/*! 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 7 In a population the initial popu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a population the initial population is \(x_{0}=100\). Suppose a population can be modelled using the differential equation $$ \frac{d X}{d t}=0.2 X-0.001 X^{2} $$ with an initial population size of \(x_{0}=100\) and a time step of 1 month. Find the predicted population after two months. (Use either an analytical solution, or a numerical solution from Maple or MATLAB.)

Short Answer

Expert verified
The predicted population after two months is 119.9.

Step by step solution

01

Understand the Model

The given differential equation is a logistic growth model defined as \( \frac{dX}{dt} = 0.2X - 0.001X^2 \). This equation represents the rate of change of the population \( X \) with respect to time \( t \). The term \( 0.2X \) signifies the growth rate, and the term \( -0.001X^2 \) represents a limiting factor as the population grows.
02

Identify the Initial Condition

The problem gives an initial population \( X_0 = 100 \). This means at time \( t = 0 \), the population \( X(0) = 100 \). This initial condition will be used to solve the differential equation explicitly or numerically.
03

Choose the Method of Solution

Since the differential equation is a nonlinear first-order equation, solving it analytically involves complex integration. For simplicity and step-by-step clarity, let's use an Euler numerical method with a time step of 1 month to predict the population after two months.
04

Apply Euler's Method for First Month

Euler's method approximates the solution by taking the initial point and predicting the next point using the derivative. At \( t = 0 \), \( X_0 = 100 \). Calculate \( \Delta X = (0.2 \cdot 100 - 0.001 \cdot (100)^2) \cdot 1 = (20 - 10) \cdot 1 = 10 \). Hence, \( X(1) = X_0 + \Delta X = 100 + 10 = 110 \).
05

Apply Euler's Method for Second Month

Using the updated population after the first month \( X(1) = 110 \), calculate for the next step. \( \Delta X = (0.2 \cdot 110 - 0.001 \cdot (110)^2) \cdot 1 = (22 - 12.1) \cdot 1 = 9.9 \). Thus, \( X(2) = X(1) + \Delta X = 110 + 9.9 = 119.9 \).
06

Conclude the Population After Two Months

After applying Euler’s method for two consecutive months, the predicted population of the model after two months is \( X(2) = 119.9 \).

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.

Differential Equations
Differential equations are mathematical formulations that describe how a quantity changes with respect to another quantity. In this context, the differential equation \[ \frac{dX}{dt} = 0.2X - 0.001X^2 \] represents how the population size changes over time. This specific equation is a type of differential equation called a logistic growth model, which is commonly used in modeling populations.

  • The term \(0.2X\) refers to the natural growth rate of the population. Here, each individual contributes to the growth.
  • The term \(-0.001X^2\) acts as a limiting factor, reducing growth as the population increases, essentially representing competition for resources.
Understanding differential equations provides insights into predicting complex systems. It's crucial for solving problems where quantities change continuously over time.
Numerical Methods
When solving differential equations, sometimes an analytic or exact solution is complex or not possible to find. Numerical methods come to the rescue in these cases. They are techniques employed to approximate solutions, especially when dealing with non-linear equations like the logistic growth model.

These methods use numerical approximations to calculate values at specific points. For our logistic growth scenario, we used Euler’s method as the numerical approach.

  • Numerical methods often trade exactness for practicality, providing an easier way to predict outcomes.
  • They are highly valuable for modeling real-world applications where precision can be compromised for the sake of understanding trends and behaviors.
In the context of this problem, a numerical method allows us to compute population forecasts with step-by-step increments.
Initial Conditions
Initial conditions are essential in solving differential equations as they specify the starting point for the simulations. In our logistic growth model, the given initial condition is the initial population size, \( X_0 = 100 \), indicating that at time \( t = 0 \), the population is 100.

These initial values anchor the solution process and allow us to determine the trajectory that the population follows over time.

  • Without such a starting point, the solution to a differential equation can be ambiguous, leading to multiple potential outcomes.
  • Initial conditions provide the necessary data to predict future states of the system accurately through numerical methods.
In population models, understanding initial conditions is crucial as they dictate the prediction's initial extrapolation.
Euler's Method
Euler’s Method is a simple yet powerful tool used for numerically solving differential equations. It's particularly straightforward to apply, making it suitable for beginning explorations in numerical solutions. In our example, Euler’s Method allows us to approximate the population size over discrete time intervals.

Steps Involved in Euler’s Method:

  • Start with the initial condition: In our problem, \( X_0 = 100 \) at \( t = 0 \).
  • Calculate the change (\(\Delta X\)): based on the slope from the differential equation for each time step.
  • Estimate the next value: Add \(\Delta X\) to the current population to find the population at the next time step.
In this logistic model, we updated the population from 100 to 110 in the first month and then to approximately 119.9 in the second month. Euler's method, by iteratively applying the differential equation, provides a step-by-step calculation paving the way for predicting outcomes over time.

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

Stability of 2-cycles. Consider the discrete logistic equation (with \(K=1\) ) $$ X_{n+1}=X_{n}+r X_{n}\left(1-X_{n}\right) $$ (a) Show that every second term in the sequence \(X_{0}, X_{1}, X_{2}, \ldots\) satisfies the difference equation $$ \begin{gathered} X_{n+2}=\left(1+2 r+r^{2}\right) X_{n}-\left(2 r+3 r^{2}+r^{3}\right) X_{n}^{2} \\ +\left(2 r^{2}+2 r^{3}\right) X_{n}^{3}-r^{3} X_{n}^{4} \end{gathered} $$ (b) For equilibrium solutions, let \(S=X_{n+2}=X_{n}\) and obtain a quartic equation (that is, an equation with the unknown raised to the fourth power, at most). Explain why \(S=0\) and \(S=1\) must be solutions of this quartic equation. Hence show that the other two solutions are $$ S=\frac{(2+r) \pm \sqrt{r^{2}-4}}{2 r} $$ [Note: Comparing with Figure \(3.10(r=2.2)\) we see that these two values of the two non-zero equilibrium solutions are the values between which the population oscillates in a two-cucle. Furthermore, when \(r\) increases to where these two equilibrium solutions become unstable this corresponds to where the two-cycle changes to a four-cycle.]

Investigating parameter change. Using Maple or MATLAB examine the effect of increasing the parameter \(r\) on the solution to the equation $$ X_{n+1}=X_{n} e^{a\left(1-X_{n} / K\right)}, \quad \text { where } \quad a=\ell \mathrm{n}(r+1) $$ Establish (roughly) for what values of \(r\) the system undergoes its first two bifurcations. (Code can be adapted from that in Section 3.6.)

Modelling insect populations with a difference equation. Many insect populations breed only at specific times of the year and all the adults die after breeding. These may be best modelled by a difference equation, such as $$ X_{n+1}=r\left(X_{n}-0.001 X_{n}^{2}\right) $$ Using Maple or MATLAB investigate what happens as the parameter \(r\) (the growth rate) is varied from \(r=0\) to \(r=3 .\) Sketch all the different types of growth patterns observed, together with the corresponding value of \(r\).

Models for the spread of technology are very similar to the logistic model for population growth. Let \(N(t)\) be the number of ranchers who have adopted an improved pasture technology in Uruguay. Then \(N(t)\) satisfies the differential equation $$ \frac{d N}{d t}=a N\left(1-\frac{N}{N^{*}}\right) $$ where \(N^{*}\) is the total population of ranchers. It is assumed that the rate of adoption is proportional to both the number who have adopted the technology and the fraction of the population of ranchers who have not adopted the technology. (a) Which terms correspond to the fraction of the population who have not yet adopted the improved pasture technology? (b) According to Banks \((1994), N^{*}=17015, a=0.490\) and \(N_{0}=141 .\) Determine how long it takes for the improved pasture technology to spread to \(80 \%\) of the population. Note: This same model can be used to describe the spread of a rumour within an organisation or population.

In view of the potentially disastrous effects of overfishing causing a population to become extinct, some governments impose quotas which vary depending on estimates of the population at the current time. One harvesting model that takes this into account is $$ \frac{d X}{d t}=r X\left(1-\frac{X}{K}\right)-h_{0} X $$ (a) Show that the only non-zero equilibrium population is $$ X_{e}=K\left(1-\frac{h}{r}\right) $$ (b) At what critical harvesting rate can extinction occur? Although extinction can occur with this model, as the harvesting parameter \(h\) increases towards the critical value the equilibrium population tends to zero. This contrasts with the constant harvesting model in Section \(3.3\) and Question 5 , where a sudden population crash (from a large population to extinction) can occur as the harvesting rate increases beyond a critical value.

See all solutions

Recommended explanations on Biology 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.