__Polar Explorations__

By: Brandie Thrasher

By exploring
Polar equations, we can take a glance and construct the Òn-Leaf RoseÓ. LetÕs
start off with the basic equation of r = b cos(kq)

Having b and k equal
1, it appears that our equation yields a circle (with theta ranging from 0 to 2p). Lets overlay a similar equation, in which a constant ÒaÓ is
introduced,

R = a + b cos (kq)

Again, we have b
and k equal to 1 as well as a. We see that we no longer have a circle. Polar
graphs usually have symmetry either at the polar axis, the pole or the line q = ^{p}/_{2}.
Here the symmetry appears to occur at the polar axis.

Lets investigate
various changes to a, b, and k, starting with k = 2.

We can see that
as k increases, we begin to see the pedals of the Òn-leafÓ. It appears that the
equation r = a + b cos (kq) has exactly 2
symmetrical sides, while the equation r = b cos (kq) has symmetry that doubles the amount of k. It still appears that
the symmetry rests along the pole. Will this pattern continue to hold true?
LetÕs increase k to 3 and compare.

This assumption
does not hold true, as both equations yield ÒpedalsÓ of the same amount as k.
LetÕs again increase k to 4 and see the results.

It appears that
with even values for k, the equation r = b cos (kq) will yield a
graph that has the pedals the amount of 2k, while our equation r = a + b cos (kq) will always have pedals in he amount of k. (k increases by 1
each graph shown)

If we change the
equations to r = b sin (kq) and r = a +
sin (kq) or line of symmetry becomes q = ^{p}/_{2 }

But just as
before, the numbers of pedals are composed by the amount multiplying data, be
it k or 2k (for odd or even).