Session 05 - Vector-Valued Functions and Derivatives

📅 Micro-Lecture

Engineering Context: Motion in Mechanical Systems

In this session, we investigate the mathematical principles of Vector-Valued Functions and Derivatives and explore how engineers apply these concepts to analyze real-world physical and structural systems.

Segment 1 of 8

Skill Block 1

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

Explanations and derivations of the core concepts.

Worked Example

Determine the domain of $\vec{r}(t) = \left\langle \sqrt{t+2}, \frac{-9}{t+3}, \frac{-7}{\sqrt{t-5}} \right\rangle$ and plot the function using Desmos.

💡 Reveal Worked Solution

Here is the step-by-step solution to the problem presented in the image.

Problem Statement

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

To find the domain of a vector-valued function, we must find the intersection of the domains of its individual component functions ( $x(t)$ , $y(t)$ , and $z(t)$ ).

1. Analyze the $x$ -component: $x(t) = \sqrt{t+2}$

The term inside a square root must be greater than or equal to zero.
Set up the inequality:

t + 2 \ge 0 \implies t \ge -2

Domain of $x(t)$ : $[-2, \infty)$

2. Analyze the $y$ -component: $y(t) = \frac{-9}{t+3}$

The denominator cannot equal zero.
Set up the equation:

t + 3 \neq 0 \implies t \neq -3

Domain of $y(t)$ : $(-\infty, -3) \cup (-3, \infty)$

3. Analyze the $z$ -component: $z(t) = \frac{-7}{\sqrt{t-5}}$

The term inside the square root must be non-negative ( $t - 5 \ge 0$ ).
Additionally, because the square root is in the denominator, it cannot be equal to zero ( $t - 5 \neq 0$ ).
Combining these restrictions yields:

t - 5 > 0 \implies t > 5

Domain of $z(t)$ : $(5, \infty)$

Finding the Intersection

Now, we intersect the three individual domains to find where all components are simultaneously defined:

$t \ge -2$
$t \neq -3$
$t > 5$

Since any value of $t$ that is strictly greater than $5$ automatically satisfies $t \ge -2$ and $t \neq -3$ , the restriction $t > 5$ dictates the overall domain.

Final Answer

Inequality Notation: $t > 5$
Interval Notation: $(5, \infty)$

Desmos Plotting Tip

To plot this 3D vector-valued function in Desmos (specifically using the Desmos 3D Grapher), you can input the curve using the parameter $t$ exactly as written:

Plaintext

r(t) = (sqrt(t+2), -9/(t+3), -7/sqrt(t-5))

Be sure to set the parameter bounds for $t$ starting from just above $5$ (e.g., 5 < t < 15) to see the curve render correctly!

Active Practice Problem

Determine the domain and plot using Desmos:

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

Workspace

Solution

Problem Statement

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

Step-by-Step Solution

To find the domain of this vector-valued function, we find the domain of each component independently and then determine where they intersect.

1. Analyze the $x$ -component: $x(t) = \sqrt{3t-4}$

The expression inside the square root must be greater than or equal to zero. $3t - 4 \ge 0 \implies 3t \ge 4 \implies t \ge \frac{4}{3}$
Domain of $x(t)$ : $\left[\frac{4}{3}, \infty\right)$

2. Analyze the $y$ -component: $y(t) = \sqrt{t+7}$

The expression inside the square root must be greater than or equal to zero. $t + 7 \ge 0 \implies t \ge -7$
Domain of $y(t)$ : $[-7, \infty)$

3. Analyze the $z$ -component: $z(t) = \frac{6}{t-1}$

The denominator cannot equal zero. $t - 1 \neq 0 \implies t \neq 1$
Domain of $z(t)$ : $(-\infty, 1) \cup (1, \infty)$

Finding the Intersection

Now, find the interval where all three conditions are satisfied at the same time:

$t \ge \frac{4}{3}$ (which is approximately $1.33$ )
$t \ge -7$
$t \neq 1$

Looking at the first condition ( $t \ge \frac{4}{3}$ ), any value in this range is automatically greater than $-7$ .
Furthermore, since $\frac{4}{3} > 1$ , the restriction $t \neq 1$ is already excluded automatically because $1$ is strictly less than $\frac{4}{3}$ .

Therefore, the condition $t \ge \frac{4}{3}$ is the limiting restriction that satisfies all components.

Final Answer

Inequality Notation: $t \ge \frac{4}{3}$
Interval Notation: $\left[\frac{4}{3}, \infty\right)$

Extra Practice & Extensions

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

Refer to the 'Core Theory' tab or review the Worked Example demonstration for this skill.

Segment 2 of 8

📅 Regroup 1

Review common misconceptions and clarify key notations.

Segment 3 of 8

Skill Block 2

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

Explanations and derivations of the core concepts.

Worked Example

Find the derivative of the vector valued function and the principal unit tangent vector:

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

💡 Reveal Worked Solution

Problem Statement

Find the derivative of the vector-valued function and the principal unit tangent vector:

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

Part 1: Find the Derivative $\vec{r}^{\prime}(t)$

To find the derivative of a vector-valued function, differentiate each component with respect to $t$ individually using the power rule:

$\hat{i}$ -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}^{\prime}(t) = -\hat{i} - 4t^{3}\hat{j} - 12t\hat{k}

Part 2: Find the Principal Unit Tangent Vector $\hat{T}(t)$

The principal unit tangent vector is found by dividing the tangent vector $\vec{r}^{\prime}(t)$ by its magnitude $|\vec{r}^{\prime}(t)|$ :

\hat{T}(t) = \frac{\vec{r}^{\prime}(t)}{|\vec{r}^{\prime}(t)|}

1. Calculate the Magnitude $|\vec{r}^{\prime}(t)|$ :

|\vec{r}^{\prime}(t)| = \sqrt{(-1)^2 + (-4t^3)^2 + (-12t)^2}

|\vec{r}^{\prime}(t)| = \sqrt{1 + 16t^6 + 144t^2}

Rearranging the terms in standard descending order:

|\vec{r}^{\prime}(t)| = \sqrt{16t^6 + 144t^2 + 1}

2. Divide $\vec{r}^{\prime}(t)$ by its magnitude:

\hat{T}(t) = \frac{-\hat{i} - 4t^{3}\hat{j} - 12t\hat{k}}{\sqrt{16t^6 + 144t^2 + 1}}

Final Answer

Derivative:
$\vec{r}^{\prime}(t) = -\hat{i} - 4t^{3}\hat{j} - 12t\hat{k}$
Principal Unit Tangent Vector:
$\hat{T}(t) = \frac{-1}{\sqrt{16t^6 + 144t^2 + 1}}\hat{i} - \frac{4t^3}{\sqrt{16t^6 + 144t^2 + 1}}\hat{j} - \frac{12t}{\sqrt{16t^6 + 144t^2 + 1}}\hat{k}$

Active Practice Problem

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:

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

Workspace

Solution

Problem Statement

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:

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

Step-by-Step Solution

First, let's rewrite the given vector-valued function $\vec{r}(t)$ and scalar function $f(t)$ in simplified power form to make differentiation easier:

\vec{r}(t) = -6t^{3}\hat{i} - 3t^{7}\hat{j} + 3t^{-4}\hat{k}

f(t) = t^{1/4}

Part 1: Calculate $\frac{d}{dt}[\vec{r}(t)]$

To find the derivative, use the power rule on each component individually:

$\hat{i}$ -component: $\frac{d}{dt}[-6t^3] = -18t^2$
$\hat{j}$ -component: $\frac{d}{dt}[-3t^7] = -21t^6$
$\hat{k}$ -component: $\frac{d}{dt}[3t^{-4}] = -12t^{-5} = -\frac{12}{t^5}$

\frac{d}{dt}[\vec{r}(t)] = -18t^{2}\hat{i} - 21t^{6}\hat{j} - \frac{12}{t^{5}}\hat{k}

Part 2: Calculate $\frac{d}{dt}[\vec{r}(f(t))]$

There are two common methods to solve this: by substitution first, or by using the Vector Chain Rule. Let's use Method 1: Direct Substitution as it is often simpler.

Step 1: 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}

Step 2: Differentiate with respect to $t$ Now, apply the power rule to each component of this composite vector function:

$\hat{i}$ -component: $\frac{d}{dt}[-6t^{3/4}] = -6 \cdot \frac{3}{4}t^{-1/4} = -\frac{9}{2}t^{-1/4}$
$\hat{j}$ -component: $\frac{d}{dt}[-3t^{7/4}] = -3 \cdot \frac{7}{4}t^{3/4} = -\frac{21}{4}t^{3/4}$
$\hat{k}$ -component: $\frac{d}{dt}[3t^{-1}] = 3 \cdot (-1)t^{-2} = -3t^{-2}$

\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}

Final Answers

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

Extra Practice & Extensions

\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 of the car, $\vec{a}(t)$ . Hint: acceleration is the derivative of velocity with respect to time.
Find the acceleration and principal unit tangent vector at $t = 20$ s.

Refer to the 'Core Theory' tab or review the Worked Example demonstration for this skill.

Segment 4 of 8

📅 Regroup 2

Reflect on the physical modeling applications and mathematical setups.

Segment 5 of 8

Skill Block 3

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

Explanations and derivations of the core concepts.

Worked Example

Evaluate the indefinite integral:

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

💡 Reveal Worked Solution

Problem Statement

Evaluate the indefinite integral:

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

To integrate a vector-valued function, integrate each component function independently with respect to $t$ . Remember to include a constant of integration for each component, which can ultimately be combined into a single constant vector $\vec{C}$ .

1. Integrate the $x$ -component: $f(t) = 5 \cos(7t)$

Using $u$ -substitution where $u = 7t$ and $du = 7\,dt$ :

\int 5 \cos(7t) \, dt = 5 \cdot \frac{1}{7} \sin(7t) + C_1 = \frac{5}{7} \sin(7t) + C_1

2. Integrate the $y$ -component: $g(t) = 3 \cos(4t)$

Similarly, using $u$ -substitution where $u = 4t$ and $du = 4\,dt$ :

\int 3 \cos(4t) \, dt = 3 \cdot \frac{1}{4} \sin(4t) + C_2 = \frac{3}{4} \sin(4t) + C_2

3. Integrate the $z$ -component: $h(t) = \frac{-9}{t}$

Using the standard logarithmic integration rule $\int \frac{1}{t} \, dt = \ln|t|$ :

\int \frac{-9}{t} \, dt = -9 \ln|t| + C_3

Combining the Components

We can write out the final vector in standard component form, merging the individual integration constants ( $C_1, C_2, C_3$ ) into a single constant vector $\vec{C} = \langle C_1, C_2, C_3 \rangle$ :

\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}

Final Answer

\left\langle \frac{5}{7} \sin(7t), \frac{3}{4} \sin(4t), -9 \ln|t| \right\rangle + \vec{C}

Active Practice Problem

Given that an object has a velocity 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 displacement vector of the object.
Use a definite integral to find the displacement vector of the object over the interval $1 \le t \le 5$ .

Practice Space Solve this problem on paper or tablet.

Refer to the 'Core Theory' tab or review the Worked Example demonstration for this skill.

Step-by-Step Execution

Problem Statement

Given that an object has a velocity 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 displacement vector of the object.
Use a definite integral to find the displacement vector of the object over the interval $1 \le t \le 5$ .

To find the displacement vector from a velocity vector function, we integrate each component function independently. For easier integration, rewrite the velocity function using fractional exponents:

\vec{v}(t) = t^{1/3}\hat{i} + \frac{1}{t+2}\hat{j} + e^{8t}\hat{k}

Part 1: Indefinite Integral for Displacement Vector $\vec{r}(t)$

Integrate each component individually with respect to $t$ :

$\hat{i}$ -component: Use the power rule ( $\int t^n dt = \frac{t^{n+1}}{n+1}$ ):
$\int t^{1/3} \, dt = \frac{t^{4/3}}{4/3} = \frac{3}{4}t^{4/3} + C_1$
$\hat{j}$ -component: Use the standard logarithmic rule:
$\int \frac{1}{t+2} \, dt = \ln|t+2| + C_2$
$\hat{k}$ -component: Use the exponential rule ( $\int e^{at} dt = \frac{1}{a}e^{at}$ ):
$\int e^{8t} \, dt = \frac{1}{8}e^{8t} + C_3$

Combine the components and merge the constants $C_1, C_2, C_3$ into a single constant vector $\vec{C}$ :

\vec{r}(t) = \left( \frac{3}{4}t^{4/3} \right)\hat{i} + \left( \ln|t+2| \right)\hat{j} + \left( \frac{1}{8}e^{8t} \right)\hat{k} + \vec{C}

Part 2: Definite Integral over the Interval $1 \le t \le 5$

To find the total displacement over the time interval $[1, 5]$ , evaluate the definite integral component-by-component using the antiderivatives found above:

\Delta\vec{r} = \int_{1}^{5} \vec{v}(t) \, dt

1. Evaluate the $\hat{i}$ -component:

\left[ \frac{3}{4}t^{4/3} \right]_{1}^{5} = \frac{3}{4}(5)^{4/3} - \frac{3}{4}(1)^{4/3} = \frac{3}{4}\left(5^{4/3} - 1\right) \approx 5.66

2. Evaluate the $\hat{j}$ -component:

\left[ \ln|t+2| \right]_{1}^{5} = \ln|5+2| - \ln|1+2| = \ln(7) - \ln(3) = \ln\left(\frac{7}{3}\right) \approx 0.847

3. Evaluate the $\hat{k}$ -component:

\left[ \frac{1}{8}e^{8t} \right]_{1}^{5} = \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}

Combine the components into the final 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}

Final Answers

Indefinite Integral Vector:
$\vec{r}(t) = \left( \frac{3}{4}t^{4/3} \right)\hat{i} + \left( \ln|t+2| \right)\hat{j} + \left( \frac{1}{8}e^{8t} \right)\hat{k} + \vec{C}$
Definite Integral (Exact Value):
$\Delta\vec{r} = \left[ \frac{3}{4}\left(5^{4/3} - 1\right) \right]\hat{i} + \left[ \ln\left(\frac{7}{3}\right) \right]\hat{j} + \left[ \frac{1}{8}\left(e^{40} - e^8\right) \right]\hat{k}$

Segment 6 of 8

📅 Regroup 3

Verify calculations and mathematical reasoning.

Segment 7 of 8

📅 Synthesis Wrap-up

Core takeaways from Session 05 and overview of homework homework assignment: Vector-Valued Functions and Space Curves Calculus of Vector-Valued Functions.

Segment 8 of 8

Vector-Valued Functions and Derivatives

Draft Handout - Content Not Finalized

📅 Micro-Lecture

Engineering Context: Motion in Mechanical Systems

Skill Block 1

Worked Example

Problem Statement

1. Analyze the xxx-component: x(t)=t+2x(t) = \sqrt{t+2}x(t)=t+2​

2. Analyze the yyy-component: y(t)=−9t+3y(t) = \frac{-9}{t+3}y(t)=t+3−9​

3. Analyze the zzz-component: z(t)=−7t−5z(t) = \frac{-7}{\sqrt{t-5}}z(t)=t−5​−7​

Finding the Intersection

Final Answer

Desmos Plotting Tip

Active Practice Problem

Workspace

Solution

Problem Statement

Step-by-Step Solution

1. Analyze the xxx-component: x(t)=3t−4x(t) = \sqrt{3t-4}x(t)=3t−4​

2. Analyze the yyy-component: y(t)=t+7y(t) = \sqrt{t+7}y(t)=t+7​

3. Analyze the zzz-component: z(t)=6t−1z(t) = \frac{6}{t-1}z(t)=t−16​

Finding the Intersection

Final Answer

Extra Practice & Extensions

📅 Regroup 1

Skill Block 2

Worked Example

Problem Statement

Part 1: Find the Derivative r⃗′(t)\vec{r}^{\prime}(t)r′(t)

Part 2: Find the Principal Unit Tangent Vector T^(t)\hat{T}(t)T^(t)

Final Answer

Active Practice Problem

Workspace

Solution

Problem Statement

Step-by-Step Solution

Part 1: Calculate ddt[r⃗(t)]\frac{d}{dt}[\vec{r}(t)]dtd​[r(t)]

Part 2: Calculate ddt[r⃗(f(t))]\frac{d}{dt}[\vec{r}(f(t))]dtd​[r(f(t))]

Final Answers

Extra Practice & Extensions

📅 Regroup 2

Skill Block 3

Worked Example

Problem Statement

1. Integrate the xxx-component: f(t)=5cos⁡(7t)f(t) = 5 \cos(7t)f(t)=5cos(7t)

2. Integrate the yyy-component: g(t)=3cos⁡(4t)g(t) = 3 \cos(4t)g(t)=3cos(4t)

3. Integrate the zzz-component: h(t)=−9th(t) = \frac{-9}{t}h(t)=t−9​

Combining the Components

Final Answer

Active Practice Problem

Step-by-Step Execution

Problem Statement

Part 1: Indefinite Integral for Displacement Vector r⃗(t)\vec{r}(t)r(t)

Part 2: Definite Integral over the Interval 1≤t≤51 \le t \le 51≤t≤5

Final Answers

📅 Regroup 3

📅 Synthesis Wrap-up

1. Analyze the $x$ -component: $x(t) = \sqrt{t+2}$

2. Analyze the $y$ -component: $y(t) = \frac{-9}{t+3}$

3. Analyze the $z$ -component: $z(t) = \frac{-7}{\sqrt{t-5}}$

1. Analyze the $x$ -component: $x(t) = \sqrt{3t-4}$

2. Analyze the $y$ -component: $y(t) = \sqrt{t+7}$

3. Analyze the $z$ -component: $z(t) = \frac{6}{t-1}$

Part 1: Find the Derivative $\vec{r}^{\prime}(t)$

Part 2: Find the Principal Unit Tangent Vector $\hat{T}(t)$

Part 1: Calculate $\frac{d}{dt}[\vec{r}(t)]$

Part 2: Calculate $\frac{d}{dt}[\vec{r}(f(t))]$

1. Integrate the $x$ -component: $f(t) = 5 \cos(7t)$

2. Integrate the $y$ -component: $g(t) = 3 \cos(4t)$

3. Integrate the $z$ -component: $h(t) = \frac{-9}{t}$

Part 1: Indefinite Integral for Displacement Vector $\vec{r}(t)$

Part 2: Definite Integral over the Interval $1 \le t \le 5$