Graphing Rose Curves, Limaçons & Lemniscates

Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

Plotting polar equations is not difficult but does require some special considerations. This lesson explores how to plot three of the more spectacular polar equations: rose curves, limacons and lemniscates.

Plotting Polar Equations

Creating a table of x and y values is a tried-and-true curve plotting strategy. Additional strategies include symmetry testing, locating maximum/minimum values and recognizing curve types. In this lesson, we use these methods to plot graphs that sound like creatures of the sea: rose curves, limaçons and lemniscates.

Useful Ideas

Before diving in with polar equations, let's visit with polar graphs and symmetry.

Polar Graphs

Some polar curves have symmetry lines. These lines act like mirrors. Points in front of the 'mirror' are imaged behind the mirror. The symmetry lines are the polar axis and the θ = π/2 line. Symmetry is also possible about the pole.

Important polar features

The pole is equivalent to the origin in rectangular coordinates. The ray starting at the pole and extending forever to the right is the polar axis corresponding to the positive x-axis. The θ = π/2 line corresponds to the y-axis.

The distance from the pole to a point in the plane is the radius r. The angle θ is the counter-clockwise rotation referenced to the polar axis. Knowing r and θ locates any point in the plane.


The number of tabulated points is reduced if there is symmetry about the polar axis, the θ = π/2 line and/or the pole. Symmetry exists:

  • about the polar axis if (r, θ) = (r, -θ) or (-r, π - θ)
  • about the θ = π/2 line if (r, θ) = (-r, -θ) or (r, π - θ)
  • about the pole if (r, θ) = (-r, θ) or (r, θ + π)

There are two tests for each type of symmetry. If one algebraic test passes, symmetry exists. However, even if both tests fail, don't rule out symmetry. Sometimes an algebraic symmetry test will not pass, yet the curve when plotted is symmetric.

Plotting Rose Curves

Rose curves resemble flowers:

12-petal rose curve

Rose curve equations have two forms: r = a cos(nθ) and r = a sin(nθ) where a ≠ 0 and n is a positive integer. Petals have length determined by a. If n is odd, the number of petals is n. However, if n is even, the number of petals is 2n. By the way, the equation for the 12-petal rose curve is r = 2 cos(6θ).

In the cosine form, r maximum occurs for nθ = 0, meaning a petal exists at θ = 0. For sine, maximum r is at θ = π/2. Thus, nθ = π/2, and a petal appears at θ = π/(2n). In both cases, the separation between petals is 360o divided by the number of petals.

Example: plot r = 2 cos(3θ)

  • a is 2 and n = 3
  • n is odd, so there are n = 3 petals
  • the separation between the petals is 360o/(number of petals) = 360o/3 = 120o.

For cosine, the first petal is at θ = 0. Is the curve symmetric about the polar axis? The test for symmetry is (r, θ) = (r, -θ) or (-r, π - θ).

(r, -θ) means r = 2 cos(3(-θ)) = 2 cos(-3θ) = 2 cos(θ) = (r, θ). Test passed. Plotting the known petal locations:

Known points

Drawing the rest of the curve:

Drawing through the points

A very nice sea urchin.

Plotting Limaçons

French for ''snail'' is ''limaçon''. Okay, not a sea creature but still a creature. The limaçon equations are r = a ± b cos θ and r = a ± b sin θ, where a > 0 and b > 0. Also, the sign between a and b can tell us were most of the limaçon will be with respect to the polar (horizontal) axis and the π/2 line (vertical axis):

Sign between a and b Type of limaçon Location
negative limaçon with cosine to the left of the vertical axis
positive limaçon with cosine to the right of the vertical axis
negative limaçon with sine below the horizontal axis
positive limaçon with sine above the horizontal axis

Let's explore the four general shapes that a limaçon can have, using the form r = a + bsin(θ):

I. Inner Loop: a/b is < 1

e.g., r = 2 + 3sin(θ) has a = 2, b = 3 and a/b = 2/3 < 1:

Inner loop

II. Cardioid (heart): a/b = 1

e.g., r = 2 + 2sin(θ) has a = b = 2 and a/b = 2/2 = 1:


III. Dimple: 1 < a/b < 2

e.g., r = 3 + 2sin(θ) has a = 3 and b = 2. Thus, a/b = 3/2 (between 1 and 2):


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