Pre-Calc Homework Solutions 445

# Pre-Calc Homework Solutions 445 - Cumulative Review 47 Let...

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Unformatted text preview: Cumulative Review 47. Let s be the rope length remaining and x be the horizontal distance from the dock. (a) x s2 52, ds dt 445 4 54. 2 x x 4 1 dx 1 2 x 4 1 dx 4 x x 1 12 5 1.5, and 8 ft, 8 64 25 5 5 8 64 s 25 s2 dx dt s s2 ds , 25 dt 7 1 dx. x 2 which means that for s speed (b) s dx dt 55. Let u 1.9 ft/sec Then ds , which for 25 dt ln x, so du dx x(ln x)2 2e e ( 1.5) u dx x(ln x)2 1 du 1 u C 1 ln x C. d s sec 1 , so dt 5 Therefore, 1 ln x 2e e 8 ft becomes ( 1.5) 0.15 rad/sec. 1 ln 2 1 1 1 ln 2 ln 2 0.409 48. (a) Let h be the level of the coffee in the pot, and let V be 56. the volume of the coffee in the pot. h dh V , so dt 16 dV/dt 16 9 16 (3 (3t 2t)i t 2)i ex 1 j dt t 3 0.179 in./min. (ln t)j 1 (ln 3)j e x dx. csc2 u cot u du (csc2 u cot u cot (e x 2 2i 2i (ln 3)j (b) Now let h be the level of the coffee in the cone, and let V be the volume of the coffee in the cone. V 1 h 2 h 3 2 dV 4 h 2 dt dV h , so dt 12 4 ( 9) 25 3 57. Let u 1, so du Use the identity cot2 u e cot (e x 2 x 1. 1) dx 4 36 25 h 2 dh dh and dt dt 1) du u 1) C (e x 1) C 0.458 in./min. Since dh is negative, the level in the cone is falling at dt Since 1 C is an arbitrary constant, we may redefine C cot (e x 1) ex C. the rate of about 0.458 in./min. 49. (a) (1)(0 (b) (1)(1.8 1 0 and write the solution as 58. Let u 1.8 6.4 6.4 12.6 ... ... 1 16.2) 0) 165 in. 165 in. 1 2 x 2 0 2 1 0 50. 2 x dx 2 x dx 0 x dx 1 2 x 2 2 1 2 ds s , so du . 2 2 ds 1 ds ds 2 2[(s/2)2 s2 4 4(s/2)2 4 1 1 s tan 1 u C tan 1 C 2 2 2 1] 1 2 du u2 1 2.5 59. Let u 2, 2 2 cos (x 3) dx 3) 3) 3) , so du ( u C 3 sin (x 1 2 u 2 3) dx. C 51. Using Number 29 in the Table of Integrals, with a 2 2 4 ( x dx ) 2 . 2 x 2 4 x 2 2 sin 1 x 2 sin (x cos3 (x ) du 1 2 cos2 (x Alternately, observe that the region under the curve and above the x-axis is a semicircle of radius 2, so the area is 1 (2)2 2 3 2 . 1 1 3 dx x x 3 26 9.765 3 3 52. 1 x2 ln 3 /4 ln x 1 9 ln 3 1 3 53. 0 sec2 x dx /4 tan x 0 1 ...
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