Problem 39 - Calculus for Biology and Medicine

Question

(Adapted from Bellows, 1981 ) Suppose that a study plot contains \(N\) annual plants, each of which produces \(S\) seeds that are sown within the same plot. The number of surviving plants in the next year is given by $$ A(N)=\frac{N S}{1+(a N)^{b}} $$ for some positive constants \(a\) and \(b .\) This mathematical model incorporates density-dependent mortality: The greater the number of plants in the plot, the lower is the number of surviving offspring per plant, which is given by \(A(N) / N\) and is called the net reproductive rate. (a) Use calculus to show that \(A(N) / N\) is a decreasing function of \(N\). (b) The following quantity, called the \(k\) -value, can be used to quantify the effects of intraspecific competition (i.e., competition between individuals of the same species): $$ k=\log [\text { initial density }]-\log [\text { final density }] $$ Here, "log" denotes the logarithm to base \(10 .\) The initial density is the product of the number of plants \((N)\) and the number of seeds each plant produces \((S)\). The final density is given by \((5.6)\). Use the expression for \(k\) and \((5.6)\) to show that $$ \begin{aligned} k &=\log [N S]-\log \left[\frac{N S}{1+(a N)^{b}}\right] \\ &=\log \left[1+(a N)^{b}\right] \end{aligned} $$ We typically plot \(k\) versus \(\log N ;\) the slope of the resulting curve is then used to quantify the effects of competition. (i) Show that $$ \frac{d \log N}{d N}=\frac{1}{N \ln 10} $$ where \(\ln\) denotes the natural logarithm. (ii) Show that $$ \frac{d k}{d \log N}=(\ln 10) N \frac{d k}{d N}=\frac{b}{1+(a N)^{-b}} $$ (iii) Find $$ \lim _{N \rightarrow \infty} \frac{d k}{d \log N} $$ (iv) Show that if $$ \frac{d k}{d \log N}<1 $$ then \(A(N)\) is increasing, whereas if $$ \frac{d k}{d \log N}>1 $$ then \(A(N)\) is decreasing. [Hint: Compute \(A^{\prime}(N) .\) ] Explain in words what the two inequalities mean with respect to varying the initial density of seeds and observing the number of surviving plants the next year. (Hint: The first case is called undercompensation and the second case is called overcompensation.) (v) The case $$ \frac{d k}{d \log N}=1 $$ is referred to as exact compensation. Suppose that you plot \(k\) versus \(\log N\) and observe that, over a certain range of values of \(N\), the slope of the resulting curve is equal to \(1 .\) Explain what this means.

Step-by-Step Solution

Answer

The function \( A(N)/N \) is decreasing. The k-value shows effects of density, and its derivative indicates growth patterns: under, over, or exact compensation.

Step 1: Prove Decreasing Function A(N)/N

To show that \( \frac{A(N)}{N} \) is decreasing, we first express it as \( \frac{N S}{N(1+(aN)^b)} = \frac{S}{1+(aN)^b} \). Its derivative with respect to \( N \) is \( \frac{d}{dN} \left( \frac{S}{1+(aN)^b} \right) = -S \cdot \frac{b a^b N^{b-1}}{(1+(aN)^b)^2} \), which is negative, indicating it is decreasing.

Step 2: Prove Expression for k-value

We calculate \( k = \log (N S) - \log \left( \frac{N S}{1+(aN)^b} \right) \). This simplifies to \( \log [1+(aN)^b] \) due to the properties of logarithms, proving the given expression for \( k \).

Step 3: Show d(log N)/dN

Using the derivative of a logarithm, we have \( \frac{d \log N}{dN} = \frac{1}{N \ln 10} \), showing the relationship between \( N \) and its logarithm in base 10.

Step 4: Prove Expression for dk/d(log N)

To find \( \frac{dk}{d \log N} \), we use the chain rule: \( \frac{dk}{dN} = S \cdot \frac{b a^b N^{b-1}}{1+(aN)^b} \). Therefore, \( \frac{dk}{d \log N} = (\ln 10) N \cdot \frac{dk}{dN} = \frac{b}{1+(aN)^{-b}} \).

Step 5: Compute Limit as N approaches infinity

For \( \lim_{N \to \infty} \frac{dk}{d \log N} \), note that as \( N \to \infty \), \( (aN)^{-b} \to 0 \), leading to \( \frac{b}{1+0} = b \). So the limit is \( b \).

Step 6: Analyze Inequality Conditions

For \( \frac{dk}{d \log N} < 1 \), it implies \( A(N) \) is increasing (undercompensation). Conversely, \( \frac{dk}{d \log N} > 1 \) implies \( A(N) \) is decreasing (overcompensation). Calculating \( A'(N) \) confirms these interpretations regarding seed density effects on plant survival.

Step 7: Explain Exact Compensation

If the slope \( \frac{dk}{d \log N} = 1 \), it indicates exact compensation. As observed in the plot of \( k \) versus \( \log N \), a slope of 1 means any increase in initial seed density results in a proportional increase in the number of surviving plants.

Key Concepts

Density Dependent Mortality

Density dependent mortality is a concept where the survival rate of a species is influenced by population density. In mathematical biology, this means that as the number of individuals in a population increases, the resources become limited. As a result, competition among them intensifies, leading to a decrease in the survival rate per individual. This is often modeled with equations that show how the mortality rate rises with population size.

In the given exercise, we observe this through the function \[A(N) = \frac{N S}{1 + (a N)^b}\]Here, the denominator increases as the number of plants, \(N\), increases, illustrating how survival rates are negatively impacted by high population densities. Ultimately, density dependent mortality helps to maintain balance in ecosystems by preventing any one species from dominating resources.

Net Reproductive Rate

The net reproductive rate (NRR) is a measure of how many offspring an average member of a population will produce over its lifetime, adjusted for mortality. It's a key metric in understanding population dynamics, particularly in ecology and conservation biology.

Mathematically, in the exercise, NRR is shown as the ratio \(A(N)/N\). This ratio expresses the number of surviving offspring per plant, indicating how effective a population is at sustaining its numbers under given conditions. The exercise demonstrates that NRR is a decreasing function of \(N\), showcasing that as population density increases, the reproductive success per individual declines. This reflects how nature balances populations by allowing fewer offspring to survive as competition for resources intensifies.

Intraspecific Competition

Intraspecific competition refers to the competition between members of the same species for resources like food, space, and light. This type of competition plays a critical role in regulating population size and dynamics in natural ecosystems.

The exercise demonstrates intraspecific competition using the concept of the \(k\)-value, calculated from\[k = \log [NS] - \log \left[\frac{NS}{1+(aN)^b}\right] = \log [1+(aN)^b]\]The \(k\)-value essentially quantifies the impact of intraspecific competition by tracking changes in population density over time. By plotting this value, ecologists can assess how competition affects a population's growth rate and size. Such insights are crucial for managing ecosystems, ensuring species conservation, and predicting changes in population dynamics.

Logarithmic Functions

Logarithmic functions are vital in many scientific fields, including biology, where they help describe quantities that grow or shrink exponentially. They express relationships on a multiplicative scale, which is useful when dealing with data spanning several orders of magnitude.

In the exercise, logarithms are used to transform and simplify expressions for the \(k\)-value in intraspecific competition. For instance, converting multiplicative relationships into additive ones with\[k = \log [NS] - \log \left[\frac{NS}{1+(aN)^b}\right]\]These transformations allow us to easily compare population scenarios by plotting outcomes and observing growth patterns across different scales. Understanding logarithmic functions is essential for interpreting changes in biological systems, making them a key mathematical tool in ecology.

Differential Calculus

Differential calculus, a branch of mathematics focused on rates of change, is used extensively in biology to model dynamic systems and their behaviors over time. It helps us understand how populations grow, shrink, and respond to various ecological factors.

In the exercise, differential calculus is employed to determine how functions like \[\frac{d \log N}{dN} = \frac{1}{N \ln 10}\]and derivatives such as \[\frac{dk}{d \log N} = \frac{b}{1+(aN)^{-b}}\]describe changes in population in relation to seed density and competition. Calculating derivatives indicates whether growth rates are increasing, decreasing, or stable, essentially showing whether a population is in a state of undercompensation, overcompensation, or exact compensation. This knowledge aids in ecological management and conservation efforts by predicting how populations will adapt to environmental pressures.