/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 In a city of half a million, the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 6

In a city of half a million, there are initially 800 cases of a particularly virulent strain of flu. The Centers for Disease Control and Prevention in Atlanta claims that the cumulative number of infections of this flu strain will increase by \(40 \%\) per week if there are no limiting factors. Make a logistic model of the potential cumulative number of cases of flu as a function of weeks from initial outbreak, and determine how long it will be before 100,000 people are infected.

Short Answer

Expert verified
About 17.7 weeks.

Step by step solution

01

Understanding the Logistic Growth Model

The logistic growth model is defined by the equation \( C(t) = \frac{L}{1 + be^{-kt}} \), where \( C(t) \) is the cumulative number of cases at time \( t \), \( L \) is the carrying capacity of the environment, \( k \) is the growth rate, and \( b \) is a constant determined by initial conditions. In this problem, \( L = 500,000 \), since that's the total population.
02

Finding the Growth Rate

The initial number of cases, \( C(0) \), is 800, and the weekly growth rate is 40\%. To turn this percentage into the growth rate \( k \), we must convert it into a continuous growth factor using the formula \( k \approx \ln(1.4) \approx 0.3365 \).
03

Determine the Constant b

Use the initial condition to find \( b \). We know \( C(0) = 800 = \frac{500,000}{1 + b} \). By solving this equation for \( b \), we find that \( 1 + b = \frac{500,000}{800} = 625 \). Therefore, \( b = 624 \).
04

Setting Up the Logistic Function

The logistic model becomes: \( C(t) = \frac{500,000}{1 + 624e^{-0.3365t}} \). This represents the potential cumulative number of flu cases over time.
05

Solving for 100,000 Infections

Set \( C(t) = 100,000 \) and solve for \( t \). Substituting gives \( 100,000 = \frac{500,000}{1 + 624e^{-0.3365t}} \). Rearranging gives \( 1 + 624e^{-0.3365t} = 5 \), leading to \( 624e^{-0.3365t} = 4 \). Solving for \( t \), we find \( e^{-0.3365t} = \frac{4}{624} = \frac{1}{156} \). Taking the natural logarithm, \( -0.3365t = \ln\left(\frac{1}{156}\right) \). Thus, \( t = \frac{-\ln\left(\frac{1}{156}\right)}{0.3365} \approx 17.68 \) weeks.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Carrying Capacity
In the context of the logistic growth model, carrying capacity, often denoted as \( L \), refers to the maximum population size that a particular environment can support. This is an important concept in understanding how populations grow and eventually stabilize. For our problem involving the flu in a city of half a million people:
  • The city's total population is 500,000.
  • This entire population represents the carrying capacity, \( L = 500,000 \), assuming that everyone could potentially be infected.
Carrying capacity does not mean that everyone will be infected, but it sets the upper limit on the potential size of the outbreak in the absence of interventions. Over time, as infections approach this ceiling, the growth rate of new flu cases will naturally slow down, even without intentional control measures.
Growth Rate
The growth rate in the logistic growth model is crucial for understanding how quickly the number of cases increases. It is often denoted by \( k \) and represents the continuous growth. Unlike a simple percentage increase, continuous growth accounts for compound effects over time.In our scenario:
  • The weekly growth rate is given as 40%.
  • To apply it in a continuous model, we use the natural logarithm: \( k = \ln(1.4) \approx 0.3365 \).
This transformation is necessary because the logistic growth equation requires \( k \) in terms of time's continuous effect. The chosen growth rate helps in setting up the logistic equation to model the outbreak properly over the weeks.
Continuous Growth
Continuous growth captures the idea of exponential increases in a population or infection cases, where each increment of time sees the current amount added to. This contrasts with simple periodic growth, where increases are at fixed amounts or percentages.For the flu outbreak:
  • Continuous growth means the infection rate compounds weekly.
  • The formula \( e^{-kt} \), used in our logistic equation, is key in adjusting for how continuous growth affects population increase.
      • This aspect of a logistic model allows us to predict accurately how fast the infection spreads over time, considering the compounded nature of epidemic data. It more closely mirrors real-world scenarios than simple percentage calculations.
Logistic Function
The logistic function is the mathematical model that integrates the concepts of carrying capacity, growth rate, and continuous growth into a comprehensive formula used to predict population dynamics under constraints.In our flu outbreak example:
  • The logistic function is given by: \[ C(t) = \frac{L}{1 + be^{-kt}} \]
  • Here, \( L \) is the carrying capacity, \( b \) is determined by initial conditions, and \( k \) is the growth rate.
  • Specific to this scenario, the initial setup provided a formula: \( C(t) = \frac{500,000}{1 + 624e^{-0.3365t}} \)
This function shows how the number of flu cases progresses over time, accounting for both the initial rapid spread and the eventual slowdown as the carrying capacity is approached. Using this model, we accurately determined that it takes about 17.68 weeks for the infection count to reach 100,000 under specified conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If we view a star now, and then view it again 6 months later, our position will have changed by the diameter of the Earth's orbit around the sun. (See Figure 5.68.) For stars within about 100 light-years of Earth, the change in viewing location is sufficient to make the star appear to be in a different location in the sky. Half of the angle from one location to the next is known as the parallax angle. Even for nearby stars, the parallax angle is very \(\operatorname{small}^{36}\) and is normally measured in seconds of arc. The distance to a star can be determined from the parallax angle. The table below gives parallax angle \(p\) measured in seconds of arc and the distance \(d\) from the sun measured in light-years. $$ \begin{array}{|l|c|c|} \hline \text { Star } & \text { Parallax angle } & \text { Distance } \\ \hline \text { Markab } & 0.030 & 109 \\ \hline \text { Al Na'ir } & 0.051 & 64 \\ \hline \text { Alderamin } & 0.063 & 52 \\ \hline \text { Altair } & 0.198 & 16.5 \\ \hline \text { Vega } & 0.123 & 26.5 \\ \hline \text { Rasalhague } & 0.056 & 58 \\ \hline \end{array} $$ a. Make a plot of \(\ln d\) against \(\ln p\), and determine whether it is reasonable to model the data with a power function. b. Make a power function model of the data for \(d\) in terms of \(p\). c. If one star has a parallax angle twice that of a second, how do their distances compare? d. The star Mergez has a parallax angle of \(0.052\) second of arc. Use functional notation to express how far away Mergez is, and then calculate that value. e. The star Sabik is 69 light-years from the sun. What is its parallax angle?

A building that is subjected to shaking (caused, for example, by an earthquake) may collapse. Failure depends both on intensity and on duration of the shaking. If an intensity \(I_{1}\) causes a building to collapse in \(t_{1}\) seconds, then an intensity \(I_{2}\) will cause the collapse in \(t_{2}\) seconds, where $$ \frac{t_{1}}{t_{2}}=\left(\frac{I_{2}}{I_{1}}\right)^{2} $$ If a certain building collapses in 30 seconds at one intensity, how long would it take the building to collapse at triple that intensity?

When seeds of a plant are sown at high density in a plot, the seedlings must compete with each other. As time passes, individual plants grow in size, but the density of the plants that survive decreases. \({ }^{33}\) This is the process of selfthinning. In one experiment, horseweed seeds were sown on October 21 , and the plot was sampled on successive dates. The results are summarized in Table \(5.8\), which gives for each date the density \(p\), in number per square meter, of surviving plants and the average dry weight \(w\), in grams, per plant. a. Explain how the table illustrates the phenomenon of self-thinning. b. Find a formula that models \(w\) as a power function of \(p\). c. If the density decreases by a factor of \(\frac{1}{2}\), what happens to the weight? d. The total plant yield y per unit area is defined to be the product of the average weight per plant and the density of the plants: \(y=w \times p\). As time goes on, the average weight per plant increases while the density decreases, so it's unclear whether the total yield will increase or decrease. Use the power function you found in part b to determine whether the total yield increases or decreases with time. Check your answer using the table. $$ \begin{array}{|l|r|c|} \hline \text { Date } & \text { Density } p & \text { Weight } w \\ \hline \text { November 7 } & 140,400 & 1.6 \times 10^{-4} \\ \hline \text { December } 16 & 36,250 & 7.7 \times 10^{-4} \\ \hline \text { January 30 } & 22,500 & 0.0012 \\ \hline \text { April 2 } & 9100 & 0.0049 \\ \hline \text { May 13 } & 4510 & 0.018 \\ \hline \text { June } 25 & 2060 & 0.085 \\ \hline \end{array} $$

A queue, or line, of traffic can form when a feeder road meets a main road with a high volume of traffic. In this exercise we assume that gaps in traffic on the main road (allowing cars to enter from the feeder road) appear randomly and that cars on the feeder road arrive at the intersection randomly also. We let \(s\) be the average rate at which a gap in traffic appears and \(a\) the average arrival rate, both per minute. Assume that \(s\) is greater than \(a\). a. The average length \(L\) of a queue is given by the rational function $$ L=\frac{a^{2}}{s(s-a)} . $$ Explain what happens to average queue length if arrival rate and gap rate are nearly the same. b. The average waiting time \(w\) (the time in minutes spent in the queue) is given by $$ w=\frac{a}{s(s-a)} . $$ If the gap rate is 3 per minute, what arrival rate will result in a driver's average waiting time of 2 minutes?

An exponential model of growth follows from the assumption that the yearly rate of change in a population is \((b-d) N\), where \(b\) is births per year, \(d\) is deaths per year, and \(N\) is current population. The increase is in fact to some degree probabilistic in nature. If we assume that population increase is normally distributed around \(r N\), where \(r=b-d\), then we can discuss the probability of extinction of a population. a. If the population begins with a single individual, then the probability of extinction by time \(t\) is given by $$ P(t)=\frac{d\left(e^{r t}-1\right)}{b e^{r t}-d} $$ If \(d=0.24\) and \(b=0.72\), what is the probability that this population will eventually become extinct? (Hint: The probability that the population will eventually become extinct is the limiting value for \(P\).) b. If the population starts with \(k\) individuals, then the probability of extinction by time \(t\) is $$ Q=P^{k}, $$ where \(P\) is the function in part a. Use function composition to obtain a formula for \(Q\) in terms of \(t, b, d, r\), and \(k\). c. If \(b>d\) (births greater than deaths), so that \(r>0\), then the formula obtained in part b can be rewritten as $$ Q=\left(\frac{d\left(1-a^{t}\right)}{b-d a^{t}}\right)^{k} $$ where \(a<1\). What is the probability that a population starting with \(k\) individuals will eventually become extinct? d. If \(b\) is twice as large as \(d\), what is the probability of eventual extinction if the population starts with \(k\) individuals? e. What is the limiting value of the expression you found in part \(\mathrm{d}\) as a function of \(k\) ? Explain what this means in practical terms.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.