EMAT 6680 :: Assignment 11
:: Problem 1 :: Clay Kitchings

Investigate

Note:

When
a and b are equal, and k is an integer, this is one textbook version of the
" n-leaf rose."

Compare
with

for
various k. What if . . . cos( ) is replaced with sin( )?

***Note: For the purposes of working this problem,
any ordered triple shall represent the values a, b, and k as (a, b, k).***

Unless otherwise noted,.

We
shall begin with perhaps the most basic case: (1, 1, 1). The resulting graph
is:

For
now, we shall leave k=1 and modify a and b to see how this affects the graph.

(same as first graph above)

(new blue graph for a=2, b=2)

(The
value k remains at 1 for now.)

It
appears that when a = b the graph has a similar pattern each time. Furthermore,
it appears that the a and b values have something to do with where the graphs
intersect the x-axis.

Observation
leads me to conjecture that a-b will be the opposite value of the x-intercept
on the negative x-axis and a+b is the value of the x-intercept of the
positive x-axis. Even if a<b, a-b will be a negative distance for the
negative x-axis which would place the x-intercept on the positive x-axis and
its opposite value (-[a-b]) will be positive. For example, if a=1 and b=2,
consider the graph below.

Values: (1, 2, 1)

So,
notice a<b in this case. a – b = -1, and its
opposite is positive 1. For the
a+b conjecture, a+b=3 and the graph intersects the positive x-axis at 3.

What
happens if a and b are both less than zero?

(-1, -2, 1)

If
a<0 and b<0, the graph is reflected over the y-axis.

If
,
then the graph does not loop within itself, but rather begins to appear more
circular as the distance between and b increases (when ).

Now,
let us modify the value of k for these kinds of graphs.

Fix
a=b=1 and vary k to obtain the following graphs:

K=2:

k=3

k=4

k=5

From
the former three observations seem to imply that for a=b=1, the k-value will correspond
to the number of leaves or pedals the graph has.

Now,
let us fix a=1, k=5 and vary the value of b. At this point, we shall also fix
a>0 and b>0

b=2

b=3

It
appears that as long as a<b, another set of pedals (corresponding to k)
forms within the outer pedals. What happens when a>b? At this point we
shall fix b=1, k=5 and vary a.

a=3

a=4

a=5

As
we observed earlier in this problem there appears to be a relationship between
a, b, and the left and right x intercepts. Notice that a-b appears to provide
the left-most intercept on the x-axis and a+b provides the right-most
x-intercept. As before, we see the a-value appears to correspond to the lower
y-intercept (-a) and the upper y-intercept (a).

What
happens if we observe the equation: ?

First
we notice that as b increases, the size of the resulting graph from the k-value
increases as well. So how do
various values of k change the graphs characteristics? For the purposes of
this investigation, we shall fix b at 3.

We
shall now compare the following graphs:

Green Red

There
appears to be a correlation between even k-value and the number of leaves or
pedals that exist in the graphs. For example, the green graph has a k-value
of 2 and contains four pedals whereas the red graph results from a k-value of
4 and has 8 pedals in it. From
observing other even values for k, we conjecture that if k is even, there are
2k pedals in the resulting graph. For negative, even values of k, we observe
identical graphs.

What
happens when k is an odd value?

Using
the same equation as the previous example, we will only modify k such that it
takes on odd values. If k is odd,
there are k leaves or pedals in the graph.

Example:
k=5 (there are five pedals/leaves):

Finally,
what might happen if we decided to use sine instead of cosine in these
equations?

First,
let us view the graphs of r = sin(bq) and r = cos(bq) in blue and red
respectively. (k=2)

It
appears that there is some kind of rotation going on, but this is only an
observation for k=2. It would make sense that this might be rotated in some way
since sine and cosine are shifts of each other.

Lets
make k=4 and try it again

I
have superimposed an angle between one pedal from cosine and one sine pedal.
This angle is representative of a possible angle of rotation. For the even
values of k, it is worth noting that this angle of rotation appears to be equivalent
to 90/k (assuming a counter-clockwise rotation). Here, 90/4 = 22.5 degrees. Why
four? Because four such superimposed angles would fit into 90 degrees.

Let
us view this with an odd value for k and see if it appears to be true there as
well.

If
indeed 90/k provides the amount of the counter-clockwise rotation, then in this
case the rotation would be 18 degrees.