# topics_15_6_Directional_Derivatives_Gradient_Vector - 15.6...

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Unformatted text preview: 15.6. Directional Derivatives and Gradient Vector Problem 6/page 956. Find the directional derivative of f ( x, y ) = x sin( xy ) at the point (2 , 0) in the direction θ = π/ 3. Solution. D u f ( a, b ) = ∇ f ( a, b ) · u = h f x ( a, b ) , f y ( a, b ) i · h u 1 , u 2 i . We compute: f x ( x, y ) = sin( xy ) + xy cos( xy ) , f y ( x, y ) = x 2 cos( xy ) f x (2 , 0) = 0 , f y (2 , 0) = 4 u = h cos( π/ 3) , sin( π/ 3) i = h 1 / 2 , √ 3 / 2 i . Then, D u f (2 , 0) = h , 4) i · h 1 / 2 , √ 3 / 2 i = 2 √ 3 . Problem 9/page 956. (a) Find the gradient of f ( x, y, z ) = xe 2 yz . (b) Evaluate the gradient at P (3 , , 2). (c) Find the rate of change of f at P (3 , , 2) in the direction u = h 2 / 3 ,- 2 / 3 , 1 / 3 i . Solution. (a) f x ( x, y, z ) = e 2 yz , f y ( x, y, z ) = 2 zxe 2 yz , f z ( x, y, z ) = 2 yxe 2 yz Then, the gradient of f at ( x, y, z ) is: ∇ f ( x, y, z ) = h f x ( x, y, z ) , f y ( x, y, z ) , f z ( x, y, z ) i = h e 2 yz , 2 zxe 2 yz , 2 yxe 2 yz i . (b) ∇ f (3 , , 2) = h f x (3 , , 2) , f y (3 , , 2) , f z (3 , , 2) i = h 1 , 12 , i . (c) The rate of change of f ( x, y, z ) at the point ( a, b, c ) in the direction of u is equal to the directional derivative of f ( x, y, z ) at the point ( a, b, c ) 1 in the direction of u . Hence, the rate of change of the given f ( x, y, z ) at the point (3 , , 2) in the direction of u = h 2 / 3 ,- 2 / 3 , 1 / 3 i is D u f (3 , , 2) = ∇ f (3 , , 2) · h 2 / 3 ,- 2 / 3 , 1 / 3 i = h 1 , 12 , i · h 2 / 3 ,- 2 / 3 , 1 / 3 i = 2 3- 8 =- 22 3 . Remark. Note that a direction can be given with a vector that is not unit (not of length 1). Then first, you have to normalize it to be unit. Remark. Given f ( x, y, z ) the rate of change of f at the point ( a, b, c ) in the direction of u is equal to the directional derivative of f ( x, y, z ) at the point ( a, b, c ) in the direction of u : D u f ( a, b, c ) = h f x ( a, b, c ) , f y ( a, b, c ) , f z ( a, b, c ) i·h u 1 , u 2 , u 3 i = ∇ f ( a, b, c ) · u = = |∇ f ( a, b, c ) || u | cos( φ ) = |∇ f ( a, b, c ) | cos( φ ) where φ is the angle in [0 , π ] between the two vectors:the gradient vector ∇ f ( a, b, c ) and the unit vector u . Then, obviously we shall have a maximum rate of change of f at ( a, b, c ) when φ = 0, i.e., in a direction of the vector ∇ f ( a , b , c ). But ∇ f ( a, b, c ) is not a unit vector so, we have to normalize it by u * = 1 |∇ f ( a, b, c ) | ∇ f ( a, b, c ) . Hence, u * is the direction of maximal change of f at ( a, b, c ). Then, D u * f ( a, b, c ) = |∇ f ( a, b, c ) | 1 |∇ f ( a, b, c ) | |∇ f ( a, b, c ) | = |∇ f ( a, b, c ) | = q [ f x ( a, b, c )] 2 + [ f y ( a, b, c )] 2 + [ f z ( a, b, c )] 2 ....
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