Deriving the formula for arc length in $\mathbb{R}^2$

I've seen the proof the arc length formula in $$\mathbb{R}^2$$ in both polar and standard cartesian coordinates before, but I saw this interesting formula while piecing through my brother's calculus textbook the other day, and I actually can't follow it.

For the proof, they construct a finite number of right-triangles along their curve (smooth-continuous), and call $$s$$ the hypotenuse, so $$s^2=x^2+y^2$$. Then, they take the differential and consider $$s$$ as $$x$$ and $$y$$ become infinitesimal. So, they say $$(ds)^2=(dx)^2+(dy)^2$$. It seems like kind of an abuse of notation, but I follow the argument thus far. Here's what I can't follow:

If $$S$$ denotes the entire arc length,

$$S=\int ds= \int \sqrt{(dx)^2+(dy)^2}=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$ which is of course the standard formula for arc length. My question is, where does that last equality come from? Specifically, how do we get

$$\int \sqrt{(dx)^2+(dy)^2}=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$

I mean, algebraically, if you treat $$(dy)^2$$ and $$(dx)^2$$ purely symbolically, like they're part of some division ring, I think you can force one expression into the other, but obviously differentials aren't just symbolic expressions. So, my question is, how do we get to that final equality following this proof?

• I think you are missing a $dx$ on the right hand side of your two last equations. Jun 22, 2023 at 18:47
• This is not a proof. It doesn't even have a definition of arclength Jun 22, 2023 at 19:01

$$\sqrt{dx^2+dy^2} = \sqrt{(x(t+h)-x(t))^2 + (y(t+h)-y(t))^2} = \sqrt{x'(t)^2 dt^2 + y'(t)^2 dt^2 } = \sqrt{ 1 + \frac{y'(t)^2}{x'(t)^2}} x'(t)^2 dt^2$$
Let $$x(t) =t$$, $$y(t) = y(x)$$
$$=\sqrt{1+y'(x)^2} dx^2$$
So when they write your equation in doubt, they reparameterize implicitly as $$x(t)=t$$ and $$y(t) = y(x)$$.
This is a definition, not a proof, however for a $$C^1$$ arc defined on $$[0,1]$$ you can show that $$\sum_{k = 1}^n \| \gamma(k/n) - \gamma((k-1)/n)\|_2 \to \int_0^1 \|\gamma'\|_2$$