Problem 37 - Calculus Single Variable

Question

In 1980 , the population density for the central core of Houston, a disk with a 6 -mile radius, was \(f(x)=5860\) \(\exp (-0.148 x) .\) What was the population of that part of Houston?

Step-by-Step Solution

Answer

The population of the central core of Houston in 1980 was approximately 186,416.

Step 1: Understand the Problem

We are tasked with finding the total population of a circular area around Houston's center based upon a given population density function. The radius is 6 miles, and the density function is given by \( f(x) = 5860 \exp(-0.148x) \).

Step 2: Set Up the Integral

To find the total population, we need to integrate the density function over the entire disk. Since the area is circular, polar coordinates are appropriate. The integral is set up as \( \int_{0}^{2\pi} \int_{0}^{6} f(r) \cdot r \, dr \, d\theta \).

Step 3: Convert Density Function to Polar Coordinates

In polar coordinates, the density function remains the same except that we replace \( x \) with \( r \), where \( r \) is the radial distance. The function is \( f(r) = 5860 \exp(-0.148r) \).

Step 4: Evaluate the Integrals

First, evaluate the inner integral over the radius from 0 to 6: \( \int_{0}^{6} 5860r \exp(-0.148r) \, dr \). Solve this integral by parts or using a substitution method.

Step 5: Solve the Inner Integral

The inner integral \( \int_{0}^{6} 5860r \exp(-0.148r) \, dr \) can be solved using substitution. Let \( u = -0.148r \), then \( dr = du / -0.148 \). The integral becomes \( -39617.57 \cdot \left[ -0.148r\exp(-0.148r) + \int \exp(-0.148r) \, dr \right]_{0}^{6} \).

Step 6: Calculate the Inner Integral Value

Through substitution or integration by parts, find the value of the inner definite integral. This evaluates to approximately \( 29670.9 \).

Step 7: Complete the Outer Integral

The result from Step 6 represents the integral result for a single angle \( \theta \). Now multiply by the full angle around the circle \( \int_{0}^{2\pi} d\theta \) which equals \( 2\pi \). Thus, the total population is \( 2\pi \times 29670.9 \).

Step 8: Calculate the Total Population

Multiply the integral result from Step 6 by \( 2\pi \) (approximated as 6.28318) to get \( 2\pi \times 29670.9 \approx 186415.6 \). This is the estimated total population.

Key Concepts

Polar Coordinates

Polar coordinates are a way of describing the position of a point based on its distance from a fixed origin and the angle from a reference direction. This system is particularly useful when dealing with circular or rotational problems, as it simplifies the integral setup.

When using polar coordinates:

• Radial coordinate ( ): Represents the distance from the origin to the point.
• Angular coordinate ( heta): Denotes the angle from the reference direction, usually the positive x-axis.

In calculations involving circular regions, such as the population density of a city core, polar coordinates allow us to express the circular area naturally. This makes them ideal for integrals describing situations with radial symmetry. The integration process typically involves evaluating a double integral, with the inner integral tackling the radial dimension and the outer one dealing with the angular sweep of the circle.

Definite Integral

Definite integrals calculate the accumulation of quantities, such as area under a curve, over a specific interval. In this exercise, it is essential for determining the total population within a given radius.

We use the definite integral because:

• Continuous population density: The density function tells us how densely populated each infinitesimal region is.
• Integration bounds: They give us the range (from the center to the outer edge of the disk) over which to integrate.

By evaluating the definite integral in polar coordinates over the circular region, we find the total population. The inner integral relates to the radius, and once completed, it is multiplied by the angle from 0 to \(2\pi\), encapsulating the entire circle. Integrating functions like \(5860 \exp(-0.148r)\) within these bounds helps to accumulate the population over the entire disk.

Population Estimation

Population estimation involves calculating the total number of people within an area based on a density function. The problem involved the population density function \(f(x) = 5860\exp (-0.148 x)\).

This concept matters because:

• Density function application: By applying a density function, we can model how population numbers vary as we move further from the center of the city.
• Integration result interpretation: The definite integral over the density function gives a summed value representing the entire population within the region.

Using the integration set-up and evaluation, the exercise arrives at a total population estimate for the central core of Houston. This approach effectively combines data analysis and mathematical modeling to make complex demographic estimates achievable.

Exponential Decay Function

An exponential decay function describes processes that decrease rapidly at first, then more slowly over time. In this exercise, the population density function is \(f(x) = 5860 \exp(-0.148x)\).

Key points about exponential decay:

• Initial value: The coefficient 5860 indicates the population density at the city center (where \(x=0\)).
• Decay rate: The constant -0.148 signifies how quickly population density decreases with distance.

This particular density function reflects that population density is highest at the core, decreasing as we move outwards. Mathematically, this function aligns with the trends of modern cities, where central regions are densely populated, and density decreases due to factors like distance from economic centers. Understanding exponential decay enables effective demographic predictions and urban planning strategies.