Unit 3 · Topic 3.14

Polar Function Graphs

A polar function r = f(θ) assigns a signed distance to every direction.

Practice 2: Multiple Representations

Part 1 · Hook

Why this matters

Look at the pickup pattern printed on a microphone box. It is drawn as a polar graph: in each direction, the distance from the center shows how well the microphone hears. The heart-shaped one is a cardioid, and its equation is a polar function, r=1+cos⁡θr = 1 + \cos\theta.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Where is the polar point (−2,π4)\left(-2, \frac{\pi}{4}\right)?

  1. Quadrant III
  2. Quadrant I
  3. Quadrant II
  4. Quadrant IV
Show the answer

The angle π4\frac{\pi}{4} points into Quadrant I, and r=−2r = -2 moves backward through the pole into Quadrant III.

  • Correct: Quadrant III:
  • Quadrant I:
  • Quadrant II:
  • Quadrant IV:

2. What is the period of y=3sin⁡(2x)y = 3\sin(2x)?

  1. π\pi
  2. 2π2\pi
  3. 33
  4. π2\frac{\pi}{2}
Show the answer

The period of sin⁡(bx)\sin(bx) is 2πb=π\frac{2\pi}{b} = \pi.

  • Correct: π\pi:
  • 2π2\pi:
  • 33:
  • π2\frac{\pi}{2}:

Part 4 · See it

See it first

The polar graph of r = 3 sin(2 theta): a rose with four petals of length 3, centered on the rays pi/4, 3 pi/4, 5 pi/4 and 7 pi/4, each starting and ending at the pole.
The rose r=3sin⁡(2θ)r = 3\sin(2\theta): four petals, each reaching distance 33 from the pole. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Let θ\theta sweep from 00 to 2π2\pithe input of r=f(θ)r = f(\theta) is the direction
  2. Read rr from the graph of rr against θ\thetathe output is the signed distance along that direction
  3. rr is 00the curve passes through the pole
  4. rr is negativethe point is drawn on the opposite ray, in the opposite quadrant

Part 6 · Key ideas

Key ideas

  • A polar function r=f(θ)r = f(\theta) gives a distance for each direction; the curve is all points (f(θ),θ)(f(\theta), \theta).
  • Each hump of the graph of rr against θ\theta draws one loop or petal; zeros of rr are visits to the pole.
  • r=acos⁡(nθ)r = a\cos(n\theta) and r=asin⁡(nθ)r = a\sin(n\theta) are roses: nn petals if nn is odd, 2n2n if nn is even.
  • r=a+bcos⁡θr = a + b\cos\theta is a limaçon: an inner loop if |a|<|b||a| < |b|, a cardioid if |a|=|b||a| = |b|, no loop if |a|>|b||a| > |b|.

Part 7 · Misconception

A common mistake

The wrong idea: Where rr is negative, the curve is drawn in the quadrant the angle points to, just closer to the pole.

What actually happens: A negative rr sends the point backward through the pole, so it is drawn in the opposite quadrant. That is how the inner loop of r=1+2cos⁡θr = 1 + 2\cos\theta ends up to the right of the pole.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

A polar rose, r=3sin⁡(2θ)r = 3\sin(2\theta), and its graph of rr against θ\theta

The first graph shows r=3sin⁡(2θ)r = 3\sin(2\theta) as a function of θ\theta on rectangular axes, for 0≤θ≤2π0 \le \theta \le 2\pi. The second shows the same function drawn in the polar plane.

0π/2π3π/22π−3−2−1123θr
The polar graph of r = 3 sin(2 theta): a rose with four petals of length 3, centered on the rays pi/4, 3 pi/4, 5 pi/4 and 7 pi/4, each starting and ending at the pole.
LevlPrep original diagram.

1. How many petals does the polar curve r=3sin⁡(2θ)r = 3\sin(2\theta) have?

  1. 44
  2. 22
  3. 88
  4. 33
Show the answer

For r=asin⁡(nθ)r = a\sin(n\theta) with nn even, the rose has 2n2n petals: here 2⋅2=42\cdot 2 = 4. On the rr against θ\theta graph there are four humps on [0,2π][0, 2\pi], two above the axis and two below, and each hump draws one petal.

  • Correct: 44: Right: n=2n = 2 is even, so 2n=42n = 4 petals.
  • 22: Each hump of the rr graph draws a petal, and there are four humps on [0,2π][0, 2\pi], including the two where r<0r < 0.
  • 88: That doubles the count twice. Four humps on [0,2π][0, 2\pi] draw four petals, each once.
  • 33: An odd number of petals comes from an odd nn; here n=2n = 2.

2. For π2<θ<π\frac{\pi}{2} < \theta < \pi, in which quadrant are the points of the curve drawn?

  1. Quadrant IV
  2. Quadrant II
  3. Quadrant III
  4. Quadrant I
Show the answer

On (π2,π)\left(\frac{\pi}{2}, \pi\right) the rr graph is below the axis, so r<0r < 0. The angle points into Quadrant II, and a negative rr sends each point backward through the pole, into Quadrant IV. That petal is the one centered on the ray 7π4\frac{7\pi}{4}.

  • Correct: Quadrant IV: Right: negative rr with an angle in Quadrant II lands in Quadrant IV.
  • Quadrant II: That would need r>0r > 0; the rr graph is negative on this interval.
  • Quadrant III: The point opposite a Quadrant II direction is in Quadrant IV, not III.
  • Quadrant I: Quadrant I points come from 0<θ<π20 < \theta < \frac{\pi}{2}, where r>0r > 0.

Graph

The polar function r=1+2cos⁡θr = 1 + 2\cos\theta

The first graph shows r=1+2cos⁡θr = 1 + 2\cos\theta against θ\theta on rectangular axes, for 0≤θ≤2π0 \le \theta \le 2\pi. The second shows the same function in the polar plane.

0π/2π3π/22π−1123θr
The polar graph of r = 1 + 2 cos theta: a limacon with a large outer loop reaching farthest to the right and a small inner loop to the right of the pole.
LevlPrep original diagram.

3. For which θ\theta in [0,2π)[0, 2\pi) does the curve r=1+2cos⁡θr = 1 + 2\cos\theta pass through the pole?

  1. 2π3\frac{2\pi}{3} and 4π3\frac{4\pi}{3}
  2. π3\frac{\pi}{3} and 5π3\frac{5\pi}{3}
  3. π\pi
  4. π2\frac{\pi}{2} and 3π2\frac{3\pi}{2}
Show the answer

r=0r = 0 when cos⁡θ=−12\cos\theta = -\frac{1}{2}: θ=2π3\theta = \frac{2\pi}{3} and 4π3\frac{4\pi}{3}. On the rr graph these are where the curve crosses the θ\theta-axis.

  • Correct: 2π3\frac{2\pi}{3} and 4π3\frac{4\pi}{3}: Right: the two zeros of rr.
  • π3\frac{\pi}{3} and 5π3\frac{5\pi}{3}: These solve cos⁡θ=+12\cos\theta = +\frac{1}{2}, where r=2r = 2.
  • π\pi: At θ=π\theta = \pi, r=−1r = -1: the curve is 11 unit from the pole.
  • π2\frac{\pi}{2} and 3π2\frac{3\pi}{2}: At π2\frac{\pi}{2} and 3π2\frac{3\pi}{2}, r=1r = 1.

4. Which interval of θ\theta draws the small inner loop of r=1+2cos⁡θr = 1 + 2\cos\theta?

  1. (2π3,4π3)\left(\frac{2\pi}{3}, \frac{4\pi}{3}\right)
  2. (π2,3π2)\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)
  3. (0,2π3)\left(0, \frac{2\pi}{3}\right)
  4. (4π3,2π)\left(\frac{4\pi}{3}, 2\pi\right)
Show the answer

The inner loop is traced while r<0r < 0, between the two zeros: on (2π3,4π3)\left(\frac{2\pi}{3}, \frac{4\pi}{3}\right) the rr graph is below the axis. The angles point left, but negative rr puts the points to the right of the pole, inside the big loop.

  • Correct: (2π3,4π3)\left(\frac{2\pi}{3}, \frac{4\pi}{3}\right): Right: the interval where rr is negative.
  • (π2,3π2)\left(\frac{\pi}{2}, \frac{3\pi}{2}\right): r=1r = 1 at π2\frac{\pi}{2} and 3π2\frac{3\pi}{2}; rr is still positive at the ends of this interval.
  • (0,2π3)\left(0, \frac{2\pi}{3}\right): There r>0r > 0: that piece is the upper half of the big loop.
  • (4π3,2π)\left(\frac{4\pi}{3}, 2\pi\right): There r>0r > 0: that piece is the lower half of the big loop.

5. Which describes the graph of r=θr = \theta for 0≤θ≤4π0 \le \theta \le 4\pi?

  1. A spiral that winds outward, two full turns
  2. A circle of radius 4π4\pi
  3. A ray from the pole at angle 4π4\pi
  4. A rose with 44 petals
Show the answer

As θ\theta sweeps around, rr keeps growing: each point is farther from the pole than the last, so the curve winds outward. From 00 to 4π4\pi the angle makes two full turns.

  • Correct: A spiral that winds outward, two full turns: Right: growing distance with a turning angle makes a spiral.
  • A circle of radius 4π4\pi: A circle needs rr constant; here rr grows with θ\theta.
  • A ray from the pole at angle 4π4\pi: A ray needs θ\theta constant; here θ\theta changes.
  • A rose with 44 petals: A rose needs rr to rise and fall back to 00; here rr only grows.

Part 9 · Summary

Summary

A polar function r=f(θ)r = f(\theta) assigns a signed distance to every direction. Read it from its graph of rr against θ\theta: zeros are visits to the pole, negative values are drawn on the opposite ray, and each hump draws a petal or loop. Circles (r=ar = a, r=acos⁡θr = a\cos\theta), limaçons and cardioids (r=a+bcos⁡θr = a + b\cos\theta) and roses (r=asin⁡(nθ)r = a\sin(n\theta)) are the families to know.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic