Problem 2 - Advanced Engineering Mathematics

Question

From Problem 1 the growth constant is \(k=\frac{1}{5} \ln 2 .\) Then \(P=P_{0} e^{(1 / 5)(\ln 2) t}\) and \(10,000=P_{0} e^{(3 / 5) \ln 2}\). Solving for \(P_{0}\) we get \(P_{0}=10,000 e^{-(3 / 5) \ln 2}=6,597.5 .\) Now $$P(10)=P_{0} e^{(1 / 5)(\ln 2)(10)}=6,597.5 e^{2 \ln 2}=4 P_{0}=26,390$$ The rate at which the population is growing is $$P^{\prime}(10)=k P(10)=\frac{1}{5}(\ln 2) 26,390=3658 \text { persons/year. }$$

Step-by-Step Solution

Answer

The initial population is 6597.5, the population at \(t=10\) is 26,390, and the growth rate at \(t=10\) is 3658 persons/year.

Step 1: Understanding the problem

The problem involves finding the initial population \(P_0\), population at \(t=10\), and the growth rate at \(t=10\) using a given growth constant \(k = \frac{1}{5} \ln 2\).

Step 2: Solving for Initial Population \(P_0\)

We start with the equation for population growth: \(10,000 = P_0 e^{(3/5) \ln 2}\). To isolate \(P_0\), apply the exponential laws to get \(P_0 = 10,000 e^{-(3/5) \ln 2}\). Simplify to find \(P_0 = 6,597.5\).

Step 3: Finding Population at \(t=10\)

Use the population growth formula \(P = P_0 e^{(1/5)(\ln 2) t}\) with \(t=10\). This becomes \(P(10) = 6597.5 e^{2 \ln 2}\). Applying the exponential property \(e^{2 \ln 2} = (e^{\ln 2})^2 = 2^2 = 4\), therefore \(P(10) = 4 \times 6597.5 = 26,390\).

Step 4: Calculating the Growth Rate at \(t=10\)

The growth rate is given by \(P'(10) = k P(10)\). Substitute the known values to get \(P'(10) = \frac{1}{5} (\ln 2) \times 26,390\), which calculates to approximately \(3658\) persons/year.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions that depict how quantities grow or decay at a constant growth rate. In simpler terms, these functions help us model situations where something is increasing by a percentage or factor over time.

The general form of an exponential function is given by:

• \( f(t) = a e^{kt} \)

Where:

• \( a \) is the initial amount;
• \( k \) is the growth or decay constant;
• \( e \) is the mathematical constant approximately equal to 2.71828;
• \( t \) is time.

When \( k \) is positive, the function describes exponential growth, where the quantity increases over time. If \( k \) is negative, it describes exponential decay, indicating that the quantity decreases over time.

These functions are fundamental in modeling real-world scenarios like population growth, radioactive decay, and interest calculations.

Population Growth

Population growth is a specific type of exponential growth where the number of individuals in a population increases over time. It's often modeled using exponential functions due to its nature of compounding growth.

In many biological systems, such as populations of humans or animals in a habitat, populations grow by a constant proportion in regular time intervals. This growth can be modeled with the formula:

• \( P(t) = P_0 e^{kt} \)

Where:

• \( P(t) \) is the population at time \( t \);
• \( P_0 \) is the initial population;
• \( k \) is the growth constant;
• \( e \) is Euler's number, approximately 2.71828.

Such growth is unsustainable indefinitely, as it assumes unlimited resources. Therefore, real-world models often adjust over time to incorporate limited resources, which leads to logistic growth models.

The exercise above uses this concept to find the population at a specific time and the growth rate, showcasing how exponential models apply to real-world population dynamics.

Differential Equations

Differential equations are mathematical equations that involve the rates of change of quantities. These equations are crucial for modeling numerous physical processes, including exponential growth discussed earlier.

For population growth, we use a differential equation to represent the rate at which a population changes over time:

• \( \frac{dP}{dt} = kP \)

Where:

• \( \frac{dP}{dt} \) represents the rate of change of the population.
• \( k \) is the constant growth rate.
• \( P \) is the population size at time \( t \).

This equation shows how the rate of population growth is proportional to the current population size, typical for exponential growth.

Solving differential equations like this one help us predict future states of systems and understand how these systems evolve over time. In the problem provided, the differential equation was solved to calculate the current population and its growth rate at a specific time. This highlights the power of differential equations in projecting and understanding complex changes in systems.