# Polar and Cartesian Coordinates

**... and how to convert between them.**

*In a hurry? Read the Summary. But please read why first:*

To pinpoint where we are on a map or graph there are two main systems:

## Cartesian Coordinates

Using Cartesian Coordinates we mark a point by **how far along** and **how far up** it is:

## Polar Coordinates

Using Polar Coordinates we mark a point by **how far away**, and **what angle** it is:

## Converting

To convert from one to the other we will use this triangle:

## To Convert from Cartesian to Polar

When we know a point in Cartesian Coordinates (x,y) and we want it in Polar Coordinates (r,*θ*) we **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):

^{2}= 12

^{2}+ 5

^{2}

^{2}+ 5

^{2})

**13**

Use the Tangent Function to find the angle:

*θ*) = 5 / 12

*θ*= tan

^{-1 }( 5 / 12 ) =

**22.6°**(to one decimal)

**Answer**: the point (12,5) is **(13, 22.6°)** in Polar Coordinates.

### What is **tan**^{-1} ?

^{-1}

It is the Inverse Tangent Function:

**Tangent**takes an angle and gives us a ratio,**Inverse Tangent**takes a ratio (like "5/12") and gives us an angle.

**Summary**: 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 we know a point in Polar Coordinates (r, *θ*), and we want it in Cartesian Coordinates (x,y) we **solve a right triangle with a known long side and angle**:

### Example: What is (13, 22.6°) in Cartesian Coordinates?

**12.002...**

**4.996...**

Answer: the point (13, 22.6°) is *almost exactly* **(12, 5)** in Cartesian Coordinates.

*θ*) to Cartesian Coordinates (x,y):

**x = r**×**cos(***θ*)**y = r**×**sin(***θ*)

### How to Remember Which is Which?

**(x,y) is alphabetical,
(cos,sin) is also alphabetical**

Also * "y and sine rhyme"* (try saying it!)

And in JavaScript (if you like coding):

function toCartesian(r, angle) { let x = r * Math.cos(angle) let y = r * Math.sin(angle) return [x, y] }

## But What About Negative Values of X and Y?

### Four Quadrants

When we include negative values, the x and y axes divide the

space up into 4 pieces:

**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°)

x = 12 × −0.9659...

x =**−11.59**to 2 decimal places - y = 12 × sin(195°)

y = 12 × −0.2588...

y =**−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 can give 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 |

(Yes, both Quadrant's II and III are "Add 180°")

### Example: P = (−3, 10)

P is in **Quadrant II**

- r = √((−3)
^{2}+ 10^{2})

r = √109 =**10.4**to 1 decimal place - θ = tan
^{-1}(10/−3)

θ = 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})

r= √89=**9.4**to 1 decimal place - θ = tan
^{-1}(−8/5)

θ = 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°