quiz9 - curve given by x = cos t , y = sin t from (1 , 0)...

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Quiz 9 Name: Math 240 - Calculus III April 14, 2009 Note: In order to receive full credit, you must show work that justifies your answer. 1. (6 points) Evaluate the integral Z C (2 x - y ) dx + ( x + 3 y ) dy where C is the path given by y = 1 - x 2 from (0 , 1) to (1 , 0). Solution : The path is parametrized by x = t , y = 1 - t 2 , for 0 t 1. Thus, Z C (2 x - y ) dx + ( x + 3 y ) dy = Z 1 0 ( 2 t - (1 - t 2 ) ) dt + ( t + 3(1 - t 2 ) ) ( - 2 t ) dt = Z 1 0 ( 6 t 3 - t 2 - 4 t - 1 ) dt = - 11 6 2. (4 points) For what value A is the integral Z C (2 x - y ) dx + ( Ax + 3 y ) dy independent of path? What is the corresponding potential function? Solution : Let P = 2 x - y and Q = Ax + 3 y . The integral is independent of path if P y = Q x , which occurs when A = - 1. The potential function is a function φ ( x, y ) such that φ x = P and φ y = Q . Examining antiderivatives, we find φ ( x, y ) = x 2 - xy + 3 2 y 2 + C for any constant C .
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Quiz 9 Name: Math 240 - Calculus III April 16, 2009 Note: In order to receive full credit, you must show work that justifies your answer. 1. (7 points) Find the work done by the force F ( x, y ) = - x 2 i + xy j acting along the
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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 h-x 2 , xy i h-sin t, cos t i dt = Z / 2 h-cos 2 t, cos t sin t i h-sin 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 = h-x 2 , xy i . Another way of saying this is that-x 2 dx + xy dy is not an exact dierential....
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quiz9 - curve given by x = cos t , y = sin t from (1 , 0)...

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