Problem 64 - Calculus for Biology and Medicine

Question

Fish Recruitment Model 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 $$ R(P)=\alpha P e^{-\beta P}, \quad P \geq 0 $$ where \(\alpha\) and \(\beta\) are positive constants. (a) Sketch the graph of the function \(R(P)\) when \(\beta=1\) and \(\alpha=2\). (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 shows a hump at P = 1; (b) Derivative is \(2e^{-P}(1-P)\); (c) Horizontal tangent at (1, 0.736).

Step 1: Understand the Recruitment Function

The given model for recruitment is \( R(P) = \alpha P e^{-\beta P} \). For part (a), we need to consider \( R(P) \) with \( \alpha = 2 \) and \( \beta = 1 \). The function becomes \( R(P) = 2P e^{-P} \). This function describes how the number of recruits \( R \) changes with respect to the size of the parent stock \( P \).

Step 2: Sketch the Graph of the Function

To sketch \( R(P) = 2P e^{-P} \), first note that when \( P = 0 \), \( R(0) = 0 \). As \( P \) increases, \( R(P) \) initially increases and then decreases because of the \( e^{-P} \) term. Calculate a few values: \( R(0.5) \approx 0.61 \), \( R(1) = 0.736 \), \( R(2) = 0.541 \). Around \( P = 1 \), the function reaches its maximum and then decreases, forming a hump-like curve.

Step 3: Differentiate the Function

To find \( \frac{dR}{dP} \), use the product rule. The function is \( R(P) = 2P e^{-P} \). Let \( u = 2P \) and \( v = e^{-P} \). The derivatives are \( \frac{du}{dP} = 2 \) and \( \frac{dv}{dP} = -e^{-P} \). Apply the product rule: \( \frac{d}{dP}(uv) = u \frac{dv}{dP} + v \frac{du}{dP} \). This gives \( \frac{dR}{dP} = 2P(-e^{-P}) + e^{-P}(2) = 2e^{-P} - 2Pe^{-P} = 2e^{-P}(1-P) \).

Step 4: Find Horizontal Tangents

Horizontal tangents occur where \( \frac{dR}{dP} = 0 \). Setting \( 2e^{-P}(1-P) = 0 \), we find \( 2e^{-P} eq 0 \), so \( 1-P = 0 \). Solving \( 1-P = 0 \) gives \( P = 1 \). Thus, the point with a horizontal tangent is \( (1, R(1)) \). Calculate \( R(1) = 2 \cdot 1 \cdot e^{-1} = \frac{2}{e} \approx 0.736 \). So, the point is \( (1, 0.736) \).

Key Concepts

Differentiation in Calculus

Differentiation is a fundamental concept in calculus that deals with the rate at which a function changes. It is essentially the process of finding the derivative of a function. The derivative tells us how the function value changes as its input changes, which is mathematically expressed as \ \( \frac{d}{dx}f(x) \ \).
For the Fish Recruitment Model described in the exercise, we want to differentiate the function \ \( R(P) = 2P e^{-P} \ \). To do this, we employ the product rule of differentiation.The product rule states that for two functions \ \( u(x) \ \) and \ \( v(x) \ \), the derivative of their product \ \( uv \ \) is:

• \ \( \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \ \)

Applying this to our function, where \ \( u = 2P \ \) and \ \( v = e^{-P} \ \), we find:- \ \( \frac{du}{dP} = 2 \ \)- \ \( \frac{dv}{dP} = -e^{-P} \ \)
Thus, \ \( \frac{dR}{dP} = 2P(-e^{-P}) + e^{-P}(2) = 2e^{-P} - 2Pe^{-P} = 2e^{-P}(1-P) \ \).
This result helps find where the tangent to the curve is horizontal, crucial for understanding phenomena such as maximum recruitment.

Exponential Functions

Exponential functions are a type of mathematical function where the variable appears in the exponent. These functions are typically in the form \ \( a^x \ \) or in our case with the fish recruitment model, \ \( e^{-P} \ \), where \ \( e \ \) is Euler's number, approximately equal to 2.718.In the recruitment model \ \( R(P) = 2P e^{-P} \ \), the \ \( e^{-P} \ \) part is an exponential decay function. Exponential decay affects the growth patterns, often causing an initial increase followed by a slowdown and eventual decrease, as the \ \( e^{-P} \ \) term makes the function decrease at larger values of \ \( P \ \).Some key characteristics of exponential functions include:

• Their rapid increase or decrease as the variable changes.
• Their role in modeling growth and decay processes in real-world scenarios.

In this fish recruitment scenario, the exponential decay represents factors that limit recruitment as the parent stock size increases, leading to a peak or maximum value before the number of recruits starts to decline.

Horizontal Tangents

A horizontal tangent to the curve of a function occurs at a point where the slope of the tangent is zero. This happens where the derivative of the function equals zero.For the fish recruitment model, the derivative is \ \( \frac{dR}{dP} = 2e^{-P}(1-P) \ \). To find points with horizontal tangents, we set this derivative equal to zero:

• \ \( 2e^{-P}(1-P) = 0 \ \)

Since \ \( 2e^{-P} \ \) is never zero, we focus on \ \( 1-P = 0 \ \), leading to \ \( P = 1 \ \).
At \ \( P = 1 \ \), the point on the graph where the horizontal tangent exists is calculated using the original function:- \ \( R(1) = 2 \cdot 1 \cdot e^{-1} = \frac{2}{e} \approx 0.736 \ \)
Thus, the horizontal tangent is at point \ \( (1, 0.736) \ \). This critical point often represents a maximum or minimum in the context of the problem, informing us of the peak recruitment rate.