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Maths Notes: Polar Coordinates and Graphs
Mar 3, 2017
2 minutes read

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.


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:

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

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