Problem 11 - Applied Calculus

Question

A curve representing the total number of people, \(P\), infected with a virus often has the shape of a logistic curve of the form $$P=\frac{L}{1+C e^{-k t}}$$ with time \(t\) in weeks. Suppose that 10 people originally have the virus and that in the early stages the number of people infected is increasing approximately exponentially, with a continuous growth rate of \(1.78 .\) It is estimated that, in the long run, approximately 5000 people will become infected. (a) What should we use for the parameters \(k\) and \(L ?\) (b) Use the fact that when \(t=0,\) we have \(P=10,\) to find \(C\) (c) Now that you have estimated \(L, k,\) and \(C,\) what is the logistic function you are using to model the data? Graph this function. (d) Estimate the length of time until the rate at which people are becoming infected starts to decrease. What is the value of \(P\) at this point?

Step-by-Step Solution

Answer

(a) \( L = 5000, k = 1.78 \); (b) \( C = 499 \); (c) \( P(t) = \frac{5000}{1 + 499e^{-1.78t}} \); (d) Rate decreases at \( t \approx 3.49 \) and \( P = 2500 \).

Step 1: Determine Parameters L and k

The long-term limit of the number of people infected is denoted by \( L \). Since it is estimated that 5000 people will eventually become infected, \( L = 5000 \).The problem states that the early growth of the infection is approximately exponential with a continuous growth rate of \( 1.78 \). In a logistic model, this corresponds to the constant \( k = 1.78 \).

Step 2: Solve for Parameter C

We know that when \( t = 0 \), \( P = 10 \). Substitute \( t = 0 \) and \( P = 10 \) into the logistic equation:\[10 = \frac{5000}{1 + C \cdot e^{-1.78 \cdot 0}}\]Since \( e^0 = 1 \), the equation simplifies to:\[10 = \frac{5000}{1 + C}\]Solve for \( C \):\[10(1 + C) = 5000 \quad \Rightarrow \quad 10 + 10C = 5000\]Subtract 10 from both sides:\[10C = 4990 \quad \Rightarrow \quad C = 499\]

Step 3: Formulate the Logistic Function

Now substitute \( L = 5000 \), \( k = 1.78 \), and \( C = 499 \) back into the logistic equation:\[P(t) = \frac{5000}{1 + 499e^{-1.78t}}\]

Step 4: Graph the Logistic Function

Plot the logistic function \( P(t) = \frac{5000}{1 + 499e^{-1.78t}} \).The graph should show a steep initial increase, representative of exponential growth, that gradually slows and approaches \( L = 5000 \) as \( t \) becomes large, forming an S-shaped curve.

Step 5: Determine When the Infection Rate Peaks

The infection rate peaks when the second derivative of \( P(t) \) changes from positive to negative, which is at the inflection point of the logistic curve. For logistic models, this occurs when \( P = \frac{L}{2} \). \( \frac{5000}{2} = 2500 \).

Step 6: Estimate the Time to Inflection Point

To find the time \( t \) when \( P(t) = 2500 \), substitute into the logistic equation and solve for \( t \):\[2500 = \frac{5000}{1 + 499e^{-1.78t}}\]Simplifying:\[1 + 499e^{-1.78t} = 2 \quad \Rightarrow \quad 499e^{-1.78t} = 1 \]Divide both sides by 499:\[e^{-1.78t} = \frac{1}{499} \quad \Rightarrow \quad -1.78t = \ln \left(\frac{1}{499}\right)\]Solve for \( t \):\[t = -\frac{\ln\left(\frac{1}{499}\right)}{1.78} \approx \frac{\ln(499)}{1.78}\]Calculate \( t \):\( t \approx \frac{6.21}{1.78} \approx 3.49 \) weeks.

Key Concepts

Exponential Growth

Exponential growth is a process where a quantity increases over time at a rate proportional to its current value. This pattern of growth is characterized by the presence of a constant growth rate that causes the quantity to rise rapidly.

• This type of growth is often observed in populations or financial investments, where the increase is continuous and accelerates the quantity swiftly.
• In the context of infectious diseases, when a virus spreads through a population, as in the problem, it initially follows an exponential growth curve because an increasing number of individuals become infected over time.
• The constant growth rate, in this case, 1.78, signifies that at every moment, the number of new infections depends significantly on how many people are already infected.

Understanding exponential growth is essential for predicting early-stage trends where the infection count increases swiftly. However, this eventually slows as resources or susceptible individuals in the population diminish.

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are used extensively to describe physical, biological, and economic phenomena.

• In the case of logistic growth, differential equations help model how the rate of change of the infected population varies with time.
• The logistic growth model itself is derived from a differential equation, demonstrating how the growth rate of a population slows as it approaches a maximum carrying capacity, denoted by the parameter \( L \).
• The importance of understanding this concept lies in its ability to show how populations evolve over time and respond to limiting factors such as resources or space.

In our scenario, solving differential equations allows us to determine constants such as \( k \), \( L \), and \( C \) and to formulate the complete logistic model.

Inflection Point

An inflection point on a graph is where the curve changes concavity, or direction, from concave up to concave down (or vice versa). In terms of population growth models, this is significant because it marks the transition from accelerating growth to decelerating growth.

• In logistic growth, the inflection point occurs precisely where the growth rate is at its maximum.
• For the logistic function described in the problem, this happens when \( P = \frac{L}{2} \), or half of the carrying capacity. Meaning, for \( L = 5000 \), the inflection point occurs at \( P = 2500 \).
• This point represents the fastest rate of infection spread, after which the rate begins to slow down.

Understanding inflection points in logistic growth models helps predict shifts in growth patterns, allowing interventions to be more effectively timed.