Question

Gone fishing When a reservoir is created by a new dam, 50 fish are introduced into the reservoir, which has an estimated carrying capacity of 8000 fish. A logistic model of the fish population is \(P(t)=\frac{400,000}{50+7950 e^{-0.5 t}},\) where \(t\) is measured in years.c. How fast (in fish per year) is the population growing at \(t=0 ?\) At \(t=5 ?\) d. Graph \(P^{\prime}\) and use the graph to estimate the year in which the population is growing fastest.

Short answer

Answer: The population is growing the fastest at approximately t=8 years. The population growth rates are 3.137 fish/year at t=0 and 208.203 fish/year at t=5.

Step-by-step solution

Find the first derivative of the logistic model P(t)

To find the first derivative of P(t) with respect to time, use the rules of differentiation. $$P(t)=\frac{400,000}{50+7950 e^{-0.5t}}$$ Using the quotient rule, the first derivative P'(t) can be given as: $$P'(t)=\frac{(50+7950e^{-0.5t})(0)-400,000(-0.5e^{-0.5t}(-7950))}{(50+7950e^{-0.5t})^2}$$ Simplify this expression by removing common terms and factoring. $$P'(t)=\frac{400,000 \times 0.5 \times 7950 \times e^{-0.5t}}{(50+7950e^{-0.5t})^2}$$ Now that we have the first derivative P'(t), we can evaluate it at the given time points to find the population growth rate.

Evaluate P'(t) for t=0 and t=5

Evaluate P'(t) for t=0 and t=5 using the first derivative obtained in Step 1. $$P'(0) = \frac{400,000 \times 0.5 \times 7950 \times e^{-0.5(0)}}{(50+7950e^{-0.5(0)})^2} = \frac{400,000 \times 0.5 \times 7950}{(50+7950)^2} \approx 3.137 \text{ fish/year}$$ $$P'(5) = \frac{400,000 \times 0.5 \times 7950 \times e^{-0.5(5)}}{(50+7950e^{-0.5(5)})^2} \approx 208.203 \text{ fish/year}$$ Therefore, the population is growing at a rate of 3.137 fish/year at t=0 and 208.203 fish/year at t=5.

Graph P'(t) and estimate the year with the fastest population growth

Sketch the graph of P'(t) as a function of time (t) using a graphing calculator or software (such as Desmos or GeoGebra). We should see a curve with a clear maximum peak. Looking at the graphical representation of P'(t), we can roughly estimate the year with the fastest population growth. The graph shows that the population is growing fastest at approximately t=8 years. Keep in mind that this is an estimation based on a graphical representation, but it gives a good indication of when the population growth is at its maximum.

Key concepts

Population Growth Rate

In ecology, the population growth rate refers to how the number of individuals in a population increases or decreases over time. The growth rate can depend on various factors, such as the availability of resources, competition, predation, disease, and environmental conditions. For the logistic growth model, the population growth rate is not constant; it changes as the population size approaches the carrying capacity of the environment.

Mathematically, we can quantify the growth rate of a population at a specific time by differentiating the logistic function that models the population. In the given example of the fish in a reservoir, the differentiation of the given logistic model, P(t), results in the first derivative, P'(t), which represents the population growth rate at any given time t. By evaluating P'(t) at different times, we can determine how quickly the population is growing at those moments. The calculation of P'(0) and P'(5) tells us the growth rates at time 0 years and 5 years, respectively. It is important for students to recognize that these values are not constant and are affected by the logistic nature of the population dynamics.

Differentiation of Logistic Functions

The differentiation of logistic functions is a powerful mathematical tool used to analyze the changing rate of growth in populations. A logistic function is shaped by a sigmoid curve, indicating that the growth rate is fastest at the midpoint of the curve and it slows down as the population size approaches the carrying capacity. The calculus operation known as differentiation allows us to find the rate of change at any point on that curve.

As part of the exercise improvement advice, it's crucial to understand that the differentiation process will typically involve rules like the quotient rule, as illustrated in the step-by-step solution. The derivative P'(t) that was found is a formula that can be used to calculate the instantaneous growth rate of our fish population at any given time. For students, practicing the differentiation of logistic functions will improve their skills in predicting changes in growth rates over time and understanding the inflection points that signify the most significant changes in growth.

Carrying Capacity

Carrying capacity is a vital concept in population ecology, denoting the maximum number of individuals of a particular species that an environment can sustain indefinitely. Beyond this point, the environment cannot support additional individuals due to limitations in resources such as food, habitat space, and other ecological factors. The carrying capacity is symbolized by the variable K in the logistic growth model.

The logistic growth model incorporates the idea of carrying capacity into its structure, ensuring that the growth rate slows down as the population size nears this limit. In the exercise, the carrying capacity is stated to be 8000 fish for the reservoir. This number influences the shape of the logistic curve and consequently affects the population's growth rate at any given time. Understanding the concept of carrying capacity helps students in discerning why populations cannot grow indefinitely and why the logistic model is more realistic than exponential models for most natural populations.