Tangents with Polar Coordinates — Question 6

Question 6 – 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 $\theta=\pi/3$, the circle $r=2$ has tangent direction $5\pi/6$.

4. The second circle has slope $1/\sqrt3$, hence tangent direction $\pi/6$.

5. The smaller angle between the tangent lines is $\boxed{\pi/3}$.