# Curvature and arc length

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 $t_0 = 0$?

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How did you get $2\sinh(t)$? Isn't $r'(t)=(1,\sinh(t),0)$? – Stefan Hansen Jan 21 '13 at 1:10
I took the magnitude of $|r'(t)|$ – Hawk Jan 21 '13 at 1:12

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: $$\frac{dr}{ds} = \frac{dr}{dt}\frac{dt}{ds} = \frac{dr}{dt}\frac{1}{\frac{ds}{dt}}=(1, \sinh t, 0) \frac{1}{\cosh t}.$$
Going from $0$ to $t$ gives half of the branch of the curve. The other half is given by the $2$ – Hawk Jan 21 '13 at 1:20
What branch? $s(t)$ measures the arc length of the (single-valued) curve $r(u)$ from $u=0$ to $u=t$. – user7530 Jan 21 '13 at 1:21
It sounds like you're measuring the arc length of the curve from $u=-t$ to $u=t$. To unit-speed reparameterize the curve you want to measure the arc length from a fixed point (say, 0) to $t$. – user7530 Jan 21 '13 at 1:33
But why choose $0$? Why not say $t = 4$? – Hawk Jan 21 '13 at 1:35