Polar graphs of the form r=a+bcos(kΘ)
Polar graphs like r=a+bcos(kΘ) have several interesting properties we can look at. First, the value of k affects the period of the graph. The idea that y=cos(kx) has values that will repeat every †is familiar. This looks somewhat different in polar coordinates. Letís look at r=3+2cos(6Θ) to see what that means.
As , r (the distance from the origin) is shrinking. As , we see r increase to its maximum again. We see the same decrease from maximum increase back to maximum pattern over the period we are used to, but the heights are now distances from the origin and the period is an angular section of the plane. Notice that if we draw the graph over six periods we return to where we started in the plane and begin to trace the same path over again.
After k repetitions we will end where we started, Θ=k . For integer k values this means the r value will be the same, and a k-leaf rose will be formed. Rational values for k result in r values not lining up at Θ=0 and Θ=. They will, however, line up at multiples of †multiplied by the denominator of the fraction k. For example r=3+2cos(Θ) repeats every †instead of every . Also note that the denominator of the fraction still gives the number of rose leafs.
We know that cos(Θ) always returns a value between -1 and 1. So, r=a+bcos(kΘ) has amaximum r value of a+b and a minimum r value of a-b.
For instance, r=3+2cos(5Θ) is always between the circles r=1 and r=5
If the a and b values are the same then the leaves meet at the minimum value r=a-a=0
If a<b we run into the situation where a-b is a negative number. On the graph this means we turn a positive angle Θ and instead of moving out along the terminal side of the angle we move along the extension of the terminal side on the other side of the origin. Letís look at r=1+2cos(5Θ) for an example.
When, we get that r=1 and r continues to be positive until †=†when r=0 (blue). The r value is then negative on << † (orange).
The graph continues in this manner, and we end with a graph with a large rose inside r=3 and a small rose inside r=-1 (note this is the same graph as r=1).
The situation is the same for r=bcos(kΘ) (a=0). The a value is less than the b value. So we sometimes get negative r values, but now the minimum r value is Ėb and the circles enclosing circles are the same. So the negative portions will overlap the positive portions of the graph. The following graph shows r= 4cos(7Θ) and its first three and a half cycles.
Notice that the graph has negative sections for half of each period and that the second half of the third period will start to retrace the previous portions of the graph. We end with a seven leaf rose that has been traced over twice as much.