← Back to the Polar Coordinates Parametric Equations worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: N-CN.B.4, F-IF.C.7 · Level: Grade 12 Precalculus
Problem 1 Polar to Rectangular
x = 6 cos 60° = 6 · 12 = 3 and y = 6 sin 60° = 6 · √32 = 3√3.
Answer: (3, 3√3)
Problem 2 A Negative Radius
x = −3 cos π4 = −3 · √22 = −3√22 and y = −3 sin π4 = −3√22.
This is the same point as (3, 5π4), in Quadrant III.
Answer: (−3√22, −3√22) ≈ (−2.12, −2.12)
Problem 3 Rectangular to Polar
r = √(27 + 9) = 6. The reference angle is tan⁻¹(33√3) = π6, and the point is in Quadrant II, so θ = π − π6 = 5π6.
Answer: (6, 5π6)
Problem 4 A Point in Quadrant III
r = √(36 + 36) = 6√2. The reference angle is π4 and the point is in Quadrant III, so θ = π + π4 = 5π4.
Answer: (6√2, 5π4)
Problem 5 Reading a Polar Grid
(a) A is on the 4th ring along the 210° ray: (4, 210°).
(b) Add a full turn: (4, 570°). Use a negative radius and add 180° (or subtract it): (−4, 30°).
(c) x = 4 cos 210° = −2√3 and y = 4 sin 210° = −2.
Answer: (4, 210°), also (4, 570°) and (−4, 30°); rectangular (−2√3, −2)
Problem 6 A Circle from a Sine
Multiply by r: r2 = 8r sin θ, so x2 + y2 = 8y. Move and complete the square: x2 + (y − 4)2 = 16.
The graph is a circle with center (0, 4) and radius 4.
Answer: x2 + (y − 4)2 = 16
Problem 7 A Line in Polar Form
Substitute x = r cos θ and y = r sin θ: 2r cos θ − 5r sin θ = 10. Factor out r: r(2 cos θ − 5 sin θ) = 10.
Answer: r = 102 cos θ − 5 sin θ
Problem 8 A Circle Through the Pole
Replace x2 + y2 with r2 and x with r cos θ: r2 + 6r cos θ = 0. Divide by r: r + 6 cos θ = 0.
Answer: r = −6 cos θ
Problem 9 Analyzing a Cardioid
(a) Since a = b = 3, it is a cardioid.
(b) r = 6 at 0, r = 3 at π2, r = 0 at π, r = 3 at 3π2.
(c) The maximum distance is 6, at θ = 0, and the graph reaches the pole at θ = π.
Answer: cardioid; values 6, 3, 0, 3; farthest 6 units, cusp at the pole at θ = π
Problem 10 An Eight-Petal Rose
Since 4 is even, the rose has 2 · 4 = 8 petals. Each petal has length |5| = 5.
The first tip occurs where sin 4θ = 1, so 4θ = π2 and θ = π8.
Answer: 8 petals, each 5 units long, first tip at θ = π8
Problem 11 A Limaçon with a Loop
Since a = 2 < b = 4, the limaçon has an inner loop. r = 0 when cos θ = −12, so θ = 2π3 and θ = 4π3.
The largest r occurs at θ = 0: r = 2 + 4 = 6.
Answer: inner loop; pole at θ = 2π3 and 4π3; maximum r = 6
Problem 12 A Three-Petal Rose
Since 3 is odd, the rose has 3 petals. Tips occur where |cos 3θ| = 1, so 3θ = 0, π, 2π and θ = 0, π3, 2π3.
At θ = 0, r = 4. At θ = π3, r = −4, which is the point (4, 4π3). At θ = 2π3, r = 4.
Answer: 3 petals; tips at distance 4 in the directions 0, 2π3 and 4π3
Problem 13 Eliminating a Parameter
Solve x = 2t + 1 for t: t = x − 12. Then y = x − 12 − 3 = x2 − 72.
At t = 0 the point is (1, −3). At t = 5 it is (11, 2). The path is the line segment between them.
Answer: y = x2 − 72, from (1, −3) to (11, 2)
Problem 14 An Elliptical Path
cos t = x − 25 and sin t = y + 13. Using cos2 t + sin2 t = 1: (x − 2)225 + (y + 1)29 = 1.
The path is an ellipse with center (2, −1), horizontal semi-axis 5 and vertical semi-axis 3.
Answer: (x − 2)225 + (y + 1)29 = 1, center (2, −1)
Problem 15 Is It a Parabola?
Multiply: r + r cos θ = 4, so r = 4 − x. Square both sides: x2 + y2 = 16 − 8x + x2.
The x2 terms cancel: y2 = 16 − 8x. This is a parabola opening to the left with vertex (2, 0).
Answer: y2 = 16 − 8x, a parabola
Problem 16 The Kicked Ball
(a) x = 9 and y = 5 + 12 − 4 = 13. The position is (9, 13).
(b) The vertex of y is at t = 2432 = 0.75: y = 5 + 18 − 9 = 14 ft.
(c) Solve −16t2 + 24t + 5 = 0: t = 24 + √89632 ≈ 1.685 s. Then x = 18(1.685) ≈ 30.3 ft.
Answer: (9, 13); 14 ft; about 30.3 ft
Problem 17 Two Boats on Radar
(a) 150° − 30° = 120°.
(b) Law of cosines: d2 = 92 + 72 − 2(9)(7) cos 120° = 81 + 49 + 63 = 193, so d = √193 ≈ 13.9 mi.
Check with coordinates: A ≈ (7.79, 4.5) and B ≈ (−6.06, 3.5). The differences are 13.85 and 1.0, and √(191.8 + 1) ≈ 13.9.
Answer: 120°; about 13.9 miles
Problem 18 Microphone Pickup Pattern
(a) a = b = 4, so the pattern is a cardioid.
(b) At θ = π2: r = 4 + 4 = 8 ft. At θ = 3π2: r = 4 − 4 = 0, so the microphone picks up nothing from directly behind.
(c) At θ = π6: r = 4 + 4 · 12 = 6 ft.
Answer: cardioid; 8 ft ahead and 0 ft behind; 6 ft at θ = π6
Problem 19 Where Two Curves Meet
Set 3 = 6 cos θ, so cos θ = 12 and θ = ±π3. The points are (3, π3) and (3, −π3).
Rectangular: x = 3 cos π3 = 32 and y = ±3 sin π3 = ±3√32.
Check: x2 + y2 = 94 + 274 = 9 and (x − 3)2 + y2 = 94 + 274 = 9.
Answer: (3, ±π3), or (32, ±3√32)
Problem 20 Ferris Wheel Motion
(a) cos(t5) = x20 and sin(t5) = y − 2420. So x2 + (y − 24)2 = 400, a circle with center (0, 24) and radius 20.
(b) At t = 5 the angle is 1 radian: x = 20 cos 1 ≈ 10.8 and y = 24 + 20 sin 1 ≈ 40.8.
(c) A full turn needs t5 = 2π, so t = 10π ≈ 31.4 s.
Answer: x2 + (y − 24)2 = 400; (10.8, 40.8); 10π ≈ 31.4 s
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