EMT 668 - ASSIGNMENT 11, #1
Polar Equations

by

Kimberly N. Bennekin


Investigate

r = a + bcos (kt)

for a=b, a<b and a>b when

k = integer values within the interval [0,2Pi],
k = 1/2, 3/2, 5/2... within the interval [0, 4Pi],
and k = 1/3, 2/3, 4/3... within the interval [0, 6Pi].


We will use Theorist to investigate this problem by graphing many variations of the polar equation:

r = a + b cos (kt)

1) Consider the polar equation for a = b = 1.

A) We will begin the investigation for integer values of k. Consider the following graphs:

r = 1 + cos (t) - black
r = 1 + cos (2 t) - red
r = 1 + cos (3 t) - green

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:

r = 1 + cos (1/2 t) - black
r = 1 + cos (3/2 t) - red
r = 1 + cos (5/2 t) - green

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:

r = 1 + cos (1/3 t) - black
r = 1 + cos (2/3 t) - red
r = 1 + cos (4/3 t) - green

The results are similar to our previous results.


2) Now let's consider the equation for a < b. Let a = 1 and b = 2.

A) Consider the equation r = 1 + 2 cos (k t) for k = 1, 2, and 3:

r = 1 + 2 cos (t) - black
r = 1 + 2 cos (2 t) - red
r = 1 + 2 cos (3 t) - green

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).


B) Consider the same equation for values of k = 1/2, 3/2 and 5/2.

r = 1 + 2 cos (1/2 t) - black
r = 1 + 2 cos (3/2 t) - red
r = 1 + 2 cos (5/2 t) - green

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."


C) Lastly, let's consider r = 1 + 2 cos (k t) for k = 1/3, 2/3, and 4/3. I predict the same results as in the explorations with r = 1 + cos (k t) for the same values of k. Consider the following graphs:

r = 1 + 2 cos (1/3 t) - black
r = 1 + 2 cos (2/3 t) - red
r = 1 + 2 cos (4/3 t) - green


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:

r = 2 + cos (t) - black
r = 2 + cos (2 t) - red
r = 2 + cos (3 t) - green

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.

r = 3 + cos (t) - black
r = 3 + cos (2 t) - red
r = 3 + cos (3 t) - green

The prediction is correct. As a increases, the more our graphs are stretched outward.


B) Let's consider the original equation for k = 1/2 3/2 and 5/2:

r = 2 + cos (1/2 t) - black
r = 2 + cos (3/2 t) - red
r = 2 + cos (5/2 t) - green

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.


C) A similar situation will occur for k = 1/3, 2/3 and 4/3:

r = 2 + cos (1/3 t) - black
r = 2 + cos (2/3 t) - red
r = 2 + cos (4/3 t) - green

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.


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