Curvature and arc length

Question

Given $r(t) = (t,\cosh(t),0)$ where $t \in \mathbb{R}$. Compute the arclength function and verify that it is unit speed, that is $|r'(s)| = 1$

Basically since cosh(t) is symmetric about $x = 0$, I used the arclength function

$$s = \int_{0}^{t}2|r'(t)| dt =2\sinh(t) $$

So $t = \log \left ( \dfrac{s + \sqrt{s^2 + 4}}{2} \right )$. I save you all the trouble from doing the horrible algebra, but in the very end, one gets

$r'(s) = (1/(s^2 + 4),s/(2\sqrt{s^2 + 4})$ and clearly $|r'(s)| = 1/2 \neq 1 $

I am guessing something went wrong from the very start. Perhaps my symmetry argument was flawed?

EDIT I was given $r(t)$ with $t \in \mathbb{R}$. So I wasn't told what is the starting point to integrate. How do I know that I should let $t_0 = 0$?

Answer 1

I'm not sure where the $2$ came from, or why you need any symmetry arguments: the arc length $s(t)$ of a curve $\gamma(t)$ is simply $\int_0^t \|\gamma'(u)\|\,du$ and so, as you've calculated, $s(t) = \sinh(t)$.

Also, notice that you don't need too much nasty algebra to confirm that $r(s)$ is unit speed: $$\frac{dr}{ds} = \frac{dr}{dt}\frac{dt}{ds} = \frac{dr}{dt}\frac{1}{\frac{ds}{dt}}=(1, \sinh t, 0) \frac{1}{\cosh t}.$$