Compartment Models - Differential Equations

1. Consider a system with two compartments, A and B, with amounts $x(t)$ and $y(t)$ and constant volumes $V_A$ and $V_B$. A substance flows from A to B at a rate of 10 L/min. Part of the substance is then recycled from B back to A at a rate of 3 L/min. An external source adds the substance to A at a rate of 7 L/min, and the substance is removed from B at a rate of 4 L/min. Which pair of terms correctly represents the transfer of the substance between A and B in the differential equation for $\frac{dx}{dt}$?
   Answer: $-10 \frac{x}{V_A} + 3 \frac{y}{V_B}$ When modeling compartment systems, you need to carefully track both flow rates and concentrations. The key insight is that substance transfer depends on concentration (amount per unit volume), not just the total amount in each compartment. For compartment A, you're looking at how the transfer between compartments affects $\frac{dx}{dt}$. Substance flows out of A to B at 10 L/min, so the rate of substance leaving A is $10 \times \frac{x}{V_A}$ (flow rate times concentration in A). This represents a loss, so it's negative: $-10\frac{x}{V_A}$. Meanwhile, substance is recycled back from B to A at 3 L/min, giving a rate of substance entering A of $3 \times \frac{y}{V_B}$ (flow rate times concentration in B). This represents a gain, so it's positive: $+3\frac{y}{V_B}$. Answer D correctly captures both terms: $-10\frac{x}{V_A} + 3\frac{y}{V_B}$. Answer A treats amounts directly without considering volumes, ignoring that flow depends on concentration, not total amount. Answer B has the wrong sign on the recycling term—it should be positive since substance flows into A from B. Answer C makes both the flow from A to B and the recycling from B to A appear as losses to compartment A, which contradicts the physical setup where recycling adds substance back to A. Remember: in compartment models, always divide amounts by volumes to get concentrations, and carefully track the direction of each flow to determine the correct signs.
2. A factory waste treatment system has four interconnected settling tanks. Contaminated water enters Tank 1 at 100 L/h with pollutant concentration 500 mg/L. Water flows from Tank 1 to Tank 2 at 80 L/h, from Tank 2 to Tank 3 at 60 L/h, from Tank 3 to Tank 4 at 40 L/h, and exits Tank 4 at 40 L/h. Each tank has overflow pipes that maintain constant volume. At steady state, if the pollutant concentration decreases by 30% in each tank due to settling, what is the pollutant removal efficiency of the entire system?
   Answer: 75.99% If each tank reduces pollutant concentration by 30%, then 70% of the pollutant concentration remains after each tank. Starting with 500 mg/L: after Tank 1, concentration is $500 \times 0.7 = 350$ mg/L; after Tank 2, concentration is $350 \times 0.7 = 245$ mg/L; after Tank 3, concentration is $245 \times 0.7 = 171.5$ mg/L; after Tank 4, concentration is $171.5 \times 0.7 = 120.05$ mg/L. The overall removal efficiency is $\frac{500 - 120.05}{500} \times 100\% = \frac{379.95}{500} \times 100\% = 75.99\%$. Alternatively, the remaining fraction is $0.7^4 = 0.2401$, so removal efficiency is $(1 - 0.2401) \times 100\% = 75.99\%$. Choice B (82.40%) would result from $(1 - 0.7^3) \times 100\% = 65.7\%$ or some other calculation error. Choice C might come from using an arithmetic average. Choice D is too high for this level of treatment.
3. The system of differential equations below models the amount of a substance in two compartments, $A$ and $B$, in milligrams (mg), over time $t$ in minutes. $\begin{cases} \frac{dx_A}{dt} = -0.4x_A + 0.1x_B + 7 \\ \frac{dx_B}{dt} = 0.4x_A - 0.3x_B \end{cases}$ What is the correct physical interpretation of the term $0.4x_A$ in this model?
   Answer: The rate at which the substance flows from compartment $A$ to compartment $B$. In compartment models, a term of the form $k x_i$ represents a rate of transfer proportional to the amount $x_i$ in compartment $i$. The term $0.4x_A$ appears with a negative sign in the equation for $\frac{dx_A}{dt}$ (as part of $-0.4x_A$) and with a positive sign in the equation for $\frac{dx_B}{dt}$. This indicates that this term corresponds to a flow of substance out of compartment $A$ and into compartment $B$. Therefore, $0.4x_A$ represents the rate at which the substance flows from compartment $A$ to compartment $B$.
4. A tank with a capacity of 500 liters initially contains 200 liters of pure water. A salt solution with concentration 0.1 kg/L is pumped in at a rate of 10 L/min. The well-mixed solution is pumped out at a rate of 5 L/min. Let $S(t)$ be the amount of salt in the tank at time $t$. Which initial value problem correctly models this process?
   Answer: $\frac{dS}{dt} = 1 - \frac{S}{40+t}, \quad S(0) = 0$ When you encounter a mixing problem like this, you need to set up a differential equation based on the principle: rate of change of salt = rate of salt entering - rate of salt leaving. Let's work through this systematically. First, determine the volume of solution in the tank at time $t$. You start with 200 L, gain 10 L/min, and lose 5 L/min, so the net gain is 5 L/min. Therefore, the volume at time $t$ is $200 + 5t$ liters. For the rate equation: Salt enters at $(0.1 \text{ kg/L})(10 \text{ L/min}) = 1$ kg/min. Salt leaves at a rate equal to the concentration times the outflow rate. The concentration at time $t$ is $\frac{S(t)}{200+5t}$ kg/L, so salt leaves at $\frac{S(t)}{200+5t} \times 5 = \frac{5S}{200+5t}$ kg/min. This gives us $\frac{dS}{dt} = 1 - \frac{5S}{200+5t}$ with $S(0) = 0$ (initially pure water). Choice A incorrectly assumes constant volume (200 L) and wrong outflow calculation. Choice C has the correct volume term but is missing from our options - wait, let me recheck. Actually, choice B simplifies correctly: $\frac{dS}{dt} = 1 - \frac{5S}{200+5t} = 1 - \frac{S}{\frac{200+5t}{5}} = 1 - \frac{S}{40+t}$. Choice D uses the wrong inflow rate (should be 1, not 0.1). Study tip: Always track the changing volume in mixing problems, and remember that concentration equals amount divided by volume at any given time.
5. A system of two compartments is described by the differential equations: $\begin{cases} \frac{dx}{dt} = 15 - 5x + y \\ \frac{dy}{dt} = 3x - 3y \end{cases}$ As $t \to \infty$, the system approaches a stable equilibrium state $(x_e, y_e)$. What is the ratio $x_e / y_e$ at this equilibrium?
   Answer: $1$ At equilibrium, the rates of change are zero: $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} = 0$. From the second equation, $3x - 3y = 0$, which simplifies to $3x = 3y$, or $x=y$. This directly implies that the ratio $x_e / y_e = 1$. We can also solve for the specific values to confirm. Substituting $x=y$ into the first equilibrium equation gives $15 - 5x + x = 0$, which means $15 - 4x = 0$, so $x_e = 15/4$. Since $y_e=x_e$, $y_e = 15/4$. The ratio is $(15/4) / (15/4) = 1$.
6. A pharmaceutical drug is administered to a patient. Let $x(t)$ be the amount of the drug in the gastrointestinal (GI) tract and $y(t)$ be the amount in the bloodstream. The drug moves from the GI tract to the blood at a rate proportional to the amount in the GI tract. It is eliminated from the blood at a rate proportional to the amount in the blood. Which system of differential equations is a plausible model for this process, assuming $k_1, k_2 > 0$ are the rate constants?
   Answer: $\begin{cases} \frac{dx}{dt} = -k_1 x \\ \frac{dy}{dt} = k_1 x - k_2 y \end{cases}$ When modeling drug kinetics with differential equations, you need to carefully track the rates of change based on the physical processes described. Think about what causes each compartment (GI tract and bloodstream) to gain or lose drug over time. For the GI tract, $x(t)$: The only process affecting it is drug moving out to the bloodstream at rate $k_1x$. Since drug is leaving, this creates a negative rate of change: $\frac{dx}{dt} = -k_1x$. For the bloodstream, $y(t)$: Two processes are occurring simultaneously. Drug enters from the GI tract at rate $k_1x$ (positive contribution) and is eliminated at rate $k_2y$ (negative contribution). This gives: $\frac{dy}{dt} = k_1x - k_2y$. Answer D correctly captures both equations: $\frac{dx}{dt} = -k_1x$ and $\frac{dy}{dt} = k_1x - k_2y$. Looking at the incorrect options: Answer A has $\frac{dx}{dt} = k_1x - k_2y$, incorrectly suggesting the GI tract gains drug and is affected by blood elimination. Answer B shows $\frac{dx}{dt} = k_1x$, implying the GI tract somehow gains drug over time, which contradicts the one-way transfer described. Answer C has $\frac{dy}{dt} = k_1x$ without the elimination term, ignoring that drug is removed from the bloodstream. Study tip: In compartmental models, always check that your signs make physical sense—what processes add to each compartment versus what processes remove from it. The conservation principle also helps: drug leaving one compartment should enter another at the same rate.
7. The general solution for the amount of a substance $x(t)$ in a certain compartment is given by $x(t) = 15e^{-0.9t} - 4e^{-0.2t} + 25$, where $t \ge 0$. What is the initial amount of the substance in the compartment, $x(0)$?
   Answer: $36$ To find the initial amount of the substance, we must evaluate the function $x(t)$ at time $t=0$. Substituting $t=0$ into the expression: $x(0) = 15e^{-0.9(0)} - 4e^{-0.2(0)} + 25$. Since $e^0 = 1$, this simplifies to: $x(0) = 15(1) - 4(1) + 25 = 15 - 4 + 25 = 11 + 25 = 36$. The value $25$ represents the long-term equilibrium amount, not the initial amount. The value $11$ is an intermediate calculation ($15-4$).
8. A medication is introduced into the bloodstream (compartment $x_1$) and subsequently absorbed by body tissues (compartment $x_2$). The system is modeled by: $\begin{cases} \frac{dx_1}{dt} = 10 - 0.5x_1 \\ \frac{dx_2}{dt} = 0.5x_1 - 0.1x_2 \end{cases}$ where amounts are in mg and time is in hours. The term $10$ represents a constant intravenous infusion rate. Let $T(t) = x_1(t) + x_2(t)$ be the total amount of medication in the body. Which differential equation governs $T(t)$?
   Answer: $\frac{dT}{dt} = 10 - 0.1x_2$ When analyzing coupled differential equation systems, you need to understand how quantities flow between compartments and what happens to the total amount in the entire system. To find the differential equation for $T(t) = x_1(t) + x_2(t)$, differentiate both sides: $\frac{dT}{dt} = \frac{dx_1}{dt} + \frac{dx_2}{dt}$ Substituting the given equations: $\frac{dT}{dt} = (10 - 0.5x_1) + (0.5x_1 - 0.1x_2)$ Notice that the $0.5x_1$ terms cancel out: $\frac{dT}{dt} = 10 - 0.1x_2$ This makes physical sense: medication enters the system at rate 10 (intravenous infusion) and leaves only from compartment $x_2$ at rate $0.1x_2$. The transfer between compartments ($0.5x_1$) doesn't affect the total amount. Answer A incorrectly assumes medication is eliminated from the total at rate $0.1T$, but elimination only occurs from compartment $x_2$, not proportionally from the total amount. Answer B uses the wrong coefficient (0.6 instead of 0.1) and incorrectly applies it to the total. Answer D simply adds both original equations without recognizing that internal transfers cancel out when considering the total. Study tip: In compartment models, when finding the rate of change of total quantity, internal transfer terms always cancel out. Only external inputs and outputs affect the total amount in the system.
9. A pharmacokinetic model describes drug absorption and elimination using two compartments: gut (G) and blood (B). Drug transfers from gut to blood at rate $k_a G$ and is eliminated from blood at rate $k_e B$. If a 100 mg dose is given orally at $t = 0$ and $k_a = 0.8 \text{ h}^{-1}$, $k_e = 0.2 \text{ h}^{-1}$, what is the time when blood concentration reaches its maximum?
   Answer: $t = \frac{\ln(4)}{0.6}$ hours The system equations are $\frac{dG}{dt} = -k_a G$ and $\frac{dB}{dt} = k_a G - k_e B$. With $G(0) = 100$ and $B(0) = 0$, we get $G(t) = 100e^{-k_a t} = 100e^{-0.8t}$. For blood: $\frac{dB}{dt} + k_e B = k_a G = 80e^{-0.8t}$. The solution is $B(t) = \frac{k_a}{k_a - k_e}(e^{-k_e t} - e^{-k_a t}) \cdot 100 = \frac{0.8}{0.6}(e^{-0.2t} - e^{-0.8t}) \cdot 100$. Maximum occurs when $\frac{dB}{dt} = 0$: $k_a G - k_e B = 0$, so $k_a e^{-k_a t} = k_e \frac{k_a}{k_a - k_e}(e^{-k_e t} - e^{-k_a t})$. Simplifying: $e^{-k_a t} = \frac{k_e}{k_a - k_e}(e^{-k_e t} - e^{-k_a t})$. After algebraic manipulation: $e^{-(k_a - k_e)t} = \frac{k_e}{k_a}$, so $t = \frac{\ln(k_a/k_e)}{k_a - k_e} = \frac{\ln(0.8/0.2)}{0.8-0.2} = \frac{\ln(4)}{0.6}$.
10. An epidemiological model tracks disease progression through three compartments: Susceptible (S), Infected (I), and Recovered (R). The infection rate is $\beta SI/N$ where $N$ is total population, and recovery rate is $\gamma I$. If $\beta = 0.5$ per day, $\gamma = 0.1$ per day, and $N = 10000$, what is the basic reproduction number $R_0$, and what does it predict about disease spread?
    Answer: $R_0 = 5$; disease spread depends on initial proportion of susceptible individuals The basic reproduction number is $R_0 = \frac{\beta}{\gamma} = \frac{0.5}{0.1} = 5$. However, $R_0 > 1$ doesn't guarantee exponential spread from any initial condition. The effective reproduction number is $R_{eff} = R_0 \cdot \frac{S}{N}$, so disease spread depends on the fraction of susceptible individuals. With $R_0 = 5$, if all individuals are initially susceptible, each infected person will infect 5 others on average. But if some are already immune or recovered, $R_{eff} < R_0$. Choice A is incorrect because it ignores the dependence on susceptible fraction. Choice C miscalculates $R_0$. Choice D incorrectly calculates $R_0 = \beta N / \gamma$.
11. A closed system consists of two compartments, with the amount of a tracer substance in each given by $x(t)$ and $y(t)$. In a closed system, the total amount of the substance $x(t) + y(t)$ remains constant over time. The dynamics are modeled by the linear system $\frac{d}{dt}\begin{pmatrix} x \\ y \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix}$. Which of the following is a necessary property of the matrix $A$?
    Answer: The sum of the entries in each column of $A$ is zero. If the total amount $T = x(t) + y(t)$ is constant, then its derivative must be zero: $\frac{dT}{dt} = \frac{dx}{dt} + \frac{dy}{dt} = 0$. Let $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$. Then $\frac{dx}{dt} = a_{11}x + a_{12}y$ and $\frac{dy}{dt} = a_{21}x + a_{22}y$. Summing these gives $\frac{dT}{dt} = (a_{11}+a_{21})x + (a_{12}+a_{22})y$. For this expression to be zero for all possible values of $x$ and $y$, the coefficients must be zero: $a_{11}+a_{21}=0$ and $a_{12}+a_{22}=0$. These are precisely the sums of the entries in each column of $A$.
12. A cascade system is modeled by: $\begin{cases} \frac{dx}{dt} = 8 - 4x \\ \frac{dy}{dt} = 4x - 2y \end{cases}$ This represents a substance flowing into compartment $X$, then to $Y$, then out of the system. Suppose the external inflow rate of $8$ is halved to $4$. How does the equilibrium amount of the substance in compartment $Y$ change?
    Answer: It is halved. First, find the original equilibrium $(x_e, y_e)$. Set the derivatives to zero: $8 - 4x = 0 \implies x_e = 2$. $4x_e - 2y = 0 \implies 4(2) - 2y = 0 \implies 8 - 2y = 0 \implies y_e = 4$. Now, find the new equilibrium $(x'_e, y'_e)$ with the new inflow rate of $4$: $4 - 4x = 0 \implies x'_e = 1$. $4x'_e - 2y = 0 \implies 4(1) - 2y = 0 \implies 4 - 2y = 0 \implies y'_e = 2$. The equilibrium amount in compartment $Y$ changes from $y_e = 4$ to $y'_e = 2$. Therefore, the amount is halved.
13. The amounts of two interacting chemical species, $A$ and $B$, in a reactor are given by $x(t)$ and $y(t)$. Species $A$ converts to species $B$ at a rate $0.2x$. Species $B$ converts back to species $A$ at a rate $0.1y$. Additionally, species $A$ and $B$ react with each other to form an inert product, consuming both at a rate proportional to the product of their amounts, with a rate constant of $k=0.01$. Which system of differential equations models this process?
    Answer: $\begin{cases} \frac{dx}{dt} = -0.2x + 0.1y - 0.01xy \\ \frac{dy}{dt} = 0.2x - 0.1y - 0.01xy \end{cases}$ For species A (amount $x$): It is consumed when converting to B (term $-0.2x$), it is produced from B (term $+0.1y$), and it is consumed in the reaction with B (term $-0.01xy$). So, $\frac{dx}{dt} = -0.2x + 0.1y - 0.01xy$. For species B (amount $y$): It is produced from A (term $+0.2x$), it is consumed when converting to A (term $-0.1y$), and it is consumed in the reaction with A (term $-0.01xy$). So, $\frac{dy}{dt} = 0.2x - 0.1y - 0.01xy$. This matches choice A.

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