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Chapter 11: Problem 34

The total number of people infected with a virus often grows like a logistic curve. Suppose that time, \(t,\) is in weeks and that 10 people originally have the virus. In the early stages, the number of people infected is increasing exponentially with \(k=1.78 .\) In the long run, 5000 people are infected. (a) Find a logistic function to model the number of people infected. (b) Sketch a graph of your answer to part (a). (c) Use your graph to estimate the length of time until the rate at which people are becoming infected starts to decrease. What is the vertical coordinate at this point?

Short Answer

Expert verified
The logistic function is \(P(t) = \frac{5000}{1 + e^{-1.78(t-2.998)}}\). Infection rate decreases at 3 weeks at 2500 infected.

Step by step solution

01

Understand the Logistic Function Form

The logistic model is given by \[ P(t) = \frac{L}{1 + e^{-k(t-t_0)}} \]where \(L\) is the carrying capacity of the population, \(k\) is the growth rate, and \(t_0\) is the midpoint in time where the population grows fastest.
02

Extract Given Information

From the problem, \(L = 5000\), \(P(0) = 10\), and \(k = 1.78\). We need to find \(t_0\).
03

Set Initial Condition

The initial number of infected people is 10, so \(P(0) = \frac{5000}{1 + e^{-1.78(0-t_0)}} = 10\). This can be rewritten as: \[ 10 = \frac{5000}{1 + e^{1.78t_0}} \]
04

Solve for \(t_0\)

Rearranging the equation:\[ 1 + e^{1.78t_0} = \frac{5000}{10} = 500 \]\[ e^{1.78t_0} = 499 \]Taking the natural logarithm:\[ 1.78t_0 = \ln(499) \]\[ t_0 = \frac{\ln(499)}{1.78} \approx 2.998 \]
05

Write the Logistic Function

Using the value of \(t_0\), the logistic function for the number of infected people is:\[ P(t) = \frac{5000}{1 + e^{-1.78(t-2.998)}} \]
06

Sketch the Graph of the Logistic Function

To sketch the graph, plot the logistic curve with the function from Step 5. It will have an S-shape, starting near zero, steeply increasing around \(t = 2.998\), and leveling off at 5000.
07

Determine When the Rate of Infection Starts to Decrease

The rate of infection starts to decrease at the midpoint \(t_0\), which we've calculated as approximately 3 weeks. At this point, the inflection happens and the curve transitions from a steep increase to a slower increase.
08

Find the Vertical Coordinate at \(t_0\)

To find \(P(t_0)\), substitute \(t_0 = 2.998\) into the logistic equation:\[ P(2.998) = \frac{5000}{1 + e^{-1.78(2.998-2.998)}} = \frac{5000}{2} = 2500 \]

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Key Concepts

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

Logistic Function
The logistic function is a fundamental concept in modeling biological growth and other processes constrained by a limiting factor. It is known for its characteristic S-shaped curve. The mathematical form of the logistic function is given by:\[ P(t) = \frac{L}{1 + e^{-k(t-t_0)}} \]where:
  • \( L \) is the carrying capacity, indicating the maximum population size that the environment can sustain.
  • \( k \) is the growth rate, determining how fast the population approaches the carrying capacity.
  • \( t_0 \) is the inflection point where the highest growth rate occurs.
The logistic function is widely used for populations that begin growing exponentially, slow down as resources become limited, and eventually level off when the carrying capacity is reached.
Exponential Growth
Exponential growth occurs when the rate of increase of a population is proportional to the current population size. For example, if a population doubles every period, it exhibits exponential growth. The formula for exponential growth is typically expressed as:\[ P(t) = P_0 e^{kt} \]where:
  • \( P_0 \) is the initial population size.
  • \( k \) is the growth rate constant.
  • \( t \) is time.
In the case of the virus infection from our exercise, the initial stages follow exponential growth, with a rate \( k = 1.78 \), meaning that the number of infected people rises rapidly. However, unlike exponential growth that continues indefinitely, the logistic growth model accounts for practicality by incorporating a carrying capacity.
Carrying Capacity
The carrying capacity \( L \) is a key element of the logistic model, defining the maximum population size that the environment can support. In the context of the virus infection model, the carrying capacity represents the maximum number of people who can become infected. In the exercise, this number is given as 5000.The carrying capacity influences how the population grows over time: as the population size nears \( L \), the growth rate decreases, leading to a leveling off of the population size. It reflects the limit imposed by factors such as available resources or space, ensuring the model remains realistic over longer periods.
Inflection Point
The inflection point \( t_0 \) is a crucial point in logistic growth, marking where the population growth transitions from increasing at an accelerating rate to increasing at a decelerating rate. It represents the point of maximum growth rate. In our exercise, we calculated \( t_0 \) to be approximately 2.998 weeks.At the inflection point, half of the carrying capacity is typically reached. For our logistic model, when \( t = t_0 \), the population size \( P(t) \) is half the carrying capacity, which is 2500 given the full capacity of 5000. This point is vital as it indicates where the intervention to control growth can be most effective.

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Most popular questions from this chapter

Give \(k, L, A,\) a formula for \(P\) as a function of time \(t,\) and the time to the peak value of \(d P / d t\) $$\frac{d P}{d t}=0.02 P-0.0025 P^{2}, \quad P_{0}=1$$

Give an explanation for your answer. For any positive values of the constant \(k\) and any positive values of the initial value \(P(0),\) the solution to the differential equation \(d P / d t=k P(L-P)\) has limiting value \(L\) as \(t \rightarrow \infty\)

Many organ pipes in old European churches are made of tin. In cold climates such pipes can be affected with tin pest, when the tin becomes brittle and crumbles into a gray powder. This transformation can appear to take place very suddenly because the presence of the gray powder encourages the reaction to proceed. The rate of the reaction is proportional to the product of the amount of tin left and the quantity of gray powder, \(p,\) present at time \(t .\) Assume that when metallic tin is converted to gray powder, its mass does not change. (a) Write a differential equation for \(p .\) Let the total quantity of metallic tin present originally be \(B\) (b) Sketch a graph of the solution \(p=f(t)\) if there is a small quantity of powder initially. How much metallic tin has crumbled when it is crumbling fastest? (c) Suppose there is no gray powder initially. (For example, suppose the tin is completely new.) What does this model predict will happen? How do you reconcile this with the fact that many organ pipes do get tin pest?

At 1: 00 pm one winter afternoon, there is a power failure at your house in Wisconsin, and your heat does not work without electricity. When the power goes out, it is \(68^{\circ} \mathrm{F}\) in your house. At 10: 00 pm, it is \(57^{\circ} \mathrm{F}\) in the house, and you notice that it is \(10^{\circ} \mathrm{F}\) outside. (a) Assuming that the temperature, \(T,\) in your home obeys Newton's Law of Cooling, write the differential equation satisfied by \(T\) (b) Solve the differential equation to estimate the temperature in the house when you get up at 7: 00 am the next morning. Should you worry about your water pipes freezing? (c) What assumption did you make in part (a) about the temperature outside? Given this (probably incorrect) assumption, would you revise your estimate up or down? Why?

The population data from another experiment on yeast by the ecologist G. F. Gause is given. $$\begin{array}{l|c|c|c|c|c|c|c}\hline \text { Time (hours) } & 0 & 13 & 32 & 56 & 77 & 101 & 125 \\\\\hline \text { Yeast pop } & 1.00 & 1.70 & 2.73 & 4.87 & 5.67 & 5.80 & 5.83 \\\\\hline\end{array}$$ (a) Do you think the population is growing exponentially or logistically? Give reasons for your answer. (b) Estimate the value of \(k\) (for either model) from the first two pieces of data. If you chose a logistic model in part (a), estimate the carrying capacity, \(L,\) from the data. (c) Sketch the data and the approximate growth curve given by the parameters you estimated.
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