# Vector-valued functions

> AP Precalculus · Functions Involving Parameters, Vectors, and Matrices
> Source: https://www.owlsprep.com/study/ap-precalculus-u4-vector-valued-functions/

This module covers 2D vector-valued functions for AP Precalculus: definitions and components, the magnitude and direction of a position vector, using average rates of change over an interval to describe planar motion, and converting between vector form and Cartesian parametric curves.

**Prerequisites:** [Basic vector operations](https://www.owlsprep.com/study/ap-precalculus-u4-vectors-in-the-plane/); [Parametric equations](https://www.owlsprep.com/study/ap-precalculus-u4-parametric-equations/); Average rate of change over an interval

## Learning objectives

- Define 2D vector-valued functions and identify their components
- Find the magnitude (distance from the origin) and direction of a position vector
- Use average rates of change of the components over an interval to describe planar motion
- Convert between vector-valued functions and Cartesian parametric curves

## Definition of a Vector-valued Function

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.

**2D Vector-valued Function** — 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, $\mathbf{r}(t)$ is called the position function for a moving object.

*Notation:* $\mathbf{r}(t) = \langle x(t), y(t) \rangle$

*Example:* $\mathbf{r}(t) = \langle t^2, 2t+1 \rangle$ has components $x(t)=t^2$, $y(t)=2t+1$

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.

## Position, Components, Magnitude, and Direction

A position vector $\mathbf{r}(t) = \langle x(t), y(t) \rangle$ 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.

**Magnitude of a Position Vector** — The magnitude of the position vector is the distance from the origin to the point $(x(t), y(t))$ at parameter $t$.

*Notation:* $|\mathbf{r}(t)| = \sqrt{x(t)^2 + y(t)^2}$

*Example:* If $\mathbf{r}(2) = \langle 6, 8 \rangle$, then $|\mathbf{r}(2)| = \sqrt{6^2 + 8^2} = 10$, 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, $\theta = \arctan\left(\frac{y}{x}\right)$ (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 $\mathbf{r}(t) = \langle 3t, 4t \rangle$ 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 $|\mathbf{r}(t)|$ is always the distance from the origin — use the Pythagorean form $\sqrt{x^2 + y^2}$, never $x + y$.

*Calculator:* allowed

## Average Rates of Change and Planar Motion

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.

**Average Velocity Components (Average Rates of Change)** — 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 $\left\langle \dfrac{\Delta x}{\Delta t}, \dfrac{\Delta y}{\Delta t} \right\rangle$.

*Notation:* $\dfrac{\Delta x}{\Delta t} = \dfrac{x(t_2)-x(t_1)}{t_2-t_1}, \qquad \dfrac{\Delta y}{\Delta t} = \dfrac{y(t_2)-y(t_1)}{t_2-t_1}$

*Example:* The average speed of the net displacement is the magnitude of the average velocity vector, $\left| \left\langle \dfrac{\Delta x}{\Delta t}, \dfrac{\Delta y}{\Delta t} \right\rangle \right|$.

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

**Worked example:** A drone's position (in meters) is $\mathbf{r}(t) = \langle 3t, -0.1t^2 + 4t \rangle$ 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 $\Delta t = t_2 - t_1$.

*Calculator:* allowed

## Vector-valued Functions and Parametric Curves

Every 2D vector-valued function $\mathbf{r}(t) = \langle x(t), y(t) \rangle$ defines a parametric curve in the $xy$-plane, where $\mathbf{r}(t)$ 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 $\mathbf{r}(t) = \langle 2\cos t, 3\sin t \rangle$ for $0 \leq t \leq 2\pi$, 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 $2\pi$, $x \in [-2, 2]$ and $y \in [-3, 3]$, so this is the full ellipse:
6. $$\frac{x^2}{4} + \frac{y^2}{9} = 1$$

**Check your understanding**

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

1. An object has position $\mathbf{r}(t) = \langle t^2, 4t \rangle$ (meters, with $t$ in seconds). What is its average velocity vector over the interval $[0, 2]$?

   - A) $\langle 2, 4 \rangle$
   - B) $\langle 4, 8 \rangle$
   - C) $\langle 2, 2 \rangle$
   - D) $\langle 1, 4 \rangle$

   *Why:* The average rate of change of each component: $\frac{\Delta x}{\Delta t} = \frac{4 - 0}{2} = 2$ and $\frac{\Delta y}{\Delta t} = \frac{8 - 0}{2} = 4$, so the average velocity vector is $\langle 2, 4 \rangle$.

> **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.

## Common pitfalls

- **Wrong:** Computing an average rate of change as $x(t_2) - x(t_1)$ without dividing by $\Delta t$.
  - Why it fails: You remember the change in position but forget that a rate divides that change by the change in $t$.
  - Correct: Always divide by $\Delta t = t_2 - t_1$ to get the average rate of change of a component.
- **Wrong:** Using the average rate of change of $y$ to describe horizontal motion (or $x$ for vertical motion).
  - Why it fails: You lose track of which component controls which direction.
  - Correct: Horizontal motion comes from $\Delta x/\Delta t$ and vertical motion from $\Delta y/\Delta t$; label each before interpreting direction.
- **Wrong:** When asked for average speed, reporting the average velocity vector instead of its magnitude.
  - Why it fails: You mix up the average velocity (a vector) with the average speed (its scalar magnitude).
  - Correct: Average speed is $\left| \left\langle \Delta x/\Delta t, \Delta y/\Delta t \right\rangle \right|$; compute the magnitude after finding the average velocity vector.
- **Wrong:** For $\mathbf{r}(t) = \langle e^t, e^{2t} + 1 \rangle$, you write the Cartesian equation as $y = x^2 + 1$, the full parabola.
  - Why it fails: You forget that $e^t = x$ is always positive, so $x > 0$ from the original function.
  - Correct: After eliminating the parameter, add any domain restrictions implied by the original domain of $t$.
- **Wrong:** You calculate magnitude as $|\langle a, b \rangle| = a + b$ instead of using the Pythagorean theorem.
  - Why it fails: You confuse component-wise addition with vector magnitude after doing other component-wise operations.
  - Correct: Always use $|\langle a, b \rangle| = \sqrt{a^2 + b^2}$ for magnitude, regardless of the components.

## Cheatsheet

| Category | Formula/Rule | Notes |
| --- | --- | --- |
| 2D Vector-valued Function | $\mathbf{r}(t) = \langle x(t), y(t) \rangle$ | Scalar input $t$ (usually time); $x(t), y(t)$ are scalar components |
| Magnitude (distance from origin) | $\|\mathbf{r}(t)\| = \sqrt{x(t)^2 + y(t)^2}$ | How far the point is from the origin at parameter $t$ |
| Average velocity component (horizontal) | $\dfrac{\Delta x}{\Delta t} = \dfrac{x(t_2)-x(t_1)}{t_2-t_1}$ | Average rate of change of $x$ over $[t_1, t_2]$ |
| Average velocity component (vertical) | $\dfrac{\Delta y}{\Delta t} = \dfrac{y(t_2)-y(t_1)}{t_2-t_1}$ | Average rate of change of $y$ over $[t_1, t_2]$ |
| Average speed (net displacement) | $\left\| \left\langle \dfrac{\Delta x}{\Delta t}, \dfrac{\Delta y}{\Delta t} \right\rangle \right\| = \sqrt{\left(\dfrac{\Delta x}{\Delta t}\right)^2 + \left(\dfrac{\Delta y}{\Delta t}\right)^2}$ | 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 |

## 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.

- [Matrices](https://www.owlsprep.com/study/ap-precalculus-u4-matrices/)
- [The inverse and determinant of a matrix](https://www.owlsprep.com/study/ap-precalculus-u4-the-inverse-and-determinant-of/)
- [Linear transformations and matrices](https://www.owlsprep.com/study/ap-precalculus-u4-linear-transformations-and-matrices/)

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