Polar Equations of the form:

r=a+bcos(kt)

Polar equations can form many different and many times very pretty figures.
One of the most well known polar equations is that of **r=a+bcos(kt)**.
This equation may be one of the most widely known of all polar equations
because of what the equation creates when **a** and **b **are equal
and **k** is an integer. This variation of the above mentioned equation
creates what is known as an 'n-leaf rose'. **Figure 1** illustrates the
'n-leaf rose' for values of **k=1, 2, 3, 4, 5,**
and **6**. These colors correspond with
the colors of the colors in the graph.

Notice that the value of **k** determines how many leaves the rose
has. There is one leaf for **k=1** and five leaves for **k=5**. This
is what makes this variation of the equation so unique.

Now let us keep **a** and **b** the same once again but this time
let **k=.5, 1.5, 2.5,
3.5, 4.5,**
and **5.5**. What would one think the graph
will look like now? Using what was just found in **Figure 1**, one may
want to hypothesize that there will only be a half of a leaf, one and half
leaves, etc. **Figure 2** will prove this hypothesis.

Only half of the first leaf of every graph is present when the values
of **k** are half odd-integers. So, if **a** and **b** are equal,
then the value of **k** will determine the number of leaves the 'n-leaf
rose' will have. For this discussion it should be called a 'k-leaf rose'.

One last modification to the 'n-leaf rose' equation. What will happen
when **a=0** and **b=1** for **k=1, 2,
3, 4, 5,**
and **6**? Or in other words, **r=bcos(kt)**.
**Figure 3** shows the graphs of these equations.

From this one can see that for odd values of **k**, there were **k**
leaves drawn and for even values of **k**, there were **2k** leaves
drawn. When **k** is negative, nothing different is formed because all
that a negative **k** means is all the angles will be negative and since
cosine is positive in the first and fourth quadrants and negative in the
second and third quadrants, the graph will just be just be traversed in
the opposite direction it was when **k** was positive.

If one was to replace **cos(kt)** with **sin(kt)**, using what
you know about the relationship between sine and cosine, what would you
expect the graphs to look like? You should know that sine and cosine are
just phase shifts of each other. From this you should be able to determine
what the graphs will look like now. If you are still not sure, take a look
at **Figure 4** for more incite.

From this figure one can see that the graphs are shifted. When **k=1**,
the graph shifted a full 90 degrees, however, all the other graphs shifted
90 divided by **k** degrees. Or 45 degrees for **k=2**, 30 degrees
for **k=3**, etc.

For the next set of investigations, let **a < b**. This variation
creates some very interesting graphs. Can you guess what they will look
like for **a=1** and **b=2** and **k=1, 2,
3, 4**? **Figure
5** demostrates what these values of **a**, **b**, and **k**
create.

When **a < b**, for each integer value of **k**, there are **2k**
leaves in the graph of which half are larger than the other half. However,
when **k** is odd, the half of the leaves that are smaller than the others
are located inside the larger ones and when **k** is even, the half of
the leaves that are smaller than the others are located between the larger
leaves not inside them. If the value of **a** and/or **b** is made
larger, the graphs become larger. This can be understood by letting **t=0**
and solving the equation. The cosine of zero is always **1**, thus if
**a** and/or **b** is larger, then the value of **r** becomes larger
and the opposite is true also. If **a** and/or **b** become smaller,
then the value of **r** becomes smaller.

Just as an addition to what was just stated about **r** becoming larger
and smaller, **Figure 6** shows graphs for **a=1**, **k=4** and
**b=1.5, 2, 2.5,** and **3**.

When **k** is rational, some very weird graphs are created. I am not
sure what can be determined from them, so if the reader has any insight,
I would appreciate hearing what you have found. **Figure 7** shows some
graphs of of the equation we have been using when **a=1**, **b=2**,
and **k** is rational.

The graph on the left for **cos(kt)** and the one on the right is
for **sin(kt)**.

Another variation of the equation comes about when **a > b**. Using
what has been discussed so far, what would you predict the graphs to look
like in this case? To narrow the possibilities down a little, consider only
those cases when **a=2**, **b=1** and **k=1, 2,
3, **and **4**
and only the graphs of **cos(kt)**. Now what do think? **Figure 8**
can be of some help to you if you need some help.

This is a quite interesting figure. Notice how none of the graphs touch
the origin, unlike all of the previous graphs discussed. If the **cos(kt)**
is replaced by **sin(kt)**, one gets the same outcome as earlier in this
discussion. The graphs of the **cos(kt)** are shifted by 90 divided by
**k** degrees to form the graphs for the **sin(kt)**. This is shown
in **Figure 9**.

Now we take a look at the graphs when **k** is a rational. The graphs
on the left in **Figure 10** are for **cos(kt)** and the graphs on
the right in **Figure 10** are for **sin(kt)**. The values of **k**
used are **k=.5, 1.5, 2.5,
**and **3.5**.

The graphs for **cos(kt)** are not closed and form some odd looking
shapes. On the other hand, the graphs for **sin(kt)** are closed and
form shapes similar to those created when **k** was an integer.

The investigations done here are only a few of the many one could do.
My knowledge of polar equations is not real strong at this time so I would
appreciate it if the reader has any questions, comments, suggestions, or
incites to contact me at **lrapley@aol.com**.