Answer
$$x = \ln 2 + \frac{1}{2}t;{\text{ }}y = {e^{ - 2}} - {e^{ - 2}}t,{\text{ }}z = 8 + 12t$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \ln t{\bf{i}} + {e^{ - t}}{\bf{j}} + {t^3}{\bf{k}};{\text{ }}{t_0} = 2 \cr
& {\text{Calculate }}{\bf{r}}\left( {{t_0}} \right) \cr
& {\bf{r}}\left( 2 \right) = \ln t{\bf{i}} + {e^{ - t}}{\bf{j}} + {t^3}{\bf{k}} \cr
& {\bf{r}}\left( 2 \right) = \ln \left( 2 \right){\bf{i}} + {e^{ - 2}}{\bf{j}} + {\left( 2 \right)^3}{\bf{k}} \cr
& {\bf{r}}\left( 2 \right) = \ln 2{\bf{i}} + {e^{ - 2}}{\bf{j}} + 8{\bf{k}} \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( {{t_0}} \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\ln t{\bf{i}} + {e^{ - t}}{\bf{j}} + {t^3}{\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = \frac{1}{t}{\bf{i}} - {e^{ - t}}{\bf{j}} + 3{t^2}{\bf{k}} \cr
& {\bf{r}}{\text{'}}\left( 2 \right) = \frac{1}{2}{\bf{i}} - {e^{ - 2}}{\bf{j}} + 3{\left( 2 \right)^2}{\bf{k}} \cr
& {\bf{r}}{\text{'}}\left( 2 \right) = \frac{1}{2}{\bf{i}} - {e^{ - 2}}{\bf{j}} + 12{\bf{k}} \cr
& {\text{Let }}{{\bf{r}}_0} = {\bf{r}}\left( {{t_0}} \right){\text{ and }}{{\bf{v}}_0}{\text{ = }}{\bf{r}}{\text{'}}\left( {{t_0}} \right){\text{ }} \cr
& {\text{The tangent line to the graph of }}{\bf{r}}\left( t \right){\text{ at }}{{\bf{r}}_0}{\text{ is given by the vector }} \cr
& {\text{equation}} \cr
& {\bf{r}} = {{\bf{r}}_0} + t{{\bf{v}}_0} \cr
& {\bf{r}} = {\bf{r}}\left( {{t_0}} \right) + t{\bf{r}}{\text{'}}\left( {{t_0}} \right){\text{ }} \cr
& {\bf{r}} = \ln 2{\bf{i}} + {e^{ - 2}}{\bf{j}} + 8{\bf{k}} + t\left( {\frac{1}{2}{\bf{i}} - {e^{ - 2}}{\bf{j}} + 12{\bf{k}}} \right) \cr
& {\bf{r}} = \ln 2{\bf{i}} + {e^{ - 2}}{\bf{j}} + 8{\bf{k}} + \frac{1}{2}t{\bf{i}} - {e^{ - 2}}t{\bf{j}} + 12t{\bf{k}} \cr
& {\bf{r}} = \left( {\ln 2 + \frac{1}{2}t} \right){\bf{i}} + \left( {{e^{ - 2}} - {e^{ - 2}}t} \right){\bf{j}} + \left( {8 + 12t} \right){\bf{k}} \cr
& {\text{Thus}},{\text{the parametric equations of the tangent line at }}t = {t_0}{\text{ are:}} \cr
& x = \ln 2 + \frac{1}{2}t;{\text{ }}y = {e^{ - 2}} - {e^{ - 2}}t,{\text{ }}z = 8 + 12t \cr} $$
