Section 10.4 – Polar
Coordinates and Polar Graphs
Introduction to Polar Curves
Parametric equations allowed us a new way to
define relations: with two equations. Parametric
curves opened up a new world of curves:
x  2 cos 2 t
y  sin 4 t
Polar coordinates will introduce a new coordinate
system.
Introduction to Polar Curves
You have only been graphing with standard Cartesian
coordinates, which are named for the French
philosopher-mathematician, Rene Descartes.
Example: Plot (−3,2)
𝑦
𝑥
Polar Coordinates
In polar coordinates we identify the origin 𝑂 as the pole and the
positive 𝑥-axis as the polar axis. We can then identify each point
𝑃 in the plane by polar coordinates (𝑟, 𝜃), where 𝑟 gives the
distance from 𝑂 to 𝑃 and 𝜃 gives the angle from the initial ray to
the ray 𝑂𝑃. By convention, angles measured in the
counterclockwise direction are positive.
Since it easier to plot
a point by starting
with the angle, polar
equations are like
inverses. 𝜃 =
independent
variable. 𝑟 =
dependent variable.
NOTE: The
origin 𝑂 has no
well-defined
coordinate. For
our purposes
the coordinates
will be (0, 𝜃) for
any 𝜃.
Example 1
Example: Plot the polar coordinates (3, 2𝜋
).
3
To plot a
point using
polar
coordinates
(𝑟, 𝜃), we
often use a
polar grid:
𝟑
𝟐𝝅
𝟑
First find the
angle 𝜃 on
the polar grid.
Now plot the
point 𝑟 units in
the direction of
the angle.
Example 2
Example: Plot the polar coordinates (−2, 5𝜋
).
3
To plot a
point using
polar
coordinates
(𝑟, 𝜃), we
often use a
polar grid:
First find the
angle 𝜃 on
the polar grid.
𝟓𝝅
𝟑
−2
Now plot the
point 𝑟 units in
the direction of
the angle. If 𝑟
is negative, the
point is plotted
𝑟 units in the
opposite
direction.
Example 3
Graph the polar curve 𝑟 = 3 cos 𝜃. Indicate the direction in
which it is traced.
Notice Polar equations are like
inverses. 𝜃 = independent
variable. 𝑟 = dependent variable.
r
𝜽
3 cos 0 = 3
0
3 cos 𝜋6 = 2.598
𝜋 6
3 cos 𝜋4 = 2.121
𝜋 4
3 cos 𝜋3 = 1.5
𝜋 3
3 cos 𝜋2 = 0
𝜋/2
3 cos 2𝜋
3 = −1.5
3 cos 3𝜋
4 = −2.1
2𝜋 3
3 cos 5𝜋
6 = −2.6
5𝜋 6
3 cos 𝜋 = −3
𝜋
3𝜋 4
The Relationships Between Polar and
Cartesian Coordinates
Find the relationships between 𝑥, 𝑦, 𝑟, & 𝜃.
(𝑟,𝜃)
𝑦
𝑟
𝑦
𝑥
Right triangles are always a
convenient shape to draw.
Using Pythagorean Theorem…
2
2
𝑥 +𝑦 =𝑟
𝑥
2
The Relationships Between Polar and
Cartesian Coordinates
Find the relationships between 𝑥, 𝑦, 𝑟, & 𝜃.
What about the angle 𝜃?
(𝑟,𝜃)
𝑟
sin 𝜃
𝑦
𝑥
You can use a reference angle to find
𝑦
a relationship but that would require an
extra step.
(cos 𝜃 , sin 𝜃)
Instead, compare the coordinates to
the unit circle coordinates.
1
𝑥
cos 𝜃
𝜃
The red and blue triangles are similar
with a scale factor of 1𝑟 = 𝑟. Thus…
𝑥 = 𝑟 cos 𝜃
𝑦 = 𝑟 sin 𝜃
The Relationships Between Polar and
Cartesian Coordinates
Find the relationships between 𝑥, 𝑦, 𝑟, & 𝜃.
(𝑟,𝜃)
𝑦 = 𝑟 sin 𝜃
𝑦
𝑟
𝑥 𝑥 = 𝑟 cos 𝜃
What about a relationship with
𝑥, 𝑦, & 𝜃?
To find the angle measure 𝜃, it is
possible to use the tangent function to
find the reference angle.
Instead investigate the tangent
function and 𝜃:
𝜃
tan 𝜃 =
sin 𝜃
cos 𝜃
𝑟 sin 𝜃
= 𝑟 cos 𝜃
Therefore:
tan 𝜃 =
𝑦
𝑥
(Remember tangent is also the slope of the radius.)
Conversion Between Polar and
Cartesian Coordinates
When converting between coordinate systems the
following relationships are helpful to remember:
𝒙 = 𝒓 𝐜𝐨𝐬 𝜽
𝒚 = 𝒓 𝐬𝐢𝐧 𝜽
𝑦
𝑥
𝒙𝟐 + 𝒚𝟐 = 𝒓𝟐
𝒚
𝐭𝐚𝐧 𝜽 =
𝒙
NOTE: Because of conterminal angles and negative values of r, there are
infinite ways to represent a Cartesian Coordinate in Polar Coordinates.
Example 1
Complete the following:
a) Convert (1, 3) into polar coordinates.
tan  
1
  tan
  3
3
1
r 1  3
2
r 4
r  2
2
3
1
2
2

2,
 3
b) Express your answer in (a) as many ways as you can.


2,

2,
 3   3  2 n 
4
4

2,


2,
 3   3  2 n 
Example 2
Find rectangular coordinates for (16,5𝜋/6).
x  16 cos 56
x  8 3
 8
y  16sin 56
y 8
3,8

NOTE: In Cartesian coordinates, every point in
the plane has exactly one ordered pair that
describes it.
Example 3
Use the polar-rectangular conversion
formulas to show that the polar graph of 𝑟 =
4 sin 𝜃 is a circle.

r  4sin    r
2
r  4r sin 
2
2
x  y  4y
A circle
2
2
x  y  4y  0
centered at (0,2)
2
2
x  y  4y  4  4
with a radius of
2
2
2 units.
x   y  2  4
Conversion Between Polar Equations
and Parametric Equations
The polar graph of 𝑟 = 𝑓(𝜃) is the curve defined
parametrically by:
The slope of
𝒙 = 𝒓 𝐜𝐨𝐬 𝒕 = 𝒇(𝒕) 𝐜𝐨𝐬 𝒕
tangent lines
𝒚 = 𝒓 𝐬𝐢𝐧 𝒕 = 𝒇(𝒕) 𝐬𝐢𝐧 𝒕
is dy/dx not
dr/dΘ.
Example: Write a set of parametric equations for the
polar curve 𝑟 = sin 6𝜃
x  r cos t  sin 6t cos t
y  r sin t  sin 6t sin t
Since we can easily convert a polar equation into parametric
equations, the calculus for a polar equation can be performed
with the parametrically defined functions.
Example
Use polar equation 𝑟 = 2 sin 3𝜃 to answer the
following questions:
(a) Find the Cartesian equation of the tangent line
at 𝜃 = 𝜋/6.
Find the slope of the tangent
line (Remember 𝑡 = 𝜃):
dy
dx t   
Parametric Equations:
x  r cos t  2sin 3t cos t
y  r sin t  2sin 3t sin t
Find dy/dx not dr/dΘ:
d 2sin 3t sin t
dy
dy / dt
dt
d 2sin 3t cos t
dx
dx / dt
dt



6cos3t sin t  2sin 3t cos t
6cos3t cos t  2sin 3t sin t
 3
6
Find the point:


6
6


6
6
x  2sin 3  cos  3
y  2sin 3  sin  1
Find the
equation:

y 1   3 x  3

Example (Continued)
Use polar equation 𝑟 = 2 sin 3𝜃 to answer the
following questions:
(b) Find the length of the arc from 𝜃 = 0 to 𝜃 = 𝜋/6.
Parametric Equations:
x  r cos t  2sin 3t cos t
y  r sin t  2sin 3t sin t
Find dy/dt and dx/dt:
dy
d
dt
dt
dx
dt
Use the Arc Length Formula:
d 

6
0
 2sin 3t sin t
 6 cos 3t sin t  2sin 3t cos t
 dtd 2sin 3t cos t
 6 cos 3t cos t  2sin 3t sin t


6
0
   
dx 2
dt
dy
dt
2
dt
 6 cos 3t cos t  2sin 3t sin t    6 cos 3t sin t  2sin 3t cos t 
2
 2.227
2
dt
Example (Continued)
Use polar equation 𝑟 = 2 sin 3𝜃 to answer the
following questions:
(c) Is the curve concave up or down at 𝜃 = 𝜋/6.
Parametric Equations:

dy
Find dy/dx:
dx
2
Find
d2y/dx2:
d y
dx
2
dy / dt
dx / dt

d
dt
x  r cos t  2sin 3t cos t
y  r sin t  2sin 3t sin t

d 2sin 3t sin t
dt
d 2sin 3t cos t
dt
 dy

6cos3t sin t  2sin 3t cos t
6cos3t cos t  2sin 3t sin t
dx 
2 4cos 2 3t  5 
 3cos3t cos t sin 3t sin t 2


dx dt
Find value of the second
d2y
derivative (Remember 𝑡 = 𝜃): dx 2
t   6
 40
Since the second
derivative is
positive, the graph
is concave up.
Alternate Formula for the Slope of a
Tangent Line of a Polar Curve
If 𝑓 is a differentiable function of 𝜃, then the
slope of the tangent line to the graph of 𝑟 =
𝑓(𝜃) at the point (𝑟, 𝜃) is:
dy
dx

dy / d
dx / d

f ( )cos( )  f '( )sin 
 f ( )sin   f '( )cos
If you do not want to easily convert a polar equation
into parametric equations, you can always memorize
another formula...
Alternate Arc Length Formula for Polar
Curves
The arc length for a polar curve 𝑟(𝜃)
between 𝜃 = 𝛼 and 𝜃 = 𝛽 is given by
L


r 
2
dr
d

2
d
If you do not want to easily convert a polar equation
into parametric equations, you can always memorize
another formula...