Answer
\[K = \frac{{\sqrt 2 }}{3}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = \frac{1}{2}{t^2}{\mathbf{i}} + t{\mathbf{j}} + \frac{1}{3}{t^3}{\mathbf{k}},{\text{ }}P\left( {\frac{1}{2},1,\frac{1}{3}} \right) \hfill \\
{\text{For }}t = 1 \hfill \\
{\mathbf{r}}\left( 1 \right) = \frac{1}{2}{\left( 1 \right)^2}{\mathbf{i}} + \left( 1 \right){\mathbf{j}} + \frac{1}{3}{\left( 1 \right)^3}{\mathbf{k}} \hfill \\
{\mathbf{r}}\left( 1 \right) = \frac{1}{2}{\mathbf{i}} + {\mathbf{j}} + \frac{1}{3}{\mathbf{k}} \hfill \\
{\text{Then }}t = 1{\text{ at the point }}P\left( {\frac{1}{2},1,\frac{1}{3}} \right) \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) = \frac{1}{2}{t^2}{\mathbf{i}} + t{\mathbf{j}} + \frac{1}{3}{t^3}{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) = t{\mathbf{i}} + {\mathbf{j}} + {t^2}{\mathbf{k}} \hfill \\
{\mathbf{r}}''\left( t \right) = {\mathbf{i}} + 0{\mathbf{j}} + 2t{\mathbf{k}} \hfill \\
{\text{At }}t = 1 \hfill \\
{\mathbf{r}}'\left( 1 \right) = {\mathbf{i}} + {\mathbf{j}} + {\mathbf{k}} \hfill \\
{\mathbf{r}}''\left( 1 \right) = {\mathbf{i}} + 0{\mathbf{j}} + 2{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( 1 \right) \times {\mathbf{r}}''\left( 1 \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
1&1&1 \\
1&0&2
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( 1 \right) \times {\mathbf{r}}''\left( 1 \right) = \left( {2 - 0} \right){\mathbf{i}} - \left( {2 - 1} \right){\mathbf{j}} + \left( {0 - 1} \right){\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( 1 \right) \times {\mathbf{r}}''\left( 1 \right) = 2{\mathbf{i}} - {\mathbf{j}} - {\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( 1 \right) \times {\mathbf{r}}''\left( 1 \right)} \right\| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} \hfill \\
\left\| {{\mathbf{r}}'\left( 1 \right) \times {\mathbf{r}}''\left( 1 \right)} \right\| = \sqrt 6 \hfill \\
and \hfill \\
\left\| {{\mathbf{r}}'\left( 1 \right)} \right\| = \left\| {{\mathbf{i}} + {\mathbf{j}} + {\mathbf{k}}} \right\| \hfill \\
\left\| {{\mathbf{r}}'\left( 1 \right)} \right\| = \sqrt 3 \hfill \\
{\text{Therefore, the curvature at the point }}P\left( {\frac{1}{2},1,\frac{1}{3}} \right){\text{ is at }}t = 1 \hfill \\
K = \frac{{\left\| {{\mathbf{r}}'\left( 1 \right) \times {\mathbf{r}}''\left( 1 \right)} \right\|}}{{{{\left\| {{\mathbf{r}}'\left( 1 \right)} \right\|}^3}}} = \frac{{\sqrt 6 }}{{{{\left( {\sqrt 3 } \right)}^3}}} \hfill \\
K = \frac{{\sqrt 6 }}{{3\sqrt 3 }} \hfill \\
K = \frac{{\sqrt 2 }}{3} \hfill \\
\end{gathered} \]
