/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Federal or state agencies contro... [FREE SOLUTION] | 91Ó°ÊÓ

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Federal or state agencies control hunting and fishing by setting a quota on how many animals can be harvested each season. Determining the appropriate quota means achieving a balance between environmental concerns and the interests of hunters and fishers. For example, when a June 8,2007 decision by the Delaware Superior Court invalidated a two-year moratorium on catching horseshoe crabs, the Delaware Department of Natural 91Ó°ÊÓ and Environmental Control imposed instead an annual quota of 100,000 on male horseshoe crabs. Environmentalists argued this would exacerbate a decrease in the protected Red Knot bird population that depends on the crab for food. For a population \(P\) that satisfies the logistic model with harvesting, $$\frac{d P}{d t}=k P\left(1-\frac{P}{L}\right)-H$$ show that the quota, \(H,\) must satisfy \(H \leq k L / 4,\) or else the population \(P\) may die out. (In fact, \(H\) should be kept much less than \(k L / 4 \text { to be safe. })\)

Short Answer

Expert verified
Quota \( H \) must satisfy \( H \leq \frac{k L}{4} \) to prevent extinction.

Step by step solution

01

Understanding the logistic model with harvesting

The given logistic model for population with harvesting is \( \frac{d P}{d t} = k P \left(1-\frac{P}{L}\right) - H \), where \( P \) is the population, \( k \) is the intrinsic growth rate, \( L \) is the carrying capacity, and \( H \) is the harvesting rate or quota. The logistic model explains how a population grows limited by its carrying capacity while harvesting reduces the growth by \( H \).
02

Analyzing equilibrium points

To find the equilibrium, set \( \frac{d P}{d t} = 0 \):\[ k P \left(1-\frac{P}{L}\right) - H = 0 \]Solving for \( P \), we get:\[ P = \frac{L}{2} \pm \sqrt{\left(\frac{L}{2}\right)^2 - \frac{H L}{k}} \]The term under the square root (the discriminant) must be non-negative for \( P \) to have real solutions, ensuring the population does not die out. This implies the condition for the discriminant is fulfilling \( \left(\frac{L}{2}\right)^2 - \frac{H L}{k} \geq 0 \).
03

Solving the inequality for the discriminant

The inequality from Step 2 is:\[ \left(\frac{L}{2}\right)^2 - \frac{H L}{k} \geq 0 \]This simplifies to:\[ \frac{L^2}{4} \geq \frac{H L}{k} \]Dividing both sides by \( L \) (assuming \( L eq 0 \)) gives:\[ \frac{L}{4} \geq \frac{H}{k} \]Thus, multiplying both sides by \( k \) results in:\[ H \leq \frac{k L}{4} \]
04

Conclusion

To ensure the population \( P \) does not die out, the harvesting quota \( H \) must satisfy \( H \leq \frac{k L}{4} \). It is recommended to keep \( H \) much less than \( \frac{k L}{4} \) to ensure the population remains safe from extinction.

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Key Concepts

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

Harvesting Quota
When managing wildlife populations, setting a harvesting quota is crucial.
The quota represents the maximum number of animals that can be harvested in a given period.
This ensures that the population remains sustainable. In the context of a logistic population model, the quota is denoted by the parameter \( H \).
This parameter directly affects the balance of the population under study.

The goal is to set this quota responsibly so that the population doesn't decline to dangerous levels. In our exercise, ensuring that \( H \) is less than \( \frac{kL}{4} \) is essential.
By doing so, the population has room to grow or stabilize without being overharvested.
Setting too high of a quota can risk the population's ability to recover, leading to possible extinction scenarios.

Considering environmental needs alongside economic and recreational interests helps maintain a balanced ecosystem.
Regulatory bodies often use models like the logistic equation to guide these decisions.
Harmonizing these interests is key to effective wildlife management.
Equilibrium Points
Equilibrium points in a population model indicate where the population neither increases nor decreases over time.
They are pivotal in understanding the population dynamics.
In logistic models, the equilibrium is found when the rate of change in population, \( \frac{dP}{dt} \), equals zero.

By solving the equation \( kP\left(1-\frac{P}{L}\right) - H = 0 \), we identify these points.
The solutions to this equation tell us where the population can stabilize.
To ensure realistic solutions, the term inside the square root must be non-negative.
This ensures there are real numbers for equilibrium points.

In practical terms, ensuring a real solution means the population doesn't die out and can maintain itself despite the harvesting.
This balance is dictated by the interplay of growth, carrying capacity, and harvesting rate.
Logistic Equation
The logistic equation is at the heart of many population dynamics studies.
It models how populations grow, taking into account a limit or carrying capacity \( L \).
This capacity is the maximum population size that the environment can sustain indefinitely.

The equation is expressed as \( \frac{dP}{dt} = kP\left(1-\frac{P}{L}\right) \), where \( k \) represents the intrinsic growth rate.
Without harvesting, the population would approach \( L \) as it grows.
However, the introduction of a harvesting quota \( H \) modifies the equation to \( \frac{dP}{dt} = kP\left(1-\frac{P}{L}\right) - H \).
The stochastic nature of the equation makes it a robust tool for studying population dynamics and planning sustainable harvesting strategies.

Through this modified equation, we understand how extraction or harvesting affects the population size and its long-term viability.
Population Growth
Population growth in natural systems is rarely unlimited due to environmental constraints.
The logistic population model provides a more realistic framework that includes these limits.
It acknowledges that as the population size \( P \) approaches the carrying capacity \( L \), the growth rate slows down and eventually reaches stability.

This growth can be affected by several factors like resources, space, and predation.
Understanding growth patterns allows us to predict changes in the ecosystem.
For management purposes, knowing how quickly a population might rebound or decline helps in making informed decisions about harvesting quotas.

Sustainable population growth ensures that species and ecosystems can thrive without depletion.
This balance is critical for long-term environmental health and human socio-economic interests.
Consciously monitoring these growth patterns allows regulatory bodies to establish quotas that prevent overharvesting and ensure population resilience.

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Most popular questions from this chapter

In the 1930 s, the Soviet ecologist G. F. Gause \(^{22}\) studied the population growth of yeast. Fit a logistic curve, \(d P / d t=k P(1-P / L),\) to his data below using the method outlined below. $$\begin{array}{l|c|c|c|c|c|c|c}\hline \text { Time (hours) } & 0 & 10 & 18 & 23 & 34 & 42 & 47 \\ \hline \text { Yeast pop } & 0.37 & 8.87 & 10.66 & 12.50 & 13.27 & 12.87 & 12.70 \\\\\hline\end{array}$$(a) Plot the data and use it to estimate (by eye) the carrying capacity, \(L\) (b) Use the first two pieces of data in the table and your value for \(L\) to estimate \(k\) (c) On the same axes as the data points, use your values for \(k\) and \(L\) to sketch the solution curve \(P=\frac{L}{1+A e^{-k t}} \quad\) where \(\quad A=\frac{L-P_{0}}{P_{0}}\)

As you know, when a course ends, students start to forget the material they have learned. One model (called the Ebbinghaus model) assumes that the rate at which a student forgets material is proportional to the difference between the material currently remembered and some positive constant, \(a\). (a) Let \(y=f(t)\) be the fraction of the original material remembered \(t\) weeks after the course has ended. Set up a differential equation for \(y .\) Your equation will contain two constants; the constant \(a\) is less than \(y\) for all \(t\). (b) Solve the differential equation. (c) Describe the practical meaning (in terms of the amount remembered) of the constants in the solution \(y=f(t)\)

Give an explanation for your answer. For any positive values of the constant \(k\) and any positive values of the initial value \(P(0),\) the solution to the differential equation \(d P / d t=k P(L-P)\) has limiting value \(L\) as \(t \rightarrow \infty\)

(a) An object is placed in a \(68^{\circ} \mathrm{F}\) room. Write a differential equation for \(H,\) the temperature of the object at time \(t.\) (b) Find the equilibrium solution to the differential equation. Determine from the differential equation whether the equilibrium is stable or unstable. (c) Give the general solution for the differential equation. (d) The temperature of the object is \(40^{\circ} \mathrm{F}\) initially and \(48^{\circ} \mathrm{F}\) one hour later. Find the temperature of the object after 3 hours.

Many organ pipes in old European churches are made of tin. In cold climates such pipes can be affected with tin pest, when the tin becomes brittle and crumbles into a gray powder. This transformation can appear to take place very suddenly because the presence of the gray powder encourages the reaction to proceed. The rate of the reaction is proportional to the product of the amount of tin left and the quantity of gray powder, \(p,\) present at time \(t .\) Assume that when metallic tin is converted to gray powder, its mass does not change. (a) Write a differential equation for \(p .\) Let the total quantity of metallic tin present originally be \(B\) (b) Sketch a graph of the solution \(p=f(t)\) if there is a small quantity of powder initially. How much metallic tin has crumbled when it is crumbling fastest? (c) Suppose there is no gray powder initially. (For example, suppose the tin is completely new.) What does this model predict will happen? How do you reconcile this with the fact that many organ pipes do get tin pest?

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