Question

A fish hatchery has 500 fish at \(t=0,\) when harvesting begins at a rate of \(b>0\) fish/year. The fish population is modeled by the initial value problem \(y^{\prime}(t)=0.01 y-b\) \(y(0)=500,\) where \(t\) is measured in years. a. Find the fish population, for \(t \geq 0,\) in terms of the harvesting rate \(b\) b. Graph the solution in the case that \(b=40\) fish/year. Describe the solution. c. Graph the solution in the case that \(b=60\) fish/year. Describe the solution.

Short answer

Answer: For a harvesting rate of 40 fish per year, the fish population is given by the function \(y(t) = (500 + \frac{40}{0.01})e^{-0.01t} - \frac{40}{0.01}\). This function shows that the fish population initially decreases, reaches a minimum, and then gradually increases back to its carrying capacity of 1000 fish.

Step-by-step solution

Solve the initial value problem

To find the general solution for the fish population, we need to first solve the given initial value problem. We are given the differential equation: \(y^{\prime}(t)=0.01y(t) - b\) with the initial condition: \(y(0)=500\) This is a first-order linear differential equation, and we can solve it using an integrating factor: \(IF(t) = e^{\int P(t) dt}\), where \(P(t) = 0.01\). So, \(IF(t) = e^{0.01t}\). Now, we multiply the differential equation by the integrating factor: \(e^{0.01t} y^{\prime}(t) + 0.01e^{0.01t} y(t) = -be^{0.01t}\) The left-hand side of this equation is the derivative of the product of the dependent variable y(t) and the integrating factor: \((y(t)e^{0.01t})^{\prime} = -be^{0.01t}\) Now we integrate both sides with respect to t: \(\int (y(t)e^{0.01t})^{\prime} dt = \int -be^{0.01t} dt\) So, \(y(t)e^{0.01t} = -\frac{b}{0.01}e^{0.01t} + C\) Now we can solve for y(t): \(y(t) = Ce^{-0.01t} - \frac{b}{0.01}\) To find the constant C, we can use the initial condition y(0)=500: \(500 = Ce^0 - \frac{b}{0.01}\) \(C = 500 + \frac{b}{0.01}\) So, the fish population is given by: \(y(t) = (500 + \frac{b}{0.01})e^{-0.01t} - \frac{b}{0.01}\)

Graph the solution for b = 40 fish/year

Now, we want to graph the solution of the initial value problem with a harvesting rate of b = 40 fish/year. We substitute b=40 into the equation for y(t): \(y(t) = (500 + \frac{40}{0.01})e^{-0.01t} - \frac{40}{0.01}\) By graphing this function with respect to time, we can observe that the fish population initially decreases and reaches a minimum before gradually increasing back to its carrying capacity of 1000 fish (given by b/0.01). This indicates that with a harvesting rate of 40 fish/year, the fish population will eventually adapt and recover.

Graph the solution for b = 60 fish/year

Now, we want to graph the solution of the initial value problem with a harvesting rate of b = 60 fish/year. We substitute b=60 into the equation for y(t): \(y(t) = (500 + \frac{60}{0.01})e^{-0.01t} - \frac{60}{0.01}\) By graphing this function with respect to time, we can observe that the fish population steadily decreases and converges to a steady state with a significantly reduced population. This indicates that with a harvesting rate of 60 fish/year, the fish population will not be able to recover and will eventually reach a new equilibrium with a smaller population.

Key concepts

Understanding Differential Equations

A differential equation is a mathematical equation that relates a function to its derivatives. It is a way of describing how a quantity changes over time or space, and is a central tool in modeling a wide range of physical and biological processes.

In this particular problem, the differential equation models the growth of a fish population over time. It takes into account the natural growth rate of the fish population and the effect of harvesting. The key part of solving such an equation is to find a function that satisfies the equation; this function describes the behavior of the fish population over time. Differential equations are categorized according to their order, which is determined by the highest derivative present in the equation. The equation in our exercise is a first-order differential equation since it involves only the first derivative of the function.

Applying the Integrating Factor

An integrating factor is a clever mathematical technique used to solve linear differential equations of the form \(y'+P(t)y=Q(t)\). It involves finding a function, typically denoted as \(IF(t)\), that when multiplied by the original equation, simplifies it to a form that can be easily integrated.

In our fish population problem, we have used the integrating factor \(e^{0.01t}\) which, when multiplied by our differential equation, allows us to write the left-hand side of the equation as the derivative of a product of functions. This greatly simplifies the problem, as we can then integrate both sides directly to find an expression for the fish population over time.

The Fish Population Model

The fish population model given in this exercise is an example of how differential equations can be used to model real-world scenarios. This model consists of two parts: a growth term that is proportional to the current fish population (denoted by \(0.01y\)), and a harvesting term (denoted by \(b\)) that represents the rate at which fish are removed from the population.

By solving the differential equation, we can predict how the fish population changes over time, given different rates of harvesting. These predictions can inform fishery management decisions, such as setting sustainable harvest levels to prevent overfishing and ensure the long-term health of the fish population.

Analyzing the Harvesting Rate

The harvesting rate, represented by the variable \(b\), refers to the number of fish removed from the population per year. It plays a crucial role in the fish population model, as it directly impacts the population's ability to sustain itself over time.

If the harvesting rate is too high (as in the case with \(b=60\) fish/year), the model predicts that the fish population will decrease and stabilize at a lower level than it would without harvesting. Conversely, a smaller harvesting rate (such as \(b=40\) fish/year) may allow the population to eventually recover to its original size, assuming that it is not too high compared to the natural growth rate of the fish population. Understanding how different harvesting rates affect the population can guide the implementation of effective conservation practices.