Arc Length with Vector Functions — Question 10

Section 13-9 · Calculus pages 2559–2560

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Question 10

Consider r→(t)=⟨t2,t3⟩\vec r(t)=\left\langle t^2,t^3\right\rangle on −1≤t≤1-1\le t\le 1. The parameterization is not regular at t=0t=0. Determine the total arc length exactly and explain why one must split the integral or preserve an absolute value.

Question 10 – Solution

Strategy The speed contains t2=|t|\sqrt{t^2}=|t|; use symmetry across the cusp.

Diagram shown in the original SVG worksheet.

Speed r→′(t)=⟨2t,3t2⟩,∥r→′(t)∥=|t|4+9t2.\vec r'(t)=\left\langle 2t,3t^2\right\rangle,\qquad \|\vec r'(t)\|=|t|\sqrt{4+9t^2}.

Length L=2∫01t4+9t2dt=227[(4+9t2)3/2]01=227(1313−8).L=2\int_0^1t\sqrt{4+9t^2}\,dt =\frac{2}{27}\left[(4+9t^2)^{3/2}\right]_0^1 =\boxed{\frac{2}{27}(13\sqrt{13}-8)}.

Why care about |t||t|? Replacing |t||t| by tt on the full symmetric interval would make the integrand odd and falsely give zero. Although velocity vanishes at the cusp, each half is smooth and their nonnegative lengths add.