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AP Calculus BC · Cram sheet

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

10–15% of the AP exam 16 key terms

● Core concept  ·  ○ Supporting concept

9.1 Defining and Differentiating Parametric Equations

Parametric equations ○ — A way of describing a curve with two functions of a parameter: x = x(t) and y = y(t). As the parameter t varies, the pair (x(t), y(t)) traces out the curve.

Derivative of a parametric function ● (core concept) — For a curve given parametrically, dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0. Its value at a point is the slope of the line tangent to the curve there.

9.2 Second Derivatives of Parametric Equations

Second derivative of a parametric function ● (core concept) — d²y/dx² = [d/dt(dy/dx)] / (dx/dt): differentiate the first parametric derivative with respect to t, then divide by dx/dt.

9.3 Finding Arc Lengths of Curves Given by Parametric Equations

Arc length of a parametric curve ● (core concept) — For x = x(t), y = y(t) from t₁ to t₂, the curve's length is the definite integral of √((dx/dt)² + (dy/dt)²) dt.

9.4 Defining and Differentiating Vector-Valued Functions

Vector-valued function ○ — A function r(t) = ⟨x(t), y(t)⟩ whose output is a vector; each component is an ordinary function of the parameter t, and it traces a curve in the plane.

Derivative of a vector-valued function ● (core concept) — Differentiate component by component: r′(t) = ⟨x′(t), y′(t)⟩. The same derivative rules used for real-valued functions apply to each component.

9.5 Integrating Vector-Valued Functions

Integral of a vector-valued function ● (core concept) — Integrate component by component, the same way real-valued functions are integrated. With a rate vector and initial conditions, this produces the particular position function of a particle.

Position from a velocity vector ● (core concept) — Given a particle's velocity vector and its initial position, its position function is the antiderivative of the velocity vector that satisfies the initial conditions — an initial value problem in the plane.

9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions

Velocity vector ● (core concept) — For a particle in planar motion, v(t) = ⟨x′(t), y′(t)⟩ — the derivative of the position vector. It gives both the rate of motion and its direction.

Speed of a particle in planar motion ● (core concept) — The magnitude of the velocity vector: |v(t)| = √((x′(t))² + (y′(t))²). Speed is a nonnegative scalar, unlike velocity.

Acceleration in planar motion ● (core concept) — a(t) = v′(t) = ⟨x″(t), y″(t)⟩ — the rate of change of the velocity vector of a particle moving in the plane.

Displacement vs. distance traveled ● (core concept) — Over a time interval, displacement (net change in position) is the definite integral of the velocity vector, while total distance traveled is the definite integral of speed.

9.7 Defining Polar Coordinates and Differentiating in Polar Form

Polar coordinates ○ — A way of locating a point by (r, θ), where r is the distance from the origin and θ is the angle from the positive x-axis. Conversions: x = r cos θ, y = r sin θ, r² = x² + y², tan θ = y/x.

Derivative in polar form ● (core concept) — For a curve given by r = f(θ), dy/dx = (dy/dθ)/(dx/dθ). Derivatives of r, x, and y with respect to θ — and first and second derivatives of y with respect to x — give information about the curve's behavior.

9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

Area of a polar region ● (core concept) — For r = f(θ) swept from θ = α to θ = β, the area is ½ times the definite integral of r² dθ — the polar analogue of accumulating area.

9.9 Finding the Area of the Region Bounded by Two Polar Curves

Area between two polar curves ● (core concept) — A = ½∫(r_outer² − r_inner²) dθ, integrated over the θ-interval where one curve lies outside the other. The curves' intersection points determine the limits of integration.