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

Question: Given a population model as a differential equation \(y'(t) = 0.008y - h\) with an initial condition \(y(0) = y_0\), calculate the following: a. The annual harvesting rate \(h\) that ensures a constant population of 2000 individuals, given the initial population \(y_0 = 2000\). b. The initial population \(y_0\) that will lead to a constant population size with a given harvesting rate of \(h=200\) per year. Answer: a. To maintain a constant population of 2000, the annual harvesting rate should be 16 individuals per year. b. To maintain a constant population with a harvesting rate of 200 individuals per year, the initial population should be 25,000 individuals.

Step-by-step solution

Set up the differential equation for a constant population of y=2000

We are given the population model \(y'(t) = 0.008y - h\), and we are to find the harvesting rate that maintains a constant population of \(y = 2000\). To maintain a constant population, the rate of change of the population (\(y'(t)\)) should be equal to 0. So, we can write this as: $$0 = 0.008(2000) - h$$

Solve for the harvesting rate h

Now, we just need to solve the equation for h: $$0 = 0.008(2000) - h$$ $$h = 0.008(2000)$$ $$h = 16$$ So, to maintain a constant population of 2000, the annual harvesting rate should be 16 individuals per year. #b. Find the initial population to maintain a constant population with a harvesting rate of 200/year#

Set up the differential equation for a constant population with h=200/year

We are given the population model \(y'(t) = 0.008y - h\), and we are given the harvesting rate of \(h = 200\) per year. To find the initial population that maintains a constant population, we again need to set the rate of change of the population (\(y'(t)\)) to 0: $$0 = 0.008y - 200$$

Solve for the initial population y_0

Now we just need to solve the equation for y: $$0 = 0.008y - 200$$ $$200 = 0.008y$$ $$y_0 = \frac{200}{0.008}$$ $$y_0 = 25000$$ So, to maintain a constant population with a harvesting rate of 200 individuals per year, the initial population should be 25,000 individuals.

Key concepts

Population Dynamics

Population dynamics refers to the study of how and why populations change over time. This includes examining factors that influence the size and composition of populations, such as birth rates, death rates, immigration, and emigration. In the context of differential equations, population dynamics can be modeled using mathematical equations that describe how a population grows or declines under various circumstances.

For instance, a simple model might assume that the population grows at a constant rate; such models can often be expressed through equations like:

• Exponential growth models where population increases rapidly over time as a function of its current size.
• Logistic growth models that consider the carrying capacity of the environment, leading to a population that increases until it levels off.

In our current exercise, the population dynamics are affected by a harvesting activity. This means while discussing population growth or decline, we also need to factor in the removal of individuals from the population due to harvesting efforts.

Thus, understanding population dynamics is critical for establishing how external factors like harvesting influence long-term population stability.

Harvesting Models

Harvesting models are essential in managing natural resources like wildlife populations and fisheries. They describe how many individuals can be removed from a population without causing long-term declines. The goal is to find a sustainable balance where the harvesting does not exceed the population’s ability to replenish itself.

In our problem, the harvesting model is given by the differential equation:

• \( y'(t) = 0.008y - h\)

Here, \(y'(t)\) represents the rate of change of the population, \(0.008y\) indicates the natural growth rate of the population, and \(h\) is the annual harvesting rate.

When the harvesting is equal to the growth term \(0.008y\), the population remains stable, or constant. Thus, solving these equations helps in deciding the optimal harvesting rate or identifying the initial population needed for sustainable management.

Harvesting models allow conservationists and resource managers to plan harvesting strategies that ensure long-term viability of species while allowing for economic utilization.

Rate of Change

In mathematics, the rate of change refers to how a quantity changes over time. It's a crucial concept in calculus and differential equations, often used to describe how one quantity varies in relation to another.

For populations, the rate of change is typically given as a derivative \(y'(t)\). This derivative shows whether the population is increasing, decreasing, or remaining constant at any given time. Analyzing the rate of change gives insights into population trends.

In the context of the textbook exercise, the rate of change is expressed as:

• \( y'(t) = 0.008y - h\)

If \(y'(t) = 0\), the population does not change over time. This situation is often desired in controlled environments, where the goal is to maintain a steady population numbers.

Knowing the rate of change is vital when adjusting the variables in the equation (such as the initial population size or the harvesting rate) to achieve desired population outcomes. It helps to ensure that the resource is managed sustainably without unintended population crashes.