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: 17

Answer

$$-1+\frac{8}{\pi}$$

Work Step by Step

Given $$ f(x, y)=\ln x+\ln y, \quad \mathbf{r}(t)=\left\langle\cos t, t^{2}\right\rangle, \quad t=\frac{\pi}{4} $$ since $\mathbf{r}(\pi/4)=\langle \frac{1}{\sqrt{2}},\frac{\pi^2}{16}\rangle$ and $$ \begin{aligned} \nabla f &=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle \\ &=\langle \frac{1}{x},\frac{1}{y}\rangle \\ \mathbf{r}^{\prime}(t) &=\left\langle -\sin t, 2t\right\rangle \end{aligned} $$ Then \begin{aligned} \left.\frac{d}{d t} f(\mathbf{r}(t))\right|_{t=\pi/4}&=\nabla f_{\mathbf{r}(\pi/4)} \cdot \mathbf{r}^{\prime}(\pi/4)\\ &=\langle\sqrt{2},\frac{16}{\pi^2}\rangle \cdot\langle -1/\sqrt{2},\pi/2\rangle \\ &=-1+\frac{8}{\pi} \end{aligned}
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