Problem 71 - Precalculus: Mathematical for Calculus

Question

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 the sequence $$P_{n}=35,000(1.02)^{n}$$ (a) Find the first five terms of the sequence. (b) Find the population in 2014 .

Step-by-Step Solution

Answer

(a) 35,000, 35,700, 36,414, 37,142, 37,885; (b) 42,665.

Step 1: Calculate First Term

To find the first term of the sequence, set \(n = 0\). This represents the year 2004, the starting point of our sequence. Using the formula \(P_{n} = 35,000(1.02)^{n}\):\[ P_{0} = 35,000(1.02)^{0} = 35,000 \times 1 = 35,000 \]Thus, the population in 2004 is 35,000.

Step 2: Calculate Second Term

For the second term, set \(n = 1\). This corresponds to the year 2005:\[ P_{1} = 35,000(1.02)^{1} = 35,000 \times 1.02 = 35,700 \]Thus, the population in 2005 is 35,700.

Step 3: Calculate Third Term

For the third term, set \(n = 2\). This corresponds to the year 2006:\[ P_{2} = 35,000(1.02)^{2} = 35,000 \times 1.0404 = 36,414 \]Thus, the population in 2006 is 36,414.

Step 4: Calculate Fourth Term

For the fourth term, set \(n = 3\). This corresponds to the year 2007:\[ P_{3} = 35,000(1.02)^{3} = 35,000 \times 1.061208 = 37,142 \]Thus, the population in 2007 is 37,142.

Step 5: Calculate Fifth Term

For the fifth term, set \(n = 4\). This corresponds to the year 2008:\[ P_{4} = 35,000(1.02)^{4} = 35,000 \times 1.08243216 = 37,885 \]Thus, the population in 2008 is 37,885.

Step 6: Calculate Population in 2014

To find the population in 2014, we need the term where \(n = 10\) since 2014 is ten years after 2004:\[ P_{10} = 35,000(1.02)^{10} = 35,000 \times 1.218994 \\approx 42,665.29 \]Thus, the population in 2014 is approximately 42,665.

Key Concepts

Population Growth

Population growth refers to the change in the number of individuals in a population over time. In many real-world scenarios, this growth can be influenced by factors such as birth rates, death rates, and migration. In our exercise, we are examining the growth of a city's population. The initial number of individuals at the time of the city's incorporation represents our starting point.

This type of problem often involves using a mathematical formula that helps estimate future populations based on current growth trends. The formula provided is a recurrence relation that predicts the population of the city each year. In the exercise, the city starts with 35,000 residents in 2004. Each subsequent year sees a population increase by 2%.

Understanding the dynamics of population growth is essential for urban planning, resource management, and policy-making. It helps cities plan for things like housing, schools, and other infrastructure needs.

Exponential Growth

Exponential growth occurs when the growth rate of a value is directly proportional to its current size, leading to a rapid increase over time. It's a common pattern in many biological, economic, and natural systems. In our exercise, the city's population grows exponentially because the population size increases by a fixed percentage each year.

The exponential growth formula is given by:\[ P_{n} = P_{0} imes (1 + r)^n \]where:

• \(P_{n}\) is the population after \(n\) years.
• \(P_{0}\) is the initial population.
• \(r\) is the growth rate (in decimal form).

In our example, \(P_{0} = 35,000\) and \(r = 0.02\), which means the formula accounts for a 2% increase every year.

Exponential growth can lead to substantial increases over time, as seen in the decades following the starting date.

Sequence Calculation

Sequence calculation is an essential component of mathematics, especially when analyzing changes over time like in population growth. In this exercise, the population growth over the years forms a geometric sequence.

A geometric sequence is a series of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Here, each year's population is the result of multiplying the previous year's population by 1.02. This constant ratio (1.02 in our case) indicates a consistent rate of growth.

To find specific terms of a geometric sequence, we use the general term formula for sequences:\[ a_n = a_1 imes r^{(n-1)} \]where:

• \(a_n\) is the \(n\)-th term of the sequence.
• \(a_1\) is the first term of the sequence.
• \(r\) is the common ratio.

By calculating these terms, students can track how the population changes each year and predict future values.