Problem 4 - Differential Equations and Linear Algebra

Question

At time \(t,\) the population \(P(t)\) of a certain city is increasing at a rate that is proportional to the number of residents in the city at that time. In January \(2000,\) the population of the city was 10,000 and by 2005 it had risen to 20,000. (a) What will the population of the city be at the beginning of the year \(2020 ?\) (b) In what year will the population reach one million?

Step-by-Step Solution

Answer

(a) The population at the beginning of 2020 will be \(160,000\) residents. (b) The population will reach one million in the year \(2033\).

Step 1: Write down the differential equation

Since the rate of population increase is proportional to the population itself, we can write a differential equation to represent this: \( \frac{dP}{dt} = kP \), where k is the proportionality constant.

Step 2: Solve the differential equation

In order to find the population function P(t), we need to solve this first-order linear differential equation. To do this, we remember that the solution of this ODE is: \( P(t) = Ce^{kt} \), where C is the constant of integration, to be determined using the given data points.

Step 3: Find the constant of integration

We are given that in January 2000, P(0) = 10000, and in 2005, P(5) = 20000. We'll use these data points to find the constants C and k. Using the data point P(0) = 10000: \( P(0) = Ce^{k(0)} = 10000 \) \( C = 10000 \) Now we have \( P(t) = 10000e^{kt} \) Next, using the data point P(5) = 20000: \( P(5) = 10000e^{k(5)} = 20000 \) \( e^{5k} = 2 \) \( 5k = \ln(2) \) \( k = \frac{\ln(2)}{5} \) Now we can write our population function as \( P(t) = 10000e^{\frac{\ln(2)}{5}t} \).

Step 4: Calculate the population at the beginning of 2020

Now that we have found the population function, we can find the population at the beginning of the year 2020 by plugging in t = 20: \( P(20) = 10000e^{\frac{\ln(2)}{5}(20)} \) \( P(20) = 10000e^{4\ln(2)} \) \( P(20) = 10000 \cdot (2^4) \) \( P(20) = 160000 \) residents The population at the beginning of 2020 will be 160,000 residents.

Step 5: Determine when the population will reach one million

To find the year when the population will reach one million, we need to solve for t when P(t) = 1000000: \( 1000000 = 10000e^{\frac{\ln(2)}{5}t} \) Divide both sides by 10000: \( 100 = e^{\frac{\ln(2)}{5}t} \) Take the natural logarithm of both sides: \( \ln(100) = \frac{\ln(2)}{5}t \) Now, solve for t: \( t = \frac{5\ln(100)}{\ln(2)} = \frac{5\ln(10^2)}{\ln(2)} = \frac{10\ln(10)}{\ln(2)} \) \( t \approx 33.219 \) Since we started in the year 2000, the population will reach one million around the year 2000 + 33.219 = 2033.219. Therefore, the population will reach one million in the year 2033.

Key Concepts

Differential Equations

Differential equations are a fundamental tool to model situations where a quantity changes at a rate dependent on its current value. In the context of population growth, this concept is expressed as a direct relationship between the growth rate and the current population size. The differential equation formulated for this is \( \frac{dP}{dt} = kP \), where

• \( P(t) \) denotes the population at time \( t \)
• \( k \) is the proportionality constant reflecting how the population grows over time.

This type of equation is known as a first-order linear differential equation, which is common in exponential growth problems. Such equations are essential in predicting future values based on present conditions.

Growth Rate Proportionality

The term 'growth rate proportionality' refers to the principle that the rate at which a population grows is directly related to its current size. This means larger populations grow more quickly than smaller ones because they have more individuals reproducing. Mathematically, this is represented by the equation \( \frac{dP}{dt} = kP \), where

• \( \frac{dP}{dt} \) is the rate of change of the population over time
• \( k \) is a constant that indicates the percentage by which the population increases each infinitesimal time increment.

By solving this equation, you can find a general expression for the population as a function of time, which is crucial for predicting future trends.

Population Modeling

Population modeling involves creating mathematical representations of how populations change over time. In our example, we use an exponential model because the growth rate is proportional to the current population size. The solution to the differential equation is:\[ P(t) = Ce^{kt} \]- \( C \) is the integration constant found using initial conditions.- \( k \) is the growth rate constant.

• This model helps us understand dynamics like how quickly a population can grow and when it will reach certain milestones.
• In practice, it allows urban planners and policymakers to make informed decisions.

Understanding population modeling is vital for resource allocation, infrastructure development, and environmental planning.

Integration Constant

In solving differential equations, particularly those concerning population growth, the integration constant \( C \) is determined using initial conditions. This constant reflects the initial number of individuals in the population at a specified starting point. For instance, if we know that in January 2000, the population was \( 10,000 \), then:\[ P(0) = C \cdot e^{k \cdot 0} = 10,000 \]Hence:

• \( C = 10,000 \)

By integrating the differential equation and applying known values, you can accurately describe the population at any given time. These initial conditions are key in shaping the population model's accuracy and relevance.

Natural Logarithm

When solving for the time at which a population reaches a particular size, the natural logarithm becomes incredibly useful. It allows us to isolate the time variable \( t \) in the exponential equation\[ P(t) = Ce^{kt} \]To solve for \( t \), you rearrange this to:

• Divide both sides to isolate the expression \( e^{kt} \), e.g., \( \frac{P(t)}{C} = e^{kt} \)
• Apply the natural logarithm to both sides, resulting in \( \ln(\frac{P(t)}{C}) = kt \)

From here, you can easily solve for \( t \). The natural logarithm is thus a fundamental tool in transitioning from exponential equations to algebraic ones, enabling us to solve complex growth-related problems efficiently.