Answer
\[\frac{4}{{{{\left( {\sqrt {17} } \right)}^3}}}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = {e^t}{\mathbf{i}} + 4t{\mathbf{j}},{\text{ }}P\left( {1,0} \right) \hfill \\
{\text{For }}t = 0 \hfill \\
{\mathbf{r}}\left( 0 \right) = {e^0}{\mathbf{i}} + 4\left( 0 \right){\mathbf{j}} \hfill \\
{\mathbf{r}}\left( 0 \right) = {\mathbf{i}} + 0{\mathbf{j}} \hfill \\
{\text{Then }}t = 0{\text{ at the point }}P\left( {1,0} \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}{\mathbf{i}} + 4t{\mathbf{j}} \hfill \\
{\mathbf{r}}'\left( t \right) = {e^t}{\mathbf{i}} + 4{\mathbf{j}} \hfill \\
{\mathbf{r}}''\left( t \right) = {e^t}{\mathbf{i}} \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}}&4&0 \\
{{e^t}}&0&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
4&0 \\
0&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{{e^t}}&0 \\
{{e^t}}&0
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{{e^t}}&4 \\
{{e^t}}&0
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = - 4{e^t}{\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = 4{e^t} \hfill \\
and \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \left\| {{e^t}{\mathbf{i}} + 4{\mathbf{j}}} \right\| \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt {{e^{2t}} + 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{{4{e^t}}}{{{{\left( {\sqrt {{e^{2t}} + 16} } \right)}^3}}} \hfill \\
{\text{At }}t = 0 \hfill \\
K = \frac{{4{e^0}}}{{{{\left( {\sqrt {{e^0} + 16} } \right)}^3}}} = \frac{4}{{{{\left( {\sqrt {17} } \right)}^3}}} \hfill \\
\end{gathered} \]
