Answer
\[K = \frac{{2\sqrt 5 }}{{{{\left( {4 + 5{t^2}} \right)}^{3/2}}}}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = 2t{\mathbf{i}} + \frac{1}{2}{t^2}{\mathbf{j}} + {t^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) = 2t{\mathbf{i}} + \frac{1}{2}{t^2}{\mathbf{j}} + {t^2}{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) = 2{\mathbf{i}} + t{\mathbf{j}} + 2t{\mathbf{k}} \hfill \\
{\mathbf{r}}''\left( t \right) = 0{\mathbf{i}} + 2{\mathbf{j}} + 2{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
2&t&{2t} \\
0&1&2
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
t&{2t} \\
1&2
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
2&{2t} \\
0&2
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
2&t \\
0&1
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = 0{\mathbf{i}} - 4{\mathbf{j}} + 2{\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = \sqrt {16 + 4} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = \sqrt {20} \hfill \\
and \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \left\| {2{\mathbf{i}} + t{\mathbf{j}} + 2t{\mathbf{k}}} \right\| \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt {4 + {t^2} + 4{t^2}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt {4 + 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 {20} }}{{{{\left( {\sqrt {4 + 5{t^2}} } \right)}^3}}} \hfill \\
{\text{Simplifying}} \hfill \\
K = \frac{{2\sqrt 5 }}{{{{\left( {4 + 5{t^2}} \right)}^{3/2}}}} \hfill \\
\end{gathered} \]
