Chapter 12 - Vector-Valued Functions - 12.5 Exercises - Page 860: 30

$K = \frac{2}{{{{\left( {4 - 3{{\cos }^2}\pi t} \right)}^{3/2}}}}$

\[\begin{align} & \mathbf{r}\left( t \right)=2\cos \pi t\mathbf{i}+\sin \pi t\mathbf{j} \\ & \text{Use the Theorem which states} \\ & \text{If }C\text{ is a smooth curve given by }\mathbf{r}\left( t \right),\text{ then the curvature }K\text{ of} \\ & 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}}} \\ & \mathbf{r}\left( t \right)=2\cos \pi t\mathbf{i}+\sin \pi t\mathbf{j} \\ & \mathbf{r}'\left( t \right)=-2\pi \sin \pi t\mathbf{i}+\pi \cos \pi t\mathbf{j} \\ & \mathbf{r}''\left( t \right)=-2{{\pi }^{2}}\cos \pi t\mathbf{i}-{{\pi }^{2}}\sin \pi t\mathbf{j} \\ & \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2\pi \sin \pi t & \pi \cos \pi t & 0 \\ -2{{\pi }^{2}}\cos \pi t & -{{\pi }^{2}}\sin \pi t & 0 \\ \end{matrix} \right| \\ & \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left| \begin{matrix} \pi \cos \pi t & 0 \\ -{{\pi }^{2}}\sin \pi t & 0 \\ \end{matrix} \right|\mathbf{i}-\left| \begin{matrix} -2\pi \sin \pi t & 0 \\ -2{{\pi }^{2}}\cos \pi t & 0 \\ \end{matrix} \right|\mathbf{j} \\ & +\left| \begin{matrix} -2\pi \sin \pi t & \pi \cos \pi t \\ -2{{\pi }^{2}}\cos \pi t & -{{\pi }^{2}}\sin \pi t \\ \end{matrix} \right|\mathbf{k} \\ & \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=\left( 2{{\pi }^{3}}{{\sin }^{2}}\pi t+2{{\pi }^{3}}{{\cos }^{2}}\pi t \right)\mathbf{k} \\ & \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=2{{\pi }^{3}}\left( {{\sin }^{2}}2\pi t+{{\cos }^{2}}2\pi t \right)\mathbf{k} \\ & \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right)=2{{\pi }^{3}}\mathbf{k} \\ & \left\| \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right) \right\|=2{{\pi }^{3}} \\ & and \\ & \left\| \mathbf{r}'\left( t \right) \right\|={{\left\| -2\pi \sin \pi t\mathbf{i}+\pi \cos \pi t\mathbf{j} \right\|}^{3}} \\ & \left\| \mathbf{r}'\left( t \right) \right\|=\sqrt{{{\left( -2\pi \sin \pi t \right)}^{2}}+{{\left( \pi \cos \pi t \right)}^{2}}} \\ & \left\| \mathbf{r}'\left( t \right) \right\|=\sqrt{4{{\pi }^{2}}{{\sin }^{2}}\pi t+{{\pi }^{2}}{{\cos }^{2}}\pi t} \\ & \left\| \mathbf{r}'\left( t \right) \right\|=\sqrt{{{\pi }^{2}}\left( 4{{\sin }^{2}}\pi t+{{\cos }^{2}}\pi t \right)} \\ & \left\| \mathbf{r}'\left( t \right) \right\|=\sqrt{{{\pi }^{2}}\left( 4\left( 1-{{\cos }^{2}}\pi t \right)+{{\cos }^{2}}\pi t \right)} \\ & \left\| \mathbf{r}'\left( t \right) \right\|=\sqrt{{{\pi }^{2}}\left( 4-4{{\cos }^{2}}\pi t+{{\cos }^{2}}\pi t \right)} \\ & \left\| \mathbf{r}'\left( t \right) \right\|=\sqrt{{{\pi }^{2}}\left( 4-3{{\cos }^{2}}\pi t \right)} \\ & \left\| \mathbf{r}'\left( t \right) \right\|=\pi \sqrt{4-3{{\cos }^{2}}\pi t} \\ & \text{Therefore}\text{,} \\ & K=\frac{\left\| \mathbf{r}'\left( t \right)\times \mathbf{r}''\left( t \right) \right\|}{{{\left\| \mathbf{r}'\left( t \right) \right\|}^{3}}}=\frac{2{{\pi }^{3}}}{{{\left( \pi \sqrt{4-3{{\cos }^{2}}\pi t} \right)}^{3}}} \\ & K=\frac{2{{\pi }^{3}}}{{{\left( \pi \sqrt{4-3{{\cos }^{2}}\pi t} \right)}^{3}}} \\ & K=\frac{2}{{{\left( 4-3{{\cos }^{2}}\pi t \right)}^{3/2}}} \\ \end{align}\]