Investigation of Polar Equations

By: Denise Natasha Brewley and Venki Ramachandran

To begin our investigation, we consider polar equations of the following form, where a, b, n are integers such that a=b >0.

*r = acos(nq)
and r=a+bcos(nq)*

For this case we consider when a=2 and b=2. Notice that for this case, as depicted below, the graph of three polar equations are given including the equation of r=2. We will first consider what happens when n=0 and how this impacts our graphs. Notice that for the n=0 case, the equations r=2cos(nq) and r=2 are the same - they overlap. This follows because cos(nq)=cos(0)=1. We have a graph of two concentric circles such that the blue circle has radius 4. This also follows because r=2+2cos(nq) simplifies to r=2+2(1) = 4.

We now want to know what happens to our graphs when a and b are fixed but n varies.

Now let us investigate the n=1 case. Notice that our purple
graph becomes r=2cosq which is
the graph of a circle.
For *q* in the
interval 0__<__*q*<2p,
we can get a full range of values for *r *given below. As
q increases from 0 to 2p,
the values of *r* on our circle start at 2 for q=0
and ends at 2 for q=2p.
This explains why the graph of r=2cosq
is translated to the right two units as opposed to centered at the origin.
Notice for this equation that the radius of this circle is 1. Lets figure
our why this is so. We can conjecture that for r=acosq,
a is the diameter of the circle. We know this because on the interval 0__<__*q*
<p/2,* *it yields
values for *r* on the interval 2__<__r__<__0. Since a=2 is the
diameter, we can conclude that the radius is 1. Another important
observation is that r=2cosq
is symmetric with respect to the Polar Axis.

q |
0 | p/6 | p/3 | p/2 | 2p/3 | 5p/6 | p | 7p/6 | 3p/2 | 11p/6 | 2p |

r |
2 | Ö3 | 1 | 0 | -1 | -Ö3 | -2 | -Ö3 | 0 | Ö3 | 2 |

The other graph that will be discussed is our blue
graph of r=2+2cos(q).
This graph is a bit different. It is the graph of a cardioid which is a
heart shaped figure. Again, for
*q* in the interval
0__<__*q*<2p,
we will look at a range of values for *r *given below. The graph of
the cardioid is also symmetric with respect to the Polar Axis. Also for 0__<__*q*<p,
we have values of *r *that range from *2*__<__r<0.

q |
0 | p/6 | p/3 | p/2 | 2p/3 | 5p/6 | p | 7p/6 | 3p/2 | 11p/6 | 2p |

r |
4 | 2+Ö3 | 3 | 2 | 1 | 2-Ö3 | 0 | 2-Ö3 | 2 | 2+Ö3 | 4 |

Now we can consider the n=2 case. Some interesting things
happen to our graphs. Let us start with the r=2cos(2q).
An observation is that we have a four pedal rose which is symmetric with respect
to both the Polar Axis and the line q=p/2.
The length of each pedal is 2 and two pedals are on the Polar Axis and two
pedals are on the line q=p/2
line, which confirms the symmetry property. The maximum value of r is 2.
Let us take a look at the table of values of *r *below. This table
helps us to construct each pedal of our rose curve.

q |
0 | p/6 | p/4 | p/3 | p/2 | 2p/3 | 3p/4 | 5p/6 | p | 7p/6 | 3p/2 | 11p/6 | 2p |

r |
2 | 1 | 0 | -1 | -2 | -1 | 0 | 1 | -2 | 1 | -2 | 1 | 2 |

Now let us look at r=2+2cos(2q).
In this case, we have a two pedal rose. Again, for this case the rose is
symmetric with respect to the Polar Axis. Furthermore, we can also look at
the range of values of *r *for this case as well. Notice that for 0<q<p/2,
the values for *r* range from 4<r<0.

q |
0 | p/6 | p/4 | p/3 | p/2 | 2p/3 | 3p/4 | 5p/6 | p | 7p/6 | 3p/2 | 11p/6 | 2p |

r |
4 | 3 | 2 | 1 | 0 | 1 | 2 | 3 | 4 | 3 | 0 | 3 | 4 |

What happens as we continue to increase the values of n ? It appears that the number of pedals can be found when n is odd and when n is even. Let see if this hypothesis holds.

Consider the n=3 case. Let us discuss what happens in general for both graphs. In the case of r=2cos(3q) and r=2+2cos(3q) we only have three pedal roses. So it appears that when n is odd for both polar equations r=2cos(nq) and r=2+2cos(nq), we have n pedals. Does this conjecture hold when n is even. Let us see.

For the n=4 case, we have eight pedals for the
polar equation r=2cos(4q)
and only four pedals for r=2+2cos(4q).
So our conclusion is a bit different than in the case of odd values of n.
We can conclude that for the polar equation of the form r=2cos(nq),
when n is even, we obtain 2n petals provided that n__>__2. But for the
equation r=2+2cos(nq),
we only have n petals.

Now what if we changed our polar equations and considered the following with the same restrictions on a, b and n.

*r = asin(nq)
and r=a+bsin(nq)*

What happens to our graph for n=0,1,2,3, and 4. Let's take a look below.

For n=0, as depicted below, the graph of three polar equations are given again including the equation of r=2. The graph of r=2sin(nq) is a point. This follows because 2sin(nq)=2sin(0)=2(0)=0. Notice also that the equations r=2+2sin(nq) and r=2 are the same. They also overlap. We have a graph of one circle such that radius is 2. This also follows because r=2+2sin(nq) simplifies to r=2+2(0) = 2.

For n=1, we have the same graphs as the cosine case but they are translated by p/2. For r=2sin(q), we have a circle that is symmetric with respect to the line q=p/2. And for r=2+2sin(q) we have a cardioid that is translated the same number of radians and also symmetric with respect to the line q=p/2.

In the case when n=2, we have the rose curves again. This
time in both cases, r=2sin(2q) and r=2+2sin(2q)
we have symmetry with respect to the Pole. Note that the conjecture that
we made for the cosine case holds here as well. That is, when n is even
for r=2sin(nq) we also have 2n petals for n__>__2.
And for r=2+2sin(nq), we have n petals as well.

Since we have generalized, we will conclude this investigation with the cases n=3 and n=4. It looks like our conjecture holds.

When n=3,

When n=4,