7.3 The Net Change Theorem & Rectilinear Particle Motion

7.3 The Net Change Theorem & Rectilinear Particle Motion

Core CLEP Concept: The Net Change Theorem is the direct physical interpretation of the Fundamental Theorem of Calculus. It asserts that integrating the rate of change of any quantity computes the net change in that quantity. In physics, engineering, and economics, this theorem enables us to compute final state values from initial conditions and rate functions. On the CLEP Calculus exam, Net Change is most frequently tested through rectilinear particle kinematics—distinguishing between signed displacement and non-negative total distance traveled across both non-calculator and graphing calculator sections.

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1. The Net Change Theorem & Future State Modeling

The formal statement of the Net Change Theorem reformulates FTC Part 2 in terms of rates of change.

Theorem (The Net Change Theorem): The definite integral of a rate of change $F'(t)$ over a time interval $[a, b]$ equals the net change in the original quantity $F(t)$: $\int_a^b F'(t) \, dt = F(b) - F(a)$

The Accumulation / Future State Formula

Rearranging the theorem isolates the final quantity $F(b)$:

$$F(b) = F(a) + \int_a^b F'(t) \, dt$$ $$\text{Final State at time } b = \text{Initial State at time } a + \text{Net Accumulated Change on } [a, b]$$

   [ Current Quantity F(b) ]  =  [ Starting Value F(a) ]  +  / b  [ Rate of Change F'(t) ] dt
                                                             / a

Real-World Applications of Net Change

1. Fluid Dynamics / Reservoir Storage: If water enters a tank at rate $R_{\text{in}}(t)$ and drains at rate $R_{\text{out}}(t)$, the net rate of change is $V'(t) = R_{\text{in}}(t) - R_{\text{out}}(t)$. The volume at time $t_2$ is: $$V(t_2) = V(t_1) + \int_{t_1}^{t_2} \left[ R_{\text{in}}(t) - R_{\text{out}}(t) \right] dt$$
2. Population Dynamics: If a population changes at rate $P'(t) = \text{Births}(t) - \text{Deaths}(t)$, the population at time $t$ is: $$P(t) = P(0) + \int_0^t P'(u) \, du$$
3. Thermal Energy / Temperature: If an object heats or cools at rate $T'(t)$, its temperature at time $t$ is: $$T(t) = T(0) + \int_0^t T'(u) \, du$$
4. Economic Total Cost from Marginal Cost: If $C'(x)$ represents marginal cost, the cost of increasing production from $x_1$ to $x_2$ units is: $$\Delta C = \int_{x_1}^{x_2} C'(x) \, dx = C(x_2) - C(x_1)$$

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2. Rectilinear Kinematics Hierarchy: Position, Velocity, Acceleration

In one-dimensional (rectilinear) particle motion along a coordinate axis, the relationship between position $s(t)$, velocity $v(t)$, speed $|v(t)|$, and acceleration $a(t)$ forms a two-way calculus hierarchy:

               DIFFERENTIATION (d/dt)                          INTEGRATION (int ... dt)
   Position s(t) ------------------> Velocity v(t) ------------------> Acceleration a(t)
                 <-----------------                <-----------------
                 Position s(t) =                   Velocity v(t) =
                 s(t_0) + int v dt                 v(t_0) + int a dt

Reconstructing Velocity and Position Initial Value Problems

Given acceleration $a(t)$ and initial velocity $v(t_0)$:

$$v(t) = v(t_0) + \int_{t_0}^t a(u) \, du$$

Given velocity $v(t)$ and initial position $s(t_0)$:

$$s(t) = s(t_0) + \int_{t_0}^t v(u) \, du$$

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3. Displacement vs. Total Distance Traveled

The distinction between displacement and total distance traveled is one of the most heavily tested concepts on the CLEP Calculus exam.

   Motion of Particle:
   Starts at s(0)=5 ---> moves right to s(1)=9 ---> reverses left to s(3)=5 ---> moves right to s(4)=9

              <--------------------------- (Moves Left: Delta s_2 = -4)
   ----+------+------+------+------+------+------+------+------+------+----> Coordinate Line s
       0      1      2      3      4      5      6      7      8      9
                                   s(0),s(3)                     s(1),s(4)
       ---------------------------------->                       ---------->
          (Moves Right: Delta s_1 = +4)                     (Moves Right: Delta s_3 = +4)

   - Displacement = s(4) - s(0) = 9 - 5 = +4  (Net shift from start to end)
   - Total Distance = |+4| + |-4| + |+4| = 4 + 4 + 4 = 12  (Total odometer mileage)

Comparative Summary Table

Fundamental Kinematic Inequality

For any motion over $[t_1, t_2]$:

$$\text{Total Distance Traveled} \ge |\text{Displacement}|$$

Equality holds ($\text{Total Distance} = |\text{Displacement}|$) if and only if the particle moves in a single direction without reversing—meaning $v(t)$ never changes sign on $[t_1, t_2]$.

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4. The 4-Step Analytical Algorithm for Total Distance (Non-Calculator)

When evaluating total distance without a graphing calculator, follow this systematic protocol:

Step 1: Find All Turning Points

Set $v(t) = 0$ and solve for all critical roots $t_c$ strictly inside the open interval $(t_1, t_2)$. These roots represent potential points where the particle changes direction.

Step 2: Partition the Domain into Subintervals

Subdivide $[t_1, t_2]$ at each real turning point:

$$[t_1, t_{c1}], \quad [t_{c1}, t_{c2}], \quad \dots, \quad [t_{cn}, t_2]$$

On each subinterval, the velocity maintains a constant sign (the particle travels purely forward or purely backward).

Step 3: Compute Sub-Displacements

Calculate the signed displacement on each subinterval using the position function:

$$\Delta s_k = s(t_{\text{end}}) - s(t_{\text{start}}) = \int_{t_{\text{start}}}^{t_{\text{end}}} v(t) \, dt$$

Step 4: Sum the Absolute Values of All Sub-Displacements

Add the absolute values of the individual segments:

$$\text{Total Distance} = |\Delta s_1| + |\Delta s_2| + \dots + |\Delta s_n|$$

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5. Comprehensive Step-by-Step Kinematic Case Studies

Case Study 1: Complete Non-Calculator Kinematics Analysis

Problem: A particle moves along a horizontal axis with velocity $v(t) = 6t^2 - 18t + 12$ meters per second for $0 \le t \le 3$. The particle's initial position at $t = 0$ is $s(0) = 4$ meters.

1. Find the position function $s(t)$.
2. Determine the displacement of the particle over $[0, 3]$.
3. Compute the total distance traveled by the particle over $[0, 3]$.
4. Determine the time intervals when the particle is speeding up and slowing down.

Part 1: Position Function $s(t)$

$$s(t) = s(0) + \int_0^t (6u^2 - 18u + 12) \, du = 4 + [2u^3 - 9u^2 + 12u]_0^t = 2t^3 - 9t^2 + 12t + 4$$

Part 2: Displacement on $[0, 3]$

$$\text{Displacement} = \int_0^3 v(t) \, dt = s(3) - s(0)$$ $$s(3) = 2(3)^3 - 9(3)^2 + 12(3) + 4 = 2(27) - 9(9) + 36 + 4 = 54 - 81 + 40 = 13$$ $$\text{Displacement} = 13 - 4 = +9 \text{ meters}$$

Part 3: Total Distance Traveled on $[0, 3]$

• Step 1 (Find turning points): Set $v(t) = 6(t^2 - 3t + 2) = 6(t - 1)(t - 2) = 0 \implies t = 1, t = 2$. Both lie in $(0, 3)$.
• Step 2 (Partition intervals): $[0, 1]$, $[1, 2]$, and $[2, 3]$.
• Step 3 (Evaluate position at all boundary points):
  • $s(0) = 4$
  • $s(1) = 2(1)^3 - 9(1)^2 + 12(1) + 4 = 2 - 9 + 12 + 4 = 9 \implies \Delta s_1 = 9 - 4 = +5$
  • $s(2) = 2(2)^3 - 9(2)^2 + 12(2) + 4 = 16 - 36 + 24 + 4 = 8 \implies \Delta s_2 = 8 - 9 = -1$
  • $s(3) = 13 \implies \Delta s_3 = 13 - 8 = +5$
• Step 4 (Sum absolute sub-displacements): $$\text{Total Distance} = |+5| + |-1| + |+5| = 5 + 1 + 5 = 11 \text{ meters}$$

Part 4: Speeding Up vs. Slowing Down Analysis

• Speeding Up: $v(t)$ and $a(t)$ have the SAME sign (both positive or both negative).
• Slowing Down: $v(t)$ and $a(t)$ have OPPOSITE signs.
• Velocity signs: $v(t) = 6(t - 1)(t - 2)$
  • Positive ($v > 0$) on $[0, 1) \cup (2, 3]$.
  • Negative ($v < 0$) on $(1, 2)$.
• Acceleration: $a(t) = v'(t) = 12t - 18 = 6(2t - 3) = 0 \implies t = 1.5$.
  • Negative ($a < 0$) on $[0, 1.5)$.
  • Positive ($a > 0$) on $(1.5, 3]$.
• Sign Comparison Table:
  • On $(0, 1)$: $v > 0, a < 0 \implies$ Slowing down
  • On $(1, 1.5)$: $v < 0, a < 0 \implies$ Speeding up
  • On $(1.5, 2)$: $v < 0, a > 0 \implies$ Slowing down
  • On $(2, 3)$: $v > 0, a > 0 \implies$ Speeding up

$$\text{Speeding Up on } (1, 1.5) \cup (2, 3) \qquad \text{Slowing Down on } (0, 1) \cup (1.5, 2)$$

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6. Graphing Calculator Protocol for CLEP Section 2

On Section 2 of the CLEP exam, you have access to the online scientific/graphing calculator (TI-84 Plus CE emulator). Use the numerical integration command fnInt:

Calculator Commands

1. Enter the Velocity Function: Press Y= and input $Y_1 = v(X)$.
2. Compute Displacement: $$\text{Displacement} = \int_{t_1}^{t_2} v(t) \, dt \implies \text{MATH} \to \text{9:fnInt} \to \text{fnInt}(Y_1, X, t_1, t_2)$$
3. Compute Total Distance Traveled: $$\text{Total Distance} = \int_{t_1}^{t_2} |v(t)| \, dt \implies \text{fnInt}(\text{abs}(Y_1), X, t_1, t_2)$$ (To access abs(), press MATH $\to$ NUM $\to$ 1:abs().
4. Compute Future Position $s(t_2)$: $$s(t_2) = s(t_1) + \text{fnInt}(Y_1, X, t_1, t_2)$$

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7. Common CLEP Pitfalls & Traps

1. Evaluating Displacement When Asked for Total Distance: Displacement ($\int v \, dt$) allows negative motion to cancel positive motion. Total distance ($\int |v| \, dt$) requires summing all path lengths as positive numbers. Always check which quantity the question requests.
2. Omitting Initial Conditions in Accumulation Problems: When asked for the particle's position $s(t)$ or tank volume $V(t)$ at a given time, adding the net change integral $\int_0^t v(u) \, du$ to $0$ instead of the initial value $s(0)$ or $V(0)$ is a standard distractor choice.
3. Equating "Speeding Up" with Positive Acceleration: A particle speeds up when its speed $|v(t)|$ increases. If $v(t) = -10$ and $a(t) = -5$, the particle is moving to the left faster and faster (speed is increasing from 10 to 15), even though acceleration is negative. Speeding up requires $v(t)$ and $a(t)$ to share the same sign.
4. Searching for Turning Points where $a(t) = 0$: Turning points (reversals of motion) occur where velocity changes sign ($v(t) = 0$), NOT where acceleration is zero.