Answer
$\kappa \left( {\frac{\pi }{4}} \right) \simeq 0.2258$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {\tan t,\sec t,\cos t} \right)$. Using Theorem 2 of Section 3.6 on page 138, we obtain
${\bf{r}}'\left( t \right) = \left( {{{\sec }^2}t,\sec t\tan t, - \sin t} \right)$
${\bf{r}}{\rm{''}}\left( t \right) = \left( {2{{\sec }^2}t\tan t,\sec t{{\tan }^2}t + {{\sec }^3}t, - \cos t} \right)$
At $t = \frac{\pi }{4}$, we get
${\bf{r}}'\left( {\frac{\pi }{4}} \right) = \left( {2,\sqrt 2 , - \frac{1}{2}\sqrt 2 } \right)$
${\bf{r}}{\rm{''}}\left( {\frac{\pi }{4}} \right) = \left( {4,3\sqrt 2 , - \frac{1}{2}\sqrt 2 } \right)$
By Eq. (3) of Theorem 1 (Section 14.4) the curvature is
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
So, at $t = \frac{\pi }{4}$, the curvature is
$\kappa \left( {\frac{\pi }{4}} \right) = \frac{{||{\bf{r}}'\left( {\frac{\pi }{4}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{4}} \right)||}}{{||{\bf{r}}'\left( {\frac{\pi }{4}} \right)|{|^3}}}$
Evaluate ${\bf{r}}'\left( {\frac{\pi }{4}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{4}} \right)$
${\bf{r}}'\left( {\frac{\pi }{4}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{4}} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
2&{\sqrt 2 }&{ - \frac{1}{2}\sqrt 2 }\\
4&{3\sqrt 2 }&{ - \frac{1}{2}\sqrt 2 }
\end{array}} \right|$
$ = \left( { - 1 + 3} \right){\bf{i}} - \left( { - \sqrt 2 + 2\sqrt 2 } \right){\bf{j}} + \left( {6\sqrt 2 - 4\sqrt 2 } \right){\bf{k}}$
$ = 2{\bf{i}} - \sqrt 2 {\bf{j}} + 2\sqrt 2 {\bf{k}}$
So,
$||{\bf{r}}'\left( {\frac{\pi }{4}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{4}} \right)|| = \sqrt {\left( {2, - \sqrt 2 ,2\sqrt 2 } \right)\cdot\left( {2, - \sqrt 2 ,2\sqrt 2 } \right)} $
$ = \sqrt {4 + 2 + 8} = \sqrt {14} $
and
$||{\bf{r}}'\left( {\frac{\pi }{4}} \right)|| = \sqrt {\left( {2,\sqrt 2 , - \frac{1}{2}\sqrt 2 } \right)\cdot\left( {2,\sqrt 2 , - \frac{1}{2}\sqrt 2 } \right)} $
$ = \sqrt {4 + 2 + \frac{1}{2}} = \sqrt {\frac{{13}}{2}} $
$||{\bf{r}}'\left( {\frac{\pi }{4}} \right)|{|^3} = {\left( {\frac{{13}}{2}} \right)^{3/2}}$
Substituting these results in $\kappa \left( {\frac{\pi }{4}} \right)$:
$\kappa \left( {\frac{\pi }{4}} \right) = \frac{{||{\bf{r}}'\left( {\frac{\pi }{4}} \right) \times {\bf{r}}{\rm{''}}\left( {\frac{\pi }{4}} \right)||}}{{||{\bf{r}}'\left( {\frac{\pi }{4}} \right)|{|^3}}}$
gives
$\kappa \left( {\frac{\pi }{4}} \right) = \sqrt {14} {\left( {\frac{2}{{13}}} \right)^{3/2}} \simeq 0.2258$
