/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 In economic theory, the followin... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 15

In economic theory, the following model is used to describe a possible capital investment policy. Let \(f(t)\) represent the total invested capital of a company at time \(t .\) Additional capital is invested whenever \(f(t)\) is below a certain equilibrium value \(E\), and capital is withdrawn whenever \(f(t)\) exceeds \(E .\) The rate of investment is proportional to the difference between \(f(t)\) and \(E\). Construct a differential equation whose solution is \(f(t)\) and sketch two or three typical solution curves.

Short Answer

Expert verified
The differential equation is \( \frac{df(t)}{dt} = k (E - f(t)) \). Typical solutions are curves that approach the equilibrium value \(E\) asymptotically.

Step by step solution

01

Identify the Proportional Relationship

The problem states that the rate of investment is proportional to the difference between the total invested capital, represented by \(f(t)\), and the equilibrium value \(E\). This can be mathematically represented as \( \frac{df(t)}{dt} \propto E - f(t) \).
02

Formulate the Differential Equation

To form the differential equation, introduce a proportionality constant \(k\). The relationship can be written as \( \frac{df(t)}{dt} = k (E - f(t)) \). Here, \(k\) is a positive constant.
03

Interpret the Differential Equation

The differential equation \( \frac{df(t)}{dt} = k (E - f(t)) \) suggests that the rate of change of capital investment depends on how far \(f(t)\) is from the equilibrium value \(E\). When \(f(t)\) is less than \(E\), \( \frac{df(t)}{dt} > 0 \) leading to an increase in investment. Conversely, when \(f(t)\) is greater than \(E\), \( \frac{df(t)}{dt} < 0 \) leading to a withdrawal of capital.
04

Solve the Differential Equation

To solve the differential equation \( \frac{df(t)}{dt} = k (E - f(t)) \), rewrite it in a separable form: \( \frac{df(t)}{E - f(t)} = k \, dt \). Integrate both sides: \( \int \frac{df(t)}{E - f(t)} = \int k \, dt \). Let \(u = E - f(t)\). Then \( \frac{-du}{-f'(t)} = du \) and \( - \ln|E - f(t)| = kt +C \), where \(C\) is the integration constant. Solving for \(f(t)\), we get: \( f(t) = E - Ce^{-kt} \).
05

Draw Typical Solution Curves

Typical solution curves for \( f(t) = E - Ce^{-kt} \) show how \( f(t) \) approaches \( E \) over time. For different initial values, the curve will start either above or below \(E\), but will always approach \(E\) asymptotically as \( t \) increases.

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Key Concepts

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

economic theory
Economic theory often seeks to describe how various factors influence financial decisions over time. In the context of capital investment, a company must decide how and when to allocate resources to maximize profitability and growth. One common model assumes that the decision to invest additional capital or to withdraw it is based on the difference between current investment levels and a target value or equilibrium value, denoted as \(E\). This model helps us understand and predict investment behaviors in response to varying conditions, providing valuable insights into financial planning and economic stability.
differential equations
A differential equation is a mathematical equation that relates some function with its derivatives. In our case, we look at how the rate of investment (a derivative) changes with respect to time. The differential equation that we derived is \( \frac{df(t)}{dt} = k (E - f(t)) \). Here, \(f(t)\) represents the investment at time \(t\), \(E\) is the equilibrium investment, and \(k\) is a proportionality constant. Differential equations are widely used in economic theory to model dynamic systems and predict how they evolve.
proportional relationships
A proportional relationship in mathematics indicates that two quantities increase or decrease uniformly. For the capital investment model, the rate at which investment changes is proportional to the difference between current investment \(f(t)\) and the equilibrium value \(E\). This relationship is expressed as \( \frac{df(t)}{dt} = k (E - f(t)) \). Here, the proportionality constant \(k\) dictates the speed at which investment reacts to deviations from the equilibrium.
equilibrium value
The equilibrium value \(E\) represents the optimal level of capital investment which the company aims to maintain. If the current investment \(f(t)\) is below \(E\), additional capital is invested to reach this value. Conversely, if \(f(t)\) is above \(E\), capital is withdrawn. The differential equation \( \frac{df(t)}{dt} = k (E - f(t)) \) ensures that, over time, the investment will approach \(E\) asymptotically. Equilibrium values are fundamental in economic theory as they represent the point where the system is in balance.
integration
Integration is the mathematical process of finding the function from its derivative, often referred to as the antiderivative. To solve the differential equation \( \frac{df(t)}{dt} = k (E - f(t)) \), we rearrange it to a separable form and integrate both sides: \( \int \frac{df(t)}{E - f(t)} = \int k \, dt \). This gives us the equation \( - \ln |E - f(t)| = kt + C \), where \(C\) is the constant of integration. By solving for \(f(t)\), we obtain \( f(t) = E - Ce^{-kt} \), showing how the capital investment approaches the equilibrium value \(E\) over time.

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Most popular questions from this chapter

One or more initial conditions are given for each differential equation in the following exercises. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solutions. Include a \(y z\) -graph if one is not already provided. Always indicate the constant solutions on the \(t y\) -graph whether they are mentioned or not. \(y^{\prime}=-\frac{1}{2} y, y(0)=-2, y(0)=0, y(0)=2\)

Solve the given equation using an integrating factor. Take \(t>0\). $$ 6 y^{\prime}+t y=t $$

The fish population in a pond with carrying capacity 1000 is modeled by the logistic equation $$ \frac{d N}{d t}=\frac{.4}{1000} N(1000-N) $$ Here, \(N(t)\) denotes the number of fish at time \(t\) in years. When the number of fish reached 250, the owner of the pond decided to remove 50 fish per year. (a) Modify the differential equation to model the population of fish from the time it reached 250 . (b) Plot several solution curves of the new equation, including the solution curve with \(N(0)=250\). (c) Is the practice of catching 50 fish per year sustainable or will it deplete the fish population in the pond? Will the size of the fish population ever come close to the carrying capacity of the pond?

Use Euler's method with \(n=4\) to approximate the solution \(f(t)\) to \(y^{\prime}=2 t-y+1, y(0)=5\) for \(0 \leq t \leq 2\) Estimate \(f(2)\).

A model that describes the relationship between the price and the weekly sales of a product might have a form such as $$ \frac{d y}{d p}=-\frac{1}{2}\left(\frac{y}{p+3}\right), $$ where \(y\) is the volume of sales and \(p\) is the price per unit. That is, at any time, the rate of decrease of sales with respect to price is directly proportional to the sales level and inversely proportional to the sales price plus a constant. Solve this differential equation. (Figure 6 shows several typical solutions.)
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