The Òn=leafÓ Rose and
Variations

By:

Alex Moore

In
this investigation we explore polar coordinates and the Òn-leafÓ rose. Before we dive into mathematics fun we
first review polar coordinates.
Just like the standard x,y-plane is a coordinate system, polar
coordinates is another coordinate system of the plane. To each point in the plane we assign an
ordered pair (r,t) where r is the distance from the point to the origin and t
is the angle of the line segment from the origin to the point, counterclockwise
with respect to the x-axis. Here
is a picture below: (we use t in place of theta for typing ease)

What
is the relationship between our beloved x,y-coordinates and polar
coordinates? Consider the point P
above. Notice the dashed lines
create right triangles. This
allows us to use our basic trigonometry knowledge! Using trig, we see the length of the leg of the triangle
along the x-axis is rcos(t) and the length of the leg of the triangle parallel
to the y-axis is rsin(t). Thus, we
have just ÒparametrizedÓ the plane:

x = rcos(t);

y = rsin(t).

For
function; we generally write r as a function of t, r(t), rather than y =
f(x). Geometrically, this means
that as we rotate counterclockwise around the origin the distance from the
origin is changing depending on t.
For example: consider the
function r=1. This function
represents the set of all points a distance 1 from the origin (as t changes, r
stays constant at r=1), therefore r=1 is the unit circle. What about r=at for any real a? This says r grows with t at a linear
rate:

Now
that we have a basic feel for this new coordinate system we would like to
investigate functions of the form r = a +b cos(n t). The simplest example is a=b=0 but this would mean r=0 and
this is just one point: the origin.
If we let b=0 then we get r=a and this is simply the circle centered at
the origin of radius a. Now let us
examine our first real example: let a=b be nonzero and let k be an
integer. What happens? This!

These
graphs are the Ò3-leafÓ and Ò6-leafÓ roses. In general we call the graph of r=a(1+cos(k t)) the Òn-leafÓ
rose, scaled by a factor of a.
Nice pictures, eh? Lets
alter this slightly. What if we
ignore the 1, that is, what about the functions of the form r=a cos(k t)? Let us see using the same values of a
and k in the previous example.

What
changed? It ÒappearsÓ that the
blue graph didnÕt change in the number of leaves and the red graph went from 6
leaves to 12 leaves. What actually
happened to the blue graph is the graph repeated. If we let t run from only 0 to pi then we get the same
graph. While the red graph needs t
from 0 to 2pi to complete its 12 leaves the red graph needs only 0 to pi.

The
reason we get nice, closed paths (leaves) is because the values for k are
Ònice.Ó These values are
integers. If we let k be rational
we get partial graphs (depending on the denominator). The most interesting case, in my opinion, is when we let k
be an irrational number. If k is
irrational then k t is a multiple of pi only when t=0. This means that r=a cos(k t) will never
have a closed path. Speaking in
topological terms, if we consider the set of all points (r,r(t)) with r then we
get a dense subset of the unit disk!
Here we graph for value of t between 0 and 100.

What
changes if we use sin(k t) rather than cos(k t)? After some algebra,

r = a + bsin(k t) = a +
bcos(k t – pi/2) = a + bcos(k(t-pi/2k)) = a + bcos(k tÕ), where
tÕ=t-pi/2k.

Using
this change of variables we see that the graphs of these two functions are just
rotations of one another! This
should not be surprising to us given the nature of sine and cosine and the
relationship between them, sin(t)=cos(t-pi/2).