Problem 12 - Applied Calculus

Question

The population of the world \(t\) years after 2010 is predicted to be \(P=6.9 e^{0.012 t}\) billion. (a) What population is predicted in \(2020 ?\) (b) What is the predicted average population between 2010 and \(2020 ?\)

Step-by-Step Solution

Answer

(a) 7.775 billion; (b) 6.3325 billion.

Step 1: Identify the Year for Part (a)

To find the population for the year 2020, determine how many years after 2010 this is. Since 2020 - 2010 = 10, the value of \( t \) is 10.

Step 2: Calculate Population for 2020

Using the formula \( P = 6.9 e^{0.012 t} \), substitute \( t = 10 \). Evaluate \( P = 6.9 e^{0.012 imes 10} \). Calculate the value: \( P = 6.9 e^{0.12} \).

Step 3: Compute the Exponential Term

Calculate \( e^{0.12} \) using a calculator. \( e^{0.12} \approx 1.1275 \).

Step 4: Determine Population in 2020

Substitute \( e^{0.12} \approx 1.1275 \) back into the equation. \( P = 6.9 imes 1.1275 \approx 7.775 \). The predicted population in 2020 is approximately 7.775 billion.

Step 5: Determine the Integral Limits for Part (b)

For the average population between 2010 and 2020, set up the integral from \( t = 0 \) to \( t = 10 \) for the function \( P(t) = 6.9 e^{0.012 t} \).

Step 6: Integrate the Population Function

Integrate \( P(t) = 6.9 e^{0.012 t} \) over the interval from \( t = 0 \) to \( t = 10 \). The indefinite integral \( \int 6.9 e^{0.012 t} \, dt \) results in \( \frac{6.9}{0.012} e^{0.012 t} + C \).

Step 7: Evaluate the Definite Integral

Calculate the definite integral \( \left[ \frac{6.9}{0.012} e^{0.012 t} \right]_{0}^{10} \). This becomes \( \left( \frac{6.9}{0.012} e^{0.12} - \frac{6.9}{0.012} e^{0} \right) \).

Step 8: Calculate the Definite Integral Result

Compute \( e^{0.12} \approx 1.1275 \) and \( e^{0} = 1 \). Thus, \( \left( \frac{6.9}{0.012} \times 1.1275 - \frac{6.9}{0.012} \times 1 \right) \). This results in \( 575.75 \times 1.1275 - 575.75 \).

Step 9: Find the Average Population

The population sum between 2010 and 2020 from the integral is \( 63.325 \). Divide by the interval length (10 years) to find the average: \( \frac{63.325}{10} = 6.3325 \).

Step 10: Conclusion for Part (b)

The predicted average population between 2010 and 2020 is approximately 6.3325 billion.

Key Concepts

World Population Prediction

Predicting the world population involves understanding exponential growth, a situation where quantities grow at a rate proportional to their size. In this exercise, we're given a formula for predicting the world population at any time after 2010, expressed as \( P=6.9 e^{0.012 t} \) billion people, where \( P \) is the population and \( t \) is the number of years since 2010.
To predict the world population in 2020, we need to identify how many years after 2010 that is, which is 10 years. Substituting \( t = 10 \) in the equation, we can compute:

• \( P = 6.9 e^{0.012 imes 10} \).
• This results in \( P = 6.9 e^{0.12} \).

Using a calculator, we find \( e^{0.12} \approx 1.1275 \). Multiply this result by 6.9 to get approximately 7.775 billion. Thus, the predicted world population in 2020 is about 7.775 billion people.

Average Population Calculation

To calculate the average world population between two dates, we must consider the growth pattern over the time interval. Here, the task is to find the average population from 2010 to 2020. This requires the use of the concept of an integral in mathematics, which helps us find accumulated quantities over an interval.
This involves integrating the population function \( P(t) = 6.9 e^{0.012 t} \) over \( t \) from 0 to 10, representing 2010 to 2020. The integral expression is:

• \( \int_{0}^{10} 6.9 e^{0.012 t} \, dt \).
• When we evaluate this definite integral, we use the antiderivative \( \frac{6.9}{0.012} e^{0.012 t} \).

This is evaluated from \( t = 0 \) to \( t = 10 \). Substituting these limits, we get: \[ \left( \frac{6.9}{0.012} e^{0.12} - \frac{6.9}{0.012} \times 1 \right) \]. Calculating this gives us about 63.325 billion. To find the average, we divide this result by 10, obtaining around 6.3325 billion as the predicted average population for the decade.

Definite Integral Application

Definite integrals are powerful tools for finding accumulated growth over a specified interval. In this context, they help us calculate how much the world population has grown or averaged over a period of time.
A definite integral provides the total sum of an exponential function between two bounds, making it ideal for population predictions and average calculations. It is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper bounds, respectively. The value of a definite integral can represent a total change, like population growth over time, or, as in this exercise, the total population over a period.
Upon computing the integral result for our population function, we find the accumulated growth from 2010 to 2020. By dividing by the number of years, we obtained what is known as the average population during this interval. This approach not only aids in projecting population changes but also helps understand how variables change across specified periods, making it a crucial concept in calculus.