Sketching Polar Curves

Let r depend on $\theta$

A polar curve gives r as a function of $\theta$, written $r = f\left(\theta\right)$. For each direction $\theta$, go out a distance r from the pole along that direction and mark the point. As $\theta$ turns, the points trace the curve.

The simplest case holds r fixed. For r = 2, every direction gets a point 2 from the pole, so a full turn of $\theta$ traces the circle of radius 2 about the pole. A circle about the origin, which needs $x^{2} + y^{2} = 4$ in Cartesian form, is just r = 2 in polar form.

A circle through the pole

Now let r change with $\theta$: $r = 2 \cos \theta$. At $\theta = 0$, r = 2. At $\frac{\pi}{6}$, $r = \sqrt{3} \approx 1.73$. At $\frac{\pi}{4}$, $r = \sqrt{2} \approx 1.41$. At $\frac{\pi}{3}$, r = 1. At $\frac{\pi}{2}$, r = 0, so the curve reaches the pole.

Below the initial line, $\cos\left(-\theta\right) = \cos \theta$, so the angles from 0 down to $- \frac{\pi}{2}$ give the mirror image. From $\theta = - \frac{\pi}{2}$ to $\frac{\pi}{2}$ the curve is traced once.

Multiplying $r = 2 \cos \theta$ by r gives $r^{2} = 2r \cos \theta$, which is $x^{2} + y^{2} = 2x$, or $\left(x - 1\right)^{2} + y^{2} = 1$. So the curve is the circle of radius 1 about (1, 0): its diameter is 2, from the pole to the point (2, 0).

The gold curve is $r = 2 \cos \theta$, traced from $\theta = - \frac{\pi}{2}$ to $\frac{\pi}{2}$: a circle of radius 1 that runs from the pole to (2, 0). The dashed curve is r = 2, the circle of radius 2 about the pole. The two meet at (2, 0), where $\theta = 0$.

The cardioid

For $r = 1 + \cos \theta$, read r at the quarter turns: 2 at $\theta = 0$, 1 at $\frac{\pi}{2}$, 0 at $\pi$, and 1 at $\frac{3\pi}{2}$. Between them, r = 1.5 at $\frac{\pi}{3}$ and 0.5 at $\frac{2\pi}{3}$.

The largest value, r = 2, is at $\theta = 0$, where $\cos \theta = 1$. The curve reaches the pole only where $\cos \theta = -1$, at $\theta = \pi$. There it comes in to a point, called a cusp.

Replacing $\theta$ by $-\theta$ leaves r unchanged, because $\cos\left(-\theta\right) = \cos \theta$, so the curve is symmetric about the initial line. Its heart shape gives it its name: a cardioid.

An arm of length $r = 1 + \cos \theta$ turns from $\theta = 0$ and lays down the cardioid behind it. At $\theta = 120^\circ$ the arm is 1 + cos 120° = 0.5 long. Turned on to 180°, the arm shrinks to 0 at the pole: the cusp.

A rose with four petals

For $r = 2 \cos 2\theta$, the angle inside the cosine turns twice as fast. r = 2 at $\theta = 0$, falls to 0 at $\frac{\pi}{4}$, and is −2 at $\frac{\pi}{2}$.

A negative r means going the distance |r| in the opposite direction, $\theta + \pi$. So at $\theta = \frac{\pi}{2}$, r = −2 puts the point 2 below the pole, at (0, −2).

Over a full turn, r = 0 at $\theta = \frac{\pi}{4}$, $\frac{3\pi}{4}$, $\frac{5\pi}{4}$ and $\frac{7\pi}{4}$: the curve returns to the pole four times. Between those returns it draws a petal, out to 2 and back. The petals along the x-axis come from the angles where r is positive, and the petals along the y-axis from those where r is negative.

$r = 2 \cos 2\theta$ over a full turn. The dots are its four tips: (2, 0) at $\theta = 0$, (0, −2) at $\theta = \frac{\pi}{2}$ with r = −2, (−2, 0) at $\theta = \pi$, and (0, 2) at $\theta = \frac{3\pi}{2}$ with r = −2.

A method for any curve

To sketch a polar curve, read r at the quarter turns $\theta = 0$, $\frac{\pi}{2}$, $\pi$ and $\frac{3\pi}{2}$ and mark those points. Find where r = 0, where the curve passes through the pole, and where r is largest. Check for symmetry. Then join the points smoothly as $\theta$ turns.

Try it on $r = 2 + \sin \theta$. At the quarter turns r is 2, 3, 2 and 1, giving the points (2, 0), (0, 3), (−2, 0) and (0, −1). The sine never goes below −1, so r is never less than 1 and the curve never reaches the pole. It is largest, 3, at $\theta = \frac{\pi}{2}$.

Replacing $\theta$ by $\pi - \theta$ leaves $\sin \theta$ unchanged, so the curve is symmetric about the y-axis. Joining the four points gives an oval, taller above the initial line than below it.

$r = 2 + \sin \theta$, through its four quarter-turn points (2, 0), (0, 3), (−2, 0) and (0, −1). It reaches 3 above the pole and only 1 below it, and never touches the pole.

The usual mistakes

Reading $r = 2 \cos \theta$ as a circle of radius 2. The 2 is the diameter: the circle has radius 1 and passes through the pole.

Putting the cardioid’s cusp at $\theta = 0$. There $\cos \theta = 1$ and r = 2, the farthest point. The cusp is at $\theta = \pi$, where r = 0.

Counting the petals of $r = 2 \cos 2\theta$ from the positive values of r only. The negative values draw the other two petals, along the y-axis.

A cardioid microphone

In the application below, a microphone’s pick-up pattern is $r = \frac{1}{2}\left(1 + \cos \theta\right)$, a cardioid half the size of $r = 1 + \cos \theta$. The direction with no pick-up is the cusp, where r = 0, and the directions where r is at most a quarter come from an inequality in $\cos \theta$.