Unit 9 · Parametric Equations, Polar Coordinates, and Vector-Valued Functions

Unit 9

Parametric Equations, Polar Coordinates, and Vector-Valued Functions

~3 weeks · March · BC only · 10–15%

Why it's here now

This is the first genuine widening of what a "function" is, and it exists to break an assumption that has been quietly limiting everything since Unit 1: that curves are graphs of y = f(x).

They aren't. Circles aren't. Orbits aren't. Anything that loops, crosses itself, stops, or reverses isn't. The machinery all still works — you just have to describe the curve differently.

It's placed here because it needs both differentiation and integration in hand, and because Unit 10 will also require the idea that functions can be built in unfamiliar ways.

Parametric curves

Parametrization Instead of y = f(x), give both coordinates as functions of a third variable: x = x(t), y = y(t). Think of t as time and the curve as the path traced by a moving point. The variable t is the parameter; it doesn't appear on the graph.

Why this is the more natural description

It's how Newton thought — kinematically, in terms of quantities flowing in time. And it's strictly more general: every function y = f(x) can be parametrized as x = t, y = f(t), but most parametric curves cannot be written as functions.

Crucially, the curve and the parametrization are different objects. The unit circle is (cos t, sin t), but it's also (cos 2t, sin 2t) — same path, traversed twice as fast. Speed, direction, and starting point are properties of the parametrization, not of the curve.

Derivatives

dy/dx = (dy/dt)/(dx/dt)

Chain rule: dy/dt = (dy/dx)·(dx/dt), so divide.

Trap — the second derivative

d²y/dx² is not (d²y/dt²)/(d²x/dt²). That's the natural guess and it's wrong.

The correct move: you already have dy/dx as a function of t. Differentiate that with respect to t, then divide by dx/dt again:

d²y/dx² = [ d/dt (dy/dx) ] / (dx/dt)

The logic: to convert any t-derivative into an x-derivative you divide by dx/dt. Do it once for the first derivative, once more for the second.

Worked — both derivatives

For x = t², y = t³ − 3t, find dy/dx and d²y/dx² at t = 2.

dx/dt = 2t, dy/dt = 3t² − 3
dy/dx = (3t² − 3)/(2t)	at t = 2: (12−3)/4 = 9/4
rewrite: dy/dx = (3/2)t − (3/2)t⁻¹	easier to differentiate in this form
d/dt(dy/dx) = 3/2 + (3/2)t⁻²	at t = 2: 3/2 + 3/8 = 15/8
d²y/dx² = (15/8)/(2·2) = 15/32	divide by dx/dt one more time

Motion, speed, and arc length

Quantity	Formula
Velocity vector	⟨dx/dt, dy/dt⟩
Speed	√((dx/dt)² + (dy/dt)²)
Acceleration vector	⟨d²x/dt², d²y/dt²⟩
Distance travelled (arc length)	∫_a^b√((dx/dt)² + (dy/dt)²) dt
Displacement	⟨∫dx/dt dt, ∫dy/dt dt⟩ — a vector
Position at time b	⟨x(a) + ∫_a^bx′dt, y(a) + ∫_a^by′dt⟩

Why the arc length formula is cleaner here

It's literally ∫ |velocity| dt — distance equals speed integrated over time, the most ordinary fact in physics.

Compare Unit 8's Cartesian version, ∫√(1 + (dy/dx)²)dx. Same Pythagorean origin (ds² = dx² + dy²), but the parametric form is symmetric in x and y and doesn't break when the curve goes vertical. The parametric formula is the real one; the Cartesian one is the special case x = t.

Trap — speed is a scalar, velocity is a vector

Speed is the magnitude of the velocity vector and is never negative. On the AP, "how fast is the particle moving" wants the scalar; "find the velocity" wants the ordered pair. And total distance is ∫speed dt, while displacement is the vector of separate integrals — they're different objects, not just different numbers.

Check — second route

Parametric slope is (dy/dt)/(dx/dt). On x = t, y = t², that gives 2t/1 = 2t = 2x — exactly what you get differentiating y = x² directly ✓.

Whenever a parametric curve can also be written as a plain function, the two routes have to agree. It is the quickest way to confirm you have the quotient the right way up.

Polar coordinates

Polar coordinates Locate a point by distance from the origin (r) and angle from the positive x-axis (θ), rather than by horizontal and vertical offsets. Conversions: x = r cos θ, y = r sin θ, and back via r² = x² + y², tan θ = y/x. A polar curve is given as r = f(θ) — the distance from the origin, as a function of direction.

Why polar exists — coordinates should match the symmetry of the problem

Some situations have rotational symmetry, and forcing them into a rectangular grid produces ugly algebra. A circle of radius 3 is x² + y² = 9 in Cartesian and simply r = 3 in polar.

The best example: Kepler's second law says a planet sweeps equal areas in equal times. In Cartesian coordinates that's a mess. In polar it's ½r²(dθ/dt) = constant — which is exactly the polar area formula differentiated, and it's conservation of angular momentum. The physics becomes visible when the coordinates match the geometry.

Area in polar

A = ½ ∫_α^β r² dθ

Derivation — why it isn't ∫y dx

slice the region into thin sectors, not rectangles	a wedge from the origin, spanning angle dθ
a sector of radius r and angle dθ has area ½r²dθ	the Unit 1 sector formula: fraction dθ/2π of the full circle πr²
sum the wedges: A = ½∫r²dθ

The whole formula is the sector-area fact from the very first unit, used as a slice. Slice shapes follow the coordinate system — rectangles in Cartesian, wedges in polar, washers for revolution. That's the unifying idea behind every area and volume formula in the course.

Trap — the limits of integration

Getting α and β right is the entire difficulty in polar area problems, and it's where nearly all the errors live.

Find where r = 0 — those angles are usually where a petal or loop begins and ends.
Watch for retracing. r = cos 2θ (a four-petal rose) draws its full picture over 0 ≤ θ ≤ 2π, but r = cos 3θ (three petals) completes over 0 ≤ θ ≤ π and then redraws. Integrating to 2π double-counts.
Exploit symmetry: compute one petal and multiply.

Sketch first, always. In polar, the algebra will not warn you that you've gone around twice.

Worked — area of one petal

Find the area of one petal of r = 2cos(3θ).

r = 0 when cos3θ = 0 → 3θ = ±π/2 → θ = ±π/6	the petal starts and ends where r hits zero
A = ½∫_−π/6^π/64cos²(3θ)dθ	r² = 4cos²(3θ)
cos²(3θ) = (1 + cos6θ)/2	the Unit 3 half-angle identity — the only way to integrate cos²
= ∫_−π/6^π/6(1 + cos6θ)dθ	the 2·½·... constants collapse
= [θ + sin(6θ)/6]_−π/6^π/6
= (π/6 + 0) − (−π/6 + 0) = π/3	sin(±π) = 0

Slope in polar

There's no separate formula worth memorizing. Convert to parametric with θ as the parameter:

x = r(θ)cos θ, y = r(θ)sin θ, then dy/dx = (dy/dθ)/(dx/dθ)

Trap — dr/dθ is not a slope

dr/dθ tells you how fast the distance from the origin changes as you sweep around. It is not the slope of the tangent line and it is not dy/dx. Students substitute one for the other constantly. If a question asks for a tangent line, you must go through x and y.

Common polar curves

Equation	Shape
r = a	circle of radius a centred at the origin
θ = c	a line through the origin
r = a cos θ or a sin θ	circle of diameter a, through the origin
r = a(1 ± cos θ)	cardioid — heart-shaped
r = a ± b cos θ	limaçon; has an inner loop when a < b
r = a cos(nθ)	rose: n petals if n is odd, 2n petals if n is even

Vector-valued functions

Just parametric equations packaged as a single object: r(t) = ⟨x(t), y(t)⟩. Differentiate and integrate componentwise.

Why this packaging matters

It's the on-ramp to multivariable calculus. Once position, velocity, and acceleration are vectors, everything in mechanics can be written in one line instead of two, and the same notation survives into three dimensions and beyond.

It also makes a distinction visible that scalars hide: acceleration can change a particle's direction without changing its speed. Uniform circular motion has constant speed and constant-magnitude acceleration pointing always toward the centre. In one dimension that's impossible; in two it's the most common motion in the universe.

One thing to show her — the brachistochrone

A bead slides under gravity from A down to a lower point B. Which curve gets it there fastest?

Not the straight line. The answer is an upside-down cycloid — the path traced by a point on the rim of a rolling wheel, which is naturally parametric: x = t − sin t, y = 1 − cos t.

Johann Bernoulli posed it as a public challenge in 1696. Newton, then 55 and running the Royal Mint, received it in the evening, solved it overnight, and published anonymously. Bernoulli recognized the author immediately: "tanquam ex ungue leonem" — one knows the lion by its claw.

The same cycloid is also the tautochrone: a bead released from any point on it reaches the bottom in the same time. Huygens used that to design a pendulum clock in 1659, before the calculus existed to explain why it worked.

Solving it properly requires the calculus of variations — optimizing over a space of functions rather than over numbers. That framework is the whole of Lagrangian mechanics, and it's what a sophomore who likes this material might chase next.

Beyond BC · what a college course does here

Polar arc length, which BC omits even though it's two lines from what she has. Treat r = f(θ) as parametric with θ as the parameter: x = r cos θ, y = r sin θ. Differentiate, square, add, and the cross terms cancel by sin² + cos² = 1:

L = ∫_α^β √( r² + (dr/dθ)² ) dθ

Check — known case

Sanity check on a circle r = a: dr/dθ = 0, so L = ∫₀^2π a dθ = 2πa. ✓

Conic sections in polar, with a focus at the origin — the form that makes orbital mechanics tractable:

r = ed / (1 + e·cos θ)

One equation, and the eccentricity e selects the shape: e = 0 circle, 0 < e < 1 ellipse, e = 1 parabola, e > 1 hyperbola. In Cartesian coordinates these are four different-looking equations; in polar they are one equation with a dial. That is the entire argument for polar coordinates, in a single formula — and it's why Newton could derive Kepler's laws at all.

Beyond BC · the Russian approach — the cycloid, done properly

The main text tells the brachistochrone story. The tradition would make her derive the curve's properties, because it's the single richest object available at this level.

A wheel of radius a rolls along the x-axis. The path of a point on its rim is

x = a(t − sin t), y = a(1 − cos t)

Everything in Unit 9 falls out of one curve:

Arc length of one arch: √(x′² + y′²) = a√(2 − 2cos t) = 2a·sin(t/2) using the half-angle identity. Integrating from 0 to 2π gives 8a — exactly four diameters, with no π in it at all, which is startling for a curve generated by a circle.
Area under one arch: ∫y dx = ∫a(1−cos t)·a(1−cos t)dt = 3πa² — exactly three times the area of the rolling circle. Galileo tried to determine this by weighing paper cutouts and got it approximately; Roberval proved it in 1634.
Cusps: at t = 0, both x′ and y′ vanish, so dy/dx is 0/0. The curve has a cusp where the rim point momentarily stops — the contact point of a rolling wheel is instantaneously at rest, which is why the bottom of a moving car's tyre isn't blurred in a photograph.

Why this is the right problem: it exercises parametric derivatives, arc length, area, and the failure mode of dy/dx, all on one curve, all with clean answers, and every answer is surprising. That's the tradition's actual pedagogy — fewer problems, each one carrying more.

Formula sheet

Unit 9 — Parametric, Polar, Vector

Parametric

dy/dx = (dy/dt)/(dx/dt)
d²y/dx² = [d/dt(dy/dx)] / (dx/dt) — not a ratio of second derivatives
speed = √((dx/dt)² + (dy/dt)²) · arc length = ∫speed dt
Horizontal tangent: dy/dt = 0 (and dx/dt ≠ 0). Vertical: dx/dt = 0 (and dy/dt ≠ 0).
The curve and the parametrization are different objects.

Polar

x = r cosθ · y = r sinθ · r² = x² + y² · tanθ = y/x
Area = ½∫_α^β r²dθ — slices are sectors, not rectangles
Limits: find where r = 0; watch for retracing; use symmetry
Slope: convert to x(θ), y(θ) and use the parametric formula. dr/dθ is not a slope.
Rose r = a cos(nθ): n petals if n odd, 2n if n even

Vector

r(t) = ⟨x(t), y(t)⟩ · v = r′ · a = r″ — all componentwise
Speed = |v| (scalar) · velocity = v (vector) · distance = ∫|v|dt · displacement = the vector of integrals

Identities you'll reach for

cos²θ = (1 + cos2θ)/2 · sin²θ = (1 − cos2θ)/2 — required for nearly every polar area

Practise Unit 9 → Fresh problems from this unit, with worked solutions.