/*! 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 14 A breeding group of foxes is int... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 14

A breeding group of foxes is introduced into a protected area and exhibits logistic population growth. After \(t\) years the number of foxes is given by $$ N(t)=\frac{37.5}{0.25+0.76^{t}} \text { foxes } . $$ a. How many foxes were introduced into the protected area? b. Calculate \(N(5)\) and explain the meaning of the number you have calculated. c. Explain how the population varies with time. Include in your explanation the average rate of increase over the first 10-year period and the average rate of increase over the second 10-year period. d. Find the carrying capacity for foxes in the protected area. e. As we saw in the discussion of terminal velocity for a skydiver, the question of when the carrying capacity is reached may lead to an involved discussion. We ask the question differently. When is \(99 \%\) of carrying capacity reached?

Short Answer

Expert verified
a. 30 foxes; b. 74 foxes after 5 years; c. Average rates: 9.175 foxes/year (0-10), 2.62 foxes/year (10-20); d. 150 foxes; e. 99% reached at 25 years.

Step by step solution

01

Determine Initial Population

To find out how many foxes were introduced, calculate \( N(0) \).\[ N(0) = \frac{37.5}{0.25 + 0.76^0} = \frac{37.5}{1.25} \]Calculating this gives:\[ N(0) = 30 \text{ foxes} \]Therefore, 30 foxes were introduced.
02

Calculate the Population at 5 Years

We need to calculate \( N(5) \) to determine the number of foxes after 5 years:\[ N(5) = \frac{37.5}{0.25 + 0.76^5} \]First, calculate \( 0.76^5 \):\[ 0.76^5 \approx 0.2541 \]Substitute back into the equation:\[ N(5) = \frac{37.5}{0.25 + 0.2541} = \frac{37.5}{0.5041} \]Calculating this gives:\[ N(5) \approx 74.37 \text{ foxes} \]This means that approximately 74 foxes are present in the protected area after 5 years.
03

Describe Population Variation Over Time

To describe how the population varies, we need to look at how it changes over time. The logistic growth model implies the population will increase rapidly at first and then slow as it approaches the carrying capacity.To find the average rate of increase over the first 10 years \((0-10)\) and the second 10 years \((10-20)\):First, calculate \( N(10) \):\[ N(10) = \frac{37.5}{0.25 + 0.76^{10}} \]Calculating \( 0.76^{10} \approx 0.058 \) and substituting:\[ N(10) = \frac{37.5}{0.308} \approx 121.75 \text{ foxes} \]Average rate for the first 10 years:\[ \text{Average rate} = \frac{N(10) - N(0)}{10} = \frac{121.75 - 30}{10} = 9.175 \text{ foxes per year} \]For the second 10 years, calculate \( N(20) \):\[ N(20) = \frac{37.5}{0.25 + 0.76^{20}} \]Calculating \( 0.76^{20} \approx 0.0033 \) and substituting:\[ N(20) = \frac{37.5}{0.2533} \approx 147.96 \text{ foxes} \]Average rate for the second 10 years:\[ \text{Average rate} = \frac{N(20) - N(10)}{10} = \frac{147.96 - 121.75}{10} = 2.62 \text{ foxes per year} \]
04

Find the Carrying Capacity

The carrying capacity is the maximum population size that the environment can sustain. It can be determined by finding the limit of \( N(t) \) as \( t \rightarrow \infty \):\[ \lim_{t \to \infty} N(t) = \frac{37.5}{0.25 + 0.76^t} \]As \( t \to \infty \), \( 0.76^t \to 0 \), so:\[ \lim_{t \to \infty} N(t) = \frac{37.5}{0.25} = 150 \]Thus, the carrying capacity is 150 foxes.
05

Determine When 99% of the Carrying Capacity is Reached

To find when 99% of carrying capacity is reached, calculate when \( N(t) = 0.99 \times 150 \).Calculating 99% of 150:\[ 0.99 \times 150 = 148.5 \]Set up the equation:\[ \frac{37.5}{0.25 + 0.76^t} = 148.5 \]Cross-multiply to solve:\[ 37.5 = 148.5(0.25 + 0.76^t) \]Divide by 148.5:\[ 0.2525 = 0.25 + 0.76^t \]This simplifies to:\[ 0.76^t = 0.0025 \]Taking the logarithm of both sides:\[ t \log(0.76) = \log(0.0025) \]Solving for \( t \):\[ t = \frac{\log(0.0025)}{\log(0.76)} \approx 24.8 \]Thus, 99% of the carrying capacity is reached after approximately 25 years.

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Key Concepts

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

Carrying Capacity
Carrying capacity refers to the maximum number of individuals that can be supported by an environment without depleting resources. It is a crucial concept in understanding logistic population growth, like the scenario of foxes introduced into a protected area. When a population grows, it initially increases rapidly due to abundant resources. However, as the population approaches the carrying capacity, resources become limited. This leads to a slowdown in growth.In our exercise, the carrying capacity can be identified by evaluating the limit of the foxes' population formula, \( N(t) = \frac{37.5}{0.25+0.76^t} \), as time \( t \) approaches infinity. Here, the carrying capacity of the fox population is calculated to be 150. This means that at maximum, the environment can sustain 150 foxes without suffering resource depletion.It's also interesting to calculate what happens when the population reaches a significant percentage of the carrying capacity. In this exercise, reaching 99% of the carrying capacity requires finding when the population will be around 148.5 foxes. This underscores the importance of carrying capacity in planning resources and maintaining ecological balance.
Logarithmic Functions
Logarithmic functions are mathematical expressions involving logs, which are used to solve equations where variables appear as exponents. These functions are integral to understanding population dynamics, as seen in the logistic growth model of the fox population.When determining when 99% of the carrying capacity is reached, the resulting equation involves logs. Given that the formula is \( 37.5 = 148.5(0.25 + 0.76^t) \), it simplifies to \( 0.76^t = 0.0025 \). To solve for \( t \), we use logarithms:
  • Apply the log to both sides: \( \log(0.76^t) = \log(0.0025) \).
  • Using properties of logs, this becomes \( t \log(0.76) = \log(0.0025) \).
  • Solve for \( t \): \( t = \frac{\log(0.0025)}{\log(0.76)} \).
This type of calculation allows us to pinpoint more precise moments in population dynamics by expressing exponential growth in terms of more easily manageable linear functions.
Average Rate of Change
The average rate of change provides insight into how quickly a population grows over a specific time period. In our scenario, we examined the average rate of change in the fox population over two distinct ten-year periods.For the first ten years (0 to 10 years), the population grows from 30 to approximately 121.75 foxes. We calculate:
  • First Interval: Average rate = \( \frac{N(10) - N(0)}{10} = \frac{121.75 - 30}{10} = 9.175 \) foxes per year.
For the second ten years (10 to 20 years), the growth slows, reaching approximately 148.96 foxes by year 20. We find:
  • Second Interval: Average rate = \( \frac{N(20) - N(10)}{10} = \frac{147.96 - 121.75}{10} = 2.62 \) foxes per year.
The decline in the average rate of change over the second period suggests that the growth rate decreases as the population nears its carrying capacity. This reflects a core characteristic of logistic growth, where the environmental limits increasingly restrict the population expansion as it approaches the maximum ecological tolerance.

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Most popular questions from this chapter

In this exercise we develop a model for the growth rate \(G\), in thousands of dollars per year, in sales of a product as a function of the sales level \(s\), in thousands of dollars. \({ }^{30}\) The model assumes that there is a limit to the total amount of sales that can be attained. In this situation we use the term unattained sales for the difference between this limit and the current sales level. For example, if we expect sales to grow to 3 thousand dollars in the long run, then \(3-s\) gives the unattained sales. The model states that the growth rate \(G\) is proportional to the product of the sales level \(s\) and the unattained sales. Assume that the constant of proportionality is \(0.3\) and that the sales grow to 2 thousand dollars in the long run. a. Find a formula for unattained sales. b. Write an equation that shows the proportionality relation for \(G\). c. On the basis of the equation from part b, make a graph of \(G\) as a function of \(s\). d. At what sales level is the growth rate as large as possible? e. What is the largest possible growth rate?

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One class of models for population growth rates in marine fisheries assumes that the harvest from fishing is proportional to the population size. For one such model, we have $$ G=0.3 n\left(1-\frac{n}{2}\right)-0.1 n \text {. } $$ Here \(G\) is the growth rate of the population, in millions of tons of fish per year, and \(n\) is the population size, in millions of tons of fish. a. Make a graph of \(G\) versus \(n\). Include values of \(n\) up to \(1.5\) million tons. b. Use functional notation to express the growth rate if the population size is \(0.24\) million tons, and then calculate that value. c. Calculate \(G(1.42)\) and explain in practical terms what your answer means. d. At what population size is the growth rate the largest?
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