# Y 0 x 2 ? 1 2 x 2 2 x which is impossible now check

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Problem 7(a) - Spring 2008 Consider the equation x 2 + y 2 / 9 + z 2 / 0 , y 0 , and z 4 = 1. Identify this quadric (i.e. quadratic surface), and graph the portion of the surface in the region x 0. Your graph should include tick marks along the three positive coordinate axes, and must clearly show where the surface intersects any of the three positive coordinate axes. Solution: This is an ellipsoid . A problem of this type will not be on this midterm.
y = 0 = x = ± 2. λ = 1 = 2 x - 2 = 2 x , which is impossible. Now check the value of T at 3 points: T (1 , 0) = - 1 , T (2 , 0) = 0 , T ( - 2 , 0) = 8 . Maximum temperature is 8 and the minimum temperature is - 1.
Problem 7(b) - Spring 2008 Consider the equation x 2 , y , z , y , z ) = x 2 + y 2 9 z x F x = 2 x = - 2 x 1 2 z = Problem 7(c) - Spring 2008 Consider the equation x 2 ) = x 2 = F ( 1 2 , 0 = = ( 2 , = 2( x - ( r , θ θ = π 2 . + y 2 e xy )( f ∂θ Determine the equation of the tangent plane to the surface at the point f θ at r 1). ) = x 2 = F ( 1 2 , 0 = = ( 2 , = 2( x - ( r , θ θ = π 2 . + y 2 e xy )( f ∂θ Problem 8(a) - Spring 2008 Given the function f ( x , y Find the linearization L ( x , y ) = 2 xy + y (0 , Let L ( x , y L ( x , y ) = f (0 , L ( x , y Calculating at ( . 1 , 4 . L ( . 1 , 4 . . Calculate f θ at r = 5 and θ = π 2 . + y 2 e xy )( f ∂θ Problem 8(c) - Spring 2008 Given the function f ( x , y ) = f ( x ( t ) , y ( t )) where f ) is the function defined above. Calculate F prime (3) , y (3)) = (0 , Solution: The Chain Rule gives d F dt = f x dx dt + f y dy dt F prime (3) = (0 + 5 2 e 0
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