MATH-182 (Calculus II) Curriculum

Session 05 - Net Change Theorem

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

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

Differentiating Between Net Change (Displacement) and Total Change (Distance Traveled)
Generalizing and Translating Mathematical Models Across Contexts

Micro-Lecture

Engineering Context: Engineering Calculus II Application

In real-world engineering, tracking rates of change is vital for monitoring and managing dynamic systems where direct measurement is impossible. Aerospace engineers apply these concepts to distinguish between a vehicle's net altitude change (displacement) and its total structural wear and fuel burn (distance traveled), while electrical engineers integrate electrical current over time to dynamically calculate a smartphone or electric vehicle's remaining battery capacity from its initial charge. Additionally, civil and chemical engineers rely on the Net Change Theorem to balance changing fluid volumes in municipal water reservoirs or to precisely quantify the total volume of oil lost during a decaying pipeline leak. By translating these calculus models across disciplines, engineers can accurately predict future system states, optimize efficiency, and prevent critical infrastructure failures.

Skill Block 1

Regroup 1

Review common misconceptions and clarify key notations.

Skill Block 2

Regroup 2

Reflect on the physical modeling applications and mathematical setups.

Skill Block 3

Active Practice Problem

Oil is leaking from a tank and engineers inspected the leak and determine that the amount of oil in the tank, $Q(t)$ , is changing such that:

Q'(t) = \frac{-8}{(t+3)^2}

where

Q

is in thousands of gallons and

t

is in hours since the leak started at midnight.

How much oil will leak from the tank between 5 and 6 AM? Give a complete sentence final answer with units.

Worked Solution:

Concept Explanation

The problem provides the rate of change of the amount of oil in the tank, $Q'(t)$ , and asks for the total quantity of oil lost over a specific time window.

According to the Net Change Theorem, the total net change of a quantity over a time interval $[a, b]$ is found by calculating the definite integral of its rate of change function:

\text{Net Change} = \int_{a}^{b} Q'(t) \, dt

Since $t$ represents the hours passed since midnight ( $t = 0$ ):

5 AM corresponds to $t = 5$
6 AM corresponds to $t = 6$

The rate function $Q'(t)$ is negative because the total volume of oil inside the tank is decreasing. To find the actual positive amount of oil that escaped the tank, we can drop the negative sign and integrate the absolute rate of leakage:

\text{Amount Leaked} = \int_{5}^{6} \frac{8}{(t+3)^2} \, dt

Step-by-Step Calculation

Step 1: Set up the definite integral

To prepare the fraction for integration, rewrite it using a negative exponent:

\text{Amount Leaked} = \int_{5}^{6} 8(t + 3)^{-2} \, dt

Step 2: Find the antiderivative

We can integrate this using the power rule ( $\int u^n \, du = \frac{u^{n+1}}{n+1}$ ), where the inner function is $u = t + 3$ (since the derivative of $t+3$ is just $1$ , no complex substitution steps are required):

Retain the constant coefficient $8$ .
Add $1$ to the exponent: $-2 + 1 = -1$ .
Divide by the new exponent: $-1$ .

The resulting antiderivative function is:

F(t) = \frac{8(t + 3)^{-1}}{-1} = -\frac{8}{t + 3}

Step 3: Evaluate the definite integral

Apply the Fundamental Theorem of Calculus by substituting the upper limit ( $t = 6$ ) and subtracting the value at the lower limit ( $t = 5$ ):

\text{Amount Leaked} = \left[ -\frac{8}{t + 3} \right]_{5}^{6}

Substitute the upper limit ( $t = 6$ ):

F(6) = -\frac{8}{6 + 3} = -\frac{8}{9}

Substitute the lower limit ( $t = 5$ ):

F(5) = -\frac{8}{5 + 3} = -\frac{8}{8} = -1

Subtract the lower limit evaluation from the upper limit evaluation:

\text{Amount Leaked} = \left( -\frac{8}{9} \right) - (-1)

\text{Amount Leaked} = -\frac{8}{9} + 1 = \frac{1}{9}

Step 4: Unit Conversion

The problem notes that the function $Q(t)$ measures oil in thousands of gallons. Therefore, our mathematical result of $\frac{1}{9}$ represents $\frac{1}{9}$ thousand gallons.

To state this in a more standard fashion, we convert it to individual gallons by multiplying by 1,000:

\frac{1}{9} \times 1000 = \frac{1000}{9} \approx 111.11 \text{ gallons}

Final Answer

Between 5 and 6 AM, $\frac{1}{9}$ thousand gallons (or approximately 111.11 gallons) of oil will leak from the tank.

Regroup 3

Verify calculations and mathematical reasoning.

Synthesis Wrap-up

Core takeaways from Session 05 and overview of homework homework assignment.