# Cardioid in coffee mug?

I've been learning about polar curves in my Calc class and the other day I saw this suspiciously $r=1-\cos \theta$ looking thing in my coffee cup (well actually $r=1-\sin \theta$ if we're being pedantic.) Some research revealed that it's called a caustic. I started working out why it would be like this, but hit a snag. Here's what I did so far:

Consider the polar curves $r=1-\cos \theta$ and $r=1$. Since for a light ray being reflected off a surface (or the inside of my cup) the $\angle$ of incidence =$\angle$ of reflection, a point on the circle $(1,\theta)\to(1,2\theta)$. It looks like this has something to do with the tangent lines, so I pretend there's an $xy$ plane centered at the pole to find the slope of the line connecting the points. Since it's the unit circle the corresponding rectangular coordinates are $(\cos \theta, \sin \theta)$ and $(\cos 2\theta, \sin 2\theta).$ So $$m={\sin 2\theta-\sin \theta \over \cos 2\theta-\cos \theta}$$ now we see if it matches up with the slope of the lines tangent to the cardioid $$\frac{dy}{dx}={r'\sin \theta+r \cos \theta \over r' \cos \theta - r \sin \theta}={\sin^2\theta - \cos^2 \theta +\cos \theta \over 2\sin \theta \cos \theta -\sin\theta}={\cos \theta - \cos 2\theta\over \sin 2\theta-\sin \theta}$$

They're similar, but not identical. In particular $\frac{dy}{dx}=-\frac1m$. What error have I made, or what have I overlooked conceptually? Thanks in advance. • physics.stackexchange.com/a/91164/31276 is related, and may be helpful. I'm not certain this is the issue, but note that the slope you want is the negative reciprocal of the slope you got. This suggests perpendicularity, which appears involved as en.wikipedia.org/wiki/Caustic_%28mathematics%29 mentions the "orthotomic". – Mark S. Nov 28 '15 at 19:36
• A little variation of your figure and you get a Pascal´s limaçon but it is would be not surprising because both are curves of equation $r=a+b\cos \theta$ and each case depends of coefficients $a$ and $b$ relying to the conchoid of a circle (even you can get a circle when $a=b$). Anyway, my opinion is that you have not a cardiod in your coffe but something like a cardioid. – Piquito Nov 28 '15 at 19:55
• @Ataulfo Probably it isn't since nothing turns out nice in the physical world, but what I'm interested in more generally is why a cardioid would appear in any situation. I was mostly intrigued by the following wikipedia claim: "The catacaustic of a circle with respect to a point on the circumference is a cardioid." en.wikipedia.org/wiki/Cardioid#Caustics – mysatellite Nov 28 '15 at 20:02
• I said you (maybe in bad English) all these curves are conchoids of a circle. I'll try to give something more later. – Piquito Nov 28 '15 at 20:09
• I get this figure that would be useful for you. See my answer (it is not possible to give it in comments) – Piquito Nov 28 '15 at 20:42

The ray, when intersecting the cardiod, must be orthogonal to the plane of incidence. It just so happens that the derivative of the cardiod's equation, as given in your question, gives the slope of the plane of incidence at that particular angle and $m \times \frac{dy}{dx} = -1$ as required!

But shouldn't the derivative of the cardiod's equation gives the slope of the tangent line? Well it does... to see this supposed discrepancy, we have to carefully assess the model that we're using.

Let's fix our point source of light at the right of the circle, as in the diagram below at $P$. Physically, $P$ is where your incoming beam, exterior to the ring, is focused. Shifting the location of the point source would disrupt the cardiod and produce some other envelope. Keeping the point source on the circumference would keep the envelope of rays in the shape of a cardiod; in response to your comment, I think that's what is meant by "the catacaustic of a circle with respect to a point on the circumference is a cardioid." (Source of the picture.)

$$m(\phi)={\sin 2\phi-\sin \phi \over \cos 2\phi-\cos \phi}$$

is slope of the path of the reflected ray. The angle $\phi$ is being measured from a coordinate system at the center of the circle, measured counterclockwise from the positive $x$-axis. This is convenient, since the law of reflection will indeed give $(r,\phi)\to(r,2\phi)$.

Now let's match the cardiod in the diagram. It looks something like$$r = a(\cos \theta -1)$$ This curve is a bit different from yours, due to how we're defining $\theta$. Anyway, we don't know $a$. It's not unit length if our circle is a unit circle. See here. Check the linked plots, and you'll probably notice that the circle isn't positioned properly. This is because $\theta$ is not measured from the center of the circle, but from the cusp of the cardiod. We're using two coordinate systems. Regardless, we can compare slopes in the two systems since they're translations. We still have

\begin{align*} \frac{dy}{dx}(\theta) = \frac{dy/ d\theta}{dx/ d\theta} & = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta} \\ \\ & = \frac{-a\sin^2 \theta +a\cos^2 \theta- a\cos \theta}{-a\sin\theta \cos \theta - a\cos\theta \sin \theta + \sin \theta} \\ \\ & = -\frac{\cos 2\theta- \cos \theta}{\sin 2\theta - \sin \theta} \\ \\ & = -\frac{1}{m(\theta)} \end{align*} For reference, I marked a sample ray in yellow. It reflects at $\phi = \frac{\pi}{2}$ and the reflected portion has slope $m(\frac{\pi}{2}) = 1$. Now, without proving it, it does appear that at $\theta = \phi$, the reflected ray and the tangent have parallel slopes. Neat. But why don't the formulas give this result? Why does $\frac{dy}{dx}(\frac{\pi}{2}) = -1$ for our ray? This is the negative reciprocal.

Recall the cardiod has formula $r = a(\cos \theta -1)$. Let's examine how one of these points is mapped. At $\theta = \frac{\pi}{2}$, we have $r = -a$. This point is on the opposing side of the cardiod! The slope of this tangent given by the derivative is actually that of the pink line below. Interestingly, the pink tangent on the opposing side of the cardiod appears to be perpendicular to the reflected ray itself. That is, it is parallel to the plane of incidence.

Thanks for the cool question.

• @Sky Ay caramba. This answer is totally wrong (and nobody caught it!). I'll edit and post a revised answer very soon. Your analysis makes sense to me, but I think there's a slight error in the model. – zahbaz Nov 29 '15 at 19:26
• Big revision. Let me know if you have questions. – zahbaz Nov 30 '15 at 0:03
• Great answer, and even better after revision. Thank you :) – mysatellite Nov 30 '15 at 0:05

Light From Infinity

Most likely, the light is from a distance that is large on the scale of the cup. Therefore, we will take the incoming rays to be parallel. If so, the caustic in the coffee cup is a nephroid.

Consider the following diagram The ray reflected at the point $(-\cos(\theta),\sin(\theta))$ is $$\frac{y-\sin(\theta)}{x+\cos(\theta)}=-\tan(2\theta)\tag{1}$$ which is $$x\sin(2\theta)+y\cos(2\theta)=-\sin(\theta)\tag{2}$$ Here is a plot of the reflected rays from $(2)$ generated by uniformly distributed parallel incoming rays Taking the derivative of $(2)$ with respect to $\theta$: $$2x\cos(2\theta)-2y\sin(2\theta)=-\cos(\theta)\tag{3}$$ Solving $(2)$ and $(3)$ simultaneously gives the envelope of the family of lines in $(1)$: \begin{align} \begin{bmatrix} x\\ y \end{bmatrix} &= \begin{bmatrix} \cos(2\theta)&-\sin(2\theta)\\ \sin(2\theta)&\cos(2\theta) \end{bmatrix}^{-1} \begin{bmatrix} -\frac12\cos(\theta)\\ -\sin(\theta) \end{bmatrix}\\[6pt] &=\frac14\begin{bmatrix} \cos(3\theta)-3\cos(\theta)\\ 3\sin(\theta)-\sin(3\theta) \end{bmatrix}\tag{4} \end{align} The curve from $(4)$ is added in red Equation $(4)$ describes a nephroid.

Light From A Point On The Circle

Since it is mentioned in a comment that a cardoid is formed by light coming from a point on the edge of the cup, we will do the same computation with light from a point on the circle.

Consider the following diagram The ray reflected at the point $(-\cos(\theta),\sin(\theta))$ is $$\frac{y-\sin(\theta)}{x+\cos(\theta)}=-\tan\left(\frac{3\theta}2\right)\tag{5}$$ which is $$y(1+\cos(3\theta))+x\sin(3\theta)=\sin(\theta)-\sin(2\theta)\tag{6}$$ Here is a plot of the reflected rays from $(6)$ reflected at uniformly spaced points on the circle Taking the derivative of $(6)$ with respect to $\theta$: $$-3y(\sin(3\theta))+3x\cos(3\theta)=\cos(\theta)-2\cos(2\theta)\tag{7}$$ Solving $(6)$ and $(7)$ simultaneously gives the envelope of the family of lines in $(5)$: \begin{align} \begin{bmatrix} x\\ y \end{bmatrix} &= \begin{bmatrix} \cos(3\theta)&-\sin(3\theta)\\ \sin(3\theta)&1+\cos(3\theta) \end{bmatrix}^{-1} \begin{bmatrix} \frac13\cos(\theta)-\frac23\cos(2\theta)\\ \sin(\theta)-\sin(2\theta) \end{bmatrix}\\[6pt] &=\frac13\begin{bmatrix} \cos(2\theta)-2\cos(\theta)\\ 2\sin(\theta)-\sin(2\theta) \end{bmatrix}\tag{8} \end{align} The curve from $(8)$ is added in red Equation $(8)$ describes a cardoid.

As I promised you in the comments, here three distinct values for $a$ and $b$ in the equation $r = a +b \cos \theta$. You can see different curves however the equation is similar.

• Yes, cardioids are of the form $r=a(1-\cos \theta)$ – mysatellite Nov 28 '15 at 20:49