MAT 131 Calculus I Practice Exam 2 Vocabulary Review

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Indefinite Integration and Power-Law Formulations

• Indefinite Integral Evaluation:

  • Problem: Evaluate the indefinite integral     $\int \left(6x^2 - \frac{4}{\sqrt{x}} + 3\right) dx$

  • Rewrite the integrand using negative rational exponents:     $6x^2 - 4x^{-1/2} + 3$

  • Apply the power rule of integration $\int x^n dx = \frac{x^{n+1}}{n+1}$ for $n \neq -1$:

  • First term:     $\int 6x^2 dx = 6 \left(\frac{x^3}{3}\right) = 2x^3$

  • Second term:     $\int -4x^{-1/2} dx = -4 \left(\frac{x^{1/2}}{1/2}\right) = -4 \cdot 2\sqrt{x} = -8\sqrt{x}$

  • Third term:     $\int 3 dx = 3x$

• Include the arbitrary constant of integration $+ C$:   $2x^3 - 8\sqrt{x} + 3x + C$

• Multiple choice classification:

  • (A) $2x^3 - 8\sqrt{x} + 3x + C$ (Correct)

  • (B) $2x^3 + 8\sqrt{x} + 3x + C$ (Incorrect sign on radical term)

  • (C) $2x^3 - 8\sqrt{x} + C$ (Omission of linear integral term $3x$)

  • (D) $2x^3 + 8\sqrt{x} + C$ (Omission of linear term and erroneous sign)

Riemann Sums and Monotonic Tabular Approximations

• Upper Riemann Sum Calculation from Tabulated Data:

  • Problem: A neighborhood extends along a road with population density $r(x)$ in residents per kilometer at distance $x$ in kilometers from the west end. On each consecutive interval, $r(x)$ is strictly increasing throughout or strictly decreasing throughout:

  • $x_0 = 0\text{ km}$, $r(0) = 900\text{ residents/km}$

  • $x_1 = 1\text{ km}$, $r(1) = 600\text{ residents/km}$

  • $x_2 = 3\text{ km}$, $r(2) = 800\text{ residents/km}$

  • $x_3 = 4\text{ km}$, $r(3) = 500\text{ residents/km}$

  • $x_4 = 6\text{ km}$, $r(4) = 700\text{ residents/km}$

  • Subinterval width and monotonicity analysis:

  • Subinterval $[0, 1]$: Width $\Delta x_1 = 1 - 0 = 1\text{ km}$. The function decreases from $900$ to $600$. The supremum occurs at the left endpoint:       $\max_{[0, 1]} r(x) = r(0) = 900\text{ residents/km}$       Contribution: $900 \cdot 1$

  • Subinterval $[1, 3]$: Width $\Delta x_2 = 3 - 1 = 2\text{ km}$. The function increases from $600$ to $800$. The supremum occurs at the right endpoint:       $\max_{[1, 3]} r(x) = r(3) = 800\text{ residents/km}$       Contribution: $800 \cdot 2$

  • Subinterval $[3, 4]$: Width $\Delta x_3 = 4 - 3 = 1\text{ km}$. The function decreases from $800$ to $500$. The supremum occurs at the left endpoint:       $\max_{[3, 4]} r(x) = r(3) = 800\text{ residents/km}$       Contribution: $800 \cdot 1$

  • Subinterval $[4, 6]$: Width $\Delta x_4 = 6 - 4 = 2\text{ km}$. The function increases from $500$ to $700$. The supremum occurs at the right endpoint:       $\max_{[4, 6]} r(x) = r(6) = 700\text{ residents/km}$       Contribution: $700 \cdot 2$

  • Formulating the Upper Sum:     $\text{Upper Sum} = 900 \cdot 1 + 800 \cdot 2 + 800 \cdot 1 + 700 \cdot 2$

  • Answer: Option (D).

Work, Vector Displacement, and Properties of Definite Integrals

• Definite Integral Additivity and Directed Orientation:

  • Problem: A continuous variable force $F(x)$ in newtons ($\text{N}$) operates along the $x$-axis ($x$ in meters). Given:     $\int_1^4 F(x) dx = 7\text{ J}$     $\int_1^6 F(x) dx = 12\text{ J}$

  • Work performed by the force as the object is displaced from $x = 6\text{ m}$ to $x = 4\text{ m}$ is given by:     $W = \int_6^4 F(x) dx$

  • Decomposition via integral additivity:     $\int_1^6 F(x) dx = \int_1^4 F(x) dx + \int_4^6 F(x) dx$     $12 = 7 + \int_4^6 F(x) dx \implies \int_4^6 F(x) dx = 5\text{ J}$

  • Inversion of limits of integration reverses sign:     $\int_6^4 F(x) dx = -\int_4^6 F(x) dx = -5\text{ J}$

  • Multiple choice classification:

  • (A) $-19\text{ joules}$

  • (B) $-5\text{ joules}$ (Correct)

  • (C) $5\text{ joules}$

  • (D) $19\text{ joules}$

Mean Value Theorem and Interpretations of Integral Rates

• Average Rate of Change versus Average Value:

  • Problem: City electricity demand is defined as $P(t)$ megawatts, where $t$ is hours past midnight. Define the contextual meaning of:     $\frac{1}{6} \int_6^{12} P'(t) dt$

  • Analytical reduction using the Fundamental Theorem of Calculus:     $\int_6^{12} P'(t) dt = P(12) - P(6)$     $\frac{1}{12 - 6} \int_6^{12} P'(t) dt = \frac{P(12) - P(6)}{12 - 6}$

  • Conceptual differentiation:

  • $\frac{1}{b - a} \int_a^b f(t) dt$ represents the average value of function $f(t)$.

  • Applying this definition to $f(t) = P'(t)$ gives the average value of the rate of change of electricity demand over $t \in [6, 12]$, which is equivalent to the average rate of change of electricity demand from 6 a.m. to noon, measured in megawatts per hour ($\text{MW/h}$.

  • Contrast with distractors:

  • Average electricity demand: $\frac{1}{6} \int_6^{12} P(t) dt$ (megawatts).

  • Total energy consumed: $\int_6^{12} P(t) dt$ (megawatt-hours).

  • Net change in demand: $\int_6^{12} P'(t) dt = P(12) - P(6)$ (megawatts).

  • Answer: Option (C).

Modeling Physical Dynamics with Differential Equations

• Oscillatory Mechanics and Restoring Acceleration:

  • Problem: A mass oscillates about equilibrium with signed displacement $y(t)$. Acceleration is proportional to displacement and points directly toward the equilibrium position ($k > 0$).

  • Mathematical representation:

  • Displacement: $y(t)$

  • Acceleration: $y''(t)$

  • Magnitude relationship: $|y''(t)| \propto |y(t)|$

  • Restoring directional dynamic: When displacement is positive ($y > 0$), acceleration acts in the opposite direction ($y'' < 0$). When displacement is negative ($y < 0$), acceleration acts in the positive direction ($y'' > 0$).

  • Differential equation governing simple harmonic motion:       $y'' = -ky$

  • Multiple choice classification:

  • (A) $y' = ky$ (First-order runaway growth)

  • (B) $y'' = ky$ (Second-order unstable exponential runaway)

  • (C) $y' = -ky$ (First-order exponential decay)

  • (D) $y'' = -ky$ (Correct oscillatory model)

Direction Fields and Geometric Differential Analysis

• Slope Field Identification for First-Order Systems:

  • Problem: Determine which differential equation generates the given directional slope field.

• Evaluation across coordinate quadrants:

  • In Quadrant I ($x > 0$, $y > 0$): Slopes are negative.

  • In Quadrant II ($x < 0$, $y > 0$): Slopes are positive.

  • In Quadrant III ($x < 0$, $y < 0$): Slopes are negative.

  • In Quadrant IV ($x > 0$, $y < 0$): Slopes are positive.

• Boundary and axis behavior:

  • As $y \rightarrow 0$, the tangent line segments become vertical ($\frac{dy}{dx} \rightarrow \pm\infty$).

  • As $x \rightarrow 0$, the tangent line segments become horizontal ($\frac{dy}{dx} \rightarrow 0$).

• Solution curves:

  • The slope segments are tangent to concentric circles centered at the origin: $x^2 + y^2 = C$.

  • Implicit differentiation of $x^2 + y^2 = C$:     $2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$

• Choice comparison:

  • (A) $\frac{dy}{dx} = \frac{x}{y}$ (Yields positive slopes in Quad I and hyperbolic trajectories)

  • (B) $\frac{dy}{dx} = -\frac{x}{y}$ (Correct)

  • (C) $\frac{dy}{dx} = \frac{y}{x}$ (Yields radial straight-line slopes from origin)

  • (D) $\frac{dy}{dx} = -\frac{y}{x}$ (Yields negative slopes in Quad I, but vertical slopes on $x = 0$ and horizontal on $y = 0$

Logarithmic Differentiation and Exponential Variables

• Derivative of Variable Base with Variable Exponent:

  • Problem: Compute $\frac{dy}{dx}$ for $y = x^{\sqrt{x}}$ on the domain $x > 0$.

  • Step 1: Take the natural logarithm of both sides:     $\ln(y) = \ln\left(x^{\sqrt{x}}\right) = \sqrt{x} \ln(x) = x^{1/2} \ln(x)$

  • Step 2: Differentiate implicitly with respect to $x$ applying the product rule:     $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}\left[x^{1/2}\right] \ln(x) + x^{1/2} \frac{d}{dx}[\ln(x)]$     $\frac{1}{y} \frac{dy}{dx} = \frac{1}{2}x^{-1/2} \ln(x) + x^{1/2} \left(\frac{1}{x}\right)$     $\frac{1}{y} \frac{dy}{dx} = \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}}$

  • Step 3: Establish a common denominator inside the factor:     $\frac{1}{y} \frac{dy}{dx} = \frac{\ln(x) + 2}{2\sqrt{x}}$

  • Step 4: Multiply through by the original function $y$:     $\frac{dy}{dx} = x^{\sqrt{x}} \left(\frac{\ln(x) + 2}{2\sqrt{x}}\right) = \frac{x^{\sqrt{x}}(\ln(x) + 2)}{2\sqrt{x}}$

  • Multiple choice classification:

  • (A) $\sqrt{x} \cdot x^{\sqrt{x}-1}$ (Faulty application of standard power rule)

  • (B) $\frac{x^{\sqrt{x}}\ln(x)}{2\sqrt{x}}$ (Omission of product rule derivative term)

  • (C) $\frac{x^{\sqrt{x}}(\ln(x) + 1)}{2\sqrt{x}}$ (Algebraic error in fraction addition)

  • (D) $\frac{x^{\sqrt{x}}(\ln(x) + 2)}{2\sqrt{x}}$ (Correct)

Price Elasticity and Marginal Demand Approximation

• Linear Approximation of Exponential Demand Changes:

  • Problem: Product demand is modeled by $q(p) = 1200(0.9)^p$ units at price $p$ dollars. At $p = 5$, evaluate the anticipated percentage shift in demand resulting from a $1\%$ increase in price.

  • Marginal derivative computation:     $q'(p) = 1200(0.9)^p \ln(0.9) = q(p) \ln(0.9)$

  • Price increment:     $\Delta p = 0.01 \times p = 0.01 \times 5 = 0.05\text{ dollars}$

  • First-order differential approximation for absolute demand change:     $\Delta q \approx q'(p) \Delta p = q(p) \ln(0.9) \Delta p$

  • Relative and percentage demand change:     $\frac{\Delta q}{q(p)} \approx \ln(0.9) \Delta p = \ln(0.9) \cdot 0.05$

  • Numerical evaluation of $\ln(0.9)$:       $\ln(0.9) \approx -0.1053605$

  • Relative change:       $\frac{\Delta q}{q} \approx -0.1053605 \times 0.05 = -0.005268$

  • Percentage change:       $\% \Delta q \approx -0.5268\% \approx -0.53\%$

  • Direct Elasticity Formulation:     $E = \frac{p}{q} \frac{dq}{dp} = p \ln(0.9)$     At $p = 5$:     $E = 5 \ln(0.9) \approx 5(-0.10536) = -0.5268$     $\% \Delta q \approx E \times (\% \Delta p) = -0.5268 \times 1\% \approx -0.53\%$

  • Answer: Option (C) (Demand decreases by approximately $0.53\%$).

Non-Uniform Linear Density and Riemann Sums of a Metal Bar

• Geometric Density Profile of a 3-Meter Metal Bar:

• Total length: $L = 3\text{ m}$ spanning $x \in [0, 3]$.

• Variable linear density: $r(x)$ in kilograms per meter ($\text{kg/m}$.

• Partition into $n = 6$ equal subintervals:     $\Delta x = \frac{b - a}{n} = \frac{3 - 0}{6} = 0.5\text{ m}$

• Coordinate grid values extracted from the density profile:

  • $r(0.0) = 2\text{ kg/m}$

  • $r(0.5) = 3\text{ kg/m}$

  • $r(1.0) = 6\text{ kg/m}$

  • $r(1.5) = 7\text{ kg/m}$ (Peak density)

  • $r(2.0) = 6\text{ kg/m}$

  • $r(2.5) = 4\text{ kg/m}$

  • $r(3.0) = 3\text{ kg/m}$

  • Part (a): Upper and Lower Riemann Sum Construction:

• Subinterval extrema identification:

  • Interval $[0, 0.5]$: $\min = r(0) = 2$, $\max = r(0.5) = 3$

  • Interval $[0.5, 1.0]$: $\min = r(0.5) = 3$, $\max = r(1.0) = 6$

  • Interval $[1.0, 1.5]$: $\min = r(1.0) = 6$, $\max = r(1.5) = 7$

  • Interval $[1.5, 2.0]$: $\min = r(2.0) = 6$, $\max = r(1.5) = 7$

  • Interval $[2.0, 2.5]$: $\min = r(2.5) = 4$, $\max = r(2.0) = 6$

  • Interval $[2.5, 3.0]$: $\min = r(3.0) = 3$, $\max = r(2.5) = 4$

• Lower Sum Calculation ($L_6$):     $L_6 = \Delta x \sum_{i=1}^6 \min_i = 0.5 \cdot (2 + 3 + 6 + 6 + 4 + 3) = 0.5 \cdot 24 = 12\text{ kg}$

• Upper Sum Calculation ($U_6$):     $U_6 = \Delta x \sum_{i=1}^6 \max_i = 0.5 \cdot (3 + 6 + 7 + 7 + 6 + 4) = 0.5 \cdot 33 = 16.5\text{ kg}$

• Physical Contextual Interpretation:

  • The definite integral $\int_0^3 r(x) dx$ denotes the total mass of the bar in kilograms.

  • The lower sum ($12\text{ kg}$) gives a guaranteed lower bound (underestimate), and the upper sum ($16.5\text{ kg}$) gives a guaranteed upper bound (overestimate) of the bar's true physical mass $M$:       $12\text{ kg} \le M \le 16.5\text{ kg}$

  • Part (b): Subinterval Refinement Effects:

• If each subinterval is divided into smaller subdivisions, the upper sum decreases (or stays constant) and the lower sum increases (or stays constant).

• Mathematical justification: Over any refined sub-partition, the supremum over smaller subsets cannot exceed the supremum over the original set, and the infimum cannot be less than the original infimum. Thus, the upper and lower sums squeeze closer together, progressively bounding the exact total mass of the bar with decreasing error.

  • Part (c): Evaluation and Interpretation of Derivative Integral:

• Analytical evaluation via the Net Change Theorem (Fundamental Theorem of Calculus Part 2):     $\int_0^3 r'(x) dx = r(3) - r(0) = 3\text{ kg/m} - 2\text{ kg/m} = 1\text{ kg/m}$

• Contextual interpretation: This represents the net change in the bar's linear density from the origin ($x = 0\text{ m}$) to its opposite end ($x = 3\text{ m}$). The bar is $1\text{ kg/m}$ denser at $x = 3\text{ m}$ than at $x = 0\text{ m}$.

Kinematics, Net Displacement, and Geometric Integration

• Velocity Profile and Geometric Trajectory:

• Motion characteristics: Cyclist moving linearly east-west over $t \in [0, 12]$ seconds.

• Conventions: Eastward velocity is positive ($v(t) > 0$); westward velocity is negative ($v(t) < 0$).

• Initial position: At $t = 0\text{ s}$, cyclist is located $5\text{ m}$ east of the reference landmark ($s(0) = +5\text{ m}$.

• Component geometric breakdown:

  • Segment 1 ($0 \le t \le 4$): Triangular region with base $4 - 0 = 4\text{ s}$ and peak height $4\text{ m/s}$ at $t = 2\text{ s}$.       $\text{Area}_1 = \frac{1}{2} \times 4 \times 4 = 8\text{ m}$

  • Segment 2 ($4 \le t \le 8$): Semicircular region beneath the $t$-axis with diameter spanning from $t = 4\text{ s}$ to $t = 8\text{ s}$ (diameter $4\text{ s}$, radius $R = 2\text{ s}$) and minimum velocity $-2\text{ m/s}$ at $t = 6\text{ s}$.       $\text{Area}_2 = \frac{1}{2} \pi R^2 = \frac{1}{2} \pi (2)^2 = 2\pi\text{ m}$

  • Segment 3 ($8 \le t \le 12$): Triangular region with base $12 - 8 = 4\text{ s}$ and peak height $2\text{ m/s}$ at $t = 10\text{ s}$.       $\text{Area}_3 = \frac{1}{2} \times 4 \times 2 = 4\text{ m}$

  • Part (a): Net Displacement Evaluation:

• Signed integral calculation:     $\int_0^{12} v(t) dt = \text{Area}_1 - \text{Area}_2 + \text{Area}_3 = 8 - 2\pi + 4 = 12 - 2\pi\text{ m} \approx 5.72\text{ m}$

• Contextual interpretation: The net change in position (displacement) of the cyclist between $t = 0\text{ s}$ and $t = 12\text{ s}$ is $12 - 2\pi\text{ meters}$ (approximately $5.72\text{ meters}$ east).

  • Part (b): Extreme Eastward Position and Temporal Location:

• Extreme value analysis on the closed interval $[0, 12]$ for position $s(t) = s(0) + \int_0^t v(\tau) d\tau$:

  • Critical points occur when $v(t) = 0$: $t = 0, 4, 8, 12$.

  • Evaluation at critical values and boundaries:

    • $s(0) = 5\text{ m}$

    • $s(4) = 5 + 8 = 13\text{ m}$

    • $s(8) = 13 - 2\pi \approx 13 - 6.28 = 6.72\text{ m}$

    • $s(12) = 6.72 + 4 = 10.72\text{ m}$ (or $17 - 2\pi\text{ m}$

• Justification via derivative signs:

  • For $t \in (0, 4)$, $v(t) > 0$ (moving east).

  • For $t \in (4, 8)$, $v(t) < 0$ (moving west).

  • For $t \in (8, 12)$, $v(t) > 0$ (moving east).

  • Because the westward distance traveled during $t \in [4, 8]$ ($2\pi \approx 6.28\text{ m}$) exceeds the subsequent eastward distance traveled during $t \in [8, 12]$ ($4\text{ m}$), the absolute maximum position is attained at the end of the first eastward stage.

• Result: The cyclist is farthest east at $t = 4\text{ seconds}$, at a location $13\text{ meters}$ east of the landmark.

  • Part (c): Total Distance Traveled:

• Total distance corresponds to the definite integral of speed ($|v(t)|$):     $\text{Total Distance} = \int_0^{12} |v(t)| dt = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 = 8 + 2\pi + 4 = 12 + 2\pi\text{ m} \approx 18.28\text{ meters}$

Periodic Temperature Modeling and Trigonometric Calculus

• Mathematical Model of Climate-Controlled Greenhouse:

  • Temperature equation: $T(t) = 20 + 4\sin\left(\frac{\pi t}{12}\right)$ for $0 \le t \le 24$

  • Units: $T$ in degrees Celsius ($^\circ\text{C}$), $t$ in hours past midnight.

• Part (a): Curve Sketching and Extrema Analysis:

  • Fundamental periodic parameters:

  • Vertical baseline (midline): $T = 20^\circ\text{C}$

  • Amplitude: $A = 4^\circ\text{C}$

  • Angular frequency: $\omega = \frac{\pi}{12}\text{ rad/h}$

  • Period: $P = \frac{2\pi}{\omega} = \frac{2\pi}{\pi / 12} = 24\text{ hours}$

  • Principal time-point values:

  • Midnight ($t = 0\text{ h}$): $T(0) = 20 + 4\sin(0) = 20^\circ\text{C}$

  • Morning ($t = 6\text{ h}$): $T(6) = 20 + 4\sin\left(\frac{\pi}{2}\right) = 24^\circ\text{C}$ (Global Maximum Temperature)

  • Noon ($t = 12\text{ h}$): $T(12) = 20 + 4\sin(\pi) = 20^\circ\text{C}$

  • Evening ($t = 18\text{ h}$): $T(18) = 20 + 4\sin\left(\frac{3\pi}{2}\right) = 16^\circ\text{C}$ (Global Minimum Temperature)

  • Next Midnight ($t = 24\text{ h}$): $T(24) = 20 + 4\sin(2\pi) = 20^\circ\text{C}$

• Part (b): Instantaneous Rate of Temperature Change:

  • Differentiation via the chain rule:     $T'(t) = 4 \cos\left(\frac{\pi t}{12}\right) \cdot \frac{d}{dt}\left[\frac{\pi t}{12}\right] = 4 \left(\frac{\pi}{12}\right) \cos\left(\frac{\pi t}{12}\right) = \frac{\pi}{3} \cos\left(\frac{\pi t}{12}\right)$

  • Evaluation at $t = 9\text{ hours}$ (9:00 a.m.):     $T'(9) = \frac{\pi}{3} \cos\left(\frac{9\pi}{12}\right) = \frac{\pi}{3} \cos\left(\frac{3\pi}{4}\right) = \frac{\pi}{3} \left(-\frac{\sqrt{2}}{2}\right) = -\frac{\pi\sqrt{2}}{6}^\circ\text{C/hour}$

  • Numerical approximation:     $T'(9) \approx -\frac{3.14159 \times 1.4142}{6} \approx -0.74^\circ\text{C/hour}$

  • Contextual interpretation: At 9:00 a.m., the greenhouse temperature is decreasing at an instantaneous rate of approximately $0.74^\circ\text{C per hour}$.

• Part (c): Average Temperature over a 12-Hour Interval:

  • Application of the mean value formulation for integrals:     $T_{\text{avg}} = \frac{1}{12 - 0} \int_0^{12} \left(20 + 4\sin\left(\frac{\pi t}{12}\right)\right) dt$

  • Antiderivative integration:     $\int \left(20 + 4\sin\left(\frac{\pi t}{12}\right)\right) dt = 20t - 4\left(\frac{12}{\pi}\right) \cos\left(\frac{\pi t}{12}\right) = 20t - \frac{48}{\pi} \cos\left(\frac{\pi t}{12}\right)$

  • Boundary evaluation from $0$ to $12$:     $\left[ 20(12) - \frac{48}{\pi}\cos(\pi) \right] - \left[ 20(0) - \frac{48}{\pi}\cos(0) \right]$     $\left( 240 - \frac{48}{\pi}(-1) \right) - \left( -\frac{48}{\pi}(1) \right) = 240 + \frac{48}{\pi} + \frac{48}{\pi} = 240 + \frac{96}{\pi}$

  • Division by interval length ($12\text{ h}$):     $T_{\text{avg}} = \frac{1}{12} \left(240 + \frac{96}{\pi}\right) = 20 + \frac{8}{\pi}^\circ\text{C} \approx 20 + 2.55 = 22.55^\circ\text{C}$

  • Contextual interpretation: The average temperature inside the greenhouse over the first 12 hours of the day (from midnight until noon) is $20 + \frac{8}{\pi}^\circ\text{C}$ (approximately $22.55^\circ\text{C}$).

First-Order Mixing Differential Equations

• Physical System Parameters:

  • Initial fluid volume: $V(0) = 5000\text{ liters}$

  • Initial dissolved solute mass: $S(0) = 20\text{ kg}$

  • Inflow brine concentration: $c_{\text{in}} = 0.03\text{ kg/L}$

  • Volumetric inflow rate: $r_{\text{in}} = 25\text{ L/min}$

  • Volumetric outflow rate: $r_{\text{out}} = 25\text{ L/min}$

  • Dynamic variable: $S(t)$ denotes the mass of salt in kilograms at time $t$ minutes.

• Part (a): Derivation of the Governing Rate Differential Equation:

  • Volumetric conservation:     $\frac{dV}{dt} = r_{\text{in}} - r_{\text{out}} = 25\text{ L/min} - 25\text{ L/min} = 0\text{ L/min}$     Because the net rate of volume change is zero, the fluid volume remains invariant at $V(t) = 5000\text{ L}$ for all $t \ge 0$.

  • Solute entry rate:     $\text{Rate}_{\text{in}} = c_{\text{in}} \times r_{\text{in}} = (0.03\text{ kg/L}) \times (25\text{ L/min}) = 0.75\text{ kg/min}$

  • Solute exit rate:

  • With continuous and thorough mixing, the instantaneous concentration of salt leaving the tank matches the uniform tank concentration:       $c_{\text{out}}(t) = \frac{S(t)}{V(t)} = \frac{S(t)}{5000}\text{ kg/L}$

  • The rate at which salt exits is:       $\text{Rate}_{\text{out}} = c_{\text{out}}(t) \times r_{\text{out}} = \left(\frac{S(t)}{5000}\text{ kg/L}\right) \times (25\text{ L/min}) = \frac{25S(t)}{5000}\text{ kg/min} = \frac{S(t)}{200}\text{ kg/min}$

  • Solute mass balance equation:     $\frac{dS}{dt} = \text{Rate}_{\text{in}} - \text{Rate}_{\text{out}} = 0.75 - \frac{S}{200}$     Both terms carry units of kilograms per minute ($\text{kg/min}$.

• Part (b): Solution Verification for Candidate Model:

  • Proposed candidate function: $S(t) = 150 - 130e^{-t/100}$

  • Verification of initial condition:     $S(0) = 150 - 130e^0 = 150 - 130 = 20\text{ kg}$     The candidate satisfies the initial state $S(0) = 20$.

  • Substitution into left-hand side ($\text{LHS}$) of differential equation:     $\frac{dS}{dt} = \frac{d}{dt}\left[150 - 130e^{-t/100}\right] = -130 \left(-\frac{1}{100}\right) e^{-t/100} = 1.3e^{-t/100}$

  • Substitution into right-hand side ($\text{RHS}$) of differential equation:     $0.75 - \frac{S}{200} = 0.75 - \frac{150 - 130e^{-t/100}}{200} = 0.75 - 0.75 + \frac{130}{200}e^{-t/100} = 0.65e^{-t/100}$

  • Equivalence check:     $\text{LHS} = 1.3e^{-t/100} \neq 0.65e^{-t/100} = \text{RHS}$

  • Conclusion: The proposed function fails to satisfy the differential equation because the decay constant in the exponent is incorrect (yielding an erroneous time scale).

• Part (c): Analytical Solution via Separation of Variables:

  • Step 1: Algebraic restructuring of the differential equation:     $\frac{dS}{dt} = \frac{150 - S}{200}$

  • Step 2: Separate variables:     $\frac{1}{150 - S} dS = \frac{1}{200} dt$

  • Step 3: Integrate both sides:     $\int \frac{1}{150 - S} dS = \int \frac{1}{200} dt$     $-\ln|150 - S| = \frac{t}{200} + C_1$     $\ln|150 - S| = -\frac{t}{200} - C_1$

  • Step 4: Exponentiate both sides:     $|150 - S| = e^{-C_1} e^{-t/200}$     $150 - S = A e^{-t/200} \quad (\text{where } A = \pm e^{-C_1})$     $S(t) = 150 - A e^{-t/200}$

  • Step 5: Enforce the initial boundary condition $S(0) = 20$:     $20 = 150 - A e^0 \implies 20 = 150 - A \implies A = 130$

  • Step 6: Formulate the particular solution:     $S(t) = 150 - 130e^{-t/200}$

  • Long-term limiting behavior:     $\lim_{t \rightarrow \infty} S(t) = 150\text{ kg}$     Consistent with terminal equilibrium: $5000\text{ L} \times 0.03\text{ kg/L} = 150\text{ kg}$.