Arc Length with Polar Coordinates — Question 10

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Question 10

Problem

Find aa so that r=aθr=a\theta, 0≤θ≤10\le\theta\le1, has length 2+arsinh⁡1\sqrt2+\operatorname{arsinh}1.

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Original worksheet page 1: question and worked solution for 3-9-010
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Question 10 – Solution

See the diagram in the original worksheet below.

Solution

  1. Differentiate the polar radius to obtain r′=dr/dθr'=dr/d\theta and choose an interval that traces the requested arc exactly once.

  2. Use the polar arc-length formula L=∫abr2+(drdθ)2dθ.L=\int_a^b\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta. Simplify the expression under the square root before evaluating or reporting the integral.

  3. Its length is |a|∫011+θ2dθ=|a|2(2+arsinh⁡1)|a|\int_0^1\sqrt{1+\theta^2}d\theta=\tfrac{|a|}{2}(\sqrt2+\operatorname{arsinh}1).

  4. Thus a=±2\boxed{a=\pm2}.

Original worksheet page 2: question and worked solution for 3-9-010

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