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Chapter 10: Problem 34

Amount of Information Learned In certain learning situations a maximum amount, \(M,\) of information can be learned, and at any time, the rate of learning is proportional to the amount yet to be learned. Let \(y=f(t)\) be the amount of information learned up to time \(t .\) Construct and solve a differential equation that is satisfied by \(f(t).\)

Short Answer

Expert verified
The amount of information learned over time is given by \( y = M(1 - e^{-kt}) \).

Step by step solution

01

Identify the proportional relationship

The rate of learning is proportional to the amount yet to be learned. This can be written as \ \ \ \[ \frac{dy}{dt} = k(M - y) \] where \(k\) is a positive constant and \(M - y\) is the amount yet to be learned.
02

Separate the variables

Rearrange the equation to separate variables \(y\) and \(t\): \[ \frac{1}{M - y} \frac{dy}{dt} = k \] or, equivalently, \[ \frac{1}{M - y} dy = k dt \]
03

Integrate both sides

Integrate both sides to solve the differential equation: \[ \int \frac{1}{M - y} dy = \int k dt \] This leads to: \[ -\ln|M - y| = kt + C \] where \(C\) is the constant of integration.
04

Solve for \( y \)

Exponentiate both sides to solve for \( y \): \[ |M - y| = e^{-kt + C} \] Let \( C' = e^C\), \[ |M - y| = C' e^{-kt} \] Removing the absolute value and solving for \( y \): \[ M - y = C' e^{-kt} \] \[ y = M - C' e^{-kt} \]
05

Determine the specific solution

Assume an initial condition where \( y(0) = 0 \): \[ 0 = M - C' \] So, \( C' = M \). Thus, the solution becomes \[ y = M(1 - e^{-kt}) \]

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Key Concepts

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

Differential Equations
Differential equations are equations that involve the rates of change of a variable and the variable itself. They are powerful tools for modeling real-world phenomena.
In our learning situation, the amount of information learned over time can be modeled using a differential equation.
Specifically, we are given that the rate of learning is proportional to the amount of information yet to be learned, which is reflected in the differential equation: \[ \frac{dy}{dt} = k(M - y) \] where \(k\) is a constant.

Differential equations like this help us understand dynamic systems where change over time is crucial. They are fundamental in physics, engineering, economics, biology, and many other fields.
Separation of Variables
Separation of Variables is a method to solve differential equations by separating the variables and integrating both sides individually. In our example, we start with: \[ \frac{dy}{dt} = k(M - y) \]
We rearrange this equation to isolate \(y\) and \(t\) on each side: \[ \frac{1}{M - y} dy = k dt \]
This makes it possible to integrate each side independently, simplifying our equation.
By separating variables, complex differential equations become simpler to handle and solve, revealing how each variable behaves over time.
Exponential Growth or Decay
Exponential functions are a key part of describing growth and decay processes. They occur in various natural phenomena, such as population growth, radioactive decay, and learning processes.
When something grows or decays exponentially, its rate of change is proportional to its current value. In our learning model, the equation: \[ y = M(1 - e^{-kt}) \]
shows how the information learned evolves over time. Here, \(e^{-kt}\) represents the decay of the amount yet to be learned.
Exponential growth or decay functions describe how quantities change rapidly at first and slow down as they approach a limit.
Integration Techniques
Integration is a fundamental concept in calculus, allowing us to find the area under curves and solve differential equations. To solve our learning rate equation, we need to integrate both sides: \[ \text{Integrate:} \int \frac{1}{M - y} dy = \int k dt \]
which results in: \[ -\text{ln}|M - y| = kt + C \]
Here, \(C\) is the constant of integration.
Integration techniques, such as substitution and integration by parts, are critical tools in calculus.
They enable us to transform and solve complex equations, offering insights into different physical and abstract systems.

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Most popular questions from this chapter

Review concepts that are important in this section. In each exercise, sketch the graph of a function with the stated properties. Domain: \(0 \leq t \leq 7 ;(0,2)\) is on the graph; the slope is always positive, the slope becomes more positive as \(t\) increases from 0 to 3, and the slope becomes less positive as \(t\) increases from 3 to 7.

A person took out a loan of \(\$ 100,000\) from a bank that charges \(7.5 \%\) interest compounded continuously. What should be the annual rate of payments if the loan is to be paid in full in exactly 10 years? (Assume that the payments are made continuously throughout the year.)

Review concepts that are important in this section. In each exercise, sketch the graph of a function with the stated properties. Domain: \(0 \leq t \leq 8 ;(0,6)\) is on the graph; the slope is always negative, the slope becomes more negative as \(t\) increases from 0 to \(3,\) and the slope becomes less negative as \(t\) increases from 3 to 8.

Find an integrating factor for each equation. Take \(t>0\). $$y^{\prime}-\frac{y}{10+t}=2$$

A certain drug is administered intravenously to a patient at the continuous rate of \(r\) milligrams per hour. The paticnt's body removes the drug from the bloodstream at a rate proportional to the amount of the drug in the blood, with constant of proportionality \(k=.5\) (a) Write a differential equation that is satisfied by the amount \(f(t)\) of the drug in the blood at time \(t\) (in hours). (b) Find \(f(t)\) assuming that \(f(0)=0 .\) (Give your answer in terms of \(r .)\) (c) In a therapeutic 2 -hour infusion, the amount of drug in the body should reach 1 milligram within 1 hour of administration and stay above this level for another hour. However, to avoid toxicity, the amount of drug in the body should not exceed 2 milligrams at any time. Plot the graph of \(f(t)\) on the interval \(1 \leq t \leq 2,\) as \(r\) varies between 1 and 2 by increments of \(.1 .\) That is, plot \(f(t)\) for \(r=1,1.1,1.2,1.3, \ldots . .2 .\) By looking at the graphs, pick the values of \(r\) that yield a therapeutic and nontoxic 2-hour infusion.
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