Tangents with Polar Coordinates — Question 7

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Question 7

Problem

For r=eθr=e^\theta, show that the angle between the radius vector and tangent is constant.

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Original worksheet page 1: question and worked solution for 3-7-007
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Question 7 – 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. In polar components, tangent has radial component r′=rr'=r and transverse component rr.

  4. Thus it makes angle arctan⁡(r/r)=π/4\arctan(r/r)=\boxed{\pi/4} with the radius vector.

Original worksheet page 2: question and worked solution for 3-7-007

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