Ch%2015%20Review%20Ex%2027%20-%2032

Ch%2015%20Review%20Ex%2027%20-%2032 - Chapter Review...

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Chapter Review Exercises 811 SOLUTION The function whose value we want to approximate is f ( x , y , z ) = q x 2 + y 2 + z We use the linear approximation at the point ( 7 , 5 , 70 ) , hence h = 7 . 1 7 = 0 . 1, k = 4 . 9 5 =− 0 . 1, and l = 70 . 1 70 = 0 . 1. We get f ( 7 . 1 , 4 . 9 , 70 . 1 ) f ( 7 , 5 , 70 ) + 0 . 1 f x ( 7 , 5 , 70 ) 0 . 1 f y ( 7 , 5 , 70 ) + 0 . 1 f z ( 7 , 5 , 70 ) (1) We compute the partial derivatives of f : f x ( x , y , z ) = 2 x 2 p x 2 + y 2 + z = x p x 2 + y 2 + z f x ( 7 , 5 , 70 ) = 7 p 7 2 + 5 2 + 70 = 7 12 f y ( x , y , z ) = 2 y 2 p x 2 + y 2 + z = y p x 2 + y 2 + z f y ( 7 , 5 , 70 ) = 5 p 7 2 + 5 2 + 70 = 5 12 f z ( x , y , z ) = 1 2 p x 2 + y 2 + z f z ( 7 , 5 , 70 ) = 1 2 p 7 2 + 5 2 + 70 = 1 24 Also, f ( 7 , 5 , 70 ) = p 7 2 + 5 2 + 70 = 12. Substituting the values in (1) we obtain the following approximation: q 7 . 1 2 + 4 . 9 2 + 70 . 1 12 + 0 . 1 · 7 12 0 . 1 · 5 12 + 0 . 1 · 1 24 = 12 1 48 12 . 020833 That is, q 7 . 1 2 + 4 . 9 2 + 70 . 1 12 . 020833 The value obtained using a calculator is 12.021647. 26. Suppose that the plane z = 2 x y 3 is tangent to the graph of z = f ( x , y ) at P = ( 2 , 4 ) . (a) Determine f ( 2 , 4 ) , f x ( 2 , 4 ) ,and f y ( 2 , 4 ) . (b) Approximate f ( 2 . 2 , 3 . 9 ) . (a) The tangent plane to the surface z = f ( x , y ) at the point ( a , b ) has the equation z = f ( a , b ) + f x ( a , b )( x a ) + f y ( a , b )( y b ) Since z = 2 x y 3 is the equation of the tangent plane at ( 2 , 4 ) , the following holds: f x ( 2 , 4 ) = 2 f y ( 2 , 4 ) 1( 1 ) The tangency point ( 2 , 4 , f ( 2 , 4 )) lies on the tangent plane, hence f ( 2 , 4 ) = 2 · 2 4 3 3 (b) We now Fnd linear approximation to f ( 2 . 2 , 3 . 9 ) using the linear approximation at ( a , b ) = ( 2 , 4 ) .S ince h = 2 . 2
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Ch%2015%20Review%20Ex%2027%20-%2032 - Chapter Review...

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