What does a negative r mean in polar coordinates?

Turn to the angle, then move that distance in the opposite direction. A point with r = −2 at some angle is the same as the point with r = 2 at that angle plus pi.

How do I tell if a polar curve is moving away from the origin?

Check whether r is moving away from 0. That happens when r is positive and increasing, or negative and decreasing.

Where does a polar curve pass through the pole?

Wherever r = 0. On the graph of r against theta, those are the places the graph crosses or touches the horizontal axis.

Why graph r against theta on rectangular axes?

It shows where r is positive, negative, increasing, and decreasing at a glance. Those four facts answer most polar questions.

Polar functions for AP Precalculus: coordinates, graphs, and distance

Updated November 10, 2026

Polar coordinates name a point by an angle and a distance instead of by $x$ and $y$ . On AP Precalculus they show up in three ways: converting between polar and rectangular, reading a polar graph, and deciding when a point on a polar curve is moving toward or away from the origin.

This guide covers all three, with worked examples. The key tool is the ordinary graph of $r$ against $\theta$ , which shows everything you need about the polar curve.

Polar coordinates and converting points

The polar point $(r, \theta)$ means: turn through the angle $\theta$ , measured counterclockwise from the positive $x$ -axis, then move a directed distance $r$ . If $r$ is negative, move $|r|$ units in the opposite direction. The origin is called the pole.

Polar to rectangular: $x = r\cos\theta$ and $y = r\sin\theta$ .
Rectangular to polar: $r^2 = x^2 + y^2$ and $\tan\theta = \frac{y}{x}$ .
Many names for one point: adding $2\pi$ to $\theta$ gives the same point, and so does changing the sign of $r$ while adding $\pi$ to $\theta$ .

One point each way

(a) Convert the polar point

\left(-4, \frac{\pi}{3}\right)

to rectangular coordinates. (b) Find polar coordinates, with

r > 0

and an angle from

0

up to but not including

2\pi

, for the rectangular point

(5\sqrt{3}, -5)

(a) $x = -4\cos\frac{\pi}{3} = -4 \cdot \frac{1}{2} = -2$ and $y = -4\sin\frac{\pi}{3} = -4 \cdot \frac{\sqrt{3}}{2} = -2\sqrt{3}$ . The angle points into Quadrant I, but $r$ is negative, so the point lands in Quadrant III.

(b) $r = \sqrt{75 + 25} = 10$ . Next, $\tan\theta = \frac{-5}{5\sqrt{3}} = -\frac{1}{\sqrt{3}}$ . Tangent is negative in Quadrants II and IV, and the point has $x > 0$ and $y < 0$ , so it is in Quadrant IV. That gives $\theta = \frac{11\pi}{6}$ .

Answer: (a) $(-2, -2\sqrt{3})$ . (b) $\left(10, \frac{11\pi}{6}\right)$

Converting equations

To turn a polar equation into a rectangular one, aim for the pieces $r\cos\theta$ , $r\sin\theta$ , and $r^2$ , then replace them with $x$ , $y$ , and $x^2 + y^2$ . Multiplying both sides by $r$ is often the first step. To go the other way, replace $x$ and $y$ and solve for $r$ .

$r = 5$ is $x^2 + y^2 = 25$ , a circle of radius $5$ centered at the pole.
$\theta = \frac{\pi}{4}$ is the line $y = x$ , since every point on it has $\frac{y}{x} = 1$ , and negative $r$ fills in the other half.

A circle and a line

(a) Rewrite

r = -10\cos\theta

in rectangular form and describe its graph. (b) Rewrite the line

x + y = 4

as a polar equation solved for

r

(a) Multiply both sides by $r$ : $r^2 = -10r\cos\theta$ . Replace the pieces: $x^2 + y^2 = -10x$ . Move $-10x$ to the left and complete the square: $x^2 + 10x + 25 + y^2 = 25$ , so $(x + 5)^2 + y^2 = 25$ .

(b) Replace $x$ and $y$ : $r\cos\theta + r\sin\theta = 4$ . Factor out $r$ : $r(\cos\theta + \sin\theta) = 4$ . Divide to solve for $r$ .

Answer: (a) $(x + 5)^2 + y^2 = 25$ , a circle with center $(-5, 0)$ and radius $5$ that passes through the pole. (b) $r = \frac{4}{\cos\theta + \sin\theta}$

Sketching a polar graph from r versus theta

Graph $r = f(\theta)$ on ordinary axes, with $\theta$ across and $r$ up. That graph shows directly where $r$ is positive, negative, zero, largest, and smallest. Then turn each piece into a piece of the polar curve.

r = 1 + 2\sin\theta

The graph of r against theta for r = 1 + 2 sin theta. The horizontal axis is theta in radians.

Read the graph one stretch at a time, and plot where the polar point is at each key angle.

$0$ to $\frac{\pi}{2}$ : $r$ grows from $1$ to $3$ . The point sweeps through Quadrant I from $(1, 0)$ out to $(0, 3)$ .
$\frac{\pi}{2}$ to $\pi$ : $r$ shrinks from $3$ to $1$ . The point moves through Quadrant II to $(-1, 0)$ .
$\pi$ to $\frac{7\pi}{6}$ : $r$ shrinks to $0$ . The point reaches the pole.
$\frac{7\pi}{6}$ to $\frac{11\pi}{6}$ : $r$ is negative, so the angle points down but the point is above the origin. It traces a small inner loop through $(0, 1)$ at $\theta = \frac{3\pi}{2}$ and back to the pole.
$\frac{11\pi}{6}$ to $2\pi$ : $r$ grows from $0$ back to $1$ , ending at $(1, 0)$ .

The result is a limaçon with an inner loop. Each crossing of the $\theta$ -axis is a pass through the pole, and each stretch below the axis is drawn on the opposite side of the origin from where the angle points.

When the distance from the origin is increasing or decreasing

The distance from the point $(r, \theta)$ to the origin is $|r|$ . So the question is never just whether $r$ is going up. It's whether $r$ is moving away from $0$ or toward $0$ .

$r > 0$ and increasing: distance increasing, moving away from the origin.
$r < 0$ and decreasing: $r$ is getting more negative, so distance increasing, moving away.
$r > 0$ and decreasing: distance decreasing, moving toward the origin.
$r < 0$ and increasing: $r$ is rising toward $0$ , so distance decreasing, moving toward.

On the graph of $r$ against $\theta$ , that's one picture: the distance is increasing wherever the graph is moving away from the $\theta$ -axis, and decreasing wherever it's moving toward the axis.

Five intervals on one curve

Let

r = 1 + 2\sin\theta

for one full turn, with

\theta

from

0

2\pi

, as graphed above. (a) On which intervals is the distance between the point on the polar curve and the origin increasing? (b) Which point on the curve is farthest from the origin? (c) Find the average rate of change of

r

over

\frac{7\pi}{6} \le \theta \le \frac{3\pi}{2}

, and say what happens to the distance on that interval.

Split at the zeros, $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ , and at the turning points, $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ .

$\left(0, \frac{\pi}{2}\right)$ : $r > 0$ and increasing, so away.
$\left(\frac{\pi}{2}, \frac{7\pi}{6}\right)$ : $r > 0$ and decreasing, so toward.
$\left(\frac{7\pi}{6}, \frac{3\pi}{2}\right)$ : $r < 0$ and decreasing, so away.
$\left(\frac{3\pi}{2}, \frac{11\pi}{6}\right)$ : $r < 0$ and increasing, so toward.
$\left(\frac{11\pi}{6}, 2\pi\right)$ : $r > 0$ and increasing, so away.

(b) The farthest point has the largest $|r|$ . The greatest value of $r$ is $3$ and the least is $-1$ , so the largest distance is $3$ , at $\theta = \frac{\pi}{2}$ . In rectangular coordinates that point is $(0, 3)$ .

(c) The average rate of change is $\frac{r\left(\frac{3\pi}{2}\right) - r\left(\frac{7\pi}{6}\right)}{\frac{3\pi}{2} - \frac{7\pi}{6}} = \frac{-1 - 0}{\frac{\pi}{3}} = -\frac{3}{\pi} \approx -0.955$ per radian. It's negative, but the distance went from $0$ to $1$ . A negative rate means $r$ decreased, not that the point got closer.

Answer: (a) $\left(0, \frac{\pi}{2}\right)$ , $\left(\frac{7\pi}{6}, \frac{3\pi}{2}\right)$ , and $\left(\frac{11\pi}{6}, 2\pi\right)$ . (b) $(0, 3)$ , at $\theta = \frac{\pi}{2}$ , distance $3$ . (c) $-\frac{3}{\pi} \approx -0.955$ per radian, while the distance increases from $0$ to $1$

r = 4\sin(3\theta)

The graph of r against theta for the three-petal rose r = 4 sin 3 theta. Between π/3 and π/2 the graph moves away from the θ-axis.

A negative stretch of a rose

For the polar curve

r = 4\sin(3\theta)

, graphed above against

\theta

, which statement describes the point on the curve on the interval

\frac{\pi}{3} < \theta < \frac{\pi}{2}

A. It moves away from the origin
B. It moves toward the origin
C. It stays the same distance from the origin
D. It moves toward the origin, then away

On that interval $3\theta$ runs from $\pi$ to $\frac{3\pi}{2}$ , where sine falls from $0$ to $-1$ . So $r$ is negative and decreasing, from $0$ to $-4$ . The distance $|r|$ grows from $0$ to $4$ .

Answer: A

Polar graphs in Desmos

1Define the function on one line: type $f(x) = 1 + 2\sin(x)$ .
2Draw the polar curve: type $r = f(\theta)\{0 \le \theta \le 2\pi\}$ , using the $\theta$ key. The restriction draws exactly one trip around.
3Draw $r$ against $\theta$ : type $y = f(x)\{0 \le x \le 2\pi\}$ . Click its zeros and its highest and lowest points to see their coordinates.
4To watch one point move, add a slider $a$ and plot $(f(a)\cos(a), f(a)\sin(a))$ next to $(a, f(a))$ .

Desmos is available on the calculator parts of the exam. The no-calculator parts still expect conversions and sign readings by hand, so practice both. More calculator moves are in how to use Desmos on the AP Precalculus exam, and how free-response answers earn points is in the free-response guide.

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Practice on exam-style questions

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Questions

What does a negative r mean in polar coordinates?: Turn to the angle, then move that distance in the opposite direction. A point with r = −2 at some angle is the same as the point with r = 2 at that angle plus pi.
How do I tell if a polar curve is moving away from the origin?: Check whether r is moving away from 0. That happens when r is positive and increasing, or negative and decreasing.
Where does a polar curve pass through the pole?: Wherever r = 0. On the graph of r against theta, those are the places the graph crosses or touches the horizontal axis.
Why graph r against theta on rectangular axes?: It shows where r is positive, negative, increasing, and decreasing at a glance. Those four facts answer most polar questions.