Unit velocity parametrization of a parabola.

Question

I have a parametrization for the parabola $y = x^2$ given by: $$x(t) = t$$ $$y(t) = t^2$$ However, this doesn't have constant unit velocity, since $$\sqrt{x'(t)^2 + y'(t)^2} = \sqrt{1 + 4t^2} \neq 1\,.$$ Is there a way to fix this so that the velocity is 1 for all $t$?

Answer 1

Yes...in theory. If you define the arclength function $$ s = f(t) = \int_{0}^{t} \sqrt{1 + 4\tau^{2}}\, d\tau = t\sqrt{1 + 4t^{2}} + \tfrac{1}{4}\ln\left(t + \sqrt{1 + 4t^{2}}\right) $$ and put $t = g(s) = f^{-1}(s)$ (which in general is "practically impossible"), then $$ \bigl(g(s), g^{2}(s)\bigr) $$ is a unit-speed parametrization. (The same trick works—subject to the same obstacle—for an arbitrary smooth curve whose speed is non-vanishing.)