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