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Fishing for sardines: This is a continuation of Example 6.10. If we take into account an annual fish harvest of \(F\) million tons of fish, then the equation of change for Pacific sardines becomes $$ \frac{d N}{d t}=0.338 N\left(1-\frac{N}{2.4}\right)-F . $$ a. Suppose that there are currently \(1.8\) million tons of Pacific sardines off the California coast and that you are in charge of the commercial fishing fleet. It is your goal to leave the Pacific population of sardines as you found it. That is, you wish to set the fishing level \(F\) so that the biomass of Pacific sardines remains stable. What value of \(F\) will accomplish this? (Hint: You want to choose \(F\) so that the current biomass level of \(1.8\) million tons is an equilibrium solution.) b. For the remainder of this exercise, take the value of \(F\) to be \(0.1\) million tons per year. That is, assume the catch is 100,000 tons per year. i. Make a graph of \(\frac{d N}{d t}\) versus \(N\), and use it to find the equilibrium solutions. ii. For what values of \(N\) will the biomass be increasing? For what values will it be decreasing? iii. On the same graph, sketch all equilibrium solutions and the graphs of \(N\) versus \(t\) for each of the initial populations \(N(0)=0.3\) million tons, \(N(0)=1.0\) million tons, and \(N(0)=2.3\) million tons. \(\rightarrow\) iv. Explain in practical terms what the picture you made in part iii tells you. Include in your explanation the significance of the equilibrium solutions.

Step by step solution

01

Find Equilibrium Solution for Part (a)

To maintain the sardine population at a stable level of 1.8 million tons, set \( \frac{d N}{d t} = 0 \) and solve the equation \[ 0.338N \left(1-\frac{N}{2.4}\right) - F = 0. \] Plug \( N = 1.8 \) into the equation and solve for \( F \):\[ 0.338 \cdot 1.8 \left(1-\frac{1.8}{2.4}\right) - F = 0. \] This simplifies to \( 0.338 \cdot 1.8 \left(1-0.75\right) = F \), which gives \( F = 0.15255 \). Thus, the fishing level \( F \) should be approximately \( 0.1526 \) million tons per year.

02

Graph dN/dt versus N for Part (b)(i)

Substitute \( F = 0.1 \) into the equation, \[ \frac{dN}{dt} = 0.338N\left(1-\frac{N}{2.4}\right) - 0.1. \] We will plot \( \frac{dN}{dt} \) as a function of \( N \) in a graphing tool. The zeros of this function, where \( \frac{dN}{dt} = 0 \), correspond to equilibrium solutions.

03

Solve for Equilibrium Solutions

Set \[ \frac{dN}{dt} = 0.338N\left(1-\frac{N}{2.4}\right) - 0.1 = 0 \] to find equilibrium solutions. Rearranging gives \[ 0.338N\left(1-\frac{N}{2.4}\right) = 0.1. \] Solve for \( N \). The simplified equation is quadratic and further simplification gives roots indicating equilibria. Through calculation, approximately \( N = 0.426 \) and \( N = 2.1 \) as equilibria.

04

Determine Increasing or Decreasing Biomass Levels

Consider the signs of \( \frac{dN}{dt} \) for different intervals of \( N \).- If \( N < 0.426 \), \( \frac{dN}{dt} > 0 \), biomass increases.- If \( 0.426 < N < 2.1 \), \( \frac{dN}{dt} < 0 \), biomass decreases.- If \( N > 2.1 \), \( \frac{dN}{dt} > 0 \), biomass increases.

05

Sketch Equilibrium and N(t) Graphs for Part (b)(iii)

Plot the equilibrium solutions from Step 3 on a new graph with the initial conditions \( N(0)=0.3 \), \( N(0)=1.0 \), and \( N(0)=2.3 \).- For each initial condition, draw the trajectory of \( N(t) \) based on increasing or decreasing behavior explained in Step 4.

06

Explain Graph Interpretation for Part (b)(iv)

The graph shows how populations evolve over time for different initial populations.- Below \( N = 0.426 \) and above \( N = 2.1 \), populations increase towards the nearest equilibrium.- Between \( N = 0.426 \) and \( N = 2.1 \), populations decrease towards equilibrium \( N = 2.1 \). This analysis signifies population stability points, with \( N = 2.1 \) as a stable equilibrium and \( N = 0.426 \) as an unstable equilibrium.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Solutions

In the context of differential equations and population modeling, an equilibrium solution is a consistent state where the population does not change over time. This occurs when the rate of change, represented by \( \frac{dN}{dt} \), is zero. For the sardine population, the equilibrium solutions are found by setting \( \frac{dN}{dt} = 0 \). This means the growth of the population is perfectly balanced by the fish harvest, resulting in a stable sardine biomass.
To find specific equilibrium solutions, we solve the equation: \[ 0.338N \left(1-\frac{N}{2.4}\right) - F = 0. \] For instance, using a fishing level \( F \), if we want the population to remain stable at 1.8 million tons, we plug \( N = 1.8 \) into the equation. This calculation determines the precise fishing level needed to keep the population constant. In practice, this helps manage resources sustainably by setting the harvest at a level that ensures the population neither declines nor increases excessively.
Equilibrium solutions are crucial because they denote points where interventions like fishing can be optimized for long-term ecological balance.

Population Modeling

Population modeling involves using mathematical equations to represent how a population changes over time. In this model, the Pacific sardine population is influenced by both biological growth and human interaction, through fishing, modeled by the equation: \[ \frac{dN}{dt} = 0.338N\left(1-\frac{N}{2.4}\right) - F, \] where \( N \) represents the biomass of the sardine population, and \( F \) is the harvest rate.
The first term, \( 0.338N\left(1-\frac{N}{2.4}\right) \), represents the logistic growth of the population, accounting for natural reproduction and the carrying capacity of the environment. This means population growth slows as it approaches the environmental limit. The second term, \( -F \), represents external removal of the population due to fishing.
To manage the sardine population effectively, understanding this model enables predictions about how changes in fishing intensity or natural conditions affect long-term population dynamics. By adjusting \( F \), stakeholders can foresee the impact of different fishing strategies, helping to maintain ecological balance.

Rate of Change

The concept of the rate of change in population modeling refers to how quickly or slowly a population increases or decreases. This is indicated by the derivative \( \frac{dN}{dt} \), representing the rate at which the sardine biomass changes with respect to time. It determines whether the population is stable, growing, or declining.
In our sardine example, the rate of change is derived from the equation: \[ \frac{dN}{dt} = 0.338N\left(1-\frac{N}{2.4}\right) - F. \]
By analyzing this equation, we can determine how different population sizes influence the increase or decrease of sardines over time. For example, for biomass \( N < 0.426 \), the rate of change \( \frac{dN}{dt} > 0 \), meaning the population is increasing. Conversely, between \( 0.426 < N < 2.1 \), \( \frac{dN}{dt} < 0 \), indicating a decline.
Understanding rate of change is essential as it helps predict future population trends under varying conditions, allowing for informed decision-making in resource management and ensuring the population remains within sustainable limits.