Answer
\[K = \frac{{36}}{{125}}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = t{\mathbf{i}} + \frac{1}{9}{t^3}{\mathbf{j}},{\text{ at }}t = 2 \hfill \\
{\text{By Theorem 12}}{\text{.8 }} \hfill \\
{\text{If }}C{\text{ is a smooth curve given by }}{\mathbf{r}}\left( t \right),{\text{ then the curvature }}K{\text{ of}} \hfill \\
C{\text{ at }}t{\text{ is }}K = \frac{{\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\mathbf{r}}'\left( t \right)} \right\|}^3}}} \hfill \\
{\mathbf{r}}\left( t \right) = t{\mathbf{i}} + \frac{1}{9}{t^3}{\mathbf{j}} \hfill \\
{\mathbf{r}}'\left( t \right) = {\mathbf{i}} + \frac{1}{3}{t^2}{\mathbf{j}} \hfill \\
{\mathbf{r}}''\left( t \right) = \frac{2}{3}t{\mathbf{j}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
1&{\frac{1}{3}{t^2}}&0 \\
0&{\frac{2}{3}t}&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\frac{1}{3}{t^2}}&0 \\
{\frac{2}{3}t}&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
1&0 \\
0&0
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
1&{\frac{1}{3}{t^2}} \\
0&{\frac{2}{3}t}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \frac{2}{3}t{\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = \frac{2}{3}t \hfill \\
\left\| {{\mathbf{r}}'\left( 2 \right) \times {\mathbf{r}}''\left( 2 \right)} \right\| = \frac{2}{3}\left( 2 \right) = \frac{4}{3} \hfill \\
{\left\| {{\mathbf{r}}'\left( t \right)} \right\|^3} = \left\| {{\mathbf{i}} + \frac{1}{3}{t^2}{\mathbf{j}}} \right\| = {\left( {\sqrt {1 + {{\left( {\frac{1}{3}{t^2}} \right)}^2}} } \right)^3} \hfill \\
{\left\| {{\mathbf{r}}'\left( 2 \right)} \right\|^3} = {\left( {\sqrt {1 + {{\left( {\frac{1}{3}{{\left( 2 \right)}^2}} \right)}^2}} } \right)^3} = {\left( {\sqrt {\frac{{25}}{9}} } \right)^3} = \frac{{125}}{{27}} \hfill \\
{\text{Therefore,}} \hfill \\
K = \frac{{\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\mathbf{r}}'\left( t \right)} \right\|}^3}}} \hfill \\
{\text{Evaluate at }}t = 1 \hfill \\
K = \frac{{4/3}}{{125/27}} \hfill \\
K = \frac{{36}}{{125}} \hfill \\
\end{gathered} \]
