19 polar equations and graphs x

Polar Coordinates & Graphs
In the last section, we tracked the location of a point
P in the plane by its polar coordinates (r, ),
P
x
y
(r, )p

P in the plane by its polar coordinates (r, ), where
r = a is the signed distance from the origin (0, 0) to P,
P
x
y
(r, )p
r

P
x
y
O
and  = a is an angle that is measured from the
positive x–axis which gives the direction to P.
(r, )p
r

Conversion Rules
Let (x, y)R and (r, )P be the
rectangular and the polar
coordinates of P,
P
x
y
O
(r, )p = (x, y)R
r

Conversion Rules
coordinates of P, then
x = r*cos()
P
x
y
O
x = r*cos()
The rectangular and polar conversions
(r, )p = (x, y)R
r

Conversion Rules
x = r*cos()
y = r*sin()
P
x
y
O
x = r*cos()
y = r*sin()
r
(r, )p = (x, y)R

Conversion Rules
x = r*cos()
y = r*sin()
r = √x2 + y2
tan() = y/x
P
x
y
O
x = r*cos()
y = r*sin()
r = √x2 + y2
(r, )p = (x, y)R

Conversion Rules
P
x
y
O
x = r*cos()
y = r*sin()
x = r*cos()
y = r*sin()
r = √x2 + y2
tan() = y/x
r = √x2 + y2
If P is in quadrants I, II or IV
then  may be extracted
by inverse trig functions. But if P is in quadrant III, then
 can’t be calculated directly by inverse trig-functions.
(r, )p = (x, y)R

Polar Equations

Polar Equations
A rectangular equation in x and y gives the relation
between the horizontal displacement x and vertical
displacement y of locations.

Polar Equations
displacement y of locations. A polar equation gives a
relation of the distance r the direction .

Polar Equations
The rectangular equation y = x
specifies that the horizontal
displacement x must be the same
as the vertical displacement y for
our points P.

Polar Equations
our points P.
y
y
x
x
P(x, y)
The graph of y = x in the
the rectangular system

Polar Equations
our points P.
y
y
x
x
P(x, y)
The polar equation r = rad says that
the distance r must be the same as
the rotational measurement  for P.
Its graph is the Archimedean spiral.

Polar Equations
our points P.
y
y
x
x
P(x, y)
x
P(r, )

r
Graph of r =  in the
polar system.
The polar equation r = rad says that
the distance r must be the same as
the rotational measurement  for P.
Its graph is the Archimedean spiral.

Let’s look at some basic examples of polar graphs.
The Constant Equations r = c &  = c

Example A. Graph the following polar
equations.
a. (r = c)
b. ( = c)

equations.
a. (r = c) The constant equation r = c
indicates that “the distance r is c, a
fixed constant” and that  may be of
any value.
b. ( = c)

equations.
any value. This equation describes the
circle of radius c, centered at (0,0).
b. ( = c)

Polar Coordinates
Example A. a. Plot the following polar coordinates
A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P.
Find their corresponding rectangular coordinates.
For A(4, 60o)P
x = r*cos()
y = r*sin()
(x, y)R = (4*cos(60⁰), 4*sin(60⁰)),
r2 = x2 + y2
tan() = y/x
x
y
60o
4
A(4, 60o)P

equations.
x
y
c
The constant
equation r = c
b. ( = c)

equations.
x
y
c
The constant
equation r = c
b. ( = c) The constant equation  = c
requires that “the directional angle  is c,
a fixed constant” and the distance r may
be of any value.

equations.
x
y
c
The constant
equation r = c
be of any value. This equation describes
the line with polar angle c.

equations.
x
y
c
The constant
equation r = c
be of any value. This equation describes
the line with polar angle c.
x
y
The constant
equation  = c
 = c

For the graphs of other polar equations we need to plot
points accordingly.

points accordingly. For this we will use the polar graph
paper such as the one shown here which is gridded in
terms of the distances and directional angles.

Example B. a. Graph r = 3cos().

Example B. a. Graph r = 3cos(). r 
3 0o
3(3/2) ≈ 2.6 30o
3(2/2) ≈ 2.1 45o
1.5 60o
0 90o
Let’s plot the
points starting
with  going
from 0 to 90o.

3 0o
3(3/2) ≈ 2.6 30o
3(2/2) ≈ 2.1 45o
1.5 60o
0 90o
Let’s plot the
points starting
with  going
from 0 to 90o. 3

3 0o
3(3/2) ≈ 2.6 30o
3(2/2) ≈ 2.1 45o
1.5 60o
0 90o
–1.5 120o
≈ –2.1 135o
≈ –2.6 150o
–3 180o
Let’s plot the
points starting
with  going
from 0 to 90o. 3
Next continue
with  from
90o to 180o as
shown in the table.

3 0o
3(3/2) ≈ 2.6 30o
3(2/2) ≈ 2.1 45o
1.5 60o
0 90o
–1.5 120o
≈ –2.1 135o
≈ –2.6 150o
–3 180o
Let’s plot the
points starting
with  going
from 0 to 90o. 3
Next continue
with  from
90o to 180o as
shown in the table.
negative so the points are in the 4th quadrant.
Note the r’s are

Continuing with  from 180o to 270o, r 
–3 180o
≈ –2.6 210o
≈ –2.1 225o
–1.5 240o
0 270o
1.5 300o
≈ 2.1 315o
≈ 2.6 330o
3 360o
again r’s are
negative, hence
the points are
located in the
1st quadrant.

–3 180o
≈ –2.6 210o
≈ –2.1 225o
–1.5 240o
0 270o
1.5 300o
≈ 2.1 315o
≈ 2.6 330o
3 360o
again r’s are
negative, hence
the points are
located in the
1st quadrant.
In fact, they
trace over the
same points as  goes from 0o to 90o

–3 180o
≈ –2.6 210o
≈ –2.1 225o
–1.5 240o
0 270o
1.5 300o
≈ 2.1 315o
≈ 2.6 330o
3 360o
again r’s are
negative, hence
the points are
located in the
1st quadrant.
3
In fact, they
trace over the

–3 180o
≈ –2.6 210o
≈ –2.1 225o
–1.5 240o
0 270o
1.5 300o
≈ 2.1 315o
≈ 2.6 330o
3 360o
again r’s are
negative, hence
the points are
located in the
1st quadrant.
3
In fact, they
trace over the
Finally as  goes from 270o to 360o we trace over the
same points as  goes from 90o to 180o in the 4th
quadrant.

–3 180o
≈ –2.6 210o
≈ –2.1 225o
–1.5 240o
0 270o
1.5 300o
≈ 2.1 315o
≈ 2.6 330o
3 360o
again r’s are
negative, hence
the points are
located in the
1st quadrant.
3
In fact, they
trace over the
Finally as  goes from 270o to 360o we trace over the
same points as  goes from 90o to 180o in the 4th
quadrant. As we will see shortly, these points form a
circle and for every period of 180o the graph of
r = cos() traverses this circle once.

b. Convert r = 3cos() to a
rectangular equation. Verify it’s
a circle and find the center and
radius of this circle.
3

Multiply both sides by r so we
have r2 = 3r*cos(). 3

have r2 = 3r*cos().
In terms of x and y, it’s
x2 + y2 = 3x
3

x2 + y2 = 3x
x2 – 3x + y2 = 0
3

x2 + y2 = 3x
x2 – 3x + y2 = 0
completing the square,
x2 – 3x + (3/2)2 + y2 = 0 + (3/2)2
3

x2 + y2 = 3x
x2 – 3x + y2 = 0
x2 – 3x + (3/2)2 + y2 = 0 + (3/2)2
(x – 3/2)2 + y2 = (3/2)2
3

3
x2 + y2 = 3x
x2 – 3x + y2 = 0
x2 – 3x + (3/2)2 + y2 = 0 + (3/2)2
(x – 3/2)2 + y2 = (3/2)2
so the points form the circle centered at (3/2, 0)
with radius 3/2.

In general, the polar
equations of the form
r = ±D*cos()
r = ±D*sin()
are circles with diameter D
and tangent to the x or y axis
at the origin. r = ±a*cos()
r = ±a*sin()
D x
y

r = ±D*cos()
r = ±D*sin()
r = ±a*sin()
D x
y
The points (r, ) and (r, –) are the
vertical mirror images of each other
across the x–axis.
x
(r, )
(r, –)
1

r = ±D*cos()
r = ±D*sin()
r = ±a*sin()
D x
y
across the x–axis. So if r = f() = f(–)
such as r = cos() = cos(–),
then its graph is symmetric with
respect to the x–axis,
x
(r, )
(r, –)
1

r = ±D*cos()
r = ±D*sin()
r = ±a*sin()
D x
y
respect to the x–axis,
x
(r, )
(r, –)
r = cos() = cos(–)
1

r = ±D*cos()
r = ±D*sin()
r = ±a*sin()
D x
y
respect to the x–axis, so r = ±D*cos()
are the horizontal circles.
x
(r, )
(r, –)
r = cos() = cos(–)
1

x
(r, )(–r, –)
y
The points (r, ) and (–r, –) are the
mirror images of each other across
the y–axis.

x
(r, )
r = sin() = –sin(–)
(–r, –)
y
the y–axis. So if r = f(–) = –f()
such as r = sin() = –sin(–),
then its graph is symmetric to the
y–axis and so r = ±D*sin()
are the two vertical circles.

x
(r, )
r = sin() = –sin(–)
(–r, –)
y
x
y
r = cos()r = –cos()
r = sin()
r = –sin()
1
1
Here they are with their orientation
starting at  = 0.
the y–axis. So if r = f(–) = –f()
such as r = sin() = –sin(–),
then its graph is symmetric to the
y–axis and so r = ±D*sin()
are the two vertical circles.

The Cardioids
r = c(1 ± cos())
r = c(1 ± sin())
The graphs of the equations of the form
are called the cardioids, or the heart shaped curves.
Example C. Graph r = 1 – cos().
The graph of r = 1 – cos() is symmetric with respect
to the x–axis because cos() = cos(–). Therefore we
will plot  from 0o to 180o and take its mirrored image
across the x–axis for the complete graph. As  goes
from 0o to 180o, cos() goes from 1 to –1, and the
expression 1 – cos() goes from 0 to 2. The table is
shown below, readers may verify the approximate
values of r’s.

The Cardioids
r = c(1 ± cos())
r = c(1 ± sin())

The Cardioids
r = c(1 ± cos())
r = c(1 ± sin())
to the x–axis because cos() = cos(–).

The Cardioids
r = c(1 ± cos())
r = c(1 ± sin())
across the x–axis for the complete graph.

The Cardioids
r = c(1 ± cos())
r = c(1 ± sin())
across the x–axis for the complete graph. As  goes
from 0o to 180o, cos() goes from 1 to –1, and the
expression 1 – cos() goes from 0 to 2.

r=1–cos() 
0 0o
≈ 0.13 30o
≈ 0.29 45o
0.5 60o
1 90o
1.5 120o
≈ 1.71 135o
≈ 1.87 150o
2 180o
x
2

r=1–cos() 
0 0o
≈ 0.13 30o
≈ 0.29 45o
0.5 60o
1 90o
1.5 120o
≈ 1.71 135o
≈ 1.87 150o
2 180o
Reflecting across the x–axis, we have the cardioid.
x
2

r=1–cos() 
0 0o
≈ 0.13 30o
≈ 0.29 45o
0.5 60o
1 90o
1.5 120o
≈ 1.71 135o
≈ 1.87 150o
2 180o
x
2

The cardioid is the
track of a point on
a circle as it
r=1–cos() 
0 0o
≈ 0.13 30o
≈ 0.29 45o
0.5 60o
1 90o
1.5 120o
≈ 1.71 135o
≈ 1.87 150o
2 180o
x
2
revolves around another circle of the same size.

We use the conversion rules to convert equations
between the rectangular and the polar coordinates.

Example C. Convert each equation into a
corresponding rectangular form.
a. r = 5

The polar equation states that the
distance from the origin to the points
on its graph is a constant 5.
a. r = 5

This is the circle of radius 5,
centered at (0, 0).
a. r = 5
r = 5 x
y

This is the circle of radius 5,
centered at (0, 0).
a. r = 5
r = 5 x
Set r = √x2 + y2 = 5 we have that
x2 + y2 = 52 or the rectangular form of a circle.
y

b. Convert r = 4cos() into the rectangular form.

Multiplying by r to both sides, we have
r2 = 4 rcos(),

r2 = 4 rcos(), in terms of x and y, we have
x2 + y2 = 4x
x2 – 4x + y2 = 0

x2 + y2 = 4x
x2 – 4x + y2 = 0 completing the square,
x2 – 4x + 4 + y2 = 4

x2 + y2 = 4x
x2 – 4x + 4 + y2 = 4 we have
(x – 2)2 + y2 = 4

2
x2 + y2 = 4x
x2 – 4x + 4 + y2 = 4 we have
(x – 2)2 + y2 = 4
x
y
This is the circle centered at (2, 0)
with radius r = 2.

Example D. Convert 2x2 = 3x – 2y2 – 8 into a polar
equation.

2x2 = 3x – 2y2 – 8
equation.

2x2 = 3x – 2y2 – 8
2x2 + 2y2 = 3x – 8
equation.
grouping the square terms,

2x2 = 3x – 2y2 – 8
2x2 + 2y2 = 3x – 8
2(x2 + y2) = 3x – 8
equation.

2x2 = 3x – 2y2 – 8
2x2 + 2y2 = 3x – 8
2(x2 + y2) = 3x – 8
2r2 = 3rcos() – 8
equation.
converting into r and ,

19 polar equations and graphs x

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