Assignment #11

Polar Equations

By: Keith Schulte

Polar equations are in the form:

Where r is the radial coordinate and theta is the polar angle. What would the generic graph of r equals theta look like, if theta is equal to 2pi?

Remember, that due to the cyclical and continuous nature of these graphs, the range of pi determines how much of the graph we actually see. If we take the same generic polar equation and let theta equal 4pi, we would get the following graph:

There are many families of graphs of polar equations. One family of graphs is of the form:

These graphs are called limacons, when k=1. Limacons are conchoids (shell forms) of a circle. When a, b and k are equal to one and theta is greater than or equal to 2pi, the limacon looks like the graph below. Also, whenever a and b are equal and k =1, (and theta is 2pi or more), it will have the same general shape, just a different size:

We can vary a, b, k and/or theta to explore this family of graphs. What happens if we change a? What does a represent? Let's change 'a' and see if we can make a hypothesis about what 'a' represents in this family of graphs. In the next graph b and k are equal to one, theta is equal to 2pi and red is a = .5, purple is a = 1, blue is a = 1.5 and green is a = 3.

So, as we change 'a', the larger we make 'a', the larger the graph gets. When 'a' is 3, the graph is circular. As we make 'a' smaller, a cusp develops on the left side of our graph. When 'a' is one, the cusp is at the origin. As 'a' gets smaller, an eye or a loop is created, with it's left edge at the origin. As we make 'a' even smaller, the loop becomes larger.

Next we will look at this family of graphs when a and k are equal to one and theta is equal to 2pi, and vary b. What do we expect to happen? Let's look at the graph with; red is b = .5, purple is b = 1, blue is b = 1.5 and green is b = 3.

So we see that as b is made larger, the size of the graph increases, so that is similar to when we changed 'a'. What about the shapes of the graph?  As b becomes smaller, the graph takes on a circular shape, as b increases, the graph develops an eye. So this is the reverse of the change we saw with a.

Now what would happen if we make k different integers, while keeping a and b equal to one? In the following graph, red is k = 2 and blue is k = 8.

We see that when k is changed (with a=b), from two to eight, we get flower shaped patterns with the number of petals equaling the value of k, when k is an integer! So with k=n, we get an "n-leaf rose".

Now we can explore some other patterns. What if we have:

for different values of k? When k is an odd integer and k= red is k = 1, purple is k = 3, and blue is k = 5 do we still get flowers with n numbered petals? Let's see!

Yes, when k is an odd integer, we get flowers with n petals. The one petal flower is a circle. What happens if we let k be an even numbered integer?  When k is: red is k = 2, blue is k = 4 we get:

So when we have r=b*cos(k*theta) and k is an even integer, we get flowers with 2n petals!

What if we replace cosine with sine in the equation?

We'll let k=1 and 3 and see what the difference is between the two equations' graphs.

When red is k = 1, purple is k = 3 the graph for our new equation gives us a similar picture, but with a different orientation.