Problem 64 - Calculus for Biology and Medicine

Question

The following model is used in the fisheries literature to describe the recruitment of fish as a function of the size of the parent stock: If we denote the number of recruits by \(R\) and the size of the parent stock by \(P\), then $$ R(P)=a P e^{-b P}, \quad P \geq 0 $$ where \(a\) and \(b\) are positive constants. (a) Sketch the graph of the function \(R(P)\) when \(b=1\) and \(a=2\). (b) Differentiate \(R(P)\) with respect to \(P .\) (c) Find all the points on the curve that have a horizontal tangent.

Step-by-Step Solution

Answer

(a) Sketch von the curve \( R(P) = 2P e^{-P} \). (b) The derivative is \( R'(P) = 2e^{-P}(1-P) \). (c) The horizontal tangent occurs at \( P = 1 \).

Step 1: Understand the Function

The function given is \( R(P) = aP e^{-bP} \), where \( a = 2 \) and \( b = 1 \) for this exercise. This is a product of the linear term \( aP \) and the exponential decay term \( e^{-bP} \). For \( b = 1 \) and \( a = 2 \), the function becomes \( R(P) = 2P e^{-P} \).

Step 2: Sketch the Graph

To sketch the graph of \( R(P) = 2P e^{-P} \), observe the function behavior:- When \( P = 0 \), \( R(P) = 0 \).- As \( P \) increases from 0, \( R(P) \) initially increases, reaches a maximum, then decreases back towards zero because of the \( e^{-P} \) factor which dominates for large \( P \).- The function has a well-defined peak where the derivative equals zero.

Step 3: Differentiate the Function

Differentiate \( R(P) = 2P e^{-P} \) using the product rule:The product rule states: \( (uv)' = u'v + uv' \), where \( u = 2P \) and \( v = e^{-P} \).- \( u' = 2 \).- \( v' = -e^{-P} \).Substitute these into the product rule:\[ R'(P) = 2 e^{-P} + 2P (-e^{-P}) = 2e^{-P} - 2Pe^{-P} = 2e^{-P}(1 - P) \].

Step 4: Find Horizontal Tangents

Horizontal tangents occur where the derivative \( R'(P) = 0 \).From \( R'(P) = 2e^{-P}(1 - P) \) set the expression equal to zero:\[ 2e^{-P}(1 - P) = 0 \].Since \( e^{-P} \) is never zero for \( P \geq 0 \), the equation \( 1 - P = 0 \) gives \( P = 1 \).Thus, the curve has a horizontal tangent at \( P = 1 \).

Key Concepts

Differentiation

Differentiation is a key concept in calculus, used extensively in analyzing functions and curves. It tells us how a function changes at any given point, essentially measuring the rate of change or the "slope" of the function.
In this exercise, we're looking at the function described by the recruitment model: \[ R(P) = 2P e^{-P} \] To find the derivative of this function with respect to \( P \), we need to apply the product rule because the function is the product of two terms: \( 2P \) and \( e^{-P} \).
The product rule states that the derivative of a product of two functions \( u \, \text{and} \, v \) is \[ (uv)' = u'v + uv' \]
Here's how it applies to our function:

• Let \( u = 2P \) and \( v = e^{-P} \).
• Then the derivatives are \( u' = 2 \) and \( v' = -e^{-P} \).

Using the product rule, the differentiation gives us:\[ R'(P) = u'v + uv' = 2e^{-P} + 2P(-e^{-P}) = 2e^{-P}(1 - P) \].
The derivative \( R'(P) \) indicates the slope of the curve at any point \( P \). To find where the slope is zero, indicating horizontal tangents, we solve \( R'(P) = 0 \), leading us to \( P = 1 \).
This means, at \( P = 1 \), the curve of the recruitment model has a horizontal tangent.

Exponential Function

The exponential function appears frequently in various fields, including mathematics, physics, and environmental science. In this recruitment model, the exponential function comes in the form of \[ e^{-P} \], behaving as a "decay" factor.
The general form of an exponential function is \( e^{x} \), and here \( x = -P \), hence the negative power indicates exponential decay.
Features to note about exponential functions:

• Exponential functions grow or decay at rates proportional to their current value.
• As \( P \) becomes large, \( e^{-P} \) decreases to zero, which explains why \( R(P) \) declines for large \( P \) even if \( P \) itself becomes large.

This exponential decay essentially balances the linear increase due to the \( 2P \) term, resulting in a function \( R(P) \) which has an increase at first, reaches a peak, and then decays.
This is why the graph will initially rise as \( P \) increases, reach a maximal point where the derivative is zero, and then decline as \( e^{-P} \) becomes the dominant factor causing the value of \( R(P) \) to decrease.
Understanding the role of the exponential decay in this model helps us predict how the number of recruits will respond to changes in the parent stock size.

Fisheries Biology

Fisheries biology involves studying and managing fish populations to ensure sustainable fishing practices. In this context, recruitment models like the one discussed are vital for estimating how new fish enter into a fishery based on the population size of the parent stock.
The recruitment model given: \[ R(P) = aP e^{-bP} \], uses factors like the size of the parent stock \( P \), a scaling constant \( a \), and a decay constant \( b \) to estimate recruit numbers. Understanding this function:

• The term \( aP \) represents growth – as parent stock increases, more recruits are expected.
• The term \( e^{-bP} \) represents density-dependent factors – recruitment decreases when the parent stock is too large due to resource limitations or other environmental factors.

The peak of the graph, determined by differentiation, represents the optimal condition where recruitment is at its highest before declining. This insight is crucial for fisheries biologists in developing strategies to manage fish populations sustainably.
By understanding these dynamics, fisheries managers can set catch limits and other control measures to prevent overfishing, ensuring a balance between fishing industry needs and conservation efforts.