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Sec 14.5-sm.dvi

# Sec 14.5-sm.dvi - S E C T I O N 15.5 The Gradient and...

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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 ( 0 , 0 ) , then there would exist a function ( h , k ) satisfying lim ( h , k ) ( 0 , 0 ) ( h , k ) = 0 such that f ( h , k ) = L ( h , k ) + ( h , k ) h 2 + k 2 in a disk containing ( 0 , 0 ) . By part (c), f x ( 0 , 0 ) = f y ( 0 , 0 ) = 0, therefore L ( h , k ) = f ( 0 , 0 ) + f x ( 0 , 0 ) h + f y ( 0 , 0 ) k = 0 + 0 h + 0 k = 0 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, 0 | R 1 ( h , k ) | | ( h , k ) | and 0 | R 2 ( h , k ) | | ( h , k ) | . Since lim ( h , k ) ( 0 , 0 ) ( h , k ) = 0, the Squeeze Theorem implies that also lim ( h , k ) ( 0 , 0 ) R 1 ( h , k ) = lim ( h , k ) ( 0 , 0 ) R 2 ( h , k ) = 0 We show that the conditions are satisfied: hR 1 ( h , k ) + kR 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 ) = hR 1 ( h , k ) + kR 2 ( h , k ). (e) Setting h = k , we get f ( h , h ) = hR 1 ( h , h ) + hR 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 0 f ( h , h ) h = lim h 0 R 1 ( h , h ) + lim h 0 R 2 ( h , h ) = 0 + 0 = 0 However by the definition of f , we have for h = 0 f ( h , h ) = 2 h 2 ( h + h ) h 2 + h 2 = 2 h f ( h , h ) h = 2 So obviously lim h 0 f ( h , h ) h = 2 = 0. We arrive at a contradiction and conclude that f is not differentiable at ( 0 , 0 ) . (f) By the Criterion for Differentiability, the continuity of f x and f y at ( 0 , 0 ) implies that f ( x , y ) is differentiable at ( 0 , 0 ) . In part (e) we showed that f is not differentiable at ( 0 , 0 ) , 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

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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 . 3. Describe the two main geometric properties of the gradient f . SOLUTION The gradient of f points in the direction of maximum rate of increase of f and is normal to the level curve (or surface) of f . 4. Express the partial derivative f x as a directional derivative D u f for some unit vector u . SOLUTION The partial derivative f x is the following limit: f x ( a , b ) = lim t 0 f ( a + t , b ) - f ( a , b ) t Considering the unit vector i = 1 , 0 , we find that f x ( a , b ) = lim t 0 f ( a + t · 1 , b + t · 0 ) - f ( a , b ) t = D i f ( a , b ) We see that the partial derivative f x can be viewed as a directional derivative in the direction of i . 5. You are standing at point where the temperature gradient vector is pointing in the northeast (NE) direction. In which direction(s) should you walk to avoid a change in temperature?
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Sec 14.5-sm.dvi - S E C T I O N 15.5 The Gradient and...

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