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