Business Calc Homework w answers_Part_90

Business Calc - 446 Cumulative Review 65 Use integration by parts cos 2x dx 1 sin 2x 2 1 x e sin 2x dx 2 60 Use integration by parts u du e x e x

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Unformatted text preview: 446 Cumulative Review 65. Use integration by parts. cos 2x dx 1 sin 2x 2 1 x e sin 2x dx 2 60. Use integration by parts. u du e x e x dv x u du x2 2x dx dv v x 2 cos x sin x dx cos x 2x cos x dx e dx v cos 2x dx 1 x e sin 2x 2 x 2 sin x dx Now let Now let u du Then e x e x dv x e dx v 1 sin 2x dx 2 1 cos 2x 4 u du x dx dv v 2 cos x dx 2 sin x 2x sin x 2x sin x 2 sin x dx 2 cos x C C x 2 sin x dx cos 2x dx 1 x e cos 2x 4 e 5 x x 2 cos x x cos x 2 1 x e sin 2x 2 1 e x cos 2x dx 4 dy dx (2 x 2) cos x 2x sin x so e 61. x x2 x The graph of the slope field of the differential equation cos 2x dx (2 sin 2x cos 2x) A x 1 x B 6 C x 2 sin x and the antiderivative y (2 x 2) cos x 2 5x 6 (x x 2 1)(x 6) 2x sin x is shown below. x 2 A (x B 6) B(x 1, B 6A 1) (A B) x (B 1 ,B 7 6A) 8 so 7 [ 5, 5] by [ 10, 10] Solving A x x2 2 yields A 2 8 1 . Then 5x 6 7(x 6) 7(x 1) x 2 8 1 dx dx x 2 5x 6 7(x 6) 7(x 1) 8 1 1 (x 6)8 ln x 6 ln x 1 C ln 7 7 7 x 1 5 [3 2 66. Use integration by parts. C 359; u du x dx xe x e x dx dv v xe x e x dx ex ex C e x(x 1) C 62. Area Volume 63. y C 64. y y y y (t 2(8.3) 359 1 2(9.9) 8975 ft 3 ... 2(8.3) 3] 25 1) x e x dx 1 2 Confirm by differentiation: 7 and y 2 1 2t e C; y (0) 2 1 2t 7 1 e . 2 2 t 1 1 C 2, so d x [e (x dx 1) C] ex (x 1)e x xe x Ce k(5). 67. (a) y 1 cos 2 sin 2 1 cos 2 sin 2 1 sin 2 cos 4 1 sin 2 cos 4 Ce kt, with 6,000 10,000 6,000 ln 3 3 5 Ce k(2) and 10,000 5 3 C1, and y 1 . 2 1 2 1 2 2 0 Then e k(5 2) , so e 3k and therefore C2, and y 4 2 0 k 0.170. 6,000 e 2k Furthermore, C 4268. The approximate number 4268e 0.170t. of bacteria is given by y (b) About 4268 Cumulative Review 68. Let t be the time in minutes where t 0 represents right 72. The region has four congruent portions, so /2 447 now, and let T(t) be the number of degrees above room temperature. Then we may write T(t) T(0) k 50 and T( 15) 13 1 ln 10 15 k(120) Area 73. Solve 5 2x 2 Area 4 0 sin 2x dx x2 x2 4 1 cos 2x 2 /2 4 0 T0e kt where 3 to find the integration limits: 65, giving T0 50 and 8x 2 2. Then x 2) 64 3 0.0175. 6.13C above room temperature. 50e kt [(5 2 (x 2 2 3)] dx (8 2 2x 2) dx (a) 50e 8x 74. Solve y y2 y 2 (b) Solving 5 gives t ln 0.1 k 2 3 x 3 2 2 131.6 minutes, 3 5 (1 (1 2 0 or about 2 hours and 12 minutes from now. dy y 0.08y 1 dx 500 500 dy 0.08 dx y(500 y) (500 y) y dy 0.08 dx y(500 y) 1 1 dy 0.08 dx y 500 y y 0y 21)/2 2 to find the integration limits: 1 2 21 . Then 3) dy 16.039. 69. Area 21)/2 [(y 1 2 2 0 2 0 2 0 2) 1 2 2 0 (y 2 75. Area 9 2 9 2 r2 d 2 cos 2 cos cos 9(1 cos )2 d (1 1 3 2 cos2 ) d 1 cos 2 2 Integrate both sides. ln y y 500 y d ln 500 y 0.08x C1 9 2 1 cos 2 2 1 sin 2 4 2 0 d C2e 0.08x C2e 0.08x 9 3 2 2 2 sin 0) 1 y (1 y 70. 1 ) 500C2e 0.08x 9 (3 2 27 2 x3 2 dx 2 42.412 1 7 x 4 7 1 1 500 Ce 0.08x 76. Volume 1 14 0.224 dy dx dy y 4 dy y 4 (y (x (x x2 2 4)(x 3) dx 3) dx 3x 2 3) 77. Solve 4x x 0 or x x2 0 to find the limit of integration: 4. By the cylindrical shell method, 4 Volume 0 2 x(4x 1 4 x 4 4 0 x 2) dx 128 3 4 2 0 (4x 2 x 3) dx ln y y 4 4 y C1 4 4 2 4 3 x 3 e C1e(x Ce(x 2 /2) 3x 134.041. 78. The average value is the integral divided by the interval length. Using NINT, /2) 3x 71. Use EULERT. x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 y 0 0.1 0.2095 0.3285 0.4568 0.5946 0.7418 0.8986 1.0649 1.2411 1.4273 average value 79. y 1 0 sin x dx 0.763 sec 2 x, so we may use NINT to obtain /4 Length /4 1 dy dt (sec 2 x)2 dx 1 2.556. 80. dx dt cos t and sin t, so we may use NINT to obtain /2 Length /2 /2 /2 cos 2 t 2 (1 sin t)2 dt 4. 2 sin t dt 448 Cumulative Review 88. (a) The work required to raise a thin disk at height y from 2 0 81. Using NINT, Length dy 82. dx dr 2 d d 0 1 2 d 6.110. the bottom is (weight)(distance) 10 Area 2 0 1 x/2 e , so we may use NINT to obtain 2 2 e x/2 2 2 e x/2 1 dx 2 0 60 y 2 dy (12 2 y). Total work 10 0 15 y 2(12 12y ) dy 2 y) dy 15 1 4 y 4 15 0 ( y 3 4y 3 10 0 e x/2 4 dy dt e x dx 8.423. sin t, so 22,500 (b) sin t) dt 3.470. 2 70,686 ft-lb. 257 sec 4 min, 17 sec 83. dx dt cos t and /2 0 0 /2 1 Area 2 sin t 2 sin t cos t 2 2 22,500 275 (1 2 sin t dt 89. The sideways force exerted by a thin disk at depth y is its edge area times the pressure, or dr 84. d 1, so we may use NINT to obtain 2 /2 1 (2 dy)(849y) H 1698 y dy. 1698 y dy 0 Area sin x 2 4 2 +1d 32.683. Total force Solve: 40,000 849 H 2, where H is depth. 85. Volume 0 1 x2 2 849 H 2 H ln x x dx x 4) dx 1 5 x 5 1 0 H 40,000 and V 849 12.166 ft 3. 1 x 1/2 (x 2x 5/2 4 7/2 x 7 0 1 2 x 4 2 9 280 90. Use l'Hpital's Rule: lim x x lim x lim x 2 x 0. 2 0.101. f (x) ln x grows slower than g (x) x. 1 t2 4 91. Use the limit comparison test with f (t) 86. Use the region's symmetry: /4 and ), g(t) (2 tan x) dx 2 Volume 2 0 1 . Since f and g are both continuous on [3, t2 f (t) g(t) /4 lim tan x dx 2 t 3 1, and t2 3 g(t) dt converges, we conclude that 4 8 8 8 8 8 87. (a) F F 0 /4 0 f (t) dt dt 3 converges. 2, 1 ln x 1 , and x (sec x /4 2 1) dx 92. Use the comparison test: for x b tan x 1 2 x 0 lim b 2 dx x x lim (ln b b 0 ln 2) e b 0 . Both integrals diverge. x 4 2 93. 5.394. k(0.8) k 250 N/m, so for 94. 1.2 0 e Since dx e x e x dx lim b 0 dx b 2 0 e e x dx b dx e x dx lim x 1, the 0 0 kx 200 300 N, x 1.2 original integral converges. 1 300 250 1.2 m. 125x 2 180 J 0 4r dr 1 r 2 b lim b1 0 4r dr 1 r 2 lim b1 4 1 r2 b 0 (b) Work 0 250x dx lim ( 4 b1 10 1 1 b2 dx 4) 10 4. The integral converges. 95. 0 dx 1 x 1 0 0 1 x 1 Since lim b1 dx 1 x lim b1 b dx 1 x b dx x 0 1 ln (1 x) 0 , the original integral diverges. Cumulative Review 2 449 96. 0 dx 3 1 dx 3 2 dx 3 101. The first six terms of the Maclaurin series are 1 1 x6 . By the Alternating 6! x7 1 Series Estimation Theorem, error 0.001. 7! 7! x 1 b 0 0 x 1 1 2 x dx lim b1 dx 3 1 x lim 1 a1 2/3 b a x2 2! x3 3! x4 4! x5 5! x 3 x lim b1 3 (x 2 1) lim 0 a1 3 (x 2 1)2/3 2 0. a The whole integral converges. 97. We know that 1 1 x 102. f (0) f (0) 1, f (0) 2 9(0 1 3(0 1)2/3 1)2/3 1 , 3 1 1 x 1 2x x x x2 x3 ... ( 1)nx n ... ... f (n)(0) ... 2 , 9 2 5 ... (3n ( 1)n 1 3n 4) , so the for 1 1 1. Substituting 2x for x yields 2x 4x 2 8x 3 ... ( 1)n2nx n Taylor series is 1 1 x 3 2 5 8 4 2 5 3 x x 4! 34 3! 33 2 5 ... (3n 4) n ... ( 1)n 1 x . n! 3n 2 x2 2! 32 2x for 1 2 1 1, so the interval of convergence is ... 1 . 2 98. (a) cos t 2 Since by the Ratio Test 1 1 (t 2)2 (t 2)4 (t 2)6 ... (t 2)2n ( 1)n 2! 4! 6! (2n)! 4n t 12 t4 t8 n t ... ( 1) 6! 2! 4! (2n)! ... lim n an 1 an 2 5 ... (3n 4)(3n (n 1)!3n 1 (3n 1)x (n 1)(3) 1)x n 1 Integrating each term with respect to t from 0 to x yields x (b) x 5(2!) 5 lim n n!3n 2 5 ... (3n 4)x n lim x 9(4!) 9 x , the radius of convergence is 1. an 1 an x 13(6!) 13 ... ( 1)n x (4n 4n 1 1)(2n)! .... n 103. Using the Ratio Test, lim n lim n 2 3 n 1 3n 2 1 , so 3 x ; Since the cosine series converges for all the series converges. 104. Note that an Test, since n 1 real numbers, so does the integrated series, by, the term-by-term integration theorem (Section 9.1, Theorem 2). 99. ln (2 ln 2 2x) x ln [2(x x 2 2 1 for every n. By the Direct Comparison n 1 2 diverges, so does . n n 1 n 105. Use the alternating series test. 1)] 3 ln 2 ... ln (x ( 1)n lim n 1) 1x n x 3 x 4 4 n n 1 ... x x , the Note that n 0 ( 1)n n 1 n 0 ( 1)nun, where un un 1 1 n 1 . an 1 Since by the Ratio Test lim an n Since each un is positive, un n for all n, and n lim un 0, the original series converges. series converges for 1 x 1. 1 does not. n 106. Using the Ratio Test, lim n n 1 ( 1)n converges, but n n 1 an 1 an lim n 3n 1 (n 1)! n! 3n lim n 3 n 1 0, and the 100. Let f (x) f (x) sin x. Then f (x) cos x, f (4)(x) cos x, f (x) sin x, 2 the series converges. sin x, and so on. At x sine terms are zero and the cosine terms alternate between 1 and (x 1, so the Taylor series is 2 )3 (x 2 )5 3! 5! (x 2 )2n 1 .... ( 1)n (2n 1)! (x 2 ) ... 450 Cumulative Review dr dt dv dt 107. (a) Using the Ratio Test, lim n 112. (a) v(t) 2) 1 n 1 ( cos t)i (sin t)i (1 (cos t)j sin t)j a n 1 an lim n (x n n (x 2)n x 1 x 2 , which 2 1, a(t) means that the series converges for or 3 x 1. Furthermore, at x (b) Using NINT, the distance traveled is 3 /2 3 /2 3, the series v(t) dt /2 3 /2 /2 ( cos t)2 4. (1 sin t)2 dt is n 1 n 1 1 , which diverges, and at x 1, the series is n ( 1)n , which converges. The interval of n 2 /2 2 sin t dt 113. Yes. The path of the ball is given by x y 100(cos 45 )t 16t 2 50 2t and 16t 2 13 5 2 13 5 2 convergence is convergence is 1. (b) 3 x 1 1 (c) At x 3 x 1 and the radius of 100(sin 45 )t 130, we have t 13 2 50 2t. When x an 1 an n ln2 n xn and so 75.92 ft, high enough 108. (a) Using the Ratio Test, lim n y 16 5 50 2 2 lim n (n (n xn 1 1)ln2 (n nx ln 2 (n) 1) ln 2 (n n n 1 to easily clear the 35-ft tree. 114. Since r cos x, r sin y, the Cartesian equation is x y 2. The graph is a line with slope 1 and y-intercept 2. x , which means x 1. At x 1, 115. 1) 1) lim n x lim n 2 ln n 1) n ln (n lim that the series converges for 1 the series converges by the Integral Test: 1 dx x(ln x)2 b lim b 2 2 lim b 1 ln b 1 ln 2 1 dx x(ln x)2 1 ln 2 1 ln x b 2 [ 3, 3] by [ 0.5, 3.5] The shortest possible -interval has length 2 . 116. x r cos cos cos 2 , . So the x 1 and the radius of convergence interval is convergence is 1. (b) 1 x 1 (c) Nowhere 109. 1 22 ( 3)2 3 1 y dy dx r sin dy/d dx/d dy : d sin cos sin sin cos sin 2 cos 2 2 sin cos 2, 3 2 13 , 3 13 Zeros of cos cos (2 cos 110. 1 cos , 1 sin 3 1 3 , 2 2 sin 2 1 cos 2 2 cos2 0 1) 4 , or 3 0 1)(cos 2 , 3 dx : d 0 2 111. dy dx t 3 /4 dy/dt dx/dt t 3 /4 1 32 42 3 sin t 4 cos t t 3 /4 4 3 , and 5 5 4 and 5 3 . The 4 0, Zeros of sin tangent vectors are 4, 3 2 sin 0 or cos cos 1 2 3 0 4 , 5 3 3 . The normal vectors are , 5 5 3 4 , . 5 5 sin 0, , or 5 , 3 2 ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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