Short on time? This page is the recap. For the explanations, the path diagram and three worked examples, use the full study guide. BC-only material. Prerequisites (vector derivatives 9.4, FTC 6.7, antiderivatives and substitution 6.8–6.9, particular solutions 7.7) are on the Calculus BC roadmap.
Recap
- You differentiate a vector one component at a time, so you integrate it one component at a time too.
- An indefinite integral of a vector has a constant vector C = ⟨C₁, C₂⟩. The two constants are usually different.
- A definite integral of a vector is a vector: one number per component.
- Being given dx/dt and dy/dt is the same as being given the vector ⟨dx/dt, dy/dt⟩.
- An initial condition (a known position or velocity at one time) fixes the constants.
Key relationships
| Idea | Formula | Note |
|---|---|---|
| Indefinite integral | ∫ ⟨f(t), g(t)⟩ dt = ⟨F(t) + C₁, G(t) + C₂⟩ | F′ = f, G′ = g |
| Definite integral | ∫ from a to b of ⟨f, g⟩ dt = ⟨∫ from a to b of f dt, ∫ from a to b of g dt⟩ | Answer is a vector |
| Net change | ∫ from a to b of r′(t) dt = r(b) − r(a) | FTC, applied to each component |
| Position from velocity | r(b) = r(a) + ∫ from a to b of v(t) dt | Add the known position |
| Velocity from acceleration | v(b) = v(a) + ∫ from a to b of a(t) dt | Then integrate again for position |
| Accumulation form | r(t) = r(t₀) + ∫ from t₀ to t of r′(s) ds | Use when no antiderivative can be written |
Assumptions
- Angles are in radians.
- Calculator integrals: keep 4 or more decimal places, round to 3 at the end.
- An earlier time works too: r(0) = r(1) − ∫ from 0 to 1 of r′(t) dt.
Mistakes to avoid
- One shared constant for both components.
- Using t = 0 when the condition is given at another time.
- Stopping at the net change and calling it the position.
- Skipping v(t₀) when working up from acceleration.
- Adding the components of a definite integral to get one number.
- Missing a chain-rule factor: ∫ sin(3t) dt = −⅓ cos(3t).
- A lost minus sign when going back in time.
Quick self-check
- Evaluate ∫ from 0 to π/2 of ⟨cos t, 2t⟩ dt. (⟨1, π²/4⟩)
- r′(t) = ⟨4t, 3⟩ and r(0) = ⟨1, −2⟩. Find r(2). (⟨9, 4⟩)
- r′(t) = ⟨1/t, 2t⟩ for t > 0 and r(1) = ⟨0, 3⟩. Find r(e). (⟨1, e² + 2⟩, since r(t) = ⟨ln t, t² + 2⟩)
Next: practice questions.