/*! 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 23 Suppose that the logistic equati... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 23

Suppose that the logistic equation \(d x / d t=k x(M-x)\) models a population \(x(t)\) of fish in a lake after \(t\) months during which no fishing occurs. Now suppose that, because of fishing, fish are removed from the lake at the rate of \(h x\) fish per month (with \(h\) a positive constant). Thus fish are "harvested" at a rate proportional to the existing fish population, rather than at the constant rate of Example 4. (a) If \(0

Short Answer

Expert verified
(a) Limit is \( M - \frac{h}{k} \) if \( h < kM \). (b) If \( h \geqq kM \), \( x(t) \rightarrow 0 \) as \( t \rightarrow +\infty \).

Step by step solution

01

Understanding the Logistic Equation

The logistic equation given is \( \frac{dx}{dt} = kx(M - x) \). This models a population where \( x(t) \) represents the fish population at time \( t \), \( k \) is the growth rate constant, and \( M \) is the carrying capacity of the lake.
02

Modify the Equation for Harvesting

With fish being removed at the rate of \( hx \), the modified logistic equation becomes \( \frac{dx}{dt} = kx(M - x) - hx \). This introduces a term \( -hx \) to account for the harvesting.
03

Simplify the Modified Equation

Factor the right-hand side of the equation: \( \frac{dx}{dt} = kx(M - x) - hx = x(kM - kx - h) = x(k(M - x) - h) \). This shows that the equation still has the form of a logistic equation with an altered growth rate.
04

Analyze the New Equilibrium Points

Set \( \frac{dx}{dt} = 0 \) to find equilibrium points: \( x(k(M - x) - h) = 0 \). This gives equilibrium solutions \( x = 0 \) and \( k(M - x) - h = 0 \), or \( x = M - \frac{h}{k} \).
05

Condition for Positive Limiting Population

For a positive equilibrium \( M - \frac{h}{k} > 0 \), the condition \( h < kM \) must hold. Thus, the population stabilizes at \( x = M - \frac{h}{k} \) if \( 0 < h < kM \).
06

Analyze \( h \geq kM \) Condition

If \( h \geq kM \), the equilibrium point \( M - \frac{h}{k} \leq 0 \), meaning the only feasible solution is \( x(t) \rightarrow 0 \) as \( t \rightarrow +\infty \). This indicates the population will eventually be fished out.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Population Dynamics
Population dynamics refers to the changes in the size and composition of a population over time. It considers how the population numbers increase or decrease due to births, deaths, and migration.
In the context of our exercise, the logistic growth model is used to understand how populations of fish behave in a lake environment. Without human intervention, fish populations grow in a predictable, non-linear fashion controlled by factors such as birth rates and the availability of resources, depicted by parameters like growth rate (\(k\)) and carrying capacity (\(M\)).
Understanding these dynamics allows us to predict how a population will change under different circumstances. The equation \( \frac{dx}{dt} = kx(M - x) \) shows this relationship clearly, where the growth changes based on how close the population is to its carrying capacity.
Equilibrium Solutions
Equilibrium solutions are important in determining the long-term behavior of a population as time goes to infinity. They represent the population sizes that don't change over time.
In logistic equations, we typically aim to find when the rate of change is zero, which is when the fish population stabilizes. For our model, this occurs when \( \frac{dx}{dt} = 0 \). By solving this equation, we find that \(x = 0\) and \(x = M - \frac{h}{k}\) are equilibria. The point \(x = 0\) represents extinction, while \(x = M - \frac{h}{k}\) represents a stable population size under the influence of both natural growth and harvesting.
  • If \(0 < h < kM\), the population stabilizes at \(x = M - \frac{h}{k}\).
  • If \(h \geq kM\), the stable solution becomes \(x = 0\).
Thus, knowing equilibrium solutions helps us predict whether the population will persist or decline to extinction over time.
Carrying Capacity
Carrying capacity is the maximum number of individuals that an environment can sustainably support without being degraded.
In the logistic growth model, this is depicted by the parameter \(M\). It acts as a natural limit to the growth of the fish population, accounting for limited resources like food and space. The closer a population comes to its carrying capacity, the slower its growth becomes until it eventually levels off.
When the equation \(M - x\) in our logistic growth model closes to zero, it indicates that the population is nearing its carrying capacity. Without harvesting (\(h = 0\)), the population will stabilize at this capacity. However, if harvesting is introduced, the effective carrying capacity becomes \(M - \frac{h}{k}\), altering the level at which the population stabilizes.
Harvesting Impact
Harvesting impact refers to how removing individuals from a population affects its dynamics and stability.
In our exercise, harvesting is represented by the term \(hx\), which is subtracted from the logistic growth model, reducing the population at a rate proportional to its size. This action decreases the overall growth rate and alters the equilibrium points.
  • If the harvest rate \(h\) is lower than \(kM\), the population can still reach a new stable state, albeit at a lower capacity: \(x = M - \frac{h}{k}\).
  • However, if harvesting is too aggressive (\(h \geq kM\)), it drives the population size to zero, indicating that the population will eventually be depleted.
Understanding harvesting impact is essential for sustainable management practices, ensuring that fishing activities do not lead to regional extinction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A computer with a printer is required for Problems 17 through 24. In these initial value problems, use the Runge-Kutta method with step sizes \(h=0.2,0.1,0.05\), and \(0.025\) to approximate to six decimal places the values of the solution at five equally spaced points of the given interval. Print the results in tabular form with appropriate headings to make it easy to gauge the effect of varying the step size \(h .\) Throughout, primes denote derivatives with respect to \(x\). $$ y^{\prime}=\sin x+\cos y, y(0)=0 ; 0 \leqq x \leqq 1 $$

This problem deals with the differential equation \(d x / d t=\) \(k x(x-M)-h\) that models the harvesting of an unsophisticated population (such as alligators). Show that this equation can be rewritten in the form \(d x / d t=k(x-\) \(H)(x-K)\), where $$ \begin{aligned} &H=\frac{1}{2}\left(M+\sqrt{M^{2}+4 h / k}\right)>0, \\ &K=\frac{1}{2}\left(M-\sqrt{M^{2}+4 h / k}\right)<0 . \end{aligned} $$ Show that typical solution curves look as illustrated

The time rate of change of an alligator population \(P\) in a swamp is proportional to the square of \(P .\) The swamp contained a dozen alligators in 1988 , two dozen in 1998 . When will there be four dozen alligators in the swamp? What happens thereafter?

A hand-held calculator will suffice for, where an initial value problem and its exact solution are given. Apply the improved Euler method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.1 .\) Construct a table showing four-decimal-place values of the approximate solution and actual solution at the points \(x=0.1,0.2,0.3,0.4\), \(0.5\). $$ y^{\prime}=y+1, y(0)=1 ; y(x)=2 e^{x}-1 $$

It is proposed to dispose of nuclear wastes - in drums with weight \(W=640 \mathrm{lb}\) and volume \(8 \mathrm{ft}^{3}\) -by dropping them into the ocean \(\left(v_{0}=0\right)\). The force equation for a drum falling through water is $$ m \frac{d v}{d t}=-W+B+F_{R} $$ where the buoyant force \(B\) is equal to the weight (at \(62.5\) \(\mathrm{lb} / \mathrm{ft}^{3}\) ) of the volume of water displaced by the drum (Archimedes' principle) and \(F_{R}\) is the force of water resistance, found empirically to be \(1 \mathrm{lb}\) for each foot per second of the velocity of a drum. If the drums are likely to burst upon an impact of more than \(75 \mathrm{ft} / \mathrm{s}\), what is the maximum depth to which they can be dropped in the ocean without likelihood of bursting?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.