The number we get in this way coincides with so

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The number we get in this way coincides with so called the directional derivative F v ( x, y, z ) = lim t 0 F ( x + tv 1 , y + tv 2 , z + tv 3 ) - F ( x, y, z ) t which is known to exist whenever F is differentiable. In particular, when v is a unit vector, we call the number F v ( x, y, z ) as the rate of change of F in the direction of the unit vector v . In general, the rate of change of F in the direction of a nonzero vector v is the rate of change of F in the direction of the vector v/ || v || . At a point ( a, b, c ) , we can ask in which direction, the rate of change of F is maximal/minimal. The answer is simple. The rate of change of F at ( a, b, c ) maximizes (i.e. F increases most) in the direction of the vec- tor F ( a, b, c ) and minimizes (i.e. F decreases most) in the direction of - F ( a, b, c ) if F ( a, b, c ) is not zero. If F ( a, b, c ) is zero, the point ( a, b, c ) is called a critical point of F and in this case, we need to know more infor- mations, namely the second derivatives F xx , F xy , F yy to say more about F . The gradient vector is extremely useful when we compute the tangent plane to the surface given by the equation S : F ( x, y, z ) = 0. 14
We can find the tangent plane to the surface S at the point ( a, b, c ) as fol- lows. First, we compute the gradient vector F ( a, b, c ) at ( a, b, c ) . Suppose F ( a, b, c ) = A, B, C is not zero vector. Then, the tangent plane to the surface S at ( a, b, c ) is given by A ( x - a ) + B ( y - b ) + C ( z - c ) = 0. Namely, it is the plane which contains ( a, b, c ) and has normal direction F ( a, b, c ) = A, B, C . Examples (work out by yourself) Compute the gradient vector for each the following functions F ( x, y, z ) . The function F ( x, y, z ) = x 2 + y 2 + z 2 . The function F ( x, y, z ) = x 2 - y 2 - z 2 . The function F ( x, y, z ) = 3x 2 - y 2 . The function F ( x, y, z ) = xyz . The function F ( x, y, z ) = x + y + z . The function F ( x, y, z ) = 2x + y + 3z . The function F ( x, y, z ) = z ( sin x + cos y ) . The function F ( x, y, z ) = e - x sin y + e x cos y . The function F ( x, y, z ) = e - x 2 - y 2 - z 2 ( x 2 + y 2 + z 2 ) . Problems and Explanations (think by yourself first before looking at explanations) 14.6.5 Problem. Let f ( x, y, z ) = yz + e xyz . Find the rate of change of f at the point P ( 0, 1, 2 ) towards the point Q (- 2, - 1, 0 ) . 15
Explanation : We compute the gradient vector f of f as f = f x , f y , f z = yze xyz , z + xze xyz , y + xye xyz which is at the point P ( 0, 1, 2 ) equal to 2, 2, 1 . We compute the direction - PQ from P ( 0, 1, 2 ) towards the point Q (- 2, - 1, 0 ) as - PQ = - 2, - 1, 0 - 0, 1, 2 = - 2, - 2, - 2 . We compute the unit vector - PQ/ || - PQ || as - PQ || - PQ || = - 2, - 2, - 2 4 + 4 + 4 = - 1, - 1, - 1 1 + 1 + 1 = 1 3 - 1, - 1, - 1 . Hence, the rate of change of f at the point P in the direction of - PQ is f · - PQ || - PQ || = 2, 2, 1 ⟩ · 1 3 - 1, - 1, - 1 = 1 3 (- 2 - 2 - 1 ) = - 5 3 . 14.6.6 Problem. The temperature at a point ( x, y, z ) in space is given by T ( x, y, z ) = xye xz . What is the rate of change of temperature at the point ( 2, 1, 1 ) in the direction toward the point ( 4, 0, 3 ) ? Explanation : We compute the gradient vector f as f = f x , f y , f z = ye xz + xyze xz , xe xz , x 2 ye xz which is at the point ( 2, 1, 1 ) equal to 3e 2 , 2e 2 , 4e 2 .

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