CHAPTER 5 Logarithmic, Eponential, and Other Transcendental Functions Section 5. The Natural Logarithmic Function: Differentiation.... 9 Section 5. The Natural Logarithmic Function: Integration...... 98 Section 5. Inverse Functions...................... 5 Section 5. Eponential Functions: Differentiation and Integration.. 59 Section 5.5 Bases Other than e and Applications............ 56 Section 5.6 Differential Equations: Growth and Deca......... 5 Section 5.7 Differential Equations: Separation of Variables...... 57 Section 5.8 Inverse Trigonometric Functions: Differentiation..... 55 Section 5.9 Inverse Trigonometric Functions: Integration....... 59 Section 5. Hperbolic Functions.................... 5 Review Eercises.............................. 58 Problem Solving.............................. 55
CHAPTER 5 Logarithmic, Eponential, and Other Transcendental Functions Section 5. Solutions to Even-Numbered Eercises The Natural Logarithmic Function: Differentiation. (a). The graphs are identical.. (a) ln 8..6 8. dt.6 t 6. (a) ln.6.58.6 dt.58 t 8. f ln. Reflection in the -ais Matches (d) f ln Reflection in the -ais and the -ais Matches (c). f ln. f ln 6. g ln Domain: > Domain: Domain: > 8. (a) (c) ln.5 ln ln ln.86. ln ln ln ln ln ln.779 ln ln ln.88 (d) ln 7 ln ln ln.765. ln z ln ln ln z. lna lna lna 6. ln e ln ln e ln 8. ln ln ln e e 9
9 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions. ln ln ln z ln ln ln z ln z. ln ln ln ln ln. ln ln ln ln ln 6. 8. lim ln6. lim ln ln 5.69 6 5 f = g 5. ln ln. ln ln At,,. At,,. 6. h ln 8. h ln d ln ln 5. ln ln 5. d f ln ln ln f 5. ht ln t t ht tt ln t t ln t t 56. lnln d ln ln 58. ln ln ln 6. f ln f
Section 5. The Natural Logarithmic Function: Differentiation 95 6. ln ln ln d Note that: Hence, d 6. 68. ln csc 66. 7. (a) csc cot csc cot ln sin ln sin 7. d sin cos sin cos sin sin ln,, 7. d When, d. Tangent line: ln sec tan d sec tan sec sec tan sec sec tan sec tan ln g l t dt sec g ln d d ln ln (Second Fundamental Theorem of Calculus) ln 5 ln ln 5 5 d d 5 d 5 5 8
96 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 76. ln ln ln ln ln 78. ln Domain: > when. 6 (, ) 6 > Relative minimum:, 8. ln Domain: e, e e, e ( ) ( ) > 6 ln ln Relative maimum: e, e Point of inflection: e, e ln when e. ln when e. 8. ln. Domain > 8. ln ln when f ln, f f ln, f f, f ln ln e P f f, P ln ln P f f f when e Relative minimum: e, 8e Point of inflection: e, e ( ) e /, e, P P, P P, P P, P ( e /, 8e ) 8 The values of f, P, P, and their first derivatives agree at. The values of the second derivatives of f and P agree at. f P P
Section 5. The Natural Logarithmic Function: Differentiation 97 86. Find such that ln. 88. f ln f n n f n f n n ln n n n n f n.6.79.69.9. Approimate root:.8 ln ln ln ln d d 9. ln ln ln d d 9. ln ln ln ln ln d d 6 6 6 9. The base of the natural logarithmic function is e. 96. g ln f, f > g f f (a) Yes. If the graph of g is increasing, then g >. Since f >, ou know that f gf and thus, f >. Therefore, the graph of f is increasing. No. Let f (positive and concave up). g ln is not concave up. 98. 5.5 t < 6.7968 ln, (a) 5 t67. ears T 67. $8,78. (c) t68.5 ears T 68.5 $8,6. (d) dt d 5.56.7968 ln 5.5 6.7968 ln When 67., dtd.65. When 68.5, dtd.585. (e) There are two obvious benefits to paing a higher monthl pament:. The term is lower. The total amount paid is lower.
98 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions. (a) 5 T p.96 p.955 p (c) T.75 deglbin T7.97 deglbin lim Tp p. ln ln ln (a) (c) lim d d When 5, d. When 9, d 99.. ln > for >. Since ln is increasing on its entire domain,, it is a strictl monotonic function and therefore, is one-to-one. 6. False is a constant. d d ln Section 5. The Natural Logarithmic Function: Integration u 5, du d. d d ln C. 5 d ln 5 C 6. d d 8. ln C u, du d d d ln C
Section 5. The Natural Logarithmic Function: Integration 99. u 9, du d 9 d 9 d 9 C. d 6 d u ln C. 8. 7 9 d d 6. 9 9 ln C d d. 6 5 ln d ln d d 5 9 5 5 d 5 9 5 ln 5 C. d ln C u du, d. d ln C d ln ln C d d d d d ln C 6. u, du d d u du d u u du u du u ln u C ln C ln C
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 8. u, du d d u du d u u u du u u u u du u u u du u u u ln u C ln ln C C. tan 5 d 5 5 sin 5 d cos 5. sec d sec d 5 ln cos 5 C ln sec tan C. u cot t, du csc t dt 6. csc t cot t dt ln cot t C sec t tan t dt lnsec t tan t lncos t C sec t tan t ln cos t C ln sec tsec t tan t C 8. d. 9 ln 9 C, : ln 9 C C ln 9 ln 9 ln 9 r sec t tan t dt ln tan t C, : ln C C r ln tan t (, ) 8 9 9 8 8 ( π, ) ln., d, (a) ln Hence, d ln C ln. ln C C
Section 5. The Natural Logarithmic Function: Integration 5. 8. d ln ln ln ln d d d ln ln 6. u ln, du d e e e ln d e ln d lnln e e ln 5... csc cot d.... csc csc cot cot d csc csc cot d cot csc 5. csc C ln csc C 5. 56. ln sin C ln ln... ln csc cot C ln csc cot csc cot csc cot C csc cot C ln csc cot C d 6 ln C ln C where C C 5. ln csc cot csc cot C 58. tan sec d ln sec tan sin C 6. sin cos cos d ln sec tan sin ln.66 Note: In Eercises 6 and 6, ou can use the Second Fundamental Theorem of Calculus or integrate the function. 6. F tan t dt 6. F tan F t dt F
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 66. 68. A 6 d d ln ln ln 8.55 A Matches (a) 9 7. tan. d ln cos. 6 ln cos. ln cos..7686 5 8 7. Substitution: u and Power Rule 7. Substitution: u tan and Log Rule 76. Answers will var. 78. Average value d ln d 8. ln ln Average value 6 ln sec tan 6 ln sec 6 d 6 ln ln ln ln.886 8. t ln T dt 5 ln lnt ln ln 5 5 ln ln ln.5 units of time 8. ds dt k t St k t dt k ln t S k ln C S k ln C Solving this sstem ields k ln and C. Thus, St ln t ln C k ln t C since t >. ln t ln.
Section 5. Inverse Functions 5 86. k : f k.5: f.5.5 8 f k.: f.. lim k f k ln 6 f.5 f. 6 8 88. False d d ln 9. False; the integrand has a nonremovable discontinuit at. Section 5. Inverse Functions. (a) f g f 8 f g f g f g 8 g. (a) f g f f g f g f g g 6. (a) f 6, g 6 f g f 6 6 6 6 6 g f g6 6 6 6 8 f 8 g 6 8. (a) f, g, < g f g f f g f g
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions. Matches. Matches (d). f 5 6. F One-to-one; has an inverse Not one-to-one; does not have an inverse 8. gt t Not one-to-one; does not have an inverse. f 5. One-to-one; has an inverse h Not one-to-one; does not have an inverse 9 9 9 6 9. f cos f sin when,,,... f is not strictl monotonic on,. Therefore, f does not have an inverse. 6. f 6 8. f for all. f is increasing on,. Therefore, f is strictl monotonic and has an inverse. f ln, > f > for >. f is increasing on,. Therefore, f is strictl monotonic and has an inverse.. f. f. f, f f f f f 5 5 5 f 5 f f f
Section 5. Inverse Functions 55 6. f, 8. f, 5 f f 5 f 5 5 86 5 86 f 5 86 6 f 8 6 f f The graphs of f and are reflections of each other across the line.. f 5. 5 5 f 5 f The graphs of f and f are reflections of each other across the line. f f f 6 6. f The graphs of f and f are reflections of each other across the line. f f 6 8 6 (, 6) (, ) (, ) 6 8 6. f k is one-to-one for all k. Since f, f k k k. 8. f on, 5. f > on, f is increasing on,. Therefore, f is strictl monotonic and has an inverse. f cot on, f csc < on, f is decreasing on,. Therefore, f is strictl monotonic and has an inverse. 5. f sec on, f sec tan > on, f is increasing on,. Therefore, f is strictl monotonic and has an inverse.
56 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 5. f on, 5 The graphs of f and f are f reflections of each other across the line. f ± ± f, < 56. (a), h h 58. (a), f f (c) h is not one-to-one and does not have an inverse. The inverse relation is not an inverse function. (c) Yes, f is one-to-one and has an inverse. The inverse relation is an inverse function. 6. f 6. Not one-to-one; does not have an inverse f a b f is one-to-one; has an inverse a b b a b a f b a, a 6. f 6 is one-to-one for. 66. is one-to-one for. f 6 6 6 6 f, f 6, 6 68. No, there could be two times t t for which ht ht. 7. Yes, the area function is increasing and hence one-to-one. The inverse function gives the radius r corresponding to the area A. 7. f 7 5 ; f 5 a. 7 f 7 5 6 f f f f 7 5 6 7
Section 5. Inverse Functions 57 7. f cos, f a f sin f f f f sin which is undefined. 76. f, f 8 a f f f f f8 8 78. (a) Domain f Domain f, Range f Range f, (c) f f (d) f,, f f f,, 5 5 f f 8. (a) Domain f,, Domain f, Range f,, Range f, (c) f f (d) f f 8, f f f, f 8. ln d d. At,, d 6 6 6. In Eercises 8 and 86, use the following. f 8 and g f 8 and g 8. g f g f g 86. g g g g g 9
58 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions In Eercises 88 and 9, use the following. f and g 5 f and g 5 88. f g f g 9. f 5 5 g f g f g 5 Hence, g f Note: g f f g 9. The graphs of f and f are mirror images with respect 9. Theorem 5.9: Let f be differentiable on an interval I. to the line. If f has an inverse g, then g is differentiable at an for which fg. Moreover, g fg, fg 96. f is not one-to-one because different -values ield the same -value. Eample: f f 5 98. If f has an inverse, then f and f are both one-to-one. Let f then f and f. Thus, f f. Not continuous at ±.. If f has an inverse and f then f f f f, f. Therefore, f is one-to-one. If f is one-toone, then for ever value b in the range, there corresponds eactl one value a in the domain. Define g such that the domain of g equals the range of f and gb a. B the refleive propert of inverses, g f.. True; if f has a -intercept.. False Let f or g. 6. From Theorem 5.9, we have: g fg g fg f gg fg fg fg fg f g fg f > < If f is increasing and concave down, then and f which implies that g is increasing and concave up.
Section 5. Eponential Functions: Differentiation and Integration 59 Section 5. Eponential Functions: Differentiation and Integration. e.5.... ln.5.69... 6. e ln ln.5... e.69... 6 8.. e 8. e 8 ln 8. e 5. e 5 ln ln.68 6 e 8 ln e e e ln.5 ±e 5 ±8. 6. ln 8. e e e.68 ln e e 6 e 6 5.9. e. e. (a) 8 8 Horizontal asmptotes: and 8 Horizontal asmptote:
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 6. Ce a 8. Horizontal asmptote: Reflection in the -ais Matches (d) C lim a C e lim C e a Horizontal asmptotes: Matches C a e C and. f e. f e. In the same wa, g ln ln g ln lim r e r for r >. 8 6 f g f 6 8 g 6. 6 7.78596 5 e.78888 e > 6 7 5 8. (a) e e e e At,,. At,,.. f e. f e e. d e e d e e e 6. gt e t 8. gt e t 6t 6 t e t ln e e 5. ln e ln e d e e e e e e ln e e lne e ln d e e e e e e 5. e e d e e 5. e e e d e e e
( Section 5. Eponential Functions: Differentiation and Integration 5 56. f e ln 58. f e ln e d 6. e 6. d e d d e e d e e g e ln g e e ln g e e e e ln e e ln 6. e cos sin e 6 sin 8 cos e cos sin e sin 5 cos 5e sin cos 5e cos sin 5e sin cos 5e 5 cos 5e cos 5e cos 5e sin cos 5e cos sin 5 Therefore, 5 5. 66. f e e f e e > (, ) f e e Point of inflection:, when. 68. g g e g e Relative maimum: Points of inflection:, e,,.99 e,, e,.,,..8 (, (, ( π e.5, π ( ( e.5 π 6 7. f e f e e e when. f e e e when. Relative maimum:, e Point of inflection:, e (, e ) (, e )
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 7. f e f e e e 6 when 5. f e 6 e 6 e 8 when. Relative maimum: 5, 96.9 Point of inflection:, 7.798 5 (, 96.9 ) ( ), 7.798.5.5 7. (a) f c f c ce c c e c c c c e e c ce c c e c (d) A f c e ee c e e ee ce c ce c (c) A 6 c e e ee (.8,.59) lim c lim c 9 The maimum area is.59 for.8 and f.57. 76. Let be the desired point on e,. e e Slope of tangent line e Slope of normal line (.6, e.6 ) e e We want, to satisf the equation: e e e e Solving b Newton s Method or using a computer, the solution is.6..6, e.6
Section 5. Eponential Functions: Differentiation and Integration 5 78. V 5,e.686t, t (a), dv dt 99e.686t (c), When t, dv 58.8. dt When t 5, dv 6.89. dt 8..56e.t cos.9t.5 ( inches equals one-fourth foot.) Using a graphing utilit or Newton s Method, we have t 7.79 seconds. 8. (a) V 686.79t,8.79 V 9.5t.t 8,.6 The slope represents the rate of decrease in value of the car., (c) (e) V,5.77.859 t,5.77e.58t dv dt For 77.e.58t t 5, dv dt For t 9, dv dt 5 dollarsear. 8 dollarsear. (d) Horizontal asmptote: lim V t t As t, the value of the car approaches. 8. f e, f f e, f f e e e, f P, P P, P P, P P, P P, P P f P The values of f, P, P and their first derivatives agree at. The values of the second derivatives of f and P agree at.
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 86. n th term is n n! in polnomial:!!! Conjecture: e!!... 9. e d e e e 9. e d e d C e 88. Let u, du d. e d e C 9. e d e d e C 96. Let u e, du e d. e e d e e d ln e C 98. Let u, du d. e d e d e e e e. Let u e e, du e e d. e e e e d lne e C. e e d e e e d 6. e e C. Let u e e, du e e d. e e e e d e e e e d e e C esec sec tan d esec C u sec, du sec tan 8. lne d d. C e e d e e d e e C. f sin e d cos e C f C C f cos e f cos e d sin e C f C C f sin e
Section 5. Eponential Functions: Differentiation and Integration 55. (a) d,, e. e. d.e..d. e. C.5e. C, :.5e C.5 C C.5e. 6. b e d a e b a ea e b 8. e d e e.9 a b. (a) e d, n...t dt Midpoint Rule: 9.898 Trapezoidal Rule: 9.87 e.t Simpson s Rule: 9.785 e. Graphing Utilit: 9.77 e d, n e.. ln ln Midpoint Rule:.96 Trapezoidal Rule:.87 ln. minutes. Simpson s Rule:.88 Graphing Utilit:.8799. t R 5 8 7 6 ln R 6.5 5.8.77.6.58 (a) ln R.655t 6.69 R e.655t6.69 8.78e.655t 5 5 (c) Rt dt 8.78e.655t dt 67. liters
56 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 6. The graphs of f ln and g e are mirror images across the line. 8. (a) Log Rule: u e Substitution: u. e ln a e ln b ea ln e b a b ln e ab a b Therefore, ln ea ln and since is one-to-one, we have ea eab ln e b eab. eb Section 5.5 Bases Other than e and Applications. t8. t5 6. log 7 9 log 7 7 At t 6, 68 At t, 5.7579 8. log a a log a log a a. (a) 7 9. (a) log 9 log 7 9 6 8 9 9 7 log 9 7 log 6 8. 6. 8. ± ± 9 9 6 6 9 8 8 6 6. (a) log 8 8 log 6 6 6 6. (a) log b 7 b 7 b log b 5 b 5 b 5
Section 5.5 Bases Other than e and Applications 57. (a) log log log OR is the onl solution since the domain of the logarithmic function is the set of all positive real numbers. log log log 9 6. 5 6 8 8. 6 ln 5 ln 8 ln 8 6 ln 5.95 5 86 5 86 ln 5 ln 86 ln 86 ln 5 ln 86.85 ln 5.. 65 65t. 65t ln. ln 65 t 65 ln ln. 65 6.9 log t.6 t.6 t.6.7. log 5. 5. 5. 5 6. 5 6. 9,78.59 6. f t.75 t 75. 8. Zero: t g log Zeros:.86,.86 5 (, ) (.86, ) 5 5 (.86, ) 5
58 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions. f g log f 9 9 9 6 f 9 9 g g 6 9. g. g ln 6 6. d ln 66 6 6 ln 6 6 ln 6 f t t t ft t ln t t t t t ln t 8. g 5 sin g 5 cos ln 5 5 sin 5. log log log d ln ln 5. h log log log log h ln ln ln ln 5. log log log d ln ln ln ln 56. f t t log t t lnt ln ft ln t t t lnt 58. ln ln d ln d ln ln 6. ln ln d ln d ln 6. 5 d 5 ln 5 C 6. 5 d d 7 5 d ln
Section 5.5 Bases Other than e and Applications 59 66. 7 d 7 d 68. sin cos d, u sin, du cos d ln 7 7 C ln sin C 7. (a) 6 d esin cos, e sin cos d e sin C, : e sin C C C e sin 7. log b ln ln b log log b 7. f log (a) Domain: > (d) If f <, then < <. log (e) f log log log f must have been increased b a factor of. (c) log log log, log If,, then f. (f) log log log n n n Thus, n n. 76. f a (a) f u v a uv a u a v f u f v f a a f 78. Vt, (a) V, 6,, 8,, t dv dt, ln t When t : When t : dv 5. dt dv 8.9 dt (c) 5 V ( ) 6 6 8 t 6 V, $,5 Horizontal asmptote: v As the car ages, it is worth less each ear and depreciates less each ear, but the value of the car will never reach $.
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 8. P $5, r 6%.6, t A 5.6 n n A 5e.6 8.9 n 65 Continuous A 87.8 855.9 86.66 875.5 899.7 8.9 8. P $5, r 7%.7, t 5 A 5.7 n 5n A 5e.75 n 65 Continuous A 7,7.6 7,9.6 8,.78 8,67.9 8,768.9 8,77. 8., Pe.6t P,e.6t t 5 P 9,76.5 5,88.6,9. 6,59.89 97.8 978.7 86., P.7 65 65t P,.7 65 65t t 5 P 9,. 9,66.86,66.,8. 68.6.75 88. Let P $, t. (a) A e.t A 8. A e.5t A 7.8.6t A = e.5t A = e.t A = e 9. (a) lim n.86 e.5n.86 or 86% P.86.5e.5n P.69 P.6 e.5n.5e.5n e.5n (c) A e.6t A. 9. (a) Limiting size:, fish (c), (d) pt, 9e t5 pt 8, 5 9e t5 t ln 9 5 t 5 ln 9.7 e t5 9e t5 9e t5 pt e t5 9e t5 9 5, 8,et5 9e t5 p.5 fishmonth p. fishmonth
Section 5.5 Bases Other than e and Applications 5 9. (a) 6.56 97.557.75 7.88 ln 99.557.56.875.7 5 (c) The slope of 6.56 is the annual rate of change in the amount given to philanthrop. (d) For 996, 6 and 6.65,.. 6.56,.969, is increasing at the greatest rate in 996. 96. 8 seems best. A d ln 6 98. ln.666 6.59.75.78.78. t 6 6 66.5 69.58 79. Ck t When t, 6 C 6. 6k t 6 66.5.5, 6 6 Let k.5. 6.5 t 69.58 79..5,.5,.5 66.5 69.58. True. f e n f e n ln e n ln e n n n. True. d n dn Ce for n,,,... 6. True. f ge g since e > for all.
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 8. sin ln ln sin sin ln At sin sin sin,, sin cos ln cos ln sin cos Tangent line: ln Section 5.6 Differential Equations: Growth and Deca.. d d C 6. d d d ln C e C Ce Ce d d C 9 C 8.. d d d ln C e C e C e Ce d d d ln C ln C e C e C e Ce
Section 5.6 Differential Equations: Growth and Deca 5. dp k t. dt dp dt dt k t dt dp k t C P k t C kl d k L d L d d k d k L C lnl k C L e k C L e C e k L Ce k 6. (a) (, ) d,, ln d C e C C e, : C e C e 8. dt t,, 5. (, ) t dt dt,, dt (, ) 5 5 t C 5 ln t C 5 C C t e tc e C e t Ce t Ce C e t. dn dt kn. dp kp dt N Ce kt (Theorem 5.6) P Ce kt (Theorem 5.6), 5: C 5, : 5e k k ln 5 ln 8 5 When t, N 5e ln85 5e ln85 5 8 5 89 5, 5: C 5, 75: 75 5e k k ln 9 When t 5, P 5e ln95 5 9 5 868.95
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 6. Ce kt,,, 5, 8. C e kt e5k k ln8 5 e.59t.59 Ce kt,,,, 5 Cek 5 Ce k Ce k 5 Cek e k e k e k k ln.6 Ce.6t 5 Ce.6 C.5.5e.6t. k. dt d when >. Quadrants I and II. d >. Since Ce ln6t, we have.5 Ce ln6 C. which implies that the initial quantit is. grams. When t,, we have.e ln6,. gram. 6. Since Ce ln57t, we have. Ce ln57, C 6.7 which implies that the initial quantit is 6.7 grams. When t, we have 6.7e ln57 5.9 grams. 8. Since Ce ln57t, we have. Ce ln57 C.6. Initial quantit:.6 grams. When t,, we have.8 grams.. Since Ce ln,6t, we have. Ce ln,6, C.5 which implies that the initial quantit is.5 gram. When t, we have.5e ln,6.5 gram.. Since d k, Cekt or e kt.. Since A,e.55t, the time to double is given b,,e.55t and we have e57k e.55t k ln 57 ln 57t.5 e ln t ln.5 57 57 ln.5 t 5,68.8 ears. ln ln.55t t ln.6 ears..55 Amount after ears: A,e.55 $,665.6
Section 5.6 Differential Equations: Growth and Deca 55 6. Since A,e rt and A, when t 5, we 8. Since A e rt and A 56.56 when t, we have the following. have the following.,,e 5r r ln 5.86.86% Amount after ears: A,e ln 5 $, 56.56 e r r ln56.56 The time to double is given b. % e.t t ln 6.9 ears.. 5. 5, P.6 5. 5, P.9 5 P 5,.5 8 $5,6. P 5,.9 $5,.9 5. (a).6 t (c).6 t ln t ln.6 t ln.9 ears ln.6.6 t.6 t ln t ln.6 t ln ln.6.58 ears (d).6 65 65t.6 65 65t ln 65t ln.6 65 t ln 65 ln.6.55 ears 65 e.6t e.6t ln.6t t ln.55 ears.6 56. (a).55 t (c).55 t ln t ln.55 t ln.95 ears ln.55.55 t.55 t ln t ln.55 t ln ln.55.6 ears (d).55 65 65t.55 65 65t ln 65t ln.55 65 t 65 e.55t e.55t ln.55t ln ln.55.6 ears 65 t ln.6 ears.55
56 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 58. P Ce kt Ce.t 6. P.6 Ce. C.965 P.965e.t P Ce kt Ce.t P.6 Ce. C.5856 P.5856e.t P 6. or 6,, people in P.5 or,5, people in 6. (a) N.596.55 t 6. Ce kt,, 7,,, 6, N when t 6. hours (graphing utilit) Analticall,.596.55 t.55 t.996.596 t ln.55 ln.996 C 7, 6, 7,e k k ln67 7,e.8t When t, $58,7..8 66. (a) e k t k ln N e.66t ln.996 6. hours. ln.55 e k ln.66 5 e.66t e.66t 6 t ln 6 9 das.66 68. S 5 e kt (a) 5 e k e k 5 ek 5 k ln.7 5 5, units lim S t 5 (c) When t 5, S.55 which is,55 units. (d) 5 8 7. (a) R 979.99.69 t 979.99e.67t I.85t.77t 9.9755t.85t 66.9 Rate of growth Rt 65.7e.67t (c) 5 (d) Pt I R
Section 5.7 Differential Equations: Separation of Variables 57 7. I 9 log log 6 I 6 7. 6.7 log I I 6.7 I 8 log log 6 I 6 8 log I I 8 Percentage decrease: 6.7 8 6.7 95% Since k 8 dt 8 k dt ln 8 kt C. When t, 5. Thus, C ln. When t,. Thus, k ln ln 8 Thus, e lnt 8. When t 5, 79.. k ln ln ln. 76. True 78. True Section 5.7 Differential Equations: Separation of Variables. Differential equation: Solution: C Check: C C C. Differential equation: Solution: C e cos C e sin Check: C C e sin C C e cos C e sin C e cos C e sin C e cos C C e sin C C e cos C e cos C e sin C C C C e sin C C C C e cos 6. e e e e e e e e e e Substituting, e.
58 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions In Eercises 8, the differential equation is 6. 8. cos. 5 ln 8 cos 6 8 cos 8 cos, Yes. 6 8 ln, No.. e sin 8e 6 sin 6 e 6 sin 6e sin, Yes In 8, the differential equation is e.. e, e e e 6. sin, cos e e e, Yes. cos sin e, No. 8. e 5, e e e e e 5 e, Yes.. A sin t. C passes through, d dt A sin t Since d dt 6, we have A sin t 6A sin t. Thus, 6 and ±. 9 6 C C Particular solution:. Differential equation: General solution: Particular solutions: C C, Point C, C, Circles 6. Differential equation: General solution: C 6 Initial condition: : 8 C Particular solution:
Section 5.7 Differential Equations: Separation of Variables 59 8. Differential equation: Initial conditions:, General solution: C C ln C, C C C C C ln C C C, C ln Particular solution: ln ln ln. Differential equation: 9 General solution: e C C e C C C e e C C C e C C C e C e C C C 9 9 Initial conditions:, e C C C C e C C C e C C C e C C e C C Particular solution: e. d. d e e d C e e d ln e C 6. d cos 8. d tan sec cos d sin C sec d tan C u, du d. 5. Let u 5, u 5, d u du d 5 d 5 u uu du u u du u u5 5 C 5 5 5 5 C
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions. 5e d 5e d 5e d e C. d d C 6. dr ds.5s 8. dr.5s ds r.5s C 5. d ln ln ln C ln C C d 6 cos 6 cos d 6 sin C sin C 5. 9 5 5. d 5 9 d 5 9 C e d e d e C 56. 58. d C C Initial condition:, 8 9 C Particular solution: 9 ln ln d ln d : C ln ln C 6. d : d 6. C C C dr ers ds e r s dr e ds e r es C r : C C e r es e r es r ln es ln es r ln e s
Section 5.7 Differential Equations: Separation of Variables 5 6. dt kt 7 dt 66. dt T 7 k dt lnt 7 k t C Initial condition: T 7 Ce kt T ; 7 7 Ce C Particular solution: T 7 7e kt, T 7 e kt d d ln ln ln C C Initial condition: 8, C8, C 8 Particular solution: 8, 68. m d d 7. f, f t, t t t t Not homogeneous ln ln C ln ln C ln C C 7. f, f t, t 7. t t t t t t t Homogeneous of degree f, tan f t, t tant t tant Not homogeneous 76. f, tan f t, t tan t t tan Homogeneous of degree 78. d v, dv v d v dv v d v d v dv v d d v d v dv d 8. v dv d v ln C ln C ln C v dv d v v v d dv v d v v v dv d lnv ln ln C ln C v C C C, v
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 8. v dv v v d dv d v dv v d ln v ln ln C ln C v C C, v 8. C C d, v v d vv d dv v dv v d v ln v ln ln C ln C C ev v C v ln C v e Ce Initial condition:, Ce C e Particular solution: e 86. d 88. Let v, dv v d. v d v dv v d v d v dv v d v dv d v v dv d ln ln v C ln ln v ln C C v C C C d C C : C C
Section 5.7 Differential Equations: Separation of Variables 5 9..5 d 8.5 d.5 d d ln 8 C e C 8 Ce 8 Ce 8 9., 9. d., 9 d 9 5 5 5 5 96. dt k, Cekt 98. k d Initial conditions:, 6 The direction field satisfies d along : Ce C Matches. 6 e k k ln 5 Particular solution: e t ln5 When 75% has been changed: 5 e t ln5 et ln5 t ln 6. hr ln5. d k. From Eercise, The direction field satisfies d along, and grows more positive as increases. Matches (d). w Ce kt, k w Ce t w w C C w w w e t
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions. Let the radio receiver be located at,. The tangent line to joins, and,. Transmitter (, ) = Radio (a) If, is the point of tangenc on the, then (c) 5 6 Then 6.5 ± 8 5 6 6 6.55 6 Now let the transmitter be located at, h. h Then, h h h h h h h h h h h ± h h h h h h h h h h h h There is a vertical asmptote at h, which is the height of the mountain. 6. Given famil (hperbolas): Orthogonal trajector: C d ln ln ln k k k 8. Given famil (parabolas): Orthogonal trajector (ellipse): C C C d K K 6 6
Section 5.8 Inverse Trigonometric Functions: Differentiation 55. Given famil (eponential functions): Ce Orthogonal trajector (parabolas): Ce d 6 6 K K. The number of initial conditions matches the number of constants in the general solution.. Two families of curves are mutuall orthogonal if each curve in the first famil intersects each curve in the second famil at right angles. 6. True d 8. True C d C K K d K K K C C C Section 5.8 Inverse Trigonometric Functions: Differentiation. arccos (a).8.6.....6.8..99..98.77.57.69.59.97.6 (c) (d) Intercepts:, and, π No smmetr.,,,, 6 6. arcsin,, 6
56 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 8. arccos. arc cot 5 6. arccos 5 6. arcsin.9. 6. arctan.5 8. (a) tan arccos tan θ θ cos arcsin 5 5. (a) θ 5 sec arctan 5 5 tan arcsin 5 6 5 θ 6 5. secarc tan. cosarccot arctan sec 6 + 6 θ arccot cos θ + 6. secarcsin 8. arcsin sec θ cos arcsin h r arcsin h r cos r h r r θ r ( h) h. 5. arctan 5 f = g 5 tan tan 5.7 5 Asmptote: arccos cos θ tan
Section 5.8 Inverse Trigonometric Functions: Differentiation 57. arccos arcsec cosarcsec ± θ 6. (a) arcsin arcsin, arccos arccos, Let arcsin. Then,. sin sin sin. Thus, arcsin arcsin. Therefore, arcsin arcsin. Let arccos. Then,. cos cos cos. arccos arccos. Thus, Therefore, arccos arccos. 8. f arctan tan Domain:, π π. f arccos cos cos (, π) π Range:, 6 6 f is the graph of arctan shifted units upward. Domain:, Range:, (, ) 6 6. f t arcsin t. t ft t f arcsec f 6. f arctan 8. f h arctan h arctan 5. f arcsin arccos f 5. lnt arctan t t t t t t t t t 5. arcsin 56. arctan ln d arctan 8 arctan
58 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 58. 5 arcsin 5 5 5 5 5 5 5 5 5 5 5 6. arctan 8 6. f arctan, a f P () π f π f P () P f f P f f f 6. f arcsin 66. f ±. f f > when or Relative minimum: f <,.68 Relative maimum:,.68 f arcsin arctan f 6 ±.68 B the First Derivative Test,.68,.7 is a relative maimum and.68,.7 is a relative minimum. 68. arctan. is not in the range of arctan. 7. The derivatives are algebraic. See Theorem 5.8. 7. (a) cot d dt arccot d 9 dt 7. cos d dt 75 s arccos 75 d ds ds dt s 75s 75 ds s dt θ s 75 If, d. radhr. dt 75 ds ss 75 dt If, d 66.667 radhr. dt A lower altitude results in a greater rate of change of.
Section 5.9 Inverse Trigonometric Functions: Integration 59 76. (a) Let arcsin u. Then Let arctan u. Then sin u cos u u d cos u u. (c) Let arcsec u. Then sec u sec tan u d d u sec tan u u u. Note: The absolute value sign in the formula for the derivative of arcsec u is necessar because the inverse secant function has a positive slope at ever value in its domain. (e) Let arccot u. Then csc u d u u d csc u. u + u u cot u u u u tan u sec u d u u. d sec u (d) Let arccos u. Then cos u sin u d (f) Let arccsc u. Then u u d sin u. csc cot u d + u u csc u d u csc cot u u u u u u u Note: The absolute value sign in the formula for the derivative of arccsc u is necessar because the inverse cosecant function has a negative slope at ever value in its domain. 78. f sin g arcsinsin (a) The range of arcsin is Maimum:. g f Minimum: 8. False The range of arcsin is,. 8. False arcsin arccos Section 5.9 Inverse Trigonometric Functions: Integration. d d arcsin C. d arcsin 6 6. 9 d 9 d arctan C 8.. d arctan C. 9 d arctan 6 d d C
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions. Let u t, du t dt. t t 6 dt t t dt 8 arctan t C 6. Let u, du d. d d arcsec C 8. Let u arccos, du d. arccos d arccos d. d arccos cos. sin d arctansin.95. Let u, du d. d d d arctan 6. 8 8. arcsin C d d d. d d d ln 5 arctan C 5. d d 7 d d. arctan arctan 9 d ln ln ln d. u, du d, d u du u du u u du arctan u C u 7 arctan C arctan C 6. d d d arcsin C 8. Let u, du d. d d C. d d arcsec C
Section 5.9 Inverse Trigonometric Functions: Integration 5. Let u, du d. 9 8 d 5 d arcsin 5 C. Let u, u, u du d d u u du u 6 6 u u du du 6 u du 6 arctan u C arctan C 6. The term is 9 9 9 9 : 8. (a) e d cannot be evaluated using the basic integration rules. e d e C, u (c) e d e C, u 5. (a) cannot be evaluated using the basic integration rules. d d d arctan C, u (c) d d ln C, u 5. (a) 5. d 6,.5.5, d 6, d 6, : arcsin C arcsin arcsin 6 sin 6 sin C C 6 6
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 56. A d arcsin 6 58. arcsin d.57 π π 6. F (a) F represents the average value of f over the interval,. Maimum at, since the graph is greatest on,. F arctan t F t dt arctan arctan when. 5 5 5 6. 6 d (a) 6 9 6 9 9 6 d d 9 arcsin C (c) 7 u, u, u du d 6u u u du 6 u du arcsin u 6 C arcsin 6 C The antiderivatives differ b a constant,. Domain:, 6 6. Let f arctan f for >. > Since f and f is increasing for >, arctan for >. Thus, > arctan >. Let g arctan 5 6 8 g for >. > Since g and g is increasing for >, arctan > for >. Thus, > arctan. Therefore, < arctan <.
Section 5. Hperbolic Functions 5 Section 5. Hperbolic Functions. (a) cosh e e sech e e.68. (a) sinh tanh 6. (a) csch ln 5.8 coth ln.7 8. cosh e e e e e e cosh. sinh cosh e e e e e e sinh. cosh cosh e e e e e e e e e e e e cosh cosh. tanh sech sech cosh coth sech Putting these in order: sinh cosh tanh csch sech coth sinh tanh cosh csch 6. coth 8. csch g lncosh g sinh tanh cosh. cosh sinh. sinh cosh cosh sinh ht t coth t ht csch t coth t
5 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions. g sech 6. g sech sech tanh 6 sech tanh f e sinh 8. sech f cosh e sinh sech tanh. f sinh cosh 6 f cosh sinh sinh cosh f for. B the First Derivative Test,, cosh,.5 is a relative minimum. 6 (,.5) 6. h tanh. (.88,.5) (.88,.5) a cosh a sinh a cosh Therefore,. Relative maimum:.88,.5 Relative minimum:.88,.5 6. f cosh f cosh f sinh f sinh f cosh f cosh P f f P f P P 8. (a) 8 5 cosh, 5 5 At ±5, 8 5 cosh 56.577. 5 At, 8 5. 8 sinh (c) At 5, sinh.75 5. 5 5. Let cosh u, du d. d cosh d sinh C. Let u cosh, du sinh d. sinh sinh d sinh d cosh sech C cosh C. Let u, du d. sech d sech d 6. Let u sech, du sech tanh d. sech tanh d sech sech tanh d tanh C sech C
Section 5. Hperbolic Functions 55 cosh 8. cosh d d sinh C 5. 5 d arcsin 5 arcsin 5 sinh C 5. d d ln C 5. Let u sinh, du cosh d. 56. cosh 9 sinh d arcsin sinh C arcsin e e 6 C tanh 58. sech cos, < < cos cos sin sin cos sin sec, cos since sin for < <. 6. csch csch csch 6. tanh ln tanh ln 6. See page, Theorem 5.. tanh tanh 66. Equation of tangent line through P, : a sech a a a Q When, a sech a a a a sech a. a P ( a, ) Hence, Q is the point, a sech a. L Distance from P to Q: d a a 68. 9 d ln C 9 d 6 ln C
56 Chapter 5 Logarithmic, Eponential, and Other Transcendental Functions 7. Let u, du d. 7. 7. d d sinh C ln C d 8 d ln C 8 d 6 d d 6 ln C 76. Let u, du d. 78. 8. 8 d d ln 8 d d d ln ln A tanh d e e e e d e e e e d lne e lne e ln ln e e C ln ln.65 8. C 5 A 6 d 6 ln 5 6 ln5 6 ln 5 6 ln 5.66 5 C 8. (a) vt t st vt dt t dt 6t C s 6 C C st 6t CONTINUED
Section 5. Hperbolic Functions 57 8. CONTINUED (c) dv kv dt dv kv dt dv kv dt Let u k v, then du k dv. k ln k v t C k v Since v, C. ln k v k v k t k v k v ek t k v e k t k v vk k e k t e k t v ek t ke k t ek t e k t k ek t e k t e k t e k t tanhk t k (d) lim t k tanhk t k. The velocit is bounded b k. (e) Since tanhct dt c ln cosh ct (which can be verified b differentiation), then st tanhk t dt k k When t, s C lncoshk t C. k When k., k lncoshk t. s t lncosh. t s t 6t. lncoshk t C k s t when t 5 seconds. s t when t 8. seconds When air resistance is not neglected, it takes approimatel. more seconds to reach the ground. 86. Let arcsintanh. Then, sin tanh e e e e tan e e sinh. and Thus, arctansinh. Therefore, arctansinh arcsintanh. e + e e e 88. 9. sech tanh cosh sech sech sinh sinh sech tanh cosh sinh sech sech