MATH-283 (Calculus III) Curriculum

Session 05 - Vector-Valued Functions and Derivatives

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Evaluate vector-valued functions $\vec{r}(t) = \langle f(t), g(t), h(t) \rangle$ , determine component and global domains via set intersection, and interpret spatial trajectories and parametric curves in $\mathbb{R}^3$ .
Differentiate vector-valued functions component-wise to compute velocity $\vec{v}(t) = \vec{r}'(t)$ , acceleration $\vec{a}(t) = \vec{r}''(t)$ , and the principal unit tangent vector $\hat{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}$ , applying the vector chain rule to composite space curves.
Integrate vector-valued functions component-wise to solve kinematic Initial Value Problems (IVPs) with constant vectors $\vec{C} = \langle C_1, C_2, C_3 \rangle$ and calculate net displacement vectors $\Delta \vec{r} = \int_a^b \vec{v}(t)\,dt$ across mechanical and aerospace systems.

Micro-Lecture

1. Parametric Space Curves: Beyond the Vertical Line Test

Mathematical Foundation: In single-variable calculus, functions $y = f(x)$ are restricted to planar paths where each input produces exactly one vertical output. In multivariable engineering, physical trajectories twist, ascend, and cross themselves in 3D space. We describe these paths using Vector Valued Functions $\vec{r}(t) = \langle f(t), g(t), h(t) \rangle = f(t)\hat{i} + g(t)\hat{j} + h(t)\hat{k}$ , where a single scalar parameter $t \in \mathbb{R}$ (typically time) governs independent Cartesian coordinate functions. The domain of $\vec{r}(t)$ requires all three component functions to be simultaneously valid: $\text{Domain}(\vec{r}) = \text{Domain}(f) \cap \text{Domain}(g) \cap \text{Domain}(h)$ .
Engineering Context: Robotic arms, CNC milling spindles, and aircraft routing systems rely on vector-valued functions to establish smooth 3D toolpaths and spatial operating boundaries.

2. Motion in Space: Velocity, Tangent Vectors, and Heading Control

Mathematical Foundation: Taking the derivative of a position vector function yields the velocity vector $\vec{v}(t) = \vec{r}'(t) = \langle f'(t), g'(t), h'(t) \rangle$ (defined in Derivatives of Vector-Valued Functions), which is strictly tangent to the space curve. Differentiating velocity produces the acceleration vector $\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t)$ . To isolate the pure directional heading of motion independent of traversal speed, we normalize velocity to form the Unit Tangent Vector $\hat{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}$ .
Engineering Context: Autonomous drones and autonomous vehicles decouple throttle speed ( $\|\vec{v}(t)\|$ ) from directional attitude control ( $\hat{T}(t)$ ) to execute stabilized flight turns and collision avoidance.

3. Kinematic Accumulation: Reconstructing Trajectories from Sensor Logs

Mathematical Foundation: Integrals of Vector-Valued Functions reverse differentiation component-by-component. Indefinite integration produces an independent integration constant for each axis, grouped into a single constant vector $\vec{C} = \langle C_1, C_2, C_3 \rangle$ . Definite integration evaluates the net displacement vector $\Delta \vec{r} = \int_{a}^{b} \vec{v}(t)\,dt = \vec{r}(b) - \vec{r}(a)$ .
Engineering Context: In GPS-denied environments (tunnels, subterranean mines, deep space), inertial navigation systems (INS) integrate onboard accelerometer logs twice over time (dead reckoning) to maintain exact real-time vehicle coordinates.

Skill Block 1

Worked Example A: Finding the Domain with Radical and Rational Components

Determine the domain of the vector-valued function:

\vec{r}(t) = \left\langle \sqrt{t+2}, \frac{-9}{t+3}, \frac{-7}{\sqrt{t-5}} \right\rangle

Worked Solution:

Analyze Component 1 ( $x(t) = \sqrt{t+2}$ ): State the Square Root Domain Condition: The radicand must be non-negative ( $u(t) \ge 0$ ).
$t + 2 \ge 0 \implies t \ge -2$ Domain of $x(t)$ : $[-2, \infty)$ .
Analyze Component 2 ( $y(t) = \frac{-9}{t+3}$ ): State the Rational Denominator Condition: The denominator cannot equal zero ( $d(t) \ne 0$ ).
$t + 3 \ne 0 \implies t \ne -3$ Domain of $y(t)$ : $(-\infty, -3) \cup (-3, \infty)$ .
Analyze Component 3 ( $z(t) = \frac{-7}{\sqrt{t-5}}$ ): State the Denominator Square Root Condition: The radicand must be strictly positive ( $u(t) > 0$ ).
$t - 5 > 0 \implies t > 5$ Domain of $z(t)$ : $(5, \infty)$ .
Intersect All Component Domains: State the Vector Domain Intersection Formula:
$\text{Domain}(\vec{r}) = \text{Domain}(x) \cap \text{Domain}(y) \cap \text{Domain}(z)$ $\text{Domain}(\vec{r}) = [-2, \infty) \cap \{t \ne -3\} \cap (5, \infty)$ Any real number satisfying $t > 5$ automatically satisfies $t \ge -2$ and $t \ne -3$ . $\text{Domain}(\vec{r}) = (5, \infty)$

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) How is the domain of a vector-valued function $\vec{r}(t) = \langle f(t), g(t), h(t) \rangle$ determined from the domains of its individual component functions?
(b) In Worked Example A, explain why the condition $t > 5$ makes the restrictions $t \ge -2$ and $t \ne -3$ redundant.
(c) In the linked engineering application Robotic Arm Workspace Domain (Mechanical & Robotics Engineering), what physical condition restricts the robot's $z$ -component domain to $t < 14$ ?

Solution Key:

(a) Vector Domain Definition: A vector-valued function is only defined at parameter values $t$ where every single component function produces a real output. Therefore, the global domain is the strict set intersection of the individual domains: $\text{Domain}(\vec{r}) = \text{Domain}(f) \cap \text{Domain}(g) \cap \text{Domain}(h)$
(b) Redundancy Analysis: The third component requires $t > 5$ . On the real number line, any number strictly greater than $5$ is already greater than or equal to $-2$ , and cannot equal $-3$ . Thus, $(5, \infty) \subset [-2, \infty) \setminus \{-3\}$ , making $t > 5$ the limiting bound.
(c) Engineering Physical Link: The denominator contains $\sqrt{14 - t}$ . For real Cartesian displacement coordinates, the radicand must be positive ( $14 - t > 0$ ), which enforces a maximum cycle cutoff time of $t < 14\text{ seconds}$ before joint limiters are breached.

Show your steps and calculations here.

Practice Problem 2

Determine the domain of the vector-valued function:

\vec{r}(t) = \left\langle \sqrt{3t-4}, \sqrt{t+7}, \frac{6}{t-1} \right\rangle

Solution Key:

Analyze Component 1 ( $x(t) = \sqrt{3t-4}$ ): Apply the Square Root Domain Condition:
$3t - 4 \ge 0 \implies 3t \ge 4 \implies t \ge \frac{4}{3}$ Domain of $x(t)$ : $[\frac{4}{3}, \infty)$ .
Analyze Component 2 ( $y(t) = \sqrt{t+7}$ ): Apply the Square Root Domain Condition:
$t + 7 \ge 0 \implies t \ge -7$ Domain of $y(t)$ : $[-7, \infty)$ .
Analyze Component 3 ( $z(t) = \frac{6}{t-1}$ ): Apply the Rational Denominator Condition:
$t - 1 \ne 0 \implies t \ne 1$ Domain of $z(t)$ : $(-\infty, 1) \cup (1, \infty)$ .
Intersect Component Domains via the Vector Domain Intersection Formula:
$\text{Domain}(\vec{r}) = \left[\frac{4}{3}, \infty\right) \cap [-7, \infty) \cap \{t \ne 1\}$ Since $\frac{4}{3} \approx 1.333 > 1 > -7$ , any $t \ge \frac{4}{3}$ is automatically greater than $-7$ and strictly greater than $1$ . $\text{Domain}(\vec{r}) = \left[\frac{4}{3}, \infty\right)$

Show your steps and calculations here.

Practice Problem 3

Determine the domain of the vector-valued function:

\vec{r}(t) = \left\langle \frac{\sqrt{t+2}}{t-1}, \sqrt{4-t}, \frac{3}{\sqrt{t-6}} \right\rangle

Solution Key:

Analyze Component 1 ( $x(t) = \frac{\sqrt{t+2}}{t-1}$ ): Numerator requires $t + 2 \ge 0 \implies t \ge -2$ . Denominator requires $t - 1 \ne 0 \implies t \ne 1$ . Domain of $x(t)$ : $[-2, 1) \cup (1, \infty)$ .
Analyze Component 2 ( $y(t) = \sqrt{4-t}$ ): Radicand must be non-negative:
$4 - t \ge 0 \implies t \le 4$ Domain of $y(t)$ : $(-\infty, 4]$ .
Analyze Component 3 ( $z(t) = \frac{3}{\sqrt{t-6}}$ ): Denominator radicand must be strictly positive:
$t - 6 > 0 \implies t > 6$ Domain of $z(t)$ : $(6, \infty)$ .
Evaluate Set Intersection: Applying the Vector Domain Intersection Formula:
$\text{Domain}(\vec{r}) = ([-2, 1) \cup (1, \infty)) \cap (-\infty, 4] \cap (6, \infty)$ Notice the direct conflict between $t \le 4$ and $t > 6$ . There exists no real number that is simultaneously $\le 4$ and $> 6$ : $(-\infty, 4] \cap (6, \infty) = \emptyset$ Therefore: $\text{Domain}(\vec{r}) = \emptyset$

Show your steps and calculations here.

Regroup 1

Review domain boundary intervals and confirm understanding of open versus closed interval notation with radical roots.
Discuss how 3D parametric graphing tools (e.g. Desmos 3D) display curves over finite parameter windows.

Skill Block 2

Worked Example A: Finding the Derivative and Principal Unit Tangent Vector

Find the derivative vector $\vec{r}'(t)$ and the principal unit tangent vector $\hat{T}(t)$ for:

\vec{r}(t) = -t\hat{i} - t^4\hat{j} - 6t^2\hat{k}

Worked Solution:

Differentiate $\vec{r}(t)$ Component-Wise via the Vector Derivative Formula: State the Vector Derivative Formula:
$\vec{r}'(t) = \frac{d}{dt}[x(t)]\hat{i} + \frac{d}{dt}[y(t)]\hat{j} + \frac{d}{dt}[z(t)]\hat{k}$
(This formula is defined in Derivatives of Vector-Valued Functions.) Apply the power rule to each component:
- $\hat{i}$ -component: $\frac{d}{dt}[-t] = -1$
- $\hat{j}$ -component: $\frac{d}{dt}[-t^4] = -4t^3$
- $\hat{k}$ -component: $\frac{d}{dt}[-6t^2] = -12t$ $\vec{r}'(t) = -\hat{i} - 4t^3\hat{j} - 12t\hat{k} = \langle -1, -4t^3, -12t \rangle$
Compute the Vector Magnitude $\|\vec{r}'(t)\|$ : State the Vector Magnitude Formula:
$\|\vec{r}'(t)\| = \sqrt{[x'(t)]^2 + [y'(t)]^2 + [z'(t)]^2}$ $\|\vec{r}'(t)\| = \sqrt{(-1)^2 + (-4t^3)^2 + (-12t)^2} = \sqrt{1 + 16t^6 + 144t^2} = \sqrt{16t^6 + 144t^2 + 1}$
Construct the Principal Unit Tangent Vector $\hat{T}(t)$ : State the Principal Unit Tangent Vector Formula:
$\hat{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}$ (This formula is defined in Unit Tangent Vector.) $\hat{T}(t) = \frac{-\hat{i} - 4t^3\hat{j} - 12t\hat{k}}{\sqrt{16t^6 + 144t^2 + 1}} = \left\langle \frac{-1}{\sqrt{16t^6 + 144t^2 + 1}}, \frac{-4t^3}{\sqrt{16t^6 + 144t^2 + 1}}, \frac{-12t}{\sqrt{16t^6 + 144t^2 + 1}} \right\rangle$

Practice Problem 4: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) State the distinct mathematical and physical definitions of velocity $\vec{v}(t)$ , scalar speed $\|\vec{v}(t)\|$ , and the unit tangent vector $\hat{T}(t)$ .
(b) In Worked Example A, explain why the principal unit tangent vector $\hat{T}(t)$ must have a magnitude of exactly 1 for all real values of $t$ .
(c) In the linked engineering application Autonomous Drone Flight Kinematics (Aerospace & Computer Engineering), how can an aircraft experience non-zero acceleration even when its scalar flight speed is constant?

Solution Key:

(a) Kinematic Definitions:
- Velocity Vector: $\vec{v}(t) = \vec{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle$ , tangent to the trajectory.
- Scalar Speed: $v(t) = \|\vec{v}(t)\| = \sqrt{[x'(t)]^2 + [y'(t)]^2 + [z'(t)]^2}$ , a positive scalar.
- Principal Unit Tangent Vector: $\hat{T}(t) = \frac{\vec{v}(t)}{\|\vec{v}(t)\|}$ , a unit vector of length 1 isolating direction.
(b) Magnitude Invariance: For any non-zero vector $\vec{w}$ , the norm of $\frac{\vec{w}}{\|\vec{w}\|}$ is $\frac{1}{\|\vec{w}\|} \|\vec{w}\| = 1$ . Hence $\|\hat{T}(t)\| = 1$ identically for all $t$ where $\vec{r}'(t) \ne \vec{0}$ .
(c) Centripetal Turning in Flight: Velocity is a vector having both magnitude and direction: $\vec{v}(t) = v(t)\hat{T}(t)$ . Differentiating yields $\vec{a}(t) = v'(t)\hat{T}(t) + v(t)\hat{T}'(t)$ . When speed is constant ( $v'(t) = 0$ ), the acceleration $\vec{a}(t) = v(t)\hat{T}'(t)$ is entirely centripetal, turning the drone's direction through 3D space.

Show your steps and calculations here.

Practice Problem 5

Given $\vec{r}(t) = -6t^3\hat{i} - 3t^7\hat{j} - \frac{-3}{t^4}\hat{k}$ and $f(t) = \sqrt[4]{t}$ , calculate:

$\frac{d}{dt}[\vec{r}(t)]$
$\frac{d}{dt}[\vec{r}(f(t))]$

Solution Key:

Compute $\frac{d}{dt}[\vec{r}(t)]$ via the Vector Derivative Formula: State the Vector Derivative Formula:
$\frac{d}{dt}[\vec{r}(t)] = \frac{d}{dt}[-6t^3]\hat{i} + \frac{d}{dt}[-3t^7]\hat{j} + \frac{d}{dt}[3t^{-4}]\hat{k}$
Apply the power rule ( $\frac{d}{dt}[t^n] = n t^{n-1}$ ):
- $\hat{i}$ -component: $-6(3t^2) = -18t^2$
- $\hat{j}$ -component: $-3(7t^6) = -21t^6$
- $\hat{k}$ -component: $3(-4t^{-5}) = -12t^{-5} = -\frac{12}{t^5}$ $\frac{d}{dt}[\vec{r}(t)] = -18t^2\hat{i} - 21t^6\hat{j} - \frac{12}{t^5}\hat{k}$
Compute $\frac{d}{dt}[\vec{r}(f(t))]$ via Parameter Substitution: Substitute $f(t) = t^{1/4}$ into $\vec{r}(t)$ :
$\vec{r}(f(t)) = -6\left(t^{1/4}\right)^3\hat{i} - 3\left(t^{1/4}\right)^7\hat{j} + 3\left(t^{1/4}\right)^{-4}\hat{k}$

$\vec{r}(f(t)) = -6t^{3/4}\hat{i} - 3t^{7/4}\hat{j} + 3t^{-1}\hat{k}$
Now differentiate component-wise with respect to $t$ :
- $\hat{i}$ -component: $\frac{d}{dt}[-6t^{3/4}] = -6\left(\frac{3}{4}\right)t^{-1/4} = -\frac{9}{2}t^{-1/4} = -\frac{9}{2t^{1/4}}$
- $\hat{j}$ -component: $\frac{d}{dt}[-3t^{7/4}] = -3\left(\frac{7}{4}\right)t^{3/4} = -\frac{21}{4}t^{3/4}$
- $\hat{k}$ -component: $\frac{d}{dt}[3t^{-1}] = 3(-1)t^{-2} = -\frac{3}{t^2}$ $\frac{d}{dt}[\vec{r}(f(t))] = -\frac{9}{2t^{1/4}}\hat{i} - \frac{21t^{3/4}}{4}\hat{j} - \frac{3}{t^2}\hat{k}$

Show your steps and calculations here.

Practice Problem 6

The velocity of a test vehicle traveling along a proving ground track is modeled by:

\vec{v}(t) = \left\langle t\sin(2t), 2t\ln(t-3), \frac{t^2}{t-2} \right\rangle \text{ m/s}

Find the acceleration vector function $\vec{a}(t) = \vec{v}'(t)$ .
Calculate the acceleration vector $\vec{a}(20)$ and the principal unit tangent vector $\hat{T}(20)$ at $t = 20\text{ seconds}$ .

Solution Key:

Differentiate Each Component to Find Acceleration $\vec{a}(t) = \vec{v}'(t)$ :
- $x$ -component ( $v_x = t\sin(2t)$ ): Apply the Product Rule: $\frac{d}{dt}[t\sin(2t)] = (1)\sin(2t) + t(\cos(2t) \cdot 2) = \sin(2t) + 2t\cos(2t)$
- $y$ -component ( $v_y = 2t\ln(t-3)$ ): Apply the Product Rule: $\frac{d}{dt}[2t\ln(t-3)] = (2)\ln(t-3) + 2t\left(\frac{1}{t-3}\right) = 2\ln(t-3) + \frac{2t}{t-3}$
- $z$ -component ( $v_z = \frac{t^2}{t-2}$ ): Apply the Quotient Rule: $\frac{d}{dt}\left[\frac{t^2}{t-2}\right] = \frac{(2t)(t-2) - (t^2)(1)}{(t-2)^2} = \frac{2t^2 - 4t - t^2}{(t-2)^2} = \frac{t^2 - 4t}{(t-2)^2}$ Combine into the full acceleration vector: $\vec{a}(t) = \left\langle \sin(2t) + 2t\cos(2t), 2\ln(t-3) + \frac{2t}{t-3}, \frac{t^2 - 4t}{(t-2)^2} \right\rangle \text{ m/s}^2$
Evaluate $\vec{a}(20)$ at $t = 20\text{ s}$ :
- $a_x(20) = \sin(40) + 40\cos(40) \approx 0.7451 + 40(-0.6669) \approx -25.93\text{ m/s}^2$
- $a_y(20) = 2\ln(17) + \frac{40}{17} \approx 2(2.8332) + 2.3529 \approx 8.02\text{ m/s}^2$
- $a_z(20) = \frac{20^2 - 4(20)}{(20-2)^2} = \frac{320}{324} \approx 0.99\text{ m/s}^2$ $\vec{a}(20) \approx \langle -25.93, 8.02, 0.99 \rangle \text{ m/s}^2$
Evaluate $\hat{T}(20)$ at $t = 20\text{ s}$ : State the Principal Unit Tangent Vector Formula:
$\hat{T}(20) = \frac{\vec{v}(20)}{\|\vec{v}(20)\|}$
Evaluate $\vec{v}(20)$ :
- $v_x(20) = 20\sin(40) \approx 20(0.7451) \approx 14.90\text{ m/s}$
- $v_y(20) = 40\ln(17) \approx 40(2.8332) \approx 113.33\text{ m/s}$
- $v_z(20) = \frac{20^2}{18} = \frac{400}{18} \approx 22.22\text{ m/s}$ $\vec{v}(20) \approx \langle 14.90, 113.33, 22.22 \rangle \text{ m/s}$ Compute the speed $\|\vec{v}(20)\|$ : $\|\vec{v}(20)\| = \sqrt{14.90^2 + 113.33^2 + 22.22^2} \approx \sqrt{222.01 + 12843.69 + 493.73} = \sqrt{13559.43} \approx 116.44\text{ m/s}$ Normalize to find $\hat{T}(20)$ : $\hat{T}(20) \approx \left\langle \frac{14.90}{116.44}, \frac{113.33}{116.44}, \frac{22.22}{116.44} \right\rangle \approx \langle 0.128, 0.973, 0.191 \rangle$

Show your steps and calculations here.

Regroup 2

Contrast the physical dimensions of velocity (m/s) versus the unit tangent vector (dimensionless unit vector).
Emphasize that evaluating trig functions in calculus must always be done with radians, not degrees.

Skill Block 3

Worked Example A: Indefinite Vector Integration with Constant Vector $\vec{C}$

Evaluate the indefinite integral:

\int \left\langle 5\cos(7t), 3\cos(4t), \frac{-9}{t} \right\rangle dt

Worked Solution:

Integrate Component-by-Component via the Indefinite Vector Integral Formula: State the Indefinite Vector Integral Formula:
$\int \vec{r}(t) \, dt = \left\langle \int x(t)\,dt, \int y(t)\,dt, \int z(t)\,dt \right\rangle$ (This formula is defined in Integrals of Vector-Valued Functions.)
Evaluate Individual Component Antiderivatives:
- $x$ -component: $\int 5\cos(7t) \, dt = \frac{5}{7}\sin(7t) + C_1$
- $y$ -component: $\int 3\cos(4t) \, dt = \frac{3}{4}\sin(4t) + C_2$
- $z$ -component: $\int \frac{-9}{t} \, dt = -9\ln|t| + C_3$
Group Constants into Vector Constant $\vec{C}$ :
$\int \left\langle 5\cos(7t), 3\cos(4t), \frac{-9}{t} \right\rangle dt = \left\langle \frac{5}{7}\sin(7t), \frac{3}{4}\sin(4t), -9\ln|t| \right\rangle + \vec{C}$ where $\vec{C} = \langle C_1, C_2, C_3 \rangle$ .

Practice Problem 7: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Why must the constant of integration for a 3D vector-valued function be written as a vector $\vec{C} = \langle C_1, C_2, C_3 \rangle$ rather than a single scalar constant $C$ ?
(b) In Worked Example A, explain how component-wise antiderivatives are determined for the individual terms before collecting the constants into $\vec{C}$ .
(c) In the linked engineering application Electric Vehicle Acceleration & Trajectory Recovery (Electrical & Automotive Engineering), how are the initial calibration vectors $\vec{v}(0)$ and $\vec{r}(0)$ used to isolate the unique vehicle trajectory during dead reckoning?

Solution Key:

(a) Vector Constant Principle: Each component represents an independent scalar differential equation ( $\frac{dx}{dt} = f(t), \frac{dy}{dt} = g(t), \frac{dz}{dt} = h(t)$ ), yielding independent constants $C_1, C_2, C_3$ . In vector arithmetic, a scalar cannot be added to a vector; the constants must be collected into $\vec{C} = \langle C_1, C_2, C_3 \rangle$ .
(b) Component-Wise Antiderivative Deconstruction: In Worked Example A:
- $\int 5\cos(7t)\,dt = \frac{5}{7}\sin(7t) + C_1$ via $u = 7t, du = 7\,dt$ .
- $\int 3\cos(4t)\,dt = \frac{3}{4}\sin(4t) + C_2$ via $u = 4t, du = 4\,dt$ .
- $\int \frac{-9}{t}\,dt = -9\ln|t| + C_3$ via $\int \frac{1}{t}dt = \ln|t|$ . Grouping $C_1\hat{i} + C_2\hat{j} + C_3\hat{k} = \vec{C}$ consolidates the individual constants into a single constant vector.
(c) Initial Condition Isolation in Dead Reckoning: Indefinite integration produces an infinite family of parallel trajectories. Evaluating the antiderivatives at $t = 0$ and equating them to the calibrated IMU sensor states $\vec{v}(0)$ and $\vec{r}(0)$ fixes $\vec{C}_1 = \vec{v}(0) - \vec{V}(0)$ and $\vec{C}_2 = \vec{r}(0) - \vec{R}(0)$ , uniquely locking the true physical vehicle path.

Show your steps and calculations here.

Practice Problem 8

An automated guided vehicle (AGV) moves through an automated logistics warehouse with acceleration vector:

\vec{a}(t) = \langle 6t, 12t^2, 4 \rangle \text{ m/s}^2

t = 0\text{ seconds}

, its initial velocity is

\vec{v}(0) = \langle 2, -1, 5 \rangle \text{ m/s}

and its initial position is

\vec{r}(0) = \langle 0, 10, -3 \rangle \text{ meters}

Find the velocity vector function $\vec{v}(t)$ .
Find the position vector function $\vec{r}(t)$ .

Solution Key:

Integrate Acceleration to Find Velocity $\vec{v}(t)$ : State the Kinematic Velocity Integral Formula:
$\vec{v}(t) = \int \vec{a}(t) \, dt = \left\langle \int 6t \, dt, \int 12t^2 \, dt, \int 4 \, dt \right\rangle = \langle 3t^2, 4t^3, 4t \rangle + \vec{C}_1$ Apply the initial velocity condition $\vec{v}(0) = \langle 2, -1, 5 \rangle$ : $\vec{v}(0) = \langle 3(0)^2, 4(0)^3, 4(0) \rangle + \vec{C}_1 = \langle 0, 0, 0 \rangle + \vec{C}_1 = \langle 2, -1, 5 \rangle \implies \vec{C}_1 = \langle 2, -1, 5 \rangle$ Therefore: $\vec{v}(t) = \langle 3t^2 + 2, 4t^3 - 1, 4t + 5 \rangle \text{ m/s}$
Integrate Velocity to Find Position $\vec{r}(t)$ : State the Kinematic Position Integral Formula:
$\vec{r}(t) = \int \vec{v}(t) \, dt = \left\langle \int (3t^2 + 2) \, dt, \int (4t^3 - 1) \, dt, \int (4t + 5) \, dt \right\rangle = \langle t^3 + 2t, t^4 - t, 2t^2 + 5t \rangle + \vec{C}_2$ Apply the initial position condition $\vec{r}(0) = \langle 0, 10, -3 \rangle$ : $\vec{r}(0) = \langle 0, 0, 0 \rangle + \vec{C}_2 = \langle 0, 10, -3 \rangle \implies \vec{C}_2 = \langle 0, 10, -3 \rangle$ Therefore: $\vec{r}(t) = \langle t^3 + 2t, t^4 - t + 10, 2t^2 + 5t - 3 \rangle \text{ meters}$

Show your steps and calculations here.

Practice Problem 9

Given that an object has a velocity vector of:

\vec{v}(t) = \left[ \sqrt[3]{t}\hat{i}, \frac{1}{t+2}\hat{j}, e^{8t}\hat{k} \right] \text{ ft/s}

Use an indefinite integral to find the general displacement vector function $\vec{r}(t)$ .
Use a definite integral to calculate the exact displacement vector of the object over the time interval $1 \le t \le 5\text{ seconds}$ .

Solution Key:

Find General Indefinite Antiderivative $\vec{r}(t)$ : State the Indefinite Vector Integral Formula:
$\vec{r}(t) = \int \vec{v}(t) \, dt = \left( \int t^{1/3} \, dt \right)\hat{i} + \left( \int \frac{1}{t+2} \, dt \right)\hat{j} + \left( \int e^{8t} \, dt \right)\hat{k}$
- $\hat{i}$ -component: $\int t^{1/3} \, dt = \frac{t^{4/3}}{4/3} + C_1 = \frac{3}{4}t^{4/3} + C_1$
- $\hat{j}$ -component: $\int \frac{1}{t+2} \, dt = \ln|t+2| + C_2$
- $\hat{k}$ -component: $\int e^{8t} \, dt = \frac{1}{8}e^{8t} + C_3$ Combine into vector form: $\vec{r}(t) = \left( \frac{3}{4}t^{4/3} \right)\hat{i} + (\ln|t+2|)\hat{j} + \left( \frac{1}{8}e^{8t} \right)\hat{k} + \vec{C}$ where $\vec{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$ .
Evaluate Definite Integral over $[1, 5]$ : State the Definite Vector Integral Formula:
$\Delta \vec{r} = \int_{1}^{5} \vec{v}(t) \, dt = \left[ \left( \frac{3}{4}t^{4/3} \right)\hat{i} + (\ln|t+2|)\hat{j} + \left( \frac{1}{8}e^{8t} \right)\hat{k} \right]_{1}^{5}$
Evaluate each component between limits $t = 1$ and $t = 5$ :
- $\hat{i}$ -component: $\frac{3}{4}(5)^{4/3} - \frac{3}{4}(1)^{4/3} = \frac{3}{4}\left(5^{4/3} - 1\right) \approx 5.662\text{ ft}$
- $\hat{j}$ -component: $\ln|5+2| - \ln|1+2| = \ln(7) - \ln(3) = \ln\left(\frac{7}{3}\right) \approx 0.847\text{ ft}$
- $\hat{k}$ -component: $\frac{1}{8}e^{8(5)} - \frac{1}{8}e^{8(1)} = \frac{1}{8}\left(e^{40} - e^8\right) \approx 2.94 \times 10^{16}\text{ ft}$ Combine into the exact displacement vector: $\Delta \vec{r} = \left[ \frac{3}{4}\left(5^{4/3} - 1\right) \right]\hat{i} + \left[ \ln\left(\frac{7}{3}\right) \right]\hat{j} + \left[ \frac{1}{8}\left(e^{40} - e^8\right) \right]\hat{k} \text{ ft}$

Show your steps and calculations here.

Regroup 3

Confirm why definite integrals eliminate the constant vector $\vec{C}$ ( $\vec{C} - \vec{C} = \vec{0}$ ).
Discuss when exact symbolic forms are required versus decimal approximations in engineering reports.

Synthesis Wrap-up

Core Takeaways:
- Vector-valued functions parameterize curves in space by mapping scalar inputs $t$ to 3D position vectors $\vec{r}(t)$ .
- The domain of $\vec{r}(t)$ is the strict intersection of the domains of its component functions.
- $\vec{r}'(t) = \vec{v}(t)$ gives velocity (tangent vector), $\|\vec{v}(t)\|$ gives scalar speed, and $\hat{T}(t) = \frac{\vec{v}(t)}{\|\vec{v}(t)\|}$ gives unit directional heading.
- Integrating $\vec{a}(t)$ and $\vec{v}(t)$ with initial condition vectors uniquely reconstructs position trajectories in dead-reckoning navigation.
Upcoming Knewton Alta Homework Modules:
- Vector-Valued Functions and Space Curves
- Derivatives of Vector-Valued Functions
- Tangent Vectors and Unit Tangent Vectors
- Integrals of Vector-Valued Functions
Next Session Preview: Session 06 - Arc Length and Curvature.

MATH-283 (Calculus III) Curriculum

Session 05 - Vector-Valued Functions and Derivatives

Micro-Lecture

Skill Block 1

Worked Example A: Finding the Domain with Radical and Rational Components

Practice Problem 1: Concept & Reading Check

Practice Problem 2

Practice Problem 3

Regroup 1

Skill Block 2

Worked Example A: Finding the Derivative and Principal Unit Tangent Vector

Practice Problem 4: Concept & Reading Check

Practice Problem 5

Practice Problem 6

Regroup 2

Skill Block 3

Worked Example A: Indefinite Vector Integration with Constant Vector C⃗\vec{C}C

Practice Problem 7: Concept & Reading Check

Practice Problem 8

Practice Problem 9

Regroup 3

Synthesis Wrap-up

Worked Example A: Indefinite Vector Integration with Constant Vector $\vec{C}$