/*! 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 17 Consider a population \(P=P(t)\)... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a population \(P=P(t)\) with constant relative birth and death rates \(\alpha\) and \(\beta,\) respectively, and a constant emigration rate \(m,\) where \(\alpha, \beta,\) and \(m\) are positive constants. Assume that \(\alpha>\beta .\) Then the rate of change of the population at time \(t\) is modeled by the differential equation $$ \frac{d P}{d t}=k P-m \quad \text { where } k=\alpha-\beta $$ (a) Find the solution of this equation that satisfies the initial condition \(P(0)=P_{0}\). (b) What condition on \(m\) will lead to an exponential expansion of the population? (c) What condition on \(m\) will result in a constant population? A population decline? (d) Is \(47,\) the population of Ireland was about \(8 \mathrm{mil}\) - lion and the difference between the relative birth and death rates was \(1.6 \%\) of the population. Because of the potato famine in the \(1840 \mathrm{s}\) and \(1850 \mathrm{s}\), about \(210,000\) potabitants per year emigrated from Ireland. Was the population expanding or declining at that time?

Short Answer

Expert verified
The population of Ireland was declining at that time.

Step by step solution

01

Understand the Differential Equation

We are given the differential equation \( \frac{dP}{dt} = kP - m \), where \( k = \alpha - \beta \). This represents the rate of change of the population where \( \alpha \) and \( \beta \) are the birth and death rates respectively, and \( m \) is the emigration rate. The initial condition is \( P(0) = P_0 \).
02

Solve the Homogeneous Equation

Consider the corresponding homogeneous differential equation \( \frac{dP}{dt} = kP \). The solution to this equation is found using separation of variables: \( \frac{1}{P} dP = k \, dt \). Integrate both sides to get \( \ln|P| = kt + C \), where \( C \) is a constant. Exponentiating both sides gives \( P = Ce^{kt} \).
03

Find Particular Solution Using Constant Coefficients

Assume a particular solution of the form \( P(t) = A \) (a constant). Substituting this into the non-homogeneous equation gives \( 0 = kA - m \), or \( A = \frac{m}{k} \).
04

Solve the Non-Homogeneous Equation

The general solution of \( \frac{dP}{dt} = kP - m \) combines the homogeneous and particular solutions: \( P(t) = Ce^{kt} + \frac{m}{k} \).
05

Apply Initial Condition

Using \( P(0) = P_0 \), substitute into the general solution: \( P_0 = Ce^{0} + \frac{m}{k} \). This simplifies to \( P_0 = C + \frac{m}{k} \), solving for \( C \) gives \( C = P_0 - \frac{m}{k} \).
06

Derive Final Solution

Substituting \( C \) back into the general solution, we have \( P(t) = \left(P_0 - \frac{m}{k}\right)e^{kt} + \frac{m}{k} \). This is the solution satisfying the initial condition \( P(0)=P_{0} \).
07

Discuss Conditions for Population Change

For exponential growth, the \( e^{kt} \) term should dominate, meaning \( k > 0 \). This occurs if \( m < \alpha P_0 \). For a constant population, the emigration \( m \) should balance the net increase due to births and deaths, so \( m = \alpha P_0 \). For a decline, \( m > \alpha P_0 \).
08

Analyze Historical Population Based on Given Data

Given \( k = 0.016 \) and using \( P_0 = 8,000,000 \), the rate of emigration was \( m = 210,000 \). Calculate \( \alpha P_0 = 0.016 \times 8,000,000 = 128,000 \). Since \( m = 210,000 > 128,000 \), the population was declining.

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Key Concepts

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

Population Dynamics
Population dynamics is an essential field in understanding how populations change over time. A population model, like the one in our exercise, considers factors such as birth rates, death rates, and emigration rates to predict population changes. Here's how these dynamics work:
- **Birth rates (\(\alpha\)):** The rate at which new individuals are added to the population.
- **Death rates (\(\beta\)):** The rate at which individuals leave the population due to death.
- **Emigration rates (\(m\)):** The number of individuals leaving the population permanently.
When analyzing a population using a differential equation, such as \(\frac{dP}{dt} = kP - m\), we assess how these elements interplay. The difference \(\alpha - \beta = k\), known as the net birth rate, indicates if the intrinsic growth is positive or negative. Understanding these dynamics is crucial for predicting future population sizes and trends.
Emigration Rates
Emigration rates play a vital role in population dynamics as they directly affect population size.In our exercise, the emigration rate \(m\) represents a constant rate at which individuals leave the population. This rate is critical in determining whether a population grows, remains stable, or declines.
- If \(m < kP_0\): Emigration is less than the potential growth, leading to a net increase.
- If \(m = kP_0\): Emigration exactly balances growth, resulting in a stable population.
- If \(m > kP_0\): Emigration exceeds growth contributions, causing a population decline.
Real-world scenarios, such as the historical context provided in the exercise, demonstrate how high emigration rates can lead to significant population shifts. Understanding these processes helps in policy-making and strategic planning.
Exponential Growth
Exponential growth occurs when a population increases rapidly over time, characterized by the equation \(P(t) = Ce^{kt} + \frac{m}{k}\) given in our exercise.
When examining this equation:
- The term \(e^{kt}\) describes exponential growth, dependent on the net growth rate \(k\).
- To observe exponential growth, it is essential that the growth component \(kP_0\) is greater than the emigration component \(m\). This condition ensures that the reproductive capacity overcomes losses.
Exponential growth is common in biological populations under favorable conditions with unlimited resources. However, such growth is unsustainable in real-world settings over the long term, usually impacted by resource limits or increased emigration.
Initial Value Problems
Initial value problems are a crucial concept in differential equations, including population models. An initial value problem involves finding a solution to a differential equation with a specified initial state. In our exercise, the condition \(P(0) = P_0\) represents the initial population size.
- This starting point ensures that the solution is tailored to a given scenario, allowing for precise modeling of population changes over time.
- Solving initial value problems involves substituting the initial condition into the general solution of the differential equation.
- The final solution \(P(t) = \left(P_0 - \frac{m}{k}\right)e^{kt} + \frac{m}{k}\) incorporates both the initial state and the ongoing dynamics governed by the differential equation.
Understanding initial value problems is fundamental for predicting future states in scientific fields and for applications such as population forecasting.

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Most popular questions from this chapter

Lynx eat snowshoe hares and snowshoe hares eat woody plants like willows. Suppose that, in the absence of hares, the willow population will grow exponentially and the lynx population will decay exponentially. In the absence of lynx and willow the hare population will decay exponentially. If \(L(t), H(t),\) and \(W(t)\) represent the populations of these three species at time \(t\) write a system of differential equations as a model for their dynamics. If the constants in your equation are all positive, explain why you have used plus or minus signs.

Solve the differential equation. $$ y^{\prime}+y=1 $$

An object with mass \(m\) is dropped from rest and we assume that the air resistance is proportional to the speed of the object. If \(s(t)\) is the distance dropped after \(t\) seconds, then the speed is \(v=s^{\prime}(t)\) and the acceleration is \(a=v^{\prime}(t) .\) If \(g\) is the acceleration due to gravity, then the downward force on the object is \(m g-c v,\) where \(c\) is a positive constant, and Newton's Second Law gives $$ m \frac{d v}{d t}=m g-c v $$ (a) Solve this as a linear equation to show that $$ v=\frac{m g}{c}\left(1-e^{-c t / m}\right) $$ (b) What is the limiting velocity? (c) Find the distance the object has fallen after \(t\) seconds.

Solve the differential equation and use a calcula tor to graph several members of the family of solutions. How does the solution curve change as C varies? $$ x y^{\prime}+2 y=e^{x} $$

(a) Program your computer algebra system, using Euler's method with step size \(0.01,\) to calculate \(y(2),\) where \(y\) is the solution of the initial-value problem \(y^{\prime}=x^{3}-y^{3} \quad y(0)=1\) (b) Check your work by using the CAS to draw the solution curve.

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