Ashley Jones

Assignment 11: Polar Equations

Investigate r = a + b cos(k*theta)

Compare with r = b cos(theta) for various values of k

Before comparing the two equations, I wanted to explore the first equation r = a + b cos(k*theta). When the parameter k is varied between -10 and 10 we get a very unique looking illustration. The graph shows us a rose with k number of leaves. As suggested from the assignment, this is often referred to as a 'n-leafed rose.' During this illustration the values of a and b for the equation remain equivalent to one another (a = b = 1). Below is the illustration for the 'rose' graph formed from the polar equation r = a + b cos(k*theta).

Next, I examined the graph of the polar equation as the parameter a varied between -10 and 10. For this illustration I kept the value of b = 1 and the value of k = 1. Below is the illustration of how the varying value of the parameter a affects the graph of the polar equation.

The shape of the graph is greatly affected by the parameter a. At the values when the parameter a = -10 and when a = 10, the graph is in a full ellipse.

Instead of varying the parameter a, I now wanted to keep the parameter a = 1 and k = 1 but vary the parameter b between -10 and 10. Below is the illustration of how the varying of the parameter b affects the graph of the polar equation.

In this variation of the polar equation, the parameter value of b causes the graph to look like a double ellipse, one inside another. The 'double ellipse' seems to disappear as the value of the parameter approaches 0. This is a very interesting illustration of the graph of this polar equation.

In the next illustration, I explored the graphs of the original polar equation in comparison with the polar equation r = b cos(k*theta). The graph of the new polar equation was very unique and showed a different sized 'rose' in comparison to the original polar equation I observed. Below is an illustration of the two polar equations graphed together (first equation is purple and second equation is red) as the value of the parameter k varies between -10 and 10.

When examining the two graphs of the polar equations I found a few differences. The length of the petals of the original rose (from r = a + b cos(k*theta)) was twice as long as the petals from the rose formed by the equation r = b cos(k*theta). I believe that this is because in the first equation we have added the parameter a which is equal to 1. Thus when b = 1 (in this particular case), the first equation forms a rose with petals that are longer than the second equation's petals. Also, when the absolute value of the parameter k was an even value, the number of petals in the rose from the second equation (r = b cos(k*theta)) doubled the number of petals in the rose from the first polar equation (r = a + b cos(k*theta)).

To compare the original polar equation I looked at, r = a + b cos(k*theta) I added another equation r = a + b sin(k*theta). So, I simply changed the cosine to a sine function to see what the change in the polar equation would do to the graph. Comparing the two graphs, as the parameter k varied between -10 and 10, of the polar equations, I saw that the sine polar equation (red) differed from the cosine polar equation. While they were the same graph shapes with the same number of petals on their roses, the sine polar equation was rotated slightly so as not too lay directly on top of the cosine polar equation. Below is the illustration that I examined for this exploration.

This exploration opened my eyes to the world of polar equations. I know that more explorations could be done with these equations, which is a great way to expand the activity for classroom use. I feel that students would enjoy seeing such creative uses of polar equations, which might make the mathematical topic more interesting to them.

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