Answer
\[K = \frac{2}{5}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = 4\cos t{\mathbf{i}} + 3\sin t{\mathbf{j}} + t{\mathbf{k}},{\text{ }}P\left( { - 4,0,\pi } \right) \hfill \\
{\text{For }}t = \pi \hfill \\
{\mathbf{r}}\left( \pi \right) = 4\cos \left( \pi \right){\mathbf{i}} + 3\sin \left( \pi \right){\mathbf{j}} + \pi {\mathbf{k}} \hfill \\
{\mathbf{r}}\left( \pi \right) = - 4{\mathbf{i}} + 0{\mathbf{j}} + \pi {\mathbf{k}} \hfill \\
{\text{Then }}t = \pi {\text{ at the point }}P\left( { - 4,0,\pi } \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) = 4\cos t{\mathbf{i}} + 3\sin t{\mathbf{j}} + t{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) = - 4\sin t{\mathbf{i}} + 3\cos t{\mathbf{j}} + {\mathbf{k}} \hfill \\
{\mathbf{r}}''\left( t \right) = - 4\cos t{\mathbf{i}} - 3\sin t{\mathbf{j}} + 0{\mathbf{k}} \hfill \\
{\text{At }}t = \pi \hfill \\
{\mathbf{r}}'\left( \pi \right) = - 4\sin \left( \pi \right){\mathbf{i}} + 3\cos \left( \pi \right){\mathbf{j}} + {\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( \pi \right) = 0{\mathbf{i}} - 3{\mathbf{j}} + {\mathbf{k}} \hfill \\
{\mathbf{r}}''\left( \pi \right) = - 4\cos \left( \pi \right){\mathbf{i}} - 3\sin \left( \pi \right){\mathbf{j}} + 0{\mathbf{k}} \hfill \\
{\mathbf{r}}''\left( \pi \right) = 4{\mathbf{i}} + 0{\mathbf{j}} + 0{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( \pi \right) \times {\mathbf{r}}''\left( \pi \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
0&{ - 3}&1 \\
4&0&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( \pi \right) \times {\mathbf{r}}''\left( \pi \right) = \left( 0 \right){\mathbf{i}} - \left( {0 - 4} \right){\mathbf{j}} + \left( {0 + 12} \right){\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( \pi \right) \times {\mathbf{r}}''\left( \pi \right) = 0{\mathbf{i}} + 4{\mathbf{j}} + 12{\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( \pi \right) \times {\mathbf{r}}''\left( \pi \right)} \right\| = \sqrt {{{\left( 0 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( {12} \right)}^2}} \hfill \\
\left\| {{\mathbf{r}}'\left( \pi \right) \times {\mathbf{r}}''\left( \pi \right)} \right\| = \sqrt {160} \hfill \\
and \hfill \\
\left\| {{\mathbf{r}}'\left( \pi \right)} \right\| = \left\| {0{\mathbf{i}} - 3{\mathbf{j}} + {\mathbf{k}}} \right\| \hfill \\
\left\| {{\mathbf{r}}'\left( \pi \right)} \right\| = \sqrt {0 + 9 + 1} = \sqrt {10} \hfill \\
{\text{Therefore, the curvature at the point }}P\left( { - 4,0,\pi } \right){\text{ is at }}t = \pi \hfill \\
K = \frac{{\left\| {{\mathbf{r}}'\left( \pi \right) \times {\mathbf{r}}''\left( \pi \right)} \right\|}}{{{{\left\| {{\mathbf{r}}'\left( \pi \right)} \right\|}^3}}} = \frac{{\sqrt {160} }}{{{{\left( {\sqrt {10} } \right)}^3}}} \hfill \\
K = \frac{{\sqrt {160} }}{{10\sqrt {10} }} \hfill \\
K = \frac{2}{5} \hfill \\
\end{gathered} \]
