Graphs of polar equations

The graph of a polar equation ${r=f(\theta)}$ is the set of all points ${(r,\theta)}$ that make it true. Some of these graphs are simple, like circles. Others are curves with names like the cardioid (heart shape), the limaçon (snail), and the rose. This page shows how to sketch them.

Graphing from a table

The basic method is the same as for ${y=f(x)}$: make a table of values, plot the points, and join them. But now each point is plotted by its angle and distance.

Example 1: Sketch the graph of ${r=2\sin\theta}$.

Solution:

Make a table for ${\theta}$ from ${0}$ to ${\pi}$:

${\theta}$${r=2\sin\theta}$
${0}$${0}$
${\dfrac{\pi}{6}}$${1}$
${\dfrac{\pi}{4}}$${\sqrt{2}\approx 1.41}$
${\dfrac{\pi}{3}}$${\sqrt{3}\approx 1.73}$
${\dfrac{\pi}{2}}$${2}$
${\dfrac{2\pi}{3}}$${\sqrt{3}\approx 1.73}$
${\dfrac{5\pi}{6}}$${1}$
${\pi}$${0}$

Plot the points, turning to each angle and walking out the distance ${r}$:

π/6π/3π/22π/35π/6π7π/64π/33π/25π/311π/612
The points from the table lie on a circle: the graph of ${r=2\sin\theta}$.

From ${\pi}$ to ${2\pi}$, ${\sin\theta}$ is negative, so ${r}$ is negative and the points land on the same circle again. So the graph is traced once for ${0\le\theta\le\pi}$.

Check with rectangular form: Multiply by ${r}$: ${r^2=2r\sin\theta}$, so ${x^2+y^2=2y}$. Completing the square gives ${x^2+(y-1)^2=1}$: the circle with center ${(0,1)}$ and radius ${1}$.

Circles

Example 1 and Example 5 of Polar coordinates show a pattern:

${r=a}$: circle of radius ${|a|}$, centered at the pole

${r=a\cos\theta}$: circle through the pole, centered on the polar axis

${r=a\sin\theta}$: circle through the pole, centered on the line ${\theta=\dfrac{\pi}{2}}$

The last two have diameter ${|a|}$.

Symmetry

Symmetry halves the work. The most useful test: if replacing ${\theta}$ by ${-\theta}$ gives the same equation, the graph is a mirror image of itself across the polar axis. This is true whenever ${r}$ depends on ${\theta}$ only through ${\cos\theta}$, because ${\cos(-\theta)=\cos\theta}$. Then you only need to plot ${\theta}$ from ${0}$ to ${\pi}$ and reflect.

Cardioids and limaçons

Example 2: Sketch the graph of ${r=1+\cos\theta}$.

Solution:

The equation uses ${\cos\theta}$, so the graph is symmetric about the polar axis. Some key values:

  • ${\theta=0}$: ${r=2}$ (the farthest point).
  • ${\theta=\dfrac{\pi}{2}}$: ${r=1}$.
  • ${\theta=\pi}$: ${r=0}$ (the curve reaches the pole).

As ${\theta}$ goes from ${0}$ to ${\pi}$, ${r}$ shrinks steadily from ${2}$ to ${0}$. Reflecting across the polar axis gives the other half:

π/6π/3π/22π/35π/6π7π/64π/33π/25π/311π/612
The cardioid ${r=1+\cos\theta}$. The orange points are at ${\theta=0}$, ${\dfrac{\pi}{2}}$, ${\pi}$, and ${\dfrac{3\pi}{2}}$.

This heart shape is called a cardioid.

The cardioid is one of the limaçons, the graphs of ${r=a+b\cos\theta}$ or ${r=a+b\sin\theta}$. Their shape depends on how ${a}$ compares with ${b}$. When ${a<b}$, ${r}$ becomes negative for some angles, and the curve has an inner loop.

Example 3: Sketch the graph of ${r=1+2\cos\theta}$.

Solution:

At ${\theta=0}$, ${r=3}$. The curve reaches the pole where ${r=0}$:

${1+2\cos\theta=0}$, so ${\cos\theta=-\dfrac{1}{2}}$

That is at ${\theta=\dfrac{2\pi}{3}}$ and ${\theta=\dfrac{4\pi}{3}}$. Between these angles, ${r}$ is negative, so those points are drawn on the opposite side. They form the small inner loop. At ${\theta=\pi}$, ${r=-1}$: one unit out in the direction ${0}$.

π/6π/3π/22π/35π/6π7π/64π/33π/25π/311π/6123
The limaçon ${r=1+2\cos\theta}$ has an inner loop. The curve passes through the pole at ${\theta=\dfrac{2\pi}{3}}$ and ${\dfrac{4\pi}{3}}$ (dashed).

Roses

The graphs of ${r=a\cos n\theta}$ and ${r=a\sin n\theta}$, for a whole number ${n\ge 2}$, are roses. Each petal is ${|a|}$ long.

If ${n}$ is odd, the rose has ${n}$ petals.

If ${n}$ is even, the rose has ${2n}$ petals.

Example 4: Sketch the graph of ${r=3\sin 2\theta}$.

Solution:

Here ${n=2}$ is even, so there are ${4}$ petals, each ${3}$ units long.

The petal tips are where ${|r|=3}$, that is, where ${\sin 2\theta=\pm 1}$: ${2\theta=\dfrac{\pi}{2}+k\pi}$, so

${\theta=\dfrac{\pi}{4}+\dfrac{k\pi}{2}}$

The curve passes through the pole where ${\sin 2\theta=0}$: at ${\theta=0}$, ${\dfrac{\pi}{2}}$, ${\pi}$, and ${\dfrac{3\pi}{2}}$.

π/4π/23π/4π5π/43π/27π/4123
The rose ${r=3\sin 2\theta}$ has four petals, each ${3}$ units long.

For ${\dfrac{\pi}{2}<\theta<\pi}$, ${r}$ is negative, so that petal is drawn in quadrant IV instead of quadrant II. In the end, every quadrant gets one petal.

A gallery of polar graphs

Here are the common shapes side by side:

Circle: ${r=a}$
Circle: ${r=a\cos\theta}$
Cardioid: ${r=a(1+\cos\theta)}$
Limaçon: ${r=a+b\cos\theta}$, ${a<b}$
Rose: ${r=a\cos 3\theta}$ (3 petals)
Rose: ${r=a\cos 2\theta}$ (4 petals)
Spiral: ${r=a\theta}$

The last one, ${r=a\theta}$, is the spiral of Archimedes: as ${\theta}$ grows, the curve keeps winding outward.

Summary