Question

Spread of Flu The spread of flu in a certain school is modeled by the equation \(P(t)=\frac{200}{1+e^{5-t}}\) where \(P(t)\) is the total number of students infected \(t\) days after the flu first started to spread. (a) Estimate the initial number of students infected with this flu. (b) How fast is the flu spreading after 4 days? (c) When will the flu spread at its maximum rate? What is that rate?

Short answer

The initial number of students infected is \(200e^{-5}\). The rate of spread after 4 days is \(-\frac{200e}{(e+1)^2}\). The maximum rate of spread of the flu is \(-200e^{-5}\), and it occurs at \(t=0\).

Step-by-step solution

Find the initial number of students infected

The initial number of students who were infected is simply \(P(0)\). So plug in \(t = 0\) into the equation \(P(t)=\frac{200}{1+e^{5-t}}\). This gives \(P(0)=\frac{200}{1+e^{5}}=200e^{-5}\).

Find the growth rate after four days

To find the growth rate, we need to calculate the derivative of the function \(P(t)\). The derivative of \(P(t)\) is \(P'(t)=-\frac{200e^{5-t}}{(e^{5-t}+1)^2}\). Plug in \(t = 4\) to find the spread rate of the flu after 4 days, which is \(P'(4)=-\frac{200e}{(e+1)^2}\).

Find the maximum growth rate and when it occurs

To find the maximum growth rate, we need to find the critical points of the derivative function \(P'(t)\). We set \(P'(t)=0\) to solve for \(t\). In this situation, it is not possible to solve for \(t\) algebraically as the equation involves a transcendental function. However, by looking at the derivative function \(P'(t)\), we can see that it is always negative because the numerator is always negative and the denominator is always positive. Therefore, the growth rate is decreasing over time. Hence, the maximum rate must occur initially at \(t=0\). Plug in \(t = 0\) into \(P'(t)\), we get \(P'(0)=-\frac{200e^{5}}{(e^{5}+1)^2}=-200e^{-5}\), as the maximum rate of spread of the flu.

Key concepts

Exponential Growth and Decay

Understanding the exponential growth and decay of a flu outbreak is fundamental in epidemiology. The spread of diseases often follows an exponential pattern in the initial stages, whereby the number of infected individuals can increase rapidly. Conversely, exponential decay models the waning phase of an outbreak, where the number of infections decreases over time.

An exponential function has the general form of \( a \times e^{kx} \), where \( a \), \( e \), and \( k \), are constants, and \( x \) is the variable. In the context of a flu's spread within a school, we model the number of infected students, \( P(t) \), as a function of time \( t \), using an exponential decay function within a logistic growth framework.

It's important to note that such models assume that everyone in the population has an equal chance of infection and that the population size is large enough for the probabilities to remain constant. In the given exercise, the initial number of students infected might seem to imply an initial rapid growth, but as \( t \) increases, the growth 'decays' due to the function's nature.This decay aspect is reflected in the increase of the denominator as time goes by, representing the slowing down of infection spread. Thus, the concept of exponential growth and decay can help us predict the progression of an epidemic and determine necessary interventions to control it.

Mathematical Modeling in Epidemiology

Mathematical models like the one described in the flu problem are powerful tools in epidemiology. They enable us to understand and predict the behavior of infectious diseases within populations. Such models can range from simple to complex, incorporating various factors that influence the course of an epidemic, such as transmission rates, recovery rates, and population contact patterns.

In the exercise provided, the model \( P(t)=\frac{200}{1+e^{5-t}} \) represents the schematic spread of flu infections over time. Physicians and public health officials can use this type of logistical model to estimate critical details like infection peaks, rates of spread, and disease longevity.

Improving this mathematical model would involve adjusting for more realistic scenarios such as the inclusion of new infections over time, recovery rates, and perhaps the effects of interventions like vaccination or quarantine. A more sophisticated model might use differential equations to represent these dynamics. Such improvements would help the model to mirror real-life scenarios more closely and assist in making more accurate public health predictions and decisions.

Derivative of an Exponential Function

The derivative of an exponential function is essential for analyzing the rate at which something like a flu virus spreads. In the exercise, the derivative of our model \( P(t) \) with respect to \( t \) provides information on how fast the number of infected students changes at any given time.

The derivative, denoted as \( P'(t) \) or \( \frac{dP}{dt} \) gives us the rate of change of the infected population with respect to time. This rate can be positive, implying an increasing number of infections, or negative, suggesting a decrease. By calculating \( P'(t) \) for the given function and observing its sign and magnitude, we can determine how quickly the infection is spreading or subsiding at a specific time.

In the given solution, \( P'(t) \) is always negative due to the structure of the exponential function within the derivative, indicating a continuous decay in the spread rate. Furthermore, analyzing \( P'(t) \) can lead to determining when the flu spread is at its maximum rate and what that rate is, which is crucial for implementing timely control measures. Understanding the derivative of an exponential function is a key step in effective epidemic modeling and the management of infectious diseases.