## Investigations of The Polar Equation r = a + bcos(kt) varying a, b, and k.

#### by Jennifer Roth

Given the equation r = a + bcos(kt), first we will examine different graphs for a = b, with k having interger values.

For a and b both equal to 1 and k = 1, we get the following graph

What happens if we change k? If we let k = 4, leaving a and b both equal to 1, we get the following graph.

Now there are 4 leaves, and remember what the value for k was!
One more graph with a and b both at 1 and a different value for k might help us make some conclusions. If w let k = 6, we get the following graph.

Now there are 6 leaves, and we set k = 6. It appears that the number of leaves of the rose is determined by k, but we need to do some further investigation letting a = b with different values for a.
If we let a = 2 = b, and let k = 1, we get the following graph.

Compare this to our first graph in which we set a = 1 =1 and let k = 1.

Now we will vary k, and let a = 2 = b. In the following graph we set k = 6.

This graph looks like the graph in which we set a = 1 = b, and let k = 6. So, it "appears" that when a and b are equal, and k is an integer we get a k-leaf rose.

Now we will set k equal to non-integer values.
If we let a = 2 = b, and let k = 1/2, we get the following graph.

How does this compare to our original graph in which a = 1 = b, and k = 1.
If we let a = 2 = b, and k = 5/2, we get the following graph.

We have a 5-leaf rose, but the leaves overlap to get two sets of 5-leaf roses. So, it appears we will get a m sets of n-leaf roses for k = n/m. Lets view more graphs to see if this holds true.
If we let a = 2 = b, and k = 1/3, we get the following graph.

And, if we let a = 2 = b, and k = 4/3 we get the following graph.

It appears that we are getting m sets of n-leaf roses for k=4/3, but what if we vary the values for a = b.
If we let a = 1/2 = b, and let k = 4/3, then we get the following graph.

It appears the same thing is going on here.

Now we will examine different graphs for a < b, with k having interger values.

If we let a = 1, b = 2, and k = 1 we get the following graph.

How does this compare to our other graph's. Let's do some more graphs and see what conclusions we can draw.

If we again let a =1, b = 2, but let k = 6, we get the following graph.

So it appears we are getting 2 sets of k-leaf roses in this case. So, where is the 2 soming from. Let's vary a and b, and see if there could be some relation between their values and the number of sets.

If we Let a = 1, b= 3, and let k = 6, we get the following graph.

It appears we are still getting 2 sets of k-leaf roses. If we do further graphs for different values of a and b as long as a < b, and k is an integer value we will get 2 sets of k-leaf roses.

Now we will look at graphs with a < b, and k set equal to a non-integer value.
If we let a = 1, b = 2, and k = 1/2 we get the following graph

If we let a = 1, b = 2, and k = 5/2 we get the following graph.

So, what is going on with these graph's that is different from the other sets of graphs?
It might be easier to see if we look at some different values for k.
If we let a = 1, b = 2, and k = 1/3 we get the following graph.

If we let a = 1, b = 2, and k = 4/3, we get the following graph

What conclusions can we draw? How many sets of roses do we have, and how many leaves does each rose have?

Now we will look at graphs with b < a , and k set equal to a non-integer value.
If we let a = 2, b = 1, and k = 1 we get the following graph

If we let a = 2, b = 1, and k = 1/2, we get the following graph.

Again we will let a = 2, and b = 1. If we let k = 5/2 we get the following graph.

Look at the other graphs in which k = 5/2. What is going on here that is different?
Let's look at some more values for k and see if we can make any conclusions.

If we again let a = 2, and b = 1, and let k = 1/3, then we get the following graph.

And, in the following graph k = 4/3

Make comparisons with this graph and the other graphs. What conclusions can be made for:
1) a = b, k integer value
2) a = b, other values for k
3) a < b, k integer value
4) a < b, other values for k
5) b < a, k integer value
6) b < a, other values for k
What kind of general conclusions can be made?