# How do you determine whether a curve uses arc length as a parameter?

I know how to find arc length because it's simply a matter of plugging in values into a formula:

s(t) = $\int_a^t |v(u)|\,du$

But given an equation r(t), how do I show whether or not the curves use arc length as a parameter? e.g) $r(t) = <2 \cos{t}, 2 \sin{t}>$ for $0 \leq t \leq 2\pi$

I did some calculating and figured this much out:

$$v(t) = \left< -2\sin{t}, 2\cos{t}\right >$$

$$|v(t)| = 2$$

$$s(t) = 2t$$

Any examples or tips?

• if $r$ is parameterized by arclength parameter $s$ then $|r'(s)|=1$ Commented Sep 4, 2013 at 17:18
• You're missing a derivative from your arclength formula. That is, you have a |v(t)| where there should be a $\|v'(t)\|$. Commented Sep 4, 2013 at 17:25
• @Omnomnomnom I don't believe that's correct. If $v(t)= r ' (t)$ then his work is good, since $v(t)$ usually means velocity. Granted he should have used $\int_a^t |r'(u)|\,du$.
– john
Commented Oct 6, 2017 at 4:59
• Also $s(t) = 2t - 2a$.
– john
Commented Oct 6, 2017 at 5:02

A parametrized $r(t)$ is parametrized by arclength when $t=s(t)$, where the arclength $s(t)$ is given by $$\int_{t_0}^t\|r'(t)\|dt$$ In your example you found that $$s(t)=2t$$ Since $t\neq 2t$, we can conclude that $r$ is not parametrized by arclength.
If, on the other hand, you were given $$r(t)=\left\langle -2\sin\frac t2, 2\cos \frac t2 \right\rangle$$ You would find $$\|r'(t)\|=1\implies s(t)=t$$ Which would mean that $r$ is parametrized by arclength.
As you can see, this also means that $r$ is parametrized by arclength whenever $\|r'(t)\|=1$ for all $t$.
• Correct me if I'm wrong, but following your reasoning a parametrized $r(t)$ is parametrized by arclength when $s(t) = t - t_0$, where the arclength $s(t)$ is given by $\int_{t_0}^t\|r'(u)\|du$. This implies $\|r'(t)\|=1$. Nice explanation.