Vector-Valued Functions — AP® Precalculus Lesson

Part 1 · Hook

Why this matters

A delivery drone reports its position as an arrow from the depot: so far east, so far north. As the minutes pass the arrow swings and stretches, and its tip draws the flight path. Comparing two arrows tells you how far the drone got and in which direction, on average, it was heading.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\left\langle 6, 9 \right\rangle - \left\langle 2, 5 \right\rangle$ ?

$\left\langle 4, 4 \right\rangle$
$\left\langle 8, 14 \right\rangle$
$\left\langle -4, -4 \right\rangle$
$\left\langle 3, 4 \right\rangle$

Show the answer

Subtract matching components: $6 - 2 = 4$ and $9 - 5 = 4$ .

Correct: $\left\langle 4, 4 \right\rangle$ :
$\left\langle 8, 14 \right\rangle$ :
$\left\langle -4, -4 \right\rangle$ :
$\left\langle 3, 4 \right\rangle$ :

2. For $x(t) = 2t$ and $y(t) = 6t - t^2$ , where is the point at $t = 3$ ?

$(6, 9)$
$(6, 27)$
$(9, 6)$
$(3, 9)$

Show the answer

$x(3) = 6$ and $y(3) = 18 - 9 = 9$ .

Correct: $(6, 9)$ :
$(6, 27)$ :
$(9, 6)$ :
$(3, 9)$ :

Part 4 · See it

See it first

The path of p(t) = <2t, 6t minus t squared> for t from 0 to 6, an arch from (0, 0) up to (6, 9) at t = 3 and down to (12, 0) at t = 6, with points at each whole t. Position vectors p(1) = <2, 5> and p(3) = <6, 9> are drawn from the origin in blue, and the displacement vector p(3) minus p(1) = <4, 4> is drawn from the tip of p(1) to the tip of p(3) in vermillion. — The position vectors $\mathbf{p}(1) = \left\langle 2, 5 \right\rangle$ and $\mathbf{p}(3) = \left\langle 6, 9 \right\rangle$ of $\mathbf{p}(t) = \left\langle 2t, 6t - t^2 \right\rangle$ , and the displacement $\mathbf{p}(3) - \mathbf{p}(1) = \left\langle 4, 4 \right\rangle$ between them. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Write a parametric function as $\mathbf{p}(t) = \left\langle x(t), y(t) \right\rangle$ each $t$ gives a position vector from the origin to the point
Let $t$ increasethe tip of the position vector traces the same path as the parametric function
Subtract $\mathbf{p}(t_1)$ from $\mathbf{p}(t_2)$ the displacement vector, from the first position to the second; its magnitude is the distance between them
Divide the displacement by $t_2 - t_1$ the average rate of change of $\mathbf{p}$ : the average rates of $x$ and $y$ , as one vector

Part 6 · Key ideas

Key ideas

A vector-valued function $\mathbf{p}(t)$ $= \left\langle x(t), y(t) \right\rangle$ $= x(t)\mathbf{i}$ $+ y(t)\mathbf{j}$ gives a vector for each input $t$ .
Drawn from the origin, $\mathbf{p}(t)$ is a position vector; its tip traces the curve $(x(t), y(t))$ .
Displacement on $[t_1, t_2]$ : $\mathbf{p}(t_2) - \mathbf{p}(t_1)$ . Distance between the two positions: its magnitude.
Average rate of change of $\mathbf{p}$ on $[t_1, t_2]$ : $\frac{\mathbf{p}(t_2) - \mathbf{p}(t_1)}{t_2 - t_1}$ , whose components are the average rates of change of $x$ and $y$ .

Part 7 · Misconception

A common mistake

The wrong idea: The magnitude of the average rate of change of $\mathbf{p}$ is the average speed along the path.

What actually happens: It is the straight-line distance between the two positions divided by the time. If the path curves or doubles back, the distance traveled along it is longer, so the average speed along the path is larger.

Part 8 · Check yourself

Check yourself

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

Table

A delivery drone

Position of a drone, in meters east (x) and north (y) of its depot, t seconds after takeoff
$t$ (seconds)	0	5	10	15
$x(t)$ (meters)	0	40	60	60
$y(t)$ (meters)	0	30	70	120

1. What is the drone's displacement from $t = 0$ to $t = 10$ ?

$\left\langle 60, 70 \right\rangle$
$\left\langle 70, 60 \right\rangle$
$\left\langle 6, 7 \right\rangle$
$\left\langle 60, 120 \right\rangle$

Show the answer

$\mathbf{p}(10) - \mathbf{p}(0)$ $= \left\langle 60 - 0, 70 - 0 \right\rangle$ $= \left\langle 60, 70 \right\rangle$ meters.

Correct: $\left\langle 60, 70 \right\rangle$ : Read both rows at $t = 10$ and subtract the start.
$\left\langle 70, 60 \right\rangle$ : The components are swapped; east comes first.
$\left\langle 6, 7 \right\rangle$ : This divides by $10$ : that is a rate, not a displacement.
$\left\langle 60, 120 \right\rangle$ : $120$ is $y(15)$ , from the wrong column.

2. What is the average rate of change of the drone's position vector on $[5, 15]$ , in meters per second?

$\left\langle 2, 9 \right\rangle$
$\left\langle 20, 90 \right\rangle$
$\left\langle 6, 12 \right\rangle$
$\left\langle 4, 3 \right\rangle$

Show the answer

$\frac{\left\langle 60, 120 \right\rangle - \left\langle 40, 30 \right\rangle}{15 - 5}$ $= \frac{\left\langle 20, 90 \right\rangle}{10}$ $= \left\langle 2, 9 \right\rangle$ .

Correct: $\left\langle 2, 9 \right\rangle$ : Displacement $\div$ $10$ seconds.
$\left\langle 20, 90 \right\rangle$ : This is the displacement; divide by the $10$ seconds.
$\left\langle 6, 12 \right\rangle$ : This divides $\mathbf{p}(15)$ by $10$ ; subtract $\mathbf{p}(5)$ first.
$\left\langle 4, 3 \right\rangle$ : This is $\mathbf{p}(5)$ divided by $10$ .

3. How far is the drone from the depot at $t = 5$ ?

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

Show the answer

$\left\| \left\langle 40, 30 \right\rangle \right\|$ $= \sqrt{1600 + 900}$ $= 50$ meters.

Answer: 50 meters

4. On $[10, 15]$ the average rate of change of the position vector is $\left\langle 0, 10 \right\rangle$ . Which statement does the table support?

At $t = 15$ the drone is due north of where it was at $t = 10$
The drone hovered in place from $t = 10$ to $t = 15$
The drone moved east and not north from $t = 10$ to $t = 15$
The drone flew a straight line due north the whole time

Show the answer

$x$ is $60$ at both times and $y$ rose from $70$ to $120$ , so the net move is due north. The table cannot show the path between readings.

Correct: At $t = 15$ the drone is due north of where it was at $t = 10$ : Same $x$ , larger $y$ : due north.
The drone hovered in place from $t = 10$ to $t = 15$ : $y$ rose by $50$ meters, so it did not stay put.
The drone moved east and not north from $t = 10$ to $t = 15$ : The east component of the rate is $0$ .
The drone flew a straight line due north the whole time: Two readings give the net move, not the route in between.

5. For a vector-valued function $\mathbf{p}(t)$ drawn from the origin, what does the tip of $\mathbf{p}(t)$ trace as $t$ increases?

The path of $(x(t), y(t))$
The graph of $x(t)$ against $t$
A circle centered at the origin
A straight line from the origin

Show the answer

The tip is at $(x(t), y(t))$ , so it traces the same curve in the same direction.

Correct: The path of $(x(t), y(t))$ : Tip at $(x(t), y(t))$ .
The graph of $x(t)$ against $t$ : The tip is a point in the $xy$ -plane, not on a $t$ -axis.
A circle centered at the origin: Only if $\left\| \mathbf{p}(t) \right\|$ stays constant.
A straight line from the origin: Only for special functions, such as $\left\langle t, 2t \right\rangle$ .

6. What does $\left\| \mathbf{p}(t) \right\| = 4$ for every $t$ tell you about $\mathbf{p}(t) = \left\langle 4\cos t, 4\sin t \right\rangle$ ?

Its tip stays $4$ units from the origin: a circle of radius $4$
Its tip moves $4$ units for each unit of time, along a straight line
Its tip stays at one point, $4$ units from the origin
Its tip traces a circle of radius $16$

Show the answer

$\sqrt{16\cos^2 t + 16\sin^2 t}$ $= 4$ : every position is $4$ units from the origin, and the direction changes with $t$ , so the tip goes around a circle.

Correct: Its tip stays $4$ units from the origin: a circle of radius $4$ : Constant distance from the origin, changing direction.
Its tip moves $4$ units for each unit of time, along a straight line: The length of $\mathbf{p}(t)$ is a distance from the origin, not a step size.
Its tip stays at one point, $4$ units from the origin: The direction changes with $t$ , so the tip moves.
Its tip traces a circle of radius $16$ : $16$ is the length squared.

Part 9 · Summary

Summary

A vector-valued function

\mathbf{p}(t) = \left\langle x(t), y(t) \right\rangle

is a parametric function written as position vectors. The displacement

\mathbf{p}(t_2) - \mathbf{p}(t_1)

says how far and which way the object moved, and dividing by

t_2 - t_1

gives the average rate of change vector, made of the average rates of

x

and

y

.

Part 10 · Up next

What comes next

Part 11 · Connections