Investigation:

1) I looked at the different graphs of the equation

I varied *a*
as follows: 1,
2, -1, 3, 1.5, -1.5, and 0.25

I noticed
that as *a* increases, the graph begins to look more and
more like a circle.

This time,
I kept *a* as a constant 1 and varied *b* as follows:

1, 2, -1, -2, 0.5, and 0.25.

I notice that
the graphs when *b*=2 and *b*=-2 are reflections of
each other along with *b* = -1 and *b* = 1.

Looking at
the four graphs in which *b* is greater than 0, we can see
this time that as *b* decreases, the graph takes on a more
circular shape.

This time,
I will fix *a *and *b* and change *k*.

Here, *k* = .05. The graph
looks like a circle, but if we look closer around theta equaling
2, we can see that the circle is not closed. There is a slight
opening.

and expanding the domain of theta to 1000, gives us the following:

.

(this was an interesting side-thought).
If we were to look again at fixed *a *and *b*, and change
*k*, we would see the same sort of pattern seen with changing
*a *and *b*. As *k *gets smaller, the graph takes
on a more circular shape.

If we look closely at the graphs
with *k* = 1, 2, and 3, we can see
a pattern with the number of petals on the graph.

For *k* = 100, we have
the following:

with 100 petals ( we can assume).

Now we can compare the following two polar equations:

and
. For fixed values of *a,b, *and
*k*, we get the following graphs:

The equation for the graph in purple is , while the equation for the graph in green is .

If we vary *k*, we can
see something very different. Above, as *k* varied in the
original equation, we saw that the number of loops in the graph
corresponded to the value of *k. *This does not appear to
be the case with .

In this case *k *= 2.

In this case *k* = 3.
If this pattern follows, the number of petals for *k*, even,
should be twice *k*, for *k* odd, the number of petals
should be equal to *k. *

* *