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Chapter 11: Problem 4

A quantity \(P\) satisfies the differential equation $$\frac{d P}{d t}=k P\left(1-\frac{P}{250}\right), \quad \text { with } k>0$$ Sketch a graph of \(d P / d t\) as a function of \(P\)

Short Answer

Expert verified
The graph of \(\frac{dP}{dt}\) vs. \(P\) is a symmetric, inverted parabola crossing the \(P\)-axis at 0 and 250.

Step by step solution

01

Understand the Differential Equation

The equation given is \(\frac{d P}{d t}=k P\left(1-\frac{P}{250}\right)\). It's a logistic differential equation where \(k > 0\) is the growth rate, and 250 is the carrying capacity.
02

Identify Key Points

Calculate \(\frac{dP}{dt}\) when \(P = 0\) and \(P = 250\), which are the equilibrium points. At \(P = 0\), \(\frac{dP}{dt} = 0\). At \(P = 250\), \(\frac{dP}{dt} = 0\). These are points where the function intersects the line \(\frac{dP}{dt} = 0\).
03

Sketch the General Shape

For values of \(0 < P < 250\), the expression \(1 - \frac{P}{250}\) is positive, so \(\frac{dP}{dt} > 0\), meaning \(P\) is increasing. For \(P > 250\), \(1 - \frac{P}{250}\) becomes negative, making \(\frac{dP}{dt} < 0\), meaning \(P\) is decreasing. Use these details to sketch the shape of the graph.
04

Plot Important Points

Plot the points \((0,0)\) and \((250,0)\) on the \(P\)-axis. The curve will start at \(0\), rise to a maximum, and then return to \(0\) at \(P = 250\). The maximum occurs at \(P = 125\) due to symmetry, where \(\frac{dP}{dt}\) is at its peak.
05

Draw the Graph

Based on the previous steps, draw the curve starting at \(P = 0\), increasing to the highest point at \(P = 125\), and then decreasing back to intersect at \(P = 250\). The shape will look like a symmetrical curve shaped like an inverted parabola.

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Key Concepts

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

Understanding Growth Rate
In the context of a logistic differential equation, the growth rate is represented by the constant \( k \). This parameter is crucial because it determines how rapidly the population \( P \) can increase when there are no constraints on the resources or space. In simple terms, imagine if the population were to grow without any hindrances; this rate defines how fast that would happen.
However, in reality, populations do not grow indefinitely, which brings us to the role of the carrying capacity in the equation. The growth rate works alongside the carrying capacity to maintain balance in the system, ensuring that growth is not unchecked.
- Think of the growth rate as the potential speed of increase.
- It's the constant that accelerates growth when resources are abundant.
- A higher value of \( k \) means the population can grow faster under optimal conditions.
The Role of Carrying Capacity
Carrying capacity is a critical concept in understanding logistic growth and is denoted by the number 250 in our equation. It represents the maximum sustainable population size that the environment can support indefinitely without being degraded.
As the population \( P \) approaches the carrying capacity, the growth rate slows down. This happens because resources such as food, space, and other essentials become limited. The term \( \left(1 - \frac{P}{250}\right) \) becomes significant here, reducing the growth rate as \( P \) gets closer to 250.
- Think of carrying capacity as the ceiling for population size.
- \( P \) grows slower as it nears 250, showing a constraint on growth.
- Environmental factors and resource availability define carrying capacity.
Analyzing Equilibrium Points
Equilibrium points in a logistic differential equation are values of \( P \) where the population stops changing, meaning \( \frac{dP}{dt} = 0 \). In our example, these points occur at \( P = 0 \) and \( P = 250 \).
At \( P = 0 \), the population is at one equilibrium point, naturally stable since no growth can occur from zero without an external factor. At \( P = 250 \), the other equilibrium point represents a stable population size, balanced by the environmental constraints.
- Equilibrium points indicate where \( P \) remains constant over time.
- They help identify the stable population sizes in different scenarios.
- The logistic equation uses these points to define where growth ceases naturally.

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Most popular questions from this chapter

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$\begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array}$$. (a) For two armies of strengths \(x\) and \(y\) fighting a conventional battle governed by Lanchester's differential equations, write a differential equation involving \(d y / d x\) and the constants of attrition \(a\) and \(b\) (b) Solve the differential equation and hence show that the equation of the phase trajectory is $$a y^{2}-b x^{2}=C$$ for some constant \(C .\) This equation is called Lanchester's square law. The value of \(C\) depends on the initial sizes of the two armies.

Give an example of: A differential equation that has a slope field with all the slopes above the \(x\) -axis positive and all the slopes below the \(x\) -axis negative.

A model for the population, \(P,\) of carp in a landlocked lake at time \(t\) is given by the differential equation $$\frac{d P}{d t}=0.25 P(1-0.0004 P)$$ (a) What is the long-term equilibrium population of carp in the lake? (b) A census taken ten years ago found there were 1000 carp in the lake. Estimate the current population. (c) Under a plan to join the lake to a nearby river, the fish will be able to leave the lake. A net loss of \(10 \%\) of the carp each year is predicted, but the patterns of birth and death are not expected to change. Revise the differential equation to take this into account. Use the revised differential equation to predict the future development of the carp population.

A bank account earns \(5 \%\) annual interest, compounded continuously. Money is deposited in a continuous cash flow at a rate of 1200 dollars per year into the account. (a) Write a differential equation that describes the rate at which the balance \(B=f(t)\) is changing. (b) Solve the differential equation given an initial balance \(B_{0}=0\) (c) Find the balance after 5 years.

In Problems \(52-54\), explain what is wrong with the statement. The solution to \(d P / d t=0.2 t\) is \(P=B e^{0.2 t}\)
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