15.5%20Ex%2034%20-%2040

15.5%20Ex%2034%20-%2040 - S E C T I O N 15.5 The Gradient...

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SECTION 15.5 The Gradient and Directional Derivatives (ET Section 14.5) 689 T ( 3 , 9 , 4 ) = e 9 4 h 1 , 3 , 3 i = e 5 h 1 , 3 , 3 i The rate of change of the bug’s temperature at the starting point P is the directional derivative D u f ( P ) =∇ T ( 3 , 9 , 4 ) · u = e 5 h 1 , 3 , 3 i · 1 3 h 2 , 2 , 1 i=− e 5 3 ≈− 49 . 47 The answer is 49 . 47 degrees Celsius per meter. 34. Suppose that f P = h 2 , 4 , 4 i .Is f increasing or decreasing at P in the direction v = h 2 , 1 , 3 i ? SOLUTION We compute the derivative of f at P with respect to v : D v f ( P ) f P · v = h 2 , 4 , 4 i · h 2 , 1 , 3 i = 4 4 + 12 = 12 > 0 Since the derivative is positive, f is increasing at P in the direction of v . 35. Let f ( x , y ) = xe x 2 y and P = ( 1 , 1 ) . (a) Calculate k∇ f P k . (b) Find the rate of change of f in the direction f P . (c) Find the rate of change of f in the direction of a vector making an angle of 45 with f P . (a) We compute the gradient of f ( x , y ) = x 2 y . The partial derivatives are f x = 1 · e x 2 y + x 2 y · 2 x = e x 2 y ³ 1 + 2 x 2 ´ f y =− x 2 y The gradient vector is thus f = ¿ f x , f y À = D e x 2 y ³ 1 + 2 x 2 ´ , x 2 y E = e 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%20Ex%2034%20-%2040 - S E C T I O N 15.5 The Gradient...

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