Tangents with Polar Coordinates — Question 9

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Question 9

Problem

A student cancels rr before using the polar slope formula at a pole. Explain the danger.

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Original worksheet page 1: question and worked solution for 3-7-009
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Question 9 – Solution

See the diagram in the original worksheet below.

Solution

  1. Write the polar curve parametrically as x(θ)=r(θ)cos⁡θ,y(θ)=r(θ)sin⁡θ.x(\theta)=r(\theta)\cos\theta, \qquad y(\theta)=r(\theta)\sin\theta. Differentiation gives dxdθ=r′cos⁡θ−rsin⁡θ,dydθ=r′sin⁡θ+rcos⁡θ.\frac{dx}{d\theta}=r'\cos\theta-r\sin\theta, \qquad \frac{dy}{d\theta}=r'\sin\theta+r\cos\theta.

  2. Wherever dx/dθ≠0dx/d\theta\ne0, compute dydx=r′sin⁡θ+rcos⁡θr′cos⁡θ−rsin⁡θ.\frac{dy}{dx}= \frac{r'\sin\theta+r\cos\theta} {r'\cos\theta-r\sin\theta}. Test numerator and denominator separately when locating horizontal or vertical tangents.

  3. At r=0r=0, direct division by rr is undefined. Valid cancellation where r≠0r\ne0 can help compute a limit, but does not by itself justify substitution at the pole.

  4. One must use a limit or inspect the parameter values producing the pole.

Original worksheet page 2: question and worked solution for 3-7-009

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