Vector Functions — Question 9

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Question 9

Construct a vector function for a helix that lies on the cylinder (x−2)2+(y+1)2=9(x-2)^2+(y+1)^2=9, begins at (5,−1,0)(5,-1,0) when t=0t=0, rises 4 units during each counterclockwise revolution, and makes two revolutions.

Original worksheet page 1: question and worked solution for 6-6-009
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Question 9 – Solution

Strategy Use the cylinder center and radius for the circular components, then choose a linear height matching the pitch.

See the diagram in the original worksheet below.

Construction Counterclockwise motion from the rightmost point is x=2+3cos⁡t,y=−1+3sin⁡t.x=2+3\cos t,\qquad y=-1+3\sin t. One revolution is Δt=2π\Delta t=2\pi and should raise zz by 4, so z=(2/π)tz=(2/\pi)t. Two revolutions require 0≤t≤4π0\le t\le 4\pi. Thus r→(t)=⟨2+3cost,−1+3sint,2t/π⟩,0≤t≤4π.\boxed{\vec r(t)=\left\langle 2+3\cos t,-1+3\sin t,2t/\pi\right\rangle},\quad 0\le t\le 4\pi.

Verification The cylinder equation is identically 9, and the final height is 8.

Original worksheet page 2: question and worked solution for 6-6-009

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