Polar Equations

Rozina Essani

Invetigate

r
= a + b cos(kQ)

Using graphing
calculator I investigated various forms of the above polar equation. As noted
in the problem that when a and b are equal, and k is
an integer, we have a Òn-leaf roseÓ.

a = b = 1

k = 1, 2, 3, 4, and 9

Keeping a=b=1, our
graphs stay within a domain and range of -2 to 2. Given an even number of k,
the petals evenly distribute on the axis, and hence show symmetry along the x
and y- axis. When we have an odd integer for integer we no longer see that
symmetry.

What happens when a is not equal to b?

I tried two cases for
this scenario.

Case 1: a = 1, b = 4,
and k = 6

Increasing the value of
b gives us two Òn-leaf rosesÓ simultaneously. This also increased the domain and range.

Case 2: a = 3, b = 1, k
= 6

Increasing the value of
a does not change the number of leaves, but instead now the curve does not hit
the origin. The graph is enveloped between two circles, the smaller with radius
2 and the larger with radius 4.

Let us compare the
graphs of r = a + bcos(kQ) to r = bcos(kQ) for various values for k.

Setting b = 4 we
enlarge the domain and range to 4 units from the origin. We see a difference in
the k = 1 graph. In this case we have a circle with radius 2 in the first and
fourth quadrants. When we have an even number for k, for instance when k = 2
and k = 6, we get Ò2n-leaf rosesÓ. The leaves then double and give us a 4-leaf
rose for k = 2 and a 12- leaf rose for k = 6.

What if we change the
formula to r = bsin(kQ)?

I set b = 4 and check
for k = 1, 2, 3, 4, and 5.

When we change to cos
to sin, for k = 1, we see that the circle is now in the first and second quadrant.
Again our even number selection for k gives us a Ò2n-leaf roseÓ. The graphs
have also rotated 45¡ counterclockwise.