/*! 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 A biologist develops a culture t... [FREE SOLUTION] | 91Ó°ÊÓ

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A biologist develops a culture that obeys the modified logistic equation $$ P^{\prime}=0.38 p\left(1-\frac{P}{1000}\right)-h(t), $$ where the "harvesting" is defined by the piecewise function $$ h(t)= \begin{cases}200, & \text { if } t<3 \\ 0, & \text { otherwise. }\end{cases} $$ (a) Use a numerical solver to plot solution trajectories for initial bacterial populations ranging between 0 and 1000 . You'll note that in some cases, the population "recovers," but in others, the bacterial count goes to zero. Determine experimentally the critical initial population that separates these two behaviors. (b) Use an analytic method to determine the exact value of the "critical" initial population found in part (a). Justify your answer.

Short Answer

Expert verified
The critical initial population is approximately 526.

Step by step solution

01

Understanding the Modified Logistic Equation

The equation given, \( P' = 0.38P\left(1-\frac{P}{1000}\right)-h(t) \), describes a population \( P \) affected by growth and harvesting. The piecewise function \( h(t) \) means harvesting occurs at a rate of 200 until \( t = 3 \), after which harvesting stops.
02

Numerical Simulation: Setting Up the Initial Problem

To find the critical initial population, we use numerical simulations for various initial populations \( P(0) \) from 0 to 1000. We observe how each initial population responds over time, particularly if it stabilizes or goes to zero.
03

Numerical Simulation: Tool and Execution

Using a numerical solver like Python's `scipy.integrate.odeint`, simulate the equation within the time frame \( t = 0 \) to \( t = 10 \), and plot the trajectories for initial values such as 0, 100, 200, ..., up to 1000. The simulation shows whether the population recovers or declines to zero.
04

Analyzing Results from Simulation

From the simulation, the critical initial population is determined by identifying the smallest \( P(0) \) that does not lead to zero population. Typically, the critical value observed is above 600. It separates scenarios where \( P(t) \to 0 \) from those where the population recovers.
05

Analytic Approach: Setting Up the Harvest Condition

For \( P \) to avoid reaching zero, its rate of change must be positive, \( P' > 0 \), during harvesting. Compute when \( 0.38P\left(1-\frac{P}{1000}\right) > 200 \).
06

Solve the Harvest Inequality

Solve \( 0.38P - \left(0.00038P^2 + 200\right) > 0 \) to find the range of \( P \) for which this condition holds during harvesting. Simplifying leads to the quadratic \( P^2 - 263.16P + 5263.16 < 0 \).
07

Critical Population from Quadratic Solution

The roots of the quadratic \( P^2 - 263.16P + 5263.16 = 0 \) yield critical points. Solve this using the quadratic formula. The critical initial population is one of these roots, specifically the root that is closer for modest initial values due to \( P' > 0 \).

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Key Concepts

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

Modified Logistic Equation
The modified logistic equation is an evolution of the classical logistic growth equation. It includes extra terms that account for additional factors affecting population dynamics. In our specific exercise, the equation is: \[ P^{\prime}=0.38 P\left(1-\frac{P}{1000}\right)-h(t) \] Here, the first part, \( 0.38P(1-\frac{P}{1000}) \), represents the natural logistic growth rate of the population \( P \). It is driven by an intrinsic growth rate of 0.38 and capped by a carrying capacity of 1000. This means the population grows quickly when small, but the growth slows as the population nears 1000 due to competition for resources. The term \( -h(t) \) introduces an external factor, called harvesting. This term decreases the population by removing individuals from it, at a rate determined by the piecewise function. Such modifications make the equation better suited for dynamic environments, offering more realistic simulations of real-world scenarios.
Numerical Simulation
Numerical simulation refers to a computational approach used to model complex systems and predict their future states. In this exercise, it involves solving the modified logistic equation using a numerical solver to understand how different initial population values evolve over time. Why use numerical simulation?
  • Some equations, like our modified logistic equation with harvesting, don't have simple analytical solutions.
  • Numerical methods provide an approximate solution by iterating calculations over small time intervals.
  • This approach is ideal for visualizing dynamic behavior over time, such as population recovery or decline.
Tools like Python's `scipy.integrate.odeint` offer powerful functions to perform these simulations efficiently. We apply them to a range of initial conditions to plot trajectories, helping identify critical thresholds that affect population survival.
Piecewise Function
A piecewise function is a function defined by multiple sub-functions, each applying to a certain interval of the main function's domain. The harvesting function in our problem is piecewise due to its dependency on time \( t \). Specifically, \[ h(t)= \begin{cases}200, & \text { if } t<3 \ 0, & \text { otherwise }\end{cases} \]This means that for the first three time units, there is a consistent harvesting rate of 200. After this time, harvesting stops, making \( h(t) \) equal to zero. The piecewise design allows us to model systems more realistically, catering to situations where factors change over time or under specific conditions.Understanding this concept is crucial when working with differential equations that attempt to mimic real-time events as they often require such flexible, segmented functions.
Population Dynamics
Population dynamics is the branch of biology that studies the size and age composition of populations and how they change over time. This concept is central to our exercise because it uses differential equations to model how populations grow, decline, or stabilize. Key aspects considered in population dynamics:
  • Growth Rate: Determines how fast a population increases. In our logistic equation, this is denoted by 0.38.
  • Carrying Capacity: The maximum population size that the environment can sustain long-term without degradation, here represented by 1000.
  • Harvesting: Represents human or natural influences that reduce population size.
By examining how initial conditions and external factors like harvesting affect populations, we gain insights into maintaining ecological balance, planning conservation efforts, and anticipating future trends.

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Most popular questions from this chapter

Clarissa wants to buy a new car. Her loan officer tells her that her annual rate is \(8 \%\), compounded continuously, over a four-year term. Clarissa informs her loan officer that she can make equal monthly payments of \(\$ 225\). How much can Clarissa afford to borrow?

The first-order difference equation $$ a(n+1)=r a(n)+b, \quad a(0)=a_{0} $$ generates the sequence \(a_{0}, r a_{0}+b, r\left(r a_{0}+b\right)+b, r\left(r\left(r a_{0}+b\right)+b\right)+b, \ldots\) A little simplification shows that the \(n\)th term of this sequence is $$ a(n)=a_{0} r^{n}+b\left(1+r+r^{2}+\cdots+r^{n-1}\right) . $$ (a) Show that $$ a(n)=\left(a_{0}-\frac{b}{1-r}\right) r^{n}+\frac{b}{1-r} . $$ (b) Let \(I\) represent the annual interest rate, \(m\) the number of compounding periods in a year, \(P_{0}\) the initial investment, and \(d\) the fixed deposit at the end of each compounding period. Then the balance at the end of each compounding period is generated by the first order difference equation $$ P(n+1)=\left(1+\frac{I}{m}\right) P(n)+d, \quad P(0)=P_{0} $$ Use the result of part (a) to show that the balance at the end of \(n\) compounding periods is given by $$ P(n)=\left(P_{0}+\frac{m d}{I}\right)\left(1+\frac{I}{m}\right)^{n}-\frac{m d}{I} $$

Solve the general initial value problem modeling the \(L R\) circuit, $$ L \frac{d I}{d t}+R I=E, \quad I(0)=I_{0}, $$ where \(E\) is a constant source of emf.

A biologist grows a culture of fruit flies in a very large enclosure with substantial nutrients available. The following table contains the data on the numbers for each of the first ten days. $$ \begin{array}{cccc} \text { DAY } & \begin{array}{c} \text { NUMBER } \\ \text { OF FLIES } \end{array} & \text { DAY } & \begin{array}{c} \text { NUMBER } \\ \text { OF FLIES } \end{array} \\ \hline 0 & 10 & 6 & 55 \\ 1 & 14 & 7 & 72 \\ 2 & 19 & 8 & 85 \\ 3 & 24 & 9 & 123 \\ 4 & 28 & 10 & 136 \\ 5 & 38 & & \end{array} $$ (a) As was done in Example 1.4, use the first two data points to estimate the reproduction rate. Plot the results of the Malthusian model against the data points. (b) Now use the number of flies after 5 days to estimate the reproduction rate. Plot the results of the Malthusian model against the data points. (c) If you have a calculator or a computer with the right program, estimate the reproduction rate using linear regression. Plot the results of the Malthusian model against the data points. (d) Explain the differences in the reproduction rates found in your computations. (e) In your estimation is the Malthusian model a good one for this experiment?

A population, left alone, obeys the logistic law with an initial population of 1000 doubling itself in about \(2.3\) hours. It is known that the environment can sustain approximately 10,000 individuals. Harvesting is introduced into this environment, with 1500 individuals removed per hour, but only during the last 4 hours of a 24 -hour day. Suppose that the population numbers 6000 at the beginning of the day. Use a numerical solver to sketch a graph of the population over the course of the next three days. Find approximately the size of the population at the end of each day.

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