Problem 71
Question
Urban population density An urban density model is a formula that relates the population density \(D\) (in thousands/ \(\mathrm{mi}^{2}\) ) to the distance \(x\) (in miles) from the center of the city. The formula \(D=a e^{-b x}\) for the central density \(a\) and coefficient of decay \(b\) has been found to be appropriate for many large U.S. cities. For the city of Atlanta in \(1970, a=5.5\) and \(b=0.10\). At approximately what distance was the population density 2000 per square mile?
Step-by-Step Solution
Verified Answer
The population density was approximately 2000/sq mile at about 10.12 miles from the city center.
1Step 1: Set Up the Equation
Given the population density equation \(D = a e^{-bx}\), we need to find the distance \(x\) where the density \(D = 2\) (since 2000 per sq mile is equivalent to 2 in thousands/sq mile). Substitute the known values \(a = 5.5\) and \(b = 0.10\) into the equation. This gives us the equation \(2 = 5.5 e^{-0.10x}\).
2Step 2: Isolate the Exponential Term
Divide both sides of the equation by 5.5 to isolate the exponential term. This results in \( \frac{2}{5.5} = e^{-0.10x} \).
3Step 3: Solve for the Exponent
To solve for \(x\), take the natural logarithm of both sides. This gives \(\ln\left(\frac{2}{5.5}\right) = -0.10x\).
4Step 4: Solve for Distance \(x\)
Calculate \(x\) by dividing both sides by -0.10. Therefore, \(x = \frac{\ln\left(\frac{2}{5.5}\right)}{-0.10}\). Compute this value using a calculator.
5Step 5: Calculate the Density Distance
Computing the right side gives \(x \approx \frac{\ln(0.3636)}{-0.10} \approx \frac{-1.0116}{-0.10} \approx 10.116\). So, the distance \(x\) is approximately 10.12 miles.
Key Concepts
Exponential Decay ModelPopulation Density FormulaNatural Logarithm CalculationsDistance from City Center
Exponential Decay Model
The concept of the exponential decay model is crucial in understanding how quantities decrease at a rate proportional to their current value. This model is especially relevant in urban studies for various metrics like population density. When you see a equation in the form of \(D = a e^{-bx}\), it follows this model.
Here:
Here:
- \(D\) represents the density, such as population, diminishing as you move away from the center.
- \(a\) is the initial or central value of density at the very core of the urban area.
- \(b\) is the decay constant, guiding how quickly the density falls with distance.
Population Density Formula
This formula is a tool to quantify how densely a space, like a city, is populated. It allows you to predict the density at any given distance from a central point in the city. The formula \(D = a e^{-bx}\) helps analyze urban planning and development policies.
The central density \(a\) represents the base population density at the city's center, and the exponential term \(e^{-bx}\) demonstrates how this density decreases.
For example, in the case of Atlanta in 1970, with \(a=5.5\) and \(b=0.10\), it tells us that the initial density is 5.5 thousand people per square mile, reducing as you move further out. This formula becomes an essential part of evaluating and understanding urban sprawl, infrastructural needs, and environmental implications.
The central density \(a\) represents the base population density at the city's center, and the exponential term \(e^{-bx}\) demonstrates how this density decreases.
For example, in the case of Atlanta in 1970, with \(a=5.5\) and \(b=0.10\), it tells us that the initial density is 5.5 thousand people per square mile, reducing as you move further out. This formula becomes an essential part of evaluating and understanding urban sprawl, infrastructural needs, and environmental implications.
Natural Logarithm Calculations
Natural logarithms, denoted as \(\ln\), are a mathematical function used to solve equations where an unknown variable appears in an exponent, just like in the exponential decay equations. They allow you to "bring down" the exponent to a more accessible form.
Say you have an equation like \(e^{y} = x\), taking the natural log on both sides transforms it to \(y = \ln(x)\). This technique simplifies equations making it easier to find unknowns such as distance in our urban density problem.
In our exercise, after isolating \(e^{-0.10x}\), using \(\ln\) helps deduce that \(x = \frac{\ln(0.3636)}{-0.10}\). This step is essential to calculate the specific point where the population density hits a threshold, showcasing the power of logarithms in unraveling exponential equations.
Say you have an equation like \(e^{y} = x\), taking the natural log on both sides transforms it to \(y = \ln(x)\). This technique simplifies equations making it easier to find unknowns such as distance in our urban density problem.
In our exercise, after isolating \(e^{-0.10x}\), using \(\ln\) helps deduce that \(x = \frac{\ln(0.3636)}{-0.10}\). This step is essential to calculate the specific point where the population density hits a threshold, showcasing the power of logarithms in unraveling exponential equations.
Distance from City Center
The concept of distance from the city center is pivotal in urban geography and city planning. It refers to how far a specific point is from the central hub, often considered zero on the distance scale.
In our urban density model, as distance \(x\) increases, population density \(D\) typically decreases, showing how density thins out as you move away from the nucleus of urban activity.
This distance factor is not just physical but also symbolizes changes in infrastructure, housing, and social dynamics. The computation of \(x\) in our exercise, coming out to be approximately 10.12 miles, practically shows how far from the city center you would find a reduced density of 2000 people per square mile. Understanding this helps urban planners make informed decisions about resource allocation and infrastructural development.
In our urban density model, as distance \(x\) increases, population density \(D\) typically decreases, showing how density thins out as you move away from the nucleus of urban activity.
This distance factor is not just physical but also symbolizes changes in infrastructure, housing, and social dynamics. The computation of \(x\) in our exercise, coming out to be approximately 10.12 miles, practically shows how far from the city center you would find a reduced density of 2000 people per square mile. Understanding this helps urban planners make informed decisions about resource allocation and infrastructural development.
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