Vector-Valued Functions

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

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.

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 $c$.

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

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
• 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.

Vector-Valued Functions — Velocity & Acceleration

Part 2 of 7 — Derivatives of Vector Functions

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

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

Velocity, Speed, and Acceleration

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 $= \sqrt{0 + 4} = 2$.

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 $x'(t) = 0$ AND $y'(t) = 0$ 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

Key Concepts

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.

Vector-Valued Functions — Integration & Displacement

Part 3 of 7 — Antiderivatives and Distance

Integration of vector functions is also done component-wise:

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

Displacement vs. Distance

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

Concept Checks

Computation

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.

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 $t = a$ to $t = b$ is:

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

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

Unit Tangent Vector

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

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

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

Concept Checks

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.

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

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 $\vec{r}(t)$.

$x(t) = \int(2t-1)\,dt = t^2 - t + C_1$, $x(0) = 3 \implies C_1 = 3$

$y(t) = \int e^{-t}\,dt = -e^{-t} + C_2$, $y(0) = 5 \implies -1 + C_2 = 5 \implies C_2 = 6$

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

(b) Speed at $t = 2$: $\vec{v}(2) = \langle 3, e^{-2} \rangle$. Speed $= \sqrt{9 + e^{-4}} \approx 3.002$.

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

Practice

Quick Checks

FRQ Practice

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.

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 1

$\vec{a}(t) = \langle 2, -6t \rangle$, $\vec{v}(0) = \langle 1, 4 \rangle$, $\vec{r}(0) = \langle 0, 0 \rangle$.

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 Checks

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.

Vector-Valued Functions — Comprehensive Review

Part 7 of 7 — Full Topic Review

Master Reference Table

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: $\vec{v}(t) = \vec{r}\,'(t)$, distance $= \int\|\vec{v}\|\,dt$

Common Mistakes:

1. Confusing displacement (vector) with distance (scalar)
2. Forgetting to check BOTH components for "at rest"
3. Using $|x'| + |y'|$ instead of $\sqrt{(x')^2+(y')^2}$ for speed
4. Not applying initial conditions after integration

Review Questions

Concept Checks

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.