1.3 Polar Coordinates

Big idea

Rectangular coordinates locate a point by two perpendicular distances, which is the right choice when the important structure in a problem runs along straight lines. A great deal of geometry does not. Circles, spirals, orbits, radiation patterns and anything with a centre are described by how far out you are and which way you are facing, and forcing those descriptions through $x$ and $y$ produces square roots and implicit equations for curves that are simple in the other language.

Polar coordinates use exactly those two quantities: a distance $r$ from a fixed origin, called the pole, and an angle $\theta$ measured from a fixed ray, called the polar axis. The unit circle, which needs two square roots to graph as a function, becomes the single statement $r = 1$. That compression is the reason to learn the system, and the trade is that some easy rectangular curves become awkward: the line $y = 3$ turns into $r = 3\csc\theta$.

The system has one genuine oddity. Rectangular coordinates name each point exactly once, and polar coordinates do not. Adding a full turn to the angle returns to the same point, and allowing $r$ to be negative, which means measuring backwards along the ray, adds infinitely many more names. The origin is the extreme case: $r = 0$ describes it for every angle. This is not a defect to be patched; it is a feature you will use when tracing curves.

The practical work of this section is twofold. You convert points and equations back and forth, which is mechanical once you have the four relations. And you learn to read a polar equation as a curve, which means building a table of angles, tracking whether $r$ is growing or shrinking, and recognizing the standard families - circles, lines, limacons and roses - on sight.

Decoder

A point has exactly one pair of rectangular coordinates but infinitely many polar pairs, since adding a full turn to the angle, or negating the radius and adding a half turn, names the same point.

The first half is why conversion one way is easy and the other way requires a decision. Given $(r,\theta)$, the point is determined and the two conversion formulas compute it. Given $(x,y)$, you must choose which of the infinitely many polar names to report, and the usual convention takes $r \ge 0$ with $\theta$ in one full turn.

The phrase negating the radius and adding a half turn is the part that surprises people. A negative $r$ does not mean an impossible distance; it means you face in the direction $\theta$ and then walk backwards. That lands you diametrically opposite, which is the same as facing $\theta + \pi$ and walking forwards. Curves like roses rely on this: half their petals are traced while $r$ is negative.

The classic mistake lives in the reverse conversion. The relation $\tan\theta = y/x$ has two solutions per turn, and an inverse tangent returns only the one between $-\pi/2$ and $\pi/2$. That is the right angle for points to the right of the $y$-axis and off by $\pi$ for points to the left. Always plot the point first, note its quadrant, and then confirm the angle you computed lands there.

Definitions and results

Polar coordinates. The pair $(r,\theta)$ names the point reached by turning through angle $\theta$ from the polar axis and moving a signed distance $r$ along that direction. Positive $\theta$ is counterclockwise.

Multiple names. For any integer $k$, the pairs $(r, \theta + 2k\pi)$ and $(-r, \theta + \pi + 2k\pi)$ all name the same point. The pole is $(0,\theta)$ for every $\theta$.

Polar to rectangular. Always valid, with no case analysis:

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

Rectangular to polar. From the same two relations,

$$ r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x} \;\; (x \neq 0) $$

Take $r = \sqrt{x^2+y^2}$ for the standard name and choose $\theta$ so the point lands in the correct quadrant.

Converting equations. Going from polar to rectangular, look for the combinations $r\cos\theta$, $r\sin\theta$ and $r^2$, creating them by multiplying both sides by $r$ when that helps. Going the other way, substitute $x = r\cos\theta$ and $y = r\sin\theta$ and simplify.

Circles. The equation $r = a$ is the circle of radius $|a|$ centred at the pole. The equations $r = 2a\cos\theta$ and $r = 2a\sin\theta$ are circles of radius $|a|$ through the pole, centred at $(a,0)$ and $(0,a)$ respectively.

Lines. The equation $\theta = \alpha$ is the full line through the pole at angle $\alpha$, since $r$ may be negative. The equations $r\cos\theta = a$ and $r\sin\theta = b$ are the vertical line $x = a$ and the horizontal line $y = b$.

Limacons and cardioids. Curves $r = a \pm b\cos\theta$ and $r = a \pm b\sin\theta$ are limacons. When $a = b$ the curve reaches the pole and has a heart-shaped dimple: a cardioid. When $a < b$ it has an inner loop, and when $a > b$ it does not reach the pole at all.

Roses. For $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$ with $n$ a positive integer, the graph is a rose with $n$ petals when $n$ is odd and $2n$ petals when $n$ is even, each petal of length $|a|$.

Symmetry tests. Replacing $\theta$ by $-\theta$ without changing the equation means symmetry about the polar axis; replacing $\theta$ by $\pi - \theta$ means symmetry about the vertical line through the pole; replacing $r$ by $-r$ means symmetry about the pole.

Worked examples

Converting a point in each direction

Convert $(r,\theta) = \left(4, \frac{2\pi}{3}\right)$ to rectangular form. Since $\cos\frac{2\pi}{3} = -\frac12$ and $\sin\frac{2\pi}{3} = \frac{\sqrt3}{2}$,

$$ x = 4\left(-\tfrac12\right) = -2, \qquad y = 4\left(\tfrac{\sqrt3}{2}\right) = 2\sqrt3 $$

Now go the other way with the point $(-3,3)$. The radius is $r = \sqrt{9+9} = 3\sqrt2$, and $\tan\theta = 3/(-3) = -1$. An inverse tangent reports $-\pi/4$, which points into the fourth quadrant, but the point is in the second, so add $\pi$ to get $\theta = 3\pi/4$ and the standard name $\left(3\sqrt2, \frac{3\pi}{4}\right)$.

Check by converting back: $3\sqrt2\cos\frac{3\pi}{4} = 3\sqrt2 \cdot \left(-\frac{\sqrt2}{2}\right) = -3$ and $3\sqrt2 \sin\frac{3\pi}{4} = 3$. Two other valid names for the same point are $\left(3\sqrt2, \frac{11\pi}{4}\right)$ and $\left(-3\sqrt2, -\frac{\pi}{4}\right)$.

Turning a polar equation into a familiar one

Identify the curve $r = 6\cos\theta$.

Nothing in this equation is one of the standard combinations yet, but multiplying both sides by $r$ creates two of them at once:

$$ r^2 = 6r\cos\theta \quad \Longrightarrow \quad x^2 + y^2 = 6x $$

Move everything to one side and complete the square in $x$:

$$ x^2 - 6x + y^2 = 0 \quad \Longrightarrow \quad (x-3)^2 + y^2 = 9 $$

This is the circle of radius $3$ centred at $(3,0)$, which passes through the origin. Check one point: at $\theta = 0$ the polar equation gives $r = 6$, the point $(6,0)$, the far side of the circle, and at $\theta = \pi/2$ it gives $r = 0$, the pole, which is on the circle as well. Multiplying by $r$ can introduce the pole as a spurious solution, so confirming it really belongs is part of the work.

Two lines, one in each direction

The equation $\theta = \frac{\pi}{4}$ describes every point whose direction is $45$ degrees, including those with negative $r$ pointing the opposite way, so it is the entire line $y = x$, not just a ray.

Going the other direction, convert the vertical line $x = 4$ by substituting $x = r\cos\theta$, which gives $r\cos\theta = 4$, that is $r = 4\sec\theta$. This is defined for every angle except $\pm\pi/2$, which are exactly the directions in which a vertical line is never met. The polar form is worse than the rectangular one here, and that is the honest general lesson: straight lines away from the pole are the case polar coordinates handle badly.

Sketching a cardioid and a rose

Graph $r = 2 + 2\cos\theta$.

Because cosine is even, replacing $\theta$ by $-\theta$ changes nothing, so the curve is symmetric about the polar axis and you only need to trace $0 \le \theta \le \pi$ and reflect. Build a short table:

$$ \theta = 0 \to r = 4, \quad \theta = \tfrac{\pi}{3} \to r = 3, \quad \theta = \tfrac{\pi}{2} \to r = 2, \quad \theta = \tfrac{2\pi}{3} \to r = 1, \quad \theta = \pi \to r = 0 $$

The radius shrinks steadily from $4$ to $0$ as the direction sweeps from east to west. Reflecting below the axis completes the heart shape, with the dimple at the pole where $r$ reaches zero. Since the constant and the coefficient are equal, this is a cardioid.

Now graph $r = 3\sin(2\theta)$. The multiplier inside is even, so this rose has $2n = 4$ petals of length $3$. The first is traced as $\theta$ runs from $0$ to $\pi/2$, with $r$ climbing to $3$ at $\theta = \pi/4$ and returning to $0$. On $\pi/2 \le \theta \le \pi$ the value of $r$ is negative, so that petal is drawn opposite the direction you are facing. Those negative-radius passes fill in two of the four petals.

Practice

First the coordinates themselves: convert points in both directions, choose a correct angle for the quadrant, and produce alternative names for the same point.

Then the curves. Convert equations between the two systems and identify the standard families from their equations.

Quiz

Five items on converting points and equations, multiple polar names, and recognizing polar graphs.