Problem 3 - Calculus

Question

A population, \(P\), in millions, is 1500 at time \(t=0\) and its growth is governed by $$\frac{d P}{d t}=0.00008 P(1900-P)$$ Use Euler's method with \(\Delta t=1\) to estimate \(P\) at time \(t=1,2,3\)

Step-by-Step Solution

Answer

At \( t=1 \), \( P \approx 6300 \); at \( t=2 \), \( P \approx -2015700 \); at \( t=3 \), \( P \approx 614748780 \).

Step 1: Identify the Differential Equation

The given differential equation is \( \frac{dP}{dt} = 0.00008 P (1900 - P) \), where \( P \) is the population in millions and \( t \) is time in years.

Step 2: Apply Euler's Method for t=0 to t=1

Euler's method states \( P_{n+1} = P_n + \Delta t \times \frac{dP}{dt} \big|_{t_n,P_n} \). At \( t=0 \), \( P_0 = 1500 \). Compute \( \frac{dP}{dt} \big|_{t=0,P=1500} = 0.00008 \times 1500 \times (1900-1500) = 4800 \). Thus, \( P_1 = 1500 + 1 \times 4800 = 6300 \).

Step 3: Apply Euler's Method for t=1 to t=2

For \( t=1 \), use \( P_1 = 6300 \). Compute \( \frac{dP}{dt} \big|_{t=1,P=6300} = 0.00008 \times 6300 \times (1900-6300) = -2016000 \). Thus, \( P_2 = 6300 - 2016000 = -2015700 \).

Step 4: Apply Euler's Method for t=2 to t=3

For \( t=2 \), use \( P_2 = -2015700 \). Compute \( \frac{dP}{dt} \big|_{t=2,P=-2015700} = 0.00008 \times (-2015700) \times (1900 - (-2015700)) = 616764480 \). Thus, \( P_3 = -2015700 + 616764480 = 614748780 \).

Key Concepts

Differential Equations

Differential equations are a powerful mathematical tool. They help us understand how different quantities change over time. In the problem given, the differential equation is expressed as \( \frac{dP}{dt} = 0.00008 P (1900 - P) \). This equation describes the rate of change of the population \( P \) over time \( t \). Here, the amount of change depends on both the current population and a factor related to the potential maximum capacity (in this case, 1900).

Differential equations can model various real-world phenomena. In our exercise, it represents how a population grows, possibly under conditions such as limited resources or competition. This particular equation is non-linear, due to the term \( P(1900 - P) \), which makes it involve complex interactions. The goal of such equations is to capture the dynamics of systems, which can be anything from populations to chemical processes.

Population Growth

Population growth is a biological process concerned with changes in the number of individuals in a population over time. In the exercise, the differential equation illustrates a logistic growth model, a common model in population dynamics. This model accounts for a scenario where an environment has a carrying capacity, here represented by the value 1900.

To understand the concept, consider that populations cannot grow indefinitely in a confined space. At low population levels, growth is typically fast, akin to bacteria in a petri dish. However, as the population approaches the carrying capacity, resources become scarce, leading to slow growth and eventually stabilizing.

• The initial rapid multiplication happens due to ample resources.
• As resources dwindle, competition increases, slowing growth.
• Finally, the population hovers around the carrying capacity, where births roughly equal deaths.

In mathematical terms, this slows the population's rate of growth, leading to a more manageable population size.

Numerical Methods

Numerical methods allow us to approximate solutions to complex mathematical problems that don't have simple algebraic solutions. Euler's Method is one such numerical approach, especially useful in solving differential equations.

Euler's Method approximates the solution by taking small steps along the tangent of the curve. For instance, starting at an initial point, it uses the differential equation to calculate the slope (rate of change), then moves from the point in the direction of the slope. The formula used is \( P_{n+1} = P_n + \Delta t \times \frac{dP}{dt} \big|_{t_n,P_n} \), where \( \Delta t \) is the step size.

• Begin at the known point \((t_0, P_0)\).
• Calculate the slope using the function \( \frac{dP}{dt} \).
• Move to the next point using the formula.

This continues step-by-step, yielding an approximated value of the function at multiple points. While not always exact, it provides insight into the behavior of the solution over time and can be remarkably close when steps are small.