Polar Coordinates

A distance and an angle

Cartesian coordinates name a point by how far across and how far up it is. Polar coordinates name it by how far it is from a fixed point, and in which direction.

The fixed point is the pole, O, at the origin, and the directions are measured from the initial line, the positive x-axis. A point P has polar coordinates $\left(r,\theta \right)$(r, θ): r is the distance OP, and $\theta$θ is the angle turned counterclockwise from the initial line to OP, in radians.

For example, the point 2 from O in the direction $\frac{\pi }{3}$π/3, which is 60° above the initial line, has r = 2 and $\theta =\frac{\pi }{3}$θ = π/3.

The point r = 2, $\theta =\frac{\pi }{3}$θ = π/3. It lies on the dashed circle of radius 2 about the pole, where the arrow from the pole at $\frac{\pi }{3}$π/3 above the initial line meets it. Straight below it, at (1, 0), is the foot of the right triangle whose legs are its x and y.

From polar to Cartesian

Drop a perpendicular from P to the x-axis. That makes a right triangle with hypotenuse r and angle $\theta$θ at the pole. The horizontal leg is adjacent to $\theta$θ and the vertical leg is opposite, so $x=r\cos \theta$x = r cos θ and $y=r\sin \theta$y = r sin θ.

For r = 2 and $\theta =\frac{\pi }{3}$θ = π/3, x = 2 × ½ = 1 and $y=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$y = 2 × √3/2 = √3, so the point is $\left(1,\sqrt{3}\right)$(1, √3).

The same formulas work in every quadrant, because $\cos \theta$cos θ and $\sin \theta$sin θ carry the signs. For r = 4 and $\theta =\frac{5\pi }{6}$θ = 5π/6, $x=4\cos \frac{5\pi }{6}=-2\sqrt{3}\approx -3.46$x = 4 cos 5π/6 = −2√3 ≈ −3.46 and $y=4\sin \frac{5\pi }{6}=2$y = 4 sin 5π/6 = 2, a point up and to the left.

From Cartesian to polar

Going back, r is the hypotenuse, so Pythagoras gives $r=\sqrt{x^2+y^2}$r = √(x² + y²). The angle satisfies $\tan \theta =\frac{y}{x}$tan θ = y/x.

The tangent repeats every half turn, so two directions share each value of $\tan \theta$tan θ. A calculator’s inverse tangent only returns angles between $-\frac{\pi }{2}$−π/2 and $\frac{\pi }{2}$π/2, which point to the right of the y-axis. So find the quadrant first, then correct the angle.

Take (−3, 4). It is $r=\sqrt{9+16}=5$r = √(9 + 16) = 5 from the pole. The inverse tangent of $\frac{4}{-3}$4/(−3) gives −0.927, which points down and to the right, but (−3, 4) is up and to the left, in the second quadrant. The acute angle with tangent $\frac{4}{3}$4/3 is 0.927, so $\theta =\pi -0.927\approx 2.214$θ = π − 0.927 ≈ 2.214. Check: 5 cos 2.214 = −3 and 5 sin 2.214 = 4.

Taking $\theta$θ between 0 and $2\pi$2π, the other quadrants work the same way. (−3, −4), in the third quadrant, has $\theta =\pi +0.927\approx 4.069$θ = π + 0.927 ≈ 4.069, and (3, −4), in the fourth, has $\theta =2\pi -0.927\approx 5.356$θ = 2π − 0.927 ≈ 5.356. Some books take $\theta$θ between $-\pi$−π and $\pi$π instead; then (3, −4) has $\theta \approx -0.927$θ ≈ −0.927.

The arrow runs from the pole to (−3, 4), 5 long at the angle 2.214. The hollow point is (3, −4), where the inverse tangent’s angle −0.927 would put a point 5 from the pole: the same tangent, the opposite direction.

There and back

r = 2 at $\theta =\frac{\pi }{3}$θ = π/3 is the point $\left(2\cos \left(\frac{\pi }{3}\right),2\sin \left(\frac{\pi }{3}\right)\right)=\left(1,\sqrt{3}\right)$(2 cos(π/3), 2 sin(π/3)) = (1, √3). Going back, $r=\sqrt{1+3}=2$r = √(1 + 3) = 2, and $\tan \theta =\sqrt{3}$tan θ = √3. The point is in the first quadrant, so $\theta =\frac{\pi }{3}$θ = π/3, and the round trip returns the same pair.

A point has more than one polar name. Adding $2\pi$2π to $\theta$θ turns the direction a full turn and back to where it was, so $\left(2,\frac{\pi }{3}\right)$(2, π/3) and $\left(2,\frac{7\pi }{3}\right)$(2, 7π/3) are the same point. The pole itself has r = 0, and any $\theta$θ names it.

Equations in polar form

A curve’s equation converts the same way. The circle $x^2+y^2=9$x² + y² = 9 becomes $r^2=9$r² = 9, that is r = 3: every point is 3 from the pole.

The circle $x^2+y^2=6x$x² + y² = 6x passes through the pole. Substituting gives $r^2=6r\cos \theta$r² = 6r cos θ, and dividing by r gives $r=6\cos \theta$r = 6 cos θ. The distance from the pole now depends on the direction: 6 along the initial line, and 0 at $\theta =\frac{\pi }{2}$θ = π/2.

The usual mistakes

Swapping the legs. $x=r\cos \theta$x = r cos θ and $y=r\sin \theta$y = r sin θ: the cosine goes with the leg next to the angle.

Trusting the inverse tangent in every quadrant. It gives −0.927 for both (−3, 4) and (3, −4), so the quadrant has to be checked.

Leaving out the square root. $x^2+y^2$x² + y² is $r^2$r², not r: for (3, 4) it is 25, while r = 5.

Adding the legs. 3 + 4 = 7 is longer than the distance 5, because a straight line is the shortest way.

A dish antenna

In the application below, the cross-section of a dish antenna is given from its receiver as $r=\frac{1.2}{1-\cos \theta }$r = 1.2/(1 − cos θ). Multiplying out gives $r-r\cos \theta =1.2$r − r cos θ = 1.2. Replacing $r\cos \theta$r cos θ by x leaves r = x + 1.2, and squaring with $r^2=x^2+y^2$r² = x² + y² turns it into a Cartesian equation.