GRAPHING IN POLAR COORDINATES
SYMMETRY
Recall from Algebra and Calculus I that the concept of symmetry was discussed using
Cartesian equations. Also remember that there are three types of symmetry - y-axis, xaxis, and origin. Do you recall how we could test the functions for symmetry? If not,
here are the tests.
1.
A graph has symmetry with respect to the y-axis if, whenever (x, y) is on the
graph, so is the point (-x, y).
2.
A graph has symmetry with respect to the origin if, whenever (x, y) is on the
graph, so is the point (-x, -y).
3.
A graph has symmetry with respect to the x-axis if, whenever (x, y) is on the
graph, so is the point (x, -y).
The big question is how do we test for symmetry of an equation in polar coordinates?
Let us look at the following diagrams to determine the answer to this question.
y-axis symmetry
x-axis symmetry
symmetry about the origin
So here are the symmetry tests for polar graphs.
1. Symmetry about the x-axis: If the point (r,  ) lies on the graph, then the point
(r, - ) or (-r,  -  ) also lies on the graph.
2. Symmetry about the y-axis: If the point (r,  ) lies on the graph, then the point (r,
 -  ) or (-r, - ) also lies on the graph.
3. Symmetry about the origin: If the point (r,  ) lies on the graph, then the point (r,  ) or (r,  +  ) also lies on the graph.
EXAMPLE 1:
Identify the symmetries of the curve r = 2 + 2 cos  and then
sketch the graph.
SOLUTION:
(r, -  )  r = 2 + 2 cos (- )  r = 2 + 2 cos 
(Remember that cosine is an even function.)
x-axis symmetry: yes
(-r, -  )  -r = 2 + 2 cos (- ) -r = 2 + 2 cos   r = -2 - 2 cos

y-axis symmetry: no
(-r,  )  -r = 2 + 2 cos   r = -2 - 2 cos 
symmetry with respect to the origin: no
Now, let us compare our findings with the graph of this function.
Notice that the graph's only symmetry is with respect to the xaxis, and this is what we determine with our testing.
EXAMPLE 2:
Identify the symmetries of the curve r = 2 + sin  and then sketch
the graph.
SOLUTION:
(r, -  )  r = 2 + sin (- )  r = 2 -sin 
(Remember that sine is an odd function.)
x-axis symmetry: no
(-r, -  )  -r = 2 + sin (- )  -r = 2 - sin   r = -2 + sin 
y-axis symmetry: no
Is this a correct answer? No! Let us look at the graph of these two
functions on the same coordinate axis.
r = 2 + sin  is the purple graph
r = - 2 + sin  is the teal graph
We have the same graph, but they start in different places.
Therefore, this function does have y-axis symmetry.
Sometimes it is best to look at the graph of the polar function
instead of trusting algebraic manipulation.
EXAMPLE 3:
Identify the symmetries of the curve r 2 = cos  and then sketch
the graph.
SOLUTION:
(r, -  )  r 2 = cos (- )  r 2 = cos 
(Remember that cosine is an even function.)
x-axis symmetry: yes
(-r, -  )  (-r) 2 = cos (- )  r 2 = cos 
y-axis symmetry: yes
(-r,  )  (-r) 2 = cos   r 2 = cos 
symmetry with respect to the origin: yes
Now, let us compare our findings with the graph of this function.
Yes this graph does fit the results that we received from algebraic
manipulation.
SLOPES
Now let us look at how to determine the slope of a polar curve r = f ( ). Remember that the
slope of any curve is given by dy/ dx not dr/ d , so we will have to derive out the formula for
dy/dx.
Let x = r cos  = f ( ) cos  and y = r sin  = f ( ) sin  . If f is a differentiable function of ,
then so is x and y. When dx/ d  0, we can find dy/ dx from the parametric formula.
EXAMPLE 4:
Find the slope of the curve r = -1 + sin  at    .
SOLUTION:
Now evaluate dy/dx at    .
EXAMPLE 5:
Find the slope of the curve r = cos 2  at    / 2.
SOLUTION:
Now evaluate dy/ dx at    / 2.
FINDING POINTS WHERE POLAR GRAPHS INTERSECT
There are two types of intersection points. They are (1) simultaneous, and (2) non-simultaneous.
Here is how your find both types of points.
To find the simultaneous intersection points, set the two equations equal to each other and solve
for  .
To find the non-simultaneous intersection points, graph both equations and determine where the
graphs cross each other.
EXAMPLE 6:
Find the points of intersection (both types) of the pair of curves r
= 1 + sin  and r = 1 - sin  .
SOLUTION:
SIMULTANEOUS INTERSECTION POINTS
1 + sin  = 1 - sin   2sin  = 0   = 0 and  = 
When  = 0, then r = 1 + sin 0 = 1  (1, 0).
When  =  , then r = 1 + sin  = 1  (1,  ).
NON-SIMULTANEOUS INTERSECTION POINTS
Let us graph both equations on the same axis.
r = 1 + sin  is in purple
r = 1 - sin  is in teal
Notice that the graphs cross each other at the point (0, 0), so this
is the non-simultaneous intersection point. This is the only one.
EXAMPLE 7:
Find the points of intersection (both types) of the pair of curves r
= cos  and r = 1 - cos  .
SOLUTION:
SIMULTANEOUS INTERSECTION POINTS
NON-SIMULTANEOUS INTERSECTION POINTS
r = cos  is in purple
r = 1 - cos  is in teal
The graphs cross each other at the origin, so the only nonsimultaneous intersection point is (0, 0).
EXAMPLE 8:
Find the points of intersection (both types) of the pair of curves r 2
= cos 2  and r 2 = sin 2  .
SOLUTION:
SIMULTANEOUS INTERSECTION POINTS
NON-SIMULTANEOUS INTERSECTION POINTS
r 2 = sin 2  is in purple
r 2 = cos 2  is in teal
The only non-simultaneous intersection point for these two graphs
is the origin, (0, 0).
I have discussed three major topics in this set of supplemental notes. The first was how to
determine the symmetry of a polar graph. When looking at some examples, we concluded that
we would sometimes have to look at the graph of the equation. The use of symmetry will be
important when we start to determine the area inside the curve. The second topic that I discussed
is the slope of a polar curve. This is an application of the derivative of a parametric curve.
Finally, I talked about how to find the two types of intersection points. This will be useful when
we start to determine the area between two curves