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Unformatted text preview: S E C T I O N 15.5 The Gradient and Directional Derivatives (ET Section 14.5) 677 (d) If f ( x , y ) were differentiable at ( , ) , then there would exist a function ( h , k ) satisfying lim ( h , k ) ( , ) ( h , k ) = such that f ( h , k ) = L ( h , k ) + ( h , k ) . h 2 + k 2 in a disk containing ( , ) . By part (c), f x ( , ) = f y ( , ) = 0, therefore L ( h , k ) = f ( , ) + f x ( , ) h + f y ( , ) k = + h + k = Therefore we get f ( h , k ) = ( h , k ) . h 2 + k 2 (1) We define the following functions: R 1 ( h , k ) = ( h , k ) h . h 2 + k 2 ; R 2 ( h , k ) = ( h , k ) k . h 2 + k 2 We now show that R 1 ( h , k ) and R 2 ( h , k ) satisfy the required properties. First notice that . . . . . h . h 2 + k 2 . . . . . . . . . h h 2 . . . . = 1 and . . . . . k . h 2 + k 2 . . . . . . . . . k k 2 . . . . = 1 Therefore,  R 1 ( h , k )   ( h , k )  and  R 2 ( h , k )   ( h , k )  . Since lim ( h , k ) ( , ) ( h , k ) = 0, the Squeeze Theorem implies that also lim ( h , k ) ( , ) R 1 ( h , k ) = lim ( h , k ) ( , ) R 2 ( h , k ) = We show that the conditions are satisfied: h R 1 ( h , k ) + k R 2 ( h , k ) = h 2 ( h , k ) . h 2 + k 2 + k 2 ( h , k ) . h 2 + k 2 = . h 2 + k 2 . ( h , k ) . h 2 + k 2 = ( h , k ) . h 2 + k 2 Combining with (1) we get f ( h , k ) = h R 1 ( h , k ) + k R 2 ( h , k ). (e) Setting h = k , we get f ( h , h ) = h R 1 ( h , h ) + h R 2 ( h , k ) or f ( h , h ) h = R 1 ( h , h ) + R 2 ( h , h ) Taking the limit as h 0 gives lim h f ( h , h ) h = lim h R 1 ( h , h ) + lim h R 2 ( h , h ) = + = However by the definition of f , we have for h = f ( h , h ) = 2 h 2 ( h + h ) h 2 + h 2 = 2 h f ( h , h ) h = 2 So obviously lim h f ( h , h ) h = 2 = 0. We arrive at a contradiction and conclude that f is not differentiable at ( , ) . (f) By the Criterion for Differentiability, the continuity of f x and f y at ( , ) implies that f ( x , y ) is differentiable at ( , ) . In part (e) we showed that f is not differentiable at ( , ) , hence f x and f y are not continuous at this point. 15.5 The Gradient and Directional Derivatives (ET Section 14.5) Preliminary Questions 1. Which of the following is a possible value of the gradient f of a function f ( x , y ) of two variables? (a) 5 (b) 3 , 4 (c) 3 , 4 , 5 678 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) SOLUTION The gradient of f ( x , y ) is a vector with two components, hence the possible value of the gradient f = . f x , f y . is (b). 2. True or false: A differentiable function increases at the rate f P in the direction of f P ? SOLUTION The statement is true. The value f P is the rate of increase of f in the direction f P ....
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This note was uploaded on 07/15/2009 for the course MATH 210 taught by Professor Hubscher during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 Hubscher
 Derivative

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