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Chapter 8: Problem 45

In this question we will interpret the recovery rate, \(c\), that appears in the model. Assume that a population of infected individuals is quarantined (that is, they are unable to transmit the disease to others, or to catch it again once they recover). (a) Explain why under these assumptions we expect: $$ \frac{d I}{d t}=-c I $$ (b) Assuming \(I(0)=I_{0}\), find \(I(t)\) by solving \((8.71)\). (c) How long will it take for the number of infected individuals to decrease from \(I_{0}\) to \(I_{0} / 2 ?\) (d) Assume that it takes 7 days for the number of infected individuals to decrease from 50 to \(25 .\) Calculate the recovery rate \(c\) for this disease.

Short Answer

Expert verified
(a) The rate of decrease is proportional to the number infected. (b) Solution: \( I(t) = I_0 e^{-ct} \). (c) Time to halve: \( t = \frac{\ln(2)}{c} \). (d) \( c \approx 0.099 \text{ per day} \).

Step by step solution

01

Understanding the Differential Equation

The differential equation given is \( \frac{dI}{dt} = -cI \). This equation models the rate of change of infected individuals as proportional to their current number, with a negative sign indicating a decrease. Under the assumptions that recovered individuals are not reinfected nor can they infect others, the equation suggests that the decrease in the number of infected individuals is directly proportional to the current number of infected individuals, scaled by the recovery rate \( c \).
02

Solving the Differential Equation

To solve \( \frac{dI}{dt} = -cI \), we can separate variables and integrate both sides. Rewriting the equation, we have \( \frac{1}{I} dI = -c dt \). Integrating both sides gives \( \ln|I| = -ct + C \), where \( C \) is the integration constant. Solving for \( I \), we find \( I(t) = I_0 e^{-ct} \) using the initial condition \( I(0) = I_0 \).
03

Finding the Time for Infected Count to Halve

We are given that the number of infected individuals decreases from \( I_0 \) to \( \frac{I_0}{2} \). Substituting into the solution \( I(t) = I_0 e^{-ct} \), set \( I(t) = \frac{I_0}{2} \). This gives the equation \( \frac{I_0}{2} = I_0 e^{-ct} \). Dividing by \( I_0 \) and taking the natural logarithm, we have \( \ln \left( \frac{1}{2} \right) = -ct \). Solving for \( t \), we find \( t = \frac{\ln(2)}{c} \).
04

Calculating the Recovery Rate

Given that it takes 7 days for the infected population to decrease from 50 to 25, use the equation \( t = \frac{\ln(2)}{c} \). Substituting \( t = 7 \), we have \( 7 = \frac{\ln(2)}{c} \). Solving for \( c \), we find \( c = \frac{\ln(2)}{7} \). Numerically, this evaluates to \( c \approx 0.099 \text{ per day} \).

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Key Concepts

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

Recovery Rate
The recovery rate, often denoted by the letter \( c \), is an important factor in understanding the dynamics of infectious diseases within a population. When dealing with differential equations in epidemiology, \( c \) represents the rate at which infected individuals recover over time.
This rate influences how quickly the number of infected individuals decreases, assuming that once someone recovers, they cannot transmit the infection further or become reinfected themselves.
It's important to note:
  • A higher recovery rate implies a faster reduction in the number of infected individuals.
  • This rate is a crucial part of the differential equation \( \frac{dI}{dt} = -cI \), which helps model the decline of infections in a quarantined population.
Understanding the recovery rate helps in designing quarantine measures and predicting the progression of an outbreak.
Infected Individuals
In epidemiology, infected individuals refer to those who have contracted the disease being studied. In the context of the differential equation \( \frac{dI}{dt} = -cI \), \( I \) denotes the number of people currently infected.
This equation emphasizes how the decrease in infected individuals is directly proportional to their current number. Simply put, more infected persons result in a greater overall drop as they recover over time.
Studying infected individuals involves:
  • Tracking how rapidly their numbers decline, which is tied to the recovery rate \( c \).
  • Using initial data to predict future disease progression.
These infected numbers serve as the basis for solving differential equations, providing key insights into disease control strategies.
Initial Condition
To solve differential equations effectively, initial conditions are essential. An initial condition provides the starting point or the initial value of the variable being studied.
In the equation \( I(t) = I_0 e^{-ct} \), \( I_0 \) represents the number of infected individuals at time \( t = 0 \). This acts as a crucial anchor for calculations.
Key aspects include:
  • Determining the initial size of the infected population ensures accurate future projections.
  • Initial conditions help set precise paths for the differential equation's solution over time.
By appropriately using initial conditions, students can predict the evolution of infectious diseases with much greater accuracy.
Solving Differential Equations
Differential equations are mathematical equations that describe how a quantity changes in relation to others. Solving them typically requires separating and integrating variables.
In this scenario, the equation \( \frac{dI}{dt} = -cI \) is solved by:
  • Separating variables: \( \frac{1}{I} dI = -c dt \).
  • Integrating both sides to find \( \ln|I| = -ct + C \), where \( C \) is a constant.
  • Using the initial condition \( I(0) = I_0 \) to solve for \( C \), resulting in \( I(t) = I_0 e^{-ct} \).
This process helps model the decline of infected individuals over time, providing a mathematical representation of disease dynamics.

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Most popular questions from this chapter

By breaking down each equation into two parts that you can sketch, determine how many equilib\mathrm{\\{} r i a ~ e a c h ~ d i f f e r e n t i a l ~ e q u a t i o n ~ h a s , ~ a n d ~ c l a s s i f y ~ t h e m ~ a s ~ s t a b l e ~ or unstable. You do not need to determine the location of the equilibria. $$ \frac{d N}{d t}=N^{2}-N+1 \quad N>0 $$

A cell constantly gains or loses small molecules to its environment because the small molecules are able to diffuse through the cell membrane. We will build a model for this process. Suppose a molecule is present in the cell at a concentration \(C(t)\), and present in its environment at a concentration \(C_{\infty}\) (you may assume \(C_{\infty}\) is a constant). One model for the diffusion of molecules across the cell membrane is that the rate at which molecules travel through the membrane is proportional to the difference in concentration between the cell and its surroundings. That is: Rate at which $$ \text { molecules flow out }=k\left(C-C_{\infty}\right) $$ of cell The constant \(k\) is known as the permeability of the membrane: \(k>0\), and \(k\) depends on the surface area of the cell and the chemistry of the membrane, as well as the type of molecule. (a) Starting with a word equation for the amount of small molecules in the cell, show, if the cell volume is \(V\), then: $$ \frac{d C}{d t}=-\frac{k}{V}\left(C-C_{\infty}\right) $$ (b) Find the equilibrium of \((8.53)\) and use a graphical analysis to determine whether it is stable or unstable. (c) Suppose that the molecule we are studying is produced within the cell. The cell produces the molecule at a rate \(r\); that is, a quantity \(r\) is produced (added to the cell) in unit time. Explain why the differential equation for the concentration of molecules in the cell should be modified to: $$ \frac{d C}{d t}=-\frac{k}{V}\left(C-C_{\infty}\right)+\frac{r}{V} $$ (d) Analyze Equation (8.54) to find the equilibrium value of the cell concentration. Is this equilibrium stable or unstable? You may use a graphical argument or calculate the eigenvalue to determine the equilibrium's stability.

To model the spread of a disease in a population of size \(N\) we derived a differential equation model: $$ \frac{d I}{d t}=(k b-c) I-\frac{k b}{N} I^{2} $$ where \(I(t)\) is the number of infected individuals at time \(t\), and \(k, b\), and \(c\) are all positive coefficients. Assuming that I ( \(t\) ) is modeled by Equation \((8.70)\), you should locate the equilibria of the model, and find which of these equilibria are stable. Draw a vector field plot for each problem. \(k=1, b=1, c=0.5, N=50 .\)

Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d N}{d t}=N(N-1)(5-N)\) (a) \(N(0)=1\), (b) \(N(0)=1 / 2\), (c) \(N(0)=3 / 2\), (d) \(N(0)=7\).

Find the equilibria of the following differential equations. $$ \frac{d y}{d t}=y^{3}+y $$
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