*This is part of my notes for the Edexcel FP2 exam, which I will be taking in
2017*

Polar coordinates are different to the Cartesian alternative. Instead of
representing a point as an x,y pair (or however many dimensions you wish), they
are written as a distance (or length) from the pole/origin and an angle -
`\((\textbf{r}, \theta)\)`

.

To make calculus involving polar coordinates easier, we only really use radians for the angles.

If one were to express a line in the Cartesian form, something like `\(y = x + 2\)`

may be used. With a polar graph, it is usually expressed in terms of **r**.

Here is an example of a graph where `\(r = 1\)`

:

Another example of a polar graph is the Archimedian spiral. These are of the
form `\(r = a + b\theta\)`

, and were (as the name suggests) discovered by
Archimedes.

Here is an example, where `\(r = \frac{\theta}{2\pi}\)`

:

Adjusting the value of *a* will turn the spiral, while the value of *b* will
change the distance between the arms - for any given spiral this remains
constant. This is a good example of a curve that is most definitely best
described by a polar equation.

## Converting Cartesian to Polar form

Finding **r** is quite easy, we just use Pythagoras’ Theorem. Given the
Cartesian coordinates `\((x, y)\)`

:

```
$$
r = \sqrt{x^2 + y^2}
$$
```

Finding the angle is almost as easy, with a small caveat. Given `\(\theta\)`

, we
know that `\(tan\theta = \frac{y}{x}\)`

. However, finding `\(\theta\)`

itself is
ever so slightly more involved.

As you can see in the above example, like a Cartesian graph, a polar graph also has four quadrants.

If we were to simply try `\(\theta = arctan\frac{y}{x}\)`

, then `\(\theta\)`

would only ever be in the top or bottom right quadrant. This is not what we
want. Because of this, it may well be necessary to add fractions of pi to the
resulting angle in order to ensure that the result is in the correct quadrant.

### Example

As an example, here is how you would convert `\((3, 4)\)`

from Cartesian to
polar form.

```
$$
r = \sqrt{3^2 + 4^2} = \sqrt{25} = 5
$$
```

```
$$
\theta = \arctan\frac{4}{5} = 0.93
$$
```

Resulting in an answer of `\((5, 0.93)\)`

## Converting Polar to Cartesian

This is actually easier than the inverse, as nothing needs adjusting in regards to the quadrants.

Given a polar coordinate `\((r, \theta)\)`

, the Cartesian equivalent is given
quite simply as:

```
$$
(r\cos\theta, r\sin\theta)
$$
```