TITLE: Why, conceptually, do limaçons $r=a+b\cos\theta$ have dimples when $|\frac{a}{b}|<2$?
QUESTION [5 upvotes]: Using calculus, I can justify that limaçons—the polar graphs of $r=a+b\cos\theta$ for various nonzero real values of $a$ and $b$—are dimpled when $|\frac{a}{b}|<2$, but that doesn't seem to yield any conceptual reason why that should be the case.  The boundary value 2 seems too nice to not have a conceptual explanation, so is there one?

The calculus:  For this sort of limaçon, with cosine, there are always vertical tangents to the graph at $\theta=k\pi$, and the dimple is characterized by a pair of vertical tangents near one of those two locations, but equally spaced before and after it, whereas a non-dimpled limaçon only has those two vertical tangents.  Vertical tangents occur when $\frac{dy}{dx}$ is undefined; for polars, that means when $\frac{dy}{d\theta}$ is undefined (for our limaçon, never) or when $\frac{dx}{d\theta}=0$.  For our limaçon, $\frac{dx}{d\theta}=-\sin\theta(a+2b\cos\theta)$, so $\frac{dx}{d\theta}=0$ implies $\sin\theta=0$ ($\theta=k\pi$) or $a+2b\cos\theta=0$.  This latter case, which can be rewritten as $\cos\theta=-\frac{a}{2b}$, has no solutions when $|\frac{a}{2b}|>1$, a single solution that is already in the solutions from $\sin\theta=0$ when $|\frac{a}{2b}|=1$ (so, no additional vertical tangents and hence no dimple when $|\frac{a}{b}|\ge 2$), and two solutions when $|\frac{a}{2b}|<1$ (so two additional vertical tangents and hence a dimple when $|\frac{a}{b}|<2$).

REPLY [2 votes]: An answer is already given in the paper "Why do certain limacons have a dimple, Teaching Mathematics and its Applications Advance Access published March 29, 2006".