Math Core

Rectangular coordinates tell you how far to walk east and then north. Sometimes it is more natural to say "face this direction and walk this far." A radar screen, a lighthouse beam, and a sprinkler all work that way. Polar coordinates describe a point by its distance from a center and the angle you turn to face it.

Locating a point with distance and angle

Start with a fixed point $O$ called the pole (it sits where the origin usually is) and a ray from $O$ pointing right called the polar axis (the positive $x$ -axis).

Definition

Polar coordinates

The point $P$ with polar coordinates $(r, \theta)$ is found by rotating the polar axis through the angle $\theta$ and then moving a directed distance $r$ along that ray.

$\theta$ is measured from the polar axis, counterclockwise for positive angles and clockwise for negative angles. Radians are standard.
$r$ is the directed distance from the pole. If $r$ is negative, you move $|r|$ units in the opposite direction.

To plot $(3, \tfrac{\pi}{4})$ , turn $\tfrac{\pi}{4}$ (a $45^\circ$ turn) and walk out 3 units. To plot $(2, \tfrac{5\pi}{6})$ , turn $150^\circ$ and walk out 2 units. The graph below shows both, along with the circles $r = 2$ and $r = 3$ that help you gauge distance.

Circles of radius 2 and 3 act like the grid lines of the polar system.Open in grapher →

On polar graph paper, the "grid lines" are circles centered at the pole (constant $r$ ) and rays leaving the pole (constant $\theta$ ).

One point, many names

In rectangular coordinates every point has exactly one address. In polar coordinates every point has infinitely many.

Adding a full turn does not move you: $(r, \theta)$ and $(r, \theta + 2\pi k)$ name the same point for any integer $k$ .
Turning halfway around and walking backward lands you in the same place: $(r, \theta)$ and $(-r, \theta + \pi)$ name the same point.
The pole is $(0, \theta)$ for every angle $\theta$ .

For example, $(4, \tfrac{\pi}{3})$ , $(4, \tfrac{7\pi}{3})$ , $(4, -\tfrac{5\pi}{3})$ and $(-4, \tfrac{4\pi}{3})$ are all the same point.

Common mistake

A negative $r$ does not mean a negative angle. $(-4, \tfrac{\pi}{3})$ means "face $\tfrac{\pi}{3}$ , then back up 4 units," which puts the point in Quadrant III, not Quadrant I or IV. When in doubt, rewrite it with a positive $r$ by adding $\pi$ to the angle: $(-4, \tfrac{\pi}{3}) = (4, \tfrac{4\pi}{3})$ .

Converting polar to rectangular

Drop a perpendicular from $P$ to the $x$ -axis. You get a right triangle with hypotenuse $r$ and angle $\theta$ , so the legs are $r\cos\theta$ and $r\sin\theta$ . These formulas also work when $r$ is negative or $\theta$ is outside the first quadrant, because the signs of cosine and sine take care of the direction.

Conversion formulas

x = r\cos\theta, \qquad y = r\sin\theta

r^2 = x^2 + y^2, \qquad \tan\theta = \dfrac{y}{x} \ (x \ne 0)

Worked example: Polar to rectangular

Convert (a) $(6, \tfrac{2\pi}{3})$ and (b) $(-2, \tfrac{\pi}{6})$ to rectangular coordinates.

Solution.

(a) $x = 6\cos\tfrac{2\pi}{3} = 6\left(-\tfrac{1}{2}\right) = -3$ and $y = 6\sin\tfrac{2\pi}{3} = 6 \cdot \tfrac{\sqrt{3}}{2} = 3\sqrt{3}$ . The point is $(-3, 3\sqrt{3})$ .

(b) $x = -2\cos\tfrac{\pi}{6} = -2 \cdot \tfrac{\sqrt{3}}{2} = -\sqrt{3}$ and $y = -2\sin\tfrac{\pi}{6} = -2 \cdot \tfrac{1}{2} = -1$ . The point is $(-\sqrt{3}, -1)$ , in Quadrant III, just as the negative $r$ predicted.

Converting rectangular to polar

Going the other way takes two steps. Finding $r$ is easy: $r = \sqrt{x^2 + y^2}$ if you want $r \ge 0$ . Finding $\theta$ needs care, because $\tan\theta = \tfrac{y}{x}$ has two solutions in $[0, 2\pi)$ , one in each of two opposite quadrants. The inverse tangent only returns angles between $-\tfrac{\pi}{2}$ and $\tfrac{\pi}{2}$ (Quadrants I and IV).

A reliable method:

Plot the point (roughly) and note its quadrant.
Find the reference angle $\alpha = \tan^{-1}\left|\tfrac{y}{x}\right|$ .
Place the angle in the correct quadrant: $\theta = \alpha$ (QI), $\pi - \alpha$ (QII), $\pi + \alpha$ (QIII), or $2\pi - \alpha$ (QIV).

Points on an axis are quick: $(0, 5)$ is $(5, \tfrac{\pi}{2})$ and $(-7, 0)$ is $(7, \pi)$ .

Worked example: Rectangular to polar

Write $(-2\sqrt{3}, -2)$ in polar form with $r > 0$ and $0 \le \theta < 2\pi$ .

Solution. $r = \sqrt{(-2\sqrt{3})^2 + (-2)^2} = \sqrt{12 + 4} = 4$ .

Both coordinates are negative, so the point is in Quadrant III. The reference angle satisfies $\tan\alpha = \dfrac{2}{2\sqrt{3}} = \dfrac{1}{\sqrt{3}}$ , so $\alpha = \tfrac{\pi}{6}$ . In Quadrant III, $\theta = \pi + \tfrac{\pi}{6} = \tfrac{7\pi}{6}$ .

The point is $(4, \tfrac{7\pi}{6})$ . Check: $4\cos\tfrac{7\pi}{6} = -2\sqrt{3}$ and $4\sin\tfrac{7\pi}{6} = -2$ .

Tip

Always check your polar answer by converting back with $x = r\cos\theta$ , $y = r\sin\theta$ . If a calculator gave you $\tan^{-1}\tfrac{y}{x}$ directly and the point is in Quadrant II or III, you need to add $\pi$ .

Distance between polar points

Two points $(r_1, \theta_1)$ and $(r_2, \theta_2)$ form a triangle with the pole. The sides from the pole have lengths $r_1$ and $r_2$ , and the angle between them is $\theta_2 - \theta_1$ . The Law of Cosines gives the third side:

d^2 = r_1^2 + r_2^2 - 2r_1 r_2 \cos(\theta_2 - \theta_1).

Worked example: Distance without converting

Find the distance between $(3, \tfrac{\pi}{6})$ and $(5, \tfrac{5\pi}{6})$ .

Solution. The angle between the rays is $\tfrac{5\pi}{6} - \tfrac{\pi}{6} = \tfrac{2\pi}{3}$ , and $\cos\tfrac{2\pi}{3} = -\tfrac{1}{2}$ .

d^2 = 9 + 25 - 2(3)(5)\left(-\tfrac{1}{2}\right) = 34 + 15 = 49,

so $d = 7$ .

Practice

Practice 1

Convert the polar point $(4, \tfrac{\pi}{3})$ to rectangular coordinates. Enter it as $(x, y)$ .

Enter a point like (2, -3)

Practice 2

Convert the polar point $(-2, \tfrac{\pi}{4})$ to rectangular coordinates.

Enter a point like (2, -3)

Practice 3

Which polar point is not the same point as $(3, \tfrac{\pi}{6})$ ?

Practice 4

Write the rectangular point $(-3, 3)$ in polar form $(r, \theta)$ with $r > 0$ and $0 \le \theta < 2\pi$ .

Enter a point like (2, -3)

Practice 5

Write the rectangular point $(1, -\sqrt{3})$ in polar form with $r > 0$ and $0 \le \theta < 2\pi$ .

Enter a point like (2, -3)

Practice 6

Find the exact distance between the polar points $(2, \tfrac{\pi}{6})$ and $(4, \tfrac{\pi}{2})$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

The rectangular point $(-5, -12)$ has polar coordinates $(13, \theta)$ with $0 \le \theta < 2\pi$ . Find $\theta$ in radians, rounded to the nearest hundredth.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.