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 802: 40

Answer

The values of $t$ are $t=0$ and $t=2$.

Work Step by Step

We are given $\nabla T = \left( {y - 4,x + 2y} \right)$ and the path ${\bf{r}}\left( t \right) = \left( {{t^2},t} \right)$. So, $T\left( {{\bf{r}}\left( t \right)} \right) = \left( {t - 4,{t^2} + 2t} \right)$. The velocity vector is ${\bf{r}}'\left( t \right) = \left( {2t,1} \right)$. By Theorem 2, the rate of change of $T\left( {{\bf{r}}\left( t \right)} \right)$ is given by $\frac{d}{{dt}}T\left( {{\bf{r}}\left( t \right)} \right) = \nabla {f_{{\bf{r}}\left( t \right)}}\cdot{\bf{r}}'\left( t \right)$ We find the values of $t$ such that $\frac{d}{{dt}}T\left( {{\bf{r}}\left( t \right)} \right) = 0$: $\frac{d}{{dt}}T\left( {{\bf{r}}\left( t \right)} \right) = \left( {t - 4,{t^2} + 2t} \right)\cdot\left( {2t,1} \right) = 0$ $2{t^2} - 8t + {t^2} + 2t = 0$ $3{t^2} - 6t = 0$ $3t\left( {t - 2} \right) = 0$ So, the values of $t$ are $t=0$ and $t=2$.
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