## Equations using Polar Co-ordinates

Equations with polar co-ordinates, like those with Cartesian ones, are a relationship between the two co-ordinates. Graphs of these equations can be draw by drawing a line through all the points that satisfy the equation.

**Note// This post assumes you can use polar co-ordinates. **

Examples of some simple polar equations are

**r=**θ

and

**r=3sin**θ

### Tips for Sketching Polar Graphs

There are number fo ways to make it easier to sketch polar graphs. Some of the basic methods are as follows:

- Plot a point for every π/6
- If the graph is a cos function it will be symmetrical about the initial line (since cos(-x) = cosx). This means you only have to calculate from 0 to π and then draw the other side using symmetry.
- If the graph is a sin function it will be symmetrical about the line θ=π/2 so you only need to calculate from -π/2 to π/2 and then flip it.

An applet for sketching these graphs is at http://www.ies.co.jp/math/java/calc/sg_kyok/sg_kyok.html however I recommend that you attempt to sketch the graphs for yourself first and then use it to check. Note that for some graphs it doesn’t seem to give you the sketch for the full range you enter.

### Conversion to and from Cartesian Form

As with points in a 2D plane polar equations can be converted to Cartesian equations and similarly Cartesian equations can be converted to polar ones. To do this we will use the same equations as were need for the conversion of points. These are:

- x = rcosθ
- y=rsinθ
- r
^{2}= x^{2}+ y^{2} - θ = tan
^{-1}(y/x)

We can then substitute the appropriate equations into the one we want to convert to find its equivlinet.

**A Cartesian to polar example**

consider the equation of a straight line in Cartesian co-ordinates

y=3+2x

To write this in polar form we can substitute in equations 1 and 2 to get

rsinθ = 3 + 2rcosθ

which can be re-arranged however you like.

**A polar to Cartesian example**

consider the polar equation

r=3sin θ

Then from 2 we know sinθ=y/r so we get

r = 3y/r

which gives

r^{2} = 3y

but from 3 we know r^{2} = x^{2} + y^{2}

so we get

x^{2} + y^{2} = 3y

## Using Polar Co-ordinates and Converting to and from Cartesian

Polar co-ordinates are a different co-ordinate scheme to the standard Cartesian co-ordinates. Again any point in a 2D plane can be located using only two numbers.

It works by taking an “initial line” (shown in red — the equivalent of the positive x axis in Cartesian co-ordinates) with the origin at one end. Then any point can be found by drawing a line from the point to the origin and quoting the length of this line (the radius) and the angle the line makes with the origin. These numbers are usually written in brackets in the same way as Cartesian co-ordinates with the radius first followed by the angle. For example look at the point (3,π/3) below.

**Note//** The angles are usually measured in radians

### Cartesian Equivalent of Polar Co-ordinates

Since both Cartesian and polar co-ordinates are a way of describing a point position in a 2D plane it is possible to convert between then. When doing this the initial line is taken as the x-axis.

To find the Cartesian co-ordinates we must use trigonometry by drawing a vertical line down from the point to the x-axis to form a right angled triangle. The length of the vertical line then gives the y- coordinate and its distance from the origin gives the x-coordinate.

To find the x and y values for the point (r,Θ) we must therefore use the equations

x = rcosΘ

and

y = rsinΘ

If we want to go in reverse to find the polar co-ordinates of the point (x,y) in a Cartesian system we must solve these equations simultaneously.

We can eliminate Θ by squaring both of the equations to obtain

x^{2} = r^{2}cos^{2}Θ

and

y^{2} = r^{2}sin^{2}Θ

and then adding these equations to get

x^{2} + y^{2} = r^{2}(cos^{2}Θ + sin^{2}Θ)

by substituting the trigonometric identity sin^{2}+cos^{2} = 1 to get

x^{2} + y^{2} = r^{2}

we have removed Θ and can therefore calculate r using

**r =√(x ^{2} + y^{2})**

To find Θ we can divide the two equations given at the start such that the r’s cancel to get

y/x = sin Θ/cos Θ

which using the identity tan Θ=sin Θ/cos Θ gives

** Θ = tan ^{-1}(y/x)**

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