EVEN05 - CHAPTER 5 Logarithmic, Exponential, and Other...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 5 Logarithmic, Exponential, and Other Transcendental Functions Section 5.1 Section 5.2 Section 5.3 Section 5.4 Section 5.5 Section 5.6 Section 5.7 Section 5.8 Section 5.9 The Natural Logarithmic Function: Differentiation . . . . 493 The Natural Logarithmic Function: Integration . . . . . . 498 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . 503 Exponential Functions: Differentiation and Integration . . 509 Bases Other than e and Applications . . . . . . . . . . . . 516 Differential Equations: Growth and Decay . . . . . . . . . 522 Differential Equations: Separation of Variables . . . . . . 527 Inverse Trigonometric Functions: Differentiation . . . . . 535 Inverse Trigonometric Functions: Integration . . . . . . . 539 Section 5.10 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . 543 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554 CHAPTER 5 Logarithmic, Exponential, and Other Transcendental Functions Section 5.1 The Natural Logarithmic Function: Differentiation Solutions to Even-Numbered Exercises 2. (a) 3 0.2 2 1 −4 x 1 −1 2 3 4 4 y (b) 3 The graphs are identical. 4. (a) ln 8.3 8.3 2.1163 2.1163 6. (a) ln 0.6 0.6 0.5108 0.5108 (b) 1 1 dt t ln x (b) 1 1 dt t ln x 8. f x 10. f x Reflection in the x-axis Matches (d) 12. f x 2 ln x 14. f x ln x 0 Reflection in the y-axis and the x-axis Matches (c) 16. g x 2 ln x Domain: x > 0 y Domain: x y 3 2 Domain: x > 0 y 2 1 3 1 x 1 −1 −2 −2 −3 2 3 4 −3 −2 −1 1 2 3 x 2 1 x 1 2 3 18. (a) ln 0.25 (b) ln 24 (c) ln (d) ln 22. ln xyz 3 ln 1 4 3 ln 2 1 3 ln 1 ln 3 2 ln 2 3.1779 ln 3 1.3862 20. ln 23 ln 23 2 3 2 ln 2 12 2 ln 2 0.8283 4.2765 24. ln a 1 e 1 ln a 1 12 1 72 ln 1 ln x ln y 3 ln 2 ln z 2 ln 3 1 2 ln a 1 26. ln 3e2 ln 3 2 ln e 2 ln 3 28. ln ln 1 ln e 1 493 494 Chapter 5 2 ln y Logarithmic, Exponential, and Other Transcendental Functions 4 ln z ln x3 ln x 3y 2 z4 x 1x x x2 1 x2 x2 1 1 3 2 30. 3 ln x ln y 2 ln z4 32. 2 ln x ln x 1 ln x 1 2 ln x 1 ln 34. 3 ln x2 2 1 ln x 1 ln x 1 3 x2 1 ln 2 x 1x 1 ln 36. 3 38. lim ln 6 x →6 x 40. lim ln x →5 x x 4 ln 5 1.6094 f=g −1 −1 5 42. y y ln x3 3 2x 2 3 ln x 2 44. y y ln x1 1 2x 2 1 ln x 2 At 1, 0 , y 46. h x hx ln 2x2 1 2x2 3 2. At 1, 0 , y 1 48. 4x 2x2 1 1 ln x2 2 x x2 4 56. ln t 1 ln t t2 y dy dx x ln x x 1 x 1 2. 1 4x ln x 1 ln x 50. y dy dx ln x2 1 2x 2 x2 4 ln t t t1 t t2 x x 4 4 52. f x fx ln 1 x 2x x x 3 1 3 ln 2x 3 xx 3 ln x 3 54. h t ht y dy dx ln ln x 1x ln x 1 x ln x 58. y y ln 3 1 1 1 x 1 ln x 3 1 1 ln x 1 2 3x 2 60. f x fx 1 ln x x 1 4 x2 1 4 4 x2 x2 1 x 4 x2 1 1 3x 1 12 3 x2 1 Section 5.1 x2 2x2 2x2 x 4 1 2 ln 4 4 4x 4 x2 x 4x x2 4 x2 2x2 1 42 1 x2 4 The Natural Logarithmic Function: Differentiation 1 ln 2 4 x x2 1 ln x 4 495 62. y dy dx x2 4 1 4x x2 4 4 4 Note that: 1 x2 Hence, 1 x2 1 2x x 1 x2 4x x 2 2 2 4 dy dx 2 4 2 2 x2 x2 4 4 12 4 4 1 4x 2 x2 x2 x2 x2 4 1 4x 4 4 12 2 2x x2 4 4 x2 4 x3 x2 4 x2 4 x3 x x2 4 1 4x x2 4 x3 x2 4 x3 64. y y ln csc x csc x cot x csc x cot x 66. y dy dx ln sec x tan x sec x tan x sec2 x sec x tan x sec x sec x tan x sec x tan x sec x 68. y dy dx ln 1 sin2 x 1 ln 1 2 ln x sin2 x 70. g x l t2 3 dt d ln x dx ln x 2 1 2 sin x cos x 2 1 sin2 x sin x cos x 1 sin2 x gx ln x 2 3 3 x (Second Fundamental Theorem of Calculus) 1 x 2 72. (a) y dy dx 4 2x 2x x2 ln 1, 1 12 0, 4 74. ln x 1 x ln xy ln y 1 dy y dx 5x 5x 5 1 dy y dx dy dx 30 30 0 1 x y x 5 5y y x 5xy 1 1 2x 1 x dy 0, dx 4 y 2 1 . 2 When x Tangent line: y 1 x 2 1 x 2 0 4 (b) 8 −4 4 −4 496 76. Chapter 5 y y x y x xy ln x Logarithmic, Exponential, and Other Transcendental Functions x ln x x x 1 x 4x ln x x ln x 4 4x x 3 3 6 ln x ln x 0 78. y Domain: x > 0 y y 1 1 x (1, 1) 0 when x 1. −1 −1 6 1 >0 x2 Relative minimum: 1, 1 ln x x −1 80. y 2 (e, e−1) (e 3 2 ,2e 3 3 −2 ) 6 Domain: x > 0 y y x1 x x2 x2 1x 1 x4 ln x 1 ln x x2 ln x 2x 1 32 0 when x 2 ln x x3 e. −4 3 0 when x e3 2. Relative maximum: e, e 3 Point of inflection: e3 2, 2 e 82. y y 1 y y x x2 ln . 4 x2 1 x Domain x > 0 2x ln x 4 x1 2 ln x 4 4e x 4 0 when 84. fx fx fx x ln x, 1 1 , x f1 f1 x 1 f1 f1 1 0 1 ln x, f1 x x 2 ln ⇒ ln 4 4 1 2 ln x 4 2x 4e 1 x 32 12 32 1 ⇒x 2 3 2 ln 12 P1 x P2 x f1x f1x 1 x 2 1 1 1 1 1 1 2, x 1, P1 1 1 0 2 0 1 f1x 2 P2 1 0 when x Relative minimum: 4e Point of inflection: 4e 4 , 8e 24e 1 3 , P1 x P2 x P2 x 1, 1 x, P1 1 x P2 1 (4e−3/2, −24e−3) x, P2 1 1 −4 8 −4 (4e−1/2, −8e−1) The values of f, P1, P2, and their first derivatives agree at x 1. The values of the second derivatives of f and P2 agree at x 1. 3 f P 2 −2 −1 P 1 4 S ection 5.1 The Natural Logarithmic Function: Differentiation 497 86. Find x such that ln x fx fx xn n xn f xn 2 0.3069 x 1 xn ln x 1 x f xn f xn 1 3 3 0 x. 88. y ln y 1 dy y dx x 1 ln x 2 1x 1 2x ln x 1 x 2 x 3 2 1 3 ln x 3 1 xn 4 ln xn 1 xn 3 2.2079 0.0000 1 1 2x 1 1 2 3x x 2 1x 12x 2y 12x 11 2x 3 11 2 2.2046 0.0049 2.208 dy dx 3x2 2 Approximate root: x x2 x2 1 ln x2 2 1 2x 2 x2 1 x2 x2 x2 x2 1 1 1 1 ln x2 2x x2 1 1 2x 1 x4 1 x2 12 3x2 12x 11 x 1x 2x 3 90. y ln y 1 dy y dx dy dx 92. 1 y ln y 1 dy y dx dy dx x x ln x 1 x y x x x2 1x 1x 1 1 2 1 1x 1x x x 2 2 ln x 1 2 4 x2 2 2 4 x 2 2 1 x 1 y ln x 1 x x2 1 2 ln x 2 6x2 12 1 x2 4 2 1 x2 1 22x 1x 6 x2 2 1x 2x 1 x2 1 12 1 2x 1 3 2 x2 6 x2 2 12x 2 94. The base of the natural logarithmic function is e. 96. g x gx ln f x , f x > 0 fx fx (b) No. Let f x x2 1 (positive and concave up). g x ln x2 1 is not concave up. (a) Yes. If the graph of g is increasing, then g x > 0. g x f x and Since f x > 0, you know that f x thus, f x > 0. Therefore, the graph of f is increasing. 5.315 , 1000 < x 6.7968 ln x 50 98. t (a) (d) dt dx 5.315 6.7968 ln x 2 1 x x 1000 0 4000 5.315 6.7968 ln x 1167.41, dt dx 0.1585. 2 (b) t 1167.41 T (c) t 1068.45 T 20 years 1167.41 20 12 30 years 1068.45 30 12 $384,642.00 $280,178.40 When x dt dx 0.0645. When x 1068.45, (e) There are two obvious benefits to paying a higher monthly payment: 1. The term is lower 2. The total amount paid is lower. 498 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 34.96 p 3.955 p 100. (a) 350 (b) T p T 10 T 70 (c) 30 4.75 deg lb in2 0.97 deg lb in2 0 0 100 0 0 100 p→ lim T p 0 102. y (a) 10 ln 20 10 100 x x2 100 x2 10 ln 10 100 x2 dy dx ln x 0 100 x2 (c) lim x → 10 0 0 10 (b) dy dx 10 x 100 x 100 x 100 x 100 x2 10 100 10 100 10 100 x2 x2 1 10 x x2 10 x 1 x x 100 x 100 10 x x2 x2 x2 10 x2 10 x 10 x 100 100 x2 x2 x2 2 100 x2 100 x2 10 x 10 x 3. When x 10 x 10 When x 104. y y ln x 100 x x2 5, dy dx 9, dy dx 19 9. 106. False is a constant. d ln dx 0 1 > 0 for x > 0. x , it is a Since ln x is increasing on its entire domain 0, strictly monotonic function and therefore, is one-to-one. Section 5.2 2. 10 dx x 10 The Natural Logarithmic Function: Integration 1 dx x 10 ln x C 4. u x x 1 5 3 x2 2 C 3 x3 5, du dx dx 5 C ln x 6. 1 3x 2 dx 1 1 3 dx 3 3x 2 1 ln 3x 3 8. u x3, du dx 3x2 dx 1 1 3 3 x3 1 ln 3 3 x3 3x2 dx C Section 5.2 The Natural Logarithmic Function: Integration 499 10. u 9 x 9 xx x3 x2, du x2 dx 2x dx 1 2 9 x2 12 2x dx 9 x2 C 12. 2 3x2 4 dx 1 3x2 6x dx 3 3x 3x2 4 1 ln x3 3 3x2 4 u C x3 3x2 4 14. 2x2 7x x2 3 dx x2 2x 11x 11 19 x 19 ln x 2 dx 2 C 16. x3 x 6x 5 20 dx x3 3 x2 5x 5x2 2 19 19x 115 dx x5 115 ln x 5 C 18. x3 3x2 x2 4x 3 9 dx 3x 3 x2 2 x x x2 1 ln x2 2 3 dx 3 C 20. 1 dx x ln x3 11 3 ln x 1 ln ln x 3 1 dx x C 22. u 1 1 x2 3 1 x1 3, du dx 1 dx 3x2 3 3 1 1 x1 3 24. 1 3x2 3 xx x 2 dx 13 x2 x x 1 x ln x 2x 1 x 13 12 dx 13 1 dx 1 dx 1 x 1 1 1 1 2 3 3 x1 3 3 dx dx 3 ln 1 x1 C x 1 2x dx C 1 26. u 1 1 1 3x 3x, du dx 2 3 3 dx ⇒ dx 2 3x 12 u u3 1 1 du 1 du u ln u 3x 3x C ln 1 2 u 3 1 du 2 u 3 2 1 3 2 3 3x 3x C1 C 2 ln 1 3 500 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 1 3x2 u u 3 3 3 3 u u u2 u3 3 x1 3 28. u x1 3 1, du x 1 dx 3 dx ⇒ dx 1 3u 1 u2 3u 3u2 2 1 3 3 3 3u 1 2 du 3 3 x 1 2 du 2u 3 3u 3 x1 3x2 2 1 du 1 du u ln u 3 C 1 2 2 3 3 x1 3 3 1 C1 ln x1 3 1 C 3 ln x1 1 3x1 x 30. tan 5 d 1 5 sin 5 d 5 cos 5 1 ln cos 5 5 C 32. sec x dx 2 2 sec x1 dx 22 x 2 tan x 2 C 2 ln sec 34. u cot t, du csc2 t dt cot t csc2 t dt ln cot t C 36. sec t tan t dt ln sec t ln tan t ln cos t C tan t C C sec t tan t cos t ln sec t sec t 38. y 2x x2 ln x2 9 dx C 9 4 C⇒C ln 9 4 ln 9 40. r sec2 t dt tan t 1 ln tan t ,4 : 4 1 ln 0 1 10 9 ln 0 9 8 C 1 4 C⇒C 4 0, 4 : 4 y ln x2 (0, 4) r ln tan t −9 9 −8 −4 −2 (π , 4) 8 42. dy dx (a) 2 1 ln x , 1, x y 2 (b) y y1 x ln x dx x 2⇒ ln x 2 2 ln x 2 2 2. 2 C ln 1 2 2 C⇒C 2 4 −1 −2 Hence, y Section 5.2 1 The Natural Logarithmic Function: Integration 1 dx x e e 2 501 44. 1 2 1 1x dx ln x ln 3 2 1 46. u ln 3 e e 2 ln x, du 1 dx x ln x ln 1 11 dx ln x x e2 ln ln x e ln 2 1 48. 0 x x 1 dx 1 1 1 0 1 dx 0 x 1 2 dx 1 1 x 2 ln x 1 0 2 ln 2 0.2 0.2 50. 0.1 csc 2 cot 2 2 d 0.1 0.2 csc2 2 2 csc2 2 0.1 2 csc 2 cot 2 2 csc 2 cot 2 0.2 cot2 2 1d 0.0024 d cot 2 csc 2 0.1 52. ln sin x C ln 1 csc x C ln csc x C 54. ln csc x cot x C ln ln csc x 1 csc x cot x csc x csc x cot x cot x C cot x C cot x ln csc2 x csc x cot2 x cot x C ln csc x C 56. 1 1 x dx x 1 4x x x 2 61 x x 4 ln 1 C where C x C1 C1 5. 4 ln 1 58. tan2 2x dx sec 2x 4 1 ln sec 2x 2 tan 2x sin 2x C 60. sin2 x cos2 x dx cos x 4 4 ln sec x ln 2 2 1 1 tan x 2 sin x 4 22 1.066 Note: In Exercises 62 and 64, you can use the Second Fundamental Theorem of Calculus or integrate the function. x x2 62. F x 0 tan t dt tan x 64. F x 1 1 dt t 2 x Fx Fx 2x x2 502 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 4 66. 2 1 y 68. A 1 x x 4 4 dx 1 1 x 4 dx x 4 1 4 ln x 4 ln 4 4 ln 4 1 8.5452 x 1 2 3 4 4 3 6 −1 −2 A3 Matches (a) 0 0 9 4 70. 1 2x tan 0.3x dx x2 16 10 ln cos 0.3x 3 10 ln cos 1.2 3 4 1 8 1 10 ln cos 0.3 3 11.7686 0 −2 5 72. Substitution: u 76. Answers will vary. x2 4 and Power Rule 74. Substitution: u tan x and Log Rule 78. Average value 1 4 4 4 4x x2 1 2 1 x dx 80. Average value 1 2 0 2 sec 0 2 x dx 6 tan ln 1 x 6 0 2 0 2 2 1 dx x2 1 x 1 4 1 4 4 2 16 x ln sec 2 6 3 1 2 1 2 1.8863 3 ln 2 ln 2 3 3 2 ln x 2 ln 4 2 ln 2 ln 2 ln 4 82. t 10 ln 2 300 250 T 1 dT 100 300 10 ln T ln 2 dS dt St S2 S4 k t k dt t k ln 2 k ln 4 100 250 10 ln 200 ln 2 ln 150 10 4 ln ln 2 3 4.1504 units of time 84. k ln t C C 200 300 C k ln t C since t > 1. Solving this system yields k St 100 ln t ln 2 100 100 ln 2 and C 100 ln t ln 2 1. 100. Thus, Section 5.3 86. k k k k→0 Inverse Functions 503 1: f1 x 0.5: f0.5 x x 1 x1 0.5 10 10 y 2 10 x 10 1 x 1 f1 8 6 0.1: f0.1 x ln x x1 0.1 f 4 2 0.5 f 2 4 6 8 lim fk x 0.1 x 10 88. False d ln x dx 1 x 90. False; the integrand has a nonremovable discontinuity at x 0. Section 5.3 2. (a) fx gx f gx gfx 4. (a) fx gx f gx f 1 gfx g1 3 Inverse Functions 3 3 4 f 3 4 4x x3 1 3 4x x (b) 8 y f x 3 3 4 3 4 3 4 x 4x 4 x −2 2 x 2 −2 4 g3 1 3 x g 8 (b) x f3 2 y 1 1 x3 1 x x 1 x 3 1 x 3 g −2 −1 −1 x 2 3 1 x3 3 x3 x (b) 20 16 y −2 6. (a) fx gx f gx 16 16 f 16 x2, x ≥ 0 x x x 16 x 16 16 x2 16 x 2 16 16 x2 x 12 8 f g x 8 12 16 20 gfx g 16 x2 1 1 1 x f 1 x x 8. (a) fx gx f gx ,x≥0 ,0<x≤1 (b) 3 y x g 2 x 1 1 x 1 1 1 x 1 1 1 x x 1 1 x x 1 x 1 f x 1 2 3 gfx g 1 1 x 1 x 1 x x 504 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 12. Matches (d) x2 x2 1 t2 1 Not one-to-one; does not have an inverse 3 10. Matches (b) 14. f x 5x 3 16. F x One-to-one; has an inverse y 4 Not one-to-one; does not have an inverse y 18. g t −2 −1 −1 x 1 2 1 1 2 −3 −3 −1 x − 1 2 1 2 3 −1 1 −2 1 −1 20. f x 5x x 1 22. h x x 4 x 4 One-to-one; has an inverse 12 Not one-to-one; does not have an inverse 9 −9 9 0 0 6 −9 24. f x fx cos 3x 2 0 when x 0, , 24 , ,. . . 33 . Therefore, f does not have an inverse. 28. f x 3x 2 2 3 3x sin 2 2 f is not strictly monotonic on 26. f x fx x3 3x2 6x2 12x 12x 12 ln x 1 x 3 3 ,x > 3 > 0 for x > 3. . Therefore, f is strictly monotonic ≥ 0 for all x. fx , . Therefore, f is strictly f is increasing on monotonic and has an inverse. f is increasing on 3, and has an inverse. x3 3 3 3 y 30. fx x y f 1 3x y 3 x 3 x 3 y 3 2 1 y 32. fx x y f 1 1 y x x 1 1 1 y 34. fx x y f 1 x2 y x x y, 0 ≤ x x x y x 5 4 3 2 −5 − 4 −3 4 f f −1 x 2345 3 2 1 f −1 f f −1 f x 1 2 3 −4 −5 1 2 3 4 x −3 −2 −3 S ection 5.3 36. fx x y f y 5 4 3 2 1 x 1 2 3 4 5 −6 Inverse Functions 505 x2 y2 x2 x2 4 4 4 y, x ≥ 2 38. fx x y f 1 3 y 5 5 2x 1 y 1 x 4, x ≥ 0 243 486 x5 243 486 x5 243 486 x 6 f −1 f −4 f f −1 8 The graphs of f and f 1 are reflections of each other across the line y x. 40. fx x y f 1 x3 y5 x5 x5 2 5 3 3 3 y 42. fx x y x x 2 y 2 x 2 x 4 2 y 44. x f 1 0 x 6 2 2 4 0 1 y x 1 1 f −1 8 6 4 2 f −1 f −3 3 (0, 6) f 1 x (2, 2) (4, 0) x 8 −2 f −6 6 2 4 6 The graphs of f and f 1 are reflections of each other across the line y x. −4 The graphs of f and f 1 are reflections of each other across the line y x. 46. f x 48. f x fx k2 x x x x 2 on 2 1 2 x3 is one-to-one for all k 2, 1 > 0 on 2, f is decreasing on 0, and has an inverse. . Therefore, f is strictly monotonic 0. Since f 1 3 2, f 2 3 k2 2 2 3 12k ⇒ k 1 4. 50. f x fx cot x on 0, csc2 x < 0 on 0, f is increasing on 2, . Therefore, f is strictly monotonic and has an inverse. 52. f x fx sec x on 0, 2 2 sec x tan x > 0 on 0, f is increasing on 0, 2 . Therefore, f is strictly monotonic and has an inverse. 506 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions The graphs of f and f 1 are reflections of each other across the line y x. 10 0 54. 2x2 x2 2 fx 3 y x y f 1 2 x2y 3 ± ± 3 x2 y on 0, 10 5 f −1 f 0 3 2 3 2 3 x x y x 2 ,x<2 56. (a), (b) −3 2 58. (a), (b) h 3 −10 10 f −1 f 10 h −1 −2 −10 (c) h is not one-to-one and does not have an inverse. The inverse relation is not an inverse function. 60. f x 3 (c) Yes, f is one-to-one and has an inverse. The inverse relation is an inverse function. 62. f x ax b Not one-to-one; does not have an inverse f is one-to-one; has an inverse ax b x y f 1 y y a x a x a b ,a 0 b b x 64. f x 16 16 16 4 4 x4 is one-to-one for x ≥ 0. x4 y y x 1 66. f x x x 3 x y f 1 3 is one-to-one for x ≥ 3. y y x x 3 3 3, x ≥ 0 y x4 x y 4 16 16 f x x 16 x, x ≤ 16 t2 for which 70. Yes, the area function is increasing and hence one-to-one. The inverse function gives the radius r corresponding to the area A. 68. No, there could be two times t1 h t1 h t2 . 72. f x fx f 1 15 x 27 1 5x4 27 11 2x3 ; f 6x2 1 ff 1 3 1 27 243 54 11 a. 1 11 f 3 5 3 4 27 6 3 2 1 17 S ection 5.3 74. fx fx f 1 Inverse Functions 507 cos 2x, f 0 2 sin 2x 1 ff x 1 2x 1 ff 1 1 1 a 1 1 4, f 8 4 2 1 f0 2 a 1 2 sin 0 1 which is undefined. 0 76. fx fx f 1 2 1 f8 1 128 , , 4 1 14 4 78. (a) Domain f (b) Range f (c) f f −1 3 2 Domain f Range f y 1 (d) fx fx f1 f 1 3 4 4 3 4 1 4 1 4 4 1 x2 x2 4x, 1, 1 1 x x 1 x , 1, 1 −5 −4 −3 −2 −1 −2 −3 −4 −5 x 23 f f 1 1 80. (a) Domain f (b) Range f (c) 4 3 2 1 y 0, , Domain f 1 1 0, 4 (d) fx fx f 1 0, 4 , Range f 0, 8x ,f 1 12 4 x 2 ,f x x 1 2 x x x2 x 1 2 f −1 f x 1 2 3 4 f 1 2 4 82. x 1 dy dx 2 ln y2 2 1 y2 3 4y 3 3 2y dy dx dy dx 16 3 16 13 . 16 y2 . At 0, 4 , In Exercises 84 and 86, use the following. fx f 84. g 1 1 1 8x 3 and g x 3 and g g 1 1 x3 x g 3 x 1 8x 3 x 0 0 86. g 1 f f 1 3 1 g 1 4 g 3 1 g 1 4 9 g 4 13 4 3 4 508 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions In Exercises 88 and 90, use the following. fx f 1 x x x 4 and g x 4 and g 1 2x x x 5 5 2 90. g f x gfx gx 2x 4 4 3 1 1 88. f 1 g 1 x f f x 1 g x 1 x 5 1 2 5 2 3 2 Note: g f f 1 5 4 2x Hence, g f x x x 2 g 3 1 92. The graphs of f and f to the line y x. 1 are mirror images with respect 94. Theorem 5.9: Let f be differentiable on an interval I. If f has an inverse g, then g is differentiable at any x for which f g x 0. Moreover, gx 1 , f gx f gx 0 96. f is not one-to-one because different x-values yield the same y-value. Example: f 3 f 4 3 3 5 98. If f has an inverse, then f and f Let f 1 1 x y then x f Thus, f 1 1 f. 1 1 are both one-to-one. y and f x y. Not continuous at ± 2. f x2 , then f 1 f x1 f 1 f x2 ⇒ x1 x2. Therefore, f is one-to-one. If f x is one-to100. If f has an inverse and f x1 one, then for every value b in the range, there corresponds exactly one value a in the domain. Define g x such that the domain of g equals the range of f and g b a. By the reflexive property of inverses, g f 1. 102. True; if f has a y-intercept. 104. False Let f x 106. From Theorem 5.9, we have: gx 1 f gx f gx 0 f gx g x f gx 2 f gx 1 f gx f gx 2 x or g x 1 x. gx f gx f gx 3 If f is increasing and concave down, then f > 0 and f < 0 which implies that g is increasing and concave up. Section 5.4 Exponential Functions: Differentiation and Integration 509 Section 5.4 2. e 2 Exponential Functions: Differentiation and Integration 0.1353. . . 2 4. e ln 0.5 0.6931. . . 0.6931. . . 1 2 6. eln 2x 2x x 12 12 6 ln 0.1353. . . 8. 4ex ex x 83 83 4 83 ln 4 3.033 10. 6 3ex 3ex ex x 8 14 14 3 ln 14 3 1.540 12. 200e e 4x 15 15 200 3 ln 40 40 1 ln 4 3 1 e e 4 e 0.680 0.648 3 40 14. ln x2 x2 x 10 e10 ± e5 ± 148.4132 4x 4x x 16. ln 4x 4x x 18. ln x x x 2 2 2 2 12 e12 e6 2 e6 405.429 2 x 20. y 1x 2e y 4 3 2 1 x 1 2 3 22. y e x2 y 4 3 2 −1 −2 −1 x 1 2 24. (a) 10 (b) 10 −8 −2 10 −8 −2 10 Horizontal asymptotes: y 0 and y 8 Horizontal asymptote: y 4 510 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions C e C e 1 26. y Ce ax 28. y 0 x→ 1 ax Horizontal asymptote: y Reflection in the y-axis Matches (d) lim lim 1 ax C 0 C and y 0 x→ C e ax Horizontal asymptotes: y Matches (b) 30. f x gx y 8 6 4 ex ln 3 32. f x 3 ln x gx y 4 3 2 1 x −1 −1 ex 1 1 34. In the same way, ln x x→ x3 lim 1 r x x er for r > 0. f g f −2 2 x 1 2 3 4 −2 g 2 4 6 8 36. 1 1 1 2 1 6 1 24 1 120 1 720 1 5040 e 2.71825396 2.718281828 1 1 2 1 6 1 24 1 120 1 720 2x e>1 1 5040 38. (a) y y e2x 2e2x 2. 42. x (b) y y e 2e 2x At 0, 1 , y 40. f x fx e1 x At 0, 1 , y y dy dx e x2 2. 44. y dy dx xe x2e x e1 2xe x2 x2e x x 2xe x e x x 2 46. g t gt e e 3 t2 48. 6t 3 y ln 1 1 ex ex ex ex 1 ln 1 ex ex ex 50. y ln ex 2 3 t2 6 t 3e3 t2 ln 1 dy dx ex 1 ln ex dy dx ex ex e2x e2x e e 1 1 e x x x ln 2 2ex 1 e2x 52. y dy dx ex 2 ex 2 e x e x 54. y dy dx xex ex ex ex x ex x 1 1 xex Section 5.4 56. f x fx e3 ln x e3 x exy x dy dx y exy 2x dy xy xe dx x2 2y y2 dy dx 10 0 yexy yexy xexy 2x 2x 2y Exponential Functions: Differentiation and Integration 58. y dy dx 62. g x gx ln ex 1 x 511 60. x 1 2x 1 4x3 ex ln x ex x 2 ex ln x xex x2 ex 2x 1 x2 ex ex x ex ln x ex ln x 2y dy dx gx 1 4x x 64. y y ex 3 cos 2x ex e x 4 sin 2x 8 cos 2x 5 cos 2x 2 sin 2x 2 ex 3 cos 2x 5ex 4 sin 2x cos 2x cos 2x 5ex 5 cos 2x 4 sin 2x 25ex cos 2x 5y 6 sin 2x 10 sin 2x 5ex 4 cos 2x 25ex cos 2x 2y 2 sin 2x y y 2y 5ex 2 sin 2x cos 2x 0. 5ex 2 sin 2x 2y 5y 5ex 3 cos 2x Therefore, y ex 2 ex 2 ex 2 e e e 5y ⇒ y x 2 66. f x fx x >0 x −3 (0, 0) 3 fx 0 when x 0. −2 Point of inflection: 0, 0 68. g x gx gx 1 e 2 1 x 2 1 2 x x 3 2 2 0.8 ( 2, e− 0.5 2π 3, 1 2π ( 4, e− 0.5 2π 3e 2x 3, 2, x 3 2 2 ( x 3 2 (( ( 6 4e 1 2 1 e 2 2 0 0 Relative maximum: Points of inflection: 3, 0.399 12 , 4, 1 e 2 12 2, 0.242 , 4, 0.242 70. f x fx fx xe x x x 2 xe e e x e x x 1 x x e x 0 when x x 2 1. −2 (1, e −1) (2, e−2) 4 e 1 1 2 0 when x 2. −2 Relative maximum: 1, e Point of inflection: 2, 2e 512 72. f x fx fx Chapter 5 2 e3x e3x 2 6 Logarithmic, Exponential, and Other Transcendental Functions 2x 2x 6x e3x 10 e3x 24 6x 18x 0 when x 0 when x 5 3. 4 3. − 0.5 0 2.5 e3x 4 100 3e3x 4 3e3x 10 5 3, 4 3, ( 5 , 96.942) 3 ( 70.798 ) 4 , 3 Relative maximum: Point of inflection: 96.942 70.798 x ex 1 1 74. (a) fc 10ce c fc 10 c cx ec x c c x x ex x xe c x (b) A x xf c x 10 e ex x ex 1 10x2 x e ex 1 x ex 2 c ec cec x x ec x (d) c 1 cex cex c c (c) A x 6 1 1 e x 0 0 4 10x2 x e x e 1 (2.118, 4.591) x →0 x→ lim c 1 0 lim c 0 0 9 The maximum area is 4.591 for x 2.547. fx 2.118 and 76. Let x0, y0 be the desired point on y y y 1 y y e x0 e x. 3 2 1 y e x e ex ex0 x x Slope of tangent line Slope of normal line x0 −1 ( 0.4263, e − 0.4263 ) x 1 2 3 −1 We want 0, 0 to satisfy the equation: e x0 x0ex0 x0e2x0 0 0.4263. 1 x0e2x0 1 Solving by Newton’s Method or using a computer, the solution is x0 0.4263, e 0.4263 Section 5.4 78. V (a) 15,000e 20,000 Exponential Functions: Differentiation and Integration 513 0.6286t , 0 ≤ t ≤ 10 (b) dV dt When t 9429e 1, 5, 0.6286t (c) 5028.84. 20,000 dV dt dV dt 0 0 10 0 10 0 When t 406.89. 80. 1.56e 0.22t cos 4.9t ≤ 0.25 (3 inches equals one-fourth foot.) Using a graphing utility or Newton’s Method, we have t ≥ 7.79 seconds. 0 2 10 −2 82. (a) V1 V2 20,000 1686.79t 109.52t2 23,181.79 3220.12t 28,110.36 (b) The slope represents the rate of decrease in value of the car. 0 0 10 (c) V3 31,450.77 0.8592 t 31,450.77e 0.1518t (d) Horizontal asymptote: lim V3 t t→ 0 As t → (e) dV3 dt For t For t 4774.2e 5, 9, dV3 dt dV3 dt ,f 0 x 2, 2 , the value of the car approaches 0. 0.1518t 2235 dollars year. 1218 dollars year. 84. fx fx fx P1 x P1 x P2 x P2 x P2 x e x2 2 1 f0 e 0 0 0 0 1 1 x 2 0 2 x2 2 2 xe 0 e x2 2 P 1 2 x2e x 2 x2 1 1 ,f 0 1 −3 f 3 P2 −2 1 0x 1, P1 0 0, P1 0 1 0x x, P2 0 1, P2 0 1 x2 ,P 0 22 1 The values of f, P1, P2 and their first derivatives agree at x f and P2 agree at x 0. 0. The values of the second derivatives of 514 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 88. Let u e x3 3! ... x 4 86. nth term is xn n! in polynomial: y4 1 x x2 2! 1 x x3 3! x2 2! x4 4! x4, du 4x dx 3 4x3 dx. e x 4 C Conjecture: ex 4 4 90. 3 e3 x dx e3 x 3 e 1 1 1 1 e 92. x2ex 3 2 dx 2 x3 e 3 2 3x2 dx 2 2 x3 e 3 2 C 94. e1 x dx x3 2 11 e 2 x2 2 dx x3 11 e 2 x2 C 96. Let u 1 1 e2x, du 2e2x dx. 1 ln 1 2 e2x C e2x dx e2x 1 2e2x dx 2 1 e2x 98. Let u 2 x2 , du 2 x2 2 dx 0 x dx. 2 xe 0 e x2 2 x dx e x2 2 0 2 1 e 1 e e 1 100. Let u ex ex ex e e x x e x, du dx ln ex ex e x e x dx. 102. Let u 2ex ex ex x e x, du 2 ex ex 2 e e e x x dx. 2 C 2e dx e x2 ex e x dx ex e2x 2ex ex 1 x C 104. dx ex ex 2x 2 e e x x dx C 106. e sec 2x sec 2x tan 2x dx u sec 2x, du 1 sec 2x e 2 C 2 sec 2x tan 2x 108. ln e2x 1 dx x2 2x x 1 dx C 110. y ex e2x 1 2x e 2 e 2 2x x 2 dx e 2x dx 2x 1 e 2 C 112. f x f0 fx fx sin x 1 1 2 e 2x dx cos x 1 ⇒ C1 2 1 1 dx C2 0 1 2x e 2 1 C1 C1 1 2x e 2 cos x cos x sin x 1 2x e 2 x 1 2x e 4 f0 fx 1 4 x C2 sin x 1 ⇒ C2 4 1 2x e 4 Section 5.4 Exponential Functions: Differentiation and Integration dy dx y 0.2x2 515 114. (a) 4 y (b) xe , 0, 3 2 1 0.4 C 2.5e0 1 e 0.2x2 xe 0.2x2 dx 0.4x dx C C⇒C 1 −4 x 4 1 e 0.4 0, y 3 : 2 2.5e 3 2 0.2x2 2.5e C 0.2x2 −4 2.5 0.2x2 b b 2 116. a e x dx e x a e a e b 118. 0 e 2x 2 dx 1 e 2 1 e 2 4 2 2x 2x 4 0 1 2 4.491 a b 4 −2 0 4 4 x 120. (a) 0 x ex dx, n 12 122. 0 0.3 e e 0.3t dt x 1 2 1 2 1 2 1 2 ln 1 2 ln 2 2.31 minutes Midpoint Rule: 92.1898 Trapezoidal Rule: 93.8371 Simpson’s Rule: 92.7385 Graphing Utility: 92.7437 2 0.3t 0 0.3x 1 0.3x e (b) 0 2xe x dx, n 12 0.3x x Midpoint Rule: 1.1906 Trapezoidal Rule: 1.1827 Simpson’s Rule: 1.1880 Graphing Utility: 1.18799 124. ln 2 0.3 t R ln R 0 425 6.052 1 240 5.481 2 118 4.771 3 71 4.263 4 36 3.584 (a) ln R R (b) 450 0.6155t e 0.6155t 6.0609 428.78e 0.6155t 6.0609 −1 0 5 4 4 (c) 0 R t dt 0 428.78e 0.6155t dt 637.2 liters 516 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions ex are mirror 128. (a) Log Rule: u (b) Substitution: u ex x2 1 126. The graphs of f x ln x and g x images across the line y x. 130. ln ln ea ea eb b ln ea a ea eb b ln eb a b Therefore, ln ln ea b and since y ln x is one-to-one, we have ea eb ea b. Section 5.5 2. y At t0 1 2 t8 Bases Other than e and Applications 4. y 1 2 t5 6. log27 9 2, y 1 2 25 log27 272 3 2 3 16, y 1 2 16 8 1 4 At t0 0.7579 8. loga 1 a loga 1 loga a 1 10. (a) 272 3 9 2 3 8 3 4 12. (a) log3 1 9 3 (b) 2 2 2 1 9 log27 9 (b) 163 4 491 7 1 2 log49 7 log16 8 14. y x y 3x 1 16. y 1 0 1 3 2x x y 2 18. y 2 16 y 3 x y x 1 1 2 3 3 9 1 2 0 1 1 2 2 16 0 1 y ±1 1 3 ±2 1 9 1 9 y 12 10 8 1 6 8 6 4 2 −2 −1 x 1 2 3 4 −2 −1 x 1 2 4 2 −1 −1 x 1 20. (a) log3 1 81 3x x x 1 81 4 x 36 2 22. (a) logb 27 b3 b (b) logb 125 b 3 3 27 3 3 125 5 (b) log6 36 6 x b x Section 5.5 24. (a) log3 x log3 x log3 x x xx x2 x x 2x 1x 2 2 2 3 3 1 1 31 0 0 3 Bases Other than e and Applications 3 log10 log10 x x x x x x 3 3 x 3 3 1 1 101 10x 9x 1 3 517 (b) log10 x 1 OR x x 3 is the only solution since the domain of the logarithmic function is the set of all positive real numbers. 26. 56x 6x ln 5 x 8320 ln 8320 ln 8320 6 ln 5 0.935 x 1 ln 5 ln ln x 1 28. 3 5x 5x 1 86 86 3 86 3 1 86 3 ln 5 ln 86 3 ln 5 x 1 3.085 30. 1 365t ln 1 0.10 365 365t 2 ln 2 t 1 365 ln 2 0.10 ln 1 365 6.932 32. log10 t t 3 3 t 2.6 102.6 3 102.6 401.107 0.10 365 34. log5 x x x 4 4 4 x 3.2 53.2 53.2 4 2 56.4 29,748.593 735.41 38. g x 1 2 log10 x x 3 56.4 36. f t Zero: t 10 300 1.007512t 10 Zeros: x 5 0.826, 3.826 (10, 0) 0 20 −5 (− 0.826, 0) 5 (3.826, 0) −10 −5 518 40. f x gx Chapter 5 3x log3 x Logarithmic, Exponential, and Other Transcendental Functions y x fx 1 9 2 1 3 1 0 1 1 3 2 9 9 6 3 f x gx 1 9 1 3 1 1 0 3 1 9 2 −3 g x 3 6 9 2 −3 42. g x gx 2 x 44. x y dy dx x6 x 6 6 2x 46. f t 2x 32t t t 2 ln 3 32t t2 32t 2t ln 3 t2 1 32t ln 2 2 2 ln 6 6 2x 2x 6 1 2x ft 2x ln 6 1 2x ln 6 50. 48. g g 5 5 2 2 sin 2 2 cos 2 1 2 y dy dx log10 2x 0 1 x ln 10 x2 x 1 1 log10 2 1 x ln 10 log10 x ln 5 5 2 sin 2 52. h x log3 xx 2 1 54. 1 1 1 ln 3 1 log3 2 0 y log10 log3 x hx 1 x ln 3 11 ln 3 x 1 log3 x 2 1 2 2x 2 1 x 1 log10 x2 dy dx x2 log10 x 1 x ln 10 1 x 1 1 2x 1 ln 10 1 2x ln 10 x2 1 x2 1 ln 10 x x2 1 3x ln 3 2x x 56. f t ft t 3 2 log2 t 1 t3 2 1 ln t 1 2 ln 2 1 58. y ln y 1 dy y dx dy dx xx x x y xx x 1 1 1 t3 2 2 ln 2 t1 3 12 t ln t 2 1 ln x 1 1 x 2 1 x ln x ln x 1 0 x x ln x 0 60. y ln y 1 dy y dx dy dx 1 x 1x 62. 5 x dx 5x ln 5 C 64. 2 3 3 5 dx 2 0 2 27 25 dx 1 ln 1 x 1 1 x1 x y 1 xx 1 1 x x x ln 1 ln x x 1 x 1 ln x x 1 1 x 1 x2 2 dx 2 0 2x 2 4 1x Section 5.5 2 Bases Other than e and Applications 519 66. 3 x7 3 x dx 1 2 23 x x7 2 3 x 2 dx 68. 2 sin x cos x dx, u C sin x, du cos x dx 1 73 2 ln 7 C 1 sin x 2 ln 2 dy dx y ,2 : 2 70. (a) 6 4 2 y (b) esin x cos x e sin x ,2 e 1 sin x cos x dx C 1 C 1 esin C⇒C x 10 −2 y esin x 72. logb x ln x ln b log10 x log10 x log10 b 74. f x (a) Domain: x > 0 (b) y 10y f (c) 1 (d) If f x < 0, then 0 < x < 1. (e) f x 1 log10 x log10 10x x must have been increased by a factor of 10. log10 10 log10 x x 10x log10 103 log10 104 3 4 x log10 1000 log1010,000 (f) log10 x1 x2 log10 x1 3n n log10 x2 2n 100n . If 1000 ≤ x ≤ 10,000, then 3 ≤ f x ≤ 4. Thus, x1 x2 76. f x (a) f u ax v au 3 4 t v 102n au av fufv (b) f 2x a2x ax 2 fx 2 78. V t (a) 20,000 V 20,000 16,000 12,000 8,000 4,000 (b) dV dt 20,000 ln dV 1: dt 4: dV dt 3 4 3 4 t (c) 4315.23 1820.49 V ′(x) x When t When t t 2 4 6 8 10 − 1000 − 2000 − 3000 − 4000 − 5000 − 6000 2 4 6 12 V2 20,000 3 4 2 $11,250 Horizontal asymptote: v 0 As the car ages, it is worth less each year and depreciates less each year, but the value of the car will never reach $0. 520 80. P A A 82. P A A Chapter 5 $2500, r 2500 1 2500e 0.06 $5000, r 5000 1 Logarithmic, Exponential, and Other Transcendental Functions 6% 0.06 n 20 20n 0.06, t 20 n A 1 8017.84 2 8155.09 4 8226.66 12 8275.51 365 8299.47 Continuous 8300.29 8300.29 0.07, t 25n 7% 0.07 n 25 n A 1 27,137.16 2 27,924.63 4 28,340.78 12 28,627.09 365 28,768.19 Continuous 28,773.01 5000e0.07 25 Pe0.06t ⇒ P 100,000e 0.06t 84. 100,000 t P 1 94,176.45 10 54,881.16 20 30,119.42 30 16,529.89 40 9071.80 50 4978.71 86. 100,000 P1 0.07 365 365t ⇒P 100,000 1 0.07 365 30 365t t P 1 93,240.01 10 49,661.86 20 24,663.01 40 6082.64 50 3020.75 12,248.11 88. Let P (a) $100, 0 ≤ t ≤ 20. A 100e0.03t 182.21 100e0.05t 0 400 90. (a) lim n→ 1 P 0.86 e 0.25n 0.86 1 0.069 0.016 0.86 or 86% 0.25n A = 100e0.06t A = 100e 0.05t A 20 (b) A A 20 (c) A A 20 (b) 0.25 e e 0.25n 0.215e 1e 0.25n 0.25n A = 100e0.03t 20 0 P3 P 10 271.83 100e0.06t 332.01 92. (a) 12,000 (d) pt 19e t5 38,000 e 5 1 ln 19 5 ln 19 t5 1 1 19e 19e t5 t53 0 0 0 40 t 5 t (b) Limiting size: 10,000 fish (c) pt pt 1 1 10,000 19e t e t5 5 14.72 19et 52 t5 19 10,000 5 38,000e 1 19e p1 p 10 t52 113.5 fish month 403.2 fish month Section 5.5 Bases Other than e and Applications 521 94. (a) y1 y2 y3 y4 (b) 150 6.0536x 100.0751 97.5571 17.8148 ln x x (c) The slope of 6.0536 is the annual rate of change in the amount given to philanthropy. (d) For 1996, x y3 6 and y1 6.0536, y2 2.9691, 99.4557 1.0506 101.2875x0.1471 6.6015, y4 3.2321. y3 is increasing at the greatest rate in 1996. 0 100 8 y3 seems best. 3 x 96. A 0 y 3 dx x 3 ln 3 3 0 26 ln 3 98. 23.666 x 1 x 1x 1 2 10 1 10 2 10 4 10 6 2.594 2.705 2.718 2.718 30 20 10 x 1 2 3 100. t y y 0 600 C kt 1 630 2 661.50 3 694.58 4 729.30 When t y 0, y 600 kt 600 ⇒ C 600. 630 661.50 1.05, 600 630 Let k 1.05. y 600 1.05 t 1.05, 694.58 661.50 1.05, 729.30 694.58 1.05 102. True. f en 1 104. True. f en ln en n 1 1 1 106. True. Cex y for n 1, 2, 3, . . . fx gx g x ex 0⇒ ln en n dy dxn n 0 since ex > 0 for all x. 522 108. y Chapter 5 x sin x ln xsin x sin x xsin x ,, 22 sin Logarithmic, Exponential, and Other Transcendental Functions ln y y y y At sin x cos x ln x ln x ln x 1 x sin x x cos x 2 y sin 2 2 2 2 2 0 1 cos 2 ln 2 Tangent line: y 2 y 1x x 2 Section 5.6 2. dy dx y 4 4 x Differential Equations: Growth and Decay 4. x dx 4x x2 2 C dy dx dy 4 1 4 ln 4 y y dy x e 4 10. x 4 dx y 6. y 3yy 3yy dx x 3y x x dx 23 x 3 C 2 dx C1 C1 y dy 4 y y 3y2 2 Ce x C1 9y2 4x3 2 Ce x 8. y 1 y 1 1 ln 1 1 y y y dx x1 x x dx x dx x2 2 ex 2 y xy y y 100x 100x x x dx x dx x2 2 x2 2 e e x2 2 C1 xy x 100 y y 100 y 100 y y dx dy y y y y C1 2 2 1 dy 100 y ln 100 1 1 100 y y y ln 100 y y C1 2 C1 C1 C1 x2 2 eC1 ex Cex 2 2 e 100 x2 2 100 Ce Section 5.6 dP dt dP dt dt dP P Differential Equations: Growth and Decay dy dx 1 L 1 L L ln L L dy y dx 523 12. k 10 k 10 k 10 2 k 10 2 t t dt t t 2 14. kx L kx kx dx kx2 2 kx2 2 e y C C dy dx y dx 1 y dy y y y y dy dx dy y L 2 C1 C1 C1 kx2 2 L e Ce C1 e kx2 2 kx2 2 16. (a) 4 y (b) xy, x dx x2 2 ex 2 0, 1 2 (0, 1 ) 2 −4 x 4 ln y −4 C 2 C y 0, C1ex 2 2 11 : 22 C1e0 ⇒ C1 1 ⇒y 2 1 x2 e 2 2 18. dy dt dy y 10 y 3 4 3 4 13 t 2 1 0 2 13 t 2 t, 0, 10 t dt 0 15 20. (0, 10) dy dt dy y ln y 3 y, 0, 10 4 3 dt 4 3 t 4 e3 4t −5 40 10 (0, 10) 5 −5 2 C C⇒C 10 −5 C1 C1 4t 32 10 y eC1 e 3 2 Ce3t 10 4 10 y Ce0 ⇒ C 10e3t 4 22. dN dt N kN Cekt (Theorem 5.16) 250 250ek ⇒ k 250e4ln 8 250 8 5 4 5 24. dP dt P kP Cekt (Theorem 5.16) 5000 5000ek ⇒ k 5000eln 19 5000 19 20 20 5 5 0, 250 : C 1, 400 : 400 When t 4, N 0, 5000 : C ln 400 250 5 ln 4 8 5 1, 4750 : 4750 When t 5, P ln 19 20 250eln 8 8192 5 3868.905 524 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 1 2 1 , 4, 5 2 26. y C y 1 2 k y Cekt, 0, 4 , 4 4ekt 4e5k 5, 28. y 1 2 5 2Ce3k Cekt, Ce3k Ce4k 1 4k Ce 5 e4k ek ln 10 3, ln 1 8 5 4e 0.4159t 0.4159 10e3k 10 k y 5 C y 2.3026 Ce2.3026t Ce2.3026 4 0.0005 0.0005e2.3026t 12 xy 2 30. y dy dt ky 32. dy dx dy > 0 when y > 0. Quadrants I and II. dx 34. Since y When t Ce ln 1 2 1620 t , we have 1.5 Ce ln 1 10,000, we have y 2.30e ln 1 2 1620 2 1620 1000 10,000 ⇒ C 2.30 which implies that the initial quantity is 2.30 grams. 0.03 gram. 36. Since y Ce ln 1 2 5730 t, we have 2.0 Ce ln 1 2 5730 10,000 ⇒ C 6.70 which implies that the initial quantity is 6.70 grams. When t 1000, we have y 6.70e ln 1 2 5730 1000 5.94 grams. 38. Since y Ce ln 1 2 5730 t, we have 3.2 Ce ln 1 2 5730 1000 ⇒C 3.61. Initial quantity: 3.61 grams. When t 10,000, we have y 1.08 grams. 40. Since y Ce ln 1 2 24,360 t, we have 0.4 Ce ln 1 2 24,360 10,000 ⇒ C 0.53 which implies that the initial quantity is 0.53 gram. When t 1000, we have y 0.53e ln 1 2 24,360 1000 0.52 gram. dy dx 42. Since ky, y y0e5730k ln 2 5730 y0e Cekt or y y0ekt. 44. Since A 40,000 2 ln 2 t 20,000e0.055t, the time to double is given by 20,000e0.055t and we have e0.055t 0.055t ln 2 0.055 12.6 years. 1 y 20 k 0.15y0 ln 0.15 t ln 2 5730 t ln 2 t 5730 5730 ln 0.15 ln 2 15,682.8 years. Amount after 10 years: A 20,000e 0.055 10 $34,665.06 Section 5.6 46. Since A 10,000ert and A have the following. 20,000 r 10,000e5r ln 2 5 0.1386 13.86% 10,000e ln 2 5 10 Differential Equations: Growth and Decay 5436.56 when t 10, we 525 20,000 when t 5, we 48. Since A 2000ert and A have the following. 5436.56 r 2000e10r ln 5436.56 2000 10 0.10 10% Amount after 10 years: A $40,000 The time to double is given by 4000 t 2000e0.10t ln 2 0.10 6.93 years. 50. 500,000 P P1 0.06 12 12 40 52. 500,000 480 P1 0.09 12 12 25 500,000 1.005 $45,631.04 P 500,000 1 $53,143.92 0.09 12 300 54. (a) 2000 2 ln 2 t (b) 2000 2 ln 2 t 1000 1 1.06t t ln1.06 ln 2 ln 1.06 1000 1 1 0.06 12 0.6 t (c) 2000 2 1000 1 1 0.06 365 0.06 365 365t 365t 11.90 years 0.06 12 12t 12t ln 2 t 365t ln 1 0.06 365 1 ln 2 0.06 365 ln 1 365 11.55 years (d) 2000 0.06 12 11.58 years 2 ln 2 t 1000e0.06t e0.06t 0.06t ln 2 0.06 11.55 years 12t ln 1 1 12 ln 2 0.06 ln 1 12 56. (a) 2000 2 ln 2 t (b) 2000 2 ln 2 t 1000 1 1.055t t ln1.055 ln 2 ln 1.055 1000 1 1 0.055 t (c) 2000 2 1000 1 1 0.055 365 365t 365t 0.055 365 12.95 years 0.055 12 12t 12t ln 2 t 365t ln 1 1 365 0.055 365 12.60 years 0.055 12 ln 2 0.055 ln 1 365 (d) 2000 0.055 12 12.63 years 2 ln 2 t 1000e0.055t e0.055t 0.055t ln 2 0.055 12.60 years 12t ln 1 1 12 ln 2 0.055 ln 1 12 526 58. P P P Chapter 5 Cekt 1 Logarithmic, Exponential, and Other Transcendental Functions 60. P 1 Ce0.031t 11.6 Ce0.031 ⇒C 11.9652 Cekt 1 Ce 3.6 0.004t P P Ce 0.004t 0.004 1 ⇒C 3.5856 11.9652e0.031t 16.31 or 16,310,000 people in 2010 t 3.5856e 3.45 y C P 10 62. (a) N (b) N P 10 64. or 3,450,000 people in 2010 100.1596 1.2455 400 when t Cekt, 0, 742,000 , 2, 632,000 742,000 742,000e2k ln 632 742 2 742,000e 4, y 0.0802t 6.3 hours (graphing utility) Analytically, 400 1.2455t t ln 1.2455 t 66. (a) 20 30e30k k N 68. S 25 1 30 1 10 ln 1 3 30 30 1 ekt ek 1 t→ 632,000 100.1596 1.2455 400 100.1596 ln 3.9936 ln 3.9936 ln 1.2455 e30k 6.3 hours. (b) e ln 3 30 e 0.0366t t k y When t 0.0802 3.9936 $538,372. 25 0.0366t 30 1 1 6 e 0.0366t 0.0366 t ln 6 0.0366 49 days (a) 4 25 1 ⇒1 25 ek 4 ⇒ ek 25 21 ⇒k 25 ln 21 25 0.1744 (b) 25,000 units lim S (c) When t 5, S 14.545 which is 14,545 units. (d) 25 0 0 8 70. (a) R I (b) 2000 979.3993 1.0694 0.1385t4 t 979.3993e0.0671t 9.9755t2 23.8513t 266.4923 2.1770t3 Rate of growth 0 0 10 Rt 65.7e0.0671t (c) 500 (d) P t 1 I R 0 0 10 0 0 10 Section 5.7 I 10 16 Differential Equations: Separation of Variables dy dt dy kt 527 72. 93 6.7 80 8 10 log10 10 log10 I 10 6.7 16 74. Since 1 y 80 ky k dt 80 log10 I ⇒ I I 10 log10 10 16 log10 I ⇒ I 10 log10 I 10 10 8 16 ln y When t 80 0, y 1, y C. 1500. Thus, C 1120. Thus, ln 1120 ln 1040 142 t ln 1420. 6.7 Percentage decrease: 10 6.7 8 10 100 95% When t k1 ln 1420 k 80 ln 1420 80. ln 104 . 142 Thus, y When t 76. True 78. True 1420e ln 104 5, y 379.2 . Section 5.7 Differential Equations: Separation of Variables 2xy x2 y2 2. Differential equation: y Solution: x 2 Check: 2x y2 2yy y y Cy Cy 2y 2y 2 2x C 2xy Cy 2y 2 2xy x2 y2 y2 2xy x2 2xy x2 y2 4. Differential equation: y Solution: y Check: C1e x 2y C2e C1 2 C1e 2C1e 2 2C1 x x 2y x 0 cos x y y sin x x C2 e sin x sin x 2C2e 2C2e x x x C1 cos x C2 e x cos x y 2y 2y sin x C2 e 2C1 cos x C1 x C1 sin x 2C2 e C2 e x cos x 2C2 2 C1e 2C1 x cos x C2e x x sin x 0 2C2 sin x 2C2 2C1 e cos x 6. y y y 2 3 2 3 2 3 e 2x ex 2x 2e 4e 2x ex ex 2y 2 3 Substituting, y 4e 2x ex 2 2 3 2e 2x ex 2ex. 528 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 16y 0. 10. y y4 48 cos 2x 0, Yes. y4 16y 5 ln x 30 x4 30 x4 80 ln x 0, No. In Exercises 8 –12, the differential equation is y 4 8. y y4 y4 16y 3 cos 2x 48 cos 2x 48 cos 2x 12. y y 4 y 4 3e2x 48e2x 24e2x 4 sin 2x 64 sin 2x 64 sin 2x 16 3e2x 4 sin 2x x3ex. 0, Yes 16y In 14–18, the differential equation is xy x 2ex, y 2y x2ex 2y x x 2ex ex x2 2xex 2x x2ex 2xex 2 ex x2 x2ex 2y 14. y xy 18. y xy 20. y d 2y dt 2 2x x3ex, 10x 2 x2ex 5x2 x3ex, Yes. 16. y xy sin x, y 2y cos x x cos x 2 sin x x3ex, No. 5x2, y x x2ex 2xex 10x Yes. 22. 2x 2 y2 16 C passes through 3, 4 C⇒C 2 y2 2 A sin t A 2 sin t 16y 0, we have 0. 29 Particular solution: 2x 2 A 2 2 Since d 2y d t 2 sin t 16 and 16A sin t ± 4. Thus, 24. Differential equation: yy General solution: x 2 Particular solutions: C C y2 x C 0 y 2 1 0, Point 1, C 4, Circles 1 2 x 26. Differential equation: 3x General solution: 3x2 6x 2 3x 3x 4yy 2yy 2yy 0 0 0 23 2 2yy 2y2 C 0 Initial condition: y1 3: 3 1 2 3 2y2 18 21 21 C Particular solution: 3x2 Section 5.7 28. Differential equation: xy General solution: y y xy 1 C2 ,y x y x C1 y 0 Differential Equations: Separation of Variables Initial conditions: y 2 0 y C1 C2 x C2 ⇒ C2 2 1, C1 ln 2 ln 2 ln x ln x 2 C2 ln 2 0, y 2 1 2 529 C2 ln x 1 C2 2 x 1 1 C2 2 C2 x x 0 1 2 Particular solution: y 30. Differential equation: 9y General solution: y y y 9y 2 2x e 3 2 2x e 3 12y 3 12y 3 4y C2 x 3 0 e2x C1 C1 2 C 31 4y C2x C2 9 C2e2x 2 Cx 32 3 e2x e2x 3 2 C 31 2 C2 2 2x e 3 3 2 Cx 32 2 C 31 12 e2x 3 3 32 C 3 2C2 2 C 31 2 Cx 32 C2 2 Cx 32 4 e2x 3 2 2x e 3 2 C 31 0 2C2 2 Cx 32 C1 C2 x 0 Initial conditions: y 0 0 4 0 e2 C1 1 C1 e2 4 3C2 4, y 3 0 ⇒ C1 3C2 ⇒ C2 e2x 3 4 4 3 4 4 x 3 dy dx y ex 1 1 ex ex ex dx ln 1 ex C Particular solution: y 32. dy dx y x3 x3 4x 4x dx x4 4 2x2 C 34. 36. dy dx y u dy dx y x cos x 2 x cos x2 dx x 2, du 2x dx 1 sin x2 2 C 38. dy dx y tan2 x sec2 x sec2 x 1 dx 1 tan x x C 40. x5 x5 x. Let u x dx 5 5 x, u2 u2 u 5 2u du x, dx 2u du 10u2 10u3 3 10 5 3 2u4 du 2u5 5 x 32 C 2 5 5 x 52 C 530 dy dx y Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions dy dx 3y 2 dy y3 x3 3 x2 2 3y 2 x2 2x 2 dx C 42. 5e 5e x2 44. dx 5 2 10e e x2 x2 x2 1 dx 2 C 46. dr ds dr r 0.05s 0.05s ds 0.025s 2 C 48. xy dy y ln y y y dx x ln x Cx ln C ln Cx 50. y dy dx 6 cos x 6 cos x dx 6 sin x sin x C1 C y dy y2 2 y2 12 52. x2 9 dy dx dy y 5x 5x dx x2 9 5 x2 0 x1 2 dx 23 x 3 C 4, 2 2 2 54. 4y dy dx 3ex 3ex dx 3ex ln x2 dy 2x dx 0 2 ln x ln x dx x ln x 2 C 2 2 4y dy C 58. 2xy 2y2 9 12 C 56. x yy y1 2 dy 23 y 3 2 C1 dy y y3 2 x3 2 C Initial condition: y 1 43 Particular solution: y3 1 x3 32 2 8 9 1 9 C y1 y 2: 2 1 ln x 2 2 60. 1 y1 y2 1 y0 1: x2 1 2y dy dx dy 12 x1 1 1 0 1 1 y2 x2 x2 12 62. x dx C 1 r0 e e dr ds e rdr e r er e 2s 2s ds 2s y2 12 C⇒C x2 1 1 e 2 0: 1 1 e 2 1 e 2 ln ln 2s 2s C 1 2 1 2 C⇒C 1 2 1 y2 r r 1 2 2s r r 1 e 2 1 1 2 2s ln 1 e 2 2s 2 e Section 5.7 Differential Equations: Separation of Variables dy dx 3 dy y 2y 3x 2 dx x ln x2 Cx 2 531 64. dT kT 70 dt dT 70 70 70 0 k dt kt Ce kt 66. T ln T T C1 ln y3 y Ce0 kt, 3 ln C Initial condition: T 0 140 Particular solution: T 140; 70 70 70 70e C 70 1 e kt Initial condition: y 8 Particular solution: 8y3 2, 23 x2, y C 82 , C 12 x 2 3 1 8 T 68. m dy y ln y y dy dx dx x ln x Cx y x 0 0 y x 70. f x, y f t x, ty x3 t3x3 3x2y2 3t4x2y2 2y2 2t2y2 Not homogeneous C1 ln x ln C ln Cx 72. f x, y f t x, t y xy x2 y2 t2 tx ty x2 t 2 y 2 t xy x2 y2 74. f x, y f t x, t y tan x tan t x y ty tan t x y Not homogeneous t xy t x2 y2 Homogeneous of degree 1 y x ty tx tan y x 2 76. f x, y f t x, t y tan tan Homogeneous of degree 0 x3 y3 xy2 x3 dy y3 dx x dv v dx x3 v3 dx dx 1 dx x ln x 3 ln x 3x3 ln x C C Cx3 v dx x3 x3 dx vx 3 78. y 80. v dx 2v dx 2v x y dv dx x2 2xy x2 1 v y2 ,y vx x y 2 dy y vx, 2 v 2 x2 2x 2 v v2 dx x ln x ln C ln C x dx x vx x dv 2x dv dv 1 1 1 x2 x4 v2 dv x v 2 dv v 2 dv v3 3 y x 3 v3 x3 dx v2 ln v 2 v2 y2 x2 y2 1 C x C x Cx y3 532 Chapter 5 2x x 2x x 2 ln 2 Logarithmic, Exponential, and Other Transcendental Functions 3y 82. v x x ln 1 1 1 y dv dx dv dx v v y x y x y ,y vx 2 3v 84. x v dx 2 2 y 2 dx x 2 2 xx y dy x dv dx x ln ln ev ey x 0, y 0 vx 3vx x v v dx 1 v v dv 2v ⇒ x2 ln C dv 1 ln v x2C 2 dx x v ln v ln x ln C1 v C1 vx C1 x C1 xv xC x2C x2C Cx 3 1 x Initial condition: y 1 Particular solution: y dy dx x y y C1 y y 1, 1 e 1 Ce yx Ce 1 ⇒C e yx 86. 2 x 2 Let y 2x 2 2x2 2 y2 dx v x, dy v 2x 2 dx xy d y x dv 0 v dx. v dx 0 0 88. x vx x dv x3v xv dv v 1 v2 1 ln 1 2 ln 1 C1 C1 C x2 dv v2 v2 v2 y2 x2 y2 12 12 12 4 2x2v 2 2v 2 dx dx dv 2 −4 −2 −2 x 2 4 2 dx x 2 ln x ln x x 2 2 −4 C1 ln C y dy y2 2 y2 y2 1 2 x2 x dx x2 2 C C1 1 x2 1 x y1 0: 1 1 x 1 C2 x x 12 C1 x2 x x2 0 ⇒C y2 y2 1 Section 5.7 dy dx dy 4 dy y 4 y Differential Equations: Separation of Variables 533 90. 0.25x 4 0.25x dx y 8 y 4 0.25x dx −4 −2 −2 x 2 4 1 x dx 4 ln y y 4 4 y 12 x 8 eC1 4 1 8 x2 C1 Ce 1 8 x2 Ce 1 8 x2 92. dy dx 2 y, y 0 9 4 94. dy dx 0.2 x 2 10 y ,y 0 9 −5 −1 5 −5 5 0 96. dy ky, y Cekt dt Initial conditions: y 0 20 16 k Particular solution: y 98. 20, y 1 Ce 0 dy dx kx 4 0 along x 4: 16 C The direction field satisfies dy dx Matches (b). 20ek ln 4 5 5 20et ln 4 When 75% has been changed: 5 1 4 t 20et ln 4 et ln 4 5 5 ln 1 4 ln 4 5 6.2 hr 100. dy dx ky2 0, and 102. From Exercise 101, w w w0 w 1200 1200 w0 1200 Ce Ce kt, t 0 along y The direction field satisfies dy dx grows more positive as y increases. Matches (d). k 1 1200 1200 C⇒C w0 e t 1200 w0 534 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions y 104. Let the radio receiver be located at x0, 0 . The tangent line to y x x2 joins 1, 1 and x0, 0 . Transmitter (−1, 1) 2 1 y=x− −1 x2 Radio x1 x0 x (a) If x, y is the point of tangency on the y then 1 x 2x2 x2 1 2x 2x 2x 2 x y 1 Then 1 0 x0 3 1 1 6 x0 (c) 10 x 1 x2, (b) Now let the transmitter be located at 1 x x2 2x2 2x 1 h 2x 2x 1 x y x x 0 2± 1 4 2 2 x2 h h 1 4 2 x0 h x0 1 2h 2h 4 2 32 h h 32 h h 2 h h h 1 x2 x x h x2 1, h . h 1 y x x 0 1 1 x2 x x 1 x2 1 2± 2 x x2 4h 1 4 33 5 3 33 x0 6 1.155 8 5 6 1 3 y x 33 11 x0 6 33 33 3 Then, 33 h 1 0 x0 h 32 32 1 2h 4 4 h 43 6 6 33 h h 1 h2 h 4 32 0.25 −2 3 There is a vertical asymptote at h height of the mountain. 106. Given family (hyperbolas): x2 2x 2y2 4yy y Orthogonal trajectory: y dy y ln y y 2 1 4, which is the C 0 x 2y 108. Given family (parabolas): 2yy y Orthogonal trajectory (ellipse): y2 2C C y 2Cx y2 1 2x y y y dy 2x y y 2x 2y x 2 dx x 2 ln x kx 2 2x dx x2 K K1 ln k k x2 4 y2 2 2x 2 y2 −3 3 −6 6 −2 −4 Section 5.8 110. Given family (exponential functions): y y Orthogonal trajectory (parabolas): y y dy y2 2 y2 x 2x Cex Cex 1 y dx K1 K y Inverse Trigonometric Functions: Differentiation 535 4 −6 6 −4 112. The number of initial conditions matches the number of constants in the general solution. 114. Two families of curves are mutually orthogonal if each curve in the first family intersects each curve in the second family at right angles. 116. True dy x dx 118. True x2 y2 dy dx x C y K 2y 1 2Cy x C x y y Kx Cy x2 y2 x2 y2 dy dx 2Kx K y x2 x2 y2 y2 2x2 2y 2 y2 x2 x2 y2 1 x 2Kx 2Cy 2x2 2y 2 Section 5.8 2. y (a) x y (b) π Inverse Trigonometric Functions: Differentiation arccos x 1 3.142 y 0.8 2.499 0.6 2.214 (c) 0.4 1.982 0.2 1.772 0 1.571 0.2 1.369 0.4 1.159 0.6 0.927 0.8 0.634 0, 1 0 and 1, 0 (d) Intercepts: 2 No symmetry −1 0 x 1 1 −1 4. , , 4 6 1, 4 3 , 3 3, 6 3 6. arcsin 0 0 3, 536 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 5 6 16. arctan 3 1.25 8. arccos 0 2 3 2 5 6 2 2 14. arcsin 0.39 10. arc cot 3 12. arccos 0.40 18. (a) tan arccos tan 4 1 2 2 (b) cos arcsin 5 13 12 13 θ 13 5 12 θ 2 20. (a) sec arctan 3 5 34 5 5 θ −3 34 (b) tan arcsin 5 6 5 11 11 11 θ 6 −5 22. y sec arc tan 4x arctan 4x 1 + 16x 2 4x θ 24. y cos arccot x arccot x θ x2 + 1 1 y sec 1 16x2 1 y cos x x2 x r h r r2 r 5 5 x 1 tan 1 tan 2 1 1 5 x h 2 x 1 h r x−h 26. y sec arcsin x arcsin x 1 1 2x 1 1 x−1 28. y θ 2x − x2 cos arcsin arcsin x θ r 2 − (x − h)2 y sec x2 y cos 30. −2 5 32. arctan 2x f=g 2 2x 1.721 −5 Asymptote: x arccos cos tan x 2 x 2 0 2 4 − x2 θ x 4 x x2 S ection 5.8 34. arccos x x x x2 x 36. (a) arcsin Let y x Thus, arcsin y x arcsec x cos arcsec x x Inverse Trigonometric Functions: Differentiation 537 1 x 1 ±1 x2 − 1 θ 1 x arcsin arcsin x, x ≤ 1. x . Then, sin y ⇒ x sin y. sin y ⇒ x arcsin x ⇒ y arcsin x. (b) arccos Let y x x arccos arccos x, x ≤ 1. x . Then, cos y ⇒ x cos y. cos y ⇒ x arcsin x. Therefore, y arccos x ⇒ y arccos x. Thus, arccos x. Therefore, arccos x x 4 y 38. f x x arctan x tan y , 2 π 40. f x x 4 x − 6 − 4 −2 x 2 4 6 arccos cos y 4 cos y y (− 4, π ) π 2 π 2 Domain: Range: 0, Domain: Range: 0, 4, 4 (4, 0) −6 −4 −2 2 4 6 x f x is the graph of arctan x shifted 42. f t ft arcsin t 2 2t 1 t4 arctan 1 1 x x 1 2x 1 2 x1 2 units upward. 44. f x fx arcsec 2x 2x 2 4x2 1 x 1 4x2 1 46. f x fx 48. h x hx x2 arctan x 2x arctan x x2 1 x2 x 50. f x fx arcsin x 0 arccos x 2 52. y y ln t2 2t t2 2t t2 4 4 4 1 2 1 t arctan 2 2 1 1 1 t2 4 t 2 2t t2 2 1 2 1 4 54. y y 1 x4 2 11 x4 22 1 2 4 x2 4 x2 x2 x2 4 arcsin 12 x 2 4 4 4 x2 x2 2 1 1 x2 2 56. y dy dx x arctan 2x 2x 4x2 1 ln 1 4 arctan 2x 4x2 1 8x 4 1 4x2 arctan 2x 2x 4 x2 1 x2 538 Chapter 5 x 5 x2 Logarithmic, Exponential, and Other Transcendental Functions x 2 1 x2 4 1 2 x2 2 58. y y 25 arcsin 5 1 1 x 25 25 x2 x2 x 1 25 2 x2 12 60. y 2x y arctan 1 21 2 x2 4 12 x 2 4 2 2 2x 25 25 x2 2x2 25 x2 62. f x fx fx P1 x P2 x arctan x, a 1 1 1 f1 f1 x2 2x x2 25 x2 25 x2 x2 25 x2 x x2 4 2 2x2 8 x x2 4 2 1 π 2 π 4 −4 −2 x 2 y P1 (x) f 2 P ( x) 2 f1x f1x 1 1 4 1 x 2 1 1 2 1 f1x 2 4 1 x 2 1 1 x 4 66. f x 1 2 64. f x fx x fx f 3 2 arcsin x 1 1 ± 2x 2 0 when 1 x2 1 or 2 arcsin x 1 1 1 x2 x4 3 x x2 2 arctan x 2 1 x2 x2 x2 21 41 0 ± 0.681 x2 fx 0 3 . 2 x x2 1 32 2x2 6x2 1 x4 >0 3 , 2 0.68 Relative minimum: f 3 2 <0 By the First Derivative Test, 0.681, 0.447 is a relative maximum and 0.681, 0.447 is a relative minimum. Relative maximum: 3 , 0.68 2 arctan x. 70. The derivatives are algebraic. See Theorem 5.18. 750 s arccos d dt 11.001 rad hr. 66.667 rad hr. d ds ds dt 750 s 1 1 750 s 2 68. arctan 0 0. is not in the range of y 72. (a) cot x 3 arccot x 3 74. cos s θ 750 (b) d dt If x If x x2 3 dx 9 dt d dt 750 ds s2 dt 10, 3, 750 ds s s2 7502 dt d dt A lower altitude results in a greater rate of change of . Section 5.9 76. (a) Let y arcsin u. Then 1 Inverse Trigonometric Functions: Integration (b) Let y arctan u. Then tan y sec2 y dy dx dy dx u y 539 1 + u2 u sin y cos y y dy dx (c) Let y u u u cos y u . 1 u2 u y 1 − u2 u u u sec2 y u 1 u2 . 1 arcsec u. Then sec y u y u2− 1 1 (d) Let y arccos u. Then 1 dy sec y tan y dx dy dx cos y sin y dy dx dy dx (f) Let y u y 1 − u2 u u sec y tan y u u sin y u . 1 u2 u y u u u u2 1 . Note: The absolute value sign in the formula for the derivative of arcsec u is necessary because the inverse secant function has a positive slope at every value in its domain. (e) Let y arccot u. Then cot y csc2 y dy dx dy dx u y 1 + u2 1 u arccsc u. Then csc y csc y cot y dy dx dy dx u u u csc y cot y 1 u2 − 1 u u csc2 y u 1 . u2 u u u2 1 Note: The absolute value sign in the formula for the derivative of arccsc u is necessary because the inverse cosecant function has a negative slope at every value in its domain. 78. f x gx sin x arcsin sin x arcsin x is 2 2 2≤y≤ −2 3 f 2 (a) The range of y (b) Maximum: Minimum: 80. False The range of y 2. g −3 82. False 2 arcsin x is ,. 22 arcsin2 0 arccos2 0 0 2 1 Section 5.9 2. 3 1 4x2 dx Inverse Trigonometric Functions: Integration 3 2 1 2 4x2 dx 3 2 1 arcsin 2x C 4. 0 1 4 x 2 dx arcsin x 2 1 0 6 6. 4 1 9x2 1 x dx 4 3 dx 3 1 9x2 4 arctan 3x 3 3 C 8. 3 1 9 x 2 dx 1 x arctan 3 3 3 3 36 10. 4 1 2 dx 1 x1 arctan 2 2 C 12. x4 x2 1 dx 1 x2 1 dx 13 x 3 x C 540 Chapter 5 t 2, du t t4 16 dt 1 2 Logarithmic, Exponential, and Other Transcendental Functions 2t dt. 1 4 2 14. Let u 16. Let u t2 2 x2, du 4 dx 2x dx. 1 2 x2 1 x2 2 22 C 2x dx 2t dt 1 arctan 8 4 t2 C 1 x x4 1 x2 arcsec 4 2 1 1 1 0 2 18. Let u 1 0 2 arccos x, du arccos x dx 1 x2 x2 dx. 20. Let u 0 3 1 x 1 x 2 x2, du dx 1 2 2x dx. 0 3 arccos x dx 1 x2 1 0 2 1 1 x2 x2 2x dx 0 1 arccos2 x 2 2 32 32 0.925 1 ln 1 2 ln 2 3 22. 1 3 2 1 x 2 2 2 dx 1 1 32 x 2 2 dx 2 1 x2 arctan 3 3 2 1 3 18 1 dx, dx 2x C C 24. 0 cos x dx 1 sin2 x arctan sin x 0 4 26. 3 2 x1 3 2u du 2 u 1 u2 x dx. u 3 du 1 x, du 2u du u2 3 arctan u 3 arctan x 28. 4x 1 x x 1 3 dx x2 2 2 2 2x 1 2x x 1 2 x2 dx 3 1 1 x2 dx 41 x2 3 arcsin x C 30. 4 dx 1 2 2 4 dx x 3 12 4 dx C 1 ln x2 2 2 2x 5 3 x1 arctan 2 2 1 x2 arctan 3 3 7 1 x dx 32. 2 x2 2x dx 4x 5 2x 2 x2 2 2 13 dx 2 x 2x dx 22 2 2x 2 2 2 2 9 dx 4 1 arctan 3 3 ln x2 2x 2 7 arctan x 1 C 34. x2 x2 1 1 2 36. 4x dx 4 4 2 arcsin x2 2 x x 2 4x 2 2 dx 2 C 4 dx 38. Let u x2 2x, du 1 2 2x x2 x2 2 dx. 2x 12 x1 dx x2 2x 2x 2 dx 2x C 40. 1 x 1 x2 2x dx x 1 1 x 1 2 1 dx arcsec x 1 C Section 5.9 42. Let u x2 x 8x2 x 2 dx 1 4, du dx 2x dx. 1 2 2x x2 dx 1 x2 4 arcsin 2 5 Inverse Trigonometric Functions: Integration 541 9 44. Let u x x x4 25 4 dx 6 3 C 2 C 2, u2 2u2 u2 2u 2 3 x, 2u du du 2u2 u2 6 du 2 du 2 6 1 u2 3 du x 3 2 C u 6 arctan 3 3 92 :x 4 2x 2 3 arctan 46. The term is 3 2 2 3x x2 3x 9 4 9 4 x 3 2 2 9 4 48. (a) (b) ex dx cannot be evaluated using the basic integration rules. xex dx 2 2 1 x2 e 2 C, u x2 (c) 1 1x e dx x2 e1 x C, u 1 x 50. (a) (b) (c) 1 1 x 1 x3 1 x4 x4 x4 dx cannot be evaluated using the basic integration rules. dx dx 1 2x dx 21 x2 2 1 4x3 dx 4 1 x4 1 arctan x2 2 1 ln 1 4 x4 C, u C, u 1 x2 x4 52. (a) 4 y 54. dy dx 2y ,y 0 16 x2 3 2 −1.25 x 1.25 −3 −4 −1 3 (b) dy dx x 16 dy 16 y2 arcsin y 4 y2, 0, x dx x2 2 2 4 x2 2 y 4 y sin 2 2 C C⇒C 0, 2 : arcsin arcsin y 4 6 6 x2 6 x2 2 6 4 sin 542 Chapter 5 1 Logarithmic, Exponential, and Other Transcendental Functions x 2 1 0 1 56. A 0 y 1 4 dx x2 arcsin 6 58. 0 arcsin x dx y 0.571 1 π 2 π 4 x 1 2 x 1 2 1 1 60. F x 1 2 x x 2 2 t2 1 dt 2 . Maximum at x 1, since the graph is (a) F x represents the average value of f x over the interval x, x 1, 1 . greatest on x 2 (b) F x Fx arctan t x arctan x 1 2 2 2 1 arctan x x2 x2 1 x2 x2 4x 5 4x 5 4x 1 x2 1 4x 0 when x 1. 1 1 x 1 x2 x2 5 62. 1 6x (a) 6x x2 x2 1 6x dx 9 dx x, 2u du 2u du x2 6x 9 dx x dx 2 6 u2 du 2 arcsin u 6 C 2 arcsin x 6 C 9 x 3 2 (c) x 3 3 C 4 y2 x2 9 3 2 arcsin y1 −1 7 (b) u x, u2 1 6u2 u4 −2 The antiderivatives differ by a constant, 2. Domain: 0, 6 64. Let f x fx Since f 0 arctan x 1 1 x2 x 1 1 1 x2 x2 x2 2 1 2x2 > 0 for x > 0. x2 5 y 0 and f is increasing for x > 0, arctan x x 1 x2 . x 1 x2 y3 > 0 for x > 0. Thus, 4 3 2 1 arctan x > Let g x gx Since g 0 Therefore, x 1 x2 x 1 y2 y1 2 4 6 8 10 arctan x 1 1 x 2 x x2 1 x2 > 0 for x > 0. arctan x > 0 for x > 0. Thus, x > arctan x. 0 and g is increasing for x > 0, x < arctan x < x. Section 5.10 Hyperbolic Functions 543 Section 5.10 2. (a) cosh 0 (b) sech 1 e0 2 2 ee Hyperbolic Functions e0 1 0.648 4. (a) sinh (b) tanh 1 1 1 0 0 0 0 6. (a) csch (b) coth 1 2 3 ln 1 2 5 0.481 1 1 4 ln 2 2 1 e2x 2 ex 2 x 2 y e 0.347 8. 1 cosh 2x 2 e 2x 2 e2x 2 4 e2x e 2x ex 2 e x2 cosh2 x x 10. 2 sinh x cosh x 2 ex 2 ex ex e x e 2 2x sinh 2x 12. 2 cosh x 2 y y2 cosh 2 2 e 2 e 4 x y2 ex x y2 e 2 e 2 x x y2 ey y e ex ey 2 e y cosh x 1 2 1 ⇒ sech2 x 1 32 1 12 2 1 2 23 3 cosh y Putting these in order: 14. 1 2 2 tanh x sech2 x cosh x coth x sinh x csch x 3 ⇒ sech x 4 3 2 sinh x cosh x tanh x 3 3 23 3 1 2 csch x sech x coth x 2 3 3 2 tanh x cosh x 1 33 3 23 3 3 3 16. y y coth 3x 3 csch2 3x 18. g x gx ln cosh x 1 sinh x cosh x t 1 coth t csch2 t coth2 t tanh x 20. y y x cosh x x sinh x sinh x cosh x cosh x x sinh x 22. h t ht 544 Chapter 5 sech2 3x Logarithmic, Exponential, and Other Transcendental Functions 26. f x fx esinh x cosh x esinh x 28. y y sech x sech x 1 1 tanh x 1 24. g x gx 2 sech 3x sech 3x tanh 3x 3 6 sech 3x tanh 3x 2 30. f x fx x sinh x x cosh x 1 1 cosh x sinh x 1 1 sinh x 1 x cosh x 1 −6 6 fx 0 for x Test, 0, cosh 0. By the First Derivative 1 0, 1.543 is a relative minimum. 6 (0, −1.543) −2 32. h x 2 tanh x 2 x 34. y y a cosh x a sinh x a cosh x y 0. (0.88, 0.53) −3 3 y (− 0.88, − 0.53) −2 Therefore, y Relative maximum: 0.88, 0.53 Relative minimum: 0.88, 0.53 3 36. f x fx fx P1 x P2 x cosh x sinh x cosh x f0 1 f1 f1 f1 f0x 0 cosh 0 sinh 0 cosh 0 1 1 0 1 −2 f P 2 P1 2 0 12 2x 38. (a) y 18 25 cosh 80 x , 25 25 ≤ x ≤ 25 (b) At x At x (c) y ± 25, y 18 18 25 25 cosh 1 43. sinh 1 56.577. 0, y sinh x . At x 25 25, y 1.175 −25 −10 25 40. Let u cosh x x x, du dx 1 dx. 2x 2 cosh x 1 2x dx 2 sinh x C 42. Let u 1 cosh x, du sinh dx sinh2 x sinh x dx. sinh x dx cosh2 x sech x C 1 cosh x C 44. Let u 2x 1, du 1 dx 2 dx. 1 sech2 2x 2 1 tanh 2x 2 1 1 2 dx C 46. Let u sech x, du sech x tanh x dx. sech2 x 1 sech3 x 3 sech x tanh x dx C sech2 2x sech3 x tanh x dx Section 5.10 1 1 x 2 1 x 2 2 x1 4x2 cosh 2x dx 2 sinh 2x 2 1 sinh 2x 4 1 1 C C 4 Hyperbolic Functions x 5 4 545 48. cosh2 x dx 50. 0 1 25 x 2 dx arcsin arcsin 0 4 5 52. dx 2 2x 2x 2 2 dx 2 ln 1 1 2x 4x2 C 54. Let u 9 sinh x, du cosh x dx sinh2 x cosh x dx. arcsin arcsin sinh x 3 e x 56. y C x tanh 1 x 2 2 y C 1 1 x2 1 2 2 4 x2 e 6 58. y sech y 1 cos 2x , 0 < x < 1 4 2 sin 2x 4. 2 sin 2x cos 2x sin 2x 2 cos 2x 2 sec 2x, cos 2x 1 cos2 2x since sin 2x ≥ 0 for 0 < x < 60. y y csch 2 csch 1 x 1 2 x 1 x 1 x2 2 csch 1 x x 1 x2 1 ln 1 2 1 62. y y x tanh x 1 1 1 x ln 1 tanh 1 x2 x x tanh x x2 1 x x2 64. See page 401, Theorem 5.22. x2 1 tanh x 66. Equation of tangent line through P y When x y a sech 0, a sech 1 1 x0, y0 : a2 x02 x x0 x0 Q y x0 a a2 x02 a P x0 a a2 x02 1 a2 x0 a . a2 x02 a sech 1 x0 . a (a, 0) x Hence, Q is the point 0, a sech Distance from P to Q: d x 9 x4 1 29 1 3 ln 12 3 L x02 2x dx x2 2 x2 x2 C x02 2 x2 x2 a 68. dx 11 3 ln 26 3 C 546 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions 3 2 2 3 70. Let u x 1 x3 2, du dx x dx. 1 1 x3 22 x3 3 2 x dx 2 sinh 3 1 x3 2 C 2 ln x3 3 2 1 x3 C 72. x 2 dx x2 4x 8 x 2 dx x x x 2 2 4 2 2 1 ln 2 1 2x2 2 2 4 C 74. x 1 4x 8 dx x 1 2 1 1 2x 1 1 2 6 dx dx 1 ln 6 3 x x 1 1 2 3 C x 1 x 1 2 3 2 76. Let u y 2x 1 , du 1 4x2 2 dx. dx 2x 1 2 3 2 x 1 8x 1 dx 2x 1 2 1 ln 3 3 2x 4x2 1 8x 1 C 78. y 1 4x ln 4x 2x dx x2 x2 4 4x 3 x ln 4 x 2x dx x2 2 2 2 2 3 x C 1 22 ln 4x 4 dx x2 3x4 ln 4 x C 2 5 80. A 0 tanh 2x dx 2 2x e 0 82. A 3 6 x2 4 dx 5 e2x 2 0 e e 1 e 2x 2x dx 2 e2x 2 2x 0 6 ln x e 2x x2 21 21 5 4 3 1 2 e2x 2x dx 6 ln 5 6 ln 5 3 6 ln 3 3.626 5 1 ln e2x 2 1 ln e4 2 ln e4 2 32t e e 4 1 ln 2 2 4 e 1.654 84. (a) v t (b) s t s0 st v t dt 16 0 16t2 2 32t dt C 400 16t2 C 400 400 ⇒ C —CONTINUED— Section 5.10 84. —CONTINUED— dv dt dv kv2 dv 32 Let u 1 k Since v 0 ln 32 32 32 32 32 v k ke 2 Hyperbolic Functions 547 (c) 32 dt dt kv2 (d) lim t→ 32 tanh k 32k t 32 32 . k k. 32 kv 2 The velocity is bounded by (e) Since tanh ct dt 1 c ln cosh ct (which can be verified by differentiation), then k dv. kv kv t C st 32 tanh k 32 k 32k t dt 32k t C. C k v, then du 1 ln 2 32 0, C kv kv kv kv kv 32k t 32 32 0. 1 ln cosh 32k 32k t 2 32k t e e 2 32k t 1 ln cosh k When t s0 32 32k t 0, C 400 ⇒ 400 1 k ln cosh 32k t . 2 32k t kv When k 1 1 1 e e 32k t e e 32k t 32k t 0.01, 400 16t 0 when t 0 when t 2 32 e 32 e ke 32 k 2 s2 t s1 t s1 t s2 t 100 ln cosh 0.32 t 400. 5 seconds. 8.3 seconds 2 2 32k t 32k t v e e 32k t 32k t 32k t 32k t When air resistance is not neglected, it takes approximately 3.3 more seconds to reach the ground. 32 tanh k 86. Let y arcsin tanh x . Then, tanh x ex 2 e x sin y tan y Thus, y ex ex e e x x and e +e x −x e −e y 2 x −x sinh x. arctan sinh x . Therefore, arcsin tanh x . y sech y sech x 1 1 sech y tanh y 1 1 arctan sinh x 88. x sech y tanh y y y 1 sech y 1 sech y 2 1 x1 x2 90. y sinh y cosh y y y sinh x 1 x 1 cosh y 1 sinh2 y 1 1 x2 1 ...
View Full Document

This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

Ask a homework question - tutors are online