Exploring Polar Equations: nLeaf Roses
By: Laura Lowe
Problem: Investigate
r = a + b cos(kq)
We begin by graphing r = a + b cos(kq) with a
= 1, b = 1, k = 1, and 0 < q < 2p.
This
is an interesting graph, but what if we hold a = 1, b =
1, and 0 < q < 2p and let k vary? (We will restrict a, b, and k to
be positive rational numbers greater than 1.)


k = 2 
k = 3 


k = 4 
k = 5 
It
appears that k, a natural number, dictates the number of loops (called
leaves)in the figure. If k = 7, there should be 7 leaves. Try it and see.
Now
that we have a conjecture when k is a natural number, what about when k is
rational? Logically, we would
expect to get partial leaves.
LetŐs try a few.


k =
½ 
k = 3/2 


k = 5/2 
k = 7/2 
\We
did get partial leaves, but it looks like the pattern may continue if we make q larger.
LetŐs try letting 0 < q < 4p.


k =
½ 
k = 3/2 


k = 5/2 
k = 7/2 
If we
suppose k = c/d where c and d
are natural numbers, it appears that the graphs have c leaves when we let 0 < q < 2dp. LetŐs
try a few.


0 < q < 6p 
0 < q < 8p 


It
appears our conjecture is true.
In fact
when a and b are equal (but not 0) these properties still
hold. The only difference is
the relative size of the rose.
Now
letŐs look at the graph when we hold a =1 and k = 1and let b vary.



b = 2 
b = 3 




b = 4 
b = 5 
It appears that as be
increases, the graph becomes a circular figure with an outer loop intersecting
the xaxis at x = 0 and x = b +1,
and the inner loop intersecting the xaxis
at x = 0 and x = b
– 1. At the same time the yintercepts remain at y = ±1. So we
would expect that when b = 6, then
the outer loop would intersect at x
= 0 and x = 7 and the inner loop
at x = 0 and x = 5. Try
this and see.
Again,
we have a conjecture when b is a
natural number, so what about when b
is rational? Since b appears only to affect the ŇsizeÓ of the loops, we
would not expect this to change when b is rational. LetŐs
try a few.


b = 3/2 
b = 5/2 
Again it appears our
conjecture is true.
Now
letŐs look at the graph when we hold b =1 and k = 1and let a vary.
a = 2 
a = 3 
a = 4
It appears that as a increases, the graph becomes more and more circular
with xintercepts x = a + 1
and x = a +1, and yintercepts
y = ±a. So
when a = 6, we would expect our xintercepts to be x = 5 and x
= 7 and yintercepts to be y = 6
and y = 6. Try this and see.
Again,
we have a conjecture when a is a natural
number, so what about when a is
rational? Since a appears only to affect the ŇsizeÓ of the circular
figure, we would not expect this to change when a is rational. LetŐs try a few.
a = 3/2 
a = 5/2 
Again it appears our
conjecture is true.
Now letŐs compare these
graphs to the graphs of r = bcos(kq). To
begin we will let b = 1, k = 1, and 0 < q < 2p.
From
our earlier explorations we can expect that if we hold k = 1 constant, and let b vary, the graph should remain a circle only bigger.
b = 2 
b = 3 
Now
what if we hold b = 1 constant and
let k vary we should expect, from
our earlier explorations, that k
will dictate the number of leaves in the rose.


k = 2 
k = 3 


k = 4 
k = 5 
This is
interesting. Now, when k is even, we have 2k leaves on the rose, but when n is odd we only have k leaves on the rose. LetŐs see what happens when k is rational.


k =
½ 
k = 3/2 


k = 5/2 
k = 7/2 
Again,
we got partial leaves. When k = c/d the graph has c – 1full leaves and 2 half leaves. As we found earlier, we can guess that if we make q larger, we will see more leaves.
Let 0 < q < 4p and graph.


k =
½ 
k = 3/2 


k = 5/2 
k = 7/2 
Now it
appears that the graphs have 2c
leaves when we let 0 < q < 2dp. LetŐs
try a few more and see if our conjecture holds.


0 < q < 6p 
0 < q < 8p 


Another
interesting result. This time,
when c is odd and d is odd there are c leaves, and when c is odd and d is even, there are 2c
leaves. What if c even and d
is odd? LetŐs try , with 0 <
q < 6p. (We will disregard the
case of c and d both even, as it it will reduce to one of the other
cases.)
So when
c is even and d is odd, there are 2c leaves.
Try more examples and see if this result holds.
What if
cos is replaced with sin in all of these?
A quick check reveals that this only rotates the graph 90ˇ counterclockwise. Why?
(Hint: Look to your unit
circle.)
Extensions:
What
happens when b or k are 0?
What
happens when we let a, b, and k be negative?