Question

Using a Logistic Differential Equation In Exercises 53 and \(54,\) the logistic differential equation models the growth rate of a population. Use the equation to (a) find the value of \(k,\) (b) find the carrying capacity, (c) graph a slope field using a computer algebra system, and (d) determine the value of \(P\) at which the population growth rate is the greatest. $$ \frac{d P}{d t}=3 P\left(1-\frac{P}{100}\right) $$

Short answer

The value of \(k\) is 3, the carrying capacity is 100, and the population growth rate is greatest when \(P\) is 50.

Step-by-step solution

Find the value of \(k\)

The logistic differential equation is normally written as \(\frac{d P}{d t}= k P\left(1-\frac{P}{N}\right)\) where \(N\) is the carrying capacity and \(k\) is the rate constant. Comparing this with the given equation \(\frac{d P}{d t}=3 P\left(1-\frac{P}{100}\right)\), we can see that the value of \(k\) is 3.

Find the carrying capacity

From the equation \(\frac{d P}{d t}=3 P\left(1-\frac{P}{100}\right)\), we can derive that the carrying capacity is the value of \(P\) when \(\frac{d P}{d t}=0\). This gives \(P=\frac{d P}{d t}=3 P\left(1-\frac{P}{100}\right) = 0 \), or \(P=100\). So the carrying capacity \(N\) is 100.

Graph a slope field

Graphing a slope field requires the use of a computer algebra system, which is beyond the abilities of this exercise. However, you can input the equation \(\frac{d P}{d t}=3 P\left(1-\frac{P}{100}\right)\) into a computer algebra system, where \(P\) is on the y-axis and \(t\) is on the x-axis.

Determine the value of \(P\) at which the population growth rate is the greatest

The rate of change of a population in a logistic model is greatest at the midpoint of the growth process, or when \(P=\frac{N}{2}\). Substituting \(N=100\) (which we determined earlier), we get \(P=50\) as the population size at which the growth rate is the greatest.

Key concepts

Population Growth Rate

Understanding how a population grows over time is crucial for predicting future trends and planning resources or interventions. The **population growth rate** is the change in the number of individuals in a population over a specified period. It is governed by both natural factors and mathematical models. In this context, the logistic differential equation provides a framework for modeling such growth.

In our logistic equation, \(\frac{dP}{dt} = 3P\left(1-\frac{P}{100}\right)\), the growth rate is influenced by \(P\), the current population size, and two critical parameters: \(k\), the rate constant, and the carrying capacity \(N\). Initially, when \(P\) is small compared to \(N\), the population grows almost exponentially. As \(P\) approaches \(N\), the growth rate slows down, eventually becoming zero when \(P = N\).

It's important to note that the peak growth rate occurs when the population is at half the carrying capacity, giving us insight into optimal conditions for resource allocation or control measures.

Carrying Capacity

In population dynamics, **carrying capacity** represents the maximum population size that an environment can sustain indefinitely given the available resources. It is a key component of the logistic growth model. The equation \(\frac{dP}{dt} = kP\left(1-\frac{P}{N}\right)\) explicitly includes \(N\), where \(N\) is the carrying capacity.

From the given logistic differential equation, \(\frac{dP}{dt} = 3P\left(1-\frac{P}{100}\right)\), it is clear that our carrying capacity \(N\) is 100. This is figured by noting that the differential \(\frac{dP}{dt}\) becomes zero when \(P = N\).

The concept of carrying capacity helps in understanding the limits of population growth. When the population reaches this capacity, it means that resources are being used to their fullest extent. At this point, factors such as food availability, habitat space, and competition curtail further growth.

Slope Field

A **slope field** is a visual representation used in differential equations to depict the slope of a function at various points. It provides insight into the behavior of solutions to differential equations without requiring an explicit solution.

To visualize how the population changes over time for the equation \(\frac{dP}{dt} = 3P\left(1-\frac{P}{100}\right)\), you can create a slope field using computational tools. This plot shows the slope of the solution at different values of \(P\) and \(t\), usually with arrows or short line segments.

Graphing the slope field helps identify patterns such as equilibria and points of rapid change. It is particularly useful for taking a holistic view of how the system evolves, offering a dynamic snapshot where trajectories of different initial populations can be traced.

Rate Constant

The **rate constant** \(k\) plays a vital role in determining how fast the population grows. In logistic growth models, it influences both the speed and pattern of population expansion.

In our equation \(\frac{dP}{dt} = 3P\left(1-\frac{P}{100}\right)\), the rate constant \(k\) is 3. It signifies how rapidly the population would grow at its fastest pace, assuming other conditions allow. Although it's a simple concept, the rate constant's impact is profound: higher values of \(k\) mean the population reaches the carrying capacity more quickly.

Importantly, \(k\) also determines the responsiveness of the population to changes in environmental conditions or resource availability. Adjustments to \(k\) in simulations allow for the exploration of different scenarios, revealing potential outcomes of changes in ecological or resource conditions.