Problem 42 - Applied Calculus

Question

A flu epidemic hits a college community, beginning with five cases on day \(t=0\). The rate of growth of the epidemic (new cases per day) is given by the following function \(r(t),\) where \(t\) is the number of days since the epidemic began. a. Find a formula for the total number of cases of flu in the first \(t\) days. b. Use your answer to part (a) to find the total number of cases in the first 20 days. \(r(t)=20 e^{0.04 t}\)

Step-by-Step Solution

Answer

The total number of cases in the first 20 days is approximately 1117.

Step 1: Understanding the function

The given rate of growth function is \( r(t) = 20 e^{0.04t} \), which represents the number of new flu cases per day. Our goal is to determine a formula for the total number of flu cases over a period of \( t \) days.

Step 2: Total cases as an integral

To find the total number of cases, we need to integrate the rate function \( r(t) \) with respect to \( t \). The integral will give us the accumulated number of cases from the starting day \( t = 0 \) to any day \( t \). Therefore, we need to evaluate \( C(t) = \int r(t) \ dt = \int 20 e^{0.04t} \ dt \).

Step 3: Integrating the rate function

To solve the integral \( \int 20 e^{0.04t} \ dt \), we use the substitution method. Let \( u = 0.04t \), then \( du = 0.04 dt \). Therefore, \( dt = \frac{du}{0.04} \). The integral becomes \( \int 20 e^{u} \times \frac{1}{0.04} \ du = 500 \int e^{u} \ du = 500 e^{u} + C \). Substituting back \( u = 0.04t \), we obtain \( C(t) = 500 e^{0.04t} + C \).

Step 4: Applying initial condition

We are given that there are 5 cases at \( t = 0 \). So using \( C(0) = 5 \), we can solve for \( C \). Plugging \( t = 0 \) into the integrated formula gives \( 500 e^{0} + C = 5 \), which simplifies to \( 500 + C = 5 \). Thus, \( C = 5 - 500 = -495 \).

Step 5: Final formula for total cases

Substituting back the value of \( C \) into our integral result, the formula for the total number of flu cases in the first \( t \) days becomes \( C(t) = 500 e^{0.04t} - 495 \).

Step 6: Calculating total cases for 20 days

To find the total cases in the first 20 days, we substitute \( t = 20 \) into the total cases formula: \( C(20) = 500 e^{0.04 \times 20} - 495 \). Calculating \( e^{0.8} \) and substituting gives \( C(20) = 500 \times 2.2255 - 495 \approx 1117 \).

Key Concepts

Exponential Growth

Exponential growth is a type of increase where the rate of growth is proportional to the current size or quantity of something. In this context, the growth of the flu epidemic is described by the function \( r(t) = 20 e^{0.04t} \). This formula represents how the number of new cases arises over time.
This kind of growth progresses rapidly because it keeps multiplying by a constant factor. It's like compounding interest in a bank account. The exponential function here, \( e^{0.04t} \), indicates that every additional day, the growth rate compounds at approximately 4%. This implies the epidemic spreads faster as more time passes.

• Exponential growth allows small initial numbers to become large.
• It models many natural phenomena, including populations and epidemics.

In the case of our flu example, initially beginning with just a handful of cases, the epidemic can grow into a large number within days due to this compounding effect.

Definite Integral

A definite integral is a calculation that helps in finding the total accumulation of quantities when presented as a rate function over a specific interval. In the flu epidemic scenario, we used a definite integral to calculate the accumulated number of flu cases over days.
The expression \( \int r(t) \ dt = \int 20 e^{0.04t} \ dt \) signifies that we're finding the total number of cases from day zero to day \( t \).

• A definite integral helps calculate the total from a rate.
• The integral is essential in understanding cumulative growth over time.

The area under the curve represented by \( r(t) \) gives us the total number of cases, which accumulates the new cases day by day. This calculation is practical for any scenario where we need to calculate the total from a variable rate, whether it’s disease spreading, rainfall, or even stock prices.

Epidemic Modeling

Epidemic modeling is crucial for understanding, projecting, and controlling the spread of diseases. By using mathematical models, we can predict how a disease like the flu spreads within a population over time. Our equation \( r(t) = 20 e^{0.04t} \) is an example of such a model.
This representation helps in:

• Predicting future outbreaks and their impact.
• Providing insights into how fast an epidemic might accelerate.

The model assists public health officials in planning and response efforts to effectively manage the epidemic's spread. Moreover, it shows us how quick, early interventions are critical to controlling fast-growing epidemics. Thus, having a reliable model not only aids in epidemiological research but also informs policy decisions during public health emergencies.