Problem 48 - Calculus for Biology and Medicine

Question

Suppose that the rate of growth of a plant in a certain habitat depends on a single resource-for instance, nitrogen. The dependence of the growth rate \(f(R)\) on the resource level \(R\) is modeled using Monod's equation $$f(R)=a \frac{R}{k+R}$$ where \(a\) and \(k\) are constants. Express the percentage error of the growth rate, \(100 \frac{\Delta f}{f}\), as a function of the percentage error of the resource level, \(100 \frac{\Delta R}{R}\).

Step-by-Step Solution

Answer

The percentage error of the growth rate is \(\frac{k}{k+R}\) times the percentage error of the resource level.

Step 1: Identify Variables and Constants

First, understand the given formula for the growth rate: \(f(R) = a \frac{R}{k+R}\), where \(a\) and \(k\) are constants, and \(R\) is the resource level.

Step 2: Differentiate Growth Rate

Differentiate the growth rate \(f(R)\) with respect to \(R\). Use the quotient rule: \(\frac{d}{dR}\left(\frac{R}{k+R}\right) = \frac{(k+R)\cdot 1 - R\cdot 1}{(k+R)^2}\), which simplifies to \(\frac{k}{(k+R)^2}\). Thus, \(\frac{df}{dR} = a \cdot \frac{k}{(k+R)^2}\).

Step 3: Apply Differential Concept

Express the change in \(f\), denoted as \(\Delta f\), using the differential approximation: \(\Delta f \approx \frac{df}{dR} \cdot \Delta R = a \cdot \frac{k}{(k+R)^2} \cdot \Delta R\).

Step 4: Calculate Percentage Errors

The percentage error in the growth rate is given by \(100 \frac{\Delta f}{f}\), and the percentage error in the resource level is \(100 \frac{\Delta R}{R}\). Substitute \(\Delta f = a \cdot \frac{k}{(k+R)^2} \cdot \Delta R\) into the percentage error formula for \(f\):\[100 \frac{\Delta f}{f} = 100 \frac{a \cdot \frac{k}{(k+R)^2} \cdot \Delta R}{a \cdot \frac{R}{k+R}}\]Simplify to get:\[100 \frac{k}{R(k+R)} \cdot \Delta R\]

Step 5: Express in Terms of Percentage Error of R

Since \(100 \frac{\Delta R}{R}\) represents the percentage error of the resource level, replace \(\Delta R\) with \(R \cdot \frac{\Delta R}{R}\) in the expression:\[100 \cdot \frac{k}{R(k+R)} \cdot R \cdot \frac{\Delta R}{R} = 100 \cdot \frac{k}{k+R} \cdot \frac{\Delta R}{R}\]Thus, the percentage error for the growth rate is \(\frac{k}{k+R}\) times the percentage error of the resource.

Key Concepts

Growth Rate

In biology, the growth rate of an organism, such as a plant, can describe how quickly it is increasing in size or biomass. The Monod equation is often used for modeling growth rates that depend on a single limiting resource. This is especially important in environments where one resource might control how fast growth can occur. With the formula \(f(R) = a \frac{R}{k+R}\), we know that:

• \(f(R)\) represents the growth rate as a function of the resource level \(R\).
• \(a\) is the maximum growth rate, which is reached when the resource is abundant.
• \(k\) is the half-saturation constant, representing the resource level at which the growth rate is half of its maximum.

As the resource \(R\) increases, the growth rate approaches \(a\). Conversely, if \(R\) is low, the growth rate can be significantly less than \(a\). Understanding the dynamics of the growth rate in relation to resource availability can help make decisions about resource management and conservation.

Percentage Error

Percentage error gives insight into how significant a measurement's error is in relation to the true value. This is often expressed as a percentage, which can be very helpful when comparing relative errors across different contexts. When referring to the Monod equation, percentage error in growth rate \(100 \frac{\Delta f}{f}\) is of interest:

• \(\Delta f\) represents the change in the growth rate.
• \(f\) is the calculated or expected growth rate.
• The formula \(100 \frac{\Delta f}{f}\) shows the percentage error related to \(f\).

Similarly, the percentage error in the resource level \(R\) is expressed as \(100 \frac{\Delta R}{R}\). It shows how much the actual resource level differs from an expected or normal level. For the Monod equation, knowing both of these percentages can help in understanding how sensitive the growth rate is to changes in resource levels. This can guide effective resource allocation.

Differential Approximation

Differential approximation is a helpful mathematical technique for estimating how slight changes in one variable affect another. In this context, we use this to determine how a small change in resource level \(\Delta R\) impacts the growth rate \(\Delta f\). It involves differentiating the function and using the derivative to approximate the change:

• The derivative \(\frac{df}{dR}\) is crucial; in this case, it used the quotient rule and simplified to \(a \cdot \frac{k}{(k+R)^2}\).
• This allows us to use \(\Delta f \approx \frac{df}{dR} \cdot \Delta R\) to estimate changes in \(f\).

By employing the differential approximation, one can effectively relate the changes in resource levels to changes in growth rate, giving us a clear tool for predicting how alterations in resources will affect growth, which is critical for managing resources and understanding biological growth patterns.