Problem 49 - Contemporary Precalculus

Question

According to data from the U.S. Department of Health and Human Services, the cumulative number \(y\) of AIDS cases (in thousands) as of year \(x\) is approximated by \(y=.004 x^{3}-1.367 x^{2}+54.35 x+569.72 \quad(0 \leq x<11)\) where \(x=0\) corresponds to 1995 (a) When did the cumulative number of cases reach \(944,000 ?\) (b) If this model remains accurate after \(2006,\) in what year will the cumulative number of cases reach 1.1 million?

Step-by-Step Solution

Answer

And in what year will the cumulative number of cases reach 1.1 million if this model remains accurate after 2006? Answer: The cumulative number of AIDS cases reached 944,000 approximately in the year 2001. If the model remains accurate after 2006, the cumulative number of cases will reach 1.1 million around the year 2004.

Step 1: (a) Find the value of x for 944,000 cases

To find the year when the cumulative number of cases reached 944,000, we first need to convert the case number to thousands since the function \(y\) is given in thousands. Thus, we get: \(944,000/1,000 = 944\). Now, plugging \(y = 944\) into the function and solving for \(x\): \(944 = .004 x^{3} - 1.367 x^{2} + 54.35x + 569.72\).

Step 2: (a) Solving the equation for x

We have to solve the following equation for \(x\): \(.004 x^{3} - 1.367 x^{2} + 54.35x + 569.72 - 944 = 0\). By using either a graphing calculator or an online solver, we find that \(x \approx 6.06.\)

Step 3: (a) Converting x to years

Since \(x=0\) corresponds to 1995, \(x=6.06\) corresponds to the year: \(1995 + 6.06 = 2001.06\). Therefore, the cumulative number of cases reached 944,000 approximately in 2001.

Step 4: (b) Find the value of x for 1.1 million cases

Now, let's find the year when the cumulative number of cases will reach 1.1 million. First, we need to convert the case number to thousands: \(1,100,000/1,000 = 1,100\). Plugging \(y = 1100\) into the function and solving for \(x\): \(1100 = .004 x^{3} - 1.367 x^{2} + 54.35x + 569.72\).

Step 5: (b) Solving the equation for x

Now, we need to find the value of \(x\) that solves: \(.004 x^{3} - 1.367 x^{2} + 54.35x + 569.72 - 1100 = 0\). Using a graphing calculator or an online solver, we find that \(x \approx 8.95\).

Step 6: (b) Converting x to years

Since \(x=0\) corresponds to 1995, \(x=8.95\) corresponds to the year: \(1995 + 8.95 = 2003.95\). Assuming this model remains accurate after 2006, the cumulative number of cases will reach 1.1 million approximately in 2004.

Key Concepts

Cubic Polynomial

A cubic polynomial is an expression of the form \( ax^3 + bx^2 + cx + d \), where \( a, b, c, \) and \( d \) are coefficients and \( a eq 0 \). This particular form is called a cubic because the highest power of \( x \) is three. Such expressions are vital in various areas of mathematics and modeling because they can capture more complex behaviors than linear or quadratic polynomials.
In our model for AIDS cases, we have a cubic polynomial

• \( y = 0.004x^3 - 1.367x^2 + 54.35x + 569.72 \).

This equation allows modeling the cumulative number of AIDS cases over time, with \( x = 0 \) corresponding to the year 1995.

Mathematical Modeling

Mathematical modeling involves using mathematical expressions to represent real-world situations. Here, we use a cubic polynomial to predict the cumulative number of AIDS cases over different years.
Models like the one we are using are crucial because they allow us to make forecasts and understand trends in data. This modeling helps in planning healthcare resources and responses.
By inserting different values of \( x \) into the polynomial, we can estimate the corresponding cumulative cases \( y \). These kinds of models depend on accurate data collection and careful formulation of the polynomial equation. In our exercise, the polynomial provides a framework to understand how the number of cases changes over time.

Solving Equations

Solving polynomial equations involves finding the value(s) of \( x \) that satisfy the equation. For a cubic polynomial, you may need advanced techniques such as graphing, recognizing patterns, or using software tools.
Here, the exercise requires determining the values of \( x \) when given specific \( y \) values (e.g., 944,000 or 1.1 million cases). First, you adjust \( y \) to match the units in the polynomial (thousands), then solve:

• For 944,000 cases, solve \( 0.004 x^3 - 1.367 x^2 + 54.35 x + 569.72 = 944 \).
• For 1.1 million cases, solve \( 0.004 x^3 - 1.367 x^2 + 54.35 x + 569.72 = 1100 \).

Such problems often require using graphing calculators or algebraic manipulation tools to derive the solution.
Once you find \( x \), convert it back to the corresponding year to provide a real-world interpretation.