Session 01 - Course Overview + Prerequisite Review

📅 Micro-Lecture

1. Trigonometric Unit Circle & Right Triangle Geometry

Mathematical Foundation: A right triangle with acute angle $\theta$ , adjacent side $x$ , opposite side $y$ , and hypotenuse $r = \sqrt{x^2 + y^2}$ defines the standard trigonometric ratios: $\sin\theta = \frac{y}{r}, \quad \cos\theta = \frac{x}{r}, \quad \tan\theta = \frac{y}{x}$ Angles are converted between radians and degrees using the ratio $\pi \text{ rad} = 180^\circ$ . The unit circle maps angles to coordinates $(x,y) = (\cos\theta, \sin\theta)$ on $x^2 + y^2 = 1$ , where sign conventions are determined by the quadrant (CAST rule).
Engineering Context: Crucial for static force decompositions in structural trusses, coordinate system rotations, robotics kinematics, and analyzing alternating current (AC) circuits.

2. Limits, Continuity, and Differentiability

Mathematical Foundation: Limits describe function behavior as the input $x$ approaches a value $c$ . Indeterminate forms like $\frac{0}{0}$ are resolved algebraically via factoring or rationalizing. A function $f(x)$ is continuous at $c$ if $\lim_{x \to c} f(x) = f(c)$ , and differentiable if the derivative limit $\lim_{x \to c} f'(x)$ exists and is continuous at $c$ .
Engineering Context: Modeling state transitions in physical processes (such as thermodynamics or fluid mechanics) requires continuous and differentiable functions to avoid non-physical shocks or discontinuities in stress, pressure, or temperature.

3. Derivatives and Optimization

Mathematical Foundation: The derivative $f'(x)$ measures the instantaneous rate of change. Critical points occur where $f'(x) = 0$ or is undefined. The second derivative $f''(x)$ dictates concavity (concave up if $f''(x) > 0$ , concave down if $f''(x) < 0$ ), and inflection points occur where concavity changes sign. Optimization uses these derivatives to locate maximum or minimum values under specific constraints.
Engineering Context: Used to minimize material cost (e.g., structural optimization), maximize structural load capacities, and determine peak operating efficiencies.

4. Definite Integrals and the Net Change Theorem

Mathematical Foundation: A definite integral $\int_a^b f(x) dx$ represents the net signed area bounded by the curve and the $x$ -axis. Total area treats all bounded regions as positive. The Net Change Theorem states that the integral of a rate of change function $F'(t)$ yields the net accumulated change: $F(b) - F(a) = \int_a^b F'(t) dt$
Engineering Context: Enables calculating total physical accumulation (e.g., total volume of water discharged, total charge accumulated, or total work done) from a time-varying rate.

Segment 1 of 8

Skill Block 1

Real-Life Engineering Applications

🏗️

Civil Engineering

Bridge Truss Force Decomposition (Civil & Mechanical Engineering)

In structural engineering, joint analysis relies on right-triangle trigonometry to decompose static force vectors into axial components along support members.

A bridge truss joint (Junction A) experiences a downward gravitational load of $150\text{ kN}$ from traffic. Two steel struts meet at this junction: Strut 1 is horizontal, and Strut 2 forms an angle of $\theta = 30^\circ$ with the horizontal strut.

Determine the following under static equilibrium conditions (where the vertical component of the force in Strut 2 must balance the $150\text{ kN}$ downward load):

Draw the force triangle representing the equilibrium state.
Determine the exact force in Strut 2.
Determine the exact force in Strut 1.

🎯 Why This Matters:

& Real-Life Application: Decomposing force vectors using trigonometry is essential for sizing steel beams and joints. Underestimating the axial forces in members can lead to structural buckling and catastrophic failure of the bridge under peak traffic loads.

💡 Show Worked Derivation

In structural engineering, joint analysis relies on right-triangle trigonometry to decompose static force vectors into axial components along support members.

A bridge truss joint (Junction A) experiences a downward gravitational load of $150\text{ kN}$ from traffic. Two steel struts meet at this junction: Strut 1 is horizontal, and Strut 2 forms an angle of $\theta = 30^\circ$ with the horizontal strut.

Determine the following under static equilibrium conditions (where the vertical component of the force in Strut 2 must balance the $150\text{ kN}$ downward load):

Draw the force triangle representing the equilibrium state.
Determine the exact force in Strut 2.
Determine the exact force in Strut 1.

Step-by-Step Solution:

Draw the Force Triangle: The downward gravitational load vector of $150\text{ kN}$ forms the vertical leg of a right-angled force triangle. The axial force in Strut 2 is along the hypotenuse, and the axial force in Strut 1 is along the horizontal leg.
Determine the Force in Strut 2 ( $F_2$ ): Using the sine trigonometric ratio:
$\sin(30^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{150}{F_2}$ $F_2 = \frac{150}{\sin(30^\circ)} = \frac{150}{1/2} = 300\text{ kN}$
Determine the Force in Strut 1 ( $F_1$ ): Using the cosine trigonometric ratio:
$\cos(30^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{F_1}{F_2}$ $F_1 = F_2 \cdot \cos(30^\circ) = 300 \cdot \frac{\sqrt{3}}{2} = 150\sqrt{3}\text{ kN}$

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

Pre-Calculus & Trigonometric Fluency

General Overview
Skill Block 1 reviews right triangle geometry, trigonometric ratios, angle conversions, and unit circle evaluations.

Conceptual Framework

Ratios in Right Triangles: Trigonometric ratios represent the ratios of the sides of a right triangle.
Unit Circle Evaluation: Evaluating trigonometric functions of any angle is accomplished by determining the quadrant, reference angle, and coordinate signs.
Angle Conversion: Converting between radians and degrees using the relationship $\pi \text{ rad} = 180^\circ$ .

Key Equations & Structural Rules

Radian to Degree: $\text{Degrees} = \text{Radians} \times \frac{180^\circ}{\pi}$
Degree to Radian: $\text{Radians} = \text{Degrees} \times \frac{\pi}{180^\circ}$
Pythagorean Theorem: $a^2 + b^2 = c^2$

Worked Example A: Unit Circle Evaluation

Evaluate the following, giving your answer in exact form:

\sin\left(\frac{3\pi}{4}\right)

💡 Reveal Worked Solution

Identify the Quadrant: The angle $\theta = \frac{3\pi}{4}$ lies in Quadrant II, where the sine function is positive.
Find the Reference Angle ( $\theta_R$ ): $\theta_R = \pi - \frac{3\pi}{4} = \frac{\pi}{4}$
Evaluate: $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ Since sine is positive in Quadrant II, the exact value is: $\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Worked Example B: Right Triangle Side Lengths

Consider triangle $ABC$ , a right triangle with a right angle at $C$ . Given the hypotenuse $c = 12$ and adjacent side $b = 9$ , find the exact length of the third side $a$ .

💡 Reveal Worked Solution

Apply the Pythagorean Theorem: $a^2 + b^2 = c^2 \implies a^2 + 9^2 = 12^2$
Isolate $a^2$ : $a^2 + 81 = 144 \implies a^2 = 63$
Solve for $a$ (exact form): $a = \sqrt{63} = \sqrt{9 \cdot 7} = 3\sqrt{7}$

Worked Example C: Right Triangle Trigonometric Ratios

Consider triangle $ABC$ , a right triangle with a right angle at $C$ . Given adjacent side $b = 6$ and opposite side $a = 6$ , find the exact values of $\sin A$ and $\cos A$ .

💡 Reveal Worked Solution

Find the Hypotenuse ( $c$ ): $c^2 = a^2 + b^2 = 6^2 + 6^2 = 36 + 36 = 72 \implies c = \sqrt{72} = 6\sqrt{2}$
Evaluate $\sin A$ : $\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
Evaluate $\cos A$ : $\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Practice Problem 1

Consider triangle $ABC$ , a right triangle with a right angle at $C$ . Given the hypotenuse $c = 18$ and angle $A = 63^\circ$ , find the length $a$ . Round your answer to 2 decimal places.

Practice Space Solve this problem on paper or tablet.

Identify the given side (hypotenuse) and angle. Choose the trigonometric ratio (sine) that relates the opposite side $a$ , the hypotenuse, and the angle $A$ .

Answer:

a \approx 16.04

Step-by-Step Execution

Identify the Given Information:
- Angle $A = 63^\circ$
- Hypotenuse $c = 18$
- Side $a$ is opposite to angle $A$ .
Apply the Sine Ratio: $\sin(A) = \frac{\text{Opposite}}{\text{Hypotenuse}} \implies \sin(63^\circ) = \frac{a}{18}$
Solve for $a$ : $a = 18 \cdot \sin(63^\circ) \approx 18 \cdot 0.8910065 \approx 16.038 \implies a \approx 16.04$

Practice Problem 2

Express $\frac{7\pi}{12}$ rad in degrees. (Enter exact values.)

Practice Space Solve this problem on paper or tablet.

Multiply the radian value by the conversion factor $\frac{180^\circ}{\pi}$ .

Answer:

105^\circ

Step-by-Step Execution

\text{Degrees} = \left(\frac{7\pi}{12}\right) \times \frac{180^\circ}{\pi} = \frac{7}{12} \times 180^\circ = 7 \times 15^\circ = 105^\circ

Practice Problem 3

Express $23^\circ$ in radians. (Enter exact values.)

Practice Space Solve this problem on paper or tablet.

Multiply the degree value by the conversion factor $\frac{\pi}{180^\circ}$ .

Answer:

\frac{23\pi}{180} \text{ rad}

Step-by-Step Execution

\text{Radians} = 23^\circ \times \frac{\pi}{180^\circ} = \frac{23\pi}{180} \text{ rad}

Practice Problem 4

A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is $6\text{ ft}$ from the ground and the angle between the ground and the ramp is to be $26^\circ$ , how long does the ramp need to be? Round your answer to four decimal places.

Practice Space Solve this problem on paper or tablet.

Model the ramp as the hypotenuse of a right triangle. The height of the staircase ( $6\text{ ft}$ ) is the opposite side to the angle $26^\circ$ .

Answer:

13.6870\text{ ft}

Step-by-Step Execution

Apply the Sine Ratio: $\sin(26^\circ) = \frac{6}{x} \implies x = \frac{6}{\sin(26^\circ)}$
Calculate: $x \approx \frac{6}{0.438371} \approx 13.68703 \implies x \approx 13.6870\text{ ft}$

Segment 2 of 8

📅 Regroup 1

Review unit circle quadrant signs and reference angles.
Review radian-degree conversion factors.

Segment 3 of 8

Skill Block 2

Real-Life Engineering Applications

🧪

Chemical Engineering

Spot Optimization (Chemical Engineering)

Continuous temperature monitoring in exothermic reactors is critical for safety. The temperature profile across reaction zones is modeled by piecewise functions, which must be continuous and differentiable to avoid physical discontinuities in heat transfer models.

An exothermic chemical reactor has a temperature profile along its central axis modeled by the piecewise function:

T(x) = \begin{cases} -x^2 + 6x + 80 & : 0 \le x < 3 \\ ax + b & : x \ge 3 \end{cases}

where

x

is the distance in meters from the inlet, and

T(x)

is the temperature in degrees Celsius.

Determine the following:

Find the values of $a$ and $b$ such that the temperature profile is continuous and differentiable at the boundary $x = 3$ meters.
Optimize the reactor design by finding the maximum temperature along the axis in the interval $0 \le x \le 5$ .

🎯 Why This Matters:

& Real-Life Application: Continuity and differentiability of thermal profiles prevent heat flux discontinuities in modeling. A smooth temperature transition ensures that the physical catalyst beds do not experience localized thermal shock, which could lead to reactor runaway.

💡 Show Worked Derivation

Continuous temperature monitoring in exothermic reactors is critical for safety. The temperature profile across reaction zones is modeled by piecewise functions, which must be continuous and differentiable to avoid physical discontinuities in heat transfer models.

An exothermic chemical reactor has a temperature profile along its central axis modeled by the piecewise function:

T(x) = \begin{cases} -x^2 + 6x + 80 & : 0 \le x < 3 \\ ax + b & : x \ge 3 \end{cases}

where

x

is the distance in meters from the inlet, and

T(x)

is the temperature in degrees Celsius.

Determine the following:

Find the values of $a$ and $b$ such that the temperature profile is continuous and differentiable at the boundary $x = 3$ meters.
Optimize the reactor design by finding the maximum temperature along the axis in the interval $0 \le x \le 5$ .

Step-by-Step Solution:

Find $a$ and $b$ for Continuity and Differentiability:
- Continuity Condition: The left-hand limit must equal the right-hand limit at $x = 3$ :
  $\lim_{x \to 3^-} (-x^2 + 6x + 80) = \lim_{x \to 3^+} (ax + b)$ $-3^2 + 6(3) + 80 = 3a + b$ $-9 + 18 + 80 = 3a + b \implies 89 = 3a + b \quad \text{--- (Equation 1)}$
- Differentiability Condition: The derivative from the left must equal the derivative from the right at $x = 3$ :
  $T'(x) = \begin{cases} -2x + 6 & : 0 \le x < 3 \\ a & : x > 3 \end{cases}$ $\lim_{x \to 3^-} (-2x + 6) = \lim_{x \to 3^+} (a)$ $-2(3) + 6 = a \implies a = 0$
- Solve for $b$ : Substitute $a = 0$ into Equation 1:
  $89 = 3(0) + b \implies b = 89$
  
  So, the parameters are $a = 0$ and $b = 89$ .
Find the Maximum Temperature on $0 \le x \le 5$ : With $a = 0$ and $b = 89$ :
$T(x) = \begin{cases} -x^2 + 6x + 80 & : 0 \le x < 3 \\ 89 & : x \ge 3 \end{cases}$

To find the maximum temperature:
- For $0 \le x < 3$ , find critical points of $T(x) = -x^2 + 6x + 80$ : $T'(x) = -2x + 6 = 0 \implies x = 3$ Since $x=3$ is the boundary, the function increases on $[0, 3)$ because $T'(x) > 0$ for $x < 3$ .
- At $x = 0$ , $T(0) = 80^\circ\text{C}$ .
- For $x \ge 3$ , the temperature is constant at $T(x) = 89^\circ\text{C}$ .
Therefore, the maximum temperature along the axis is $89^\circ\text{C}$ , reached at $x \ge 3$ .

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

Calculus 1 Review (Limits, Continuity, & Derivatives)

General Overview
Skill Block 2 establishes fluency in limit evaluation, piecewise continuity/differentiability, curve sketching (concavity, inflection points), and optimization.

Conceptual Framework

Indeterminate Limits: Resolving $\frac{0}{0}$ limits using algebraic factoring or rationalizing.
Continuity and Differentiability of Piecewise Functions: Ensuring a piecewise function is smooth at its boundary by setting the values and derivatives of each piece equal at that boundary.
Second Derivative Tests: Concavity and inflection points are found using $f''(x)$ .
Applied Optimization: Formulating equations for a system under constraints and finding the extreme values using derivatives.

Key Equations & Structural Rules

Continuity at $x = c$ : $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$
Differentiability at $x = c$ : $\lim_{x \to c^-} f'(x) = \lim_{x \to c^+} f'(x)$
Optimization: Solve $f'(x) = 0$ to find critical points.

Worked Example A: Algebraic Limit Evaluation

Evaluate the following limit:

\lim_{x \to -2} \frac{-x^2 + 5x + 14}{-5x^2 - 2x + 16}

💡 Reveal Worked Solution

Check for Indeterminate Form: Substituting $x = -2$ gives $\frac{0}{0}$ .
Factor Numerator and Denominator:
- Numerator: $-x^2 + 5x + 14 = -(x - 7)(x + 2)$
- Denominator: $-5x^2 - 2x + 16 = (x + 2)(-5x + 8)$
Cancel Common Factor $(x + 2)$ : $\lim_{x \to -2} \frac{-(x - 7)}{-5x + 8}$
Evaluate by Substitution: $\frac{-(-2 - 7)}{-5(-2) + 8} = \frac{9}{18} = \frac{1}{2}$

Worked Example B: Piecewise Continuity and Differentiability

Find the values of $a$ and $b$ that make the following piecewise defined function both continuous and differentiable everywhere:

f(x) = \begin{cases} 3x + 4, & x < -3 \ 2x^2 + ax + b, & x \ge -3 \end{cases}

💡 Reveal Worked Solution

Apply Continuity Condition at $x = -3$ : $3(-3) + 4 = 2(-3)^2 + a(-3) + b \implies -5 = 18 - 3a + b \implies -23 = -3a + b \quad \text{--- (Eq. 1)}$
Apply Differentiability Condition at $x = -3$ : $f'(x) = \begin{cases} 3, & x < -3 \ 4x + a, & x > -3 \end{cases}$ $3 = 4(-3) + a \implies 3 = -12 + a \implies a = 15$
Solve for $b$ : $-23 = -3(15) + b \implies -23 = -45 + b \implies b = 22$

Worked Example C: Norman Window Optimization

A Norman window is constructed by adjoining a semicircle to the top of a rectangular window. What is the maximum possible area, in square meters, of a Norman window with a perimeter of $26$ meters?

💡 Reveal Worked Solution

Set Up Perimeter and Area Equations: Let the rectangle have width $y$ and height $x$ . The semicircle has radius $r = y/2$ .
- Perimeter: $P = 2x + y + \pi(y/2) = 26 \implies 2x = 26 - y(1 + \pi/2)$
- Area: $A = xy + \frac{1}{2}\pi(y/2)^2 = xy + \frac{\pi y^2}{8}$
Substitute $x$ to express $A$ in terms of $y$ : $A = y\left(13 - \frac{y}{2} - \frac{\pi y}{4}\right) + \frac{\pi y^2}{8} = 13y - \frac{y^2}{2} - \frac{\pi y^2}{8} = 13y - y^2\left(\frac{4 + \pi}{8}\right)$
Maximize Area by Taking the Derivative: $\frac{dA}{dy} = 13 - y\left(\frac{4 + \pi}{4}\right) = 0 \implies y = \frac{52}{4 + \pi}$
Calculate Maximum Area: $A = 13\left(\frac{52}{4 + \pi}\right) - \left(\frac{52}{4 + \pi}\right)^2\left(\frac{4 + \pi}{8}\right) = \frac{338}{4 + \pi} \text{ m}^2$

Practice Problem 5

Evaluate:

\lim_{x \to -5} \frac{\sqrt{-x-2} - \sqrt{3}}{-x-5}

Practice Space Solve this problem on paper or tablet.

Substitute $x = -5$ to check for the indeterminate form $\frac{0}{0}$ . Rationalize the numerator by multiplying the numerator and denominator by the conjugate $\sqrt{-x-2} + \sqrt{3}$ .

Answer:

\frac{\sqrt{3}}{6}

Step-by-Step Execution

Multiply by the Conjugate: $\lim_{x \to -5} \frac{(\sqrt{-x-2} - \sqrt{3})(\sqrt{-x-2} + \sqrt{3})}{(-x-5)(\sqrt{-x-2} + \sqrt{3})} = \lim_{x \to -5} \frac{-x - 5}{(-x - 5)(\sqrt{-x-2} + \sqrt{3})}$
Cancel Common Factor $(-x - 5)$ and Evaluate: $\lim_{x \to -5} \frac{1}{\sqrt{-x-2} + \sqrt{3}} = \frac{1}{\sqrt{3} + \sqrt{3}} = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6}$

Practice Problem 6

Consider a function $f(x)$ with critical points at $-2$ , $-\frac{3}{2}$ , $-\frac{3}{5}$ , and $-\frac{1}{2}$ . Based on the sign table for $f'(x)$ below, on which of the intervals is $f(x)$ increasing?

Interval	$(-\infty, -2)$	$\left(-2, -\frac{3}{2}\right)$	$\left(-\frac{3}{2}, -\frac{3}{5}\right)$	$\left(-\frac{3}{5}, -\frac{1}{2}\right)$	$\left(-\frac{1}{2}, \infty\right)$
Sign of $f'(x)$	$+$	$-$	$+$	$+$	$+$

Practice Space Solve this problem on paper or tablet.

A function is increasing on an interval where its derivative is positive ( $f'(x) > 0$ ).

Answer:

(-\infty, -2), \ \left(-\frac{3}{2}, -\frac{3}{5}\right), \ \left(-\frac{3}{5}, -\frac{1}{2}\right), \ \left(-\frac{1}{2}, \infty\right)

Step-by-Step Execution

Read the table and select the intervals with a positive ( $+$ ) derivative sign. All intervals except $\left(-2, -\frac{3}{2}\right)$ have $f'(x) > 0$ , indicating the function is increasing.

Practice Problem 7

The graph of the derivative $f'(x)$ is given below. On what interval(s) is the function $f(x)$ concave up?

Practice Space Solve this problem on paper or tablet.

The function $f(x)$ is concave up when its second derivative $f''(x) > 0$ . On the graph of the first derivative $f'(x)$ , this corresponds to intervals where the slope of $f'(x)$ is positive (i.e., where the graph of $f'(x)$ is increasing).

Answer:

(-6, -4), (0, \infty)

Step-by-Step Execution

Identify the intervals where the graph of $f'(x)$ has a positive slope (is rising):

Rising from $x = -6$ to $x = -4$
Rising from $x = 0$ to $x = \infty$ Thus, the function $f(x)$ is concave up on $(-6, -4) \cup (0, \infty)$ .

Practice Problem 8

Consider the function $f(x) = 4x^5 - 7x^3 - 5x + 1$ . Find the $x$ -values of all inflection points.

Practice Space Solve this problem on paper or tablet.

Find the second derivative $f''(x)$ , set it to zero, and solve for $x$ . Ensure that the sign of $f''(x)$ changes across these points.

Answer:

x = 0, \ \pm\frac{\sqrt{210}}{20}

Step-by-Step Execution

Find $f'(x)$ and $f''(x)$ : $f'(x) = 20x^4 - 21x^2 - 5 \implies f''(x) = 80x^3 - 42x$
Solve $f''(x) = 0$ : $2x(40x^2 - 21) = 0 \implies x = 0 \quad \text{or} \quad x^2 = \frac{21}{40} \implies x = \pm \sqrt{\frac{21}{40}} = \pm \frac{\sqrt{210}}{20}$ Since $f''(x)$ changes signs at each of these roots, they are all inflection points.

Practice Problem 9

Consider the function $f(x) = \frac{x^4}{4} + \frac{22x^3}{3} + \frac{153x^2}{2} + 324x + 3$ . Over what open interval(s) is the function increasing and concave up?

Practice Space Solve this problem on paper or tablet.

Find the intervals where $f'(x) > 0$ (increasing) and $f''(x) > 0$ (concave up), then find the intersection of these intervals.

Answer:

(-4, \infty)

Step-by-Step Execution

Find Increasing Intervals ( $f'(x) > 0$ ): $f'(x) = x^3 + 22x^2 + 153x + 324 = (x + 4)(x + 9)^2 = 0 \implies \text{Increasing on } (-4, \infty)$
Find Concave Up Intervals ( $f''(x) > 0$ ): $f''(x) = 3x^2 + 44x + 153 = (3x + 17)(x + 9) = 0 \implies \text{Concave up on } (-\infty, -9) \cup \left(-\frac{17}{3}, \infty\right)$
Find the Intersection: The intersection of $(-4, \infty)$ and $(-\infty, -9) \cup \left(-\frac{17}{3}, \infty\right)$ is $(-4, \infty)$ .

Practice Problem 10

A cell phone plan charges $30 per month for unlimited calls and texts, and each gigabyte (GB) of data is charged at $10 per GB (prorated). If a user uses more than $15\text{ GB}$ of data, the cost is capped at the cost for $15\text{ GB}$ . Write a piecewise-defined function for the cost $C$ as a function of data $d$ in GB.

Practice Space Solve this problem on paper or tablet.

For $0 \le d \le 15$ , the cost is the base fee plus $\$10$ times $d$ . For $d > 15$ , the cost is capped at the value of $C(15)$ .

Answer:

C(d) = \begin{cases} 30 + 10d, & 0 \le d \le 15 \\ 180, & d > 15 \end{cases}

Step-by-Step Execution

For $0 \le d \le 15$ : $C(d) = 30 + 10d$
For $d > 15$ : Cost is capped at $C(15) = 30 + 10(15) = 180$ . $C(d) = \begin{cases} 30 + 10d, & 0 \le d \le 15 \\ 180, & d > 15 \end{cases}$

Segment 4 of 8

📅 Regroup 2

Review the definition of derivative, concavity, and critical points.
Discuss standard optimization setup (objective function and constraint).

Segment 5 of 8

Skill Block 3

Real-Life Engineering Applications

🏗️

Civil Engineering

Hydroelectric Reservoir Net Change (Civil & Environmental Engineering)

Tracking net water volume changes in hydroelectric reservoirs requires integrating the net flow rate (inflow minus spillway discharge) over an operational time period.

Water discharges through a spillway of a reservoir at a time-varying rate of $r(t) = 18 - \frac{t^2}{2}$ cubic meters per minute for $t \ge 0$ , where $t$ is measured in minutes. Initially, the reservoir has $65$ cubic meters of active storage.

Determine the following:

Write the definite integral representing the total volume of water discharged (lost) from the reservoir between $t = 0$ and $t = 3$ minutes.
Evaluate this integral and determine how much water is remaining in the reservoir after $3$ minutes.

🎯 Why This Matters:

& Real-Life Application: Applying the Net Change Theorem allows engineers to monitor reservoir reserves and predict power generation capacities. Integrating flow rates ensures that water levels remain above minimum hydraulic intake heights to prevent turbine cavitation.

💡 Show Worked Derivation

Tracking net water volume changes in hydroelectric reservoirs requires integrating the net flow rate (inflow minus spillway discharge) over an operational time period.

Water discharges through a spillway of a reservoir at a time-varying rate of $r(t) = 18 - \frac{t^2}{2}$ cubic meters per minute for $t \ge 0$ , where $t$ is measured in minutes. Initially, the reservoir has $65$ cubic meters of active storage.

Determine the following:

Write the definite integral representing the total volume of water discharged (lost) from the reservoir between $t = 0$ and $t = 3$ minutes.
Evaluate this integral and determine how much water is remaining in the reservoir after $3$ minutes.

Step-by-Step Solution:

Write the Definite Integral: The rate of water leaving the reservoir is given by $r(t) = 18 - \frac{t^2}{2}$ . The total volume of water discharged over the interval $[0, 3]$ is the definite integral:
$\text{Volume Discharged} = \int_{0}^{3} \left(18 - \frac{t^2}{2}\right) dt$
Evaluate the Integral and Calculate Remaining Water: First, find the antiderivative of the rate function:
$\int \left(18 - \frac{t^2}{2}\right) dt = 18t - \frac{t^3}{6}$

Evaluate this antiderivative from $t = 0$ to $t = 3$ :
$\left[18(3) - \frac{3^3}{6}\right] - \left[18(0) - \frac{0^3}{6}\right]$ $= \left[54 - \frac{27}{6}\right] - 0$ $= 54 - 4.5 = 49.5\text{ cubic meters}$

Subtract the discharged volume from the initial $65\text{ cubic meters}$ to find the remaining volume:
$\text{Volume Remaining} = 65 - 49.5 = 15.5\text{ cubic meters}$

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

Calculus 2 Review (Integration & Net Change)

General Overview
Skill Block 3 covers definite integrals evaluated using geometric formulas, net signed area, and the Net Change Theorem.

Conceptual Framework

Geometric Definite Integrals: Using known areas of circles, semicircles, triangles, and trapezoids to evaluate integrals without antiderivatives.
Net Signed Area: Area above the $x$ -axis is positive; area below is negative.
The Net Change Theorem: The definite integral of a rate of change function $F'(t)$ gives the total change in the quantity $F(t)$ over that interval.

Key Equations & Structural Rules

Semicircle Equation: $y = \sqrt{R^2 - (x - x_0)^2} \implies (x - x_0)^2 + y^2 = R^2$
Net Change Theorem: $F(b) = F(a) + \int_a^b F'(t) dt$

Worked Example A: Geometric Definite Integral (Semicircle)

Evaluate the definite integral using geometric formulas:

\int_{-7}^{-3} \left(\sqrt{-(x+5)^2 + 4}\right) dx

💡 Reveal Worked Solution

Identify the Curve: The integrand $y = \sqrt{4 - (x+5)^2}$ represents a semicircle centered at $(-5, 0)$ with radius $R = \sqrt{4} = 2$ , sitting above the $x$ -axis.
Check the Bounds: The bounds $[-7, -3]$ represent the full domain of the semicircle (from $-5 - 2$ to $-5 + 2$ ).
Calculate Semicircle Area: $\text{Area} = \frac{1}{2}\pi R^2 = \frac{1}{2}\pi (2)^2 = 2\pi$

Worked Example B: Net Signed Area

Find the net signed area between the graph of $f(x) = 4 - \frac{5x}{4}$ and the $x$ -axis over the interval $\left[\frac{6}{5}, \frac{46}{5}\right]$ .

💡 Reveal Worked Solution

Set Up the Definite Integral: $\text{Net Signed Area} = \int_{6/5}^{46/5} \left(4 - \frac{5x}{4} \right) dx$
Find the Antiderivative: $F(x) = 4x - \frac{5x^2}{8}$
Evaluate at the Bounds:
- Upper bound: $F(46/5) = 4(46/5) - \frac{5(46/5)^2}{8} = \frac{184}{5} - \frac{529}{10} = -\frac{161}{10}$
- Lower bound: $F(6/5) = 4(6/5) - \frac{5(6/5)^2}{8} = \frac{24}{5} - \frac{9}{10} = \frac{39}{10}$
Subtract: $\text{Net Signed Area} = -\frac{161}{10} - \frac{39}{10} = -\frac{200}{10} = -20$

Worked Example C: Net Change from a Graph

A function $F(x)$ has an initial value $F(0) = -1$ . The graph of its derivative $F'(x)$ is a line passing through $(0,0)$ with a constant slope of $-2$ . What is the value of $F(3)$ ?

💡 Reveal Worked Solution

Apply the Net Change Theorem: $F(3) = F(0) + \int_{0}^{3} F'(x) dx$
Calculate the Area under $F'(x)$ Geometrically: From $x = 0$ to $x = 3$ , the graph of $F'(x) = -2x$ forms a right triangle below the $x$ -axis.
- Base = 3
- Height = $F'(3) = -6 \implies$ vertical length is 6.
- Area (below $x$ -axis) = $-\frac{1}{2} \times 3 \times 6 = -9$ .
Solve for $F(3)$ : $F(3) = -1 + (-9) = -10$

Practice Problem 11

Evaluate the definite integral using geometric formulas:

\int_{1}^{7} (3 - |-4 + x|) dx

Practice Space Solve this problem on paper or tablet.

Identify the shape under the curve. The absolute value function creates a triangular region. Find the base and height of this triangle.

Answer:

9

Step-by-Step Execution

Find Triangle Dimensions:
- Base: from $x = 1$ to $x = 7 \implies b = 6$ .
- Height: peak is at $x = 4 \implies h = f(4) = 3 - 0 = 3$ .
Calculate Area: $\text{Area} = \frac{1}{2} \times 6 \times 3 = 9$

Practice Problem 12

The piecewise function $f(x)$ is graphed below. Use geometric formulas to find the total area between the graph of the function and the $x$ -axis.

Practice Space Solve this problem on paper or tablet.

"Total Area" treats all bounded regions as positive. Split the shape into two triangles: one above the $x$ -axis ( $[0,4]$ ) and one below ( $[4,8]$ ).

Answer:

8

Step-by-Step Execution

Calculate Area of Triangle 1 (above axis):
- Base $b_1 = 4$ , Height $h_1 = 1 \implies A_1 = \frac{1}{2} \times 4 \times 1 = 2$
Calculate Area of Triangle 2 (below axis):
- Base $b_2 = 4$ , Height $h_2 = 3 \implies A_2 = \frac{1}{2} \times 4 \times 3 = 6$
Sum the Areas: $\text{Total Area} = A_1 + A_2 = 2 + 6 = 8$

Practice Problem 13

Water leaks out of a tank at a rate of $r(t) = 18 - \frac{t^2}{2}$ for $t \ge 0$ , measured in gallons per minute. Initially the tank has $65$ gallons of water in the tank. How much water is left in the tank after $3$ minutes?

Practice Space Solve this problem on paper or tablet.

The total water lost is the integral of the rate function $\int_{0}^{3} r(t) dt$ . Subtract this lost volume from the initial volume.

Answer:

15.5 \text{ gallons}

Step-by-Step Execution

Set Up the Equation: $\text{Water Left} = 65 - \int_{0}^{3} \left(18 - \frac{t^2}{2}\right) dt$
Integrate: $\int_{0}^{3} \left(18 - \frac{t^2}{2}\right) dt = \left[18t - \frac{t^3}{6}\right]_0^3 = 18(3) - \frac{27}{6} = 54 - 4.5 = 49.5\text{ gallons lost}$
Subtract: $\text{Remaining Water} = 65 - 49.5 = 15.5\text{ gallons}$

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📅 Regroup 3

Discuss the difference between net signed area and total area.
Review the Net Change Theorem.

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📅 Synthesis Wrap-up

Re-emphasize key calculus prerequisites.
Discuss how limits, derivatives, optimization, and integration are generalized to multi-dimensional surfaces in Calculus 3.

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