1. What polar coordinates are
A point in the plane is named by an ordered pair $(r, \theta)$ where $r$ is its distance from the origin (the pole) and $\theta$ is the angle that the ray from the origin to the point makes with the positive $x$-axis, measured counterclockwise.
The two pieces of information are the two you actually want when you're thinking radially. "How far am I from here, and which way am I facing?" — the polar pair answers exactly that question. The rectangular pair answers a different question: "How many steps east, and how many steps north?"
Both questions describe the same point. They're equally valid; they're just different vocabularies.
2. Converting points between polar and rectangular
Drop the point $P = (x, y) = (r\cos\theta, r\sin\theta)$ onto the plane. The right triangle whose hypotenuse is the segment from the origin to $P$ has horizontal leg $x$, vertical leg $y$, and hypotenuse $r$ at angle $\theta$ from the $x$-axis. Two pieces of basic trig fall out immediately.
Polar → rectangular
$$ x = r\cos\theta \qquad y = r\sin\theta $$These are just the definitions of sine and cosine on a circle of radius $r$. You already know them from the unit circle, scaled up.
Rectangular → polar
$$ r = \sqrt{x^2 + y^2} \qquad \theta = \arctan\!\frac{y}{x} \;\;(\text{with quadrant care}) $$The first is the Pythagorean theorem. The second is the trig identity solved for the angle — but with a catch we'll address in a moment.
Worked: polar to rectangular
Convert $(r, \theta) = (4, \pi/3)$:
$$ x = 4\cos(\pi/3) = 4 \cdot \tfrac{1}{2} = 2 \qquad y = 4\sin(\pi/3) = 4 \cdot \tfrac{\sqrt{3}}{2} = 2\sqrt{3} $$So $(4, \pi/3) \longleftrightarrow (2, 2\sqrt{3})$.
Worked: rectangular to polar — and the quadrant trap
Convert $(x, y) = (1, 1)$. Easy: $r = \sqrt{1+1} = \sqrt{2}$, and $\theta = \arctan(1) = \pi/4$. The point is in Quadrant I and $\arctan$ returns $\pi/4$. Everything matches.
Now try $(x, y) = (-1, -1)$. By the same formula: $r = \sqrt{2}$ and $\arctan(-1/-1) = \arctan(1) = \pi/4$. But $(-1, -1)$ is in Quadrant III, nowhere near $\pi/4$. The arithmetic of $-1/-1$ throws away the sign information, and $\arctan$ only ever returns angles in $(-\pi/2, \pi/2)$ — it has no way to know which quadrant you started in.
The fix: look at the original signs of $x$ and $y$ and correct $\theta$ by hand.
| Quadrant | Signs of $(x, y)$ | $\theta$ from $\arctan(y/x)$ |
|---|---|---|
| I | $(+, +)$ | Use directly |
| II | $(-, +)$ | Add $\pi$ |
| III | $(-, -)$ | Add $\pi$ (or subtract $\pi$) |
| IV | $(+, -)$ | Use directly (negative result) |
For $(-1, -1)$: Quadrant III, so $\theta = \pi/4 + \pi = 5\pi/4$ (equivalently $-3\pi/4$). Sanity check: $r\cos(5\pi/4) = \sqrt{2} \cdot (-\sqrt{2}/2) = -1$ ✓.
Every programming language has a function called atan2(y, x) that does the quadrant correction for you. It takes both $y$ and $x$ as separate arguments — so it can see the signs — and returns the right angle in $(-\pi, \pi]$. If you ever need to do this conversion in code, reach for atan2 and forget the table above.
3. Non-uniqueness of polar names
Here's a thing rectangular coordinates do but polar doesn't: uniquely name each point. Every $(x, y)$ pair names exactly one point and every point has exactly one $(x, y)$ name. Polar plays by looser rules.
Consider the point at distance $2$ along the angle $\pi/4$. All of these name the same point:
$$ (2,\, \pi/4) \quad=\quad (2,\, \pi/4 + 2\pi) \quad=\quad (2,\, 9\pi/4) \quad=\quad (-2,\, \pi/4 + \pi) \quad=\quad (-2,\, 5\pi/4) $$Three things are going on:
- Adding $2\pi$ to $\theta$ spins all the way around and lands on the same spot.
- Flipping the sign of $r$ and adding $\pi$ to $\theta$ also lands on the same spot — "go $|r|$ units in the opposite direction" is identical to "go $|r|$ units this way."
- The origin is $(0, \theta)$ for every $\theta$. There's no angle attached when $r = 0$.
The redundancy is annoying for bookkeeping but it's also useful. The "negative $r$" convention is what lets a single equation like $r = 1 + 2\cos\theta$ trace out a curve with an inner loop without piecewise definitions — values where $r$ goes negative simply draw on the opposite side of the origin, and the curve stays smooth.
4. Converting equations
Converting a single point is mechanical. Converting an equation — a whole curve, viewed through a different vocabulary — is where polar starts to pay off. The two directions use the same four substitutions, applied with judgment.
| Substitution | Use when |
|---|---|
| $x = r\cos\theta$ | Polar → rect: replace $r\cos\theta$ with $x$. Rect → polar: replace $x$. |
| $y = r\sin\theta$ | Same idea, vertical. |
| $r^2 = x^2 + y^2$ | The most useful trick — gets rid of square roots. |
| $\tan\theta = y/x$ | When the polar equation involves $\theta$ alone. |
Rectangular → polar: a circle
$x^2 + y^2 = 9$ is the rectangular form of a circle of radius $3$ at the origin. Substituting directly:
$$ r^2 = 9 \quad\Longrightarrow\quad r = 3 $$One symbol versus three. Polar form is a single equation: "distance from origin equals $3$."
Polar → rectangular: an off-center circle
The polar equation $r = 2\cos\theta$ doesn't look like a circle at first glance. The standard trick: multiply both sides by $r$ to get into shape for our substitutions.
$$ r = 2\cos\theta $$ $$ r^2 = 2r\cos\theta $$ $$ x^2 + y^2 = 2x $$ $$ x^2 - 2x + y^2 = 0 $$ $$ (x - 1)^2 + y^2 = 1 $$It's a circle of radius $1$ centered at $(1, 0)$ — it just happens to pass through the origin. More generally, $r = 2a\cos\theta$ is a circle of radius $|a|$ centered at $(a, 0)$. That's a fact worth memorizing; it shows up everywhere.
Whenever you see $r$ alone, multiplying both sides by $r$ converts it to $r^2$ and a term like $r\cos\theta$ becomes $x$ — which exposes the rectangular form. The cost: you might pick up the spurious solution $r = 0$ (the origin), but the origin almost always already satisfies the original anyway.
5. Why polar is natural for rotation
The whole point of having a second coordinate system isn't variety — it's that some problems are structurally radial. Forcing them into Cartesian form is fighting their geometry.
Examples where polar makes everything easier:
- Orbits. A planet's orbit is naturally described by its distance from the Sun ($r$) and the angle it has swept out ($\theta$). Kepler's first law states this orbit is an ellipse with the Sun at one focus, and the polar equation $r = p / (1 + e\cos\theta)$ states the same thing in two short symbols. The rectangular form takes a quarter of a page.
- Antenna and microphone patterns. "How sensitive is this microphone at angle $\theta$?" is a question about a function of $\theta$ alone. The cardioid pickup pattern $r = 1 + \cos\theta$ — most sensitive in front, dead behind — is one curve in polar; in rectangular it's a mess.
- Radar and sonar. The native data is range $r$ and bearing $\theta$. You only convert to $(x, y)$ when you want to do vector arithmetic — and you often don't bother.
- Anything radially symmetric. The temperature distribution around a hot wire, the gravitational field of a planet, the diffraction pattern from a circular aperture: all functions of $r$ alone, and they look trivial in polar.
The picture and the equation should agree on what's hard. If your picture is rotational, your equation should also be rotational.