Problem 3 - Advanced Engineering Mathematics

Question

Let \(P=P(t)\) be the population at time \(t .\) Then \(d P / d t=k P\) and \(P=c e^{k t} .\) From \(P(0)=c=500\) we see that \(P=500 e^{k t} .\) since \(15 \%\) of 500 is \(75,\) we have \(P(10)=500 e^{10 k}=575 .\) Solving for \(k,\) we get \(k=\frac{1}{10} \ln \frac{575}{500}=\frac{1}{10} \ln 1.15 .\) When \(t=30\), $$P(30)=500 e^{(1 / 10)(\ln 1.15) 30}=500 e^{3 \ln 1.15}=760 \text { years }$$ and $$P^{\prime}(30)=k P(30)=\frac{1}{10}(\ln 1.15) 760=10.62 \text { persons/year. }$$

Step-by-Step Solution

Answer

Population at \( t = 30 \) is 760 and changing at 10.62 persons/year.

Step 1: Understanding the Differential Equation

The equation \( \frac{dP}{dt} = kP \) is a basic differential equation describing exponential growth. The general solution to this equation is \( P(t) = c e^{kt} \), where \( c \) is the initial population value at \( t = 0 \).

Step 2: Plugging Initial Conditions

Given \( P(0) = c = 500 \), we substitute \( c \) into the general solution. This gives us \( P(t) = 500 e^{kt} \).

Step 3: Calculate Population at t = 10

We know \( P(10) = 500e^{10k} = 575 \). This tells us that the population increases by 75 from its initial value at \( t = 10 \).

Step 4: Solve for Growth Rate Constant k

Using \( 500e^{10k} = 575 \), we solve for \( k \) by simplifying: \[ e^{10k} = \frac{575}{500} \]\[ k = \frac{1}{10} \ln \left( \frac{575}{500} \right) = \frac{1}{10} \ln 1.15 \].

Step 5: Calculate Population at t = 30

To find \( P(30) \), use \( P(t) = 500 e^{kt} \) with the value of \( k \) previously found:\[ P(30) = 500 e^{\left(\frac{1}{10} \ln 1.15\right) 30} = 500 e^{3 \ln 1.15} \].

Step 6: Simplify Population Calculation

Use the property \( e^{a \ln b} = b^a \) to simplify:\[ P(30) = 500 (e^{\ln 1.15})^3 = 500 (1.15)^3 = 760 \].

Step 7: Compute Rate of Change of Population at t = 30

The derivative \( \frac{dP}{dt} \) represents the rate at which the population changes. We compute:\[ P'(30) = k P(30) = \left(\frac{1}{10} \ln 1.15\right) \times 760 = 10.62 \text{ persons/year} \].

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. This type of equation is fundamental in expressing how quantities change over time and are widely used in various fields, such as physics, engineering, and economics. In our problem, the differential equation \( \frac{dP}{dt} = kP \) describes exponential growth. Here, \( P \) represents the population as a function of time \( t \), and \( k \) is a proportional constant known as the growth rate.

A solution to this differential equation is given by the exponential function \( P(t) = ce^{kt} \), where \( c \) is the initial population at \( t=0 \). This implies that if we know the initial population and the growth rate, we can predict future population sizes.

In growth models, understanding differential equations helps us uncover dynamic behavior, predict future trends, and manage resources effectively.

Population Modeling

Population modeling involves creating mathematical equations to represent the growth or decline of populations over time. A common model for population growth is based on the exponential growth equation, which assumes that the growth rate of the population is proportional to the current population size.

In the example exercise, the population at time \( t \) follows the equation \( \frac{dP}{dt} = kP \), indicating that the population growth rate is proportional to the population size. This is a characteristic of unrestricted exponential growth, where resources are assumed to be unlimited.

Through population modeling, we can:

• Predict future population sizes.
• Understand the effect of changing growth rates.
• Plan resource allocation efficiently.

Population modeling provides essential insights for resource management, urban planning, and environmental conservation.

Exponential Functions

Exponential functions play a central role in modeling growth processes because of their unique property of increasing at a rate proportional to their current value. The general form of an exponential function is \( f(t) = ce^{kt} \), where:

• \( c \) is the initial amount (at \( t = 0 \)).
• \( k \) is the constant growth rate.
• \( t \) is time.

In exponential growth, the function \( e^{kt} \) determines how the quantity grows over time.

Exponential functions are notably characterized by:

• Rapid increase or decrease.
• Constant percentage growth rate.
• Applications in various domains like finance, biology, and physics.

This mathematical representation allows flexible adaptation to different growth scenarios and facilitates the calculation of future values with ease.

Growth Rate Calculation

The growth rate is a crucial parameter in exponential growth models, influencing how quickly a population or a process expands over time. In our example, we need to calculate the growth rate constant \( k \) from given data points.

Given the initial condition \( P(0) = 500 \) and the fact that \( P(10) = 575 \), we deduced that:
\[ e^{10k} = \frac{575}{500} \]
From this expression, we solve for \( k \):
\[ k = \frac{1}{10} \ln \left( \frac{575}{500} \right) = \frac{1}{10} \ln 1.15 \]

Knowing \( k \) allows us to predict future population sizes and understand the system's dynamics effectively. In practical applications, calculating the growth rate helps in:

• Forecasting population or production levels.
• Assessing the timing of reaching specific thresholds.
• Implementing strategies to manage growth sustainably.

Precise growth rate calculations ensure that models align closely with real-world behavior.