As k increases, the number of "petals" in our graph increases.
B) Let's investigate the same equation for k = 1/2, 3/2 and 5/2:
For k = 1/2, we get 1/2 of a "petal" of r = 1 + cos (t).
For k = 3/2, we get the graph of r = 1 + cos (2 t) minus 1/2 of a
"petal." Similarly, for k = 5/2, we get the graph of r = 1
+ cos (3 t) minus 1/2 of a "petal."
Let's investigate k = 1/3, 2/3 and 4/3 within the same equation and see
if we get the same results:
The results are similar to our previous results.
Multiplying the parameter t by 2 doubles the number of "petals" which appear in the graph of r = 1 + cos (k t). I predict if b = 3, it would triple the number of "petals" in the graph of r = 1 + cos (k t).
This had the same effect as in the previous exploration of r = 1 +
cos (k t) for k = 1/2, 3/2 and 5/2.
For k = 1/2, we get 1/2 of a large "petal" of r = 1 + 2 cos
(t). For k = 3/2, we get the graph of
r = 1 + 2 cos (2 t) minus 1/2 of a large "petal." Similarly,
for k = 5/2, we get the graph of r = 1 + 2 cos (3 t) minus 1/2 of
a large "petal."
3) Lastly, let us consider the equation for a > b. Let a = 2 and
b = 1.
A) Consider the equation r = 2 + cos (k t) for k = 1, 2, and 3:
This has a much different affect. It seems that increasing a stretches
the graph outward. To be sure, let us consider the same set of graphs for
a = 3 and b = 1. Consider the graphs below.
The prediction is correct. As a increases, the more our graphs are stretched outward.
Again, with a slight distortion, this had the same effect as in the previous exploration of r = 1 + cos (k t) for k = 1/2, 3/2 and 5/2 by taking away 1/2 a "petal" from each graph.
These are just a few of the many explorations that can be easily made
using Theorist. Additonal topics may include replacing the cosine
function with the sine function or maybe eliminating a and considering
r = b cos (k t). Explore some of these topics for yourself.