How (and why) would I reparameterize a curve in terms of arclength?

Intuitively you can think of a parameterized curve as describing how a particle moves through space as a function of time, and you can think of a curve as the path that a particle took as it moved through space. The extra data attached to the curve when it is given a parameterization is, more or less, the velocities of the particle at each point.

But sometimes you only want to think about the intrinsic geometric properties of the curve, such as its curvature. To compute things like the curvature it's convenient to parameterize the curve, but that attaches extra velocity data to the curve which has nothing to do with things like the curvature, so you don't have a guarantee that things you compute from a parameterization are actually properties of the curve. You can guarantee this if you pick a special parameterization, the arc-length parameterization. Intuitively it corresponds to having velocity a unit vector everywhere, and things that you compute from the arc-length parameterization really are properties of the curve and not properties of a particular choice of velocities.

The basic explanation for the idea is that classical (and modern!) differential geometers are interested in intrinsic, rather than extrinsic, properties of smooth manifolds. For example, we would like to understand the geometric properties of curves and surfaces that do not depend on their embedding in Euclidean space.

A curve in space is a smooth map $\gamma:I\to \mathbb{R}^3$ with a non-zero derivative; where, $I$ is an open interval in $\mathbb{R}$. The condition that the derivative of $\gamma$ be non-zero is purely technical. Of course, a curve is a function and thus we lose information if we identify a curve with its image.

For example, the images of the curves $f:(0,2\pi)\to \mathbb{R}^2$ and $g:(0,\pi)\to \mathbb{R}^2$ given by the rules $f(t)=(\cos t,\sin t)$ and $g(t)=(\cos 2t, \sin 2t)$ are equal but the curves $f$ and $g$ (i.e., the functions $f$ and $g$) are not equal. Thus, their domains are different but physically the curve $g$ is a curve with twice the angular velocity (and, hence, twice the speed) of the curve $f$.

The curves $f$ and $g$ in the example above are said to be the reparametrizations of one another. Formally, if $\gamma:I\to \mathbb{R}^3$ is a curve and if $p:I'\to I$ is a smooth map with non-zero derivative, then we say that the $\underline{\textit{function composition}} \,\,\,\gamma\circ p:I'\to \mathbb{R}^3$ is a reparametrization of $\gamma:I\to \mathbb{R}^3$. The condition that the derivative of $p$ be non-zero is required so that the reparametrization of a curve is a curve (under our earlier definition of "curve").

A basic question we may ask is: which properties of curves do not depend on the parametrization? Indeed, as far as intrinsic geometry is concerned, we only care about the image of the curve rather than the curve itself. For example, the velocity of a curve depends on its parametrization (by the chain rule) as does the speed of a curve (i.e., the norm of the velocity).

The following exercise provides an answer to this question and I encourage you to attempt the exercise:

Exercise 1: The length of a curve $\gamma:(a,b)\to \mathbb{R}^3$ is the integral $\int_{a}^{b} \left|\gamma'(t)\right|dt$. Prove that the length of a curve is independent of its parametrization. (Hint: use the chain rule in differential calculus and the change of variables theorem in integral calculus.)

We deduce that the length of a curve is an intrinsic property of the curve. Of course, there are other properties of curves that do not depend on their parametrizations. The curvature is one such property and if you are familiar with this notion, then I encourage you to prove this on your own.

Exercise 2: If $\gamma:I\to \mathbb{R}^3$ is a curve, then the arclength parametrization of $\gamma$ is given by the change of parameter $p^{-1}:I'\to I$ where $p(t)=\int_{a}^{t} \left|\gamma'(t)\right|dt$ and $I'=p(I)$. Prove that the arclength parametrization is indeed a reparametrization of $\gamma$. (The set $I'\subseteq \mathbb{R}$ is an open interval by the inverse function theorem if you wish to be technical.)

Exercise 3: Determine the arclength parametrization for the curve of your question (this should be a simple integral computation).

The crucial observation is that the arclength paramatrization is intrinsic to the curve, i.e., the arclength parametrization does not depend on the initial parametrization of the curve with which we began (this should be obvious but, if not, prove it using Exercise 1 and Exercise 2).

In modern differential geometry, people are interested in those properties of curves and surfaces which depend only on the induced Riemannian metric from the Euclidean space. For example, such properties include the arclength, curvature, torsion, bi-torsion etc.

A famous theorem in differential geometry is the Theorema Egregium of Gauss which asserts that the Gaussian curvature of a surface (i.e., the product of the principal curvatures of the surface) does not depend on the embedding of the surface in Euclidean space but only on the induced Riemannian metric. The analogue of the arclength parametrization in higher dimensions (i.e., the analogue of the arclength parametrization for higher dimensional Riemannian manifolds) is given by the exponential map and normal coordinates.

I hope this helps!