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I apologize if this is trivial but I have not been able to figure it out. For a curve $\sigma(t)$, I have a definition for arc length:

$$s(t)=\int_{t_0}^t |\sigma'(t)|dt$$

We reparameterize a curve $\sigma$ by observing that $s$ has an inverse, $t(s)$, and the resulting reparameterized curve has underlying assignment $s \mapsto t \mapsto \sigma(t)$.

A curve is a unit speed curve if

$$\forall t\qquad |\sigma'(t)|=1$$

Thanks

Zain Patel
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A__A__0
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    This is answered here http://math.stackexchange.com/questions/1634974/prove-that-for-any-piecewise-smooth-curve-it-is-possible-to-find-the-parametrisa/1635008#1635008, as a byproduct to the answer. – Thomas Jul 07 '16 at 16:10
  • Thank you, I had not seen that answer. – A__A__0 Jul 07 '16 at 18:02

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If $t=s$ your definition of arc-length becomes $$ s=\int_{s_0}^s|\sigma'(u)|\,du $$ Differentiating w.r.t $s$ we get $$ 1=|\sigma'(s)| $$ which shows when parametrized using $s$, the curve becomes unit speed.

Jack's wasted life
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  • Thank you for your response. I think this makes sense, but I am confused why you wrote $u$ for the variable - shouldn't it be s? My understanding is that the integration that defines the arc length function should always be with respect to a variable that represents the domain of the function. Then again, maybe that is what your $u$ does, and you chose to write $u$ since $s$ was already bound as one of the limits of integration. – A__A__0 Jul 07 '16 at 18:02
  • @A__A__0 $u$ is a dummy variable. $\int_0^xf(x),dx$ is considered an abuse of notation, it should be written as $\int_0^xf(u),du$. $u$ can be replaced with any dummy variable of your choice. – Jack's wasted life Jul 07 '16 at 18:16
  • Sorry this is probably something I've forgotten from basic calculus but why is $\frac{d}{ds} \int |\frac{d}{ds}\sigma(u)|du=|\frac{d}{ds}\sigma{u}$? I thought the varable we're deriving with respect to would have to be the same as the one in the integral for them to cancel like this. – A__A__0 Jul 08 '16 at 00:26
  • Is the point of the change to $u$ in this case that the variable $u$ is in fact 'really' $s$ but that we change it so that the bindings are correct? – A__A__0 Jul 08 '16 at 00:27
  • Also I do not understand why we can assume $t=s$. By the reparameterization we know that we are precomposing a function which maps $s\mapsto t$. Sorry for the trouble. – A__A__0 Jul 08 '16 at 00:53
  • @A__A__0 Your "Is the point..." comment is correct. $t$ was an arbitrary parameter. I just for the sake of simplicity assumed $t=s$. If you want $s$ to be a function of $t$, you can do that. Let $s=f(t)\implies\int_{s_0}^s|{d\over du}\sigma(f^{-1}(u))|du\implies 1=|{d\over ds}\sigma(f^{-1}(s))|$ which was to be shown since with arclength as parameter the curve is $\sigma\cdot f^{-1}$. – Jack's wasted life Jul 08 '16 at 03:22
  • wait, int $s$ a function of... itself? i mean $s(s)$ somehow, since we get $s$ as a parameter on the right side. – Joaquin Brandan Apr 21 '20 at 17:50