Problem 79 - Precalculus: Mathematics for Calculus

Question

Population of a City \(\quad\) A city was incorporated in 2004 with a population of \(35,000 .\) It is expected that the population will increase at a rate of \(2 \%\) per year. The population \(n\) years after 2004 is given by $$P_{n}=35,000(1.02)^{n}$$ (a) Find the first five terms of the sequence. (b) Find the population in 2014.

Step-by-Step Solution

Answer

The first five terms are 35,000; 35,700; 36,414; 37,142; and 37,884. The population in 2014 is approximately 42,664.

Step 1: Understanding the Formula

We are given the formula for the population of the city as \( P_n = 35,000 (1.02)^n \). This formula calculates the population \( n \) years after 2004, starting with a base population of 35,000 and increasing at a rate of 2% per year.

Step 2: Finding the First Five Terms (a)

To find the first five terms, we will calculate \( P_n \) for \( n = 0, 1, 2, 3, \) and \( 4 \). 1. \( n = 0 \): \( P_0 = 35,000(1.02)^0 = 35,000 \)2. \( n = 1 \): \( P_1 = 35,000(1.02)^1 = 35,000 \times 1.02 = 35,700 \)3. \( n = 2 \): \( P_2 = 35,000(1.02)^2 = 35,000 \times 1.0404 = 36,414 \)4. \( n = 3 \): \( P_3 = 35,000(1.02)^3 = 35,000 \times 1.061208 = 37,142 \approx 37,142 \)5. \( n = 4 \): \( P_4 = 35,000(1.02)^4 = 35,000 \times 1.08243216 = 37,885 \approx 37,884 \) Thus, the first five terms of the sequence are: 35,000; 35,700; 36,414; 37,142; and 37,884.

Step 3: Calculate Population in 2014 (b)

To find the population in 2014, we note that 2014 is 10 years after 2004. Thus, \( n = 10 \).Calculate \( P_{10} \):\[ P_{10} = 35,000(1.02)^{10} \]First find \( (1.02)^{10} \):\( (1.02)^{10} \approx 1.21899 \)Calculate:\[ P_{10} = 35,000 \times 1.21899 \approx 42,664 \]Thus, the population in 2014 is approximately 42,664.

Key Concepts

Population Model

A population model is a mathematical way to predict changes in a population over time. This particular model calculates the growth of a city's population that started with 35,000 inhabitants in 2004. The model is structured around a consistent growth rate of 2% per year. The formula given, \[ P_n = 35,000(1.02)^n \]models the population for any year \( n \) after 2004. It's essential to break this down:

• The initial population is 35,000. This is sometimes referred to as the base or starting value.
• The term \( 1.02^n \) accounts for the consistent yearly increase of 2%.

To forecast the population for any year after 2004, you insert the year difference from 2004 into \( n \). For example, to find the population in 2014, \( n \) would be 10, since 2014 is ten years after 2004.

Geometric Sequence

A geometric sequence is a series of numbers where each term is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In this exercise, the city's changing population forms a geometric sequence.Here's how the sequence is built:

• The first term, \( P_0 \), is the initial population, 35,000.
• Each following term is found by multiplying the previous term by 1.02, the common ratio. This reflects a 2% increase annually.

The first five terms of this sequence are calculated by plugging in values \( n = 0, 1, 2, 3, 4 \) into the formula:

• \( n=0 \), \( P_0 = 35,000 \)
• \( n=1 \), \( P_1 = 35,000 \times 1.02 = 35,700 \)
• \( n=2 \), \( P_2 = 35,000 \times 1.04 = 36,414 \)
• \( n=3 \), \( P_3 = 35,000 \times 1.061208 = 37,142 \)
• \( n=4 \), \( P_4 = 35,000 \times 1.08243216 = 37,884 \)

This sequence helps visualize the steady growth of the population over the first few years.

Compound Growth Rate

The compound growth rate refers to the annual increase rate applied to a quantity over multiple periods. In this model, a 2% growth rate compounds yearly, meaning each year's population is increased by 2% of the current year's population, not the original.This compounding effect is crucial:

• Every year’s increase builds on the previous year’s new total, resulting in exponential growth rather than simple linear growth.
• The formula \( (1.02)^n \) indicates that the growth rate is compounded annually, meaning that each year, the entire population from the year before is increased by 2%.

The power of compounding makes a significant difference in long-term predictions. For instance, at a compound growth rate of 2% per year, the population predicted for 2014 (ten years after 2004) becomes:\[P_{10} = 35,000(1.02)^{10} \approx 42,664\]This implies a substantial increase over 10 years, primarily due to the compounding effect.