MATHEMATICS / COORDINATES

Polar coordinates: turn, then travel

A lighthouse beam, a radar sweep, or a planet held near the Sun by gravity is described more naturally by a direction and a distance from a centre than by how far right and how far up it is. To describe a point that way, we turn through an angle and then move out a distance. The unit circle page uses the same turning idea for points on a circle, reading off their $x$ - and $y$ -coordinates. Here we use it for the whole plane.

Changing coordinates does not move the point. We are describing the same position: $(x,y)$ records horizontal and vertical distances, while the new description records a direction and a distance from the origin. Put the two descriptions of one point beside each other so that each number has somewhere to point. Once that is familiar, we can use the angle as an input and let a rule choose the distance, leaving a whole curve behind the moving point. If radians, sine or cosine feel unfamiliar, the circle below rebuilds them; for now, keep track of the turn and the distance outwards.

01Two ways to give a point an address

In Cartesian coordinates, the pair $(x,y)$ locates a point: $x$ gives its horizontal position and $y$ gives its vertical position. Together they determine one point. Polar coordinates describe the same plane using two different questions. Standing at the origin, how far away is the point, and in which direction?

Choose a positive distance $r$ . All points at that distance from the origin form a circle of radius $r$ . Knowing $r$ therefore leaves a whole circle of possible points. To choose one, we also give an angle $\theta$ , measured anticlockwise from the positive $x$ -axis. That direction meets the circle at one point, which is why we write a polar coordinate pair as $(r,\theta)$ .

Why does a full turn have another number?

A degree divides one full turn into $360$ equal parts. We can also measure an angle using lengths on a circle. Start at the right-hand point, turn anticlockwise, and follow the gold arc. The radius is the distance from the centre to the circle. Let $R$ be that radius and $s$ the length travelled along the arc. Both lengths use the same unit.

Arc length alone cannot tell us the angle: the same turn on a larger circle travels farther. But if we enlarge the circle by a factor of $2$ , both the radius and this arc length double, leaving $(2s)/(2R)=s/R$ . The ratio removes the circle size. We define the angle in radians to be $\theta=s/R$ . One radian therefore means that the arc length equals the radius.

A full turn follows the circumference: the distance all the way around the circle. The number $\pi$ is the ratio of a circle circumference to its diameter. The diameter is $2R$ , so the circumference is $2\pi R$ . A full turn follows that entire circumference, giving $\theta=(2\pi R)/R=2\pi$ radians. Half a turn is $\pi$ , and a quarter is $\pi/2$ .

Circle radius

R=2

Angle in radians

Keep the angle fixed and change the radius: the gold arc and green radius grow together

\theta=1,\qquad s=R\theta=2\times(1)\approx2.000

The radius is

2

and the arc length is also

2

, so

s/R=2/2=1

radian. Changing the circle size preserves that ratio.

For an angle of $d$ degrees, the fraction of a full turn is $d/360$ . Multiply that fraction by the full turn in radians: $\theta=(d/360)\times2\pi=d\pi/180$ . For example, $90^\circ$ gives $90\pi/180=\pi/2$ . This is why the same direction can carry a degree label or a radian label.

Cartesian coordinates:

x

says how far across,

y

says how far up. These two directions together pin down exactly one point.

Polar coordinates: a positive distance

r

leaves a circle of possible points. An angle

\theta

selects one of them. If

r

is zero, we are already at the origin, whatever angle we choose.

Choose $r$ , then $\theta$

(r,\theta)=(2.5,\ 40^\circ)

Distance r2.5

Angle θ40° = 0.70 rad

Change $r$ first and watch the dashed circle: every point on it is $r$ away from the origin, so $r$ alone cannot tell them apart. Then turn $\theta$ , measured anticlockwise from the positive $x$ -axis, and the point moves around that same circle until it stops at one place. You can also drag on the dial. At $r=0$ the circle has shrunk to one point; changing $\theta$ keeps that point at the origin.

How do the two addresses describe the same point?

We can locate a point by turning and travelling, but later formulas will also use its horizontal and vertical coordinates. We need to see where those two numbers come from. The reference circle below has radius $1$ ; a circle of that radius is called a unit circle. At the chosen angle, call its horizontal coordinate cosine, written $\cos\theta$ , and its vertical coordinate sine, written $\sin\theta$ .

A horizontal coordinate tells us how far right or left the point lies: right is positive, left negative. A vertical coordinate records up as positive and down as negative. On the unit circle these definitions work all the way round, including directions where there is no nonzero right triangle to draw.

Now keep the angle and enlarge or shrink the unit circle to radius $r$ . Every horizontal and vertical displacement is multiplied by the same factor $r$ . So the new horizontal coordinate is $x=r\cos\theta$ , and the new vertical coordinate is $y=r\sin\theta$ . The controls below reveal that scaling before drawing the two displacements.

Target distance after scaling

r=2

Direction

90^\circ

Dashed circle: radius 1 reference. Green: the current radius. Blue: horizontal; orange: vertical. The gold point is the current point.

Stage 1 of 3

Start on a circle of radius

1

. The point keeps the chosen angle; its horizontal coordinate defines

\cos\theta

, and its vertical coordinate defines

\sin\theta

Circle	Horizontal coordinate	Vertical coordinate
Unit circle	$\cos\theta\approx0$	$\sin\theta\approx1.000$

The coordinate readouts are rounded to three decimal places; the equations describe the exact scaling.

At $90^\circ$ , the unit-circle point is at the top: it is $0$ units sideways and $1$ unit up. Thus $\cos90^\circ=0$ and $\sin90^\circ=1$ by the coordinate definitions. Scaling to $r=2$ gives $x=2\times0=0$ and $y=2\times1=2$ . At $0^\circ$ , the point is on the right instead, so the same reasoning gives $x=2$ and $y=0$ .

At the origin, $r=0$ makes $x=0\cos\theta=0$ and $y=0\sin\theta=0$ for every angle. Thus the origin has no single direction attached to it. Try keeping the point at the origin and changing the angle: the unit-circle reference moves, but the actual point stays put. Away from the origin, restricting the distance to be positive and choosing an angle from $0$ inclusive to $2\pi$ exclusive gives one address; allowing extra full turns gives additional addresses for the same point.

02One point has many polar addresses

There is something worth noticing before we draw anything. Turning is different from walking across: if we keep turning, we eventually come back to the direction we started from. What does that mean for addresses?

Turning once more lands on the same point

(2.5,\ \tfrac{\pi}{6})

Press “Add a full turn” and watch the arm go all the way round. A full turn is $2\pi$ , which is $360^\circ$ , and after it we are indeed facing exactly the same way as before, so the point finishes where it started; it moves during the turn, then returns. Therefore one point has infinitely many polar addresses, namely $(r,\theta+2k\pi)$ for every integer $k$ :

(2.5,\ -\tfrac{11\pi}{6})=(2.5,\ \tfrac{\pi}{6})=(2.5,\ \tfrac{13\pi}{6})=(2.5,\ \tfrac{25\pi}{6})=\cdots

This is the main difference from Cartesian coordinates, where $(x,y)$ is unique. There is one more convention you will meet in curves: a negative $r$ means “face the direction $\theta$ , then walk backwards”. Press “Use $-r$ ” to see it: the point ends up on the opposite side, so $(-r,\theta)$ is the same point as $(r,\theta+\pi)$ .

03Plotting a polar rule point by point

For $y=2x+1$ , an input $x$ gives an output $y$ , so we can plot the point $(x,y)$ . Repeating this gives samples of the graph; the rule also determines the points between them. We can sample a polar curve in a similar way. The input is now the angle $\theta$ , and the output is $r$ . The rule $r=\theta$ tells us how far from the origin to place a point in each direction. Choose nine angles from 0 to $2\pi$ , calculate each $r$ , and plot the pairs $(r,\theta)$ .

r=\theta

$\\theta$	$r$	point
$0$	0	…
$\tfrac{\pi}{4}$	0.79	…
$\tfrac{\pi}{2}$	1.57	…
$\tfrac{3\pi}{4}$	2.36	…
$\pi$	3.14	…
$\tfrac{5\pi}{4}$	3.93	…
$\tfrac{3\pi}{2}$	4.71	…
$\tfrac{7\pi}{4}$	5.50	…
$2\pi$	6.28	…

Progress0%

For each row, the arm turns to the angle $\theta$ . The pen moves along that direction by $r=r(\theta)$ , leaving a point where it stops.

04Why does $r=2$ give a circle?

Now let’s use the same method on the rule $r=2$ . This time $r$ does not depend on $\theta$ at all: whichever angle we choose, the point is 2 away from the origin. Play it and compare with $r=\theta$ .

r=2

$\\theta$	$r$	point
$0$	2.00	…
$\tfrac{\pi}{4}$	2.00	…
$\tfrac{\pi}{2}$	2.00	…
$\tfrac{3\pi}{4}$	2.00	…
$\pi$	2.00	…
$\tfrac{5\pi}{4}$	2.00	…
$\tfrac{3\pi}{2}$	2.00	…
$\tfrac{7\pi}{4}$	2.00	…
$2\pi$	2.00	…

Progress0%

For each row, the arm turns to the angle $\theta$ . The pen moves along that direction by $r=r(\theta)$ , leaving a point where it stops.

In the first quadrant, the perpendicular distances to the $x$ - and $y$ -axes are $k\sin\theta$ and $k\cos\theta$ . In other quadrants, distances remain nonnegative, so they are $|k\sin\theta|$ and $|k\cos\theta|$ . The signed coordinates are still $y=k\sin\theta$ and $x=k\cos\theta$ : down gives a negative $y$ and left gives a negative $x$ . The Pythagorean calculation uses their squares, so it works in every quadrant.

The two perpendiculars and the line from the origin form a right triangle, so Pythagoras’ theorem gives the distance from the origin. Inside the square root we can take out $k^2$ as a common factor, and then we use $\sin^2\theta+\cos^2\theta=1$ . Recall where that comes from: on the unit circle, the point at angle $\theta$ is $(\cos\theta,\sin\theta)$ and it is 1 away from the origin, so by Pythagoras $\cos^2\theta+\sin^2\theta=1^2=1$ . Finally $\sqrt{k^2}=k$ because $k$ is a positive distance. Drag $\theta$ and see that the readout stays equal to $k$ .

Every point is $k$ away from the origin

Distance k2.0

Angle θ50°

signed y-coordinate $k\sin\theta \approx 1.53$ signed x-coordinate $k\cos\theta \approx 1.29$ distance $\sqrt{x^2+y^2}=k=2.00$

\begin{aligned} r &= \sqrt{(k\sin\theta)^2 + (k\cos\theta)^2} \\ &= \sqrt{k^2\left(\sin^2\theta + \cos^2\theta\right)} \\ &= \sqrt{k^2\cdot 1} = k \end{aligned}

Whatever $\theta$ we choose, the distance comes out as $k$ . Therefore every point with $r=k$ lies on the circle of radius $k$ , and every angle gives one of its points, which leads to the whole circle. In other words, $r=k$ is a circle of radius $k$ centred at the origin.

05Other curves from the same idea

Now that we know how a rule $r(\theta)$ becomes a curve, let’s look at some well-known ones. Choose a family and press “Trace the curve”: the orange arm turns steadily through $\theta$ while the rule sets $r$ , and the shaded slice shows the part just swept. Notice how differently the pen behaves, even though the method is always the same.

r=2.6\cos(2\theta)

Size a2.6

$r=a\cos(k\theta)$ swings between $a$ and $-a$ as $\theta$ turns. Each time $\cos(k\theta)$ passes through zero the pen returns to the origin, forming the petals. Negative $r$ puts the pen on the opposite side. For odd $k$ , the $k$ petals are traced once from $0$ to $\pi$ ; for even $k$ , the $2k$ petals need $0$ to $2\pi$ . The current value of $k$ is 2 gives 4 petals.

Why are there

k

2k

petals?

A circle keeps the pen at one distance. Here we let a rule change that distance while the angle turns: r=a cos(kθ), with a>0 and k a positive integer. Choose Rose and use Trace the curve. Between two successive zeros of the cosine, the pen leaves the origin, reaches distance a and returns. That outward-and-back part makes one petal. A negative radius draws it on the opposite side of the chosen direction.

Let us count distinct petals rather than just counting repeated trips. Positive maxima occur when kθ=2jπ. Negative maxima occur when kθ=(2j+1)π, but the pen then points an extra π away. Here j runs through 0,1,…,k−1. The two lists of petal directions are therefore:

\begin{aligned}\phi_+&=\frac{2j\pi}{k}\\\phi_-&=\frac{(2j+1)\pi}{k}+\pi\\&=\frac{(2j+k+1)\pi}{k}\end{aligned}

When k is odd, k+1 is even, so every negative-radius direction is already in the positive list, allowing full turns to identify the same direction. There are k distinct petals. When k is even, 2j+k+1 is odd, so none of these directions equals an even-numbered direction in the positive list. We then have k+k=2k petals. Compare k=3 and k=4 in the model.

This also explains the drawing range. Adding π to θ reverses the direction. For odd k it reverses the sign of cos(kθ) too, so both reversals return the pen to the same point; [0,π] traces the complete curve. For even k the radius keeps its sign, so the second half-turn visits the opposite petals and [0,2π] is needed.

Model scope: ideal plane geometry with a fixed unit length. Angles are in radians in every formula; the dial marks the quarter turns. Rose curves use whole numbers $k$ .

OpenStax · Polar coordinates OpenStax · Polar area