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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!

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

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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.

polar_graph_of_cardioid

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