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: 24



Work Step by Step

Given $$ f(x, y)=\sin (x-y), \quad \mathbf{v}=\langle 1,1\rangle, \quad P=\left(\frac{\pi}{2}, \frac{\pi}{6}\right) $$ Since the direction of $\mathbf{v}$ is \begin{align*} \mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\ &=\frac{\langle 1,1\rangle}{\sqrt{1+1}}\\ &=\left\langle\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle \end{align*} and \begin{align*} \nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\\ &=\left\langle \cos (x-y), -\cos (x-y)\right\rangle \\ \nabla f_{(\pi/2,\pi/6)}&=\left\langle \frac{1}{2}, \frac{-1}{2}\right\rangle \end{align*} Then the directional derivatives are given by \begin{align*} D_{\mathbf{u}} f(\pi/2,\pi/6)&=\nabla f_{(\pi/2,\pi/6)} \cdot \mathbf{u}\\ &=\left\langle \frac{1}{2}, \frac{-1}{2}\right\rangle \cdot\left\langle\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle\\ &=0 \end{align*}
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