Mathematics Education

EMAT 6680, Professor Wilson

Exploration 11, Polar Equations by Ursula Kirk

For this exploration, I will investigate and compare the following polar equations:

When a and b are equal and k is an integer for various values of k

CASE 1

For the first graph a = b = k = 1 and for the second graph a= b=2, k = 1

In this case, our graphs are on the shape of a bean and they have one petal which is symmetric about the x-axis. In the first picture, the graph crosses the x-axis at 2 when the values of a and b are equal to 1. Because k = 1 we have one only petal. However in the second picture since a and b are 2 the graph crosses the x-axis at 4; but since k = 1 we still have one only petal.

For the first graph a = b =1, k = 2 and for the second graph a= b=2, k = 2

In this case, our graphs have two petals which are symmetric about the y-axis and the x-axis. In the first picture, the graph crosses the x-axis at 2 when a and b are equal to 1. Because k = 2 we have two petals. However in the second picture since a and b are 2 the graph crosses the x-axis at 4; but since k = 2 we still have two petals.

Therefore, I can conjecture that k determines the number of petals while a and b determine where the graph intercepts the x-axis. It seems that when I double the value of a and b we also double the size of the graph. Let’s check my conjecture looking at other cases.

It seems that my conjecture is right for the cases above. The value of k determines the number of petals. When k is 3, I have 3 petals, when k is 4, I have 4 petals and so on.  The values of a and b determine the x-intercept. When a=b=1, the intercept is at 2, but when a=b=2 the intercept is at 4 and the graph has doubles in size

CASE 2

For the first graph b = k = 1 and for the second graph b = 2, k = 1

In this case, our graphs are circular. In the first picture, the graph crosses the x-axis at 1 when the value b is equal to 1. Because k = 1 we have only one circular “petal”. However in the second picture since b is 2 the graph crosses the x-axis at 2; but since k = 1 we still have one only circular petal.

In this case, our graphs have four petals. In the first picture, the graph crosses the x-axis and the y-axis at 1 when the value of b is equal to 1. Because k = 2 we have a four petal “flower”. However in the second picture since b is 2 the graph crosses the x-axis at 4; but since k = 2 we still have a four petal flower.

Now, my conjecture is that if k is even I will end up with 2k petals, but if k is odd, I will end up with k petals. However, the x-intercept will be the same than the x-intercept.