Section 13.4: Polar Coordinates
In this section, we describe a coordinate system introduced by Newton, called the polar
coordinate system.
Choose a point in the plane that is called the pole (origin) and labeled O. If P is any other
point in the plane, let r be the distance from O to P and let θ be the angle (in radians)
between the positive x-axis and the line OP as in Figure 1. Then the point P can be
represented by the ordered pair (r, θ) and r, θ are called the polar coordinates of P .
Figure 1: A point expressed in polar coordinates.
Note: We use the convention that an angle is positive if measured in the counterclockwise
direction from the positive x-axis and negative in the clockwise direction. The meaning of
polar coordinates (r, θ) can be extended to the case in which r is negative by taking the
convention that (−r, θ) = (r, θ + π).
The connection between Cartesian and polar coordinates can be seen in Figure 2. If the
point P has Cartesian coordinates (x, y) and polar coordinates (r, θ), then
x = r cos θ
y = r sin θ
and
r=
p
x2 + y 2
tan θ =
y
.
x
Figure 2: Relationship between Cartesian and polar coordinates.
2π
Example: Convert the point 2,
from polar to Cartesian coordinates.
3
Since r = 2 and θ =
2π
,
3
2π
1
x = r cos θ = 2 cos
=2 −
= −1,
3
2
√ !
√
2π
3
y = r sin θ = 2 sin
=2
= 3.
3
2
√
Thus, the point is (−1, 3) in Cartesian coordinates.
√
Example: Convert the point (2 3, −2) from Cartesian to polar coordinates.
√
Since x = 2 3 and y = −2,
r=
p
√
x2 + y 2 = 12 + 4 = 4,
y
1
tan θ = = − √ .
x
3
√
11π
. Thus, the point
Since the point (2 3, −2) lies in the fourth quadrant, we choose θ =
6
11π
is 4,
in polar coordinates.
6
Example: Sketch the region in the plane consisting of points whose polar coordinates satisfy
0 ≤ r ≤ 4 and π/4 ≤ θ ≤ 3π/4.
Figure 3: Polar region 0 ≤ r ≤ 4,
π
3π
≤θ≤
.
4
4
Definition: The graph of a polar equation r = f (θ) is the set
{(r, θ)|r = f (θ)}.
Example: Find a Cartesian equation for the curve r = 2 sin θ and identify the graph of the
equation.
Using the relation y = r sin θ, it follows that sin θ = y/r and so
r = 2 sin θ
2y
r =
r
r2 = 2y
x2 + y 2 = 2y
x2 + (y 2 − 2y) = 0
x2 + (y − 1)2 = 1.
The graph is a circle of radius 1 centered at (0, 1).
Example: Find a polar equation for the curve 2xy = 1.
Using the relations x = r cos θ and y = r sin θ, it follows that
2xy = 1
2r sin θ cos θ = 1
r2 sin(2θ) = 1
2
1
sin 2θ
= csc 2θ.
r2 =
r2