Arc Length with Vector Functions — Question 8

Section 13-9 · Calculus pages 2555–2556

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Question 8

Suppose q→(u)=r→(ϕ(u))\vec q(u)=\vec r(\phi(u)) traces the same regular curve, where ϕ\phi is continuously differentiable and strictly monotone. Prove that the arc length is unchanged by this reparameterization, including the case in which orientation is reversed.

Question 8 – Solution

Strategy Apply the chain rule to speed, then use substitution while carefully retaining the absolute value.

Diagram shown in the original SVG worksheet.

Transformed speed q→′(u)=r→′(ϕ(u))ϕ′(u),∥q→′(u)∥=∥r→′(ϕ(u))∥|ϕ′(u)|.\vec q'(u)=\vec r'(\phi(u))\phi'(u),\qquad \|\vec q'(u)\|=\|\vec r'(\phi(u))\|\,|\phi'(u)|.

Increasing case If ϕ′≥0\phi'\ge 0 and ϕ(α)=a,ϕ(β)=b\phi(\alpha)=a,\phi(\beta)=b, substitution gives ∫αβ∥q→′(u)∥du=∫ab∥r→′(t)∥dt.\int_\alpha^\beta\|\vec q'(u)\|\,du=\int_a^b\|\vec r'(t)\|\,dt.

Decreasing case If ϕ′≤0\phi'\le 0, then ϕ(α)=b\phi(\alpha)=b and ϕ(β)=a\phi(\beta)=a. Since |ϕ′|=−ϕ′|\phi'|=-\phi', reversing the substituted limits again produces ∫ab∥r→′(t)∥dt\int_a^b\|\vec r'(t)\|dt. Thus length depends on the traced curve, not its orientation or timing.