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Recall that the length of a curve $\alpha : [a,b] \rightarrow \mathbb{R}^3$ is given by $L(\alpha) = \int |\alpha'(t)| dt$. Let $\beta(r): [c,d] \rightarrow \mathbb{R}^3$ be a reparametrization of $\alpha$ defined by taking a map $h: [c,d] \rightarrow [a,b]$ with $h(c) = a, h(d)=b$ and $h'(r) \geq 0$ for all $r \in [c,d]$. Show that the arclength does not change under this type of reparametrization.

I believe I have to use the chain rule. I initially set $\beta(r) = \alpha(h(r))$. Then using the chain rule, I get $\beta '(r) = \alpha '(h(r)) \cdot \frac{dh}{ds} (r)$. Does this imply that the magnitudes are the same as well? Because then that would be sufficient to prove that the arclength is the same.

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  • $\begingroup$ Try something like changing variable in $L(\alpha) = \int |\alpha'(t)| dt$ by changing $t$ to $h(r)$ and then apply the reverse chain rule $\endgroup$ Jan 22 '14 at 15:46
  • $\begingroup$ Well the Inverse function theorem plays a role as you can probably identify. look at $\frac{dh}{ds}(r)$ $\endgroup$
    – Vishesh
    Jan 22 '14 at 16:23
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Something like :
$L(\alpha) = \int |\alpha'(t)| dt \overset{t \rightarrow h(r)}{=} \int |\alpha'(h(r))| h'dr \overset{h'(r)\geqslant 0}{=} \int |\alpha'(h(r))h'| dr \overset{inverse\ chain\ rule}{=} \int |\beta'(r)| dr = L(\alpha \circ h)$

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