Transcript Document

Graphs of Polar
Equations
Using Polar Grids to Graph Polar
Equations
Recall that a polar equation is an equation whose variables are r and θ. The
graph of a polar equation is the set of all points whose polar coordinates
satisfy the equation. We use polar grids like the one shown to graph polar
equations. The grid consists of circles with centers at the pole. This polar grid
shows five such circles. A polar grid also shows lines passing through the pole,
In this grid, each fine represents an angle for which we know the exact values
of the trigonometric functions.
2
3 3
4
5
6

2

4 
˝
6

7
6
5
4 4
3

3
2
3
2
4
0
11 
6
7
5 4
3
Text Example
• Graph the polar equation r = 4 cos  with  in radians.
Solution
We construct a partial table of coordinates using multiples of 6.
Then we plot the points and join them with a smooth curve, as shown.

r = 4 cos 
(r, )
0
4 cos 0 = 4 • 1 = 4
(4, 0)
 /6
4 cos  /6 = 4 •3/2=2 3=3.5
(3.5,  /6)
 /3
4 cos  /3 = 4 • 1/2 = 4
(2,  /3)
 /2
4 cos  /2 = 4 • 0 = 0
(0,  /2)
2 /3
4 cos2  /3 = 4(- 1/2) = -2
(-2, 2  /3)
5/6
4 cos5  /6 = 4(- 3/2)=-2 3=-3.5
(-3.5,5 /6)
˝
4 cos ˝  = 4(-1) = -4
(-4, )
(2,  /3)
(0,  /2)
2
3 3
4
5
6

2

3
˝
4 
6
(3.5,  /6)
0
˝
2 4
(4, 0) or (-4, )
7
11 
6
6
5
7
(-2, 2  /3)
4 4
5 4
3

3
3
2
(-3.5, 5 /6)
Circles in Polar Coordinates
The graphs of
r = a cos  and
r = a sin 
Are circles.
r = a cos 
r = a sin 
 /2
/2
a
˝
0
˝
0
a
3  /2
3  /2
Text Example
• Check for symmetry and then graph the polar equation: r =
1 - cos .
Solution
We apply each of the tests for symmetry.
Polar Axis: Replace  by -  in r = 1 - cos  :
r = 1 - cos (- )
Replace  by -  in r = 1 - cos  .
r = 1 - cos 
The cosine function is even: cos (-  ) = cos  .
Because the polar equation does not change when  is replaced by - , the
graph is symmetric with respect to the polar axis.
Text Example cont.
Solution
The Line  = 2: Replace (r, ) by (-r, - ) in r = 1 - cos  :
-r = 1 - cos(-)
Replace r by -r and  by – in -r = 1 - cos(-  ).
-r = 1 – cos 
cos(-  ) = cos  .
r = cos  - 1
Multiply both sides by -1.
Because the polar equation r = 1 - cos  changes to r = cos  - 1 when (r, ) is
replaced by (-r, - ), the equation fails this symmetry test. The graph may of
may not be symmetric with respect to the line  = 2.
The Pole: Replace r by -r in r = 1 - cos  :
-r = 1 – cos 
Replace r by –r.
r = cos  - 1
Multiply both sides by -1.
Because the polar equation r = 1 - cos  changes to r = cos  - 1 when r is
replaced by -r, the equation fails this symmetry test. The graph may or may
not be symmetric with respect to the pole.
Text Example cont.
Solution
Now we are ready to graph r = 1 - cos . Because the period of the
cosine function is 2r, we need not consider values of  beyond 2. Recall that
we discovered the graph of the equation r = 1 - cos  has symmetry with
respect to the polar axis. Because the graph has symmetry, we may be able to
obtain a complete graph without plotting points generated by values of  from
0 to 2. Let's start by finding the values of r for values of  from 0 to .

r
0
0
 /6
 /3
2
2  /3
0.13
0.50
1.00
1.50
5  /6

1.87
2
The values for r and  are in the
table. Examine the graph. Keep in mind
that the graph must be symmetric with
respect to the polar axis.

2
3 3
4

˝
2
3
4
5
6

6

7
6
5
4 4
3

1
0
2
11 
6
3
2
7
5 4
3
Text Example cont.
Solution
Thus, if we reflect the graph from the last slide about the polar axis, we will
obtain a complete graph of r = 1 - cos , shown below.

2
3 3
4

2
3
4
5
6

6

7
6
5
4 4
3

1
0
2
11 
6
3
2
7
5 4
3
Limacons
The graphs of
r = a + b sin , r = a - b sin ,
r = a + b cos , r = a - b cos , a > 0, b > 0
are called limacons. The ratio ab determines a limacon's shape.
Inner loop if ab < 1
Heart shaped if ab = 1 Dimpled with no inner
loop if 1< ab < 2
and called cardiods


2

3
2


2
0
3
2
loop if ab  2.

2
0
No dimple and no inner

2
0
3
2

0
3
2
Example
• Graph the polar equation
y= 2+3cos
Example
• Graph the polar equation
y= 2+3cos
Solution:
Rose Curves
The graphs of
r = a sin n
and
r = a cos n, a does not equal 0,
are called rose curves. If n is even, the rose has 2n petals. If n is odd, the
rose has n petals.
r = a sin 2
r = a cos 3
r = a cos 4
r = a sin 5
Rose curve
with 4 petals
Rose curve
with 3 petals
Rose curve
with 8 petals

Rose curve
with 5 petals



2
2
2
2
n=4
a
a
n=3

0

0
a

0

0
a
n=2
3
2
3
2
3
2
a
3
2
n=5
Example
• Graph the polar equation y=3sin2
Example
• Graph the polar equation y=3sin2
Solution:
Lemniscates
• The graphs of r2 = a2 sin 2 and r2 = a2
cos 2 are called lemniscates
Lemniscate:
r2 = a2 cos 2