SAT / PSAT
SAT / PSAT Prep
History & Social Science
AP World History AP US History AP European History AP Human Geography AP US Government & Politics AP Psychology AP Macroeconomics AP Microeconomics
English
AP English Language & Composition AP English Literature & Composition
Math & Computer Science
AP Calculus AB/BC AP Precalculus AP Statistics AP Computer Science A AP Computer Science Principles
Sciences
AP Biology AP Chemistry AP Environmental Science AP Physics 1 AP Physics 2
World Languages & Arts
AP Spanish Language AP Art History AP Music Theory Start studying →
Unit 9 · Parametric, Polar & Vector (BC) Flashcards Cheat Sheet Essentials Visual Review MC Practice FRQ Practice

AP Calculus AB/BC Unit 9 Visual Review

A topic-by-topic visual walkthrough of Unit 9 (BC): Parametric, Polar & Vector-Valued Functions — differentiating and integrating them, plus polar area.

← Back to Unit 9 hub
TOPIC 9.1 BC ONLY Differentiating Parametric Equations dy/dx = (dy/dt) / (dx/dt) Worked example x = t², y = t³ dx/dt = 2t, dy/dt = 3t² dy/dx = 3t²/2t = (3/2)t Divide the two time-derivatives — no need to eliminate the parameter. Tangent lines & special points HORIZONTAL tangent where dy/dt = 0 (and dx/dt ≠ 0). VERTICAL tangent where dx/dt = 0 (and dy/dt ≠ 0). Evaluate dy/dx at a t-value for a slope there. Parametric slope: dy/dx = (dy/dt)/(dx/dt). The Review Hub · AP Calculus AB/BC Unit 9 TOPIC 9.2 BC ONLY Second Derivatives (Parametric) d²y/dx² = [ d/dt (dy/dx) ] / (dx/dt) Differentiate dy/dx, divide by dx/dt The second derivative is NOT the ratio of the second time-derivatives. Take d/dt of your dy/dx expression, then divide once more by dx/dt. d²y/dx² ≠ (d²y/dt²)/(d²x/dt²) Worked example From 9.1: dy/dx = (3/2)t d/dt(dy/dx) = 3/2 d²y/dx² = (3/2)/(2t) = 3/(4t) Its sign gives concavity along the curve. Concave up where d²y/dx² > 0. d²y/dx² = [d/dt(dy/dx)] / (dx/dt) — differentiate the slope, then divide again. The Review Hub · AP Calculus AB/BC Unit 9 TOPIC 9.3 BC ONLY Parametric Arc Length L = ∫_t₁^t₂ √( (dx/dt)² + (dy/dt)² ) dt Integrate the speed The integrand √((x′)²+(y′)²) is the SPEED of the particle. Arc length = total distance the particle travels from t₁ to t₂. Usually evaluated with a calculator. Worked example x = cos t, y = sin t, 0 ≤ t ≤ 2π: √(sin²t + cos²t) = 1 L = ∫₀^2π 1 dt = 2π The circumference of the unit circle — as expected. Parametric arc length = ∫√((x′)²+(y′)²) dt — integrate the speed. The Review Hub · AP Calculus AB/BC Unit 9 TOPIC 9.4 BC ONLY Differentiating Vector Functions r(t) = ⟨x(t), y(t)⟩ → r′(t) = ⟨x′(t), y′(t)⟩ Differentiate componentwise A vector-valued function packages the position; differentiate each component. r′(t) = VELOCITY vector ⟨x′, y′⟩. r″(t) = ACCELERATION vector ⟨x″, y″⟩. SPEED = |r′(t)| = √((x′)²+(y′)²). Worked example r(t) = ⟨t², sin t⟩ r′(t) = ⟨2t, cos t⟩ r″(t) = ⟨2, −sin t⟩ Velocity points along the direction of motion at time t. Differentiate componentwise: r′=⟨x′,y′⟩=velocity; speed = |r′|. The Review Hub · AP Calculus AB/BC Unit 9 TOPIC 9.5 BC ONLY Integrating Vector Functions ∫ r(t) dt = ⟨ ∫x(t) dt, ∫y(t) dt ⟩ Integrate componentwise Antidifferentiate each component separately. Recover VELOCITY from acceleration, or POSITION from velocity. Add a vector constant, found from initial data. Position from velocity x(t) = x(t₀) + ∫_t₀^t vₓ dt y(t) = y(t₀) + ∫_t₀^t v_y dt Use the initial position to pin down the constant of integration in each component. Displacement = ∫ velocity (a vector). ∫ Integrate each component separately — recover velocity or position. The Review Hub · AP Calculus AB/BC Unit 9 TOPIC 9.6 BC ONLY Planar Motion Problems QuantityHow to find it velocity vector⟨x′(t), y′(t)⟩ speed√((x′)² + (y′)²) total distance on [a,b]∫ₐᵇ √((x′)²+(y′)²) dt position at time tinitial + ∫ velocity dt Same ideas as straight-line motion, in 2D Differentiate for velocity/acceleration; integrate speed for distance; use initial conditions for position. In 2D motion: velocity=⟨x′,y′⟩, speed=|v|, distance=∫speed dt. The Review Hub · AP Calculus AB/BC Unit 9 TOPIC 9.7 BC ONLY Differentiating in Polar Form x = r cos θ, y = r sin θ, with r = f(θ) dy/dx = (dy/dθ) / (dx/dθ) Treat polar as parametric in θ Convert to x(θ) and y(θ) using r = f(θ), then differentiate each with respect to θ. dx/dθ = r′cos θ − r sin θ, dy/dθ = r′sin θ + r cos θ The slope of the tangent line is the ratio dy/dθ ÷ dx/dθ. dr/dθ tells you how fast the radius grows as the angle sweeps. Polar: with x=r cosθ, y=r sinθ, use dy/dx = (dy/dθ)/(dx/dθ). The Review Hub · AP Calculus AB/BC Unit 9 TOPIC 9.8 BC ONLY Area of a Polar Region r = f(θ) sweep θ from α to β A = ½ ∫_α^β r² dθ Sum of thin circular sectors Each slice is a tiny wedge of area ½r²dθ. r = 2: A = ½∫₀^2π 4 dθ = 4π (matches πr² = π·2² = 4π for a circle.) α and β are the angles bounding the region. Find them where r = 0 or where the curve closes. Polar area = ½∫r² dθ — sum of thin circular sectors. The Review Hub · AP Calculus AB/BC Unit 9 TOPIC 9.9 BC ONLY Area Between Two Polar Curves A = ½ ∫_α^β ( [r_outer]² − [r_inner]² ) dθ Outer minus inner (squared) Like the washer method: subtract the inner curve's r² from the outer curve's r². Find the INTERSECTION angles first by setting the two r-expressions equal. Those angles become α and β. Watch which curve is outer Over the interval, the curve with the LARGER r is the outer boundary. If they swap, split the integral at the crossing angle. Sketch both curves to identify the region. Between two polar curves: A = ½∫(r_out² − r_in²)dθ. The Review Hub · AP Calculus AB/BC Unit 9
1 / 9

How to use the visual review

Spend 30 seconds per slide before clicking next. Look at the diagram, then ask yourself: "Could I set up this derivative or integral from the picture?"

Use the fullscreen button () on desktop for the best experience. Use arrow keys to navigate. Tap "Show all slides" to jump around.

This is great for review the night before the exam — fast, visual, and covers every idea you need to recognize in Unit 9.