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: 13

Answer

$$5\cos (1)$$

Work Step by Step

Given $$ f(x, y)=\sin (x y), \quad \mathbf{r}(t)=\left\langle e^{2 t}, e^{3 t}\right\rangle, \quad t=0$$ Since $ \mathbf{r}(0)=\langle 1,1 \rangle$ and \begin{align*} \nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\\ &=\langle y\cos (xy), x\cos (xy)\rangle\\ \mathbf{r}^{\prime}(t)&=\langle 2e^{2 t}, 3e^{3 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 \cos (1), \cos (1)\rangle\cdot\langle 2e^{2 t}, 3e^{3 t} \rangle \end{align*} Hence \begin{align*} \frac{d}{d t} f(\mathbf{r}(t))\bigg|_{t=\pi/2} &= \langle \cos (1), \cos (1)\rangle\cdot \langle2,3\rangle \\ &= 5\cos (1) \end{align*}
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