#### Answer

$\dfrac{1}{2}$

#### Work Step by Step

The curvature $\kappa$ for a plane curve system is:
$\kappa (t)= \dfrac{||r'(t) \times r''(t)||}{||r'(t)||^3}$
We have:
$r'(t) =\lt \sinh t, \cosh t, 1 \gt$ and $r''(t) =\lt \cosh t, \sinh t, 0 \gt$
Thus,
$\kappa(t) = \dfrac{||-\sinh ti+\cosh t j-k||}{||\lt \sinh t, \cosh t, 1 \gt||^3} \\=\dfrac{\sqrt {\sinh^2 t+\cosh^2 t+1}}{[\sinh^2t+\cosh^2 t+1]^{3/2}}=\dfrac{1}{\cosh (2t)+1}$
Now, we will compute $\kappa (t)$ at $t=0$
Therefore,
$\kappa(0)= \dfrac{1}{\cosh 2(0)+1}=\dfrac{1}{2}$