Tangents with Polar Coordinates — Question 3

PDF ↗

Question 3

Problem

A student claims that θ=0\theta=0 gives the rightmost point and a vertical tangent of r=1+sin⁡θr=1+\sin\theta. Test both parts of the claim.

See the diagram in the original worksheet below.

Original worksheet page 1: question and worked solution for 3-7-003
Show solutionHide solution

Question 3 – 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 θ=0\theta=0, the point is (1,0)(1,0) and the slope is 11, so the tangent is not vertical.

  4. Also x′(0)=1>0x'(0)=1>0, so xx is still increasing; the point is not rightmost.

  5. Both parts are false.

Original worksheet page 2: question and worked solution for 3-7-003

Original worksheet layout. Use Enlarge or open the PDF for a closer view.