Answer
The derivative of ${\bf{r}}\left( t \right)\cdot{\bf{a}}\left( t \right)$ at $t=2$ is
$\frac{d}{{dt}}\left( {{\bf{r}}\left( t \right)\cdot{\bf{a}}\left( t \right)} \right){|_{t = 2}} = 13$
Work Step by Step
By Eq. (4) of Theorem 3,
$\frac{d}{{dt}}\left( {{\bf{r}}\left( t \right)\cdot{\bf{a}}\left( t \right)} \right) = \left[ {{\bf{r}}'\left( t \right)\cdot{\bf{a}}\left( t \right)} \right] + \left[ {{\bf{r}}\left( t \right)\cdot{\bf{a}}'\left( t \right)} \right]$
The derivative of ${\bf{r}}\left( t \right)\cdot{\bf{a}}\left( t \right)$ at $t=2$ is
$\frac{d}{{dt}}\left( {{\bf{r}}\left( t \right)\cdot{\bf{a}}\left( t \right)} \right){|_{t = 2}} = \left[ {{\bf{r}}'\left( 2 \right)\cdot{\bf{a}}\left( 2 \right)} \right] + \left[ {{\bf{r}}\left( 2 \right)\cdot{\bf{a}}'\left( 2 \right)} \right]$
We have ${\bf{r}}\left( t \right) = \left( {{t^2},1 - t,4t} \right)$, ${\bf{a}}\left( 2 \right) = \left( {1,3,3} \right)$ and ${\bf{a}}'\left( 2 \right) = \left( { - 1,4,1} \right)$.
So,
${\bf{r}}\left( 2 \right) = \left( {4, - 1,8} \right)$
${\bf{r}}'\left( t \right) = \left( {2t, - 1,4} \right)$, ${\ \ }$ ${\bf{r}}'\left( 2 \right) = \left( {4, - 1,4} \right)$
Substituting the corresponding values in $\frac{d}{{dt}}\left( {{\bf{r}}\left( t \right)\cdot{\bf{a}}\left( t \right)} \right){|_{t = 2}}$, that is
$\frac{d}{{dt}}\left( {{\bf{r}}\left( t \right)\cdot{\bf{a}}\left( t \right)} \right){|_{t = 2}} = \left[ {{\bf{r}}'\left( 2 \right)\cdot{\bf{a}}\left( 2 \right)} \right] + \left[ {{\bf{r}}\left( 2 \right)\cdot{\bf{a}}'\left( 2 \right)} \right]$
gives
$\frac{d}{{dt}}\left( {{\bf{r}}\left( t \right)\cdot{\bf{a}}\left( t \right)} \right){|_{t = 2}} = \left[ {\left( {4, - 1,4} \right)\cdot\left( {1,3,3} \right)} \right] + \left[ {\left( {4, - 1,8} \right)\cdot\left( { - 1,4,1} \right)} \right]$
$\frac{d}{{dt}}\left( {{\bf{r}}\left( t \right)\cdot{\bf{a}}\left( t \right)} \right){|_{t = 2}} = 13 + 0 = 13$
