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I am reading on Wikipedia that

''...Any regular curve may be parametrized by the arc length (the natural parametrization) and...''

I know that if $a(t) = (x(t),y(t),z(t))$ is a curve (say, smooth) then it is regular iff for all $t$: $a' (t) \neq 0$. I also know the definition of arc length:

The arc length of a curve $a$ between $t_0$ and $t$ is defined as

$$ l = \int_{t_0}^t |a'(t)|dt$$

But what is the parametrization of $a$ using its arc lenght?

student
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  • Arc length parametrization just means unit speed parametrization. The former is used for some dumb historical reasons, I think the latter is far clearer. – Yasha May 26 '25 at 22:58

3 Answers3

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Simply if $$\alpha: I\to \mathbb{R}^3$$ is a regular curve and the arc length is $$s(t)=\int_{t_0}^t |\alpha'|dt$$ Now we solve for $t$ as $t=t(s)$ to get the function $$t:J\to I$$So we define a new curve $\beta(s)=\alpha\circ t=\alpha(t(s))$ where $$\beta: J\to I \to \mathbb{R}^3 \\ |\beta'(s)|=|\frac{d\beta}{ds}|=|\frac{d\beta}{dt}. \frac{dt}{ds}|=|\frac{d\alpha(t(s))}{dt}. \frac{1}{\alpha'(t)}|=1$$

Arvin Rasoulzadeh
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Semsem
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    Why founding that the arc lenght parametrization gives speed $1$ implies? Why is it important? – Poperton Apr 13 '16 at 21:29
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    @GuerlandoOCs one good reason would appear at a later stage, when someone studies analysis on manifolds where the theorems there are mainly valid for structures known as smooth manifolds and its sub-manifolds. One of the conditions that sub-manifolds have to fulfill is to be the image of a map with a constant rank derivation (equal to the dimension of the domain). Hence having a regular curve, by knowing the arclength parameterization, you can fulfill such a condition and by providing the other conditions you can treat it as a sub-manifold where needed. – Arvin Rasoulzadeh Feb 10 '19 at 17:21
  • How do we know that it is necessarily possible to solve for t to get our function t(s) from J to I? – Sachin Jan 24 '25 at 19:14
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"Parameterization by arclength" means that the parameter $t$ used in the parametric equations represents arclength along the curve, measured from some base point. One simple example is $$ x(t) = \cos(t) \quad ; \quad y(t) = \sin(t) \quad (0 \le t \le 2\pi) $$ This a parameterization of the unit circle, and the arclength from the start of the curve to the point $(x(t), y(t))$ is $t$.

In most cases, it's not possible to find simple formulas that give arclength parameterizations, so the whole approach is somewhat academic, in my view.

bubba
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We may have a good notion of arc length even for some nonregular (or not even differentiable) curves; all we need (by definition) is that the curve $a\colon[0,b]\to \mathbb R^n$ is rectifiable, that is: $$ l(a):=\sup\{\,d(a(0),a(t_1))+d(a(t_1),a(t_2))+\ldots+d(a(t_{n}),a(b))\mid 0<t_1<\ldots<t_n<b\,\}$$ exists (is finite). For the case you mention, this supremum definition coincides with the integral definition, of course.

This $l$ gives us a map $\ell\colon[0,b]\to[0,\infty)$ given by $\ell(t)=l(a|_{[0,t]})$ (because automatically all these restrictions are also rectifiable). Then $\ell(0)=0$, $\ell(b)=l(a)$, $\ell$ is continuous and strictly increasing. Thus $\ell^{-1}\colon[0,l(a)]\to[0,b]$ exists and allows us to reparametrize our curve as $\hat a=a\circ\ell^{-1}\colon[0,l(a)]\to\mathbb R^n$. This is reparametrization by arc length. With this the arc length from $\hat a(t_1)$ to $\hat a(t_2)$ is always $t_2-t_1$ for $0\le t_1\le t_2\le l(a)$.

Hagen von Eitzen
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  • Why does the supremum definition coincide with the integral definition, in this case? – étale-cohomology Jul 25 '17 at 17:38
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    @Hagen von Eitzen Suppose that the curve is constant in some interval $[t_1 , t_2]$. Then I don't understand how one can claim that the map $\mathscr{l}$ is strictly increasing on $[t_1 , t_2]$. In that case $\ell(t_1)=l(a|{[0,t_1]} = l(a|{[0,t_2]} ) =\ell(t_2)$. It will be very helpful if you give me some reference to understand it properly. – Anindya Biswas Mar 23 '19 at 04:30