Problem 35 - Applied Calculus

Question

Soybean production, in millions of tons $$\begin{array}{c|c|c|c|c|c|c} \hline \text { Year } & 2000 & 2001 & 2002 & 2003 & 2004 & 2005 \\ \hline \text { Production } & 161.0 & 170.3 & 180.2 & 190.7 & 201.8 & 213.5 \\ \hline \end{array}$$ Whooping cough was thought to have been almost wiped out by vaccinations. It is now known that the vaccination wears off, leading to an increase in the number of cases. w, from 1248 in 1981 to 18,957 in 2004 (a) With \(t\) in years since \(1980,\) find an exponential function that fits this data. (b) What does your answer to part (a) give as the average annual percent growth rate of the number of cases? (c) On May \(4,2005,\) the Arizona Daily Star reported (correctly) that the number of cases had more than doubled between 2000 and \(2004 .\) Does your model confirm this report? Explain.

Step-by-Step Solution

Answer

Exponential function: \( w(t) = 1248 \cdot e^{0.113t} \), with a growth rate of 11.97%. Yes, the model confirms the number of cases more than doubled between 2000 and 2004.

Step 1: Understand the Problem

This problem is about modeling the increase in whooping cough cases with an exponential function. We have data from 1981 to 2004 and need to find the function and determine growth characteristics. Additionally, we need to confirm a statement made by the Arizona Daily Star regarding case numbers doubling between 2000 and 2004.

Step 2: Define Exponential Growth Function

The exponential growth function can be written as \( w(t) = w_0 \cdot e^{kt} \), where \( w_0 \) is the initial amount of cases, \( t \) is the time in years since 1980, and \( k \) is the growth rate. We know \( w(1) = 1248 \) and \( w(24) = 18957 \). We start by solving for \( k \) using the formula: \( w(t) = 1248 \cdot e^{kt} \).

Step 3: Calculate Growth Constant k

Using the known values, set up the equation \( 18957 = 1248 \cdot e^{24k} \). Solving for \( k \) involves dividing both sides by 1248, then taking the natural logarithm to find \( k \): $$ e^{24k} = \frac{18957}{1248} $$ $$ \ln(e^{24k}) = \ln\left(\frac{18957}{1248}\right) $$ $$ 24k = \ln\left(15.1947\right) \approx 2.72 $$ $$ k \approx \frac{2.72}{24} \approx 0.113 $$

Step 4: Exponential Function and Average Growth Rate

Using the calculated \( k \), the exponential function is \( w(t) = 1248 \cdot e^{0.113t} \). The average annual percent growth rate is found by converting \( k \) into a percentage: \( e^{0.113} - 1 \times 100\). Calculating: \[ \text{Growth Rate} = (e^{0.113} - 1) \times 100 \approx 11.97\% \]

Step 5: Confirm Daily Star Report

Determine the number of cases in 2000 by setting \( t = 20 \): \[ w(20) = 1248 \cdot e^{0.113 \times 20} \approx 5171 \] Determine the number of cases in 2004 by setting \( t = 24 \): \[ w(24) = 1248 \cdot e^{0.113 \times 24} \approx 18957 \] Since 5171 doubled is 10342 and \( w(24) = 18957 \), the cases more than doubled from 2000 to 2004, confirming the report.

Key Concepts

Exponential Growth Rate

In the context of exponential growth, the growth rate is a crucial concept. It describes how quickly the value of something increases over time in a multiplying manner. For the whooping cough data, the growth rate is expressed through the constant \( k \) in the exponential function \( w(t) = w_0 \cdot e^{kt} \). Here, \( w_0 \) represents the initial number of cases, and \( t \) stands for the time in years since 1980.

To find \( k \), use the data points from 1981 and 2004: 1,248 cases in 1981 and 18,957 cases in 2004. When solving for \( k \), you convert it into an annual growth rate like this:

• Start with the equation: \( 18957 = 1248 \cdot e^{24k} \)
• Solve for \( k \) by isolating \( e^{24k} \) and using natural logarithms.
• Compute \( k \approx 0.113 \), representing an annual multiplier of change.

Ultimately, this tells us that the average annual percent growth rate is about \( 11.97\% \). This fast growth implies that the number of cases was growing exponentially each year, a very steep rate of growth in a real-world context like disease spread.

Data Analysis

Data analysis is the process of evaluating data using analytical and statistical tools to discover useful information. In this case, it involves examining the whooping cough cases in terms of time to understand the trend over the years.

The observation of data from 1981 to 2004 shows an apparent upward trend. Here, the initial step is to outline clearly the specific data points – in this case, the annual whooping cough cases. Critical points include:

• Initial data point in 1981 with 1,248 cases
• Ending data point in 2004 with 18,957 cases
• The process of determining points in between, such as for the year 2000

Conducting such analysis systematically involves recognizing that data points are interconnected by some trend, such as exponential growth. You take the raw data, apply mathematical models (like the exponential function), and verify real-world consistency as with the Arizona Daily Star's claim. The analysis not only helps in model verification but also in making future predictions possible.

Mathematical Modeling

Mathematical modeling involves creating a mathematical representation of a real-world phenomena. It uses formulas and computations to simulate and study behaviors and trends. In this exercise, an exponential model was used to understand and represent the increase in whooping cough cases.

The model chosen was \( w(t) = 1248 \cdot e^{0.113t} \), which represents the exponential growth of cases over time. Some essential parts of model formation include:

• Choosing the right type of function (exponential) to reflect the nature of the data.
• Identifying parameters like initial values and coefficients (initial cases \( w_0 \) and growth rate \( k \)) by fitting the model to actual data.
• Calculating specific values using the model to predict future points, such as cases in 2000 or 2004.

Such models are invaluable because they allow predictions beyond immediate observations, providing insights into potential future trends and helping inform responses. They link theoretical computations with practical decisions, like confirming news reports with actual trends.