Problem 545 - Precalculus

Question

For the following exercises, use this scenario: The equation \(N(t)=\frac{1200}{1+199 e^{-0.625 t}}\) models the number of people in a school who have heard a rumor after \(t\) days. What is the carrying capacity?

Step-by-Step Solution

Answer

The carrying capacity is 1200.

Step 1: Understanding the Scenario

The equation provided, \(N(t)=\frac{1200}{1+199 e^{-0.625 t}}\), is a logistical growth model used to describe how the number of people who have heard a rumor in a school changes over time. The parameter we are asked to find is the carrying capacity.

Step 2: Identifying the Logistic Growth Model Components

A logistic growth model is generally given by the form \(N(t)=\frac{L}{1+be^{-kt}}\), where \(L\) is the carrying capacity of the system. In this model, \(L\) represents the maximum number of people who can hear the rumor, given as \(1200\) in the equation \(N(t)=\frac{1200}{1+199 e^{-0.625 t}}\).

Step 3: Determining the Carrying Capacity

In the equation \(N(t)=\frac{1200}{1+199 e^{-0.625 t}}\), the carrying capacity \(L\) is the value that appears as the numerator of the logistic function. Thus, the carrying capacity \(L\) is \(1200\).

Key Concepts

Carrying Capacity

Carrying capacity is a vital concept in understanding logistic growth models. It represents the maximum population size that a particular environment, system, or situation can sustain. In the logistic growth model equation \(N(t)=\frac{1200}{1+199 e^{-0.625 t}}\), the carrying capacity is the value of \(L\), which is the numerator of the equation. Here, \(L=1200\), indicating that a maximum of 1,200 people can hear the rumor. This limit reflects the point at which the growth rate will slow down and stabilize because all or most members of the population have heard the rumor.

In logistic models:

• \(L\) is a limiting factor that halts indefinite growth.
• Environmental resources often influence carrying capacity.

For the rumor, this could include limited social interactions or information channels.

Growth Rate

The growth rate in a logistic growth model describes how quickly a population approaches its carrying capacity. It's determined by the exponential component of the equation. In our example, the term \(199 e^{-0.625 t}\) contains both the growth rate and decay factor. The growth rate constant is embedded in the exponent, which in this case is \(-0.625\).

How does this growth rate affect rumor spread?

• Higher growth rates mean faster information spread.
• Initially, the number of people hearing the rumor increases rapidly.

However, as the population nears the carrying capacity, the increase in growth becomes slower, balancing eventually to reflect the maximum reach or limitation that the carrying capacity presents.

Exponential Decay

Exponential decay plays a critical role in understanding growth dynamics in a logistic model. While exponential growth suggests rapid increase without constraints, exponential decay accounts for the gradual slowdown as a population maxes out its resources or limits, like in our rumor scenario. In \(N(t)=\frac{1200}{1+199 e^{-0.625 t}}\), \(e^{-0.625 t}\) represents the decay component that tempers the initial growth.

Key aspects of exponential decay in logistic models:

• It establishes a decelerating growth curve.
• As time passes, the impact of the decay increases.
• This ensures the population increment becomes less frequent as it approaches \(L\).

Exponential decay ensures that not all individuals hear about the rumor instantly, allowing the model to simulate realistic spreading scenarios.