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M25008SelSol2 - Mathematics 250 Selected Solutions Fall...

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Mathematics 250 Fall 2008 Selected Solutions Assignment #2 MM, Exercise 3.2.1: The equation of the tangent plane to the graph of a scalar-valued function f at a given point a is Df ( a )( x - a ) = 0. b) If f ( x y ) = x 2 + y 2 , then Df = [ 2 x 2 y ]. At the point - 1 2 , the equation of the tangent space is [ - 2 4 ] ( x y - - 1 2 ) = 0 , or - 2 x + 4 y - 10 = 0, or - x + 2 y - 5 = 0. d) If f ( x y ) = 4 - x 2 - y 2 , then Df = - x 4 - x 2 - y 2 - y 4 - x 2 - y 2 . At the point 1 1 , the equation of the tangent plane is - 1 2 - 1 2 ( x y - 1 1 ) = 0 , or - 1 2 ( x + y - 2) = 0, or x + y - 2 = 0. e) if f ( x y z ) = xyz , then Df = [ yz xz xy ], At the point 1 2 3 , the equation of the tangent plane is [ 6 3 2 ] (( x y z - 1 2 3 ) = 0 , or 6 x + 3 y + 2 z - 18 = 0. MM, Exercise 3.2.3: Compute the Jacobian or derivative matrices of the following functions. a) f ( x y ) = ( xy x 2 + y 2 ). c) f ( s t ) = ( s cos t s sin t ). e) f ( x y ) = ( x cos y x sin y y . The derivative of Jacobian matrix is the matrix of partial derivatives of the function. By differentiation, we find the following Jacobians: a) y x 2 x 2 y b) cos t - s sin t sin t s cos t c) cos y - x sin y sin y x cos y 0 1 MM, Exercise 3.2.5: The area of a triangle with sides of lengths a and b , and included angle θ is A = 1 2 ab sin θ . The derivative of this is DA = 1 2 [ b sin θ a sin θ ab cos θ ] . 1
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For a = 3, b = 4, and θ = π 3 , we have DA = 3 3 3 4 3 . The largest of these components is the third one, so A is most sensitive to a small change in θ . AS2.1 : In this exercise, we are to make various computations about the derivatives of the function g ( x y ) = xy 3 x 2 + y 6 when x y = 0 0 , and g ( 0 0 ) = 0 a) Partial derivatives at points outside the origin: Using the usual rules for differentiating, especially
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