Problem 40 - Applied Calculus

Question

The population of a city is 50,000 in 2008 and is growing at a continuous yearly rate of \(4.5 \%\). (a) Give the population of the city as a function of the number of years since 2008 . Sketch a graph of the population against time. (b) What will be the city's population in the year 2018 ? (c) Calculate the time for the population of the city to reach \(100,000 .\) This is called the doubling time of the population.

Step-by-Step Solution

Answer

(a) \( P(t) = 50000 e^{0.045t} \); (b) approximately 78,400; (c) approximately 15.4 years.

Step 1: Understand the Growth Formula

The population grows continuously according to the formula \( P(t) = P_0 e^{rt} \), where \( P_0 \) is the initial population, \( r \) is the growth rate, and \( t \) is the time in years since the initial measurement. Given that \( P_0 = 50000 \) and \( r = 0.045 \) (as a decimal), we can form the function describing the population over time.

Step 2: Write the Population Function

Using the formula from Step 1, plug in the known values: \( P(t) = 50000 e^{0.045t} \). This represents the population \( P \) as a function of time \( t \) in years since 2008.

Step 3: Graph the Function (Hypothetical)

To sketch the graph of \( P(t) = 50000 e^{0.045t} \), plot it with respect to \( t \). The graph is an exponential curve starting at 50000 when \( t = 0 \) and increasing as \( t \) increases, concave upwards.

Step 4: Calculate the Population in 2018

To find the population in 2018, calculate \( P(10) \) since 2018 is 10 years after 2008. Substitute \( t = 10 \) into the equation: \( P(10) = 50000 e^{0.045 \times 10} \). Simplifying this, \( P(10) = 50000 e^{0.45} \). Calculate the value of \( e^{0.45} \) and finally find \( P(10) \).

Step 5: Numerical Calculation for 2018

Approximate \( e^{0.45} \approx 1.568 \). Therefore, \( P(10) = 50000 \times 1.568 \approx 78400 \). Thus, the population in 2018 is approximately 78,400.

Step 6: Find the Doubling Time

Doubling time occurs when \( P(t) = 100000 \). Using the formula \( P(t) = 50000 e^{0.045t} = 100000 \), solve for \( t \). Divide both sides by 50000 to get \( e^{0.045t} = 2 \). Take the natural logarithm of both sides to solve for \( t \): \( 0.045t = \ln(2) \).

Step 7: Solve for Doubling Time

Calculate \( \ln(2) \approx 0.693 \). Then \( t = \frac{0.693}{0.045} \approx 15.4 \). Thus, the doubling time for the population to reach 100,000 is approximately 15.4 years.

Key Concepts

Population Modeling

Population modeling is a way to predict future population sizes based on current data and assumptions about growth. These models are incredibly helpful for urban planning, environmental studies, and resource management.

In the given exercise, we deal with exponential growth, which means a population increases at a rate proportional to its current size. This is modeled using the formula: \( P(t) = P_0 e^{rt} \) where:

• \( P(t) \) is the population at time \( t \).
• \( P_0 \) is the initial population.
• \( r \) is the rate of growth.
• \( t \) is the time in years since the start point.

In our example, the city’s population in 2008 is 50,000, and it grows at a rate of 4.5% per year. Using the formula, we create a model \( P(t) = 50000 e^{0.045t} \), predicting the population over time.

This model assumes continuous, unrestricted growth, which means it can predict future population sizes assuming current trends continue.

Continuous Growth Rate

Continuous growth rate is a concept where the growth happens consistently over time, rather than in discrete steps. In our city population model, the annual growth rate is 4.5%, which is considered continuous.

This concept is expressed mathematically using the number 'e', the base of natural logarithms, which approximately equals 2.71828. The formula used to model this growth is \( P(t) = P_0 e^{rt} \), where 'e' calculates the effect of continuous compounding.

Continuous growth is more realistic for natural processes, such as populations or investments, which change more fluidly compared to processes calculated at intervals. Using 'e' helps simulate this continuous nature, ensuring predictions closely reflect reality.

• The growth is proportional to the current size, meaning larger populations grow faster.
• The exponent \( rt \) represents the proportionate growth over time \( t \).

Doubling Time

Doubling time refers to the amount of time it takes for a population to double in size at a constant growth rate. It's a practical way to understand the implications of exponential growth. In the example of the city's population, the doubling time is when the population goes from 50,000 to 100,000.

To find the doubling time, we use the equation:\[ 2 = e^{rt} \]Solving this involves taking the natural logarithm of both sides: \( ln(2) = rt \).

The formula \( t = \frac{ln(2)}{r} \) gives us the doubling time when the growth rate \( r \) is known. In our exercise, substituting the rate of 4.5%, we calculate:\( ln(2) \approx 0.693 \), so:\( t \approx \frac{0.693}{0.045} \approx 15.4 \) years.

This means, at the rate given, the population will double in about 15.4 years. Knowing this helps in planning for future needs and challenges of growth.