Answer
At $t=3$ the location of the particle is ${\bf{r}}\left( 3 \right) = \left( {\frac{{45}}{4},5} \right)$.
Work Step by Step
We have $\frac{{d{\bf{r}}}}{{dt}} = \left( {2t - \frac{1}{{{{\left( {t + 1} \right)}^2}}},2t - 4} \right)$ and ${\bf{r}}\left( 0 \right) = \left( {3,8} \right)$.
We find the position vector by integration:
${\bf{r}}\left( t \right) = \mathop \smallint \limits_{}^{} \left( {2t - \frac{1}{{{{\left( {t + 1} \right)}^2}}},2t - 4} \right){\rm{d}}t$
${\bf{r}}\left( t \right) = \left( {{t^2} + \frac{1}{{t + 1}},{t^2} - 4t} \right) + {\bf{c}}$,
where ${\bf{c}}$ is a constant vector.
Using the initial condition ${\bf{r}}\left( 0 \right) = \left( {3,8} \right)$, we obtain
${\bf{r}}\left( 0 \right) = \left( {1,0} \right) + {\bf{c}} = \left( {3,8} \right)$
Therefore, ${\bf{c}} = \left( {2,8} \right)$. So, the position vector with the given initial condition is
${\bf{r}}\left( t \right) = \left( {{t^2} + \frac{1}{{t + 1}},{t^2} - 4t} \right) + \left( {2,8} \right)$
${\bf{r}}\left( t \right) = \left( {2 + {t^2} + \frac{1}{{t + 1}},8 + {t^2} - 4t} \right)$
At $t=3$ the location of the particle is ${\bf{r}}\left( 3 \right) = \left( {\frac{{45}}{4},5} \right)$.
