Review Eercises for Chapter 6. r v 0 sin. Let f, 00, d 0.6. v 0 00 ftsec changes from 0 to dr 00 cos d 6 0 d 0 r dr 80 00 6 96 feet 80 cos 0 96 feet 8080 f f fd d f 99. 00 0.6 9.97 00 Using a calculator: 99. 9.9699 7. Let f, 6, d. f f f d d f 6 6.998 00 6 9. Let f,, d 0.0, f. Then f.0 f fd.0 0.0 0.0. Using a calculator, 6.9980.. In general, when 0, d approaches.. True. True Review Eercises for Chapter. A number c in the domain of f is a critical number if fc 0 or is undefined at c. f f c is undefined. f c = 0. g cos, 0, g sin 8 6.8, 7.7 0 when sin..7, 0.88 Critical numbers: 0.,.7 Left endpoint: 0, Critical number: 0.,. Critical number:.7, 0.88 Minimum Right endpoint:, 7.7 Maimum
6 Chapter Applications of Differentiation. Yes. f f 0. f is continuous on,, differentiable on,. 7. f a f 0 for. c satisfies fc 0. 6 6 0 6 f f 7 0 b f is not differentiable at. 9. f b f a b a f, 8. f 8 7 fc c 7 c 9 7.76 79 f b f a b a f sin fc sin c c 0 f cos,. f A B C f A B f f A B A B fc Ac B A B Ac A c Midpoint of,. f f 7 Critical numbers: and 7 Interval: < < < < 7 7 < < Sign of f: f > 0 f < 0 f > 0 Conclusion: Increasing Decreasing Increasing 7. h Domain: 0, Interval: Sign of h: 0 < < h < 0 < < h > 0 h Conclusion: Decreasing Increasing Critical number:
Review Eercises for Chapter 6 9. ht t 8t ht t 8 0 when t. Relative minimum:, Test Interval: < t < < t < Sign of ht: ht < 0 ht > 0 Conclusion: Decreasing Increasing. cost sint v sint cost a When t inch and v inches/second. 8, b sint cost 0 when sint Therefore, sint and cost The maimum displacement is. c Period: inch. Frequenc: 6 6 6 cost tant.. f cos, 0 f sin f cos 0 when,. Points of inflection:,,, Test Interval: 0 < < Sign of f: f < 0 f > 0 f < 0 Conclusion: Concave downward Concave upward Concave downward < < < <. g g g Critical numbers: 0, ±, 0, 0, g0 > 0 g ± 8 < 0 Relative minimum at 0, 0 Relative maimums at ±, 7. 7 6, f, f 6, 0 0, 0 7 9. The first derivative is positive and the second derivative is negative. The graph is increasing and is concave down.
66 Chapter Applications of Differentiation. a b D 0.00t 0.t.9t 0.86t 9.0 69 0 0 9 c Maimum at.9, 9. 99 Minimum at.6, 69.6 97 d Outlas increasing at greatest rate at the point of inflection 9.8, 7.7 979. lim lim cos. lim 0, since cos. 7. h 9. f Discontinuit: Discontinuit: 0 lim lim lim Vertical asmptote: Horizontal asmptote: Vertical asmptote: 0 Horizontal asmptote:. f. f Relative minimum:, 08 Relative maimum:, 08 Relative minimum: 0.,.077 Relative maimum:., 0.077 00 0. 00 Vertical asmptote: 0. Horizontal asmptote: 0. f Domain:, ; Range:,, f 0 when. f Therefore,, is a relative maimum. Intercepts: 0, 0,, 0
Review Eercises for Chapter 67 7. f 6, Domain:,, Range: 8, 8 Domain:, ; Range: 8, 8 f 6 0 when ± and undefined when ±. 6 f 6, 8 8 6, 0, 0 8 6 6 8 0, 0 8, 8 f > 0 Therefore,, 8 is a relative minimum. f < 0 Therefore,, 8 is a relative maimum. Point of inflection: 0, 0 Intercepts:, 0, 0, 0,, 0 Smmetr with respect to origin 9. f Domain:, ; Range:, f 0 when,,. f 0 when, ±6. f > 0 Therefore,, 0 is a relative minimum. f < 0 Therefore,, 6 is a relative maimum. Points of inflection:, 0, 6 Intercepts: 0, 9,, 0,, 0, 0.60, 6, 0.6, 0,..69, 0.6, 0 6.7, 0.60. f Domain:, ; Range:, f f 0 when and undefined when, 0. is undefined when 0,. B the First Derivative Test, 0 is a relative maimum and is a relative minimum. 0, 0 is a point of inflection.,, 0,.9 0, 0 Intercepts:, 0, 0, 0
68 Chapter Applications of Differentiation. f Domain:,,, ; Range:,,, f < 0 if. f Horizontal asmptote: Vertical asmptote: Intercepts:, 0, 0,. f Domain:, ; Range: 0, f 8 0 when 0. f 8 0 when ±., 0,, f0 < 0 Therefore, 0, is a relative maimum. Points of inflection: ±, Intercept: 0, Smmetric to the -ais Horizontal asmptote: 0 7. f 0 Domain:, 0, 0, ; Range:, 6, 6,, 6 f 0 when ±., 6 0 f 6 8 6 8 0 f < 0 Therefore,, 6 is a relative maimum. f > 0 Therefore,, 6 is a relative minimum. Vertical asmptote: 0 Smmetric with respect to origin
Review Eercises for Chapter 69 9. f 9 Domain:, ; Range: 0, f 9 9 f 9 is undefined at ±. 9 f0 < 0 Therefore, 0, 9 is a relative maimum. 0 when 0 and is undefined when ±. 0, 0 0, 9, 0 Relative minima: ±, 0 Points of inflection: ±, 0 Intercepts: ±, 0, 0, 9 Smmetric to the -ais 6. f cos, Domain: 0, ; Range:, f sin 0, f is increasing. f cos 0 when,. Points of inflection: Intercept: 0,,,,, 0,, 6. 6 0 a 6 The graph is an ellipse: Maimum:, Minimum:, b 6 0 8 d d 6 d d 0 d 8 6 d d d 8 6 8,, The critical numbers are and. These correspond to the points,,,,,, and,. Hence, the maimum is, and the minimum is,.
70 Chapter Applications of Differentiation 6. Let t 0 at noon. L d 00 t 0t 0,000 00t t dl 00 00 88t 0 when t.9 hr. dt 6 Ship A at 0.98, 0; Ship B at 0, 9.8 d 0,000 00t t 098.6 when t.9 : P.M.. d 6 km 0, 0 B d 0, 0 t 00 t, 0 A 00, 0 67. We have points 0,,, 0, and, 8. Thus, Let m 8 0 f L 8. 0 8 or 8 f 8 6 0. 6 0 when 0, minimum. Vertices of triangle: 0, 0,, 0, 0, 0 0 0 8 6 0,, 8, 0 6 8 0 69. A Average of basesheight s s s see figure da d s s s s s s s s 0 when s. s s A is a maimum when s. s s s+ s s s s 7. You can form a right triangle with vertices 0, 0,, 0 and 0,. Assume that the hpotenuse of length L passes through, 6. Let m 6 0 6 0 6 or f L 6. f 7 0 when 0 or. L.0 feet 0
Review Eercises for Chapter 7 7. csc csc L L L L 6 csc 9 csc dl 6 csc cot 9 sec tan 0 d L 6 or L 6 csc see figure 9 or L 9 csc L π θ θ 9 6 csc 9 sec L θ 6 tan tan sec csc tan sec tan L 6 9 Compare to Eercise 7 using a 9 and b 6. ft.07 ft 7. Total cost Cost per hournumber of hours T v 600 0 v 0 v 60 v dt 0 dv 60 v v,000 60v 0 when v 000 00.8 mph. d T 00 so this value ields a minimum. dv v > 0 when v 00 77. f From the graph ou can see that f has three real zeros. f n n f n f n f n f n n f n f n.000 0.0.700 0.0.. 0.009.00 0.00. n n f n f n f n f n n f n f n 0.000 0.70.00 0.667 0. 0.07.6667 0.09 0. 0.7 0.7 0.000.68 0.000 0.7 n n f n f n f n f n n f n f n.9000 0.90 7.800 0.00.8797.8797 0.00 7.998 0.000.879 The three real zeros of f are., 0.7, and.879.
7 Chapter Applications of Differentiation 79. Find the zeros of f. f From the graph ou can see that f has two real zeros. f changes sign in,. n n f n f n f n f n n f n f n.000 0.76 7.90 0.06.6.6 0.000 7. 0.00.60 On the interval, :.6. f changes sign in,. n n f n f n f n f n n f n f n.000 0.6.000 0.00.0.0 0.068. 0.00.6.6 0.000.60 0.0000.6 On the interval, ;.. 8. cos cos 8. d sin cos d d sin cos d S r. dr r ±0.0 ds 8r dr 89±0.0 ±.8 square cm ds S 00 8r dr dr r 00 00 r ±0.0 00 ±0.6% 9 V r dv r dr 9 ±0.0 ±8. cubic cm dv V 00 r dr dr 00 00 r r ±0.0 00 ±0.8% 9 Problem Solving for Chapter. Assume < d <. Let g f d a. g is continuous on a, b and therefore has a minimum c, gc on a, b. The point c cannot be an endpoint of a, b because ga fa d d < 0 gb fb d d > 0 Hence, a < c < b and gc 0 fc d.
Problem Solving for Chapter 7. a For a,,, 0, p has a relative maimum at 0, 0. For a,,, p has a relative maimum at 0, 0 and relative minima. b p a a 0 0, ± a p a a For 0, p0 < 0 p has a relative maimum at 0, 0. c If a > 0, ± are the remaining critical numbers. a p ± a a > 0 p has relative minima for a > 0. d 0, 0 lies on. Let ± Then a. p a a 6 a 9 8 a a 9 a. Thus, 9 is satisfied b all the relative etrema of p. a ± a a = a = a = a = 8 a = a = a = 0. p a a p a a p 6 a For a 0, there is one relative minimum at 0, 0. b For a < 0, there is a relative maimum at 0,. c For there are two relative minima at ± a < 0, a. d There are either or critical points. The above analsis shows that there cannot be eactl two relative etrema. a = a = a = 8 7 6 a = 0 a = a = a = 7. f c f c 0 c c c f c If c 0, f has a relative minimum, but no relative maimum. If c > 0, c is a relative minimum, because If c < 0, c is a relative minimum too. Answer: all c. f c > 0.
7 Chapter Applications of Differentiation 9. Set fb fa fab a b a k. Define F f f a fa a k a. Fa 0, Fb f b f a fab a kb a 0 F is continuous on a, b and differentiable on a, b. There eists c, a < c < b, satisfing Fc 0. F f fa k a satisfies the hpothesis of Rolle s Theorem on a, c : Fa 0, Fc 0. There eists c, a < c < c satisfing Fc 0. Finall, F f k and Fc 0 implies that k fc. Thus, k fb fa fab a fc f b f a fab a fc b a. b a. E tan 0. tan 0. tan E 0 tan 0 sec tan sec 0 tan tan 0 sec 0 0 tan 0 tan 0 sec tan sec 0 tan tan 0 sec 0 sec tan sec 00 tan sec 0 tan sec 0 tan 0 tan 0 tan tan 0 0 tan ± 00 0 0 tan tan 0 tan 0.9099,.099 Using the positive value, 0.76, or.. 00 tan sec 0 tan sec. v 00 sin v 00 cos 0 n, n, n an integer
Problem Solving for Chapter 7. The line has equation or. Rectangle: Area A. A 8 0 8 Dimensions: Calculus was helpful. Circle: The distance from the center r, r to the line must be r: 0 r r r 9 6 r 7r 7r 7r r or r 6. Clearl, r. Semicircle: The center lies on the line and satisfies r. r Thus No calculus necessar. r 7 r r 7. 7. : + + + + + + 0 0 ± ± The tangent line has greatest slope at, and least slope at,. 9. a 0. 0. 0. 0. 0..0 sin 0.0998 0.9867 0.9 0.89 0.79 0.87 sin b Let f sin. Then f cos and on 0, ou have b the Mean Value Theorem, fc cosc sin Hence, f f 0, 0 < c < 0 sin cosc sin sin