Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.5 The Gradient and Directional Derivatives - Exercises - Page 801: 32


$$\frac{15}{ \sqrt{14}}$$

Work Step by Step

Given $$ f(x, y)=xy+ z^{3}, \quad P=(3,-2,-1) $$ Since $$PO= (-3,2,1) $$ Then the direction of $\mathbf{v}$ is \begin{align*} \mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\ &=\frac{\langle -3,2,1\rangle}{\sqrt{9+4+1}}\\ &=\left\langle\frac{-3}{\sqrt{14}},\frac{ 2}{\sqrt{14}},\frac{ 1}{\sqrt{14}}\right\rangle \end{align*} and \begin{align*} \nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right\rangle\\ &=\left\langle y ,x, 3z^2\right\rangle \\ \nabla f_{(3,-2,-1)}&=\left\langle -2,3,3\right\rangle \end{align*} Then the directional derivatives are given by \begin{align*} D_{\mathbf{u}} f(3,-2,-1)&=\nabla f_{(3,-2,-1)} \cdot \mathbf{u}\\ &=\left\langle -2,3,3\right\rangle \cdot\left\langle\frac{-3}{\sqrt{14}},\frac{ 2}{\sqrt{14}},\frac{ 1}{\sqrt{14}}\right\rangle\\ &=\frac{15}{ \sqrt{14}} \end{align*}
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