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 “n-leaf rose”.

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 “n-leaf roses” 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 “2n-leaf roses”. 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 “2n-leaf rose”. The graphs have also rotated 45° counterclockwise.