Tangents with Polar Coordinates — Question 8

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Question 8

Problem

Find a value of aa so that r=a+2cos⁡θr=a+2\cos\theta has tangent slope 11 at θ=π/2\theta=\pi/2.

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Original worksheet page 1: question and worked solution for 3-7-008
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Question 8 – Solution

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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 π/2\pi/2, the slope is 2/a2/a.

  4. Setting 2/a=12/a=1 gives a=2\boxed{a=2}.

Original worksheet page 2: question and worked solution for 3-7-008

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