Suppose the curve $C$ is given by the vector function
r$(t)=(\sinh(t),\cosh(t),t)$
where $t\geq-1$. Find the exact curvature of $C$ at the point $(0,1,0)$.
My idea isto find the length of T$'(t)$ and divide that by the length of r$'(t)$. So I started off by finding $|$r$'(t)|$ and i got:
r$'(t)=(\cosh(t),\sinh(t),1)$
$|$r$'(t)|$ = $\sqrt{2\sinh(t)^2+2}$
Now should I take r$'(t)$ and divide it by $|$r$'(t)|$, then take that whole value and divide it by $|$r$'(t)|$?
At first you compute the Expression $r''(t)=\frac{dr'(t)}{dt}$. Since it holds $\frac{ds}{dt}=|r'(t)|$ and $\frac{dr'(t)}{ds}=\kappa n$ for the arclength $s$ and the unit normal vector $n$, you arrive at $\kappa = |\frac{dr'(t)}{ds}| = |\frac{dr'(t)}{dt} \frac{dt}{ds}|=|\frac{dr'(t)}{|r'(t)|dt}|=\frac{|r''(t)|}{|r'(t)|}$. You get the curvature of the curve $\kappa$ as a function of $t$. Find the value $t$ which corresponds to the Point $C$ and you are done.