Assignment 11

Polar Equations

by Rives Poe

In this investigation we will look at some different parametric equations, varying the values of the coefficients. Let's see if what we find!

The graph below shows five different functions:

Purple: r=2a cos(k*theta) / Red: r=2a sin(k*theta)

Blue: r=2a sin(k*theta)+b / Green: r=2a cos(k*theta)+b

Light blue: r=(c)/(a cos(k*theta+b sin(k*theta))

The red and the purple functions create perfect circles, when a and k are equal to 1. The dark blue and green function have a small "loop" in the graph. The light blue function is the formula of a line. For each of these functions above, a,b,c, and k are all equal to 1. Let's look at what happens when we change those values.

What do you think will happen if we change the value of a? I am going to set a equal to 4.

I did not change the functions, the colors still represent the same functions as above. It seems that a changes the size of the graph and for the linear funtion(the light blue) a changes the slope of the line. The x-intercepts remain the same for all of the functions, however the y-intercepts are changing.
Okay, moving on. Let's see how b affects the way these functions act. Be sure to make a conjecture, then scroll down to see if you are right!

For this picture above, I set b = 4 and a,k, and c equal to 1.

Both r=2a sin(k*theta)+b and r=2a cos(k*theta)+b have lost their inner "loop" it seems! The x-intercepts have changed and the slope of the light blue line has decreased.

If I continue to enlarge the value of b, the blue and green functions appear to become closer to circles and the light blue line seems to approach the x-axis. (Our red and purple circles are not affected by this change, since they do not carry the value of b.)

Below is a graph where b = 15.

Okay, let's see how c affects the r=(c)/(a cos(k*theta+b sin(k*theta)). This function is the only of the five that has c.

It seems as though, c changes the x and y intercepts. When c was equal to 1, the line crossed the x and y axes at (0,1) and (1,0). Now, in the graph above, where c=3, the line crosses the axes at (0,3) and (3,0). Again, I have left a,b, and k equal to 1.

I have saved the best for last! For the big finale, how do you think k will affect all of these functions? Any ideas? You might want to scroll up to the top of the page to see what the picture looked like when k(and a,b, and c) were equal to 1.

When k = 2, the functions look like this:

In this picture it is too busy to see exactly what is happening. So, let's look at the functions seperately below.

on the left k=2 / on the right k=3.

The two functions r=2a cos(k*theta) and r=2a sin(k*theta) were circles, but have now been pulled into the center, making a flower-like shape.

For the folowing 2 pictures, I am only showing the function: r=2a cos(k*theta), because it is difficult to see what is happening with both of the functions on the screen.

when k = 6 , there are 12 "petals" (to the left below) and when k = 7, there are 7 "petals"(to the right below)

So, let's say that for the two functions r=2a cos(k*theta) and r=2a sin(k*theta) that k affects the shape by creating "petals". When k is even, the number of "petals" is twice the value of k and when k is odd, the number of "petals" is the same as the value of k.

Let's continue this investigation, by looking at r=2a sin(k*theta)+b and r=2a cos(k*theta)+b for k =2.

***remember that these two functions looked like this before:

:

Okay, let's see what they look like when k=3:

(I am only going to show r=2a sin(k*theta)+b , so that we have a better view of what is happening. The graphs are almost identical(just rotations of each other).)

Alright, I think I am ready to state what is happening! Are you? It seems that for these two functions k is doing close to the same thing as before, however the even and odd values do not effect the graph in the same way. If the value of k is even, the small, inner "petals" alternate with the larger "petals" and if k is odd, the small, inner "petals" are inside the larger "petals".

K creates "petals" and the value of k determines the number of petals.

Now, one more function to explore: r=(c)/(a cos(k*theta+b sin(k*theta)). This function is a linear function. Let's see what happens when k = 2.

It seems as though we have 4 assymptotes, creating 2 parabolas.

When k=5:

Now, we have 5 assymptotes and 5 parabolas. Let's look at one more graph when k=10:

Above there are 10 sets of assymptotes (20 in all) and 20 parabolas, which is double the value of k. Thus, k affects this linear function, by constructing assymptotes for the value of k. If k is an odd value, then there are as many assympotes as the value of k. If k is an even value, then there are twice as many assymptotes and twice as many parabolas.

This concludes the investigation of Polar Equations. I hope that this will help us understand how different constants, a,b,c, and k will affect Polar Equations. And if you ever want to construct flowers in a mathematical way, you should have no problem creating exactly what you want!!!!