Answer
$$\kappa \left( t \right) = \frac{1}{{2{{\cosh }^2}t}}$$
Work Step by Step
$$\eqalign{
& x = \cosh t,\,\,\,\,y = \sinh t,\,\,\,\,z = t \cr
& {\text{Let }}{\bf{r}}\left( t \right) = x\left( t \right){\bf{i}} + y\left( t \right){\bf{j}} \cr
& {\text{Then}} \cr
& {\bf{r}}\left( t \right) = \cosh t{\bf{i}} + \sinh t{\bf{j}} + t{\bf{k}} \cr
& {\text{Calculate the derivatives }}{\bf{r}}'\left( t \right){\text{ and }}{\bf{r}}{\text{''}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\cosh t} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {\sinh t} \right]{\bf{j}} + \frac{d}{{dt}}\left[ t \right]{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = \sinh t{\bf{i}} + \cosh t{\bf{j}} + \left( 1 \right){\bf{k}} \cr
& and \cr
& {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ {\sinh t} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {\cosh t} \right]{\bf{j}} + \frac{d}{{dt}}\left[ 1 \right]{\bf{k}} \cr
& {\bf{r}}''\left( t \right) = \cosh t{\bf{i}} + \sinh t{\bf{j}} \cr
& \cr
& {\text{Calculate the cross product }}{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right) \cr} $$
\[\begin{gathered}
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{\sinh t}&{\cosh t}&1 \\
{\cosh t}&{\sinh t}&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\cosh t}&1 \\
{\sinh t}&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{\sinh t}&1 \\
{\cosh t}&0
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{\sinh t}&{\cosh t} \\
{\cosh t}&{\sinh t}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left( { - \sinh t} \right){\mathbf{i}} - \left( { - \cosh t} \right){\mathbf{j}} + \left( {{{\sinh }^2}t - {{\cosh }^2}t} \right){\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = - \sinh t{\mathbf{i}} + \cosh t{\mathbf{j}} - {\mathbf{k}} \hfill \\
\end{gathered} \]
$$\eqalign{
& {\text{Use the formula }}\left( 3 \right)\,\,\,\kappa \left( t \right) = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}}.{\text{ Thus}}{\text{,}} \cr
& \kappa \left( t \right) = \frac{{\left\| { - \sinh t{\bf{i}} + \cosh t{\bf{j}} - {\bf{k}}} \right\|}}{{{{\left\| {\sinh t{\bf{i}} + \cosh t{\bf{j}} + {\bf{k}}} \right\|}^3}}} \cr
& \kappa \left( t \right) = \frac{{\sqrt {{{\sinh }^2}t + {{\cosh }^2}t + 1} }}{{{{\left( {\sqrt {{{\sinh }^2}t + {{\cosh }^2}t + 1} } \right)}^3}}} \cr
& \kappa \left( t \right) = \frac{1}{{{{\sinh }^2}t + {{\cosh }^2}t + 1}} \cr
& \kappa \left( t \right) = \frac{1}{{{{\cosh }^2}t - 1 + {{\cosh }^2}t + 1}} \cr
& \kappa \left( t \right) = \frac{1}{{2{{\cosh }^2}t}} \cr} $$
