# Why, conceptually, do limaçons $r=a+b\cos\theta$ have dimples when $|\frac{a}{b}|<2$?

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 $|\frac{a}{b}|<2$).

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The following doesn't answer your question, but it might be interesting anyway: If you take the geometric inverse of the limaçon with respect to the unit circle you get a conic section (hyperbola, parabola, or ellipse, depending on the values of $a$ and $b$) with a focus at the origin. The $|a| = 2|b|$ case is an ellipse in which each focus bisects the semimajor axis it's lying on. –  Mike Spivey Oct 9 '11 at 22:42
@Mike's observation corresponds to the fact that limaçons and conics are in fact inverse curves. That is, the inverse of a conic with respect to one of its foci is a limaçon. :) (Incidentally, due to this, one can form a correspondence between methods for drawing parabolas and methods for drawing cardioids.) –  Guess who it is. Oct 10 '11 at 3:49

Here is a geometric way of looking at it. For the purposes of this answer, I shall be reparametrizing the limaçon as

$$r=2a(1-h\cos\,\theta)$$

(I consider $h$ here to be positive; negative values of $h$ correspond to a congruent curve reflected about the vertical axis.) I have chosen this form because this allows for a convenient phrasing of one of the definitions of a limaçon:

The limaçon is the locus of a point attached to a circle of radius $a$ rolling around the outside of a fixed circle of radius $a$, with the tracing point at a distance $ah$ from the center of the rolling circle.

In other words, the limaçon is a special case of the epitrochoid.

Your border case corresponds to $h=\frac12$. Interpreted from the epitrochoid viewpoint, this means that the tracing point is halfway between the circumference and the center of the rolling circle:

For $0 < h < \frac12$, the tracing point is near the center of the rolling circle, and thus the limaçon can be expected to be convex (with the extreme $h=0$ case giving the circle whose radius is twice the original). For $h > \frac12$, the tracing point "juts out" a bit, and thus we see the dimples in the limaçon:

For the cardioid case ($h=1$) in particular, the cusp corresponds to where the tracing point coincides with the point of contact of the fixed and rolling circles:

while for $h > 1$, the self-intersection point is expected due to the tracing point jutting out of the rolling circle.

As seen in this book, there is more than one way to define the limaçon geometrically, and I suppose those other definitions will also shed light on the differences between dimpled and undimpled limaçons. I think that treating the limaçon as an epitrochoid is the easiest to understand, though.

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This makes it quite clear. Nice answer! –  Mike Spivey Oct 11 '11 at 15:26

Somewhat related to Robert's and Rahul's answers: it might be more illustrative to consider the curvature of the limaçon. Without loss of generality, consider the case $b=1$ and positive $a$ (all other cases are related by scaling and reflection about the horizontal axis). The curvature function is

$$\frac{a^2+3a\cos\,\theta+2}{\sqrt{(a^2+2a\cos\,\theta+1)^3}}$$

For $a=2$ (your boundary), a plot of the curvature function shows that its only zero is at $\theta=\pi$, and that point is a tangency point:

(Remember that a point of zero curvature corresponds to your curve being "locally straight").

For $a > 2$ or for $a < 1$, the curvature function is entirely positive (no "straight sections"). The cardioid case, $a=1$, yields a curvature function that is singular at $\theta=\pi$ (corresponding to the cusp). Finally, for the case $1 < a < 2$, the curvature becomes zero at the points $\theta=\arccos\dfrac{-a^2-2}{3a}$ and $\theta=2\pi-\arccos\dfrac{-a^2-2}{3a}$, which exactly corresponds to the case where one starts to see dimples...

A similar analysis can be seen in Gibson's Elementary Geometry of Differential Curves: An Undergraduate Introduction.

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To put it in a wider context: the planar parametric curve ${\bf p}= {\bf p(t)}$ with velocity vector ${\bf v}$ and acceleration ${\bf a}$ is "turning to the left" if $v_1 a_2 - v_2 a_1 > 0$ and "turning to the right" if $v_1 a_2 - v_2 a_1 < 0$. For a polar curve, taking $t = \theta$, ${\bf p}(t) = r(t) [\cos(t), \sin(t)]$, $$v_1 a_2 - v_2 a_1 = r(t)^2 + 2 r'(t)^2 - r(t) r''(t)$$ In the case of your limaçons, with $r(t) = a + b \cos(t)$, this turns out to be $2 b^2+3 a b \cos(t)+a^2$. The minimum, when $\cos(t) = \pm 1$ (depending on sign of $ab$) is $2 b^2 - 3 |a b| + a^2 = (|a|-|b|)(|a| - 2 |b|)$. For this to be negative, producing your "dimple", we need $|b| < |a| < 2 |b|$.

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If I understand correctly, the left/right test is essentially equivalent to looking at whether the cross-product of velocity and acceleration is pointing up or down (relative to the plane) as a way of determining if acceleration is to the right or left of velocity (which is nice and something I hadn't thought about). But I don't see any more insight into why the boundary value is 2 than I got from looking at the vertical tangents. –  Isaac Oct 9 '11 at 19:50
Let's say $a > b > 0$. At $t = \pi$, which is the centre of the "dimple" if there is any, $r' = 0$ by symmetry and the condition becomes $r(\pi) - r''(\pi) = (a - b) - b < 0$. Does that help? –  Robert Israel Oct 9 '11 at 23:57

Let a particle move along the limaçon at a constant rate of $\dot\theta = 1$, and consider on the instant when $\theta = \pi$ and $r = a - b$.

If it were moving at a constant radius, its inward acceleration would be $r\dot\theta^2 = a - b$. But since its radius is also changing, you get an additional outward acceleration of $\ddot r = b$. These cancel out when $a = 2b$.

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The invariant description of what you are looking for is that in a family of smooth curves, the appearance and disappearance of a "dimple" happens at curves where a tangent line has contact of order 4 with (at least one point of one branch of) the curve.

This is a relatively high order of local calculation and might not be amenable to a simple geometric argument about rolling circles. The acceleration and curvature calculations are equivalent (because the two quantities are proportional to each other), and show contact of order at least 3. The reflection symmetry of the curves enhances this to order 4; an even analytic function vanishing to $O(t^3)$ is $O(t^4)$. Or one checks the signs of the curvature near the critical point to show it is a minimum. In all these approaches some algebra is necessary to get the answer.

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When you say the tangent line has contact of order 4 with the curve, is that detectable in my calculus exploration? I could see order 3 from my work (one from the sine branch of the solution and a double from the cosine branch), but it's not clear to me whether that's related to what you're talking about. –  Isaac Oct 12 '11 at 21:52
Two inflection points come together, hence order 4. Your exploration used symmetry in assuming that the tangent line at the critical point (where dimple birth/death occurs) would be vertical, and in the presence of symmetry the order 3 vanishing implies order 4. –  zyx Oct 12 '11 at 21:59
zyx: I though the epitrochoid argument took care of the "intuition" being asked for by @Isaac. –  Guess who it is. Oct 13 '11 at 1:12
@J.M., I may have read your answer too quickly, but thought it was illustrating (not proving) that $h = 1/2$ is the critical case and that additional argument is needed. I will look more carefully to see how it compares with the analytic order of contact arguments. (Actually my main thought was "how did he produce those animations?") –  zyx Oct 13 '11 at 1:48
Actually, I think I have a geometric solution right in the alley of what you seem to want to see (there really is more than one way to skin a cat!)... I'll need to figure out how to assemble it properly, though. –  Guess who it is. Oct 13 '11 at 1:50