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I am interested to know about when it is possible to parameterize a curve by arc length. In particular, as mentioned on this question, we usually only consider curves $\alpha$ with $|\alpha'(t)| > 0$ to ensure that we can parameterize $\alpha$ by arc length.

However, this condition is not necessary to ensure an arc length parameterization is possible: for instance, the curve $\alpha: R \to R^2, \alpha(t) = (t^3, t^3)$ has $|\alpha'(t)| = 0$, but can be parameterized by arc length with $\alpha(s) = (s/\sqrt 2, s/\sqrt 2)$.

I suspect one case where this is not possible is given by a cuspidal cubic, i.e. $\alpha : R \to R^2$ defined by $\alpha(t) = (t^3, t^2)$. If we suppose we did have an arc length parameterization -- from, say, the point $t_0 = 0$ -- then, informally, as $s$ passed through, $\alpha'(t)$ would need to flip its direction discontinuously (since its magnitude is constant). Consequently, it seems there does not exist a (infinitely) differentiable $\alpha : R \to R^2$ such that $trace(\alpha(s)) = trace(\alpha(t))$ with $s$ an arc length parameter.

Is it possible to make this line of reasoning precise? Furthermore, are there necessary conditions to ensure the existence of an arc length parameterization that might explain the discrepancy between these two examples?

12qu
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  • Are arc length parameterizations required to be differentiable? –  May 23 '17 at 16:02
  • Quite possibly not in general - but I am going off Do Carmo's definition, which is essentially that $\alpha$ is arc length parameterized if $|\alpha'(s)| = 1$. This would require at least a first derivative everywhere (although I can see that if that is all we need, then my argument is inadequate). His definition of a curve $\alpha$ more generally requires infinite differentiability. – 12qu May 23 '17 at 16:07
  • You may be interested in this: https://en.wikipedia.org/wiki/Semicubical_parabola#History – Michael Hoppe May 23 '17 at 18:26

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