/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Refer to Exercise \(20 .\) A pop... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Refer to Exercise \(20 .\) A population of 20 goldfish was introduced into a pond that has an estimated carrying capacity of 200 fish. After 1 month, the population of goldfish had grown to \(80 .\) If the pattern of growth of the population followed the Gompertz curve, how many goldfish were in the pond after 3 months?

Short Answer

Expert verified
After following the step-by-step solution, we find that the population of goldfish in the pond after 3 months, following the Gompertz curve, is: \[P(3) = 200 \cdot e^{-(\ln{10}) e^{-(3) \cdot \ln{\frac{-\ln\frac{2}{5}}{\ln{10}}}}}\]

Step by step solution

01

Identify the given information

We are given the following information: - Initial population (\(P_0\)) = 20 goldfish - Carrying capacity (\(K\)) = 200 goldfish - Population after 1 month (\(P(1)\)) = 80 goldfish - Time for which the population is asked (\(t\)) = 3 months
02

Set up equations to find parameters \(b\) and \(g\)

Use the provided information to formulate two Gompertz equations for the different time points: 1. Population at \(t = 0\): \(P(0) = 20 = 200 \cdot e^{-b e^{0 \cdot g}}\) 2. Population at \(t = 1\): \(P(1) = 80 = 200 \cdot e^{-b e^{-1 \cdot g}}\)
03

Solve for parameter \(b\) from the initial population equation

Using the first Gompertz equation from Step 2, we can solve for \(b\): \[20 = 200 \cdot e^{-b}\] \[b = - \ln{\frac{20}{200}} = \ln{10}\]
04

Solve for parameter \(g\) using parameter \(b\) and the population at \(t = 1\)

Substitute the value of \(b = \ln{10}\) in the second Gompertz equation from Step 2: \[80 = 200 \cdot e^{-(\ln{10})e^{-g}}\] Divide by 200: \[\frac{80}{200} = e^{-(\ln{10})e^{-g}}\] Now, take the natural logarithm of both sides: \[\ln{\frac{2}{5}} = -(\ln{10})e^{-g}\] Now, divide by \(-\ln{10}\) and invert both sides: \[e^{g} = \frac{-\ln\frac{2}{5}}{\ln{10}}\] Now, take the natural logarithm of both sides one more time to find \(g\): \[g = \ln{\frac{-\ln\frac{2}{5}}{\ln{10}}}\]
05

Find the population after 3 months using parameters \(b\) and \(g\)

Now that we have found \(b\) and \(g\), we can use them to find the population size at \(t = 3\) months. From the Gompertz equation: \[P(3) = 200 \cdot e^{-(\ln{10}) e^{-(3) \cdot \ln{\frac{-\ln\frac{2}{5}}{\ln{10}}}}}\] Plug in the values for \(b\) and \(g\) to evaluate the expression and find the population after 3 months.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carrying Capacity
Understanding the concept of carrying capacity is crucial when examining population dynamics. In ecological terms, carrying capacity refers to the maximum number of individuals or species that an environment can sustain indefinitely without being degraded. It signifies the upper limit of a population an environment can hold, which is determined by factors like food availability, habitat space, water, and other resources.

For instance, a pond can only support a limited number of fish before resources become too strained to support any more. In the case of our goldfish example, the pond's carrying capacity is estimated to be 200 fish. This does not mean that the population will always reach this number; various factors including competition, disease, and predation might prevent it from doing so. Also, carrying capacity can change over time due to environmental changes, such as a decrease in water quality or space.
Population Dynamics
Population dynamics examines how populations of organisms change over time and space, considering the growth rates, reproduction, death rates, and migration. It is a fundamental principle in ecology that helps scientists and researchers understand the complex interactions between species and their environments.

The study of population dynamics involves several models that predict population sizes under different conditions. The Gompertz growth model, used in our exercise, is a type of sigmoid function that describes growth as quick at first and then slowing down as the population approaches its carrying capacity. This model is particularly useful in situations where a population initially grows exponentially but its growth rate decreases as resource limitations become more significant.

Applying Population Dynamics to Goldfish

With our goldfish scenario, the initial introduction of fish leads to rapid population growth due to the lack of competition and ample resources. However, as the population grows, resources such as food and space become limited, slowing down the growth rate and leveling out as the carrying capacity is reached.
Exponential Growth
Exponential growth occurs when the growth rate of the value of a mathematical function (e.g., a population) is proportional to the function's current value. In simpler terms, it means that as the population grows, it does so faster and faster, assuming that resources are unlimited. The formula for exponential growth is often written as \( P(t) = P_0e^{rt} \) where \( P(t) \) is the population at time \( t \) , \( P_0 \) is the initial population, \( r \) is the growth rate, and \( e \) is Euler's number (approximately 2.71828).

In real-world ecosystems, though, unlimited growth is unsustainable, hence why models like Gompertz are used—they integrate a slowing of growth as the population nears the environment's carrying capacity. The initial stages of the Gompertz model mimic exponential growth, which is why we see the goldfish population leap from 20 to 80 in the first month. However, as the carrying capacity is neared, the growth slows down, and it's this change that differentiates models like Gompertz from pure exponential growth.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Von Bertalanffy Growth Model The von Bertalanffy growth model is used to predict the length of commercial fish. The model is described by the differential equation $$ \frac{d x}{d t}=k(L-x) $$ where \(x(t)\) is the length of the fish at time \(t, k\) is a positive constant called the von Bertalanffy growth rate, and \(L\) is the maximum length of the fish. a. Find \(x(t)\) given that the length of the fish at \(t=0\) is \(x_{0}\). b. At the time the larvae hatch, the North Sea haddock are about \(0.4 \mathrm{~cm}\) long, and the average haddock grows to a length of \(10 \mathrm{~cm}\) after 1 year. Find an expression for the length of the North Sea haddock at time \(t\). c. Plot the graph of \(x\). Take \(L=100(\mathrm{~cm})\). d. On average, the haddock that are caught today are between \(40 \mathrm{~cm}\) and \(60 \mathrm{~cm}\) long. What are the ages of the haddock that are caught?

Gompertz differential equation, a model for restricted population growth, is obtained by modifying the logistic differential equation and is given by $$ \frac{d P}{d t}=c P \ln \left(\frac{L}{P}\right) $$ where \(c\) is a constant and \(L\) is the carrying capacity of the environment. a. Find the equilibrium solution of the differential equation. b. Illustrate graphically the solutions of the equation with initial conditions \(P(0)=P_{0}\), where (i) \(P_{0}>L\), (ii) \(P_{0}=L\), and (iii) \(0

Spread of Disease A simple mathematical model in epidemiology for the spread of a disease assumes that the rate at which the disease spreads is jointly proportional to the number of infected people and the number of uninfected people. Suppose that there are a total of \(N\) people in the population, of whom \(N_{0}\) are infected initially. Show that the number of infected people after \(t\) weeks, \(x(t)\), is given by $$ x(t)=\frac{N}{1+\left(\frac{N-N_{0}}{N_{0}}\right) e^{-k N t}} $$ where \(k\) is a positive constant.

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. If \(y_{1}\) is a solution of the homogeneous equation \(y^{\prime}+P y=0\) associated with the nonhomogeneous equation \(y^{\prime}+P y=f\) and \(y_{2}\) is a solution of the nonhomogeneous equation, then \(y=c y_{1}+y_{2}\) is a solution of the nonhomogeneous equation, where \(c\) is any constant.

Lambert's Law of Absorption According to Lambert's Law of Absorption, the percentage of incident light \(L\), absorbed in passing through a thin layer of material \(x\), is proportional to the thickness of the material. For a certain material, if \(\frac{1}{2}\) in. of the material reduces the light to half of its intensity, how much additional material is needed to reduce the intensity to one fourth of its initial value?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.