This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: curve given by x = cos t , y = sin t from (1 , 0) to (0 , 1). Hint : Work = R C F d r Solution : The path is r ( t ) = h cos t, sin t i for 0 t 2 . Z C F d r = Z / 2 hx 2 , xy i hsin t, cos t i dt = Z / 2 hcos 2 t, cos t sin t i hsin t, cos t i dt = Z / 2 ( sin t cos 2 t + cos 2 t sin t ) dt = 2 Z / 2 sin t cos 2 t dt = 2 3 . 2. (3 points) Is the integral you evaluated above independent of path? How do you know? Solution : No, the integral is not independent of path because the vector eld given by F ( x, y ) is not the gradient of any function. We know this because P y 6 = Q x , where F ( x, y ) = h P, Q i = hx 2 , xy i . Another way of saying this is thatx 2 dx + xy dy is not an exact dierential....
View
Full
Document
This note was uploaded on 09/09/2009 for the course MATH 240 taught by Professor Cremins during the Spring '07 term at Maryland.
 Spring '07
 Cremins
 Calculus, Linear Algebra, Algebra

Click to edit the document details