Assignment XI: "N-leaf" Rose

by

Rui Kang

In this investigation, we look at the equation

Note:

(1) We only look at the case when a and b are equal, and k is an integer. This is one textbook version of the "n-leaf rose",

(2) Then, we compare the above equation with

for various k.

(3) Finally, we examine what if ...co() is replaced with sin()?

(1) First, let's fix a=b=1, and let k=1, we obtain a one-leave rose shown below:

By changing the values of a and b, we find that as long as a=b, no matter what value they take, the graph remains a one-leaf rose except that the size of the rose changes.

Since this investigation mainly concerns the shape of the graph, in the following investigations, we fix a, b at 1.

Next, let's change the value of k, let's k=2, 3, 4, 5

**Conclusion 1: In general, if a and b are equal and k is a positive integer, the graph of **

** **

**is a k-leaf rose which is symmetric with respect to the x-axis. **

**If k is even, then the graph is symmetric with respect to both the x-axis and the y-axis. If k is odd, the graph is symmetric with respect to only the x-axis.**

The next question is then what happens if k is a negative number or zero?

**Conclusion 2: When k=0, the graph is a circle. **

**When k is an negative number, the graph remains that same as those when k is a positive number. **

**For example, when k = -5, the graph is the same as when k=5. When k=-2, the graph is the same as when k=2.**

Note that this conclusion should not suprise us because we know that

(2) Now we are ready to compare the two polar equations:

and

We let k takes on values from 1 to 5 as we did above.

**Conclusion 3: The graph of **

**is a k-leaf rose when k is odd; **

**however, when k is even, the graph of **

**is a 2k-leaf rose. **

**In comparison to the graphs of **

**, **

**when k is odd, the graphs of the two polar equations are of the same shape.**

**When k is even, the graphs of **

**have twice as many rose leaves as those of **

**.**

(3) Finally, we are ready to investigate what happens if we change *cos *in the equations to *sin*?

We first try to compare

and

**Conclusion 4: **

** **

**is also a k-leaf rose, and in fact, if we rotate the graph of **

** **

**by a certain degree, we will get **

**. **

**Regardless of the value of k, the graphs for these two polar equations are always congruent despite the fact that their leaves point to different directions. **

We now try to compare

and

**Conclusion 5: By examininh the above set of graphs, we find that **

** **

**is a k-leaf rose when k is odd, and a 2k-leaf rose when k is even. **

**In addition, if we roate the corresponding graph of **

**by a certain degree, we will get **

** and vice versa. **

**Regardless of the value of k, the graphs for these two polar equations are always congruent despite the fact that their leaves point to different directions.**