5 47. Accoring Use te totable te peak of values oil teor, of fto first propose etermine wic 956 b of geopsicist A or B inm. Figure Hubbert, 7 is te total grapamount of f. of crue Eplain. oil Qt prouce worlwie up to time t as a grap like tat in Figure 6. a Sketc te erivative Q t 0for 900 0.5 t 50..5 Wat oes.5q t 3represent? 3.5 4 b In wic ear approimatel oes Q t take on its maimum value? f 0 55 98 39 77 0 37 c Wat is L lim Qt? An wat is its interpretation? t 57 68 Wat is te value of lim t Q t? A 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. 49. Compute te erivatives, were c is a constant. Let R be a variable an r a constant. Compute te erivatives: a a R R b R r c t ct3 b 9c 3 4c c z 5z + R r R 4cz 3 a t ct3 3ct. b z 5z + 4cz 5 + 8cz. c 9c 3 4c 7c. 5. Fin te points on te grap of + 3 7 at Fin te points on te grap of f 3 wic te slope of te tangent line is equal to 4. were te tangent line is orizontal. Let + 3 7. Solving / + 3 4 iels. 53. Determine a an b suc tat p + a 3..5 Fin te values of were 3 + b satisfies an p 0 an p 4. + 5 ave parallel tangent lines. 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. 55. Let f 3 3 +. Sow tat f 3 for all an Fin all values of suc tat te tangent line to 4 tat, for ever m> 3, tere are precisel two points were f m. Inicate te position of tese points an te corresponing + + is tangent steeper lines tan for te one tangent value line of m to in a sketc 3. of te grap of f. Let P a, b be a point on te grap of f 3 3 +. Te erivative satisfies f 3 3 3 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 + 3 m + 3 > 0 an tus a ±. 3 3 Te two parallel tangent lines wit slope are sown wit te grap of fere. B 4 57. Compute te erivative of f 3/ Sow tat te tangent lines to using te 3 3 limit efinition. Hint: Sow tat at a an at b are parallel if a b or a + b. f+ f + 3 3 + 3 + 3 Ma 5, 0
6 CHAPTER 3 DIFFERENTIATION Once we ave te ifference of square roots, we multipl b te conjugate to solve te problem. f + 3/ 3/ + lim lim 3 3 + 3 + 3 0 0 + 3 + 3 + 3 3 lim 0 + 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/ 3 3 3 /. 59. Let f e. Use te limit efinition to compute f 0, an fin te equation of te tangent line at 0. Use te limit efinition of mb to approimate m4. Ten estimate te slope of te tangent line to 4 at 0 an Let f. e. Ten f0 0, an f f0 + f0 e 0 0 lim lim lim 0 0 0 e. Te equation of te tangent line is f 0 0 + f0 0 + 0. 6. Biologists ave observe tat te pulse rate P in beats per minute in animals is relate to bo mass in kilograms Te average spee in meters per secon of a gas molecule is 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 8RT tat represent goat m 33 an man m 68. v avg 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, wicguinea as apigmolar mass of 0.03 kg/mol. Pulse beats/min 00 00 Goat Man 00 00 300 FIGURE 8 Cattle 400 500 Mass kg P /m 50m 5/4. For m>0, P /m 50m 5/4. P /m 0asm gets 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, P33 83.445 beats per minute bpm an P m 5033 5/4 0.636 bpm/kg. Hence, 0.636m 33 + 83.445. For a man m 68 kg, we ave P68 69.647 bpm an P m 5068 5/4 0.5606 bpm/kg. Hence, te tangent line as formula 0.5606m 68 + 69.647. 63. Te Clausius Claperon Law relates te vapor pressure of water P in atmosperes to te temperature T in kelvins: 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 0.85. Calculate P K/m at m 68. Ten calculate te erivative wit respect to m of te relative kine-to-mass ratio K/m at m T 68. 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 Ma 5, 0
SECTION 3. Te Derivative as a Function 7 T K 93 303 33 33 333 343 353 P atm 0.078 0.048 0.0808 0.3 0.067 0.373 0.4754 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 P 333 P 343 P33 P93 0 P33 P303 0 P333 P33 0 P343 P33 0 P353 P333 0 0.0808 0.078 0 0.3 0.048 0 0.067 0.0808 0 0.373 0.3 0 0.4754 0.067 0 0.0065 atm/k; 0.00445 atm/k; 0.00695 atm/k; 0.0093 atm/k; 0.03435 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 303 33 33 333 343 T P P T 5047.59 505.76 5009.54 4994.57 498.45 Tese values are rougl constant, suggesting tat te Clausius Claperon law is vali, an tat k 5000. 65. In te setting of Eercise 64, sow tat te point of tangenc is te mipoint of te segment of L ling in te first Let L be te tangent line to te perbola at a, were a>0. Sow tat te area of te triangle quarant. boune b L an te coorinate aes oes not epen on a. 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. 67. Make a roug sketc of te grap of te erivative of te function in Figure 0A. Matc functions A C wit teir erivatives I III in Figure 9. 3 3 4 0 3 4 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: 3 4 Ma 5, 0
8 CHAPTER 3 DIFFERENTIATION 69. Sketc te grap of f. Ten sow tat f 0 eists. Grap te erivative of te function in Figure 0B, omitting points were te erivative is not efine. For <0, f, an f. For >0, f, an f.at 0, we fin an f0 + f0 lim lim 0+ 0+ 0 f0 + f0 lim lim 0 0 0. 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 In Eercises Determine 7 76, te fin values te points of at c wic if an te suc function tat f in c Figure oes notis: eist. a iscontinuous, an b nonifferentiable. 7. f.5 0.5 3 Here is te grap of f. Its erivative oes not eist at. At tat value of tere is a sarp corner. 73. f 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..5 75. f f 3/ Here is te grap of f. Its erivative oes not eist at orat. At tese values of, te grap as sarp corners. 3 f Ma 5, 0
SECTION 3. Te Derivative as a Function 9 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. 0. 0. 0. 0. 0. 0. 0.3 79. f 3 /3, f 3 5/3 a 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..9.95 3 3.05 3. 8. f sin, f sin /3 a 0, 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. 0. 0.08 0.04 0. 0.05 0.05 0. 83. Plot te erivative f f sin, a of f 0 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 6 + 0. 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. 400 800 300 600 00 400 00 00 4 6 8 00 4 6 8 Fin te coorinates of te point P in Figure at wic te tangent line passes troug 5, 0. Ma 5, 0 FIGURE Grap of f 9.
0 CHAPTER 3 DIFFERENTIATION Eercises 85 88 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 3 85. Calculate te subtangent of f + 3 at Let f + 3. Ten f + 3, an te equation of te tangent line at is f + f 7 + 0 7 4. Tis line intersects te -ais at 4 7. Tus Q as coorinates 4 7, 0, R as coorinates, 0 an te subtangent is 4 7 0 7. 87. Prove in general tat te subnormal at P is Sow tat te subtangent of f e f f. is everwere equal to. 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. 89. Prove te following teorem of Apollonius Sow tat PQas lengt f of Perga te Greek matematician born in 6 bce wo gave te parabola, ellipse, an perbola teir names: Te subtangent + f 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 0 9. Sow Formulate tat te subtangent an prove to a generalization 3 at of a is Eercise equal to 90 for 3 a. 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. Ma 5, 0
SECTION 3. Te Derivative as a Function Furter Insigts an Callenges 93. A vase is forme b rotating aroun te -ais. If we rop in a marble, it will Two small arces ave te sape of parabolas. Te first is given b f eiter touc te bottom point of te vase or be suspene for an te secon b g 4 4 above te bottom b toucing te sies Figure 5. How small must te marble be to touc te bottom? for 6. A boar is place on top of tese arces so it rests on bot Figure 9. Wat is te slope of te boar? Hint: Fin te tangent line to ftat intersects g in eactl one point. 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. 95. Negative Eponents Let n be a wole number. Use te Power Rule for n to calculate te erivative of f n b sowing Let tat 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 tef+ left if c>0 f or rigt if c<0. + n n n + n Let f n were n is a positive integer. Te ifference quotient for f is f+ f + n n + n n + n n n + n. n + n n + n Terefore, f f+ f + n n lim lim 0 0 n + n lim 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 n is a negative integer an we ave for negative integers k. k n n n k k ; i.e. k k k 97. Infinitel Rapi Oscillations Define 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. sin 0 f 0 0 Sow tat fis continuous at 0 but f 0 oes not eist see Figure 4. Ma 5, 0
CHAPTER 3 DIFFERENTIATION { sin if 0 Let f.as 0, 0 if 0 f f0 sin 0 sin 0 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. 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. 3.3 Prouct an Quotient Rules sin 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. Eercises In Eercises 6, use te Prouct Rule to calculate te erivative.. f 3 + Let f 3 +. Ten f 3 + + + 3 3 4 + + 3 0 4 + 3. 3. f e f 3 5 3 Let f e. Ten f e + e e + e e +. f 94e + Ma 5, 0
5. s, s s / + s7 s s4 Let s s / + s7 s. Ten s s / + s 7 s + 7 s s s / + ss + 7 s s 3/ + SECTION 3.3 Prouct an Quotient Rules 3 s / + s 7 s 3/ + 3 s 5/ + 4. Terefore, s 7 s4 4 3/ + 3 4 5/ + 4 87 64. In Eercises 7, use te Quotient t, t 8t Rule e t + t to calculate te erivative. 7. f t Let f. Ten f g 9. t f, gt + 4 t + t + + t Let gt t + t. Ten g t Terefore,. t t t + t + t t t t t t + t t. g w + z, e w z z9 z + z Let g + e. Ten g t 4 t 8 9. g + e + e + e + e 0 e e + e + e. 4t t. In Eercises 3 6, f e 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. 5. t f t 3 + 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, Hence t for t. t t t + t t +. Ma 5, 0
4 CHAPTER 3 DIFFERENTIATION In Eercises 7 38, g 3 calculate + te + 3 erivative. 7. f 3 + 5 3 + + Let f 3 + 5 3 + +. Ten f 3 + 53 + + 3 + + 3 6 5 + 4 3 + 8 + 5. 9. f, 4e 3 + 3 + 0 Let +0. Using te quotient rule: + 00 + 0 + 0. Terefore, 3 3 + 0 69.. f + z Let f, z 3 +. Multipling troug first iels f for 0. Terefore, + f for 0. If we carr out te prouct rule on f / + /, weget f / + / + / / + / + /. 3. f 95/, 4 4 5 Let 4 4 5. Ten Terefore, 5 4 3 5 z 5., z f 4 + e 3 + + Let z 3. Using te quotient rule: + 4 4 5 0 3 + 8 5. 5 0 3 + 8 5 80. z 3 + 0 3 3 + 3 3 +. Terefore, z 3 3 + 3 4. t 7. t f t + 33 t + + t Let t t + t + t t 3 + t + t +. Ten t 3 + t + t + t 3t + t + t t 3 + t + t + t3 t + t 3 + t + t +. 9. ft f 3/ 3/ 5 / 4 3 + / Let ft 3 5. Ten f t 0, since ftis a constant function! Ma 5, 0
SECTION 3.3 Prouct an Quotient Rules 5 3. f + π 3 5 Let f + 3 5. Using te Prouct Rule insie te Prouct Rule wit a first factor of + 3 an a secon factor of 5, wefin f + 3 + 5 + 5 3 6 3. Alternativel, f + 3 6 + 5 3 3 3 + 5. Terefore, f 3 6 3. 33. f f e + e + + 4 Let f e e + +. Ten f e + + e e e + + + e e e e + + e + +. z g e+ 4 z 35. gz + e Hint: Simplif first. z z + e + Let z 4 z z + z gz z z + z for z an z. Ten, t 4 37. t t a + bab + a,b constants constant Let ft t 4 t. Using te quotient rule: z + z z + g z z + + z z. z z + f t t t 4t t t t + 8t t t + 8t t. In Eercises 39 4, a + b calculate te erivative using te values: a, b, c, constants c + f4 f 4 g4 g 4 39. fg 4 an f/g 4. 0 5 Let fg an H f/g. Ten fg + gf an H gf fg. Finall, an 4 f4g 4 + g4f 4 0 + 5 0, H 4 g4f 4 f4g 4 g4 g 5 0 5 0. 4. G 4, F were G g 4, were F. f. Let G g gg. Ten G gg + gg gg, an G 4 g4g 4 5 0. Ma 5, 0
6 CHAPTER 3 DIFFERENTIATION 43. Calculate F 0, were H 4, were H gf. F 9 + 8 + 4 5 7 4 3 + + 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 an let Ten F f g. Now, f 9 + 8 + 4 5 7 g 4 3 + +. f 9 8 + 8 7 + 0 4 7 an g 4 3 6 +. Moreover, f0 0, f 0 7, g0, an g 0. Using te quotient rule: F 0 g0f 0 f0g 0 g0 7 0 7. 45. UseProcee te Prouct as inrule Eercise to calculate 43 to calculate F 0, were e. Note tat e e e. Terefore F + + 4/3 + 5/3 35 + 5 4 + 5 + 8 9 7 4 + e e e e e + e e e. 47. Plot f / in a suitabl Plot te erivative of f / boune viewing bo. Use te plot to etermine weter f is positive + over [ 4, 4]. Use te grap to etermine te intervals on wic f or negative on > 0 an f its omain { : ±}. Ten compute f an confirm < 0. Ten plot fan escribe ow te sign of f our conclusion algebraicall. is reflecte in te grap of f. 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 +, 5 4 3 3 4 5 49. Fin a>0 suc Let P V tat te tangent R/R + r line to te grap of as in Eample 7. Calculate P/r, assuming tat r is variable an R is constant. f e at a passes troug te origin Figure 4. f e a FIGURE 4 Ma 5, 0
SECTION 3.3 Prouct an Quotient Rules 7 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. 5. Te revenue per mont earne b te Couture cloting cain at time t is Rt NtSt, were Ntis te number Current I amperes, voltage V volts, an resistance R oms in a circuit are relate b Om s Law, I V/R. of stores an St is a Calculate I 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 R ifmont, V is constant an B wit to use value avertising V 4. to increase average revenue per store at a rate of $0,000 per mont. Assume tat R6 b Calculate V N0 50 an S0 $50,000. a Sow tat total revenue will R if I is increase constant atwit te rate value I 4. R6 R 5St + 0,000Nt t 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 NtS t + StN t. t We are tol tat N t 5 stores per mont an S t 0,000 ollars per mont. Terefore, R 5St + 0,000Nt. t b Using part a an te given values of N0 an S0, wefin R t 550,000 + 0,00050,50,000. 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. 53. Te Te curve tip spee / ratio + of a is turbine calle Figure te witc 5 is of te Agnesi ratio R Figure T/W, 6 after were te T Italian is te spee matematician of te tip of Maria a blae Agnesi an 78 799, W is te spee wo wrote of te one win. of te Engineers first books ave on foun calculus. empiricall Tis strange tat name a turbine is te wit result n blaes of a mistranslation etracts maimum of te Italian power wor from la versiera, te win meaning wen R tat π/n. wic Calculate turns. Fin R/t equations t in minutes of te tangent if W lines 35 at km/ ±. an W ecreases at a rate of 4 km/ per minute, an te tip spee as constant value T 50 km/. 3 FIGURE 6 Te witc of Agnesi. Let f +. Ten f + 0 + +. At, te tangent line is 3 f + + f + + +. Ma 5, 0
8 CHAPTER 3 DIFFERENTIATION At, te tangent line is f + f + +. 55. Use te Prouct Rule to sow tat f Let f g. Sow tat f/g ff f. /g. Let g f ff. Ten g f ff ff + ff ff. Furter Sow Insigts tat f 3 an 3f Callenges f. 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. 59. Derivative of te Reciprocal Use te limit efinition to prove Prove te Quotient Rule using te limit efinition of te erivative. f f f Hint: Sow tat te ifference quotient for /f is equal to f f+ f f + Let g f. We ten compute te erivative of g using te ifference quotient: g g + g lim lim 0 0 f+ f f+ lim 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. 6. Use te limit efinition of te erivative to prove te following special case of te Prouct Rule: Prove te Quotient Rule using Eq. 7 an te Prouct Rule. f f + f 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 lim lim + f+ 0 0 f+ f lim + lim f+ 0 0 f + f. 63. Te Power Rule Revisite If ou are familiar wit proof b inuction, use inuction to prove te Power Rule for Carr out Maria Agnesi s proof of te Quotient Rule from er book on calculus, publise in 748: Assume tat all wole numbers n. Sow tat te Power Rule ols for n ; ten write f, g, an f/g are ifferentiable. Compute te erivative of g f using n as te n an use te Prouct Rule. Prouct Rule, an solve for. Let k be a positive integer. If k, ten k. Note tat 0. 7 Ma 5, 0
SECTION 3.4 Rates of Cange 9 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. 65. Use Eercise 64 to etermine weter c is a multiple root: Sow tat c is a multiple root of fif an onl if c is a root of bot fan f. a 5 + 4 4 3 8 + b 4 + 3 5 3 + a To sow tat is a multiple root of f 5 + 4 4 3 8 +, it suffices to ceck tat f f 0. We ave f + + 4 8 + + 0 an f 5 4 + 8 3 6 f 5 8 + 6 0 b Let f 4 + 3 5 3 +. Ten f 4 3 + 3 0 3. Because f 5 + 3 + 0 but f 4 + 3 + 0 3 6 0, it follows tat is a root of f, but not a multiple root. 67. Accoring to Eq. 6 in Section 3., Figure 7 is te grap of a polnomial b mb b wit roots. Use te Prouct Rule to sow tat mab ma + mb. at A, B, an C. Wic of tese is a multiple root? Eplain our reasoning using Eercise 64. 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? Since bot trains travel te same istance in te same amount of time, te ave te same average velocit: 90 mp. 3. Estimate f6, assuming tat f5 43, f 5 0.75. f f5 + f 5 5, sof6 43 + 0.756 5 43.75. 4. Te population Ptof Freeonia in 009 was P009 5 million. a Wat is te meaning of P 009? b Estimate P00 if P 009 0.. 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 009 0., ten P009 5. million. Ma 5, 0
30 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. 3. Cube root 3 wit respect to wen, 8, 7. Volume of a cube wit respect to its sie s wen s 5. 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 3 3 8 f 8 3 8 /3 3 4 7 f 7 3 7 /3 3 9 7 5. Te iameter of a circle wit respect to raius. Te reciprocal / wit respect to wen,, 3. 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. 7. Volume V of a cliner wit respect to raius if te eigt is equal Surface area A of a spere wit respect to raius r A 4πr to te raius.. Te volume of te cliner is V πr πr 3. Tus V /r 3πr. In Eercises 9, refer to Figure 0, te grap of istance st from te origin as a function of Spee of soun v in m/s wit respect to air temperature T in kelvins, were v 0 time for a car trip. T. 50 00 50 Distance km 0.5.0.5 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 50 0 00 km/our. 0.5 0 b Te average velocit over te interval [0.5, ] is 00 50 0.5 c Te average velocit over te interval [,.5] is 00 00.5 Te average velocit over te interval [, ] is 50 00 At wat time is velocit at a maimum?. Matc te escriptions i iii wit te intervals a c. i Velocit increasing ii Velocit ecreasing iii Velocit negative.0.5 3.0 00 km/our. 0km/our. 50 km/our. t Ma 5, 0
SECTION 3.4 Rates of Cange 3 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. 3. Use Figure 3 from Eample to estimate te instantaneous rate of cange of Martian temperature wit respect to Use te ata from Table in Eample to calculate te average rate of cange of Martian temperature T wit time in egrees Celsius per our at t 4 am. respect to time t over te interval from 8:36 am to 9:34 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 75 70 3..565 C/our. 5. Te velocit in cm/s of Te temperature in bloo molecules flowing troug a capillar of raius C of an object at time t in minutes is Tt 8 3 0.008 cm is v 6.4 0 8 0.00r, were r is te istance from te molecule to te center of te capillar. Fin te t rate 5t of cange + 80 for of velocit 0 t wit 0. At respect wat to r wen rate is r te object 0.004 cm. cooling at t 0? Give correct units. Te rate of cange of te velocit of te bloo molecules is v r 0.00r. Wen r 0.004 cm, tis rate is 8 0 6 /s. 7. Use Figure to estimate T / at 30 an 70, were T is atmosperic temperature in egrees Celsius an Figure isplas te voltage V across a capacitor as a function of time wile te capacitor is being carge. is altitue in kilometers. Were is T / equal to zero? 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 Temperature quickl or more slowl as time goes on? Eplain in terms of tangent lines. C 50 00 50 00 50 0 50 00 Tropospere Stratospere Mesospere Altitue km Termospere 0 50 00 50 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 0 50 40 3 50 7.94; 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 00 0 88 58 00 3.33; 30 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 90. 9. Calculate te rate of cange of escape velocit v esc.8 Te eart eerts a gravitational force of Fr.99 0 6 0 7 r /r / m/s wit respect to istance r from te center of te eart. 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 tat escape velocit canges is v esc r.4 07 r 3/. Ma 5, 0