Question

Population Growth \(\ln\) Exercises 57 and 58, use a graphing utility to graph the growth function. Use the table of integrals to find the average value of the growth function over the interval, where \(N\) is the size of a population and \(t\) is the time in days. $$ N=\frac{375}{1+e^{4.20-0.25 t}}, \quad[21,28] $$

Short answer

The average value of the function \(N=\frac{375}{1+e^{4.20-0.25 t}}\) over the interval \([21,28]\) should be determined by applying the formula for average value of a function using integration, and performing necessary calculations. This exercise calls upon calculus skills, focusing particularly on integration of functions involving natural logarithms and exponential growth.

Step-by-step solution

Understand the Growth Function

The given population function is \(N=\frac{375}{1+e^{4.20-0.25 t}}\). This function describes an exponential decay, where \(N\) is the size of the population and \(t\) is the time in days. Notice that the function has a horizontal asymptote at \(N=375\). This is the population size the model approaches as time \(t\) increases.

Set Up the Average Value Function

The average value of a function \(f(x)\) over the interval \([a, b]\) can be calculated using the formula: \(\frac{1}{b-a} \int_{a}^{b} f(x) dx\). Here, \(a\) and \(b\) are the limits of the interval over which we are trying to find the average value. In the exercise, these are 21 and 28 respectively. Substitute these values and the population growth function \(N\) into the average value formula.

Calculate the Integral

Now, calculate the integral in the average value function. This requires the application of integration techniques like the power rule, chain rule, and potentially integration by parts or substitution. The integral in this case will yield an expression containing \(t\), which will then be evaluated at the limits of 28 and 21.

Complete the Calculation

Substitute \(t=28\) and \(t=21\) in the evaluated integral, and perform the subtraction. This will yield a single numerical value. Finally, divide this value by the difference of the interval limits, \((28 - 21)\), to find the average value of the function over the interval \([21,28]\).

Key concepts

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. This often results in a rapid decline initially, which slows down over time. In the growth function \(N = \frac{375}{1+e^{4.20-0.25 t}}\), exponential decay is evident in the term \(e^{4.20-0.25t}\). As \(t\) (time in days) increases, the term \(e^{4.20-0.25t}\) grows smaller, meaning the denominator of the function increases. Hence, the rate of increase of \(N\) approaches zero. Eventually, the population size \(N\) approaches the horizontal asymptote at 375, indicating that the population stabilizes at this maximum value. This represents the carrying capacity of the environment in the model.

Average Value of a Function

The average value of a function over a given interval provides us with an overall measure of the function’s performance or behavior over that span. Mathematically, for a continuous function \(f(x)\), the average value from \([a,b]\) is determined by the formula \(\frac{1}{b-a} \int_{a}^{b} f(x)\, dx\).

In simple terms, this formula computes the area under the curve of \(f(x)\) from \(a\) to \(b\), then divides by the interval length \(b-a\) to yield an average value. This is particularly useful in understanding complex behaviors over time, especially in population modeling where observation over multiple days can vary greatly.

Definite Integral

A definite integral calculates the accumulated area under a curve between two defined limits, \(a\) and \(b\). It is expressed as \(\int_{a}^{b} f(x)\, dx\). This tool is incredibly helpful in determining the total quantity in situations where change is continuous, such as the accumulation of population over time.

In the context of our exercise, finding the definite integral of the growth function \(N = \frac{375}{1+e^{4.20-0.25 t}}\) from \(t = 21\) to \(t = 28\) provides an overall measure of the population's total growth in this interval. The integration involves techniques such as substitution, which simplifies its evaluation, and leads to understanding complex dynamics efficiently.

Population Dynamics

Population dynamics focuses on how populations change over time in response to factors like birth rates, death rates, and the carrying capacity of the environment. The given growth function models these dynamics by showing how population changes over days with a limiting factor introduced by the environment.

This model is particularly adept at indicating exponential growth and eventual stabilization, revealing insights into real-world ecological systems. By exploring population dynamics through such functions, we can anticipate future trends, assess ecological impacts, and make informed decisions in conservation and resource management. The analysis of these trends over intervals, using integrals, ensures a comprehensive understanding of the growth rate and constraints faced by the population that allow it to reach equilibrium or stability.