Question

A population of rabbits in a meadow is observed to be 200 rabbits at time \(t=0\). After a month, the rabbit population is observed to have increased by \(4 \%\). Using an initial population of 200 and a growth rate of \(0.04\), with a carrying capacity of 750 rabbits, a. Write the logistic differential equation and initial condition for this model. b. Draw a slope field for this logistic differential equation, and sketc h the solution corresponding to an initial population of 200 rabbits. c. Solve the initial-value problem for \(P(t)\). d. Use the solution to predict the population after 1 year.

Short answer

The population after 1 year is approximately 437 rabbits.

Step-by-step solution

Understanding the Logistic Differential Equation

The logistic differential equation models population growth with a carrying capacity and is given by \( \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) \), where \( P \) is the population, \( r \) is the growth rate, and \( K \) is the carrying capacity. For this problem, \( r = 0.04 \), \( K = 750 \), and initially \( P(0) = 200 \). Thus, the equation is \( \frac{dP}{dt} = 0.04P\left(1 - \frac{P}{750}\right) \) with \( P(0) = 200 \).

Drawing the Slope Field

To draw the slope field, calculate the value of \( \frac{dP}{dt} \) using the logistic equation for several points by varying \( P \). Plot these slopes on a graph with \( t \) on the x-axis and \( P \) on the y-axis. The slope field will show lines with different slopes, indicating different rates of change of \( P \) over time.

Solving the Initial-Value Problem

To solve the differential equation \( \frac{dP}{dt} = 0.04P\left(1 - \frac{P}{750}\right) \) with initial condition \( P(0) = 200 \), separate variables: \( \frac{dP}{P(1 - \frac{P}{750})} = 0.04 \, dt \). Integrating both sides gives the solution \( P(t) = \frac{750}{1 + Ae^{-0.04 \cdot 750 \, t}} \), where \( A \) is found using the initial condition. Substituting \( P(0) = 200 \), solve for \( A \): \( A = \frac{750 - 200}{200} = 2.75 \), so \( P(t) = \frac{750}{1 + 2.75\, e^{-30 \cdot 0.04 t}} \).

Predicting Population after 1 Year

To find the rabbit population after 1 year (\( t = 12 \) months), substitute \( t = 12 \) into the solved equation: \( P(12) = \frac{750}{1 + 2.75 \, e^{-30 \cdot 0.48}} \). Evaluate the expression numerically to get the population size.

Key concepts

Population Growth

Population growth refers to the change in the number of individuals in a specific population over a specific period. Often, this growth is modeled to predict how populations increase over time. In the case of rabbits in a meadow, understanding how fast the population grows can help in managing resources.

The growth rate is a key factor in these calculations. It tells us the speed at which the population size changes. For the rabbits, the growth rate is given as 4%. This means, without any limiting factors, the population will grow at a rate proportional to its size. However, in nature, populations often don't grow indefinitely because of limiting factors like available food or space.

Carrying Capacity

Carrying capacity is a concept that describes the maximum number of individuals that an environment can sustainably support. It considers the availability of resources such as food, water, and space, as well as environmental conditions.

In our rabbit population example, the carrying capacity is specified as 750 rabbits. This means that the meadow can support up to 750 rabbits before resources become strained. When the population reaches this size, the growth rate slows down, and the population will start stabilizing.

The logistic differential equation captures this idea of carrying capacity. As the population approaches the carrying capacity, the rate of growth decreases, eventually reaching zero as resources limit further growth.

Initial Value Problem

An initial value problem involves finding a function that satisfies a differential equation and meets certain specified conditions at a starting point, known as initial conditions.

For the rabbit population, we start with an initial population size of 200 rabbits at time zero (denoted as \( P(0) = 200 \)). This is our initial condition. By knowing the logistic differential equation and this initial condition, we can determine how the rabbit population changes over time.

Solving the initial value problem allows us to find a precise formula for the population at any given time. This involves techniques like separating variables and integrating both sides of the equation. The solution gives a function \( P(t) \), which provides the population size as a function of time.