Problem 32 - Applied Calculus

Question

The population of the United States is predicted to be \(P(t)=310 e^{0.0073 t}\) million, where \(t\) is the number of years after the year \(2010 .\) Predict the average population between the years 2010 and \(2060 .\)

Step-by-Step Solution

Answer

The average population between 2010 and 2060 is approximately 372.90 million.

Step 1: Identify the Function

The population function is given as \( P(t) = 310 e^{0.0073t} \), where \( t \) represents the number of years after 2010.

Step 2: Define the Interval

We need to find the average population between 2010 and 2060. So, \( t \) starts at 0 (for year 2010) and ends at 50 (for year 2060).

Step 3: Set Up the Average Population Formula

The average value of a function \( f(t) \) over the interval \([a, b]\) is given by \( \frac{1}{b-a} \int_{a}^{b} f(t) \, dt \). Here, \( a = 0 \) and \( b = 50 \), so the average population \( \bar{P} \) is \( \frac{1}{50 - 0} \int_{0}^{50} 310 e^{0.0073t} \, dt \).

Step 4: Compute the Integral

Calculate the integral \( \int_{0}^{50} 310 e^{0.0073t} \, dt \). Using the formula for exponential integrals, \( \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \), we get: \( \int 310 e^{0.0073t} \, dt = \frac{310}{0.0073} e^{0.0073t} \) evaluated from 0 to 50.

Step 5: Evaluate the Integral

Substitute the limits into the antiderivative: \[ \frac{310}{0.0073} (e^{0.0073 \times 50} - e^0) \]. Calculate: \( e^{0.0073 \times 50} = e^{0.365} \). Use approximate value: \( e^{0.365} \approx 1.440 \). Then, it becomes \[ \frac{310}{0.0073} (1.440 - 1) \].

Step 6: Calculate the Average Population

Calculate \( \frac{310}{0.0073} \approx 42465.75 \) and then multiply by \(0.440\) to get the final integral result. The result is \( \approx 18644.93 \). Thus, the average population is \( \frac{1}{50} \times 18644.93 = 372.90 \text{ million} \).

Key Concepts

Understanding Exponential Growth

Exponential growth occurs when the growth rate of a mathematical function is constantly proportional to its current value. This produces a curve on a graph that rapidly increases over time.
It's like observing how a small stream might quickly swell into a massive river. In our exercise, the population grows exponentially, as denoted by the function:

• Given as: \( P(t) = 310 e^{0.0073t} \)
• An exponential term \( e^{0.0073t} \) captures the growth factor.

Since the base of natural logarithms, \( e \), is a fundamental constant, any exponentiation with it results in a form of growth.
The growth rate here is defined by the exponent \( 0.0073t \), which indicates how fast or slow the population will swell over time.

The Role of a Population Function

A population function serves as a mathematical representation to model how a population changes over time. This is particularly useful when predicting future populations or understanding past trends.In the exercise framework, we have:

• The initial value, \( P(0) = 310 \) million, represents the population at the year 2010.
• Function of time: \( P(t) = 310 e^{0.0073t} \), translating how the population evolves.

The constant \( 310 \) is the initial population in millions, while \( e^{0.0073t} \) indicates the multiplication out of each point on the timeline.
This setup implies, as time \( t \) increases, the function predicts more accurately how the US population might grow.

Using Definite Integrals to Find Total Change

Definite integrals can be used to determine the total accumulation of a quantity, especially when dealing with a variable that changes over time—like population. In our case, we seek the average population over a specific time span, from 2010 to 2060.
This involves calculating:

• The integral \( \int_{0}^{50} 310 e^{0.0073t} \, dt \) to find the area under the population curve.
• This area represents the aggregate population over the 50-year interval.

Such an integral takes into account all the continuous changes within the timeline.
The integral's result, filled with detailed values, then divides over the interval range for simplification.

Calculating the Average Value of a Function

Averages are a familiar concept, but calculating the average of a function over an interval is slightly more involved. Here, we use the formula for the average value of a continuous function:\( \bar{P} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt \) to find:

• The formula evaluates to \( \frac{1}{50} \int_{0}^{50} 310 e^{0.0073t} \, dt \)
• The factor \( \frac{1}{50} \) accounts for the interval length.

Through this approach, we're able to determine an equitable average over the decades despite the variations in annual growth rates.
This calculated result, \( 372.90 \) million, offers a simple summary of the otherwise complex growth pattern observed.