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. $$ N=\frac{5000}{1+e^{4.8-1.9 t}}, \quad[0,2] $$

Short answer

The graph of the growth function \(N = \frac{5000}{1 + e^{4.8 - 1.9t}}\) shows the exponential growth of the population over time in the interval [0, 2]. The average value of the growth function in the interval [0, 2] may be obtained by calculating the integral of the function over the interval divided by the interval length. The actual numerical values would depend on the result of the integral calculation.

Step-by-step solution

Graph the growth function

To plot the graph, replace \(t\) by values from the interval [0, 2] in the function \(N = \frac{5000}{1 + e^{4.8 - 1.9t}}\) and compute corresponding \(N\) values. A graph can then be plotted with the obtained data points. Connect the points to visualize the growth over time.

Calculate the integral of the function over the interval

The integral of the function in the interval [0, 2] can be calculated by following formula:\[ \int_{0}^{2} N(t) \, dt = \int_{0}^{2} \frac{5000}{1 + e^{4.8 - 1.9t}} \, dt \]Use the common integral rule for \(\int f(x) \cdot e^{g(x)} \, dx\) .

Calculate the Average Value

The average value of the function over the interval [0, 2] can be calculated with the formula \(A = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx\). Substitute \(a = 0, b = 2\), and the integral \(\int_{0}^{2} N(t) \, dt\). Solve the equation to get the average value.

Key concepts

Differential Equations

At the heart of many natural phenomena like population growth lies a mathematical tool called a differential equation. A differential equation describes the relationship between a function and its derivative, capturing the essence of change within the system. In the context of population growth, a differential equation would express how the population size changes over time, taking into consideration birth rates, death rates, immigration, and emigration.

In simple terms, if you were to imagine a growing population of animals in a forest, a differential equation would allow us to predict the number of animals at any future time based on their current population and their rates of reproduction and death. These calculations can be quite complex and often require sophisticated mathematical techniques to solve. However, by solving these equations, we can make powerful predictions about the world around us.

Exponential Functions

Exponential functions are essential in modeling growth processes that increase at a rate proportional to the current size, which is exactly what happens in many real-world population growth scenarios. The function often takes the form of \( N(t) = N_0e^{rt} \), where \( N(t) \) is the population at time \( t \), \( N_0 \) is the initial population, \( r \) is the growth rate, and \( e \) is the base of natural logarithms, an irrational and transcendental number approximately equal to 2.71828.

In our exercise, though, we have a more complex form due to the presence of a logistic model which accounts for limited resources or other constraints on growth. Nonetheless, at its core, the function still relies on the exponential mechanism to describe how the population grows over time. It's the exponential nature of the growth that leads to the rapid increase of population, an important concept for students to grasp when considering the long-term implications of population dynamics.

Integration Calculus

Integration calculus is like the flip side of differentiation; it deals with accumulation rather than rates of change. In our problem, integration is used to calculate the total population growth over a given time interval and to find the average population size within that interval. The act of integrating the growth function over the interval from \( a \) to \( b \) can be thought of as adding up tiny slices of population growth, each slice representing a moment in time, to get the whole picture of growth across the interval.

The average value of the function over an interval is a particularly useful concept in understanding overall trends, as it smooths out the fluctuations that might occur over short periods. The formula for the average value of a continuous function over an interval \( [a, b] \) is given by \( A = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx \). In population studies, this can provide insight into the 'typical' population over a period, allowing researchers and policymakers to make informed decisions about resource allocation, conservation efforts, or infrastructure development.