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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 \(t \rightarrow \infty\)

Short Answer

Expert verified
The solution is \(L\), as \( P(t) \rightarrow L \) as \( t \rightarrow \infty \).

Step by step solution

01

Understanding the Differential Equation

The given differential equation is \( \frac{dP}{dt} = kP(L-P) \), known as the logistic growth equation. Here, \(P(t)\) represents the population at time \(t\), \(k\) is the growth rate, and \(L\) is the carrying capacity.
02

Identifying Equilibrium Points

The solutions to this differential equation possess equilibrium points found by setting \( \frac{dP}{dt} = 0 \). Thus, solving \( kP(L-P) = 0 \) gives the equilibrium points \( P = 0 \) and \( P = L \).
03

Analyzing Stability of Equilibrium Points

To determine the stability of the equilibrium points, we analyze small perturbations around these points. If \( P < L \), \( \frac{dP}{dt} > 0 \), indicating that the population grows towards \( L \). If \( P > L \), \( \frac{dP}{dt} < 0 \), indicating that the population decreases back towards \( L \). This shows that \( P = L \) is a stable equilibrium.
04

Determining the Limiting Population

As \( t \rightarrow \infty \), the population \( P(t) \) progresses towards the stable equilibrium point \( P = L \). Hence, regardless of the initial condition \( P(0) > 0 \), the limiting value of \( P(t) \) as \( t \rightarrow \infty \) is \( L \).

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Key Concepts

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

Differential Equations
Differential equations are mathematical expressions that relate a function to its derivatives. They are powerful tools used to model how a quantity changes over time.
In our problem, we deal with a specific type called a logistic growth equation: \[ \frac{dP}{dt} = kP(L-P) \] Here, \(P(t)\) represents a population at time \(t\), \(k\) is a constant growth rate, and \(L\) is the carrying capacity—the maximum population that the environment can support.
This equation is termed 'logistic' because it involves a limiting factor, \((L-P)\), which slows down the growth as \(P\) approaches \(L\). At the start, when \(P\) is small, the growth is nearly exponential. As \(P\) increases, the growth rate decreases and eventually halts when \(P=L\).
This behavior is commonly seen in biological systems, where resources are limited. Understanding this principle is crucial for analyzing real-world population dynamics.
Equilibrium Points
Equilibrium points occur where the population stops changing, meaning \( \frac{dP}{dt} = 0 \). In logistic growth, equilibrium points can be found from the equation:\[ kP(L-P) = 0 \] Solving this gives us two possible solutions: \( P = 0 \) and \( P = L \).
\( P = 0 \) represents extinction, where there is no population present. However, \( P = L \) is a state where the population perfectly matches the carrying capacity.
In terms of stability, an equilibrium point can either be stable or unstable. A stable equilibrium suggests that if a population is slightly displaced from it, it will tend to move back to the equilibrium point. Conversely, an unstable equilibrium suggests that small disturbances grow, leading the population away from the point.
In our logistic growth case, \( P = L \) is a stable equilibrium while \( P = 0 \) is unstable. Thus, populations tend to move towards carrying capacity over time.
Carrying Capacity
Carrying capacity, denoted as \(L\) in the logistic growth model, is an important concept in ecology and population dynamics.
It represents the maximum population size that the environment can sustainably support, given resources such as food, habitat, and water. As the population \(P\) approaches \(L\), growth slows and eventually ceases.
The carrying capacity is a critical factor for long-term sustainability. If a population exceeds \(L\), negative feedback mechanisms invlove a decrease in population size due to resource limits.
In the logistic differential equation \( \frac{dP}{dt} = kP(L-P) \), the term \((L-P)\) acts as a regulatory factor. When \(P\) nears \(L\), the term diminishes, slowing growth significantly. This dynamic tends to age the population near a stable equilibrium, aligning with \(L\) in the long term.
Understanding carrying capacity helps in managing ecosystem resources effectively, ensuring sustainability for different species over time.

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Most popular questions from this chapter

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 \(y=f(t)\)

(a) Sketch the slope field for \(y^{\prime}=x / y\) (b) Sketch several solution curves. (c) Solve the differential equation analytically.

In Problems \(55-58,\) give an example of: A differential equation all of whose solutions form the family of functions \(f(x)=x^{2}+C\).

(a) Define the variables. (b) Write a differential equation to describe the relationship. (c) Solve the differential equation. Nicotine leaves the body at a rate proportional to the amount present, with constant of proportionality 0.347 if the amount of nicotine is in \(\mathrm{mg}\) and time is in hours. The amount of nicotine in the body immediately after smoking a cigarette is \(0.4 \mathrm{mg}\).

A rumor spreads among a group of 400 people. The number of people, \(N(t),\) who have heard the rumor by time \(t\) in hours since the rumor started is approximated by $$N(t)=\frac{400}{1+399 e^{-0.4 t}}$$ (a) Find \(N(0)\) and interpret it. (b) How many people will have heard the rumor after 2 hours? After 10 hours? (c) Graph \(N(t)\) (d) Approximately how long will it take until half the people have heard the rumor? 399 people? (e) When is the rumor spreading fastest?

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