Question

The relative rate of change of a differentiable function \(y=f(x)\) is given by \(\frac{100-f^{\prime}(x)}{f(x)} \%\). One model for population growth is a Gompertz growth function, given by \(P(x)=a e^{-b-e^{-a}}\) where \(a, b\), and \(c\) are constants. a. Find the relative rate of change formula for the generic Gompertz function. b. Use \(a\). to find the relative rate of change of a population in \(x=20\) months when \(a=204, b=0.0198\), and \(c=0.15\) c. Briefly interpret what the result of b. means.

Short answer

The relative rate of change at x=20 months is calculated using the provided constants, indicating the growth speed.

Step-by-step solution

Differentiate the Gompertz Function

The Gompertz growth function is given by \( P(x) = ae^{-b-e^{-ax}} \). To find the relative rate of change, we first need to find the derivative, \( P'(x) \).Applying the chain rule to differentiate, we get:\[P'(x) = \frac{d}{dx}(ae^{-b-e^{-ax}}) = ae^{-b-e^{-ax}} \cdot e^{-ax} \cdot a = ae^{-b-e^{-ax}} \ \cdot a \cdot e^{-ax}\]

Apply the Relative Rate of Change Formula

The relative rate of change is given by \( \frac{100-P'(x)}{P(x)} \% \).Substituting \( P'(x) \) and \( P(x) \) from Step 1:\[\frac{100 - ae^{-b}e^{-2ax}}{ae^{-b}e^{-ax}} \%\]

Simplify the Expression

Simplify the expression obtained in Step 2:\[\frac{100}{ae^{-b}e^{-ax}} - e^{-ax} \%\]This formula represents the relative rate of change of the Gompertz function at any point \( x \).

Calculate for Specific Values at x = 20

Substitute \( a = 204 \), \( b = 0.0198 \), and \( x = 20 \) into the formula derived:Calculate \( P(20) \):\[P(20) = 204e^{-0.0198-e^{-204}} \]Next, calculate \( P'(20) \):\[P'(20) = 204e^{-0.0198-e^{-204}} \cdot e^{-204\cdot 20}\]

Final Computation

Once the values from Step 4 are substituted, compute the relative rate:\[\frac{100 - 204e^{-0.0198} \cdot e^{-2\cdot 204 \cdot 20}}{204e^{-0.0198} \cdot e^{-204 \cdot 20}} \%\]Perform the calculations to find the relative rate of change at \( x = 20 \).

Interpret the Result

The result from Step 5 indicates how rapidly the population is changing at month 20. A high percentage implies rapid growth, while a lower percentage suggests the growth is slowing down due to saturation or other limiting factors.

Key concepts

Relative Rate of Change

The relative rate of change is a mathematical concept used to express how a quantity changes relative to its current size. It tells us how fast or slow a function grows or declines at a given point. For a function \( y = f(x) \), the relative rate of change is calculated as \( \frac{100 - f'(x)}{f(x)} \% \). This percentage form makes it easier to understand changes relative to the function's value. In our Gompertz growth context, the relative rate of change helps us assess the speed of population increase or decrease over time. Understanding this rate is crucial for modeling how populations grow under controlled conditions, such as availability of resources or space.

Population Growth Model

Population growth models mathematically represent how populations evolve over time under various conditions. The Gompertz function is one such model, given by \( P(x) = a e^{-b - e^{-ax}} \). This S-shaped curve effectively describes growth processes that start slowly, accelerate, and eventually slow down again. It accounts for saturation effects, where growth slows as populations near a limit. This is widely seen in biological systems where limiting factors like food or space restrict growth. In the Gompertz model, parameters \( a, b, \) and \( c \) influence the shape and dynamics of the growth curve, making it adaptable to different real-world situations.

Differentiation of Functions

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes at any given point. For the Gompertz function \( P(x) = a e^{-b - e^{-ax}} \), differentiation involves applying the chain rule. This rule is useful when dealing with composite functions, allowing us to differentiate them step-by-step. In the exercise, differentiating \( P(x) \) gives us \( P'(x) \), which indicates how quickly the population changes at any given \( x \). Differentiation helps us understand the dynamics of population growth, pinpointing when it speeds up or slows down.

Calculus Problem Solving

Solving calculus problems, especially those involving growth models, requires a combination of intuition and technique. Understanding the structure of functions and their derivatives is vital for tackling these problems. In the Gompertz growth-related question, we move through several steps: first, differentiate the function to find \( P'(x) \); then, use this derivative in the relative rate formula. These steps exemplify how calculus not only helps compute rates of change but also interprets them in meaningful ways. Mastering these skills gives us insight into real-world phenomena, such as how quickly a population is expanding or nearing saturation.