Is there a correct mathematical way to prove the arc length integral from the pythagorean theorem?

Question

It might not be clean what do I mean by the question, but I'm trying to explain it:

By the pythagorean theorem: $$\mathrm{d}S=\sqrt{(\mathrm{d}x)^2+(\mathrm{d}y)^2}$$ If we factor out d$x$: $$\mathrm{d}S=\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\mathrm{d}x$$ So the arc length is: $$S=\int\limits_a^b \sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\mathrm{d}x$$

But it does not seem like a real proof.

Answer 1

Let $a=x_0< x_1 < \cdots < x_n =b$ be a partition of $[a,b]$.

Define arc length as;

\begin{equation} S_n = \sum_{i=1}^n \sqrt{(x_i-x_{i-1})^2+(f(x_i)-f(x_{i-1}))^2} \end{equation}

Recall Mean Value Theorem; \begin{equation} \frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}} = f'(\tilde{x_i}) \iff f(x_i)-f(x_{i-1}) = f'(\tilde{x_i})(x_i-x_{i-1}) \end{equation} for some $x_{i-1}\leq\tilde{x_i}\leq x_i$. Now, rewrite $S_n$; \begin{equation} S_n = \sum_{i=1}^n \sqrt{ (x_i-x_{i-1})^2+f(\tilde{x_i})^2(x_i-x_{i-1})^2} =\sum_{i=1}^n \sqrt{1+f'(\tilde{x_i})^2}(x_i-x_{i-1}) \end{equation} Which is a Riemann Sum. Now take limit; \begin{equation} \lim_{n\to\infty}S_n = \int_a^b\sqrt{1+f'(x)}dx \end{equation}

Note that I replaced $y$ with $f(x)$. Also, when taking limit, I identify partition with number of elements $n$, but we actually require $(x_i-x_{i-1})\to 0$.

Answer 2

This is the heuristic way of doing things---which nonetheless helps a lot in some cases. The problem here is that you are considering $dx$ and $dy$ as 'infinitesimal' quantities, which is not so rigorous. Some problems: suppose $dx$ to be positive. Then the integral is actually a finite sum and the 'length' of the curve is only approximated by the sum of the lengths of straight segments. Then $dx$ should be something positive but less than all the positive numbers, which gives a contradiction. Hence $dx = 0$, which does not allow you to divide by $dx$. In all cases something goes wrong.

In general the formula you state is formalized using a metric and the concept of derivative, which involves a limit and is formal from a topological point of view. If $\gamma \colon [a,b] \rightarrow \mathbb{R}^n$ is (at least) a $C^1$ curve, then its length is given by $$\mathcal{L}_a^b(\gamma):=\int_a^b ||\gamma'(t)||_g\ dt.$$ The metric used does not need to be the standard one, that is why I put a $g$ as a lower script---it denotes a generic metric on $\mathbb{R}^n$. In the particular case where $g$ is standard, $n=2$ and $\gamma(t) = (t,f(t)), \ f\colon [a,b] \rightarrow \mathbb{R}$ (as in your example) you get your formula $$\int_a^b \sqrt{1+f'(t)}\ dt.$$ About technical details: https://en.wikipedia.org/wiki/Arc_length

Answer 3

$dS=\lim_{x\rightarrow x_o}(\sqrt{(x-x_o)^2+(y(x)-y(x_o))^2})$

$dS=\lim_{x\rightarrow x_o}((x-x_o)\sqrt{1+(\frac{y(x)-y(x_o)}{x-x_o})^2})$

$dS=\lim_{x\rightarrow x_o}(x-x_o)\lim_{x\rightarrow x_o}\sqrt{1+(\frac{y(x)-y(x_o)}{x-x_o})^2}$

$dS=dx\lim_{x\rightarrow x_o}\sqrt{1+(\frac{y(x)-y(x_o)}{x-x_o})^2}$

$dS=dx\sqrt{1+(\frac{dy}{dx})^2}$

Then integrate both sides

$S-0=S=\int_a^b{\sqrt{1+(\frac{dy}{dx})^2}}dx$