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# final_fall2005solutions - Math 192 Final Exam Solutions Dec...

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Math 192 Final Exam Solutions. Dec 8 , 2005 1. f ( x, y ) = 1 - x 2 2 - y 2 16 f x = - x 2 1 - x 2 2 - y 2 16 - 1 2 , f y = - y 16 1 - x 2 2 - y 2 16 - 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 C ( F · T ) ds = D [1 - ( - 1)] dA = 2(Area of D ) . b) F · dr dt = ( b sin t )( a sin t ) + ( a cos t )( b cos t ) = ab. Therefore, C ( F · T ) ds = 2 π 0 ( F · dr dt ) dt = 1 2 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 = 5 π 6 π 6 2 sin θ 1 rdrdθ = 1 2 5 π 6 π 6 (4 sin 2 θ - 1) = 1 2 5 π 6 π 6 (1 - 2 cos 2 θ ) = 1 2 [ θ - sin 2 θ ] 5 π 6 π 6 = π 3 + 3 2 . 4. Since z = ρ cos φ , the plane and sphere intersect when 1 = 2 cos φ φ = π 3 . Also, ρ = sec φ every- where on the plane. Therefore, Volume = 2 π 0 π 3 0 2 sec φ ρ 2 sin

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final_fall2005solutions - Math 192 Final Exam Solutions Dec...

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