Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Two kilos of rotenone are mixed into a lake which 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. Fifteen percent of the rotenone decomposes every day. a. Write a mathematical model that describes the daily change in the amount of rotenone in the lake. b. Let \(R_{0}, R_{1}, R_{2}, \cdots\) denote the amounts of rotenone in the lake, \(R_{t}\) being the amount of rotenone 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 \(R_{0}\) ? Compute \(R_{1}\) from your dynamic equation. Compute \(R_{2}\) from your dynamic equation. d. Find a solution equation for your dynamic equation.

Step-by-Step Solution

Answer

The amount of rotenone reduces daily as 2 kg initially, then follows a geometric progression with (0.75)^t factor in each day.

Step 1: Understand the Problem

We are tasked with modeling the decrease of rotenone in a lake with inflow and outflow of water. The rotenone decomposes daily, and clean water flows in and out, affecting concentration.

Step 2: Develop the Mathematical Model

Let \( R_t \) be the amount of rotenone at the start of day \( t \). The daily decrease is due to decomposition and water replacement. The daily change is given by:\[R_{t+1} = 0.85R_t - \left(\frac{R_t}{4000}\right) \times 1000\]Where the first term accounts for decomposition (15% is lost, so 85% remains), and the second term addresses outflow, based on current concentration.

Step 3: Initial Condition and Calculate R_0

We begin with 2 kg of rotenone, so initially:\[ R_0 = 2 \text{ kg} \]

Step 4: Calculate R_1 Using the Dynamic Equation

Substitute \( R_0 \) into the dynamic equation:\[R_1 = 0.85 \times 2 - \left(\frac{2}{4000}\right) \times 1000\]Simplifying this gives:\[R_1 = 1.7 - 0.5 = 1.2 \text{ kg}\]

Step 5: Calculate R_2 Using the Dynamic Equation

Substitute \( R_1 \) into the dynamic equation for the second day:\[R_2 = 0.85 \times 1.2 - \left(\frac{1.2}{4000}\right) \times 1000\]Simplifying this gives:\[R_2 = 1.02 - 0.3 = 0.72 \text{ kg}\]

Step 6: Develop the General Solution Equation

The equation or model showing how \(R_t\) changes over time can be generalized by recognizing a pattern:\[R_{t+1} = 0.75R_t\]This simplifies to a geometric progression that can be solved as:\[R_t = R_0 \times (0.75)^t = 2 \times (0.75)^t\]

Key Concepts

Dynamic Equations

Dynamic equations are a powerful tool to describe how quantities change over time. They are particularly useful in situations involving processes that evolve step by step, such as the concentration of a substance in a liquid. These equations allow mathematicians and scientists to predict future states from known conditions.

In our exercise, we deal with a dynamic equation that portrays the decrease in rotenone in a lake. This involves multiple factors—decomposition and water flow—that cause changes in rotenone levels. The dynamic equation given as:

• \[ R_{t+1} = 0.85R_t - \left(\frac{R_t}{4000}\right) \times 1000 \]

contains two crucial terms:

• The first part, \(0.85R_t\), represents the rotenone amount after 15% decomposes daily, leaving 85% on each following day.
• The second part, \(\left(\frac{R_t}{4000}\right) \times 1000\), accounts for the concentration of rotenone changing due to the inflow and outflow of water.

This is a classic example of a dynamic equation at work.

Mathematical Modeling

Mathematical modeling involves creating representations of real-world phenomena through mathematical constructs, making complex systems easier to analyze and predict. In this case, the situation in our exercise—a controlled environment with specific parameters like lake size, flow rates, and decomposition rates—is converted into a mathematical form.

The mathematical model for our rotenone problem is the dynamic equation itself, as it quantifies the daily change in rotenone levels. This model is built based on assumptions:

• Immediate mixing of inflow and outflow water ensures uniform concentration.
• Daily decomposition follows a consistent rate.

These assumptions help simplify the model while providing a useful tool to predict how rotenone decays over time. Through mathematical modeling, we can explore different scenarios by altering parameters, such as changing flow rates or decomposition percentages.

Geometric Progression

A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In our problem, the solution equation derived shows a geometric progression reflecting how rotenone decreases over each day.

The process can be summarized in the formula:

• \[ R_t = R_0 \times (0.75)^t \]

Here, \(R_0\) represents the initial amount of rotenone (2 kg), and the common ratio is 0.75, derived from the fact that 25% of the substance diminishes overall daily due to combined effects of decomposition and water flow. This ratio highlights the exponential decay pattern—a characteristic behavior in many natural and chemical processes.

Recognizing this geometric pattern, we can easily compute rotenone amounts for any future day, enabling effective environmental management and planning.