Differential Equations Quiz: Compartment Models

What this quiz covers

This quiz focuses on Compartment Models, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The rate at which the substance flows from compartment $A$ to compartment $B$. (correct answer)
2. The rate at which the substance flows from compartment $B$ to compartment $A$.
3. The rate at which the substance in compartment $A$ is eliminated from the system.
4. The total rate of change of the substance in compartment $A$ at equilibrium.

Explanation: 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 $B$.

Question 2

1. $\begin{cases} \frac{dx}{dt} = -0.2x + 0.1y - 0.01xy \\ \frac{dy}{dt} = 0.2x - 0.1y - 0.01xy \end{cases}$ (correct answer)
2. $\begin{cases} \frac{dx}{dt} = -0.2x + 0.1y + 0.01xy \\ \frac{dy}{dt} = 0.2x - 0.1y + 0.01xy \end{cases}$
3. $\begin{cases} \frac{dx}{dt} = -0.2x + 0.1y - 0.01(x+y) \\ \frac{dy}{dt} = 0.2x - 0.1y - 0.01(x+y) \end{cases}$
4. $\begin{cases} \frac{dx}{dt} = 0.2x - 0.1y - 0.01xy \\ \frac{dy}{dt} = -0.2x + 0.1y - 0.01xy \end{cases}$

Explanation: 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.

Question 3

1. $25$
2. $11$
3. $36$ (correct answer)
4. $0$

Explanation: 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 ($15-4$).

Question 4

1. It is quartered.
2. It is halved. (correct answer)
3. It does not change.
4. It decreases by 2 units.

Explanation: 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.

Question 5

1. $-10 x + 3 y$
2. $+10 \frac{x}{V_A} - 3 \frac{y}{V_B}$
3. $-10 \frac{x}{V_A} - 3 \frac{y}{V_B}$
4. $-10 \frac{x}{V_A} + 3 \frac{y}{V_B}$ (correct answer)

Explanation: 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.

Question 6

1. $\frac{dT}{dt} = 10 - 0.1T$
2. $\frac{dT}{dt} = 10 - 0.6(x_1+x_2)$
3. $\frac{dT}{dt} = 10 - 0.1x_2$ (correct answer)
4. $\frac{dT}{dt} = 10 - 0.5x_1 - 0.1x_2$

Explanation: 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.

Question 7

1. $\begin{cases} \frac{dx}{dt} = k_1 x - k_2 y \\ \frac{dy}{dt} = k_2 y \end{cases}$
2. $\begin{cases} \frac{dx}{dt} = k_1 x \\ \frac{dy}{dt} = -k_1 x + k_2 y \end{cases}$
3. $\begin{cases} \frac{dx}{dt} = -k_1 x - k_2 y \\ \frac{dy}{dt} = k_1 x \end{cases}$
4. $\begin{cases} \frac{dx}{dt} = -k_1 x \\ \frac{dy}{dt} = k_1 x - k_2 y \end{cases}$ (correct answer)

Explanation: 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.

Question 8

1. $1$ (correct answer)
2. $4/5$
3. $5/4$
4. $3$

Explanation: 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 $(15/4) / (15/4) = 1$.

Question 9

1. The sum of the entries in each column of $A$ is zero. (correct answer)
2. The sum of the entries in each row of $A$ is zero.
3. $A$ is a symmetric matrix, i.e., $A = A^T$.
4. The diagonal entries of $A$ must be zero.

Explanation: 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 $A$.

Question 10

1. $\frac{dS}{dt} = 1 - \frac{S}{200}, \quad S(0) = 0$
2. $\frac{dS}{dt} = 1 - \frac{S}{40+t}, \quad S(0) = 0$ (correct answer)
3. $\frac{dS}{dt} = 1 - \frac{5S}{200+5t}, \quad S(0) = 0$
4. $\frac{dS}{dt} = 0.1 - \frac{S}{40+t}, \quad S(0) = 0$

Explanation: 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.

Question 11

1. $t = \frac{\ln(4)}{0.6}$ hours (correct answer)
2. $t = \frac{\ln(4)}{0.8}$ hours
3. $t = \frac{\ln(0.25)}{0.6}$ hours
4. $t = \frac{\ln(2)}{0.6}$ hours

Explanation: 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 $t = \frac{\ln(k_a/k_e)}{k_a - k_e} = \frac{\ln(0.8/0.2)}{0.8-0.2} = \frac{\ln(4)}{0.6}$.

Question 12

1. 75.99% (correct answer)
2. 82.40%
3. 78.15%
4. 85.20%

Explanation: 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.

Question 13

1. $R_0 = 5$; disease will spread exponentially from any initial infected individual
2. $R_0 = 5$; disease spread depends on initial proportion of susceptible individuals (correct answer)
3. $R_0 = 0.05$; disease will die out regardless of initial conditions
4. $R_0 = 50$; disease will spread exponentially with very high transmission rate

Explanation: 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 $R_0 = \beta N / \gamma$.