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Vector-Valued Functions

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Vector-Valued Functions - Complete Interactive Lesson

Part 1: Core Concepts

Vector-Valued Functions

Part 1 of 7 — Introduction & Position Vectors

A vector-valued function describes a curve using a position vector:

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

This is the natural extension of parametric equations — same information, vector notation.

Parametric vs. Vector Form

Parametric	Vector
$x = f(t),\; y = g(t)$	$\vec{r}(t) = \langle f(t), g(t) \rangle$

Key Fact: The AP BC exam uses both notations interchangeably. Be fluent in both.

Examples of Vector-Valued Functions

Example 1. Circular motion:
$\vec{r}(t) = \langle \cos t,\, \sin t \rangle$ traces the unit circle counterclockwise.

Practice Problems

Concept Checks

Computation

Summary

Vector-valued functions: $\vec{r}(t) = \langle x(t), y(t) \rangle$
Equivalent to parametric equations in vector notation

Part 2: Worked Examples

Vector-Valued Functions — Velocity & Acceleration

Part 2 of 7 — Derivatives of Vector Functions

The derivative of a vector-valued function is taken component-wise:

$\vec{r}\,'(t) = \langle x'(t),\, y'(t) \rangle$

Part 3: Problem-Solving Patterns

Vector-Valued Functions — Integration & Displacement

Part 3 of 7 — Antiderivatives and Distance

Integration of vector functions is also done component-wise:

$\int_a^b \vec{r}\,'(t)\,dt = \vec{r}(b) - \vec{r}(a) = \text{displacement}$

Part 4: Graphs and Interpretation

Vector-Valued Functions — Arc Length & Unit Tangent

Part 4 of 7 — Arc Length and the Unit Tangent Vector

Arc Length

The arc length of $\vec{r}(t) = \langle x(t), y(t) \rangle$ from to is:

Part 5: Applications

Vector-Valued Functions — Motion & FRQ Strategies

Part 5 of 7 — AP Free-Response Motion Problems

Motion problems with vector-valued functions are one of the most common BC FRQ topics. Here's the typical structure:

Common FRQ Parts

Part	They Ask	You Do
(a)	Position at time $t$	Integrate $\vec{v}$ , apply

Part 6: Exam Strategy

Vector-Valued Functions — Workshop

Part 6 of 7 — Problem-Solving Workshop

Mixed problems covering position, velocity, acceleration, arc length, and motion analysis.

Workshop Problems

Problem	Skills Tested
1	Position from velocity + ICs
2	Speed and direction analysis
3	Arc length computation

Problem 1

$\vec{a}(t) = \langle 2, -6t \rangle$ , , .

Part 7: Mixed Review

Vector-Valued Functions — Comprehensive Review

Part 7 of 7 — Full Topic Review

Master Reference Table

Concept	Formula
Position	$\vec{r}(t) = \langle x(t), y(t) \rangle$

Example 2. Line through $(1, 3)$ with direction $\langle 2, -1 \rangle$ :
$\vec{r}(t) = \langle 1 + 2t,\, 3 - t \rangle$

Example 3. Parabolic path:
$\vec{r}(t) = \langle t,\, t^2 \rangle$

Domain and Continuity

$\vec{r}(t)$ is continuous at $t = c$ if both component functions $x(t)$ and $y(t)$ are continuous at $c$ .

$\lim_{t \to c} \vec{r}(t) = \left\langle \lim_{t \to c} x(t),\; \lim_{t \to c} y(t) \right\rangle$

Limits and continuity are evaluated component-wise

The path is traced by the tip of the position vector

$\boxed{\vec{r}(t) = \langle x(t),\, y(t) \rangle}$

Next: Part 2 — Velocity, speed, and acceleration vectors.

⟨x′(t),y′(t)⟩

Velocity, Speed, and Acceleration

Quantity	Definition	Formula
Position	$\vec{r}(t)$	$\langle x(t), y(t) \rangle$
Velocity	$\vec{v}(t) = \vec{r}\,'(t)$
Acceleration	$\vec{a}(t) = \vec{v}\,'(t)$
Speed	$\\|\vec{v}(t)\\|$	$\sqrt{[x'(t)]^2 + [y'(t)]^2}$

Key Fact: Velocity is a vector (has direction). Speed is a scalar (magnitude only).

Example

Let $\vec{r}(t) = \langle t^3 - 3t,\, t^2 \rangle$ .

Velocity: $\vec{v}(t) = \langle 3t^2 - 3,\, 2t \rangle$

Acceleration: $\vec{a}(t) = \langle 6t,\, 2 \rangle$

Speed at $t = 1$ :
$\vec{v}(1) = \langle 0, 2 \rangle$ , so speed .

Direction of motion at $t = 1$ : Purely vertical (upward) since $v_x = 0$ .

When is the particle at rest?

The particle is at rest when $\vec{v}(t) = \vec{0}$ , meaning AND simultaneously.

$3t^2 - 3 = 0 \implies t = \pm 1$ , and $2t = 0 \implies t = 0$ .

No value satisfies both — the particle is never at rest (it's always moving in at least one direction).

Practice Problems

Speed Computation

Summary

$\vec{v}(t) = \vec{r}\,'(t)$ — velocity is the derivative of position
$\vec{a}(t) = \vec{v}\,'(t) = \vec{r}\,''(t)$ — acceleration is the second derivative
Speed $= \|\vec{v}(t)\| = \sqrt{[x'(t)]^2 + [y'(t)]^2}$
Particle at rest: $\vec{v}(t) = \vec{0}$ (both components zero)

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

Next: Part 3 — Integration of vector functions and displacement.

Displacement vs. Distance

Concept	Formula	Type
Displacement	$\int_a^b \vec{v}(t)\,dt = \langle \int_a^b x'\,dt,\, \int_a^b y'\,dt \rangle$	Vector
Total distance	$\int_a^b \\|\vec{v}(t)\\|\,dt = \int_a^b \sqrt{(x')^2 + (y')^2}\,dt$

AP Tip: "How far" = total distance (scalar). "Net change in position" = displacement (vector). The exam is precise about this distinction.

Example

A particle has velocity $\vec{v}(t) = \langle 2t, 3 \rangle$ and initial position $\vec{r}(0) = \langle 1, -2 \rangle$ .

Position function: $\vec{r}(t) = \int \vec{v}(t)\,dt = \langle t^2 + C_1,\, 3t + C_2 \rangle$

Apply ICs: $\vec{r}(0) = \langle C_1, C_2 \rangle = \langle 1, -2 \rangle$

$\boxed{\vec{r}(t) = \langle t^2 + 1,\, 3t - 2 \rangle}$

Displacement from $t=0$ to $t=2$ : $\vec{r}(2) - \vec{r}(0) = \langle 5, 4 \rangle - \langle 1, -2 \rangle = \langle 4, 6 \rangle$

Total distance from $t=0$ to $t=2$ : $\int_0^2 \sqrt{4t^2 + 9}\,dt$ This requires trig sub or a calculator. On the AP exam, a calculator-active section would provide a numerical answer.

Practice Problems

Summary

Integrate vector functions component-wise
Displacement $= \int_a^b \vec{v}\,dt$ (vector)
Total distance $= \int_a^b \|\vec{v}\|\,dt$ (scalar)
Use initial conditions to find constants of integration

$\boxed{\text{Distance} = \int_a^b \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt}$

Next: Part 4 — Arc length and the unit tangent vector.

$s = \int_a^b \|\vec{r}\,'(t)\|\,dt = \int_a^b \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$

Note: this is identical to the total distance formula — arc length = distance traveled.

$\hat{T}(t) = \frac{\vec{r}\,'(t)}{\|\vec{r}\,'(t)\|}$

The unit tangent points in the direction of motion with magnitude 1.

Vector	Formula	Meaning
$\vec{r}\,'(t)$	$\langle x', y' \rangle$	Tangent (velocity)
$\hat{T}(t)$	$\vec{r}\,'/\\|\vec{r}\,'\\|$
$\hat{N}(t)$	$\hat{T}\,'/\\|\hat{T}\,'\\|$

Example

$\vec{r}(t) = \langle 3\cos t, 3\sin t \rangle$ , $0 \le t \le 2\pi$ .

$\vec{r}\,'(t) = \langle -3\sin t, 3\cos t \rangle$

$\|\vec{r}\,'(t)\| = \sqrt{9\sin^2 t + 9\cos^2 t} = 3$

Arc length: $s = \int_0^{2\pi} 3\,dt = 6\pi$ ✓ (circumference of circle of radius 3)

Unit tangent: $\hat{T}(t) = \frac{1}{3}\langle -3\sin t, 3\cos t \rangle = \langle -\sin t, \cos t \rangle$

AP Tip: Arc length and total distance are computed with the same integral. The only difference is interpretation: arc length describes the curve, distance describes the motion.

Practice Problems

Arc Length Computation

Summary

Arc length $= \int_a^b \|\vec{r}\,'(t)\|\,dt$ (same as total distance)
Unit tangent vector: $\hat{T}(t) = \vec{r}\,'/\|\vec{r}\,'\|$
$\hat{T}$ gives direction of motion, $\|\vec{r}\,'\|$ gives speed

$\boxed{s = \int_a^b \|\vec{r}\,'(t)\|\,dt \qquad \hat{T} = \frac{\vec{r}\,'}{\|\vec{r}\,'\|}}$

Next: Part 5 — Motion problems and free-response strategies.

Scoring: Show all setup. Even with a calculator problem, write the integral before evaluating.

Full FRQ Practice

A particle moves in the $xy$ -plane with velocity $\vec{v}(t) = \langle 2t - 1,\, e^{-t} \rangle$ for $t \ge 0$ . At $t = 0$ , the particle is at $(3, 5)$ .

(a) Find $\vec{r}(t)$ .

$x(t) = \int(2t-1)\,dt = t^2 - t + C_1$ ,

$y(t) = \int e^{-t}\,dt = -e^{-t} + C_2$ ,

$\vec{r}(t) = \langle t^2 - t + 3,\, -e^{-t} + 6 \rangle$

(b) Speed at $t = 2$ : $\vec{v}(2) = \langle 3, e^{-2} \rangle$ . Speed .

(c) Total distance from $t = 0$ to $t = 3$ : $\int_0^3 \sqrt{(2t-1)^2 + e^{-2t}}\,dt \approx 4.512$ (calculator).

(d) $\vec{a}(t) = \langle 2, -e^{-t} \rangle$ . At : .

Summary

AP FRQs follow a predictable pattern: position → speed → distance → acceleration
Always show integral setup before calculator evaluation
"At rest" means $\vec{v} = \vec{0}$ (both components zero)
Direction changes when individual velocity components change sign

Next: Part 6 — Problem-Solving Workshop.

\vec{v}(0) = \langle 1, 4 \rangle

\vec{r}(0) = \langle 0, 0 \rangle

Find $\vec{r}(t)$ .

$\vec{v}(t) = \int\langle 2, -6t \rangle\,dt = \langle 2t + C_1, -3t^2 + C_2 \rangle$

$\vec{v}(0) = \langle C_1, C_2 \rangle = \langle 1, 4 \rangle$

$\vec{v}(t) = \langle 2t + 1, -3t^2 + 4 \rangle$

$\vec{r}(t) = \int\vec{v}\,dt = \langle t^2 + t + D_1, -t^3 + 4t + D_2 \rangle$

$\vec{r}(0) = \langle D_1, D_2 \rangle = \langle 0, 0 \rangle$

$\boxed{\vec{r}(t) = \langle t^2 + t,\, -t^3 + 4t \rangle}$

Workshop Questions

Workshop Computation

Workshop Summary

Integrate acceleration → velocity → position (apply ICs at each step)
Speed at a point: evaluate $\|\vec{v}(t_0)\|$
Direction: analyze signs of $x'(t)$ and $y'(t)$
Arc length of non-circular curves often requires calculator

Next: Part 7 — Comprehensive Review.

AP Tip: This table covers everything you need for vector motion questions. Memorize it.

Key Connections

Vectors ↔ Parametric: Same math, different notation. $\vec{r}(t) = \langle x(t), y(t) \rangle$ is the same as $x = x(t), y = y(t)$ .

Vectors ↔ 1D Motion: Extends AB motion concepts:

AB: $v(t) = s'(t)$ , distance $= \int|v|\,dt$
BC: , distance

Common Mistakes:

Confusing displacement (vector) with distance (scalar)
Forgetting to check BOTH components for "at rest"
Using $|x'| + |y'|$ instead of $\sqrt{(x')^2+(y')^2}$ for speed

Review Questions

Final Computation

Topic Complete!

You've mastered vector-valued functions:

Position, velocity, acceleration — component-wise derivatives
Speed vs. velocity vs. displacement vs. distance
Integration with initial conditions
Arc length and unit tangent vectors
AP FRQ strategies for motion problems

$\boxed{\vec{v}(t) = \vec{r}\,'(t) \qquad \text{Distance} = \int_a^b \|\vec{v}(t)\|\,dt}$

Up next: Arc Length & Surface Area — extending these ideas to general curves.

[x^{'} (t)]^{2} + [y^{'} (t)]^{2}

= \sqrt{0 + 4} = 2

[x′(t)]2+[y′(t)]2

∫ab(x′)2+(y′)2dt

[x′(t)]2+[y′(t)]2

x(0) = 3 \implies C_1 = 3

y(0) = 5 \implies -1 + C_2 = 5 \implies C_2 = 6

= \sqrt{9 + e^{-4}} \approx 3.002

\vec{a}(1) = \langle 2, -e^{-1} \rangle

\vec{v}(t) = \vec{r}\,'(t)

= \int\|\vec{v}\|\,dt

(x^{'})^{2} + (y^{'})^{2}

Not applying initial conditions after integration