Polar equations

Investigation 1, conducted with Theorist, involves graphing the following:
r = a + b cos(kt)
for a=b, a<b, and a>b;
using integer values for k from 0 to 6.

For the first case, when a=b, and k is an integer, the result is the "n-leaf rose"; where the number of leaves is the value of k.

For k=5:

This graph holds for both odd and even values of k. The following is k=6:

Returning the k value to 5, we can change the relationship of a and b. In the case where a>b, the "rose" no longer has a center. Because the graph does not go into the origin, the "rose" now looks more like a "starfish".

The last relationship with a and b is a<b. In this case we get a smaller petal within each petal.

When we go back to a=b, and change the value of k to a rational number such as 5/2, we get an incomplete "rose":

In order to complete this "rose" change the values for t from 0...2 Pi to 0...4.Pi.

When the value of k is a rational number with a different denominator, the value of needs to equal 2k Pi.

An interesting graph results when a>b, and k is 5/2:

We can now change the function to see if the same effects hold for changes in b, and k.

The function under investigation is:

r = b cos(kt)

The resulting graphs are much smaller than the previous ones and the number of pedals equals 2k for even values of k; 1k if k is odd. The size of the graph can be changed by increasing the values of b.

The following graph is of this function when k is 7, and b is 2.

The changing from cos to sin merely changes the orientation of the graph's pedals.

The investigation leads into changingthe function to:

r = 2a sin (kt)
varying the values of a, and k

to get the resulting graphs. The a value determined the size of the "rose" and the k value determined the number of pedals. This graph is when k= 7/2 and the range of t is expaned to 0...4Pi.

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