ps6sol

# ps6sol - 18.02 Fall 2008 Problem Set 6 Part B Solutions x =...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 18.02 Fall 2008 Problem Set 6, Part B Solutions x = x0 + (v1 /v3 )z y = y0 + (v2 /v3 )z 1. a) z = z0 + z provided v3 = 0 (z can't be constant). b) y/z = v2 /v3 , and, since this is constant, the limit is the same, dy/dz = v2 /v3 . c) The derivative is calculated along a line: the intersection of the planes u = a and v = b for some constants a, b. The line of intersection is in the direction of 1, 1, 1 2, -1, 3 = 4, -1, -3 . Using the result of (b), along this line we have (y/z)u,v = dy/dz = 1/3. We get the same answer at every point because the lines are all parallel to each other (all in the direction of 4, -1, -3 ). 1 x 2 2-x 2. dy dx: I = R (x2 + y 2 ) dy dx = 0 0 (x2 + y 2 ) dy dx + 1 0 3 (x2 + y 2 ) dy dx - x) . 6 y=x Inner: x y + Outer: I = 2 1 3 x 3y 0 = 4 3 3x and x y + 2 1 3 2-x 3y 0 = 2x - x + 2 1 2 3 1 3 (2 q (1, 1) y =2-x 1 4 1 3x 0 2 + ( 3 x3 - 1 x4 - 4 1 12 (2 1 - x)4 ) 2-y = 1 3 4 + (4 - 1) = 3. 3 3 q (0, 0) qy=0 (2, 0) dx dy: I = R (x2 + y 2 ) dx dy = 0 y 2-y y (x2 + y 2 ) dx dy. Inner: 1 3 3x + y2 x 1 7 = ( 1 (2 - y)3 + 2y 2 - y 3 ) - 4 y 3 = 3 (2 - y)3 + 2y 2 - 3 y 3 . 3 3 7 4 1 12 y 0 1 = (- 12 + 2 3 2 1 Outer: I = - 12 (2 - y)4 + 3 y 3 - - 7 12 ) + 16 12 = 4. 3 a 3. a) 1 1 e-xy dy = - e-xy x y=a y=1 e-x - e-ax 1 . = - (e-ax - e-x ) = x x a a b) 0 e-x - e-ax dx = x e-xy dy dx = 0 1 1 a 0 e-xy dx dy 1 - e-xy y x= a = 1 dy = x=0 1 dy = ln a. y 1 ...
View Full Document

## This note was uploaded on 04/28/2009 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

Ask a homework question - tutors are online