Graphing Polar Equations

Graphing polar equations means plotting every point (r, \theta) that makes an equation like r = 2 + 2 \cos \theta true. Make a table of \theta and r values, plot the points on a polar grid, and connect them in order. Learn the common shapes: circles, rose curves, cardioids and limaçons.

Before this lesson: You should know Polar Coordinates, The Unit Circle and Graphs of Sine and Cosine.

How Polar Graphs Work

In a polar equation, r depends on \theta. As the angle sweeps around from 0 to 2\pi, the distance from the pole grows and shrinks. The curve traces where those points land.

If r comes out negative, plot the point in the opposite direction. For example, r = -3 at \theta = \frac{\pi}{2} lands 3 units straight down, not up.

Three polar graphs on grids with circles of radius 2 and 4. Left, solid blue: the rose r = 3 cos 2θ with four petals of length 3 along the axes. Middle, dashed orange: the cardioid r = 2 + 2 cos θ, a heart shape that reaches 4 on the right and touches the pole on the left. Right, dotted green: the limaçon r = 1 + 2 sin θ, reaching 3 at the top with a small inner loop inside it, just above the pole.
Three common polar shapes: a four-petal rose, a cardioid and a limaçon with an inner loop.

Common Polar Graphs

EquationShapeKey facts
r = aCircle at the poleRadius \lvert a \rvert
\theta = \alphaLine through the poleMakes angle \alpha with the polar axis
r = a \cos \theta or r = a \sin \thetaCircle through the poleDiameter \lvert a \rvert
r = a \cos n\theta or r = a \sin n\thetaRose curven petals if n is odd, 2n if n is even; petal length \lvert a \rvert
r = a + a \cos \theta (or with \sin, or minus)CardioidHeart shape that touches the pole
r = a + b \cos \theta (or with \sin)LimaçonInner loop if \lvert a \rvert \lt \lvert b \rvert; dimple if \lvert b \rvert \lt \lvert a \rvert \lt 2\lvert b \rvert; no dimple if \lvert a \rvert \ge 2\lvert b \rvert

Symmetry shortcuts

  • Equations with \cos \theta are symmetric about the polar axis (the x-axis), because \cos(-\theta) = \cos \theta.
  • Equations with \sin \theta are symmetric about the line \theta = \frac{\pi}{2} (the y-axis).
  • Use symmetry to plot half the points, then mirror them.

Steps for Graphing a Polar Equation

  1. Identify the shape from the table above, if you can.
  2. Make a table of values for \theta from 0 to 2\pi, using unit circle angles.
  3. Plot each point (r, \theta). For a negative r, go the opposite way.
  4. Connect the points in order of \theta, and use symmetry to check.

Worked Examples

Example 1: A circle through the pole

Graph r = 4 \sin \theta.

Make a table for 0 \le \theta \le \pi: at \theta = 0, r = 4 \sin 0 = 0

r = 4 \cdot \frac{1}{2} = 2

At \theta = \frac{\pi}{6}.

r = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \approx 3.46

At \theta = \frac{\pi}{3}.

r = 4 \cdot 1 = 4

At \theta = \frac{\pi}{2}.

At \theta = \frac{5\pi}{6} and \theta = \pi, the values repeat in reverse: r = 2, then r = 0

Plot and connect: the points form a circle through the pole, reaching up to (4, \frac{\pi}{2})

Answer: a circle with diameter 4, centered at (0, 2) in rectangular coordinates. From \pi to 2\pi, r is negative and the graph traces the same circle again.

Example 2: A cardioid

Graph r = 2 + 2 \cos \theta.

Identify the shape: a = b = 2, so it is a cardioid, symmetric about the polar axis

r = 2 + 2(1) = 4

At \theta = 0.

r = 2 + 2 \cdot \frac{1}{2} = 3

At \theta = \frac{\pi}{3}.

r = 2 + 2(0) = 2

At \theta = \frac{\pi}{2}.

r = 2 + 2(-\frac{1}{2}) = 1

At \theta = \frac{2\pi}{3}.

At \theta = \pi: r = 2 + 2(-1) = 0, so the curve touches the pole

Mirror these points across the polar axis for \pi \lt \theta \lt 2\pi

Answer: a heart shape that reaches 4 on the right and has its point at the pole, as in the middle graph above.

Example 3: A rose curve

Graph r = 3 \cos 2\theta.

Identify the shape: a rose with n = 2. Since n is even, there are 2n = 4 petals, each 3 long

At \theta = 0: r = 3 \cos 0 = 3, the tip of a petal on the polar axis

r = 3 \cos \frac{\pi}{3} = 1.5

At \theta = \frac{\pi}{6}.

r = 3 \cos \frac{\pi}{2} = 0

At \theta = \frac{\pi}{4} (back at the pole).

At \theta = \frac{\pi}{2}: r = 3 \cos \pi = -3. Plot it 3 units in the opposite direction, straight down

Keep going around: the petals point along \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}

Answer: a four-petal rose with petal tips 3 units out on each axis, as in the left graph above.

Example 4: A limaçon with an inner loop

Graph r = 1 + 2 \sin \theta.

Identify the shape: a = 1 and b = 2, and 1 \lt 2, so it is a limaçon with an inner loop, symmetric about \theta = \frac{\pi}{2}

Find the largest r: at \theta = \frac{\pi}{2}, r = 1 + 2(1) = 3

Find where it passes through the pole: 1 + 2 \sin \theta = 0, so \sin \theta = -\frac{1}{2} and \theta = \frac{7\pi}{6}, \frac{11\pi}{6}

Between those angles, r is negative. At \theta = \frac{3\pi}{2}: r = 1 + 2(-1) = -1, which plots 1 unit straight up

r = 1 + 2(0) = 1

At \theta = 0 and \theta = \pi.

Answer: a large loop reaching 3 at the top, with a small inner loop that reaches 1 above the pole, as in the right graph above.

Example 5: Word problem with a microphone

A cardioid microphone picks up a singer within r = 5 + 5 \cos \theta feet, where \theta = 0 points straight out the front. How far away can the singer be when standing in front, to the side (\theta = \frac{\pi}{2}), behind, and at \theta = \frac{\pi}{3}?

r = 5 + 5(1) = 10

In front, \theta = 0 (feet).

r = 5 + 5(0) = 5

To the side, \theta = \frac{\pi}{2} (feet).

r = 5 + 5(-1) = 0

Behind, \theta = \pi (feet).

r = 5 + 5 \cdot \frac{1}{2} = 7.5

At \theta = \frac{\pi}{3} (feet).

Answer: 10 ft in front, 5 ft to the side, 7.5 ft at \frac{\pi}{3}, and none from directly behind. That is why these microphones block noise from the back.

Common Mistakes

MistakeExample of the errorFix
Plotting negative r on the same rayPlotting r = -3 at \theta = \frac{\pi}{2} as 3 units upNegative r goes the opposite way: 3 units down.
Wrong number of petalsSaying r = 3 \cos 2\theta has 2 petalsIf n is even, there are 2n petals: 4.
Too few pointsPlotting only \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}Use steps of \frac{\pi}{6} or smaller so you see the shape.
Mixing up the limaçon typesCalling r = 1 + 2 \sin \theta a cardioidA cardioid needs \lvert a \rvert = \lvert b \rvert. Here 1 \lt 2, so there is an inner loop.
Stopping at \piGraphing a cardioid only for 0 \le \theta \le \piMost curves need 0 \le \theta \lt 2\pi. Use symmetry to fill in the rest.

Related Methods

Convert to rectangular form

Circles are easy to check by converting. Multiply r = 4 \sin \theta by r to get x^2 + y^2 = 4y, a circle, as shown in polar coordinates.

Use a graphing calculator

Set the calculator to polar mode, enter r as a function of \theta, and set \theta from 0 to 2\pi. Compare the screen with your hand-drawn sketch.

Think of it like a sine graph

The table of r values follows the same ups and downs as the sine and cosine graphs. Picture r against \theta first, then wrap that graph around the pole.

Practice Problems

  1. Name the graph of r = 5.
  2. Name the graph of \theta = \frac{\pi}{3}.
  3. Name the graph of r = 6 \cos \theta and give its center and radius in rectangular coordinates.
  4. How many petals does r = 4 \sin 3\theta have, and how long is each?
  5. How many petals does r = 2 \cos 4\theta have?
  6. Name the graph of r = 3 - 3 \sin \theta. Find its largest r and the angle where it happens.
  7. Name the graph of r = 2 + 4 \cos \theta and find where it passes through the pole on [0, 2\pi).
  8. Name the graph of r = 4 + 2 \cos \theta. Find its largest and smallest r.
  9. For r = 2 \sin 2\theta, find r at \theta = \frac{\pi}{12}, \frac{\pi}{4}, \frac{5\pi}{12}, \frac{\pi}{2}.
  10. A microphone picks up sound within r = 4 + 4 \cos \theta feet. How far away can a singer be at \theta = \frac{2\pi}{3}? At \theta = \frac{\pi}{4}?
Show answers

1.

r is always 5, at every angle
Answer: a circle centered at the pole with radius 5

2.

Every point has angle \frac{\pi}{3}, with any r, positive or negative
Answer: a line through the pole at angle \frac{\pi}{3}

3.

r^2 = 6r \cos \theta

Multiply by r.

x^2 + y^2 = 6x

Replace r squared with x squared plus y squared, and r cos θ with x.

x^2 - 6x + 9 + y^2 = 9

Complete the square.

(x - 3)^2 + y^2 = 9

Factor the perfect square.

Answer: a circle through the pole, center (3, 0), radius 3

4.

n = 3 is odd, so there are 3 petals
Petal length is \lvert a \rvert = 4
Answer: 3 petals, each 4 long

5.

n = 4 is even, so there are 2n = 2 \cdot 4 = 8 petals
Answer: 8 petals, each 2 long

6.

a = b = 3, so it is a cardioid, symmetric about \theta = \frac{\pi}{2}

r = 3 - 3(-1) = 6

r is largest when \sin \theta = -1, at \theta = \frac{3\pi}{2}.

Answer: a cardioid with largest r = 6 at \theta = \frac{3\pi}{2}

7.

\lvert a \rvert = 2 \lt \lvert b \rvert = 4, so it is a limaçon with an inner loop

2 + 4 \cos \theta = 0

Set r = 0.

\cos \theta = -\frac{1}{2}

Subtract 2 and divide by 4.

Answer: an inner-loop limaçon through the pole at \theta = \frac{2\pi}{3} and \frac{4\pi}{3}

8.

\lvert a \rvert = 4 \ge 2\lvert b \rvert = 4, so it is a limaçon with no dimple
Largest: \theta = 0 gives r = 4 + 2(1) = 6
Smallest: \theta = \pi gives r = 4 + 2(-1) = 2
Answer: a limaçon with no dimple, r from 2 to 6

9.

\theta = \frac{\pi}{12}: r = 2 \sin \frac{\pi}{6} = 2 \cdot \frac{1}{2} = 1
\theta = \frac{\pi}{4}: r = 2 \sin \frac{\pi}{2} = 2
\theta = \frac{5\pi}{12}: r = 2 \sin \frac{5\pi}{6} = 1
\theta = \frac{\pi}{2}: r = 2 \sin \pi = 0
Answer: r = 1, 2, 1, 0, one petal in Quadrant I

10.

r = 4 + 4(-\frac{1}{2}) = 4 - 2 = 2

Substitute \frac{2\pi}{3}. The distance is in feet.

r = 4 + 4 \cdot \frac{\sqrt{2}}{2} = 4 + 2\sqrt{2} \approx 6.83

Substitute \frac{\pi}{4}. The distance is in feet.

Answer: 2 ft and about 6.83 ft

FAQ

How do you graph a polar equation?

Make a table of \theta and r values, plot each point on a polar grid, and connect them in order. Knowing the basic shapes helps you check the result.

How do you know how many petals a rose curve has?

For r = a \cos n\theta or r = a \sin n\theta, there are n petals when n is odd and 2n petals when n is even.

What is the difference between a cardioid and a limaçon?

Both have the form r = a + b \cos \theta or r = a + b \sin \theta. A cardioid is the special limaçon with \lvert a \rvert = \lvert b \rvert, so it just touches the pole.

What does a negative r mean on a polar graph?

It means plot the point in the opposite direction from the angle. That is how a rose gets petals on the other side and how a limaçon gets its inner loop.

Summary

  • Graph by making a table of \theta and r, plotting, and connecting in order.
  • r = a is a circle; \theta = \alpha is a line; r = a \cos \theta and r = a \sin \theta are circles through the pole.
  • Roses r = a \cos n\theta: n petals if n is odd, 2n if even.
  • Cardioids have \lvert a \rvert = \lvert b \rvert; limaçons with \lvert a \rvert \lt \lvert b \rvert have an inner loop.
  • Cosine graphs are symmetric about the polar axis; sine graphs about \theta = \frac{\pi}{2}.

Previous lesson: Polar Coordinates

Next lesson: Trigonometric Form of Complex Numbers

Back to the Trigonometry Index