Aim for FiveCalculus BC

Unit 9 · Topic 9.5

BC only

9.5 Integrating Vector-Valued Functions

BC only. You integrate a vector-valued function one component at a time, with a separate constant for each component. Given a velocity vector and a starting position, this lets you find the position at any later time.

Key terms

  • antiderivative
  • component
  • constant of integration
  • initial condition

Antiderivatives component by component

This whole unit is BC only. ∫ ⟨f(t), g(t)⟩ dt = ⟨∫ f(t) dt, ∫ g(t) dt⟩ = ⟨F(t) + C₁, G(t) + C₂⟩.

Each component gets its own constant, because the x- and y-motions are independent. You can also write the constant as a single vector C = ⟨C₁, C₂⟩.

Position from velocity

If v(t) = ⟨x′(t), y′(t)⟩ and you know the position at time a, then for each component:

x(b) = x(a) + ∫ₐᵇ x′(t) dt and y(b) = y(a) + ∫ₐᵇ y′(t) dt.

In vector form: r(b) = r(a) + ∫ₐᵇ v(t) dt. The integral of the velocity vector is the displacement vector, the net change in position. It's the same net change idea as in 8.2, done twice.

Two ways to use the initial condition

  • Find the general antiderivative with C₁ and C₂, then plug in the known time and position to solve for each constant. This gives a formula for r(t) at all times.
  • Use definite integrals from the known time, as above. This is the fastest route when you need the position at one time, and it's the only route when the integral needs a calculator.

Velocity from acceleration

The same idea works one level down: v(b) = v(a) + ∫ₐᵇ a(t) dt, component by component. If you start from acceleration and want position, integrate twice, using an initial velocity and an initial position for the two sets of constants.

Reading the displacement vector

The vector ∫ₐᵇ v(t) dt tells you how far the particle ended up from where it started, in each direction. If ∫₀³ v(t) dt = ⟨4, −2⟩, the particle finished 4 units to the right of its starting point and 2 units below it, whatever path it took. The straight-line distance between the start and end points is √(4² + 2²) = √20, but the distance traveled along the path is usually longer (9.6).

A quick check

Differentiate the position you found. You should get back the velocity you started with, and at the starting time your position should match the given point. Doing this for each component catches the most common slip: a wrong or missing constant.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1

    Position formula from velocity

    A particle has velocity v(t) = ⟨2t, 3t² − 1⟩ and position r(0) = ⟨1, 4⟩. Find r(t) and r(2).

    Show the solution
    1. Step 1: Integrate each component: x(t) = t² + C₁ and y(t) = t³ − t + C₂.
    2. Step 2: At t = 0: x(0) = C₁ = 1 and y(0) = C₂ = 4.
    3. Step 3: So r(t) = ⟨t² + 1, t³ − t + 4⟩.
    4. Step 4: r(2) = ⟨4 + 1, 8 − 2 + 4⟩ = ⟨5, 10⟩.

    Answer: r(t) = ⟨t² + 1, t³ − t + 4⟩; r(2) = ⟨5, 10⟩

  2. Example 2Calculator allowed

    Calculator: one component has no formula

    A particle has velocity v(t) = ⟨cos(t²), e^(t/2)⟩ and is at (3, −1) when t = 0. Find its position at t = 2.

    Show the solution
    1. Step 1: x(2) = 3 + ∫₀² cos(t²) dt. There's no elementary antiderivative, so use the calculator: ∫₀² cos(t²) dt ≈ 0.461. So x(2) ≈ 3.461.
    2. Step 2: y(2) = −1 + ∫₀² e^(t/2) dt = −1 + [2e^(t/2)] from 0 to 2 = −1 + 2e − 2 = 2e − 3 ≈ 2.437.
    3. Step 3: Position ≈ (3.461, 2.437).

    Answer: About (3.461, 2.437)

  3. Example 3

    Trap: one constant for both components

    A student integrates v(t) = ⟨cos t, e²ᵗ⟩ and writes r(t) = ⟨sin t, ½e²ᵗ⟩ + C, then uses r(0) = ⟨2, 5⟩ to say C = 2. What's wrong?

    Show the solution
    1. Step 1: The constant is a vector, ⟨C₁, C₂⟩, not one number.
    2. Step 2: x(0) = sin 0 + C₁ = 2, so C₁ = 2.
    3. Step 3: y(0) = ½e⁰ + C₂ = 5, so C₂ = 5 − ½ = 9/2.
    4. Step 4: Correct: r(t) = ⟨sin t + 2, ½e²ᵗ + 9/2⟩.

    Answer: r(t) = ⟨sin t + 2, ½e²ᵗ + 9/2⟩

Common mistakes

  • Using one constant for both components.
  • Forgetting to add the starting position: ∫₀² v(t) dt is the displacement, not the position.
  • Plugging the starting time into the wrong component.
  • Rounding the x-integral early and carrying the error into later parts.

On the exam

  • A typical BC free-response part: “The particle is at (3, −1) at t = 0. Find its position at t = 2.” Write x(2) = 3 + ∫₀² x′(t) dt and y(2) = −1 + ∫₀² y′(t) dt, then the values.
  • If a component has a simple antiderivative, you can still use the calculator for it on calculator-active questions, but show the setup.

Connected topics

Videos

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  • ❖ Integrating a Vector Function to Find Position ❖

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Check yourself

4 questions on 9.5 Integrating Vector-Valued Functions. Pick an answer to see if you got it, and why.

Question 1 of 4

A particle moves in the xy-plane with velocity vector v(t) = ⟨6t², 2e²ᵗ⟩. At time t = 0, the particle is at the point (1, 3). What is the position vector of the particle at t = 1?

Question 2 of 4

∫ from 0 to π/2 of ⟨cos t, 2t⟩ dt =

Question 3 of 4

A particle moves in the xy-plane with acceleration vector a(t) = ⟨6t, −2⟩. At t = 0 its velocity is ⟨1, 4⟩ and its position is ⟨0, 0⟩. What is the position of the particle at t = 2?

Question 4 of 4

A particle moves in the xy-plane with velocity vector v(t) = ⟨2t, 1/(t + 1)⟩ for t ≥ 0. At time t = 0, the particle is at the point (3, −2). What is the position vector of the particle at time t = e − 1?

0 of 4 answered