Lesson 8: Parametric Equations and Polar Coordinates

Cartesian equations describe *where* a curve is but not *how it is travelled*. Parametric equations add the clock: x and y are each functions of a parameter t, so a single equation set captures position, direction and speed. Polar coordinates change the ruler instead: a point is located by its distance r from the origin and angle θ, which turns rotations and circles into simple equations. This lesson builds both systems; Lesson 18 puts them to work on motion and area.

Parametric equations and eliminating the parameter

A parametric curve is x = f(t), y = g(t). The table of values *is* the graph, drawn in order of increasing t — the orientation. Worked example: x = 2t − 1, y = t^2 + 1. Values: t = −1 → (−3, 2); t = 0 → (−1, 1); t = 1 → (1, 2); t = 2 → (3, 5). To find the Cartesian equation, solve for t in the simpler equation: t = (x + 1)/2, and substitute: y = ((x + 1)/2)^2 + 1 — a parabola with vertex (−1, 1), traced left to right as t increases. But the Cartesian form has forgotten something: if t is restricted, say t ≥ 0, the parametric curve is only the right half of the parabola, starting at (−1, 1). Eliminating the parameter gains simplicity and loses information — always carry the restriction across.

Parametric motion: position as a function of time

Because t is usually time, parametric equations describe motion directly: x(t) is the horizontal position, y(t) the vertical. A projectile with launch speed v at angle α (gravity g) follows x = v cos α · t, y = v sin α · t − (1/2) g t^2 — Lesson 18 exploits this fully. Even without physics, reading a parametric graph means reading three things at once: the path (from the Cartesian form), the timing (from t-values marked on the path), and the direction (the orientation arrows). Two different parametrisations can share a path while describing completely different journeys along it.

Polar coordinates: converting points

In polar form a point is (r, θ): r is the signed distance from the pole (origin), θ the angle from the polar axis (positive x-axis). Conversion formulas: x = r cos θ, y = r sin θ, r^2 = x^2 + y^2, tan θ = y/x. Worked examples. (r, θ) = (4, 2π/3): x = 4 cos(2π/3) = −2, y = 4 sin(2π/3) = 2·sqrt(3) ≈ 3.46, so the point is (−2, 2·sqrt(3)). In reverse, (3, sqrt(3)): r = sqrt(9 + 3) = 2·sqrt(3) ≈ 3.46, tan θ = sqrt(3)/3 = 1/sqrt(3) with the point in quadrant I, so θ = π/6. Polar coordinates are not unique: (4, 2π/3), (4, 2π/3 + 2π) and (−4, 5π/3) are the same point. Negative r means "walk backwards": (−4, 5π/3) points at angle 5π/3 but 4 units in the opposite direction, landing at angle 2π/3, distance 4 ✓.

Polar equations and classic curves

Converting equations uses the same formulas. Worked example: r = 6 cos θ. Multiply by r: r^2 = 6r cos θ, i.e. x^2 + y^2 = 6x, i.e. (x − 3)^2 + y^2 = 9 — a circle of radius 3 centred at (3, 0), traced once for θ in [0, π]. The classic polar families:

• Circles: r = 2a cos θ or r = 2a sin θ (through the pole, diameter 2a)
• Cardioid: r = a(1 + cos θ) — heart-shaped, one dimple-free loop; e.g. r = 2 + 2 cos θ passes through (4, 0°), (2, 90°), (0, 180°)
• Limaçon with inner loop: r = 1 + 2 sin θ (the inner loop appears because r goes negative)
• Rose: r = 3 sin(2θ) — for sin(nθ)/cos(nθ), if n is even there are 2n petals, if n odd, n petals; r = 3 sin(2θ) has 4 petals of length 3

Symmetry tests speed up sketching: replace θ by −θ (x-axis symmetry), θ by π − θ (y-axis symmetry), or r by −r (origin symmetry), and check whether the equation is unchanged.

Common mistakes

• Losing the parameter restriction. y = ((x + 1)/2)^2 + 1 alone claims the whole parabola; with t ≥ 0 the curve starts at (−1, 1). Graders and engineers both care about the difference.
• Quadrant errors in θ. For the point (−1, −1), tan θ = 1 suggests θ = π/4, but the point is in quadrant III: θ = 5π/4. arctan gives a quadrant-I/IV answer; adjust by the point's actual position.
• Assuming polar coordinates are unique. When a problem states a polar point, equivalent forms with θ shifted by 2π or with r negated and θ shifted by π describe the same location — especially when checking whether a point lies on a curve.
• Graphing r = 3 sin 2θ with 2 petals. n = 2 is even, so the rose has 2n = 4 petals. The odd/even rule is about whether the "backward" tracing retraces petals or draws new ones.
• Trying to plot polar curves point-by-point in Cartesian style. Use the special values (θ = 0, π/2, π, 3π/2) and symmetry; a cardioid is four computed points plus a smooth join, not forty calculator plots.

Exam Experience

1. The polar point (5, 7π/6) has rectangular coordinates

• (A) (5·sqrt(3)/2, 5/2)
• (B) (−5·sqrt(3)/2, −5/2)
• (C) (−5/2, −5·sqrt(3)/2)
• (D) (5/2, −5·sqrt(3)/2)

Answer: (B). x = 5 cos(7π/6) = 5(−sqrt(3)/2) = −5·sqrt(3)/2; y = 5 sin(7π/6) = 5(−1/2) = −5/2. The angle is in quadrant III, so both coordinates are negative — (A) is the quadrant-I reflex answer.

2. The polar equation r = 2 sin θ represents

• (A) a circle centred at (0, 1) with radius 1
• (B) a circle centred at (1, 0) with radius 1
• (C) a cardioid
• (D) a line through the origin

Answer: (A). Multiply by r: x^2 + y^2 = 2y, i.e. x^2 + (y − 1)^2 = 1 — centre (0, 1), radius 1. The sin θ version centres on the y-axis; the cos θ version (B) centres on the x-axis.

3. For the parametric curve x = 3t, y = t^2 − 4, the Cartesian equation is

• (A) y = x^2/9 − 4
• (B) y = x^2 − 4
• (C) y = 9x^2 − 4
• (D) x = 3·sqrt(y + 4)

Answer: (A). t = x/3, so y = (x/3)^2 − 4 = x^2/9 − 4. (B) comes from substituting t = x instead of t = x/3 — the scaling in x = 3t must be undone, not ignored.

Constructed response. Consider the cardioid r = 2 + 2 cos θ. (a) Complete a table of r for θ = 0, π/3, π/2, 2π/3, π, and use it to state the maximum and minimum values of r. (b) Convert the equation to Cartesian form. (c) State the symmetry of the curve and the point where it meets the pole.

Scoring guide. (a) 2 points: r = 4, 3, 2, 1, 0 respectively (cos values 1, 1/2, 0, −1/2, −1) — 1 point for all five values; maximum r = 4 (at θ = 0), minimum r = 0 (at θ = π) — 1 point. (b) 2 points: r^2 = 2r + 2r cos θ → x^2 + y^2 = 2·sqrt(x^2 + y^2) + 2x, equivalently x^2 + y^2 − 2x = 2·sqrt(x^2 + y^2) — 1 point for multiplying through by r, 1 point for the substitutions r^2 = x^2 + y^2 and r cos θ = x (leaving sqrt(x^2 + y^2) for the standalone r is correct and expected). (c) 1 point: symmetric about the polar axis (x-axis), since cos(−θ) = cos θ; it meets the pole where r = 0, at θ = π.

Going deeper: the cycloid and the limaçon

The cycloid is the path of a point on a rolling wheel's rim: x = t − sin t, y = 1 − cos t (unit wheel, t the rotation angle). Table: t = 0 → (0, 0); t = π/2 → (0.57, 1); t = π → (3.14, 2) — the top, at height equal to the diameter; t = 3π/2 → (5.71, 1); t = 2π → (6.28, 0) — one full arch spanning the circumference 2π. The curve cannot be written as a single y = f(x) worth using (eliminating t is hopeless in elementary functions), which is the deepest reason parametric form exists: some curves are *about* the parameter. The cycloid is also the brachistochrone — the curve of fastest descent — a calculus-of-variations celebrity hiding in a rolling wheel.

The limaçon r = 1 + 2 cos θ shows why polar plotting needs care with negative r. Values: θ = 0 → r = 3; θ = π/2 → r = 1; θ = 2π/3 → r = 0 (the curve passes through the pole); θ = π → r = −1, plotted one unit in the direction *opposite* π, i.e. at (1, 0°) — completing an inner loop. The loop exists because |constant term| < |coefficient| (1 < 2): the curve crosses the pole, and negative r wraps part of the graph back inside. Compare the cardioid r = 2 + 2 cos θ (equal terms: touches the pole, no loop) and r = 3 + 2 cos θ (constant larger: dimpled or convex, never reaches the pole). One family, three shapes, decided by a single ratio — pattern recognition worth more than any individual sketch.

Workshop: reading polar and parametric motion

Circular motion, parametrically. x = 3 cos t, y = 3 sin t traces the circle of radius 3 (since x^2 + y^2 = 9), starting at (3, 0), reaching (0, 3) at t = π/2 — anticlockwise, one revolution per 2π of t. Reversing the second sign, x = 3 cos t, y = −3 sin t, traverses the *same circle* clockwise: identical Cartesian equation, opposite orientation. Orientation is invisible to the Cartesian form and obvious in the parametric — the recurring theme of Lesson 8.

Negative r, concretely. The polar point (−3, π/4) and (3, 5π/4) coincide: both give rectangular coordinates (−3·sqrt(2)/2, −3·sqrt(2)/2) ≈ (−2.12, −2.12). When checking whether a point lies on a polar curve, failing with one representation proves nothing — try the equivalent forms (r negated, θ shifted by π or 2π) before concluding.