Polar Function Graphs

In Topic 3.13 a single pair $(r, \theta)$ named a single point. Now let $r$ depend on $\theta$ . As the direction $\theta$ sweeps around the pole, the distance $r = f(\theta)$ changes, and the points trace a curve. This page shows how to read and sketch such curves from the ordinary graph of $r$ against $\theta$ , and names the families you will see.

A function of the angle

A polar function $r = f(\theta)$ has the angle as its input and a signed distance as its output. Its graph is the set of points with polar coordinates $(f(\theta), \theta)$ . You can also graph $r$ against $\theta$ on ordinary axes, with $\theta$ across and $r$ up. The two pictures carry the same information:

On the graph of $r$ against $\theta$	On the polar curve
moving right (input increases)	the direction turns counterclockwise
height $r$ (output)	signed distance along the current ray
a zero, $r = 0$	the curve passes through the pole
$r < 0$ (below the axis)	the point is on the opposite ray
a peak or valley of $\|r\|$	a point farthest from the pole nearby, such as a petal tip

Sketching from the graph of $r$

Worked example 1. Sketch $r = 3\sin(2\theta)$ (Figure 1).

Key values. $r = 0$ at $\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ ; $r = 3$ at $\frac{\pi}{4}$ and $\frac{5\pi}{4}$ ; $r = -3$ at $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$ .

First quarter turn, $0 \le \theta \le \frac{\pi}{2}$ : $r$ rises from $0$ to $3$ and falls back to $0$ , all positive. The points leave the pole, reach distance $3$ along the ray $\frac{\pi}{4}$ , and return: one petal in Quadrant I.

Second quarter turn, $\frac{\pi}{2} \le \theta \le \pi$ : $r$ is negative. The angle points into Quadrant II, so the points land in Quadrant IV: a petal there, with its tip at $\left(-3, \frac{3\pi}{4}\right)$ , which is the same point as $\left(3, \frac{7\pi}{4}\right)$ .

The next two quarter turns draw petals in Quadrants III and II. Four humps, four petals.

The polar graph of r = 3 sin(2 theta): a rose with four petals of length 3, centered on the rays pi/4, 3 pi/4, 5 pi/4 and 7 pi/4, each starting and ending at the pole. — **Figure 1.** $r = 3\sin(2\theta)$ for $0 \le \theta \le 2\pi$ . LevlPrep original diagram.

The families

Equation	Name	Shape
$r = a$	circle	radius $\|a\|$ , centered at the pole
$r = a\cos\theta$ , $r = a\sin\theta$	circle	diameter $\|a\|$ , through the pole, centered on an axis
$r = a + b\cos\theta$ (or $\sin$ ), $\|a\| < \|b\|$	limaçon with inner loop	passes through the pole twice
$\|a\| = \|b\|$	cardioid	heart shape, touches the pole once
$\|a\| > \|b\|$	limaçon without a loop	never reaches the pole
$r = a\cos(n\theta)$ , $r = a\sin(n\theta)$	rose	$n$ petals if $n$ is odd, $2n$ if $n$ is even, each of length $\|a\|$

Cosine versions are symmetric about the polar axis; sine versions are symmetric about the vertical line $\theta = \frac{\pi}{2}$ .

The inner loop

Worked example 2. Where does $r = 1 + 2\cos\theta$ form its inner loop?

$r = 0$ where $\cos\theta = -\frac{1}{2}$ : $\theta = \frac{2\pi}{3}$ and $\frac{4\pi}{3}$ . Between them, $r < 0$ ; at $\theta = \pi$ , $r = -1$ .

On $\left(\frac{2\pi}{3}, \frac{4\pi}{3}\right)$ the angle points left, but the negative $r$ puts the points to the right of the pole. The small loop runs from the pole out to $(1, 0)$ in rectangular coordinates and back. Outside that interval $r > 0$ and the big loop is drawn, reaching $(3, 0)$ at $\theta = 0$ .

Common slips

Plotting a negative $r$ in the quadrant the angle points to. Reverse through the pole.
Counting petals as $n$ when $n$ is even. Even $n$ gives $2n$ .
Confusing $r = 4\cos\theta$ (a circle through the pole) with $r = 4$ (a circle around the pole).