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Theorem:

let $X:I\to \mathbb R^n$ be a curve whose speed $v(t)$ is $>0$ for all $t$ in the interval of definition, Let $\text{ (1)}$ $$s(t)=\int_a^tv(u)\,du \text{ (2)}$$ and $t=f(s)$ be the inverse function of $s(t)$, then the curve given by $\text{ (3)}$ $$s\to Y(s)=X(f(s)) \text{ (4)}$$ is parametrized by arc length, and $Y'(s)$ is perpendicular to $Y''(s)$ for each value of $s$ $\text{ (5)}$

I'm stuck with this theorem for $2$ days, and honestly, I wanna just skip it.

First, I want to know if there exists a well-known name to this theorem so I can search for it somewhere.

I understand the first and the second line, but for the third line, I don't know what is going on in the notation for the inverse function, and for the fourth line I don't know why the author decomposed the curve $X$ with $f(s)$, (I mean what is the reason?)

I'm okay with the last line.

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  • $\begingroup$ FYI you can use $$[...]\tag{1}$$ to get $(1)$ to appear at the end of an equation line. $\endgroup$ – DMcMor Apr 19 at 19:06
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You have a curve $X(t)$ parametrized by the variable $t$ in the interval $I = [a,b]$, but you want to reparametrize the curve, using a new parameter $s$, which lives in a new interval $J$, so that if changing $s$ by 1 unit, moves along the curve by 1 distance unit (measured by the arc length formula).

This theorem is maybe known as the "arc-length parametrization" for curves.

So I want to have a different parametrization $Y(s)$, which traces over the same curve $X$, and has a different domain $s\in J$.

The author defines a map $$s:I\to J \\ t\mapsto s(t).$$ The inverse of this map can be written as $$s^{-1}:J\to I \\s\mapsto t(s)$$ but instead of writing $t(s)$, which is maybe a shorthand, you can write explicitly that $t\in I$ is some function of $s\in J$, and call that function $f(s)$. In other words $$s\mapsto f(s) = t(s) = s^{-1}(s)$$

Finally put this together to get a curve $$Y: J \xrightarrow{f} I\xrightarrow{X}\mathbb R^n \\ Y = X\circ f \\ Y(s) = X(f(s)).$$

Does that help clarify what's going on?

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  • $\begingroup$ good explanation thank you @Keshav $\endgroup$ – Yassir Apr 19 at 20:08

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