Problem 25 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

2 kilos of a fish poison, 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. What is the amount \(\left(p_{t}\right.\) for discrete time or \(P(t)\) for continuous time) of poison in the lake at time \(t\) days after the poison is applied? a. Treat the problem as a discrete time problem with one-day time intervals. Solve the difference equation $$ p_{0}=2 \quad p_{t+1}-p_{t}=-\frac{1000}{4000} p_{t} $$ b. Let \(t\) denote continuous time and \(P(t)\) the amount of poison in the lake at time \(t\). Let \([t, t+\Delta t]\) denote a short time interval (measured in units of days). An equation for the mathematical model is $$ P(t+\Delta t)-P(t)=-\frac{P(t)}{4000} \Delta t 1000 $$ Show that the units on the terms of this equation balance. c. Argue that $$ P(0)=2, \quad P^{\prime}(t)=-0.25 P(t) $$ d. Compute the solution to this equation. e. Compare the solution to the discrete time problem, \(p_{t},\) with the solution to the continuous time problem, \(P(t)\). f. For what value of \(k\) will the solution, \(Q(t),\) to $$ Q(0)=2, \quad Q^{\prime}(t)=k Q(t) \quad \text { satisfy } $$ \(Q(t)=p_{t}, \quad\) for \(\quad t=0,1,2, \cdots ?\) g. Which of \(P(t)\) and \(Q(t)\) most accurately estimates rotenone levels? h. On what day, \(\bar{t}\) will \(P(\bar{t})=4 \mathrm{~g}\) ?

Step-by-Step Solution

Answer

The poison amount decreases exponentially. At 24 days, the continuous model \(P(t)\) predicts 4 g of rotenone.

Step 1: Formulate the Discrete Time Problem

For the discrete time problem, we apply the given difference equation: \( p_{t+1} - p_t = -\frac{1000}{4000} p_t \). The initial condition is \( p_0 = 2 \). Simplifying the difference equation, we have: \( p_{t+1} = p_t (1 - 0.25) = 0.75p_t \). This shows that each day, the amount of poison is multiplied by 0.75.

Step 2: Solve the Discrete Time Equation

Using \( p_{t+1} = 0.75p_t \), we recursively get: - \( p_1 = 0.75 \times 2 = 1.5 \) - \( p_2 = 0.75 \times 1.5 = 1.125 \)- Generalizing, \( p_t = 2 \times (0.75)^t \).

Step 3: Formulate the Continuous Time Problem

For the continuous time problem, we use the given model: \( P(t+\Delta t) - P(t) = -\frac{P(t)}{4000} \Delta t \times 1000 \), which simplifies to \( P(t+\Delta t) - P(t) = -0.25 P(t) \Delta t \).

Step 4: Confirm Unit Balance for Continuous Model

Checking unit balance: - \( P(t) \) is in kilograms - \( \Delta t \) is in days - Hence, both sides of \( P(t+\Delta t) - P(t) = -0.25 P(t) \Delta t \) equate to a volume per time unit, verifying the equation is balanced.

Step 5: Solve the Continuous Time Differential Equation

Given \( P(0) = 2 \) and \( P'(t) = -0.25 P(t) \), this is a first-order linear differential equation. The solution is \( P(t) = 2e^{-0.25t} \).

Step 6: Compare Discrete and Continuous Solutions

For the discrete solution: \( p_t = 2 \times (0.75)^t \) and continuous solution: \( P(t) = 2e^{-0.25t} \). Both models show exponential decay, but with different decay rates.

Step 7: Determine \( k \) for \( Q(t) \) to Equal \( p_t \)

To have the form \( Q(t) = 2 \times (0.75)^t \), the new differential equation \( Q'(t) = kQ(t) \) must satisfy \( k = \ln(0.75) \).

Step 8: Identify Most Accurate Estimation

\( P(t) \), the continuous model, often provides a closer approximation of real-world processes due to its smooth nature versus discrete steps in \( Q(t) \).

Step 9: Find Day \( \bar{t} \) for \( P(\bar{t}) = 4 \text{ g} \)

Set \( P(t) = 2e^{-0.25t} = 4 \times 10^{-3} \). Solving \( 0.002 = e^{-0.25t} \) gives \( \bar{t} = -\frac{\ln(0.002)}{0.25} \), approximately 24 days.

Key Concepts

Difference Equation

A difference equation is like a recipe that helps us understand how a quantity changes from one time point to another. In this exercise, we are dealing with the amount of poison in a lake over discrete time intervals, specifically one-day intervals. The difference equation we're focusing on is given by \( p_{t+1} = 0.75p_t \). This equation tells us that each day, the amount of poison is 75% of what it was the previous day.

Here is how you can interpret the equation:

• \( p_{t+1} \) is the amount of poison the next day.
• \( p_t \) is the amount of poison today.
• The factor 0.75 means that the poison decreases by 25% each day.

By solving this difference equation, we calculated \( p_t = 2 \times (0.75)^t \), where \( t \) is the number of days after the poison was initially added. This shows how the poison amount decreases exponentially with time in a discrete manner.

Exponential Decay

Exponential decay is a process where a quantity decreases at a rate proportional to its current value. In this problem, the poison in the lake decreases according to the models we established for both discrete and continuous scenarios.

For the continuous model, we have the differential equation \( P'(t) = -0.25 P(t) \). This equation signifies that the rate of change of the poison is proportional to the amount of poison present, with a rate factor of -0.25. The negative sign indicates decay.

Solving this differential equation gives us \( P(t) = 2e^{-0.25t} \). This equation confirms exponential decay, as the term \( e^{-0.25t} \) reflects an exponential function where \( e \) is the base of the natural logarithm and \( -0.25 \) is the decay rate.

• Initial amount: 2 kg of poison
• Decay factor: -0.25, indicating 25% decay continuously per day

Whether in discrete or continuous form, exponential decay elegantly captures how the quantity lessens over time. The amount diminishes more slowly than linear decrease, allowing us to model gradual reduction accurately.

Continuous vs. Discrete Models

Understanding the key differences between continuous and discrete models is crucial when dealing with real-world situations. In this exercise, we examined how rotenone behaves in both scenarios:

The discrete model uses a difference equation. It breaks time into distinct phases, in this case, days. Every day, the poison multiplies by a constant factor (0.75 here), effectively using steps to depict how things change over time. This method simplifies calculations and fits many practical applications with intervals, such as business quarters or school grades.

The continuous model, contrastingly, employs a differential equation to present time as a smooth, ongoing variable. The continuous equation \( P'(t) = -0.25 P(t) \) provides a refined understanding of changes without the restrictions of stepping through time abruptly. Thus, it suits phenomena with constant fluctuation, such as chemical reactions or population growth.

• Discrete models: easier calculations, ideal for specific, non-constant intervals.
• Continuous models: smoother representation, best for processes with uninterrupted change.

Comparing them, we notice continuous models usually mirror real-life gradual processes more accurately. They reduce approximation errors common in the blocky nature of discrete models.