Polar equations
are composed of the polar coordinates
where
r is the directed distance
of a point P on the polar curve to the origin; and 
is
the angle measure, in radians, of the point P from the x-axis.
Therefore, the coordinates 
determine
the location of the point P. It is possible to rewrite an equation
that is in polar coordinates as rectangular coordinates using
the following coversion identities:
R 
R 
R
.
The graph appears to be an ellipse. But, how can we know for sure? If it is an ellipse, how can we determine, algebraically, information about the center and the major and minor axes? Merely, by observation, we can estimate the following information :
R Major Axis: 5
R Minor Axis: 4
R Center: (3,0)
We could, therefore,
guess that the equation for this figure, in rectangular coordinates
is: 
.This information, however, would
be more difficult to observe for various figures.
As you will
notice, this equation is very similar to the previous one. The
only difference between the two is the angular measure of which
the cosine is being taken for each .
verses 
?
Let's take a look at the graphs of these two polar equations together
:
in blue; 
in green.In rectangular
coordinates, cos(x-2) is the translation of the function cos(x)
2 units to the right. Similarly, in polar coordinates,
is the rotation of about the orgin
by 45 degrees in the positive direction.
Let's take this
knowledge about the cosine function as we look at the polar equation
of real interest: and how it compares
with: .
in
red; in
blue.Again, it seems as though the blue graph is a rotation of the red about the origin. Does it hold that the rotation angle is by 45 degrees as we saw in the graphs of cosine?
as
well as as increases
.|
for |
for |
|
|
|
 ,5.5581) |
,8) |
 ,3.2) |
 ,5.5581) |
 , 5.5581) |
 ,3.2) |
By looking at
this table of values, you can see that indeed
is the a or 45 degrees rotation of . Click here for a Graphing Calculator
2.2 animation demonstrating how this fact.
THIRDLY, let's investigate the
function .
The following is a graph of the function:
Why is it that we get an assymptote at x=2 and y=-2? It is much easier to answer these questions by looking at the equation in its rectangular coordinates.
is equivalent
to 
in rectangular coordinates. Click
here
to verify this.
It is easy to
see that 
has a vertical assymptote at x=(-2) because this value of
x gives a zero in the denominator of the function.
The horizontal
assymptote can be found by taking the limit, L, of 
as x approaches infinity. Since the highest power of x in the
numerator is equivalent to the highest power of x in the denominator,
L = (-2/1) = -2. Therefore, a horizontal assymptote occurs at y=-2.
Therfore, because
the vertical assymptote is determined by the "2" in
the original polar equation, we see that whatever constant, c,
appears in the numerator x= -c is the vertical assymptote. To
test this out, let's try graphing 
Notice, however,
that by altering the numerator, you also alter the location of
the horizontal asssymptote as well. By reverting to our rectangular
coordinates, we will find that the new horizontal assymptote lies
as y = - 5.
IN SUMMARY, we find that it is fairly easy (dispite
slightly messy Algebra) to converty from polar equations to rectangular
equations. Through this fairly simple conversion we learn some
things about our graphs that are not so obvious from the polar
form of the equation. It is very useful to be able to do such
conversions in order to show students the relationships between
the polar and rectangular coordinate sysytem. This conversion
process can help students understand that one graph can be represented
in numerous ways, one of which is its polar equation.
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