Arc Length with Parametric Equations — Question 9

PDF ↗

Question 9

Problem

A particle goes around the unit circle twice. Is its path length 2π2\pi or 4π4\pi? Justify using x=cos⁡2t,y=sin⁡2tx=\cos2t,y=\sin2t, 0≤t≤2π0\le t\le2\pi.

See the diagram in the original worksheet below.

Original worksheet page 1: question and worked solution for 3-4-009
Show solutionHide solution

Question 9 – Solution

See the diagram in the original worksheet below.

Solution

  1. Differentiate: x′(t)=−2sin⁡2t,y′(t)=2cos⁡2t.x'(t)=-2\sin2t, \qquad y'(t)=2\cos2t.

  2. The speed is constant: x′2+y′2=4sin⁡22t+4cos⁡22t=2.\sqrt{x'^2+y'^2} =\sqrt{4\sin^22t+4\cos^22t} =2.

  3. Integrate over the full parameter interval: L=∫02π2dt=4π.L=\int_0^{2\pi}2\,dt =\boxed{4\pi}.

  4. The angle of the point is 2t2t, which increases from 00 to 4π4\pi. Therefore, the unit circle is traversed twice. Its geometric circumference is 2π2\pi, but the particle travels that circumference twice, producing total traveled length 4π4\pi.

Original worksheet page 2: question and worked solution for 3-4-009

Original worksheet layout. Use Enlarge or open the PDF for a closer view.