Assignment 11

Polar Equations

by Jeff Hall

Problem 1

Investigate the following equation:

It produces this graph:

When a and b are equal, and k is an integer, this equation

generates one textbook version of the "n-leaf rose."

In this example, a, b, and k all equal 1, and t goes from -pi to pi.

How does this equation compare to a more basic equation:

When k=1, we get this graph.

When k=2, we get this graph:

At this point, it looks like the number of blue petals is determined by k-squared.

Let's try k=3 just to be sure:

Hmm.... Now the number of blue petals equals k.

Let's try k=4:

The number of petals now equals k times 2.

Let's try 5:

Okay, now we are starting to see a more defined pattern.
When k is odd, the number of petals equals k.

Now let's try one more even number...try k=6:

Since there are 12 petals, we can safely safe that k*2 works when k is even.

Let's change the original equation's k value, as well. Let k=2 for both:

Now try k=3 for both:

This time, the number of red petals equals k every time.

Let's try k=4 just to be sure:

That's it!

What would happen if we changed these equations from cos to sin?

We get this when k=1:

It rotated the image 90 degrees counterclockwise.
Problem 2

Investigate the following equations varying a,b,c, and k:


When a,b,c, and k=1, we get this graph:

Let's manipulate k first. Let k=2:

Now let k=3:

And k=4:

Now let's put k back to 1 and try a=2:

The variable a seems to determine the size of each circle.

Let's try a=3 to be sure:

Yes, that's what is happening.

Notice that the red and blue circles have a as their radius.

Also note that the black line is getting a steeper slope.

Now put a back to 1 and change b to equal 2:


Now try b=3:

And finally, try b=4:

Changing b obviously enlarges the green and purple shapes,

but notice the distortion in their shapes, as well.

They are becoming more circular.

Also note that the black line's slope is becoming more horizontal.

Now let's manipulate c. Change b back to 1 and let c=2:

Now try c=3:

Finally, try c=4:

As you can see, changing the c only moves the black line.

Notice the x- and y- intercepts equal c.
Problem 3

Investigate the following equations for different values of p:

The parameter k is called the "eccentricity" of these conics.

When p=1 and k=1, we get this:

When p=2, we get this:

When p=3, we get this:

As you can see, each parabola passes through -p and p on either the x- or y-axis.

Now let k=2 and p=1:

Now try p=2:

It seems that when k>1, the parabolas become hyperbolic figures.

Increasing the value of p just widens these figures apart. Notice that some of

these figures now cross the x- and y-axes at -2p and +2p.

Now let's try k<1. Change p back to 1 and let k=0.5

Now change p=2:

Finally, change p=3:

As you can see, when k<1, the parabolic figures begin to look

like circles. Increases in p make the figures grow larger.

What happens when k=0? Absolutely nothing, since each equation is r=0.

What happens when k<0? The graphs look identical to the respective positive k graph.

The figures may be inverted, but you can't tell by the way they look on a graph.
Problem 4

Investigate the following equations:


These equations generate the following graph:

Notice that the blue figure is rotated in comparison to the red figure.

The purple equation generated a hyperbola.

Now let's see what happens when you place a 2 in front of each theta:

Changing the cos to sin in the original first two equations gives us this:

Again, the red and blue figures appear to be similar, just rotated differently.

Problem 5

Try some interesting graphs.

Try this one when a is small, such as a=0.1:

We get this graph:

Now put a=1 and add these equations:

Since this graph has become cluttered (a hazard of geometry), let's try

graphing these equations on a brand new graph: