15.5%20Prel%20Q%20and%20Ex%201%20-%2012

15.5%20Prel%20Q%20and%20Ex%201%20-%2012 - S E C T I O N...

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SECTION 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 ²( h , k ) satisfying lim ( h , k ) ( 0 , 0 ) h , k ) = 0 such that f ( h , k ) = L ( h , k ) + h , k ) p 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 ) p h 2 + k 2 (1) We deFne the following functions: R 1 ( h , k ) = h , k ) · h p h 2 + k 2 ; R 2 ( h , k ) = h , k ) · k p h 2 + k 2 We now show that R 1 ( h , k ) and R 2 ( h , k ) satisfy the required properties. ±irst notice that ¯ ¯ ¯ ¯ ¯ h p h 2 + k 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ h h 2 ¯ ¯ ¯ ¯ = 1a n d ¯ ¯ ¯ ¯ ¯ k p 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 satisFed: hR 1 ( h , k ) + kR 2 ( h , k ) = h 2 h , k ) p h 2 + k 2 + k 2 h , k ) p h 2 + k 2 = ³ h 2 + k 2 ´ h , k ) p h 2 + k 2 = h , k ) p h 2 + k 2 Combining with (1) we get f ( h , k ) = 1 ( h , k ) + 2 ( h , k ). (e) Setting h = k ,weget f ( h , h ) = 1 ( h , h ) + 2 ( h , k ) or f ( h , h ) h = R 1 ( h , h ) + R 2 ( h , h ) Taking the limit as h 0gives 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 deFnition of f ,wehavefor h 6= 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) h 3 , 4 i (c) h 3 , 4 , 5 i
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678 CHAPTER 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 = D f x , f y E is (b). 2. True or false: A differentiable function increases at the rate k∇ f P k in the direction of f P ? The statement is true. The value k∇ f P k istherateofincreaseof f in the direction f P . 3. Describe the two main geometric properties of the gradient f . 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 . 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 = h 1 , 0 i ,weFndthat 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
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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.

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15.5%20Prel%20Q%20and%20Ex%201%20-%2012 - S E C T I O N...

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