Vector-valued functions — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Definitions of 2D vector-valued functions, limits and continuity of vector-valued functions, derivatives of vector-valued functions, velocity, speed, and conversion between vector form and Cartesian parametric curves.

You should already know: Basic vector operations, parametric equations, limits and derivatives of single-variable functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Vector-valued functions?

A vector-valued function (often shortened to vector function) is a function that 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, which follow the standard notation $\mathbf{r}(t)=⟨x(t),y(t)⟩$, where $x(t)$ and $y(t)$ are scalar-valued functions called the components of $\mathbf{r}(t)$. When the input represents time, $\mathbf{r}(t)$ is often called the position function, as it gives the position of a moving object at time $t$. According to the AP Precalculus Course and Exam Description (CED), this topic accounts for approximately 1-2% of the total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam. It is most commonly tested as a tool for modeling motion, or as a connection between parametric equations and vector concepts. Many AP questions combine vector-valued functions with prior knowledge of parametric curves and single-variable calculus, making it a key unifying topic in Unit 4.

2. Limits and Continuity of Vector-valued Functions

All operations on vector-valued functions in AP Precalculus work component-wise: that is, we can apply the same rules we use for single-variable functions to each component of the vector function separately. For limits, this means that the limit of a vector-valued function as $t$ approaches a constant $a$ is just the vector of the limits of its components, as long as both component limits exist: $$\underset{t\to a}{\lim }\mathbf{r}(t)=⟨\underset{t\to a}{\lim }x(t),\underset{t\to a}{\lim }y(t)⟩$$ Continuity follows from the same definition as single-variable functions: $\mathbf{r}(t)$ is continuous at $t=a$ if and only if ${\lim }_{t\to a}\mathbf{r}(t)=\mathbf{r}(a)$. This means both $x(t)$ and $y(t)$ must be continuous at $t=a$, and $\mathbf{r}(a)$ must be defined. The intuition here is simple: a vector is just two scalars packaged together, so there is no new rule for limits or continuity beyond what you already know for single-variable functions.

Worked Example

Find ${\lim }_{t\to 2}\mathbf{r}(t)$ for $\mathbf{r}(t)=⟨\frac{t^2-4}{t-2},\frac{3t^2-2t-8}{t-2}⟩$, and determine if $\mathbf{r}(t)$ 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 $t\ne 2$ to get $x(t)=t+2$, so ${\lim }_{t\to 2}x(t)=2+2=4$.
2. Evaluate the limit of the $y$-component: factor the numerator: $3t^2-2t-8=(3t+4)(t-2)$. Cancel $(t-2)$ for $t\ne 2$ to get $y(t)=3t+4$, so ${\lim }_{t\to 2}y(t)=3(2)+4=10$.
3. Combine the component limits: ${\lim }_{t\to 2}\mathbf{r}(t)=⟨4,10⟩$.
4. Check continuity: $\mathbf{r}(2)$ is undefined because the denominator of both components is zero at $t=2$, so ${\lim }_{t\to 2}\mathbf{r}(t)\ne \mathbf{r}(2)$, meaning $\mathbf{r}(t)$ 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 of Vector-valued Functions, Velocity, and Speed

The derivative of a vector-valued function is defined identically to the single-variable derivative, as the limit of the difference quotient: $${\mathbf{r}}^{\prime }(t)=\underset{\Delta t\to 0}{\lim }\frac{\mathbf{r}(t+\Delta t)-\mathbf{r}(t)}{\Delta t}$$ Just like limits, the derivative simplifies to component-wise differentiation: if $\mathbf{r}(t)=⟨x(t),y(t)⟩$, then ${\mathbf{r}}^{\prime }(t)=⟨x^{\prime }(t),y^{\prime }(t)⟩$, as long as both components are differentiable at $t$. Geometrically, ${\mathbf{r}}^{\prime }(t)$ is the tangent vector to the parametric curve defined by $\mathbf{r}(t)$ at parameter $t$, pointing in the direction of increasing $t$. Physically, when $\mathbf{r}(t)$ is the position function of a moving object, ${\mathbf{r}}^{\prime }(t)$ is called the velocity vector, and the magnitude of the velocity vector $|{\mathbf{r}}^{\prime }(t)|$ is called speed, a non-negative scalar that measures how fast the object is moving regardless of direction. The second derivative ${\mathbf{r}}^{\prime \prime }(t)=⟨x^{\prime \prime }(t),y^{\prime \prime }(t)⟩$ is the acceleration vector.

Worked Example

Given position function $\mathbf{r}(t)=⟨t^2+2t,\cos (2t)⟩$ for $t\ge 0$, find ${\mathbf{r}}^{\prime }(t)$ and the speed of the object at $t=\pi /4$.

1. Differentiate the $x$-component: $x(t)=t^2+2t$, so $x^{\prime }(t)=2t+2$.
2. Differentiate the $y$-component using the chain rule: $y(t)=\cos (2t)$, so $y^{\prime }(t)=-2\sin (2t)$. This gives ${\mathbf{r}}^{\prime }(t)=⟨2t+2,-2\sin (2t)⟩$.
3. Evaluate ${\mathbf{r}}^{\prime }(\pi /4)$: $x^{\prime }(\pi /4)=2(\pi /4)+2=\pi /2+2$, and $y^{\prime }(\pi /4)=-2\sin (2(\pi /4))=-2\sin (\pi /2)=-2$.
4. Calculate speed as the magnitude of ${\mathbf{r}}^{\prime }(\pi /4)$: $$\text{Speed}=\sqrt{{\left(\frac{\pi }{2}+2\right)}^2+(-2{)}^2}=\sqrt{\frac{{\pi }^2}{4}+2\pi +8}$$

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

Every 2D vector-valued function $\mathbf{r}(t)=⟨x(t),y(t)⟩$ 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 for any input $t$. To find the Cartesian equation of the curve (an equation in $x$ and $y$ without $t$), you eliminate the parameter $t$ using the same techniques you learned for parametric equations: solve one component for $t$ and substitute into the other component, or use a trigonometric identity for trigonometric components. 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 give a portion of the full implicit curve.

Worked Example

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

1. Set $x=2\cos t$ and $y=3\sin t$, then rearrange to isolate the trigonometric terms: $\cos t=\frac{x}{2}$ and $\sin t=\frac{y}{3}$.
2. Use the Pythagorean identity ${\cos }^{2}t+{\sin }^{2}t=1$.
3. Substitute the expressions for $\cos t$ and $\sin t$ into the identity: ${\left(\frac{x}{2}\right)}^2+{\left(\frac{y}{3}\right)}^2=1$, which simplifies to $\frac{x^2}{4}+\frac{y^2}{9}=1$.
4. Since $t$ ranges from $0$ to $2\pi$, $\cos t$ ranges from $-2$ to $2$ and $\sin t$ ranges from $-3$ to $3$, so this is the full ellipse centered at the origin with a vertical major axis.

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

5. Common Pitfalls (and how to avoid them)

• Wrong move: After canceling $(t-a)$ to find ${\lim }_{t\to a}\mathbf{r}(t)$, you conclude $\mathbf{r}(t)$ is continuous at $t=a$. Why: You confuse existence of the limit with the requirement that the limit equals the function's value at $t=a$ for continuity. Correct move: After finding the limit, explicitly evaluate $\mathbf{r}(a)$ and confirm it equals the limit before concluding continuity.
• Wrong move: You differentiate $y(t)=\sin (3t)$ as $y^{\prime }(t)=\cos (3t)$ when finding ${\mathbf{r}}^{\prime }(t)$, 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 $\mathbf{r}(t)=⟨e^t,e^{2t}+1⟩$, you write the Cartesian equation as $y=x^2+1$, the full parabola. Why: You forget that $e^t=x$ is always positive, so $x>0$. 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 $|⟨a,b⟩|=\sqrt{a^2+b^2}$ for magnitude, regardless of the components.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following is the velocity vector and speed of an object moving with position function $\mathbf{r}(t)=⟨t^3-4t,\ln (t+1)⟩$ at $t=1$? A) Velocity $⟨-1,1/2⟩$, speed $\sqrt{5}/2$ B) Velocity $⟨-1,1/2⟩$, speed $-1/2$ C) Velocity $⟨-1,1⟩$, speed $\sqrt{5}/2$ D) Velocity $⟨1,1/2⟩$, speed $\sqrt{5}/2$

Worked Solution: Velocity is the component-wise derivative of the position function. Differentiate the $x$-component: $\frac{d}{dt}(t^3-4t)=3t^2-4$, which evaluates to $3(1{)}^2-4=-1$ at $t=1$. Differentiate the $y$-component: $\frac{d}{dt}\ln (t+1)=\frac{1}{t+1}$, which evaluates to $\frac{1}{1+1}=1/2$ at $t=1$. Speed is the magnitude of the velocity vector: $\sqrt{(-1{)}^2+(1/2{)}^2}=\sqrt{5/4}=\sqrt{5}/2$. This matches option A. Correct answer: A.

Question 2 (Free Response)

Let $\mathbf{r}(t)=⟨\frac{t^2-9}{t-3},2t+1⟩$ for all real $t$. (a) Find ${\lim }_{t\to 3}\mathbf{r}(t)$. (b) Is $\mathbf{r}(t)$ continuous at $t=3$? Justify your answer. (c) Find the derivative ${\mathbf{r}}^{\prime }(t)$ for $t\ne 3$, and find the magnitude of ${\mathbf{r}}^{\prime }(4)$.

Worked Solution: (a) Simplify the $x$-component for $t\ne 3$: $\frac{t^2-9}{t-3}=\frac{(t-3)(t+3)}{t-3}=t+3$. The limit of $x(t)$ as $t\to 3$ is $3+3=6$. The limit of $y(t)=2t+1$ as $t\to 3$ is $2(3)+1=7$. So ${\lim }_{t\to 3}\mathbf{r}(t)=⟨6,7⟩$. (b) $\mathbf{r}(t)$ is not continuous at $t=3$. The $x$-component is undefined at $t=3$ (denominator equals zero), so $\mathbf{r}(3)$ does not exist. For continuity, ${\lim }_{t\to 3}\mathbf{r}(t)$ must equal $\mathbf{r}(3)$, which is impossible here. (c) For $t\ne 3$, $x(t)=t+3$ so $x^{\prime }(t)=1$, and $y(t)=2t+1$ so $y^{\prime }(t)=2$. Thus ${\mathbf{r}}^{\prime }(t)=⟨1,2⟩$ for all $t\ne 3$. At $t=4$, the magnitude is $|⟨1,2⟩|=\sqrt{1^2+2^2}=\sqrt{5}$.

Question 3 (Application / Real-World Style)

A small drone flying over a field has position at time $t$ seconds (for $0\le t\le 10$) given by $\mathbf{r}(t)=⟨3t,-0.1t^2+4t⟩$, where $x$ and $y$ are measured in meters from the starting point at the origin. Find the velocity vector and speed of the drone at $t=5$ seconds, and interpret the velocity components in context.

Worked Solution: Differentiate component-wise to get velocity: $\mathbf{v}(t)={\mathbf{r}}^{\prime }(t)=⟨3,-0.2t+4⟩$. Evaluate at $t=5$: $\mathbf{v}(5)=⟨3,-0.2(5)+4⟩=⟨3,3⟩$ meters per second. Speed is the magnitude: $|\mathbf{v}(5)|=\sqrt{3^2+3^2}=3\sqrt{2}\approx 4.24$ meters per second. In context, the $x$-component of 3 m/s means the drone's eastward position is increasing at 3 meters per second, and the $y$-component of 3 m/s means the drone's northward position is also increasing at 3 meters per second at $t=5$ seconds.

7. Quick Reference Cheatsheet

8. 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.

• Vectors in the plane
• Parametric equations
• Matrix transformations of vectors