Topic 9.5 Notes – Integrating Vector-Valued Functions

What Integrating a Vector-Valued Function Means

A vector-valued function describes motion in the plane or space:

$$\mathbf{r}(t)=\langle x(t), y(t) \rangle \quad \text{or} \quad \langle x(t), y(t), z(t) \rangle$$

Think of it as tracking a particle’s coordinates over time.

You already know:

• Velocity: $\mathbf{v}(t)=\mathbf{r}'(t)$
• Acceleration: $\mathbf{a}(t)=\mathbf{v}'(t)=\mathbf{r}''(t)$

Integration moves you backward through that chain:

• Integrate acceleration → get velocity
• Integrate velocity → get position

The key fact:

$$\int \mathbf{v}(t)\,dt = \left\langle \int v_x(t)\,dt,\; \int v_y(t)\,dt,\; \int v_z(t)\,dt \right\rangle$$

You integrate each component separately. Nothing fancy beyond that.

Indefinite and Definite Integrals of Vectors

Indefinite Integrals

If
$\mathbf{v}(t)=\langle f(t), g(t) \rangle$,

then

$$\int \mathbf{v}(t)\,dt = \langle \int f(t)\,dt,\; \int g(t)\,dt \rangle$$

Each component gets its own constant:

$$\mathbf{r}(t)=\langle F(t)+C_1,\; G(t)+C_2 \rangle$$

In 3D, you’ll have three constants.

Students often forget this and write just one $C$. On a free-response question, that costs points.

Definite Integrals

$$\int_a^b \mathbf{v}(t)\,dt = \left\langle \int_a^b v_x(t)\,dt,\; \int_a^b v_y(t)\,dt \right\rangle$$

This represents displacement, meaning:

$$\mathbf{r}(b)-\mathbf{r}(a)$$

It tells you the net change in position, not how far the particle traveled.

That distinction shows up a lot on quizzes and the AP exam.

Solving an Initial Value Problem for Motion

This is the main skill for this topic.

You’re typically given:

• A rate vector (velocity or acceleration)
• An initial condition like $\mathbf{r}(0)=\langle 2,-1\rangle$

The Process

1. Identify what you're given

   • Acceleration → integrate to get velocity.
   • Velocity → integrate to get position.

2. Integrate component-wise

   • Add constants $C_1, C_2, (C_3)$.

3. Use the initial condition

   • Plug in the given time.
   • Solve for each constant separately.

4. Write the particular solution

   • No constants left.

Quick Example

Suppose
$\mathbf{v}(t)=\langle 4t, \cos t \rangle$
and $\mathbf{r}(0)=\langle 3,2\rangle$.

Integrate:

$$\mathbf{r}(t)=\langle 2t^2 + C_1,\; \sin t + C_2 \rangle$$

Apply $\mathbf{r}(0)=\langle 3,2\rangle$:

• $2(0)^2 + C_1 = 3 \Rightarrow C_1=3$
• $\sin 0 + C_2 = 2 \Rightarrow C_2=2$

Final answer:

$$\mathbf{r}(t)=\langle 2t^2 + 3,\; \sin t + 2 \rangle$$

That’s the complete particular position function.

Position vs Displacement vs Distance

These get mixed up constantly.

Position

$\mathbf{r}(t)$
The location at time $t$.

Displacement

$$\int_a^b \mathbf{v}(t)\,dt$$

A vector from starting point to ending point.

Total Distance Traveled

This uses speed, which is the magnitude of velocity:

$$|\mathbf{v}(t)|=\sqrt{(v_x)^2+(v_y)^2}$$

$$\text{Distance}=\int_a^b |\mathbf{v}(t)|\,dt$$

Notice the absolute value bars. That makes it a scalar.

If the problem says “how far,” they usually mean total distance, not displacement.

Visualizing the Difference

Here’s the idea geometrically.

Distance versus displacement

The curved path length is total distance.
The straight arrow is displacement.