AP Precalculus: Polar Functions

The average rate of change of a polar function r = f(θ) over an interval [θ₁, θ₂] is (f(θ₂) − f(θ₁)) / (θ₂ − θ₁), the change in the radius divided by the change in the angle. Its units are radius units per radian. Geometrically it is the slope of the secant line on the θr-graph. It describes how quickly the radius changes as the point turns, not how fast the point moves in the plane. A positive average rate of change means r increased over the interval, a negative one means r decreased. Remember that r itself can be negative: a negative rate of change with r positive means the point is moving closer to the pole, but with r negative the same sign would mean the point is getting farther from the pole (more negative). Always use the exact value of r at the endpoints, and keep angles in radians.

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1. Polar Coordinates and Plotting Points

Rectangular coordinates locate a point by how far to go right and up. Polar coordinates locate it by turn, then walk. A fixed point called the pole (the origin) and a ray from it called the polar axis (the positive x-axis) set up the system. A point is written (r, θ), where θ is a directed angle measured from the polar axis and |r| is a distance along the terminal side. A positive θ turns counterclockwise and a negative θ turns clockwise. If r is positive, the point is r units out along the terminal side of θ. If r is negative, first picture the terminal side, then go |r| units in the opposite direction, straight through the pole. Unlike rectangular coordinates, a point has many polar names: you can add any multiple of 2π to θ, and you can switch the sign of r while adding or subtracting π to θ. So (r, θ), (r, θ + 2kπ), and (−r, θ + π + 2kπ) all name the same point, where k is any integer. The pole itself is (0, θ) for every θ.

Worked Example: Plot A(4, 2π/3), B(3, −π/4), and C(−2, π/3). Then give each one a name with r > 0 and 0 ≤ θ < 2π.

A: θ = 2π/3 is a counterclockwise turn of 120° (Quadrant II). Walk 4 units out. It already has r > 0 and θ in range, so A = (4, 2π/3).

B: θ = −π/4 turns 45° clockwise (Quadrant IV). Walk 3 units. To get a positive angle, add 2π: −π/4 + 2π = 7π/4, so B = (3, 7π/4).

C: r is negative. Draw the terminal side of π/3, then walk 2 units the opposite way, which lands in Quadrant III. To make r positive, add π to the angle: π/3 + π = 4π/3, so C = (2, 4π/3).

Three points on a polar grid. For C(−2, π/3), the red segment shows the terminal side of π/3, and the point is plotted 2 units in the opposite direction, at the same place as (2, 4π/3).

1. Which point is the same as P(−3, π/4)?

Explanation: A negative r reverses direction, which is the same as adding π to the angle: (−3, π/4) = (3, π/4 + π) = (3, 5π/4). Choice (3, π/4) is the point on the opposite side of the pole.

2. Which of these is NOT another name for (2, π/6)?

Explanation: 13π/6 = π/6 + 2π and −11π/6 = π/6 − 2π, which are full turns. (−2, 7π/6) works because the angle changed by π while r changed sign. (−2, π/6) only flips r and does not change the angle, so it is a different point.

3. Where is the point (4, −π/2)?

Explanation: A clockwise quarter turn from the polar axis points straight down. With r = 4 > 0, the point is 4 units down.

4. Write (−5, π/3) with r > 0 and 0 ≤ θ < 2π.

Explanation: Make r positive by adding π to the angle: π/3 + π = 4π/3. The result is in range, so (5, 4π/3).

2. Converting Between Polar and Rectangular Coordinates

Put the pole on the origin and the polar axis on the positive x-axis, and the two systems describe the same plane. A right triangle with hypotenuse r and angle θ gives x = r cos θ and y = r sin θ for polar-to-rectangular. These formulas work for any r, positive or negative, so there is nothing special to remember for negative r. Going the other way, r = √(x² + y²) (taking r > 0) and θ satisfies tan θ = y/x. The arctangent only returns angles in (−π/2, π/2), so you must place θ in the right quadrant. Start from the reference angle α = arctan(|y|/|x|): for x > 0, y > 0 use θ = α; for x < 0, y > 0 use θ = π − α; for x < 0, y < 0 use θ = π + α; for x > 0, y < 0 use θ = 2π − α. Points on an axis have quadrantal angles (0, π/2, π, 3π/2), so check the signs of x and y before using a formula. Always sketch the point to confirm the quadrant.

Worked Example: (a) Convert (6, 2π/3) to rectangular. (b) Convert (−2, 3π/4) to rectangular. (c) Convert (−2√3, 2) to polar with r > 0 and 0 ≤ θ < 2π. (d) Convert (0, −7) to polar.

(a) x = 6 cos(2π/3) = 6(−1/2) = −3, and y = 6 sin(2π/3) = 6(√3/2) = 3√3. The point is (−3, 3√3).

(b) x = −2 cos(3π/4) = −2(−√2/2) = √2, and y = −2 sin(3π/4) = −2(√2/2) = −√2. The point is (√2, −√2).

(c) r = √(12 + 4) = 4. The point is in Quadrant II, and the reference angle is arctan(2/(2√3)) = arctan(1/√3) = π/6, so θ = π − π/6 = 5π/6. The polar form is (4, 5π/6).

(d) r = 7. The point is on the negative y-axis, a quadrantal angle, so θ = 3π/2. The polar form is (7, 3π/2).

The point (6, 2π/3) has x = 6 cos(2π/3) = −3 and y = 6 sin(2π/3) = 3√3 ≈ 5.20. The dashed lines are the legs of the right triangle whose hypotenuse is r = 6.

1. Convert (8, π/3) to rectangular coordinates.

Explanation: x = 8 cos(π/3) = 8(1/2) = 4 and y = 8 sin(π/3) = 8(√3/2) = 4√3. Be careful not to swap cosine and sine.

2. Convert (−2, π/2) to rectangular coordinates.

Explanation: x = −2 cos(π/2) = 0 and y = −2 sin(π/2) = −2. A negative r sends the point to the opposite side of the pole, 2 units below it.

3. Convert (−3, −3) to polar with r > 0 and 0 ≤ θ < 2π.

Explanation: r = √(9 + 9) = 3√2. The point is in Quadrant III with reference angle arctan(1) = π/4, so θ = π + π/4 = 5π/4. Arctan alone gives π/4, which is the wrong quadrant.

4. Convert (√3, 1) to polar with r > 0 and 0 ≤ θ < 2π.

Explanation: r = √(3 + 1) = 2. In Quadrant I, θ = arctan(1/√3) = π/6.

3. Polar Form of a Complex Number

A complex number a + bi is plotted as the point (a, b) in the complex plane, with the real part a on the horizontal axis and the imaginary part b on the vertical axis. Because every point has polar coordinates, a + bi can be written using a distance and an angle. From a = r cos θ and b = r sin θ we get the polar (trigonometric) form z = r(cos θ + i sin θ), sometimes written r cis θ. Here r = √(a² + b²) is the modulus of z and θ is the argument, found exactly as in the previous section with tan θ = b/a and the quadrant taken from the signs of a and b. The argument is not unique, since adding 2π gives the same number, so unless told otherwise give θ in [0, 2π). To go back, evaluate cos θ and sin θ and distribute r. Every complex number has a polar form. Even z = 0 has modulus 0, and any angle works because 0 · (cos θ + i sin θ) = 0; the angle only becomes meaningful when r > 0. Numbers on the axes use quadrantal angles: positive real → 0, positive imaginary → π/2, negative real → π, negative imaginary → 3π/2.

Worked Example: (a) Write 1 − √3 i in polar form. (b) Write −4i in polar form. (c) Write 6(cos π/3 + i sin π/3) as a + bi. (d) Write k + ki, where k > 0, in polar form.

(a) a = 1 and b = −√3, so r = √(1 + 3) = 2. The point is in Quadrant IV, with reference angle arctan(√3) = π/3, so θ = 2π − π/3 = 5π/3. Then z = 2(cos 5π/3 + i sin 5π/3).

(b) a = 0 and b = −4, so r = 4. The point is on the negative imaginary axis, so θ = 3π/2 and z = 4(cos 3π/2 + i sin 3π/2).

(d) r = √(k² + k²) = k√2, and tan θ = k/k = 1 in Quadrant I gives θ = π/4. So k + ki = k√2 (cos π/4 + i sin π/4).

The complex number 1 − √3 i is the point (1, −√3). Its distance from the origin (the modulus) is 2, and its angle (the argument) is 5π/3. The real axis is the horizontal axis and the imaginary axis is the vertical axis.

1. Write −3i in polar form.

Explanation: The number is on the negative imaginary axis, so r = 3 and θ = 3π/2. Check: cos 3π/2 = 0 and sin 3π/2 = −1, so 3(0 − i) = −3i. The choice with π/2 would give +3i.

2. Write 2(cos π/6 + i sin π/6) in the form a + bi.

Explanation: cos π/6 = √3/2 and sin π/6 = 1/2. Multiply by 2: √3 + 1i. Do not forget to multiply both parts by the modulus.

3. For k > 0, which is the polar form of k + ki?

Explanation: The modulus is √(k² + k²) = k√2, not k. The angle satisfies tan θ = 1 in Quadrant I, so θ = π/4.

4. Write −1 + √3 i in polar form.

Explanation: r = √(1 + 3) = 2. The point is in Quadrant II with reference angle arctan(√3) = π/3, so θ = π − π/3 = 2π/3.

4. Graphing Basic Polar Equations

A polar graph of r = f(θ) is the set of all points whose polar coordinates satisfy the equation. The input θ is an angle and the output r is a distance, so changes in the input turn the point around the pole while changes in the output move it toward or away from the pole. Two simple equations show this at once. r = k (a constant) keeps the distance fixed while the angle is free: it is a circle centered at the pole with radius |k|. θ = c (a constant angle) keeps the direction fixed while r is free: because r can be negative, it is the full line through the pole making angle c with the polar axis, with slope tan c. If the domain is restricted to r ≥ 0, only the ray in the direction c is drawn. Next comes r = aθ, a spiral: as θ grows the radius grows too. Any restriction on θ or r chosen in the problem must be honored, so read the domain before you draw. A handy plan for other equations is to build a table of (θ, r) pairs and plot them, but a table alone can mislead; the next section shows a better method.

Worked Example: Describe and graph (a) r = 3, (b) θ = π/6, (c) r = θ for 0 ≤ θ ≤ 2π.

(a) Every point is 3 units from the pole, so the graph is a circle with center at the pole and radius 3. In rectangular form, x² + y² = 9.

(b) Every point lies on the line making an angle of π/6 with the polar axis. For r > 0 the points are in Quadrant I along that direction, and for r < 0 they are on the extension in Quadrant III. The graph is the line y = (tan π/6)x = x/√3 through the origin.

(c) At θ = 0, r = 0 (the pole). At θ = π/2, r ≈ 1.57; at θ = π, r ≈ 3.14; at θ = 2π, r ≈ 6.28. The radius grows steadily while the point turns counterclockwise: an outward spiral.

The graph of r = 3 is a circle of radius 3 around the pole. The graph of r = θ for 0 ≤ θ ≤ 2π is a spiral that starts at the pole and widens counterclockwise as the radius grows to about 6.28.

For r = θ with θ < 0, the angle turns clockwise and r is negative, so each point lands on the opposite side. The result is the mirror image of the first spiral across the y-axis.

1. What is the graph of r = 5?

Explanation: r is constant, so every point is 5 units from the pole, whatever the angle. That is a circle of radius 5 centered at the pole.

2. What is the graph of θ = π/4 with r ≥ 0?

Explanation: The angle is fixed at 45°. Restricting r ≥ 0 keeps only the ray starting at the pole and heading into Quadrant I. Without the restriction, negative r would add the opposite ray and the whole line y = x.

3. What does the graph of r = θ for θ ≥ 0 look like?

Explanation: As θ increases, the point turns counterclockwise, and r = θ also increases, so the distance from the pole grows. That produces an outward spiral.

4. The point (2, π/3) lies on which of these graphs? r = 2; θ = π/3; r = θ

Explanation: The point has r = 2 and θ = π/3, so it satisfies both r = 2 and θ = π/3. For r = θ, we would need r = π/3 ≈ 1.05, but r = 2, so it is not on that spiral.

5. Using the θr-Plane to Graph Polar Curves

A table of values often produces repeated points and hides how the curve moves. A stronger method uses the ordinary graph of r against θ as a guide. Step 1: graph r = f(θ) on a regular coordinate plane with θ on the horizontal axis (the θr-plane). Step 2: read it as instructions: the height of the graph at θ is how far to walk, and θ is the direction to face. Move along the guide from left to right, one interval at a time, and trace the matching piece in the polar plane. When the guide is positive the point is on the terminal side of θ. When it is negative the point is on the opposite side, so a stretch that would be in Quadrant II is actually drawn in Quadrant IV. When r = 0, the curve passes through the pole. The curve begins to retrace itself once the pairs (|r|, direction) repeat, which tells you the interval that gives the whole graph. The equations r = a sin θ and r = a cos θ are circles of diameter |a|: sine circles lie on the y-axis (center (0, a/2)) and cosine circles lie on the x-axis (center (a/2, 0)). Only half a turn (0 ≤ θ ≤ π) is needed to draw them.

Worked Example: Graph r = 4 sin θ for 0 ≤ θ ≤ 2π.

Make a table: θ = 0 gives r = 0; π/4 gives 2√2 ≈ 2.83; π/2 gives 4; 3π/4 gives 2√2; π gives 0; 3π/2 gives −4. In the polar plane there are only a few distinct points, so use the guide graph (below) to see the motion.

From 0 to π/2, r grows from 0 to 4, so the curve leaves the pole along the polar axis direction and swings counterclockwise to the point (4, π/2) on the positive y-axis. From π/2 to π, r shrinks back to 0, so the curve continues around through Quadrant II and returns to the pole. From π to 2π, r is negative, so each point is plotted on the opposite side of its terminal ray, which is exactly the same circle again. The graph is a circle of diameter 4 centered at (0, 2).

Step 1: graph r = 4 sin θ as an ordinary function of θ (horizontal axis) with the output r on the vertical axis. It rises to r = 4 at π/2, returns to 0 at π, and is negative from π to 2π.

Step 2: use the heights of the first graph as radii. As θ goes from 0 to π, r rises 0 → 4 → 0 and the point sweeps out the full circle of radius 2 centered at (0, 2). For π < θ < 2π, r is negative and the point retraces the same circle.

1. The graph of r = 4 cos θ is

Explanation: Cosine circles lie on the x-axis. The diameter is |a| = 4 along the polar axis, so the center is (2, 0) and the radius is 2.

2. For r = 4 sin θ, which interval of θ draws the complete circle exactly once?

Explanation: On [0, π], r goes 0 → 4 → 0, covering the whole circle. On [π, 2π] r is negative and the same circle is traced a second time.

3. For r = 6 sin θ, the point at θ = 7π/6 is the same as which point?

Explanation: r = 6 sin(7π/6) = 6(−1/2) = −3, so the point is (−3, 7π/6). Adding π to the angle and flipping the sign gives (3, 7π/6 − π) = (3, π/6).

4. What is the greatest distance from the pole of any point on r = 5 cos θ?

Explanation: |cos θ| ≤ 1, so |r| ≤ 5. The maximum distance 5 occurs at θ = 0 (and θ = π, where r = −5 gives the same point).

6. Families of Polar Curves: Circles, Cardioids, Limaçons, and Roses

Certain polar equations keep producing the same shapes, and recognizing them saves time. Circles: r = a sin θ or r = a cos θ. Cardioids: r = a(1 ± sin θ) or r = a(1 ± cos θ), heart-shaped with a cusp at the pole; the farthest point is 2a from the pole. Limaçons: r = a ± b sin θ or r = a ± b cos θ with a, b > 0 and a ≠ b. Compare a and b: if a < b there is an inner loop (and r becomes negative for part of the turn); if b < a < 2b the curve is a dimpled limaçon with a flattened side; if a ≥ 2b it is convex (an oval). Roses: r = a sin nθ or r = a cos nθ (n ≥ 2) have petals of length |a|. If n is odd there are n petals; if n is even there are 2n petals. Sine versions are symmetric about the y-axis (the first petal points along the y-axis for odd n), and cosine versions are symmetric about the x-axis. Sine versions of the other shapes point up or down, while cosine versions point right or left, and a minus sign flips the curve. The maximum |r| in each case is the largest value the formula can reach.

Worked Example: Identify each curve: (a) r = 3 + 3 cos θ, (b) r = 1 + 2 cos θ, (c) r = 3 + 2 sin θ, (d) r = 4 sin 3θ, (e) r = 3 cos 2θ.

(a) The form a(1 + cos θ) with a = 3: a cardioid opening to the left with its cusp at the pole; the farthest point is 6 units right at θ = 0.

(b) a = 1 < b = 2: a limaçon with an inner loop. The inner loop appears where r < 0, which happens when cos θ < −1/2.

(d) n = 3 is odd, so there are 3 petals of length 4.

(e) n = 2 is even, so there are 2 × 2 = 4 petals of length 3, with petal tips on the x- and y-axes.

Cardioid: r = 3 + 3 cos θ

Limaçon with an inner loop: r = 1 + 2 cos θ

Rose with 3 petals: r = 4 sin 3θ

Rose with 4 petals: r = 3 cos 2θ

1. How many petals does the graph of r = 3 sin 5θ have?

Explanation: For r = a sin nθ, n odd gives n petals. Here n = 5, so 5 petals.

2. How many petals does r = 2 cos 6θ have?

Explanation: n = 6 is even, so there are 2n = 12 petals.

3. Which describes r = 1 + 2 cos θ?

Explanation: The form is a + b cos θ with a = 1 and b = 2. Since a < b, the curve has an inner loop. A cardioid would need a = b.

4. What is the greatest distance from the pole of any point on r = 3 + 2 cos θ?

Explanation: The largest value of r is 3 + 2(1) = 5, reached at θ = 0. The smallest is 3 − 2 = 1, so the curve never touches the pole.

7. Distance from the Pole: Increasing, Decreasing, and Extrema

The behavior of r = f(θ) in the θr-plane tells you what the curve is doing in the polar plane, but you must account for the sign of r. What matters for the picture is |r|, the distance from the pole. The distance is increasing when r is positive and increasing, or negative and decreasing (becoming more negative). The distance is decreasing when r is positive and decreasing, or negative and increasing (moving toward 0). If f changes from increasing to decreasing, or from decreasing to increasing, the function has a relative extremum there. This corresponds to the curve being relatively farthest from or closest to the pole. The sign of r also tells you which side of the terminal ray the point is on, but it does not change how far away it is. Reading the guide graph interval by interval is a systematic way to describe the whole curve: state where r increases or decreases, where r = 0 (the curve is at the pole), and where r reaches its greatest and least values.

Worked Example: Describe the behavior of r = 2 + 2 sin θ on [0, 2π].

From the guide graph: on [0, π/2], r is positive and increasing from 2 to 4, so the distance is increasing and the curve sweeps up from the point (2, 0) to the top at (4, π/2). On [π/2, 3π/2], r is positive and decreasing from 4 to 0, so the distance is decreasing and the curve swings around the left side and returns to the pole. On [3π/2, 2π], r is positive and increasing from 0 to 2, so the distance increases and the curve leaves the pole and comes back to (2, 0).

Extrema: r has a relative maximum of 4 at θ = π/2 (farthest from the pole) and a relative minimum of 0 at θ = 3π/2 (at the pole). The curve is a cardioid.

The guide graph of r = 2 + 2 sin θ. It increases on [0, π/2], decreases on [π/2, 3π/2], and increases on [3π/2, 2π]. Here r is never negative.

The matching cardioid r = 2 + 2 sin θ. Its farthest point is 4 units above the pole (θ = π/2, a relative maximum of r) and it touches the pole at θ = 3π/2 (a relative minimum).

1. For r = 5 cos θ on [0, π/2], what is happening to the distance from the pole?

Explanation: r goes from 5 down to 0. It is positive and decreasing, so the point gets closer to the pole until it reaches it at θ = π/2.

2. For r = −2θ on (0, π), what is happening to the distance from the pole?

Explanation: r = −2θ is negative and becomes more negative as θ increases, so |r| = 2θ grows. A negative r reflects the side the point is on but not its distance.

3. Where on [0, 2π] is r = 3 − 3 cos θ farthest from the pole?

Explanation: r = 3 − 3 cos θ is largest when cos θ = −1, at θ = π, where r = 6. At θ = 0 the curve is at the pole.

4. For r = cos 2θ on [π/4, π/2], what is happening to the distance from the pole?

Explanation: On this interval 2θ runs from π/2 to π, so r goes from 0 down to −1. It is negative and decreasing, so |r| increases from 0 to 1.

8. Average Rate of Change in Polar Functions

Worked Example: For r = 5 + 3 cos θ, find the average rate of change over (a) [0, π/3], (b) [π/3, π], and (c) [π, 3π/2].

(a) r(0) = 8 and r(π/3) = 5 + 3(1/2) = 6.5, so the rate is (6.5 − 8)/(π/3 − 0) = −1.5/(π/3) = −4.5/π ≈ −1.43 radius units per radian.

(b) r(π) = 2, so (2 − 6.5)/(π − π/3) = −4.5/(2π/3) = −6.75/π ≈ −2.15.

The signs match the picture: the curve moves toward the pole on (a) and (b) and away from it on (c), as r is positive on the whole interval.

r = 5 + 3 cos θ as a function of θ. Marked points: (0, 8), (π/3, 6.5), (π/2, 5), (π, 2). The average rate of change of r over an interval is the slope of the line joining the endpoints on this graph.

1. For r = 3 + 2 sin θ, what is the average rate of change over [0, π/2]?

Explanation: r(0) = 3 and r(π/2) = 5, so (5 − 3)/(π/2) = 2/(π/2) = 4/π.

2. For r = 2θ, what is the average rate of change over [1, 4]?

Explanation: r(1) = 2 and r(4) = 8, so (8 − 2)/(4 − 1) = 6/3 = 2. For r = kθ, the rate is constant and equal to k.

3. For r = 6 cos θ, what is the average rate of change over [0, π/3]?

Explanation: r(0) = 6 and r(π/3) = 3. The rate is (3 − 6)/(π/3) = −3 · 3/π = −9/π.

4. For r = 1 + 2 cos θ, what is the average rate of change over [π/2, π]?

Explanation: r(π/2) = 1 and r(π) = 1 − 2 = −1. The rate is (−1 − 1)/(π/2) = −4/π. Note that r passes through 0 and becomes negative on this interval, so the point passes through the pole.

Common Mistakes to Avoid

Treating the order as (θ, r). Polar coordinates are always written (r, θ), radius first.
Plotting a negative r on the terminal side of θ. A negative r goes in the opposite direction, through the pole.
Believing each point has one polar name. Add 2π to θ, or flip the sign of r and add π, to get other names for the same point.
Using arctan(y/x) alone for θ and ignoring the quadrant. Find the reference angle, then place it in the right quadrant (π − α, π + α, or 2π − α).
Forgetting the modulus when writing r(cos θ + i sin θ), or computing r for k + ki as k instead of k√2.
Using the wrong quadrantal angle for numbers on an axis, for example θ = π/2 for −3i (it should be 3π/2).
Counting petals incorrectly: for r = a sin nθ or a cos nθ, odd n gives n petals and even n gives 2n petals.
Reading r negative and decreasing as “distance decreasing.” If r is negative and decreasing, |r| grows and the distance increases.
Computing the average rate of change as a change in the point’s position. It is (change in r)/(change in θ), in radius units per radian.

Frequently Asked Questions

Why can the same point have different polar coordinates?

Because angles repeat every 2π and because a negative radius reverses direction. For example (3, π/4), (3, 9π/4), and (−3, 5π/4) are all the same point.

How do I decide the angle when converting to polar form?

Find the reference angle α = arctan(|y|/|x|), then use the quadrant: Quadrant I gives α, II gives π − α, III gives π + α, IV gives 2π − α. For points on an axis, use 0, π/2, π, or 3π/2.

Can every complex number be written in polar form?

Yes. Nonzero numbers have modulus r = √(a² + b²) and an argument θ as described in the lesson. Zero has modulus 0 and any angle works, so it is the only one where the angle is not determined.

What do the names cardioid and limaçon mean?

A cardioid is a heart-shaped curve (from the Greek for heart), and a limaçon is a looped or oval curve (French for snail). They are the family r = a ± b cos θ or r = a ± b sin θ, with a = b giving the cardioid.

Do I need a calculator for polar graphs?

It is useful for checking, but the exam expects you to be able to reason from the θr-graph and from the equation type. Use the calculator in POLAR mode with the angle step small enough that the curve looks smooth, and use radians.

Where can I practice more problems like these?

The AP Precalculus Trigonometric and Polar Functions Quiz 2 in the AP Precalculus QBank gives you more practice on this topic, and you can browse every other topic in the full AP Precalculus QBank.