Isn't Santa at the North Polar?

by

Mike Rosonet

This exercise will investigate polar equations of the form

and how the value of

kaffects the graph. To begin with, what is a polar equation?

A polar equation is a function expressed in polar coordinates with the radius *r* expressed as a function of the angle .

Consider the equation r = a. This is a transformation of the original equation of this investigation with b = 0. If this equation is graphed in the polar coordinate plane, it looks like this:

Graphs of polar equations with differing values of a

Thus, the polar equation r = a is a circle with a radius of a. The reason for this is because every point in the polar coordinate system is determined by its distance r from the origin (0, 0) and the measure of the angle constructed by the segment from the point to the origin and the initial side of the angle, the x-axis.

The parameter *a.*

What does changing the value of a do to the graph of the given polar equation?

graph of the inital equation with a = 1Let

b= 4 andk= 1. Thus, the equation isr =. Allowing the value ofa+ 4cosato vary from 0 to 10, the graph changes like so:

changing the value of a from 0 to 10When

a= 0, the graph is of a circle with a diameter equal to 4 with endpoints at the points (0, 0) and (4, 0) and center at (2, 0). As the value ofaincreases from 0 to 4, the circle expands to reveal that it is actually two circles, and while one is expanding into a bud shape, the other is shrinking and looks like a tear drop. Whena= 0, the tear drop can no longer be seen. Asaincreases from 4 to 10, the tear drop circle pushes out along the x-axis, increasing the circumference of the larger bud-shape as it expands to look more and more like a perfect circle.

The parameter *b.*

What does changing the parameter of

bdo to the graph of the given polar equation?Here, let

a= 4 andk= 1. Thus, the equation isr = 4 +. Allowing the value ofbcosbincrease from 0 to 8:

*changing the value of b from 0 to 8*

When

b= 0, the graph is of a circle with center (0, 0) and a radius of 4. Asbincreases from 0 to 4, the center of the circle shifts to the right and the left side of the circle begins to indent while the circle's circumference expands. Asbincreases from 4 to 8, a loop forms inside the circle; its circumference increases while the circumference of the circle increases. The inner loop approaches the outer circle but never reaches it asbincreases.

The relationship between *a* and *b*

The parameters

aandbhave inverse effects on the polar equation and its graph. Whena= 0, it graphs as a perfect circle tangent to the y-axis. Asaincreases to infinity, the circle separates into two loops, one inside the other, until the inner loop is joined with the outer loop ata=b. From that point to infinity, the graph expands in all directions and approaches the shape of a perfect circle. Asbincreases to infinitiy, the circle indents on its left side as it shifts to the right. The dent forms a loop inside the growing circle and both the inner loop and the outer circle expand asbincreases to infinity.One thing worth noticing is that the graph when

b= 0 is the same as the graph asaapproaches infinity. Conversely, the graph asbapproaches infinity grows very close to the graph whena= 0. The neutral case is whena=b. Each of these findings, however, depend on the parameterk= 1.

The parameter *k*.

The parameter

kaffects the graph of the polar equation in a totally different manner from those effects made by changing the values of parametersaandb. It is important, however, to look at three different cases,a<b,a=b, anda>b.

I.a<.bFor this case, the base equation will be r = a + bcos(k ). The general form, with k = 1, is below.

graph of the base equation with a = 1, b = 2 so that a < b,and k = 1How will this graph change as the value of

kincreases?

*graphs of the equation with different values for k, with a < b (a = 1, b = 2)*

From these examples, it is clear to see that when

kis an even integer, there will be a number of loops equal tokoutside of the larger loops. Likewise, whenkis an odd integer, there will be a nmber of loops equal tokinside the larger loops.

II.a = b.For this case, the base equation will be r = 2 + 2cos(k ). The general form, with k = 1, is below.

How will this graph change as the value of

kincreases to 2, 3, 4, and 5, respectively?

*graphs of the equation with different values of k, and a = b (a = b = 2)*

From these examples, it is plain to see that, when

a=b, the value ofkequals the number of loops, or, what look like flower petals in these examples. Notice that each petal begins and ends at the origin and each petal is the same size as the other petals in each graph.

III. a > b.For this case, the base equation will be r = a + bcos(k ). The general form, with k = 1, is below.

*the graph of the equation with a > b (a = 2, b = 1), k = 1*

How will this graph change as the value of

kincreases?

*graph of the equation with increasing values of k, and a > b (a = 2, b = 1)*

From these examples, it can be seen that the value of

kcorresponds to the number of petal-like divisions that occur. The center of each graph is the origin, and each petal is the same size as the other petals in each graph.

**Conclusion**

The behavior of the polar equation r = a + bcos(k ) is predictable, as shown by each of these investigations. By changing the values of the parameters

a,b, ork, the graph of the equation can take on many different shapes. By lettingk= 1, base equations and graphs can be used to determine the effects of changingaandb. Then, changing the value ofkwill divide the graph intokequal sections.

Extension 1

What happens to the graphs when cos ( ) is replaced with sin ( )?

* The theme of this page was inspired by my wife, who loves Christmas.