**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.