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Table 11.7 gives values for a logistic function \(P=f(t)\) (a) Estimate the maximum rate of change of \(P\) and estimate the value of \(t\) when it occurs. (b) If \(P\) represents the growth of a population, estimate the carrying capacity of the population. $$\begin{array}{c|c|c|c|c|c|c|c|c}\hline t & 0 & 10 & 20 & 30 & 40 & 50 & 60 & 70 \\ \hline P & 120 & 125 & 135 & 155 & 195 & 270 & 345 & 385 \\\\\hline\end{array}$$

Short Answer

Expert verified
(a) Maximum rate of change is 75 at around \(t = 45\). (b) Carrying capacity is approximately 385.

Step by step solution

01

Understand the Logistic Function

A logistic function generally represents growth that starts exponentially but then slows down and approaches a maximum limit, known as the carrying capacity. In this logistics model, the rate of change is initially increasing, reaching a peak, and then decreases as it approaches the carrying capacity.
02

Identify Maximum Rate of Change

To find the maximum rate of change, we analyze the differences in consecutive values of \(P\) over time. Calculate \(\Delta P\) for each consecutive pair: \(\Delta P = P_{t+1} - P_{t}\).\(\Delta P:\)- \(P(10) - P(0) = 125 - 120 = 5\)- \(P(20) - P(10) = 135 - 125 = 10\)- \(P(30) - P(20) = 155 - 135 = 20\)- \(P(40) - P(30) = 195 - 155 = 40\)- \(P(50) - P(40) = 270 - 195 = 75\)- \(P(60) - P(50) = 345 - 270 = 75\)- \(P(70) - P(60) = 385 - 345 = 40\)The maximum \(\Delta P\) is 75, occurring between \(t = 40\) and \(t = 60\), indicating the maximum rate of change around \(t = 45\).
03

Estimate Carrying Capacity

The carrying capacity is the value of \(P\) as \(t\) approaches a large number, where \(P\) levels off. From the table, the highest value given is \(P(70) = 385\). Since \(P\) is slowing down as it approaches this value, the carrying capacity appears to be around 385.

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Key Concepts

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

Rate of Change
The rate of change in a logistic function is a vital concept to grasp. It shows how fast the population (or whatever is being measured) is growing or shrinking over time.

In our logistic model, the rate of change can be computed by analyzing the differences in consecutive values of the function. You do this by subtracting the previous value from the next. This difference tells us how much the population has grown in each time step.

For example, when the difference between populations at consecutive times is biggest, that's where the population is growing the fastest. In the given exercise, the biggest difference, or maximum rate of change, happens between the 5th and 6th period (between time 40 and 60). This tells us the population is growing at a maximum rate around the middle of this interval, approximately at time 45.

This peak in the rate of change is important because it represents the point of most rapid growth in the cycle of this particular population dynamics.
Population Growth
Population growth in logistic functions begins with a slow increase, speeds up to reach a maximum growth rate, and then gradually slows as it approaches its upper limits called carrying capacity. Understanding this process is crucial for predicting how populations behave over time.

As you analyze population growth, it’s essential to note the stages in a logistic growth model:
  • Initial Phase: Growth begins slowly.
  • Exponential Phase: Growth rate increases rapidly.
  • Deceleration Phase: Growth slows down as resources become limited.
  • Stable Phase: Population levels off near the carrying capacity.
The logistic function used in the exercise is typical of biological populations, where resources like food or space are finite.

Recognizing where the population is on this curve helps in understanding its dynamics. In the given exercise, we see that the growth is starting to slow down as it nears a carrying capacity of about 385.
Carrying Capacity
Carrying capacity is a key concept in logistic growth, representing the maximum population size an environment can sustain indefinitely.

A population grows until it nears carrying capacity, at which point limiting factors like food, space, and nutrient availability slow down growth. In our logistic function model, this capacity is where the population stabilizes after its period of rapid growth.

Estimating this capability in real-world scenarios can help biologists and ecologists make important decisions regarding conservation and resource management.

In the exercise, the population levels off at around 385. This suggests that 385 individuals are the environment's carrying capacity for this particular population under the conditions measured. It's where growth ceases because the resources available can no longer support additional increase.

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Most popular questions from this chapter

Federal or state agencies control hunting and fishing by setting a quota on how many animals can be harvested each season. Determining the appropriate quota means achieving a balance between environmental concerns and the interests of hunters and fishers. For example, when a June 8,2007 decision by the Delaware Superior Court invalidated a two-year moratorium on catching horseshoe crabs, the Delaware Department of Natural 91Ó°ÊÓ and Environmental Control imposed instead an annual quota of 100,000 on male horseshoe crabs. Environmentalists argued this would exacerbate a decrease in the protected Red Knot bird population that depends on the crab for food. For a population \(P\) that satisfies the logistic model with harvesting, $$\frac{d P}{d t}=k P\left(1-\frac{P}{L}\right)-H$$ show that the quota, \(H,\) must satisfy \(H \leq k L / 4,\) or else the population \(P\) may die out. (In fact, \(H\) should be kept much less than \(k L / 4 \text { to be safe. })\)

Decide whether the statement is true or false. Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=2 x-y .\) Justify your answer. If \(f(a)=b,\) the slope of the graph of \(f\) at \((a, b)\) is \(2 a-b\)

A drug is administered intravenously at a constant rate of \(r\) mg/hour and is excreted at a rate proportional to the quantity present, with constant of proportionality \(\alpha>0\) (a) Solve a differential equation for the quantity, \(Q,\) in milligrams, of the drug in the body at time \(t\) hours. Assume there is no drug in the body initially. Your answer will contain \(r\) and \(\alpha .\) Graph \(Q\) against \(t\) What is \(Q_{\infty},\) the limiting long-run value of \(Q ?\) (b) What effect does doubling \(r\) have on \(Q_{\infty} ?\) What effect does doubling \(r\) have on the time to reach half the limiting value, \(\frac{1}{2} Q_{\infty} ?\) (c) What effect does doubling \(\alpha\) have on \(Q_{\infty} ?\) On the time to reach \(\frac{1}{2} Q_{\infty} ?\)

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 1 million barrels of oil in the well; six years later 500,000 barrels remain. (a) At what rate was the amount of oil in the well decreasing when there were 600,000 barrels remaining? (b) When will there be 50,000 barrels remaining?

The radioactive isotope carbon- 14 is present in small quantities in all life forms, and it is constantly replenished until the organism dies, after which it decays to stable carbon- 12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5730 years. Suppose \(C(t)\) is the amount of carbon- 14 present at time \(t.\) (a) Find the value of the constant \(k\) in the differential equation \(C^{\prime}=-k C.\) (b) In 1988 three teams of scientists found that the Shroud of Turin, which was reputed to be the burial cloth of Jesus, contained \(91 \%\) of the amount of carbon-14 contained in freshly made cloth of the same material. \(^{8}\) How old is the Shroud of Turin, according to these data?

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