Polar Graphs of Limacons and Cardioids, precalculus lesson

A cosine wave such as $r = 1 + 2\cos\theta$ rises to $3$ , falls to $-1$ and rises again. Wrap that wave around the pole, so that $\theta$ is a direction instead of a position along a horizontal axis, and you get a limaçon. Depending on how the constant compares with the amplitude, the curve looks like an oval, a dented oval, a heart, or a loop with a smaller loop inside it. Two numbers in the equation tell you which before you plot a single point.

Key terms

Symmetry tests

Definition A polar graph is symmetric about a line or a point if the mirror image of each of its points is again on the graph. For an equation in $r$ and $\theta$ , each replacement in the table produces a polar name of the mirror image of the point $(r, \theta)$ . So if a replacement turns the equation into an equivalent equation, the graph has that symmetry.

Symmetric about	Replace $(r, \theta)$ by	Or replace $(r, \theta)$ by
the polar axis, the $x$ -axis	$(r, -\theta)$	$(-r, \pi - \theta)$
the line $\theta = \frac{\pi}{2}$ , the $y$ -axis	$(r, \pi - \theta)$	$(-r, -\theta)$
the pole	$(-r, \theta)$	$(r, \theta + \pi)$

The tests are sufficient but not necessary. If a replacement changes the equation, the graph may still have the symmetry, so a failed test proves nothing. To show that a symmetry is missing, find one point of the graph whose mirror image is not on the graph.

In plain words Each replacement is another way to name the reflected point. If the equation does not notice the change, every reflected point is on the graph too, and you only need to plot half of it.

Example Test $r = 1 - 2\sin\theta$ . Replacing $\theta$ by $\pi - \theta$ leaves it unchanged, because $\sin(\pi - \theta) = \sin\theta$ , so the graph is symmetric about the $y$ -axis. Replacing $\theta$ by $-\theta$ gives $r = 1 + 2\sin\theta$ , a different equation, so that test is silent. A point settles the question. At $\theta = \frac{3\pi}{2}$ the radius is $3$ , which gives the point $(0, -3)$ . Its mirror image in the $x$ -axis is $(0, 3)$ . A point of the graph other than the pole lies on the $y$ -axis only for the angles $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ , up to full turns, and these angles give $r = -1$ and $r = 3$ , the points $(0, -1)$ and $(0, -3)$ . So $(0, 3)$ is not on the graph, and the graph is not symmetric about the $x$ -axis.

Limaçon

Definition Let $a > 0$ and $b > 0$ . The graphs of

\begin{aligned} r &= b + a\cos\theta, \\[0.4em] r &= b - a\cos\theta, \\[0.4em] r &= b + a\sin\theta, \\[0.4em] r &= b - a\sin\theta \end{aligned}

are limaçons. Each one is drawn completely as $\theta$ runs through $[0, 2\pi)$ , and apart from the pole no point is drawn twice. The cosine forms are symmetric about the $x$ -axis and the sine forms about the $y$ -axis. All four are turned copies of $r = b + a\cos\theta$ , because replacing $\theta$ by $\theta - \alpha$ turns a polar graph counterclockwise through $\alpha$ , and

\begin{aligned} b - a\cos\theta &= b + a\cos(\theta - \pi), \\ b + a\sin\theta &= b + a\cos\left(\theta - \tfrac{\pi}{2}\right), \\ b - a\sin\theta &= b + a\cos\left(\theta - \tfrac{3\pi}{2}\right). \end{aligned}

The shape depends only on the ratio $\frac{b}{a}$ of the constant term to the coefficient of the trigonometric term. Some books swap the two letters, so compare those two numbers whatever they are called.

Ratio	Shape
$\frac{b}{a} < 1$	inner loop
$\frac{b}{a} = 1$	cardioid
$1 < \frac{b}{a} < 2$	dimpled
$\frac{b}{a} \ge 2$	convex

In plain words The constant $b$ sets the typical distance from the pole. The term with $a$ pushes the curve out on one side and pulls it in on the other, and the larger $a$ is compared with $b$ , the harder the pull.

Example For $r = 3 - \sin\theta$ the ratio is $3$ , so the curve is a convex limaçon. Four angles give its points on the axes.

$\theta$	$0$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$
$r$	$3$	$2$	$3$	$4$
Point	$(3, 0)$	$(0, 2)$	$(-3, 0)$	$(0, -4)$

The minus sign in front of $\sin\theta$ makes the curve bulge downward, toward $(0, -4)$ , and flatten on top.

Limaçon with an inner loop

Definition Let $0 < b < a$ . The limaçon $r = b + a\cos\theta$ passes through the pole exactly at the angles in $[0, 2\pi)$ with

\cos\theta = -\frac{b}{a},

which are $\theta_1 = \arccos(-\frac{b}{a})$ , between $\frac{\pi}{2}$ and $\pi$ , and $\theta_2 = 2\pi - \theta_1$ . For $\theta_1 < \theta < \theta_2$ the radius is negative, and these angles draw the inner loop. Its points lie on the opposite side of the pole from the directions $\theta$ , so the small loop sits inside the large one, on the right. Apart from the pole it meets the $x$ -axis only at $(a - b, 0)$ , drawn at $\theta = \pi$ , while the large loop reaches out to $(b + a, 0)$ .

In plain words When $a$ is larger than $b$ , the term $a\cos\theta$ can outweigh $b$ and push the radius below zero. A negative radius is drawn on the far side of the pole, and that stretch of the curve is the small loop.

Example For $r = 1 + 2\cos\theta$ the radius is zero where $\cos\theta = -\frac{1}{2}$ , at $\theta = \frac{2\pi}{3}$ and $\theta = \frac{4\pi}{3}$ .

$\theta$	$r$	Point $(x, y)$
$0$	$3$	$(3, 0)$
$\frac{\pi}{3}$	$2$	$(1, \sqrt{3})$
$\frac{\pi}{2}$	$1$	$(0, 1)$
$\frac{2\pi}{3}$	$0$	$(0, 0)$
$\pi$	$-1$	$(1, 0)$

As $\theta$ runs from $\frac{2\pi}{3}$ to $\frac{4\pi}{3}$ , the point leaves the pole, swings through the fourth quadrant to $(1, 0)$ and comes back through the first quadrant. That is the inner loop. The angles from $\frac{4\pi}{3}$ to $2\pi$ draw the lower half of the large loop, the mirror image of the upper half drawn by the angles from $0$ to $\frac{2\pi}{3}$ .

Cardioid

Definition A cardioid is a limaçon with $a = b$ , so one of

\begin{aligned} r &= a(1 + \cos\theta), \\[0.4em] r &= a(1 - \cos\theta), \\[0.4em] r &= a(1 + \sin\theta), \\[0.4em] r &= a(1 - \sin\theta), \end{aligned}

with $a > 0$ . Its radius is never negative, and it equals $0$ at exactly one angle in $[0, 2\pi)$ . There the curve reaches the pole and turns back the way it came, forming a cusp. The farthest point is at distance $2a$ from the pole, at the angle half a turn away from the angle of the cusp.

In plain words The cardioid is the borderline case between an inner loop and no loop at all. The radius shrinks to zero for an instant and grows again, so the curve comes to a point at the pole instead of passing through it. The name means heart shaped.

Example For $r = 2 + 2\cos\theta$ ,

$\theta$	$0$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\frac{2\pi}{3}$	$\pi$
$r$	$4$	$3$	$2$	$1$	$0$

The points are $(4, 0)$ , $(\frac{3}{2}, \frac{3\sqrt{3}}{2})$ , $(0, 2)$ , $(-\frac{1}{2}, \frac{\sqrt{3}}{2})$ and the pole. Reflecting them in the $x$ -axis gives the lower half. The result is a heart lying on its side. Its cusp is a dent at the pole on the left side, and its far point $(4, 0)$ is on the right.

Dimpled and convex limaçons

Definition Let $b > a > 0$ . Then $r = b + a\cos\theta \ge b - a > 0$ for every $\theta$ , so the curve never reaches the pole and has no loop. If $b < 2a$ the limaçon is dimpled, which means it bends inward near the point $(-(b - a), 0)$ nearest the pole. If $b \ge 2a$ it is convex, which means the region it encloses contains the whole segment between any two of its points. The other three forms behave the same way after turning.

In plain words Once $b$ is larger than $a$ , the curve stays away from the pole. If $b$ is only a little larger, the near side is pulled in and shows a dent. Once $b$ is at least twice $a$ , the dent is gone and the curve is an egg shaped oval.

Example Compare the left ends of $r = 3 + 2\cos\theta$ and $r = 4 + 2\cos\theta$ . On the first curve the $x$ -coordinate is

\begin{aligned} x &= r\cos\theta \\[0.4em] &= 3\cos\theta + 2\cos^2\theta \\[0.4em] &= 2\left(\cos\theta + \frac{3}{4}\right)^2 - \frac{9}{8}. \end{aligned}

At $\theta = \pi$ this is $-1$ , but where $\cos\theta = -\frac{3}{4}$ it is $-\frac{9}{8}$ , farther left. The two points above and below the axis stick out past the axis point, and that dent is the dimple. On the second curve

\begin{aligned} x &= 4\cos\theta + 2\cos^2\theta \\[0.4em] &= 2(\cos\theta + 1)^2 - 2, \end{aligned}

which is smallest at $\cos\theta = -1$ . The leftmost point is $(-2, 0)$ on the axis, and there is no dent.

Maximum r

Definition On the graph of $r = f(\theta)$ , the point $(f(\theta), \theta)$ is at distance $\lvert f(\theta) \rvert$ from the pole, so the farthest points come from the angles where $\lvert f(\theta) \rvert$ is largest. That largest value is the larger of the maximum of $f$ and the negative of its minimum. For the limaçon $r = b + a\cos\theta$ with $a > 0$ and $b > 0$ , the bounds $-1 \le \cos\theta \le 1$ give

b - a \le r \le b + a.

The maximum $r = b + a$ occurs at $\theta = 0$ , and since $\lvert b - a \rvert < b + a$ , the point $(b + a, 0)$ is the farthest from the pole. The minimum $r = b - a$ occurs at $\theta = \pi$ . If $b > a$ it gives the nearest point of the curve, if $b = a$ it is the cusp at the pole, and if $b < a$ it gives the tip of the inner loop, at distance $a - b$ . The other three forms reach their maximum where their trigonometric term equals $+a$ .

In plain words Read $r$ as an ordinary wave with midline $r = b$ and amplitude $a$ . Its highest value $b + a$ is the far edge of the curve, and its lowest value $b - a$ is either the near edge or, when it is negative, the tip of the inner loop.

Example For $r = 2 - 3\sin\theta$ the midline is $r = 2$ and the amplitude is $3$ , so $r$ runs from $-1$ to $5$ . The maximum $r = 5$ occurs at $\theta = \frac{3\pi}{2}$ , where $\sin\theta = -1$ , and gives the point $(0, -5)$ , the farthest from the pole. The minimum $r = -1$ occurs at $\theta = \frac{\pi}{2}$ and names the point $(0, -1)$ , the tip of the inner loop. The inner loop lies below the pole, and so does most of the big loop.

Check yourself

Classify $r = 3 - 3\sin\theta$ , name a symmetry it has, and find its farthest point from the pole.

Show the answer

Here $a = b = 3$ , so it is a cardioid. Replacing $\theta$ by $\pi - \theta$ leaves the equation unchanged, so it is symmetric about the $y$ -axis. The maximum $r = 6$ occurs at $\theta = \frac{3\pi}{2}$ , which gives the point $(0, -6)$ . The cusp is at the pole, at $\theta = \frac{\pi}{2}$ .

For $r = 2 + 4\cos\theta$ , find the angles in $[0, 2\pi)$ where the graph passes through the pole, and the point other than the pole where the inner loop meets the $x$ -axis.

Show the answer

The radius is zero where $\cos\theta = -\frac{1}{2}$ , at $\theta = \frac{2\pi}{3}$ and $\theta = \frac{4\pi}{3}$ . At $\theta = \pi$ the radius is $-2$ , which names the point $(2, 0)$ , the tip of the inner loop. The large loop reaches $(6, 0)$ at $\theta = 0$ .

Is $r = 5 + 2\sin\theta$ dimpled or convex? Find its largest and smallest distances from the pole.

Show the answer

The ratio is $\frac{5}{2} \ge 2$ , so it is convex. The largest distance is $7$ , at $\theta = \frac{\pi}{2}$ , the point $(0, 7)$ . The smallest is $3$ , at $\theta = \frac{3\pi}{2}$ , the point $(0, -3)$ .

Polar Graphs of Limacons and Cardioids

Key terms

Symmetry tests

Limaçon

Limaçon with an inner loop

Cardioid

Dimpled and convex limaçons

Maximum r

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus