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

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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 ...
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