Tangents with Polar Coordinates — Question 4

Question 4

Problem

The spiral $r=\theta$ reaches $\theta=\pi$. Find its tangent slope there.

Question 4 – 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. With $r'=1$, the slope is $(\sin\pi+\pi\cos\pi)/(\cos\pi-\pi\sin\pi)=\boxed{\pi}$.