/*! 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 6 Sprinkler irrigation in Nebraska... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 6

Sprinkler irrigation in Nebraska: Logistic growth can be used to model not only population growth but also economic and other types of growth. For example, the total number of acres \(A=A(t)\), in millions, in Nebraska that are being irrigated by modern sprinkler systems has shown approximate logistic growth since 1955 , closely following the equation of change $$ \frac{d A}{d t}=0.15 A\left(1-\frac{A}{3}\right) \text {. } $$ Here time \(t\) is measured in years. a. According to this model, how many total acres in Nebraska can be expected eventually to be irrigated by sprinkler systems? (Hint: This corresponds to the carrying capacity in the logistic model for population growth.) b. How many acres of land were under sprinkler irrigation when sprinkler irrigation was expanding at its most rapid rate?

Short Answer

Expert verified
a. 3 million acres; b. 1.5 million acres.

Step by step solution

01

Understanding Logistic Growth Model

The logistic growth model describes how a population grows rapidly at first and then slows as it approaches a maximum capacity. This is represented by the equation \(\frac{dA}{dt} = 0.15A\left(1 - \frac{A}{3}\right)\), where \(A\) is the number of acres (in millions) and \(t\) is time in years. The term \(1 - \frac{A}{K}\) represents the environmental carrying capacity \(K\).
02

Identifying the Carrying Capacity

In a logistic growth model, the carrying capacity is the maximum stable population size that can be sustained. In the given equation, \(K = 3\) millions of acres is the limiting factor, meaning this is the maximum number of acres that can be eventually irrigated by sprinklers.
03

Determining the Point of Maximum Growth

The rate \(\frac{dA}{dt}\) reaches its maximum when \(A = \frac{K}{2}\). This is derived by finding the critical point of \(\frac{dA}{dt} = 0.15A(1-\frac{A}{3})\). At maximum growth, \(A = \frac{3}{2} = 1.5\) million acres, which represents the number of acres irrigated when growth is most rapid.

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 91影视!

Key Concepts

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

Carrying Capacity
In the context of logistic growth, carrying capacity is a fundamental concept. It represents the upper limit that a population or any growing quantity can sustain in a given environment. For instance, the logistic model can describe not only population dynamics but also economic growth, such as the acreage used for irrigation. Think of carrying capacity as the maximum supportable level鈥攍ike how many people can fit in a room without overcrowding. This limit, in our specific scenario, is determined by resources and environmental constraints, such as water availability and infrastructure for irrigation. In the equation provided, \[ \frac{dA}{dt} = 0.15A \left(1 - \frac{A}{3}\right), \]The term \( K = 3 \) (in millions of acres) signifies that the environment can eventually support up to 3 million acres of sprinkler-irrigated land. Beyond this point, the growth rate slows and stabilizes, preventing the number of acres from exceeding this carrying capacity.
Population Growth
Population growth, often modeled using differential equations, generally undergoes different phases.
  • Initial Phase: Growth begins slowly.
  • Exponential Phase: Growth accelerates rapidly.
  • Stationary Phase: Growth rate decreases and stabilizes due to limiting factors.
The logistic growth model encapsulates these phases by showing how a growing population (or acreage of irrigated land, in this case) might initially expand quickly, only to slow as it nears its environmental limit, the carrying capacity. In our example of sprinkler irrigation expansion in Nebraska, the use of such a model helps illustrate not only the extent of expansion (up to 3 million acres) but also the grace of nature鈥檚 balance where resources dictate the ceiling of growth.
Differential Equations
Differential equations are mathematical tools used to describe changes in quantities and their rates over time. They allow us to encapsulate dynamic processes like population growth or the expansion of irrigated areas.The logistic growth model is a form of differential equation that incorporates both the current growth rate and the influence of existing limitations (i.e., carrying capacity). The given equation, \[ \frac{dA}{dt} = 0.15A \left(1 - \frac{A}{3}\right), \]describes the change in the number of acres (\( A \)) under irrigation over time (\( t \)). Here,
  • \( \frac{dA}{dt} \) refers to the rate of change in irrigated land.
  • \( 0.15A \) suggests initial exponential growth, driven by the existing number of acres.
  • \( 1 - \frac{A}{3} \) modifies growth, indicating how close the current acreage is to the carrying capacity of 3 million acres.
Through differential equations, we can model and predict how systems evolve over time, providing valuable insight into both natural and man-made systems.

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

Making up a story about a car trip: You begin from home on a car trip. Initially your velocity is a small positive number. Shortly after you leave, velocity decreases momentarily to zero. Then it increases rapidly to a large positive number and remains constant for this part of the trip. After a time, velocity decreases to zero and then changes to a large negative number. a. Make a graph of velocity for this trip. b. Discuss your distance from home during this driving event, and make a graph. c. Make up a driving story that matches this description.

The rock with a formula: If from ground level we toss a rock upward with a velocity of 30 feet per second, we can use elementary physics to show that the height in feet of the rock above the ground \(t\) seconds after the toss is given by \(S=30 t-16 t^{2}\). a. Use your calculator to plot the graph of \(S\) versus \(t\). b. How high does the rock go? c. When does it strike the ground? d. Sketch the graph of the velocity of the rock versus time.

Competition between bacteria: Suppose there are two types of bacteria, type \(A\) and type \(B\), in a place with limited resources. If the bacteria had unlimited resources, both types would grow exponentially, with exponential growth rates \(a\) for type \(A\) and \(b\) for type \(B\). Let \(P\) be the proportion of type \(A\) bacteria, so \(P\) is a number between 0 and 1. For example, if \(P=0\), then there are no type \(A\) bacteria and all are of type \(B\). If \(P=0.5\), then half of the bacteria are of type \(A\) and half are of type \(B\). Each population of bacteria grows in competition for the limited resources, so the proportion \(P\) changes over time. The function \(P\) is subject to the equation of change $$ \frac{d P}{d t}=(a-b) P(1-P) \text {. } $$ Suppose \(a=2.3\) and \(b=1.7\). a. Does \(P\) have a logistic equation of change? b. What happens to the populations if initially \(P=0\) ? c. What happens to \(P\) in the long run if \(P(0)\) is positive (but not zero)? In this case, does it matter what the exact value of \(P(0)\) is?

Chemical reactions: In a second-order reaction, one molecule of a substance \(A\) collides with one molecule of a substance \(B\) to produce a new substance, the product. If \(t\) denotes time and \(x=x(t)\) denotes the concentration of the product, then its rate of change \(\frac{d x}{d t}\) is called the rate of reaction. Suppose the initial concentration of \(A\) is \(a\) and the initial concentration of \(B\) is \(b\). Then, assuming a constant temperature, \(x\) satisfies the equation of change $$ \frac{d x}{d t}=k(a-x)(b-x) $$ for some constant \(k\). This is because the rate of reaction is proportional both to the amount of \(A\) that remains untransformed and to the amount of \(B\) that remains untransformed. Here we study a reaction between isobutyl bromide and sodium ethoxide in which \(k=0.0055, a=51\), and \(b=76\). The concentrations are in moles per cubic meter, and time is in seconds. 10 a. Write the equation of change for the reaction between isobutyl bromide and sodium ethoxide. b. Make a graph of \(\frac{d x}{d t}\) versus \(x\). Include a span of \(x=0\) to \(x=100\). c. Explain what can be expected to happen to the concentration of the product if the initial concentration of the product is 0 .

Water in a tank: Water is leaking out of a tank. The amount of water in the tank \(t\) minutes after it springs a leak is given by \(W(t)\) gallons. a. Explain what \(\frac{d W}{d t}\) means in practical terms. b. As water leaks out of the tank, is \(\frac{d W}{d t}\) positive or negative? c. For the first 10 minutes, water is leaking from the tank at a rate of 5 gallons per minute. What do you conclude about the nature of the function W during this period? d. After about 10 minutes, the hole in the tank suddenly gets larger, and water begins to leak out of the tank at 12 gallons per minute. i. Make a graph of W versus t. Be sure to incorporate linearity where it is appropriate. ii. Make a graph of dW dt versus t.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation
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.