/*! 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 19 The amount of growth of plants i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The amount of growth of plants in an ungrazed pasture is a function of the amount of plant biomass already present and the amount of rainfall. \({ }^{18}\) For a pasture in the arid zone of Australia, the formula $$ \begin{aligned} Y=&-55.12-0.01535 N \\ &-0.00056 N^{2}+3.946 R \end{aligned} $$ gives an approximation of the growth. Here \(R\) is the amount of rainfall (in millimeters) over a 3-month period, \(N\) is the plant biomass (in kilograms per hectare) at the beginning of that period, and \(Y\) is the growth (in kilograms per hectare) of the biomass over that period. (For comparison, 100 millimeters is about \(3.9\) inches, and 100 kilograms per hectare is about 89 pounds per acre.) a. Solve Equation (2.6) for \(R\). b. Ecologists are interested in the relationship between the amount of rainfall and the initial plant biomass if there is to be no plant growth over the period. Put \(Y=0\) in the equation you found in part a to get a formula for \(R\) in terms of \(N\) that describes this relationship. c. Use the formula you found in part b to make a graph of \(R\) versus \(N\) (again with \(Y=0\) ). Include values of \(N\) from 0 to 800 kilograms per hectare. This graph is called the isocline for zero growth. It shows the amount of rainfall needed over the 3 -month period just to maintain a given initial plant biomass. d. With regard to the isocline for zero growth that you found in part c, what happens to \(R\) as \(N\) increases? Explain your answer in practical terms. e. How much rainfall is needed just to maintain the initial plant biomass if that biomass is 400 kilograms per hectare? f. A point below the zero isocline graph corresponds to having less rainfall than is needed to sustain the given initial plant biomass, and in this situation the plants will die back. A point above the zero isocline graph corresponds to having more rainfall than is needed to sustain the given initial plant biomass, and in this situation the plants will grow. If the initial plant biomass is 500 kilograms per hectare and there are \(40 \mathrm{mil}-\) limeters of rain, what will happen to the plant biomass over the period?

Short Answer

Expert verified
No growth. 38.22 mm of rain maintains 400 kg/ha. 500 kg/ha with 40 mm results in decrease.

Step by step solution

01

Solve the Equation for R

Starting with the given equation for plant growth: \[ Y = -55.12 - 0.01535N - 0.00056N^2 + 3.946R \] we need to solve for \( R \). Rearrange the equation: \[ 3.946R = Y + 55.12 + 0.01535N + 0.00056N^2 \] Divide both sides by 3.946 to isolate \( R \): \[ R = \frac{Y + 55.12 + 0.01535N + 0.00056N^2}{3.946} \] This equation expresses \( R \) in terms of \( Y \) and \( N \).
02

Set Y to Zero for Zero Growth Isocline

Now, we are interested in the situation where there is no growth, meaning \( Y = 0 \). Substitute \( Y = 0 \) into the equation from Step 1: \[ R = \frac{0 + 55.12 + 0.01535N + 0.00056N^2}{3.946} \] Simplify to obtain \( R \) in terms of \( N \): \[ R = \frac{55.12 + 0.01535N + 0.00056N^2}{3.946} \] This is the formula for \( R \) when there is no plant growth.
03

Graph R vs. N for Zero Growth Isocline

Use the equation \( R = \frac{55.12 + 0.01535N + 0.00056N^2}{3.946} \) to plot \( R \) versus \( N \) for values of \( N \) from 0 to 800 kilograms per hectare. This graph shows the required rainfall to maintain plant biomass without growth.
04

Analyze the Effect of Increasing N on R

Examine the formula \( R = \frac{55.12 + 0.01535N + 0.00056N^2}{3.946} \). As \( N \) increases, the terms \( 0.01535N \) and \( 0.00056N^2 \) both increase, causing \( R \) to increase as well. In practical terms, more initial biomass requires more rainfall to maintain it without growth.
05

Calculate Rainfall for N = 400

Insert \( N = 400 \) into the equation \( R = \frac{55.12 + 0.01535N + 0.00056N^2}{3.946} \): \[ R = \frac{55.12 + 0.01535(400) + 0.00056(400)^2}{3.946} \] Simplifying: \[ R = \frac{55.12 + 6.14 + 89.6}{3.946} = \frac{150.86}{3.946} \] Compute the result: \( R \approx 38.22 \) millimeters. Thus, about 38.22 millimeters of rain are needed.
06

Determine Outcome for Given Scenario

With an initial biomass of 500 kg/hectare and 40 mm of rain, use the calculated isocline equation to find the needed rainfall: \[ R = \frac{55.12 + 0.01535(500) + 0.00056(500)^2}{3.946} \] Simplifying, \[ R = \frac{200.87}{3.946} \approx 50.89 \] millimeters are needed. Since 40 mm is less than 50.89 mm, the plant biomass will decrease.

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.

Plant Biomass
Plant biomass refers to the total mass of living plants in a particular area, usually measured in kilograms per hectare. To understand plant growth, we first need to comprehend how plant biomass functions. In an ungrazed pasture, biomass is critical because it provides the starting point for potential growth over a given period. Initial plant biomass, symbolized as \(N\), is a significant determinant in models predicting plant growth. The formula used in the exercise showcases how the amount of biomass influences growth calculations. As biomass increases, certain elements of the formula adjust, specifically through terms like \(0.01535N\) and \(0.00056N^2\). Recognizing the significance of these terms helps us understand that larger biomass requires adjustments in resources like nutrients and water to sustain or boost growth. Understanding plant biomass is essential for ecosystem management, as it affects not only plant development but also the environment, influencing soil conditions and habitat support.
Rainfall Impact
Rainfall plays a vital role in determining plant growth, especially in arid zones like parts of Australia. In the context of our exercise, rainfall, denoted by \(R\), is one of the key variables influencing growth outcomes. The formula used demonstrates that increased rainfall usually promotes better plant growth but only up to a certain point.The coefficient associated with rainfall in the formula, \(3.946R\), indicates how significantly rainfall impacts growth compared to other factors. This is particularly useful when assessing the sustainability of plant populations in dry regions. By understanding the necessary rainfall for maintaining or increasing biomass, farmers and ecologists can make more informed decisions on water usage and conservation strategies.Where rainfall is less than required, growth stalls or declines, posing risks to the local ecosystem and agricultural productivity. Conversely, optimizing rainfall impact through irrigation or predicting weather patterns enhances resilience against periods of drought.
Mathematical Modeling
Mathematical modeling is a powerful tool used to represent complex biological processes through equations and formulas. In our exercise, the model provided is a function that describes plant growth based on two primary variables: initial plant biomass and rainfall. Creating mathematical models begins with identifying critical factors that influence the system. In this case, the choice of \(N\) and \(R\) allows us to establish a relationship between these variables and plant growth \(Y\). The terms in the formula, such as \(-0.01535N\) and \(3.946R\), were derived to quantify the impact each component has on the outcome. Given the nonlinear nature of biological systems, quadratic terms like \(0.00056N^2\) are included to represent how factors might change over different levels of biomass.These models are invaluable because they allow predictions and analyses of scenarios under varying conditions. When refined and validated, such models can guide decisions in agriculture, conservation, and resource management, helping to achieve sustainability goals.
Isocline Graphing
Isocline graphing is a mathematical technique used to analyze conditions required to maintain equilibrium, such as zero growth in plant biomass. In our example, the isocline is a two-dimensional graph plotting \(R\) versus \(N\) for cases where \(Y = 0\). This graph highlights the conditions under which the system remains stable without any net increase or decrease in biomass.To create this graph, we solve the equation for \(R\) when \(Y = 0\), resulting in a plot that helps visualize how much rainfall is needed to sustain a given biomass level without change. The relationship shows that as \(N\) increases, \(R\) must also increase, reflecting the need for more resources to maintain larger biomass.Graphs like these are essential for understanding resource needs and planning for sustainable management practices. They inform stakeholders on how to balance growth and resource allocation, ultimately aiding in maintaining ecological balance and operational efficiency in agricultural practices.

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

F. E. Smith has reported on population growth of the water flea. \({ }^{24}\) In one experiment, he found that the time \(t\), in days, required to reach a population of \(N\) is given by the relation $$ e^{0.44 t}=\frac{N}{N_{0}}\left(\frac{228-N_{0}}{228-N}\right)^{4.46} $$ Here \(N_{0}\) is the initial population size. If the initial population size is 50 , how long is required for the population to grow to 125 ?

An important model for commercial fisheries is that of Beverton and Holt. \({ }^{31}\) It begins with the study of a single cohort of fishthat is, all the fish in the study are born at the same time. For a cohort of the North Sea plaice (a type of flatfish), the number \(N=N(t)\) of fish in the population is given by $$ N=1000 e^{-0.1 t} $$ and the weight \(w=w(t)\) of each fish is given by $$ w=6.32\left(1-0.93 e^{-0.095 t}\right)^{3} $$ Here \(w\) is measured in pounds and \(t\) in years. The variable \(t\) measures the so-called recruitment age, which we refer to simply as the age. The biomass \(B=B(t)\) of the fish cohort is defined to be the total weight of the cohort, so it is obtained by multiplying the population size by the weight of a fish. a. If a plaice weighs 3 pounds, how old is it? b. Use the formulas for \(N\) and \(w\) given above to find a formula for \(B=B(t)\), and then make a graph of \(B\) against \(t\). (Include ages through 20 years.) c. At what age is the biomass the largest? d. In practice, fish below a certain size can't be caught, so the biomass function becomes relevant only at a certain age. i. Suppose we want to harvest the plaice population at the largest biomass possible, but a plaice has to weigh 3 pounds before we can catch it. At what age should we harvest? ii. Work part i under the assumption that we can catch plaice weighing at least 2 pounds.

The gross domestic product is used by economists as a measure of the nation's economic position. The gross domestic product \(P\) is calculated as the sum of personal consumption expenditures \(C\), gross private domestic investment \(I\), government consumption expenditures and gross investment \(G\), and net exports \(E\) of goods and services (exports minus imports). All of these are measured in billions of dollars. a. Find a formula that gives the gross domestic product \(P\) in terms of \(C, I, G\), and \(E\). b. In 2002, personal consumption expenditures was \(7303.7\) billion dollars, gross private domestic investment was \(1593.2\) billion dollars, government consumption expenditures and gross investment was \(1972.9\) billion dollars, and net exports of goods and services was \(-423.6\) billion dollars. i. Which was larger in 2002 , U.S. imports or exports? ii. Calculate the gross domestic product for \(2002 .\) c. Solve the equation in part a for \(E\). d. Suppose that in another year the values for \(C, I\), and \(G\) remain the same as the 2002 figures, but the gross domestic product is 10,886 billion dollars. What is the net export of goods and services for this year?

Two friends enjoy competing with each other to see who has the best time in running a mile. Initially (before they ever raced each other), the first friend runs a mile in 7 minutes, and for each race that they run, his time decreases by 13 seconds. Initially, the second friend runs a mile in 7 minutes and 20 seconds, and for each race that they run, his time decreases by 16 seconds. Which will be the first race in which the second friend beats the first?

In fish management it is important to know the relationship between the abundance of the spawners (also called the parent stock) and the abundance of the recruits - that is, those hatchlings surviving to maturity. \({ }^{32}\) According to the Ricker model, the number of recruits \(R\) as a function of the number of spawners \(P\) has the form $$ R=A P e^{-B P} $$ for some positive constants \(A\) and \(B\). This model describes well a phenomenon observed in some fisheries: A large spawning group can actually lead to a small group of recruits. \({ }^{33}\) In a study of the sockeye salmon, it was determined that \(A=4\) and \(B=0.7\). Here we measure \(P\) and \(R\) in thousands of salmon. a. Make a graph of \(R\) against \(P\) for the sockeye salmon. (Assume there are at most 3000 spawners.) b. Find the maximum number of salmon recruits possible. c. If the number of recruits \(R\) is greater than the number of spawners \(P\), then the difference \(R-P\) of the recruits can be removed by fishing, and next season there will once again be \(P\) spawners surviving to renew the cycle. What value of \(P\) gives the maximum value of \(R-P\), the number of fish available for removal by fishing?

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.