/*! 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 9 A population is modeled by the d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A population is modeled by the differential equation $$\frac{d P}{d t}=1.2 P\left(1-\frac{P}{4200}\right)$$ (a) For what values of \(P\) is the population increasing? (b) For what values of \(P\) is the population decreasing? (c) What are the equilibrium solutions?

Short Answer

Expert verified
(a) Population increases when \(0 < P < 4200\). (b) It decreases when \(P > 4200\). (c) Equilibrium solutions are \(P = 0\) and \(P = 4200\).

Step by step solution

01

Understanding the Differential Equation

The differential equation given is \(\frac{dP}{dt} = 1.2P\left(1 - \frac{P}{4200}\right)\). It represents the rate of change of the population \(P\) over time \(t\). The expression is a logistic growth model.
02

Identifying When Population is Increasing

The population is increasing where the rate of change \(\frac{dP}{dt} > 0\). Thus, we solve the inequality: \[ 1.2P \left(1 - \frac{P}{4200}\right) > 0 \] Factoring, we find that \(P > 0\) and \(1 - \frac{P}{4200} > 0\), leading to the condition \(P < 4200\). Therefore, the population is increasing when \(0 < P < 4200\).
03

Identifying When Population is Decreasing

The population is decreasing where \(\frac{dP}{dt} < 0\). Solving \(1.2P\left(1 - \frac{P}{4200}\right) < 0\), we find it occurs when \(P > 4200\). Thus, the population decreases when \(P > 4200\).
04

Finding Equilibrium Solutions

Equilibrium solutions occur where \(\frac{dP}{dt} = 0\). Solving \[ 1.2P \left(1 - \frac{P}{4200}\right) = 0 \]gives \(P = 0\) or \(P = 4200\). These are the equilibrium solutions where population does not change.

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 91Ó°ÊÓ!

Key Concepts

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

Logistic Growth Model
The logistic growth model is a fundamental concept in understanding how populations evolve over time. At its heart, this model describes how a population expands quickly when numbers are low, slows down as it approaches a maximum limit, and eventually levels off.
This modeling is useful because it considers both the potential for population growth and the limitations of the environment which restrict that growth.

In mathematical terms, the logistic growth model is described by a differential equation. For our example, it is given by:
  • \[ \frac{dP}{dt} = 1.2P \left(1 - \frac{P}{4200}\right) \]
The equation consists of two main parts:
  • The term \(1.2P\), representing the intrinsic growth rate of the population \(P\).
  • The expression \(1 - \frac{P}{4200}\), indicating how growth is constrained as the population nears the carrying capacity of 4200.
This carrying capacity reflects the maximum sustainable population size over time, shaped by resource availability, environmental conditions, and other limiting factors.
Population Dynamics
Population dynamics examines how populations change over time, considering birth rates, death rates, immigration, and emigration. In the context of the logistic model, it focuses particularly on the growth and decline of populations with respect to their size.

Our model indicates:
  • The population increases when the growth rate is positive. This happens when:\[ \frac{dP}{dt} = 1.2P \left(1 - \frac{P}{4200}\right) > 0 \]Hence, the population is increasing as long as \(0 < P < 4200\).
  • The population decreases when the growth rate becomes negative. This is observed:\[ 1.2P \left(1 - \frac{P}{4200}\right) < 0 \]which occurs when \(P > 4200\).
Thus, populations grow in number when within a healthy range under carrying capacity but start to diminish when numbers exceed environmental limits.
Equilibrium Solutions
Equilibrium solutions in differential equations indicate steady states where a system does not change over time. In our population model, these points occur when the growth rate is zero, \[ \frac{dP}{dt} = 0 \]
Solving this for the given equation yields:
  • \( P = 0 \)
  • \( P = 4200 \)
At these equilibrium points:
  • When \(P = 0\), the population does not exist, implying potential extinction or absence of initial population.
  • When \(P = 4200\), the population has reached the carrying capacity of the environment, maintaining a stable state without growth or decline.
The equilibrium solutions guide us in understanding the fixed points of population size, either at zero or at the maximum sustainable limit.

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

When a raindrop falls, it increases in size and so its mass at time \(t\) is a function of \(t, m(t) .\) The rate of growth of the mass is \(k m(t)\) for some positive constant \(k\) . When we apply Newton's Law of Motion to the raindrop, we get \((m v)^{\prime}=g m\) where \(v\) is the velocity of the raindrop (directed downward) and \(g\) is the acceleration due to gravity. The terminal velocity of the raindrop is lim, \(_{t \rightarrow \infty} v(t) .\) Find an expression for the terminal velocity in terms of \(g\) and \(k .\)

Use Euler's method with step size 0.5 to compute the approximate \(y\) -values \(y_{1}, y_{2}, y_{3},\) and \(y_{4}\) of the solution of the initial-value problem \(y^{\prime}=y-2 x, y(1)=0\)

A glucose solution is administered intravenously into the bloodstream at a constant rate \(r .\) As the glucose is added, it is converted into other substances and removed from the blood stream at a rate that is proportional to the concentration at that time. Thus a model for the concentration \(C=C(t)\) of the glucose solution in the bloodstream is $$\frac{d C}{d t}=r-k C$$ where \(k\) is a positive constant. (a) Suppose that the concentration at time \(t=0\) is \(C_{0}\) . Determine the concentration at any time \(t\) by solving the differential equation. (b) Assuming that \(C_{0}

Suppose you have just poured a cup of freshly brewed coffee with temperature \(95^{\circ} \mathrm{C}\) in a room where the temperature is \(20^{\circ} \mathrm{C}\) . (a) When do you think the coffee cools most quickly? What happens to the rate of cooling as time goes by? Explain. (b) Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newton's Law of Cooling for this particular situation. What is the initial condition? In view of your answer to part (a), do you think this differential equation is an appropriate model for cooling? (c) Make a rough sketch of the graph of the solution of the initial-value problem in part (b).

Find the solution of the differential equation that satisfies the given initial condition. $$\frac{d y}{d x}=\frac{y \cos x}{1+y^{2}}, \quad y(0)=1$$

See all solutions

Recommended explanations on Math Textbooks

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.