Problem 44 - Precalculus

Question

Explain the mistake that is made. The city of San Antonio, Texas, has a population that is growing at \(5 \%\) a year, compounding continuously. If there were 1.3 million people in the greater San Antonio area in \(2006,\) approximately how many people will there be in \(2016 ?\) Apply the formula \(N=N_{0} e^{\pi},\) where \(N\) represents the number of people. Solution: Use the population growth model. \(N=N_{0} e^{r t}\). Let \(N_{0}=1.3, r=5,\) and \(t=10\) \(N=1.3 e^{(5)(10)}\). Approximate with a calculator. \(6.7 \times 10^{21}\). This is incorrect. What mistake was made?

Step-by-Step Solution

Answer

The wrong formula and growth rate as a percentage were used. Correcting them gives a population of approximately 2.143 million in 2016.

Step 1: Identify the Correct Formula

The given formula in the problem is incorrect. The correct formula for continuous compound growth is \(N = N_{0} e^{rt}\), where \(r\) is the growth rate in decimal form, and \(t\) is the time in years.

Step 2: Convert Percentage to Decimal

Convert the annual growth rate from a percentage to a decimal by dividing by 100. So, 5% becomes \(0.05\).

Step 3: Plug Values into the Formula

Use the correct formula: \(N = N_{0} e^{rt}\). Substitute \(N_{0} = 1.3\) million, \(r = 0.05\), and \(t = 10\). Thus, the expression becomes \(N = 1.3 e^{0.05 \times 10}\).

Step 4: Simplify the Exponent

Calculate the exponent: \(0.05 \times 10 = 0.5\). So, the formula becomes \(N = 1.3 e^{0.5}\).

Step 5: Calculate Using a Calculator

Use a calculator to find the value of \(e^{0.5}\), which approximates to 1.64872. Then multiply this by 1.3 million to find \(N\).

Step 6: Final Calculation

Calculate \(N = 1.3 \times 1.64872 \approx 2.143\) million. Thus, the population in 2016 is approximately 2.143 million.

Key Concepts

Continuous Compounding

Continuous compounding is a powerful concept in exponential growth, particularly when it comes to financial and population models. It allows us to calculate the growth of a quantity that compounds instanteously over a period of time. The formula used for continuous compounding is

• \[ N = N_0 e^{rt} \] where:
• \( N \) is the final amount after growth.
• \( N_0 \) is the initial amount.
• \( r \) is the growth rate as a decimal.
• \( t \) is the time period over which growth occurs.

The function \( e \) is the base of the natural logarithm, approximately equal to 2.71828. Using continuous compounding helps in projecting future scenarios by considering all at once growth that accrues continually over time. This model is preferred over simple or discrete compounding in many natural and financial processes, as it provides a more accurate representation by considering compounding at every possible fraction of time.

Population Growth

Population growth describes how the number of individuals in a population increases over time, and it is often modeled using exponential functions. For growing populations, continuous compounding gives a realistic analysis because growth isn't just an end-of-period calculation; it happens constantly as people are born or immigrate.In our exercise example, San Antonio's population growth is assumed to follow this continuous model. Initially, we have:

• An initial population, \( N_0 = 1.3 \) million.
• A growth rate, \( r = 0.05 \), representing a 5% annual increase.
• A time duration of 10 years, \( t = 10 \).

By accurately applying the formula \( N = N_0 e^{rt} \), we are able to predict that the population grows to approximately 2.143 million from 2006 to 2016, capturing the effect of these continuous inflows. Understanding population growth helps in urban planning, resource allocation, and policy making.

Mathematical Errors

Understanding mathematical errors is crucial for solving problems accurately. In this exercise, we saw how a small mistake in formula or calculation can lead to wildly inaccurate results.Initially, an incorrect expression was used and that led to an exceedingly large and incorrect value:

• The error came from misapplying constants and mistaking the growth function \( N = N_0 e^{\pi} \) rather than \( N = N_0 e^{rt} \).
• The growth rate wasn't converted to its decimal form before substituting into the formula.
  This led to the erroneous multiplication of whole numbers instead of decimals.

Knowing how to identify these errors involves checking your formulae carefully and ensuring that the right conversions and arithmetic operations are used. If unfamiliar results are obtained, re-evaluate each step for possible errors. Correcting these minimizes inaccuracies and enhances the reliability of your computations whether in academic or professional settings.