TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Every circle has circumference 2 pi r and circumference-to-diameter ratio pi

Statement

For every centre $c \in R^{2}$ and radius $r > 0$ , the once-traversed circle has circumference

$C (r) = 2 π r .$

Since its diameter is $d (r) = 2 r$ , one has $C (r) / d (r) = π$ .

Facts & Assumptions

Given: A centre $c \in R^{2}$ , a real $r > 0$ , and the once-around path $γ (t) = c + r (\cos t, \sin t)$ on $[0, 2 π]$ .

[L1]

Circumference is the length of this once-around path, and diameter is $d (r) = 2 r$ (Circular arcs, circumference as arc length, and diameter).

[L2]

Vector differentiation is componentwise (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[L3]

$(\sin t)^{'} = \cos t$ , $(\cos t)^{'} = - \sin t$ , and $\sin^{2} t + \cos^{2} t = 1$ (The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine).

[L4]

A $C^{1}$ path has length equal to the integral of its speed (If $γ : [a, b] \to R^{n}$ is continuous, differentiable on $(a, b)$ , and $γ^{'}$ extends continuously to $[a, b]$ , then $L (γ) = \int_{a}^{b} ∥ γ^{'} (t) ∥_{2} d t$ ).

[L5]

The integral of a constant $q$ on $[a, b]$ is $q (b - a)$ (If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ ).

[L6]

Every once-traversed unit semicircle has length $π$ (The arc length of a unit semicircle is pi).

Proof

technique · direct

1.1

By [L2] and [L3], $γ^{'} (t) = r (- \sin t, \cos t)$ and $∥ γ^{'} (t) ∥_{2} = r$ , since $r > 0$ .

givenL2L3algebra

2.1

By [L1], [L4], and [L5], $C (r) = L (γ) = \int_{0}^{2 π} r d t = 2 π r$ .

step 1.1L1L4L5

3.1

Because $r > 0$ , the diameter $d (r) = 2 r$ is nonzero, and step 2.1 gives $C (r) / d (r) = (2 π r) / (2 r) = π$ .

givenstep 2.1L1algebra

4.1

At $r = 1$ , step 2.1 gives circumference $2 π$ , agreeing with the sum of the two semicircle lengths from [L6].

step 2.1L6algebra∎

Depends on

Used by

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53 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis II, section 11.4.3 (standard reference, not scraped)