This is an investigation of the equation r=a+bcos(kt) for different constant values of a, b, and k.

Let's start with a and b being equal and k an integer value between 0 and 2 pi. First we graph

r=1+1cos(1t).

It's a cardiod extending from 0 to 2. Now we'll try k = 2 and see what happens.

This certainly looks different. Let's try k = 3.

Now there are three leaves. Is it time for a conjecture? Will k=4 have 4 leaves? The graph is looking a little crowded, so let's use a new one.

Yes, it's a four-leaved rose. Just to be sure let's try k=6.Notice that although this graph is larger; it's radius is 2 whereas the one above is 4.

Just as expected - a six-leaved rose. What will happen if a and b are greater than 1? Let's try 3.

Now this looks almost the same size as the one above, but look at the markings on the radius. The leaves are three times as long as the one in which a=b=1. Leet's try decreasing a and b, say a=b=!/2.

The leaves on this one extend only to the 1 radius mark. This is half the the radius of a=b=1. Can we assume that the magnitude of a and b directly affect the radius of the figure when a = b ?

Now let's see what happens if k is a fraction, say 1/2.

Well, that's different. Let's try some more halves: say 5/2 and 13/2. We'll show the plots a little bigger on the graph.

It certainly appears as if there is half a leaf for every half graphed. Let's check a couple more to get k up to 4 pi. Let's try 17/2 and 25/2 which is only slightly less than 12.56 or 4 pi.

So there is surely a half a leaf for every 1/2 in the equation. Let's take a look at thirds. What if k= 1/3 or 2/3.

Well, these look different, sort of like snails. Maybe they are spirals. Let's try a few more to see what the pattern is. It looks like the grid needs to be made a little larger. Let's try 5/3 and 8/3. It certainly looks as if there is 1 1/3 leaves on the left and more than two leaves on the right which corresponds with the above fractions.

It appears that the number of leaves graphed corresponds to the fraction that is k.

Now we"ll look at the same equation with a< b. Let's start with a=1, b=2, and k =1.Then on the right let's have a=2, b=5, and k=3.

Wow, what happened here? Let's go back to a=1,b=2, and let k=2. Also k=4.

It seems that all the graphs with a<b have one small leaf for each larger leaf. And it looks as if the odd numbered ones have the extra leaf inside and the even ones have the extra leaf outside. Let's try two more to support this idea. How about a=1, b=2, k=5 and k=6?

This certainly seems to be true. Let's check what happens if k is 1/2 and k=1/3. Remember, we still have a<b, so let's stick with a=1 and b=2.

The one on the left does look as if it might be half of the cos(1t) and the one on the right could possibly be a third of cos(1t), although that is not so clear. Let's look at some larger fractions: say k=3/2 and 7/3.

Yes, the fractions make that part of the leaf with the smaller leaves, too.

Now we'll look at the graphs of this equation when b>a. The simplest one is a=2, and b=1, so let's try that first with k as integers, say k=1 and k=2.

Here we have an "almost" cardioid and an "almost " two leaf rose. Let's try k=3 and k=4 to see if the pattern holds.

These are certainly similar to the leaves obtained from a=b, but they are all wide open in the middle. Let's check a couple of fractions just to be complete. Let's try a=3, b=2, and k = 1/2 and k=51/3.

The left one seems to be a large half; the right one is also large. Note that the right graph has a longer radius, so it will fit in the space. Fifty-one thirds gets close to 6 pi and it seems that 6 pi might make this graph look more complete.Let's check.

Fifty-seven thirds is 19 and the graph does look more complete.

We have examined quite a few graphs of the form r = a + bcos(kt) and gotten some interesting results as well as some expected ones. But this is just the beginning of the possibilities of exploring polar equations. Check below for some more possible explorations.

Explorations: What is the difference in the above situations when k = b cos (kt)? What if all the graphs are done with sine instead of cosine? Try other polar equations such as: r =2a sin(kt), r=pe/1--e cos(t) for different values of p, and r=at. Make up some equations and try variations.

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