# Changing parametrization with regard to arc length - is it worth it?

If you have found a parametrization $\vec \alpha(s)$ of a curve $C$ for which $\int \lVert \vec \alpha\,'(s)\rVert \mathrm ds$ cannot be expressed in terms of elementary functions, does it make sense to search for a different parametrization?

This question was inspired by this specific example. I feel like the answer is no, and here is a sketchy proof:

Suppose we have two parametrizations of the curve $C$, say $\vec \alpha (s)$ and $\vec \mu (t)$ valid on $s\in[a,b]$ and $t\in[c,d]$, with continuous derivatives. Then there is some function $f$ such that $\vec \alpha(s)=\vec\mu(f(s))$. Now if $\vec \alpha\,'(s)$ and $\vec \mu \, '(t)$ are never zero (which I believe we can assume without loss of generality) then we have $\vec \alpha \, ' (s)=\vec \mu\,'(f(s))f'(s)$ (the existence of $f'(s)$ is guaranteed in this case).

Furthermore, since $f$ is differentiable it is continuous. Now the fact that $\vec \alpha$ and $\vec \mu$ are differentiable implies continuity of each function $\alpha_i$, where $\vec \alpha(s)=(\alpha_1,\,\alpha_2,\,\dots,\alpha_n)$. Similarly, each $\mu_i$ is continuous, and for all $t$ at least one of $\mu_i(t)\ne 0$. Hence we have that $f'(s)=\frac {\alpha_i'(s)}{\mu_i'(f(s))}$.

There must be some overlap in non-zero functions, however. I'm not sure how to write this formally, but for example: if we have that $\mu_1'(f(s))$ is the only nonzero $\mu_i'(f(s))$ on some interval $(x,y)$, then $f'(s)=\frac{\alpha_1'(s)}{\mu_1'(f(s))}$ on $(x,y)$. Now suppose $\mu_1'(f(s))=0$ outside this interval. Then on $[y,z)$ there must be some $i=k$ such that $\mu_k'(f(s))$ is nonzero. Since $\mu_k'(f(s))$ is continuous, it must be nonzero on an open interval (including $y$), and therefore $\mu_1'(f(s))$ is not the only nonzero function on $(x,y)$.

This overlap allows us to conclude that $f'(s)$ must be continuous (that was the whole point of the above argument) and nonzero. Thus we either have that $f'(s)>0$ on $[a,b]$ or $f'(s)<0$ on $[a,b]$. Either way, we have that $$\int \lVert \vec \alpha\,'(s)\rVert \mathrm ds =\pm\int \lVert \vec \mu\,'(f(s))\rVert f'(s)\mathrm ds=\pm\int\lVert\vec\mu\,'(t)\rVert\mathrm dt$$ and so if the arc length using the parametrization $\vec \alpha$ cannot be expressed in terms of elementary functions, the new parametrization cannot either.

I'm not positive we can assume that $\vec\alpha\,'(s)$ and $\vec \mu\,'(t)$ are never zero without loss of generality, but it seems to make sense. Actually, looking at the result I believe we could argue that it wouldn't make a difference if they were equal to zero at some points.

It also occurred to me that the argument yielding continuity of $f'(s)$ was not really needed, it was obvious that $f'(s)$ was at least piecewise continuous and we can reach our conclusion with that weaker result as well.

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You shouldn't be able to do that without assuming that $\alpha$ and $\mu$ are themselves expressed in terms of elementary functions (otherwise there are trivial examples), and I don't see where you're using such an assumption in your sketch. – Henning Makholm Sep 3 '12 at 14:21
@HenningMakholm I will add this assumption - for my purposes, it would be necessary. – process91 Sep 3 '12 at 14:22
The greater problem, as far as I can see, is that you don't seem to be using the assumption -- therefore your argument cannot be valid. In fact, I don't see how you get from the body of the arguments (which seems to be entirely about continuity and derivatives) to the end conclusion that some particular function is not elementary. There's a coordinate change from $s$ to $t$ hidden in your last equals sign, and a priori it might be that this coordinate change just happens to correct for the non-elementariness of the length as a function of $s$. – Henning Makholm Sep 3 '12 at 14:29
@HenningMakholm The assumption is that $\int \lVert \vec \alpha ' (s) \rVert ds$ is not elementary. I conclude that for any other parametrization $\vec \mu$ we have that $\int \lVert \vec \alpha ' (s) \rVert \mathrm ds = \pm\int \lVert \vec \mu ' (t) \rVert \mathrm du$ and so if the latter is elementary then the former would be as well. Hence $\int \lVert \vec \mu ' (t) \rVert \mathrm du$ cannot be expressed using elementary functions. My interest is in the integrals, not the initial parametrizations. – process91 Sep 3 '12 at 14:33
Let's put $A=\int\|\alpha'(s)\|\,ds$ and $M=\int\|\mu'(t)\|\,dt$. Then its true enough that $M(t)=A(f^{-1}(t))$. But just because $A$ is non-elementary doesn't mean that its composition with $f^{-1}$ is also non-elementary. You haven't argued whether $f$ or $f^{-1}$ are elementary or not, but that doesn't even matter, because it is possible for the composition of a non-elementary function and an elementary one to be elementary. – Henning Makholm Sep 3 '12 at 14:43