Vector-valued functions

AP Precalculus· AP Precalculus CED — Functions Involving Parameters, Vectors, and Matrices· 14 min read

1. Definition of a Vector-valued Function★☆☆☆☆⏱ 3 min

A vector-valued function (often shortened to vector function) takes a single scalar input (most commonly time $t$ in AP Precalculus problems) and outputs a vector. In AP Precalculus, we almost exclusively work with 2-dimensional vector-valued functions.

📘 Definition

2D Vector-valued Function

$r (t) = ⟨ x (t), y (t)⟩$

A function with scalar input $t$ and 2D vector output, where $x (t)$ and $y (t)$ are scalar-valued component functions. When $t$ represents time, $r (t)$ is called the position function for a moving object.

Example:

$r (t) = ⟨ t^{2}, 2 t + 1 ⟩$ has components $x (t) = t^{2}$ , $y (t) = 2 t + 1$

Note that Unit 4 is not assessed on the AP Precalculus Exam — the College Board CED limits the exam to Units 1–3 — so this topic is covered at teacher discretion for enrichment. It ties together prior knowledge of parametric equations and vectors to model motion.

2. Position, Components, Magnitude, and Direction★★☆☆☆⏱ 4 min

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A position vector $r (t) = ⟨ x (t), y (t)⟩$ points from the origin to the location $(x (t), y (t))$ of an object at parameter $t$ . The component $x (t)$ gives the horizontal position and $y (t)$ gives the vertical position.

📘 Definition

Magnitude of a Position Vector

$∣ r (t) ∣ = x (t)^{2} + y (t)^{2}$

The magnitude of the position vector is the distance from the origin to the point $(x (t), y (t))$ at parameter $t$ .

Example:

If $r (2) = ⟨ 6, 8 ⟩$ , then $∣ r (2) ∣ = 6^{2} + 8^{2} = 10$ , so the object is $10$ units from the origin.

The direction of the position vector is the angle it makes with the positive $x$ -axis, $θ = arctan (\frac{y}{x})$ (adjusted for the correct quadrant). As $t$ changes, both the magnitude (distance from the origin) and the direction can change.

📐 Worked Example

A drone's position (in meters) is $r (t) = ⟨ 3 t, 4 t ⟩$ for $t \geq 0$ . Find its distance from the origin at $t = 2$ and the direction of the position vector.

1
Evaluate the position at $t = 2$ :
$\mathbf{r}(2) = \langle 3(2), 4(2) \rangle = \langle 6, 8 \rangle$
2
Magnitude = distance from the origin:
$|\mathbf{r}(2)| = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \text{ m}$
3
Direction (angle above the positive $x$ -axis):
$\theta = \arctan\left(\frac{8}{6}\right) = \arctan\left(\frac{4}{3}\right) \approx 53.1^\circ$

Exam tip:

The magnitude $∣ r (t) ∣$ is always the distance from the origin — use the Pythagorean form $x^{2} + y^{2}$ , never $x + y$ .

3. Average Rates of Change and Planar Motion★★★☆☆⏱ 4 min

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To describe how an object's position changes over a time interval $[t_{1}, t_{2}]$ , use the average rate of change of each component — a secant-line slope computed directly from the endpoints.

📘 Definition

Average Velocity Components (Average Rates of Change)

$\frac{Δ x}{Δ t} = \frac{x ( t _{2} ) - x ( t _{1} )}{t _{2} - t _{1}}, \frac{Δ y}{Δ t} = \frac{y ( t _{2} ) - y ( t _{1} )}{t _{2} - t _{1}}$

The average rate of change of $x$ over $[t_{1}, t_{2}]$ is the average horizontal velocity, and the average rate of change of $y$ is the average vertical velocity. Together they form the average velocity vector $⟨ \frac{Δ x}{Δ t}, \frac{Δ y}{Δ t} ⟩$ .

Example:

The average speed of the net displacement is the magnitude of the average velocity vector, $⟨ \frac{Δ x}{Δ t}, \frac{Δ y}{Δ t} ⟩$ .

The signs of the average rates of change give the net direction: $Δ x /Δ t > 0$ is net rightward motion and $Δ y /Δ t > 0$ is net upward motion over the interval.

📐 Worked Example

A drone's position (in meters) is $r (t) = ⟨ 3 t, - 0.1 t^{2} + 4 t ⟩$ for $0 \leq t \leq 10$ . Find the average velocity vector and the average speed over the interval $[0, 10]$ .

1
Evaluate the position at the endpoints:
$\mathbf{r}(0) = \langle 0, 0 \rangle, \quad \mathbf{r}(10) = \langle 30, 30 \rangle$
2
Average rate of change of each component over $[0, 10]$ :
$\frac{\Delta x}{\Delta t} = \frac{30 - 0}{10} = 3, \quad \frac{\Delta y}{\Delta t} = \frac{30 - 0}{10} = 3 \text{ (m/s)}$
3
Average velocity vector:
$\left\langle 3, 3 \right\rangle \text{ m/s}$
4
Average speed = magnitude of the average velocity vector:
$\sqrt{3^2 + 3^2} = 3\sqrt{2} \approx 4.24 \text{ m/s}$
5
Both components are positive, so the net motion over $[0, 10]$ is to the right and upward.

Exam tip:

Average rate of change is measured over an interval, not at an instant. Evaluate the position at both endpoints, subtract, and divide by $Δ t = t_{2} - t_{1}$ .

4. Vector-valued Functions and Parametric Curves★★☆☆☆⏱ 3 min

Every 2D vector-valued function $r (t) = ⟨ x (t), y (t)⟩$ defines a parametric curve in the $x y$ -plane, where $r (t)$ is the position vector from the origin to the point $(x (t), y (t))$ on the curve. To convert to a Cartesian equation (an equation in $x$ and $y$ without $t$ ), eliminate the parameter $t$ using the same techniques as for standard parametric equations.

It is critical to note any restrictions on $t$ from the original function, because these translate to restrictions on the domain/range of the Cartesian curve. A restricted $t$ will only produce a portion of the full implicit curve.

📐 Worked Example

Find the Cartesian equation of the curve defined by $r (t) = ⟨ 2 cos t, 3 sin t ⟩$ for $0 \leq t \leq 2 π$ , and identify the type of curve.

1
Set $x = 2 cos t$ and $y = 3 sin t$ , then rearrange to isolate the trigonometric terms:
2
$\cos t = \frac{x}{2}, \quad \sin t = \frac{y}{3}$
3
Use the Pythagorean identity $cos^{2} t + sin^{2} t = 1$ and substitute:
4
$\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1$
5
Simplify and confirm the range: since $t$ spans $0$ to $2 π$ , $x \in [- 2, 2]$ and $y \in [- 3, 3]$ , so this is the full ellipse:
6
$\frac{x^2}{4} + \frac{y^2}{9} = 1$

✓ Quick check

Test your understanding with this AP-style multiple choice question:

An object has position $r (t) = ⟨ t^{2}, 4 t ⟩$ (meters, with $t$ in seconds). What is its average velocity vector over the interval $[0, 2]$ ?
- A) $⟨ 2, 4 ⟩$
- B) $⟨ 4, 8 ⟩$
- C) $⟨ 2, 2 ⟩$
- D) $⟨ 1, 4 ⟩$
Reveal answer
A —
The average rate of change of each component: $\frac{Δ x}{Δ t} = \frac{4 - 0}{2} = 2$ and $\frac{Δ y}{Δ t} = \frac{8 - 0}{2} = 4$ , so the average velocity vector is $⟨ 2, 4 ⟩$ .

Exam tip:

If $t$ is restricted, always write the domain restriction for $x$ (and $y$ , if needed) next to your Cartesian equation; AP multiple-choice questions often include an unrestricted full curve as a distractor.

5. Common Pitfalls

Wrong move:

Computing an average rate of change as $x (t_{2}) - x (t_{1})$ without dividing by $Δ t$ .

Why:

You remember the change in position but forget that a rate divides that change by the change in $t$ .

Correct move:

Always divide by $Δ t = t_{2} - t_{1}$ to get the average rate of change of a component.

Wrong move:

Using the average rate of change of $y$ to describe horizontal motion (or $x$ for vertical motion).

Why:

You lose track of which component controls which direction.

Correct move:

Horizontal motion comes from $Δ x /Δ t$ and vertical motion from $Δ y /Δ t$ ; label each before interpreting direction.

Wrong move:

When asked for average speed, reporting the average velocity vector instead of its magnitude.

Why:

You mix up the average velocity (a vector) with the average speed (its scalar magnitude).

Correct move:

Average speed is $∣ ⟨ Δ x /Δ t, Δ y /Δ t ⟩ ∣$ ; compute the magnitude after finding the average velocity vector.

Wrong move:

For $r (t) = ⟨ e^{t}, e^{2 t} + 1 ⟩$ , you write the Cartesian equation as $y = x^{2} + 1$ , the full parabola.

Why:

You forget that $e^{t} = x$ is always positive, so $x > 0$ from the original function.

Correct move:

After eliminating the parameter, add any domain restrictions implied by the original domain of $t$ .

Wrong move:

You calculate magnitude as $∣ ⟨ a, b ⟩ ∣ = a + b$ instead of using the Pythagorean theorem.

Why:

You confuse component-wise addition with vector magnitude after doing other component-wise operations.

Correct move:

Always use $∣ ⟨ a, b ⟩ ∣ = a^{2} + b^{2}$ for magnitude, regardless of the components.

6. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
2D Vector-valued Function	$r (t) = ⟨ x (t), y (t)⟩$	Scalar input $t$ (usually time); $x (t), y (t)$ are scalar components
Magnitude (distance from origin)	$∣ r (t) ∣ = x (t)^{2} + y (t)^{2}$	How far the point is from the origin at parameter $t$
Average velocity component (horizontal)	$\frac{Δ x}{Δ t} = \frac{x ( t _{2} ) - x ( t _{1} )}{t _{2} - t _{1}}$	Average rate of change of $x$ over $[t_{1}, t_{2}]$
Average velocity component (vertical)	$\frac{Δ y}{Δ t} = \frac{y ( t _{2} ) - y ( t _{1} )}{t _{2} - t _{1}}$	Average rate of change of $y$ over $[t_{1}, t_{2}]$
Average speed (net displacement)	$⟨ \frac{Δ x}{Δ t}, \frac{Δ y}{Δ t} ⟩ = (\frac{Δ x}{Δ t})^{2} + (\frac{Δ y}{Δ t})^{2}$	Non-negative scalar magnitude of the average velocity
Cartesian equation from vector function	Eliminate $t$ from $x = x (t), y = y (t)$	Add domain restrictions from the original $t$ domain

Going deeper

Official SyllabusAP Precalculus CEDOfficial course requirements

What's Next

This topic is the foundation for vector modeling of motion, which you will extend when studying matrix transformations of vectors and parametric motion in the remainder of Unit 4. Mastering the magnitude and direction of a position vector, along with average rates of change of the components, lets you analyze transformed parametric curves and solve planar-motion problems. This topic also builds a bridge between parametric equations and vectors, and prepares you for AP Calculus AB/BC, where motion along a curve is studied with the tools of calculus.