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Chapter 3: Problem 2

Modelling the spread of technology. 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 $$ \frac{d N}{d t}=a N\left(1-\frac{N}{N_{T}}\right) $$ where \(N_{T}\) 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), \(N_{T}=17,015, a=0.490\) and \(N_{0}=141\). Determine how long it takes for the improved pasture technology to spread to \(80 \%\) of the population.

Short Answer

Expert verified
(a) \(1 - \frac{N}{N_{T}}\); (b) Solve for \(t\) when \(N(t) = 13,612\).

Step by step solution

01

Identifying the Fraction Not Adopted

The logistic differential equation given is \( \frac{d N}{d t} = a N \left(1 - \frac{N}{N_{T}}\right) \). The term that represents the fraction of the population who have not yet adopted the improved pasture technology is \( 1 - \frac{N}{N_{T}} \). This term decreases as more of the population adopts the technology.
02

Setting Up the Equation for 80% Adoption

We aim to find the time \( t \) when 80% of the population has adopted the technology. This means \( N(t) = 0.8 N_{T} \). Substituting the known values: \( N_{T} = 17,015 \), therefore \( N(t) = 0.8 \times 17,015 = 13,612 \).
03

Solving the Logistic Equation

The differential equation \( \frac{d N}{d t} = a N \left(1 - \frac{N}{N_{T}}\right) \) is in the standard logistic form. The general solution is: \[ N(t) = \frac{N_{T}}{1 + C e^{-a N_{T} t}} \] where \( C \) is a constant. We need to find \( C \) using the initial condition \( N(0) = N_{0} = 141 \).
04

Calculating the Constant C

Substitute the initial condition into the general solution: \( N(0) = \frac{N_{T}}{1 + C} = 141 \). Solving for \( C \), we have: \[ 141 = \frac{17,015}{1 + C} \] \[ C = \frac{17,015}{141} - 1 \].
05

Substituting Values for Time Calculation

With \( C \) calculated, substitute \( N(t) = 13,612 \) into the logistic equation: \[ 13,612 = \frac{17,015}{1 + C e^{-0.490 \times 17,015 \times t}} \]. Solve for \( t \) when \( N(t) = 13,612 \).
06

Solving for Time t

Re-arranging the equation: \[ C e^{-0.490 \times 17,015 \times t} = \frac{17,015}{13,612} - 1 \]. Solve the equation for \( t \) to determine how long it takes for the improved technology to spread to 80% of the population. Use logarithms to solve for \( t \).

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Key Concepts

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

Logistic Differential Equation
The logistic differential equation is a mathematical model often used to describe the growth of a population or the spread of a technology. At its core, it is represented as:
  • \( \frac{dN}{dt} = aN \left(1 - \frac{N}{N_T}\right) \)
This represents the rate of change of a quantity \(N\) over time \(t\). Here:
  • \(N\) is the current population or number of adopters.
  • \(N_T\) is the carrying capacity, or the total potential population/adopters.
  • \(a\) is the growth rate constant.
The term \( \left(1 - \frac{N}{N_T}\right) \) signifies the fraction of the population who have not adopted the technology or who are still susceptible to adopting it. As \(N\) approaches \(N_T\), this fraction decreases, reflecting the slowing rate of adoption as saturation is approached. The logistic equation thus captures both the initial rapid growth phase and the eventual slowdown as limits are approached.
Technology Adoption
Technology adoption can be modeled using the logistic differential equation due to similar dynamics with population growth.
  • The adoption rate is initially slow, but as more individuals start using the technology, the rate increases.
  • Eventually, as the majority adopts the technology, growth slows down approaching a plateau, much like population growth in a fixed habitat.
In our example, we model the spread of an agricultural technology among ranchers. As adoption begins, the number of ranchers using the technology rises quickly. However, as most ranchers adopt it, those remaining represent a smaller portion, which reduces the rate of adoption.
  • This mirrors human behavior in the marketplace where early adopters inspire others, but once a certain market saturation is reached, growth plateaus.
Realizing the eventual adoption peak helps industries plan better for product launches, marketing strategies, and resource allocation.
Population Dynamics
Population dynamics refer to the study of how and why populations change over time in terms of size and structure. It is essential in various fields like ecology, biology, and socio-economics. In the context of modeling using a logistic equation:
  • Population dynamics describe growth patterns that are not exclusively linear.
  • They account for limits, such as resource availability, that cap growth.
In both technology spread and population growth, the logistic model illustrates that:
  • Initial rapid growth is curbed by resource limitations or market saturation, leading to a leveling off as the population or adoption nears carrying capacity \(N_T\).
This model helps predict timelines and maximum capacities, which is critical for resource planning, marketing, and understanding ecological impacts that accompany population changes.

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Most popular questions from this chapter

Chemostat. A chemostat is used by microbiologists and ecologists to model aquatic environments, or waste treatment plants. It consists of a tank filled with a mixture of some medium and nutrients, which microorganisms require to grow and multiply. A fresh nutrient-medium mixture is pumped into the tank at a constant rate \(F\) and the tank mixture is pumped from the tank at the same rate. In this way the volume of liquid in the tank remains constant. Let \(S(t)\) denote the concentration of the nutrient in the tank at time \(t\), and assume the mixture in the tank is well stirred. Let \(x(t)\) denote the concentration of the microorganism in the tank at time \(t .\) (a) Draw a compartmental diagram for the amount of nutrient. (b) In the absence of the organism, suggest a model for the rate of change of \(S(t)\). (c) If the microorganisms' per-capita uptake of the nutrient is dependent on the amount of nutrient present and is given by \(p(S)\), and the per-capita reproduction rate of the microorganism is directly proportional to \(p(S)\), extend the model equation above to include the effect of the organism. (The per-capita uptake function measures the rate at which the organism is able to absorb the nutrient when the nutrient's concentration level is \(S .)\) (d) Now develop an equation describing the rate of change of the concentration of the live organism \(\left(x^{\prime}\right)\) in the tank to derive the second equation for the system. (e) The nutrient uptake function \(p(S)\) can be shown experimentally to be a monotonically increasing function bounded above. Show that a Michaelis-Menten type function $$ p(S)=\frac{m S}{a+S} $$ with \(m\) and a positive, non-zero constants, satisfies these requirements. What is the maximum absorption rate? And why is a called the half-saturation constant? (Hint: The maximum absorption rate is the maximum reached by \(p(S) .\) For the second part, consider \(p(a) .)\)

Predicting population size. 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 2 months. (Use either an analytical solution or a numerical solution from Maple or MATLAB.)

Density-dependent births. Many animal populations have decreasing per-capita birth rates when the population density increases, as well as increasing per- capita death rates. Suppose the density-dependent per-capita birth rate \(B(X)\) and density-dependent death rate \(A(X)\) are given by $$ B(X)=\beta-(\beta-\alpha) \delta \frac{X}{K}, \quad A(X)=\alpha+(\beta-\alpha)(1-\delta) \frac{X}{K} $$ where \(K\) is the population carrying capacity, \(\beta\) is the intrinsic per- capita birth rate, \(\alpha\) is the intrinsic per-capita death rate and \(\delta\), where \(0 \leq \delta \leq 1\), is a parameter describing the extent that density dependence is expressed in births or deaths. Show that this still gives rise to the standard logistic differential equation $$ \frac{d X}{d t}=r X\left(1-\frac{X}{K}\right) $$

Linear differential-delay equation. Consider the linear differential-delay equation $$ \frac{d X}{d t}=X(t-1), \quad X(0)=1 $$ Look for an exponential solution of the form \(X(t)=C e^{m t}\), where \(m\) is a constant to be determined and \(C\) is an arbitrary constant.

Fishing with quotas. In view of the potentially disastrous effects of overfishing causing a population to become extinct, some governments impose quotas that vary depending on estimates of the population at the current time. One harvesting model that takes this into account is $$ \frac{d X}{d t}=r X\left(1-\frac{X}{K}\right)-h X $$ (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 Exercise \(3.7\), where a sudden population crash (from a large population to extinction) can occur as the harvesting rate increases beyond a critical value.
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