Polar Coordinates

Revision cards for FP2 topic 7

HideShow resource information
  • Created by: Emily
  • Created on: 15-06-12 09:46

Polar and Cartesian coordinates

Polar coordinates are an alternative way of describing the position of something (e.g. a point). 

The positive x - axis is called the initial line. Polar coorinates are written as (r, θ).

x = rcosθ

y = rsinθ

r² = x² + y²

θ = arctan (y/x)

Using the origin as a starting point, the distance is measured from the origin to the point in question: this distance is r. The angle θ is the angle between the line from the origin to the point and the initial line. 

1 of 5

Converting between polar and Cartesian equations (

Cartesian - > Polar

Substitute rsinθ for any values of y: substitue rcosθ for any values of x. Then rearrange.

E.g. y² = 5x

(rsinθ)² = 5(rcosθ)

rsin²θ = 5cosθ

-> r = 5cosθ/sin²θ

      = 5cotθcosecθ


2 of 5

Converting between polar and Cartesian equations (

Polar - > Cartesian

This involves recognising terms from the first card, and/or rearranging to make substitutions easier

E.g. r = 10

r² = 100

- > x² + y² = 100    [Since r² = x² + y²]

So r = 10 is a circle centre O and radius 10

3 of 5

Sketching curves given polar equations

I always find this really tricky; look up diagrams on the internet to see how they come out. If you are fortunate enough to have a graphical calculator: use it.

Polar equations of the form r = a give a circle centre O, radius a

Polar equations of the form Θ = α give a half line throuhg O at an angle α to the initial line

Polar equations of the form r = aΘ give a spril starting at O

Other sketches are not so easily learned: try drawing up a table of values and carefuly plotting the shapes.

4 of 5

Fining tangents to a curve

For tangents that are parallel to the initial line:

dy/dΘ = 0

For tangents that are perpendicular to the initial line:

dx/dΘ = 0

5 of 5

Comments

Josh

Sorry, but theta is NOT always arctan(y/x).  For example, consider z = -1 -i.  Clearly through a plot of z, arg(z) = -0.75pi, however performing arctan (-1/-1) will give you the principle value of 0.25pi.  Instead of worrying about this, I find it easier to just take the |y/x|, arctan that, then by a quick sketch or in your head identify the quadrant and use simple angle geometry to get the right angle.  That way, you can never go wrong - good notes otherwise though :)

Similar Further Maths resources:

See all Further Maths resources »See all resources »