Parametrising smooth regular curves by arclength
Back to Curves and Surfaces homepage
Recall from the previous lecture that we defined what it means to reparametrise a smooth path and studied some examples. We then finished with the definition of a smooth regular curve. Today we will discuss the most important example of reparametrisation, which is reparametrising smooth regular curves by arclength.
Definition. Let \(p : I \rightarrow \mathbb{R}^3\) be a smooth path, and let \(t_0, t_1 \in I\) with \(t_0 < t_1\). The arclength from \(p(t_0)\) to \(p(t_1)\) is given by the formula
\[
s(t_1, t_0) = \int_{t_0}^{t_1} \left| p'(u) \right| \, du
\]
We can also fix \(t_0 \in I\) and measure the arclength along the path \(p\) from the point \(p(t_0)\) as a function of \(t\).
\[
s(t, t_0) = \int_{t_0}^t \left| p'(u) \right| \, du
\]
Arclength does not depend on the parametrisation
It is important to note that the arclength is a property of the track of the curve and does not depend on the choice of parametrisation. To see this, let \(p : I \rightarrow \mathbb{R}^3\) and \(q : J \rightarrow \mathbb{R}^3\) be smooth paths, and let \(t : J \rightarrow I\) be an orientation-preserving reparametrisation. Let \(u_1, u_2 \in J\) and define \(t_1 = t(u_1)\) and \(t_2 = t(u_2)\). Then
\[
\begin{aligned}
\int_{u_1}^{u_2} \left| q'(u) \right| \, du & = \int_{u_1}^{u_2} \left| \frac{d}{dt} p(t(u)) \right| \, du \\
& = \int_{u_1}^{u_2} \left| p'(t(u)) \frac{dt}{du} \right| \, du \quad \text{(by the Chain Rule)} \\
& = \int_{u_1}^{u_2} \left| p'(t(u)) \right| \frac{dt}{du} \, du \quad \text{(since the reparametrisation is orientation-preserving)} \\
& = \int_{t_1}^{t_2} \left| p'(t) \right| \, dt . \quad \text{(by the change of variables formula for integration)}
\end{aligned}
\]
Parametrising a path by its arclength
Now we can define what it means to parametrise a path by its arclength. It turns out (see the note below) that an arclength parametrisation is always regular.
Definition. A smooth path \(p : I \rightarrow \mathbb{R}^3\) is an arclength parametrisation if and only if \(|p'(t)| = 1\) for all \(t \in I\).
Equivalently,
\[
t-t_0 = s(t, t_0) = \int_{t_0}^t \left| p'(u) \right| \, du .
\]
If this is true, then the arclength from \(p(t_0)\) to \(p(t)\) is equal to \(t-t_0\).
Note. Since \(|p'(t)| = 1\) for all \(t \in I\), then an arclength parametrisation is always regular.
Example 1. (The circle of radius r)
In contrast to the earlier lecture, this time we will parametrise the circle by \(p : \mathbb{R} \rightarrow \mathbb{R}^3\) such that
\[
\begin{aligned}
p(t) & = \left( r \cos \left( \frac{t}{r} \right), r \sin \left( \frac{t}{r} \right), 0 \right) \\
\Rightarrow \quad p'(t) & = \left( - \sin \left( \frac{t}{r} \right), \cos \left( \frac{t}{r} \right), 0 \right) \\
\Rightarrow \quad \left| p'(t) \right| & = \sqrt{ \sin^2 \left( \frac{t}{r} \right) + \cos^2 \left( \frac{t}{r} \right)} = 1 .
\end{aligned}
\]
Therefore \(p : \mathbb{R} \rightarrow \mathbb{R}^3\) is an arclength parametrisation.
Example 2. (Helix)
Now we will parametrise the helix by \(p : \mathbb{R} \rightarrow \mathbb{R}^3\) given by the following formula (again, you might like to compare this with our previous parametrisation from the first lecture.
\[
\begin{aligned}
p(t) & = \left( r \cos \left( \frac{t}{\sqrt{r^2 + c^2}} \right), r \sin \left( \frac{t}{\sqrt{r^2+c^2}} \right), \frac{ct}{\sqrt{r^2+c^2}} \right) \\
\Rightarrow \quad p'(t) & = \left( -\frac{r}{\sqrt{r^2+c^2}} \sin \left( \frac{t}{\sqrt{r^2+c^2}} \right), \frac{r}{\sqrt{r^2+c^2}} \cos \left( \frac{t}{\sqrt{r^2+c^2}} \right), \frac{c}{\sqrt{r^2+c^2}} \right) \\
\Rightarrow \left| p'(t) \right| & = \frac{1}{\sqrt{r^2+c^2}} \sqrt{r^2 + c^2} \\
& = 1
\end{aligned}
\]
Therefore \(p\) is an arclength parametrisation of the helix.
Reparametrising by arclength
Given that the above parametrisations of the circle and the helix look different to those from our first lecture, it is natural to ask whether there is a systematic way to take a given parametrisation and reparametrise it so that the new parametrisation is by arclength.
It turns out that this is always possible if the original parametrisation is regular. We will prove this below, but first we will study the following motivating example.
Example. (Reparametrising the circle)
Consider the circle again, but this time with our original parametrisation \(p : \mathbb{R} \rightarrow \mathbb{R}^3\) given by
\[
p(t) = (r \cos t, r \sin t, 0) .
\]
Then the tangent vector is
\[
\begin{aligned}
p'(t) & = (-r \sin t, r \cos t, 0) \\
\Rightarrow \quad \left| p'(t) \right| & = r .
\end{aligned}
\]
The arclength from \(p(0)\) to \(p(t)\) is then
\[
s(t, 0) = \int_0^t \left| p'(u) \right| \, du = \int_0^t r \, du = rt .
\]
This determines \(s = s(t, 0)\) as a function of \(t\). We can rearrange this formula to solve for \(t\) as a function of \(s\)
\[
t(s) = \frac{s}{r} .
\]
Now define a new parametrisation of the circle \(q : \mathbb{R} \rightarrow \mathbb{R}^3\) by
\[
q(s) = p(t(s)) = \left( r \cos \left( \frac{s}{r} \right), r \sin \left( \frac{s}{r} \right) , 0 \right) .
\]
Then \(q(s) = p\left( \frac{s}{r} \right) = p(t(s))\) and so \(q\) is a reparametrisation of \(p\) (see the lecture on reparametrisations if you need to remind yourself of the definition). Moreover, we also have
\[q'(s) = \left( -\sin\left(\frac{s}{r} \right), \cos \left( \frac{s}{r} \right), 0 \right), \quad \text{therefore} \quad \left| q'(s) \right| = 1\]
and so \(q\) is an arclength parametrisation of the circle.
In summary, we started with a parametrisation \(p(t)\), we solved for the arclength \(s = s(t)\) as a function of \(t\), then inverted the formula to find \(t = t(s)\) as a function of \(s\), and then reparametrised to get \(q(s) = p(t(s))\). We will now show that if the original parametrisation \(p(t)\) is regular, then this process always gives us an arclength parametrisation.
Lemma. (Reparametrising by arclength)
Let \(p : I \rightarrow \mathbb{R}^3\) be a smooth regular path. Then there exists an interval \(J\) and a reparametrisation \(t : J \rightarrow I\) such that \(q : J \rightarrow \mathbb{R}^3\) given by
\[
q(s) = p(t(s))
\]
is a parametrisation by arclength.
Proof.
Choose a point \(t_0 \in I\). Then the arclength from \(p(t_0)\) to \(p(t)\) is given by
\[
s(t) = \int_{t_0}^t \left| p'(u) \right| \, du .
\]
Since \(p\) is regular, then \(|p'(t)| > 0\) for all \(t \in I\), therefore
\[
s'(t) = \frac{d}{dt} \int_{t_0}^t \left| p'(u) \right| \, du = \left| p'(t) \right| > 0
\]
by the Fundamental Theorem of Calculus. Therefore the image of \(s\) is an interval \(J \subset \mathbb{R}\) and so the arclength defines an increasing function \(s : I \rightarrow J\).
The Inverse Function Theorem then guarantees the existence of a smooth inverse \(t : J \rightarrow I\) with
\[
\frac{dt}{ds} = \frac{1}{\frac{ds}{dt}} = \frac{1}{|p'(t)|} > 0 .
\]
Now define \(q : J \rightarrow I\) by
\[
q(s) = p(t(s)) .
\]
Therefore \(q : J \rightarrow \mathbb{R}^3\) is an orientation-preserving reparametrisation of \(p : I \rightarrow \mathbb{R}^3\). Moreover, we have
\[
q'(s)= \frac{d}{ds} p(t(s)) = \frac{dp}{dt} \frac{dt}{ds} = \frac{p'(t)}{|p'(t)|}
\]
therefore \(|q'(s)| = \frac{|p'(t)|}{|p'(t)|} = 1\), and so \(q\) is parametrised by arclength. q.e.d.
Next time
We will do some more examples of reparametrising by arclength.
