Unit-speed Reparametrization of a circle

Question

I am studying some differential Geometry from Shifrin's differential Geometry notes.

From what I understand so far it seems like it is possible to unit-speed reparametrize a curve whenever the curve is regular.

In the notes that I am reading the parametrization of the circle is: $\alpha = (a\cos{t},a\sin{t}), \implies \lVert \alpha '(t)\rVert = a$

And then it is mentioned that if we reparametrize the curve by

$\beta(s) = (a\cos(s/a),a\sin(s/a)) \implies \lVert\beta'(s)\rVert=1$

So my question how did the author derive that this exact reparametrization have unit speed?, He could impossibly just have guessed it..

And is there any General way to find a unit speed reparametrization of a parametrized curve?

Answer 1

The answer to your question is at the top of p. 8 of my notes. In the case of the circle as originally parametrized, the arclength, starting at $t=0$, is $s(t)=at$. So $t=s/a$. Thus, $\beta(s) = \alpha(s/a) = \big(a\cos(s/a),a\sin(s/a)\big)$ is a reparametrization by arclength. You can immediately check that $\|\beta'(s)\|=1$, but the general argument is in the notes there.

Answer 2

The speed of a parametrization $r$ is given by $|r'|$ by definition. Intuitively you can think of it in this way: if you visualize your curve in space and assume $t$ to represent time then $r(t+ \Delta t)-r(t)$ is the vector that takes the points on the curve $r(t)$ at time $t$ to the point on the curve $\Delta t$ time later, which is $r(t+ \Delta t)$. If you divide this by $\Delta t$ you obtain a vector $\frac{r(t+ \Delta t)-r(t)}{\Delta t}$ whose norm is just "distance" between two points divided by the time it takes to go from one to anoher. By making $\Delta t$ very small and by taking the norm one obtains the so called speed of the parametrization.

You can look for unitary speed parametrized curves at https://en.wikipedia.org/wiki/Differential_geometry_of_curves#Length_and_natural_parametrization