## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 15 - Differentiation in Several Variables - 15.5 The Gradient and Directional Derivatives - Exercises - Page 801: 14

#### Answer

$$15\sin(6)$$

#### Work Step by Step

Given $$f(x, y)=\cos (y-x), \quad \mathbf{r}(t)=\left\langle e^{t}, e^{2 t}\right\rangle, \quad t=\ln 3$$ Since $\mathbf{r}(\ln 3)=\langle 3,9 \rangle$ and \begin{align*} \nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\\ &=\langle \sin (y-x),- \sin(y-x) \rangle\\ \mathbf{r}^{\prime}(t)&=\langle e^t,2e^{2 t} \rangle \end{align*} Then \begin{align*} \frac{d}{d t} f(\mathbf{r}(t))&=\nabla f_{\mathbf{r}(t)} \cdot \mathbf{r}^{\prime}(t)\\ &=\langle \sin (-6), -\sin (-6)\rangle\cdot\langle e^t,2e^{2 t} \rangle \end{align*} Hence \begin{align*} \frac{d}{d t} f(\mathbf{r}(t))\bigg|_{t=\ln 3} &= \langle - \sin (6), \sin (6)\rangle\cdot \langle3,18\rangle \\ &=15\sin(6) \end{align*}

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