1-ans

1-ans - an integral for the area enclosed by the curves x =...

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

View Full Document Right Arrow Icon
Ma 116 First Quiz September 21, 2000 Problem I. A particle moves according to E r ( t ) = ( t 3 - 3 t ) E i + ( t 2 - 2 t ) E j . 1. Does the particle ever come to a stop? If so, when and where? Answer. The particle comes to a stop when E r 0 ( t ) = ( 3 t 2 - 3 ) E i + ( 2 t - 2 ) E j = E 0. 3 t 2 - 3 = 0 when t = ± 1, and 2 t - 2 = 0 when t = 1. Thus the particle stops when t = 1, and its position is E r ( 1 ) = - 2 E i - E j . 2. Is the particle ever moving straight up or down? If so, when and where? Answer. Write E r ( t ) = x ( t ) E i + y ( t ) E j . The particle moves straight up or down when d dt x ( t ) = 0 and d dt y ( t ) 6= 0. From Part 1 this happens when t = - 1, and at that time the particle is at E r ( - 1 ) = 2 E i + 3 E j Problem II. Set up but do not evaluate
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: an integral for the area enclosed by the curves x = y 2 , y = x + 5, y = 2, and y = 0 (see the diagram). y l1 l2 c x Answer. R 2 y 2-( y-5 ) dy Problem III. Sketch the region bounded by y = x 4 and y = 1. Set up but do not evaluate an integral for the area generated by revolving this region around the line y = 2. Answer. By the washer method the volume is π R 1-1 (( 2-x 4 ) 2-1 2 ) dx . By the shell method it is 2 π R 1 ( 2-y )( 2 y 1 / 4 ) dy ....
View Full Document

This note was uploaded on 04/17/2008 for the course MA 124 taught by Professor N/a during the Spring '08 term at BU.

Ask a homework question - tutors are online