Unit 9.4 – Defining and Differentiating Vector-Valued Functions BC ONLY

AP® Calculus BC | Vectors and Motion

Why This Matters: Vector-valued functions combine parametric equations into a single vector notation! Instead of writing \(x = f(t)\) and \(y = g(t)\) separately, we write \(\vec{r}(t) = \langle f(t), g(t) \rangle\). This elegant notation is essential for describing motion in physics and appears on virtually every BC exam. Master this and parametric motion problems become much clearer!

📚 What is a Vector-Valued Function?

VECTOR-VALUED FUNCTION DEFINED

A vector-valued function is a function whose output is a vector:

\[ \vec{r}(t) = \langle x(t), y(t) \rangle = x(t)\vec{i} + y(t)\vec{j} \]

Component Form vs. Unit Vector Form:

Component form: \(\vec{r}(t) = \langle x(t), y(t) \rangle\)
Unit vector form: \(\vec{r}(t) = x(t)\vec{i} + y(t)\vec{j}\)
Both represent the same thing!
\(\vec{i} = \langle 1, 0 \rangle\) and \(\vec{j} = \langle 0, 1 \rangle\) are unit vectors

📝 Connection to Parametric: A vector-valued function \(\vec{r}(t) = \langle x(t), y(t) \rangle\) is exactly the same as parametric equations \(x = x(t)\), \(y = y(t)\)—just written in vector form!

🔄 Derivative of Vector-Valued Functions

The Derivative Formula

THE FORMULA:

For \(\vec{r}(t) = \langle x(t), y(t) \rangle\):

\[ \vec{r}'(t) = \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle = x'(t)\vec{i} + y'(t)\vec{j} \]

Differentiate component-wise!

Key Insight:

The derivative of a vector-valued function is found by differentiating each component separately. It's that simple!

🚗 Position, Velocity, and Acceleration

MOTION VECTORS

Position Vector:

\[ \vec{r}(t) = \langle x(t), y(t) \rangle \]

The location of a particle at time \(t\)

Velocity Vector:

\[ \vec{v}(t) = \vec{r}'(t) = \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle \]

The rate of change of position (direction and rate of motion)

Acceleration Vector:

\[ \vec{a}(t) = \vec{v}'(t) = \vec{r}''(t) = \left\langle \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2} \right\rangle \]

The rate of change of velocity

⚡ Speed vs. Velocity

Speed Formula

SPEED (Scalar):

Speed is the magnitude of velocity:

\[ \text{Speed} = \left|\vec{v}(t)\right| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \]

Velocity vs. Speed:

Velocity: A vector (has direction and magnitude)
Speed: A scalar (magnitude only, always ≥ 0)
Speed = \(|\vec{v}(t)|\) = length of velocity vector

➡️ Unit Tangent Vector

Unit Tangent Vector

THE FORMULA:

\[ \vec{T}(t) = \frac{\vec{r}'(t)}{\left|\vec{r}'(t)\right|} = \frac{\vec{v}(t)}{\left|\vec{v}(t)\right|} \]

Unit vector in the direction of motion

📝 Properties: The unit tangent vector \(\vec{T}(t)\) always has magnitude 1 and points in the direction of motion (same direction as velocity).

📋 Differentiation Rules for Vectors

Rules of Differentiation

1. Constant Multiple:

\[ \frac{d}{dt}[c\vec{r}(t)] = c\vec{r}'(t) \]

2. Sum/Difference:

\[ \frac{d}{dt}[\vec{u}(t) \pm \vec{v}(t)] = \vec{u}'(t) \pm \vec{v}'(t) \]

3. Scalar Function Product:

\[ \frac{d}{dt}[f(t)\vec{r}(t)] = f'(t)\vec{r}(t) + f(t)\vec{r}'(t) \]

(Product rule!)

4. Chain Rule:

\[ \frac{d}{dt}[\vec{r}(f(t))] = \vec{r}'(f(t)) \cdot f'(t) \]

📖 Comprehensive Worked Examples

Example 1: Finding Velocity and Acceleration

Problem: Given \(\vec{r}(t) = \langle t^2, t^3 \rangle\), find velocity and acceleration at \(t = 1\).

Solution:

Step 1: Find velocity

\[ \vec{v}(t) = \vec{r}'(t) = \langle 2t, 3t^2 \rangle \]

At \(t = 1\):

\[ \vec{v}(1) = \langle 2, 3 \rangle \]

Step 2: Find acceleration

\[ \vec{a}(t) = \vec{v}'(t) = \langle 2, 6t \rangle \]

At \(t = 1\):

\[ \vec{a}(1) = \langle 2, 6 \rangle \]

ANSWER: \(\vec{v}(1) = \langle 2, 3 \rangle\), \(\vec{a}(1) = \langle 2, 6 \rangle\)

Example 2: Finding Speed

Problem: For \(\vec{r}(t) = \langle \cos t, \sin t \rangle\), find the speed at any time \(t\).

Find velocity:

\[ \vec{v}(t) = \langle -\sin t, \cos t \rangle \]

Find speed:

\[ \text{Speed} = |\vec{v}(t)| = \sqrt{(-\sin t)^2 + (\cos t)^2} = \sqrt{\sin^2 t + \cos^2 t} = 1 \]

The particle moves at constant speed 1!

Example 3: Unit Tangent Vector

Problem: Find the unit tangent vector for \(\vec{r}(t) = \langle 3t, 4t \rangle\) at \(t = 2\).

Step 1: Find \(\vec{r}'(t)\)

\[ \vec{r}'(t) = \langle 3, 4 \rangle \]

Step 2: Find magnitude

\[ |\vec{r}'(t)| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \]

Step 3: Calculate unit tangent

\[ \vec{T}(t) = \frac{\langle 3, 4 \rangle}{5} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \]

This is constant for all \(t\) (straight line motion).

Example 4: Product Rule with Vectors

Problem: Find \(\frac{d}{dt}[t^2\vec{r}(t)]\) where \(\vec{r}(t) = \langle \sin t, \cos t \rangle\).

Use product rule:

\[ \frac{d}{dt}[t^2\vec{r}(t)] = (t^2)'\vec{r}(t) + t^2\vec{r}'(t) \]

\[ = 2t\langle \sin t, \cos t \rangle + t^2\langle \cos t, -\sin t \rangle \]

\[ = \langle 2t\sin t + t^2\cos t, 2t\cos t - t^2\sin t \rangle \]

📊 Complete Formula Reference

Vector-Valued Function Formulas
Concept	Formula	Notes
Position	\(\vec{r}(t) = \langle x(t), y(t) \rangle\)	Location vector
Velocity	\(\vec{v}(t) = \vec{r}'(t)\)	Rate of change of position
Acceleration	\(\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t)\)	Rate of change of velocity
Speed	\(\|\vec{v}(t)\| = \sqrt{(dx/dt)^2 + (dy/dt)^2}\)	Magnitude of velocity
Unit Tangent	\(\vec{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}\)	Direction of motion

💡 Essential Tips & Strategies

✅ Success Strategies:

Differentiate component-wise: Each component separately
Vector vs. scalar: Velocity is vector, speed is scalar
Notation matters: \(\vec{v}\) vs. \(|\vec{v}|\)
Product rule applies: For scalar times vector
Chain rule applies: For composition
Speed = magnitude: Always non-negative
Unit vectors have magnitude 1: Check your unit tangent!
Position → velocity → acceleration: Derivative chain

🔥 Common Vector Functions:

Circular motion: \(\vec{r}(t) = \langle r\cos t, r\sin t \rangle\)
Linear motion: \(\vec{r}(t) = \langle x_0 + at, y_0 + bt \rangle\)
Projectile motion: \(\vec{r}(t) = \langle v_0t\cos\theta, h_0 + v_0t\sin\theta - \frac{1}{2}gt^2 \rangle\)

❌ Common Mistakes to Avoid

Mistake 1: Confusing velocity vector with speed (vector vs. scalar)
Mistake 2: Forgetting to differentiate both components
Mistake 3: Not using angle brackets \(\langle \, \rangle\) for vectors
Mistake 4: Forgetting magnitude formula for speed
Mistake 5: Not squaring components when finding magnitude
Mistake 6: Forgetting product rule when scalar multiplies vector
Mistake 7: Not simplifying unit tangent vector
Mistake 8: Sign errors in derivatives of trig functions
Mistake 9: Mixing up position, velocity, and acceleration
Mistake 10: Not checking if unit vector has magnitude 1

📝 Practice Problems

Solve these:

Find \(\vec{v}(t)\) and \(\vec{a}(t)\) for \(\vec{r}(t) = \langle e^t, e^{-t} \rangle\).
Find speed at \(t = 2\) for \(\vec{r}(t) = \langle t^2, 2t \rangle\).
Find unit tangent vector for \(\vec{r}(t) = \langle 2\cos t, 3\sin t \rangle\) at \(t = \pi/2\).
Find \(\frac{d}{dt}[\sin t \cdot \vec{r}(t)]\) where \(\vec{r}(t) = \langle t, t^2 \rangle\).

Answers:

\(\vec{v}(t) = \langle e^t, -e^{-t} \rangle\), \(\vec{a}(t) = \langle e^t, e^{-t} \rangle\)
Speed = \(\sqrt{80} = 4\sqrt{5}\)
\(\vec{T}(\pi/2) = \langle 0, 1 \rangle\)
\(\langle \cos t \cdot t + \sin t, \cos t \cdot t^2 + 2t\sin t \rangle\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Use proper notation: \(\langle \, \rangle\) or \(\vec{i}, \vec{j}\) form
Show component-wise differentiation: Each component separately
For speed: Show magnitude calculation
For unit tangent: Show normalization step
Label vectors: Use arrow notation or bold
State what you're finding: Position, velocity, or acceleration
Include units: If given in problem
Simplify answers: Especially for unit vectors

💯 Exam Strategy:

Identify what's given: position, velocity, or acceleration?
Determine what to find
Apply appropriate formula
Differentiate component-wise
For magnitude: square root of sum of squares
For unit vector: divide by magnitude
Simplify answer
Check: Does it make sense?

⚡ Quick Reference Guide

VECTOR-VALUED FUNCTIONS ESSENTIALS

Position Vector:

\[ \vec{r}(t) = \langle x(t), y(t) \rangle \]

Velocity Vector:

\[ \vec{v}(t) = \vec{r}'(t) = \langle x'(t), y'(t) \rangle \]

Acceleration Vector:

\[ \vec{a}(t) = \vec{r}''(t) = \langle x''(t), y''(t) \rangle \]

Speed (Scalar):

\[ |\vec{v}(t)| = \sqrt{[x'(t)]^2 + [y'(t)]^2} \]

Remember:

Differentiate each component!
Velocity is a vector, speed is scalar
Magnitude uses square root of sum of squares

Master Vector-Valued Functions! A vector-valued function outputs a vector: \(\vec{r}(t) = \langle x(t), y(t) \rangle\). This is just parametric equations in vector form! Derivative: differentiate each component—\(\vec{r}'(t) = \langle x'(t), y'(t) \rangle\). For motion: position \(\vec{r}(t)\), velocity \(\vec{v}(t) = \vec{r}'(t)\), acceleration \(\vec{a}(t) = \vec{r}''(t)\). Speed is magnitude of velocity: \(|\vec{v}(t)| = \sqrt{(dx/dt)^2+(dy/dt)^2}\) (scalar, not vector!). Unit tangent vector: \(\vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|}\) (magnitude 1, direction of motion). Product rule applies: \(\frac{d}{dt}[f(t)\vec{r}(t)] = f'(t)\vec{r}(t) + f(t)\vec{r}'(t)\). Common error: confusing velocity (vector) with speed (scalar). This is fundamental BC content—appears on every exam! Practice until automatic! 🎯✨