/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 The concentration of the car exh... [FREE SOLUTION] | 91Ó°ÊÓ

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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. \(^{67}\) At what distance from the road is the concentration of \(\mathrm{NO}_{2}\) half what it is on the road?

Short Answer

Expert verified
The concentration of \(\mathrm{NO}_2\) is half at approximately 27.28 meters from the road.

Step by step solution

01

Understanding Exponential Decay

The problem states that the concentration of \(\mathrm{NO}_2\) decays exponentially at a rate of \(2.54\%\) per meter. This means that if \(C_0\) is the initial concentration of \(\mathrm{NO}_2\) at the road, the concentration \(C(d)\) at a distance \(d\) meters from the road is given by \(C(d) = C_0 e^{-0.0254d}\).
02

Setting Up the Equation

To find the distance \(d\) at which the concentration is half of the initial value, set \(C(d) = 0.5C_0\). Substituting this into the exponential decay equation, we get: \[ 0.5C_0 = C_0 e^{-0.0254d}. \]
03

Simplifying the Equation

Cancel \(C_0\) from both sides of the equation \(0.5C_0 = C_0 e^{-0.0254d}\) to get: \[ 0.5 = e^{-0.0254d}. \]
04

Solving for the Distance \(d\)

Take the natural logarithm of both sides to solve for \(d\): \[ \ln(0.5) = -0.0254d. \] Divide by \(-0.0254\) to isolate \(d\): \[ d = \frac{\ln(0.5)}{-0.0254}. \]
05

Calculating the Distance

Compute the value of \(d\) using the expression \(d = \frac{\ln(0.5)}{-0.0254}\). Using a calculator, \(\ln(0.5) \approx -0.6931\). Therefore, \[ d = \frac{-0.6931}{-0.0254} \approx 27.28 \text{ meters}. \]

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Key Concepts

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

Exponential Functions
Exponential functions describe processes that grow or decay at a constant percentage rate. In this context, they capture the behavior of concentrations of substances like nitrous oxide \( (\text{NO}_2) \) in the air. The basic form of an exponential decay formula is \( C(t) = C_0 e^{-kt} \), where:
  • \( C(t) \) is the concentration at time \( t \) or at a certain distance in our situation
  • \( C_0 \) is the initial concentration
  • \( k \) is the decay constant
  • \( e \) is the base of the natural logarithm, approximately 2.718
In this scenario, \( C(d) = C_0 e^{-0.0254d} \) illustrates how the concentration of NOâ‚‚ decreases exponentially as you move away from the road. It's important to remember that the concentration drops in a specific way, where every meter contributes to a 2.54% reduction in concentration. This property is what distinguishes exponential decay from other types of mathematical functions. It's not just a simple linear decrease but rather a multiplicative drop over distance.
Calculus
Calculus often helps us analyze changes and patterns in a system over a continuous range. When dealing with exponential decay like the one described in the exercise, calculus provides tools to examine how functions behave instantaneously. Through calculus, we can derive exponential decay formulas by integrating different rates of change.
In the solution step, we use the natural logarithm as a key tool to "undo" the exponential component and solve for the distance \( d \). The process involves simplifying the equation by canceling terms and applying the natural logarithm operation:
  • Take the logarithm of both sides of the equation to manage the exponential function
  • This step allows isolation of the \( d \) value, revealing how far one must be from the road for the concentration to reach half its original value
So, calculus not only aids in crafting these functions but also in extracting vital insights from them, thereby helping us solve real-world problems effectively.
Environmental Science
Environmental science explores how substances like \( \text{NO}_2 \) impact our surroundings and health. The exponential decay model provided in this exercise reflects real-life pollutant behavior as distance increases from a source, in this case, a busy road.
Understanding the decay of such concentrations is crucial for evaluating air quality and public health risks.
By using mathematical models, environmental scientists can predict pollutant levels at varying distances, assess exposure risks, and propose strategic interventions to minimize these impacts.This exercise simulates common environmental scenarios, helping illustrate how fast pollutants diminish in concentration as they disperse into the broader environment. Furthermore, these models provide insights into urban air pollution dynamics, assisting in policy-making for better urban planning and road system designs to mitigate pollution effects.

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Most popular questions from this chapter

A quantity \(P\) is an exponential function of time \(t .\) Use the given information about the function \(P=P_{b} a^{t}\) to: (a) Find values for the parameters \(a\) and \(P_{0}\). (b) State the initial quantity and the percent rate of growth or decay. \(P=1600\) when \(t=3\) and \(P=1000\) when \(t=1\)

What continuous percent growth rate is equivalent to an annual percent growth rate of \(10 \% ?\)

School organizations raise money by selling candy door to door. When the price is \(\$ 1\) a school organization sells 2765 candies and when the price goes up to \(\$ 1.25\) the quantity of sold candy drops down to 2440 (a) Find the relative change in the price of candy. (b) Find the relative change in the quantity of candy sold. (c) Find and interpret the ratio \(\frac{\text { Relative change in quantity }}{\text { Relative change in price }}\)

The functions in Problems \(17-20\) represent exponential growth or decay. What is the initial quantity? What is the growth rate? State if the growth rate is continuous. $$P=7.7(0.92)^{t}$$

Hydroelectric power is electric power generated by the force of moving water. The table shows the annual percent change in hydroelectric power consumption by the US industrial sector. $$\begin{array}{c|c|c|c|c|c} \hline \text { Year } & 2005 & 2006 & 2007 & 2008 & 2009 \\ \hline \text { \% growth over previous yr } & -1.9 & -10 & -45.4 & 5.1 & 11 \\ \hline \end{array}$$ (a) According to the US Department of Energy, the US industrial sector consumed about 29 trillion BTUs of hydroelectric power in \(2006 .\) Approximately how much hydroelectric power (in trillion BTUs) did the US consume in \(2007 ?\) In \(2005 ?\) (b) Graph the points showing the annual US consumption of hydroelectric power, in trillion BTUs, for the years 2004 to \(2009 .\) Label the scales on the horizontal and vertical axes. (c) According to this data, when did the largest yearly decrease, in trillion BTUs, in the US consumption of hydroelectric power occur? What was this decrease?

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