Polar and Cartesian Coordinates
... and how to convert between them.
To pinpoint where you are on a map or graph there are two main systems:
Cartesian Coordinates
Using Cartesian Coordinates you mark a point by how far along and how far up it is:
Polar Coordinates
Using Polar Coordinates you mark a point by how far away, and what angle it is:
Converting
To convert from one to the other, you need to solve the triangle:
To Convert from Cartesian to Polar
When you know a point in Cartesian Coordinates (x,y) and want it in Polar Coordinates (r,θ) you solve a right triangle with two known sides.
Example: What is (12,5) in Polar Coordinates?
Use Pythagoras Theorem to find the long side (the hypotenuse):
Use the Tangent Function to find the angle:
What is tan^{-1} ? It is the Inverse Tangent Function.
- Tangent takes an angle and gives you a ratio,
- Inverse Tangent takes a ratio (like "5/12") and gives you an angle.
Answer: the point (12,5) is (13, 22.6°) in Polar Coordinates.
So to convert from Cartesian Coordinates (x,y) to Polar Coordinates (r,θ):
- r = √ ( x^{2} + y^{2 })
- θ = tan^{-1 }( y / x )
Note: Calculators may give the wrong value of tan^{-1 }() when x or y are negative ... see below for more.
To Convert from Polar to Cartesian
When you know a point in Polar Coordinates (r, θ), and want it in Cartesian Coordinates (x,y) you solve a right triangle with a known long side and angle:
Example: What is (13, 22.6°) in Cartesian Coordinates?
Use the Cosine Function for x: | cos( 22.6 °) = x / 13 | |
Rearranging and solving: | x = 13 × cos( 22.6 °) = 13 × 0.923 = 12.002... | |
Use the Sine Function for y: | sin( 22.6 °) = y / 13 | |
Rearranging and solving: | y = 13 × sin( 22.6 °) = 13 × 0.391 = 4.996... |
Answer: the point (13, 22.6°) is almost exactly (12, 5) in Cartesian Coordinates.
So, to convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) :
- x = r × cos( θ )
- y = r × sin( θ )
But What About Negative Values of X and Y?
Four QuadrantsWhen we include negative values, the x and y axes divide the Quadrants I, II, III and IV (They are numbered in a counter-clockwise direction) |
When converting from Polar to Cartesian coordinates it all works out nicely:
Example: What is (12, 195°) in Cartesian coordinates?
r = 12 and θ = 195°
- x = 12 × cos(195°) = 12 × -0.9659... = -11.59 to 2 decimal places
- y = 12 × sin(195°) = 12 × -0.2588... = -3.11 to 2 decimal places
So the point is at (-11.59, -3.11), which is in Quadrant III
But when converting from Cartesian to Polar coordinates,
... the calculator may give you the wrong value of tan^{-1}
It all depends what Quadrant the point is in! Use this to fix things:
Quadrant | Value of tan^{-1} |
I | Use the calculator value |
II | Add 180° to the calculator value |
III | Add 180° to the calculator value |
IV | Add 360° to the calculator value |
Example: P = (-3, 10)
P is in Quadrant II
- r = √((-3)^{2} + 10^{2}) = √109 = 10.4 to 1 decimal place
- θ = tan^{-1}(10/-3) = tan^{-1}(-3.33...)
The calculator value for tan^{-1}(-3.33...) is -73.3°
So the Polar Coordinates for the point (-3, 10) are (10.4, 106.7°)
Example: Q = (5, -8)
Q is in Quadrant IV
- r = √(5^{2} + (-8)^{2}) = √89 = 9.4 to 1 decimal place
- θ = tan^{-1}(-8/5) = tan^{-1}(-1.6)
The calculator value for tan^{-1}(-1.6) is -58.0°
So the Polar Coordinates for the point (5, -8) are (9.4, 302.0°)
Summary
To convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) :
- x = r × cos( θ )
- y = r × sin( θ )
To convert from Cartesian Coordinates (x,y) to Polar Coordinates (r,θ):
- r = √ ( x^{2} + y^{2 })
- θ = tan^{-1 }( y / x )
The value of tan^{-1}( y/x ) may need to be adjusted:
- Quadrant I: Use the calculator value
- Quadrant II: Add 180°
- Quadrant III: Add 180°
- Quadrant IV: Add 360°