"Go 3 blocks east and 4 blocks north" — that's rectangular thinking: (3, 4). "Head 53° north of east and walk 5 blocks" — that's polar thinking: (5, 53°). Same destination, two addressing systems.
Radar screens, sonar, hurricanes, flower petals, orbit diagrams — anything organized around a center point speaks polar natively. The math cost is a pair of conversion formulas you already own: they're just cos and sin from the unit circle, scaled by r.
A point's polar coordinates (r, θ): distance r from the pole (origin), at angle θ from the polar axis (positive x-axis). Conversions:
polar → rectangular: x = r cos θ, y = r sin θ
rectangular → polar: r² = x² + y², tan θ = y/x (place θ by quadrant!)
The quadrant warning is real: for (−3, 3), tan θ = −1, but the point is in QII, so θ = 3π/4 (not −π/4, which is QIV). Locate the point first, then choose the angle.
Polar addresses aren't unique: (r, θ) = (r, θ + 2π) = (−r, θ + π). A negative r means "walk backwards": face direction θ, step |r| the opposite way. So (−2, π/6) is the same point as (2, 7π/6).
| equation | graph |
|---|---|
| r = a | circle, radius |a|, centered at the pole |
| θ = c | line through the pole at angle c |
| r = 2a cos θ | circle of diameter 2a through the pole, centered (a, 0) on the x-axis |
| r = 2a sin θ | circle through the pole, centered (0, a) on the y-axis |
| r = a cos(nθ) or a sin(nθ) | rose: n petals if n odd, 2n petals if n even; petal length |a| |
| r = a ± b cos θ (or sin) | limaçon: inner loop if b > a; cardioid if a = b; dimpled if a > b < 2a; convex if a ≥ 2b |
Circle check for r = 4cos θ: multiply by r → r² = 4r cos θ → x² + y² = 4x → (x − 2)² + y² = 4 — circle center (2, 0), radius 2 ✓. That multiply-by-r conversion trick handles most equation conversions.
Cos versions are symmetric about the x-axis (polar axis); sin versions about the y-axis (θ = π/2 line).
[GRAPH: Four mini-panels. (1) r = 4cos θ: circle through the pole centered (2, 0). (2) r = 3cos(2θ): four-petal rose, petals along both axes, length 3. (3) r = 2 + 3cos θ: limaçon whose outer loop reaches (5, 0) rectangular and whose small inner loop, produced by the negative-r sweep near θ = π, lies along the positive x-axis inside it. (4) r = 4 + 4sin θ: cardioid, heart shape with cusp at the pole pointing downward, top at (0, 8) rectangular.]
A polar graph is generated by sweeping θ and plotting the signed distance r(θ). Key behaviors:
Problem: (a) Convert polar (4, π/3) to rectangular. (b) Convert rectangular (−3, 3) to polar with r > 0, 0 ≤ θ < 2π.
Solution: (a) x = 4cos(π/3) = 2; y = 4sin(π/3) = 2√3 → (2, 2√3). (b) r = √(9 + 9) = 3√2. The point is in QII with reference angle π/4 → θ = 3π/4 → (3√2, 3π/4).
Interpretation: Polar→rectangular is pure plug-in. Rectangular→polar needs the quadrant decision — the formula tan θ = y/x can't see signs the way you can.
Problem: Convert r = 6 sin θ to rectangular form and identify the graph.
Solution: Multiply both sides by r: r² = 6r sin θ → x² + y² = 6y → x² + (y − 3)² = 9. Circle, center (0, 3), radius 3 — through the pole, sitting on the y-axis (as the sin version should).
Interpretation: The multiply-by-r move converts r² and r·(trig) into rectangular pieces simultaneously. Completing the square names the circle.
Problem: For r = 3 cos(2θ): number of petals, petal length, and the angles where the curve passes through the pole.
Solution: n = 2 (even) → 2n = 4 petals, each of length 3. Pole crossings: 3cos(2θ) = 0 → 2θ = π/2 + kπ → θ = π/4 + kπ/2 (π/4, 3π/4, 5π/4, 7π/4) — the four angles separating the petals. Petal tips where |cos 2θ| = 1: θ = 0, π/2, π, 3π/2 — along both axes.
Interpretation: Everything about a rose is encoded: a = tip length, n = petal count (doubled when even), r = 0 solutions = the seams between petals.
Problem: Classify r = 2 + 3 cos θ and find its maximum |r| and its pole crossings.
Solution: b = 3 > a = 2 → limaçon with an inner loop. Max r = 2 + 3 = 5 at θ = 0. Pole crossings: 2 + 3cos θ = 0 → cos θ = −2/3 → θ = arccos(−2/3) ≈ 2.30 and 2π − 2.30 ≈ 3.98 — between those angles r is negative and the curve traces its inner loop.
Interpretation: Compare a vs. b before anything else: it names the shape. Then r's max/min and zeros give the dimensions.
1. (A). (4cos π/3, 4sin π/3) = (2, 2√3). (B) swaps cos and sin.
2. (B). r = 3√2; QII → θ = 3π/4. (A) takes arctan's raw −π/4 and drops the sign; (D) lands in QIV.
3. (C). All points at distance 6: circle radius 6 at the pole. (A) halves unnecessarily — that rule is for r = a cos θ.
4. (D). Fixed angle, any r (including negative): the full line at 45°, i.e. y = x through the origin.
5. (B). r² = 4r cos θ → x² + y² = 4x → (x − 2)² + y² = 4: radius 2, center (2, 0). (A) misreads the coefficient as the radius.
6. (D). n = 2 is even → 2n = 4 petals. (A) is the odd-n rule misapplied.
7. (A). n = 3 odd → 3 petals. (B) doubles when it shouldn't.
8. (C). b = 3 > a = 2 → inner loop. Cardioid needs a = b; dimple needs a > b.
9. (B). a = b = 4 → cardioid. (D) would be r = 4 sin θ without the constant.
10. (A). (−r, θ) = (r, θ + π): (−2, π/6) = (2, 7π/6). (B)/(C) reflect instead of reversing.
11. (C). cos θ peaks at 1: r = 3 + 2 = 5 (at θ = 0). (A) is the midline value; (D) is the minimum.
12. (FRQ-style, 6 points) (i) [2 pts] r² = 4r sin θ → x² + y² = 4y → x² + (y − 2)² = 4: circle, center (0, 2), radius 2 (through the pole). (ii) [2 pts] r = 4 sin(π/6) = 2; x = 2cos(π/6) = √3, y = 2sin(π/6) = 1 → (√3, 1). (Check: on the circle? √3² + (1−2)² = 3 + 1 = 4 ✓) (iii) [2 pts] For π < θ < 2π, sin θ < 0, so r is negative — the point plots on the opposite ray, landing back on the same upper circle already traced (e.g., θ = 7π/6, r = −2 is the point (2, π/6)). The circle is complete after θ = π; the second half retraces it.
12. (FRQ-style) 🚫 Consider the polar curve r = 4 sin θ, 0 ≤ θ ≤ π. (i) Convert the equation to rectangular form and identify the graph precisely. (ii) Find the rectangular coordinates of the point at θ = π/6. (iii) Explain why the interval 0 ≤ θ ≤ π traces the entire graph (what happens for π < θ < 2π?).
The polar function r(θ) = 2 + 2 sin θ is graphed for 0 ≤ θ < 2π.
(a) (i) Classify the curve. (ii) Find the maximum value of r and the θ at which it occurs, and the point(s) where the curve meets the pole.
(b) (i) Compute r at θ = 0, π/2, π, 3π/2. (ii) Using those values, describe how the distance from the pole changes as θ sweeps from 0 to π/2 and from π/2 to π.
(c) The curve is symmetric about the line θ = π/2 (the y-axis). Justify this symmetry using the behavior of sin θ.
(a) [2 pts] (i) [1 pt] a = b = 2 → cardioid. (ii) [1 pt] Max r = 4 at θ = π/2; meets the pole where 2 + 2sin θ = 0 → sin θ = −1 → θ = 3π/2 (the cusp).
(b) [2 pts] (i) [1 pt] r(0) = 2, r(π/2) = 4, r(π) = 2, r(3π/2) = 0. (ii) [1 pt] As θ goes 0 → π/2, r increases 2 → 4: the curve moves away from the pole. As θ goes π/2 → π, r decreases 4 → 2: it draws back toward the pole (continuing to 0 at 3π/2).
(c) [2 pts] sin(π − θ) = sin θ [1 pt identity], so r(π − θ) = r(θ): the angles θ and π − θ (mirror images across the vertical line) are assigned equal distances, making the curve symmetric about θ = π/2 [1 pt conclusion].
1. (A). (4cos π/3, 4sin π/3) = (2, 2√3). (B) swaps cos and sin.
2. (B). r = 3√2; QII → θ = 3π/4. (A) takes arctan's raw −π/4 and drops the sign; (D) lands in QIV.
3. (C). All points at distance 6: circle radius 6 at the pole. (A) halves unnecessarily — that rule is for r = a cos θ.
4. (D). Fixed angle, any r (including negative): the full line at 45°, i.e. y = x through the origin.
5. (B). r² = 4r cos θ → x² + y² = 4x → (x − 2)² + y² = 4: radius 2, center (2, 0). (A) misreads the coefficient as the radius.
6. (D). n = 2 is even → 2n = 4 petals. (A) is the odd-n rule misapplied.
7. (A). n = 3 odd → 3 petals. (B) doubles when it shouldn't.
8. (C). b = 3 > a = 2 → inner loop. Cardioid needs a = b; dimple needs a > b.
9. (B). a = b = 4 → cardioid. (D) would be r = 4 sin θ without the constant.
10. (A). (−r, θ) = (r, θ + π): (−2, π/6) = (2, 7π/6). (B)/(C) reflect instead of reversing.
11. (C). cos θ peaks at 1: r = 3 + 2 = 5 (at θ = 0). (A) is the midline value; (D) is the minimum.
12. (FRQ-style, 6 points) (i) [2 pts] r² = 4r sin θ → x² + y² = 4y → x² + (y − 2)² = 4: circle, center (0, 2), radius 2 (through the pole). (ii) [2 pts] r = 4 sin(π/6) = 2; x = 2cos(π/6) = √3, y = 2sin(π/6) = 1 → (√3, 1). (Check: on the circle? √3² + (1−2)² = 3 + 1 = 4 ✓) (iii) [2 pts] For π < θ < 2π, sin θ < 0, so r is negative — the point plots on the opposite ray, landing back on the same upper circle already traced (e.g., θ = 7π/6, r = −2 is the point (2, π/6)). The circle is complete after θ = π; the second half retraces it.
🎯 Exam tip: Build a four-shape flashcard set: r = a (circle at pole), r = 2a cos/sin θ (circle through pole on an axis), roses (count petals by n's parity), limaçons (classify by a vs. b). Every polar-graph MC on the exam is one of these four families wearing different numbers.