AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/3/3-14)
Unit 3 · Topic 3.14
3.14 Polar Function Graphs
A polar function r = f(θ) gives a radius for each angle, and plotting the points (f(θ), θ) traces curves like circles, roses and limaçons. Graphing r against θ on ordinary axes first makes the polar graph much easier to sketch.
Key terms
- polar function
- rose curve
- limaçon
- cardioid
- signed radius
How a polar graph is built
For a polar function r = f(θ), the input is an angle and the output is a signed radius. Each pair gives the point (f(θ), θ) in polar coordinates.
As θ increases, imagine a ray sweeping counterclockwise from the polar axis. At each angle, mark the point that is r units out along the ray, or |r| units the opposite way if r is negative.
Changes in θ mean changes in direction. Changes in r mean changes in signed distance from the origin.
Use the rectangular graph as a guide
Sketch y = f(θ) with θ on the horizontal axis first. It shows at a glance where r is positive, negative, zero, largest and smallest.
Where r = 0, the polar curve passes through the origin. Where r is negative, that part of the polar curve appears on the opposite side of the origin. Where |r| is largest, the curve is farthest from the origin.
To draw only part of a curve, restrict θ to the interval whose angles and radii match that part.
Common polar curves
Here a and b are nonzero constants and n is a positive integer greater than 1.
| Equation | Shape | Notes |
|---|---|---|
| r = a | Circle centered at the origin | Radius equal to the absolute value of a |
| r = a cos θ or r = a sin θ | Circle through the origin | Diameter equal to the absolute value of a |
| r = a cos(nθ) or r = a sin(nθ) | Rose | n petals if n is odd, 2n if even; petal length equal to the absolute value of a |
| r = a + b cos θ or a + b sin θ | Limaçon | Shape depends on how the sizes of a and b compare |
Limaçons and symmetry
A limaçon r = a + b cos θ (or a + b sin θ) changes shape depending on how the sizes of a and b compare. If the size of a is less than the size of b, r becomes negative for some angles and the curve has a small inner loop. If they are equal, the curve is a cardioid that just touches the origin. If the size of a is greater, r is never 0 and there's no loop.
Circles and limaçons written with cos θ are symmetric over the polar axis (the x-axis), because cos(−θ) = cos θ. Those written with sin θ are symmetric over the y-axis, because sin(π − θ) = sin θ.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
A cardioid
Sketch r = 2 + 2 cos θ for 0 ≤ θ ≤ 2π by plotting key points.
Show the solutionHide the solution
- Step 1: θ = 0: r = 4, the point (4, 0) in rectangular coordinates.
- Step 2: θ = π/2: r = 2, the point (0, 2).
- Step 3: θ = π: r = 0, the origin.
- Step 4: θ = 3π/2: r = 2, the point (0, −2). Then θ = 2π returns to (4, 0).
- Step 5: r is never negative and touches 0 only at θ = π. Here |a| = |b| = 2, so this is a cardioid: a heart shape whose pointed tip (the cusp) is at the origin, with the rounded part reaching right as far as (4, 0).
Answer: A cardioid through (4, 0), (0, 2), the origin and (0, −2), symmetric over the x-axis.
- Example 2
A four-petal rose
Describe the graph of r = 3 sin(2θ). Which values of θ trace the petal in quadrant I?
Show the solutionHide the solution
- Step 1: n = 2 is even, so there are 2n = 4 petals, each 3 units long.
- Step 2: On 0 ≤ θ ≤ π/2, 2θ runs from 0 to π, so r = 3 sin(2θ) ≥ 0. The curve leaves the origin, reaches r = 3 at θ = π/4, and returns to the origin at θ = π/2. That's the quadrant I petal.
- Step 3: On π/2 < θ < π, r is negative, so those points are plotted in the opposite direction. That traces the petal in quadrant IV, not quadrant II.
Answer: A four-petal rose with petals of length 3; the quadrant I petal is traced for 0 ≤ θ ≤ π/2, with its tip at (3, π/4).
- Example 3
Trap: where the inner loop comes from
For r = 1 + 2 cos θ, where is r negative on [0, 2π], and what does that part of the graph look like?
Show the solutionHide the solution
- Step 1: r < 0 when cos θ < −1/2, which is 2π/3 < θ < 4π/3.
- Step 2: On that interval the points are plotted opposite the direction of θ. For example, at θ = π, r = −1, which is the point (1, 0), to the right of the origin, even though θ = π points left.
- Step 3: These points form a small inner loop inside the big loop. Here |a| = 1 < |b| = 2, which matches the rule for a limaçon with an inner loop.
Answer: r is negative for 2π/3 < θ < 4π/3; that part forms the inner loop, plotted on the opposite side of the origin.
Common mistakes
- Plotting a negative r in the direction of θ. Go the opposite way.
- Counting rose petals wrong. An odd n gives n petals; an even n gives 2n.
- Skipping the rectangular graph of r against θ. It's the quickest way to see where r is zero, negative or largest.
On the exam
- Expect to match a polar equation to its graph, or to say which interval of θ traces a certain part of a polar curve.
- Questions often give the graph of r = f(θ) on rectangular axes and ask about the polar curve. Read off the zeros and sign changes first.
Connected topics
Videos
Check yourself
4 questions on 3.14 Polar Function Graphs. Pick an answer to see if you got it, and why.
Which of the following describes the graph of the polar function r = 3 sin(2θ) for 0 ≤ θ ≤ 2π?
For the polar function r = 2 − 2 cos θ, what is the greatest distance from the origin of any point on the graph, and at what value of θ in [0, 2π) does it occur?
Which of the following describes the graph of the polar function r = 4 cos θ?
The polar function r = f(θ) is given by f(θ) = 1 + 2 cos θ for 0 ≤ θ ≤ 2π.
Which of the following best describes the graph of r = f(θ) in the polar coordinate system?
0 of 4 answered