Pre-Calc Homework Solutions 163

Pre-Calc Homework Solutions 163 - Section 4.4 (b) The...

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: Section 4.4 (b) The distance between the particles is the absolute value of f (t) sin t 3 163 42. The trapezoid has height (cos ) ft and the trapezoid bases measure 1 ft and (1 V( ) 1 (cos )(1 2 sin t 3 2 cos t 1 sin t. Find 2 2 sin ) ft, so the volume is given by 1 2 sin )(20) the critical points in [0, 2 ]: f (t) 3 2 sin t 1 cos t 2 3 sin t 2 0 1 cos t 2 1 20(cos )(1 sin ). 2 Find the critical points for 0 V( ) 20(cos )(cos ) 20 cos2 20(1 sin2 ) 20(1 20 sin 20 sin : 0 0 0 0 0 1 6 sin )( sin ) 20 sin2 20 sin2 sin 1)(sin 1) 1) tan t 3 5 11 The solutions are t and t , so the critical 6 6 5 11 , 1 and , 1 , and the interval points are at 6 6 3 3 , and 2 , . The particles endpoints are at 0, 2 2 5 11 sec and at t sec, and are farthest apart at t 6 6 20(2 sin2 20(2 sin sin 1 or sin 2 the maximum distance between the particles is 1 m. The critical point is at (c) We need to maximize f (t), so we solve f (t) f (t) 3 2 0. 6 , 15 3 . Since V ( ) 6 2 0 for 0 cos t 1 sin t 2 1 sin t 2 6 and V ( ) 0 for , the critical point 0 3 2 corresponds to the maximum possible trough volume. The cos t volume is maximized when 43. (a) R D 8.5 y Q C S 6 . This is the same equation we solved in part (a), so the solutions are t 3 sec and t 4 sec. 3 For the function y , 1 and f (t), the critical points occur at y L x x A P B 4 , 1 , and the interval endpoints are at 3 3 1 1 0, and 2 , . 2 2 4 Thus, f (t) is maximized at t and t . But 3 3 these are the instants when the particles pass each other, so the graph of y points and f (t) has corners at these Sketch segment RS as shown, and let y be the length of segment QR. Note that PB 8.5 x, and so QB x 2 (8.5 x)2 8.5(2x 8.5). Also note that triangles QRS and PQB are similar. QR RS PQ QB x 8.5(2x 8.5) x2 8.5(2x 8.5) 8.5x 2 2x 8.5 d f (t) is undefined at these instants. We dt cannot say that the distance is changing the fastest at any particular instant, but we can say that near t t 3 4 the distance is changing faster than at any other 3 or y 8.5 y2 8.52 time in the interval. y2 L2 L2 L2 L2 x2 x2 x 2(2x y2 8.5x 2 2x 8.5 8.5) 8.5x 2 2x 8.5 2x 3 2x 8.5 ...
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