Answer
\[K = \frac{{7\sqrt {26} }}{{676}}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = t{\mathbf{i}} + {t^2}{\mathbf{j}} + \frac{{{t^3}}}{4}{\mathbf{k}},{\text{ }}P\left( {2,4,2} \right) \hfill \\
{\text{For }}t = 2 \hfill \\
{\mathbf{r}}\left( 2 \right) = 2{\mathbf{i}} + {2^2}{\mathbf{j}} + \frac{{{2^3}}}{4}{\mathbf{k}} \hfill \\
{\mathbf{r}}\left( 2 \right) = 2{\mathbf{i}} + 4{\mathbf{j}} + 4{\mathbf{k}} \hfill \\
{\text{Then }}t = 2{\text{ at the point }}P\left( {2,4,2} \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) = t{\mathbf{i}} + {t^2}{\mathbf{j}} + \frac{{{t^3}}}{4}{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) = {\mathbf{i}} + 2t{\mathbf{j}} + \frac{{3{t^2}}}{4}{\mathbf{k}} \hfill \\
{\mathbf{r}}''\left( t \right) = 2{\mathbf{j}} + \frac{{3t}}{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}} \\
1&{2t}&{\frac{{3{t^2}}}{4}} \\
0&2&{\frac{{3t}}{2}}
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{2t}&{\frac{{3{t^2}}}{4}} \\
2&{\frac{{3t}}{2}}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
1&{\frac{{3{t^2}}}{4}} \\
0&{\frac{{3t}}{2}}
\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) = \frac{3}{2}{t^2}{\mathbf{i}} - \frac{{3t}}{2}{\mathbf{j}} + 2{\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = \sqrt {\frac{9}{4}{t^4} + \frac{9}{4}{t^2} + 4} \hfill \\
and \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \left\| {{\mathbf{i}} + 2t{\mathbf{j}} + \frac{{3{t^2}}}{4}{\mathbf{k}}} \right\| \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt {1 + 4{t^2} + \frac{{9{t^4}}}{{16}}} \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 {\frac{9}{4}{t^4} + \frac{9}{4}{t^2} + 4} }}{{{{\left( {\sqrt {1 + 4{t^2} + \frac{{9{t^4}}}{{16}}} } \right)}^3}}} \hfill \\
{\text{At }}t = 2 \hfill \\
K = \frac{{\sqrt {\frac{9}{4}{{\left( 2 \right)}^4} + \frac{9}{4}{{\left( 2 \right)}^2} + 4} }}{{{{\left( {\sqrt {1 + 4{{\left( 2 \right)}^2} + \frac{{9{{\left( 2 \right)}^4}}}{{16}}} } \right)}^3}}} = \frac{7}{{{{\left( {\sqrt {26} } \right)}^3}}} \hfill \\
K = \frac{7}{{26\sqrt {26} }} \hfill \\
{\text{Rationalizing}} \hfill \\
K = \frac{{7\sqrt {26} }}{{676}} \hfill \\
\end{gathered} \]
