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?
Best Answer
$$\sqrt{dx^2+dy^2} = \sqrt{(x(t+h)-x(t))^2 + (y(t+h)-y(t))^2} = \sqrt{x'(t)^2 dt^2 + y'(t)^2 dt^2 } = \sqrt{ 1 + \frac{y'(t)^2}{x'(t)^2}} x'(t)^2 dt^2$$
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 $y(t) = y(x)$.