/*! 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 5 The population of a town in the ... [FREE SOLUTION] | 91Ó°ÊÓ

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The population of a town in the south of Bangladesh has been growing exponentially. However, recent flooding has alarmed residents and people are leaving the town at a rate of \(N\) thousand people per year, where \(N\) is a constant. The rate of change of the population of the town can be modeled by the differential equation $$ \frac{d P}{d t}=0.02 P-N $$ where \(P=P(t)\) is the number of people in the town in thousands. (a) If \(P(0)=100\), what is the largest yearly exodus rate the town can support in the long run? (b) How big must the population of the town be in order to support the loss of 1000 people per year?

Short Answer

Expert verified
The largest yearly exodus rate the town can support in the long run is 2 thousand people per year. The minimum population required to support the loss of 1000 people per year is 50 thousand.

Step by step solution

01

Identify the steady state condition

To find the maximum exodus rate , you need to understand when the population remains steady \(i.e., \frac{d P}{d t}=0\). The equation for this scenario would be \(0.02 P=N\). Here, \(P\) denotes population and \(N\) denotes the exodus rate.
02

Solve for the exodus rate

For the town's population to remain steady, we substitute the given initial population \(P=100\) into the equation \(0.02 P=N\) to find \(N\). So, we have \(N = 0.02*100 = 2\). This tells us that the maximum exodus rate the town can support in the long run is 2 thousand people per year, given the initial population was 100 thousand.
03

Calculate minimum population

To find the minimum population required to support a loss of 1000 people per year, you substitute \(N = 1000/1000 = 1\) (since \(N\) is in thousands) into the equation \(0.02 P=N\). Solving for \(P\), you get \(P = \frac{1}{0.02} = 50\). This means that in order to support a loss of 1000 people per year, the population must be at least 50 thousand.

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Key Concepts

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

Exponential Growth
Exponential growth is a fascinating concept often observed in populations, where the number of individuals changes at a rate proportional to its current size.
It is like a snowball effect: as more individuals are present, the rate at which they can multiply also increases. This creates a rapid expansion over time, which can be modeled using differential equations.
In mathematical terms, the differential equation representing exponential growth is usually given by \( \frac{dP}{dt} = rP \), where \( P \) is the current population, and \( r \) is the growth rate.
  • The larger the growth rate \( r \), the faster the population will increase.
  • Exponential growth cannot continue indefinitely, as resources such as food or space become limited.
When external factors like emigration are involved, as with the population leaving the town in the exercise, the differential equation becomes \( \frac{dP}{dt} = rP - N \). Here, \( N \) accounts for the rate at which people are leaving, which tempers the growth.
Understanding how these components interact is crucial in modeling real-world population dynamics.
Population Dynamics
Population dynamics are driven by various factors that affect the size and growth of biological populations.
These can include birth rates, death rates, immigration, and emigration, all of which contribute to changes in population size over time.
In the context of the exercise, the key factor is emigration, quantified by the constant \( N \), representing the number of people leaving the town each year.
This affects the overall growth rate of the population and can lead to interesting dynamics where, despite a natural tendency to grow, the population may remain stable or even decrease.
  • If \( N \) is too high, it counteracts growth completely, resulting in a net decrease.
  • Consideration of external environmental factors, like natural disasters, can further complicate dynamics.
The interplay between natural growth and external influences is captured by the differential equation \( \frac{dP}{dt} = 0.02P - N \).
It highlights how constant emigration can impact a population's ability to maintain its numbers.
Understanding these concepts helps us to foresee potential issues in real-life situations, such as managing urban populations or wildlife conservation.
Steady-State Condition
The steady-state condition is an important concept in understanding population dynamics, especially when considering whether a population will grow, shrink, or stabilize over time.
A steady state occurs when the rate of growth is exactly balanced by the exodus rate, leading to no net change in the population size over time (\( \frac{dP}{dt} = 0 \)).
To achieve this, the differential equation \(0.02 P = N\) must hold true. This equation describes a situation where every individual leaving is exactly replaced by one born, keeping the population stable.
  • A higher initial population means a higher exodus rate \( N \) can be tolerated without decrease.
  • Achieving steady-state can be critical for sustainability, especially in small communities.
In the exercise, calculating the steady-state helps determine how many people the town can afford to lose annually, without affecting its long-term population size.
Similarly, it determines the minimum population required to withstand a certain level of emigration. It's a valuable tool for policymakers in creating sustainable population management strategies.

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Most popular questions from this chapter

\(P(t)=C e^{k t}+\frac{E}{k}\), where \(C\) is a constant, is the general solution to the differential equation \(\frac{d P}{d t}=k P-E .\) Below is the slope eld for \(\frac{d P}{d t}=2 P-6\). (a) i. Find the particular solution that corresponds to the initial condition \(P(0)=2\). ii. Sketch the solution curve through \((0,2)\). (b) i. Find the particular solution that corresponds to the initial condition \(P(0)=3\). ii. Sketch the solution curve through \((0,3)\). (c) i. Find the particular solution that corresponds to the initial condition \(P(0)=4\). ii. Sketch the solution curve through \((0,4)\).

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There are many places in the world where populations are changing and immigration and/or emigration play a big role. People may move to nd food, or to nd jobs, or to ee political or religious persecution. Pick a situation that interests you. You could look at the number of Tibetans in Tibet, or the number of Tibetans in India, or the number of lions in the Serengeti, or the number of tourists in Nepal. Get some data and try to model the population dynamics using a differential equation. What simplifying assumptions have you made?

Each function below is a solution to one of the second orderdifferential equations listed. To each function match the appropriate differential equation. \(C_{1}\) and \(C_{2}\) are constants. Differential Equations I. \(\frac{d^{2} x}{d t^{2}}-9 x=0\) II. \(\frac{d^{2} x}{d t^{2}}+9 x=0\) III. \(\frac{d^{2} x}{d t^{2}}=3 x\) Solution Functions (a) \(x(t)=5 e^{3 t}\) (b) \(x(t)=-2 e^{\sqrt{3} t}\) (c) \(x(t)=7 \sin 3 t\) (d) \(x(t)=C_{1} \sin 3 t+C_{2} \cos 3 t\) (e) \(x(t)=C_{1} e^{\sqrt{3} t}+C_{2} e^{-\sqrt{3} t}\)

When a population has unlimited resources and is free from disease and strife, the rate at which the population grows in often modeled as being proportional to the population. Assume that both the bee and the mosquito populations described below behave according to this model. In both scenarios described below you are given enough information to nd the proportionality constant \(k\). In one case the information allows you to nd \(k\) solely using the differential equation, without requiring that you solve it. In the other scenario you must actually solve the differential equation in order to nd \(k\). (a) Let \(M=M(t)\) be the mosquito population at time \(t, t\) in weeks. At \(t=0\) there are 1000 mosquitoes. Suppose that when there are 5000 mosquitoes the population is growing at a rate of 250 mosquitoes per week. Write a differential equation re ecting the situation. Include a value for \(k\), the proportionality constant. (b) Let \(B=B(t)\) be the bee population at time \(t, t\) in weeks. At \(t=0\) there are 600 bees. When \(t=10\) there are 800 bees. Write a differential equation re ecting the situation. Include a value for \(k\), the proportionality constant.

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