ODD02 - CHAPTER 2 Differentiation Section 2.1 Section 2.2...

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Unformatted text preview: CHAPTER 2 Differentiation Section 2.1 Section 2.2 Section 2.3 Section 2.4 Section 2.5 Section 2.6 The Derivative and the Tangent Line Problem . . . 53 Basic Differentiation Rules and Rates of Change . 60 The Product and Quotient Rules and Higher-Order Derivatives . . . . . . . . . . . . . . 67 The Chain Rule . . . . . . . . . . . . . . . . . . . 73 Implicit Differentiation . . . . . . . . . . . . . . . 79 Related Rates . . . . . . . . . . . . . . . . . . . . 85 . . . . . . . . . . . . . . . . . . . . . . . . . 92 Review Exercises Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . 98 CHAPTER Differentiation Section 2.1 2 The Derivative and the Tangent Line Problem Solutions to Odd-Numbered Exercises 1. (a) m (b) m 0 3 3. (a), (b) (c) y f4 4 3 x 3 1x f1 x 1 1 1 2 2 1 f 1) y y 6 5 4 3 2 1 f )4) 4 f )1) )x 1 1) f )1) x 1 f )4) 5 )4, 5) x f )4) f )1) 3 1 f )1) )1, 2) 2 x 1 2 3 4 5 6 5. f x 3 2x is a line. Slope 2 7. Slope at 1, 3 lim lim g1 1 1 x x→0 x x 2 g1 4 x x x 2 2 3 1 x→0 lim 2x 2x x→0 lim 2 x→0 9. Slope at 0, 0 lim lim f0 3t t t t t t 3 f0 2 t→0 11. f x 0 fx 3 lim fx 3 x 0 3 x→0 t→0 x x fx lim 3 t→0 lim x→0 lim 0 x→0 13. f x fx 5x lim fx 5x 5 5 x→0 15. h s x x x x fx hs 5x 3 lim 2 s 3 hs 2 s 3 s→0 s s hs 2 s 3 lim lim x→0 3 lim s→0 s s 3 x→0 2 s 3 lim s→0 s 2 3 53 54 Chapter 2 Differentiation 17. f x fx 2x2 lim lim x→0 x fx 2x 2x2 4x x 1 x x x 2 fx x 2x 2 x x 2 1 x x x 2x2 1 x 1 2x2 x 1 x→0 lim lim 19. f x fx x3 4x x 2x x x→0 x x→0 lim 4x x→0 2x 1 4x 1 12x fx x x3 x x x 3 lim lim lim fx 12 x x 3x x 2 2 x→0 x x x 3 x3 3 12x 12x 12 x x3 12x x→0 3x2 x x→0 x 3x2 x 3x x 3x x 12 x 3x2 12 x x 2 lim x→0 lim 3x2 x→0 12 21. f x fx 1 x lim x→0 1 fx 1 x x x 1 x x x x x 1 2 fx 1 x 1 lim x x→0 lim lim lim x xx xx x 1 1 x→0 x 1x x 1x 1 1 1 x→0 x→0 1 1x 1 x 23. f x fx x lim x→0 1 fx x x x x 1 1 x x x x x x x x 1 1 1 fx 1 x 1 1 x 1 2x x x x 1 1 1 1 1 x x x x 1 1 x x 1 1 lim lim lim x→0 x→0 x→0 x Section 2.1 The Derivative and the Tangent Line Problem 55 25. (a) f x fx x2 lim x→0 1 fx x x x 2 17. (b) fx −5 8 (2, 5) 5 −2 lim x 2x x x→0 1 x x 2 x2 1 lim x→0 x x 2x lim 2x x→0 At 2, 5 , the slope of the tangent line is m 22 4. The equation of the tangent line is y y 5 5 y 4x 4x 4x x3 lim fx x 3x2 x x→0 2 8 3. 27. (a) f x fx 18. (b) x x x3 x fx −5 10 (2, 8) 5 −4 lim lim x3 2 x→0 x→0 3x x x 3x x x x 2 3 lim 3x2 x→0 3x2 32 2 At 2, 8 , the slope of the tangent is m The equation of the tangent line is y 8 y 29. (a) f x fx x lim lim lim fx x x x x x 1 x x→0 12. 12 x 12x 2 16. 18. (b) x x x x x x x fx −1 3 (1, 1) 5 −1 x x x x→0 x x x x x x x→0 lim x→0 1 2x At 1, 1 , the slope of the tangent line is m 1 21 1 . 2 The equation of the tangent line is y 1 y 1 x 2 1 x 2 1 1 . 2 56 Chapter 2 Differentiation 31. (a) f x fx 4 x lim fx x→0 (b) x x x fx 4 x x xx x3 xx 2x 2 x 2 10 (4, 5) − 12 12 x lim lim lim lim lim x2 x2 x→0 x x x 4 x x 4x 4x x −6 x→0 4x x 2 x x xx x x3 x x→0 x x2 x xx 4x x x2 x xx x2 x x→0 x xx x2 xx 4 1 xx x 4 x2 4 x→0 At 4, 5 , the slope of the tangent line is m 1 4 16 3 4 The equation of the tangent line is y y 5 3 x 4 3 x 4 2 3x2. Since the 35. Using the limit definition of derivative, fx 1 2x x . 1 2, 4 33. From Exercise 27 we know that f x slope of the given line is 3, we have 3x2 x 3 ± 1. Since the slope of the given line is 1 2x x x 1. 1 2 we have Therefore, at the points 1, 1 and 1, 1 the tangent lines are parallel to 3x y 1 0. These lines have equations y 1 y 3x 3x 1 2 and y 1 y 3x 3x 1 2. Therefore, at the point 1, 1 the tangent line is parallel to x 2y 6 0. The equation of this line is y y 1 1 y 1 x 2 1 x 2 1 x 2 1 1 2 3 . 2 37. g 5 g5 2 because the tangent line passes through 5, 2 2 5 0 9 2 4 1 2 39. f x x⇒f x 1 b S ection 2.1 The Derivative and the Tangent Line Problem 57 41. f x x ⇒ f x matches (a) 43. 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 y decreasing slope as x → x 1 2 3 4 Answers will vary. Sample answer: y 45. (a) If f c (b) If f c 3 and f is odd, then f 3 and f is even, then f c c fc fc 3 3 4 2x. The slope of the x 47. Let x0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, f x y line through 2, 5 and x0, y0 equals the derivative of f at x0: 5 2 5 5 4x0 y0 x0 y0 x02 0 0 7 4 2 8 x0 2 2x0 x0 4 8x0 4x0 1 x0 2x0 2x02 3 3 ⇒ x0 1, 3 −2 6 5 4 3 2 1 x 1 2 3 6 (2, 5) (3, 3) (1, 3) x0 Therefore, the points of tangency are 1, 3 and 3, 3 , and the corresponding slopes are 2 and lines are y 5 y 49. (a) g 0 (b) g 3 0 8 3, 2. The equations of the tangent 2x 2x 3 2 1 y 5 y 2x 2x 2 9 (c) Because g 1 (d) Because g 4 g is decreasing (falling) at x g is increasing (rising) at x 1. 4. g 4 > 0. 2. 7 3, (e) Because g 4 and g 6 are both positive, g 6 is greater than g 4 , and g 6 (f) No, it is not possible. All you can say is that g is decreasing (falling) at x 51. f x 13 4x 32 4x . 2 By the limit definition of the derivative we have f x x fx fx 2 2 3 1.5 27 32 27 16 −2 2 1 1 4 3 4 0.5 1 32 3 16 0 0 0 0.5 1 32 3 16 1 1 4 3 4 1.5 27 32 27 16 2 2 3 −2 58 Chapter 2 Differentiation 53. g x fx 2x 3 0.01 0.01 0.01 fx x 0.01 2 55. f 2 2x x2 100 f2 24 3.99 2.1 2 4 2 4, f 2.1 2.1 4 2.1 0 3.99 0.1 Exact: f 2 g f −2 −1 4 The graph of g x is approximately the graph of f x . 57. f x 1 x 5 and f x 1 . 2x3 2 As x → 5 f −2 , f is nearly horizontal and thus f 0. f′ −5 59. f x S x 4 x x f2 4 2 3 2 x x x x x f2 3 2 x 3 2 x x 3 2 3 f2 2 1 3 x 2 19 x 10 4 5 −2 3 1 x x 1 2 x 2 3 x 5 2x 2 3 (a) x x x 1: S 0.5: S 0.1: S x 2 2 3 S 0.1 x 3 x 2 19 x 10 S1 S 0.5 − 1 f 7 x (b) As x → 0, the line approaches the tangent line to f at 2, 3 . 61. f x f2 x2 x→2 1, c fx x 2x2 x→ 2 2 f2 2 1, c fx x 0 g0 0 x→0 x→2 lim lim x2 x 1 2 3 x →2 lim x 2x x2 2 x→2 lim x 2 4 63. f x f 65. g x g0 2 x3 2 f2 2 x→ 2 lim lim x2 x x 2 2 x→ 2 lim x3 2x2 1 x2 67. f x 1 x→ 2 lim x2 23 4 6 x ,c x→0 x x→6 6 ,c f6 6 lim gx x x x x x lim x x . Does not exist. f6 lim lim fx x x As x → 0 , As x → 0 , 1 → x 1 x → x→6 623 x6 1 6 13 0 x→6 lim x Does not exist. S ection 2.1 The Derivative and the Tangent Line Problem 59 69. h x h 5 x 5 ,c x→ 5 5 h 5 5 x 5 5 5 0 5 lim hx x x x x 71. f x is differentiable everywhere except at x (Sharp turn in the graph.) 3. x→ 5 lim x→ 5 lim Does not exist. 73. f x is differentiable everywhere except at x (Discontinuity) 77. f x is differentiable on the interval 1, (At x 1 the tangent line is vertical) 81. f x x 1 . 1. 75. f x is differentiable everywhere except at x (Sharp turn in the graph) 79. f x is differentiable everywhere except at x (Discontinuity) 83. f x x x 1 3, x ≤ 1 1 2, x > 1 3. 0. The derivative from the left is x→1 The derivative from the left is x x 1 1 0 1. x→1 lim fx x f1 1 x→1 lim lim fx x f1 1 x→1 lim x x 1 1 3 0 0. 1 2 The derivative from the right is x→1 x→1 lim x lim fx x f1 1 x→1 lim x x 1 1 0 1. The derivative from the right is x→1 The one-sided limits are not equal. Therefore, f is not differentiable at x 1. lim fx x f1 1 x→1 lim x x 1 1 2 0 0. 1 x→1 lim x These one-sided limits are equal. Therefore, f is differentiable at x 1. f 1 0 85. Note that f is continuous at x 2. f x fx x fx x x2 4x f2 2 f2 2 1, x ≤ 2 3, x > 2 x→2 The derivative from the left is lim x→2 lim lim x2 x 4x x 1 2 3 2 5 5 x→2 lim x lim 4 4 y 2 4. 4. The derivative from the right is lim x→2 x→2 x→2 The one-sided limits are equal. Therefore, f is differentiable at x 87. (a) The distance from 3, 1 to the line mx d Ax1 A m3 By1 2 2. f 2 y 4 0 is C B 2 3 2 11 m2 1 4 3m m2 3 1 . 1 x (b) 5 1 2 3 4 The function d is not differentiable at m 1. This corresponds to the line y x 4, which passes through the point 3, 1 . −4 −1 4 60 Chapter 2 Differentiation 89. False. the slope is lim f2 x→0 x x f2 . 91. False. If the derivative from the left of a point does not equal the derivative from the right of a point, then the derivative does not exist at that point. For example, if f x x , then the derivative from the left at x 0 is 1 and the derivative from the right at x 0 is 1. At x 0, the derivative does not exist. x sin 1 x , 0, x x 0 0 0. Thus, lim x sin 1 x x→0 93. f x x ≤ x sin 1 x ≤ x , x Using the Squeeze Theorem, we have x 0. Using the alternative form of the derivative we have x→0 0 f 0 and f is continuous at lim fx x f0 0 x→0 lim x sin 1 x x0 0 x→0 lim sin 1 . x 0. Since this limit does not exist (it oscillates between gx x2 sin 1 x , x 0, x 0 0 1 and 1), the function is not differentiable at x Using the Squeeze Theorem again we have x2 ≤ x2 sin 1 x ≤ x2, x 0. Thus, lim x2 sin 1 x x→0 ous at x 0. Using the alternative form of the derivative again we have x→0 0 f 0 and f is continu- lim fx x f0 0 x→0 lim x2 sin 1 x x0 0 x→0 lim x sin 1 x 0. Therefore, g is differentiable at x 0, g 0 0. Section 2.2 1. (a) y y y1 x1 2 Basic Differentiation Rules and Rates of Change (b) y y y1 x6 6x5 y 7x 8 x3 2 (c) y y y1 1 x7 x2 2x 2 (d) y y y1 5 x3 3x2 3 x1 45 5 1 12 2x 1 2 3 12 2x 3 2 3. y y 8 0 5. y y 7. y x 7 9. y 7 x8 y x 1 x 5 1 5x4 5 11. f x fx x 1 1 13. f t fx 2t2 4t 3t 3 6 15. g x gx x2 2x 4x3 12x 2 17. s t st 1 x 1 x2 t3 3t2 2t 2 4 19. y y 2 2 sin cos cos sin 21. y y x2 2x 1 cos x 2 1 sin x 2 23. y y 3 sin x 3 cos x Section 2.2 Basic Differentiation Rules and Rates of Change 61 Function 25. y 5 2x 2 3 2x x x 3 x2 6x 6 Rewrite y 5 x 2 3 x 8 x 2 Derivative y 5x 9 x 8 1 x 2 3 Simplify y 5 x3 9 8x4 1 2x3 73 x, 5 0, 2 27. y 3 y y 3 y 4 y 29. y 12 y 32 y 1 2 21 2 x 5 0 31. f x 3x 2, 1, 3 3 33. f x fx f0 1 2 35. y 2x 4x2 1 2, 0, 1 4x 4 1 fx f1 6 x3 y y0 8x 4 37. f f f0 4 sin 4 cos 41 x3 1 3x2 x2 8 x3 x 1 x 2 12 , 0, 0 1 3 4 x3 x3 3 39. f x fx x2 2x 5 6x 3x 3 2 41. g t 6 x3 gt t2 4 t3 12t t2 4t 3 2x 2t 4 2t 12 t4 43. f x fx 47. f x fx x 8 3 4x 2 45. y y x x2 3x2 1 1 x3 x 1 6 x x1 23 2 6x1 1 2x 3 49. h s 2 x2 3 h (s s4 4 s 5 5 s2 45 3 2x 2 s 3 13 4 5s1 5 2 3s1 3 51. f x fx 6x 3x 12 5 cos x 5 sin x 6x1 3 2 5 cos x 5 sin x x 53. (a) y y x4 4x3 3x2 6x 2 55. (a) f x fx 1 4 2 x3 3 x 2 2x 74 34 At 1, 0 : y Tangent line: 41 3 61 0 2 0 2. 2x 3 2x7 4 3 2 y y 2 3 x 2 7 0 3 x 2 7 2 1 y 2x y At 1, 2 , f 1 Tangent line: (b) 3 −2 −1 2 (1, 0) 3x (b) 5 2y (1, 2) −2 −1 7 62 Chapter 2 Differentiation 57. y y x4 4x3 4x x2 4x x 8x2 16x 4 2x 2 59. y y 2 1 x2 2x x 3 2 2 cannot equal zero. x3 Therefore, there are no horizontal tangents. y 0⇒x 0, ± 2 14 , 2, 14 63. x 2 2x kx k 4x 4 2x 4x 9 Equate functions Equate derivatives 4 and 4x 9⇒ x2 3, k 9⇒x 10. ± 3. Horizontal tangents: 0, 2 , 2, 61. y y cos x At x x 1 sin x, 0 ≤ x < 2 cos x 0 1⇒x ,y . , Hence, k x 2 2x 3, k Horizontal tangent: For x 2 and for x 65. k x k x2 3 x 4 3 4 3 Equate functions Equate derivatives 32 x 32 4 x and 4 x Hence, k 3 x 4 3 3⇒ x 4 3 x 4 3 3⇒ x 2 (c) 3⇒x 2⇒k 3. 67. (a) The slope appears to be steepest between A and B. (b) The average rate of change between A and B is greater than the instantaneous rate of change at B. y f BC A D E x 69. g x fx 6⇒g x fx 71. 3 y f 1 3 2 1 2 1 f x 2 3 If f is linear then its derivative is a constant function. fx fx ax a b S ection 2.2 Basic Differentiation Rules and Rates of Change 63 73. Let x1, y1 and x2, y2 be the points of tangency on y these functions are y m 2x ⇒ m 2x1 x1 Since y1 m y2 x2 x22 x22 6x2 2 x2 and y 2x2 x2 6. 6x 5, respectively. The derivatives of 2x1 2x2 x2 3 and 6 y 2x 6⇒m x1 and y2 y1 x1 6x2 x2 5 x22 x22 6x2 x2 6x2 5 x1 x2 3 6x2 12x2 3 5, 5 y x12 2 4 2x2 6 6 2x2 18x2 6. 3 2 1 )2, 3) )1, 1) x 2 3 5 x2 x22 2x22 2x2 2 x2 2 2x2 2x2 4x22 0 0 1 or 2 −1 9 14 4 1 x2 3 18 y 6x2 2 x2 5 4 3 2 1 )2, 4) x2 1 ⇒ y2 0, x1 2 and y1 4 x −1 −2 )1, 0) 2 3 Thus, the tangent line through 1, 0 and 2, 4 is y x2 0 2 ⇒ y2 4 2 0 x 1 3, x1 1 ⇒y 1 and y1 1 4x 4. Thus, the tangent line through 2, 3 and 1, 1 is y 1 3 2 1 x 1 1 ⇒y 2x 1. 75. f x fx 1 2x 4 4 4 x x x x 1 x 2 x, 12 4, 0 1 2x 77. f 1 3.64 1 0y 4x 2 xy 2xx 2x 4, y 2 0.77 3.33 1.24 The point 4, 2 is on the graph of f. Tangent line: y 4y 2 8 0 x x 02 x 44 4 4y 4 4 64 Chapter 2 Differentiation 79. (a) One possible secant is between 3.9, 7.7019 and 4, 8 : y y y (b) f x Tx 8 8 Sx 31 x 2 3x 8 4 7.7019 x 3.9 4 4 −2 20 (4, 8) 12 −2 2.981 x 2.981x 3.924 2 ⇒f 4 4 8 3 2 2 3x 4 3 S x is an approximation of the tangent line T x . (c) As you move further away from 4, 8 , the accuracy of the approximation T gets worse. 20 f −2 −2 T 12 (d) x f4 T4 x x 3 1 1 2 2.828 2 1 5.196 5 0.5 6.548 6.5 0.1 7.702 7.7 0 8 8 0.1 8.302 8.3 0.5 9.546 9.5 1 11.180 11 2 14.697 14 3 18.520 17 81. False. Let f x fx gx 85. True. If g x x2 and g x 2x, but f x 3f x , then g x x2 4. Then gx. 3f x . 83. False. If y 2, then dy dx 0. 2 is a constant. 87. f t ft 2t 2 7, 1, 2 89. f x fx 1 x2 1 , 1, 2 x Instantaneous rate of change is the constant 2. Average rate of change: f2 2 f1 1 22 7 1 21 7 2 Instantaneous rate of change: 1, 2, 1 ⇒f 1 1 2 ⇒f 2 1 1 4 (These are the same because f is a line of slope 2.) Average rate of change: f2 2 f1 1 12 2 1 1 1 2 S ection 2.2 Basic Differentiation Rules and Rates of Change 65 91. (a) s t vt (b) s2 2 16t2 32t s1 1 st 1362 93. st 4.9t2 4.9t2 v0t 120t 120 120 s0 1298 32t 1: v 1 2: v 2 1346 48 ft sec vt v5 v 10 9.8t 9.8 5 9.8 10 71 m sec 22 m sec (c) v t 120 When t When t (d) 16t2 t2 (e) v 32 ft sec. 64 ft sec. 0 1362 4 1362 4 295.242 ft sec 97. v 2 3 1362 1362 ⇒t 16 32 9.226 sec 1362 4 8 1362 95. 60 v 40 mph 2 3 mi min 4 mi Distance (in miles) 10 8 s Velocity (in mph) 50 40 30 20 10 t 2 4 6 8 10 mi min 6 min 0 mph v 0 mi min 0 mi (10, 6) 6 (6, 4) 4 0 mi min 2 min v 60 mph Time (in minutes) (8, 4) 2 t 2 4 6 8 10 1mi min (0, 0) 1 mi min 2 min 2 mi Time (in minutes) (The velocity has been converted to miles per hour) 99. (a) Using a graphing utility, you obtain R (c) T dT (e) dv For v For v For v R 0.167v B 0.02. 0.1431v 0.44 (d) 60 (b) Using a graphing utility, you obtain B 0.00586v2 T 0.0239v 0.46. 0.00586v2 0.1431 0.01172v 40, T 40 80, T 80 B R 0.612. 1.081. 1.315. 0 0 100 100, T 100 (f) For increasing speeds, the total stopping distance increases. 1,008,000 Q 1,008,000 Q2 5083.095 dC dQ \$1.93. 101. A s2, dA ds 2s 4 m, 103. C dC dQ C 351 When Q C 350 350, 6.3Q 6.3 5085 \$1.91 When s dA ds 8 square meters per meter change in s. 105. (a) f 1.47 is the rate of change of the amount of gasoline sold when the price is \$1.47 per gallon. (b) f 1.47 is usually negative. As prices go up, sales go down. 66 Chapter 2 Differentiation 107. y ax2 bx c Since the parabola passes through 0, 1 and 1, 0 , we have 0, 1 : 1 1, 0 : 0 Thus, y y 1 1 a b ax2 2ax 2a 1 a 2 a 1 2x 2 a0 a1 2 2 b0 b1 a 1x 1 a 1 c⇒c 1⇒b 1 a 1. x 1, we know that the derivative is 1 at the point 1, 0 . 1. From the tangent line y a 1 3 3x 1. Therefore, y 109. y y x3 3x2 9x 9 Tangent lines through 1, y x3 9x 9 9 0 x 3x2 3x3 2x3 9: 9x 3x2 3x2 3 2 3 2, 81 8 1 9x x2 2x 9 3 0 or x The points of tangency are 0, 0 and Tangent lines: y 0 y 9x y 0 x≤2 x>2 9x 9x 9x 0 and . At 0, 0 the slope is y 0 9. At 3 2, 81 8 the slope is y 3 2 9 4. y 81 8 9 4 x 3 2 27 4 y 4y 27 0 9 4x 111. f x ax3, x2 b, f must be continuous at x x→2 x→2 2 to be differentiable at x 8a b 4 b 8a 8a 4 4 b 2. b lim f x x→2 x→2 lim ax 3 lim f x lim x2 x<2 x>2 fx 3ax2, 2x, For f to be differentiable at x 3a 2 2 2, the left derivative must equal the right derivative. 22 4 1 3 12a a b 8a 4 4 3 Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 67 113. Let f x fx cos x. lim lim fx x→0 x x fx sin x sin x x 1 sin x x→0 cos x cos x cos x cos x x sin x 1 cos x sin x x x→0 lim 0 x→0 lim sin x Section 2.3 1. g x gx x2 x 2 The Product and Quotient Rules and Higher-Order Derivatives 1 x2 1 2x 2x 2 6x 2 2x 2 2x 2x 2 2 x 2 3. h t 2x 2x 2x3 4x 2 ht 3 t t2 3 4 t2 t2 4 3t2 3 t1 32 t 4 23 t1 2t 3 2x3 4x3 1 4t 3 2t 4 7t2 4 3t2 3 5. f x fx x 3 cos x x3 2 7. f x cos x 3x 2 x3 sin x fx x x2 x2 1 11 x 2x x2 1 2 1 x2 x2 12 sin x 3x cos x 3 9. h x x3 x3 x 1 x1 3 x3 1 1 1 x 2 3 x1 3 x3 1 2 3 11. g x 3x 2 gx sin x x2 x 2 cos x x2 sin x 2x 2 x cos x x3 2 sin x hx x3 1 x 9x 2 2 3 x3 3x 12 1 3x2 3 x3 8x 3 1 2 13. f x fx x3 x3 10x4 3x 2x2 3x 4x 12x3 3x 3 3x2 5 2x2 18x 3x 15 5 3x2 3 15. f x fx x2 x x 4 3 3 2x x x2 3 2 41 2x 2 x2 x 6x x 2 x 32 6x 3 4 2 4 f0 15 1 1 6 3 4 2 f1 1 4 17. fx fx f x cos x x 2 2 sin x 2 2 cos x 1 2 4 8 cos x x sin x 4 4 68 Chapter 2 Differentiation Function 19. y 21. y x2 3 7 3x3 4x3 x 2 Rewrite 2x y 12 x 3 7 x 3 3 Derivative 2 x 3 y 2 x 3 7x 2x 2 3 4 Simplify y 2x 3 7 x4 2 x 2 y y y y 23. y 25. f x fx y 2x x x2 1 1 2 4 x, x > 0 12 y 3 x2 2 2 2 2x x2 2x x2 1 3 1 2 2x x2 2x 2x2 4x x2 1 2 x 27. f x fx x1 1 x 1 2, 12 12 x 4 x 3 x 4x x 3 x2 x2 x 6x x 6x 3 2 29. f x 9 32 3 12 fx 2x x x 12 5 2x1 5 x 2 2 5x x 12 3 4 4x 1 x 32 32 32 x 5 2 2x 5 2x3 2 2x 5 2x x 31. h s hs s3 6s5 1 x 3 2 2 s6 4s3 4 2 12s2 6s2 s3 2 33. f x fx x x2 2x xx 3x 2 x2 2x 3x 1 3 2x x2 1 3x 3 2x 2 6x 4x 2 x 2 3x 8x 2 2x 1 2x 3x 2 3 3 2x 2 x2 35. f x fx 3x3 9x2 9x2 9x 4 15x 4 x2 x2 x2 c2 c2 2 2x 2 2x 3 x2 x 3 2 5x 1 1 5 16x 32x 3x3 3x 4 20 20 4x 1 x 3x 3 6x 4 4x 2 12x 3 1 4x 3x3 3x 4 8x 2 4x x 15x3 16x 51 4x2 20x 4x x 4x 4 x2 36x3 48x 3 5x 4x 41x 2 33x 2 37. f x fx 39. f x x2 c2 2 c2 2x ft t2 sin t t2 cos t t t cos t 2t sin t 2 sin t c2 2x x2 4xc c2 2 2 x2 S ection 2.3 The Product and Quotient Rules and Higher-Order Derivatives 69 41. f t ft 45. g t gt cos t t t sin t t2 4 43. f x cos t t sin t t2 t1 4 x 1 tan x sec2 x tan2 x cos t fx t 34 8 sec t 8 sec t 1 4t3 4 47. y 8 sec t tan t y 3 1 sin x 2 cos x 3 sec x tan x 2 3 sec x tan x 2 3 sec x 2 sec2 x tan2 x tan x 3 sec x tan x 2 1 sec x 1 t 4 8 sec t tan t 49. y y csc x csc x cot x cos x sin2 x sin x cos x 51. f x fx x2 tan x x2 sec2 x xx sec2 x 2x tan x 2 tan x cos x 1 cos x csc2 x cos x cot2 x 53. y y 2x sin x 2x cos x 4x cos x 57. g g x 2 cos x 2 sin x 2 sin x x2 x2 sin x sin x 2x cos x 55. g x gx x x 2x2 x 1 2x 2 8x 2 1 2 5 (form of answer may vary) 1 1 sin sin sin 1 1 1 csc x csc x csc x 22 3 1 22 csc x cot x 1 csc x csc x cot x 1 csc x 2 43 2 csc x cot x 1 csc x 2 cos 1 2 (form of answer may vary) 59. y y y 6 61. ht ht sec t t t sec t tan t t2 sec t t tan t t2 1 1 1 2 sec t 1 h sec tan 2 70 Chapter 2 Differentiation 63. (a) f x fx x3 x3 4x3 3x 3x 6x2 1x 11 6x 2, x 5 3. 1x 1, 3 3 53. (b) − 10 10 2 3x2 (1, − 3) 10 f1 1 slope at 1, 3 − 10 Tangent line: y 65. (a) f x fx f2 x x x 1 1 ⇒y x 2 51. (b) , 2, 2 x1 2 6 11 x1 1 1 2 x 1 1 1 (2, 2) 2 −3 6 2 slope at 2, 2 . 1x 2 ⇒y x 4 −3 Tangent line: y 67. (a) fx fx f 4 tan x, sec2 x 2 2 ,1 4 55. (b) 4 (( π ,1 4 − slope at 4 ,1 . −4 Tangent line: y y 4x 2y x2 x x x2 x fx 1 1 2x x1 2x 12 x2 1 2 1 1 2x 2x 2 4 2 0 x x 5x x 2 3 3x 1 x 22 25 x 4 2 x 5x 22 3x 2 6 x 41 x 2x x 4 2 2 2 69. f x fx 71. f x gx gx 2. 6 2 2 xx x 2 12 fx 2 0 when x 0 or x f and g differ by a constant. Horizontal tangents are at 0, 0 and 2, 4 . x n sin x x n cos x x When n When n When n When n n 1 73. f x fx 75. Area nx n 1 At 2 31 t 2 2 2 2t 1 1 t 2 12 t 2t3 2 t1 2 sin x At sin x. 2 sin x . 3 sin x . 4 sin x . n sin x . x cos x n sin x x cos x x x cos x x2 x cos x x3 x cos x xn 1 1: f x 2: f x 3: f x 4: f x 3t1 1 t 2 12 6t 1 2 cm sec 2t For general n, f x x cos x S ection 2.3 The Product and Quotient Rules and Higher-Order Derivatives 71 77. C dC dx 100 100 200 x2 400 x3 10: 15: 20: x x x dC dx dC dx dC dx 30 , 1≤x 79. Pt Pt 500 1 500 500 2000 50 200 50 50 50 4t 50 t2 t2 4 4t 2t 50 t2 2 4t 2 t2 2 t2 t2 2 30 30 2 (a) When x (b) When x (c) When x \$38.13. \$10.37. P2 \$3.80. 31.55 bacteria per hour As the order size increases, the cost per item decreases. 81. (a) sec x d sec x dx (b) csc x d csc x dx (c) cot x d cot x dx 1 cos x d 1 dx cos x 1 sin x d1 dx sin x cos x sin x d cos x dx sin x sin x sin x cos x cos x sin x 2 sin2 x cos2 x sin2 x x x x 2 x 1 3 cos x 0 1 cos x 2 sin x sin x cos x cos x 1 cos x sin x cos x sec x tan x sin x 0 sin x 1 cos x 2 cos x sin x sin x 1 sin x cos x sin x csc x cot x 1 sin2x csc2 x 83. f x fx fx 4x3 6x1 3x 2 2 85. f x fx x fx 1 11 x1 x1 2 12 3 x 1 1 2 87. f x fx fx 3 sin x 3 cos x 3 sin x 89. f x fx x2 2x 91. f x f 4 2x 1 2x 2 12 x 1 x 93. 4 3 2 y 95. f x fx f2 2g x 2g x 2g 2 2 2 hx hx h2 4 97. f x fx f2 gx hx hxg x hx h2g 2 h2 1 10 2 1 2 gxh x 2 g2h 2 2 1 x 1 2 3 4 0 34 f2 0 x 2 2. One such function is f x 72 Chapter 2 Differentiation 99. f′ 2 1 y 101. v t at f x 36 2t t2, 0 ≤ t ≤ 6 v3 2 27 m sec 6 m sec 2 1 1 a3 The speed of the object is decreasing. f″ It appears that f is cubic; so f would be quadratic and f would be linear. 103. v t at 100t 2t 15 2t 15 100 100t 2 2t 15 2 1500 15 2 (a) a 5 (b) a 10 (c) a 20 1500 25 15 2 2.4 ft sec2 1.2 ft sec2 0.5 ft sec2 1500 2 10 15 1500 2 20 15 2 2t 2 105. f x (a) gxhx fx fx gxh x gxh x gxh x fx gxh gxh f 4 hxg x g xh x 2g x h x g xh x 3g x h x g xh x g 4 hxg x hxg x 2g x h x h xg x x x 4 2g x h x g xhx hxg x h xg x 3g x h x 3g x h x x gxh g x 3g x h x 3g x h x 3g x h x xh x xhx 6g x h x 4g x h x 1 g x h4 x (b) f n 4g x h x nn 1n 1n 1n g4 x h x x nn 1 n 2 . . . 2 1 g x hn 2 1 n 2 n 3...2 1 3 2 x g x hn x 2...2 1 g x hn 2...2 1 x nn 1 n 2 . . . 2 1 g x hn 3 2 1 n 3 n 4...2 1 nn 1 n 2 . . . 2 1 gn n 1 n 2...2 1 1 g x hn x n! 1! n n Note: n! nn 1 . . .3 1! g x hn 1 1 1 x ... xh x g x hn gn x h x 2 x n! 2! n 2! x ... n! gn 1 !1! 2 xh x gn x h x 1 (read “n factorial.”) Section 2.4 The Chain Rule 73 107. f x fx fx (a) P1 x P2 x cos x sin x cos x fax 1 f ax 2 1 x 4 3 a a 2 f f f 3 3 3 fa cos 3 3 3 1 2 3 2 1 2 3 fa 1 2 1 2 (b) P2 − 2 sin cos 3 x 2 fax a P1 2 f 2 3 x 2 −2 3 (c) P2 is a better approximation. (d) The accuracy worsens as you move farther away from x a 3. 111. True hc f cg c fc 0 0 gcf c gc 0 113. True 109. False. If y dy dx f x g x , then gxf x. f xg x 115. f x xx x2, if x ≥ 0 x2, if x < 0 2x fx fx 2x, if x ≥ 0 2x, if x < 0 2, if x > 0 2, if x < 0 f 0 does not exist since the left and right derivatives are not equal. Section 2.4 The Chain Rule y 1. y 3. y 5. y 7. y y 11. f x fx f gx 6x x2 csc3 x 2x 3 2x 9 2 9 3 7 3 2 u 5 4 gx 6x x2 csc x 5 1 y y y y fu u4 u u3 9. g x 34 12 4 1 1 1 2 t 9x 4 3 u u u 1 7 x2 2 6 2x 7 2 gx 13. f t 9x 12 9 108 4 9x 3 23 x2 13 2x 39 4x x2 13 ft t 12 1 1 21 t 74 Chapter 2 Differentiation 15. y y 9x2 12 9x 3 4 13 17. y 23 24 2 x2 14 4 18x 6x 9x2 4 23 y 1 4 4 x 4 x2 34 2x 4 t 2t 3 x2 3 19. y y 23. y dy dx x 12 x 2 1 21. f t 2 2 x 2 12 1 x 1 2 2 ft 3 3 t 4 2 3 3 25. f x 32 x2 x x2 4 x 2x x 2x x 2 1 x 2 2 1 2x 2 32 fx 2 2 2 3 3 3 1 x 2 x 2 2 4 2x 2x 3x 27. y y x1 x 1 1 2 x2 1 1 1 x 2 x2 x2 x2 12 x1 12 x2 2x 1 1 12 29. y 1 x2 12 x x2 x 1 x x2 1 1 32 32 1 2x x2 x2 12 1 y 12 x 2 x2 x2 x2 1 1 12 1 12 1 12 2 x2 1 2 x 2 32 x x2 1 x2 1 1 x2 2x2 1 x2 x x2 2 x x2 5 2 2 32 3 31. g x gx 33. f v 5 2 x2 10x 23 2 x2 x2 x 2 5 2x 2 1 1 3 1 1 2v v fv 2v v 2 1 2 v 1 2 v 2 1 2v 2x 52 x2 1 1 4x3 2 12 9 1 2v 1 v4 37. g t gt 3t 2 2t 35. y y x x 2 t2 1 1 3x2 2 x x2 3t t2 3t 2 t2 2t 1 3 2 The zero of y corresponds to the point on the graph of y where the tangent line is horizontal. 2 The zeros of g correspond to the points on the graph of g where the tangent lines are horizontal. 24 y −1 5 g′ −5 −2 g 3 y′ −2 S ection 2.4 The Chain Rule 75 39. y y x x 1 41. s t st 22 t 1 t 3 t 1 t x 1x 2x x 1 y has no zeros. 4 The zero of s t corresponds to the point on the graph of s t where the tangent line is horizontal. 3 y −5 4 s′ −3 6 y′ −2 s −3 43. y dy dx cos x x x sin x sin 1 3 y x x2 x cos cos x2 x x 1 1 −5 5 y′ −3 The zeros of y correspond to the points on the graph of y where the tangent lines are horizontal. 45. (a) y y y0 sin x cos x 1 (b) y y y0 sin 2x 2 cos 2x 2 1 cycle in 0, 2 2 cycles in 0, 2 The slope of sin ax at the origin is a. 47. y dy dx cos 3x 3 sin 3x sin cos x 2 49. g x gx 3 tan 4x 12 sec2 4x 51. y y 53. h x hx sin 2 2x2 x2 2 x 2 2 x cos 22 x sin 2x cos 2x sin 2x 2 cos2 2 sin 2x 2x 2 sin2 cos 2x 2 cos 2x 2x 55. f x fx cot x sin x sin2 x cos x sin2 x sin x cos x 2 sin x cos x sin4 x 1 cos2 x sin3 x 2 cos 4x. Alternate solution: hx hx 1 sin 4x 2 1 cos 4x 4 2 2 cos 4x sin2 x 2 cos2 x sin3 x 76 Chapter 2 Differentiation 57. y y 4 sec2 x 8 sec x sec x tan x 8 sec2 x tan x 59. f f 1 4 sin2 2 1 4 1 4 sin 2 2 2 sin 2 cos 2 1 2 2 sin 4 sin 2 cos 2 61. f x ft 3 sec2 6 sec 6 sec 2 t t t 1 1 sec 1 tan t t 1 tan 1 t 1 1 1 63. y x x 1 sin 2x 4 1 sin 4x2 4 12 2 6 sin t cos3 t dy dx 1 x 2 1 cos 4x2 8x 4 2x cos 2x 2t 2t t 1 2t 8 12 2 1 2x 65. y dy dx sin cos x cos cos x sin x 67. st st t2 12 t 2 t2 s2 69. fx fx f 73. 1 y y 37 3 3 x3 4 3 x3 9 25 sec3 2x , sec2 0, 36 3 x3 4 2 , 12 2, 4 2t 2 8 sin x cos cos x 3 4 3t t t 5 8 4 3x2 1, 1, 9x2 x3 4 3 5 2 71. f t ft f0 2 , 1 13 t 0, 2 3t 12 21 t 5 1 2 2x 2 sec(2x tan 2x 6 sec3 2x tan 2x y0 75. (a) f x fx 0 3x2 12 3x 2 3x 3x2 f3 9 5 2, 2 3, 5 12 77. (a) f x fx f sin 2x, 2 cos 2x 2 ,0 6x 2 Tangent line: y 63. (b) 2 2x ⇒ 2x y 2 0 Tangent line: y (b) 5 7 9 x 5 0 (π , 0) 2 3 ⇒ 9x 5y 2 0 −2 (3, 5) −5 5 −3 S ection 2.4 The Chain Rule 77 79. f x fx 2 x2 6 x2 12x x 4 12x5 1 1 3 2 81. f x 2x 1 12x 12 x2 1 fx fx sin x 2 2x cos x 2 2x 2x 2 cos x2 sin x2 2 cos x2 2x2 24x3 72x2 1 2x2 sin x2 fx 60x 4 12 5x2 83. f′ 3 2 1 y 85. 3 2 y f x 2 2 2 3 3 3 1 2 x f f′ The zeros of f correspond to the points where the graph of f has horizontal tangents. 87. g x gx 89. (a) f x fx f5 f 3x f 3x 3 ⇒ g x gxhx gxh x 3 2 g xhx 63 24 3 f 3x The zeros of f correspond to the points where the graph of f has horizontal tangents. 73. (b) f x fx f5 ghx g hx h x g3 2 2g 3 Need g 3 to find f 5 . 73. (c) f x fx f5 gx hx hxg x hx 36 3 2 73. (d) f x gxh x 2 gx 3gx 3 3 3 2g 2 fx f5 x 162 6 3 2 12 9 4 3 (b) f 132,400 331 v 1 91. (a) f f 132,400 331 v 1 1 132,400 331 132,400 331 v 2 v 2 1 f 1 132,400 331 132,400 331 v 2 v 2 1 When v 93. 0.2 cos 8t 30, f 1.461. 95. 0.2 (since S dS dt When v C R2 C 2R 30, f r2 dR dt 2r dr dt 1.016. The maximum angular displacement is 1 ≤ cos 8t ≤ 1 . d dt When t second. 0.2 3, d 8 sin 8t dt 1.6 sin 8t 1.6 sin 24 Since r is constant, we have dr dt dS dt 1.76 4.224 105 2 1.2 10 2 0 and 2 1.4489 radians per 10 10 5 0.04224. 78 Chapter 2 Differentiation 97. (a) x (b) C 1.6372t3 60x 60 1350 19.3120t2 0.5082t 0.6161 1.6372t3 4.9116t2 19.3120t2 38.624t 2317.44t 0.5082t 0.5082 30.492 0.6161 1350 dC dt 60 294.696t2 The function 99. f x dC is quadratic, not linear. The cost function levels off at the end of the day, perhaps due to fatigue. dt 101. (a) r x cos x 2 3 4 2 sin x f gx g x f g1 g 1 4 and f 4 0. Thus, r 1 5 6 0 0 2 5 4 (a) f x fx f f (b) f x (c) f f 2k 2k 1 r1 sin x cos x Note that g 1 Also, g 1 x 4 sin x 2 fx 1 1 k k sin x 2 sin x 0 (b) s x s4 g fx f x g f4 f 4 5 5 ,g 2 2 6 6 4 2 1 and 2 x x 2k 1 sin x 2k 1 cos x Note that f 4 f4 5 . 4 Thus, s 4 15 24 5 . 8 103. g xx x2 n nx nx 12 105. g x gx 2x n 2x 2 2x 2x 3 3 , 3 x 3 2 dg dx 12 x 2 2x n 2 x2 nx 2x n 2 xx n x a g x xx n2 n 107. h x hx x cos x x sin x x cos x, x x 0 S ection 2.5 Implicit Differentiation 79 109. (a) f x fx fx P1 x P2 x (b) 2 tan x 4 sec2 x 4 x 4 1 1 2 f1 f1 tan x 44 2 x 1 1 f1 1. f1 1 2 21 4 2 4 8 2 4 sec2 f1x 1 x 24 f1 f1x 8 x 1 2 2 x 1 1 P1 P2 f 0 0 3 (c) P2 is a better approximation than P1 (d) The accuracy worsens as you move away from x 111. False. If y 1 x 12 c 1. 113. True , then y 1 2 1 x 12 1. Section 2.5 1. 2x x2 y2 2yy y 36 0 x y Implicit Differentiation 3. 1 x 2 12 x1 2 y1 1 2y 2 9 0 x y y y x 1 1y y 12 12 1 y 2 y y x 5. 3x2 xy x3 y xy y2 2yy xy y 4 0 y 3x2 7. 3x3y2y x3y3 3x2y3 3x3y2 0 0 1 3x2y3 2y y 3x2 2y x 11. 2y2 2y 2 0 sin x cos x 2cos 2y 4 sin 2y y y 1 0 1 3x2y3 3x3y2 1 9. x3 3x 2 4xy y1 3x 2 3x 2y 2xy 2 6xy 12 4xyy 3x 2 3x 2 y 6xy 4xy 3x 2 6xy 2y 2 3x 2 cos x 4 sin 2y 80 Chapter 2 Differentiation 13. sin x cos x y x1 x sec2 tan y yy 1 1 tan y 1 15. y y y x cos xy y y sin xy xy y cos xy cos x tan y x sec2 y y cos xy 1 y cos xy x cos xy 17. (a) x2 y2 y2 y 16 16 ± (b) x2 16 x2 −6 −2 6 y y= 16 − x 2 2 x 2 6 −6 y=− 16 − x 2 (c) Explicitly: dy dx ± (d) Implicitly: x2 12 1 16 2 x 16 2x x 16 x2 x y (b) 2x 2yy y 0 x y x2 9x2 ± 19. (a) 16y2 y2 y 144 1 144 16 ± y 6 9x2 x2 9 16 16 x2 −6 −2 4 2 y= 3 4 16 − x2 3 4 x 2 6 16 −4 −6 3 y=−4 16 − x2 (c) Explicitly: dy dx ± (d) Implicitly: x2 12 3 16 8 3x 4 16 2x 3x 44 3y 9x 16y 18x 32yy y 0 9x 16y x2 21. xy xy y1 xy y At 4, 4 0 y y x 1: y 1 4 23. y2 2yy 2yy y x2 x2 x2 4 4 4 2x x2 2 x2 42 4 2x 16x x2 4 8x y x2 4 2 At 2, 0 , y is undefined. S ection 2.5 Implicit Differentiation 81 25. 2 x 3 13 x2 3 y2 1 3y 3 5 0 x y 13 3 13 27. 1 y x tan x y sec2 x y y y x 1 sec2 x y sec2 x y tan2 x y 2 tan x y 1 1 x2 x2 1 2 y 3 y At 8, 1 : y 1 . 2 sin2 x y At 0, 0 : y 0. 29. x2 4y x2 4y y 2x y 8 0 x2 2xy 4 2x 8 x2 4 x2 4 16x x2 4 2 31. 2 x2 4x3 4x2yy 4x2yy x2 y2 2x 4xy2 4y3y y3 y2 2yy 4y3y 4x2y x2 y 2 4x2y 4x2y 4x2y 8xy 4 2xy 2xy x2y y 8x 8xy 4x3 x3 x3 y3 4xy2 xy2 xy2 x2 4y x2y At 2, 1 : y 32 64 1 2 8 x2 4 35. At 1, 1 : y 0. Or, you could just solve for y: y 33. tan y y sec2 y y sec2 y y x 1 1 sec2 y 1 1 1 x2 2x x2 y2 2yy y x 1 1 yy yy y2 yy y x y 2 2 x2 2x y2 2yy y 36 0 x y x y cos2 y, 1 2 x2 <y< 2 tan2 y y y 1 xy y2 x3 3x2 3x2 2y 2x 3y 4x2 2x 3 3y 2x 4x2 3x 4y 3x2 2y xy xy 3y 2 y x y2 y2 x2 y3 36 y3 37. 16 0 x y 0 0 0 x2 y2 y 3 39. y2 2yy y y 3y 2x x3 y2 3y 2x 6y y3y y x2 16 y3 3y 4x2 82 82 Chapter 2 Differentiation 41. 1 x 2 y x 12 y 1 y 2 y x 4 1 2y 9 0 (9, 1) −1 −1 14 At 9, 1 , y Tangent line: y 1 3 1 y 1 x 3 1 x 3 0 9 4 x 43. x2 y2 y At 4, 3 : 3y 12 25 x y 6 Tangent line: y Normal line: y At 3, 4 : Tangent line: y Normal line: y 3 3 4 x 3 3 x 4 4 ⇒ 4x 4 ⇒ 3x 4y (4, 3) 3y 0. 25 0 −9 9 −6 6 4 4 3 x 4 4 x 3 (−3, 4) 3 ⇒ 3x 3 ⇒ 4x 4y 3y 25 0. 0 −9 9 −6 45. 2x x2 y2 2yy y y x r2 0 x y slope of tangent line slope of normal line Let x0, y0 be a point on the circle. If x0 0, then the tangent line is horizontal, the normal line is vertical and, hence, passes through the origin. If x0 0, then the equation of the normal line is y y0 y y0 x x0 y0 x x0 x0 which passes through the origin. S ection 2.5 Implicit Differentiation 83 47. 25x2 16y2 50x 200x 32yy 160y 200 400 160y y 0 0 200 160 4: 50x 32y (− 8, 5) y (− 4, 10) 10 6 (0, 5) 4 Horizontal tangents occur when x 25 16 16y2 200 4 160y (− 4, 0) −10 − 8 − 6 − 4 x −2 2 400 10 0 0⇒y 0, 10 yy Horizontal tangents: 4, 0 , 4, 10 . 5: 400 8 Vertical tangents occur when y 25x2 400 200x 800 25x x Vertical tangents: 0, 5 , 8, 5 . 0 0⇒x 0, 8 49. Find the points of intersection by letting y2 2x2 4x 6 and x 3x 1 4x in the equation 2x2 0 y2 6. The curves intersect at 1, ± 2 . Ellipse: 4x y 2yy 2x y 0 Parabola: −6 4 y 2 = 4x (1, 2) 6 2yy y 2 y 4 2x 2 + y 2 = 6 −4 (1, − 2) At 1, 2 , the slopes are: y At 1, y 1 2 , the slopes are: 1 y 1. y 1. Tangents are perpendicular. 51. y x and x sin y −6 4 x = sin y Point of intersection: 0, 0 y y x: 1 x 1 y At 0, 0 , the slopes are: y 1 y 1. sin y: y cos y sec y (0, 0) 6 −4 x+y=0 Tangents are perpendicular. 53. xy xy y y C 0 y x 2x x2 y2 2yy y K 0 −3 2 2 C=4 C=1 K = −1 −2 3 −3 3 x y K=2 −2 At any point of intersection x, y the product of the slopes is yx xy 1. The curves are orthogonal. 84 84 Chapter 2 Differentiation 55. 2y2 3x4 0 12x3 4yy y 0 12x3 12x3 4y x 1 3 cos x y 0 3 cos x sin y (b) sin y dy dt 3 cos sin dx dt dy y dt x 0 3 cos x dx dt 3x3 y y (b) 4y dy dt 12x3 dx dt 0 3x3 dx dt (a) 4yy dy dt 57. cos (a) y 3 sin sin yy 59. A function is in explicit form if y is written as a function of x: y f x . For example, y x3. An implicit equation is not in the form y f x . For example, x 2 y 2 5. 61. (a) x4 4y2 y2 y 4 4x2 16x2 4x2 ± y2 x4 − 10 10 10 14 x 4 − 10 4x2 14 x 4 14 x 4 x4 (b) y 3⇒9 36 4x 2 16x2 0 16 ± 28 x4 16x2 36 x2 256 2 144 8± 1± 28 7 2. Note that x2 8± 8±2 7 Hence, there are four values of x: 1 7, 1 7, 1 8x 1 3 7, 1 x3 ⇒ y 7 7 7 7 x 8 x2 . 23 To find the slope, 2yy For x y1 For x y2 For x y3 For x y4 1 1 3 1 3 1 7 7, y 7x 7, y 7 7x 7, y 7 7, y 7 7x 7x 1 3 7 , and the line is 3 1 3 1 7 7x 87 23 . 1 1 3 7 , and the line is 7 3 1 3 1 1 3 7 7x 23 8 7. 1 1 3 7 7 7 7 , and the line is 3 1 3 1 1 3 7 7x 23 8 7. − 10 10 7 , and the line is 10 1 7 3 1 3 7 7x 87 23 . y1 y3 − 10 y2 y4 —CONTINUED— Section 2.6 Related Rates 85 61. —CONTINUED— (c) Equating y3 and y4, 1 3 7 7x 7 7x 7 7 1 7x 7x 1 7 7 7 77 16 7 x If x 87 , then y 7 3 1 3 7 7x 14x 87 7 87 ,5 . 7 1. The derivative of 7 7x 7 7x 1 7 7x 1 7 7 77 7 3 5 and the lines intersect at xn xp q, where p and q are nonzero integers and q > 0. First consider the case where p 63. Let f x 1 q is given by fx x d1 x dx where t ft t q lim fx x→0 x x fx lim t→x ft t fx x x fx x x. Observe that t1 q t t1 q t1 1q x1 x q t1 q t1 q q t1 2q 1q x1 q x1 q t1 q t1 x1 q t1 2q q x1 q x1 q ... 2q t1 qx1 x1 2q x1 1q x1 q 1 ... 1q t1 qx1 1q . As t → x, the denominator approaches qx1 d1 x dx q . That is, 1 qx1 1q 11 x q q q 1. Now consider f x fx 1p x q 1q xp 1 xp 1 q. From the Chain Rule, 1q 1pxp 1 dp x dx 1p x q p x q pq p p 1 pp x q q 1 nxn 1 n p . q Section 2.6 1. y dy dt dx dt x 1 dx 2 x dt 2x dy dt Related Rates 3. x dy dt y xy dx dt dy dt 4 and dx dt 1 3 24 3 . 4 2, 3, dx dt (a) When x dy dt (b) When x dx dt 4 0 y dx x dt x dy y dt 8, y 1 2, and dx dt 5 . 8 6, 10, (a) When x dy dt (b) When x dx dt 25 and dy dt 2 25 2 20. 12 10 8 1, y 1 4 6 4, and dy dt 3 . 2 86 Chapter 2 Differentiation 5. y dx dt dy dt x2 2 2x 1 7. y dx dt tan x 2 sec2 x dx dt 3, 2 2 dx dt 1, 2 0, 20 2 1, 21 2 4 cm sec. 0 cm sec. 12 4 cm sec. dy dt (a) When x dy dt (b) When x dy dt (c) When x dy dt 9. (a) (a) When x dy dt (b) When x dy dt (c) When x dy dt 0, 1 2 2 4, 8 cm sec. 2 2 2 4 cm sec. 2 2 cm sec. dy dt a dx . dt dy dx negative ⇒ positive dt dt dy dx positive ⇒ negative dt dt x2 2 14 x 2 r2 3 dr dt 6, 2 63 36 cm2 min. 3x2 1 12 11. Yes, y changes at a constant rate: No, the rate (b) 13. D dx dt dD dt 15. A dr dt dA dt dy dx is a multiple of . dt dt y2 x2 x2 1 2 x4 3x2 1 4x3 6x dx dt 2x3 3x dx x4 3x2 1 dt 4x3 6x x4 3x2 1 17. (a) sin cos 2 2 A 1 2b ⇒b s h ⇒h s 1 bh 2 s cos 2s sin 2 2 s cos 2r 1 2s sin 2 2 2 (a) When r dA dt (b) When r dA dt s2 2 sin cos 2 2 2 s2 sin 2 θ 24, 2 24 3 144 cm2 min. dA dt s h s b (b) s2 d d where cos 2 dt dt dA 6 dt , s2 2 3 2 1 rad min. 2 1 2 3s 2 8 When When (c) If d dA s2 1 1 s2 3 dt 222 8 dt is constant, dA dt is proportional to cos . , S ection 2.6 Related Rates 87 19. V dV dt dr dt 4 3 dV r, 3 dt 4 r2 dr dt 800 21. s dx dt ds dt 6x2 3 12x dx dt 1, 12 1 3 10, 12 10 3 360 cm2 sec. 36 cm2 sec. 1 dV 4 r 2 dt 30, 60, 1 800 4 r2 dr dt dr dt 1 4 30 1 4 60 2 (a) When x 800 800 2 cm min. 9 1 cm min. 18 ds dt (b) When x ds dt (a) When r (b) When r 2 23. V 12 rh 3 33 h 4 1 3 92 hh 4 since 2r 3h h dV dt dV dt 10 r dh 9 2 dh h ⇒ 4 dt dt 15, dh dt 12 6 4 dV dt 9 h2 4 10 15 1 2 When h 25. 9 8 ft min. 405 3 1 (a) Total volume of pool Volume of 1m. of water 1 2 12 6 2 1 166 2 1 6 12 18 m3 144 m3 2 h=1 12 (see similar triangle diagram) % pool filled 18 144 100% 12.5% 6h, you have 3 6h h 1 144h 18h2 1 144 1 1 m min. 144 b=6 (b) Since for 0 ≤ h ≤ 2, b V dV dt 1 bh 6 2 36h dh dt 3bh 1 dh ⇒ 4 dt 88 Chapter 2 Differentiation 27. 2x dx dt x2 y2 2y dy dt 252 0 y 25 dy dt x y dx dt 2x dx since y dt 2. x (a) When x When x When x 7, y 15, y 24, y 576 400 7, dy dt 24, dy dt dy dt 27 24 2 15 20 7 ft sec. 12 3 ft sec. 2 (b) A dA dt 1 xy 2 1 dy x 2 dt y dx dt 7, y 24, dx dt 2, 20, 2 24 7 48 ft sec. 7 From part (a) we have x and dy dt 7 . 12 1 7 2 527 24 7 12 dA Thus, dt 24 2 21.96 ft2 sec. (c) sec2 tan d dt d dt Using x x y 1 y dx dt 1 y 24, 122 y 2 x y2 dx dt dx dt 62 dy dt x y2 2, dy dt dy dt 7 and cos 12 d 24 , we have 25 dt 24 25 2 θ y 25 cos2 x 7, y 1 2 24 7 24 2 7 12 1 rad sec. 12 29. When y s x2 108 6, x 12 36 x2 12 y y 122 dy dt y x x dx y dt 6 3, and 12 − y s x ( x, y ) 12. y 1 12 2 y 12 s2 2s s ds dt 2x dx dt x 2 12 dx dt y2 2y dy dt dy dt ds dt Also, x2 2x dx dt dx dt 0⇒ 12 dy dt y x dx . dt s ds dt sy 12x ds dt 12 6 12 6 3 0.2 1 53 3 15 m sec (horizontal) Thus, x x dx y dt s dx x dt dy dt 12x y ds dx ⇒ dt dt 3 15 63 6 1 m sec (vertical). 5 S ection 2.6 Related Rates 89 31. (a) s2 dx dt dy dt ds 2s dt ds dt When x ds dt x2 y2 450 in e( mi les ) y 600 dx 2x dt x dx dt s 150 and y 150 1 hr 3 x2 30 ft Di s c 200 tan s x 100 200 100 dy 2y dt y dy dt 200, s 250 and 600 750 mph. Distance (in miles) 450 200 250 (b) t 33. s2 x dx dt 2s ds dt 250 750 902 30 28 2x 20 min 2nd 3rd x s 1st ds dx ⇒ dt dt 30, 902 302 28 x s dx dt 90 ft Home When x s ds dt 35. (a) 15 6 y dx dt dy dt (b) 30 10 28 10 15x 6y 8.85 ft sec. 30 30 10 y y 5 x 3 5 5 3 dx dt dy dt x ⇒ 15y 15 6 5 5 3 dx dt 25 ft sec 3 25 3 5 10 ft sec 3 x y dy x dt 90 Chapter 2 Differentiation 37. x t 1 t sin , x 2 2 6 2 6 1 ,y 2 y2 1 39. Since the evaporation rate is proportional to the surface area, dV dt k 4 r 2 . However, since V 4 3 r 3, we have dV dt 1 2 2 (a) Period: (b) When x 12 seconds 12 0, 3 2 1 1 4 12 2 4 r2 dr . dt 3 m. 2 Therefore, k 4 r2 4 r2 dr ⇒k dt dr . dt Lowest point: (c) When x dx dt x2 2x Thus, dy dt dx dt 1 ,y 4 15 and t 4 t 6 1 1 t cos 26 6 y2 1 2y dy dt cos 0⇒ dy dt y x dx . dt 14 cos 6 15 4 12 1 3 1 24 15 12 2 5 5 120 pV1.3 k 0 0 V dp dt 5 m sec 120 5 120 . Speed 41. 1.3 pV 0.3 dV dt dV dt V1.3 V 1.3p y 30 dp dt V 0.3 1.3p dp dt dV dt 43. tan dy dt sec2 d dt d dt When y d dt 30, y 3 m sec. 1 dy 30 dt 1 cos2 30 dy dt 4 and cos 1 rad sec. 20 2 2. Thus, y θ x 30 11 3 30 2 S ection 2.6 Related Rates 91 45. tan dx dt sec2 d dt d dt y ,y x 5 L y=5 600 mi hr 5 x2 cos2 52 L2 dx dt 5 dx x2 dt 1 dx 5 dt d dt d dt d dt 120 4 120 3 4 x2 L2 sin2 5 dx x2 dt 1 5 600 120 sin2 θ x (a) When (b) When (c) When 47. d dt 30 , 60 , 75 , 30 rad hr 1 rad min. 2 3 rad min. 2 1.87 rad min. 90 rad hr 120 sin2 75 111.96 rad hr 10 rev sec 2 rad rev x 30 d dt 1 dx 30 dt d dt 20 rad sec P 30 (a) cos sin dx dt (b) 2000 θ x x 30 sin 30 sin 20 600 sin 0 4π − 2000 (c) dx dt 600 sin is greatest when sin n or n 180 . 600 600 1 2 3 2 1⇒ 2 n or 90 n 180 dx dt is least when (d) For For 30 , 60 , dx dt dx dt 600 sin 30 600 sin 60 300 cm sec. 300 3 cm sec 49. tan x ⇒x 50 dx dt 2 50 tan 50 sec2 50 sec2 ≤ ≤ d dt d dt 4 d dt 1 cos2 , 25 4 92 Chapter 2 Differentiation 51. x2 y2 25; acceleration of the top of the ladder dx dt x dx dt d 2x dt 2 dy dt dy dt dx dt d 2y dt 2 First derivative: 2x 2y y 0 0 dx dt y d 2y dt 2 dy dt dy dt d 2y dt 2 0 1 y x d 2x dt 2 dx dt 2 Second derivative: x dy dt 2 When x d 2y dt 2 7, y 1 24 24, 70 dy dt 2 2 7 dx , and 12 dt 7 12 2 2 (see Exercise 27). Since 1 24 4 49 144 1 24 29.10s dx d 2x is constant, 2 dt dt 625 144 206.2 0. 0.1808 ft sec2 53. (a) Using a graphing utility, you obtain m s (b) dm dt dm ds ds dt 1.762s 29.10 ds dt 1.2. 0.881s2 (c) If t Thus, s 1995 , then s dm dt 15.5 and ds dt 1.762 15.5 29.10 1.2 2.15 million. Review Exercises for Chapter 2 1. f x fx x2 lim x→0 2x fx x x2 3 x x x 2 fx 2x x x x 2 2 lim x x 2x 3 x2 2x x 2x 3 3 x2 2x 3 x→0 lim 2x x x→0 lim 3. f x fx x lim 2x x 2x x→0 lim 2x x→0 x 2 2x 2 1. 1 fx x x x x x x 1 x x x x x x x x x fx 1 x x x x 1 2x x x x 1 x x x x 5. f is differentiable for all x x→0 lim lim lim x→0 x→0 x→0 lim x→0 R eview Exercises for Chapter 2 93 7. f x 4 x 2 2. 2 because of the sharp turn 9. Using the limit definition, you obtain g x At x 1, g 1 4 3 1 6 3 2 (a) Continuous at x 4 x 3 1 . 6 (b) Not differentiable at x in the graph. y 7 6 5 4 3 2 x 123456 −1 −2 −3 11. (a) Using the limit defintion, f x At x y 2 y (b) −4 3x 2. 13. g 2 x→2 lim gx x x2 x x x3 x x g2 2 1 2 x2 2 2 x2 x x2 x 2 8 2 4 4 1, f 3x 3x 0 1 3. The tangent line is 1 1 x→2 2 x→2 lim lim (− 1, − 2) x→2 lim −4 x→2 lim x 2 15. 2 y 17. y f′ f 25 0 19. f x fx x8 8x7 y 1 x −1 1 21. h t ht 3t 4 12t 3 33 x x 23 23. f x fx 6x1 2 x3 3x2 2 t 3 2 3x2 6x 3x x 2 25. h x hx 6x 3x 12 3x1 x 3 27. g t gx 31. f f 3 3 1 x2 4 t 3 3 cos 3 4 3t3 sin 4 cos 4 29. f f 2 2 3 sin 3 cos 3 sin 94 Chapter 2 Differentiation 33. F Ft 200 T 100 T 4, F 4 9, F 9 50 vibrations/sec/lb. 331 vibrations/sec/lb. 3 35. st s 9.2 s0 16t2 16 9.2 1354.24 s0 2 s0 0 (a) When T (b) When T 37. (a) 15 y The building is approximately 1354 feet high (or 415 m). (c) Ball reaches maximum height when x (d) y y y0 x 20 40 60 25. x 1 1 0.6 0 0.02x2 0.04x 10 5 y 10 y 25 Total horizontal distance: 50 (b) 0 0 39. x t (a) v t at (c) v t x 3 2 x x1 t2 0.02x2 x implies x 50 2 2t 2 3 2. 3 2 y 30 y 50 50. 1 (e) y 25 0 0.2 1 3t xt vt 0 for t 2 t 3 2t 3 (b) v t < 0 for t < 2 . (d) x t v1 1 2 1 2 1 4 0 for t 21 22 1, 2. 3 3 1 1 1 v2 The speed is 1 when the position is 0. 3x2 3x2 2 45. f x fx 6x3 x 2x 7 x2 7 2x 9x2 2 41. f x fx 2x 2 16x 3 x2 7 2x 3 6x 43. h x hx x sin x 1 sin x 2x x2 x2 x2 x2 x2 1 x 1 1 2x 1 2 x1 2 sin x x cos x 2x 2 47. f x 21 1 x3 fx 1 1 x2 x2 12 x 1 2x 3 2 x3 1 x3 49. f x fx 4 4 3x 2 sec x 3x2 3x2 1 51. y 2 x2 cos x cos x 2x cos2 x tan x x sec2 x tan x x2 x sin x 2x cos x x 2 sin x cos2 x 6x 4 6x 3x2 2 y 55. y 53. y y 3x 2 sec x tan x 6x sec x y R eview Exercises for Chapter 2 95 57. y y x cos x x sin x sin x cos x cos x x sin x 59. g t gt gt t3 3t2 6t y y 3t 3 2 61. f f f 3 tan 3 sec2 6 sec sec tan 6 sec 2 63. 2 sin x 2 cos x 2 sin x 2 sin x 0 3 cos x 3 sin x 3 cos x 3 cos x 2 sin x 3 cos x tan y y y 65. f x fx 1 1 1 2 x3 12 67. h x 12 x x2 2 x x2 3 1 2 x3 3x2 hx 3 1 3 x2 x2 x2 1 3 11 x2 6x 1 x 12 3 2x 3x2 2 1 x3 s2 s2 s s2 52 52 2x 69. f s fs 1 1 s3 3s2 3s 5 s3 s2 1 3s 5 5 25 5 2 71. y s2 5 1 32 3 cos 3x 9 sin 3x 1 1 2s y 1 1 32 32 s3 s s2 8s3 73. y y 1 csc 2x 2 1 2 csc 2x cot 2x 2 csc 2x cot 2x 75. y y x 2 1 2 1 1 2 sin 2x 4 1 cos 2x 2 4 cos 2x sin2 x 77. y y 2 32 sin x 3 sin1 2 x cos x cos x cos3 x 2 72 sin x 7 sin5 2 x cos x sin2 x 79. y y sin x x2 x 2 cos x x 22 sin x sin x 1 sin x 1 1 4 5 81. f t ft t2 t tt 83. g x 2 gx 2x x x x 1 1 2 32 12 7t The zeros of f correspond to the points on the graph of f where the tangent line is horizontal. 0.1 g does not equal zero for any value of x in the domain. The graph of g has no horizontal tangent lines. 4 f′ − 0.1 1.3 f − 0.1 −2 g′ 7 g −2 96 Chapter 2 t 6t 1 5 1 Differentiation 12 85. f t ft t 1 13 t 1 56 87. y y tan 1 x 16 sec2 1 x 21 x f does not equal zero for any x in the domain. The graph of f has no horizontal tangent lines. 5 y does not equal zero for any x in the domain. The graph has no horizontal tangent lines. 5 f y − 20 7 2 f′ −2 1 y′ −4 89. y y y 2x2 4x 4 sin 2x 2 cos 2x 4 sin 2x 91. f x fx f cot x csc2 x 2 csc x csc x cot x 93. f t ft ft t 1 t 1 2t 1 t 2 1 t3 2 t4 2 csc2 x cot x 95. g g g 97. T T tan 3 3 sec2 3 sin cos 1 1 sin 1 18 sec2 3 tan 3 700 t2 t2 4t 10 1 1400 t 2 4t 10 2 1, 1400 1 2 4 10 2 5, 1400 5 30 3xy 3y 3x 2 10 y3 3y2y y2 y y 2 (a) When t T 1 (b) When t 18.667 deg hr. T (d) When t 3.240 deg hr. T 9 3, 1400 3 2 12 10 2 10, 2 10 2 7.284 deg hr. (c) When t T 25 x2 2x 3xy 1400 10 100 40 0.747 deg hr. 99. 10 0 2x 2x 3x yx 3y 3y y2 xy y1 2 101. y 1 x 2 12 16 0 y y 2x x1 2y x 1 y 2 12 y x x y 2y 2 xy x y 2y y 2 xy y 2x 2 xy y 2x 2y 2 xy x 2y x 2x y yy xx R eview Exercises for Chapter 2 97 103. x cos y y y x cos y x sin y sin y cos x y y cos x y sin x y sin x y cos x sin y 105. 2x x2 y2 2yy y 20 0 −9 6 (2, 4) 9 y sin x sin y cos x x cos y x y −6 At 2, 4 : y Tangent line: y x Normal line: y 2x 1 2 4 2y 4 y 1 x 2 10 2x 0 0 2 2 107. y dy dt dy dt x 2 units sec dx 1 dx ⇒ dt 2 x dt 1 dx , 2 dt 1, 4, dx dt dx dt 2x dy dt 4x (a) When x (b) When x (c) When x 2 2 units/sec. 4 units/sec. 8 units/sec. 109. s h s dV dt 12 2 1 h 4 1 111. s t st s 35 60 9.8t 60 4.9t2 s (t) 4.9t2 25 5 4.9 st xt 3s t 3 9.8 5 4.9 30˚ x(t ) 4.9t2 t 1 h 4 5 8 4 4 2 hh h tan 30 1 3 xt dx dt 2 1 2 Width of water at depth h: w V dV dt dh dt When h 2 2s 5 2 2 5 4 2 4 2 h dh dt 2 h 2 h 3 ds dt 1 2 38.34 m sec 2 2 dV dt 54 h 1, dh dt 2 m min. 25 s h 2 ...
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