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 $x (t) = t^{2}$ , $y (t) = 2 t + 1$

This topic accounts for ~1-2% of the total AP Precalculus exam score, appearing in both multiple-choice and free-response sections. It is most commonly tested for modeling motion, and unifies prior knowledge of parametric equations and single-variable calculus.

2. Limits and Continuity of Vector-valued Functions★★☆☆☆⏱ 4 min

🚫 No Calculator

All operations on vector-valued functions in AP Precalculus work component-wise: we apply the same rules for single-variable functions to each component separately. For limits, this gives the rule:

lim_{t \to a} \mathbf{r}(t) = \left\langle \lim_{t \to a} x(t), \lim_{t \to a} y(t) \right\rangle

Continuity follows the same definition as single-variable functions: $r (t)$ is continuous at $t = a$ if and only if $lim_{t \to a} r (t) = r (a)$ . This means both components must be continuous at $a$ , and $r (a)$ must be defined.

📐 Worked Example

Find $lim_{t \to 2} r (t)$ for $r (t) = ⟨ \frac{t ^{2} - 4}{t - 2}, \frac{3 t ^{2} - 2 t - 8}{t - 2} ⟩$ , and determine if $r (t)$ is continuous at $t = 2$ .

1
Evaluate the limit of the $x$ -component first. Factor the numerator: $t^{2} - 4 = (t - 2) (t + 2)$ . Cancel $(t - 2)$ for $t \neq = 2$ to get $x (t) = t + 2$ .
2
$\lim_{t \to 2} x(t) = 2 + 2 = 4$
3
Evaluate the limit of the $y$ -component. Factor the numerator: $3 t^{2} - 2 t - 8 = (3 t + 4) (t - 2)$ . Cancel $(t - 2)$ for $t \neq = 2$ to get $y (t) = 3 t + 4$ .
4
$\lim_{t \to 2} y(t) = 3(2) + 4 = 10$
5
Combine the component limits to get the overall limit:
6
$\lim_{t \to 2} \mathbf{r}(t) = \langle 4, 10 \rangle$
7
Check continuity: $r (2)$ is undefined because the denominator of both components is zero at $t = 2$ . Since $lim_{t \to 2} r (t) \neq = r (2)$ , $r (t)$ is not continuous at $t = 2$ .

Exam tip:

Always confirm that both component limits exist before concluding the vector limit exists; if even one component has no limit, the entire vector limit does not exist.

3. Derivatives, Velocity, and Speed★★★☆☆⏱ 4 min

✓ Calculator OK

The derivative of a vector-valued function is defined identically to the single-variable derivative, as the limit of the difference quotient, and simplifies to component-wise differentiation:

\mathbf{r}'(t) = \lim_{\Delta t \to 0} \frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{\Delta t} = \langle x'(t), y'(t) \rangle

Geometrically, $r^{'} (t)$ is the tangent vector to the parametric curve at parameter $t$ , pointing in the direction of increasing $t$ . Physically, when $r (t)$ is the position function of a moving object:

$r^{'} (t)$ = velocity vector (vector quantity with direction and magnitude)
$∣ r^{'} (t) ∣$ = speed (non-negative scalar, magnitude of velocity)
$r^{''} (t)$ = acceleration vector (derivative of velocity)

📐 Worked Example

Given position function $r (t) = ⟨ t^{2} + 2 t, cos (2 t)⟩$ for $t \geq 0$ , find $r^{'} (t)$ and the speed of the object at $t = π /4$ .

1
Differentiate the $x$ -component: $x (t) = t^{2} + 2 t$ , so
2
$x'(t) = 2t + 2$
3
Differentiate the $y$ -component using the chain rule: $y (t) = cos (2 t)$ , so
4
$y'(t) = -2\sin(2t)$
5
This gives the velocity vector $r^{'} (t) = ⟨ 2 t + 2, - 2 sin (2 t)⟩$ . Evaluate at $t = π /4$ :
6
$x'\left(\frac{\pi}{4}\right) = \frac{\pi}{2} + 2, \quad y'\left(\frac{\pi}{4}\right) = -2$
7
Calculate speed as the magnitude of the velocity vector:
8
$\text{Speed} = \sqrt{\left(\frac{\pi}{2} + 2\right)^2 + (-2)^2} = \sqrt{\frac{\pi^2}{4} + 2\pi + 8}$

📐 Worked Example

A drone has position at time $t$ (0 ≤ t ≤ 10, units in meters) given by $r (t) = ⟨ 3 t, - 0.1 t^{2} + 4 t ⟩$ . Find the velocity vector and speed at $t = 5$ seconds, and interpret the components.

1
Differentiate component-wise to get velocity:
2
$\mathbf{v}(t) = \mathbf{r}'(t) = \langle 3, -0.2t + 4 \rangle$
3
Evaluate at $t = 5$ :
4
$\mathbf{v}(5) = \langle 3, 3 \rangle \text{ m/s}$
5
Calculate speed as the magnitude:
6
$|\mathbf{v}(5)| = \sqrt{3^2 + 3^2} = 3\sqrt{2} \approx 4.24 \text{ m/s}$
7
Interpretation: The $x$ -component of 3 m/s means the drone's eastward position increases at 3 m/s, and the $y$ -component of 3 m/s means northward position also increases at 3 m/s at $t = 5$ .

Exam tip:

Always check if the question asks for velocity or speed: velocity is a vector, speed is a scalar magnitude. AP exams regularly test this terminology distinction, so circle the key term before you start working.

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:

Which of the following is the velocity vector and speed of an object moving with position function $r (t) = ⟨ t^{3} - 4 t, ln (t + 1)⟩$ at $t = 1$ ?
- A) Velocity $⟨ - 1, 1/2 ⟩$ , speed $5 /2$
- B) Velocity $⟨ - 1, 1/2 ⟩$ , speed $- 1/2$
- C) Velocity $⟨ - 1, 1 ⟩$ , speed $5 /2$
- D) Velocity $⟨ 1, 1/2 ⟩$ , speed $5 /2$
Reveal answer
A —
Velocity is the component-wise derivative: $x^{'} (t) = 3 t^{2} - 4 = - 1$ at $t = 1$ , $y^{'} (t) = 1/ (t + 1) = 1/2$ at $t = 1$ . Speed is the magnitude: $(- 1)^{2} + (1/2)^{2} = 5 /2$ , which matches option A.

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:

After canceling $(t - a)$ to find $lim_{t \to a} r (t)$ , you conclude $r (t)$ is continuous at $t = a$ .

Why:

You confuse existence of the limit with the requirement for continuity that the limit equals the function's value at $t = a$ .

Correct move:

After finding the limit, explicitly evaluate $r (a)$ and confirm it equals the limit before concluding continuity.

Wrong move:

You differentiate $y (t) = sin (3 t)$ as $y^{'} (t) = cos (3 t)$ when finding $r^{'} (t)$ , forgetting the chain rule for the inner function.

Why:

You get comfortable with component-wise differentiation and skip checking composite functions.

Correct move:

After differentiating all components, double-check every composite function to confirm you applied the chain rule.

Wrong move:

When asked for speed, you output the velocity vector instead of its magnitude.

Why:

You mix up the definitions of velocity (vector) and speed (scalar).

Correct move:

Circle the key term "speed" in the question, and always compute the magnitude after finding the 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
Limit of Vector-valued Function	$lim_{t \to a} r (t) = ⟨ lim_{t \to a} x (t), lim_{t \to a} y (t) ⟩$	Exists only if both component limits exist
Continuity of Vector-valued Function	$r (t)$ continuous at $a$ iff $lim_{t \to a} r (t) = r (a)$	Requires $r (a)$ defined and both components continuous at $a$
Derivative (Velocity)	$r^{'} (t) = ⟨ x^{'} (t), y^{'} (t)⟩$	Tangent vector to parametric curve; velocity vector if $r (t)$ is position
Speed	$Speed = ∣ r^{'} (t) ∣ = (x^{'} (t))^{2} + (y^{'} (t))^{2}$	Non-negative scalar; always output magnitude for speed
Acceleration	$r^{''} (t) = ⟨ x^{''} (t), y^{''} (t)⟩$	Derivative of velocity, second derivative of position
Cartesian Equation from Vector Function	Eliminate $t$ from $x = x (t), y = y (t)$	Add domain restrictions from original $t$ domain

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · AP Precalculus
MCQ speed calculation
2023 · AP Precalculus
FRQ continuity check

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 projectile motion in the remainder of Unit 4. Without mastering component-wise operations and interpretation of vector-valued functions, you will not be able to correctly analyze transformed parametric curves or solve motion problems that are common on the AP exam. This topic also builds a critical bridge between parametric equations, single-variable calculus, and vector concepts, and prepares you for more advanced study of vector-valued functions in AP Calculus AB/BC, where you will integrate vector functions to find displacement and arc length.