Session 05 - Vector-Valued Functions and Derivatives

📅 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.

Segment 1 of 8

Skill Block 1

Real-Life Engineering Applications

⚙️

Mechanical Engineering

Robotic Arm Workspace Domain (Mechanical & Robotics Engineering)

In advanced manufacturing and robotics, automated multi-axis robotic manipulators trace complex 3D paths for precision spot welding and component assembly. Mechanical and robotics engineers must verify that the mathematical model of the robot's toolpath is valid across the operating time envelope, ensuring that physical stroke limits, pneumatic pressure lines, and safety enclosures are never breached.

An automated robotic welding tip moves through Cartesian coordinates $(x, y, z)$ in meters as a function of operating cycle time $t$ (in seconds) according to the vector-valued position function:

\vec{r}(t) = \left\langle \sqrt{2t - 6}, \frac{15}{t - 8}, \frac{4}{\sqrt{14 - t}} \right\rangle

Determine the allowable operating domain for the $x$ -component function $x(t) = \sqrt{2t - 6}$ based on the physical requirement that joint extension values must remain real numbers.
Determine the restriction on parameter $t$ for the $y$ -component function $y(t) = \frac{15}{t - 8}$ where the actuator encounters a singularity.
Determine the allowable operating domain for the $z$ -component function $z(t) = \frac{4}{\sqrt{14 - t}}$ .
Calculate the overall valid operating cycle time domain for the entire robotic toolpath $\vec{r}(t)$ in interval notation.

🎯 Why This Matters:

& Real-Life Application: The robotic toolpath is physically and mathematically valid between $t = 3\text{ s}$ and $t = 14\text{ s}$ , excluding the instant $t = 8\text{ s}$ . In a production line programmable logic controller (PLC), engineers program an operational safety cycle within a continuous sub-interval (e.g., $3 \le t \le 7.5\text{ s}$ or $8.5 \le t \le 13.5\text{ s}$ ) to prevent singularity lockup and ensure continuous, smooth kinematic execution.

💡 Show Worked Derivation

In advanced manufacturing and robotics, automated multi-axis robotic manipulators trace complex 3D paths for precision spot welding and component assembly. Mechanical and robotics engineers must verify that the mathematical model of the robot's toolpath is valid across the operating time envelope, ensuring that physical stroke limits, pneumatic pressure lines, and safety enclosures are never breached.

An automated robotic welding tip moves through Cartesian coordinates $(x, y, z)$ in meters as a function of operating cycle time $t$ (in seconds) according to the vector-valued position function:

\vec{r}(t) = \left\langle \sqrt{2t - 6}, \frac{15}{t - 8}, \frac{4}{\sqrt{14 - t}} \right\rangle

Determine the allowable operating domain for the $x$ -component function $x(t) = \sqrt{2t - 6}$ based on the physical requirement that joint extension values must remain real numbers.
Determine the restriction on parameter $t$ for the $y$ -component function $y(t) = \frac{15}{t - 8}$ where the actuator encounters a singularity.
Determine the allowable operating domain for the $z$ -component function $z(t) = \frac{4}{\sqrt{14 - t}}$ .
Calculate the overall valid operating cycle time domain for the entire robotic toolpath $\vec{r}(t)$ in interval notation.

Step-by-Step Solution:

Analyze $x$ -component ( $x(t) = \sqrt{2t - 6}$ ): The term inside the square root must be non-negative for real coordinates:
$2t - 6 \ge 0 \implies 2t \ge 6 \implies t \ge 3$ Domain of $x(t)$ : $[3, \infty)$
Analyze $y$ -component ( $y(t) = \frac{15}{t - 8}$ ): The denominator cannot equal zero, representing an asymptotic singularity in actuator control:
$t - 8 \ne 0 \implies t \ne 8$ Domain of $y(t)$ : $(-\infty, 8) \cup (8, \infty)$
Analyze $z$ -component ( $z(t) = \frac{4}{\sqrt{14 - t}}$ ): The radicand is located in the denominator, so it must be strictly positive:
$14 - t > 0 \implies 14 > t \implies t < 14$ Domain of $z(t)$ : $(-\infty, 14)$
Calculate Global Domain via Set Intersection: Intersect the three individual component domains:
$\text{Domain}(\vec{r}) = [3, \infty) \cap \{t \ne 8\} \cap (-\infty, 14)$ Combining these restrictions: $[3, 8) \cup (8, 14)$

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Vector-Valued Functions, Space Curves & Domain Analysis

General Overview
Skill Block 1 covers evaluating 3D vector-valued functions, determining the mathematical domain through component-wise interval intersections, identifying empty-set domain classifications, and parameterizing spatial curves.

Conceptual Framework

Vector-Valued Function Form: $\vec{r}(t) = \langle f(t), g(t), h(t) \rangle = f(t)\hat{i} + g(t)\hat{j} + h(t)\hat{k}$ Maps a scalar parameter $t \in \mathbb{R}$ to a 3D position vector whose terminal point sweeps out a space curve (see Vector Valued Functions).
Domain Intersection Principle: A vector-valued function is defined only where all three component functions exist simultaneously: $\text{Domain}(\vec{r}) = \text{Domain}(f) \cap \text{Domain}(g) \cap \text{Domain}(h)$
Algebraic Restriction Rules:
- Even roots ( $\sqrt{u(t)}$ ): Radicand must satisfy $u(t) \ge 0$ .
- Fractional denominators ( $\frac{1}{d(t)}$ ): Denominator must satisfy $d(t) \ne 0$ .
- Roots in denominators ( $\frac{1}{\sqrt{u(t)}}$ ): Radicand must satisfy $u(t) > 0$ .
- Logarithms ( $\ln(u(t))$ ): Argument must satisfy $u(t) > 0$ .
Empty Set Domain ( $\emptyset$ ): If two or more component restrictions produce contradictory inequalities with no overlapping real numbers $t$ , the domain is the empty set $\emptyset$ , meaning the curve does not exist in real physical space.

Key Equations & Structural Rules

Vector-Valued Function: $\vec{r}(t) = \langle x(t), y(t), z(t) \rangle$ (Defined in Vector Valued Functions.)
Component Domain Intersection: $\text{Domain}(\vec{r}) = \{t \in \mathbb{R} \mid t \in \text{Domain}(x) \text{ and } t \in \text{Domain}(y) \text{ and } t \in \text{Domain}(z)\}$

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

💡 Reveal 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 $t < 14$ ?

Practice Space Solve this problem on paper or tablet.

Review the Domain Intersection Principle in Vector Valued Functions and inspect the step-by-step domain analysis in Worked Example A and the Robotic Arm problem.

Answer: * (a) The domain is the set intersection of the three individual component domains: $\text{Domain}(\vec{r}) = \text{Domain}(f) \cap \text{Domain}(g) \cap \text{Domain}(h)$ .

(b) Because the interval $(5, \infty)$ is a strict subset of $[-2, \infty)$ and does not contain the excluded point $-3$ , satisfying $t > 5$ automatically satisfies all three conditions simultaneously.
(c) The physical stroke envelope of the actuator is governed by $z(t) = \frac{4}{\sqrt{14 - t}}$ , requiring radicand $14 - t > 0 \implies t < 14\text{ s}$ to prevent square roots of negative numbers.

Step-by-Step Execution

(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.

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

Practice Space Solve this problem on paper or tablet.

State the Square Root Domain Condition ( $u(t) \ge 0$ ) for components 1 and 2, and the Rational Denominator Condition ( $d(t) \ne 0$ ) for component 3. Intersect all three intervals.

Answer:

\left[\frac{4}{3}, \infty\right)

Step-by-Step Execution

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 $1$ . $\text{Domain}(\vec{r}) = \left[\frac{4}{3}, \infty\right)$

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

Practice Space Solve this problem on paper or tablet.

State the domain conditions for each component independently. Check whether any real number $t$ can simultaneously satisfy $t \le 4$ and $t > 6$ .

Answer:

\emptyset \quad (\text{Empty Set; the function is nowhere defined})

Step-by-Step Execution

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: $\text{Domain}(\vec{r}) = \emptyset$

Segment 2 of 8

📅 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.

Segment 3 of 8

Skill Block 2

Real-Life Engineering Applications

🚀

Aerospace Engineering

Autonomous Drone Flight Kinematics (Aerospace & Computer Engineering)

Autonomous quadcopter drones navigate urban delivery corridors by evaluating onboard sensor feedback and tracking pre-computed 3D parametric spline trajectories. Aerospace and computer engineering teams decouple the drone's instantaneous flight speed from its 3D directional heading vector using the principal unit tangent vector $\hat{T}(t)$ , which feeds directly into flight control algorithms for gyroscopic attitude correction.

An autonomous quadcopter executes an ascending spiral corridor inspection with 3D position vector (in meters) given as a function of time $t$ (in seconds):

\vec{r}(t) = \langle 4\cos(t), 4\sin(t), 3t \rangle

Calculate the instantaneous velocity vector $\vec{v}(t) = \vec{r}'(t)$ and the acceleration vector $\vec{a}(t) = \vec{r}''(t)$ of the quadcopter.
Determine the scalar flight speed $\|\vec{v}(t)\|$ of the drone and verify whether it travels at constant speed.
Compute the principal unit tangent vector $\hat{T}(t)$ defining the drone's directional flight heading.
Evaluate the velocity $\vec{v}(\pi)$ , acceleration $\vec{a}(\pi)$ , and unit tangent heading $\hat{T}(\pi)$ at time $t = \pi\text{ seconds}$ .

🎯 Why This Matters:

& Real-Life Application: Even though the drone maintains a constant speed of $5\text{ m/s}$ , its acceleration vector is non-zero ( $\|\vec{a}(t)\| = 4\text{ m/s}^2$ , pointing centripetally toward the vertical central $z$ -axis of the cylinder). The unit tangent vector $\hat{T}(t)$ supplies the drone's flight computer with the exact unit vector along which thrust and camera gimbals must align, while $\vec{a}(t)$ governs rotor torque control for centripetal turning.

💡 Show Worked Derivation

Autonomous quadcopter drones navigate urban delivery corridors by evaluating onboard sensor feedback and tracking pre-computed 3D parametric spline trajectories. Aerospace and computer engineering teams decouple the drone's instantaneous flight speed from its 3D directional heading vector using the principal unit tangent vector $\hat{T}(t)$ , which feeds directly into flight control algorithms for gyroscopic attitude correction.

An autonomous quadcopter executes an ascending spiral corridor inspection with 3D position vector (in meters) given as a function of time $t$ (in seconds):

\vec{r}(t) = \langle 4\cos(t), 4\sin(t), 3t \rangle

Calculate the instantaneous velocity vector $\vec{v}(t) = \vec{r}'(t)$ and the acceleration vector $\vec{a}(t) = \vec{r}''(t)$ of the quadcopter.
Determine the scalar flight speed $\|\vec{v}(t)\|$ of the drone and verify whether it travels at constant speed.
Compute the principal unit tangent vector $\hat{T}(t)$ defining the drone's directional flight heading.
Evaluate the velocity $\vec{v}(\pi)$ , acceleration $\vec{a}(\pi)$ , and unit tangent heading $\hat{T}(\pi)$ at time $t = \pi\text{ seconds}$ .

Step-by-Step Solution:

Calculate Velocity $\vec{v}(t)$ and Acceleration $\vec{a}(t)$ : Differentiating component-wise:
$\vec{v}(t) = \vec{r}'(t) = \left\langle \frac{d}{dt}[4\cos(t)], \frac{d}{dt}[4\sin(t)], \frac{d}{dt}[3t] \right\rangle = \langle -4\sin(t), 4\cos(t), 3 \rangle \text{ m/s}$ Differentiating velocity to find acceleration: $\vec{a}(t) = \vec{v}'(t) = \langle -4\cos(t), -4\sin(t), 0 \rangle \text{ m/s}^2$
Compute Scalar Flight Speed $\|\vec{v}(t)\|:$
$\|\vec{v}(t)\| = \sqrt{(-4\sin(t))^2 + (4\cos(t))^2 + 3^2} = \sqrt{16\sin^2(t) + 16\cos^2(t) + 9}$ Using $\sin^2(t) + \cos^2(t) = 1$ : $\|\vec{v}(t)\| = \sqrt{16(1) + 9} = \sqrt{25} = 5\text{ m/s}$ The scalar flight speed is strictly constant at $5\text{ m/s}$ everywhere along the trajectory.
Compute Principal Unit Tangent Vector $\hat{T}(t)$ : Normalize velocity by dividing by speed:
$\hat{T}(t) = \frac{\vec{v}(t)}{\|\vec{v}(t)\|} = \frac{\langle -4\sin(t), 4\cos(t), 3 \rangle}{5} = \left\langle -\frac{4}{5}\sin(t), \frac{4}{5}\cos(t), \frac{3}{5} \right\rangle$
Evaluate at $t = \pi\text{ s}$ :
- Velocity: $\vec{v}(\pi) = \langle -4\sin(\pi), 4\cos(\pi), 3 \rangle = \langle 0, -4, 3 \rangle \text{ m/s}$
- Acceleration: $\vec{a}(\pi) = \langle -4\cos(\pi), -4\sin(\pi), 0 \rangle = \langle 4, 0, 0 \rangle \text{ m/s}^2$
- Unit Tangent Heading: $\hat{T}(\pi) = \left\langle 0, -\frac{4}{5}, \frac{3}{5} \right\rangle$

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Differentiation, Kinematics, Unit Tangent Vectors & Chain Rule

General Overview
Skill Block 2 covers differentiating vector-valued functions component-wise, calculating velocity and acceleration vectors, determining scalar speed, computing the principal unit tangent vector $\hat{T}(t)$ , and applying the Vector Chain Rule to composite trajectories.

Conceptual Framework

Vector Derivative Definition: Differentiating a vector-valued function is executed component-wise (see Derivatives of Vector-Valued Functions): $\vec{r}'(t) = \frac{d\vec{r}}{dt} = \langle f'(t), g'(t), h'(t) \rangle = f'(t)\hat{i} + g'(t)\hat{j} + h'(t)\hat{k}$
Physical Kinematics:
- Position Vector: $\vec{r}(t) = \langle x(t), y(t), z(t) \rangle$
- Velocity Vector (Tangent Vector): $\vec{v}(t) = \vec{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle$
- Scalar Speed: $\|\vec{v}(t)\| = \|\vec{r}'(t)\| = \sqrt{[x'(t)]^2 + [y'(t)]^2 + [z'(t)]^2}$
- Acceleration Vector: $\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t) = \langle x''(t), y''(t), z''(t) \rangle$
Principal Unit Tangent Vector $\hat{T}(t)$ : The unit tangent vector isolates directional heading by dividing the velocity vector by its scalar speed (see Unit Tangent Vector): $\hat{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}$
Vector Chain Rule: For a vector function parameterized by a scalar function $u = f(t)$ : $\frac{d}{dt}[\vec{r}(f(t))] = \vec{r}'(f(t)) \cdot f'(t)$

Key Equations & Structural Rules

Vector Derivative Formula: $\vec{r}'(t) = \langle f'(t), g'(t), h'(t) \rangle$ (Defined in Derivatives of Vector-Valued Functions.)
Principal Unit Tangent Vector Formula: $\hat{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}$ (Defined in Unit Tangent Vector.)
Vector Chain Rule: $\frac{d}{dt}[\vec{r}(f(t))] = \vec{r}'(f(t)) \cdot f'(t)$

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}

💡 Reveal 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?

Practice Space Solve this problem on paper or tablet.

Review the definitions in Derivatives of Vector-Valued Functions and Unit Tangent Vector, and review the circular/helical motion analysis in the Autonomous Drone problem.

Answer: * (a) $\vec{v}(t) = \vec{r}'(t)$ is the vector rate of change of position; $\|\vec{v}(t)\|$ is the scalar magnitude of velocity (speed); $\hat{T}(t) = \frac{\vec{v}(t)}{\|\vec{v}(t)\|}$ is the dimensionless unit vector pointing along the direction of motion.

(b) By definition, dividing any non-zero vector by its own scalar magnitude yields a vector of length $\frac{\|\vec{v}\|}{\|\vec{v}\|} = 1$ .
(c) Acceleration represents the rate of change of the velocity vector. Even if speed (magnitude) is constant, changing the direction of travel creates centripetal acceleration ( $\vec{a}(t) \ne \vec{0}$ ).

Step-by-Step Execution

(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.

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:

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

Practice Space Solve this problem on paper or tablet.

Rewrite $\vec{r}(t) = -6t^3\hat{i} - 3t^7\hat{j} + 3t^{-4}\hat{k}$ and $f(t) = t^{1/4}$ . For part 2, substitute $f(t)$ into $\vec{r}(t)$ before applying the power rule, or use the Vector Chain Rule $\frac{d}{dt}[\vec{r}(f(t))] = \vec{r}'(f(t)) \cdot f'(t)$ .

Answer: 1. $\frac{d}{dt}[\vec{r}(t)] = -18t^2\hat{i} - 21t^6\hat{j} - \frac{12}{t^5}\hat{k}$ 2. $\frac{d}{dt}[\vec{r}(f(t))] = -\frac{9}{2t^{1/4}}\hat{i} - \frac{21t^{3/4}}{4}\hat{j} - \frac{3}{t^2}\hat{k}$

Step-by-Step Execution

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: $\frac{d}{dt}[3t^{-1}] = 3(-1)t^{-2} = -\frac{3}{t^2}$ $\frac{d}{dt}[\vec{r}(f(t))] = -\frac{9}{2t^{1/4}}\hat{i} - \frac{21t^{3/4}}{4}\hat{j} - \frac{3}{t^2}\hat{k}$

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 $t = 20\text{ seconds}$ .

Practice Space Solve this problem on paper or tablet.

State the Product Rule ( $\frac{d}{dt}[uv] = u'v + uv'$ ) for the $x$ and $y$ components and the Quotient Rule ( $\frac{d}{dt}[\frac{u}{v}] = \frac{u'v - uv'}{v^2}$ ) for the $z$ component. For $\hat{T}(20)$ , evaluate $\vec{v}(20)$ and normalize by its magnitude.

Answer: 1. $\vec{a}(t) = \left\langle \sin(2t) + 2t\cos(2t), 2\ln(t-3) + \frac{2t}{t-3}, \frac{t^2 - 4t}{(t-2)^2} \right\rangle \text{ m/s}^2$ 2. $\vec{a}(20) \approx \langle -25.93, 8.02, 0.99 \rangle \text{ m/s}^2$ and $\hat{T}(20) \approx \langle 0.128, 0.973, 0.191 \rangle$

Step-by-Step Execution

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 $\hat{T}(20)$ : $\hat{T}(20) \approx \left\langle \frac{14.90}{116.44}, \frac{113.33}{116.44}, \frac{22.22}{116.44} \right\rangle \approx \langle 0.128, 0.973, 0.191 \rangle$

Segment 4 of 8

📅 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.

Segment 5 of 8

Skill Block 3

Real-Life Engineering Applications

⚡

Electrical Engineering

Electric Vehicle Acceleration & Trajectory Recovery (Electrical & Automotive Engineering)

In electric and autonomous vehicle engineering, onboard inertial measurement units (IMUs) log high-frequency 3-axis accelerometer data. When GPS satellite signals are occluded (such as inside tunnels or subterranean parking garages), vehicle stability control systems reconstruct the vehicle's instantaneous velocity and position trajectory by performing successive vector integrations over time (dead reckoning).

An autonomous electric delivery vehicle enters an underground facility at $t = 0\text{ seconds}$ . The onboard IMU records a 3-axis acceleration vector (in $\text{m/s}^2$ ) modeled by:

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

At the moment of entering the facility (

t = 0

), the vehicle's initial velocity is

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

and its initial position vector relative to the entrance gate is

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

.

Use indefinite vector integration and the initial velocity condition to derive the vehicle's 3D velocity vector function $\vec{v}(t)$ .
Use indefinite vector integration and the initial position condition to derive the vehicle's 3D position trajectory function $\vec{r}(t)$ .
Calculate the net displacement vector $\Delta \vec{r}$ of the vehicle during its first 3 seconds inside the facility ( $0 \le t \le 3$ ) using a definite integral.

🎯 Why This Matters:

& Real-Life Application: Over the 3-second interval, the vehicle undergoes a 3D spatial translation of $\langle 33, 78, 33 \rangle\text{ meters}$ , achieving a direct linear displacement magnitude of $\|\Delta \vec{r}\| = \sqrt{33^2 + 78^2 + 33^2} = \sqrt{1089 + 6084 + 1089} = \sqrt{8262} \approx 90.9\text{ meters}$ . Autonomous vehicle systems use these continuous vector integral calculations to update local obstacle maps and prevent collisions when GPS signals are unavailable.

💡 Show Worked Derivation

In electric and autonomous vehicle engineering, onboard inertial measurement units (IMUs) log high-frequency 3-axis accelerometer data. When GPS satellite signals are occluded (such as inside tunnels or subterranean parking garages), vehicle stability control systems reconstruct the vehicle's instantaneous velocity and position trajectory by performing successive vector integrations over time (dead reckoning).

An autonomous electric delivery vehicle enters an underground facility at $t = 0\text{ seconds}$ . The onboard IMU records a 3-axis acceleration vector (in $\text{m/s}^2$ ) modeled by:

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

At the moment of entering the facility (

t = 0

), the vehicle's initial velocity is

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

and its initial position vector relative to the entrance gate is

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

.

Use indefinite vector integration and the initial velocity condition to derive the vehicle's 3D velocity vector function $\vec{v}(t)$ .
Use indefinite vector integration and the initial position condition to derive the vehicle's 3D position trajectory function $\vec{r}(t)$ .
Calculate the net displacement vector $\Delta \vec{r}$ of the vehicle during its first 3 seconds inside the facility ( $0 \le t \le 3$ ) using a definite integral.

Step-by-Step Solution:

Derive Velocity Vector Function $\vec{v}(t)$ : Integrate acceleration component-by-component:
$\vec{v}(t) = \int \vec{a}(t) \, dt = \left\langle \int 6t \, dt, \int 12t^2 \, dt, \int 4 \, dt \right\rangle = \langle 3t^2, 4t^3, 4t \rangle + \vec{C}_1$ Apply initial velocity $\vec{v}(0) = \langle 2, -1, 5 \rangle$ : $\vec{v}(0) = \langle 0, 0, 0 \rangle + \vec{C}_1 = \langle 2, -1, 5 \rangle \implies \vec{C}_1 = \langle 2, -1, 5 \rangle$ Therefore: $\vec{v}(t) = \langle 3t^2 + 2, 4t^3 - 1, 4t + 5 \rangle \text{ m/s}$
Derive Position Vector Function $\vec{r}(t)$ : Integrate velocity component-by-component:
$\vec{r}(t) = \int \vec{v}(t) \, dt = \left\langle \int (3t^2 + 2) \, dt, \int (4t^3 - 1) \, dt, \int (4t + 5) \, dt \right\rangle = \langle t^3 + 2t, t^4 - t, 2t^2 + 5t \rangle + \vec{C}_2$ Apply initial position $\vec{r}(0) = \langle 0, 10, -3 \rangle$ : $\vec{r}(0) = \langle 0, 0, 0 \rangle + \vec{C}_2 = \langle 0, 10, -3 \rangle \implies \vec{C}_2 = \langle 0, 10, -3 \rangle$ Therefore: $\vec{r}(t) = \langle t^3 + 2t, t^4 - t + 10, 2t^2 + 5t - 3 \rangle \text{ meters}$
Calculate Net Displacement $\Delta \vec{r}$ over $[0, 3]$ : Evaluate the definite integral of velocity:
$\Delta \vec{r} = \int_{0}^{3} \vec{v}(t) \, dt = \vec{r}(3) - \vec{r}(0)$
Evaluating $\vec{r}(3)$ :
- $x(3) = (3)^3 + 2(3) = 27 + 6 = 33$
- $y(3) = (3)^4 - (3) + 10 = 81 - 3 + 10 = 88$
- $z(3) = 2(3)^2 + 5(3) - 3 = 18 + 15 - 3 = 30$ $\vec{r}(3) = \langle 33, 88, 30 \rangle \text{ m}$ Subtracting initial position $\vec{r}(0) = \langle 0, 10, -3 \rangle$ : $\Delta \vec{r} = \langle 33 - 0, 88 - 10, 30 - (-3) \rangle = \langle 33, 78, 33 \rangle \text{ meters}$

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Integration of Vector-Valued Functions & Kinematic Accumulation

General Overview
Skill Block 3 covers indefinite and definite integration of vector-valued functions, combining integration constants into a single constant vector $\vec{C}$ , solving multi-stage Initial Value Problems (IVPs), and calculating net displacement vectors over time intervals.

Conceptual Framework

Indefinite Vector Integration: Evaluated component-by-component, producing three scalar constants that form a single constant vector (see Integrals of Vector-Valued Functions): $\int \vec{r}(t) \, dt = \left\langle \int f(t)\,dt, \int g(t)\,dt, \int h(t)\,dt \right\rangle = \langle F(t), G(t), H(t) \rangle + \vec{C}$ where $\vec{C} = \langle C_1, C_2, C_3 \rangle$ .
Definite Vector Integration (Fundamental Theorem of Calculus): Evaluates net displacement over an interval $[a, b]$ : $\Delta \vec{r} = \int_{a}^{b} \vec{v}(t) \, dt = \vec{r}(b) - \vec{r}(a)$
Initial Value Problems (IVP): Given acceleration $\vec{a}(t)$ and initial vectors $\vec{v}(0)$ and $\vec{r}(0)$ : $\vec{v}(t) = \int \vec{a}(t)\,dt + \vec{C}_1, \quad \vec{r}(t) = \int \vec{v}(t)\,dt + \vec{C}_2$

Key Equations & Structural Rules

Indefinite Vector Integral: $\int \vec{r}(t) \, dt = \vec{R}(t) + \vec{C}, \quad \vec{C} = \langle C_1, C_2, C_3 \rangle$ (Defined in Integrals of Vector-Valued Functions.)
Definite Vector Integral (Net Displacement): $\Delta \vec{r} = \int_{a}^{b} \vec{v}(t) \, dt = \left\langle \int_a^b v_x(t)\,dt, \int_a^b v_y(t)\,dt, \int_a^b v_z(t)\,dt \right\rangle$

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

💡 Reveal 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?

Practice Space Solve this problem on paper or tablet.

Review the constant vector definition and initial value problem framework in Integrals of Vector-Valued Functions, and examine the step-by-step antiderivative setup in Worked Example A and the Electric Vehicle application.

Answer: * (a) Because integration is performed on three independent component functions ( $x, y, z$ ), each generating its own independent scalar constant ( $C_1, C_2, C_3$ ). Adding a scalar to a vector is algebraically undefined.

(b) Each component is integrated separately with standard scalar integration rules ( $u$ -substitution for sinusoidal terms, standard logarithmic rule for reciprocal terms), and their respective constants are combined into $\vec{C} = \langle C_1, C_2, C_3 \rangle$ .
(c) Evaluating the general antiderivative vectors at $t = 0$ allows engineers to equate the expressions directly to the measured IMU vectors, solving for the exact offset vectors $\vec{C}_1$ and $\vec{C}_2$ .

Step-by-Step Execution

(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.

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

At

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 $\vec{r}(t)$ .

Practice Space Solve this problem on paper or tablet.

State the Indefinite Vector Integral Formula for velocity $\vec{v}(t) = \int \vec{a}(t)\,dt + \vec{C}_1$ and apply $\vec{v}(0)$ to determine $\vec{C}_1$ . Then integrate velocity $\vec{r}(t) = \int \vec{v}(t)\,dt + \vec{C}_2$ and apply $\vec{r}(0)$ to determine $\vec{C}_2$ .

Answer: 1. $\vec{v}(t) = \langle 3t^2 + 2, 4t^3 - 1, 4t + 5 \rangle \text{ m/s}$ 2. $\vec{r}(t) = \langle t^3 + 2t, t^4 - t + 10, 2t^2 + 5t - 3 \rangle \text{ meters}$

Step-by-Step Execution

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: $\vec{r}(t) = \langle t^3 + 2t, t^4 - t + 10, 2t^2 + 5t - 3 \rangle \text{ meters}$

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 $1 \le t \le 5\text{ seconds}$ .

Practice Space Solve this problem on paper or tablet.

Rewrite $\sqrt[3]{t} = t^{1/3}$ . State the Power Rule for Integration ( $\int t^n dt = \frac{t^{n+1}}{n+1}$ ), the Logarithmic Rule ( $\int \frac{1}{t+a}dt = \ln|t+a|$ ), and the Exponential Rule ( $\int e^{at}dt = \frac{1}{a}e^{at}$ ). For part 2, evaluate $\Delta \vec{r} = \vec{r}(5) - \vec{r}(1)$ .

Answer: 1. $\vec{r}(t) = \left( \frac{3}{4}t^{4/3} \right)\hat{i} + (\ln|t+2|)\hat{j} + \left( \frac{1}{8}e^{8t} \right)\hat{k} + \vec{C}$ 2. $\Delta \vec{r} = \left[ \frac{3}{4}\left(5^{4/3} - 1\right) \right]\hat{i} + \left[ \ln\left(\frac{7}{3}\right) \right]\hat{j} + \left[ \frac{1}{8}\left(e^{40} - e^8\right) \right]\hat{k} \text{ ft}$

Step-by-Step Execution

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: $\Delta \vec{r} = \left[ \frac{3}{4}\left(5^{4/3} - 1\right) \right]\hat{i} + \left[ \ln\left(\frac{7}{3}\right) \right]\hat{j} + \left[ \frac{1}{8}\left(e^{40} - e^8\right) \right]\hat{k} \text{ ft}$

Segment 6 of 8

📅 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.

Segment 7 of 8

📅 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.

Segment 8 of 8

Vector-Valued Functions and Derivatives

📅 Micro-Lecture

Skill Block 1

Real-Life Engineering Applications

Robotic Arm Workspace Domain (Mechanical & Robotics Engineering)

Step-by-Step Solution:

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

Practice Problem 1: Concept & Reading Check

Step-by-Step Execution

Practice Problem 2

Step-by-Step Execution

Practice Problem 3

Step-by-Step Execution

📅 Regroup 1

Skill Block 2

Real-Life Engineering Applications

Autonomous Drone Flight Kinematics (Aerospace & Computer Engineering)

Step-by-Step Solution:

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

Practice Problem 4: Concept & Reading Check

Step-by-Step Execution

Practice Problem 5

Step-by-Step Execution

Practice Problem 6

Step-by-Step Execution

📅 Regroup 2

Skill Block 3

Real-Life Engineering Applications

Electric Vehicle Acceleration & Trajectory Recovery (Electrical & Automotive Engineering)

Step-by-Step Solution:

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

Practice Problem 7: Concept & Reading Check

Step-by-Step Execution

Practice Problem 8

Step-by-Step Execution

Practice Problem 9

Step-by-Step Execution

📅 Regroup 3

📅 Synthesis Wrap-up

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