Calculus (3rd Edition)

Published by W. H. Freeman

Chapter 15 - Differentiation in Several Variables - 15.5 The Gradient and Directional Derivatives - Exercises - Page 801: 33

Answer

$$\frac{-e^5}{3}$$

Work Step by Step

Given $$T(x, y, z)=x e^{y-z}$$ Consider $P=(3,9,4) \text { and } Q=(5,7,3)$ Since $$\mathbf{v}=\overrightarrow{P Q}=\langle 5-3,7-9,3-4\rangle=\langle 2,-2,-1\rangle$$ Then the direction of $\mathbf{v}$ is \begin{align*} \mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\ &=\frac{\langle 2,-2,-1\rangle}{\sqrt{4+4+1}}\\ &=\left\langle\frac{2}{3},\frac{- 2}{3},\frac{- 1}{3}\right\rangle \end{align*} and \begin{align*} \nabla T&=\left\langle\frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z}\right\rangle\\ &=\left\langle e^{y-z}, x e^{y-z},-x e^{y-z}\right\rangle\\ \nabla T_{(3,9,4)}&=\left\langle e^5 ,3e^5, -3e^5\right\rangle \\ \end{align*} Then the directional derivatives are given by \begin{align*} D_{\mathbf{u}} f(3,9,4)&=\nabla f_{(3,9,4)} \cdot \mathbf{u}\\ &=\left\langle e^5 ,3e^5, -3e^5\right\rangle \cdot\left\langle\frac{2}{3},\frac{- 2}{3},\frac{- 1}{3}\right\rangle\\ &=\frac{-e^5}{3} \end{align*}

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