Vector Functions β€” Question 3

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Question 3

A particle follows 𝒓(t)=⟨2cost,2sint,t/Ο€βŸ©,0≀t≀4Ο€.\mathbf r(t)=\left\langle 2\cos t,\ 2\sin t,\ t/\pi\right\rangle,\qquad 0\le t\le 4\pi. Without calculus:

  1. Describe the trace geometrically and state its axis and radius.

  2. Find every point where the particle meets the plane z=1z=1.

  3. Determine the vertical rise during one complete revolution and the total number of revolutions.

Original worksheet page 1: question and worked solution for 1-6-003
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Question 3 – Solution

Strategy. Use the identity cos⁡2t+sin⁡2t=1\cos^2t+\sin^2t=1 and compare the angular parameter with the height.

See the diagram in the original worksheet below.

Step 1: Cylindrical constraint x2+y2=4cos⁡2t+4sin⁡2t=4.x^2+y^2=4\cos^2t+4\sin^2t=4. Thus the trace lies on the cylinder of radius 22 about the zz-axis. Since z=t/Ο€z=t/\pi steadily increases while (x,y)(x,y) rotates counterclockwise, the curve is a circular helix.

Step 2: Plane intersection Setting z=1z=1 gives tΟ€=1β‡’t=Ο€.\frac{t}{\pi}=1\quad\Longrightarrow\quad t=\pi. This value is in the interval, and therefore the only point is 𝒓(Ο€)=βŸ¨βˆ’2,0,1⟩.\boxed{\mathbf r(\pi)=\left\langle -2,0,1\right\rangle}.

Step 3: Pitch and turns One revolution changes tt by 2Ο€2\pi, so Ξ”z=t+2Ο€Ο€βˆ’tΟ€=2.\Delta z=\frac{t+2\pi}{\pi}-\frac{t}{\pi}=2. The interval length 4Ο€4\pi contains 4Ο€/(2Ο€)=24\pi/(2\pi)=2 revolutions. Hence the rise per turn is 2\boxed{2} and the curve makes 2\boxed{2} complete turns.

Original worksheet page 2: question and worked solution for 1-6-003

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