HW6 even solutions

HW6 even solutions - Homework 6 Solutions 14.7.2...

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Unformatted text preview: Homework 6 Solutions 14.7.2 Calculating the partial derivatives: f x = 2( x + y ) , 2 f x 2 = 2 . Therefore, we get f y = 2( x + y ) , 2 f y 2 = 2 , 2 f yx = 2 , 2 f xy = 2 . 14.7.4 Since f ( x,y ) = xe y , the partial derivatives are f x = e y , f y = xe y , f xx = 0 , f yy = xe y , f xy = f yx = e y . 14.7.6 Since f ( x,y ) = ( x + y ) e y = xe y + ye y , the partial derivatives are f x = e y , f y = (1) e y + ( x + y ) e y = (1 + x + y ) e y , f xx = 0 , f yy = (1) e y + (1 + x + y ) e y = (2 + x + y ) e y , f xy = f yx = e y . 14.7.8 Since f ( x,y ) = sin( x/y ) , the partial derivatives are f x = cos x y 1 y , f y = cos x y - x y 2 , f xx =- sin x y 1 y 1 y f yy =- sin x y - x y 2 - x y 2 + cos x y 2 x y 3 f xy = f yx =- sin x y - x y 2 1 y + cos x y - 1 y 2 14.7.12 The quadratic Taylor expansion about (0 , 0) is given by f ( x,y ) Q ( x,y ) = f (0 , 0)+ f x (0 , 0) x + f y (0 , 0) y + 1 2 f xx (0 , 0) x 2 + 1 2 f yy (0 , 0) y 2 + f xy (0 , 0) xy. First, we find all relevant partial derivatives and their values at (0 , 0) : f ( x,y ) = ( y- 1)( x + 1) 2 , f (0 , 0) =- 1 f x ( x,y ) = 2( y- 1)( x + 1) , f x (0 , 0) =- 2 f y ( x,y ) = ( x + 1) 2 , f y (0 , 0) = 1 f xx ( x,y ) = 2( y- 1) , f xx (0 , 0) =- 2 f yy ( x,y ) = 0 , f yy (0 , 0) = 0 f xy ( x,y ) = 2( x + 1) , f xy (0 , 0) = 2 1 Substituting these values into the formula, we get Q ( x,y ) =- 1- 2 x + y- x 2 + 2 xy. Notice that this is the same as what you get if you expand ( y- 1)( x + 1) 2 and keep only the terms of degree 2 or less. 14.7.14 The quadratic Taylor expansion about (0 , 0) is given by f ( x,y ) Q ( x,y ) = f (0 , 0)+ f x (0 , 0) x + f y (0 , 0) y + 1 2 f xx (0 , 0) x 2 + 1 2 f yy (0 , 0) y 2 + f xy (0 , 0) xy. We find all relevant partial derivatives and their values at (0 , 0) : f ( x,y ) = (1 + 2 x- y )- 1 , f (0 , 0) = 1 f x ( x,y ) = (- 1)(1 + 2 x- y )- 2 (2) , f x (0 , 0) =- 2 f y ( x,y ) = (- 1)(1 + 2 x- y )- 2 (- 1) , f y (0 , 0) = 1 f xx ( x,y ) = (- 1)(- 2)(1 + 2 x- y )- 3 (2)(2) , f xx (0 , 0) = 8 f yy ( x,y ) = (- 1)(- 2)(1 + 2 x- y )- 3 (- 1)(- 1) , f yy (0 , 0) = 2 f xy ( x,y ) = (- 1)(- 2)(1 + 2 x- y )- 3 (2)(- 1) , f xy (0 , 0) =- 4 Substituting these values into the formula, we get Q ( x,y ) = 1- 2 x + y + 4 x 2 + y 2- 4 xy. Notice that this is the same as what you get if you substitute u = 2 x- y in the quadratic approximation Q ( u ) = 1 + u + u 2 for the function 1 / (1- u 2 ) . 14.7.16 The quadratic Taylor expansion about (0 , 0) is given by f ( x,y ) Q ( x,y ) = f (0 , 0)+ f x (0 , 0) x + f y (0 , 0) y + 1 2 f xx (0 , 0) x 2 + 1 2 f yy (0 , 0) y 2 + f xy (0 , 0) xy....
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HW6 even solutions - Homework 6 Solutions 14.7.2...

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