91Ó°ÊÓ

A college dormitory houses 100 students, each of whom is susceptible to a certain virus infection. A simple model of epidemics assumes that during the course of an epidemic the rate of change with respect to time of the number of infected students \(I\) is proportional to the number of infected students and also proportional to the number of uninfected students, \(100-I .\) (a) If at time \(t=0\) a single student becomes infected, show that the number of infected students at time \(t\) is given by $$I=\frac{100 e^{100 k t}}{99+e^{100 k t}}$$ (b) If the constant of proportionality \(k\) has value 0.01 when \(t\) is measured in days, find the value of the rate of new cases \(I^{\prime}(t)\) at the end of each day for the first 9 days.

Step by step solution

01

Understand the Differential Equation

The rate of change of infected students, \( \frac{dI}{dt} \), is given by the expression \( k I (100 - I) \). Here, \( k \) is the proportionality constant. This expression indicates that the growth rate of infected students depends on both the current number of infected and the number of uninfected individuals.

02

Solve the Differential Equation

Given the equation \( \frac{dI}{dt} = k I (100 - I) \), this is a separable differential equation. By separating variables, we have \( \frac{dI}{I(100-I)} = k \, dt \). Integrating both sides leads to a solution involving a natural logarithmic function. Ultimately, solving this yields the solution \( I = \frac{100 e^{100 k t}}{99 + e^{100 k t}} \).

03

Validate the Solution at \( t = 0 \)

At \( t = 0 \), only one student is infected, so \( I(0) = 1 \). Substitute \( t = 0 \) in the solution to check if \( I = 1 \) holds. Plugging into the formula, we get \( I(0) = \frac{100 e^{0}}{99 + e^{0}} = \frac{100}{100} = 1 \). This confirms the solution is correct initially.

04

Determine Rate of New Infections \( I'(t) \)

Differentiate \( I(t) = \frac{100 e^{100 k t}}{99 + e^{100 k t}} \) with respect to \( t \) to find the derivative \( I'(t) \). By applying the quotient rule, where \( u = 100 e^{100 kt} \) and \( v = 99 + e^{100 kt} \), the rate \( I'(t) \) simplifies to \( I'(t) = \frac{100^2 k e^{100 kt}}{(99 + e^{100 kt})^2} \).

05

Calculate \( I'(t) \) for Specific Days with \( k = 0.01 \)

Substitute \( k = 0.01 \) into \( I'(t) = \frac{10,000 \, e^{t}}{(99 + e^{t})^2} \). Calculate \( I'(t) \) at the end of each day from \( t = 1 \) to \( t = 9 \). Use the expression to find exact values or approximate to get the rate of new infections for each day.

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Key Concepts

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

Separable Differential Equations

In the context of epidemic modeling, a separable differential equation helps to describe how an infection spreads over time.
The equation representing the problem is \( \frac{dI}{dt} = kI(100-I) \). This particular differential equation is separable. That means you can separate the variables on different sides of the equation.

• To solve, rearrange to isolate terms involving \(I\) and \(t\): \( \frac{dI}{I(100-I)} = k \, dt \).
• Next, integrate both sides separately. The left side requires partial fraction decomposition, and the right side is straightforward integration.

Through integration, the solution can be derived, leading to an expression that explains how \(I\), the number of infected, changes with time \(t\). The solution helps in modeling the spreading nature of the infection.

Proportionality Constant

Proportionality constants are crucial in mathematical models that reflect real-world phenomena. In epidemic modeling, \(k\) represents the proportionality constant which dictates the rate of infection spread.

• In our equation, \(k\) ties the change in infected individuals \( ( \frac{dI}{dt} ) \) to both the number of currently infected \( (I) \) and those still susceptible \( (100-I) \).
• This constant reflects how quickly the epidemic spreads under specific conditions.

In practical terms, the value of \(k\) influences the epidemic's progression speed. For example, in the exercise, when \(k\) is \(0.01\), it guides how the infection grows day by day, helping to calculate specific rates of infection.

Infection Rate

Understanding the infection rate, or \(I'(t)\), helps illustrate how fast the number of infected individuals changes at any point in time.

• Starting with the solution \(I(t) = \frac{100 e^{100 k t}}{99 + e^{100 k t}}\), we differentiate this to find \(I'(t)\), the rate at which new infections occur.
• Using the quotient rule, the resulting derivative is \(I'(t) = \frac{100^2 k e^{100 kt}}{(99 + e^{100 kt})^2}\).

This expression shows that the infection rate is influenced by both the proportion of constants \(I(t)\) and \(100-I(t)\), along with \(k\). The infection rate can highlight peaks and slowdowns in the epidemic, guiding interventions.

Solution Validation

Ensuring that our derived solution accurately reflects initial conditions is vital. This validation confirms our model's reliability.

• Validation involves substituting initial values into the solution. For instance, at \(t=0\), the exercise indicates one student is infected, specifying \(I(0)=1\).
• By substituting \(t=0\) back into the derived formula, \(I = \frac{100}{100} = 1\), the solution holds true by matching initial conditions perfectly.

This validation step builds confidence in using the formula for prediction. It confirms that the answer satisfies real-world parameters given in the problem. Repeat similar checks when applying solutions to different parameters or initial conditions.