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In differential geometry, you are often asked to reparameterize a curve using arc-length. I understand the process of how to do this, but I don't understand what we are reparameterizing from.

What is the curve originally parameterized by (before we REparameterize it by arc-length)?

mmaclaren
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2 Answers2

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It is parameterized by the parameter that appears in the parametric equations given to you !

It is often the case that there is a natural way to define a curve, be it from its geometric definition or from a convenient coordinate system, and the parameter stands out.

In the case of trajectories, it can be time.

Let's take the case of Viviani's curve: the intersection of a sphere and a cylindre of half the diameter, tangent to it.

In Cartesian coordinates, we have an implicit system of equations

$$\begin{cases}x^2+y^2+z^2=4,\\(x-1)+y^2=1.\end{cases}$$

Then letting $x-1=\cos t,y=\sin t$ by educated guess, we have

$$z=\pm\sqrt{4-x^2-y^2}=\pm\sqrt{2-2\cos t}=2\sin\frac t2.$$

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It's completely arbitrary.

For example the "standard" parameterization of the unit circle is $\gamma(t) = [\cos(t),\sin(t)]$. But you could also sweep out a circle using the curves

  • $\gamma(t) = [\cos(t^2), \sin(t^2)]$, which "speeds up" as you draw it;
  • $\gamma(t) = [\cos(\sqrt{t}), \sin(\sqrt{t})]$, which "slows down" as you draw it;
  • $\gamma(t) = [t, \sqrt{1-t^2}]$, which moves at constant speed in the $x$ direction while speeding up and slowing down in the $y$, etc etc etc.

If you have an arbitrary curve $\gamma(t):\mathbb{R}\to\mathbb{R}^n$, nothing is generally known in advance about the parameterization; the "speed" at which changing $t$ sweeps out the curve, measured by $\|\gamma'(t)\|$, could be arbitrary fast in some places in arbitrarily slow in others. The only assumption that's relatively common is to assume that the curve is regular: $\|\gamma'(t)\|>0$, which rules out some pathologies like cusps in the curve.

user7530
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