Problem 2 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Two kilos of a fish poison that does not decompose are mixed into a lake that has a volume of \(100 \times 20 \times 2=4000\) cubic meters. A stream of clean water flows into the lake at a rate of 1000 cubic meters per day. Assume that it mixes immediately throughout the whole lake. Another stream flows out of the lake at a rate of 1000 cubic meters per day. a. Write a mathematical model that describes the daily change in the amount of poison in the lake. b. Let \(P_{0}, P_{1}, P_{2}, \cdots\) denote the amounts of poison in the lake, \(P_{t}\) being the amount of poison in the lake at the beginning of the \(t \underline{h}\) day after the poison is administered. Write a dynamic equation representative of the mathematical model. c. What is \(P_{0} ?\) Compute \(P_{1}\) from your dynamic equation. Compute \(P_{2}\) from your dynamic equation. d. Find a solution equation for your dynamic equation.

Step-by-Step Solution

Answer

The amount of poison in the lake decreases exponentially, with \( P_t = 2 \left(\frac{3}{4}\right)^t \).

Step 1: Understanding the System

We have a lake with 4000 cubic meters of water initially containing 2 kg of poison. Fresh water flows in and out at the same rate of 1000 cubic meters per day, ensuring the volume of the lake remains constant. This setup indicates that the concentration of poison dilutes over time as water flows out.

Step 2: Writing the Mathematical Model

Let \( P_t \) be the amount of poison in kilograms in the lake at day \( t \). Each day, 1000 cubic meters of water flows in and out. The concentration of poison exiting daily is \( \frac{P_t}{4000} \). The change in poison every day is then: \[ \Delta P_t = 0 - 1000 \cdot \frac{P_t}{4000} \] Simplifying gives: \[ \Delta P_t = -\frac{P_t}{4} \] This is the daily change in poison.

Step 3: Dynamic Equation Formulation

The dynamic equation expressing the poison amount at the end of day \( t \) is given by: \[ P_{t+1} = P_t - \frac{P_t}{4} = \frac{3}{4}P_t \] This shows that each day, 1/4th of the poison exits the lake with the outflowing water.

Step 4: Initial Condition \(P_0\) and Calculating \(P_1\) and \(P_2\)

Initially, there are 2 kg of poison, so \( P_0 = 2 \). Calculating \( P_1 \) using the dynamic equation: \[ P_1 = \frac{3}{4} \times 2 = 1.5 \] Calculating \( P_2 \) iteratively: \[ P_2 = \frac{3}{4} \times 1.5 = 1.125 \] Thus, \( P_1 = 1.5 \) kg and \( P_2 = 1.125 \) kg.

Step 5: Deriving Solution Equation

Given \( P_{t+1} = \frac{3}{4} P_t \), we can express \( P_t \) in terms of \( P_0 \): \[ P_t = \left(\frac{3}{4}\right)^t P_0 \] Substituting \( P_0 = 2 \), the solution equation becomes: \[ P_t = 2 \left(\frac{3}{4}\right)^t \] This equation gives the amount of poison at any day \( t \).

Key Concepts

Dynamic Systems

Dynamic systems refer to systems that change over time, often described by mathematical equations. They can be used to model anything from population growth to chemical reactions, and in this case, the dilution of poison in a lake.

Key characteristics of dynamic systems include:

• State Variables: These variables capture the system's state at any point in time. For the lake scenario, the state variable is the amount of poison in the lake, denoted as \(P_t\).
• Equilibrium: A point where the system reaches a steady state. In the lake example, this would be when the amount of poison stays constant over time.
• Feedback Loops: Processes where outputs of a system loop back as inputs. Here, the outflow of poisoned water reduces subsequent poison levels.

Dynamic systems can be linear or nonlinear, but many like this lake model, use linear approximations for simplicity and solution ease.

Differential Equations

Differential equations are mathematical tools used to describe the relationships involving rates of change. In modeling a dynamic system like the lake, differential equations help describe how and why the amount of poison changes each day.

In the poison in the lake scenario, the daily rate of change of poison is expressed as:\[ \Delta P_t = -\frac{P_t}{4} \]This equation tells us the rate at which poison is being removed from the lake. Derived from the principle that 1000 cubic meters of water (carrying a certain concentration of poison) exits the lake each day, the equation captures how the poison concentration decreases.

Differential equations come in various forms, but one common type used is

• Linear Differential Equations: These equations, like the one provided, are simpler to solve and analyze, often because they provide a direct proportionality and superposition of their terms.

Being able to set up and interpret such equations is crucial for effectively modeling dynamic processes.

Initial Value Problems

Initial value problems involve finding a function that satisfies a given differential equation and meets specified initial conditions. This concept is essential in predicting future states of a dynamic system, starting from a known original condition.

For our lake model, the problem starts with an initial condition: \( P_0 = 2 \) kg. This is the amount of poison present on day zero when the poison was first introduced. From this point, using the dynamic equation:\[ P_{t+1} = \frac{3}{4}P_t \]we can compute subsequent poison amounts, like \( P_1 = 1.5 \) kg and \( P_2 = 1.125 \) kg.

Key aspects of initial value problems include:

• Starting Condition: The known initial state, necessary for solving the rest of the problem.
• Solution Iteration: Continuously applying the solution formula to predict future states from the starting point and then from each subsequent state.

Solving initial value problems enables us to outline the trajectory of dynamic processes and determine future behavior based on initial circumstances.