Problem 10 - Calculus for Biology and Medicine

Question

In our compartment model we assumed that inflows and outflows are matched at \(q\) to keep the volume of water in the tank constant. It's often useful when modeling, for example, the flow of pollutant into a pristine environment, to consider what can occur if the inflows and outflows do not match. Let's assume that the tank initially contains a volume \(V_{0}\) of water. Water flows into the tank at rate \(q_{\mathrm{in}}\), and out of the tank at rate \(q_{\text {out. }}\) (You may assume \(q_{\text {in }}>q_{\text {out } .}\) ) Suppose that the water flowing into the tank contains a concentration \(C_{I}\) of solute. As usual we write \(C(t)\) for the concentration in the tank. (a) Show that the concentration in the tank can be modeled using a differential equation: $$ \frac{d}{d t}(C V)=q_{\text {in }} C_{I}-q_{\text {out }} C $$ (b) Previously we were able to treat \(V\) as a constant. Now \(V\) changes with time. Derive a formula for \(V(t)\). (c) By substituting your formula for \(V(t)\) into (a), derive a differential equation for \(C(t)\). (d) In general we cannot analyze the behavior of the solution \(C(t)\) using techniques from Section \(8.2 .\) Why not? (e) Let's assume \(C_{l}=0\). Then show that your equation from (c) can be written as: $$ \frac{d C}{d t}=\frac{-q_{\mathrm{in}} C}{V_{0}+\left(q_{\mathrm{in}}-q_{\mathrm{out}}\right) t} $$ (f) Assume some definite values for the constants in \((8.57):\) \(q_{\mathrm{in}}=2, q_{\text {out }}=1\), and \(V_{0}=20 .\) Assuming \(C(0)=1\), solve \((8.57)\) to find \(C(t) .\) Show that \(\lim _{t \rightarrow \infty} C(t)=0\).

Step-by-Step Solution

Answer

Concentration decreases to zero over time as inflow dilutes the tank.

Step 1: Setup the Differential Equation for Concentration

The problem statement gives us the diagram of water flow in and out of the tank, with different concentrations. We need to develop a differential equation using the concentration change rate formula: \[ \frac{d}{dt}(C V) = q_{\text{in}} C_{I} - q_{\text{out}} C \] This equation essentially states that the rate of change of the total amount of solute in the tank depends on the difference between the solute entering and the solute leaving.

Step 2: Derive Formula for V(t)

Since the inflow and outflow rates differ, the tank volume changes over time. The rate of change of volume \( V(t) \) is the difference between inflow and outflow: \[ \frac{dV}{dt} = q_{\text{in}} - q_{\text{out}} \] Integrating this with respect to time, we get:\[ V(t) = V_{0} + (q_{\text{in}} - q_{\text{out}}) t \]

Step 3: Substitute V(t) into the Differential Equation

Substitute the expression for \( V(t) \) from Step 2 into the differential equation from Step 1.\[ \frac{d}{dt}(C(V_{0} + (q_{\text{in}} - q_{\text{out}}) t)) = q_{\text{in}} C_{I} - q_{\text{out}} C \] This will give the new differential equation for the concentration \( C(t) \).

Step 4: Explain why Analysis of C(t) is Not Straightforward

Section 8.2 likely deals with methods for solving differential equations where the coefficients are constants or other easily manageable forms. The term \[ V_{0} + (q_{\text{in}} - q_{\text{out}}) t \] introduces complexity, making traditional separation or integration techniques insufficient for a straightforward solution.

Step 5: Set CI=0 to Rewrite Differential Equation

Under the assumption \( C_{I} = 0 \), rewrite the simplified form of the differential equation for \( C(t) \):\[ \frac{dC}{dt} = \frac{-q_{\text{in}} C}{V_{0} + (q_{\text{in}} - q_{\text{out}}) t} \]

Step 6: Solve the Differential Equation with Given Parameters

With \( q_{\text{in}} = 2 \), \( q_{\text{out}} = 1 \), and \( V_{0} = 20 \), substitute these into the equation from Step 5.\[ \frac{dC}{dt} = \frac{-2C}{20 + (2-1)t} = \frac{-2C}{20 + t} \]Separate variables and integrate to solve for \( C(t) \): \[ \int \frac{1}{C}dC = -2 \int \frac{1}{20 + t} dt \]This yields:\[ \ln |C| = -2 \ln |20 + t| + k \]Convert to exponential form:\[ C(t) = \frac{A}{(20+t)^2} \]Use the initial condition \( C(0) = 1 \) to find \( A \).\[ C(0) = \frac{A}{20^2} = 1 \Rightarrow A = 400 \]Thus,\[ C(t) = \frac{400}{(20+t)^2} \]

Step 7: Evaluate Limit of C(t) as t approaches Infinity

Check the behavior as \( t \to \infty \):\[ \lim_{t \to \infty} C(t) = \lim_{t \to \infty} \frac{400}{(20+t)^2} = 0 \] This confirms that the concentration approaches zero over a long period.

Key Concepts

Compartment Model

A compartment model is a simplified representation of a system. In our case, it describes how substances flow in and out of compartments, like a tank. This tank receives water with a certain solute concentration, while different amounts flow in and out. Inflows and outflows affect the overall dynamics of concentration within the compartment.

In such models, compartments can include different elements or environments, like liquids or gases. By analyzing the fluid in each compartment, we aim to understand how substances transfer and change over time. The primary goal of a compartment model is tracking these movements and understanding factors like concentration and volume changes, crucial when predicting outcomes such as pollution dispersion.

Concentration Dynamics

Concentration dynamics explore how solute concentration changes over time in a container. In our exercise, this is represented by the differential equation \( \frac{d}{dt}(C V) = q_{\text{in}} C_{I} - q_{\text{out}} C \). This means how fast the solute's concentration changes depends on how much solute enters minus how much leaves.

Understanding concentration dynamics involves working with variables like the initial concentration, influx, and outflux of the solute. It accounts for changes in context, like water levels varying with time, and influences the resulting concentration.

These equations help model real-world scenarios, such as determining how a contaminant spreads in the environment, based on initial and changing conditions.

Initial Value Problem

An initial value problem in differential equations is one where we need an initial condition to solve the differential equation completely. In this exercise, the initial condition provided is \( C(0) = 1 \), meaning the solute concentration is 1 at time zero.

This initial condition allows us to integrate and find the specific solution to our equation, such as how \( C(t) \) changes over time. Without this constraint, the problem has infinite solutions, and it won't provide meaningful insight into the system's behavior.

Initial conditions are vital in making predictions and understanding the system's response to changing conditions. By incorporating these in the model, we can track how the concentration evolves over time and under varying conditions.

Volume Change with Time

Volume change with time refers to how the tank's volume changes because the inflow and outflow are different. For example, in our example, this is given by \( V(t) = V_{0} + (q_{\text{in}} - q_{\text{out}}) t \). It shows volume increases or decreases based on the inflow-outflow difference.

Analyzing volume change helps understand how concentration varies in non-equilibrium conditions where \( q_{\text{in}} eq q_{\text{out}} \). This is important because it influences the concentration dynamics in the system.

In practical scenarios, if more water enters than exits, it dilates the concentration; conversely, more outflow than inflow concentrates the substance. Hence, understanding these dynamics is crucial for accurate modeling in environmental studies and similar applications.