Playing around with a rational rose

A “rose” in mathematics is typically a curve with polar equation

r = cos(kθ)

where k is a positive integer. If k is odd, the resulting graph has k “petals” and if k is even, the plot has 2k petals.

Sometimes the term rose is generalized to the case of non-integer k. This is the sense in which I’m using the phrase “rational rose.” I’m not referring to an awful piece of software by that name [1]. This post will look at a particular rose with k = 2/3.

My previous post looked at

r = cos(2θ/3)

and gave the plot below.

Plot of r = abs(cos(2 theta / 2) )

Unlike the case where k is an integer, the petals overlap.

In this post I’d like to look at two things:

  1. The curvature in the figure above, and
  2. Differences between polar plots in Python and Mathematica


The graph above has radius 1 since cosine ranges from −1 to 1. The curve is made of arcs that are approximately circular, with the radius of these approximating circles being roughly 1/2, sometimes bigger and sometimes smaller. So we would expect the curvature to oscillate roughly around 2. (The curvature of a circle of radius r is 1/r.)

The curvature can be computed in Mathematica as follows.

    numerator = D[x[t], {t, 1}] D[y[t], {t, 2}] - 
                D[x[t], {t, 2}] D[y[t], {t, 1}]
    denominator = (D[x[t], t]^2 + D[y[t], t]^2)^(3/2)
    Simplify[numerator / denominator]

This produces

\kappa = \frac{3 \sqrt{2} \left(5 \cos \left(\dfrac{4 \theta}{3}\right)+21\right)}{\left(5 \cos \left(\dfrac{4 \theta}{3}\right)+13\right)^{3/2}}

A plot shows that the curvature does indeed oscillate roughly around 2.

The minimum curvature is 13/9, which the curve takes on at polar coordinate (1, 0), as well as at other points. That means that the curve starts out like a circle of radius 9/13 ≈ 0.7.

The maximum curvature is 3 and occurs at the origin. There the curve is approximately a circle of radius 1/3.

Matplotlib vs Mathematica

To make the plot we’ve been focusing on, I plotted

r = cos(2θ/3)

in Mathematica, but in matplotlib I had to plot

r = |cos(2θ/3)|.

In both cases, θ runs from 0 to 8π. To highlight the differences in the way the two applications make polar plots, let’s plot over 0 to 2π with both.

Mathematica produces what you might expect.

    PolarPlot[Cos[2 t/3], {t, 0, 2 Pi}]

Mathematica plot of Cos[2 theta/3]

Matplotlib produces something very different. It handles negative r values by moving the point r = 0 to a circle in the middle of the plot. Notice the r-axis labels at about 22° running from −1 to 1.

    theta = linspace(0, 2*pi, 1000)
    plt.polar(theta, cos(2*theta/3))

Python plot of cos( 2 theta / 3 )

Note also that in Mathematica, the first argument to PolarPlot is r(θ) and the second is the limits on θ. In matplotlib, the first argument is θ and the second argument is r(θ).

Note that in this particular example, taking the absolute value of the function being plotted was enough to make matplotlib act like I expected. That’s only happened true when plotted over the entire range 0 to 8π. In general you have to do more work than this. If we insert absolute value in the plot above, still plotting from 0 to 2π, we do not reproduce the Mathematca plot.

    plt.polar(theta, abs(cos(2*theta/3)))

More polar coordinate posts

[1] Rational Rose was horribly buggy when I used it in the 1990s. Maybe it’s not so buggy now. But I imagine I still wouldn’t like the UML-laden style of software development it was built around.

2 thoughts on “Playing around with a rational rose

  1. Rational Rose RT was used by an infamous class in my university. We had to build an entire multiplayer game using UML and a bit of C++ in the transitions, combined with a Java Swing GUI. It was awful and had a tendency to crash and delete half your project.

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