15.4%20Ex%2016%20-%2020

15.4%20Ex%2016%20-%2020 - S E C T I O N 15.4...

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

Page1 / 2

15.4%20Ex%2016%20-%2020 - S E C T I O N 15.4...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online