Answer
\[\frac{{\sqrt 6 }}{{{{\left( {\sqrt 3 } \right)}^3}}}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = {e^t}\cos t{\mathbf{i}} + {e^t}\sin t{\mathbf{j}} + {e^t}{\mathbf{k}},{\text{ }}P\left( {1,0,1} \right) \hfill \\
{\text{For }}t = 0 \hfill \\
{\mathbf{r}}\left( 0 \right) = {e^0}\cos \left( 0 \right){\mathbf{i}} + {e^0}\sin \left( 0 \right){\mathbf{j}} + {e^0}{\mathbf{k}} \hfill \\
{\mathbf{r}}\left( 0 \right) = {\mathbf{i}} + 0{\mathbf{j}} + {\mathbf{k}} \hfill \\
{\text{Then }}t = 0{\text{ at the point }}P\left( {1,0,1} \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) = {e^t}\cos t{\mathbf{i}} + {e^t}\sin t{\mathbf{j}} + {e^t}{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) = \left( {{e^t}\cos t - {e^t}\sin t} \right){\mathbf{i}} + \left( {{e^t}\sin t + {e^t}\cos t} \right){\mathbf{j}} + {e^t}{\mathbf{k}} \hfill \\
{\mathbf{r}}''\left( t \right) = - 2{e^t}\sin t{\mathbf{i}} + 2{e^t}\cos t{\mathbf{j}} + {e^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}} \\
{{e^t}\cos t - {e^t}\sin t}&{{e^t}\sin t + {e^t}\cos t}&{{e^t}} \\
{ - 2{e^t}\sin t}&{2{e^t}\cos t}&{{e^t}}
\end{array}} \right| \hfill \\
= \left( {\sin t - \cos t} \right){e^{2t}}{\mathbf{i}} - \left( {\sin t + \cos t} \right){e^{2t}}{\mathbf{j}} + 2{e^{2t}}{\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = {e^{2t}}\sqrt {1 - 2\sin t\cos t + 1 + 2\sin t\cos t + 4} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = \sqrt 6 {e^{2t}} \hfill \\
and \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \left\| {\left( {{e^t}\cos t - {e^t}\sin t} \right){\mathbf{i}} + \left( {{e^t}\sin t + {e^t}\cos t} \right){\mathbf{j}} + {e^t}{\mathbf{k}}} \right\| \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt {{e^{2t}}\left( {1 - 2\sin t\cos t} \right) + {e^{2t}}\left( {1 + 2\sin t\cos t} \right) + {e^{2t}}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = {e^t}\sqrt {1 - 2\sin t\cos t + 1 + 2\sin t\cos t + 1} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt 3 {e^t} \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 6 {e^{2t}}}}{{{{\left( {\sqrt 3 {e^t}} \right)}^3}}} \hfill \\
{\text{At }}t = 0 \hfill \\
K = \frac{{\sqrt 6 {e^0}}}{{{{\left( {\sqrt 3 {e^0}} \right)}^3}}} = \frac{{\sqrt 6 }}{{{{\left( {\sqrt 3 } \right)}^3}}} \hfill \\
\end{gathered} \]
