14.6 - 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR...

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14.6: DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR KIAM HEONG KWA 1. Directional Derivatives The directional derivative of a function f ( x, y ) at ( x 0 , y 0 ) in the di- rection of a unit vector u = h a, b i is defined by D u f ( x 0 , y 0 ) = lim h 0 f ( x 0 + ha, y 0 + hb ) - f ( x 0 , y 0 ) h or, in vector notation, by D u f ( x 0 ) = lim h 0 f ( x 0 + h u ) - f ( x 0 ) h , where x 0 = h x 0 , y 0 i , provided the limit exists. It indicates the rate of change of f in the direction of u . In particular, note that D i f ( x 0 , y 0 ) = f x ( x 0 , y 0 ) and D j f ( x 0 , y 0 ) = f y ( x 0 , y 0 ) . Theorem 1. If f is differentiable, then it has a directional derivative in the direction of any unit vector u = h a, b i . Furthermore, D u f ( x, y ) = f x ( x, y ) a + f y ( x, y ) b. In vector notation, (1.1) D u f ( x ) = f ( x ) · u , where x = h x, y i and (1.2) f ( x ) = f x ( x ) i + f y ( x ) j , the latter of which is called the gradient of f and also denoted by grad f . If the unit vector u makes an angle α with the positive x -axis, so that u = h cos α, sin α i , then (1.3) D u f ( x, y ) = f x ( x, y ) cos α + f y ( x, y ) sin α. Date : September 26, 2010. 1
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2 KIAM HEONG KWA Example 1 (Problem 4 in the text) . Find the directional derivative of f ( x, y ) = x 2 y 3 - y 4 at the point (2 , 1) in the direction indicated by the angle α = π/ 4 . Solution. The direction is determined by the unit vector u = < cos α, sin α > = D cos π 4 , sin π 4 E = * 2 2 , 2 2 + . Thus the directional derivative is D u f ( x, y ) = f x ( x, y ) 2 2 + f y ( x, y ) 2 2 = 2 2 (2 xy 3 + 3 x 2 y 2 - 4 y 3 ) . At the point (2 , 1) , one has D u f (2 , 1) = 2 2 [2(2)(1) 3 + 3(2) 2 (1) 2 - 4(1) 3 )] = 6 2 . Example 2. Find the gradient of f ( x, y ) = y ln( x ) at the point P (1 , - 3) and the rate of change of f at P in the direction of u = - 4 5 , 3 5 . Solution. The gradient of f is given by f ( x, y ) = < f x ( x, y ) , f y ( x, y ) > = D y x , ln x E .
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