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Chapter 1: Problem 30

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.

Short Answer

Expert verified
The exponential model is \( w(t) = 1248 \cdot e^{0.1403t} \). The annual growth rate is about 15.04%. Yes, the model confirms the doubling reported by 2004.

Step by step solution

01

Establish the Exponential Growth Equation

The general form of an exponential function is \( w(t) = w_0 \cdot e^{kt} \), where \( w_0 \) is the initial value, \( k \) is the growth rate, and \( t \) is the time in years. In this problem, \( w_0 = 1248 \) at \( t = 1 \) (1981) and \( w(t) = 18,957 \) at \( t = 24 \) (2004).
02

Find the Growth Rate Constant

We relate \( w(t) \) at \( t = 24 \) to \( w_0 \) using the equation \( w(t) = w_0 \cdot e^{kt} \). Therefore, we have:\( 18,957 = 1248 \cdot e^{24k} \).We solve for \( k \) by dividing both sides by 1248 and taking the natural logarithm:\( \ln\left(\frac{18,957}{1248}\right) = 24k \).This simplifies to:\( k = \frac{\ln\left(\frac{18,957}{1248}\right)}{24} \approx 0.1403 \).
03

Write the Exponential Function

Substituting \( k = 0.1403 \) back into the exponential function, we have:\( w(t) = 1248 \cdot e^{0.1403t} \).This equation models the growth of whooping cough cases over time.
04

Calculate the Average Annual Growth Rate

The average annual growth rate is given by the formula \( 100 \times (e^k - 1) \). Calculating this gives:\( 100 \times (e^{0.1403} - 1) \approx 15.04\% \).This represents the average percent increase in the number of cases per year.
05

Confirm the Doubling of Cases

According to the report, the number of cases doubled from 2000 to 2004. Calculating from the model, in 2000 (\( t = 20 \)), the cases were \( w(20) = 1248 \cdot e^{0.1403 \times 20} \), and in 2004 (\( t = 24 \)), \( w(24) = 1248 \cdot e^{0.1403 \times 24} = 18,957 \).Now from the computed \( k \):\( w(20) \approx 1248 \cdot e^{2.806} \approx 8,600 \).This shows the cases more than doubled from approximately 8,600 to 18,957 between 2000 and 2004, confirming the report.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Function
An exponential function is a mathematical expression used to model situations where growth or decay happens at a rapid rate. Such functions take the form \( w(t) = w_0 \cdot e^{kt} \), where:
  • \( w(t) \) represents the value we want to find, such as the number of cases at a given time.
  • \( w_0 \) is the initial value鈥攊n this case, the number of whooping cough cases at the start of the observational period.
  • \( e \) is the base of the natural logarithm, approximately 2.718.
  • \( k \) is the growth rate constant, a key factor that defines how quickly or slowly the function value changes.
  • \( t \) represents time, often measured in years or another unit conducive to the context.
In situations like the resurgence of whooping cough, using an exponential function allows us to model how quickly the cases increase over time. By substituting known values, such as \( w_0 = 1248 \) cases and \( w(24) = 18,957 \) cases in 24 years, we can determine the growth rate \( k \) and predict future trends.
Growth Rate Calculation
To understand how fast something grows exponentially, we need to compute the growth rate constant, \( k \). This is achieved by using the exponential function formula set by the real data points.
  • First, we align the final observed value with the initial value and time gap: \( 18,957 = 1248 \cdot e^{24k} \).
  • Next, we solve for \( k \) by isolating it: divide both sides by \( 1248 \) and take the natural logarithm, \( \ln\left(\frac{18,957}{1248}\right) = 24k \).
  • Finally, solve for \( k \) by rearranging the equation: \( k = \frac{\ln\left(\frac{18,957}{1248}\right)}{24} \approx 0.1403 \). This gives the growth rate constant, hinting at how rapidly the cases rise over time.
Calculating \( k \) allows us to derive meaningful insights, such as the average annual increase, which turns out to be approximately 15.04% per year when we compute \( 100 \times (e^{0.1403} - 1) \). This percentage represents how intensely the number of cases grows each year, highlighting the challenging nature of combating the disease resurgence.
Model Verification
Once we have developed an exponential model, verifying it against known information ensures its accuracy and reliability. Using our model \( w(t) = 1248 \cdot e^{0.1403t} \) to check if past behavior matches the data is crucial.
  • For instance, the model projects \( w(20) \), which is the number of cases in the year 2000, to be approximately \( 8,600 \) cases by substituting \( t = 20 \).
  • The real-world report states that cases more than doubled from 2000 to 2004, where our projection of \( w(24) \) yields \( 18,957 \) cases for 2004.
  • Comparing\( 8,600 \) cases to \( 18,957 \), we indeed find more than a double increase, verifying the accuracy of our model.
Model verification like this is essential to establishing the trustworthiness of mathematical predictions, ensuring they align with historical records and real-world observations.

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Most popular questions from this chapter

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