Integrating Vector-Valued Functions

AP Calculus BC · Parametric, Polar & Vector-Valued Functions

Integrating Vector-Valued Functions

Two antiderivatives, two constants — and one initial condition that settles both.

key idea$\int\vec{r}(t)\,dt=\Big\langle \int x(t)\,dt,\ \int y(t)\,dt\Big\rangle+\vec{C}$

The rule

Integration, like differentiation, acts on each component separately. The only feature with no scalar counterpart is the constant: it is a vector $\vec{C}=\langle C_1,C_2\rangle$ with one unknown per component, and an initial condition supplies both at once.

Indefinite vector integral

\[\int\big\langle x(t),y(t)\big\rangle dt=\Big\langle \int x\,dt+C_1,\ \int y\,dt+C_2\Big\rangle\]

Two independent antiderivatives, each with its own constant. Writing a single scalar $C$ is the characteristic error — it forces the two components to share an offset they have no reason to share.

Definite vector integral

\[\int_a^b\vec{r}(t)\,dt=\Big\langle \int_a^b x\,dt,\ \int_a^b y\,dt\Big\rangle\]

No constant, and the result is a vector. Applied to velocity it gives the displacement over $[a,b]$ — a vector, not a distance.

You are given	You want	Compute
$\vec{v}(t)$ and $\vec{r}(a)$	$\vec{r}(b)$	$\vec{r}(a)+\displaystyle\int_a^b\vec{v}\,dt$
$\vec{a}(t)$ and $\vec{v}(a)$	$\vec{v}(b)$	$\vec{v}(a)+\displaystyle\int_a^b\vec{a}\,dt$
$\vec{v}(t)$	displacement on $[a,b]$	$\displaystyle\int_a^b\vec{v}\,dt$ — a vector
$\vec{v}(t)$	distance travelled	$\displaystyle\int_a^b\left\lVert\vec{v}\right\rVert dt$ — a scalar
$\vec{v}(t)$	the path's length	the same integral, over one traversal

Rows 3 and 4 are the pair that decides most questions. Displacement integrates the vector; distance integrates its magnitude — and the magnitude must go inside the integral, never outside.

Displacement$\displaystyle\int\vec{v}\,dt$ — a vector

Distance$\displaystyle\int\left\lVert\vec{v}\right\rVert dt$ — a number

The constantis a vector, one per component

Pattern recognition

four reminders

Antidifferentiate each component on its own; nothing couples them.
The constant of integration is $\vec{C}=\langle C_1,C_2\rangle$ — two unknowns.
One initial condition, being a vector equation, determines both constants simultaneously.
$\left\lVert\displaystyle\int\vec{v}\right\rVert\neq\displaystyle\int\left\lVert\vec{v}\right\rVert$ whenever the direction changes.

Worked examples

EX 1

Position from velocity

one initial condition, two constants

A particle has $\vec{v}(t)=\left\langle 2t,\,\cos t\right\rangle$ and $\vec{r}(0)=\langle1,\,3\rangle$. Find $\vec{r}(t)$.

Antidifferentiate componentwise

$\displaystyle\int 2t\,dt=t^{2}$ and $\displaystyle\int\cos t\,dt=\sin t$.

Attach a vector constant

$\vec{r}(t)=\left\langle t^{2}+C_1,\ \sin t+C_2\right\rangle$.

Apply the initial condition

$\vec{r}(0)=\langle C_1,\,C_2\rangle=\langle1,3\rangle$.

Read both constants

$C_1=1$ and $C_2=3$ — one vector equation, two scalar facts.

Write the answer

$\vec{r}(t)=\left\langle t^{2}+1,\ \sin t+3\right\rangle$.

EX 2

Displacement versus distance

the same interval, two answers

For $\vec{v}(t)=\langle\cos t,\,\sin t\rangle$ on $[0,2\pi]$, find the displacement and the distance travelled.

Integrate the vector

$\displaystyle\int_{0}^{2\pi}\cos t\,dt=0$ and $\displaystyle\int_{0}^{2\pi}\sin t\,dt=0$.

State the displacement

$\langle0,0\rangle$ — the particle returns exactly to where it started.

Compute the speed

$\left\lVert\vec{v}\right\rVert=\sqrt{\cos^{2}t+\sin^{2}t}=1$.

Integrate the speed

$\displaystyle\int_{0}^{2\pi}1\,dt=2\pi$.

Compare

Zero displacement, $2\pi$ of distance. The magnitude must be taken before integrating, or the travel disappears entirely.

EX 3

Climbing two levels

acceleration to position

A particle has $\vec{a}(t)=\langle6t,\,-2\rangle$, $\vec{v}(0)=\langle-1,\,4\rangle$ and $\vec{r}(0)=\langle0,\,0\rangle$. Find $\vec{r}(t)$.

Integrate the acceleration

$\vec{v}(t)=\left\langle3t^{2}+C_1,\ -2t+C_2\right\rangle$.

Use the velocity condition

$\vec{v}(0)=\langle C_1,C_2\rangle=\langle-1,4\rangle$, so $\vec{v}(t)=\left\langle3t^{2}-1,\ -2t+4\right\rangle$.

Integrate again

$\vec{r}(t)=\left\langle t^{3}-t+D_1,\ -t^{2}+4t+D_2\right\rangle$.

Use the position condition

$\vec{r}(0)=\langle D_1,D_2\rangle=\langle0,0\rangle$.

Write the answer

$\vec{r}(t)=\left\langle t^{3}-t,\ -t^{2}+4t\right\rangle$. Each integration needed its own vector constant and its own condition.

EX 4

Position at a later time

definite form

A particle at $\langle2,\,-1\rangle$ when $t=1$ has $\vec{v}(t)=\left\langle\dfrac{1}{t},\ 4t\right\rangle$. Find its position at $t=3$.

Choose the definite form

$\vec{r}(3)=\vec{r}(1)+\displaystyle\int_{1}^{3}\vec{v}(t)\,dt$ — no constants needed.

Integrate the first component

$\displaystyle\int_{1}^{3}\dfrac{dt}{t}=\ln3$.

Integrate the second

$\displaystyle\int_{1}^{3}4t\,dt=\left[2t^{2}\right]_{1}^{3}=18-2=16$.

Add to the starting position

$\vec{r}(3)=\langle2+\ln3,\ -1+16\rangle$.

Simplify

$\langle2+\ln3,\ 15\rangle\approx\langle3.10,\,15\rangle$. The definite form avoids solving for constants entirely.

Watch out

The constant of integration is a vector

$\vec{C}=\langle C_1,C_2\rangle$, with two unknowns. A single scalar $C$ would force both components to shift by the same amount.

One vector condition fixes both constants

$\vec{r}(0)=\langle a,b\rangle$ is two equations at once. You do not need a second condition.

Take the magnitude inside the integral

Distance is $\int\left\lVert\vec{v}\right\rVert dt$, not $\left\lVert\int\vec{v}\,dt\right\rVert$. The second is the length of the displacement, which can be zero for a long journey.

Displacement is a vector; report it as one

$\int_a^b\vec{v}\,dt$ has two components. Collapsing it to a number answers a question that was not asked.

Course map. Topic 9.5 · LO FUN-8.A — determine a particular solution given a rate vector and initial conditions. BC-only.