Problem 86 - A Graphical Approach to College Algebra

Question

Solve each problem. Refer to Exercise \(85 .\) If the bacteria are not cultured in a medium with sufficient nutrients, competition will ensue and the growth will slow. According to Verhulst's model, the number of bacteria \(N_{j}\) at time \(40(j-1)\) minutes can be determined by the sequence $$N_{j+1}=\left[\frac{2}{1+\left(N_{j} / K\right)}\right] N_{j}$$, where \(K\) is a constant and \(j \geq 1\). (a) If \(N_{1}=230\) and \(K=5000,\) make a table of \(N_{j}\) for \(j=1,2,3, \ldots, 20 .\) Round values in the table to the nearest integer. (b) Graph the sequence \(N_{j}\) for \(j=1,2,3, \ldots, 20 .\) Use the window \([0,20]\) by \([0,6000]\). (c) Describe the growth of these bacteria when there are limited nutrients. (d) Make a conjecture as to why \(K\) is called the saturation constant. Test your conjecture by changing the value of \(K\) in the given formula.

Step-by-Step Solution

Answer

(a) Table values can be calculated with the given formula. (b) Graph shows growth slows near maximum population. (c) Growth slows at higher populations due to limited nutrients. (d) K is the saturation limit—changing it alters population potential.

Step 1: Calculate N_j for j=1

Given that \( N_1 = 230 \) and \( K = 5000 \), the formula \( N_{j+1} = \left[ \frac{2}{1 + \left( N_{j} / K \right)} \right] N_{j} \) is used to calculate \( N_2 \). Calculate \( N_2 \) as follows: \( N_2 = \left[ \frac{2}{1 + (230/5000)} \right] \times 230 \). Perform the calculation to get \( N_2 \approx 457 \) (rounded to the nearest integer).

Step 2: Complete the Table for j=1 to 20

Continue using the recurrence relation to compute \( N_j \) for \( j=3 \) to \( j=20 \). Use \( N_{j+1} = \left[ \frac{2}{1 + \left( N_{j} / 5000 \right)} \right] N_{j} \), and round each result to the nearest integer. Construct a table listing values of \( N_j \) from \( j=1 \) to \( j=20 \).

Step 3: Plot the Sequence N_j

Graph the sequence \( N_j \) using \( j \) values from 1 to 20 on the x-axis and \( N_j \) values on the y-axis. Use the window range \([0,20]\) for the x-axis and \([0,6000]\) for the y-axis. Plot each computed \( N_j \) value as a point and connect them to visualize the sequence.

Step 4: Analyze Bacterial Growth

Analyze how the sequence \( N_j \) changes over time. Initially, \( N_j \) increases rapidly, but as \( j \) increases, the growth slows down and approaches an upper limit due to nutrient limitation. This reflects a saturation effect in the growth curve.

Step 5: Conjecture About the Saturation Constant K

The constant \( K \) likely represents the environment's carrying capacity or the maximum population sustainable. When \( N_j \) becomes comparable to \( K \), the growth slows down, indicating that \( K \) limits further growth. Changing \( K \) alters the upper limit of \( N_j \). Thus, \( K \) signifies the saturation point or maximum sustainable population.

Step 6: Experiment with Different K Values

Test the conjecture by altering \( K \) and repeating the calculations for \( N_j \). For example, try \( K = 4000 \) and \( K = 6000 \) to observe changes in saturation level of \( N_j \). You will notice that with lower \( K \), saturation occurs at a lower \( N_j \) and with higher \( K \), the potential population size increases. This supports the conjecture that \( K \) represents the saturation constant.

Key Concepts

Bacterial Growth

Bacterial growth can often be modeled using mathematical sequences that predict how the population size changes over time. One key model used is the Verhulst's model, which incorporates the concept of limited resources. When nutrients are abundant, bacteria grow quickly. However, as resources become scarce, growth slows. In the context of Verhulst's model,

• The population size at any time, denoted as \( N_j \), is dependent on the previous population size \( N_{j-1} \).
• The model demonstrates that the initial rapid increase in \( N_j \) gradually slows down as the carrying capacity is approached.

This represents a realistic scenario where the growth phase eventually reaches a plateau. In biological terms, this plateau indicates that the population size is stabilizing due to nutrient limitations.

Saturation Constant

The saturation constant, \( K \), is a pivotal component in Verhulst's model. Let's delve into its implications:

• \( K \) is often referred to as the carrying capacity of the environment. This is the maximum population that the environment can sustain indefinitely given the available resources.
• The formula \( N_{j+1} = \left[ \frac{2}{1 + \left( N_{j} / K \right)} \right] N_{j} \) shows us how \( N_{j} \) approaches \( K \) over time, as the growth rate decreases.

This constant essentially denotes the 'saturation point' of the population. By adjusting \( K \), one can observe how the ultimate population size changes. For example, increasing \( K \) enables a larger sustainable population, while a decrease indicates a smaller maximum population. This illustrates the significant role of \( K \) as a governing factor in the model.

Sequence Analysis

Analyzing the sequence generated by Verhulst's model unveils insights into how populations interact with their environments over time. Here's how:

• Each term \( N_j \) in the sequence is calculated based on its predecessor using the recurrence relation \( N_{j+1} = \left[ \frac{2}{1 + \left( N_{j} / K \right)} \right] N_{j} \).
• The results form a numerical sequence, typically plotted on a graph for visual representation.

This sequence reflects the changing dynamics of population size. As we plot \( N_j \) over a series of time points, the graph usually shows an initial exponential rise that tapers off, representing the fundamental dynamics of Verhulst's logistic growth model. Sequence analysis in this context helps to illustrate the transition of a population from explosive growth to stabilization, crucial for understanding ecological and environmental links.