# Tangent at the pole for the equation $r = 2(1 - \sin\theta)$

I was asked to find the tangents at the pole for the following equation: $$r=2(1-\sin\theta)$$.

I understand that the requirements for tangency at the pole are $$f(\theta)=0$$ and $$f'(\theta) \neq 0$$. I set $$0=2(1-\sin\pi)$$ and got $$\theta= \frac{\pi}{2}$$. But when I plugged that into the derivative, I got $$f'(\theta)=0$$. Why is that? Am I solving it wrong? (By the way, my $$f'$$ was $$f'(\theta)=-2\cos\theta$$). Thank You!

• What is f?????? Mar 30, 2019 at 2:26
• This tutorial explains how to typeset mathematics on this site. Mar 30, 2019 at 9:10
• You meant to write $2(1 - \sin\theta) = 0 \implies \theta = \frac{\pi}{2}$. Mar 30, 2019 at 9:23

You calculated the derivative incorrectly. The derivative is $$\frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}$$ where \begin{align*} x & = r\cos\theta\\ y & = r\sin\theta \end{align*} By the product rule, \begin{align*} \frac{dx}{d\theta} & = r'\cos\theta - r\sin\theta\\ \frac{dy}{d\theta} & = r'\sin\theta + r\cos\theta \end{align*} where $$r' = \frac{dr}{d\theta}$$ Hence, $$\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$$ We were given the function $$r(\theta) = 2(1 - \sin\theta)$$. At the pole, $$r = 0$$, so we obtain \begin{align*} 2(1 - \sin\theta) & = 0\\ 1 - \sin\theta & = 0\\ 1 & = \sin\theta\\ \frac{\pi}{2} & = \theta \end{align*} Notice that at the pole, since $$r = 0$$, \begin{align*} \frac{dy}{dx} & = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}\\ & = \frac{r'\sin\theta}{r'\cos\theta}\\ & = \tan\theta \end{align*} Observe that \begin{align*} \lim_{\theta \to \frac{\pi}{2}^-} \tan\theta & = \infty\\ \lim_{\theta \to \frac{\pi}{2}^+} \tan\theta & = -\infty \end{align*} Thus, the function $$r(\theta) = 2(1 - \sin\theta)$$ has a cusp at the pole, as can be seen from viewing its graph, which is a cardioid. 