Investigation on Polar Equations



Larousse Charlot


To have polar equations, one needs polar coordinates.  The polar coordinates are in the form of parametric equations.

For instance,

x = r cos q

y = r sin q

At times, one might have coordinates, but not polar coordinates.  So, the need to know how to acquire polar coordinate is rather essential.


Obtaining Polar coordinates

Let us say we are given rectangular coordinates such as (1,1), how does one go about converting that into polar coordinates.  Since, the polar coordinates are in the form of polar function, let us use a circle on the Descartes plane to help us geometrically.

The reason for polar coordinates is that circle does not quite translate well in x-y coordinate system, rectangular coordinates.  The alternative description involves of specifying the distance form the origin and the angle measure the positive x-axis counterclockwise to the ray connecting the point and the origin.  For example,

rectangular coordinates

Polar coordinates, on the other hand, indicate the distance and the angle.  For, if speaking of circle, we must have angle.

polar coordinates

In polar coordinates, to find the angle q, by the definition of sine function, we have

hence, the angle

If given a rectangular coordinate (3, 4), we now can convert that into polar coordinates.

Since r is the distance from the origin to the coordinate,

in this case,

and now to find q,

Now, we can talk about polar equations. A polar equation is in the form of r = f(q). r = f(q) is the set of all points (x,y) for which x = r cos q ,y = sin q and r = f(q).  The graph of a polar equation is a graph in the Descartes plane of all those points whose polar coordinates satisfy the given equation.


Let us investigate r = a + b cos (kq).

If we graph this,

a = b = k 1                 0 q 2p


we get a cardiod.  A cardiod is of the class of lamiVon, which is defined by the equation of

r = a b sin q and r = a b cos q.  A cardiod is a lamiVon with the constants a = b. 


When the constant k alters, we have

a = b         purple: k = 2 red: k = 3 blue: k = 4 green: k = 5  0 q 2p

an n-leaf rose.  What we observe here is that the value of k indicates the number of leaf rose in the graph.  What happens to the graph as k approaches infinity?


Now, if we have the equation

r = b cos (kq)

would you say that this equation is of the class of lamiVon?  The answer is yes; find why.

If compare the graphs of the two equations,

a = b = 1          k = 2    0 q 2p

we have to cardiods, one of which is four leaf rose.  What we see here, is that with the equation

r = b cos(kq) the number of leafs is twice the value of k.  But wait, what happens when k = 3?

a = b = 1          k = 3    0 q 2p

Oh my! Our hypothesis does not hold.  For when k = 3, we only have a three leaf rose.  Observe the following graphs.

purple & red: k = 3                  blue: k = 4                0 q 2p

Here, we observe that when k is odd the amount of leaf equals to k, and when k is even the number of leaf is 2k.  Well, is this observation true for all k?

purple & red: k = 3                  blue: k = 4                green: k = 5              black: k = 6

With the observation of this graph here, our theory holds.  For the equation r = b cos (kq), when k is odd the number of leaf equals to k, and k is even the number of leaf is 2k, for any integer k Z.

Using your conclusion, what would happen if the equations were to be in sine instead of cosine?

Well, Ill tell you that our theory holds still.  The graph would only change to fit the definition of sine functions.  Find out the reason.  Play with the graphs