Problem 36 - Calculus

Question

The concentration of the car exhaust fume nitrous oxide, \(\mathrm{NO}_{2},\) in the air near a busy road is a function of distance from the road. The concentration decays exponentially at a continuous rate of \(2.54 \%\) per meter. \(^{24}\) At what distance from the road is the concentration of \(\mathrm{NO}_{2}\) half what it is on the road?

Step-by-Step Solution

Answer

The concentration is half at approximately 27.25 meters from the road.

Step 1: Identify the Given Information

We're given that the concentration of \(\mathrm{NO}_2\) decays at a continuous rate of \(2.54\%\) per meter. We need to find the distance at which the concentration is half of what it is at the road.

Step 2: Use the Exponential Decay Formula

The formula for exponential decay is given by \( C(x) = C_0 e^{-kx} \), where \(C_0\) is the initial concentration, \(x\) is the distance in meters, and \(k\) is the decay rate. Here, \(k = 2.54\% = 0.0254\) per meter.

Step 3: Set up the Equation for Half Concentration

To find the distance where the concentration is half, we set \( C(x) = \frac{1}{2}C_0 \). Substituting this into the decay formula gives us: \( \frac{1}{2}C_0 = C_0 e^{-0.0254x} \).

Step 4: Simplify the Equation

Divide both sides by \(C_0\) to get \( \frac{1}{2} = e^{-0.0254x} \).

Step 5: Solve for x Using Logarithms

Take the natural logarithm of both sides: \( \ln{\frac{1}{2}} = \ln{e^{-0.0254x}} \). This simplifies to \( -\ln{2} = -0.0254x \). Thus, \( x = \frac{\ln{2}}{0.0254} \).

Step 6: Calculate the Distance

Calculate \( x \) using the expression from the previous step: \( x = \frac{\ln{2}}{0.0254} \approx 27.25 \). So, the distance is approximately 27.25 meters.

Key Concepts

Continuous Decay Rate

Exponential decay is a mathematical concept that describes how a quantity decreases at a rate proportional to its current value over time or distance. In the context of environmental studies, it can be used to model how pollutants, such as nitrous oxide \( (\text{NO}_2) \), decrease as you move away from the pollution source, such as a busy road.

The continuous decay rate is an important factor in this process. It is expressed as a percentage that indicates how quickly a substance falls off over a particular distance—in this case, \(2.54\%\) per meter. This means that for every meter you move away from the road, the concentration of \(\text{NO}_2\) decreases by \(2.54\%\) of its present value.

It’s vital when calculating exponential decay to consider the constant decay rate, as it determines how rapidly or slowly the concentration drops. This rate is expressed in natural units, making it compatible with computations involving exponential functions and natural logarithms.

Natural Logarithm

The natural logarithm is a fundamental mathematical function that is intimately connected with exponential growth and decay processes. It is denoted as \( \ln(x) \) and is the inverse of the exponential function. Logarithms are very useful for solving equations where variables are in the exponents, like in our exponential decay problem.

• When you take the natural logarithm of an exponential function, it helps to isolate the variable, making it easier to solve for unknowns.
• In our example, after setting up the equation for half concentration, we reduce the problem to solving \( \ln{(\frac{1}{2})} = -0.0254x \).

This transformation is crucial because logarithms turn multiplicative relationships into additive ones, simplifying the math involved. Because the base of the natural logarithm is \( e \), where \( e \approx 2.718 \), it neatly complements exponential functions where the constant rate of change (decay rate \( k \)) is expressed in these natural units.

Thus, by applying \( \ln \) to both sides of the equation, we isolate \( x \), the distance, allowing students to comprehend how logarithmic functions operate as a tool to reverse exponential expressions and find meaningful solutions in real-world scenarios.

Distance Calculation

Distance calculation in the context of exponential decay involves the task of finding how far you need to travel, or the length along your path, to observe a particular level of decay. For our nitrous oxide problem, this means determining how far from a road one needs to be for the concentration to drop to half of its original value.

The decay is represented by the formula: \[ \frac{1}{2} = e^{-0.0254x} \]To solve for \( x \), the distance, follow these steps using logarithms:

• First, apply the natural logarithm to both sides, yielding \( \ln\left(\frac{1}{2}\right) = -0.0254x \).
• Then, rearrange to solve for \( x \): \( x = \frac{\ln(2)}{0.0254} \).

After performing these calculations, you will find that \( x \) is approximately 27.25 meters. This intuitively means that you have to move around 27 meters away from the road for the concentration of \( \text{NO}_2 \) to halve, illustrating the practical implications of exponential decay in pollution dispersion. Understanding how to calculate distance in such scenarios helps to evaluate environmental impacts and plan mitigation strategies effectively.