Answer
\[K = \frac{1}{4}\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = 4\cos 2\pi t{\mathbf{i}} + 4\sin 2\pi t{\mathbf{j}} \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 2\pi t{\mathbf{i}} + 4\sin 2\pi t{\mathbf{j}} \hfill \\
{\mathbf{r}}'\left( t \right) = 4\left( { - 2\pi \sin 2\pi t} \right){\mathbf{i}} + 4\left( {2\pi \cos 2\pi t} \right){\mathbf{j}} \hfill \\
{\mathbf{r}}'\left( t \right) = - 8\pi \sin 2\pi t{\mathbf{i}} + 8\pi \cos 2\pi t{\mathbf{j}} \hfill \\
{\mathbf{r}}''\left( t \right) = - 16{\pi ^2}\cos 2\pi t{\mathbf{i}} - 16{\pi ^2}\sin 2\pi t{\mathbf{j}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{ - 8\pi \sin 2\pi t}&{8\pi \cos 2\pi t}&0 \\
{ - 16{\pi ^2}\cos 2\pi t}&{ - 16{\pi ^2}\sin 2\pi t}&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{8\pi \cos 2\pi t}&0 \\
{ - 16{\pi ^2}\sin 2\pi t}&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{ - 8\pi \sin 2\pi t}&0 \\
{ - 16{\pi ^2}\cos 2\pi t}&0
\end{array}} \right|{\mathbf{j}} \hfill \\
+ \left| {\begin{array}{*{20}{c}}
{ - 8\pi \sin 2\pi t}&{8\pi \cos 2\pi t} \\
{ - 16{\pi ^2}\cos 2\pi t}&{ - 16{\pi ^2}\sin 2\pi t}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left( {128{\pi ^3}{{\sin }^2}2\pi t + 128{\pi ^3}{{\cos }^2}2\pi t} \right){\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = 128{\pi ^3}\left( {{{\sin }^2}2\pi t + {{\cos }^2}2\pi t} \right){\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = 128{\pi ^3}{\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = 128{\pi ^3} \hfill \\
and \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \left\| { - 8\pi \sin 2\pi t{\mathbf{i}} + 8\pi \cos 2\pi t{\mathbf{j}}} \right\| \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( { - 8\pi \sin 2\pi t} \right)}^2} + {{\left( {8\pi \cos 2\pi t} \right)}^2}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right)} \right\| = \sqrt {64{\pi ^2}{{\sin }^2}2\pi t + 64{\pi ^2}{{\cos }^2}2\pi t} = 8\pi \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{{128{\pi ^3}}}{{{{\left( {8\pi } \right)}^3}}} \hfill \\
K = \frac{1}{4} \hfill \\
\end{gathered} \]
