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Chapter 4: Problem 4

A rumor spreads among a group of 400 people. The number of people, \(N(t)\), who have heard the rumor by time \(t\) in hours since the rumor started to spread can be approximated by a function of the form $$ N(t)=\frac{400}{1+399 e^{-0.4 t}} $$ (a) Find \(N(0)\) and interpret it. (b) How many people will have heard the rumor after 2 hours? After 10 hours? (c) Graph \(N(t)\). (d) Approximately how long will it take until half the people have heard the rumor? Virtually everyone? (e) Approximately when is the rumor spreading fastest?

Short Answer

Expert verified
(a) 1 person at t=0. (b) Use the function to calculate. (d) Solve for t when N(t) = 200.

Step by step solution

01

Evaluate N at t=0

To find \(N(0)\), substitute \(t=0\) into the function \(N(t) = \frac{400}{1+399e^{-0.4t}}\). This gives us: \[N(0)=\frac{400}{1+399e^{0}} = \frac{400}{400} = 1.\] This means only 1 person has heard the rumor at the start.
02

Evaluate N at t=2

Substitute \(t=2\) into the function to find \(N(2)\): \\[N(2) = \frac{400}{1+399e^{-0.8}}.\] \Compute \(e^{-0.8}\) and plug it in to find \(N(2)\).
03

Evaluate N at t=10

Substitute \(t=10\) into the function to find \(N(10)\): \\[N(10) = \frac{400}{1+399e^{-4}}.\] \Compute \(e^{-4}\) and plug in the value to find \(N(10)\).
04

Graphing N(t)

Plot the function \(N(t) = \frac{400}{1+399e^{-0.4t}}\) over a suitable domain, for example \(t\) from 0 to around 20, to observe how the rumor spreads over time.
05

Solve for Half the Population

To find when the rumor has reached 200 people, set \(N(t) = 200\) and solve: \\[200 = \frac{400}{1+399e^{-0.4t}}.\] \Solve for \(t\) by isolating \(e^{-0.4t}\) and use logarithms to find the approximate value of \(t\).
06

Estimate when Nearly Everyone Heard the Rumor

To find when almost everyone has heard the rumor, approximate it for \(N(t) = 400\). However, since \(N(t)\) approaches 400 asymptotically, it may be practical to find when \(N(t)\) is 390 or greater and solve for \(t\).
07

Determine Fastest Spread

The rumor spreads fastest when the rate of change is maximum, which is the inflection point of \(N(t)\). \ To find this, differentiate \(N(t)\), set the second derivative to zero, and solve for \(t\).

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Key Concepts

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

Exponential Growth
In population dynamics, **exponential growth** refers to the increase in a population where the growth rate is proportional to the population size. This concept is often used to model scenarios where resources are unlimited.

In the context of the rumor spreading, if only one person initially knows the rumor and shares it with more people, the number of those who know begins to increase rapidly. The equation for exponential growth is generally given by \[ P(t) = P_0 e^{rt} \]where:
  • \( P(t) \) is the population at time \( t \)
  • \( P_0 \) is the initial population size
  • \( r \) is the growth rate, and \( e \) is the base of the natural logarithm.

When we look at the initial stages of rumor spreading, it resembles exponential growth as the increase is rapid when more people start sharing the rumor, but with time, the growth stabilizes.
Logistic Model
The **logistic model** offers a more realistic approach to understanding population dynamics since it incorporates the concept of carrying capacity. This is the maximum population size an environment can sustain indefinitely.

For the rumor model given by \[ N(t) = \frac{400}{1+399e^{-0.4t}} \]the logistic model is evident. Here:
  • The carrying capacity is 400, representing the total number of people that can eventually hear the rumor.
  • The growth of the rumor initially accelerates, but it slows down as the number of people nearing 400 increases.
  • The expression \(1 + 399e^{-0.4t}\) is the denominator that hinders endless exponential growth, representing resistance as more people hear the rumor.
This model is more suitable than a simple exponential growth model because it realistically predicts a leveling-off as the rumor saturates the population.
Differential Equations
**Differential equations** play a crucial role in describing how rumorous information spreads over time. In the given scenario, the function governing the rumor's spread is a solution to a differential equation. This differential equation models not just the start of the rumor but its entire lifecycle.

The differential equation associated with our logistic function might take a form similar to:\[ \frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right) \]where:
  • \( \frac{dN}{dt} \) shows how the number of people who have heard the rumor changes with time \( t \).
  • \( r \) is the intrinsic growth rate of the rumor.
  • \( K \) is the carrying capacity, here 400 people.

Solving differential equations like this one allows us to predict how a rumor will spread and change over time, capturing both the rapid spread at the beginning and the eventual saturation at the maximum population.
Calculus Applications
**Calculus**, through differentiation and integration, provides tools to explore even more in the context of rumor spreading. This helps to identify important points such as when the rumor spreads quickest or when saturation occurs.

By differentiating the function \( N(t) = \frac{400}{1+399e^{-0.4t}} \), we can find the rate at which the rumor is spreading at any given time. We look for the time \( t \) when this rate is maximum, known as the point of inflection. Here the spread changes from increasing to decreasing.
  • The first derivative \( \frac{dN}{dt} \) gives the rate of the spread.
  • Setting the second derivative to zero \( \frac{d^2N}{dt^2} = 0 \) helps identify when this change in spread rate is fastest.
  • This moment typically occurs roughly when half the people have heard the rumor, a characteristic of the logistic curve.
Understanding these calculus applications is a powerful way to get insight not just into rumors, but into natural processes exhibiting logistic growth behavior.

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Most popular questions from this chapter

Dwell time, \(t\), is the time in minutes that shoppers spend in a store. Sales, \(s\), is the number of dollars they spend in the store. The elasticity of sales with respect to dwell time is 1.3. Explain what this means in simple language.

A company can produce and sell \(f(L)\) tons of a product per month using \(L\) hours of labor per month. The wage of the workers is \(w\) dollars per hour, and the finished product sells for \(p\) dollars per ton. (a) The function \(f(L)\) is the company's production function. Give the units of \(f(L) .\) What is the practical significance of \(f(1000)=400 ?\) (b) The derivative \(f^{\prime}(L)\) is the company's marginal product of labor. Give the units of \(f^{\prime}(L) .\) What is the practical significance of \(f^{\prime}(1000)=2 ?\) (c) The real wage of the workers is the quantity of product that can be bought with one hour's wages. Show that the real wage is \(w / p\) tons per hour. (d) Show that the monthly profit of the company is $$ \pi(L)=p f(L)-w L $$ (e) Show that when operating at maximum profit, the company's marginal product of labor equals the real wage: $$ f^{\prime}(L)=\frac{w}{p} $$

Sketch the graph of a function on the inter val \(0 \leq x \leq 10\) with the given properties. Has local and global minimum at \(x=3\), local and global maximum at \(x=8\).

A grapefruit is tossed straight up with an initial velocity of \(50 \mathrm{ft} / \mathrm{sec}\). The grapefruit is 5 feet above the ground when it is released. Its height at time \(t\) is given by $$y=-16 t^{2}+50 t+5$$ How high does it go before returning to the ground?

There are many brands of laundry detergent. Would you expect the elasticity of demand for any particular brand to be high or low? Explain.
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