Problem 3 - Active Calculus

Question

The total number of people infected with a virus often grows like a logistic curve. Suppose that 20 people originally have the virus, and that in the early stages of the virus (with time, \(t,\) measured in weeks), the number of people infected is increasing exponentially with \(k=2 .\) It is estimated that, in the long run, approximately 6000 people become infected. (a) Use this information to find a logistic function to model this situation. \(P=\) (b) Sketch a graph of your answer to part (a). 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? vertical coordinate =

Step-by-Step Solution

Answer

The logistic function is \( \frac{6000}{1 + 299e^{-2t}} \). The rate at which people are becoming infected starts to decrease at \( t \approx 2.85 \) weeks, with a vertical coordinate of approximately 3163 people.

Step 1: Understand the logistic function formula

The general form of the logistic function is given by \[ P(t) = \frac{L}{1 + Ae^{-kt}} \] where:- \(L\) is the carrying capacity of the population, the maximum number of people who can be infected,- \(k\) is the growth rate, and,- \(A\) is a constant to be determined based on initial conditions.

Step 2: Identify given values

From the problem, we know:- \(L = 6000\) (the carrying capacity)- Initially, 20 people are infected (\(P(0) = 20\))- Growth rate \(k = 2\)

Step 3: Use initial condition to find A

Substitute the initial condition into the logistic function: \[ 20 = \frac{6000}{1 + Ae^{0}} \] This simplifies to: \[ 20 = \frac{6000}{1 + A} \] Solving for \(A\): \[ 20(1 + A) = 6000 \] \[ 20 + 20A = 6000 \] \[ 20A = 5980 \] \[ A = 299 \]

Step 4: Write the logistic function

Using the values of \(L = 6000\), \(k = 2\), and \(A = 299\), we have: \[ P(t) = \frac{6000}{1 + 299e^{-2t}} \]

Step 5: Determine the point of maximum growth rate

For a logistic function, the maximum growth rate (inflection point) occurs at \(t = \frac{1}{k} \ln(A)\). Using \(k = 2\) and \(A = 299\): \[ t = \frac{1}{2} \ln(299) \] \( \ln(299) \approx 5.70 \) \[ t \approx \frac{5.70}{2} \] \[ t \approx 2.85 \] (weeks)

Step 6: Find the vertical coordinate at the point of maximum growth

Substitute \(t = 2.85\) back into the logistic function: \[ P(2.85) = \frac{6000}{1 + 299e^{-2 \times 2.85}} \] \( e^{-5.70} \approx 0.003 \) \[ P(2.85) = \frac{6000}{1 + 299 \times 0.003} \] \[ P(2.85) \approx \frac{6000}{1 + 0.897} \] \[ P(2.85) \approx \frac{6000}{1.897} \] \[ P(2.85) \approx 3163 \]

Key Concepts

logistic function

The logistic function models situations where growth is initially exponential but slows as it approaches a maximum limit. This limit is known as the carrying capacity. The function is expressed as \[ P(t) = \frac{L}{1 + Ae^{-kt}} \] where:

• \(L\) is the carrying capacity or the maximum number of people who can be infected.
• \(k\) is the growth rate.
• \(A\) is a constant determined by initial conditions.

This function starts rapidly but slows as \(t\) increases, creating an S-shaped curve.

exponential growth rate

Exponential growth occurs when the rate of population increase is proportional to the current population size. In the formula \( P(t) = \frac{L}{1 + Ae^{-kt}} \), the term \(e^{-kt}\) represents the exponential aspect of the logistic function. Early in the growth phase, the number of infected people increases rapidly. For example, if the growth rate \(k\) is high, the population grows quickly at first, but this rate decreases as more individuals become infected and resources or space limit further growth.

inflection point analysis

In the context of logistic growth, the inflection point is where the rate of growth switches from increasing to decreasing. This point occurs at \( t = \frac{1}{k} \ln(A) \), where the growth is at its highest. Using the given growth rate \( k = 2 \) and the constant \( A = 299 \), we find the inflection point: \[ t = \frac{1}{2} \ln(299) \approx 2.85 \text{ weeks} \] At this moment, the population infected is \( P(2.85) \approx 3163 \). This calculation uses the S-shaped curve characteristics where growth slows after reaching this point.

carrying capacity

The carrying capacity \( L \) is the maximum population size that an environment can sustain. In our problem, it is given as 6000 people. This value caps the logistic function, meaning the infection growth will stop increasing once the total infected population hits 6000. The logistic function ensures that no matter how high the exponential growth rate \( k \) is, the population will approach but never exceed this carrying capacity.