Petal Curves
by
Michael McCallum
Introduction: Graphs of equations in polar coordinates
of the form 
are commonly called petal curves because
of their shapes of their graphs. The number and sizes of the petals
and the petal arrangements are dependent upon the values of a,
b and k. This exploration will demonstrate the effects of varying
these parameters upon the graphs of the equations.
Let's begin with a = b and vary the value of k. The simplest way to start is with a = b = 1 and k = 1. The graph of this equation is shown below.
This is not even a petal curve. It turns out to be a cardiod that is symmetric about the x - axis. Let's make k = 2 and see what happens.
Now we have the most basic petal curve, a two petal curve with x and y - axis symmetry. The petals lay on the x - axis. Let's increase k to 3 and see what happens.
Now we have three petals and symmetry about the x - axis only. Further investigation shows that as long as a = b the number of petals will be equal to k. Try it yourself. Notice that the x - intercepts in the graphs above were at x = a + b = 2 for each graph and also at x = -a + (-b) for the case where k = 2. If you were to graph the case where k = 4, you would find that the y - intercepts were at +-(a + b) also. This is interesting, but no real surprises for those of you who have worked with petal curves in polar coordinates before.
Now let's look at varying the parameter k when a < b. Let a = 0.5 and b = 1. The graph when k=1 is shown below.
Again we have a cardiod. However, this cardiod has a loop inside it. Notice that the loop intercepts the x - axis at 0.5 and the cardiod intercepts the x - axis at 1.5. This equates to b - a for the inner loop and b + a for the outer loop. Lets reduce a to 0.1 and see if b - a is correct for the x - intercept of the inner loop. SEE BELOW.
Yes, b - a was correct. Let's reset a to 0.5 and change the value of k to 2.
This is different! We now have two large petals along the x - axis and two small petals along the y - axis. The two large petals are three times the length of the two small petals. Does the value of b control the length of the large petals and the value of a control the small petals? Let's change a to 0.75 and see.
No! Apparently the small petal has length b - a and the large petals have lenght b + a. Let's reset a to 0.5 and change the value of k to 3.
Now the graph has three large petals with three smaller petals each inside a larger petal. Notice that the sizes of the petals are again a - b for the small petals and a + b for the large petals. Further investigation will show you that when k is even, the petals will be arranged with each small petal between two large petals, and when k is odd, the small petals will each be inside a large petal. Look at the graph for k = 4 below.
Just as predicted.
What about the graphs where a > b? Do you think they are similar to the graphs with a < b but with the roles of a and b reversed? Let's see. Let a = 2 and b = 1 and k = 1. The graph is shown below.
What have we here? A cardiod just like the other times when k = 1, but this cardiod doesn't intersect the origin like the others. The x - axis intercepts are at -1 and 3. This is still b - a and b + a. Let's make k = 2 and see what happens.
Hey, this looks like a peanut. Now the x - intercepts are a + b and - (a + b), and the y - intercepts are a - b and - a + b. So, do we still think the roles of a and b are reversed? Let's make b = 0.5 and find out for sure.
Well, obviously we were wrong. Let's reset b to 1 and make k = 3.
What would you call this shape? It is like a three petal curve, but there is no intercept at the origin. Further investigation will show you that all of the curves will look like this when a > b. The only differences will be the number of lobes, which is determined by the value of k.
When a = 0 the polar equation becomes 
.
What are the effects of varying the parameters b and k? Are the
effects similar to the effects above? Let's see. Let b = 1 and
k = 1.
We get a circle with radius 0.5 centered at (0.5, 0). A circle is a special case of a cardiod in polar coordinates, so this is not unexpected. What happens when we make k = 2?
We have a four petal curve with all four petals of size b. Since a = 0, we should have expected this. Remember that when a was less than b we got a 2k petal curve with the smaller petals of size b - a and the larger petals of size b + a. What happens when we make k = 3?
Now, where are the 2k petals? Remember again that when a was less than b and k was odd, we had petals inside of petals. Well now the petals are the same size so that they overlay one another. You can prove this to yourself by reducing the range of theta to 0 to pi. You will still get the same trace as above when a = 0. However, when a < b, you will not get all of the petals. This same pattern persists as the value of k is increased in integer increments.
Another possible interesting question is, what is the effect of adding phase shift on the graphs? Let's look at the case where a = 0.5, b = 1, k = 3 and we add phase shift in increments of pi/8. First, let the phase shift be pi/8. (By phase shift, we mean adding a fixed value to the argument of the cosine function.)
The curve is rotated in a clockwise direction exactly pi/8 radians. Further investigation will show you that this is always true. As a matter of fact, if you have a graphing program that is capable of animation, you can vary the phase shift systematically and have the petal curves rotate around the origin. Try it!
I hope that this investigation has caused enough interest that you wish to look further on your own. You can have a lot of fun with these petal curves. Try overlaying several different curves with different parameters. You can get some very interesting patterns. Good luck.