solnshw3 - a Section 4.5 2 y = lnx5/2 4 y = In x”2 5 l...

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Section 4.5 2. y = lnx5/2 4. y = In x”2 5 l β€”21nx β€”21nx [β€”3 ,/_i 3 2x ) 2x I __Β§ / _l y(l)β€”2 )(U-z 3 10. y = ln(1 - x)”2 = Eln(l - x) v,_Β§(β€”l)_ 3 y 21-x 2(x-l) 14. y: 7 x- v,_ .rzil/x! β€” 2xlnx : xβ€” 2xlnx : l - 21nx 4 x4 18. y = in X3 22. y = ln(x\/4 + x2) = lnx +1n(4 + x2)”1 / = l + x _ Β£2.53. y x 4+x2 x(4+x2) + .T 26. f(x) = 1n]1 _ :x = 1n(l + e-") β€” ln(l β€” e") ’ _ ex _ ._e,\’ f(x) 1 + eβ€˜ l - e'β€œ = e~"(l β€” eβ€œ) + e-β€˜(l + e") z 28'" (l + e-β€˜)(l β€” e") l - elβ€œ 1 30. log3x = β€œ111.: Derivatives of Logarithmic Functions Section 4.5 Derivatives of Logarithmic Functions 149 6. f(x) = In 2): 8. y = ln(l ~ x3) , . ~ 3 _ l V, _ _ 2x f('x)_2x_x β€œ - l-x2 : 2x x2 - 1 12. y = (In x32 = (2 In x)2 = 4(1n x)2 v’ = 4[2(lnx)(l)] = 81"" x x x '7 . = : β€” - + 16 y lnx2 +1 lnx ln(x 1) ,_l_ l _i_ 2x __ lβ€˜X2 β€˜β€”x 12+l(2x)_ x2+l_x(x2+l) 20 i=ln erl=l[ln( +l)β€”1n( β€”1)] . \ x _ l 2 x x yl[1_1]=-1=1 β€˜ 2 x+1xβ€”1 xZβ€”I 1β€”x2 24. f(x) = xlneβ€˜2 = x062) = x3 f'(x) = 3x2 28. 3x = e"β€œ"” 32. log512 = film 12 =3 1.544 (calculator) In 32 . l 7 . = z - . 8 36 ogh/3 32 Ina/3) 854 42. y = 65'" 1 y’ = ln6 - 65-β€˜(5) = 51n6 - 65" 150 Chapter 4 Exponential and Logarithmic Functions W 44. f(x) = 10β€˜: f’(x) = (ln lO)lOβ€œβ€˜(2x) = 2x(ln IONO": 48. v,:x(]/x)7β€”lnx= l β€”jnx xβ€˜ x~ v’(e) = 0 1 . β€” - = 0(x - e) 6 l . y = ; Tangent lme 52. In xy + 5x = 30 lnx+lny+5x=30 l Idy β€”+-β€”β€”+ = x ydx 5 0 dv l _= ___s dx (x )V z_x_5y=_y(l+5x) x (I ~ _ _2 f(x)_ x2 60. T = 87.97 + 34.961np + 7.91 J; d_T = 34.96 + 7.91 dP p 2J5 dT 34.96 7.91 At =60,β€”=-β€”-+ p @ a) zjfi 62. f(x) = Zlnx3 = 6lnx I _ 9 f (x) β€” x At (e, 6). the slope of the tangent line is f’(e) = 6/e. . 6 Tangent lme: y β€” 6 = :(x β€” e) 6 )7 : β€”x e 46. y = 23β€œβ€œ y’ = 3m + x3~+l In 3 = 3WD + xln 3] 50. g(x) = log2(3x β€” I) = In 2 t ___3__ gmβ€˜ax- l)ln2 3 321n2 g’(11)= y~5= 54. 4x)! + ln(x2y) = 7 ll \1 4xy+ 2lnx+ lny 2 y, 4g,'+4y+_+._ x y II 0 <4x + l)y’ = -4y β€” 2- y x y : -~ 1.093 degrees per pounds per square inch. ln(3x β€” 1) I M=Lm ’\ 4x +(1/y) 4x2y + x 58. f(x) = logmx fl") = 111β€”110”2 2 x21_n]10 64. f(x) = ln(x\/m) = lnx + %1n(x + 3) At (1.20, 0.90), the slope of the tangent line is f’(l.20) = Tangent line: y β€” 0.90 = 52β€”?(x ~ 1.20) 21y β€” 18.90 = 20x β€” 24 0 = 20x β€” 21y β€” 5.10 0 = 200x -β€” 210y β€” 51 20 1. Section 4.5 Derivatives of Logarithmic Functions 151 7 .l" , ,lnx 68. v = β€” β€” in r 66. . =.β€˜-l .'=.'-β€” - ' f(r) l og3t In 3 2 ,._l' _β€˜ \,/=x_1_x2-l f (.x) β€” E0 + 21 lnx) . x A. l M = = β€˜f,(l):___ ) 0whenx 1. ln 3 I .2 + ] 0 l y”=l+;:lxz β€˜ ’ ln 3 U β€” I) x I [Notez x = ~ 1 is not in the domain of the function] y = _ Since y”(l) = 2 > 0, there is a relative minimum at ’ (l, Moreover, since y” > 0 on (0. 00), it follows that the graph is concave upward in its domain and there are no inflection points. 5 :I. 4 3 1 2 I A -1 . 70. y = xlnx 72. y = (In x)2 l 1 2 l y’ = x(-) + (ln.r)(l) = 1 + lnx y’ = 2(ln x)<β€”) = nx x x x y’=0whenx=e". y’=0whenx= l. ',,_ l 7”: x(2/x) β€” 21nx)(l)_ 2(1 -β€” lnx) .\ x r " x3 X: Since y”(eβ€˜ ’) > 0, there is a relative minimum at Since y”(1) > 0. there is a relative minimum at (l , 0). (eβ€œ, β€”eβ€œ). Moreover. since y” > 0 on (0. 00), it fol- Since y” = 0 when x = e, it follows that there is an lows that the graph is concave upward in its domain and inflection point at (e. I). there are no inflection points. β€˜ \ > Β»\ '1': imaammm 74. x = 1000 β€” plnp 76. x = 300 β€” 501n(lnp) (ix 1 dx [/11 50 β€”. = _ _ + = _ _. : _ __ = _ dp 0 Mp) (In/0(1)] (l + lnp) dp 0 solnp plnp dx dx β€”50 : β€” = β€” + z β€” _ _ = ' β€”β€”- = z β€” _ _ When p 10,dp (1 In 10) 3 3 When [1 10 dp 101nm 217l 152 Chapter 4 Exponential and Logarithmic Functions W 500 78. xβ€”m 80. Cβ€”100+25xβ€”1201nx,)t21 β€” C 100 lnx 500 : 'β€”β€”- = -β€” β€”β€” β€”__ ln(p3+I)=T (a)C x X +25 120x [,2 + I = eSOO/x (b) 6,, : β€”x1200 _ I2O[x(l/x)x2β€” In x] P = e500x _ 1 = -100 Β£0 _ I... .311 = %(e500/x _ l)β€”i/2(esoo/_t~)< _ :- x- S tt’ C’ = 0, -250 esno/x 6 mg 2 x2 ' e500} _ I 100 = -120β€œ β€” In x) Whenp=10. ~2=1nx*1 500 dp = z .3 β€” = β€” . . x mum) 108 4anddx 0215 16_l=lnx Note that dp/dx = 1 /[dx/dp] from Exercise 75. Minimum of 5.81 at x = 6.25. 5.315 82. I=m, x> 1000 (a) 5?. 2000 0 (b) Ifx = 1167.41. 1β€˜ = 20 and the total amount is (1 167.41)(20)(12) z $280,178. (c) Ifx = 1068.45,t = 30 and the total amount is (1068.45)(30)(12) z $384,642. (d) [I β€” β€˜ (lnx β€” 6.7968)2 ~5.315/x (In x β€” 6.7968)2 2 β€”5.315 x(lnx β€” 6.7968)2 When x = 1167.41, the instantaneous rate of change is β€”5.315 ll67.41(ln 1167.41 β€” 6.7968)2 == β€” 0.064. When x = 1068.45, the instantaneous rate of change is | β€˜5'315 a: β€”0.158. 1068.45(1n 1068.45 β€” 6.7968)2 (e) For a higher monthly payment, the term is shorter and the overall payment total smaller. Section 4.6 Exponential Growth and Decay 153 Section 4.6 Exponential Growth and Decay 2. Since 3 = %whent = 0. it follows that C = Moreover. 4. Since y = 2 when t = 0, it follows that C = 2. Moreover. since y = 5 when! = 5. we have 5 = Β§e5k and since y = 1 when! = 5, we have 1 = 225* and k=%ln]0’~*0.4605. Clintz .0386. Thus, v = Β£10460β€œ. Thus, v = zeβ€”onxhi 6. Using the fact that y = Γ©when t = 3 and y = 5 when t = 4. we have% = Ceβ€œ and 5 = Ceβ€œ. From these two equations we have 5 % = $6β€œ = 5eβ€œ. and e" = 10- Thus, k = In 10 and we have y = 02”" 1". Since 5 = Ceβ€˜β€œ" '0, it follows that S l C = e41n 10 : Therefore, y 2 1 e:ln 10 Q, ] €2.3026! ” 2000 2000 β€˜ dv 2 ' d3, . β€”e = β€”β€” . + = 5. i 8 dt 3y 10 dt 2" y = 20 when t = 0: y = 2050/” _v = 18 when! = 0: y = 18e5-2’ dv 2 β€”1Β°/3)i . 2 42/1): 2 d." s, q2 β€”'= -- - =β€”- - =β€”-v --=l . ~-’=5.21~'=5.2 dt 20( 3)e 3[20e ] 3) dt 8(5 2)e ( 8e ) y Exponential decay Exponential growth 12. From Example 1 we have 14. Since y = Ce"! = 3eDW/2V57'51', we have V = Celt! = Ce[ln(l/2)/1599]l t = β€˜000: y = 3e[ln(1/2)/5715)1000 z 166 grams 1.5 = Ce['β€œ(β€˜/2Vβ€˜599]'°°° = C ~ 2.314. t = 10.000: y = 3e[β€˜β€œβ€œ/2V-β€˜7β€˜5]β€œ"°°° *2 0.89 grams. The initial quantity is 2.314 grams. Whent = 10,000, y = 2.314e['β€œβ€˜β€œ21/599119000 z 0.03 grams. 16. Since y 2 Ce’" = Ce['"β€œ/2)/24β€œβ€˜β€œβ€™]’, we have 18. 0.9957C = Ceβ€˜β€œβ€œ/2V" 0.4 = Ce['""/2V24β€œ0Β°]'0~0m => C a 0.533. when t = 1 and h is the half-life The initial quantity is 0.533 grams. 0.9957 = emH/ZW = 1 1 When t 1000, β€œ09957) : n(2) V = 0β€˜533eflnl|/2)/24.l()0]10()0 m 0518- h ln(i) h : ln(0.9957) h = 160.85 years 20. 0.30C = Ce[ln1I/2)/5715]i ln(1/2) ’5715 = 5715 In 0.30 ln(l/2) In 0.30 = 1β€˜ z 9927 years 154 Chapter 4 Exponential and Logarithmic Functions 1 22. (0. 8). (20,-) 24. (a) Let t = 0 represent 1960. 2 VI : 8eky’ 'V = C6,β€œ = 2.3,β€œ 12 _ = 1140) = __ g %: 86:01:”l =21_01n(%)= 151016z 411386 12 2.3e => 40k 111(2'3) => k 0,0413 y = 2.3e"1β€œβ€˜β€œ3’ Exponential model VI ._. 86β€”().1386t 1970: y(10) = 2.3e0-04'3W z 3.5 million .Vz : 80-)” l : 8(2)20k1 => k = .Llo = β€˜ln 16 z _02 β€˜9803 YOU} ’~' 5.3 m1llion 2 2 20 g2 16 20 In 2 . 1900: y(30) z 7.9 million ."2 = 8(2)β€œ"2’ (b) 24 = 2_3e().04131 kl=(ln2)k2 _ 1 (3Β£>~ ’7 0.0413 1" 2.3 56'8 years (c) Increasing by 0.0413, or 4.13% each year. 26. Since A = 20.000e"~β€œβ€™5’. the time to double is given by the following. 40,000 = 20000120105! 28. Since A = 10,000e" and A = 20.000 when t = 5, we have the following. 20,000 = 10,000e5’ 1β€œ 2 = 0105' r = 1β€”"3β€”2- -~ 0.1386 = 13.86% In 2 I β€” 0.105 ~ 6601 years Amount after 10 years: Amount after 10 years: A = 10.0009”ln 2” 51"") = $40,000 A = 20.000e0"Β°5β€œΒ°) ~ $57,153.02 Amount after 25 years: Amount after 25 years: A = 10.000eflIn W512” = $320,000 A = 20,000e0-105β€˜25’ z $276,091.48 30. SinceA = 2000e" and A = 6008.33 whent = 25, we 32. (a) By equating A = P0 + i)’ and A = Pe" we have the have the following. following. 6008.33 = 2000e25’ 1n 3.004165 = 25r P(l + i)’ = Fe” (1 + i1r= (er)! r z 0.044 = 4.4% The time to double is given by 4000 = 2000:3004β€œ 1n 2 = 0.044t t z 15.753 years. Amount after 10 years: A = 2000eΒ°'Β°44('°’ z $3105.41 1 + i = e’ Therefore,i = e’ β€” 1. (b) lfr = 0.06. theni = e006 - 1 z 0.0618 or 6.18%. 36. 38. 42. 4 n=4: i=(l β€” l z0.077l4=7.7l4% .5” n = l2: 1': (I + β€” I % 0.07763 g 7.763% 0.075 = : '= +β€”β€”-β€”β€”- n 365 1 (l 365 Continuous: i = elm” β€” l 2 0.07788 *~v 7.788% 7.7l4% 7.763% 7.788% 365 > - I z 0.07788 a 7.788% Section 4.6 Exponential Growth and Decay 7.788% (a) For r = 10%, the approximate time necessary for the investment to double is % = 7 years. (b) For r = 7%, the approximate time necessary for the investment to double is Q = 10 years. (a) Let t = 0 correspond to 1990. Date points: (0, 150), (10. 1074) y = 150(1.21756)’ = lSOe""%35’ Exponential models ;:: 924t+β€”150 (c) 2006: y = 92,406) + 150 -~ $16284 million ((1) 2000 0 20 0 Answers will vary. β€˜ (a) Since N = 20 when t = 30, we have 20 = 30(1 β€” e30") _ ln(l/3) β€œ m N = 300 _ 9-0.03661). (b) 25 = 30(1 _ eβ€”(unsm) t = ln( 1 /6) ~0.0366 k a β€”0.0366 ~ 49 days Linear model (b) 2006: y = 150eβ€œ-β€œβ€™Β°"5"β€œβ€™ 3 $3499 million 40. S = 30(1 β€” 3’”) (a) Since S = 5 when t = l. we have 5=30(lβ€”e") e"=lβ€”% k=mg [Notez S is in thousands of units] Therefore, S = 30(1 β€” el'"β€˜5/")3’). (mummUβ€”wmmo=m I630 The saturation point for the market is 30,000 units. (c) When I = 5. we have S=3m1β€”AW%WW~11%4 Thus. S ~ 17,944 units. ((0 1 5 5 156 Chapter 4 Exponential and Logarithmic Functions 44. (21) Since p = Ceβ€œ where p = 45 when x = 1000 and p = 40 when x = 1200. we have the following. 45 : CeIOOOk and : CeIZUOk In 45 = In C + lOOOk ln40 = lnC +1200k In 45 β€” In 40 = β€”200k _ ln(45/40) ~ _ β€” _ 200 0.0005889 Therefore, we have 45 = Ce”’°°"Β°-Β°"05"89) which implies that C m 81.0915 and p = 8l.0915eβ€œ"-"°°5β€œ9*". k (b) Since R = xp = 81.0915xeβ€˜0'0"05839-", we have the following. R’ = 8l.09]5[__0.0005889xeβ€”0.0005889); + eβ€˜0.0005889.t] = 8l.0915eβ€œ)"mΒ°5389β€˜[1 β€” 0.0005889x] = 0 Since R’ = 0 when x = I/0.0005889 z l698 units. we have 17 = 81.0915eβ€œ0β€˜0Β°"5β€œ39β€œ698’ ~ $29.83. 46. A = Veβ€˜O-β€˜lβ€˜l' 48. (a) If 10 = 1, then we have : 075$ β€”0.04z 100.000e e R: In] andgj: lnl' : 100,000e(0.75./}β€”0.041) I 10 1" 10 0 75 Therefore, I = e331β€œ β€˜0 ~ 199.526.2315. A [m = 100,0β€œ)(7 _ 0.04)e(o.75/1β€”0.04n : 0 In I 2 ’ - (b) 2R = In 10 implies that 1 = eZR'n '0 = (eRβ€™β€œ '0)? 3β€”7/5. = .04 The intensity is squared if R is doubled. t dR l 1 l 0.75 (C) β€” = (β€”) = =β€”__=.5 d] 1101 1110 x/i (0.040(2) 937 n n t = 87.89 -Β« 88 The timber should be harvested in 2078 to maximize the present value. Review Exercises for Chapter 4 , 3 _ ~i/3 V3 3 2. 10V- : (JR) = 43 = 64 4. Q) = (%) = 3 6. (9V3)(3'/3) = (9 - 3W3 = 271/3 = 3 8. 56)” = 41(2)3 = % = 2 10. 32β€œ = 23-"+1 12. x5/2 = 243 = 35 (25)-.\β€˜ = 23x+l J; = ~5x = 3x + l x : 8x = -l l x = β€”Β§ 14. eβ€˜5 2 elβ€˜β€œ 16. 4x2 = e5 β€”5 = 2x + l , l i x- = 26* x = -3 e.5/2 x β€” :t 2 ...
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This note was uploaded on 05/25/2011 for the course MATH 16b taught by Professor Chuchel during the Winter '08 term at UC Davis.

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solnshw3 - a Section 4.5 2 y = lnx5/2 4 y = In x”2 5 l...

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