11.7  Polar Equations

By now you've seen, studied, and graphed many functions and equations - perhaps all of them in Cartesian coordinates. Sometimes it is more convenient to use polar equations: perhaps the nature of the graph is better described that way, or the equation is much simpler. Examples of polar equations are:

Graphing a Polar Equation

The graph of a polar equation is the set of all points in the plane whose polar coordinates (at least one representation) satisfy the equation. The graph of the polar equation r = 1 consists of those points in the plane whose distance from the pole is 1. That is the circle of radius 1 centered at the pole.
The graph of = /4 is the set of all points which are at an angle of /4 to the polar axis. In other words, it is a straight line passing through the pole at an angle of /4 to the polar axis. However, not all the graphs of polar equations are so easy to describe. We will restrict our discussion to polar equations of the form r = f(), i.e. r is a function of . Our main goal is to write polar equations for conic sections in Section 11.8. In this section we will take a brief look at graphing in general and desmonstrate with a few specific examples.

One method we use to sketch the graph of a polar equation is to plot points. In graphing a polar equation of the form r = f() we treat as the independent variable and r as the dependent variable. We select several values of , calculate the corresponding value of r, then plot the points (r,). Through these points we draw a smooth curve. Let's work an example.

Example: Sketch the graph of r = 1 + sin().
Solution: We make a table of values by selecting several values of and computing values of r. To aid graphing we use approximate values of r most of the time. Also note that 1 + sin() is periodic of period 2 so that we only need choose values of the angle in one revolution.

0 /6 /4 /3 /2 2/3 3/4 5/6 5/4 3/2 7/4
r 1 1.5 1.71 1.87 2 1.87 1.71 1.5 1 0.29 0 0.29


Our next step is to plot these points in the polar plane. To help us place points, we have drawn in several rays from the pole at intervals of /4. We have also drawn four concentric circles centered at the pole of diameters 1 through 4, respectively. After plotting the twelve points from our table of values, we obtain the following picture.

Finally, through these points we draw a curve. This graph is called a cardioid.

There are some other techniques we can use to help sketch polar graphs. One is to recognize certain forms of polar equations and the corresponding graphs. As in the above example, the graph of
r = a (1 f()),
where f is the sine or cosine function, and a ≠ 0, is a cardioid. The value of a scales the curve, and the choice of f determines the orientation.
Cardioids are special cases of curves called limaçons (pronounced lee-ma-son) which are equations of the form
r = a bf()), a, b ≠ 0.
In the following demonstration you can investigate cardioids and limaçons. You will be observing the graph of r = a + bsin() or r = a + bcos(), where a and b are non-zero real numbers. You can select the values of a and b within certain ranges. The resulting graph will be displayed as you change the values. Initially, the polar equation uses the sine function. Press the button marked "sin" to change to the cosine function. Press the button labeled "b > 0" to change to negative values of b.

Cardioids and Limaçons

Another technique used to help sketch the graph of a polar equation is to look for symmetry. There are three types of symmetry we look for; symmetry about the polar axis, symmetry about the pole, and symmetry about the line = /2. (These are analogous to symmetry about the x-axis, origin, and y-axis, respectively, in the Cartesian plane.)

The graph will be symmetric about the polar axis if whenever (r,) lies on the graph so too does (r,−). This means that f() = f(−). For example, consider the polar equation r = cos(). Now, f(−) = cos(−) = cos() = f(). So its graph is symmetric about the polar axis.


The graph will be symmetric about the pole if whenever (r,) lies on the graph so too does (r,+). This means that f() = f(+). For example, consider the polar equation r = sin(2). Now, f(+) = sin(2(+)) = sin(2+2) = sin(2) = f().


The graph will be symmetric about the line = /2 if whenever (r,) lies on the graph so too does (r,). This means that f() = f(). For example, consider the polar equation r = 2 + sin(). Now, f() = 2 + sin() = 2 + sin()cos() − cos()sin() = 2 + (0)cos() − (−1)sin() = 2 + sin() = f().

Cartesian Equations and Polar Equations

When we want to reference points in a plane with both Cartesian coordinates and polar coordinates, we superimpose the planes so that the polar axis coincides with the positive direction of the x-axis, and the pole corresponds to the origin. This allows us to more easily rewrite a Cartesian equation as a polar equation and vice versa.

Our first example shows how to convert a Cartesian equation to a polar equation. Conversion in this direction is fairly straightforward as we merely have to substitute for x and y as demonstrated below. There is no guarantee, however, that the resulting polar equation can be easily simplified.

Example: Find a polar equation of the circle with radius |a| centered at (a,0).
Solution: In Cartesian coordinates, the equation of this circle is (xa)2 + y2 = a2. In polar coordinates we may express this relationship rather nicely as follows:
(xa)2 + y2 = a2
(rcos() − a)2 + (rsin())2 = a2
r2cos2() − 2arcos() + a2 + r2sin2() = a2
r2(cos2() + sin2()) − 2arcos() = 0

Now since cos2() + sin2() = 1, we have:
r2 − 2arcos() = 0
r(r − 2acos()) = 0
r = 0,  r − 2acos() = 0
The solution r = 0 just says that the graph should contain the pole, so the equation of the circle is:

r = 2acos()

Now plot some values of r and and check that you indeed get points on the circle shown in the sketch.

What would the polar equation be if the circle had been centered at (0,a)? Carry out a similar calculation, and then click here to check your answer.

Converting a polar equation to Cartesian is not so straightforward in general. There are some polar equations in which we can algebraically make r2 show up and that can be an advantage. The next example shows such a case.

Example: Convert r = 7 to a Cartesian equation.
Solution: Squaring both sides produces r2 = 49 which is x2 + y2 = 49, a circle of radius 7.
It is true this example worked out very quickly but consider an equation such as r = 15 + 9 + 22cos(). It is likely that converting this to Cartesian form would produce something quite complicated and no easier to understand than its current form. Fortunately, as we will see, the polar forms of conics sections presented in Section 11.8 can be converted easily to Cartesian equations, thus verifying the equivalence of the two versions of the conic sections.