9.4 Defining and Differentiating Vector-Valued Functions

A vector-valued function pairs parametric functions into one vector, usually written $r(t)=\langle f(t),g(t)\rangle$. To differentiate it, take the derivative of each component separately, so $r'(t)=\langle f'(t),g'(t)\rangle$. For AP Calculus BC, component-wise differentiation connects position, velocity, acceleration, and speed.

Why This Matters for the AP Calculus Exam

Vector-valued functions are part of Unit 9, a BC-only unit that carries 11-12% of the AP Calculus BC exam. The big idea here is that the rules you already know for derivatives extend directly to vectors, so this is a familiar method in a vector format. You differentiate each component using the power rule, chain rule, and trig derivatives just like before.

This skill sets up later topics in the unit: once you can find $r'(t)$, you can find velocity, then $r''(t)$ for acceleration, and the magnitude $|r'(t)|$ for speed. Expect to use this on both multiple-choice and free-response questions involving planar motion, where careful notation and correct chain rule work matter for clear exam answers.

Key Takeaways

• A vector has both direction and magnitude; magnitude is its length.
• Write a vector-valued function as $r(t)=\langle f(t), g(t)\rangle$, where each component is a function of $t$.
• Differentiate component by component: $r'(t)=\langle f'(t), g'(t)\rangle$.
• The same derivative rules apply, so watch for the chain rule inside trig and composite components.
• When $r(t)$ is position, $r'(t)$ is the velocity vector and $r''(t)$ is the acceleration vector.
• The magnitude of a vector $\langle a, b\rangle$ is $\sqrt{a^2+b^2}$, found with the Pythagorean theorem.

---

↖️ Vector Review

A vector is a quantity that has both direction and magnitude. Magnitude refers to the length of a vector. All vectors have horizontal and vertical components, and they are written based on those two components. For example, a vector with a horizontal component of 5 and a vertical component of 4 is written as $\langle 5, 4\rangle$. This bracket notation is the one used most often on the AP Calculus BC exam.

Every vector has a head and a tail. The tail is where the vector starts. If no other information is given, you can assume the tail sits at the origin $(0,0)$. The head is the final point of the vector, drawn with an arrowhead in diagrams.

You can find the magnitude from the horizontal and vertical components. Take the vector $\langle 5, 4\rangle$. Using the Pythagorean theorem $a^2 + b^2 = c^2$ with $a = 5$ and $b = 4$, you get $c = \sqrt{41}$. So the magnitude is $\sqrt{41}$, written as $\|v\|$.

You can also find the direction from the components. Since $\tan(\theta) = \frac{\text{vertical component}}{\text{horizontal component}}$, this vector gives $\tan(\theta) = \tfrac{4}{5}$. Taking the inverse tangent gives $\theta \approx 0.675$ radians.

The reverse works too: given a magnitude and direction, you can recover the horizontal and vertical components.

---

📐 Vector-Valued Functions

Think of $r(t)$ as a vector built from two parametric functions, $f(t)$ and $g(t)$. You write it as $r(t)=\langle f(t), g(t)\rangle$. To differentiate, take the derivative of each component on its own:

$$r'(t) = \langle f'(t), g'(t)\rangle$$

The normal derivative rules still apply, including the power rule, trig derivatives, and the chain rule. Nothing about being a vector changes how you differentiate each piece.

✏️ Vector-Valued Functions Walkthrough

Try a quick example.

$$r(t) = \langle 4t^2, 7t^7\rangle$$

Find $r'(4)$.

Start by finding $r'(t)$, taking the derivative of each component. For the horizontal component, the derivative of $4t^2$ with respect to $t$ is $8t$.

For the vertical component, the derivative of $7t^7$ with respect to $t$ is $49t^6$.

Put it together:

$$r'(t) = \langle 8t, 49t^6\rangle$$

Substitute $t = 4$:

$$r'(4) = \langle 8(4), 49(4)^6\rangle$$

Simplify:

$$r'(4) = \langle 32, 200704\rangle$$

---

🤔 Practice Problems with Vector-Valued Functions

💭 Vector Practice Problem 1

Find $r'(2)$ given that

$$r(t) = \langle 4\sin(5t), 7\cos(2t^2)\rangle$$

Find the derivative of each component. For the horizontal component, the derivative of $4\sin(5t)$ with respect to $t$ is $20\cos(5t)$ (chain rule brings out the factor of 5).

For the vertical component, the derivative of $7\cos(2t^2)$ with respect to $t$ is $-28t\sin(2t^2)$ (chain rule brings out $4t$, and $7 \cdot 4t = 28t$).

Put it together:

$$r'(t) = \langle 20\cos(5t), -28t\sin(2t^2)\rangle$$

Substitute $t = 2$:

$$r'(2) = \langle 20\cos(10), -56\sin(8)\rangle$$

Simplify (in radians):

$$r'(2) = \langle -16.781, -255.404\rangle$$

💭 Vector Practice Problem 2

A particle moves along a path so that its position at any time $t \geq 0$ is given by $(4\sin(t), 4\cos(t))$. Find the particle's velocity vector at $t = \tfrac{\pi}{4}$.

A particle's velocity vector is the first derivative of its position vector, so $v(t)=\langle \tfrac{dx}{dt}, \tfrac{dy}{dt}\rangle$. Start by differentiating each component of the position vector.

For the horizontal component, the derivative of $4\sin(t)$ with respect to $t$ is $4\cos(t)$.

For the vertical component, the derivative of $4\cos(t)$ with respect to $t$ is $-4\sin(t)$.

Put it together:

$$v(t) = \langle 4\cos(t), -4\sin(t)\rangle$$

Substitute $t = \tfrac{\pi}{4}$:

$$v\left(\tfrac{\pi}{4}\right) = \langle 4\cos(\tfrac{\pi}{4}), -4\sin(\tfrac{\pi}{4})\rangle$$

Simplify:

$$v\left(\tfrac{\pi}{4}\right) = \langle 2\sqrt{2}, -2\sqrt{2}\rangle$$

---

How to Use This on the AP Calculus Exam

Problem Solving

• Differentiate one component at a time. Treat the horizontal and vertical pieces as separate single-variable derivatives.
• Watch for the chain rule inside trig, exponential, or composite components. A missing inner derivative is the most common point loss here.
• Keep your calculator in radians for trig problems. Position and velocity components are almost always in radians on BC.

Common Trap

• Notation matters for clear exam work. Write your answer as a vector with brackets, like $\langle 32, 200704\rangle$, not as two loose numbers.
• When a question asks for velocity, find $r'(t)$. When it asks for acceleration, find $r''(t)$. When it asks for speed, find the magnitude $|r'(t)| = \sqrt{(f'(t))^2 + (g'(t))^2}$, which is a single number, not a vector.

---

Common Misconceptions

• Differentiating a vector is not one big operation. You do not combine the components first. Differentiate each component separately, then write the results back as a vector.
• Velocity and speed are different. The velocity vector $r'(t)$ has direction and magnitude. Speed is just the magnitude $|r'(t)|$, a single nonnegative number.
• The chain rule does not disappear inside a vector. Components like $\sin(5t)$ or $\cos(2t^2)$ still need the chain rule. Bracket notation does not let you skip it.
• A vector-valued function and a parametric pair are the same idea in different clothes. $r(t)=\langle f(t), g(t)\rangle$ carries the same information as the parametric equations $x=f(t)$, $y=g(t)$, so the differentiation logic carries straight over.
• Magnitude is not just adding the components. Use $\sqrt{a^2+b^2}$, not $a+b$, to get the length of a vector.

Related AP Calculus Guides

• 9.1 Defining and Differentiating Parametric Equations
• Unit 9 Overview: Parametric Equations, Polar Coordinates, and Vector-Valued Functions
• 9.2 Second Derivatives of Parametric Equations
• 9.3 Finding Arc Lengths of Curves Given by Parametric Equations
• 9.7 Defining Polar Coordinates and Differentiating in Polar Form
• 9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve