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Suppose that $P:[a,b]\to \mathbb{R}^n$ and $Q:[c,d]\to \mathbb{R}^n$ be two parameterizations of the same continuously differentiable curve $\Gamma$. Can some one give a hint on how to prove that the length of the curve $\Gamma$ is independent of both $P$ or $Q$? If you can also point out the important intrinsic things one should remember when dealing with curves, it would be great. Thanks.

Kumara
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    its a substitution in the integral – Dominic Michaelis Mar 06 '13 at 15:00
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    @DominicMichaelis: $P$ is not necessary a reparametrization of $Q$. In fact, it is not obvious; the only reference I have is in French, I will try to sum up the main idea later. – Seirios Mar 06 '13 at 15:53
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    @DominicMichaelis: If I make the substutuition $t=P^{-1}(Q(s))$ in the integral $\int_a^b|P'(t)|dt$, I would get $\int_c^d|P'(P^{-1}(Q(s)))|(P^{-1})'(Q(s))Q(s)ds$. How can I prove the above equal to $\int_c^d|Q'(t)|dt$? Am I proceeding in the right direction? Or do you mean some other substituition? – Kumara Mar 07 '13 at 13:10

1 Answers1

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Let $\Gamma$ be a curve. Define the length $\ell g(\Gamma)$ of $\Gamma$ as the supremum of $\displaystyle \sum\limits_{i=1}^n d(A_i,A_{i+1})$ over the partition $A_1,...,A_{n+1}$ of $\Gamma$ (in particular, the sequence $(A_i)_i$ is monotonic along $\Gamma$).

Now, if $\varphi$ is a regular parametrization of $\Gamma$, you can show that $\displaystyle \ell g(\Gamma)= \int_0^1 ||\varphi'(t)||dt$.

Therefore, the integral does not depend on the regular parametrization you choose, since $\ell g(\Gamma)$ is here geometrically defined.

Seirios
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