Tangents with Polar Coordinates — Question 2

Question 2 – 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. The numerator $r'\sin\theta+r\cos\theta=\cos\theta+\cos2\theta$ vanishes.

4. This gives $\cos\theta=1/2$ or $-1$, and excluding the pole leaves $\boxed{\theta=\pi/3,5\pi/3}$. At these angles $dx/d\theta\ne0$, and the points are $(3/4,\pm3\sqrt3/4)$. The tangent lines are $\boxed{y=\pm3\sqrt3/4}$.