Explain why a population that fits the logistic growth model increases more rapidly at intermediate size than at relatively small and large sizes.

Short Answer

Expert verified

A population that fits the logistic growth model increases more quickly at intermediate size because relatively few individuals create offspring when population size is small. However, the carrying capacity and per capita growth rate are low in large population sizes due to limited available resources.

Step by step solution

01

Logistic growth model

Logistic model of population determines that the population grows more slowly as it nears its carrying capacity (K),the maximum population size the environment can support.

It explains that every individual inside a population will have equal access to resources and equal survival chances.

02

Logistic growth curve

The steepest part of the logistic growth curve corresponds to a population with several substantial reproducing individuals but not near carrying capacity. The rapid growth at intermediate size is due to decreasing death rates faster than the birth rate.

The population growth decreases as population size reaches the carrying capacity according to the logistic growth.

03

Exponential growth

Exponential growth allows for a rise in population size without requiring an increase in population growth rate.In exponential growth, the per capita of a population remains constant regardless of population size.

It will result in rapid population expansion due to the enormous population size. Thus, a population that fits the logistic growth curve will grow exponentially at the intermediate size.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

  1. Find the intervals of increase or decrease.
  2. Find the local maximum and minimum values.
  3. Find the intervals of concavity and the inflection points.
  4. Use the information from parts (a)-(c) to sketch the graph. You may want to check your work with a graphing calculator or computer.

48. \(g\left( x \right) = 200 + 8{x^3} + {x^4}\)

Assuming that r= 1.0 and K= 1,500, calculate the population growth rate for four cases where population size (N) is greater than carrying capacity (K): N= 1,510, 1,600, 1,750, and 2,000 individuals. To do this, first write the equation for the population growth rate given in Table 53.2. Plug in the values for each of the four cases, starting with N= 1,510, and solve the equation for each

one. Which population size has the highest growth rate?

  1. Find the intervals of increase or decrease.
  2. Find the local maximum and minimum values.
  3. Find the intervals of concavity and the inflection points.
  4. Use the information from parts (a)-(c) to sketch the graph. You may want to check your work with a graphing calculator or computer.

53. \(F\left( x \right) = x\sqrt {6 - x} \)

Contrast the selective pressures operating in high-density populations (those near the carrying capacity, K) versus low-density populations.a

(a) Find the intervals of increase or decrease.

(b) Find the local maximum and minimum values.

(c) Find the intervals of concavity and the inflection points.

(d) Use the information from parts (a)–(c) to sketch the graph.

You may want to check your work with a graphing calculator or computer.

58. \(S\left( x \right) = x - \sin x,{\rm{ 0}} \le x \le 4\pi \)

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free