Tangents with Polar Coordinates — Question 3

Question 3

Problem

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

Question 3 – Solution

Solution

1. Write the polar curve parametrically as $$x(\theta)=r(\theta)\cos\theta, \qquad y(\theta)=r(\theta)\sin\theta.$$ Differentiation gives $$\frac{dx}{d\theta}=r'\cos\theta-r\sin\theta, \qquad \frac{dy}{d\theta}=r'\sin\theta+r\cos\theta.$$

2. Wherever $dx/d\theta\ne0$, compute $$\frac{dy}{dx}= \frac{r'\sin\theta+r\cos\theta} {r'\cos\theta-r\sin\theta}.$$ Test numerator and denominator separately when locating horizontal or vertical tangents.

3. At $\theta=0$, the point is $(1,0)$ and the slope is $1$, so the tangent is not vertical.

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

5. Both parts are false.