Fun with Basic Polar Equations

r = a + b*cos(kq)

If a and b are equal and k is an integer, what do these graphs look like? First, let's look at a = b = +/-0.1, a = b = +/-0.2,... a = b = +/-1, a = b = +/-1.1.

Let's separate these two graphs into +a = +b and -a = -b.

If a and b are negative, it just flips the graph of a and b being positive. It also looks like the graph crosses the y-axis at +a and -a and crosses the x-axis at 2*a and 0 when a = b. This is called the "n-leaf rose" from Dr. Wilson's home page. Now, is the a really necessary in the equation? What happens when the a is removed? This time, I let b = +/- 0.2, 0.4, ..., 1.8, 2.0.

r = b*cos(kq)

It is now very apparent that the graph crosses the x-axis at 0 and b. The graph only crosses the y-axis at 0 so the a in the previous equation did tell us where the graph would cross the y-axis (-a and a).

For all of the graphs thus far, k = 1. What if k = 2 or -2 or 1/2? I am going to look at the original equation first where a = b = 1.

r = a + b*cos(kq)

k = 2

k = -2 is really boring -- it looks just like the graph of k = 2. Let's assume that -k looks the same as +k. This is not hard to believe since cos q = cos (-q)!

k = 3

k = 4

It seems that our "flower" will have as many petals as there are k's. So, if k = 10, will we have 10 petals?

YES! What if k is not an integer? Let's look at k = 1.0, 1.1, 1.2, ..., 1.8, 1.9, 2.0.

Wow! That is one crazy looking graph!

Now, what happens when the a is removed for different integer values of k where b = 1?

r = b*cos(kq)

k = 2

k = 3

k = 4

k = 5

I think that if k is odd, there would be k petals in the flower and if k was even, there would be 2k petals in the flower. Just to make sure, let's look at k = 10. If my conjecture is correct, there will be 20 petals on the flower.

Yea! There are 20 petals! What would happen to my graphs if b was changed to 2?

All that changed was that the length of the petals changed to 2. Therefore, the length of the petals equals b.

What if the cos is replaced with sin?

r = a + b*sin(kq)

If a and b are equal and k is an integer, what do these graphs look like? First, let's look at a = b = 0.1, 0.2,..., 1, 1.1.

I think that it is safe to say that if a = b = -value, the graph would be reflected about the x-axis like it was reflected about the y-axis for cos. Will the graph rotate 90 degrees for other equations of sin? What if a = b = 1 and k = 2?

It is definitely not a 90 degree rotation, but a rotation none the less. What if a = b = 1 and k = 3?

What if a = b = 1 and k = 4?

What if a = b = 1 and k = 10?

Changing the cosine to sine in the equation is not that exciting in my opinion. It only rotates the graph about 45 degrees (?).

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