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