Problem 56 - Calculus Early Transcendentals

Question

Suppose that the growth of a population \(y=y(t)\) is given by the logistic equation $$ y=\frac{1000}{1+999 e^{-0.9 t}} $$ (a) What is the population at time \(t=0 ?\) (b) What is the carrying capacity \(L ?\) (c) What is the constant \(k ?\) (d) When does the population reach \(75 \%\) of the carrying capacity? (e) Find an initial-value problem whose solution is \(y(t)\)

Step-by-Step Solution

Answer

(a) 1; (b) 1000; (c) 0.9; (d) \(t \approx 6.374\); (e) \( \frac{dy}{dt} = 0.9y(1-\frac{y}{1000}),\ y(0)=1 \)."}

Step 1: Population at Time t=0

To find the population at time \(t=0\), substitute \(t=0\) into the given logistic equation: \[ y = \frac{1000}{1+999 e^{-0.9 \cdot 0}} = \frac{1000}{1+999} = \frac{1000}{1000} = 1.\]So, the population at \(t=0\) is \(1\).

Step 2: Determine the Carrying Capacity L

The carrying capacity \(L\) in a logistic equation of the form \[ y = \frac{L}{1+be^{-kt}} \] is the maximum value that \(y\) approaches as \(t\to \infty\). In this equation, \(L = 1000\).

Step 3: Identify the Growth Rate k

The constant \(k\) in the logistic equation represents the growth rate. In the given equation \[ y = \frac{1000}{1+999 e^{-0.9 t}}, \] the value of \(k\) is \(0.9\).

Step 4: Solve for the Time When Population is 75% of Carrying Capacity

75% of the carrying capacity \(L = 1000\) is \[ 0.75 \times 1000 = 750. \]Set \(y(t) = 750\) and solve for \(t\): \[ 750 = \frac{1000}{1+999 e^{-0.9 t}}. \]Rearranging, \[ 1+999 e^{-0.9 t} = \frac{1000}{750} = \frac{4}{3}. \]Therefore, \[ 999 e^{-0.9 t} = \frac{4}{3} - 1 = \frac{1}{3}. \]Divide by 999: \[ e^{-0.9 t} = \frac{1}{3 \times 999} = \frac{1}{2997}. \]Taking natural log on both sides, \[ -0.9 t = \ln \left( \frac{1}{2997} \right). \]\[ t = -\frac{1}{0.9} \ln \left( \frac{1}{2997} \right). \]Calculate to find \(t \approx 6.374\).

Step 5: Formulate the Initial-Value Problem

A logistic growth rate model is given by \[ \frac{dy}{dt} = ky\left(1-\frac{y}{L}\right), \]where \(k = 0.9\) and \(L = 1000\). With initial condition \(y(0) = 1\), the initial-value problem is: \[ \frac{dy}{dt} = 0.9y\left(1-\frac{y}{1000}\right), \quad y(0) = 1. \]

Key Concepts

Population Dynamics

Population dynamics focuses on how populations change over time. In the context of logistic growth, populations do not grow indefinitely. Instead, they follow a pattern where growth accelerates rapidly at first, then slows down as the population nears its maximum carrying capacity.
This S-shaped curve, called a logistic curve, represents the impeding limits set by environmental factors, available resources, and space constraints on population size.

• Initially, growth is exponential, driven by the abundance of resources.
• As the population reaches a significant portion of the carrying capacity, growth rate diminishes.
• The population size eventually stabilizes, fluctuating around the carrying capacity.

Understanding population dynamics is crucial to anticipating changes in species abundance and distribution, impacting ecological management and conservation efforts.

Carrying Capacity

Carrying capacity, often denoted as "L" in the logistic growth equation, is the maximum population size that an environment can sustain indefinitely.
It acts as an upper boundary, determined by the availability of resources such as food, habitat space, and water. For the logistic equation used in the problem, the carrying capacity is 1000.

• Beyond this limit, competition for resources leads to a balance where birth rates equal death rates.
• Environmental changes can alter the carrying capacity temporarily or permanently.
• Understanding carrying capacity is essential for managing sustainable wildlife populations and agricultural systems.

As a population approaches its carrying capacity, the growth rate slows down significantly, paving the way for population stability or decline, depending on resource availability.

Initial-Value Problem

An initial-value problem in the context of differential equations involves finding a function that satisfies both a differential equation and an initial condition.
For the logistic growth model provided, we seek a solution to the differential equation with a specific initial condition. Here it is: \[ \frac{dy}{dt} = 0.9y\left(1-\frac{y}{1000}\right), \ y(0) = 1. \]

• The differential equation models how the population changes at any point in time.
• The initial condition specifies the starting size of the population at time \(t=0\).
• Effective modeling requires both a well-defined equation and starting criteria.

Solving these problems allows us to determine future population behavior based on current conditions and provides insights into possible growth scenarios.

Growth Rate

The growth rate, denoted as "k" in the logistic growth equation, describes how quickly a population increases in size. In logistic models, it is part of the expression that adjusts the rate of growth as the population size changes.
From the exercise, the growth rate is given as 0.9.

• A higher growth rate implies faster initial growth but also quicker saturation near the carrying capacity.
• The growth rate influences how swiftly a population can rebound from declines.
• Environmental factors and biological characteristics strongly determine the growth rate of a population.

Monitoring the growth rate helps in forecasting population trends, assessing ecosystem health, and developing strategies for wildlife conservation and resource management.