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Give an example of: A logistic differential equation for a quantity \(P\) such that the maximum rate of change of \(P\) occurs when \(P=75\)

Short Answer

Expert verified
\(\frac{dP}{dt} = kP\left(1 - \frac{P}{150}\right)\) with \(P=75\) for maximum rate of change.

Step by step solution

01

Understand Logistic Differential Equation

A logistic differential equation models population growth with a limiting factor. The general form is \(\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)\), where \(k\) is the growth rate constant and \(L\) is the carrying capacity of the environment. The maximum rate of change occurs when \(P = \frac{L}{2}\).
02

Set up the Condition for Maximum Rate of Change

The maximum rate happens at \(P = 75\). According to logistic growth, this is when \(P = \frac{L}{2}\). Therefore, \(\frac{L}{2} = 75\), which gives \(L = 150\).
03

Write the Logistic Differential Equation

With \(L = 150\) and the general logistic form, the differential equation can be written as \(\frac{dP}{dt} = kP\left(1 - \frac{P}{150}\right)\), where \(k\) is the growth rate constant to be determined by specific environmental conditions or data.

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Key Concepts

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

Population Growth
Population growth is an important biological process that can be modeled with different mathematical equations. One of the most significant models is the logistic differential equation, which describes how a population grows over time with a limiting factor.
Initially, populations tend to grow exponentially, meaning that the larger the population, the faster it grows. This is because each individual organism has the potential to reproduce multiple times, leading the population size to double, triple, or grow even more rapidly over time.
However, this exponential growth cannot continue indefinitely in a real-world environment, as resources such as food, space, and water become limiting factors. As a result, the growth rate slows down and eventually stabilizes, reaching a plateau. This is the essence of logistic growth.
  • The logistic growth is represented by the equation: \[ \frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right) \], where:
    • \( P \) is the population size or quantity at a given time.
    • \( k \) is the intrinsic growth rate constant.
    • \( L \) is the carrying capacity or maximum population an environment can sustain.
Understanding population dynamics through logistic equations enables better prediction and management of biological resources.
Carrying Capacity
Carrying capacity is a crucial concept when analyzing how populations grow in nature. It refers to the maximum population size of a species that an environment can sustain indefinitely. In the logistic differential equation, it plays a pivotal role in predicting how and when a population will stabilize.
The carrying capacity is determined by various environmental factors, including available resources, habitat space, and competition among individuals. When the population reaches this capacity, the growth rate decreases until it approaches zero, as the resources are just enough to keep the population stable.
  • In the logistic differential equation \( \frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right) \), the term \( \left(1 - \frac{P}{L}\right) \) indicates that as \( P \) approaches \( L \), the growth rate diminishes.
The environment can impose a natural limit through factors such as predation, disease, and resource availability. By understanding carrying capacities, ecologists can suggest sustainable practices for resource usage, ensuring that populations remain balanced over time.
Maximum Rate of Change
In the context of logistic growth, the maximum rate of change of a population occurs not at the beginning, but rather when the population is halfway to its carrying capacity. This is a key insight that distinguishes logistic growth from purely exponential growth.
Mathematically, this condition is realized when the population size \( P \) is equal to \( \frac{L}{2} \), where \( L \) is the carrying capacity. At this point, the force driving the growth is strongest, but it starts declining as the population continues to grow closer to \( L \).
  • Using the logistic equation: \[ \frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right) \], when \( P = \frac{L}{2} \), the derivative \( \frac{dP}{dt} \) reaches its maximum value.
  • For instance, if the carrying capacity \( L \) is 150, the maximum rate of population change occurs at \( P = 75 \).
This point of maximum change reflects how quickly a population can grow under optimal conditions—and how that growth rate slows as limits set by the environment become more significant.

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Most popular questions from this chapter

A model for the population, \(P,\) of carp in a landlocked lake at time \(t\) is given by the differential equation $$\frac{d P}{d t}=0.25 P(1-0.0004 P)$$ (a) What is the long-term equilibrium population of carp in the lake? (b) A census taken ten years ago found there were 1000 carp in the lake. Estimate the current population. (c) Under a plan to join the lake to a nearby river, the fish will be able to leave the lake. A net loss of \(10 \%\) of the carp each year is predicted, but the patterns of birth and death are not expected to change. Revise the differential equation to take this into account. Use the revised differential equation to predict the future development of the carp population.

The systems of differential equations model the interaction of two populations \(x\) and \(y .\) In each case, answer the following two questions: (a) What kinds of interaction (symbiosis, \(^{30}\) competition, predator-prey) do the equations describe? (b) What happens in the long run? (For one of the systems, your answer will depend on the initial populations.) Use a calculator or computer to draw slope fields. $$\begin{array}{l} \frac{1}{x} \frac{d x}{d t}=y-1-0.05 x \\ \frac{1}{y} \frac{d y}{d t}=1-x-0.05 y \end{array}$$

Explain what is wrong with the statement. At a time when a bank balance \(B \text{dollars},\) which satisfies \(d B / d t=0.08 B-250,\) is \(5000 \text{dollars},\) the balance is going down.

Is the statement true or false? Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=g(x) .\) If the statement is true, explain how you know. If the statement is false, give a counterexample. If \(g(x)\) is increasing for \(x>0,\) then so is \(f(x).\)

Sketch solution curves with a variety of initial values for the differential equations. You do not need to find an equation for the solution. \(\frac{d w}{d t}=(w-3)(w-7)\)

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