Lesson 17: Parametric Equations, Polar Coordinates, and Calculus

Some curves are badly served by y = f(x): a projectile's path is fine, but a planet's orbit, a rolling wheel's rim point, or a cardioid needs either a parameter (position as x(t), y(t) driven by a third variable) or polar coordinates (position by distance r and angle θ). Calculus extends to both: derivatives become ratios of rates, and areas become integrals of (1/2) r^2. This lesson develops the two toolkits in parallel, with the same concepts-first discipline — every formula is traced to the slicing or rate idea behind it.

Parametric curves: slope as a ratio of rates

If x = x(t) and y = y(t), then dy/dx = (dy/dt)/(dx/dt) — the chain rule (Lesson 5) rearranged: the vertical rate per unit of horizontal rate. Horizontal tangents occur where dy/dt = 0 (and dx/dt ≠ 0); vertical tangents where dx/dt = 0 (and dy/dt ≠ 0). Speed along the curve is sqrt( (dx/dt)^2 + (dy/dt)^2 ), and arc length integrates the speed.

Worked Example 1. Let x = t^2 − 1, y = t^3 − 3t. Then dy/dx = (3t^2 − 3)/(2t). Horizontal tangents: 3t^2 − 3 = 0 gives t = ±1, at points (0, −2) and (0, 2). Vertical tangent: 2t = 0 gives t = 0, at (−1, 0). At t = 2 the point is (3, 2) with slope 9/4. One parameter value, complete local information — this is the working style of parametric analysis.

Worked Example 2 (arc length). The circle x = 3 cos t, y = 3 sin t, 0 ≤ t ≤ 2π: dx/dt = −3 sin t, dy/dt = 3 cos t, speed = sqrt(9 sin^2 t + 9 cos^2 t) = 3, constant. Length = ∫ from 0 to 2π of 3 dt = 6π — the circumference, as it must be. When the machinery reproduces a known answer exactly, trust in its unknown answers is earned.

Polar curves: area as slices of pie

In polar form r = r(θ), a small sector between θ and θ + dθ is approximately a triangle/wedge of area (1/2) r^2 dθ — the sector formula from geometry, with dθ in place of the angle. Hence A = (1/2) ∫ r^2 dθ over the angle range that traces the region once.

Worked Example 3 (cardioid). r = 1 + cos θ, traced once for 0 ≤ θ ≤ 2π:

A = (1/2)∫ from 0 to 2π of (1 + cos θ)^2 dθ = (1/2)∫ (1 + 2 cos θ + cos^2 θ) dθ
   = (1/2)( 2π + 0 + π ) = 3π/2 ≈ 4.71

(using ∫ cos θ = 0 over a full period and ∫ cos^2 θ dθ = π). Worked Example 4 (check against geometry). r = 2 sin θ for 0 ≤ θ ≤ π is the circle of radius 1 centred at (0, 1). Its area by the formula: (1/2)∫ from 0 to π of 4 sin^2 θ dθ = (1/2)(4)(π/2) = π — matching π·1^2 exactly ✓. And slopes in polar: convert via x = r cos θ, y = r sin θ and differentiate with respect to θ, then take the ratio. For the cardioid at θ = π/2: r = 1, dr/dθ = −1; dx/dθ = −1, dy/dθ = −1, so the tangent slope is 1 at the point (0, 1).

Application: projectiles and Kepler

A projectile launched at 25 m/s... in components: x(t) = 20t, y(t) = 15t − 4.9t^2 (launch velocity split into 20 horizontal, 15 vertical). Parametric thinking separates the motions: flight time from y = 0 is t = 15/4.9 ≈ 3.06 s; range = 20 × 3.06 ≈ 61.2 m; peak at t = 15/9.8 ≈ 1.53 s with height 15(1.53) − 4.9(1.53)^2 ≈ 11.5 m. Eliminating t gives the parabola y = (3/4)x − (4.9/400)x^2 — the trajectory as a single equation, for when you need it. For polar: a planet sweeps area at rate dA/dt = (1/2) r^2 (dθ/dt), and Kepler's second law says this rate is constant — equal areas in equal times. That constancy is conservation of angular momentum in disguise, and it is why planets hurry near the Sun (small r forces large dθ/dt). The area formula of this lesson is the mathematical core of an orbital law.

Common mistakes

• Polar area without the 1/2 or the square. Wrong: A = ∫ r dθ. Right: (1/2)∫ r^2 dθ — each slice is a sector, and sector area is (1/2) r^2 θ. The formula's shape comes from geometry; recall the geometry and the formula follows.
• Tracing the region twice (or half). r = 2 sin θ traces its circle once for θ in [0, π], not [0, 2π]; integrating to 2π doubles the area. Always ask: over what θ-range is the curve traced exactly once?
• Parametric slope as dy/dt alone. Slope is vertical rate per horizontal rate; forgetting to divide by dx/dt confuses steepness with speed. A curve can have dy/dt enormous and slope zero — if dx/dt is more enormous still.
• Arc length by ∫ dy or by the area integrand. Length integrates speed sqrt( (dx/dt)^2 + (dy/dt)^2 ); area integrates (1/2) r^2 (polar) or height (Cartesian). Three different slices for three different quantities — mixing them is the most common error in this lesson.

Exam Experience

1. If x = t^3 and y = t^2, then dy/dx at t = 2 equals

• (A) 2/3
• (B) 1/3
• (C) 3
• (D) 4/3

Answer: (B). (dy/dt)/(dx/dt) = 2t/(3t^2) = 2/(3t) = 2/6 = 1/3 at t = 2.

2. The area enclosed by the polar curve r = 2 (a circle of radius 2), computed with the polar area formula, is

• (A) 2π
• (B) 4π
• (C) 8π
• (D) π

Answer: (B). (1/2)∫ from 0 to 2π of 4 dθ = (1/2)(4)(2π) = 4π = π·2^2 ✓. The formula reproduces elementary geometry where it should.

3. A particle moves with x = cos t, y = sin t. Its speed at any time is

• (A) 0
• (B) t
• (C) 1
• (D) 2π

Answer: (C). sqrt( sin^2 t + cos^2 t ) = 1 — unit circle, unit speed: one radian per second means one unit of arc per second.

Free-response question. Consider the cardioid r = 1 + cos θ. (a) Find the area enclosed by the cardioid. (b) Find the slope of the tangent line at θ = π/2 and the equation of that line. (c) Set up, and evaluate, the integral for the total arc length of the cardioid. (Hint: r^2 + (dr/dθ)^2 = 2 + 2 cos θ = 4 cos^2(θ/2).)

Scoring guideline.

• (a) 2 points. 1 point for the setup (1/2)∫ from 0 to 2π of (1 + cos θ)^2 dθ. 1 point for the evaluation 3π/2 (expanding, using ∫ cos = 0 and ∫ cos^2 = π over [0, 2π]).
• (b) 1 point. Point (0, 1); parametric derivatives dx/dθ = −1, dy/dθ = −1 at θ = π/2 give slope 1; line y = x + 1.
• (c) 1 point. L = ∫ from 0 to 2π of sqrt(2 + 2 cos θ) dθ = ∫ from 0 to 2π of 2|cos(θ/2)| dθ = 8. The absolute value (splitting at θ = π) must appear in the working for full credit; the integrand set-up alone earns the point if the evaluation slips.

Going deeper: area between two polar curves, and curvature of a parametric path

Area between polar curves follows the same top-minus-bottom logic as Lesson 10, with (1/2)(R^2 − r^2): the area inside r = 2 cos θ (a circle of radius 1) but outside r = 1: the curves meet where 2 cos θ = 1, θ = ±π/3, and

A = (1/2) ∫ from −π/3 to π/3 of (4 cos^2 θ − 1) dθ = ∫ from 0 to π/3 of (4 cos^2 θ − 1) dθ
  = [ θ + sin(2θ) ] from 0 to π/3 = π/3 + sqrt(3)/2 ≈ 1.913

(using 4 cos^2 θ = 2 + 2 cos(2θ)). Symmetry halved the work — always check for it before integrating in polar form, where symmetric limits are common and endpoints at multiples of π invite it. Second derivatives, parametrically. d^2y/dx^2 is not (d^2y/dt^2)/(d^2x/dt^2); it is the derivative of dy/dx with respect to x, computed as (d/dt (dy/dx))/(dx/dt). For Worked Example 1's curve (x = t^2 − 1, y = t^3 − 3t): dy/dx = (3t^2 − 3)/(2t) = (3/2)(t − 1/t); differentiating in t gives (3/2)(1 + 1/t^2), which at t = 2 is (3/2)(5/4) = 15/8; dividing by dx/dt = 2t = 4 gives d^2y/dx^2 = 15/32 > 0 — at the point (3, 2) the path is concave up, information the slope alone (9/4) could not give.

One more petal, and the area bookkeeping of roses

For r = cos(2θ), one petal is traced as θ runs from −π/4 to π/4 (where cos(2θ) ≥ 0). Its area:

A = (1/2) ∫ from −π/4 to π/4 of cos^2(2θ) dθ = π/8

(evaluating the integral: cos^2 averages 1/2 over the symmetric interval of length π/2, giving (1/2)(1/2)(π/2) = π/8). The full four-petal rose has area 4 × π/8 = π/2 — compare the enclosing unit circle's π: the rose fills exactly half of it, a memorable calibration. Rose bookkeeping generalises: r = cos(kθ) has k petals for odd k and 2k for even k, and integrating (1/2) r^2 over the full tracing range counts each petal once only if you track the tracing — polar area problems reward a plotted curve more than any other topic in this lesson.