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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?

Short Answer

Expert verified
Equilibrium points are \( N = 0 \) and \( N = 177 \). For \( N(0) = 6.21 / (0.035 + 0.45^0) \), the starting value is 6.

Step by step solution

01

Understand the equation of change

The equation of change for logistic growth is given by \( \frac{dN}{dt} = 0.8N\left(1-\frac{N}{177}\right) \). This represents the growth rate of a population \( N \) over time, taking into account the carrying capacity (177 in this case), which is the maximum population size the environment can sustain.
02

Plot \( \frac{dN}{dt} \) versus \( N \)

First, create a graph with \( N \) on the x-axis and \( \frac{dN}{dt} \) on the y-axis. The function \( \frac{dN}{dt} = 0.8N\left(1-\frac{N}{177}\right) \) is a parabola opening downwards, crossing the \( N \)-axis at 0 and 177. These points (0 and 177) are the equilibrium solutions where \( \frac{dN}{dt} = 0 \).
03

Interpret equilibrium solutions

Equilibrium solutions occur where \( \frac{dN}{dt} = 0 \). This implies no change in population size. At \( N = 0 \), the population is extinct, and at \( N = 177 \), the population is at carrying capacity. These points indicate stable and unstable equilibria, respectively.
04

Sketch \( N \) versus \( t \)

In a second graph, plot \( N \) on the y-axis and \( t \) on the x-axis. For \( N(0) = 10 \), the population will increase toward the carrying capacity of 177. For \( N(0) = 225 \), which is above carrying capacity, the population will decrease towards 177.
05

Find the starting value for given \( N \) formula

For \( N = \frac{6.21}{0.035+0.45^t} \), set \( t = 0 \) to find the initial population value. Substitute into the equation: \( N(0) = \frac{6.21}{0.035+0.45^0} = \frac{6.21}{0.035+1} = 6 \).
06

Verify the correspondance

Check that the initial condition above matches the solution. The derived \( N = 6 \) indeed corresponds to the solution for \( N \) at \( t = 0 \) in the provided logistic equation formula.

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Key Concepts

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

Equilibrium Solutions in Logistic Growth
Equilibrium solutions are key points in the context of population dynamics where the population does not change over time. When we look at the logistic growth equation, \[ \frac{dN}{dt} = 0.8N \left(1-\frac{N}{177}\right) \],in this scenario, equilibrium occurs at values of \( N \) where \( \frac{dN}{dt} = 0 \).
Equilibrium solutions can be found by setting the growth rate equation to zero: \( 0.8N \left(1-\frac{N}{177}\right) = 0 \). Solving this reveals two equilibria, \( N = 0 \) and \( N = 177 \).
  • At \( N = 0 \), the equilibrium is stable. It signifies the extinction point where the population ceases to exist.
  • Conversely, \( N = 177 \) represents the carrying capacity, acting as an unstable equilibrium, at which the population stabilizes if initially precisely at that number.
Understanding these equilibria is critical, as it helps predict how populations behave when they are not at these fixed points. Populations tend to grow towards the carrying capacity if below it and decline towards it if above.
Understanding Population Dynamics
Population dynamics in ecological studies refer to how and why the number of individuals in a population changes over time. The logistic growth model provides a clear picture of this process by describing how a population grows rapidly when numbers are low, then slows as it approaches a maximum limit – known as the carrying capacity.
As depicted in the equation \[ \frac{dN}{dt} = 0.8N \left(1-\frac{N}{177}\right) \],populations do not grow indefinitely. Initially, when a population is far below its carrying capacity, growth is nearly exponential. However, as \( N \) nears 177, growth rate declines, a phenomenon driven by increasing competition for limited resources.
  • Starting with \( N(0) = 10 \), the population size \( N \) will increase up to the carrying capacity, displaying an S-shaped growth curve.
  • If \( N(0) = 225 \), the population will decrease towards 177, showing how exceeding the carrying capacity leads to population decline through negative growth.
These dynamics underscore nature's self-regulating mechanisms, where limiting factors like food, space, and competition restrict unchecked growth.
The Role of Carrying Capacity in Ecosystems
The carrying capacity is a central concept in ecology, representing the maximum number of individuals an environment can sustainably support without degrading over time. It is this limit that defines the ceiling for population growth in the logistic model.
In our logistic growth equation, the carrying capacity \( K \) is 177, which symbolizes the maximum sustainable population size for the deer on the George Reserve.
  • At carrying capacity, resources such as food, water, and shelter are optimally utilized by the population.
  • Any population value above 177 leads to a decline back toward \( K \), driven by negative feedback that reduces birth rates or increases death rates.
This concept illustrates the balance in ecosystems and emphasizes the importance of maintaining populations within an environment's carrying capacity. When nature's limits are exceeded, it can lead to resource depletion, habitat destruction, and eventual population collapse, echoing the broader themes of sustainability.

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Most popular questions from this chapter

Experimental determination of the drag coefficient: When retardation due to air resistance is proportional to downward velocity \(V\), in feet per second, falling objects obey the equation of change $$ \frac{d V}{d t}=32-r V $$ where \(r\) is known as the drag coefficient. One way to measure the drag coefficient is to measure and record terminal velocity. a. We know that an average-size man has a terminal velocity of 176 feet per second. Use this to show that the value of the drag coefficient is \(r=0.1818\) per second. (Hint: To say that the terminal velocity is 176 feet per second means that when the velocity \(V\) is 176 , velocity will not change. That is, \(\frac{d V}{d t}=0\). Put these bits of information into the equation of change and solve for \(r\).) b. An ordinary coffee filter has a terminal velocity of about 4 feet per second. What is the drag coefficient for a coffee filter?

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?)

Hiking: You are hiking in a hilly region, and \(E=\) \(E(t)\) is your elevation at time \(t\). a. Explain the meaning of \(\frac{d E}{d t}\) in practical terms. b. Where might you be when \(\frac{d E}{d t}\) is a large positive number? c. You reach a point where \(\frac{d E}{d t}\) is briefly zero. Where might you be? d. Where might you be when \(\frac{d E}{d t}\) is a large negative number?

7\. A population of bighorn sheep: A certain group of bighorn sheep live in an area where food is plentiful and conditions are generally favorable to bighorn sheep. Consequently, the population is thriving. There are initially 30 sheep in this group. Let \(N=N(t)\) be the population \(t\) years later. The population changes each year because of births and deaths. The rate of change in the population is proportional to the number of sheep currently in the population. For this particular group of sheep, the constant of proportionality is \(0.04\). a. Express the sentence "The rate of change in the population is proportional to the number of sheep currently in the population" as an equation of change. (Incorporate in your answer the fact that the constant of proportionality is \(0.04\).) b. Find a formula for \(N\). c. How long will it take this group of sheep to grow to a level of 50 individuals?

A yam baking in the oven: A yam is placed in a preheated oven to bake. An application of Newton's law of cooling gives the temperature \(Y\), in degrees, of the yam \(t\) minutes after it is placed in the oven as $$ Y=400-325 e^{-t / 50} $$ a. Make a graph of the temperature of the yam at time \(t\) over 45 minutes of baking time. b. Calculate \(\frac{d Y}{d t}\) at the time 10 minutes after the yam is placed in the oven. c. Calculate \(\frac{d Y}{d t}\) at the time 30 minutes after the yam is placed in the oven. d. Explain what your answers in parts \(b\) and \(c\) tell you about the way the yam heats over time.

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