15.4Ex16-18

# 15.4Ex16-18 - S E C T I O N 15.4 Differentiability Linear...

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SECTION 15.4 Differentiability, Linear Approximation, and Tangent Planes (ET Section 14.4) 665 We substitute these values in (1) to obtain the following equation of the tangent plane: z = 5 + 4 ( r 2 ) 5 ( s 1 ) = 4 r 5 s + 2 . 16. H ( s , t ) = se st , ( 0 , 0 ) SOLUTION The equation of the tangent plane at ( 0 , 0 ) is z = f ( 0 , 0 ) + f s ( 0 , 0 ) s + f t ( 0 , 0 ) t (1) We compute the following values: f ( s , t ) = f ( 0 , 0 ) = 0 f s ( s , t ) = 1 · e + s · te = e ( 1 + ) f s ( 0 , 0 ) = 1 f t ( s , t ) = s 2 e f t ( 0 , 0 ) = 0 Substituting in (1) gives the following equation of the tangent plane: z = 0 + 1 · s + 0 · t = s That is, z = s . 17. f ( x , y ) = e x ln y , ( 0 , 1 ) The equation of the tangent plane at ( 0 , 1 ) is z = f ( 0 , 1 ) + f x ( 0 , 1 ) x + f y ( 0 , 1 )( y 1 ) (1) We compute the values of f and its partial derivatives at ( 0 , 1 ) : f ( x , y ) = e x ln yf ( 0 , 1 ) = 0 f x ( x , y ) = e x ln y f x ( 0 , 1 ) = 0 f y ( x , y ) = e x y f y ( 0 , 1 ) = 1 Substituting in (1) gives the following equation of the tangent plane: z = 0 + 0 x + 1 ( y 1 ) = y 1 That is, z = y 1. 18. f ( x , y ) = ln ( x 2 + y 2 ) , ( 1 , 1 ) The equation of the tangent plane at ( 1 , 1 ) is z = f ( 1 , 1 ) + f x ( 1 , 1 )( x 1 ) + f y ( 1 , 1 )( y 1 ) (1) We compute the following values, using the Chain Rule: f ( x , y ) = ln ( x 2 + y 2 ) f ( 1 , 1 ) = ln 2 f x ( x , y ) = 2 x x 2 + y 2 f x ( 1 , 1 ) = 1 f y ( x , y ) = 2 y x 2 + y 2 f y ( 1 , 1 ) = 1 Substituting in (1) gives the following equation of the tangent plane:
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