I am trying to understand and distinguish the difference between arc length and arc length parameterisation.

The first thing, how do denote the $\text{arc length}$ and $\textit{arc length parametrisation of a curve}$ in the same text?

Second thing. I have a unit circle $C$ at the origin. So I choose the following parametrisation of the unit circle,

$$C(\varphi) = \{(\cos \varphi, \sin \varphi): \varphi \in (-\pi,\pi]\}$$

(Note the chosen range of $\varphi$, for my specific purposes). Then using the formula $$s(\varphi) = \int \sqrt{(x'(\varphi)^{2} + y'(\varphi)^{2}} $$ the $\textit{arc length parametrisation}$ for the unit circle is just

$$s(\varphi)= \varphi,$$

However, the $\textit{arc length}$ $s$ is

$s = \varphi$ if $\varphi \in (0,\pi]$ and $s = \pi+ \varphi$, if $\varphi \in (-\pi,0)$. Is this entirely correct?

Thirdly. Suppose I want to start measuring the arc length in the opposite direction. So above, I measured arc length anticlockwise from $(1,0)$. Now I start at $(-1,0)$ and travel along the circle clockwise. Naturally I expect the arc length to be

$$s(\varphi) = \pi - \varphi$$

How do I obtain this formula while keeping the above chosen parametrisation of a circle by $\varphi$? Is it mathematically correct to use $s = \pi - \varphi$ but to keep the initial above parametrisation? Is it necessary for me to choose a different parameter a priori if I want to traverse along a curve in the opposite sense?

Thank you.


1 Answer 1


Let $\alpha\colon\mathbb{R}^2\to\mathbb{R}^2$ be a piecewise-smooth curve, $\alpha(t)=(x(t),y(t))$.

The arclength of a piecewise-smooth, parameterized curve is $s(t)=\int_{t_0}^{t}\sqrt{x'(t)^2+y'(t)^2}\,dt$ for $t_0$ and $t$ in the same differentiable piece, $t_0$ your "starting point". So $s$ is a function from the "parameterized space", $I$, say, to the "arclength space" $J$, say. So $s\colon I\to J$. Suppose further that $s$ is invertible on this piece ($s$ typically is because one usually assumes that $x'$ and $y'$ are never simultaneously zero and hence that $s(t)$ is a strictly increasing function). By abuse of notation, let's define a function $t\colon J\to I$ which is simply $t=s^{-1}$, the inverse of $s$. Then $\alpha$ can be parameterized by arclength by noting that $t\in I$ can be written as a function of $s\in J$ as $t(s)$. Hence, abusing notation further, $\alpha(t)=(\alpha\circ t)(s)$, which is $\alpha$ parameterized by arclength.

When you measure arclength as you did, you're making two assumptions. (1) That you're starting from $(1,0)$ and (2) that, starting from $(1,0)$, the arclength to the point $(\cos(3\pi/4),\sin(3\pi/4))=(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$ is $\pi-3\pi/4=\pi/4$ which makes no sense geometrically since one must first pass by $(0,-1)$ to reach that point while remaining in the domain of $\alpha(\varphi):=C(\varphi)$. I'm not sure how you arrived at that? Measuring from $0$ to $-|\varphi|$ measures the arclength of $\alpha([-|\varphi|,0])$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .