Arc Length with Vector Functions — Question 9

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Question 9

A wire is wound exactly NN times around a cylinder of radius RR, rising uniformly through total height HH. Model its centerline by 𝒓(t)=⟨Rcost,Rsint,H2πNt⟩,0≤t≤2πN.\mathbf r(t)=\left\langle R\cos t,R\sin t,\frac{H}{2\pi N}t\right\rangle, \qquad 0\le t\le 2\pi N. Tasks

  1. Derive a formula for the wire length in terms of R,H,NR,H,N.

  2. For R=3R=3, H=24H=24, and length 30π30\pi, determine N>0N>0.

  3. Verify whether the resulting value represents a whole number of turns.

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

Strategy. The model has constant speed; after integration, solve the design equation by squaring positive quantities.

See the diagram in the original worksheet below.

Step 1: General formula 𝒓′(t)=⟨−Rsint,Rcost,H2πN⟩,\mathbf r'(t)=\left\langle -R\sin t,R\cos t,\frac{H}{2\pi N}\right\rangle, so ∥𝒓′(t)∥=R2+(H2πN)2.\|\mathbf r'(t)\|=\sqrt{R^2+\left(\frac{H}{2\pi N}\right)^2}. Multiplying this constant speed by the parameter length 2πN2\pi N gives L=(2πNR)2+H2.\boxed{L=\sqrt{(2\pi NR)^2+H^2}}.

Step 2: Apply the data With R=3R=3, H=24H=24, and L=30πL=30\pi, (30π)2=(6πN)2+242.(30\pi)^2=(6\pi N)^2+24^2. Therefore 36π2N2=900π2−576,N2=25−16π2.\begin{align*} 36\pi^2N^2&=900\pi^2-576,\\ N^2&=25-\frac{16}{\pi^2}. \end{align*} Since N>0N>0, N=25−16π2.\boxed{N=\sqrt{25-\frac{16}{\pi^2}}}.

Step 3: Feasibility Numerically, N≈4.835N\approx 4.835, which is not an integer. Thus the given continuous model has a positive solution, but the specifications are incompatible with a whole number of complete turns. This is a design-consistency conclusion, not an algebraic failure.

Original worksheet page 2: question and worked solution for 1-9-009

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