Arc Length and Surface Area Revisited — Question 10

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Question 10

Problem

A numerical system can evaluate either a Cartesian or parametric arc-length integral. Give one reason to prefer each form.

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Original worksheet page 1: question and worked solution for 3-11-010
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Question 10 – Solution

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Solution

  1. Choose a representation and interval that trace the desired curve exactly once. The equivalent arc-length formulas are L=∫1+(dydx)2dx,L=∫(dxdt)2+(dydt)2dt,L=∫r2+(drdθ)2dθ.\begin{aligned} L&=\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx,\\ L&=\int\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}\,dt,\\ L&=\int\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{aligned}

  2. For a surface of revolution, multiply the appropriate arc-length element by 2π2\pi times the nonnegative distance to the axis. Check the tracing interval to prevent geometric double-counting.

  3. Cartesian is convenient for a single-valued smooth graph with simple derivative.

  4. Parametric form is preferable for loops, vertical tangents, motion data, or when it avoids an infinite derivative.

Original worksheet page 2: question and worked solution for 3-11-010

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