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Chapter 8: Problem 51

The graph of a solution to the logistic equation is known as a logistic curve, and if \(y_{0}>0\), it has one of four general shapes, depending on the relationship between \(y_{0}\) and \(L\). In each part, assume that \(k=1\) and use a graphing utility to plot a logistic curve satisfying the given condition. (a) \(y_{0}>L\) (b) \(y_{0}=L\) (c) \(L / 2 \leq y_{0}

Short Answer

Expert verified
Each graph illustrates how the population approaches the carrying capacity \( L = 10 \) differently based on the initial condition \( y_0 \).

Step by step solution

01

Understanding the Logistic Equation

The logistic equation is given by \( y(t) = \frac{L}{1 + \, e^{-k(t - t_0) + C}} \) where \( L \) is the carrying capacity, \( k \) is the growth rate, \( t_0 \) is the initial time, and \( C \) is a constant determined by the initial condition \( y_0 = y(t_0) \). In this problem, \( k = 1 \) and different conditions on \( y_0 \) are considered.
02

Plotting for Condition (a): \( y_0 > L \)

When \( y_0 > L \), the solution starts above the carrying capacity \( L \) and approaches \( L \) from above as time increases. You can use a graphing tool to plot the curve for, say, \( L = 10 \) and \( y_0 = 15 \). The graph will show the curve decreasing and asymptotically approaching \( L = 10 \) from above.
03

Plotting for Condition (b): \( y_0 = L \)

When \( y_0 = L \), the logistic curve is a constant horizontal line at \( y = L \). This is because the initial condition matches the carrying capacity, implying the population remains at equilibrium. Use a graphing tool to plot this as a horizontal line at \( y = L = 10 \).
04

Plotting for Condition (c): \( \frac{L}{2} \leq y_0 < L \)

For this range, the initial value is below \( L \) but not less than half. Use a graphing tool to plot the curve, taking, for instance, \( L = 10 \) and \( y_0 = 8 \). The graph shows growth towards \( L = 10 \), increasing from the initial condition. The curve will rise over time toward \( L \) but never exceed it.
05

Plotting for Condition (d): \( 0 < y_0 < \frac{L}{2} \)

Here, the initial condition is less than half the carrying capacity. Choose \( L = 10 \) and \( y_0 = 3 \) to graph the logistic curve. The curve will initially increase rapidly and continue to approach \( L = 10 \), with growth slowing as it nears the carrying capacity.

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Key Concepts

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

Logistic Curve
The logistic curve is the graphical representation of a solution to the logistic equation. It models how a quantity, such as population size, grows in a system where there are limitations, like resources or space. The curve typically begins with an initial input value, known as the initial condition \( y_0 \), and evolves over time towards a maximum that it can never exceed. This maximum is called the carrying capacity (\( L \)).

Logistic curves have distinctive S-shaped, or "sigmoidal," forms:
  • For \( y_0 > L \), the curve begins above \( L \) and asymptotically approaches \( L \) from above as time goes on, indicating that the population or quantity decreases at the start.
  • If \( y_0 = L \), then the curve is simply a horizontal line at \( L \), signifying that the system remains stable at the carrying capacity.
  • When \( \frac{L}{2} \leq y_0 < L \), the curve starts below \( L \) and rises over time toward \( L \), reflecting that the population or quantity increases slowly initially, then speeds up as it approaches the limit.
  • For \( 0 < y_0 < \frac{L}{2} \), the curve exhibits rapid initial growth, as the gap between \( y_0 \) and \( L \) allows for substantial acceleration in growth until approaching the limit, where it slows.
Such curves are crucial in fields such as ecology, economics, and biology, where they help visualize how systems approach equilibrium over time.
Carrying Capacity
Carrying capacity, denoted by \( L \) in the logistic equation, is a fundamental concept in understanding logistic growth. It refers to the maximum population size or quantity that an environment can sustain indefinitely without degradation. This is determined by factors such as availability of resources, space, and other environmental limits.

In the context of the logistic curve:
  • The carrying capacity is the ultimate upper bound that the curve will approach but never exceed; it represents a stable equilibrium point.
  • Whether the initial condition \( y_0 \) is less than, equal to, or greater than \( L \), the curve will trend towards \( L \) over time, indicating a balancing act of growth and resistance.
  • Ecological models use it to predict the sustainability of species populations within ecosystems, while other disciplines use it to project limits on growth impacted by finite resources.
This concept helps in planning and managing resources, ensuring that growth does not surpass ecological or infrastructural limits, which can lead to depletion and collapse.
Growth Rate
The growth rate in the logistic equation is symbolized by \( k \). It indicates how quickly the population or quantity grows or shrinks relative to its current size. In the logistic model, the growth rate is not constant; instead, it varies depending on how close the current population is to the carrying capacity \( L \).

Here's how it affects the logistic curve:
  • When \( y_0 \) is much less than \( L \), the growth rate is initially high, reflecting rapid growth as resources are abundant relative to the population size.
  • As the population size increases, reaching closer to \( L \), the growth rate slows down due to increasing resistance from environmental limits.
  • This dynamic nature of \( k \) in logistic growth ensures that the system stabilizes and avoids runaway growth, which is unsustainable.
Understanding \( k \) is crucial for any analyses involving population dynamics, resource management, and sustainable development, as it highlights the balance between growth potential and environmental constraints.

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Most popular questions from this chapter

A bullet of mass \(m\), fired straight up with an initial velocity of \(v_{0}\), is slowed by the force of gravity and a drag force of air resistance \(k v^{2}\), where \(k\) is a positive constant. As the bullet moves upward, its velocity \(v\) satisfies the equation $$ m \frac{d v}{d t}=-\left(k v^{2}+m g\right) $$ where \(g\) is the constant acceleration due to gravity. (a) Show that if \(x=x(t)\) is the height of the bullet above the barrel opening at time \(t\), then $$ m v \frac{d v}{d x}=-\left(k v^{2}+m g\right) $$ (b) Express \(x\) in terms of \(v\) given that \(x=0\) when \(v=v_{0}\). (c) Assuming that $$ \begin{aligned} &v_{0}=988 \mathrm{~m} / \mathrm{s}, \quad g=9.8 \mathrm{~m} / \mathrm{s}^{2} \\\ &m=3.56 \times 10^{-3} \mathrm{~kg}, \quad k=7.3 \times 10^{-6} \mathrm{~kg} / \mathrm{m} \end{aligned} $$ use the result in part (b) to find out how high the bullet rises. [Hint: Find the velocity of the bullet at its highest point.

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x\) $$ y^{\prime}=-x y $$

(a) Make a conjecture about the effect on the graphs of \(y=y_{0} e^{k t}\) and \(y=y_{0} e^{-k t}\) of varying \(k\) and keeping \(y_{0}\) fixed. Confirm your conjecture with a graphing utility. (b) Make a conjecture about the effect on the graphs of \(y=y_{0} e^{k t}\) and \(y=y_{0} e^{-k t}\) of varying \(y_{0}\) and keeping \(k\) fixed. Confirm your conjecture with a graphing utility.

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x\) $$ y-\frac{d y}{d x} \sec x=0 $$

(a) Suppose that a quantity \(y=y(t)\) increases at a rate that is proportional to the square of the amount present, and suppose that at time \(t=0\), the amount present is \(y_{0} .\) Find an initial-value problem whose solution is \(y(t)\) (b) Suppose that a quantity \(y=y(t)\) decreases at a rate that is proportional to the square of the amount present, and suppose that at a time \(t=0\), the amount present is \(y_{0}\). Find an initial-value problem whose solution is \(y(t)\).
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