Problem 5 - Calculus: Early Transcendentals

Question

The Pacific halibut fishery has been modeled by the differential equation \( \frac {dy}{dt} = ky (1 - \frac {y}{M}) \) where \( y(t) \) is the biomass (the total mass of the members of the population) in kilograms at time \( t \) (measured in years), the carrying capacity is estimated to be \( M = 8 \times 10^7 \) kg, and \( k = 0.71 \) per year. (a) If \( y(0) = 2 \times 10^7 \) kg, find the biomass a year later. (b) How long will it take for the biomass to reach \( 4 \times 10^7 \) kg?

Step-by-Step Solution

Answer

(a) Biomass after one year: \( 3.24 \times 10^7 \) kg. (b) Time to reach \( 4 \times 10^7 \) kg: approximately 1.548 years.

Step 1: Set Up the Logistic Growth Equation

Given the differential equation \( \frac{dy}{dt} = ky(1 - \frac{y}{M}) \), we need to solve this for the given conditions. Here, we are given \( M = 8 \times 10^7 \) kg, and \( k = 0.71 \). The initial condition is \( y(0) = 2 \times 10^7 \).

Step 2: Solving the Differential Equation

The equation is a logistic growth model. We know the general solution is \( y(t) = \frac{M}{1 + Ae^{-kt}} \). We need to find the constant \( A \) using the initial condition \( y(0) = 2 \times 10^7 \).

Step 3: Apply Initial Condition

Substitute \( t = 0 \) and \( y(0) = 2 \times 10^7 \) into the equation: \[ 2 \times 10^7 = \frac{8 \times 10^7}{1 + A} \]Solving this, we get \( 1 + A = 4 \), so \( A = 3 \).

Step 4: Write the Specific Solution

The specific solution becomes \[ y(t) = \frac{8 \times 10^7}{1 + 3e^{-0.71t}} \].

Step 5: Calculate Biomass After One Year

Substitute \( t = 1 \) into the specific solution: \[ y(1) = \frac{8 \times 10^7}{1 + 3e^{-0.71 \times 1}} \].Calculate the exponent value and evaluate to find the biomass after 1 year.

Step 6: Evaluate Exponential and Compute Mass

Compute \( e^{-0.71} \approx 0.49 \), thus \[ y(1) = \frac{8 \times 10^7}{1 + 3 \times 0.49} = \frac{8 \times 10^7}{2.47} \approx 3.24 \times 10^7 \text{ kg} \].

Step 7: Determine Time for Biomass to Reach Threshold

Set \( y(t) = 4 \times 10^7 \), and solve for \( t \):\[ 4 \times 10^7 = \frac{8 \times 10^7}{1 + 3e^{-0.71t}} \].Simplify to give \( 1 + 3e^{-0.71t} = 2 \), leading to \( 3e^{-0.71t} = 1 \), and \( e^{-0.71t} = \frac{1}{3} \).

Step 8: Solve for Time \( t \)

Taking the natural logarithm:\[ -0.71t = \, \ln \left( \frac{1}{3} \right) \approx -1.0986 \].Solving gives: \[ t = \frac{1.0986}{0.71} \approx 1.548 \text{ years} \].

Key Concepts

Differential Equation

A differential equation is a mathematical equation that involves functions and their derivatives. In the context of the Logistic Growth Model, the differential equation we are dealing with is:

• \( \frac{dy}{dt} = ky \left( 1 - \frac{y}{M} \right) \)

This equation describes how the biomass of a population, denoted by \( y(t) \), changes over time. Here, \( dy/dt \) represents the rate of change of biomass with respect to time.
The parameter \( k \) is a constant that affects the rate of growth. This type of equation is known as a "logistic differential equation," which models growth that is initially exponential but slows down as the population approaches a maximum size. This represents a more realistic model for growth in constrained environments.

Carrying Capacity

Carrying capacity is a core concept in population biology and ecology that refers to the maximum population size that an environment can sustainably support. In the logistic growth differential equation, carrying capacity is represented by the symbol \( M \).
In our example, the carrying capacity is given as \( M = 8 \times 10^7 \) kg. This value indicates the upper limit of the biomass that can be supported by the environment.

• The carrying capacity is essential because it influences how fast the population grows and stabilizes.
• When the population size \( y \) is much smaller than \( M \), growth is approximately exponential.
• As \( y \) approaches \( M \), growth slows due to limited resources and other ecological factors.

Exponential Growth

Exponential growth occurs when the rate of population increase is proportional to the existing population. In our logistic growth model, this behavior is most evident at the beginning of population growth.
Initially, when a population is far from its carrying capacity \( M \), the term \( 1 - \frac{y}{M} \) approaches 1, leading the differential equation to approximate exponential growth:

• \( \frac{dy}{dt} \approx ky \)

In this scenario, the population grows at a constant proportional rate \( k \), which in the exercise, is 0.71 per year. However, exponential growth cannot continue indefinitely, as it would soon outstrip available resources.

Initial Condition

Initial conditions in differential equations help us determine the particular solution for a given scenario. An initial condition provides specific information about the state of a system at the beginning of observation. In this exercise, the initial condition is \( y(0) = 2 \times 10^7 \) kg.

• This means that at time \( t = 0 \), the biomass of the population starts at \( 2 \times 10^7 \) kg.
• Using this initial condition helps to calculate the constant \( A \) in the logistic growth solution:

The general solution for a logistic growth equation is given as:

• \( y(t) = \frac{M}{1 + Ae^{-kt}} \)

By substituting the initial condition, we find \( A = 3 \), resulting in a specific solution that helps predict the population at any given future time.