Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
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For a C1 path the arc-length accumulation function has derivative equal to speed

Statement

Let a<b, let γ:[a,b]→Rn be C1, and let

sγ(t):=L[a,t](γ∣[a,t]).

Then sγ is differentiable on [a,b] in the relative sense and

sγ′(t)=∥γ′(t)∥2.

At a and b these are the relative one-sided derivatives.

Facts & Assumptions

Given: The C1 path in the Statement.

[L1]

The arc-length function is sγ(t)=L[a,t](γ∣[a,t]), with sγ(a)=0 (The arc-length function sγ(t)=L(γ∣[a,t]) of a rectifiable path).

[L2]

A C1 path has length equal to the integral of its continuous speed, including on a singleton interval where both values are zero (If γ:[a,b]→Rn is continuous, differentiable on (a,b), and γ′ extends continuously to [a,b], then L(γ)=∫ab∥γ′(t)∥2 dt).

[L3]

The integral function of an integrable function is differentiable at every point where the integrand is continuous, with derivative equal to the integrand; at endpoints this means the relative one-sided derivative (The first fundamental theorem: if f is integrable on [a,b] and continuous at c, then F′(c)=f(c); in particular a continuous f has F as a primitive).

Proof

technique · direct
1.1

The speed v(t):=∥γ′(t)∥2 is continuous because γ′ and the Euclidean norm are continuous.

givenalgebra
2.1

By [L1] and [L2], for every t∈[a,b], sγ(t)=∫atv(u) du, including t=a.

L1L2step 1.1
3.1

Apply [L3] to step 2.1. It gives sγ′(t)=v(t)=∥γ′(t)∥2 throughout [a,b], with the asserted endpoint interpretation.

step 1.1step 2.1L3∎

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