Tangents with Polar Coordinates — Question 8

Question 8

Problem

Find a value of $a$ so that $r=a+2\cos\theta$ has tangent slope $1$ at $\theta=\pi/2$.

Question 8 – 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 $\pi/2$, the slope is $2/a$.

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