In trying to write a nice proof of the derivatives of $\sin(x)$ and $\cos(x)$, I encountered a serious problem, namely that I have never seen a proper definition of the notion of arc length. Based on visual intuition (for whatever that means), I tried to argue as follows:
Consider the following diagram:
The chord $AC$ is shorter than the red arc which is again (by visual intuition) shorter than the path $ABC$. This means that $$2s<arc<2d$$
Note that $|OD|=\sqrt{1-s^2}$ by the Pythagorean theorem. Now, since $\Delta ABD$ and $\Delta OAD$ are similar, we see that $$\frac sd=\frac{\sqrt{1-s^2}}1$$ so dividing the chord length $2s$ by $2s,arc$ and $2d$ considering inequalities from before we then have $$1>\frac{2s}{arc}>\frac{2s}{2d}=\sqrt{1-s^2}$$ and it follows that $$\frac{2s}{arc}=\frac{chord}{arc}\longrightarrow 1\quad\text{when}\quad chord,arc\longrightarrow 0$$
Problem: Since the inequality $arc<2d$ was based solely on intuition, I could just as well have claimed that $\frac{chord}{arc}\longrightarrow 1$ by intuition in the first place anyway. Perhaps my intution about the inequality is stronger than my intuition about the limit, but that does not make it more rigorous …
Question: How can we define the notion of arc length and based on that show rigorously that $arc<2d$?
Best Answer
One way to define the arc length of a curve $\gamma:[a,b]\to\mathbb R^2$ is by considering partitions of $[a,b]$, as in Riemann integration (read this Wikipedia article first): if $P(x,t)$ is a tagged partition of $[a,b]$, then $L_\gamma(P)$, the length of $\gamma$ with respect to the partition $P$, is $\sum_{i=1}^{n-1}|\gamma(t_i)-\gamma(t_{i-1})|$.
Definition: If $\underset{P}{\lim \sup} L_\gamma(P)$ exists, then $L_\gamma = \underset{P}{\lim \sup} L_\gamma(P)$ is the length of the curve.
Now consider the portion of the red arc between the point $A$ and the line $BD$. Using the above definition, we can see that the length of this arc lies between $AD$ and $AB$: if $\gamma(t_{i-1})$ and $\gamma(t_i)$ are points on the curve, we can project them perpendicular to $AD$ onto the points $r_{i-1}$ and $r_i$ on $AD$, and $s_{i-1}$ and $s_i$ on $AB$. And then we have $|r_i-r_{i-1}| < |\gamma(t_i)-\gamma(t_{i-1})| < |s_i-s_{i-1}|$, by simple geometry.
Therefore $L_{AD}(P) < L_\gamma(P) < L_{AB}(P)$ (taking some liberties with the notation). And in the limit we get:
$$\underset{P}{\lim \sup}L_{AD} \le \underset{P}{\lim \sup}L_\gamma(P) \le \underset{P}{\lim \sup}L_{AB}$$
In other words: $$AD \le L_\gamma\le AB$$
To show strict inequality would not be too difficult $-$ for instance, if we are more than halfway to the line $BD$, then there is a constant $\mu < 1$ such that $|\gamma(t_i)-\gamma(t_{i-1})| < \mu|s_i-s_{i-1}|$. This would give $L_\gamma(P) < \frac12(1+\mu)L_{AB}(P)$, so $L_\gamma \le \frac12(1+\mu)AB$.