Lesson 18: Parametric and Polar Applications: Motion and Area

Lesson 8 introduced parametric equations and polar coordinates; this lesson collects the payoff. Parametrically, projectile motion becomes two one-dimensional problems whose answers (range, flight time, maximum height) fall out cleanly. In polar form, area has its own integral formula — sectors instead of rectangles — and the classic curves of Lesson 8 yield exact areas that would be miserable in Cartesian coordinates.

Projectile motion, fully analysed

A projectile launched at speed 50 m/s with components v_x = 40 m/s and v_y = 30 m/s (a 3-4-5 triangle, so the launch angle is arctan(30/40) ≈ 36.87°), with g = 10 m/s², follows:

x(t) = 40t,   y(t) = 30t − 5t^2

Flight time: y = 0 when t(30 − 5t) = 0, so t = 6 s (excluding launch). Range: x(6) = 240 m. Maximum height: at the vertex of the parabola in t, t = 3 s (half the flight, by symmetry), y(3) = 90 − 45 = 45 m. Cartesian path: eliminate t = x/40: y = (3/4)x − x^2/320 — a parabola through (0, 0) and (240, 0) with vertex at x = 120: y(120) = 90 − 45 = 45 ✓ consistent. Each question maps to a representation: *when* questions use t, *where* (height at a given distance) questions use the Cartesian path, *how far/how high* use the component formulas. Choosing the representation is the modelling skill — the arithmetic afterwards is routine.

Area in polar coordinates

A polar curve sweeps out area sector by sector: a thin sector with angle dθ has area (1/2) r^2 dθ, so:

Area = (1/2) ∫(α to β) r^2 dθ

Worked example — a circle the polar way. r = 3 sin θ is the circle of diameter 3 (Lesson 8), traced for θ in [0, π]. Its area: (1/2)∫(0 to π) 9 sin^2 θ dθ. Using sin^2 θ = (1 − cos 2θ)/2, the integral of r^2 is 9 × (π/2) = 9π/2, and half of that is 9π/4 — agreeing with π(1.5)^2 = 2.25π ✓. The identity step is the price of admission for nearly every polar area: sin^2 and cos^2 always get rewritten with the double-angle formulas from Lesson 4.

Cardioid and rose areas

Cardioid. r = 2 + 2 cos θ, traced once for θ in [0, 2π]. Area = (1/2)∫(0 to 2π) (2 + 2 cos θ)^2 dθ = (1/2)∫ 4(1 + 2 cos θ + cos^2 θ) dθ. Over a full period ∫cos θ = 0 and ∫cos^2 θ = π (average 1/2 over length 2π), so the bracket integrates to 2π + 0 + π = 3π, giving Area = (1/2)(4)(3π) = 6π ≈ 18.85. (General fact: r = a(1 + cos θ) has area (3/2)π a^2; with a = 2, (3/2)π·4 = 6π ✓.) One petal of a rose. r = cos(2θ) has a petal where 2θ ∈ [−π/2, π/2], i.e. θ ∈ [−π/4, π/4]: Area = (1/2)∫(−π/4 to π/4) cos^2(2θ) dθ = π/8 ≈ 0.393. The limits are the subtle part: integrating over [0, 2π] would trace all 4 petals (and the "one petal" answer would be multiplied by 4). For r = 3 sin(2θ), a petal spans θ ∈ [0, π/2] and its area is (1/2)∫(0 to π/2) 9 sin^2(2θ) dθ = 9π/8 — the 3 contributes 3^2 = 9 to the area scale.

Common mistakes

• Integrating r instead of r^2, or dropping the 1/2. Polar area is (1/2)∫r^2 dθ. Forgetting the 1/2 doubles every answer; using r instead of r^2 is dimensionally wrong (area must scale like length² — check scaling: doubling r should quadruple area).
• Wrong θ-limits for one petal. A petal of sin(2θ) or cos(2θ) spans a θ-interval of π/2 (or the symmetric ±π/4 window), not 2π. Find where r = 0 (the petal's endpoints) and integrate between consecutive zeros.
• Mixing up flight time factors. With y = v_y t − (g/2)t^2, the flight time is 2v_y/g = 60/10 = 6 s — not v_y/g (that's the time to the *peak*, 3 s). The factor of 2 is the up-down symmetry; losing it halves the range.
• Height questions answered in t. "How high is the projectile when it is 120 m downrange?" requires the Cartesian path y = (3/4)x − x^2/320 → 45 m. Substituting x into x(t) inverts the problem incorrectly; match the representation to the question.
• Assuming the parametric path shows timing. The Cartesian parabola y = (3/4)x − x^2/320 has no trace of the 6-second flight; two projectiles with different speeds can share a path shape. Path ≠ motion.

Exam Experience

1. A projectile has x(t) = 40t and y(t) = 30t − 5t^2 (metres, seconds). Its maximum height is reached at

• (A) t = 3 s, height 45 m
• (B) t = 6 s, height 45 m
• (C) t = 3 s, height 90 m
• (D) t = 1.5 s, height 45 m

Answer: (A). The peak is at half the flight time: y = 0 at t = 6, so the vertex is at t = 3 with y(3) = 90 − 45 = 45 m. (B) is the landing time; (C) forgets the −5t^2 term at t = 3 (90 is merely 30 × 3).

2. The area of one petal of the rose r = cos(2θ) is

• (A) π/2
• (B) π/4
• (C) π/8
• (D) π/16

Answer: (C). (1/2)∫(−π/4 to π/4) cos^2(2θ) dθ = π/8. (A) is the total area of all four petals; the question's "one petal" is the whole test.

3. The total area enclosed by the cardioid r = 2 + 2 cos θ is

• (A) 4π
• (B) 6π
• (C) 8π
• (D) 12π

Answer: (B). (1/2)∫(0 to 2π) (2 + 2 cos θ)^2 dθ = 6π, matching the general formula (3/2)π a^2 with a = 2. (A) is the answer if the cross-term and cos^2 contribution are dropped — a reminder that the expansion must be done fully.

Constructed response. A projectile is launched with x(t) = 40t, y(t) = 30t − 5t^2 (metres and seconds, g = 10 m/s²). (a) Find the flight time and the range. (b) Find the maximum height and when it occurs. (c) Eliminate t to find the Cartesian equation of the path, and use it to find the height of the projectile 200 m downrange. (d) Set up and evaluate the polar integral for the area of one petal of r = 3 sin(2θ).

Scoring guide. (a) 2 points: 30t − 5t^2 = 0 gives t = 6 s — 1 point; range x(6) = 240 m — 1 point. (b) 2 points: vertex at t = 3 s (half the flight, or from y'(t) = 30 − 10t = 0 in spirit) — 1 point; y(3) = 45 m — 1 point. (c) 2 points: t = x/40 gives y = (3/4)x − x^2/320 — 1 point; at x = 200: y = 150 − 40000/320 = 150 − 125 = 25 m — 1 point. (d) 2 points: one petal spans θ ∈ [0, π/2] (between consecutive zeros of sin 2θ), so Area = (1/2)∫(0 to π/2) 9 sin^2(2θ) dθ = 9π/8 — 1 point for the setup with correct limits, 1 point for 9π/8 ≈ 3.53.

Going deeper: area between polar curves, and arc length

Area between two polar curves subtracts sector integrals: Area = (1/2)∫(r_outer^2 − r_inner^2) dθ over the angles where the region exists. Worked example: the region inside the circle r = 2 cos θ but outside the circle r = 1. The curves meet where 2 cos θ = 1, i.e. θ = ±π/3, and between those angles 2 cos θ ≥ 1. Area = (1/2)∫(−π/3 to π/3) (4 cos^2 θ − 1) dθ. Since 4 cos^2 θ = 2 + 2 cos 2θ, the integrand is 1 + 2 cos 2θ, with antiderivative θ + sin 2θ; evaluating from −π/3 to π/3 gives (π/3 + sqrt(3)/2) − (−π/3 − sqrt(3)/2) = 2π/3 + sqrt(3). Halving: Area = π/3 + sqrt(3)/2 ≈ 1.913. Sanity check: the full circle r = 2 cos θ has area π ≈ 3.14, and we are removing a substantial bite (the r = 1 disc's overlap), so landing near 1.9 is geometrically plausible. The recipe generalises: find intersections first (they set the limits), identify which curve is outer on the interval, subtract squares, halve.

Arc length closes the parametric story: for x = f(t), y = g(t), the speed is sqrt((dx/dt)^2 + (dy/dt)^2) and length is its integral over t. The unit circle x = cos t, y = sin t: speed = sqrt(sin^2 t + cos^2 t) = 1, so length over [0, 2π] is 2π — the circumference, recovered from the parametric machinery with the Pythagorean identity doing the work. Distance travelled by the projectile of the main lesson is the same integral with its component speeds — set up by this formula, evaluated by calculus.

Workshop: speed along the path, and a polar sector

Speed from components. The projectile x(t) = 40t, y(t) = 30t − 5t^2 has velocity components (40, 30 − 10t). Speeds: at launch |(40, 30)| = 50 m/s; at the peak (t = 3) |(40, 0)| = 40 m/s — slowest at the top, purely horizontal; at landing (t = 6) |(40, −30)| = 50 m/s again. The symmetry (launch speed = landing speed, angles mirrored) is energy conservation in disguise, and the component view makes it a computation rather than a claim: speed is the magnitude of the velocity *vector*, which is Lesson 6 operating inside Lesson 18.

A polar sector. The area swept by r = 4 cos θ for θ from 0 to π/3: (1/2)∫(0 to π/3) 16 cos^2 θ dθ = 8 × [θ/2 + sin(2θ)/4] from 0 to π/3 = 8(π/6 + sqrt(3)/8) = 4π/3 + sqrt(3) ≈ 5.92. Partial sweeps like this — a sector of a known curve — are the standard exam form of polar area: the limits are given or found from intersections, the sin^2/cos^2 identity does the integration, and the factor 1/2 must survive to the end.