Problem 65 - Calculus and its applications

Question

Before \(1859,\) rabbits did not exist in Australia. That year, a settler released 24 rabbits into the wild. Without natural predators, the growth of the Australian rabbit population can be modeled by the uninhibited growth model \(d P / d t=k P,\) where \(P(t)\) is the population of rabbits \(t\) years after \(1859 .\) (Source: www dpi.vic.gov.au/agriculture.) a) When the rabbit population was estimated to be \(8900,\) its rate of growth was about 2630 rabbits per year. Use this information to find \(k,\) and then find the particular solution of the differential equation. b) Find the rabbit population in \(1900(t=41)\) and the rate at which it was increasing in that year. c) Without using a calculator, find \(P^{\prime}(41) / P(41)\).

Step-by-Step Solution

Answer

The constant \( k = 0.2955 \), the population was about 2.900 million in 1900, and growing at that year by similar rate in relation. \( P'(41)/P(41) = 0.2955 \).

Step 1: Write the Given Information

We are given the differential equation for the rabbit population growth: \( \frac{dP}{dt} = kP \). We know that when the population \(P = 8900\), the growth rate \( \frac{dP}{dt} = 2630 \). We need to find \(k\).

Step 2: Solve for k using Given Condition

Substitute the values \( P = 8900 \) and \( \frac{dP}{dt} = 2630 \) into the differential equation: \( 2630 = k \times 8900 \). Solve for \(k\) to get \( k = \frac{2630}{8900} \approx 0.2955 \).

Step 3: Solve the Differential Equation

The differential equation \( \frac{dP}{dt} = kP \) with \( k = 0.2955 \) is a first-order linear equation. The general solution is \( P(t) = Ce^{kt} \).

Step 4: Find the Particular Solution

Given that initially the population \( P(0) = 24 \), substitute into the general solution: \( 24 = Ce^{0} \) which gives \( C = 24 \). Therefore, the particular solution is \( P(t) = 24e^{0.2955t} \).

Step 5: Compute the Population in 1900 (t=41)

Substitute \( t = 41 \) into the particular solution: \( P(41) = 24e^{0.2955 \times 41} \). Calculate this to find \( P(41) \).

Step 6: Calculate the Growth Rate in 1900

The rate of growth is given by \( \frac{dP}{dt} = kP(t) \). Substitute \( k = 0.2955 \) and \( P(41) \) from Step 5 to find the rate of increase in 1900.

Step 7: Find \(P^{\prime}(41)/P(41)\)

Using the relationship \( \frac{dP}{dt} = kP \), it follows that \( \frac{P^{\prime}(41)}{P(41)} = k = 0.2955 \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They describe how a particular quantity changes over time. In our exercise, the change in rabbit population is modeled by such an equation: \( \frac{dP}{dt} = kP \).
This specific equation is called a first-order linear differential equation, meaning it involves the first derivative of the function but no higher derivatives.

In this context, \(P\) represents the population of rabbits at a given time, \(t\). The derivative, \(\frac{dP}{dt}\), represents the rate of change of the population.

• "\(kP\)" suggests that the rate of growth is proportional to the size of the population. The bigger the population, the faster it grows.
• This equation forms the foundation for understanding how different initial conditions influence the rate of change of populations.

A particular solution to the differential equation is obtained by solving it, considering initial conditions, such as the initial population size. This allows us to predict future changes. In our exercise, solving the equation gives us the model \( P(t) = Ce^{kt} \).

Population Dynamics

Population dynamics explores how populations change over time and space. It is a central concept in ecology and involves the study of various factors affecting the size and structure of populations.

In this exercise, we focus on the population dynamics of rabbits introduced into a new environment, which is Australia. Several factors contribute to how rabbit numbers grow:

• Initial Population: The starting number of rabbits significantly influences future growth, as seen with the initial 24 rabbits.
• Growth Rate: Determined by the constant \(k\), it is crucial in deciding how quickly the population expands. It often reflects biological rates like birth rates minus death rates.
• Unlimited Resources: In the "uninhibited growth model", or exponential growth scenario presented here, we assume resources like food and space are available without limit.

Population dynamics involve critical thinking about how these factors interplay over time. In the real world, other considerations such as competition, predation, and environmental changes also play roles. However, for educational purposes, this basic model helps in understanding fundamental growth patterns.

Uninhibited Growth Model

The uninhibited growth model describes populations that grow without restrictions from environmental factors. In the exercise, the rabbit population is modeled using this concept, expressed mathematically by the equation \( \frac{dP}{dt} = kP \).

This model, also known as "exponential growth", assumes that conditions are ideal, meaning there are ample resources for the population to grow indefinitely without competition or predation.

• Exponential Growth: When resources are unlimited, populations can grow logarithmically over time, exemplified by the function \( P(t) = Ce^{kt} \).
• Constant Growth Rate: The rate \(k\) remains the same over time, allowing for smooth, continuous growth.
• Practical Applications: Though idealized, this model offers insight into how populations might expand in real scenarios if limiting factors are removed temporarily.
  

Real-life populations eventually face limitations that slow growth. In educational contexts, the uninhibited growth model helps students grasp the potential of population expansion and understand fundamental principles of ecological modeling.