# Can anyone explain to me this theorem? (Related to the chain rule for vector functions)

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.

• FYI you can use $$[...]\tag{1}$$ to get $(1)$ to appear at the end of an equation line. – DMcMor Apr 19 at 19:06

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?

• good explanation thank you @Keshav – Yassir Apr 19 at 20:08