15.4Ex9-14

# 15.4Ex9-14 - S E C T I O N 15.4 SOLUTION Differentiability...

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SECTION 15.4 Differentiability, Linear Approximation, and Tangent Planes (ET Section 14.4) 663 SOLUTION The linear approximation to f at the point ( 2 , 1 , 2 ) is: f ( x , y , z ) f ( 2 , 1 , 2 ) + f x ( 2 , 1 , 2 )( x 2 ) + f y ( 2 , 1 , 2 )( y 1 ) + f z ( 2 , 1 , 2 )( z 2 ) (1) We compute the values of f and its partial derivatives at ( 2 , 1 , 2 ) : f ( x , y , z ) = xy z f ( 2 , 1 , 2 ) = 1 f x ( x , y , z ) = y z f x ( 2 , 1 , 2 ) = 1 2 f y ( x , y , z ) = x z f y ( 2 , 1 , 2 ) = 1 f z ( x , y , z ) =− z 2 f z ( 2 , 1 , 2 ) 1 2 We substitute these values in (1) to obtain the following linear approximation: z 1 + 1 2 ( x 2 ) + 1 · ( y 1 ) 1 2 ( z 2 ) z 1 2 x + y 1 2 z 10. Assume that f ( 1 , 0 , 0 ) 3, f x ( 1 , 0 , 0 ) 2, f y ( 1 , 0 , 0 ) = 4, and f z ( 1 , 0 , 0 ) = 2. Use the linear approximation to estimate f ( 1 . 02 , 0 . 01 , 0 . 03 ) . The linear approximation at ( 1 , 0 , 0 ) is f ( 1 + h , k , l ) f ( 1 , 0 , 0 ) + f x ( 1 , 0 , 0 ) h + f y ( 1 , 0 , 0 ) k + f z ( 1 , 0 , 0 ) l (1) We substitute h = 0 . 02, k = 0 . 01, l 0 . 03 and the given values to obtain the following estimation: f ( 1 . 02 , 0 . 01 , 0 . 03 ) ≈− 3 + ( 2 ) · 0 . 02 + 4 · 0 . 01 + 2 ( 0 . 03 ) 3 . 06 That is, f ( 1 . 02 , 0 . 01 , 0 . 03 ) 3 . 06 . In Exercises 11–18, Fnd an equation of the tangent plane at the given point.

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## This note was uploaded on 02/13/2010 for the course MATH MATH 32A taught by Professor Park during the Fall '09 term at UCLA.

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15.4Ex9-14 - S E C T I O N 15.4 SOLUTION Differentiability...

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