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The bean curve is defined as $\left(x^{2}+y^{2}\right)^{2}=x^{3}+y^{3}$ in cartesian coordinates, $r=\left(\cos\left(\theta\right)\right)^{3}+\left(\sin\left(\theta\right)\right)^{3}$ in polar coordinates and $$\gamma(t) = \left(\left(\left(\cos\left(t\right)\right)^{3}+\left(\sin\left(t\right)\right)^{3}\right)\cos\left(t\right),\left(\left(\cos\left(t\right)\right)^{3}+\left(\sin\left(t\right)\right)^{3}\right)\sin\left(t\right)\right)$$ as parametric equation for $0\le t \le \pi$. My aim is to partition the interior in smaller bean curves(just like in the figure). For instance, in the case of unit circle centred at origin, every point in the interior can be written uniquely as $(s\cos(\theta),s\sin(\theta))$ where $s$ is the distance of the point from origin and $\theta$ denotes the angle from the $x$-axis. Thus the interior in this case is the union of smaller circles centred at origin. Is it possible to do the same for the bean curve?

                                                              enter image description here

A few possibilities that don't work are $s\gamma(t)$ and $s\gamma(t)+\frac{s(1-s)}{2}(1,1)$, because for every $0\le s \le1$, $(0,0) = s\times \gamma(3\pi/4)$ and the same problem with the second one. For the same point there are multiple bean curves.

Sam
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  • I am sorry for my bad drawing, I hope the question is clear. – Sam Jul 19 '21 at 13:26
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    Does Wikipedia's "Parallel curve" entry help? – Blue Jul 19 '21 at 13:32
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    What about moving the center then, like this https://www.desmos.com/calculator/2ewqt8pabe ? – zwim Jul 19 '21 at 13:51
  • @Blue, thank you, I appreciate the suggestion. I will try to see if it works. – Sam Jul 19 '21 at 13:52
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    are you looking for progressively scaled-down curves, or for parallel curves (constant distance) – G Cab Jul 19 '21 at 13:52
  • @zwin, wow, it looks beautiful. How did you manage to find this balance? – Sam Jul 19 '21 at 13:54
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    I supposed, the center of gravity of the bean was $(\frac 14,\frac 14)$ (visually), but it probably work evezn better with the real center of gravity, but you need to calculate the double integral for this. If you put it in $(\frac 15,\frac 15)$ this work too, but not the same balance, so I should not be too far off. – zwim Jul 19 '21 at 13:56
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    Your second example should work, possibly corrected to $s\gamma(t)+\frac{1-s}{2}(1,1)$. – Intelligenti pauca Jul 19 '21 at 14:15
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    @Intelligentipauca No $(\frac12,\frac12)$ does not work, there are multiple points near the concave part of the bean. Visually it seems we cannot go beyond $0.488$ along the $y=x$ diagonal. See my modified graph https://www.desmos.com/calculator/k7so5zzojv – zwim Jul 19 '21 at 16:20

1 Answers1

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The comment by @Blue is correct. I've had experience with these parallel curves and plotted them up for the bean curve in the first figure below. The first algorithm in the Wikipedia article was used, but be advised that the algorithm breaks down if the $|d|$ is too large or too small. Here is what I did: I first centered the bean curve at its circumcenter. Then I normalized it to fit in a unit circle. For the parallel curves I took $d\in [-.3,.3]$ in $0.1$ increments. In the complex plane, let's call the bean curve $z(t)$ and the parallel curves $z_d(t)$. The we have

$$ z(t)=(\cos^3t+\sin^3t)e^{it}\\ z_d(t)=z(t)-d\frac{i\dot z}{|\dot z|} $$

where $\dot z=\dot x+i\dot y$. The Cartesian equations are given in the referenced Wiki article.

It's interesting to note the circumcircles of all the parallel curves are concentric (they share the same center) but that the circumcenters of al the beans are in different relative positions. This is demonstrated in the second figure below.

Bean curves Bean curves with circumcircles

Cye Waldman
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