Polar Coordinates and Graphs — AP® Calculus BC Notes

Rectangular coordinates say "go $x$ units right and $y$ units up". Polar coordinates say "turn to an angle, then go a distance". The 2025 exam's polar question had the lowest average of any question that year, and the readers' advice was simple: review the precalculus first. This page is that review.

Naming a point with $(r, \theta)$

Start at the pole (the origin). The angle $\theta$ is measured from the positive $x$ -axis, counterclockwise, in radians. Turn to the ray at angle $\theta$ , then go $r$ units along it. That point is $(r, \theta)$ .

A polar grid with circles of radius 1 to 4 and rays every π/6. Point P has r = 3 and θ = π/3: its legs are x = r cos θ = 1.5 and y = r sin θ = 3√3/2, about 2.598. Point Q has r = −2 and θ = π/4: it lies 2 units from the pole on the ray opposite θ = π/4, at x = −√2, y = −√2. A side panel lists x = r cos θ, y = r sin θ and r² = x² + y². — **Figure 1.** $P = (3, \frac{\pi}{3})$ and $Q = (-2, \frac{\pi}{4})$ , with the legs that give $x$ and $y$ . LevlPrep original diagram.

Two facts make polar different from rectangular:

$r$ can be negative. Then you go $|r|$ units backward, through the pole, onto the opposite ray. The figure shows $Q = (-2, \frac{\pi}{4})$ , which sits in the third quadrant.
A point has many names. $(2, \frac{\pi}{4})$ , $(2, \frac{9\pi}{4})$ and $(-2, \frac{5\pi}{4})$ are the same point. Adding $2\pi$ to $\theta$ changes nothing; adding $\pi$ and flipping the sign of $r$ changes nothing.

Converting between polar and rectangular

The segment from the pole to $(r, \theta)$ is the hypotenuse of a right triangle with legs $x$ and $y$ :

$x = r\cos\theta, \qquad y = r\sin\theta, \qquad r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x}.$

Worked example 1. Convert $(4, \frac{5\pi}{6})$ to rectangular, then convert $(0, -5)$ to polar.

$x$ $= 4\cos\frac{5\pi}{6}$ $= 4 \cdot \left(-\frac{\sqrt{3}}{2}\right)$ $= -2\sqrt{3}$ and $y$ $= 4\sin\frac{5\pi}{6}$ $= 4 \cdot \frac{1}{2}$ $= 2$ . The point is $(-2\sqrt{3}, 2)$ .

For $(0, -5)$ : $r = \sqrt{0 + 25} = 5$ . The point is straight down, on the ray $\theta = \frac{3\pi}{2}$ . So $(5, \frac{3\pi}{2})$ , or equally $(-5, \frac{\pi}{2})$ .

Do not trust $\tan\theta = \frac{y}{x}$ alone: $(1, -1)$ and $(-1, 1)$ have the same tangent. Always check the quadrant.

The same formulas convert equations. To rewrite $r = 6\cos\theta$ , multiply by $r$ : $r^2 = 6r\cos\theta$ , so $x^2 + y^2 = 6x$ , which is $(x - 3)^2 + y^2 = 9$ : a circle of radius 3 through the pole.

Plotting a polar curve from a table

A polar curve $r = f(\theta)$ gives one distance for each angle. To sketch one, make a table, then sweep around like a lighthouse beam, marking each distance on its ray.

Worked example 2. Sketch $r = 1 + 2\cos\theta$ for $0 \le \theta \le 2\pi$ .

$\theta$	$0$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\frac{2\pi}{3}$	$\pi$	$\frac{4\pi}{3}$	$\frac{3\pi}{2}$	$2\pi$
$r$	$3$	$2$	$1$	$0$	$-1$	$0$	$1$	$3$

From $\theta = 0$ to $\frac{2\pi}{3}$ the distance shrinks from 3 to 0: a big arc into the pole. From $\frac{2\pi}{3}$ to $\frac{4\pi}{3}$ , $r$ is negative, so the rays point left but the points land on the right: a small inner loop. From $\frac{4\pi}{3}$ to $2\pi$ , $r$ grows back to 3 below the axis.

Two readings matter most for calculus: where $r = 0$ (the curve passes through the pole) and where $r$ changes sign (the curve switches to the opposite ray).

Shapes to recognize

Three polar graphs drawn from their equations. Left: the cardioid r = 2 + 2 cos θ, a heart shape reaching 4 units right of the pole and touching the pole once, at θ = π. Middle: the limaçon r = 1 + 2 cos θ, a large loop with a small inner loop, in vermillion, traced while r is negative for 2π/3 < θ < 4π/3. Right: the rose r = 3 sin(2θ), four petals of length 3, one for every π/2 of θ. — **Figure 2.** A cardioid, a limaçon with an inner loop, and a four-petal rose. LevlPrep original diagram.

Equation	Shape	What to read off
$r = a$	circle centered at the pole	radius $\|a\|$
$\theta = c$	line through the pole	slope $\tan c$
$r = a\cos\theta$ , $r = a\sin\theta$	circle through the pole	diameter $\|a\|$ , on the $x$ -axis or $y$ -axis
$r = a + a\cos\theta$	cardioid	touches the pole once
$r = a + b\cos\theta$ , $\|a\| < \|b\|$	limaçon with an inner loop	loop where $r < 0$
$r = a\sin(n\theta)$ , $r = a\cos(n\theta)$	rose	$n$ petals if $n$ is odd, $2n$ if $n$ is even; length $\|a\|$

A rose $r = 3\sin(2\theta)$ has four petals (the figure): one for each stretch of $\theta$ of length $\frac{\pi}{2}$ between zeros of $r$ . On a graphing calculator, use polar mode and radians, and set the $\theta$ -window to $0 \le \theta \le 2\pi$ to see the whole curve.

Why calculus cares

Writing $x = f(\theta)\cos\theta$ and $y = f(\theta)\sin\theta$ turns a polar curve into a parametric curve with parameter $\theta$ . Everything from Topics 9.1 to 9.3 then applies: slopes in 9.7, and areas built from thin sectors in 9.8 and 9.9.