We are going to investigate the equation:
r = a + b cos (kθ)
We will graph the equation where a, b and k are all held constant, or equal to 1; and theta is between 0 and 2 ¹.
We will now hold a and b constant and vary the values of k. For example,when k=2, we get the following graph.
As you can see, when k=2, there are 2 leaves on the graph. The leaves cross the x-axis at the origin and at the points (-2,0) and(2,0).
Now lets see what happens if we hold a and b constant and let k=3?
Now there are 3 leaves in the graph with the intersection at the origin.
HmmÉ what about when k=10?
Imagine that, there are 10 leaves. It appears that k determines the number of leaves for each polar function.
Just to be sure, we can graph the function when k=100.
There appears to be 100 leaves, although they are very small. Therefore, when a and b are equal, and k is an integer, the function forms an "n-leaf rose".
Now, we want to investigate the polar equation without a.
We will look at several equations where b is held constant at 1 and k is altered.
Notice that there is only one leaf. What does this mean?It could imply that k also determines the number of leaves for the function:
LetÕs investigate a little moreÉ
Were you expecting two leaves? Then how come there are four? Maybe the number of leaves is equal to 2k.
Let's try k=3,
So, you thought you had it figured out huh? Were you expecting six leaves? Then how come we have three? Possibly the number of leaves is equal to 2k for even numbers and k for odd numbers. Let's try this for a larger even and odd number.
When k=10, we have an even number and 2(10)=20. Are there 20 leaves?
YES!! There are 20 leaves.
What about when k=11? Since 11 is an odd number, we must use
In the last part of our investigation, we will take the equation and replace the cos with sin.
Once again, we will hold the b constant and vary the values of k. First, graph the function when k=1.
Here, there is one leaf. My first assumption is that k=number of leaves....However, letÕs explore some more and see what happens when k=2.
I bet you were expecting 2 leaves, then why are there four? Now, I suspect that we will have the same properties as before when using the cosine function. When k is even, the number of leaves is equal to 2k. When k is odd, the number of leaves is equal to k. Is this true?
Let's set k=5. Five is an odd number, so there should be five leaves. Are there five?
Yes, there are five leaves.
Now let's try k=6. Six is an even number and so we should expect to have 12 leaves. Are there 12 leaves?
YES! There are 12 leaves.
Hope you enjoyed my exploration of polar equations. Feel free to explore more on your own.
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