Answer
At $t=4$ the location of the particle is ${\bf{r}}\left( 4 \right) = \left( {8,27,62} \right)$.
Work Step by Step
We have $\frac{{d{\bf{r}}}}{{dt}} = \left( {2{t^{ - 1/2}},6,8t} \right)$ and ${\bf{r}}\left( 1 \right) = \left( {4,9,2} \right)$.
The velocity at $t=4$ is $\frac{{d{\bf{r}}}}{{dt}}{|_{t = 4}} = \left( {1,6,32} \right)$.
We find the position vector by integration:
${\bf{r}}\left( t \right) = \mathop \smallint \limits_{}^{} \left( {2{t^{ - 1/2}},6,8t} \right){\rm{d}}t$,
${\bf{r}}\left( t \right) = \left( {4{t^{1/2}},6t,4{t^2}} \right) + {\bf{c}}$,
where ${\bf{c}}$ is a constant vector.
Using the initial condition ${\bf{r}}\left( 1 \right) = \left( {4,9,2} \right)$, we obtain
${\bf{r}}\left( 1 \right) = \left( {4,6,4} \right) + {\bf{c}} = \left( {4,9,2} \right)$
Therefore, ${\bf{c}} = \left( {0,3, - 2} \right)$. So, the position vector with the given initial condition is
${\bf{r}}\left( t \right) = \left( {4{t^{1/2}},6t,4{t^2}} \right) + \left( {0,3, - 2} \right)$
${\bf{r}}\left( t \right) = \left( {4{t^{1/2}},6t + 3,4{t^2} - 2} \right)$
At $t=4$ the location of the particle is ${\bf{r}}\left( 4 \right) = \left( {8,27,62} \right)$.
