Problem 36

Question

After work a person wishes to sit in a large park along a path 300 meters long. At the ends of the path there are two construction sites, one of which is 8 times as noisy as the other. In order to have the quietest repose, how far from the quieter site should the person sit? (Hint: The intensity of noise where the person sits is directly proportional to the intensity of noise at the source and is inversely proportional to the square of the distance from the source.)

Step-by-Step Solution

Verified
Answer
The person should sit 100 meters from the quieter site.
1Step 1: Define Variables
Let the distance of the person from the quieter site be \( x \). The distance from the louder site would therefore be \( 300 - x \). Let the noise intensity of the quieter site be \( I \). The intensity of the noisier site will be \( 8I \).
2Step 2: Express Noise Intensities at Point
The noise intensity from the quieter site at distance \( x \) is \( \frac{I}{x^2} \), and from the noisier site at distance \( 300 - x \) is \( \frac{8I}{(300-x)^2} \).
3Step 3: Formulate Total Noise Intensity Equation
The total noise intensity \( N \) at the sitting point is the sum of intensities from the quieter and noisier sites: \[ N = \frac{I}{x^2} + \frac{8I}{(300-x)^2} \].
4Step 4: Simplify the Equation
Since \( I \) appears in both terms of the equation, cancel out \( I \): \[ N = \frac{1}{x^2} + \frac{8}{(300-x)^2} \].
5Step 5: Find the Minimum Total Noise Intensity
To find the minimum noise intensity, take the derivative of \( N \), set it to zero, and solve for \( x \). Let \( f(x) = \frac{1}{x^2} + \frac{8}{(300-x)^2} \), compute \( f'(x) \) and set \( f'(x) = 0 \).
6Step 6: Differentiate and Solve
Compute the derivative: \[ f'(x) = \frac{-2}{x^3} + \frac{16}{(300-x)^3} = 0 \].Set it to zero and solve for \( x \): \[ \frac{2}{x^3} = \frac{16}{(300-x)^3} \].Simplifying gives: \[ \frac{1}{x^3} = \frac{8}{(300-x)^3} \],leading to: \[ (300-x)^3 = 8x^3 \].Take the cube root: \[ 300-x = 2x \],which simplifies to: \( 3x = 300 \).
7Step 7: Solve for x
Divide both sides by 3: \( x = 100 \). Thus, the person should sit 100 meters from the quieter site.

Key Concepts

Noise IntensityInverse Square LawMathematical Derivatives
Noise Intensity
Noise intensity refers to the level of sound energy reaching a specific point from a source. In our exercise, we have two construction sites emitting noise. One site is 8 times louder than the other.

When considering noise intensity, it's crucial to understand how noise dissipates over distance. Noise becomes less intense as it spreads, or radiates, from its source. The idea is simple: the farther you are from a noise source, the less noise you hear.

In mathematical terms, the noise intensity is directly proportional to the noise produced by the source (whether quiet or loud) and inversely proportional to the square of the distance from that source. This is important for setting up our noise intensity equation and solving where the quietest spot in a park might be given noise from different distances.
Inverse Square Law
The Inverse Square Law plays a crucial role in understanding how noise intensity decreases with distance.

This law states that the intensity of a sound or any wave phenomenon decreases in proportion to the square of the distance from its source. For instance, if you double the distance from a noise source, the noise intensity will drop to a quarter of its original intensity.

In our exercise, each site's noise intensity is calculated by dividing the source intensity by the square of the distance. If one site generates noise intensity of \( I \), the perceived noise at a distance \( x \) is \( \frac{I}{x^2} \).
  • For the quieter site, this intensity is \( \frac{I}{x^2} \).
  • For the noisier site, emitting 8 times the noise, it is \( \frac{8I}{(300-x)^2} \).
Understanding this law helps predict how changing the position will alter the noise levels the person experiences in the park.
Mathematical Derivatives
Mathematical derivatives are essential tools for finding the optimal solution in problems involving rates of change or optimization, like our noise exercise. A derivative represents the rate at which a function is changing at any given point, and it's crucial when determining extremes like maxima or minima.

In this exercise, our goal is to find the point where the noise intensity is minimized. Once we established the total noise intensity equation \( N = \frac{1}{x^2} + \frac{8}{(300-x)^2} \), we used calculus to optimize it.

Taking the derivative of the total noise intensity function, \( f'(x) = \frac{-2}{x^3} + \frac{16}{(300-x)^3} \), we set it equal to zero to locate the turning points. Solving \( f'(x) = 0 \) helps us find where the noise is at its lowest possible level.
  • This involves finding the values where the slope of the noise intensity curve becomes flat.
  • We determined that the person should sit 100 meters from the quieter noise source.
Derivatives are powerful in revealing these hidden extremums in complex functions, guiding our solutions.