Tangents with Polar Coordinates — Question 9

Question 9

Problem

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

Question 9 – 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 $r=0$, direct division by $r$ is undefined. Valid cancellation where $r\ne0$ 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.