Sketching Polar Curves

Curves traced by r as the angle turns.

Let r depend on θ

A polar curve gives r as a function of θ, written r = f(θ). For each direction θ, go out a distance r from the pole along that direction and mark the point. As θ 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 θ traces the circle of radius 2 about the pole. A circle about the origin, which needs x² + y² = 4 in Cartesian form, is just r = 2 in polar form.

A circle through the pole

Now let r change with θ: r = 2 cos θ. At θ = 0, r = 2. At π/6, r = √3 ≈ 1.73. At π/4, r = √2 ≈ 1.41. At π/3, r = 1. At π/2, r = 0, so the curve reaches the pole.

Below the initial line, cos(−θ) = cos θ, so the angles from 0 down to −π/2 give the mirror image. From θ = −π/2 to π/2 the curve is traced once.

Multiplying r = 2 cos θ by r gives r² = 2r cos θ, which is x² + y² = 2x, or (x − 1)² + y² = 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).

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The gold curve is r = 2 cos θ, traced from θ = −π/2 to π/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 θ = 0.

The cardioid

For r = 1 + cos θ, read r at the quarter turns: 2 at θ = 0, 1 at π/2, 0 at π, and 1 at 3π/2. Between them, r = 1.5 at π/3 and 0.5 at 2π/3.

The largest value, r = 2, is at θ = 0, where cos θ = 1. The curve reaches the pole only where cos θ = −1, at θ = π. There it comes in to a point, called a cusp.

Replacing θ by −θ leaves r unchanged, because cos(−θ) = cos θ, so the curve is symmetric about the initial line. Its heart shape gives it its name: a cardioid.

r = 1r = 2θ = 120°r = 0.5

the arm has length r = 1 + cos θ and lays the curve down as θ turns

Sweep θ to the cusp, where r = 0

An arm of length r = 1 + cos θ turns from θ = 0 and lays down the cardioid behind it. At θ = 120° 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 2θ, the angle inside the cosine turns twice as fast. r = 2 at θ = 0, falls to 0 at π/4, and is −2 at π/2.

A negative r means going the distance |r| in the opposite direction, θ + π. So at θ = π/2, r = −2 puts the point 2 below the pole, at (0, −2).

Over a full turn, r = 0 at θ = π/4, 3π/4, 5π/4 and 7π/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.

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r = 2 cos 2θ over a full turn. The dots are its four tips: (2, 0) at θ = 0, (0, −2) at θ = π/2 with r = −2, (−2, 0) at θ = π, and (0, 2) at θ = 3π/2 with r = −2.

A method for any curve

To sketch a polar curve, read r at the quarter turns θ = 0, π/2, π and 3π/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 θ turns.

Try it on r = 2 + sin θ. 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 θ = π/2.

Replacing θ by π − θ leaves sin θ 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.

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r = 2 + sin θ, 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 θ 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 θ = 0. There cos θ = 1 and r = 2, the farthest point. The cusp is at θ = π, where r = 0.

Counting the petals of r = 2 cos 2θ 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 = ½(1 + cos θ), a cardioid half the size of r = 1 + cos θ. 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 θ.

Worked example: A Cardioid Microphone on a Stage: The Direction It Does Not Hear, and Where the Monitor Speaker Can Stand

Question A cardioid microphone stands at the pole, pointing along the initial line. For a sound arriving from the direction θ, measured in radians counterclockwise from the initial line, the microphone's output is r = 12(1 + cos θ) times its output for the same sound arriving straight ahead, so the polar graph of r is the microphone's pick-up pattern. (a) Find the one direction from which the microphone picks up nothing, and show that the pattern is symmetric about the initial line. (b) The singer's monitor speaker must stand in a direction where the microphone's output is no more than a quarter of its output straight ahead. Find the range of directions that meet this condition, and the angle that range spans.

  1. 1.The output is zero where r = 0: 12(1 + cos θ) = 0 gives cos θ = −1, and over one full turn that happens only at θ = π. Everywhere else cos θ > −1, so r > 0.

    facingr = 1micr = 0 at θ = pir = 0 only where cos θ = −1, at θ = pi
    facingr = 1micr = 0 at θ = pir = 0 only where cos θ = −1, at θ = pi
    r = 0 needs cos θ = −1, which over one turn happens only at θ = π, directly behind the microphone.

  2. 2.(a) The microphone picks up nothing from θ = π, directly behind it. For the symmetry, replace θ by −θ: since cos(−θ) = cos θ, r(−θ) = r(θ), so the point of the pattern at angle −θ is the mirror image in the initial line of the point at angle θ.

    facingr = 1micr = 0 at θ = pipi/3−pi/3r = 0 only where cos θ = −1, at θ = pi(a) cos(−θ) = cos θ, so r(−θ) = r(θ)
    facingr = 1micr = 0 at θ = pipi/3−pi/3r = 0 only where cos θ = −1, at θ = pi(a) cos(−θ) = cos θ, so r(−θ) = r(θ)
    (a) cos(−θ) = cos θ, so the points at π3 and −π3 are mirror images in the initial line.

  3. 3.For the monitor the condition is 12(1 + cos θ) ≤ 14. Multiplying both sides by 2 gives 1 + cos θ ≤ 12, and subtracting 1 gives cos θ ≤ −12.

    facingr = 1pi/3−pi/3r = 1/4r = 0 only where cos θ = −1, at θ = pi(a) cos(−θ) = cos θ, so r(−θ) = r(θ)r at most 1/4 means cos θ at most −1/2
    facingr = 1pi/3−pi/3r = 1/4r = 0 only where cos θ = −1, at θ = pi(a) cos(−θ) = cos θ, so r(−θ) = r(θ)r at most 1/4 means cos θ at most −1/2
    The monitor needs 12(1 + cos θ) ≤ 14: inside the dashed circle r = 14, which is cos θ ≤ −12.

  4. 4.Over one turn, cos θ = −12 at θ = 2π3 and at θ = 4π3. Between those two angles the cosine is smaller still, reaching −1 at θ = π, and outside them it is larger. So the condition holds for 2π3 ≤ θ ≤ 4π3.

    facingr = 1pi/3−pi/32pi/34pi/3r = 0 only where cos θ = −1, at θ = pi(a) cos(−θ) = cos θ, so r(−θ) = r(θ)r at most 1/4 means cos θ at most −1/2cos θ = −1/2 at 2pi/3 and at 4pi/3
    facingr = 1pi/3−pi/32pi/34pi/3r = 0 only where cos θ = −1, at θ = pi(a) cos(−θ) = cos θ, so r(−θ) = r(θ)r at most 1/4 means cos θ at most −1/2cos θ = −1/2 at 2pi/3 and at 4pi/3
    cos θ = −12 at 2π3 and 4π3, and it is smaller between them: the shaded band of directions.

  5. 5.(b) The monitor can stand in any direction from θ = 2π3 round to θ = 4π3, a range spanning 4π3 − 2π3 = 2π3 radians, which is 120°. Check: at θ = 2π3, r = 12(1 − 12) = 14 exactly, and by the symmetry of part (a) the range is centered on θ = π.

    facingr = 1pi/3−pi/32pi/34pi/3120 degr = 0 only where cos θ = −1, at θ = pi(a) cos(−θ) = cos θ, so r(−θ) = r(θ)r at most 1/4 means cos θ at most −1/2cos θ = −1/2 at 2pi/3 and at 4pi/3(b) span 2pi/3 radians = 120 deg
    facingr = 1pi/3−pi/32pi/34pi/3120 degr = 0 only where cos θ = −1, at θ = pi(a) cos(−θ) = cos θ, so r(−θ) = r(θ)r at most 1/4 means cos θ at most −1/2cos θ = −1/2 at 2pi/3 and at 4pi/3(b) span 2pi/3 radians = 120 deg
    (b) The band runs from 2π3 to 4π3, spanning 2π3 radians, or 120°, centered on the null.

Answer: (a) θ = π (about 3.142 radians), directly behind the microphone; the pattern is symmetric about the initial line because cos(−θ) = cos θ, so r(−θ) = r(θ); (b) 2π/3 ≤ θ ≤ 4π/3 (from about 2.094 to 4.189 radians), a range spanning 2π/3 radians, or 120°

Common mistakes

  • Solving cos θ ≤ −12 as θ ≤ 2π3, as if the inequality could be undone like an equation. That range points toward the front of the microphone, where the output is at its greatest. The cosine falls from 1 to −1 and rises again over one turn, so the angles where it is at most −12 form one band around θ = π.
  • Stopping at θ = π and giving the span as π − 2π3 = π3 radians. The pattern is symmetric about the initial line, so the band carries on past θ = π to θ = 4π3 on the other side, and spans 2π3 radians.

More polar curves problems, worked step by step →

Practice Sketching Polar Curves in the app