Answer
\[K = \frac{{\sqrt 5 }}{{{{\left( {\sqrt {1 + 5{t^2}} } \right)}^3}}}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = t{\mathbf{i}} + {t^2}{\mathbf{j}} + \frac{{{t^2}}}{2}{\mathbf{k}} \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}} + {t^2}{\mathbf{j}} + \frac{{{t^2}}}{2}{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) = {\mathbf{i}} + 2t{\mathbf{j}} + t{\mathbf{k}} \hfill \\
{\mathbf{r}}''\left( t \right) = 2{\mathbf{j}} + {\mathbf{k}} \hfill \\
\hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
1&{2t}&t \\
0&2&1
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{2t}&t \\
2&1
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
1&t \\
0&1
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
1&{2t} \\
0&2
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left( {2t - 2t} \right){\mathbf{i}} - \left( {1 - 0} \right){\mathbf{j}} + \left( {2 - 0} \right){\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = - {\mathbf{j}} + 2{\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = \sqrt {{1^2} + {2^2}} = \sqrt 5 \hfill \\
and \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \left\| {{\mathbf{i}} + 2t{\mathbf{j}} + t{\mathbf{k}}} \right\| \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt {1 + 4{t^2} + {t^2}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt {1 + 5{t^2}} \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}}} = \frac{{\sqrt 5 }}{{{{\left( {\sqrt {1 + 5{t^2}} } \right)}^3}}} \hfill \\
K = \frac{{\sqrt 5 }}{{{{\left( {\sqrt {1 + 5{t^2}} } \right)}^3}}} \hfill \\
\end{gathered} \]
