Answer
\[K = \frac{5}{{29}}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = 2t{\mathbf{i}} + 5\cos t{\mathbf{j}} + 5\sin t{\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}} + 5\cos t{\mathbf{j}} + 5\sin t{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) = 2{\mathbf{i}} - 5\sin t{\mathbf{j}} + 5\cos t{\mathbf{k}} \hfill \\
{\mathbf{r}}''\left( t \right) = 0{\mathbf{i}} - 5\cos t{\mathbf{j}} - 5\sin t{\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&{ - 5\sin t}&{5\cos t} \\
0&{ - 5\cos t}&{ - 5\sin t}
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{ - 5\sin t}&{5\cos t} \\
{ - 5\cos t}&{ - 5\sin t}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
2&{5\cos t} \\
0&{ - 5\sin t}
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
2&{ - 5\sin t} \\
0&{ - 5\cos t}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left( {25{{\sin }^2}t + 25{{\cos }^2}t} \right){\mathbf{i}} + 10\sin t{\mathbf{j}} - 10\cos t{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = 25{\mathbf{i}} + 10\sin t{\mathbf{j}} - 10\sin t{\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = \sqrt {{{\left( {25} \right)}^2} + {{\left( {10\sin t} \right)}^2} + {{\left( { - 10\cos t} \right)}^2}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = \sqrt {625 + 100} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = \sqrt {725} \hfill \\
and \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \left\| {2{\mathbf{i}} - 5\sin t{\mathbf{j}} + 5\cos t{\mathbf{k}}} \right\| \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt {4 + 25{{\sin }^2}t + 25{{\cos }^2}t} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt {29} \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 {725} }}{{{{\left( {\sqrt {29} } \right)}^3}}} \hfill \\
{\text{Simplifying}} \hfill \\
K = \frac{{5\sqrt {29} }}{{29\sqrt {29} }} \hfill \\
K = \frac{5}{{29}} \hfill \\
\end{gathered} \]
