Problem 39 - Calculus Concepts

Question

AIDS Cases Records of the number of AIDS cases diagnosed in the United States between 1988 and 1991 indicate that the rate at which the rate of change in the number of cases was changing was constant at -2099 cases per year squared. The number of AIDS cases diagnosed in 1988 was 33,590 , and the number of cases was increasing at the rate of 5988.7 cases per year in 1988 . (Source: Based on data appearing in HIVIAIDS Surveillance \(1992,\) year-end edition) a. Write a differential equation for the rate of change in the rate of change of the number of AIDS cases diagnosed in year \(t,\) where \(t\) is the number of years after 1988 . b. Find a particular solution to the differential equation in part \(a\). c. Estimate how rapidly the number of AIDS cases diagnosed was changing in 1991 and the number of AIDS cases that were diagnosed in that year.

Step-by-Step Solution

Answer

Rate of change in 1991: -308.3 cases/year. Estimated cases in 1991: 41,107.

Step 1: Define the Variables

Let the number of AIDS cases diagnosed in year \( t \), where \( t \) is the number of years after 1988, be represented by \( N(t) \). The rate of change of the number of cases is \( N'(t) \), and the rate of change of this rate of change is \( N''(t) \).

Step 2: Write the Differential Equation

Since the rate of change of the rate of change is constant at \(-2099\) cases per year squared, the differential equation for \( N''(t) \) is simply:\[ N''(t) = -2099 \]

Step 3: Find the General Solution for the Differential Equation

To find \( N'(t) \), integrate \( N''(t) = -2099 \):\[ \int N''(t) \, dt = \int -2099 \, dt \]This gives:\[ N'(t) = -2099t + C \]where \( C \) is a constant of integration.

Step 4: Utilize Initial Conditions for Specific Solution

Use the initial condition given for 1988: \( N'(0) = 5988.7 \). This means:\[ 5988.7 = -2099(0) + C \]Solving gives \( C = 5988.7 \). Therefore, the specific solution for \( N'(t) \) is:\[ N'(t) = -2099t + 5988.7 \]

Step 5: Integrate to Find the Number of Cases Function

Integrate \( N'(t) \) to find \( N(t) \):\[ \int N'(t) \, dt = \int (-2099t + 5988.7) \, dt \]This results in:\[ N(t) = -1049.5t^2 + 5988.7t + K \]where \( K \) is another constant of integration.

Step 6: Apply Initial Condition for Number of Cases

Given that \( N(0) = 33590 \), use this to find \( K \):\[ 33590 = -1049.5(0)^2 + 5988.7(0) + K \]So, \( K = 33590 \). Thus, the particular solution for \( N(t) \) is:\[ N(t) = -1049.5t^2 + 5988.7t + 33590 \]

Step 7: Estimate the Number of Cases in 1991

First, calculate the rate of change of cases in 1991 (when \( t = 3 \)) using \( N'(t) \):\[ N'(3) = -2099(3) + 5988.7 = -6297 + 5988.7 = -308.3 \]Next, find the number of cases in 1991 using \( N(t) \):\[ N(3) = -1049.5(3)^2 + 5988.7(3) + 33590 \]Calculate to get:\[ N(3) = -10449 + 17966.1 + 33590 = 41107.1 \]

Step 8: Summary

The estimated rate of change of AIDS cases in 1991 was \(-308.3\) cases per year, and the estimated number of cases diagnosed that year was approximately \( 41107 \).

Key Concepts

Rate of Change

In the realm of calculus, the rate of change is a fundamental concept that helps us understand how a quantity evolves over time. When dealing with the number of AIDS cases diagnosed, "rate of change" refers to how quickly the number of cases increases or decreases each year.

In this exercise, we defined the function \( N(t) \) to represent the number of AIDS cases diagnosed in year \( t \), where \( t \) measures years after 1988. The first derivative, \( N'(t) \), indicates how the number of cases changes annually. For example, if \( N'(t) \) is positive, the number of cases is increasing, and if negative, it is decreasing.

A unique detail here is the rate at which \( N'(t) \), the rate of change itself, is altering. This is given by the second derivative, \( N''(t) \), which was constant at -2099 cases per year squared. It tells us that the pace at which the rate increases or decreases is consistent over the years. Such insights are critical for predicting future trends in data.

Initial Conditions in Calculus

Calculus problems involving differential equations often require initial conditions to find a specific solution, not just the general form. These initial conditions are values known at the start of the observation period, and they help us determine constants that appear after integration.

In our exercise, two key initial conditions were provided for the year 1988: the number of AIDS cases diagnosed, \( N(0) = 33590 \), and the rate of change, \( N'(0) = 5988.7 \) cases/year. Applying these conditions enabled us to find the constants \(C\) and \( K \) to obtain particular solutions for \( N'(t) \) and \( N(t) \).

These specific solutions give us actual values rather than abstract equations, allowing for real-world predictions and analyses, such as estimating changes in the number of cases diagnosed by 1991.

Integral Calculus

Integral calculus is all about finding the original quantity from its rate of change, and in this context, it means finding the function \( N(t) \) from \( N'(t) \). Integration is the mathematical operation used to reverse differentiation, essentially "undoing" the process to reach the function you started with.

In our step-by-step approach, after establishing the differential equation for the rate of change of the change rate \( N''(t) = -2099 \), we integrated to find \( N'(t) \). This gave an expression representing how the number of cases was trending every year. Another integration step from \( N'(t) \) landed us at \( N(t) \), the number of AIDS cases function, with constants found using initial conditions.

By integrating \( N'(t) \), we construct a specific mathematical model of how cases were possibly distributed over time. This not only aids in understanding past data but is also crucial for forecasting and planning in public health scenarios.