EVNREV02 - Review Exercises for Chapter 2 367 Review...

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Unformatted text preview: Review Exercises for Chapter 2 367 Review Exercises for Chapter 2 2. f x x x lim 1 1 fx x x2 x xx x x fx 1x xx x 1 x x lim x x x x 1 1 x x x 1 x x x 1 2 1 1 1 fx x→0 x→0 lim lim lim 4. f x fx 2 x lim x→0 x x 1x 1x 1 x xx 1 1 x lim x→0 x→0 x2 x x 1x 1 x x x→0 xx x 2x 1x 2 1x 1 x 2 6. f is differentiable for all x fx 2 x x x x 2x xx 2x xx 2 x 4x 4x xx 2, x2, xx 2 x2 if x < if x ≥ 2 2 2. 2 x 2x xx fx 3. x→0 lim lim x 2x x→0 x→0 lim lim x2 1 x→0 x→0 8. f x 10. Using the limit defintion, you obtain h x At x 2, h 2 3 8 4 2 67 . 8 3 8 4x. (a) Nonremovable discontinuity at x (b) Not differentiable at x discontinuous there. y 5 4 2 because the function is 1 −5 −4 −2 −1 −1 −2 x 1 2 12. (a) Using the limit definition, f x At x y 2 y 0, f 0 2x 2x 0 2 x 2 . 12 (b) −6 4 (0, 2) 2. The tangent line is 6 −4 368 Chapter 2 Differentiation 14. f 2 x→2 lim fx x 1 x x 3 x x 1 f2 2 1 3 2 x 2x 1 13 1 13 1 9 20. g x gx x12 12x11 x1 1 x 2 2 16. 1 y f x→2 lim − π 2 π 2 x x→2 lim −1 f′ x→2 lim 18. y y 0 12 22. f t ft 8t 5 40t 4 2 x 9 2 24. g s gs 4s4 16s3 5s2 10s 26. f x fx 6 x 12 12 28. h x 32 1 x 2 x1 2x3 2 5 sin 3 5 cos 3 2 2 hx 4 x 9 3 4 9x3 30. g g 4 cos 4 sin 32. g g 34. s 16t2 s0 First ball: 16t2 t Second ball: 16t2 t2 75 75 16 0 53 4 2.165 seconds to hit ground 100 100 16 0 10 4 2.5 seconds to hit ground Since the second ball was released one second after the first ball, the first ball will hit the ground first. The second ball will hit the ground 3.165 2.5 0.665 second later. 36. s t 16t2 t 16t2 14,400 30 sec 1 6 14,400 0 Since 600 mph mi/sec, in 30 seconds the bomb will move horizontally 1 6 30 5 miles. R eview Exercises for Chapter 2 369 38. y v02 64 ( v02 128 v02 v02 , 64 128 ) ( v32 , 0 ) x 2 0 (a) y x 32 2 x v02 x1 32 x v02 v02 . 32 v02 32. v02 64. (b) y 1 64 x v02 v02 ,y 64 1 64 v02 v02 64 0. 0 if x 0 or x When x Projectile strikes the ground when x Projectile reaches its maximum height at x (one-half the distance) (c) y x 32 2 x v02 x1 32 x v02 0 (d) v0 70 ft sec v02 32 70 2 32 v02 128 153.125 ft 70 2 128 38.28 ft when x 0 and x x02 32. Therefore, the range is x v02 32. When the initial velocity is doubled the range is x 2v0 32 2 Range: x 4v0 32 2 Maximum height: y 50 or four times the initial range. From part (a), the maximum height occurs when x v02 64. The maximum height is 0 v02 y 64 v02 64 32 v02 v02 64 2 v02 64 v02 128 v02 . 128 0 160 If the initial velocity is doubled, the maximum height is y 2v0 64 2 2v0 2 128 4 v02 128 or four times the original maximum height. 40. (a) y 0.14x2 4.43x 58.4 (b) 320 0 0 60 (c) 12 (d) If x 65, y 362 feet. 0 0 60 (e) As the speed increases, the stopping distance increases at an increasing rate. 370 Chapter 2 Differentiation 42. g x gx x3 x3 x3 4x3 x x x 1 1 3x x 3x 1 3x 6x 2 3x3 2 x 2 3x 2 6x 2 6 3x 3 6 44. f t ft t 3 cos t t3 sin t t3 sin t cos t 3t2 3t2 cos t 6x 46. f x fx 48. f x x 1 2 6x x2 x2 23 5 1 16 x2 5x x2 1 2 11 x 2 1 2 11 fx 6x 12 3x2 5 2x x 50. f x fx 9 3x2 9 3x2 2x 2x 1 52. y 2 sin x x2 x2 cos x x4 1 1 1 sin x sin x sin x cos x 1 2 cos x sin x 12x1 3x 4 2 6x 2 18 1 3x2 3x 2x 2 y 56. y sin x 2x x cos x x3 2 sin x 54. y y 2x 2 x 2 tan x x 2 sec2 x 2x tan x y 1 sin x sin x 2 cos x 1 58. v t at v4 a4 62. h t ht ht 36 vt 36 16 t2, 0 ≤ t ≤ 6 2t 20 m sec 60. f x fx fx 34 8 m sec 4 sin t 4 cos t 4 sin t 5 cos t 5 sin t 5 cos t xy 9 x 4 74 9 4x7 4 64. xy y y cos x sin x xy xy y 1 x 1 x 5 10 10 0 sin x sin x cos x x y y y sin x 66. f x fx x2 12 x 3 1 13 68. f x 23 x2 5 x2 1 2x 2x 4 fx 2x 1 x2 3 x2 1 23 R eview Exercises for Chapter 2 371 70. h h 72. y 1 1 1 3 1 cos 2x 2 cos2 x 4 cos x sin x 4 sin x cos x y 3 2 sin 2x 31 1 6 2 1 2 1 1 4 2 2 sin x cos x 0 2 1 6 3 1 74. y y csc 3x cot 3x 76. y 3 csc2 3x csc 3x y sec7 x 7 sec5 x 5 sec4 x sec x tan x 1 3 csc 3x cot 3x 3 csc 3x cot 3x sec6 x sec x tan x sec5 x tan x sec2 x sec5 x tan3 x 78. f x 3x x2 1 3 x2 1 12 80. y 3x x2 12 x 2 1 1 12 cos x 1 x1 x 1 x 1 2 2x y 1 sin x x x 1 1 2 cos x 11 fx 3 x2 1 x2 1 3 x2 82. f x fx x 4 x3 4x 1 32 3x 2 32 1 sin x 1 cos x 1 2x 3x2 2x 4 6x 2 x2 8 4 2x 8 2 100 f′ −7 f 5 1x The zeros of f correspond to the points on the graph of f where the tangent line is horizontal. − 60 84. g x gx x x2 2x2 x2 1 1 1 12 86. y y 3x x 3x 2 3 2 2 7x 2 3x 2 g does not equal zero for any value of x. The graph of g has no horizontal tangent lines. 5 y does not equal zero for any x in the domain. The graph has no horizontal tangent lines. 75 y′ g′ −6 6 −3 − 25 y 3 g −3 372 Chapter 2 Differentiation 88. y y 2 csc3 3 x x csc3 x cot x −1 10 y y′ 8 −4 The zero of y corresponds to the point on the graph of y where the tangent line is horizontal. 90. y y y x 1 tan x 2 3 92. y y y sin2 x 2 sin x cos x 2 cos 2x sin 2x 94. g x gx gx 98. v dv dh 2gh 4 h 9, 4, y2 0 dv dh dv dh x3 x3 3x2 3x2 3x2 x2 2 32 h 6x x2 2 5 1 3x2 x2 5x 12 3 5 x 2x 2 x3 sec2 x 2 sec x sec x tan x 2 sec2 x tan x x x2 2x2 x2 x x2 2x2 1 1 1 1 3 32 2 6x3 15x2 18x x2 1 3 96. h x hx hx 8h (a) When h (b) When h 4 ft sec. 3 2 ft sec. x2y x2y x2y 2xy xy xy 2xy y y2 2y2 xy y 4yy 100. 2x x2 9y2 18yy 4x 4 3y 3y 1y y 0 0 4 2x 102. 3 6y 0 x2 x 4y y y 4 2x 3 6y 1 2xy y x 4y 10 104. 1 cos x y sin x y sin x y y y y x 1 1 1 sin x y 106. 2x x2 y2 2yy y 16 0 x y 5 3 3 3y 3 5y 5 x 3 16 3 x 5 30 5 0 5 0 −10 10 sin x y sin x y 1 1 −10 At 5, 3 : y Tangent line: y csc x 5x Normal line:y 3x 108. Surface area dx dt da dt 5 12x dx dt 12 4.5 5 270 cm2 sec A 6x 2, x length of edge. Problem Solving for Chapter 2 110. tan d dt sec2 d dt dx dt When x x 32 dx dt tan2 1 dx , 2 dt 6 16 1 4 1 6 x2 1 450 km hr. rad min 1 373 θ x 15 km min 2 Problem Solving for Chapter 2 2. 10 8 6 4 −8 −6 −4 −2 −4 −6 x 2 4 6 8 10 y Let a, a2 and b, For y x 2, y b2 2b 5 be the points of tangency. x2 b b 5 2 2 2x and for y 2⇒a 5 b2 2b 4b2 0 0 1 0 1 2b 6b 2x 5, y 1 2b b 2 2x 2. 2b Thus, 2a tangent line is a2 1 b2 2b ab 2b 4b 2b b 2b 1 2 b2 1 6 4 1, or a b2 b 2b b. Furthermore, the slope of the common 5 1 2b 2 ⇒ ⇒ 2b2 ⇒ 2b2 ⇒ b2 ⇒b b For b y For b y 4 2, a 1 1, a 4x 1 2x 1 b 2, 1 and the points of tangency are 1 ⇒y b 2x 1 1, 1 , 2, 5 . The tangent line has slope 2: 2 and the points of tangency are 2, 4 and 4x 4. 1, 8 . The tangent line has slope 4: 2 ⇒y 4. (a) y x 2, y 2x. Slope 4 y 4 at 2, 4 . 4x 4x 1 4. 1 4 Tangent line: y 2 4 (b) Slope of normal line: Normal line: y 4 y y x 2, 1 4x 9 4. 9 2 x 2 9 2 1 4x x 2 ⇒ 4x 2 x 18 0 ⇒ 4x 9 81 4 , 16 9x 2 0 Second intersection point: 0 0 (c) Tangent line: y Normal line: x —CONTINUED— 374 Chapter 2 Differentiation 4. —CONTINUED— (d) Let a, a2 , a 0, be a point on the parabola y x 2. Tangent line at a, a2 is y 1 y xa a2. To find points of intersection, solve 2a x2 x2 x2 1 x 2a x x x x 1 4a 1 4a a a 1 x 2a 1 16a2 1 4a 2 2a x a a2. Normal line at a, a2 is 1 x 2a a2 a2 a ±a a a2 1 2 1 2 1 4a 1 16a2 2 1 4a 1 4a a Point of tangency a 1 2a 2a2 1 2a 2a2 1 . 2a 1 ⇒x 4a 1 ⇒x 4a The normal line intersects a second time at x 6. f x fx a b cos cx bc sin cx b 1 b cos bc sin c 4 c 4 1 3 2 1 Equation 1 Equation 2 Equation 3 b 1 c c sin 4 b cos cos c 4 c 4 3 ⇒ 2 1 2 b b cos c 4 1 2 At 0, 1 : a At 3 , :a 42 From Equation 1, a From Equation 3, b b. Equation 2 becomes 1 1 1 . Thus c c c sin c sin 4 4 1 c c sin 2 4 1 c c sin 2 4 1 ,a 2 3 ⇒f x 2 cos 3 2 1 cos c 4 Graphing the equation g c One answer: c 2, b c 4 1, you see that many values of c will work. 1 cos 2x 2 P roblem Solving for Chapter 2 8. (a) b2y 2 y2 x3 a 3 375 x ; a, b > 0 (b) a determines the x-intercept on the right: a, 0 . b affects the height. xa x b2 x3 a b x and y2 x3 a b x Graph y1 (c) Differentiating implicitly. 2b 2 y y y ⇒ 3ax 2 3a x b2y 2 y2 3a 4 3x 2 a x x3 0 3ax 2 4x 3 3ax 2 4x3 2b2y 4x3 4x 3a . 4 3 a 3a 4 ± 27a3 1 a 64 4 3 3a2 16b 3 3a2 16b 27a 4 ⇒y 256b 2 Two points: 3a 3 3a2 3a , , , 4 16b 4 dy dt dx dt 1 x 3 dx dt 10. (a) y 1 dx dt (b) D x1 3 ⇒ 1 8 3 23 23 12 cm sec x2 y2 ⇒ dD dt 12 x 2 x y 2 2x dx dt 2y dy dt dx dy y dt dt x2 y2 98 68 49 cm sec. 17 8 12 21 64 4 y ⇒ sec2 x 68 θ 8 2 (c) tan d dt x dy dt x2 y dx dt From the triangle, sec 68 d . Hence 8 dt 81 2 12 68 64 64 16 68 4 rad sec 17 376 Chapter 2 Ex Differentiation x x Ex Ex 1 x Ex x Ex x E0 lim Ex x 1 x→0 12. E x lim lim x→0 x→0 ExE x x Ex lim E x x→0 E x lim But, E 0 Thus, E x 1 x→0 lim x→0 1. ExE 0 e x. E x exists for all x. For example: E x 27 t 5 14. (a) v t at (b) v t S5 27 ft sec 27 ft sec2 5 27 t 5 27 5 10 27 2 0⇒ 27 5 27 t 5 6 27 ⇒ t 73.5 feet 5 seconds (c) The acceleration due to gravity on Earth is greater in magnitude than that on the moon. ...
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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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