Problem 26 - 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 tank 1 and flushes solute from tank 1 into tank 2 . Now assume that \(C_{1}(0)=1\) and \(C_{2}(0)=0 .\) If \(q=1, V_{1}=1\), and \(V_{2}=2\), solve the pair of differential equations to find \(C_{1}(t)\) and \(C_{2}(t) .\) Sketch both functions of time.

Step-by-Step Solution

Answer

The solutions are \(C_{1}(t) = e^{-t}\) and \(C_{2}(t) = e^{-t/2} - e^{-t}\).

Step 1: Simplify the Equations

We start with the given differential equations:\[ \frac{d C_{1}}{dt} = \frac{q}{V_{1}}(C_{\infty} - C_{1}) \] \[ \frac{d C_{2}}{dt} = \frac{q}{V_{2}}(C_{1} - C_{2}) \] Substitute the given values \(C_{\infty} = 0\), \(q = 1\), \(V_{1} = 1\), and \(V_{2} = 2\):\[ \frac{d C_{1}}{dt} = -C_{1} \] \[ \frac{d C_{2}}{dt} = \frac{1}{2}(C_{1} - C_{2}) \]

Step 2: Solve for C1(t)

The differential equation for \(C_{1}\) is:\[ \frac{d C_{1}}{dt} = -C_{1} \] This is a simple first-order linear differential equation with the solution:\[ C_{1}(t) = C_{1}(0) e^{-t} \] Given \(C_{1}(0) = 1\), we find:\[ C_{1}(t) = e^{-t} \]

Step 3: Solve for C2(t)

Substitute \(C_{1}(t) = e^{-t}\) into the equation for \(C_{2}\):\[ \frac{d C_{2}}{dt} = \frac{1}{2}(e^{-t} - C_{2}) \] This is a linear first-order differential equation. Apply the method of integrating factors with \(\mu(t) = e^{t/2}\):\[ \frac{d}{dt}[C_{2}(t) e^{t/2}] = \frac{1}{2} e^{-t} e^{t/2} \]Integrating both sides with respect to \(t\):\[ C_{2}(t) e^{t/2} = \frac{1}{2} \int e^{-t/2} \ dt + C \]Solve this integral to obtain:\[ C_{2}(t) e^{t/2} = -e^{-t/2} + C \]Thus,\[ C_{2}(t) = C e^{-t/2} - e^{-t} \]Apply the initial condition \(C_{2}(0) = 0\) to solve for \(C\):\[ 0 = C - 1 \rightarrow C = 1 \] Thus, the solution for \(C_{2}(t)\) is:\[ C_{2}(t) = e^{-t/2} - e^{-t} \]

Step 4: Sketch the Functions C1(t) and C2(t)

Now, sketch the functions \(C_{1}(t) = e^{-t}\) and \(C_{2}(t) = e^{-t/2} - e^{-t}\). - The function \(C_{1}(t)\) is an exponential decay function starting from 1, decreasing towards 0.- The function \(C_{2}(t)\) starts from 0 and initially increases as \(t\) increases, then gradually approaches a non-zero steady state as the influence of \(e^{-t}\) diminishes faster than \(e^{-t/2}\). Both functions exhibit exponential characteristics over time.

Key Concepts

Differential Equations

Differential equations are mathematical tools used to describe how quantities change over time. In the context of this two-compartment model, we use differential equations to represent the flow of liquid between two tanks in terms of concentration. We have two differential equations for this model:

• \( \frac{d C_{1}}{dt} = -C_{1} \) represents how the concentration in the first tank changes over time.
• \( \frac{d C_{2}}{dt} = \frac{1}{2}(C_{1} - C_{2}) \) models the concentration change in the second tank based on the change in the first tank.

These equations are solved to determine the concentrations at any given time, \( C_{1}(t) \) and \( C_{2}(t) \). Solving these equations involves finding functions that describe how \( C_{1} \) and \( C_{2} \) evolve over time based on given initial conditions and parameters. The task predominantly involves solving these ordinary differential equations using techniques like separation of variables and integrating factors.

Exponential Decay

Exponential decay is a process where quantities decrease at a rate proportional to their current value. In the two-compartment model, the concentration in the first tank, \( C_{1}(t) \), exhibits exponential decay. The equation \( \frac{d C_{1}}{dt} = -C_{1} \) describes this decay, indicating that the rate of change of \( C_{1} \) is proportional to \( C_{1} \) itself, with a negative sign specifying a decrease. The solution to this equation yields the function \( C_{1}(t) = e^{-t} \), showing that \( C_{1} \) decreases exponentially over time starting from an initial concentration of 1. Graphically, exponential decay functions start at a certain value, and then drop off quickly at first, gradually slowing down as it approaches zero. This concept is prevalent in natural processes such as radioactive decay, population decline, and so on, making it a pivotal concept in understanding changes in dynamic systems like our two-tank model.

Integrating Factors

The method of integrating factors is a powerful technique used to solve linear first-order differential equations, especially when the variables are not easily separable. In the context of our two-compartment model, we apply this method to find \( C_{2}(t) \).For the differential equation \( \frac{d C_{2}}{dt} = \frac{1}{2}(e^{-t} - C_{2}) \), the integrating factor \( \mu(t) \) is \( e^{t/2} \). Multiplying the entire differential equation by this integrating factor transforms it into a form that can be solved through integration:

• \( \frac{d}{dt}[C_{2}(t) e^{t/2}] = \frac{1}{2} e^{-t/2} \)

After integrating both sides with respect to \( t \), we use the initial condition \( C_{2}(0) = 0 \) to solve for the constant of integration and find that \( C_{2}(t) = e^{-t/2} - e^{-t} \). This result incorporates the balancing effect of the decrease from \( C_{1}(t) \) and provides us with a comprehensive solution of the two-compartment dynamics.