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Chapter 3: Problem 1

The number \(N(t)\) of supermarkets throughout the country that are using a computerized checkout system is described by the initial-value problem $$\frac{d N}{d t}=N(1-0.0005 N), N(0)=1.$$ (a) Use the phase portrait concept of Section 2.1 to predict how many supermarkets are expected to adopt the new procedure over a long period of time. By hand, sketch a solution curve of the given initial-value problem. (b) Solve the initial-value problem and then use a graphing utility to verify the solution curve in part (a). How many companies are expected to adopt the new technology when \(t=10 ?\)

Short Answer

Expert verified
2000 supermarkets will adopt it long-term; at \( t = 10 \), nearly 2000 have adopted.

Step by step solution

01

Understand the Differential Equation

We start with the differential equation \( \frac{dN}{dt} = N(1 - 0.0005N) \), and the initial condition \(N(0) = 1\). This is a logistic differential equation modeling population growth with carrying capacity.
02

Find Equilibrium Points

Set \( \frac{dN}{dt} = 0 \) to find equilibrium points: \( N(1 - 0.0005N) = 0 \). This gives \( N = 0 \) or \( N = 2000 \) as equilibrium points. These points are where growth stops eventually.
03

Draw the Phase Portrait

The equilibrium point \( N = 0 \) is unstable, and \( N = 2000 \) is stable as \( \frac{dN}{dt} > 0 \) for \( 0 < N < 2000 \) and \( \frac{dN}{dt} < 0 \) for \( N > 2000 \). Thus, over time, \( N \) will approach 2000.
04

Solve the Differential Equation

Separate variables and integrate: \( \int \frac{dN}{N(1 - 0.0005N)} = \int dt \). The left side uses partial fraction decomposition. Solve to find \[ N(t) = \frac{2000}{1 + 1999e^{-t}} \].
05

Verify and Sketch the Solution

At \( t = 0 \), \( N(0) = 1 \), matching our initial condition with \( N(t) = \frac{2000}{1 + 1999e^{-t}} \). Over time, \( N(t) \rightarrow 2000 \). Graphing confirms \( N(t) \) approaches 2000, starting steeply and flattening out.
06

Calculate \( N(10) \)

Use the solution \( N(t) = \frac{2000}{1 + 1999e^{-t}} \) to evaluate at \( t=10 \): \( N(10) = \frac{2000}{1 + 1999e^{-10}} \). The exponential term is small, thus \( N(10) \approx 2000 \).

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Key Concepts

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

Equilibrium Points
In the context of logistic differential equations, equilibrium points are special values where the rate of change of the population becomes zero. These points mark the conditions under which the population remains constant over time. For the equation \( \frac{dN}{dt} = N(1 - 0.0005N) \), finding these points involves setting \( \frac{dN}{dt} = 0 \), which leads to the equation \( N(1 - 0.0005N) = 0 \). Solving this, we find two equilibrium points: \( N = 0 \) and \( N = 2000 \).

These two points represent very different scenarios in terms of stability:
  • \( N = 0 \): This is called an unstable equilibrium because any small change in \( N \) will lead to further changes away from zero.
  • \( N = 2000 \): This is a stable equilibrium, meaning that if the population reaches this point, it will tend to stay around 2000 due to the limiting nature of the carrying capacity.
Recognizing equilibrium points helps predict long-term outcomes in population models.
Phase Portraits
Phase portraits are graphical representations that help visualize how the solutions of a differential equation behave over time. They are essential when analyzing the stability and behavior of equilibrium points. For the logistic differential equation, the phase portrait illustrates how the system evolves as time progresses.

In our example:
  • The phase portrait shows arrows indicating the direction of change in \( N \) (population size).
  • When the population is less than the stable equilibrium point \( N = 2000 \), the arrows point upwards, showing an increase toward the equilibrium.
  • Beyond \( N = 2000 \), the arrows point downwards, indicating a decrease towards 2000.
Thus, the phase portrait confirms that the population will stabilize at around 2000 over time, thoroughly communicating the role of phase portraits in understanding the possible solution trajectories.
Population Growth Modeling
Population growth modeling using differential equations like the logistic model provides insights into how populations expand and stabilize over time. The logistic model is a significant improvement over exponential models because it considers resource limitations, reflected in its carrying capacity.

In our logistic differential equation \( \frac{dN}{dt} = N(1 - 0.0005N) \):
  • The equation starts as an exponential growth model, but the \( 1 - 0.0005N \) term limits growth as \( N \) increases.
  • This term simulates natural processes where larger populations face greater competition, thus slowing down the growth rate.
This biological realism makes logistic growth models particularly useful for applications like supermarket expansion, as they reflect real-world growth limitations.
Carrying Capacity
Carrying capacity refers to the maximum population level an environment can sustain due to limited resources. It is an essential concept in logistic growth models, embedded within the differential equation through the parameter limiting growth.

In the equation \( \frac{dN}{dt} = N(1 - 0.0005N) \), the term \( 0.0005N \) models the decrease in growth rate as \( N \) approaches the carrying capacity, which in this case is \( N = 2000 \). This means:
  • We expect the population of supermarkets using the checkout system to level off around 2000, as seen from solving the equation.
  • Once the population hits or approaches 2000, the growth will effectively stop, stabilizing the number of supermarkets incorporating the technology.
This concept is hugely relevant in ecological studies as well as urban planning, where understanding the limits of systems is crucial.

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Most popular questions from this chapter

The population of a town grows at a rate proportional to the population present at time \(t .\) The initial population of 500 increases by \(15 \%\) in 10 years. What will be the population in 30 years? How fast is the population growing at \(t=30 ?\)

Suppose a small cannonball weighing 16 pounds is shot vertically upward, as shown in Figure \(3.1 .13,\) with an initial velocity \(v_{0}=300 \mathrm{ft} / \mathrm{s}\). The answer to the question "How high does the cannonball go?" depends on whether we take air resistance into account. (a) Suppose air resistance is ignored. If the positive direction is upward, then a model for the state of the cannonball is given by \(d^{2} s / d t^{2}=-g\) (equation (12) of Section 1.3 ). since \(d s / d t=v(t)\) the last differential equation is the same as \(d v / d t=-g,\) where we take \(g=32 \mathrm{ft} / \mathrm{s}^{2} .\) Find the velocity \(v(t)\) of the cannonball at time \(t\) (b) Use the result obtained in part (a) to determine the height \(s(t)\) of the cannonball measured from ground level. Find the maximum height attained by the cannonball.

Consider the competition model defined by $$\begin{aligned}&\frac{d x}{d t}=x(2-0.4 x-0.3 y)\\\&\frac{d y}{d t}=y(1-0.1 y-0.3 x),\end{aligned}$$ where the populations \(x(t)\) and \(y(t)\) are measured in thousands and \(t\) in years. Use a numerical solver to analyze the populations over a long period of time for each of the following cases: (a) \(x(0)=1.5, \quad y(0)=3.5\) (b) \(x(0)=1, \quad y(0)=1\) (c) \(x(0)=2, \quad y(0)=7\) (d) \(x(0)=4.5, \quad y(0)=0.5\)

A tank in the form of a right circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3 when friction and contraction of water at the hole are ignored, the height \(h\) of water in the tank is described by $$\frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h},$$ where \(A_{w}\) and \(A_{h}\) are the cross-sectional areas of the water and the hole, respectively. (a) Solve the DE if the initial height of the water is \(H . \mathrm{By}\) hand, sketch the graph of \(h(t)\) and give its interval \(I\) of definition in terms of the symbols \(A_{w}, A_{h},\) and \(H\) Use \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) (b) Suppose the tank is 10 feet high and has radius 2 feet and the circular hole has radius \(\frac{1}{2}\) inch. If the tank is initially full, how long will it take to empty?

A mathematical model for the rate at which a drug disseminates into the bloodstream is given by $$\frac{d x}{d t}=r-k x$$ where \(r\) and \(k\) are positive constants. The function \(x(t)\) describes the concentration of the drug in the bloodstream at time \(t\) (a) since the DE is autonomous, use the phase portrait concept of Section 2.1 to find the limiting value of \(x(t)\) as \(t \rightarrow \infty\) (b) Solve the DE subject to \(x(0)=0 .\) Sketch the graph of \(x(t)\) and verify your prediction in part (a). At what time is the concentration one-half this limiting value?
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