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15.5.Ex34-36

# 15.5.Ex34-36 - 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) 689 T ( 3 , 9 , 4 ) = e 9 4 1 , 3 , 3 = e 5 1 , 3 , 3 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 1 , 3 , 3 · 1 3 2 , 2 , 1 = − e 5 3 ≈ − 49 . 47 The answer is 49 . 47 degrees Celsius per meter. 34. Suppose that f P = 2 , 4 , 4 . Is f increasing or decreasing at P in the direction v = 2 , 1 , 3 ? SOLUTION We compute the derivative of f at P with respect to v : D v f ( P ) = ∇ f P · v = 2 , 4 , 4 · 2 , 1 , 3 = 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 f P . (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 . SOLUTION (a) We compute the gradient of f ( x , y ) = xe x 2 y . The partial derivatives are f x = 1 · e x 2 y + xe x 2 y · 2 x = e x 2 y 1 + 2 x 2
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