Question

In Exercises \(23-26,\) the logistic equation describes the growth of a population \(P,\) where \(t\) is measured in years. In each case, find (a) the carrying capacity of the population, (b) the size of the population when it is growing the fastest, and (c) the rate at which the population is growing when it is growing the fastest. $$\frac{d P}{d t}=0.0002 P(1200-P)$$

Short answer

The carrying capacity of the population is 1200. The size of the population when it is growing the fastest is 600, and the rate at which the population is growing the fastest is 72 individuals per year.

Step-by-step solution

Find the Carrying Capacity

To find the carrying capacity (denoted as \(K\)), we set the rate of change \(\frac{d P}{d t}\) to 0 and solve for \(P.\) This gives us \[\frac{d P}{d t} = 0 = 0.0002 P(1200 - P)\] Solving for \(P\), we get \(P = 0\) and \(P = 1200\). Since the population \(P\) cannot be zero in this context, the carrying capacity \(K\) is 1200.

Find the Size of the Population When It is Growing the Fastest

The population is growing the fastest at halfway to the carrying capacity. This is because in the logistic growth model, growth is slow when the population is small and when the population is large, but fastest when the population is at half the carrying capacity. So, the population size when it is growing at its fastest is \[P_{\text{fastest}} = \frac{K}{2} = \frac{1200}{2} = 600\].

Find the Rate of Growth When Population is Growing the Fastest

Using \(P_{\text{fastest}}\) in our initial equation for the rate of change of population gives us the fastest rate of growth (\(\frac{d P}{d t}_{\text{fastest}}\)). So, we have \[\frac{d P}{d t}_{\text{fastest}} = 0.0002 \cdot P_{\text{fastest}} \cdot (1200 - P_{\text{fastest}}) = 0.0002 \cdot 600 \cdot (1200 - 600) = 72\]. Thus, the population is growing at the fastest rate of 72 individuals per year when the population size is 600.

Key concepts

Carrying Capacity

Understanding the carrying capacity is fundamental in studying population dynamics. It refers to the maximum population size an environment can sustain indefinitely without being degraded. The carrying capacity, denoted as \( K \), is determined by various limiting factors such as availability of food, water, habitat space and the competition within the population.

In the case of the logistic growth model, the carrying capacity can be established algebraically by setting the population growth rate to zero and solving for \( P \). For example, in the given logistic equation \( \frac{dP}{dt} = 0.0002 P(1200 - P) \), when the rate of change of \( P \) (population) equals zero, the values of \( P \) that satisfy this condition give the possible population limits. In this scenario, the value of \( P \) comes out to be 0 and 1200. However, logically, a population of zero would not grow, hence the actual carrying capacity \( K \) is 1200.

Population Growth Rate

The population growth rate is the speed at which the number of individuals in a population increases in a given time period as a fraction of the initial population. To visualize this, consider a population that starts with 100 individuals and increases to 110 individuals within a year, the growth rate is \( \frac{110 - 100}{100} = 0.1 \) or 10% per year.

In logistic growth, the growth rate is not constant and depends on both the size of the population and the carrying capacity \( K \). As the population size \( P \) approaches \( K \), the growth rate decreases due to the increase in competition for limited resources. This model realistically represents many natural populations, where resources are finite and excessive population growth is checked by increased mortality and/or decreased birth rates, leading to a sigmoidal or 'S-shaped' growth curve.

Differential Equations

Differential equations are mathematical tools that describe relationships involving rates of change. They are especially valuable in modeling natural processes where something is changing continuously. A simple physical representation might be the rate at which a car's speed changes over time.

In biology, differential equations are used to model the rate at which a population changes over time. The equation \( \frac{dP}{dt} = 0.0002 P(1200 - P) \) is a differential equation. It expresses the rate of change of the population \( P \) with respect to time \( t \). Here, the term \( 0.0002 P(1200 - P) \) encapsulates how the growth rate changes with population size -- initially growing when the population is small, then slowing down as the population nears the carrying capacity.

Logistic Equation

The logistic equation models the growth of populations and is one of the most recognized differential equations in population ecology. It is a refinement of the earlier Malthusian growth model and can be used to describe how a population evolves over time considering the limit imposed by the carrying capacity.

The logistic equation is typically given by \( \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) \), where \( P \) is the population size, \( r \) is the maximum per capita growth rate, and \( K \) is the carrying capacity of the environment. The term \( 1 - \frac{P}{K} \) acts to decrease the growth rate as \( P \) gets closer to \( K \), thus reflecting the reduced growth potential in densely populated conditions. The exercise provided confronts us with a specific form of the logistic equation and challenges us to apply its principles to identify key population parameters.