Problem 32 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises 31 and \(32,\) a population function is given. (a) Show that the function is a solution of a logistic differential equation. Identify \(k\) and the carrying capacity. (b) Writing to Learn Estimate \(P(0)\) . Explain its meaning in the context of the problem. Spread of Measles The number of students infected by measles in a certain school is given by the formula \(P(t)=\frac{200}{1+e^{5.3-t}}\) where \(t\) is the number of days after students are first exposed to an infected student.

Step-by-Step Solution

Answer

The given population function can indeed be put into logistic form with a carrying capacity \(K = 200\) and growth rate \(k = 1\). The quantity \(P(0)=\frac{200}{1+e^{5.3}}\) gives the initial infected population at the start of the infection spread.

Step 1: Verify the Logistic Equation Form

The general form of a logistic function is given by \(P(t) = \frac{K}{1+ C e^{(-kt)}}\), where K represents the carrying capacity, and k is the growth rate. We will arrange the given function \(P(t)=\frac{200}{1+e^{(5.3-t)}}\) to this format and prove it to be a logistic function.

Step 2: Identify the Parameters

Comparing the given function with the generic form, it can be seen that the carrying capacity \(K = 200\). The growth rate can be found as \(k = 1\), as it is the coefficient of t in the exponent.

Step 3: Evaluate for P(0)

If we plug t = 0 into the given function, it gives the initial population who are infected by measles, that is, \(P(0)=\frac{200}{1+e^{5.3}}\).

Step 4: Interpret the meaning of P(0)

Now, \(P(0)\) stands for the infected population just when the students are first exposed to an infected student. In the context of this problem, it is crucial as it provides the initial state or the starting point of the disease spread.

Key Concepts

Logistic Growth Model

The logistic growth model is a mathematical way to describe how populations grow over time. It is especially useful when modeling constrained growth, where a population grows rapidly in the beginning but levels off as resources become limited. This behavior creates an S-shaped curve known as the logistic curve.
The logistics model is encapsulated in the differential equation \[ \frac{dP}{dt} = rP \left(1 - \frac{P}{K}\right) \]
where:

• \( r \) is the growth rate.
• \( P \) is the population size at time \( t \).
• \( K \) is the carrying capacity — the maximum population size that the environment can sustain.

A logistic function often takes the form\[ P(t) = \frac{K}{1 + Ce^{-kt}} \]where \( C \) is a constant derived from initial conditions. As time progresses, the term \( Ce^{-kt} \) diminishes, pointing the population towards the carrying capacity.
In this exercise, the provided function is similar to a standard logistic equation, helping us understand the dynamics of a measles outbreak modeled over time.

Carrying Capacity

Carrying capacity is a core concept in the logistic growth model. It refers to the maximum number of individuals of a species that an environment can support indefinitely.
This limit is dictated by resources like food, habitat, and other essentials needed for survival. Within the logistic function \[ P(t) = \frac{K}{1 + Ce^{-kt}} \],\( K \) stands for the carrying capacity.
For the given problem, the carrying capacity is identified as \( K = 200 \), which means the maximum number of students that can get measles in the school, based on environmental conditions and constraints.
Understanding carrying capacity guides us in envisioning how populations fluctuate and stabilize, offering insights into managing animal populations or combating diseases like measles.

Initial Population

The initial population is the starting number of individuals present at the beginning of observation, which plays a crucial role in understanding population dynamics. In the logistic growth equation, the initial population can be obtained by setting the time \( t = 0 \) in the function \( P(t) = \frac{K}{1 + Ce^{-kt}} \).
For the formula \( P(t) = \frac{200}{1 + e^{5.3-t}} \), substituting \( t = 0 \) results in \( P(0) = \frac{200}{1 + e^{5.3}} \).This represents the number of infected students immediately after exposure in the context of the problem.
The initial population is crucial as it sets the tone for predicting how quickly and to what extent the infection will spread under the given model. Recognizing this figure helps in planning interventions or resource allocation in response to the disease spread.