Assignment 11 Write-up

Billy Poe

Investigate r = a + b cos (kq)

When a and b are equal and k is an integer, the "n-leaf rose" is formed. Here a = b = 2 and k = 3.

Let's see what happens when a and b are not equal. We'll let a = 5 and b = 3.

As you can see, the leaves don't meet in the center when the numbers aren't equal. Let's see what happens if b = 5 and a = 3.

Wow, did you expect this to happen? Not only have the leaves met in the center again, but there are 3 smaller leaves imbedded in the original ones. Let's see what happens if we just change the value of k to 5.

OK, now I'm beginning to make a few connections. Are you? Have you noticed the only thing that changed about this graph is the amount of leaves? How many small ones are there and how many large ones are there? That's right. . .5! The same number that k is! Also, have you noticed yet that there is always one leaf that is symmetrical around the positive x axis? Let's try another one. What if k = -5?

Did you expect this? There is no change! Do you think that it's because changing k from positive to negative simply makes the graph be drawn in a clockwise direction instead of a counter-clockwise direction? It makes sense to me; every point would still be plotted, but in "reverse order"! What if a and b are both negative? Try a = -3 and b = -5.

Aha! Notice that there is no longer a leaf symmetrical across the positive x-axis, but the negative x-axis! The whole graph seems to have flipped over the y-axis! Let's see what happens if we completely take out the a term and leave b = -5.

How about that! It looks like the smaller leaves combined with the larger leaves to make "medium-sized" leaves!

There are many other ways to play with polar equations. Feel free to explore opportunities that I may have left out!