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11\. A population of bighorn sheep: There is an effort in Colorado to restore the population of bighorn sheep. Let \(N=N(t)\) denote the number of sheep in a certain protected area at time \(t\). a. Explain the meaning of \(\frac{d N}{d t}\) in practical terms. b. A small breeding population of bighorn sheep is initially introduced into the protected area. Food is plentiful and conditions are generally favorable for bighorn sheep. What would you expect to be true about the sign of \(\frac{d N}{d t}\) during this period? c. This summer a number of dead sheep were discovered, and all were infected with a disease that is known to spread rapidly among bighorn sheep and is nearly always fatal. How would you expect an unchecked spread of this disease to affect \(\frac{d N}{d t}\) ? d. If the reintroduction program goes well, then the population of bighorn sheep will grow to the size the available food supply can support and will remain at about that same level. What would you expect to be true of \(\frac{d N}{d t}\) when this happens?

Short Answer

Expert verified
\( \frac{dN}{dt} \) describes the rate of population change; expect positive during growth, negative with disease, and zero at equilibrium.

Step by step solution

01

Understanding the Concept of Derivative

The expression \( \frac{dN}{dt} \) represents the rate of change of the population \( N \) with respect to time \( t \). In practical terms, it tells us how the population of bighorn sheep is increasing or decreasing over time.
02

Predicting Population Growth

When a small breeding population is introduced into a favorable environment, we expect the population to grow due to sufficient food and good conditions. Consequently, \( \frac{dN}{dt} \) will be positive during this period, indicating an increase in the population.
03

Considering Impact of Disease

The discovery of a fatal and rapidly spreading disease among the sheep would likely lead to population decline. Therefore, \( \frac{dN}{dt} \) would become negative as the disease reduces the number of sheep over time.
04

Reaching Carrying Capacity

Once the population reaches the size that the food supply can support, the population growth will stabilize. At this equilibrium point, \( \frac{dN}{dt} \) would be zero, indicating no net change in population size as births are balanced by deaths and other losses.

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Key Concepts

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

Derivative
In the context of population dynamics, the derivative \( \frac{dN}{dt} \) is a powerful tool to measure how the population of bighorn sheep changes over time. This concept originates from calculus and helps us calculate the rate at which the population size changes. Essentially, it tells us whether the population is increasing, decreasing, or staying the same at any given moment.

A positive derivative indicates that the population is growing because more sheep are being added than removed at any given time. Conversely, a negative derivative suggests a decline in the population, revealing that more sheep are leaving the population than joining. If the derivative is zero, it reflects a stable population with no net change, meaning the number of sheep being born equals those dying. This dynamic view provides a clear snapshot of the population's health and prospects.

Understanding the derivative in practical terms is crucial for assessing changes in populations over time, recognizing trends, and helping establish conservation or management strategies.
Rate of Change
The rate of change in population dynamics, represented by \( \frac{dN}{dt} \), helps us interpret how quickly the number of sheep increases or decreases over time. It is like pressing fast forward, normal speed, or rewind on a video player of the population's timeline. By focusing on the rate of change, we can draw meaningful forecasts on population trends.

For instance, if the \( \frac{dN}{dt} \) is a high positive number, this means the sheep population is increasing rapidly, possibly due to favorable conditions such as abundant food or ideal weather. Conversely, a high negative rate signals rapid population decline, potentially due to external factors like diseases or food scarcity.

Examining the rate of change is critical to any conservation effort, as it enables predictions about population stability or threats along with necessary interventions before critical levels are reached.
Carrying Capacity
Carrying capacity is a significant concept that defines the maximum population size that the environment can sustainably support. When the population of bighorn sheep reaches this level, it has reached its carrying capacity for that habitat. At this stage, the birth and death rates align, resulting in the derivative \( \frac{dN}{dt} \) becoming zero.

Reaching the carrying capacity means that resources such as food, water, and space have reached their limit in supporting the population size. In this balanced state, any additional sheep born must be offset by an equivalent number leaving or dying due to factors like starvation or natural predators.

Understanding carrying capacity is essential in the management of wildlife populations, as exceeding this limit can lead to resource depletion and could potentially drive the population into decline due to famine and poor health.
Disease Impact
Disease impact on wildlife populations, especially those like bighorn sheep, can be profound. A disease that spreads rapidly and is fatal, as described in the problem statement, poses a grave threat to population stability. With the outbreak of such a disease, \( \frac{dN}{dt} \) would likely turn negative.

This signifies a population decrease, as the rate at which sheep die due to disease outpaces the birth and survival rate. Certain diseases can spread rapidly within a population, particularly when the individuals are in close proximity or when the population density is high.
  • Rapid spread of disease implies a sharp decline in population.
  • Monitoring health and disease can help in implementing control measures.
While disease is a natural part of ecosystems, understanding its impact helps in establishing management strategies. Vaccination, isolating infected individuals, or reducing population density may help mitigate this impact and stabilize the population. Managing disease effectively ensures the long-term sustainability of the bighorn sheep population.

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Most popular questions from this chapter

An investment: You open an account by investing \(\$ 250\) with a financial institution that advertises an APR of \(5.25 \%\), with continuous compounding. What account balance would you expect 1 year after making your initial investment?

Growing child: A certain girl grew steadily between the ages of 3 and 12 years, gaining \(5 \frac{1}{2}\) pounds each year. Let \(W\) be the girl's weight, in pounds, as a function of her age \(t\), in years, between the ages of \(t=3\) and \(t=12\). a. Is \(W\) a linear function or an exponential function? Be sure to explain your reasoning. b. Write an equation of change for \(W\). c. Given that the girl weighed 30 pounds at age 3 , find a formula for \(W\).

Looking up: The constant \(g=32\) feet per second per second is the downward acceleration due to gravity near the surface of the Earth. If we stand on the surface of the Earth and locate objects using their distance up from the ground, then the positive direction is up, so down is the negative direction. With this perspective, the equation of change in velocity for a freely falling object would be expressed as $$ \frac{d V}{d t}=-g $$ (We measure upward velocity \(V\) in feet per second and time \(t\) in seconds.) Consider a rock tossed up- ward from the surface of the Earth with an initial velocity of 40 feet per second upward. a. Use a formula to express the velocity \(V=V(t)\) as a linear function. (Hint: You get the slope of \(V\) from the equation of change. The vertical intercept is the initial value.) b. How many seconds after the toss does the rock reach the peak of its flight? (Hint: What is the velocity of the rock when it reaches its peak?) c. How many seconds after the toss does the rock strike the ground? (Hint: How does the time it takes for the rock to rise to its peak compare with the time it takes for it to fall back to the ground?)

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?

Equation of change for logistic growth: The logistic growth formula \(N=6.21 /\left(0.035+0.45^{t}\right)\) that we used in Chapter 2 for deer on the George Reserve actually came from the following equation of change: $$ \frac{d N}{d t}=0.8 N\left(1-\frac{N}{177}\right) \text {. } $$ a. Plot the graph of \(\frac{d N}{d t}\) versus \(N\), and use it to find the equilibrium solutions. Explain their physical significance. b. In one plot, sketch the equilibrium solutions and graphs of \(N\) versus \(t\) in each of the two cases \(N(0)=10\) and \(N(0)=225 .\) c. To what starting value for \(N\) does the solution \(N=6.21 /\left(0.035+0.45^{t}\right)\) correspond?

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