Question

Let \(y(t)\) be the population of a species that is being harvested, for \(t \geq 0 .\) Consider the harvesting model \(y^{\prime}(t)=0.008 y-h, y(0)=y_{0},\) where \(h\) is the annual harvesting rate, \(y_{0}\) is the initial population of the species, and \(t\) is measured in years. a. If \(y_{0}=2000,\) what harvesting rate should be used to maintain a constant population of \(y=2000,\) for \(t \geq 0 ?\) b. If the harvesting rate is \(h=200 /\) year, what initial population ensures a constant population?

Short answer

In summary, to maintain a constant population of y = 2000 with an initial population of y(0) = 2000, the appropriate harvesting rate is 16 units per year. On the other hand, to maintain a constant population with a harvesting rate of 200 units per year, the initial population should be 25,000 individuals.

Step-by-step solution

Find the derivative when the population is constant

Since the goal is to maintain a constant population in both cases, it means that the rate of change of the population y'(t) must be zero. So, set y'(t)=0

Solve for the harvesting rate (a)

Given y'(t)=0, y(t)=2000, and y(0)=2000, find the harvesting rate h: 0=0.008*2000 - h Now solve for h: h=0.008*2000 h=16 The appropriate harvesting rate to maintain a constant population of y=2000 for t≥0 with an initial population of y(0)=2000 is 16 units per year.

Solve for the initial population (b)

Given y'(t)=0 and the harvesting rate h=200/year, find the initial population y(0) to maintain a constant population: 0=0.008*y(0) - 200 Now, solve for y(0): 200=0.008*y(0) y(0)= 200/0.008 y(0)= 25000 The initial population needed to maintain a constant population, given a harvesting rate of h=200/year, is y(0)=25000.

Key concepts

Population Dynamics

Population dynamics is a fascinating area of study that involves understanding how populations of organisms change over time. This can include changes due to birth rates, death rates, immigration, and emigration. In mathematical modeling, these factors are often represented by differential equations. These equations help us make predictions and understand the long-term behavior of the population.

In the context of the given exercise, the population dynamics of a species being harvested is described by the differential equation \(y'(t) = 0.008y - h\). Here, \(y(t)\) is the population at time \(t\), and \(h\) is the rate at which the population is being harvested per year. This model helps us investigate how different harvesting strategies affect the population size.

By setting the derivative \(y'(t) = 0\), we're specifically interested in scenarios where the population stays constant over time. This means we're looking for an equilibrium state where growth exactly balances harvesting.

Harvesting Models

Harvesting models are crucial tools in resource management, especially for species conservation and sustainable economic use. These models help determine the impact of removing resources from the environment.

In our example, the harvesting model used is \(y'(t) = 0.008y - h\), representing a continuous harvesting process. Here is how this model works:

• \(0.008y\) represents the natural growth of the population without any human interference.
• \(- h\) represents the reduction in the population due to harvesting.

The challenge lies in determining a rate \(h\) that ensures the population doesn’t decline to unsustainable levels. By structuring the harvesting strategy effectively, it’s possible to maintain a population at a constant level, which is essential for species that may be endangered or economically valuable.

Initial Value Problems

Initial value problems are a fundamental concept in differential equations where the solution is determined based on an initial condition. In this scenario, it's given as \(y(0) = y_0\). This represents the known population of the species at the beginning of the observation period (at \(t = 0\)).

Why is this important? The initial condition allows us to solve the differential equation uniquely because it gives us a specific starting point.

In our case, part of the problem involves finding what initial population \(y_0\) is needed to maintain a constant population given a fixed harvesting rate. This involves manipulating the differential equation to solve for the initial population value that balances out the harvesting effect. This kind of problem is common in ecological models where a balanced state is required for effective resource management.

Constant Solutions

Constant solutions in differential equations are scenarios where the solution does not change over time. This implies that the rate of change of the population is zero, keeping the population steady.

In the harvesting model, the constant solution is achieved by setting \(y'(t) = 0\). This simplifies the equation to a balance between growth \(0.008y\) and harvesting \(h\).

• In exercise part (a), when \(y_0 = 2000\) and \(y(t) = 2000\), solving \(y'(t) = 0\) gives us the harvesting rate \(h = 16\). This ensures that the population remains at 2000.
• In part (b), given \(h = 200\), the required initial population \(y_0\) is calculated to be 25000, to keep the population constant.

These constant solutions are essential for designing sustainable practices when harvesting natural resources. By correctly tuning the parameters, a stable ecosystem can be maintained, preventing over-exploitation and ensuring availability for future generations.