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How do you integrate a vector-valued function to recover velocity and position?

Topic 9.5 Integrating Vector-Valued Functions: integrate a vector-valued function component by component to recover velocity from acceleration and position from velocity, using initial conditions (BC).

A focused answer to AP Calculus BC Topic 9.5, integrating a vector-valued function component by component to recover velocity from acceleration and position from velocity, applying initial conditions, with worked examples.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. Component-wise integration and the constants
  3. A worked recovery of velocity and position
  4. Displacement by definite integration
  5. Distance traveled versus displacement, again
  6. A worked displacement

What this topic is asking

The College Board (Topic 9.5, BC only) reverses Topic 9.4: instead of differentiating position to get velocity and acceleration, you integrate acceleration to recover velocity, and velocity to recover position. As with all antidifferentiation, each integration introduces a constant, which an initial condition pins down, here a constant in each component.

Component-wise integration and the constants

A worked recovery of velocity and position

Displacement by definite integration

A faster route to a later position uses the definite-integral form of the Fundamental Theorem in each component. If you know r(t0)\mathbf{r}(t_0) and the velocity, then

r(t1)=r(t0)+∫t0t1v(t) dt,\mathbf{r}(t_1) = \mathbf{r}(t_0) + \int_{t_0}^{t_1} \mathbf{v}(t)\,dt,

where the definite integral ∫t0t1v(t) dt=⟨∫t0t1x′(t) dt, ∫t0t1y′(t) dt⟩\int_{t_0}^{t_1}\mathbf{v}(t)\,dt = \langle \int_{t_0}^{t_1} x'(t)\,dt,\ \int_{t_0}^{t_1} y'(t)\,dt\rangle is the displacement vector. This avoids solving for constants when only the final position is wanted: integrate velocity over the interval and add the starting position. The two components are handled independently, exactly as in one-dimensional motion, but bundled into a vector.

Distance traveled versus displacement, again

Integrating the velocity vector gives displacement, the net change in position, which can be small or zero even for a long journey. To get the total distance traveled, you integrate the speed (the magnitude of velocity), a scalar:

distance=∫t0t1∣v(t)∣ dt=∫t0t1(x′(t))2+(y′(t))2 dt.\text{distance} = \int_{t_0}^{t_1} |\mathbf{v}(t)|\,dt = \int_{t_0}^{t_1}\sqrt{(x'(t))^2 + (y'(t))^2}\,dt.

This is precisely the parametric arc-length integral of Topic 9.3. So the two integrals answer two different questions: integrate the vector for where the particle ends up (displacement, then add to start for position), and integrate the magnitude for how far it travelled (distance). Mixing them, for instance integrating speed and calling it the new position, is a classic error.

A worked displacement

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (BC, style)1 marksSection I (multiple choice, no calculator). A particle has velocity ⟨2t,3⟩\langle 2t, 3\rangle and position ⟨1,0⟩\langle 1, 0\rangle at t=0t = 0. Its position at t=1t = 1 is (A) ⟨2,3⟩\langle 2, 3\rangle (B) ⟨1,3⟩\langle 1, 3\rangle (C) ⟨2,0⟩\langle 2, 0\rangle (D) ⟨0,3⟩\langle 0, 3\rangle
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The correct answer is (A), ⟨2,3⟩\langle 2, 3\rangle.

x(1)=1+∫012t dt=1+[t2]01=1+1=2x(1) = 1 + \int_0^1 2t\,dt = 1 + [t^2]_0^1 = 1 + 1 = 2. y(1)=0+∫013 dt=0+3=3y(1) = 0 + \int_0^1 3\,dt = 0 + 3 = 3. Position is ⟨2,3⟩\langle 2, 3\rangle.

AP 2024 (BC, style)4 marksSection II (free response). A particle has acceleration ⟨6t,−2⟩\langle 6t, -2\rangle, velocity ⟨1,4⟩\langle 1, 4\rangle at t=0t = 0, and position ⟨0,0⟩\langle 0, 0\rangle at t=0t = 0. (a) Find the velocity vector. (b) Find the position vector.
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A 4-point antiderivative-with-initial-conditions problem.

(a) (2 points) v(t)=⟨∫6t dt,∫−2 dt⟩=⟨3t2+C1,−2t+C2⟩\mathbf{v}(t) = \langle \int 6t\,dt, \int -2\,dt\rangle = \langle 3t^2 + C_1, -2t + C_2\rangle. Using v(0)=⟨1,4⟩\mathbf{v}(0) = \langle 1, 4\rangle: C1=1C_1 = 1, C2=4C_2 = 4, so v(t)=⟨3t2+1,−2t+4⟩\mathbf{v}(t) = \langle 3t^2 + 1, -2t + 4\rangle.
(b) (2 points) r(t)=⟨t3+t+D1,−t2+4t+D2⟩\mathbf{r}(t) = \langle t^3 + t + D_1, -t^2 + 4t + D_2\rangle. Using r(0)=⟨0,0⟩\mathbf{r}(0) = \langle 0, 0\rangle: D1=D2=0D_1 = D_2 = 0, so r(t)=⟨t3+t,−t2+4t⟩\mathbf{r}(t) = \langle t^3 + t, -t^2 + 4t\rangle.

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