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: 11

Answer

$$-3$$

Work Step by Step

Given $$ f(x, y)=x^{2}-3 x y, \quad \mathbf{r}(t)=\langle\cos t, \sin t\rangle, \quad t=0$$ Since $ \mathbf{r}(0)=\langle 1,0 \rangle$ and \begin{align*} \nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\\ &=\langle 2x-3y, -3x\rangle\\ \mathbf{r}^{\prime}(t)&=\langle -\sin t, \cos 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 2\cos t-3\sin t, -3\cos t\rangle\cdot\langle -\sin t, \cos t \rangle \end{align*} Hence \begin{align*} \frac{d}{d t} f(\mathbf{r}(t))\bigg|_{t=0} &= \langle 2,-3\rangle\cdot \langle0, 1\rangle \\ &= -3 \end{align*}
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