Problem 27 - Calculus for Biology and Medicine

Question

In Problems 25-28 consider the two-compartment model for two tanks with respective volumes \(V_{1}\) and \(V_{2}\). $$ \begin{array}{l} \frac{d C_{1}}{d t}=\frac{q}{V_{1}}\left(C_{\infty}-C_{1}\right) \\ \frac{d C_{2}}{d t}=\frac{q}{V_{2}}\left(C_{1}-C_{2}\right) \end{array} $$ where \(C_{1}(t)\) is the concentration in the first tank and \(C_{2}(t)\) is the concentration in the second tank, and \(q\) is the volume of water flowing between the two tanks in one unit of time. Let \(C_{\infty}=0\), so that the fresh water is pumped into \(\operatorname{tank} 1\) and flushes solute from tank 1 into tank 2 . Now assume that \(C_{1}(0)=1\) and \(C_{2}(0)=0\). If \(q=1, V_{1}=3\), and \(V_{2}=1\), solve the pair of differential equations to find \(C_{1}(t)\) and \(C_{2}(t)\), and sketch both functions of time.

Step-by-Step Solution

Answer

\(C_1(t) = e^{-\frac{t}{3}}\); \(C_2(t) = \frac{3}{2} e^{-\frac{t}{3}} - \frac{3}{2} e^{-t}\).

Step 1: Understanding the Initial Conditions and System Setup

Given a two-compartment model, we start with initial concentrations \( C_1(0) = 1 \) and \( C_2(0) = 0 \). The parameters provided include \( q = 1 \), \( V_1 = 3 \), \( V_2 = 1 \), and \( C_{\infty} = 0 \). Fresh water flows into the first tank, lowering its concentration and transferring solute to the second tank.

Step 2: Set Up the Differential Equations

The system is governed by the following differential equations:\[ \frac{dC_1}{dt} = \frac{1}{3}(0 - C_1) = -\frac{1}{3}C_1 \]\[ \frac{dC_2}{dt} = \frac{1}{1}(C_1 - C_2) = C_1 - C_2 \]

Step 3: Solve for \(C_1(t)\)

We solve \( \frac{dC_1}{dt} = -\frac{1}{3}C_1 \). This is a first-order linear differential equation with a general solution:\[ C_1(t) = C_1(0) e^{-\frac{1}{3}t} \]With \( C_1(0) = 1 \), the specific solution is:\[ C_1(t) = e^{-\frac{t}{3}} \]

Step 4: Solve for \(C_2(t)\) Using an Integrating Factor

Given \( \frac{dC_2}{dt} = C_1 - C_2 \) and knowing \( C_1(t) = e^{-\frac{t}{3}} \), we rewrite:\[ \frac{dC_2}{dt} + C_2 = e^{-\frac{t}{3}} \]Using the integrating factor \( \mu(t) = e^{\int 1 \, dt} = e^t \), the equation becomes:\[ e^t \frac{dC_2}{dt} + e^t C_2 = e^{t - \frac{t}{3}} \] which simplifies to:\[ \frac{d}{dt}(e^t C_2) = e^{\frac{2t}{3}} \]

Step 5: Integrate and Solve for \(C_2(t)\)

Integrate both sides of \( \frac{d}{dt}(e^t C_2) = e^{\frac{2t}{3}} \):\[ e^t C_2 = \int e^{\frac{2t}{3}} \, dt = \frac{3}{2}e^{\frac{2t}{3}} + C \]Divide by \(e^t\):\[ C_2(t) = \frac{3}{2} e^{-\frac{t}{3}} + Ce^{-t} \]Using \( C_2(0) = 0 \), solve for \(C\):\[ 0 = \frac{3}{2} + C \, \Rightarrow C = -\frac{3}{2} \]Thus,\[ C_2(t) = \frac{3}{2} e^{-\frac{t}{3}} - \frac{3}{2} e^{-t} \]

Step 6: Sketch the Functions of Time

Draw the exponential decay of \( C_1(t) = e^{-\frac{t}{3}} \), which decreases from 1 to 0 over time.For \( C_2(t) = \frac{3}{2} e^{-\frac{t}{3}} - \frac{3}{2} e^{-t} \), this starts at 0, increases as the concentration from tank 1 transfers, then decreases as it approaches 0. \( C_2(t) \) will have an initial increase followed by decay due to the second negative exponential term.

Key Concepts

Two-Compartment Model

The two-compartment model is a type of mathematical representation used to describe the movement of substances between two separate compartments or "tanks." Each tank has its own volume, and substances can flow between them.In the context of the exercise, we have:

• Tank 1 with volume \( V_1 = 3 \)
• Tank 2 with volume \( V_2 = 1 \)

The equations provided describe how the concentration of a substance evolves over time in each tank. Tank 1 receives fresh water, which then affects the concentration of solute, while Tank 2 receives the outflow from Tank 1. This setup helps us understand how solute concentration changes, whether it be a chemical, drug, or any other substance being analyzed.

Initial Conditions

Initial conditions are essential for solving differential equations because they provide the baseline concentrations or values at the starting point of our analysis. In this problem, we start with:

• \( C_1(0) = 1 \): Initial concentration in Tank 1 is 1 unit.
• \( C_2(0) = 0 \): Initial concentration in Tank 2 is 0, meaning it starts with no solute.

These conditions allow us to solve the equations specifically for this scenario. They determine the constants of integration when solving the differential equations, which means the solutions we find are unique to these starting values.

Integrating Factor

An integrating factor is a mathematical tool used to solve certain types of first-order linear differential equations. It simplifies these equations to make solving them more straightforward.For the differential equation concerning \( C_2(t) \):\[ \frac{dC_2}{dt} + C_2 = e^{-rac{t}{3}} \]We use the integrating factor \( \mu(t) = e^t \), calculated from \( e^{\int 1 \, dt} \). This transforms the equation into a convenient form:\[ \frac{d}{dt}(e^t C_2) = e^{\frac{2t}{3}} \]This form allows us to directly integrate both sides easily, helping us find the solution for \( C_2(t) \). Without applying the integrating factor, solving this equation analytically would be much more complex.

Exponential Decay

Exponential decay describes the process where a quantity decreases at a rate proportional to its current value. In the two-compartment model, both \( C_1(t) \) and \( C_2(t) \) exhibit exponential decay, but the mechanisms vary slightly.For \( C_1(t) \):- The decay is represented by \( C_1(t) = e^{-\frac{t}{3}} \).- This means the concentration decreases rapidly at first and then more slowly as time goes on.For \( C_2(t) \):- The function is \( C_2(t) = \frac{3}{2} e^{-\frac{t}{3}} - \frac{3}{2} e^{-t} \).- It shows an initial rise as Tank 2 starts filling from Tank 1, followed by decay as the process stabilizes and solute reduces in both tanks.Understanding exponential decay is crucial in many fields, like chemistry, biology, and finance, as it describes natural decrease patterns due to proportional loss over time.