Honors Pre-Calculus
Handout for 10.5 (1)
Polar Coordinates and Equations
Boudreau
Today we start two days of lessons for polar coordinates. We will consider the basics of
polar coordinates, then convert points and equations from rectangular notation to polar
notation. We’ll also look at polar equations.
The Basics of the Polar Coordinate System
Rather than operating off of a pair of perpendicular axises, we use a system much like
what we used for rotating around the unit circle: go out on the positive horizontal axis by
the given length, and rotate counter-clockwise by the given angle.
The coordinates are not given as (x, y), but as (r, θ).
Point P (r, θ)
Unlike the x-y coordinate plane,
polar coordinates are not unique.
r
θ
pole
Think about it—how many
coterminal angles would θ have?
polar axis
    13   11 
 3,    3,
   3,
, etc.
6  
6 
 6 
Not only do we have overlapping points because of coterminal angles, but we can also go
out a negative value on that horizontal axis and rotate from there—another way to wind
up at the same place.
 
5 
 2 

So, the point  5,
 lands in the same place as   5,  or   5,  .
3 
3 
 3 

If we were to place this rotation on an x-y plane, we would put it in standard position.
Point P (r, θ)
θ
pole
polar axis
1
Conversions of coordinates
In this position, it may be clear how to convert between the rectangular grid we’re used to
and the polar coordinate system.
Point P is in the same position as (r, cos θ) and (r, sin θ).
1. So, x  r cos and y  r sin  .
2. Applying the Pythagorean Identity, x 2  y 2  r 2 . (Look familiar?)
3. For the rotation, remember tan  
y
y
; so   tan 1 .
x
x
Much of this should be familiar after all the trig work.
Try it.
 5 
1. Convert  5,  to rectangular coordinate (x, y) notation. Is it in the right quadrant?
 6 

3 5 3
 5 

x  5 cos   5 

2
2
 6 


7

2. Convert   6,
3

 5 
1 5
y  5 sin    5  
 6 
2 2

 to rectangular coordinate notation. Is it in the right quadrant?

 3
 7 
  3 3
y  6 sin 
  6

 3 
 2 
 7 
1
x  6 cos
  6   3
 3 
2


3. Convert 3 3 ,3 to polar coordinates, with r > 0, and 0    2 . (Note how those
restrictions must be specified, since we deal with non-unique coordinates.)
 
2
Use the x and y values to find r first: r 2  x 2  y 2  3 3  3 2  9  3  9  36 .
So, r  6 , but we’ve specified that r has to be positive, so r = 6.
2
Use the x and y values to find θ:
 3   1 
tan   
  

3 3  3
 1  
 
 3 6
  tan 1 
This makes sense, since the original point lies in the first quadrant.
 
The final coordinates are  6,  .
 6
4. Convert (2, -2) to polar coordinates. This time r < 0 and θ lies in one rotation.
Find r first:
r 2  x 2  y 2  2 2  (2) 2  4  4  8
r   8  2 2
Use the negative value: r  2 2 .
Then find θ:
tan  
2
 1
2
  tan 1 (1)  

4
But wait a minute. The original point (2, -2) is in the fourth quadrant. If r
is negative, we have to rotate from that negative x axis and end up in the
3
fourth quadrant. That means we rotate by
.
4
3 

So, the polar coordinates are   2 2 , .
4 

7 

If we had left r as a positive value, the coordinates would be  2 2 ,
.
4 

3
Now, polar equations:
Like coordinates, polar equations are given in r and θ, usually in the form
r = (some trig operation on) θ.
Note: The order we work here is a switch in order from x-y equations. Here, we plug in
θ first, but it’s the second variable we graph.
The solution(s) is/are given as (a, b), where a is the r distance and b is the θ rotation.
The graph of the polar equation is the set of all points in the r-θ plane that make the
statement true (like every other graphing we’ve done). We can graph a polar equation
using an r-θ chart (rather like using an x-y chart). The main difference is that, while we
plug in  first, it is the second variable we graph. We’ll see how this works in a
moment.
Consider two simple graphs.
5.
What would the graph of r  a look like?
The graph is a circle with radius a. No matter the rotation, the points lie a distance of
a out from the pole. In other words, all the points have coordinates a,  .
6.
What would the graph of   a look like?
The graph is a line through the pole, with a slope of tan(a). No matter the distance
from the pole, the points are rotated by a from the polar axis.
Set a = 5 and graph either one on the calculator. Be sure and set the window for θ
large enough to get a full rotation.
4
Conversions from rectangular equations to polar equations:
We can use the earlier conversion statements to rewrite polar equations.
7. Finding a polar equation from a hyperbola: Convert the conic equation
x 2  y 2  16 . Remember cos 2   sin 2   cos 2 .
What should this look like? Graph it to check.
x 2  y 2  16
(r cos  ) 2  (r sin  ) 2  16
r 2 cos 2   r 2 sin 2   16
For cos 2  0 .
r 2 (cos 2   sin 2  )  16
r 2 cos 2  16
16
r2 
cos 2
Here is the graph in rectangular form:
And then in polar form:
5
8. Switch from polar to rectangular: Convert r  a sin  to an equation in x-y form.
What does the curve look like? Maybe multiply each side by r to start.
Remember x 2  y 2  r 2 and y  r sin  .
r  a sin 
r 2  ra sin 
x 2  y 2  ay
x 2  y 2  ay  0
a2
a2
x  ( y  ay  ) 
4
4
a
a
x2  ( y  )2  ( )2
2
2
2
2
a
 a
This is a circle centered on  0,  with a radius of . Graph r  8 sin  ; what are
2
 2
its center and radius?
9. Switch from polar to rectangular: Convert r  a cos to an equation in x-y form.
How does it compare to the curve in #8?
r  a cos 
r 2  ra cos 
x 2  y 2  ax
x 2  ax  y 2  0
a2
a2
 y2 
4
4
a
a
(x  )2  y 2  ( )2
2
2
x 2  ax 
a
a 
This is a circle again, this time centered on  ,0  with a radius of . When
2
2 
a  0 , and a  0 the circles are side by side. Graph r  6 cos ; what are its
center and radius? Look on page 833 to see where the circle is located when
a > 0, and when a < 0, and for sine and cosine operations.
6
10. Coterminal points. Are we getting a sense of what polar equations for circles are
like? What would r  4 sin  be?
Graph on your calculator to check. This time put it in POL mode with the
following window. (Radian or degree?) Set θ step = π/24.
0  
4 x4
2 y  4
The graph will be a circle with center (0, 2) and a radius of 2.
Important point: Once you graph that, change the window to 0    2 . Is
there any difference? With the first window, we get a full circle, even though the
angle stops at π. Why do we get the same circle as we swing past π?
7
As an illustration of the answer, plug in an angle greater than π. Set  
:
6
7
 1
r  4 sin
 4    2 .
6
 2
7 

 
So, we have the point   2,
 , and this point is the same as  2,  . Once we
6 

 6
get to angles greater than π, r becomes negative and the angles get us back into
upper territory.
7 
 

We can trace the point on the graph and see how the points  2,  and   2,

6 
 6

are located compared to each other.
7 

  2,

6 

 
 2, 
 6
7
We can also use an r-θ chart to see the same thing. Remember to start with the
values for θ.
_r___θ_
2.8 π/4
r  4 sin(  / 4)  2 2  2.8
3.5
4
π/3
π/2
r  4 sin(  / 3)  2 3  3.5
r  4 sin(  / 2)  4
3.5
0
2 π/3
π
r  4 sin( 2 / 3)  2 3  3.5
r  4 sin   0
Do you see where these points are on the graph above? Notice how we swing
around, the distances lengthen and shorten—the result is a circle.
We could also do this as a piecewise function. Hmm, do we have time?
11. Finding a polar equation from a line: Convert the linear equation ax  by  c  0
into a polar equation (in the form r = __________).
Use these conversion units:
x  r cos 
.
y  r sin 
ax  by  c  0
ar cos   br sin   c
For a cos  b sin   0 .
r (a cos   b sin  )  c
r
c
a cos   b sin 
8