Assignment 11
Polar
Equations
by Rives Poe

In this
investigation we will look at some different parametric equations,
varying the values of the coefficients. Let's see if what we find!
The graph
below shows five different functions:
Purple:
r=2a cos(k*theta)
**/ **Red:
r=2a sin(k*theta)
Blue:
r=2a sin(k*theta)+b **/**
Green:
r=2a cos(k*theta)+b
Light
blue: r=(c)/(a cos(k*theta+b sin(k*theta))
The red and the purple
functions create perfect circles, when *a* and *k* are
equal to 1. The dark
blue and green function have a small "loop"
in the graph. The light
blue function
is the formula of a line. For each of these functions above, *a,b,c,*
and *k *are all equal to 1. Let's look at what happens when
we change those values.

What do
you think will happen if we change the value of *a*? I am
going to set *a* equal to 4.
I did not
change the functions, the colors still represent the same functions
as above. It seems that *a *changes the size of the graph
and for the linear funtion(the light blue) *a* changes the
slope of the line. The x-intercepts remain the same for all of
the functions, however the y-intercepts are changing.

Okay,
moving on. Let's see how b affects the way these functions act.
Be sure to make a conjecture, then scroll down to see if you are
right!
For this
picture above, I set *b *= 4 and *a,k*, and *c *equal
to 1.
Both r=2a sin(k*theta)+b
and r=2a cos(k*theta)+b
have lost their
inner "loop" it seems! The x-intercepts have changed
and the slope of the light
blue line has
decreased.
If I continue
to enlarge the value of b, the blue
and green functions appear to become
closer to circles and the light blue line seems to approach the x-axis. (Our red and purple
circles are not affected by this change, since they do not carry
the value of *b*.)
Below is
a graph where b = 15.

Okay, let's
see how *c* affects the r=(c)/(a cos(k*theta+b sin(k*theta)). This function is the
only of the five that has* c*.
It seems
as though, *c* changes the x and y intercepts. When *c*
was equal to 1, the line crossed the x and y axes at (0,1) and
(1,0). Now, in the graph above, where *c*=3, the line crosses
the axes at (0,3) and (3,0). Again, I have left *a,b*, and*
k *equal to 1.

I have
saved the best for last! For the big finale, how do you think
*k* will affect all of these functions? Any ideas? You might
want to scroll up to the top of the page to see what the picture
looked like when* k*(and *a,b*, and *c*) were equal
to 1.
When*
k* = 2, the functions look like this:
In this
picture it is too busy to see exactly what is happening. So, let's
look at the functions seperately below.
on the
left k=2 / on the right k=3.
The two
functions r=2a
cos(k*theta)
and r=2a sin(k*theta) were circles, but have
now been pulled into the center, making a flower-like shape.
For the folowing
2 pictures, I am only showing the function: r=2a cos(k*theta), because it is difficult to see what
is happening with both of the functions on the screen.

when k
= 6 , there are 12 "petals" (to the left below) and
when k = 7, there are 7 "petals"(to the right below)
So, let's
say that for the two functions r=2a cos(k*theta) and r=2a
sin(k*theta)
that *k*
affects the shape by creating "petals". When *k*
is even, the number of "petals" is twice the value of
*k* and when* k* is odd, the number of "petals"
is the same as the value of *k*.
Let's continue
this investigation, by looking at r=2a sin(k*theta)+b and r=2a cos(k*theta)+b for *k* =2.
***remember
that these two functions looked like this before:
:
Okay, let's
see what they look like when *k*=3:
(I am only
going to show r=2a
sin(k*theta)+b ,
so that we
have a better view of what is happening. The graphs are almost
identical(just rotations of each other).)
What about
when *k*=6?
Alright,
I think I am ready to state what is happening! Are you? It seems
that for these two functions k is doing close to the same thing
as before, however the even and odd values do not effect the graph
in the same way. If the value of *k* is even, the small,
inner "petals" alternate with the larger "petals"
and if *k* is odd, the small, inner "petals" are
inside the larger "petals".
*K *creates "petals"
and the value of *k* determines the number of petals.
Now, one
more function to explore: r=(c)/(a cos(k*theta+b sin(k*theta)). This function is a linear function. Let's
see what happens when *k* = 2.
It seems
as though we have 4 assymptotes, creating 2 parabolas.
When *k*=5:
Now, we
have 5 assymptotes and 5 parabolas. Let's look at one more graph
when *k*=10:
Above there
are 10 sets of assymptotes (20 in all) and 20 parabolas, which
is double the value of *k*. Thus, k affects this linear function,
by constructing assymptotes for the value of *k*. If *k*
is an odd value, then there are as many assympotes as the value
of *k*. If *k* is an even value, then there are twice
as many assymptotes and twice as many parabolas.

This concludes
the investigation of Polar Equations. I hope that this will help
us understand how different constants, *a,b,c*, and *k*
will affect Polar Equations. And if you ever want to construct
flowers in a mathematical way, you should have no problem creating
exactly what you want!!!!

To do some of your
own investigations with these function, click __here.__

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