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Chapter 2: Problem 2

The number \(N(t)\) of people in a community who are exposed to a particularadvertisement is governed by the logisticequation. Initially \(N(0)=500\), and it is observed that \(N(1)=1000\). Solve for \(N(t)\) if it is predicted that the limiting number of people in the community who will see the advertisement is 50,000 .

Short Answer

Expert verified
The logistic growth model is \( N(t) = \frac{50000}{1 + 99e^{-0.69t}} \).

Step by step solution

01

Understand the Logistic Equation

The logistic equation is given by the model: \( N(t) = \frac{K}{1 + Ae^{-rt}} \). In this formula, \( K \) is the carrying capacity or the limiting number of people who will eventually be exposed to the advertisement, \( r \) is the growth rate, and \( A \) is a constant that can be determined from initial conditions.
02

Identify the Known Values

From the problem statement, we have the following values: \( N(0) = 500 \), \( N(1) = 1000 \), and \( K = 50000 \). These will help us determine the constants A and r in the logistic equation.
03

Calculate Constant A Using Initial Condition

Substitute \( N(0) = 500 \) into the logistic equation:\(500 = \frac{50000}{1 + A} \).Solving for A gives:\(1 + A = \frac{50000}{500} \rightarrow A = 99.\)
04

Calculate Growth Rate r Using Second Condition

Substitute \( N(1) = 1000 \) into the logistic equation and use the known value of A:\[1000 = \frac{50000}{1 + 99e^{-r}}. \]Rearranging for \( e^{-r} \) gives:\[1 + 99e^{-r} = \frac{50000}{1000} \rightarrow e^{-r} = \frac{50 - 1}{99}. \]Thus, \( e^{-r} = \frac{49}{99}. \)
05

Solve for r

Take the natural logarithm of both sides to solve for \( r \):\[-r = \ln\left(\frac{49}{99}\right) \rightarrow r = -\ln\left(\frac{49}{99}\right).\]Compute this value of \( r \) using a calculator, resulting in \( r \approx 0.69 \).
06

Write the Logistic Equation

Substitute \( K = 50000 \), \( A = 99 \), and \( r \approx 0.69 \) back into the logistic model:\[N(t) = \frac{50000}{1 + 99e^{-0.69t}}.\]This is the final form of the logistic equation that describes the number of people exposed over time.

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Key Concepts

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

Carrying Capacity in Logistic Growth
The carrying capacity in a logistic growth model is a fundamental concept. It represents the maximum population size that an environment can support indefinitely. In this context, it refers to the maximum number of people who will eventually see the advertisement, denoted as \( K \). This figure is crucial in predicting how the exposure will level off over time.

The carrying capacity effectively caps growth after a certain point, preventing the population from reaching unsustainable levels. Think of it as a limit beyond which no further growth occurs due to resource constraints or market saturation. In the given exercise, the carrying capacity is 50,000, meaning that no matter how long the advertisement runs, only up to 50,000 people are expected to see it.
Understanding Growth Rate Calculation
Calculating the growth rate \( r \) in a logistic model helps us understand how quickly the number of people exposed to the advertisement grows initially. This rate is not constant, as it diminishes as the population approaches the carrying capacity. When finding \( r \), we rely on specific initial conditions to find how fast or slowly exposure increases.

The process involves using the logistic equation: \[ N(t) = \frac{K}{1 + Ae^{-rt}}. \]
By substituting known values, such as \( N(1) = 1000 \), \( K = 50000 \), and the calculated constant \( A = 99 \), we solve for \( r \). We found \( r \approx 0.69 \), indicating how transitional the growth is from the initial 500 people to the eventual saturation at 50,000. This calculation taps into the unique feature of logistic curves: their initial rapid growth that gradually slows as the environment's carrying capacity is approached.
Initial Conditions and Their Impact
Initial conditions in any mathematical model set the stage for predictions and calculations. They help us determine the constants and parameters within equations such as those in the logistic growth model.

For the problem at hand, the initial number of people exposed, \( N(0) = 500 \), and the number of people after one time unit, \( N(1) = 1000 \), are critical. These values help solve for unknown constants like \( A \) and \( r\).
To find \( A \), we started with \( N(0) = 500 \) to get a simple equation:
\[ 500 = \frac{50000}{1 + A}. \]
This allowed us to find \( A = 99 \), a crucial part of the model.
Similarly, using \( N(1) = 1000 \) facilitated the calculation of \( r \). Initial conditions bridge the gap between theoretical models and real-world applications, ensuring calculations reflect actual scenarios.

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Most popular questions from this chapter

Solve the given differential equation by using an appropriate substitution. $$ t^{2} \frac{d y}{d t}+y^{2}=t y $$

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