Answer
$${\bf{r}} = \left( {4{\bf{i}} + {\bf{j}}} \right) + t\left( { - 4{\bf{i}} + {\bf{j}} + 4{\bf{k}}} \right)$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = {t^2}{\bf{i}} - \frac{1}{{t + 1}}{\bf{j}} + \left( {4 - {t^2}} \right){\bf{k}};{\text{ }}{P_0}\left( {4,1,0} \right) \cr
& {\text{Let }}t = - 2,{\text{ then }}{\bf{r}}\left( { - 2} \right) \cr
& {\bf{r}}\left( { - 2} \right) = {\left( { - 2} \right)^2}{\bf{i}} - \frac{1}{{ - 2 + 1}}{\bf{j}} + \left( {4 - {{\left( { - 2} \right)}^2}} \right){\bf{k}} \cr
& {\bf{r}}\left( { - 2} \right) = 4{\bf{i}} + {\bf{j}} + 0{\bf{k}} \cr
& t = - 2{\text{ at }}{P_0} \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( { - 2} \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{t^2}{\bf{i}} - \frac{1}{{t + 1}}{\bf{j}} + \left( {4 - {t^2}} \right){\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = 2t{\bf{i}} + \frac{1}{{{{\left( {t + 1} \right)}^2}}}{\bf{j}} - 2t{\bf{k}} \cr
& {\bf{r}}'\left( { - 2} \right) = - 4\sin \left( {\frac{\pi }{3}} \right){\bf{i}} - 3{\bf{j}} \cr
& {\bf{r}}'\left( { - 2} \right) = 2\left( { - 2} \right){\bf{i}} + \frac{1}{{{{\left( { - 2 + 1} \right)}^2}}}{\bf{j}} - 2\left( { - 2} \right){\bf{k}} \cr
& {\bf{r}}'\left( { - 2} \right) = - 4{\bf{i}} + {\bf{j}} + 4{\bf{k}} \cr
& {\text{The vector equation is}} \cr
& {\bf{r}} = \left( {4{\bf{i}} + {\bf{j}}} \right) + t\left( { - 4{\bf{i}} + {\bf{j}} + 4{\bf{k}}} \right) \cr} $$
