nLab circumference of a circle

Idea

The circumference of a circle (the length of the curve that is the circle) as discussed in geometry.

Definition/proposition and proof

Proposition

Depending on which circle constant you use, given a radius $r$ of a circle $\mathcal{C}$ in the Euclidean plane $\mathbb{R}^2$, the circumference of a circle is either $C(r) = \tau r$ or $C(r) = 2 \pi r$.

Proof by integration

Proof

In this proof, we are using the circle constant $\tau = 2 \pi$.

Given any Euclidean plane $\mathbb{R}^2$, one could select an orthonormal basis on $\mathbb{R}^2$ by postulating an origin $0$ at the center of the circle $\mathcal{C}$ and two orthonormal vectors $\hat{i}$ and $\hat{j}$. The circle $\mathcal{C}$ could be parameterized by a function $\overrightarrow{r}:[0, \tau] \to \mathbb{R}^2$ defined as

$$\overrightarrow{r}(\theta) \coloneqq r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j}$$

Then the circumference of $\mathcal{C}$ is given by the following integral:

$$C(r) = \int_{0}^{\tau} \vert \overrightarrow{r}(\theta) \vert d \theta$$

which evaluates to

$$C(r) = \int_{0}^{\tau} \vert r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j} \vert d \theta = \int_{0}^{\tau} r((\cos(\theta))^2 + (\sin(\theta))^2) d \theta = \int_{0}^{\tau} r d \theta = \tau r$$

Proof by limits of regular polygons

Proof

In this proof, we are using the circle constant $\tau = 2 \pi$.

The perimeter of a regular polygon $\mathcal{P}_n$ (strictly speaking, a regular polygonal line) with $n$ sides and circumradius $r$ is given by the sequence of functions $P_\mathcal{P}:\mathbb{N} \to (\mathbb{R} \to \mathbb{R})$

$$P_\mathcal{P}(n)(r) = r (2 n) \sin\left(\frac{\tau}{2n}\right)$$

which embeds in the $\mathbb{R}_+$-action $P_\mathcal{P}^\prime:\mathbb{R}_+ \to (\mathbb{R} \to \mathbb{R})$, defined as

$$P_\mathcal{P}^\prime(n)(r) = r (2 n) \sin\left(\frac{\tau}{2 n}\right)$$

The limit of $P^\prime$ as $n$ goes to infinity is the circumference of a circle with radius $r$:

$$C(r) = \lim_{n \to \infty} P_\mathcal{P}^\prime(n)(r) = \lim_{n \to \infty} r (2 n) \sin\left(\frac{\tau}{2 n}\right) = r \lim_{m \to 0} \frac{\sin(\tau m)}{m} = r \lim_{m \to 0} \frac{\partial_m \sin(\tau m)}{\partial_m m} = r \lim_{m \to 0} \frac{\tau \cos(\tau m)}{1} = \tau r$$

See also

• pi

• circle

• unit circle

• arc length

• area of a circle

• Euclidean space