EVEN02 - 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 . . 330 Basic Differentiation Rules and Rates of Change 338 The Product and Quotient Rules and Higher-Order Derivatives . . . . . . . . . . . . . 344 The Chain Rule . . . . . . . . . . . . . . . . . . 350 Implicit Differentiation . . . . . . . . . . . . . . 356 Related Rates . . . . . . . . . . . . . . . . . . . 361 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 367 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . 373 CHAPTER Differentiation Section 2.1 2 The Derivative and the Tangent Line Problem Solutions to Even-Numbered Exercises 2. (a) m (b) m 1 4 1 4. (a) f4 4 f4 4 Thus, f1 1 f3 3 f4 4 5 3 5 2 4.75 1 1 0.25 f3 3 f4 f1 > 1 4 (b) The slope of the tangent line at 1, 2 equals f 1 . This slope is steeper than the slope of the line through 1, 2 and 4, 5 . Thus, f4 4 6. g x 3 x 2 1 is a line. Slope 3 2 f1 < f 1. 1 lim lim g2 5 5 4 2 x 4 4x x x 4 x 2 8. Slope at 2, 1 x→0 x x x g2 2 1 1 x→0 lim x→0 lim 10. Slope at 2, 7 lim lim h 2 4 4 4t t t 4 2 t t t 2 x→0 h 3 t 2 2 7 4 t→0 12. g x gx 5 lim gx 5 x 0 x 1 x 2 fx 9 1 2 x x 12 x 1 2 fx x x 9 1 2x 0 x→0 t→0 t x x 5 gx lim lim lim lim t→0 x→0 t→0 x→0 14. f x fx 3x lim x→0 2 fx 3x 3x x 3 x x x fx 16. f x fx 2 x 3x 2 9 lim x→0 lim lim x→0 lim lim x→0 x→0 x→0 lim 3 x→0 330 S ection 2.1 18. f x fx 1 lim lim lim x→0 The Derivative and the Tangent Line Problem 331 x2 fx 1 1 x x2 2x x x x2 fx x x3 x x x 3 x x x fx 2 1 x x 2 x2 1 x2 x→0 x 2x x x 2 x→0 lim 20. f x fx x3 lim lim x→0 lim x→0 2x x 2x fx x x x 2 2 x→0 x3 3 x2 x2 x 2 x→0 lim lim 3x2 x 3x x 2 x x 3 2x x x 2 x x3 x2 x→0 3x2 x 3x x 3x x 2x x x x→0 x x 2 lim 3x2 x→0 2x 3x2 2x 22. f x fx 1 x2 lim fx 1 lim lim lim lim x x2 xx 2x x xx 2x x x→0 x→0 24. f x x x x2 x x fx 1 x2 x2 xx 22 4 x lim fx 4 lim lim lim lim x x→0 x→0 fx x x x x fx 4 x x x 4x xx x x 4 x 2 xx x x x x x x x x x x x x→0 x→0 4x 4x xxx 4x xxx xx 4 x x→0 x2 xx 22 x→0 x→0 x x 2x2 x→0 2x x4 2 x3 xx 332 Chapter 2 Differentiation 26. (a) f x fx x2 lim x→0 2x fx x 2x x 1 x x x 2 18. (b) fx 2x x x x 2 5 (−3, 4) lim x x 2x 2x 2 1 x2 2x 1 −6 −1 3 x→0 lim x→0 x→0 lim 2x At 2 3, 4 , the slope of the tangent line is m 2 3 2 4. The equation of the tangent line is y 4 y 28. (a) f x fx x3 lim lim lim x→0 4x 4x 1 fx x x3 x 3x 2 3 8. 18. (b) x x 3 4 fx −6 (1, 2) 6 x→0 1 x x x3 3x x x 2 2 1 −4 x x 3x 2 3 1 x3 1 x→0 lim 3x 2 x→0 3x x At 1, 2 , the slope of the tangent line is m The equation of the tangent line is y 2 y 30. (a) f x fx lim lim x lim x→0 31 2 3. 3x 3x 1 fx x x x x x x x 1 1. (b) x x 1 x 1 x 1 1 1 x fx x x x 1 1 1 1 1 2x 1 x x x x 1 1 x x 1 1 −2 4 (5, 2) 10 x −4 x→0 x→0 lim x→0 At 5, 2 , the slope of the tangent line is m 1 25 1 1 4 The equation of the tangent line is y 2 y 1 x 4 1 x 4 5 3 4 S ection 2.1 The Derivative and the Tangent Line Problem 333 32. (a) f x fx 1 x lim 1 fx 1 x 1 xx x 1 x 1 2 x→0 (b) x x 1 x x x 1 x 1x 1 fx 1 x x 1x 1 1 1 (0, 1) −6 3 3 lim lim lim x x −3 x→0 x→0 x→0 At 0, 1 , the slope of the tangent line is m 0 1 1 2 1. x 1. 3x 2. Since 36. Using the limit definition of derivative, f x Since the slope of the given line is 1 1 . The equation of the tangent line is y 34. Using the limit definition of derivative, f x the slope of the given line is 3, we have 3x 2 x2 3 1⇒x ± 1. 2x 32 1 , we have 2 1, 1 the tangent Therefore, at the points 1, 3 and lines are parallel to 3x y 4 0. These lines have equations y 3 y 3x 3x 1 and y 1 y 3x 3x 1 4 2x 1 1 32 1 2 x x 1 32 1 1 1⇒x 2 At the point 2, 1 , the tangent line is parallel to x 2y 7 0. The equation of the tangent line is y 1 y 1 x 2 1 x 2 2 2 2x (d) 38. h 1 h 4 because the tangent line passes through 1 6 3 4 1 2 4 1 2 1, 4 40. f x x2 ⇒ f x 42. f does not exist at x 0. Matches (c) 44. 4 3 2 1 −4 −3 −2 −2 −3 −4 y Answers will vary. Sample answer: y x 1 2 3 4 x 334 Chapter 2 fx Differentiation 2x 2Dx fx fx x x fx 46. (a) Yes. lim x→0 lim x→0 fx (b) No. The numerator does not approach zero. (c) Yes. lim fx x x→0 fx 2x x lim fx fx x x 2x fx x→0 fx 2x fx 2 x x fx fx x lim fx x→0 1 fx 2 (d) Yes. lim fx x→0 1 fx 2 fx x x fx fx 48. Let x0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, f x y through 1, 3 and x0, y0 equals the derivative of f at x0: 3 y0 1 x0 3 3 x02 x0 2x0 3 x0 y0 x02 3 1 10 2x. The slope of the line 2x0 1 2x0 0 0 ⇒ x0 3, 1 x0 2x0 2x02 (−1, 1) 8 6 4 (3, 9) −8 −6 −4 −2 −2 −4 x 2 4 6 (1, −3) Therefore, the points of tangency are 3, 9 and ( 1, 1 , and the corresponding slopes are 6 and the tangent lines are y 3 y 6x 6x 1 9 y 3 y 2x 2x 1 1 gx x x3 x x x3 x 3x2 x 3x x x 3x x 2. The equations of 50. (a) f x fx x2 lim lim lim fx x x2 x→0 (b) g x x x x2 x 2x x x x 2x x x 1 2x 1 2x 2 and the tangent line is or y 2x 1. 0. 2x 1. At x y At x y x fx lim gx x3 3x x x 2 x→0 lim x2 lim x 2 x→0 x→0 x 2 3 x3 x→0 x2 lim x→0 x→0 x→0 x 3x2 x x 2 lim x→0 x→0 lim 3x2 1, g 1 3x 0, g 0 1, g 1 1 3x 2 3x2 lim 2x At x y At x At x 1 0, f 0 1, f 1 2 1 1 3 and the tangent line is or y 3x 2. 0. 1, f At x 0 and the tangent line is y 3 and the tangent line is 1 or y 3x 2. 0 and the tangent line is y 2 and the tangent line is y −3 3 −3 3 −3 −2 For this function, the slopes of the tangent lines are always distinct for different values of x. For this function, the slopes of the tangent lines are sometimes the same. S ection 2.1 The Derivative and the Tangent Line Problem 335 52. f x 12 2x 3 By the limit definition of the derivative we have f x x fx fx fx 3x 8 x. 1 0.5 1 1.5 1.125 1.5 2 2 2 13 2 4 −2 −1 2 2 2 2 1.5 1.125 1.5 0.01 0.01 0.01 fx 1 0.5 1 0.5 0.125 0.5 0 0 0 0.5 0.125 0.5 54. g x 56. f 2 f2 2, f 2.1 2.31525 3.1525 Exact: f 2 3 3 3 100 2.31525 2 2.1 2 f g −1 −1 8 The graph of g x is approximately the graph of f x . x3 4 f′ −9 9 58. f x 3x and f x 6 32 x 4 3 f −6 60. f x x 1 x f2 22 x x x2 22 x S x x f2 2 x 5 x 6 4 x 5 16 x 21 2 x 52 x 2 2 2 5 2 5 2 5 2 2 x f2 x 5 x 6 4 x 5 16 x 21 2 5 6 9 10 41 42 5 2 x 1 2 x 2x 22 x 3 x x 5 2 x 2 5 2 5 2 2 (a) x x x 1: S 0.5: S 0.1: S 4 x −6 S0.1 S0.5 S1 f −4 6 x (b) As x → 0, the line approaches the tangent line to f at 2, 5 . 2 62. g x g1 xx lim 1 x2 g1 1 1 x, c x→1 1 x2 x x 1 0 x→1 gx x→1 x x→1 lim lim xx x 1 1 lim x 336 Chapter 2 Differentiation 64. f x f1 x3 x→1 2x, c fx x 3 fx x 3 x→ 1 3, 1 f1 1 x→1 lim lim x3 x 2x 1 3 x →1 lim x 1 x2 x x1 3 x→1 lim x2 x 3 5 66. f x f3 68. g x g 3 1 ,c x x→3 lim x f3 3 c g 3 3 x→3 lim 1x x 13 3 x→3 lim 3 3x x x 1 3 x→3 lim 1 3x 1 9 lim gx 3 x 3 x→ 3 lim x 313 x3 0 x→ 3 lim x 1 3 23 Does not exist. 70. f x f4 x x→4 4 ,c fx x 4 f4 4 x→4 lim lim x x 4 4 0 x→4 lim x x 4 4 Does not exist. 72. f x is differentiable everywhere except at x 74. f x is differentiable everywhere except at x 76. f x is differentiable everywhere except at x ± 3. (Sharp turns in the graph.) 1. (Discontinuity) 0. (Sharp turn in the graph) ± 2. (Discontinuities) 78. f x is differentiable everywhere except at x 80. f x is differentiable everywhere except at x 82. f x 1 x2 1. (Discontinuity) The derivative from the left does not exist because lim fx x f1 1 lim 1 x2 x1 0 lim 1 x x2 1 1 1 x2 x2 lim 1 1 x x2 . (Vertical tangent) 1. x→1 x→1 x→1 x→1 The limit from the right does not exist since f is undefined for x > 1. Therefore, f is not differentiable at x 84. f x x, x ≤ 1 x2, x > 1 The derivative from the left is x→1 lim fx x f1 1 x→1 lim x x 1 1 x→1 lim 1 1. The derivative from the right is x→1 lim fx x f1 1 x→1 lim x2 x 1 1 x→1 lim x 1 2. 1. These one-sided limits are not equal. Therefore, f is not differentiable at x S ection 2.1 The Derivative and the Tangent Line Problem 337 86. Note that f is continuous at x The derivative from the left is x→2 2. f x 1 2x 2x, 1, x < 2 x≥2 lim fx x f2 2 x→2 lim 1 2x 1 x 2 2 x→2 lim 1 2 x x 2 2 1 . 2 The derivative from the right is x→2 lim fx x f2 2 x→2 lim 2x 2 x2 x 2x 2 4 2x 2x 2x 2 2 2 x→2 x→2 lim lim x 2x 2 2 2x 2. f 2 2 1 2 x→2 lim 2 2x 2 1 . 2 The one-sided limits are equal. Therefore, f is differentiable at x 88. (a) f x x2 and f x 2x 5 y 72. (b) g x x3 and g x 3x2 3 y f 4 3 2 1 1 x 1 2 3 4 −2 −1 −1 −3 1 2 g′ 2 g x −4 −3 −2 −1 f' 72. (c) The derivative is a polynomial of degree 1 less than the original function. If h x 72. (d) If f x x4, then f x lim lim lim lim fx x x4 x→0 x n, then h x nx n 1. x x x4 x 4x3 x fx x4 6x2 x 2 x→0 4x x x 2 3 x 3 4 x4 x→0 x 4x3 6x2 x x 6x2 x 4x x 4x x 2 x x 3 x→0 x→0 lim 4x3 4x3 Hence, if f x x4, then f x 4x3 which is consistent with the conjecture. However, this is not a proof, since you must verify the conjecture for all integer values of n, n ≥ 2. 90. False. y x 2 is continuous at x 2, but is not differentiable at x 2. (Sharp turn in the graph) 92. True—see Theorem 2.1 94. 3 −3 −1 3 As you zoom in, the graph of y1 x2 1 appears to be locally the graph of a horizontal line, whereas the graph of y2 x 1 always has a sharp corner at 0, 1 . y2 is not differentiable at 0, 1 . 338 Chapter 2 Differentiation Section 2.2 12 Basic Differentiation Rules and Rates of Change (b) y y y1 6. y y x8 8x7 y 8x 9 2. (a) y y y1 x x 1 (c) 2 y y y1 x 32 (d) y y y1 x 2 1 32 2x 1 2 x 1 3 52 2x 3 2 2x 2 4 3 4. f x fx 0 2 8. y 1 x8 x 8 10. y 8 y x 34 x1 4 x9 x3 1 x 4 1 4x3 4 12. g x gx 20. g t gt 3x 3 1 14. y y t2 2t 2t 2 3 16. y y 24. y y 8 18. f x fx 2 cos x 4 2x3 6x2 x2 2x 3x 3 3x 2 5 2x 5 8 5 x 8 3 cos t sin t 22. y y 5 sin x 3 2 cos x 15 8x4 2 sin x cos x 3x 2 sin x Function 26. y 2 3x2 2 Rewrite y 2 x 3 x 2 Derivative y 4 x 3 3 Simplify y 3 4 3x3 2 9x3 12x2 x 2, 5, 0 30x 30 75 28. y 3x 4 x ft 3 y 2 9 y 2 x 9 12x2 y 30. y 32. y 3 3 , 5t 3 ,2 5 4x3 34. y y y2 y 3x3 9x2 36 y 6, 2, 18 36. f x 35 3x2 ft f 38. 3 5 gt gt g 3 5t2 5 3 2 3 cos t, 3 sin t 0 2x2 , 1 fx f5 6x 0 40. f x fx x2 2x 2x 3x 3 3 3x 6x 6 x3 2 3 42. f x fx x 1 1 x 2x 2 x3 2 3 44. h x hx 48. f x fx 3x x 1 x2 1 2x 3 x 1 46. y y 3x 6x 36x 5x 2 45x 2 18x 2 15x3 2 3 2x2 1 x2 x 1 x 5 x1 3 x 23 5 x1 5 50. f t 3 t2 2 t 3 3 t1 13 3 4 23 1 x 3 45 1 3x 2 1 5x 4 5 ft 1 t 3 2 3t1 3 1 3t2 3 Section 2.2 Basic Differentiation Rules and Rates of Change 339 52. f x fx 2 3 x 3 cos x 43 2x 13 3 cos x 3 sin x 2 x 3 x3 3x2 1, 3 sin x 2 3x 4 3 54. (a) y y At x 1 2: y 3 y 4x y 1 2x 2 31 2 (b) −5 5 5 1 2 2 2 1 4x 0 4. 1 −7 Tangent line: 56. (a) y x2 x3 2x x 3x2 6x (b) 12 y 3x2 At 1, 6 : y Tangent line: y 61 11 x 11x 2 1 y 5 11. −3 −2 3 6 0 58. y y x3 3x2 x 1 > 0 for all x. 60. y y At x x2 2x 1 0⇒x 1. 0 Therefore, there are no horizontal tangents. 0, y Horizontal tangent: 0, 1 2 cos x, 0 ≤ x < 2 2 sin x 3 ⇒x 2 3 ,y 3 3 23 3 3 , 0 3 or 3 . 3 . 3 3 3 , 223 , 3 3 3 2 3 64. k x2 2x Hence, x 4x 4 2 and k 7 Equate functions Equate derivatives 4 8 7⇒k 3 62. y y sin x At x At x 3x 3 2 ,y 3 Horizontal tangents: 66. k x k 2x x 1 4 Equate functions Equate derivatives Hence, k 2x x 2 x and x 4 ⇒ 2x x 4⇒x 4⇒k 4 68. The graph of a function f such that f > 0 for all x and the rate of change the function is decreasing i.e. f < 0 would, in general, look like the graph at the right. y x 340 Chapter 2 Differentiation 70. g x 5f x ⇒ g x 5f x 72. 2 1 −2 −1 y f x 1 3 4 f′ −3 −4 If f is quadratic, then its derivative is a linear function. fx fx ax2 2ax bx b c 74. m1 is the slope of the line tangent to y y x⇒y 1 ⇒ m1 1 and y x and y ± 1. x. m2 is the slope of the line tangent to y 1 ⇒y x 1 x are 1 ⇒ m2 x2 1 . x2 1 x. Since The points of intersection of y x At x 76. f x fx 1 ⇒ x2 x ± 1, m2 1⇒x 1. Since m2 1 m1, these tangent lines are perpendicular at the points intersection. 78. f 4 1 16 2 , 5, 0 x 2 x2 2 x2 0 5 y x x2y x2 2x 10 5 ,y 2 54 2, 5 −10 −1 19 10 10 10 2x 2x 2x 4x x 2 x 4 5 The point is on the graph of f. The slope of the 5 2 8 25 . tangent line is f Tangent line: y 25y 8x 25y 4 5 20 40 0 8 x 25 8x 20 5 2 S ection 2.2 Basic Differentiation Rules and Rates of Change 341 80. (a) Nearby point: 1.0073138, 1.0221024 Secant line: y 1 y (Answers will vary.) −3 (b) f x Tx 1 3x2 3x 1 1 3x 2 1.0221024 1.0073138 3.022 x 1 1 x 1 1 2 (c) The accuracy worsens at you move away from 1, 1 . 2 (1, 1) (1, 1) 3 −2 −3 3 f T −2 (d) x fx Tx 3 8 8 2 1 5 1 0 2 0.5 0.125 0.5 0.1 0.729 0.7 0 1 1 0.1 1.331 1.3 0.5 3.375 2.5 1 8 4 2 27 7 3 64 10 x3 2. 1 1 1 . The accuracy decreases more rapidly than in Exercise 59 because y 82. True. If f x 86. False. If f x 88. f t ft t2 2t gx 1 xn 3, 2, 2.1 c, then f x x n, then f x gx nx 0 n 1 x3 is less “linear” than y x 1 g x. n xn 1 84. True. If y x, then dy dx 90. f x fx sin x, 0, cos x 6 Instantaneous rate of change: 2, 1 ⇒ f 2 2.1, 1.41 ⇒ f 2.1 Average rate of change: f 2.1 2.1 f2 2 1.41 1 0.1 4.1 22 4.2 4 Instantaneous rate of change: 0, 0 ⇒ f 0 1 , 62 ⇒f 6 1 3 2 0.866 Average rate of change: f 6 6 4.9t2 4.9t2 s0 220 4.9t 2 f0 0 v0t s0 4.9 6.8 12 6 s0 0 0 3 0.955 92. st vt v3 st 16t2 32t 22t 22 220 94. s t 0 when t 2 6.8. 118 ft sec 16t2 22t 226.6 m 112 height after falling 108 ft 16t2 2t 22t 2 8t 108 27 t v2 0 0 2 32 2 22 86 ft sec 342 Chapter 2 Differentiation 96. 50 v 98. This graph corresponds with Exercise 75. s 10 Velocity (in mph) 40 Distance (in miles) 30 20 10 t 2 4 6 8 10 8 6 4 2 (10, 6) (4, 2) (6, 2) t 4 6 8 10 Time (in minutes) (0, 0) 2 Time (in minutes) (The velocity has been converted to miles per hour) 12 at 2 100. s t c and s t s t0 t0 t t at. s t0 t0 t t 1 2 a t02 2at0 t 2t at0 s t0 Instantaneous velocity at t t0 2t0 t 1 2 a t0 t 2 Average velocity: c 2t t 2 1 2 a t0 1 2 a t02 t 2 c t 2 2t0 t 2t 102. V s3, dV ds 3s2 4 cm, dV ds 48 cm2. When s 104. C gallons of fuel used cost per gallon 15,000 1.25 x 18,750 x dC dx x C dC dx 18,750 x2 10 1875 187.5 15 1250 83.333 20 537.5 46.875 25 750 30 30 625 20.833 35 535.71 15.306 40 468.75 11.719 The driver who gets 15 miles per gallon would benefit more from a 1 mile per gallon increase in fuel efficiency. The rate of change is larger when x 15. 106. dT dt KT Ta S ection 2.2 Basic Differentiation Rules and Rates of Change 343 108. y y 1 ,x>0 x 1 x2 y 2 At a, b , the equation of the tangent line is y 1 a 1 x a2 a or y x a2 2 . a 1 (a, b) = a, a 1 () x 2 3 1 The x-intercept is 2a, 0 . The y-intercept is 0, 2 . a 1 bh 2 1 2 2a 2 a 2. The area of the triangle is A 110. y y x2 2x (a) Tangent lines through 0, a : y x2 a a a ± 2x x 2x2 x2 x 0 a The points of tangency are ± y a 2 a. Tangent lines: y a y 2 2 ax ax a, a . At a a a, and y a the slope is y a y 2 2 ax ax a a a 2 a. At a, a the slope is Restriction: a must be negative. (b) Tangent lines through a, 0 : y 0 x2 0 2x x 2x2 x2 a 2ax 2ax xx 2a 0. At 2a, 4a2 the slope is y 2a 4a. The points of tangency are 0, 0 and 2a, 4a2 . At 0, 0 the slope is y 0 Tangent lines: y 0 y 0x 0 0 and y 4a2 y 4a x 4ax 2a 4a2 Restriction: None, a can be any real number. 112. f1 x f2 x sin x is differentiable for all x sin x is differentiable for all x n , n an integer. 0. You can verify this by graphing f1 and f2 and observing the locations of the sharp turns. 344 Chapter 2 Differentiation Section 2.3 2. f x fx 6x 6x 18x3 24x3 The Product and Quotient Rules and Higher-Order Derivatives 5 x3 5 3x 2 15x 2 15x2 2 x3 6x3 12 26 12 4 2s1 8. g t sin x 1 2x x cos x 1 sin x 2x gt 12. f t s 1 2 4. g s gs s4 s1 2 s2 2s 5s 2 2 s1 4 2 4 s2 12 1 s2 s 2 2s 3 2 4 s2 2 s1 2 6. g x gx x sin x x cos x s s s 1 11 s s 1 s 1 s 2 12 t2 2t 2t cos t t3 t3 2 7 7 2t 2t t2 7 2 22 2t2 2t 14t 7 4 2 10. h s ft sin t cos t 3t 2 t3 2 t sin t t4 3 cos t hs 1 s 2 2 1 1 x3 1 3x2 2 2 1 s s 2 1 2 14. f x fx x2 x2 3x2 x x 2x 2x 1 1 2 16. f x x3 1 2x x2 x 1 2 fx x x x x 1 1 11 x 1 x 2 x 1 2 2 1 2 2 x 1 2 11 2x 2x 1 2 2 5x2 f1 0 x1 12 f2 18. fx fx sin x x x cos x x2 x cos x sin x x2 f 6 6 33 2 2 sin x 1 32 2 36 18 12 3 3 2 6 Function 20. y 5x 2 4 3 Rewrite y 52 x 4 3 4 Derivative y 10 x 4 Simplify y 5x 2 S ection 2.3 The Product and Quotient Rules and Higher-Order Derivatives 345 Function 22. y 4 5x2 3x2 7 5 Rewrite y 4 x 5 32 x 7 2 Derivative y 5 7 8 x 5 6x 7 3 Simplify y 8 5x3 6 x 7 24. y y y y 26. f x fx x3 x2 x2 x4 3x 1 2 3 x2 3 x3 12 3x 2 2x 28. f x fx x4 1 x4 2x3 x 2 x 1 x x 1 2 1 x 1 2 2 x4 1 x x 1 1 x x 1 4x3 1 1 3x2 6x2 x2 4x 12 3 12 2x2 x 1 2 30. f x fx 3 x 3 x 1 x 2 16 x1 x1 23 3 x1 2 3 1 x 3 23 32. h x hx x2 4x3 x4 4x x2 2x2 1 1 x1 2 3 4x 5 x 6 5 6x1 x 1 x2 3 6 Alternate solution: fx 3 x x 3x 16 13 3 x fx 56 5 x 6 5 6x1 x 1 x2 3 1 x 23 6 34. g x gx x2 2 2 x x 1 2x x2 x 1 2 x2 2x x 1 1 2 1 2x x 2 1 x 12 1 x2 1 x2 x3 3x 3 x2 2x x2 x 2x 1 2 2 36. f x fx x2 2x 2x 2x 5 x x2 1 x2 1 x4 x 4 x x 2x2 1 1 x 2x 5 x2 1 2x 2 x 2x x2 x2 x 2x3 2x 5 x 2x2 4 1 2x x 2 x2 x x2 x2 x 1 2x x2 1 2x 1 x 2x3 x 1 1 x x 3 6x5 c2 c2 c2 4x3 x2 x2 x2 4xc2 x2 3x2 38. f x fx 40. f f 2x c2 2 1 cos 1 cos sin 1 sin cos 1 c2 x2 2 x2 2x c2 346 Chapter 2 Differentiation 42. f x fx 46. h s hs sin x x x cos x sin x x2 1 s 1 s2 10 csc s 10 csc s cot s 44. y y x 1 cot x csc2 x cot2 x 48. y y sec x x x sec x tan x x2 sec x x tan x x2 sec x 1 50. y y x sin x x cos x cos x sin x sin x x cos x 52. f x fx sin x cos x sin x cos 2x sin x cos x cos x 54. h h 5 sec 5 sec tan tan 5 sec sec2 tan 56. f x fx 2 x2 x2 x5 x 1 2x3 x2 3 x2 2x2 12 1 x 2 1 (form of answer may vary) 58. f f 1 sin cos 1 1 cos 1 1 cos 2 60. f x fx f1 tan x cot x 0 0 cos (form of answer may vary) 62. fx fx sin x sin x sin x cos x sin x cos x sin 2x f 4 sin 2 x x 2 cos x sin x sin2 x sin x cos x cos x cos2 x sin x cos x cos 2x cos 2 1 64. (a) f x fx f0 1 x2 1 2x 2 , 0, 2 x2 21 3x2 2x 2 51. (b) −4 4 4 slope at 0, 2 . 2 2, 1 3 x 1 2 Tangent line: y 66. (a) f x fx f2 x x x 2 9 1 , 1 2x ⇒ y 2x 2 54. (b) −4 4 11 x 11 x 1 . 3 2 ⇒y 2 1 2 −3 6 slope at 2, 1 3 −4 Tangent line: y 2 x 9 2 x 9 1 9 S ection 2.3 The Product and Quotient Rules and Higher-Order Derivatives 347 68. (a) fx fx f 3 sec x, 3 ,2 . (b) 6 sec x tan x − 23 slope at 3 ,2 . −2 Tangent line: y 6 3x 70. f x fx fx x2 x2 x2 1 1 2x x2 2x 2 2 x 1 0. 2x x2 1 2 2 3y 2 3x 6 3 0 72. f x gx gx x cos x x cos x sin x x 2x 3 x2 2 x2 sin x 3x x 5x sin x 2x 1 sin x 3x 1 x cos x sin x x2 x cos x sin x x2 fx 5 23 0 when x Horizontal tangent is at 0, 0 . f and g differ by a constant. 1 2 74. f x fx cos x xn x x n n x sin x 1 n cos x nx n 1 76. V cos x r 2h 13 t 2 2 t 2t1 2 2 2 t x sin x n cos x Vt x sin x n cos x xn 1 When n When n When n When n 1: f x 2: f x 3: f x 4: f x x sin x x2 x sin x x3 x sin x x4 x sin x x5 4 cos x 3 cos x cos x . . . . 78. 1 31 t 22 k V k V2 t 12 3t 2 cubic inches sec 4t1 2 P dP dV 2 cos x For general n, f x x sin x n cos x . xn 1 80. fx gx fx sec x csc x, gx 1 sin x cos x cos x 1⇒ 1 cos x sin x sin x 1 0, 2 sec x tan x sin3 x cos3 x x 37 , 44 sec x tan x csc x cot x ⇒ csc x cot x 1 ⇒ tan3 x 1 ⇒ tan x 1⇒ 348 Chapter 2 Differentiation 82. (a) n t vt (b) A vt nt 9.6643t2 276.4643t2 90.7414t 77.5029 1809.9714 84. f x fx fx x 1 192 x4 2987.6929t 32 x2 64 x3 276.46t2 2987.69t 1809.97 9.66t2 90.74t 77.50 A represents the average retail value (in millions of dollars) per 1000 motor homes. (c) A t 86. f x fx fx x2 40.46 x 2 2.09x 17.83 x 2 9.39x 8.02 2 2x x 1 x2 2 x3 2 2x 2x 2 1 1 x 2 1 x 88. f x fx fx sec x sec x tan x sec x sec2 x sec x sec2 x tan x sec x tan x tan2 x 1 90. f x f x 92. f f f 4 5 6 x x x 2x 2 0 1 2 x2 94. The graph of a differentiable function f such that f > 0 and f < 0 for all real numbers x would in general look like the graph below. y f x 96. f x fx f2 4 hx hx h2 4 98. f x fx f2 gxhx gxh x g2h 2 34 14 hxg x h2g 2 1 2 100. 3 y f′ f f ′′ x −2 −1 −1 −2 2 3 4 It appears that f is quadratic; so f would be linear and f would be constant. 102. s t vt at t sec st vt at ft s t ft sec v t ft sec 2 8.25t2 16.50t 16.50 66t 66 Average velocity on: 0, 1 is 0 0 66 16.5 1 57.75 49.5 16.5 2 99 33 16.5 3 123.75 16.5 16.5 4 132 0 16.5 3, 4 is 2, 3 is 1, 2 is 57.75 0 10 99 2 57.75 1 57.75. 41.25. 24.75. 8.25. 123.75 99 32 132 4 123.75 3 S ection 2.3 The Product and Quotient Rules and Higher-Order Derivatives 349 104. (a) fx fn x xn nn 1n 2 ... 2 1 n! 86. (b) f fx n 1 x 1 n x nn 1n xn 1 2 ... 2 1 1 nn! xn Note: n! 106. xf x xf x xf x nn xf x xf x xf x n 1 1 ...3 fx fx fx xf f f f a a 2 n 2 1 (read “n factorial.”) fx 2f x x 1 0 1 fa nf xf x xf n 1 2f x x 3f x In general, xf x 108. f x fx fx (a) P1 x P2 x sin x cos x sin x fax x. 2 2 2 1 f ax 2 1 1 x 2 fax a fa 1 2 2 1x 2 1 − 2 2 −2 2 (c) P2 is a better approximation than P1. (d) The accuracy worsens as you move farther away from x 110. True. y is a fourth-degree polynomial. dny dx n 116. (a) fg 0 when n > 4. fg fg fg (b) fg fg fg fg fg fg fg 2f g fg fg fg False fg fg fg fg True fg a 2 . 114. True. If v t at vt c then 0. 112. True ( 3 2 0x 2 1 1 (b) 2 (π , 1 2 350 Chapter 2 Differentiation Section 2.4 The Chain Rule y 2. y 4. y f gx 1 x 3 tan 3x 2 1 2 u u u gx x x2 3x 2 1 y y y fu u 12 1 x2 3 tan u 6. y 8. y y 12. f t ft cos u y cos u 10. y 34 15 4 5 1 5 2 32 3 2 4 5t 15 t 3 3 4 2x3 2 2x3 9t 2 9t 3 x2 x2 5 1 6x 2 2 23 12x2 2x3 1 y 14. g x x2 3x 3x 2x 5 12 30x 4 3x 3 12 x2 2 13 9 3 6 9t x 1 gx 2 2 3 25 3x 16. g x gx 20. s t st t2 2x 1 x 1 18. f x fx 9x 9x 14 1, x > 1 1, x < 1 3t 3t 3 1 12 2 34 9 42 27 9x 34 1 1 22. y 2 3 1 t2 t2 2t 3t 2 1 2t 3 y t 15 3 4 24. g t gt t2 12 t 2 26. f x 32 x 3x x 3 3x 3x 27 x 9 3 2 2t t t2 2 32 fx 9 9 2 2 3 3x 3 3x 9 9 3 1 9x 2 3 4x 28. y y 12 x 16 2 121 x 16 22 x3 2 16 x 2 x 3x 2 2 16 x2 x2 12 30. y 2x x2 x 16 x2 12 x x4 x4 4 4 12 1 y x 16 x 4 4 2x 4 x4 4 3 2 1 x x4 2 x4 4 4 x 4 4 12 4x 3 32 x2 2 x4 43 2 32. h t ht 2 t2 t3 t3 4t 2 t3 2 t2 2 t3 t4 3 2 2t t2 3t2 3 2 t 2 2t3 4 t3 t3 2 3 2t2 S ection 2.4 The Chain Rule 351 34. g x gx 3 3x 2 2x 3x 2 2x 2 3 2 3 2 3 2 2x 3 6x 2x 4 3x 2 32 6 3x 2 22 2 2 3x 2 2x 3 9x 4 3 3x 2 6x 2 18x 2x 3 4 2 2 36. y y 2x x 1 2x x 1 32 1 38. f x fx x x2 x 2 2 5x 2x 2 y has no zeros. 7 The zeros of f correspond to the points on the graph of f where the tangent lines are horizontal. 4 y y′ −6 −1 −2 6 −3 f′ f 6 40. y y t2 9 t 2 42. g x gx x 1 2x 1 1 x 1 2x 1 1 5t2 8t 9 2t 2 The zero of y corresponds to the point on the graph of y where the tangent line is horizontal. 15 g has no zeros. 6 g y′ −3 y 6 −2 −2 g′ 10 − 15 44. y dy dx x2 tan 1 x sec2 1 x −4 6 y 1 2x tan x 5 y′ −6 The zeros of y correspond to the points on the graph of y where the tangent lines are horizontal. 46. (a) y y y0 sin 3x 3 cos 3x 3 (b) y sin x 2 y y0 1 2 1 x cos 2 2 3 cycles in 0, 2 Half cycle in 0, 2 The slope of sin ax at the origin is a. 352 Chapter 2 Differentiation 48. y dy dx sin cos cos 1 x x 2x 2 2 50. h x hx sec x2 2x sec x2 tan x2 52. y y 54. g g cos 1 21 1 2 1 2 1 2 1 2 sin v sin v 2x 2x 2 2 sin 1 sec sec 1 2 1 2 2x 41 2x sin 1 2x 2 tan sec2 tan 1 2 1 2 sec 1 2 tan 1 2 1 2 1 1 sec 2 2 56. g v gv 58. y y cos v csc v sec2 cos v tan2 cos v cos v 2 tan3 x 6 tan2 x sec2 x sin v cos2 v sin2 v cos 2v 60. g t gt 5 cos 2 10 cos 10 t t sin 5 cos sin t t 5 sin 2 t t 2 t cos x 2 62. h t ht 2 cot2 4 cot t t 2 2 t csc2 2 csc2 t t 2 2 64. y 3x 3x 5 cos 5 cos 5 sin 10 3x3 2x 2 2 x2 4 cot dy dx 3 3 x2 2 2 2x sin x 1 5, 45 66. y y sin x1 cos x1 3 sin x 1 x 3 3 23 13 68. 1 sin x 3 3 23 y y 4x 2, 2 9x2 4 3 cos x 13 3x 5 4x 1 cos x1 3 x2 3 cos x sin x 2 9x2 4 5 3x3 4x 4 y2 1 2 x 2x 2x 5 1 , 3 5 70. f x fx f4 74. y y 1 x 1 x2 2 x2 5 32 cos x, 1 x2 2 3x 2 x2 3 3x 3 2, 4, 1 16 72. f x fx f2 2, 3 x 3 2 3x 2x 2 2x 3 x2 3x 3 31 2x 12 2x 5 3 2 , 2 sin x 2 cos x y 2 is undefined. S ection 2.4 The Chain Rule 353 76. (a) f x fx 1 x x2 3 112 x x 32 x2 3 x2 5, 5 2, 2 12 78. (a) 2x 5 12 x 3 5 12 fx fx f 4 tan2 x, 4 ,1 2 tan x sec2 x 21 2 4 5 1 3 3 1 3 x2 13 9 Tangent line: y 64. (b) 1 4x 4 f2 4 33 4 ⇒ 4x y 1 0 Tangent line: y 62. (b) −9 2 6 13 x 9 2 ⇒ 13x 9y 8 0 − −4 9 −6 80. f x fx fx 82. f x fx x x 2x sec2 2 sec 2 2 2 2 x x 1 2 x 3 1 2 2 2 3 2 x sec x tan x sec2 x x 4 sec2 x tan x x 2 x sec2 sec2 2 2 2 fx 2 2 2 2 2 tan sec2 2 x x tan2 tan2 2 x x x 2 sec2 x tan x sec4 sec2 sec2 x x3 sec2 84. f 3 2 1 −3 −2 −1 y 86. f′ f 4 3 2 y f x 4 x −1 −2 −3 1 2 3 −1 −2 f′ f′ −3 −4 , 1 so f must be negative f is decreasing on there. f is increasing on 1, so f must be postive there. 88. g x gx f x2 f x 2 2x ⇒ g x 2x f x 2 The zeros of f correspond to the points where the graph of f has horizontal tangents. 354 Chapter 2 Differentiation 90. (a) g x gx (b) tan2 x gx sin2 x cos 2 x 1⇒g x 2 cos x sin x 0 0 92. y v 1 3 cos 12t 1 3 1 4 sin 12t 1 4 2 sin x cos x 1 1 sec2 x fx y 12 sin 12t 3 cos 12t 12 cos 12t 4 sin 12t When t 8, y 0.25 feet and v 4 feet per second. Taking derivatives of both sides, gx f x. Equivalently, f x 2 sec x sec x tan x and gx 2 tan x sec2 x, which are the same. 94. y A cos t 3.5 2 1.75 (b) v y 1.75 0.35 5 sin sin t 5 t 5 (a) Amplitude: A y Period: 10 ⇒ y 1.75 cos t 2 10 t 5 (c) T t 5 1.75 cos 96. (a) Using a graphing utility, or by trial and error, you obtain a model of the form Tt (b) 100 22.15 cos 11.60 cos t 6 t 6 1 1 6 64.18 22.15 sin t 6 1 20 0 13 − 20 0 0 13 (d) The temperature changes most rapidly when t 4.1 (April) and t 10.1 (October). The temperature changes most slowly T t 0 when t 1.1 (January) and t 7.1 (July). 98. (a) g x (b) h x (c) r x fx 2⇒g x fx 2f x f 3x 3x . 1 3 x fx 3 3f 3x gx hx rx 12 2 2. sx 2 4 4 8 2 3 2 3 4 3 1 0 1 3 1 3 2 3 1 1 1 2 2 2 2 4 3 4 4 8 2f x ⇒ h x f 3x ⇒ r x Hence, you need to know f r0 r (d) s x 1 fx 3f 0 3f 3 2 ⇒s x 3 3 1 4 12 1 3 1 2 4 1 fx Hence, you need to know f x s 2 f0 1 3, etc. S ection 2.4 The Chain Rule 355 100. f x p f x for all x. 102. If f x d f dx x 1 x f x , then d x fx fx f x. f x , which shows that f is (a) Yes, f x p periodic as well. f 2x , so g x (b) Yes, let g x Since f is periodic, so is g . 2 f 2x . f x f Thus, f x is even. 104. u d u dx 106. f x fx x2 2x d dx 4 x2 x2 4 ,x 4 ±2 u2 u2 12 u 2 12 2uu uu u2 u u ,u u 0 108. f x fx sin x cos x sin x ,x sin x k 110. (a) f x fx fx sec 2x 2 sec 2x tan 2x 2 2 sec 2x tan 2x tan 2x 4 sec 2x tan2 2x sec3 2x 2 sec 2x sec2 2x 2 (b) 6 P 2 P1 f 0 0 0.78 f f f 6 6 6 sec 3 3 2 tan 23 43 56 2 2 2 sec 423 4 3x 3 P1 x P2 x 6 1 56 x 2 28 x 6 6 2 4 3x 4 3x 6 6 2 2 (c) P2 is a better approximation than P1. 112. False. If f x sin2 2x, then f x 2 sin 2x 2 cos 2x . (d) The accuracy worsens as you move away from x 6. 114. False. First apply the Product Rule. 356 Chapter 2 Differentiation Section 2.5 2. 2x x2 y2 2yy y 16 0 x y Implicit Differentiations x3 3x2 y3 3y2y y x2y y2x 2yxy 2xy y y 0 y2 2xy 8. 1 xy 2 12 4. 8 0 x2 y2 xy 12 6. x2y 2xy 2 x 1 1 2y 2y 2y 0 0 0 0 2 xy 4 xy x cos y x y 2 y2 x2 xy y y 2 xy y y 2x x x 2y x y 2 xy xy y 2 xy 4 xy y y 10. 2 sin x sin y y 2 sin x cos y cos y cos x y 1 0 cos x cos y sin x sin y cot x cot y 12. 2 sin x cos y cos cos sin x x 2 0 0 cos sin x y sin y y sin yy y 14. cot y csc2 y y y x 1 1 y y 1 csc2 y 1 cot 2 y tan2 6y y y y 9 3 3 2 2 16. x sec 1 y 1 y y 1 1 sec tan y2 y y y2 sec 1 y tan 1 y y x 1 2 3 4 5 y2 cos 1 1 cot y y 18. (a) x2 4x 4 x 2 y2 2 9 4 9 (b) −1 −2 4 (Circle) 4 3± x 4 2 2 2 y = −3 + 4 − (x − 2) 2 y (c) Explicitly: dy dx 1 ±4 2 x 4 x ± x 2 −3 −4 −5 x 2 x 2 x x 4 2 3 2 2 12 2x 2 y = −3 − 4 − (x − 2)2 (d) Implicitly: 2 2 2x 2 2yy 4 2y 6y 6y y 0 2x x y 2 2 3 4 3± x y 2 x 2 2 2 3 S ection 2.5 Implicit Differentiation 357 20. (a) 9y 2 y2 y x2 x2 9 ± 9 1 x2 3 x2 9 9 12 x 2 2 (b) 9 −6 4 6 −4 dy (c) Explicitly: dx (d) Implicitly: 9y 2 ± 9 3 12 2x ±x 3x 2 9 ±x 3 ± 3y x 9y x 9 0 2x 2x 18y x 9y 24. x3 3x2y x 3xy2 3x2y x2y x 2y 2xy 2xyy x2 y 3 18yy 2x 18yy y 22. 2x x2 y3 3y2y y 0 0 2x 3y2 2 . 3 x3 x3 0 0 0 y2 y3 y3 y3 3xy2 xy2 y2 At 1, 1 : y 2xy y y 2xy 2x 2y yy xx At 26. 3x 2 3y 2 x3 y3 3y 2y 4x y y At 2, 1 , y 4xy 4xy 4y 4y 3y 2 4 3 12 8 1 4y 3x 2 3x 2 4x 8 5 28. x 1, 1 : y 1. x cos y 1 0 cos y x sin y 1 cot y x cot y x y sin y cos y y At 2, x3 3x2 3x2 2y 4 y2 x 3 :y x3 1 . 23 y3 6xy 6xy 6y 6x y 0 0 6y 6y 3y2 16 9 83 3x2 3x2 6x 32 40 2y y2 4 . 5 x2 2x 30. 4 x 2yy 4 y2 x y2 1 y 32. 3x2 3y2y y 3y2 At 2, 2 : y 2. At 48 , :y 33 16 3 64 9 358 Chapter 2 Differentiation 34. cos y sin y y y sin2 y cos2 y sin2 y x 1 1 ,0 < y < sin y 1 1 1 1 1 x2 cos2 y cos2 y , 1 x2 sin y y 36. 1<x<1 x2y2 2x2yy x2yy 2xy2 xy2 2x 2 1 y 3 0 0 1 0 0 0 0 2x2y4 2x2y4 2xy2 2xy2 x4y3 40. y2 2yy 4x 4 2 y 2y 2y 2 y2 2 y 4 y3 1 1 xy2 x2y 2xyy x2 y 4xyy 2 x2yy x2 y 2 2xyy x2yy x2yy x 4y3y y2 y2 y2 x2y4 x 4y3y y 4 4xy2 4x2y4 4xy x 1 2 1 xy2 2 x2y2 x2y4 2xy2 38. 1 y xy xy y y y x x x 1 0 0 x x2 x2 x2 1 1 y 1 1 x 1 y y 42. y2 2yy 11 x2 1 2x2 x2 1 2 x 12 2x 1 2x y At 2, 1 2x x2 2y x2 1 2 5 :y 5 1 4 4 1 2 2 5 54 5 5 10 8 1 x 10 5 x 0 2 1 . 10 5 −1 1 (2, ) 5 5 5 Tangent line: y 10 5y x 10 5y 2 −1 S ection 2.5 Implicit Differentiation 359 44. x2 y2 y 9 x y −6 4 (0, 3) At 0, 3 : Tangent line: y Normal line: x At 2, 5: 5 2 x 5 5 x 2 2 ⇒ 2x 2⇒ 5x 5y 9 0 3 0. 6 −4 4 (2, 5 ) −6 6 Tangent line: y Normal line: y 46. y2 2yy y 4x 4 2 y 1 at 1, 2 5 2y 0. −4 Equation of normal at 1, 2 is y 2 1x a distance of 4 units from 1, 2 . Therefore, x 1 2 1,y 3 x. The centers of the circles must be on the normal and at 3 x 2x 2 1 2 2 16 16 1 ± 2 2. 2 2 and 1 2 2 22 22 2 2 x Centers of the circles: 1 Equations: x x 48. 4x2 8x y2 1 1 22 22 2 2, 2 2 2 2 2, 2 16 16 22 y y 8x 2yy 4y 8 4 4y y 0 0 8 8x 2y 4 1: 0 0⇒y 4. 2: 4 2 2. 0 0⇒x 0, 2 0, 4 4 y 4x 2 −1 −1 y (1, 0) 1 2 3 4 x (0, − 2) −3 −4 −5 (2, − 2) Horizontal tangents occur when x 41 2 (1, − 4) y2 y2 81 4y 4y yy 4 4 Horizontal tangents: 1, 0 , 1, Vertical tangents occur when y 4x2 2 2 8x 8x 4 2 4x x 4x2 Vertical tangents: 0, 2 , 2, 360 Chapter 2 Differentiation 50. Find the points of intersection by letting y2 2x2 3x3 5 1. and 3x3 2x2 5 x3 in the equation 2x2 0 3y2 5. 2 2x2 + 3y2 = 5 (1, 1) Intersect when x Points of intersection: 1, ± 1 y2 2yy y x3: 3x2 3x2 2y 2x 2 4x y 3y 2 6yy 2x 3y 5: 0 −2 4 (1, − 1) −2 y 2= x 3 At 1, 1 , the slopes are: y At 1, y 3 2 1 , the slopes are: 3 2 y 2 . 3 y 2 . 3 Tangents are perpendicular. 52. Rewriting each equation and differentiating, x3 y y 3y x 3 3 1 1 x 3y 29 y y 3 13 3x 1 . x2 29 x (3y − 29) = 3 15 x 3 = 3y − 3 x2 −15 −3 12 For each value of x, the derivatives are negative reciprocals of each other. Thus, the tangent lines are orthogonal at both points of intersection. 54. x2 2x y2 2yy y C2 0 x y y y Kx K −3 2 2 K=1 3 −3 K = −1 C=1 3 C=2 −2 −2 At the point of intersection x, y the product of the slopes is xy K x Kx K 1. The curves are orthogonal. 56. x2 3xy2 3y2 y3 10 6xyy 6xy 3y2y 3y2 y y 58. (a) 4 sin x cos y 4 sin x y 1 4 cos x cos y 0 0 3y2 3y2 3y2 2x 2x 6xy (a) 2x (b) 2x dx dt 3y 2 dx dt 6xy dy dt 2x 3y2 3y2 dy dt dx dt 0 6xy 3y2 dy dt (b) 4 sin x sin y dx dt sin y y dy dt 4 cos x dx cos y dt 0 cos x cos y sin x sin y cos x cos y sin x sin y dy dt Section 2.6 Related Rates 361 60. Given an implicit equation, first differentiate both sides with respect to x. Collect all terms involving y on the left, and all other terms to the right. Factor out y on the left side. Finally, divide both sides by the left-hand factor that does not contain y . 62. 1671 18 00 B 1994 A 18 00 Use starting point B. 64. 1 2x x 1 y 0 c dy 2 y dx dy dx y x Tangent line at x0, y0 : y y0 y0 x0 x x0 x0 y0, 0 x0 y0 x-intercept: x0 y-intercept: 0, y0 Sum of intercepts: x0 x0 y0 y0 x0 y0 x0 2 x0 y0 y0 x0 y0 2 c 2 c. Section 2.6 2. y dy dt dx dt 2 x2 4x 1 4x 3x 6 dx dt Related Rates 4. 2x dx dt x2 2y y2 dy dt dy dt dx dt dy dt 2, dy dt dx dt 43 1 41 6 62 5 12 5 2 dx dt (a) When x dy dt (b) When x dx dt 3, y 3 8 4 4, y 3 4 2 25 0 x dx y dt y dy x dt 4, and dx dt 6 3, and dy dt 3 . 2 2, 8, dy 6 dt 3 and (a) When x (b) When x 1 and 5, 362 Chapter 2 Differentiation 6. y dx dt dy dt 1 1 2 2x x2 dx dt 2, 2 0, 0 cm sec. 2, 22 2 25 8 cm sec. 25 22 25 8 cm sec. 25 x2 8. y dx dt dy dt sin x 2 cos x dx dt 6, cos 6 2 3 cm sec. 1 2 (a) When x dy dt (b) When x dy dt (c) When x dy dt 10. (a) (b) 14. D dx dt dD dt 16. A dA dt 2 12 x 2 r2 2r dr dt (a) When x dy dt (b) When x dy dt (c) When x dy dt 4, cos 4 2 2 cm sec. 3, cos 3 2 1 cm sec. dx dy negative ⇒ negative dt dt dx dy positive ⇒ positive dt dt x2 y2 x2 sin2 x 12. Answers will vary. See page 145. sin2 x 12 2x 2 sin x cos x dx dt x sin x cos x dx x2 sin2 x dt 18. V dr dt dV dt 2 2 sin x cos x x2 sin2x 43 r 3 2 4 r2 dr dt 6, dV dt dV dt 4 4 6 2 If dr dt is constant, dA dt is not constant. dA dr depends on r and . dt dt (a) When r When r 2 2 288 in3 min. 2 4608 in3 min. 24, 24 (b) If dr dt is constant, dV dt is proportional to r2. x3 3 3x 2 dx dt 1, 31 2 20. V dx dt dV dt (a) When x dV dt (b) When x 3 9 cm3 sec. dV dt 10, 3 10 2 3 900 cm3 sec. S ection 2.6 Related Rates 363 22. V dr dt dV dt 12 rh 3 2 3 r2 dr dt 12 r 3r 3 r3 24. V 12 rh 3 1 3 25 3 h 144 r 5 25 h3 3 144 5 h. 12 By similar triangles, dV dt 6, 3 6 2 h ⇒r 12 10 25 2 dh dh h ⇒ 144 dt dt 8, dh dt 5 (a) When r dV dt (b) When r dV dt dV dt 2 216 in3 min. 144 dV 25 h 2 dt 9 ft min. 10 When h 24, 3 24 2 144 10 25 64 2 3456 in3 min. r 12 h 26. V (a) 1 bh 12 2 dV dt 12h 6bh 6h2 since b 1 dV 12h dt 2, dh dt dV dt h 28. 2x dx dt x2 2y y2 dy dt dx dt 25 0 y x dy dt 0.15y dy since x dt 0.15. dh dh ⇒ dt dt 1 and dV dt When h (b) If dh dt 1 2 12 1 12 2 3 8 1 ft min 6 When x 9 ft3 min. y 3 and h 8 2.5, 18.75, dx dt 18.75 0.15 2.5 0.26 m sec 2, then 12 ft 3 ft 3 ft h ft y 5 x 30. Let L be the length of the rope. (a) L2 dL 2L dt dx dt When L x dx dt L2 4 13 5 144 x2 dx 2x dt L dL x dt 13, 4 ft/sec (b) If dx dt 4, and L x dx L dt 5 13 4 13, 4L dL since x dt 4 ft sec. dL dt 144 52 5 169 144 5 13 ft 12 ft 20 ft sec 13 As L → 0, dL increases. dt 10.4 ft sec. Speed of the boat increases as it approaches the dock. 364 Chapter 2 Differentiation 32. x2 2x dx dt y2 0 dx dt s2 2s ds dt since dy dt 0 34. s2 x dx dt 75 160 3 53 277.13 mph. ds dt 902 60 28 x s dx dt x2 2nd 30 ft x s ds x dt 10, x 240 100 480 3 25 3rd s 90 ft Home 1st When s dx dt y 10 53 When x s ds dt 60, 902 602 30 13 56 13 15.53 ft sec. 60 28 30 13 x 5 mi s x 36. (a) 20y 20 6 20x 14y y dx dt dy dt 5 10 dx 7 dt y y 6y 20 x 20x 10 x 7 x 6 y 10 7 dy dt dx dt y2 5 50 ft sec 7 50 7 1 (c) When x 3 ,y 10 3 sin 5 3 5 y2 dx dt dy dt cos 1 2y dy dt 0⇒ 3 5 dy dt x dx . y dt 1 t ⇒ sin t t 1 4 2 (b) dy x dt 3 sin 5 2 5 50 7 35 7 15 ft sec 7 38. x t t, x2 (a) Period: (b) When x 2 seconds 3 ,y 5 4 0, 5 1 3 5 2 15 and 4 1 6 4 m. 5 3 10 dx dt x2 2x Thus, 1 ⇒t 2 Lowest point: 3 10 15 4 cos 6 9 25 5 Speed 95 . 125 95 125 0.5058 m sec S ection 2.6 Related Rates 365 40. 1 R dR1 dt dR2 dt 1 R2 dR dt 1 R1 1 1.5 1 R12 1 R2 42. rg tan 32r tan 32r sec2 d dt dv dt v2 v2, r is a constant. 2v dv dt d dt dv . dt 16r sec2 v v cos2 16r dR1 dt 1 R22 75, dR2 dt Likewise, d dt When R1 R dR dt 50 and R2 30 30 2 1 50 2 1 1 75 2 1.5 0.6 ohms sec. 44. sin dx dt cos d dt d dt 10 x 1 ft sec x 10 x2 dx dt 10 θ 10 dx sec x 2 dt 10 252 1 25 252 102 10 1 25 5 21 2 25 21 2 21 525 0.017 rad sec 46. tan d dt sec2 d dt dx dt (a) When (c) When x 50 30 2 1 dx 50 dt 50 sec2 30 , 70 , x y x y2 x y y Police 60 rad min rad sec θ 50 ft d dt 200 ft sec. 3 427.43 ft sec. (b) When 60 , dx dt x dx dt dx dt 200 ft sec. 48. sin 22 0 dx dt 50. (a) dy dt 3 dx dt means that y changes three times as fast as x changes. dy dt 1 y dx dt 89.9056 mi hr (b) y changes slowly when x 0 or x L. y changes more rapidly when x is near the middle of the interval. dy dt sin 22 240 x 22˚ 366 Chapter 2 Differentiation 52. L2 144 x2; acceleration of the boat dL dt dL dt d 2L dt 2 dx dt d 2x . dt 2 First derivative: 2L L 2x x dx dt dL dt dL dt d 2x dt 2 x d 2x dt 2 1 x dL dt 2 Second derivative: L dx dt L d 2L dt 2 dx dt dL dt 2 dx dt 2 When L d 2x dt 2 13, x 1 13 0 5 1 16 5 5, dx dt 4 2 10.4, and 10.4 1 5 4 (see Exercise 30). Since d 2L dL is constant, 2 dt dt 0. 108.16 20 92.16 18.432 ft sec2 y 54. yt dy dt y1 y1 4.9t2 9.8t 4.9 9.8 20 15.1 20 y 12 x 20 By similar triangles, x 20x When y 240 y x xy. 240 x 15.1 240 240 . 4.9 12 (0, 0) x 15.1, 20x 20 15.1 x x 20x 240 20 dx dt dx dt xy x dy dt x 20 y dx dt dy y dt 9.8 97.96 m sec. At t 1, dx dt 240 4.9 20 15.1 ...
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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.

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