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 $t=2$.
