Calculate length of involute without calculus

Question

Is it possible to calculate the length of an involute of a circle without using calculus, and how?

Say we’re interested in the involute length for $\theta$ between $0$ and $2 \pi$.

Answer 1

I assume that the generating circle has radius $1$.

The idea, as illustrated here that I reproduce for a fixed $n$ :

is to replace the circle by an insribed $n$-gon. In this way, (refering to the image of the tethered cow), you can approximate an involute of circle by the union of circular arcs with radii $r_k$ in arithmetic progression where

$$r_k = k \times \text{length of the side of the n-gon} = k \ 2 \sin \tfrac{\pi}{n}$$

The length of the $k$-th circular arc is obtained from $r_k$ by multiplying it by angle $\frac{2 \pi}{n}$.

Therefore, the total length is :

$$\sum_{k=1}^n k (2 \sin \tfrac{\pi}{n}) \frac{2 \pi}{n}$$

$$=(2 \sin \tfrac{\pi}{n})\frac{2 \pi}{n}\sum_{k=1}^n k$$

$$=2 \underbrace{\sin \tfrac{\pi}{n}}_{\text{equiv. to }\tfrac{\pi}{n}} \frac{2 \pi}{n}\frac{n(n+1)}{2} $$

whose limit when $n \to \infty$ is $2 \pi^2$, as awaited.

Remark : we have a curve "dissected" into $n$ curves ; we have postulated that the limit of sum of lengths is the length of the limit curve. This is not a theorem. Here, we are in a favorable case. It may not be the case in other situations : there are known "paradoxes" built on this "non-theorem".