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Assignment 11: Polar Fearful Symmetry

by R. Adam Molnar

In the dim light of the computer screen, it seemed otherworldly.

In trigonometry and precalculus classes, polar coordinates often appear near parametric equations. For instance, Georgia covers them in Math 4, generally taken by 12th graders. These two concepts often seem tossed together with little coherence. However, they can be connected. We can plot parametric curves in the polar system, changing values of the angle \theta and radius r.

The last assignment dealt with parametric curves, changing in time t. Here, we'll examine changing angle \theta as we draw a classic parametric shape, the polar rose. These shapes have been studied for at least 300 years. The assignment title comes from the pictures we'll see. A wide variety of symmetries will arise from this basic equation, where a, b, and c are constants.

r = a + b \cos (c \theta)

Here are three examples. Sometimes we get simple symmetric objects, like the purple circle. Sometimes the symmetry is more complex and a little strange, like the four big loops and four small loops of the blue trace. And sometimes we get a single symmetry, like the red bump.

[three sample graphs]

There's a very, very large field of possibilities from this sample equation, and we can't cover them all in one write up. In this assignment, we'll restrict a, b, and c to small integer values. Furthermore, we won't even talk much about a. There will still be plenty of complexity.

Parameterization: More basic than the need to be known

Before we look at changing values, let's consider how these curves are drawn. Convention places the parametric variable on the right hand side, so we are traveling around the space defined by angle \theta . Typically, we start at 0 and end at \theta = 2 \pi , one rotation around a circle.

Let's look at three points on the blue curve r = 2 + 3 \cos (4 \theta) . Starting at angle zero, we get r = 2 + 3 (1) = 5. We plot the green point ( r = 5, \theta = 0 ). (In Graphing Calculator, the 2-vector will let you plot the point, if you place r on top.) If we move to \theta = \pi / 12, ~ 4 \theta = \pi / 3 , ~ r = 2 + 3 \cos (\pi/3) = 2 + 3 (0.5) = 3.5 . We plot the purple point ( r = 3.5, \theta = \pi / 12 ) . Let's look at one more, where \theta = 3 \pi / 4, ~ r = 2 + 3 \cos (3 \pi) = 2 - 3 = -1 . Because r is negative, the red dot is not in the upper left quadrant where the angle 3 \pi / 4 lies on the circle. It's in the lower right quadrant. ( r = -1, \theta = 3 \pi / 4 ) = ( r = +1, \theta = 7 \pi / 4 ) . In general, symmetry around the circle gives us the relationship ( r , \theta ) = ( -r , \theta + \pi ) .

[three sample points]

If you want to further experiment with the parameterization, and have access to Graphing Calculator Viewer, open the asmt11Blue.gcf file. When you double click anywhere on the blue rose, the coordinates x , y, r, \theta will appear in the upper left corner. Dragging the light blue box around the shape will trace the travel.

B is for Bigger

Of the three variables, b has the most straightforward effect. It affects the size of the loop. Larger values make the traversal longer. Negative values send the trip in the opposite direction. The graph below shows in purple r = \cos ( \theta ) , in red r = -2 \cos ( \theta ) , and in blue r = 5 \cos ( \theta ) .

[varying values of b in b cos theta]

What about the other side?

Something is interesting about the three circles in the last picture. They lie entirely on one side of the vertical axis. The purple and blue circles, with positive b, have no negative x values. The red circle with negative b has no positive x values. We know that the angle goes completely around the plane, including the missing quadrants. What happened?

The answer is negative radius r. When cosine is negative, from \pi / 2 to 3 \pi / 2 in the second and third quadrants, r becomes negative. When we have negative r, we head into the opposite quadrant using the relationship ( -r , \theta ) = ( r , \theta + \pi ) .. So, for instance, r = \cos ( \theta ) begins in the upper right quadrant I. When cosine enters the upper left quadrant II, the negative radius moves us into lower right quadrant IV. Negative cosine quadrant III moves into quadrant I, while quadrant IV remains positive. Our circles are actually being drawn twice, once with \theta values from 0 to \pi and then redrawn from there to 2 \pi .

C is for Cotyledon

Variable c, inside the cosine, affects speed and symmetry. Because \cos ( k \pi / 2) = 0 for all odd numbered k, such as \cos ( 1 \pi / 2) , \cos ( 3 \pi / 2) , \cos ( -9 \pi / 2) , the journey from zero to 2 ~ \pi will lead to (2c) crossings of zero. For instance, if c = 3, we have r = 0 at \theta = \pi / 6 , 3 \pi / 6 , 5 \pi / 6 , 7 \pi / 6 , 9 \pi / 6 , 11 \pi / 6 . From zero to the first crossing at \pi / 6 radius is positive, keeping the graph in Quadrant I. Here, the white square indicates \theta = 0.360 . I added the arrows to show the direction of travel.

[Tri leaf, first quadrant]

Once we pass zero, we move into negative values for r and thus head to the lower left Quadrant III. The white box now signifies the point at roughly ( r = -0.425 , \theta = 0.670 ) .

[Tri leaf, third quadrant]

The track continues back to the center, loops around the upper right second quadrant, heads back to zero, and enters the fourth quadrant. The fourth quadrant uses negative values of r. Eventually, we reach the point ( r = -1 , \theta = \pi ) = ( r = 1 , \theta = 0 ) . On our second pass around, we travel via the general symmetry ( r , \theta ) = ( -r , \theta + \pi ) .

In this case, we set c = 3 and saw three leaves or petals. For any odd positive integer c, we will see c leaves. This makes c the "leaf control" value. The word cotyledon means seed leaf, an appropriate term to describe this variable. The graphs can get busy. The purple graph has c = 5, while the blue graph has c = 11.

[Five and Eleven leaves]

What about even c?

If we set c = 2 we might expect two leaves. But Graphing Calculator gives us four leaves, not two. What happened?

[c=2, four leaves]

It's not a cause for panic. The difference is that we don't have the symmetric repeat. We travel through the first quadrant and enter the third quadrant, as before. Things change, as we reach the point \theta = \pi / 2 = 1.5708 where \cos ( \pi ) = -1 . marked with a white box at the bottom of the graph. We then enter Quadrant II for the angle with negative radius, putting us in the lower right corner of the graph. Once we reach \theta = 3 \pi / 4 , we again have positive values for r.

At the halfway point for odd c, \theta = \pi , we had r = -1 and so ( r = -1 , \theta = \pi ) = ( r = 1 , \theta = 0 ) . That's not true for even c. Since \cos ( 2 \pi ) = +1 , we are now at the point ( r = +1 , \theta = \pi ) on the left side of the graph. We don't have a repeat.

[c=2, first half only]

Continuing the parameterization from \theta = \pi to \theta = 2 \pi will draw another full loop and two half loops, completing the four leaf picture shown above. If you'd like to trace the picture, you can open the asmt11cos2theta.gcf file. When you double click anywhere on the purple rose, the coordinates x , y, r, \theta will appear. Dragging the light blue box around the shape will trace the travel.

Finally (at least for this assignment, since there are many more things we could do), let's look at larger even c. Because of the lack of repetition, an even value for c will lead to (2 c) petals in the rose. For instance, the graph below shows c = 6 with 12 symmetric petals. While the symmetry may be stark, by looking at polar parameterization it's not as mad as it might have seemed.


The title and first two section headings are adapted from Audrey Niffenegger's 2009 book Her Fearful Symmetry, the followup to The Time Traveler's Wife, since this is a follow up to Assignment 10.

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