HW6 even solutions - Homework 6 Solutions 14.7.2...

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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 ∂y∂x = 2 , 2 f ∂x∂y = 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. We find all relevant partial derivatives and their values at (0 , 0) : f ( x, y ) = cos( x + 3 y ) , f (0 , 0) = 1 f x ( x, y ) = - sin( x + 3 y ) , f x (0 , 0) = 0 f y ( x, y ) = - sin( x + 3 y )(3) , f y (0 , 0) = 0 f xx ( x, y ) = - cos( x + 3 y ) , f xx (0 , 0) = - 1 f yy ( x, y ) = - cos( x + 3 y )(3)(3) , f yy (0 , 0) = - 9 f xy ( x, y ) = - cos( x + 3 y )(3) , f xy (0 , 0) = - 3 Substituting these values into the formula, we get Q ( x, y ) = 1 - 1 2 x 2 - 9 2 y 2 - 3 xy. Notice that this is the same as what you get if you substitute x + 3 y for u in the single variable quadratic approximation Q ( u ) = 1 - u 2 / 2 for cos u . 2
14.7.33 The graph of f is concave up as we move parallel to the x -axis from the point (0 , 0) , so f xx (0 , 0) is positive. The graph of f is concave down as we move parallel to the y -axis from the point (0 , 0) , so f yy (0 , 0) is negative. 14.7.38 (a) Note that a = f ( x 1 , y 1 ) and k = f ( x 2 , y 2 ) . Since P and Q lie on the same level curve, we have a = k .

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