/*! 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 12 Let \(w\) be the number of worms... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(w\) be the number of worms (in millions) and \(r\) the number of robins (in thousands) living on an island. Suppose \(w\) and \(r\) satisfy the following differential equations, which correspond to the slope field in Figure \(10.42\). $$ \frac{d w}{d t}=w-w r, \quad \frac{d r}{d t}=-r+w r $$ Assume \(w=2\) and \(r=2\) when \(t=0 .\) Do the numbers of robins and worms increase or decrease at first? What happens in the long run?

Short Answer

Expert verified
Initially, worms decrease while robins increase. In the long run, neither settles at equilibrium due to instability.

Step by step solution

01

Analyze Initial Conditions

Given the initial conditions where \(w = 2\) and \(r = 2\) at \(t = 0\), we substitute these values into the differential equations to determine the initial changes in the populations of worms and robins.
02

Substitute into Worms' Equation

Substitute \(w = 2\) and \(r = 2\) into the differential equation for worms:\[\frac{d w}{d t} = w - w r = 2 - 2 \times 2 = 2 - 4 = -2\]This indicates that initially, the number of worms decreases.
03

Substitute into Robins' Equation

Substitute \(w = 2\) and \(r = 2\) into the differential equation for robins:\[\frac{d r}{d t} = -r + w r = -2 + 2 \times 2 = -2 + 4 = 2\]This indicates that initially, the number of robins increases.
04

Analyze Equilibrium Points

The equilibrium points occur where both \(\frac{d w}{d t} = 0\) and \(\frac{d r}{d t} = 0\). Solve the system of equations:1. \(w - wr = 0\) implies \(w(1 - r) = 0\) giving \(w = 0\) or \(r = 1\).2. \(-r + wr = 0\) implies \(r(w-1) = 0\) giving \(r = 0\) or \(w = 1\).The equilibrium points are \((0, 0)\) and \((1, 1)\).
05

Long-term Behavior Analysis

The long-term behavior depends on the stability of these equilibrium points. Linearize the system near each equilibrium point using Jacobian matrices, showing the stability characteristics:1. At \((0, 0)\), the system is unstable.2. At \((1, 1)\), analyze eigenvalues to find it is a saddle point, indicating unstable behavior.Thus, the system oscillates around the equilibrium but doesn't settle down at one due to inherent instability.

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.

Population Dynamics
Population dynamics is a branch of life sciences that studies the size and age composition of populations as dynamic systems. It explores how these populations change over time and what factors drive these changes. In our scenario involving worms and robins, population dynamics helps understand how quantities of each species interact and fluctuate.
For example, individuals in one population might affect the birth and death rates of another. Here, we have differential equations capturing these interactions, where the number of worms and robins influence each other reciprocally.
  • The worms' equation, \( \frac{dw}{dt} = w - wr \), implies worms proliferate unless limited by robins.
  • The robins' equation, \( \frac{dr}{dt} = -r + wr \), suggests that robins decline unless they benefit from eating worms.

The dynamic interactions captured here are a simple representation of predator-prey relationships, demonstrating the underlying principles of how species affect each other's growth through competing or symbiotic relationships.
Equilibrium Points
Equilibrium points in a differential system refer to the states where populations remain constant over time. Mathematically, these are points where the rates of change are zero.
Solving our given equations for equilibrium, we seek points where:
  • \( \frac{dw}{dt} = 0 \), meaning worms neither increase nor decrease.
  • \( \frac{dr}{dt} = 0 \), meaning robins neither increase nor decrease.
Solving \( w - wr = 0 \) gives us \( w = 0 \) or \( r = 1 \), and solving \( -r + wr = 0 \) gives \( r = 0 \) or \( w = 1 \). The intersection provides equilibrium points at \((0, 0)\) and \((1, 1)\).
Analyzing these points shows us scenarios where both populations might "balance" out; however, it's essential to study their stability to understand the system's long-term behavior fully.
Jacobian Matrix
The Jacobian matrix is a mathematical tool used to study the local behavior of dynamical systems near equilibrium points. It provides a way to linearize a system of differential equations, offering insights into the system's stability.
Given our model equations:
  • \( \frac{dw}{dt} = w - wr \)
  • \( \frac{dr}{dt} = -r + wr \)
we can form a Jacobian matrix by taking partial derivatives of these expressions with respect to \(w\) and \(r\).
The resulting Jacobian is:
\[\begin{bmatrix}\frac{\partial (w - wr)}{\partial w} & \frac{\partial (w - wr)}{\partial r} \\frac{\partial (-r + wr)}{\partial w} & \frac{\partial (-r + wr)}{\partial r} \\end{bmatrix} = \begin{bmatrix}1 - r & -w \ r & w - 1 \\end{bmatrix}\]
Evaluating this matrix at equilibrium points like \((0, 0)\) and \((1, 1)\) helps determine system stability. For example, calculations might reveal that \((1, 1)\) is a saddle point, indicating unstable behavior. This information is pivotal for predicting whether populations will stabilize, oscillate, or diverge over time in the real biological context.

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

Let \(I\) be the number of infected people and \(S\) be the number of susceptible people in an outbreak of a disease. Explain why it is reasonable to model the interaction between these two groups by the differential equations $$\begin{array}{l} \frac{d S}{d t}=-a S I\\\ \frac{d I}{d t}=a S I-b I \quad \text { where } a, b \text { are positive constants. } \end{array}$$ Why have the signs been chosen this way? Why is the constant \(a\) the same in both equations?

A country's infrastructure is its transportation and communication systems, power plants, and other public institutions. The Solow model asserts that the value of national infrastructure \(K\) increases due to investment and decreases due to capital depreciation. The rate of increase due to investment is proportional to national income, \(Y\). The rate of decrease due to depreciation is proportional to the value of existing infrastructure. Write a differential equation for \(K\).

Money in an account earns interest at a continuous rate of \(8 \%\) per year, and payments are made continuously out of the account at the rate of \(\$ 5000\) a year. The account initially contains \(\$ 50,000\). Write a differential equation for the amount of money in the account, \(B\), in \(t\) years. Solve the differential equation. Does the account ever run out of money? If so, when?

In some chemical reactions, the rate at which the amount of a substance changes with time is proportional to the amount present. For example, this is the case as \(\delta\) glucono-lactone changes into gluconic acid. (a) Write a differential equation satisfied by \(y\), the quantity of \(\delta\) -glucono-lactone present at time \(t\). (b) If 100 grams of \(\delta\) -glucono-lactone is reduced to \(54.9\) grams in one hour, how many grams will remain after 10 hours?

Give the rates of growth of two populations, \(x\) and \(y\), measured in thousands. (a) Describe in words what happens to the population of each species in the absence of the other. (b) Describe in words how the species interact with one another. Give reasons why the populations might behave as described by the equations. Suggest species that might interact in that way. $$ \begin{array}{l} \frac{d x}{d t}=0.01 x-0.05 x y \\ \frac{d y}{d t}=-0.2 y+0.08 x y \end{array} $$

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.