Pre-Calc Homework Solutions 448

# Pre-Calc Homework Solutions 448 - 448 Cumulative Review...

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Unformatted text preview: 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. ...
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