3.14 - Polar Function Graphs

Overview

A polar function

r = f(\theta)

takes an angle as its input and gives a radius as its output. Plotting all of those

(r, \theta)

pairs draws curves that would be messy to write with

x

and

y

, like circles through the origin, roses, and limaçons. The key skill is connecting the ordinary graph of

r

versus

\theta

to the curve it draws in the polar plane.

What the graph of a polar function shows

The graph of

r = f(\theta)

is every point with polar coordinates

(f(\theta), \theta)

. Each input is an angle measured from the polar axis, and each output is a signed radius.

Picture a ray spinning counterclockwise around the origin as

\theta

increases. At each angle,

f(\theta)

tells you where the point sits on the line through that ray:

f(\theta)

units out along the ray if it's positive, or

|f(\theta)|

units in the opposite direction if it's negative.

A change in the input is a change in angle, so the point moves to a different ray.

A change in the output is a change in signed radius, so the point moves toward or away from the origin along the line.

An output of 0 puts the point at the origin, whatever the angle.

Example 1: Graphing $r = 4\cos\theta$ from a table

Make a table of values for

0 \le \theta \le \pi

and plot the points.

$\theta$	$r = 4\cos\theta$	Where the point is
$0$	$4$	4 units along the polar axis: $(4, 0)$
$\frac{\pi}{4}$	$2\sqrt{2} \approx 2.83$	on the $\frac{\pi}{4}$ ray: $(2, 2)$
$\frac{\pi}{2}$	$0$	the origin
$\frac{3\pi}{4}$	$-2\sqrt{2} \approx -2.83$	opposite the $\frac{3\pi}{4}$ ray, in Quadrant IV: $(2, -2)$
$\pi$	$-4$	opposite the $\pi$ ray, back at $(4, 0)$

For

0 \le \theta \le \frac{\pi}{2}

r

is positive and the point sweeps along the top half of a circle, from

(4, 0)

into the origin (black). For

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

r

is negative, so each point lands on the opposite side of the origin, and the bottom half gets drawn (red).

The graph is a circle with center

(2, 0)

and radius 2, traced once for

0 \le \theta \le \pi

. As

\theta

continues from

\pi

2\pi

, the same circle is traced again. (Multiplying both sides by

r

gives

r^2 = 4r\cos\theta

, which is

x^2 + y^2 = 4x

, or

(x - 2)^2 + y^2 = 4

in rectangular form.)

From the graph of $r$ versus $\theta$ to the polar graph

AP questions often show the graph of

r = f(\theta)

on ordinary axes, with

\theta

on the horizontal axis and

r

on the vertical axis, then ask about the polar curve. Read that graph from left to right. At each angle, ask two questions: how big is

r

, and is it positive or negative?

Example 2: The limaçon $r = 1 + 2\cos\theta$ (AP-style)

The figure shows

r = 1 + 2\cos\theta

graphed both ways, with matching points labeled. The red parts are where

r < 0

Point	$\theta$	$r$	Location in the plane
$A$	$0$	$3$	$(3, 0)$ , on the polar axis
$B$	$\frac{\pi}{2}$	$1$	$(0, 1)$
$C$	$\frac{2\pi}{3}$	$0$	the origin
$D$	$\pi$	$-1$	$(1, 0)$ , since a negative $r$ points opposite the $\pi$ ray
$E$	$\frac{4\pi}{3}$	$0$	the origin
$F$	$\frac{3\pi}{2}$	$1$	$(0, -1)$
$G$	$2\pi$	$3$	$(3, 0)$ again

Now follow the curve in three stages:

A to C ( $0 \le \theta \le \frac{2\pi}{3}$ ): $r$ is positive and shrinks from 3 to 0. The point swings from $(3, 0)$ up through Quadrants I and II and into the origin. This is the top of the big loop.

C to E ( $\frac{2\pi}{3} < \theta < \frac{4\pi}{3}$ ): $r$ is negative. The rays point into Quadrants II and III, but the points are plotted on the opposite side, in Quadrants IV and I. This traces the small inner loop, and $D = (1, 0)$ is its point farthest from the origin.

E to G ( $\frac{4\pi}{3} \le \theta \le 2\pi$ ): $r$ is positive again and grows from 0 to 3, tracing the bottom of the big loop through Quadrants III and IV back to $(3, 0)$ .

This curve is called a limaçon with an inner loop. The inner loop exists only because

r

goes negative.

Restricting the domain

To draw only part of a polar graph, choose an interval of

\theta

whose endpoints are the angles where that part starts and stops. Find those angles from the radius you want at each end. Often the part starts or stops at the origin, so you solve

f(\theta) = 0

Example 3: Tracing only the inner loop

For

r = 1 + 2\cos\theta

, give an interval of

\theta

that traces just the inner loop.

The inner loop starts and ends at the origin, where

r = 0

1 + 2\cos\theta = 0 \quad\Rightarrow\quad \cos\theta = -\frac{1}{2} \quad\Rightarrow\quad \theta = \frac{2\pi}{3} \text{ or } \frac{4\pi}{3}

Between these angles

r

is negative (for example,

r = -1

\theta = \pi

), which is exactly the inner loop. Answer:

\frac{2\pi}{3} \le \theta \le \frac{4\pi}{3}

Rose curves

Graphs of

r = a\sin(n\theta)

and

r = a\cos(n\theta)

, where

n \ge 2

is a whole number, are called roses. Each petal has length

|a|

, the largest value of

|r|

If $n$ is odd, there are $n$ petals, and the whole rose is traced once for $0 \le \theta \le \pi$ .

If $n$ is even, there are $2n$ petals, and it takes $0 \le \theta \le 2\pi$ to trace them all.

r = 3\sin 2\theta

has

n = 2

, so it has 4 petals of length 3.

r = 2\cos 3\theta

has

n = 3

, so it has 3 petals of length 2, with tips in the directions

\theta = 0

\frac{2\pi}{3}

, and

\frac{4\pi}{3}

Example 4: Which angles trace which petal?

For

r = 3\sin 2\theta

, find an interval of

\theta

that traces only the petal in Quadrant II.

Check the sign of

r

on each quarter turn:

Interval	Sign of $r$	Petal traced
$0 \le \theta \le \frac{\pi}{2}$	positive	Quadrant I
$\frac{\pi}{2} \le \theta \le \pi$	negative	Quadrant IV (opposite Quadrant II)
$\pi \le \theta \le \frac{3\pi}{2}$	positive	Quadrant III
$\frac{3\pi}{2} \le \theta \le 2\pi$	negative	Quadrant II (opposite Quadrant IV)

Answer:

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

. The obvious guess,

\frac{\pi}{2} \le \theta \le \pi

, traces the Quadrant IV petal instead, because

r

is negative there.

Common polar graphs

The CED doesn't require these names, but recognizing the shapes helps you check your work quickly.

Equation	Graph
$r = a$	circle centered at the origin with radius $\|a\|$
$r = a\cos\theta$	circle through the origin, centered at $\left(\frac{a}{2}, 0\right)$
$r = a\sin\theta$	circle through the origin, centered at $\left(0, \frac{a}{2}\right)$
$r = a\sin(n\theta)$ or $r = a\cos(n\theta)$	rose: $n$ petals if $n$ is odd, $2n$ petals if $n$ is even
$r = a + b\cos\theta$ or $r = a + b\sin\theta$ , with $a, b > 0$	limaçon: inner loop if $a < b$ , cardioid (heart shape) if $a = b$ , dimpled if $b < a < 2b$ , convex if $a \ge 2b$

The four limaçons above all have

b = 2

and are drawn on the same scale. The cosine versions are symmetric about the x-axis. The sine versions are the same shapes turned to be symmetric about the y-axis.

Common mistakes

Plotting a negative $r$ along the ray. A negative radius goes the opposite way, through the origin.

Assuming you always need $0 \le \theta \le 2\pi$ . Some graphs, like $r = 4\cos\theta$ and $r = 2\cos 3\theta$ , are fully traced by $0 \le \theta \le \pi$ , and going further just retraces them.

Miscounting rose petals. An even $n$ gives $2n$ petals, not $n$ .

Treating the graph of $r$ versus $\theta$ as a picture of the curve. It looks nothing like the polar graph. It's a tool for reading $r$ at each angle.

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