Problem 109 - A Graphical Approach to Precalculus with Limits

Question

The number of species in a sample is approximated by $$S(n)=a \ln \left(1+\frac{n}{a}\right)$$ where \(n\) is the number of individuals in the sample and \(a\) is a constant that indicates the diversity of species in the community. If \(a=0.36,\) find \(S(n)\) for each value of \(n .\) (Hint: \(S(n)\) must be a whole number.) (a) 100 (b) 200 (c) 150 (d) 10

Step-by-Step Solution

Answer

Use the formula provided, compute \( S(n) \) for each, and round to get whole numbers.

Step 1: Understand the Formula

We are given the formula for the number of species as \( S(n) = a \ln \left(1+\frac{n}{a}\right) \). Here, \( n \) is the number of individuals, and \( a \) is a constant that represents species diversity.

Step 2: Set the Constant Value

We are told that \( a = 0.36 \). We need to substitute this value into the species diversity formula whenever it appears in the calculation of \( S(n) \).

Step 3: Calculate S(n) for n = 100

Substitute \( n = 100 \) and \( a = 0.36 \) into the formula: \[ S(100) = 0.36 \ln \left(1 + \frac{100}{0.36}\right) \]. Calculate the inner expression \( \frac{100}{0.36} \), then add 1 and finally, multiply by 0.36 after taking the natural logarithm.

Step 4: Calculate S(n) for n = 200

Substitute \( n = 200 \) and \( a = 0.36 \) into the formula: \[ S(200) = 0.36 \ln \left(1 + \frac{200}{0.36}\right) \]. Similar to Step 3, perform the division, add 1, compute the natural logarithm, and then multiply by 0.36.

Step 5: Calculate S(n) for n = 150

Substitute \( n = 150 \) and \( a = 0.36 \) into the formula: \[ S(150) = 0.36 \ln \left(1 + \frac{150}{0.36}\right) \]. Follow the same procedure as previous steps: calculate, add, take the logarithm, and multiply by 0.36.

Step 6: Calculate S(n) for n = 10

Substitute \( n = 10 \) and \( a = 0.36 \) into the formula: \[ S(10) = 0.36 \ln \left(1 + \frac{10}{0.36}\right) \]. Again, follow the same calculation steps to determine \( S(10) \).

Step 7: Round to Whole Numbers

After computing the expressions for each value of \( n \), round \( S(n) \) to the nearest whole number as specified in the question.

Key Concepts

Species Diversity

Species diversity is an essential concept in ecology. It refers to the variety and abundance of different species living within a specific ecosystem or community. In our problem, the parameter "a" represents the diversity within the community under study. A higher value of "a" indicates a more diverse and complex ecological community. In simple terms, it is a measure of how many different types of species are present and how evenly the individuals are distributed among those species. More diversity often means greater ecological stability. In the given formula, this diversity constant "a" directly affects the calculation of the total number of species in a sample using the given approximation formula.

Sample Size

Sample size, denoted by "n" in our formula, represents the number of individual organisms sampled from the community. This concept is crucial because it impacts the estimation accuracy of the number of species present. The larger the sample size, generally, the more reliable the estimation will be. However, it is essential to choose a sample size that provides a good balance between accuracy and feasibility in data collection. A large sample is beneficial but can be costly and time-consuming. In the problem, you are required to calculate the number of species for different values of "n" using the formula, demonstrating the impact of changing the sample size on species diversity estimation.

Approximation Formula

The approximation formula provided uses natural logarithm to estimate the number of species in a sample:

• The formula is: \( S(n) = a \ln \left(1+\frac{n}{a}\right) \).
• Natural logarithms (\( \ln\)) help model phenomena related to growth and decay, which is suitable for ecological calculations.

By calculating \( S(n) \), the estimator for the number of species, we utilize the natural logarithm to factor in the ratio of the sample size "n" to the diversity constant "a". The multiplicative constant \( a \) moderates the impact of the natural logarithm, ensuring the output remains realistic. This approach is used because modeling the complexity of ecosystems exactly is challenging, and this formula provides a simplified and useful estimation. The key here is that the result should be a whole number, making it practical for real-world applications.