Problem 44 - Applied Calculus

Question

For children and adults with diseases such as asthma, the number of respiratory deaths per year increases by \(0.33 \%\) when pollution particles increase by a microgram per cubic meter of \operatorname{air}^{66}. (a) Write a formula for the number of respiratory deaths per year as a function of quantity of pollution in the air. (Let \(Q_{0}\) be the number of deaths per year with no pollution.) (b) What quantity of air pollution results in twice as many respiratory deaths per year as there would be without pollution?

Step-by-Step Solution

Answer

Formula: \(D(Q) = Q_0 \times (1.0033)^Q\); Approx. 211 micrograms doubles deaths.

Step 1: Understand the Increase Rate

The problem states that the number of respiratory deaths increases by \(0.33\%\) for every microgram per cubic meter increase in pollution particles. This rate will form part of our exponential growth model.

Step 2: Define the Formula with Exponential Growth

The formula for exponential growth can be written as \(D(Q) = Q_0 \times (1 + 0.0033)^Q\), where \(D(Q)\) represents the number of deaths in terms of pollution \(Q\), \(Q_0\) is the number of deaths with no pollution, and \(1.0033\) indicates the death increase per unit increase in pollution.

Step 3: Set Up Equation for Double the Deaths

To find when the deaths double, set \(D(Q) = 2Q_0\). Substituting into the formula, we have \(2Q_0 = Q_0 \times (1.0033)^Q\).

Step 4: Simplify and Solve for Q

Divide both sides by \(Q_0\) to get \(2 = (1.0033)^Q\). Take the natural logarithm of both sides to get \(\ln(2) = Q \cdot \ln(1.0033)\). Now solve for \(Q\) by dividing: \(Q = \frac{\ln(2)}{\ln(1.0033)}\).

Step 5: Calculate the Solution

Using a calculator, find \(\ln(2) \approx 0.6931\) and \(\ln(1.0033) \approx 0.003296\). Solve for \(Q\): \(Q = \frac{0.6931}{0.003296} \approx 210.31\). Thus, it takes approximately 211 micrograms per cubic meter to double the respiratory deaths.

Key Concepts

Exponential Function

An exponential function is a mathematical expression in which a constant base is raised to a variable exponent. It is commonly used to describe situations where quantities grow or decay at a constant relative rate. In the context of pollution and respiratory deaths, an exponential function helps model the increase in deaths as the pollution level rises.

**Understanding Exponential Growth**
- Exponential growth occurs when the growth rate of a value is proportional to its current value.
- In our problem, the number of deaths increases by 0.33% for each additional microgram of pollution.
- This percentage increase translates into the exponential growth factor of 1.0033 when modeling the situation mathematically.

Therefore, our exponential function can be written as: \[ D(Q) = Q_0 \times (1 + 0.0033)^Q \] where:
- \(D(Q)\) is the number of deaths when the pollution quantity is \(Q\)
- \(Q_0\) is the death count with no pollution.

Pollution Impact

Pollution can have a severe impact on human health, particularly through respiratory diseases like asthma. In the exercise, pollution is directly correlated with the number of respiratory deaths.

**Relationship Between Pollution and Health**
The task at hand reveals how even a small change in pollution levels can significantly affect mortality rates among vulnerable populations. By understanding this model, we grasp the potential health consequences of environmental changes.
- Just a 1 microgram increase in pollution leads to a notable increase in adverse health outcomes.
- Our formula predicts the death rate as pollution increases, demonstrating the exponential increase in health risks as pollution levels rise.

Understanding this impact helps emphasize the importance of reducing pollution to improve public health. Awareness of such exponential relationships is essential in environmental policy-making and public health planning, as even minor improvements in air quality can lead to significant reductions in health risks.

Logarithms

Logarithms play a critical role in solving exponential equations, especially when finding specific values of the variable exponent. In our pollution example, we used logarithms to determine the point at which the respiratory deaths double.

**Using Logarithms for Calculations**
- To solve the equation for when deaths double, set \( D(Q) = 2Q_0 \), which leads to the equation \(2 = (1.0033)^Q\).
- Logarithms help simplify this equation by converting multiplication to addition, making it easier to isolate \(Q\).
- Taking the natural logarithm of both sides, we get: \[ \ln(2) = Q \cdot \ln(1.0033) \] - Solving for \(Q\) involves simple division: \[ Q = \frac{\ln(2)}{\ln(1.0033)} \]

These steps are crucial in breaking down complex mathematical expressions into understandable parts, making the problem easier to solve. Logarithms are thus valuable tools in calculus for dealing with exponential growth issues.

Calculus Problem Solving

Calculus is often utilized to solve problems involving rates of change and can be particularly beneficial when dealing with exponential growth scenarios, such as pollution's impact on health. Here, we applied calculus principles to address how the death rate changes with pollution.

**Applying Calculus Concepts**
- Calculus helps us comprehend the sensitivity of respiratory deaths concerning changes in pollution levels.
- By forming an exponential model, we can not only understand but predict how future changes in pollution affect health.
- Key calculus techniques, like differentiating exponential functions, allow us to investigate how the rate of deaths changes as pollution varies.

Using calculus, we transitioned from understanding the problem to setting up and solving it mathematically. This process exemplifies how calculus helps dissect and analyze real-world phenomena, leading to meaningful solutions in public health.