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Polar Co-ordinates
A polar coordinate system, gives the co-ordinates of a point with reference to a point O and a half line or ray starting at the point O. We will look at polar coordinates for points in the xy -plane, using the origin (0, 0) and the positive x-axis for reference.
A point P in the plane, has polar coordinates (r , θ), where r is the distance of the point from the origin and θ is the angle that the ray |OP| makes with the positive x-axis.
Example 1
Example 1 Plot the points whose polar coordinates are given by
π 4
π 4
7 π 4
5 π 2
Example 1
Note the representation of a point in polar coordinates is not unique. For instance for any θ the point (0, θ) represents the pole O. We extend the meaning of polar coordinate to the case when r is negative by agreeing that the two points (r , θ) and (−r , θ) are in the same line through O and at the same distance |r | but on opposite side of O. Thus
(−r , θ) = (r , θ + π)
Example 2 Plot the point (− 3 , 3 π 4 )
Example 1
Note the representation of a point in polar coordinates is not unique. For instance for any θ the point (0, θ) represents the pole O. We extend the meaning of polar coordinate to the case when r is negative by agreeing that the two points (r , θ) and (−r , θ) are in the same line through O and at the same distance |r | but on opposite side of O. Thus
(−r , θ) = (r , θ + π)
Example 2 Plot the point (− 3 , 3 π 4 )
0
Π 4
Π 2 3 Π 4
Π
5 Π 4 3 Π 2
7 Π 4
Polar to Cartesian coordinates
To convert from Polar to Cartesian coordinates, we use the identities:
x = r cos θ, y = r sin θ
Example 3 Convert the following (given in polar co-ordinates) to Cartesian coordinates (2, π 4 ) and (3, − π 3 ) I (^) For (2, π 4 ), we have r = 2, θ = π 4. In Cartesian co-ordinates, we get x = r cos θ = 2 cos(π/4) = 2 √^12 =
we get y = r sin θ = 2 sin(π/4) = 2 √^12 =
Polar to Cartesian coordinates
To convert from Polar to Cartesian coordinates, we use the identities:
x = r cos θ, y = r sin θ
Example 3 Convert the following (given in polar co-ordinates) to Cartesian coordinates (2, π 4 ) and (3, − π 3 ) I (^) For (2, π 4 ), we have r = 2, θ = π 4. In Cartesian co-ordinates, we get x = r cos θ = 2 cos(π/4) = 2 √^12 =
we get y = r sin θ = 2 sin(π/4) = 2 √^12 =
I (^) For (3, − π 3 ), we have r = 3, θ = − π 3. In Cartesian co-ordinates, we get x = r cos θ = 3 cos(− π 3 ) = 3 12 = (^32) we get y = r sin θ = 3 sin(− π 3 ) = 3 −
√ 3 2 =^
− 3 √ 3 2
Cartesian to Polar coordinates
To convert from Cartesian to polar coordinates, we use the following identities
r 2 = x^2 + y 2 , tan θ =
y x When choosing the value of θ, we must be careful to consider which quadrant the point is in, since for any given number a, there are two angles with tan θ = a, in the interval 0 ≤ θ ≤ 2 π. Example 3 Give polar coordinates for the points (given in Cartesian co-ordinates) (2, 2), (1, −
3), and (− 1 ,
I (^) For (2, 2), we have x = 2, y = 2. Therefore r 2 = x^2 + y 2 = 4 + 4 = 8, and r =
We have tan θ = yx = 2/2 = 1. Since this point is in the first quadrant, we have θ = π 4. Therefore the polar co-ordinates for the point (2, 2) are (
8 , π 4 ).
Cartesian to Polar coordinates
To convert from Cartesian to polar coordinates, we use the following identities
r 2 = x^2 + y 2 , tan θ =
y x When choosing the value of θ, we must be careful to consider which quadrant the point is in, since for any given number a, there are two angles with tan θ = a, in the interval 0 ≤ θ ≤ 2 π. Example 3 Give polar coordinates for the points (given in Cartesian co-ordinates) (2, 2), (1, −
3), and (− 1 ,
I (^) For (2, 2), we have x = 2, y = 2. Therefore r 2 = x^2 + y 2 = 4 + 4 = 8, and r =
We have tan θ = yx = 2/2 = 1. Since this point is in the first quadrant, we have θ = π 4. Therefore the polar co-ordinates for the point (2, 2) are (
8 , π 4 ). I (^) For (1, −
3), we have x = 1, y = −
- Therefore r 2 = x^2 + y 2 = 1 + 3 = 4, and r = 2. We have tan θ = yx = −
Since this point is in the fourth quadrant, we have θ = − 3 π. Therefore the polar co-ordinates for the point (1, −
- are (2, − 3 π ).
Example 3
Example 3 Give polar coordinates for the point (given in Cartesian co-ordinates) (− 1 ,
I (^) For (− 1 ,
3), we have x = −1, y =
- Therefore r 2 = x^2 + y 2 = 1 + 3 = 4, and r = 2. We have tan θ = yx = −
Since this point is in the second quadrant, we have θ = 23 π. Therefore the polar co-ordinates for the point (− 1 ,
- are (2, 23 π ).
Graphing Equations in Polar Coordinates
The graph of an equation in polar coordinates r = f (θ) or F (r , θ) = 0 consists of all points P that have at least one polar representation (r , θ) whose coordinates satisfy the equation.
Example 4 Graph the following equations r = 5, θ = π 4
Graphing Equations in Polar Coordinates
The graph of an equation in polar coordinates r = f (θ) or F (r , θ) = 0 consists of all points P that have at least one polar representation (r , θ) whose coordinates satisfy the equation. II (^) Lines: A line through the origin (0, 0) has equation θ = θ 0 I (^) Circle centered at the origin: A circle of radius r 0 centered at the origin has equation r = r 0 in polar coordinates. Example 4 Graph the following equations r = 5, θ = π 4
Graphing Equations in Polar Coordinates
The graph of an equation in polar coordinates r = f (θ) or F (r , θ) = 0 consists of all points P that have at least one polar representation (r , θ) whose coordinates satisfy the equation. II (^) Lines: A line through the origin (0, 0) has equation θ = θ 0 I (^) Circle centered at the origin: A circle of radius r 0 centered at the origin has equation r = r 0 in polar coordinates. Example 4 Graph the following equations r = 5, θ = π 4 I (^) The equation r = 5 describes a circle of radius 5 centered at the origin. The equation θ = π 4 describes a line through the origin making an angle of π 4 with the positive x axis.
Example 5
Example 5 Graph the equation r = 6 sin θ and convert the equation to an equation in Cartesian coordinates.
θ r 0
π 4 π 2 3 π 4
π
5 π 4
2 π
Example 5
Example 5 Graph the equation r = 6 sin θ and convert the equation to an equation in Cartesian coordinates.
θ r 0
π 4 π 2 3 π 4
π
5 π 4
2 π
II (^) We find the value of r for specific values of θ in order to plot some points on the curve. We note that it is enough to sketch the graph on the interval 0 ≤ θ ≤ 2 π, since (r (θ), θ) = (r (θ + 2π), θ + 2π).
