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Chapter 2: Problem 21

F. E. Smith has studied population growth for the water flea. \({ }^{35}\) Let \(N\) denote the population size. In one experiment, Smith found that \(G\), the rate of growth per day in the population, can be modeled by $$ G=\frac{0.44 N(228-N)}{228+3.46 N} $$ a. Draw a graph of \(G\) versus \(N\). Include values of \(N\) up to 350 . b. At what population level does the greatest rate of growth occur? c. There are two values of \(N\) where \(G\) is zero. Find these values of \(N\) and explain what is occurring at these population levels. d. What is the rate of population growth if the population size is 300 ? Explain what is happening to the population at this level.

Short Answer

Expert verified
The greatest rate of growth occurs near \( N = 114 \). Growth is zero at \( N = 0 \) and \( N = 228 \). At \( N = 300 \), the population is declining due to negative growth rate.

Step by step solution

01

Plot the Graph

First, we need to plot the graph of the function \( G = \frac{0.44 N(228-N)}{228+ 3.46 N} \) against \( N \) for values of \( N \) ranging from 0 to 350. Use a graphing tool or software that can handle rational functions efficiently. Analyze the shape of the graph to identify maximum or minimum points, intercepts, and asymptotic behavior.
02

Determine Maximum Rate of Growth

To find the population level where the greatest rate of growth occurs, we can look at the peak of the graph plotted in Step 1 or find the derivative of \( G \) with respect to \( N \), set it to zero, and solve for \( N \). This will give us the critical points, which can be further analyzed using the second derivative test to confirm maxima. The population level at this point is where the greatest rate of growth occurs.
03

Find Where Growth Rate is Zero

To find the values of \( N \) where \( G \) is zero, set the numerator in the formula \( 0.44 N(228-N) \) equal to zero and solve for \( N \). This gives us \( N = 0 \) and \( N = 228 \). These are the points where population growth halts. At \( N = 0 \), the population is extinct, and at \( N = 228 \), the population reaches carrying capacity.
04

Calculate Growth Rate at N = 300

Substitute \( N = 300 \) into the growth rate formula \( G = \frac{0.44 \cdot 300 (228 - 300)}{228 + 3.46 \cdot 300} \). Calculate the numerator and denominator separately, and then divide them to find \( G \). At \( N = 300 \), if \( G \) is negative, the population is declining at this level.

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Key Concepts

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

Understanding the Rate of Growth in Population Models
Population growth can often be understood by examining the rate of growth, generally represented as the function of population size. In our context, the rate of growth, denoted as \( G \), shows how fast the water flea population is increasing or decreasing per day. This concept helps us understand how swiftly a population changes at different sizes.
  • The rate of growth function \( G = \frac{0.44 N(228-N)}{228 + 3.46 N} \) is a rational function. It represents how the growth rate varies with the population size \( N \).
  • A positive \( G \) value indicates an increasing population, while a negative \( G \) signifies a declining one.
  • The rate of growth being zero means no change in the population size at that specific point.
To analyze \( G \), we plot its graph and observe its behavior across different population values. The shape and peak of the graph indicate how the rate changes, showing us critical dynamics of the population's behavior at various sizes.
Exploring the Concept of Carrying Capacity
The carrying capacity in a population model refers to the maximum population size that an environment can sustain indefinitely. It is a crucial concept that describes ecological limits.
  • In the given model, the carrying capacity is shown where the growth rate \( G \) is zero at \( N = 228 \).
  • This means that at \( N = 228 \), the environment is unable to support further growth of the water flea population, thus balancing the birth and death rates.
  • Beyond this point, any increase in population would result in scarcity of resources, preventing further population growth effectively.
Understanding carrying capacity helps us manage populations sustainably, ensuring that we don't overstep ecological limits, which is vital for conservation efforts.
Analyzing Critical Points for Maximum and Zero Growth
In any population growth model, determining the critical points is key to understanding the dynamics. Critical points are population sizes at which the rate of growth changes direction.
  • The critical points can be found where the derivative of the growth function \( G \) with respect to \( N \) equals zero.
  • For maximum growth, we look for the peak determined by setting the first derivative to zero and using the second derivative to confirm a maximum rather than a minimum.
  • The extreme points include when \( G = 0 \), such as at \( N = 0 \) (extinct population) and \( N = 228 \) (carrying capacity).
Identifying and analyzing critical points guides us in discovering optimal and limiting factors in population management plans.
The Role and Structure of Rational Functions
Rational functions are an essential component of understanding complex systems in mathematics and science. A rational function is the ratio of two polynomials, providing a robust tool for modeling various real-world phenomena.
  • In our model \( G = \frac{0.44 N(228-N)}{228 + 3.46 N} \), the function combines polynomial expressions for the numerator and the denominator, accounting for intricate population dynamics.
  • The structure of a rational function allows for phenomena such as asymptotes, which are values that the function approaches but never reaches.
  • This modeling method is ideal for showcasing relative growth rates, such as how carrying capacity impacts populations.
Understanding rational functions aids in deciphering complex biological and ecological interactions, allowing us to predict and respond to changes effectively.

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Most popular questions from this chapter

Ohm's law says that when electric current is flowing across a resistor, then the voltage \(v\), measured in volts, is the product of the current \(i\), measured in amperes, and the resistance \(R\), measured in ohms. That is, \(v=i R\). a. What is the voltage if the current is 20 amperes and the resistance is 15 ohms? b. Find a formula expressing resistance as a function of current and voltage. Use your function to find the resistance if the current is 15 amperes and the voltage is 12 volts. c. Find a formula expressing current as a function of voltage and resistance. Use your function to find the current if the voltage is 6 volts and the resistance is 8 ohms.

The background for this exercise can be found in Exercises 11, 12, 13, and \(14 \mathrm{in} \mathrm{Sec}\) tion 1.4. A manufacturer of widgets has fixed costs of \(\$ 700\) per month, and the variable cost is \(\$ 65\) per thousand widgets (so it costs \(\$ 65\) to produce 1 thousand widgets). Let \(N\) be the number, in thousands, of widgets produced in a month. a. Find a formula for the manufacturer's total cost \(C\) as a function of \(N\). b. The highest price \(p\), in dollars per thousand widgets, at which \(N\) can be sold is given by the formula \(p=75-0.02 \mathrm{~N}\). Using this, find a formula for the total revenue \(R\) as a function of \(N\). c. Use your answers to parts a and \(b\) to find \(a\) formula for the profit \(P\) of this manufacturer as a function of \(N\). d. Use your formula from part c to determine the two break-even points for this manufacturer. Assume that the manufacturer can produce at most 500 thousand widgets in a month.

You are hosting a convention for a charitable organization. You pay a rental fee of \(\$ 25,000\) for the convention center, plus you pay the caterer \(\$ 12\) for each person who attends the convention. Suppose you just want to break even. a. Use a formula to express the amount you should charge per ticket as a function of the number of people attending. Be sure to explain the meaning of the letters you choose and the units. b. Make a graph of the function that gives the amount you should charge per ticket. Include attendance sizes up to 1000 . c. How many must attend if you are to break even with a ticket price of \(\$ 50\) ?

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?

The resale value \(V\), in dollars, of a certain car is a function of the number of years \(t\) since the year 2008 . In the year 2008 the resale value is \(\$ 18,000\), and each year thereafter the resale value decreases by \(\$ 1700\). a. What is the resale value in the year 2009 ? b. Find a formula for \(V\) as a function of \(t\). c. Make a graph of \(V\) versus \(t\) covering the first 4 years since the year 2008 . d. Use functional notation to express the resale value in the year 2011 , and then calculate that value.
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