Answer
$$K = 0$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 3t{\bf{i}} + 2t{\bf{j}} \cr
& {\text{By Theorem 12}}{\text{.8 }} \cr
& {\text{If }}C{\text{ is a smooth curve given by }}{\bf{r}}\left( t \right),{\text{ then the curvature }}K{\text{ of}} \cr
& C{\text{ at }}t{\text{ is }}K = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}} \cr
& {\bf{r}}\left( t \right) = 3t{\bf{i}} + 2t{\bf{j}} \cr
& {\bf{r}}'\left( t \right) = 3{\bf{i}} + 2{\bf{j}} \cr
& {\bf{r}}''\left( t \right) = 0 \cr} $$
\[\begin{gathered}
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
3&2&0 \\
0&0&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = 0 \hfill \\
\end{gathered} \]
$$\eqalign{
& {\text{Therefore,}} \cr
& K = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}} = \frac{0}{{{{\left\| {3{\bf{i}} + 2{\bf{j}}} \right\|}^3}}} \cr
& K = 0 \cr} $$
