Problem 48 - Calculus for Biology and Medicine

Question

Model Suppose that the rate of growth of a plant in a certain habitat depends on a single resource \(-\) for instance, nitrogen. Assume that the growth rate \(f(R)\) depends on the resource level \(R\) in accordance with the formula $$f(R)=a \frac{R}{k+R}$$ 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 in growth rate is 100 \( \frac{k}{k+R} \) times the percentage error in resource level.

Step 1: Understand the Relationship Between Growth Rate and Resource Level

The given formula is \( f(R) = a \frac{R}{k+R} \), which describes how the growth rate \( f(R) \) depends on the resource level \( R \). Our goal is to express the percentage error of the growth rate in terms of the percentage error of the resource level.

Step 2: Define the Small Changes in Growth Rate and Resource Level

Let \( \Delta f \) be a small change in \( f(R) \) and \( \Delta R \) be a small change in \( R \). The percentage changes are given by \( 100 \frac{\Delta f}{f} \) and \( 100 \frac{\Delta R}{R} \). We need to relate these two expressions.

Step 3: Use the Derivative for Small Changes

For small changes, \( \Delta f \approx \frac{df}{dR} \Delta R \). Compute \( \frac{df}{dR} \) using the given formula \( f(R) = a \frac{R}{k+R} \).

Step 4: Differentiate the Growth Rate Formula

Perform the differentiation:\[ \frac{df}{dR} = a \left( \frac{(k+R) \cdot 1 - R \cdot 1}{(k+R)^2} \right) = a \frac{k}{(k+R)^2} \]This is the derivative of the growth rate with respect to resource level.

Step 5: Relate Derivatives to Percentage Change

The small change in growth rate can be expressed as \( \Delta f = \left( a \frac{k}{(k+R)^2} \right) \Delta R \). Thus, the percentage error in growth rate becomes:\[ 100 \frac{\Delta f}{f} = 100 \cdot \frac{\frac{df}{dR} \Delta R}{f(R)} = 100 \cdot \frac{a \frac{k}{(k+R)^2} \Delta R}{a \frac{R}{k+R}} \]

Step 6: Simplify the Expression

Simplify the expression:\[ 100 \frac{\Delta f}{f} = 100 \cdot \frac{k \Delta R \cdot (k+R)}{R(k+R)^2} \]\[ = 100 \cdot \frac{k \Delta R}{R(k+R)} \]\[ = 100 \frac{k}{k+R} \cdot \frac{\Delta R}{R} \]

Step 7: Conclude with the Relationship Between Percentage Errors

The percentage error of the growth rate is:\[ 100 \frac{\Delta f}{f} = 100 \frac{k}{k+R} \cdot \frac{\Delta R}{R} \]This shows that the percentage error in growth rate is a proportion of the percentage error in resource level, scaled by \( \frac{k}{k+R} \).

Key Concepts

Growth rate models

Growth rate models in biology, such as the one described in the problem, help us understand how different factors affect the growth of organisms. In our example, the growth rate of a plant is modeled based on the availability of a resource, like nitrogen, using the formula \( f(R) = a \frac{R}{k+R} \). Here, \( R \) is the resource level, and both \( a \) and \( k \) are constants. These constants allow for flexible modeling to fit various conditions or species.

Such models are crucial in predicting how plants respond to varying resource levels.

• If \( R \) is very small, \( f(R) \approx 0 \) since the plant lacks sufficient resources.
• When \( R \) is very large, \( f(R) \approx a \) indicating that the plant grows at its maximum rate, unaffected by additional resources.

Understanding these models enables biologists to make informed decisions about resource management and farming practices. By calculating and analyzing growth rates, farmers and ecologists can optimize conditions to increase yield or ensure sustainability.

Differentiation

Differentiation is a fundamental mathematical concept used in calculus to determine the rate of change of a variable. In the context of the growth rate model, differentiation helps us understand how the growth rate \( f(R) \) changes as the resource level \( R \) changes.

When differentiating the given formula \( f(R) = a \frac{R}{k+R} \), we derive \( \frac{df}{dR} = a \frac{k}{(k+R)^2} \). This derivative tells us how sensitive the growth rate is to changes in the resource level and is instrumental for calculating small adjustments in \( f(R) \) when \( R \) changes slightly.

The process of differentiation involves applying the quotient rule, which is a technique used to differentiate expressions that are ratios of two functions. This allows us to extract precise rates of change, important for theoretical and applied problems in biology, aiding in the interpretation of how minimal changes in resource availability affect overall growth.

Percentage error calculation

Percentage error calculation provides an understanding of how errors in measurement or estimation affect the results. In the given exercise, our task was to express the percentage error in growth rate \( 100 \frac{\Delta f}{f} \), in terms of the percentage error in the resource level \( 100 \frac{\Delta R}{R} \).

Starting from the derivative \( \Delta f \approx \frac{df}{dR} \Delta R \), and using the earlier formula for \( \frac{df}{dR} \), we arrive at the expression \[ 100 \frac{\Delta f}{f} = 100 \frac{k}{k+R} \cdot \frac{\Delta R}{R} \].This tells us that the percentage error in growth rate is a scaled version of the percentage error in the resource level. The scale factor \( \frac{k}{k+R} \) adjusts the sensitivity depending on the existing resource conditions.

• When the resource availability is low, the effect of a change is magnified because reactions tend to be more sensitive under scarcity.
• Conversely, for high resource levels, the sensitivity is reduced as the system might already be operating near its limits.

Understanding this relationship is vital in biology, ensuring precise adjustments and predictions in biological processes.