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.
