Integrating Vector-Valued Functions

Part 1 · Hook

Why this matters

A ship's navigator without GPS knows where the ship started and logs its velocity east and north every hour. From those two facts alone she can say where the ship is now: start plus the accumulated change in each direction. That is dead reckoning, and it is exactly how you integrate a velocity vector.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $\frac{dy}{dt} = 6t$ and $y(0) = 5$ , then $y(t) =$

$3t^2 + 5$
$3t^2$
$6t + 5$
$3t^2 - 5$

Show the answer

Antidifferentiate, $3t^2 + C$ , then $y(0) = C = 5$ .

Correct: $3t^2 + 5$ :
$3t^2$ :
$6t + 5$ :
$3t^2 - 5$ :

2. A particle on a line has velocity $v(t)$ and position $s(0) = 2$ . Its position at $t = 3$ is

$2 + \int_0^3 v(t)\,dt$
$\int_0^3 v(t)\,dt$
$\int_0^3 v(t)\,dt - 2$
$2 + v(3)$

Show the answer

Position at a time is the starting position plus the accumulated change.

Correct: $2 + \int_0^3 v(t)\,dt$ :
$\int_0^3 v(t)\,dt$ :
$\int_0^3 v(t)\,dt - 2$ :
$2 + v(3)$ :

3. If $\mathbf{r}(t) = \left\langle t^3, 4t \right\rangle$ , then $\mathbf{r}'(t) =$

$\left\langle 3t^2, 4 \right\rangle$
$\left\langle 3t^2, 4t \right\rangle$
$\left\langle \frac{t^4}{4}, 2t^2 \right\rangle$
$\left\langle 6t, 0 \right\rangle$

Show the answer

Differentiate each component.

Correct: $\left\langle 3t^2, 4 \right\rangle$ :
$\left\langle 3t^2, 4t \right\rangle$ :
$\left\langle \frac{t^4}{4}, 2t^2 \right\rangle$ :
$\left\langle 6t, 0 \right\rangle$ :

Part 4 · See it

See it first

Two paths with the same velocity v(t) = 1, 2t - 2 for t from 0 to 2.8. The emerald path starts at (0, 0) and the blue path starts at (2, 1); the blue path is the emerald one shifted 2 right and 1 up. At t = 2 both carry the same vermillion velocity vector. The antiderivative of v is t, t squared minus 2t, plus a constant vector C: C = 0, 0 for the emerald path and C = 2, 1 for the blue one. The initial position picks C: r(t) = r(0) plus the integral of v. — The velocity $\mathbf{v}(t) = \left\langle 1, 2t - 2 \right\rangle$ has the antiderivatives $\left\langle t, t^2 - 2t \right\rangle + \mathbf{C}$ : one path for each constant vector. The starting point, here $(0, 0)$ or $(2, 1)$ , decides which path the particle follows. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The velocity vector is $\left\langle x'(t), y'(t) \right\rangle$ antidifferentiate each component, adding a constant vector $\mathbf{C} = \left\langle C_1, C_2 \right\rangle$
An initial position $\mathbf{r}(0)$ is knownit fixes $\mathbf{C}$ and picks out the one path the particle follows
The integral of a rate over an interval is the net change $\mathbf{r}(b)$ $= \mathbf{r}(a)$ $+ \int_a^b \mathbf{v}(t)\,dt$ , one component at a time
The velocity is known only from a tableapproximate each component's integral with a Riemann or trapezoidal sum, then add the start

Part 6 · Key ideas

Key ideas

Integrate a vector one component at a time: $\int \left\langle f(t), g(t) \right\rangle\,dt$ $= \left\langle F(t), G(t) \right\rangle$ $+ \mathbf{C}$ .
Position from velocity always adds the start: $x(b)$ $= x(a)$ $+ \int_a^b x'(t)\,dt$ and $y(b)$ $= y(a)$ $+ \int_a^b y'(t)\,dt$ .
From acceleration, integrate twice: $\mathbf{v}(0)$ fixes the first constant and $\mathbf{r}(0)$ the second.
$\int_a^b y'(t)\,dt$ is the change in $y$ from $t = a$ to $t = b$ , not $y(b)$ and not a distance.

Part 7 · Misconception

A common mistake

The wrong idea: $\mathbf{r}(0) = \left\langle 2, 5 \right\rangle$ and $\int_0^3 \mathbf{v}(t)\,dt$ $= \left\langle 4, -1 \right\rangle$ , so $\mathbf{r}(3)$ $= \left\langle 4, -1 \right\rangle - \left\langle 2, 5 \right\rangle$ $= \left\langle 2, -6 \right\rangle$ .

What actually happens: The integral is the change in position, and a change is added to where you started: $\mathbf{r}(3)$ $= \left\langle 2, 5 \right\rangle + \left\langle 4, -1 \right\rangle$ $= \left\langle 6, 4 \right\rangle$ . Subtracting the start, or leaving it out, is the most common lost point on this topic.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A boat's velocity components and its starting point

A remote-controlled boat moves on a lake. Its velocity components are differentiable functions of time

t

, in minutes, and at

t = 0

the boat is at the point

(3, -1)

. Distances are in meters.

Velocity components of the boat at selected times
$t$ (minutes)	$x'(t)$ (meters per minute)	$y'(t)$ (meters per minute)
$0$	$4$	$1$
$2$	$6$	$-2$
$5$	$3$	$0$
$8$	$-1$	$5$

1. Use a trapezoidal sum with the three subintervals in the table to approximate $x(8)$ , the boat's $x$ -coordinate at $t = 8$ . Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$\int_0^8 x'(t)\,dt$ $\approx \frac{4 + 6}{2}(2)$ $+ \frac{6 + 3}{2}(3)$ $+ \frac{3 + (-1)}{2}(3)$ $= 10$ $+ 13.5$ $+ 3$ $= 26.5$ . Then $x(8)$ $= x(0)$ $+ \int_0^8 x'(t)\,dt$ $\approx 3$ $+ 26.5$ $= 29.5$ meters.

Answer: 29.5 meters

2. Use a left Riemann sum with the three subintervals in the table to approximate $y(8)$ , the boat's $y$ -coordinate at $t = 8$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Left sum: $1(2) + (-2)(3) + 0(3)$ $= -4$ . So $y(8)$ $= y(0)$ $+ \int_0^8 y'(t)\,dt$ $\approx -1$ $+ (-4)$ $= -5$ meters.

Answer: -5

3. What does $\int_2^5 y'(t)\,dt$ represent?

The boat's $y$ -coordinate at $t = 5$ , in meters
The change in the boat's $y$ -coordinate, in meters, from $t = 2$ to $t = 5$
The boat's average vertical velocity, in meters per minute, from $t = 2$ to $t = 5$
The distance, in meters, that the boat travels from $t = 2$ to $t = 5$

Show the answer

The integral of a rate over an interval is the net change of the quantity: $\int_2^5 y'(t)\,dt$ $= y(5) - y(2)$ , in meters.

The boat's $y$ -coordinate at $t = 5$ , in meters: To get $y(5)$ you would add a known value, such as $y(2)$ , to the change.
Correct: The change in the boat's $y$ -coordinate, in meters, from $t = 2$ to $t = 5$ : Right: $y(5) - y(2)$ .
The boat's average vertical velocity, in meters per minute, from $t = 2$ to $t = 5$ : An average would divide by $5 - 2 = 3$ minutes.
The distance, in meters, that the boat travels from $t = 2$ to $t = 5$ : Distance along the path uses the speed $\sqrt{(x')^2 + (y')^2}$ , not $y'$ alone.

4. Using one trapezoid on $[0, 2]$ for each component, which point best approximates the boat's position at $t = 2$ ?

$(10, -1)$
$(7, 0)$
$(13, -2)$
$(11, 1)$

Show the answer

$x(2)$ $\approx 3 + \frac{4 + 6}{2}(2)$ $= 13$ and $y(2)$ $\approx -1 + \frac{1 + (-2)}{2}(2)$ $= -2$ .

$(10, -1)$ : These are the changes in $x$ and $y$ ; add the starting point $(3, -1)$ .
$(7, 0)$ : This subtracts the starting point from the changes; the start is added.
Correct: $(13, -2)$ : Right: start plus the trapezoid approximation of each change.
$(11, 1)$ : This uses the left-end rates only, $4(2)$ and $1(2)$ ; a trapezoid averages both ends.

5. A particle has velocity $\mathbf{v}(t) = \left\langle 6t^2, \cos t \right\rangle$ and $\mathbf{r}(0) = \left\langle 1, 2 \right\rangle$ . Find $x(t)$ .

Type an expression for $x(t) =$ .

Show the answer

$x(t)$ $= x(0)$ $+ \int_0^t 6s^2\,ds$ $= 1$ $+ 2t^3$ . (Likewise $y(t) = 2 + \sin t$ .)

Answer: $x(t) =$ $2t^{3} + 1$

6. Find $\int \left\langle e^{2t}, \frac{1}{t} \right\rangle\,dt$ .

$\left\langle 2e^{2t}, -\frac{1}{t^2} \right\rangle + \mathbf{C}$
$\left\langle \frac{1}{2}e^{2t}, \ln|t| \right\rangle + \mathbf{C}$
$\left\langle e^{2t}, \ln|t| \right\rangle + \mathbf{C}$
$\left\langle \frac{1}{2}e^{2t}, -\frac{1}{t^2} \right\rangle + \mathbf{C}$

Show the answer

Antidifferentiate each component: $\int e^{2t}\,dt = \frac{1}{2}e^{2t}$ and $\int \frac{1}{t}\,dt = \ln|t|$ . The constant is a vector $\mathbf{C} = \left\langle C_1, C_2 \right\rangle$ , one constant per component.

$\left\langle 2e^{2t}, -\frac{1}{t^2} \right\rangle + \mathbf{C}$ : These are derivatives of the components; the integral reverses differentiation.
Correct: $\left\langle \frac{1}{2}e^{2t}, \ln|t| \right\rangle + \mathbf{C}$ : Right: one antiderivative per component, plus a constant vector.
$\left\langle e^{2t}, \ln|t| \right\rangle + \mathbf{C}$ : Differentiating $e^{2t}$ gives $2e^{2t}$ , so the antiderivative needs the factor $\frac{1}{2}$ .
$\left\langle \frac{1}{2}e^{2t}, -\frac{1}{t^2} \right\rangle + \mathbf{C}$ : $-\frac{1}{t^2}$ is the derivative of $\frac{1}{t}$ ; its antiderivative is $\ln|t|$ .

7. A particle starts at $\mathbf{r}(0) = \left\langle 1, 1 \right\rangle$ , and $\int_0^4 \mathbf{v}(t)\,dt$ $= \left\langle 5, -2 \right\rangle$ . Where is the particle at $t = 4$ ?

$\left\langle 5, -2 \right\rangle$
$\left\langle 6, -1 \right\rangle$
$\left\langle 4, -3 \right\rangle$
$\left\langle 1, 1 \right\rangle$

Show the answer

The integral of velocity is the change in position: $\mathbf{r}(4)$ $= \mathbf{r}(0)$ $+ \int_0^4 \mathbf{v}(t)\,dt$ $= \left\langle 1 + 5, 1 - 2 \right\rangle$ $= \left\langle 6, -1 \right\rangle$ .

$\left\langle 5, -2 \right\rangle$ : That is the change in position from $t = 0$ to $t = 4$ ; add it to the start.
Correct: $\left\langle 6, -1 \right\rangle$ : Right: start plus change.
$\left\langle 4, -3 \right\rangle$ : The start is added to the change, not subtracted from it.
$\left\langle 1, 1 \right\rangle$ : That is where the particle started; the integral says it moved.

Part 9 · Summary

Summary

To integrate a vector-valued function, integrate each component; an indefinite integral carries a constant vector

\mathbf{C}

. An initial condition fixes it: the position at time

b

is the starting position plus the integral of the velocity,

\mathbf{r}(b)

= \mathbf{r}(a)

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

. From a table, approximate each component's integral with a sum, then add the start.

Part 10 · Up next

What comes next

Part 11 · Connections