Answer
\[K = \frac{{160}}{{91\sqrt {91} }}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = \left\langle {5\cos t,4\sin t} \right\rangle ,{\text{ at }}t = \frac{\pi }{3} \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) = \left\langle {5\cos t,4\sin t} \right\rangle \hfill \\
{\mathbf{r}}'\left( t \right) = \left\langle { - 5\sin t,4\cos t} \right\rangle \hfill \\
{\mathbf{r}}''\left( t \right) = \left\langle { - 5\cos t, - 4\sin t} \right\rangle \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{ - 5\sin t}&{4\cos t}&0 \\
{ - 5\cos t}&{ - 4\sin t}&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{4\cos t}&0 \\
{ - 4\sin t}&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{ - 5\sin t}&0 \\
{ - 5\cos t}&0
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{ - 5\sin t}&{4\cos t} \\
{ - 5\cos t}&{ - 4\sin t}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left( {20{{\sin }^2}t + 20{{\cos }^2}t} \right){\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = 20 \hfill \\
\left\| {{\mathbf{r}}'\left( {\frac{\pi }{3}} \right) \times {\mathbf{r}}''\left( {\frac{\pi }{3}} \right)} \right\| = 20 \hfill \\
{\left\| {{\mathbf{r}}'\left( t \right)} \right\|^3} = \left\| {\left\langle { - 5\sin t,4\cos t} \right\rangle } \right\| = {\left( {\sqrt {25{{\sin }^2}t + 16{{\cos }^2}t} } \right)^3} \hfill \\
{\left\| {{\mathbf{r}}'\left( {\frac{\pi }{3}} \right)} \right\|^3} = {\left( {\sqrt {25{{\sin }^2}\left( {\frac{\pi }{3}} \right) + 16{{\cos }^2}\left( {\frac{\pi }{3}} \right)} } \right)^3} = \frac{{91\sqrt {91} }}{8} \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}}} \hfill \\
{\text{Evaluate at }}t = \frac{\pi }{3} \hfill \\
K = \frac{{20}}{{\frac{{91\sqrt {91} }}{8}}} \hfill \\
K = \frac{{160}}{{91\sqrt {91} }} \hfill \\
\end{gathered} \]
