Answer
(1) the curvature $\kappa$ at $t = \frac{\pi }{3}$ is $ \simeq 4.54$
(2) the curvature $\kappa$ at $t = \frac{\pi }{2}$ is $ \simeq 0.2$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {2\sin t,\cos 3t,t} \right)$. So, the derivatives are
${\bf{r}}'\left( t \right) = \left( {2\cos t, - 3\sin 3t,1} \right)$
${\bf{r}}{\rm{''}}\left( t \right) = \left( { - 2\sin t, - 9\cos 3t,0} \right)$
By Eq. (3) of Theorem 1, the curvature is given by
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
(1) The curvature at $t = \frac{\pi }{3}$
Evaluate $||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||$ at $t = \frac{\pi }{3}$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{2\cos t}&{ - 3\sin 3t}&1\\
{ - 2\sin t}&{ - 9\cos 3t}&0
\end{array}} \right|$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left( {9\cos 3t} \right){\bf{i}} - \left( {2\sin t} \right){\bf{j}}$
$ + \left( { - 18\cos t\cos 3t - 6\sin t\sin 3t} \right){\bf{k}}$
${\bf{r}}'\left( {\frac{\pi }{3}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{3}} \right) = - 9{\bf{i}} - \sqrt 3 {\bf{j}} + 9{\bf{k}}$
$||{\bf{r}}'\left( {\frac{\pi }{3}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{3}} \right)|{|^2} = 81 + 3 + 81 = 165$
$||{\bf{r}}'\left( {\frac{\pi }{3}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{3}} \right)|| = \sqrt {165} $
Evaluate $||{\bf{r}}'\left( t \right)||$ at $t = \frac{\pi }{3}$
$||{\bf{r}}'\left( t \right)|{|^2} = \left( {2\cos t, - 3\sin 3t,1} \right)\cdot\left( {2\cos t, - 3\sin 3t,1} \right)$
$||{\bf{r}}'\left( t \right)|{|^2} = 4{\cos ^2}t + 9{\sin ^2}3t + 1$
$||{\bf{r}}'\left( {\frac{\pi }{3}} \right)|{|^2} = 4\cdot\frac{1}{4} + 9\cdot0 + 1 = 2$
$||{\bf{r}}'\left( {\frac{\pi }{3}} \right)|| = \sqrt 2 $
So, the curvature $\kappa$ at $t = \frac{\pi }{3}$ is
$\kappa \left( {\frac{\pi }{3}} \right) = \frac{{||{\bf{r}}'\left( {\frac{\pi }{3}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{3}} \right)||}}{{||{\bf{r}}'\left( {\frac{\pi }{3}} \right)|{|^3}}}$
$\kappa \left( {\frac{\pi }{3}} \right) = \frac{{\sqrt {165} }}{{{2^{3/2}}}} \simeq 4.54$
(2) The curvature at $t = \frac{\pi }{2}$
Evaluate $||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||$ at $t = \frac{\pi }{2}$
We have from earlier result:
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left( {9\cos 3t} \right){\bf{i}} - \left( {2\sin t} \right){\bf{j}}$
$ + \left( { - 18\cos t\cos 3t - 6\sin t\sin 3t} \right){\bf{k}}$
So,
${\bf{r}}'\left( {\frac{\pi }{2}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{2}} \right) = - 2{\bf{j}} + 6{\bf{k}}$
$||{\bf{r}}'\left( {\frac{\pi }{2}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{2}} \right)|{|^2} = 4 + 36 = 40$
$||{\bf{r}}'\left( {\frac{\pi }{2}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{2}} \right)|| = 2\sqrt {10} $
Evaluate $||{\bf{r}}'\left( t \right)||$ at $t = \frac{\pi }{2}$
We have from earlier result:
$||{\bf{r}}'\left( t \right)|{|^2} = 4{\cos ^2}t + 9{\sin ^2}3t + 1$
So,
$||{\bf{r}}'\left( {\frac{\pi }{2}} \right)|{|^2} = 0 + 9\cdot1 + 1 = 10$
$||{\bf{r}}'\left( {\frac{\pi }{2}} \right)|| = \sqrt {10} $
So, the curvature $\kappa$ at $t = \frac{\pi }{2}$ is
$\kappa \left( {\frac{\pi }{2}} \right) = \frac{{||{\bf{r}}'\left( {\frac{\pi }{2}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{2}} \right)||}}{{||{\bf{r}}'\left( {\frac{\pi }{2}} \right)|{|^3}}}$
$\kappa \left( {\frac{\pi }{2}} \right) = \frac{{2\sqrt {10} }}{{{{10}^{3/2}}}} \simeq 0.2$
