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An introduction to polar coordinates, including how to convert between polar and Cartesian coordinates and graph equations in polar coordinates. It includes examples and formulas for converting polar to Cartesian coordinates, converting Cartesian to polar coordinates, graphing lines and circles in polar coordinates, and finding tangents to polar curves.
What you will learn
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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 Plot the points whose polar coordinates are given by
π 4
π 4
7 π 4
5 π 2
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 )
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
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 =
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
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 ).
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 = −
Since this point is in the fourth quadrant, we have θ = − 3 π. Therefore the polar co-ordinates for the point (1, −
Example 3 Give polar coordinates for the point (given in Cartesian co-ordinates) (− 1 ,
I (^) For (− 1 ,
3), we have x = −1, y =
Since this point is in the second quadrant, we have θ = 23 π. Therefore the polar co-ordinates for the point (− 1 ,
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
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
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 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 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).