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In Cartesian coordinates, every point in the plane is described by (x, y). Polar coordinates provide an alternative description using a distance and an angle: (r, theta). This system is particularly natural for curves with rotational symmetry and is a major topic in AQA Further Mathematics.
A fixed point O is chosen as the pole (origin), and a fixed ray from O (usually the positive x-axis) is the initial line.
A point P is described by:
We write the coordinates as (r, theta).
| Property | Detail |
|---|---|
| r >= 0 | Distance is non-negative (though negative r is sometimes used) |
| theta | Usually measured in radians |
| Convention | Anticlockwise is positive |
Given (r, theta):
x = r cos theta
y = r sin theta
Given (x, y):
r = sqrt(x^2 + y^2)
theta = arctan(y/x) (with adjustment for the correct quadrant)
Warning: The formula theta = arctan(y/x) only gives the correct angle when x > 0. For points where x < 0, you must add pi. For points on the y-axis, theta = pi/2 or -pi/2. Always check the quadrant.
Convert the polar point (4, pi/3) to Cartesian coordinates.
Solution:
x = 4 cos(pi/3) = 4 * (1/2) = 2
y = 4 sin(pi/3) = 4 * (sqrt(3)/2) = 2 sqrt(3)
The Cartesian coordinates are (2, 2 sqrt(3)).
Convert the Cartesian point (-3, 3) to polar coordinates.
Solution:
r = sqrt((-3)^2 + 3^2) = sqrt(9 + 9) = sqrt(18) = 3 sqrt(2)
arctan(3/(-3)) = arctan(-1) = -pi/4
But (-3, 3) is in the second quadrant, so theta = pi - pi/4 = 3pi/4.
The polar coordinates are (3 sqrt(2), 3pi/4).
Convert the Cartesian equation x^2 + y^2 = 25 to polar form.
Solution:
Since x^2 + y^2 = r^2, we get r^2 = 25, so r = 5.
This is a circle of radius 5 centred at the origin.
Convert the Cartesian equation y = x to polar form.
Solution:
r sin theta = r cos theta
sin theta = cos theta (dividing by r, assuming r is not 0)
tan theta = 1
theta = pi/4
This is a straight line through the origin at 45 degrees.
Convert x^2 + y^2 = 4x to polar form.
Solution:
r^2 = 4 r cos theta
r = 4 cos theta (dividing by r)
This is a circle of diameter 4, centred at (2, 0).
A key feature of polar coordinates is that the same point can have infinitely many representations:
(r, theta) = (r, theta + 2n pi) for any integer n.
Also, (-r, theta) = (r, theta + pi) -- a negative r means you go in the opposite direction.
This is important when finding intersection points of polar curves, as curves may intersect at points where they have different theta values.
| Equation | Curve |
|---|---|
| r = a | Circle of radius a, centred at O |
| theta = alpha | Straight line through O at angle alpha |
| r = a cos theta | Circle of diameter a, centred at (a/2, 0) |
| r = a sin theta | Circle of diameter a, centred at (0, a/2) |
| r = a(1 + cos theta) | Cardioid |
| r = a(1 - cos theta) | Cardioid (reflected) |
| r = a theta | Archimedean spiral |
When r < 0 for a given theta, the point is plotted in the opposite direction -- that is, at angle theta + pi with distance |r|.
For example, if r = -2 when theta = 0, the point is actually at (2, pi), which is (-2, 0) in Cartesian coordinates.
This is particularly relevant when sketching curves like r = a cos(n theta) (rose curves), where r is negative for certain ranges of theta.
Convert r = 2 sec theta to Cartesian form.
Solution:
r = 2/cos theta
r cos theta = 2
But r cos theta = x, so x = 2.
This is the vertical line x = 2.
Express the polar equation r = 6 sin theta in Cartesian form.
Solution:
Multiply both sides by r:
r^2 = 6 r sin theta
x^2 + y^2 = 6y
x^2 + y^2 - 6y = 0
Complete the square in y:
x^2 + (y - 3)^2 = 9
This is a circle with centre (0, 3) and radius 3.
| Conversion | Formula |
|---|---|
| Polar to Cartesian | x = r cos theta, y = r sin theta |
| Cartesian to Polar | r = sqrt(x^2 + y^2), theta = arctan(y/x) (check quadrant) |
| x^2 + y^2 = r^2 | Fundamental identity |
AQA Exam Tip: Conversion questions are common. Always show your working and check the quadrant when finding theta. For converting equations, use the identities x = r cos theta, y = r sin theta, and x^2 + y^2 = r^2 as building blocks.