Pre-Calc Homework Solutions 404

206 150 sin 20 area 0 2a 0 y dx a1 2 cos ta1 2 cos t t

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Unformatted text preview: cos t t 2 cos t) dt x cos t) dt 1 sin 2t 4 2 0 75 sin 20 8 150 cos 20 , y (75 sin 20 )/8 0 32t (150 sin 20 461.749 ft 32t)2 a 2 (1 0 2 Length 3 a2 (150 cos 20 )2 a2 t dx 32. dt 2 sin t cos t), so 2 dt which, using NINT, evaluates to (b) The maximum height of the projectile occurs when a(1 y cos t)] a(1 3 cos t 2 0, 75 75 sin 20 , y sin 20 16 16 Volume 0 [a(1 a 3 3 2 cos t) dt 2 so t 3 41.125 ft (1 0 3 cos t 3 sin 2t 4 2 0 cos t) dt 36. (a) (b) 37. (a) (b) 641.236 ft 5625 64 a t 3 sin t sin t 3 t 2 1 sin3 3 87.891 ft t 840.421 ft 16,875 64 5 2 3 a 263.672 ft 33. (a) QP has length t, so P can be obtained by starting at Q and moving t sin t units right and t cos t units downward. (If either quantity is negative, the corresponding direction is reversed.) Since Q (cos t, sin t), the coordinates of P are x cos t t sin t and y sin t t cos t. y Q 38. (a) It is not necessary to use NINT. 75/8 Length 5625 8 0 (150 32t) dt 150t 16t 2 75/8 0 703.125 ft 351.5625 b (b) t sin t t t t t cos t P(x, y) x 5625 16 39. In the integral a 2 y dx 2 dt dy 2 dt, replace t with x dt and y with f (x). Then dx dx becomes dt dx (1, 0) 1....
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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