Assignment 10: Parametric Equations

By Krista Floer

For various a and b, investigate

Before we change numbers around, we need to see what they original graph would look like.

We have just an average circle, nothing to special about it. Let's first look at changing the value of a for the equation for x. I first noticed that the graph looked better when a was odd rather than even.

When a is even, it seems like the spiral is only halfway completed. When a is odd, it seems like the spiral finishes the spiral. Here is a movie file that I thought was interesting.

From this we can see that when a is even, it is incorrect to assume that the spiral isn't finished. It is just that the reflection over the y-axis that we see when a is odd is just lying on top of the original graph. In other words, the other half of the spiral in concurrent with the half that we see. This concurrency only happens when a is an even number.

When a is **ODD**, the number of the sideways u shapes on one side of the y-axis is equal to a. Consider this graph:

As we can see in the graph above, there are 5 sideways u shapes on the right or left of the y-axis. Since a = 5, we have found the correlation.

But what about when a is **EVEN**?

It is easy to look at a graph and tell a is even because it looks incomplete. Also, it is not symmetric about the y-axis. Two ways exist to tell how many sideways u shapes are graphed. The first way is to see if a is divisible by 4 or 2. If it is divisible by 4, then there will be more shapes to the left of the y-axis. If a is divisible by 4, there will always be one more on the left than the right. If a is only divisible by 2, there will be more shapes on the right of the y-axis, with the total being one more on the right. Now that we know which side has more shapes, we need to know how many are on each side. If it a divisible by 4, then divide a by 2 and that will be the number of shapes on the left. Subtract one from that number and that will be the total on the right. If a is only divisible by 2, the same process will work, but the sides will be switched. Consider these two graphs while reading through this paragraph to help clarify:

The second way to predict how many shapes there will be is to subtract 1 from a. Then if a is divisible by only 2 not 4, then there will be more shapes on the right. If a is divisible by 4, then there will be more shapes on the left.

Something I noticed while writing the above paragraph is that there will always be a shape pointing to the right at 1 on the x-axis. It doesn't matter if a is odd or even, it will hold true. It also looks like the shape in the middle on the right side will always be the shape with the root at 1 pointing to the right. The domain and range is (-1, 1) for all the graphs. I think that it would hold true for any number that was put in for a. I know that the range of the cosine function is (-1, 1), so maybe because the cosine function defines the horizontal values in our parametric equation, then that explains why the domain is (-1, 1).

Now moving on to how b affects the graph. Consider these two graphs:

I thought the graphs would be similar to the graphs when a was changed, but would extend along the x-axis. This seems true when b is odd, but it is an entirely different story when b is even. When b is even, we can see that no part of the graph is concurrent with itself except for intersection points.

This is a movie that shows a constant change of b from -6 to 6. I noticed two things is this movie. The first is that when n went from 0 to 1, the procession of graphs looked similar to the parametric equation when the equation for y is changed. I go into this in more detail in a problem for the final exam. Click HERE to see that investigation. The second thing I noticed is that when n is a whole number, the domain is divided into a parts.

Looking at the domain and range of the functions, we see that each is (-1, 1), much like the domain and range for when we changed a. I think that since sine defines our y values, then the range must be (-1, 1).