## Calculus (3rd Edition)

$v(0)= \sqrt 2$.
Using the facts that $(\sinh t)'=\cosh t$ and $(\cosh t)'=\sinh t$, we have $r'(t) =\lt \sinh t, \cosh t,1\gt$. Hence the speed $v(t)$ is given by $$v(t)=\|r'(t)\|=\sqrt{ \sinh^2 t+\cosh^2t+1 } .$$ At $t=0$, $v(0)=\sqrt{0+1+1}=\sqrt 2$.