Pre-Calc Homework Solutions 404

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Ask a homework question - tutors are online