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Eagles: In an effort to restore the population of bald eagles, ecologists introduce a breeding group into a protected area. Let \(N=N(t)\) denote the population of bald eagles at time \(t\). Over time, you observe the following information about \(\frac{d N}{d t}\). \- Initially, \(\frac{d N}{d t}\) is a small positive number. \- A few years later, \(\frac{d N}{d t}\) is a much larger positive number. \- Many years later, \(\frac{d N}{d t}\) is positive but near zero. Make a possible graph of \(N(t)\).

Short Answer

Expert verified
A graph of \(N(t)\) starts with a gentle slope, steepens, and eventually flattens.

Step by step solution

01

Understand the Problem

The exercise provides information on the rate of change \(\frac{dN}{dt}\) of the population \(N(t)\) over time. We need to construct a graph of \(N(t)\) that aligns with the behavior of \(\frac{dN}{dt}\).
02

Analyze Initial Condition

Initially, \(\frac{dN}{dt}\) is a small positive number. This implies that the population is growing slowly at this stage. On the graph, \(N(t)\) should start with a gentle upward slope.
03

Consider Intermediate Growth

A few years later, \(\frac{dN}{dt}\) becomes a much larger positive number. This indicates accelerated population growth. The slope of \(N(t)\) should steepen as \(N\) increases more rapidly during this period.
04

Evaluate Long-Term Behavior

Many years later, \(\frac{dN}{dt}\) is positive but near zero. This suggests that population growth is approaching a stable point where changes are minimal, so \(N(t)\) should flatten out at a higher value, indicating it is nearing its carrying capacity.
05

Graph Construction

Based on these observations, the graph of \(N(t)\) should start with a slow upward curve, progress to a steep slope during the intermediate phase, and then flatten out over time as the rate of increase diminishes to near zero.

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Key Concepts

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

Rate of Change
Understanding the rate of change of a population is crucial in predicting its future behavior. In our exercise, the rate of change is represented by the derivative \(\frac{dN}{dt}\), which informs us about how the population \(N(t)\) changes with time \(t\). Initially, \(\frac{dN}{dt}\) is a small positive number, meaning the population is growing, but at a slow pace. Over time, this rate increases drastically, indicating a period of rapid growth. Finally, as \(\frac{dN}{dt}\) approaches zero, the population growth slows, suggesting the system is stabilizing. Understanding this rate of change:
  • Helps in determining the growth phase the population is experiencing.
  • Informs the likely future behavior, whether accelerating, stagnating, or declining.
Grasping these changes can aid ecologists in making informed decisions about conservation efforts and resource allocation.
Graphing Functions
Graphing functions such as \(N(t)\) allows us to visualize the population dynamics over time. In the context of our exercise, plotting \(N(t)\) based on the given rate of change helps us understand the historical and future trends of the eagle population. Initially, with a small positive \(\frac{dN}{dt}\), the graph starts with a gentle upward slope. As \(\frac{dN}{dt}\) becomes much larger, the slope of the graph steeply increases, visually depicting the accelerated growth phase of the population.Eventually, as \(\frac{dN}{dt}\) again becomes smaller and approaches zero, the graph will flatten out, indicating that the population is reaching an equilibrium state.
  • Visualizing these changes helps in identifying critical phases in growth.
  • Provides insights into factors affecting the population.
Through these phases, the graph of \(N(t)\) concisely encapsulates the story of the eagle population's journey toward stabilization.
Modeling Populations
Modeling populations involves creating mathematical representations to simulate and predict changes in population sizes. In our exercise, \(N(t)\) models the population of bald eagles, showing how it evolves over time based on the changing rate of change \(\frac{dN}{dt}\). Initially, the population models a slow growth phase which accelerates substantially and then stabilizes.The significance of accurate population modeling includes:
  • Predicting future population sizes and trends.
  • Assisting in the management of wildlife and conservation plans.
  • Establishing the impact of external factors like habitat change and climate conditions.
By building these models, ecologists can anticipate challenges and design effective strategies to ensure the survival and growth of species, like the bald eagles in our example.

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Most popular questions from this chapter

Making up a story about a car trip: You begin from home on a car trip. Initially your velocity is a small positive number. Shortly after you leave, velocity decreases momentarily to zero. Then it increases rapidly to a large positive number and remains constant for this part of the trip. After a time, velocity decreases to zero and then changes to a large negative number. a. Make a graph of velocity for this trip. b. Discuss your distance from home during this driving event, and make a graph. c. Make up a driving story that matches this description.

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 \(W\).

Growth of fish: Let \(w=w(t)\) denote the weight of a fish as a function of its age \(t\). For the North Sea cod, the equation of change $$ \frac{d w}{d t}=2.1 w^{2 / 3}-0.6 w $$ holds. Here \(w\) is measured in pounds and \(t\) in years. b. Make a graph of \(\frac{d w}{d t}\) against \(w\). Include weights up to 45 pounds. c. What is the weight of the cod when it is growing at the greatest rate? d. To what weight does the cod grow?

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?

The cannon with a different muzzle velocity: If the cannonball from Example 6.7 is fired with a muzzle velocity of 370 feet per second, it will follow the graph of $$ h=x-32\left(\frac{x}{370}\right)^{2} $$ where distances are measured in feet. a. Plot the graph of the flight of the cannonball. b. Find the height of the cannonball 3000 feet downrange. c. By looking at the graph of \(h\), determine whether \(\frac{d h}{d x}\) is positive or negative at 3000 feet downrange. d. Calculate \(\frac{d h}{d x}\) at 3000 feet downrange and explain what this number means in practical terms.

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