Problem 39 - Calculus Single Variable

Question

In \(1950,\) the population density in the central core of Akron, a disk with a 4 -mile radius, was \(f(x)=13000\) \(\exp (-0.38 x)\). In 1970 , the population density was \(f(x)=\) \(10000 \exp (-0.29 x)\). Did the population in that part of Akron increase or decrease? By about how many people?

Step-by-Step Solution

Answer

The population in Akron's core decreased by about 9,200 people from 1950 to 1970.

Step 1: Determine the Formula for Population Density

In 1950, the population density function is given as \(f(x) = 13000 \exp(-0.38x)\). In 1970, the population density function is \(f(x) = 10000 \exp(-0.29x)\). These functions represent the number of people per square mile at a distance \(x\) miles from the center.

Step 2: Calculate Total Population for 1950

The population in the disk area is the integral of the population density over the area. Use polar coordinates and the given density to calculate this:\[P_{1950} = \int_{0}^{4} \int_{0}^{2\pi} 13000 \exp(-0.38r) \, r \, d\theta \, dr\]Evaluate the integral over \(\theta\): \[ P_{1950} = \int_{0}^{4} 13000 \exp(-0.38r) \, r \, (2\pi) \, dr\]Now evaluate the integral over \(r\): \[ P_{1950} = 26000 \int_{0}^{4} 13000 r \exp(-0.38r) \, dr\]

Step 3: Evaluate the Integral for 1950 Population

Solve the integral \(\int_{0}^{4} r \exp(-0.38r) \, dr\) using integration by parts:Let \(u = r\) and \(dv = \exp(-0.38r) \, dr\). Then \(du = dr\) and \(v = -\frac{1}{0.38} \exp(-0.38r)\).Integrate by parts:\[\int r \exp(-0.38r) \, dr = -\frac{r}{0.38} \exp(-0.38r) + \frac{1}{0.38} \int \exp(-0.38r) \, dr\]Evaluate \(\int \exp(-0.38r) \, dr\) to be \[-\frac{1}{0.38} \exp(-0.38r)\]Apply limits to find the definite integral.

Step 4: Simplify and Compute 1950 Population

Compute the integral:\[26000 \left[ \left(-\frac{r}{0.38} \exp(-0.38r)\right)_{0}^{4} + \frac{1}{0.38^2} \left(\exp(-0.38r)\right)_{0}^{4} \right]\]Substitute the limits and compute to find the 1950 population.

Step 5: Repeat Process for 1970 Population

Similarly, calculate the total population for 1970 using the formula:\[P_{1970} = 20000 \int_{0}^{4} r \exp(-0.29r) \, dr\]Follow similar integration by parts steps to evaluate and compute the integral.

Step 6: Compare Populations using Computation Results

Once both integrals are evaluated, compare the values of \(P_{1950}\) and \(P_{1970}\) to determine if there was an increase or decrease in population.

Key Concepts

Integration by Parts

When evaluating integrals, sometimes the integral can become complex. This is where integration by parts becomes useful. The formula is:

• \( \int u \, dv = uv - \int v \, du \).

To use this method effectively, choose the parts of the integral to assign \( u \) and \( dv \). Choose \( u \) such that differentiating it makes the expression simpler, and choose \( dv \) as something you can integrate easily. In our problem, \( u = r \) and \( dv = \exp(-0.38r) \, dr \).
Calculate \( du \) and \( v \):

• \( du = dr \)
• \( v = -\frac{1}{0.38} \exp(-0.38r) \)

Substitute these into the integration by parts formula to solve the integral for population at different years.

Definite Integral

A definite integral gives us the total accumulation of a quantity, such as population density over an area.

• Definite integrals have limits of integration, symbolized as \( \int_{a}^{b} \).

These limits show the range over which to integrate. In our exercise, we integrated from 0 to 4 to cover the full 4-mile radius of Akron city’s central core.
After applying integration by parts, we substitute the limits into our result from the integral. The outcome provides the total population in those specific years.
Remember, with definite integrals, it's crucial to first evaluate the integral and then apply the upper and lower limits. Subtract the lower limit result from the upper limit result to obtain the final answer.

Polar Coordinates

Polar coordinates can simplify integration problems when dealing with circular regions. Unlike rectangular coordinates \((x, y)\), polar coordinates use \( (r, \theta) \), where \( r \) is the distance from the origin, and \( \theta \) is the angle.

• The transformation to polar coordinates uses \( x = r\cos\theta \), and \( y = r\sin\theta \).
• In integration, the area element \( dA \) becomes \( r \, dr \, d\theta \).

In our exercise, integrating over the disk means setting \( r \) limits from 0 to 4 and \( \theta \) limits from 0 to \( 2\pi \). Integrating the population density in polar coordinates accurately models the changing population across the circular area.
This method adjusts our integration for circular symmetry and helps account for variations in density as distance from the center increases.