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Chapter 5: Problem 5

There are 3600 commercial bee hives in a region threatened by African bees. Today African bees have taken over 50 hives. Experience in other areas shows that, in the absence of limiting factors, the African bees will increase the number of hives they take over by \(30 \%\) each year. Make a logistic model that shows the number of hives taken over by African bees after \(t\) years, and determine how long it will be before 1800 hives are affected.

Short Answer

Expert verified
It will take approximately 10.42 years for 1800 hives to be affected.

Step by step solution

01

Understanding the Logistic Growth Model

The logistic growth model is used for populations in environments where resources are limited. The formula is \( P(t) = \frac{K}{1 + \frac{K - P_0}{P_0} \cdot e^{-rt}} \), where:- \( P(t) \) is the population at time \( t \).- \( K \) is the carrying capacity of the environment (3600 hives in this case).- \( P_0 \) is the initial population (50 hives currently taken over).- \( r \) is the intrinsic rate of increase (30% or 0.3 per year).This model will help us understand the growth of affected hives over time.
02

Assign Known Values

We are given that:- \( K = 3600 \) hives,- \( P_0 = 50 \) hives,- \( r = 0.3 \).
03

Set Up The Equation

The logistic growth equation becomes:\[ P(t) = \frac{3600}{1 + \frac{3600 - 50}{50} \cdot e^{-0.3t}} \]
04

Solve for 1800 Hives

We need to find \( t \) when \( P(t) = 1800 \). Substitute into the logistic equation:\[ 1800 = \frac{3600}{1 + \frac{3550}{50} \cdot e^{-0.3t}} \]
05

Simplify and Solve for \( t \)

Rearrange and solve for \( t \):- Divide both sides by 3600: \[ \frac{1}{2} = \frac{1}{1 + 71 \cdot e^{-0.3t}} \]- Cross-multiply to get: \[ 1 + 71 \cdot e^{-0.3t} = 2 \]- Rearrange: \[ 71 \cdot e^{-0.3t} = 1 \]- Solve for \( e^{-0.3t} \): \[ e^{-0.3t} = \frac{1}{71} \]- Take the natural logarithm: \[ -0.3t = \ln\left( \frac{1}{71} \right) \]- Calculate \( t \): \[ t = -\frac{1}{0.3} \ln\left( \frac{1}{71} \right) \approx 10.42 \] years.
06

Conclusion

Thus, it will take approximately 10.42 years for the African bees to take over 1800 hives.

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Key Concepts

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

Intrinsic Rate of Increase
The intrinsic rate of increase is a key concept in understanding population growth. It represents the potential growth rate of a population in ideal conditions, without the influence of limiting factors like food or space. In our exercise, the intrinsic rate of increase is 30% per year for the African bees. This means that if conditions are perfect, the number of hives they take over will grow by 30% annually. This rate is typically denoted by the symbol \( r \).

The intrinsic rate of increase is vital to predicting how fast a population can expand over time, especially within the logistic growth model. This rate provides a benchmark for the potential *explosive* or *rampant* growth a population could have if nothing were to limit its expansion.
Carrying Capacity
Carrying capacity defines the maximum population size an environment can sustainably support. For the African bees and the bee hives in this exercise, the carrying capacity is 3600 hives. This concept is crucial because it establishes a boundary for the logistic growth model.

As populations reach carrying capacity, growth rates slow down and stabilizes, contrasting the exponential growth observed when a population is far below this threshold. Whether it's food, space, or other resources, once these become limited, reaching carrying capacity becomes inevitable.
  • This concept helps us understand real-world limitations on population dynamics.
  • It adds realism to predictions made by the logistic growth model.
Population Dynamics
Population dynamics involves understanding how populations change over time, including growth, stability, decline, and the factors that influence these changes. In our scenario, it particularly describes how the number of bee hives dominated by African bees will evolve.

Logistic models incorporate various aspects of population dynamics by considering initial populations, growth rates, and carrying capacities. It focuses on how these variables interact over time to model realistic population scenarios. Observing such dynamics allows us to predict and possibly manage the future growth of a population, applying these insights to a range of ecological and conservation efforts.
Exponential Growth
Exponential growth refers to a growth pattern where a quantity, such as a population, increases by a constant percentage over equal time intervals. Initially, it yields rapid increases because the population base grows continuously. This phase can be seen in the early stages of population growth before resources start becoming limiting.

With the African bee problem, the exponential growth describes the period when the number of hives taken over increases rapidly each year. However, exponential growth cannot go on indefinitely due to natural constraints, which implies a shift to logistic growth as carrying capacity becomes a dominant factor. While initial growth follows an exponential path, understanding the transition to logistic growth is key to grasping complete population dynamics.

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Most popular questions from this chapter

The rate of growth \(G\), in thousands of dollars per year, in sales of a certain product is a function of the current sales level \(s\), in thousands of dollars, and the model uses a quadratic function: $$ G=1.2 s-0.3 s^{2} . $$ The model is valid up to a sales level of 4 thousand dollars. a. Draw a graph of \(G\) versus \(s\). b. Express using functional notation the rate of growth in sales at a sales level of \(\$ 2260\), and then estimate that value. c. At what sales level is the rate of growth in sales maximized?

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The following table shows the cost \(C\) of traffic accidents, in cents per vehicle-mile, as a function of vehicular speed \(s\), in miles per hour, for commercial vehicles driving at night on urban streets. \({ }^{65}\) $$ \begin{array}{|l|c|c|c|c|c|c|c|} \hline \text { Speed } s & 20 & 25 & 30 & 35 & 40 & 45 & 50 \\ \hline \text { Cost C } & 1.3 & 0.4 & 0.1 & 0.3 & 0.9 & 2.2 & 5.8 \\ \hline \end{array} $$ The rate of vehicular involvement in traffic accidents (per vehicle-mile) can be modeled \({ }^{66}\) as a quadratic function of vehicular speed \(s\), and the cost per vehicular involvement is roughly a linear function of \(s\), so we expect that \(C\) (the product of these two functions) can be modeled as a cubic function of \(s\). a. Use regression to find a cubic model for the data. (Keep two decimal places for the regression coefficients written in scientific notation.) b. Calculate \(C(42)\) and explain what your answer means in practical terms. c. At what speed is the cost of traffic accidents (for commercial vehicles driving at night on urban streets) at a minimum? (Consider speeds between 20 and 50 miles per hour.)

An exponential model of growth follows from the assumption that the yearly rate of change in a population is \((b-d) N\), where \(b\) is births per year, \(d\) is deaths per year, and \(N\) is current population. The increase is in fact to some degree probabilistic in nature. If we assume that population increase is normally distributed around \(r N\), where \(r=b-d\), then we can discuss the probability of extinction of a population. a. If the population begins with a single individual, then the probability of extinction by time \(t\) is given by $$ P(t)=\frac{d\left(e^{r t}-1\right)}{b e^{r t}-d} $$ If \(d=0.24\) and \(b=0.72\), what is the probability that this population will eventually become extinct? (Hint: The probability that the population will eventually become extinct is the limiting value for \(P\).) b. If the population starts with \(k\) individuals, then the probability of extinction by time \(t\) is $$ Q=P^{k}, $$ where \(P\) is the function in part a. Use function composition to obtain a formula for \(Q\) in terms of \(t, b, d, r\), and \(k\). c. If \(b>d\) (births greater than deaths), so that \(r>0\), then the formula obtained in part b can be rewritten as $$ Q=\left(\frac{d\left(1-a^{t}\right)}{b-d a^{t}}\right)^{k} $$ where \(a<1\). What is the probability that a population starting with \(k\) individuals will eventually become extinct? d. If \(b\) is twice as large as \(d\), what is the probability of eventual extinction if the population starts with \(k\) individuals? e. What is the limiting value of the expression you found in part \(\mathrm{d}\) as a function of \(k\) ? Explain what this means in practical terms.

Binary stars are pairs of stars that orbit each other. The period \(p\) of such a pair is the time, in years, required for a single orbit. The separation \(s\) between such a pair is measured in seconds of arc. The parallax angle \(a\) (also in seconds of arc) for any stellar object is the angle of its apparent movement as the Earth moves through one half of its orbit around the sun. Astronomers can calculate the total mass \(M\) of a binary system using $$ M=s^{3} a^{-3} p^{-2} . $$ Here \(M\) is the number of solar masses. a. Alpha Centauri, the nearest star to the sun, is in fact a binary star. The separation of the pair is \(s=17.6\) seconds of arc, its parallax angle is \(a=0.76\) second of arc, and the period of the pair is \(80.1\) years. What is the mass of the Alpha Centauri pair? b. How would the mass change if the separation angle were doubled but parallax and period remained the same as for the Alpha Centauri system? c. How would the mass change if the parallax angle were doubled but separation and period remained the same? d. How would the mass change if the period doubled but parallax angle and separation remained the same?
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