*Polar equation*s

by

**Ana Kuzle**

The purpose of this assignment is to explore the polar equation

.

Let's begin with our investigation by changing parameters a, b and k and see how they influence the graph of given polar equation.

I. I decided to fix the value of *a=b*(=1 in my case) and change the value of *k* from 1 to 4.

From the pictures we can tell that parameter *k* is actually the number of leaves. As *k* increases, the number of leaves increases as well.

Therefore, on the picture there are 27 leaves. Hence, k=27.

In the video bellow, *k* increases from 0-20.

What if a=b is negative? Let's take a look when *a*=*b*= –1 and *k*=1.

How is this different from a=b=k=1? If we compare the two possibilities we can see that new graph is the reflection of the original one. Hence, the new graph is the same as the old one but just rotated by 180° with respect to origin.

What happens when *a*=*b*=–1 but *k*=2? Well, we get the same graph because the graph is symmetrical with both axes so *a* and *b* being being negative dose not affect graph of the curve.

For *a*=*b*=–1 and *k*=3 we can see that the new graph can be obtained from the old one by rotating it by 60° with respect to origin.

II. Let us now consider the case when *a* and *b* are equal but changing their value and *k* fixed.

Thus, as k=1 we got only one leaf, but as a and b are getting greater, the inside area of the curve is changing. It is getting bigger.

The same is the case with the pictures bellow.

As *a*=*b* are getting bigger, the inside area of the curve gets bigger as well what can be seen in the movie bellow likewise. Parameter *k* determines number of leaves.

III. What happens if we take *a* and *b* such that *a*<*b*?

This is what I got for *a*=1, *b*=1, 2, 3, 4 and *k*=1.

By comparing these graphs with the graphs in I., we can see that when *a*<*b*, a new set of smaller leaves appears and the size of both sets of leaves increases by one.

For *k*=2 I got this.

If *a*>*b*, what happens? Let's take a look of situation when *b*=1, *k*=3 and *a* varies from 1 to 4.

Changing the value of *a* influenced the form of the curve. By increasing parameter *a*, leaves lose their form and get further from the origin of the coordinate system.

IV. We have already mentioned that parameter *k* is equal to the number of leaves. So far, we have only taken look of situation when *k* was an integer. (The case when k is negative we actually discussed. Keep in mind that cosine is an even function therefore taking negative *k* does not change nor the equation of the polar equation nor the graph).

Let *k* be any real number. I took k=3.14, 1.618 and 2.5. The picture bellow shows what I got.

VI. Let us now consider the equation for *a*=0. Hence, the equation

.

I set up *b* to be 1 and *k* ranging from 1 to 5.

From the picture we can see that number of leaves differs when *k* is even or odd.

When *k* is an odd number, the number of leaves is equal to *k*. But, when *k* is an even number, number of leaves is equal to 2*k*.

VII. If we replace cosine with the sine in the original polar equation, we get

.

The following graph is obtained for *a*=*b*=1 and *k*=1.

We can see these graphs are the same except the sine graph is rotated by 90°.

Similarly for *a*=*b*=1 and *k*=2 I got graphs that are the same with sine graph rotated by 45°.

As, in cosine case we can take a look what happens if *a*>*b *and* a<b.*

In case *a<b, *I got these graphs.

As in the case of cosine a new set of leaves appeared but because of the sine the graph is rotated by 45°.

In case *a>b*, the following graph is obtained.

Similarly as in cosine case, the form of the curve loses as a is increased and the curve is furthering from the origin. The difference is again in the rotation by 45°.

VIII. For the end, let's take a look of equation

.