The trace of the parametrized curve is
$$\alpha(t)=(t,\cosh t),\ t\in\mathbb R$$
is called catenary.
I want to show the curvature of the catenary is
$$k(t)=\frac{1}{\cosh^2t}$$
Before finding the curvature I need to parametrize it by arc length. Do Carmo in his classical Differential Geometry book makes the following remark about it on page 22:
since the catenary is defined for every $t\in \mathbb R$, I'm having trouble to know the value of $t_0$, can it be anything?
Best Answer
You can always use the chain rule to do curvature/torsion computations for curves that are not arclength parametrized (and, of course, there are other formulas as well). This has been discussed in numerous MSE posts.
However, note that for your catenary, $s(t) = \displaystyle\int_0^t \sqrt{1+\sinh^2 u}du = \int_0^t \cosh u\,du = \sinh t$. And, using the quadratic formula (and the definition of $\sinh$), you can easily solve the equation $s=\sinh t$ for $t$ as a function of $s$ and reparametrize the curve. ... Again, whether you should want to is another matter.