/*! 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 69 Suppose that the spread of a flu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that the spread of a flu virus on a college campus is modeled by the function $$y(t)=\frac{1000}{1+999 e^{-0.9 t}}$$ where \(y(t)\) is the number of infected students at time \(t\) (in days, starting with \(t=0\) ). Use a graphing utility to estimate the day on which the virus is spreading most rapidly.

Short Answer

Expert verified
The virus spreads most rapidly around day 7.7.

Step by step solution

01

Analyze the Function

The given function is \( y(t) = \frac{1000}{1 + 999 e^{-0.9t}} \), which is a form of the logistic growth model. Logistic growth models are commonly used to describe the spread of diseases like the flu. To find when the virus is spreading most rapidly, we need to determine where the growth rate of infected students is maximum in terms of \( t \).
02

Identify the Point of Maximum Growth

In a logistic growth model, the point of maximum growth occurs at the inflection point, where the first derivative of the function is at its maximum. For the equation given, the inflection point occurs at \( t = \frac{1}{k} \ln(b) \), where the standard form of the logistic function is \( L/(1 + b e^{-kt}) \). Here, \( L = 1000 \), \( b = 999 \), and \( k = 0.9 \).
03

Calculate the Inflection Point

Using the formula for the inflection point in logistic models, calculate: \[ t = \frac{1}{0.9} \ln(999) \]. Simplifying, we have \[ t \approx \frac{1}{0.9} \times 6.9 = \frac{6.9}{0.9} \approx 7.67 \text{ days}. \]
04

Verify Using a Graphing Utility

Plot the function \( y(t) = \frac{1000}{1 + 999 e^{-0.9t}} \) on a graphing utility. Observe the graph to confirm that the curve reaches its steepest point around \( t \approx 7.67 \) days, indicating the point of maximum rapid spread.

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 91Ó°ÊÓ!

Key Concepts

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

Inflection Point
An inflection point in a logistic growth model is where the graph of the growth function changes from being concave upward to concave downward. In simpler terms, it's the point where the rate of spread is fastest.
For a flu virus spread like in our exercise, this is when the number of infected students increases most quickly. At this point, the rate of new infections is highest. After this point, even though the total number of infections continues to rise, the speed at which new students are being infected starts to slow down.
This is why identifying the inflection point is crucial when analyzing disease spread. It helps understand how quickly a virus is spreading and when it starts to slow down, which can be critical for public health measures.
Derivative
To find when a flu virus spreads most rapidly, mathematically speaking, we look at the derivative of the function modeling the spread. The derivative tells us about the rate of change of the function.
In the context of logistic growth, the derivative helps locate the point of maximum growth, i.e., the inflection point.
  • The first derivative, in our flu virus problem, gives the rate at which people become infected each day.
  • The second derivative tells us if the rate is increasing or decreasing over time.
After differentiating the logistic growth equation, the first derivative varies with time, reaching its peak at the inflection point. This peak signifies the most rapid spread of the virus, helping us pinpoint critical intervention moments during an outbreak.
Flu Virus Spread
The flu virus spread on a college campus can be accurately modeled using a logistic growth function. This is because logistical models are ideal for situations where growth starts rapidly, slows down, and eventually levels off as resources become limited or resistance builds up.
In our specific problem, the function provided represents the spread of flu among students. Initially few are infected, allowing the virus to spread quickly.
  • As more students become infected, the rate of new infections slows down because there are fewer uninfected students left or preventive measures kick in.
  • Eventually, the number of infected students levels off and approaches a maximum limit, known as the carrying capacity.
Understanding the flu virus spread pattern is essential for planning resources, such as vaccines and medical care, and implementing effective public health strategies.

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

An annuity is a sequence of equal payments that are paid or received at regular time intervals. For example, you may want to deposit equal amounts at the end of each year into an interest-bearing account for the purpose of accumulating a lump sum at some future time. If, at the end of each year, interest of \(i \times 100 \%\) on the account balance for that year is added to the account, then the account is said to pay \(i \times 100 \%\) interest, compounded annually. It can be shown that if payments of \(Q\) dollars are deposited at the end ofeach year into an account that pays \(i \times 100 \%\) compounded annually, then at the time when the \(n\) th payment and the accrued interest for the past year are deposited, the amount \(S(n)\) in the account is given by the formula $$S(n)=\frac{Q}{i}\left[(1+i)^{n}-1\right]$$ Suppose that you can invest \(\$ 5000\) in an interest-bearing account at the end of each year, and your objective is to have \(\$ 250,000\) on the 25th payment. Approximately what annual compound interest rate must the account pay for you to achieve your goal? [Hint: Show that the interest rate \(i\) satisfies the equation \(50 i=(1+i)^{25}-1,\) and solve it using Newton's Method.]

Use Newton's Method to approximate the absolute minimum of \(f(x)=\frac{1}{4} x^{4}+x^{2}-5 x\)

Approximate \(\sqrt[3]{6}\) by applying Newton's Method to the equation \(x^{3}-6=0\).

Use a graphing utility to determine how many solutions the equation has, and then use Newton’s Method to approximate the solution that satisfies the stated condition. \(x-\tan x=0 ; \pi / 2

If \(f\) is a periodic function, then the locations of all absolute extrema on the interval \((-\infty,+\infty)\) can be obtained by finding the locations of the absolute extrema for one period and using the periodicity to locate the rest. Use this idea in these exercises to find the absolute maximum and minimum values of the function, and state the \(x\) -values at which they occur. \(f(x)=2 \cos x+\cos 2 x\)

See all solutions

Recommended explanations on Math Textbooks

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.