Answer
The curvature is $\kappa \left( t \right) = csch {\ }t$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t - \tanh t,secht} \right)$. So, the derivatives are
${\bf{r}}'\left( t \right) = \left( {1 - sec{h^2}t, - secht\tanh t} \right)$
Since ${\tanh ^2}t + sec{h^2}t = 1$, so
${\bf{r}}'\left( t \right) = \left( {{{\tanh }^2}t, - secht\tanh t} \right)$
${\bf{r}}{\rm{''}}\left( t \right) = \left( {2sec{h^2}t\tanh t, - sec{h^3}t + secht{{\tanh }^2}t} \right)$
By Eq. (3) of Theorem 1, the curvature is given by
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
To apply Theorem 1, we treat ${\bf{r}}'\left( t \right)$ and ${\bf{r}}{\rm{''}}\left( t \right)$ as vectors in ${\mathbb{R}^3}$ by setting the $z$-components equal to zero. Thus,
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)$
$ = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{{{\tanh }^2}t}&{ - secht\tanh t}&0\\
{2sec{h^2}t\tanh t}&{ - sec{h^3}t + secht{{\tanh }^2}t}&0
\end{array}} \right|$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = ({\tanh ^2}t\left( { - sec{h^3}t + secht{{\tanh }^2}t} \right)$
$ + \left( {2sec{h^2}t\tanh t} \right)\left( {secht\tanh t} \right)){\bf{k}}$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = ( - {\tanh ^2}tsec{h^3}t + secht{\tanh ^4}t$
$ + 2sec{h^3}t{\tanh ^2}t){\bf{k}}$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = \left( {secht{{\tanh }^4}t + sec{h^3}t{{\tanh }^2}t} \right){\bf{k}}$
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = secht{\tanh ^2}t\left( {{{\tanh }^2}t + sec{h^2}t} \right){\bf{k}}$
Since ${\tanh ^2}t + sec{h^2}t = 1$, so
${\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right) = secht{\tanh ^2}t{\bf{k}}$
So, $||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)|| = secht{\tanh ^2}t$.
Next, we evaluate $||{\bf{r}}'\left( t \right)||$
$||{\bf{r}}'\left( t \right)|{|^2} = \left( {{{\tanh }^2}t, - secht\tanh t} \right)\cdot\left( {{{\tanh }^2}t, - secht\tanh t} \right)$
$||{\bf{r}}'\left( t \right)|{|^2} = {\tanh ^4}t + sec{h^2}t{\tanh ^2}t$
$||{\bf{r}}'\left( t \right)|{|^2} = {\tanh ^2}t\left( {{{\tanh }^2}t + sec{h^2}t} \right)$
Since ${\tanh ^2}t + sec{h^2}t = 1$, so
$||{\bf{r}}'\left( t \right)|{|^2} = {\tanh ^2}t$
So, $||{\bf{r}}'\left( t \right)|| = \tanh t$.
Substituting these values in
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
gives
$\kappa \left( t \right) = \frac{{secht{{\tanh }^2}t}}{{{{\tanh }^3}t}} = \frac{{secht}}{{\tanh t}} = csch{\ }t$
