final_fall2005solutions - Math 192 Final Exam Solutions....

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Math 192 Final Exam Solutions. Dec 8 , 2005 1. f ( x, y ) = q 1 - x 2 2 - y 2 16 f x = - x 2 ± 1 - x 2 2 - y 2 16 ² - 1 2 , f y = - y 16 ± 1 - x 2 2 - y 2 16 ² - 1 2 , f x (1 , 2) = - 1 , f y (1 , 2) = - 1 4 , f (1 , 2) = 1 2 a) Equation of tangent plane: - 1( x - 1) - 1 4 ( y - 2) - ( z - 1 2 ) = 0 . b) Linearization: L ( x, y ) = f (1 , 2) + f x (1 , 2)( x - 1) + f y (1 , 2)( y - 2) = 1 2 - ( x - 1) - 1 4 ( y - 2) , L (0 . 8 , 2 . 4) = 0 . 6 2. a) Since the components of ~ F and their partial derivatives are continuous everywhere, we may apply Green’s Theorem with M = - y and N = x to get Z C ( ~ F · ~ T ) ds = Z Z D [1 - ( - 1)] dA = 2(Area of D ) . b) ~ F · d~ r dt = ( b sin t )( a sin t ) + ( a cos t )( b cos t ) = ab. Therefore, R C ( ~ F · ~ T ) ds = R 2 π 0 ( ~ F · d~ r dt ) dt = 1 2 R 2 π 0 ( ab ) dt = 2 πab. Thus, area of ellipse = 1 2 (2 πab ) = πab. 3. The circles intersect when 2 sin θ = 1 sin θ = 1 2 θ = π 6 , 5 π 6 . Therefore, area = R 5 π 6 π 6 R 2 sin θ 1 rdrdθ = 1 2 R 5 π 6
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final_fall2005solutions - Math 192 Final Exam Solutions....

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