Problem 66 - A Graphical Approach to Precalculus with Limits

Question

Insulation workers who were exposed to asbestos and employed before 1960 experienced an increased likelihood of lung cancer. If a group of insulation workers have a cumulative total of \(100,000\) years of work experience, with their first date of employ. ment \(x\) years ago, then the number of lung cancer cases occurring within the group can be modeled by $$N(x)=0.00437 x^{3.2}$$ (Source: Walker, A... Observation and Inference: An Introduction to the Methods of Epidemiology, Epidemiology Resources, Inc.) (a) Calculate \(N(x)\) when \(x=5,10,\) and \(20 .\) What happens to the likelihood of cancer as \(x\) increases? (b) If \(x\) doubles, does the number of cancer cases also double?

Step-by-Step Solution

Answer

As \(x\) increases, \(N(x)\) significantly increases; doubling \(x\) more than quadruples \(N(x)\).

Step 1: Calculate N(5)

Using the formula \(N(x) = 0.00437 x^{3.2}\), calculate \(N(5)\).Plug \(x = 5\) into the equation:\[N(5) = 0.00437 \times 5^{3.2} \]First, calculate \(5^{3.2}\):\[5^{3.2} \approx 172.4667\]Now substitute back into the equation:\[N(5) = 0.00437 \times 172.4667 \approx 0.7535\]

Step 2: Calculate N(10)

Using the same formula, calculate \(N(10)\).Plug \(x = 10\) into the equation:\[N(10) = 0.00437 \times 10^{3.2} \]First, calculate \(10^{3.2}\):\[10^{3.2} \approx 1584.8932\]Now substitute back into the equation:\[N(10) = 0.00437 \times 1584.8932 \approx 6.9267\]

Step 3: Calculate N(20)

Using the same formula, calculate \(N(20)\).Plug \(x = 20\) into the equation:\[N(20) = 0.00437 \times 20^{3.2} \]First, calculate \(20^{3.2}\):\[20^{3.2} \approx 14573.9645\]Now substitute back into the equation:\[N(20) = 0.00437 \times 14573.9645 \approx 63.667\]

Step 4: Analyze the trend of N(x) as x increases

Compare the values of \(N(x)\):- For \(x=5\), \(N(5) \approx 0.7535\)- For \(x=10\), \(N(10) \approx 6.9267\)- For \(x=20\), \(N(20) \approx 63.667\)As \(x\) increases, \(N(x)\) increases significantly, indicating that the likelihood of cancer increases as the years since first employment increases.

Step 5: Determine the effect of doubling x

Let us check if doubling \(x\) also doubles \(N(x)\):Calculate \(N(2x)\) and compare with \(2 \times N(x)\):\[N(2x) = 0.00437 (2x)^{3.2} = 0.00437 \times 2^{3.2} \times x^{3.2}\]\[2 \times N(x) = 2 \times 0.00437 x^{3.2} = 0.00874 x^{3.2}\]Calculate \(2^{3.2}\):\[2^{3.2} \approx 9.189\]Substitute into the equation:\[N(2x) = 0.00437 \times 9.189 \times x^{3.2} \approx 0.0402 x^{3.2}\]As \(N(2x) \approx 4.62 \times N(x)\), doubling \(x\) more than doubles \(N(x)\) by a factor of approximately 4.62.

Key Concepts

Exponential Growth

Exponential growth is a rapid increase in number or size where the rate of growth is proportional to the current size. Think of it like a snowball rolling down a hill, getting bigger as it gathers more snow. In the context of epidemiology, and specifically in this exercise, exponential growth explains how the number of lung cancer cases increases as the number of years since first exposure to asbestos increases.
The model given for this exercise, \(N(x) = 0.00437 x^{3.2}\), is a classic example of exponential growth. As the years \(x\) increase, the exponent 3.2 dictates how fast the number of cases grows. For instance:

• When \(x = 5\), \(N(x)\) is around 0.75.
• When \(x = 10\), \(N(x)\) jumps to approximately 7.
• And at \(x = 20\), \(N(x)\) soars to about 63.7.

This pattern reveals that not only does the number of cases increase with time, but it does so in a manner that accelerates quickly. Therefore, exponential growth captures the escalating threat that longer asbestos exposure times bring in terms of health risk.

Mathematical Analysis

Mathematical analysis provides the toolkit to evaluate and understand models like the one in this exercise. It equips us with the ability to predict outcomes and analyze trends through precise calculations. In this problem, the function \(N(x) = 0.00437 x^{3.2}\) is analyzed by inputting different values of \(x\) to see how they affect the outcome.
By analyzing this function:

• We calculated \(N(5)\) to be around 0.7535, \(N(10)\) to be 6.9267, and \(N(20)\) to be 63.667.
• We observed a pattern illustrating how the number of cases rises dramatically, confirming the non-linear nature of exponential growth.

Mathematical analysis further explores if changes in \(x\) proportionately affect \(N(x)\). When \(x\) is doubled, mathematical calculations reveal that \(N(x)\) more than doubles; in fact, it increases by about 4.62 times. This illustrates the power of using mathematical tools to uncover dynamic relationships in data and trends.

Epidemiological Modeling

Epidemiological modeling involves creating mathematical representations to simulate the spread of diseases or health outcomes, enabling public health professionals to predict and manage health risks effectively. In this exercise, the model \(N(x) = 0.00437 x^{3.2}\) is fashioned to evaluate how asbestos exposure correlates with lung cancer cases.
This model highlights fundamental epidemiology principles:

• Exposure over time: The variable \(x\) represents the years since exposure, allowing the model to project cancer risk as time progresses.
• Quantifying risk: Through calculated projections, it estimates the number of cases based on years exposed, useful for assessing long-term health impacts.

Through such models, epidemiologists can visualize potential health crises, prioritize resources efficiently, and develop strategies to minimize exposure risks. This particular model emphasizes the devastating long-term impact of asbestos, reflecting the importance of timely intervention and regulation to protect workers' health.