Assignment 11


Polar Equations


Laura Kimbel






Let's explore how a, b, and k affect the graph above.

By fixing a = b = 1, and letting k vary, we produce the following graph:


The movie above shows various graphs of an "n-leaf rose." As you can see, our k variable affects the number of leaves in our rose as it varies from 0 to 10. All graphs go through the origin except for values of k between 0 and 1. It seems though that if we were to draw a circle that surrounded our graph above, the radius of that circle would not change as k varies. Perhaps a and b might have an effect on the radius of that circle. Let's look at separate graphs as we manually change k.

Beginning with even values of k:

The graphs above are the graphs of k = 2, k = 4, and k = 6, respectively. In each of the graphs, k is the number of leaves. If we exclude the origin, each of these graphs with even values of k, cross the x-axis in 2 places. For this case (a = 1 and b = 1), it is always at x = -2 and x = 2. The leaves are all the same length.

For odd values of k:

Above, we have the graphs for k = 1, k = 3, and k = 5, respectively. As you would expect we have 1, 3, and 5 leaves. This time, if we exclude the origin, our graphs only cross the x-axis in 1 place, x = 2.


Now, let's fix a = 1, k = 2 and let b vary.

For b ranging from 0 to 1:

Above, we can see that b is affecting the length of our leaves, but in this case, our leaves don't appear to actually form leaves until b = 1.

For b ranging from 1 to 5:

When b > 1, we can see our leaves are formed. Here we have 4 leaves for most of the range of b but it does appear to collapse to 2 leaves when b = 1. Since a and b are only equal when b = 1, that is when k determines the number of leaves. It is also interesting here to notice that the leaves are different lengths, yet it is still symmetrical.



Now, let's fix b = 1 and k = 2 and let a vary.

The above video is when a varies from 0 to 1. It appears that we have a similar case here to when b varied from 1 to 5.

Let's see what happens when a varies from 1 to 5.

When a varies from 1 to 5, we have a case similar to when b varied from 0 to 1. It seems as though a and b behave similarly, yet under opposite conditions.



Now let a = 0 and let's examine

Our value for b represents the length of the leaves, so let's let b = 2 and let k vary.

While this video does create a cool arrangement of graphs, it might be easier to analyze this if we consider distinct graphs for the different values of k. Since b affects the length of our leaves, we will just leave b = 2, as it is above. Let's look at k = 1, k = 3, k = 5, and k = 7.

You can see that we produce similar graphs to the ones above when we still had a as a constant. Now let's see what happens for even values of k.

For k = 2, k = 4, and k = 6, we have the following:

These graphs have twice as many leaves for each factor of k than our graphs in part I did. However, the other properties are very similar. They all have even number of leaves, and cross the x-axis twice (excluding the origin).



What if we look at instead of ?

(The purple graph represents cos, while the red graph represents sin.)

When k = 1, it appears that our circle was rotated counterclockwise by 90 degrees.

When k = 2, our graph was rotated but what seems to be 45 degrees or 90 / 2.

When k = 3, our graph was rotated by 90/3 = 30 degrees

Our last example shows that when k = 4, our graph was rotated by 90/4 = 22.5 degrees.



By replacing cosine with sine, it appears that the same graphs are produced but they are rotated by a certain amount. That amount is related to our factor, k. This makes sense since, in the x-y coordinate plane, the graph of sin(x) looks the same as the graph of cos(x), only translated by pi / 2. Since sin(x) and cos(x) are graphed in the x-y plane, their shift is horizontal. Since our investigation above is in the polar coordinate plane, it makes sense that the shift is generated by a rotation.