Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

A pristine lake of volume \(1,000,000 \mathrm{~m}^{3}\) has a river flowing through it at a rate of \(10,000 \mathrm{~m}^{3}\) per day. A city built beside the river begins dumping \(1000 \mathrm{~kg}\) of solid waste into the river per day. 1\. Write a derivative equation that describes the amount of solid waste in the lake \(t\) days after dumping begins. 2\. What will be the concentration of solid waste in the lake after one year?

Step-by-Step Solution

Answer

The concentration of solid waste in the lake after one year is approximately 0.1 kg/m³.

Step 1: Identify Variables and Constants

Let \( W(t) \) be the amount of solid waste in the lake in kg at time \( t \) days. The flow rate of the river is \( 10,000 \ \mathrm{m^3/day} \), and the volume of the lake is \( 1,000,000 \ \mathrm{m^3} \). The city adds \( 1000 \ \mathrm{kg/day} \) of waste.

Step 2: Formulate the Derivative Equation

The change in the amount of waste in the lake is the rate at which waste enters minus the rate at which it leaves. Hence, the derivative equation is: \[ \frac{dW}{dt} = 1000 - \frac{10,000}{1,000,000}W(t) = 1000 - 0.01W(t) \]

Step 3: Solve the Differential Equation

The equation \( \frac{dW}{dt} = 1000 - 0.01W(t) \) can be solved using the integrating factor method. The solution is: \[ W(t) = Ce^{-0.01t} + 100,000 \] where \( C \) is a constant determined by initial conditions.

Step 4: Apply Initial Condition

Assume initially, \( W(0) = 0 \). Applying this, we get: \( W(0) = Ce^{0} + 100,000 = 0 \). Thus, \( C = -100,000 \). Therefore, \( W(t) = -100,000e^{-0.01t} + 100,000 \).

Step 5: Determine Solid Waste After One Year

Replace \( t \) by 365 days to find \( W(365) \): \[ W(365) = -100,000e^{-0.01 \times 365} + 100,000 \] Calculate this to get \( W(365) \approx 100,000 \).

Step 6: Calculate Concentration After One Year

The concentration of waste \( C(t) \) at \( t = 365 \) is calculated by dividing the amount of waste by the lake's volume: \[ C(365) = \frac{W(365)}{1,000,000} \] \( C(365) \approx \frac{100,000}{1,000,000} = 0.1 \ \mathrm{kg/m^3} \).

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates a function to its derivatives. In the context of life sciences, it often represents how specific quantities change over time.
A simple way to think about it:

• It tells us how fast something is growing or shrinking.
• It connects the rate of change of a quantity with the amount already present.

In the problem we examined, a differential equation is derived to describe the change in the amount of solid waste in a lake over time. This equation highlights how the added waste and outflow from the lake influence the overall waste concentration.
The specific differential equation derived is: \[\frac{dW}{dt} = 1000 - 0.01W(t)\] Here:

• \( \frac{dW}{dt} \) is the rate of change of waste in the lake.
• "1000" indicates the city dumps 1000 kg of waste daily.
• "0.01W(t)" represents how much waste leaves the lake daily due to the outflow.

Integrating Factor Method

The Integrating Factor Method is a useful technique for solving first-order linear differential equations. It simplifies the process of finding a solution by making the equation more straightforward to integrate.
Here's how it works in a nutshell:

• The method finds a special function (the integrating factor) that turns the differential equation into a simpler form.
• By multiplying the entire differential equation by this integrating factor, it allows us to easily integrate both sides.

In the example, the differential equation is: \[\frac{dW}{dt} = 1000 - 0.01W(t)\]To solve it, we use the integrating factor, which in this case is \( e^{0.01t} \). This turns the differential equation into an easier form to integrate, leading to the solution: \[W(t) = Ce^{-0.01t} + 100,000\]Where \( C \) is a constant determined by initial conditions. In real applications, the integrating factor method is a favorite because it transforms a potentially complex problem into a more manageable one. It is highly practical for life science models involving growth, decay, or mixing processes.

Rate of Change

Rate of change is a key concept in calculus that tells us how much a quantity changes as some other quantity, often time, changes. In life sciences, understanding rates can help model and predict behaviors like growth, decay, or reaction speeds.
Think of it like this:

• It's similar to "speed" but can apply to any changing quantity, like concentrations or populations.
• Provides a dynamic "snapshot" of how fast or slow things are changing at any given moment.

In the problem at hand, the rate of change is the focus of the differential equation:

• The incoming waste impacts the system immediately as 1000 kg/day.
• The outgoing waste depends on the current waste level, leaving at a rate of 0.01 times the waste present, \( 0.01W(t) \).

This concept of rate of change allows us to model how the lake's waste content evolves over time, giving insights into environmental impacts and the effectiveness of potential interventions. By understanding these rates, life scientists can better predict and manage changes in diverse ecosystems.