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When forgetfulness is taken into account, the rate of memorization of a subject is given by $$\frac{d A}{d t}=k_{1}(M-A)-k_{2} A$$ where \(k_{1}>0, k_{2}>0, A(t)\) is the amount memorized in time \(t\) \(M\) is the total amount to be memorized, and \(M-A\) is the amount remaining to be memorized. (a) since the DE is autonomous, use the phase portrait concept of Section 2.1 to find the limiting value of \(A(t)\) as \(t \rightarrow \infty\) Interpret the result. (b) Solve the DE subject to \(A(0)=0 .\) Sketch the graph of \(A(t)\) and verify your prediction in part (a).

Short Answer

Expert verified
As \( t \to \infty \), \( A(t) \to \frac{k_1 M}{k_1 + k_2} \) which means this is the amount retained long-term.

Step by step solution

01

Analyze the Differential Equation

The differential equation given is \( \frac{d A}{d t} = k_1(M-A) - k_2 A \), which simplifies to \( \frac{d A}{d t} = k_1 M - (k_1 + k_2)A \). This is an autonomous equation because it only depends on \( A \) and not explicitly on \( t \). Identifying this helps us later when we use phase portraits.
02

Identify the Equilibrium Point

To find the equilibrium point where \( \frac{d A}{d t} = 0 \), set \( k_1 M - (k_1 + k_2)A = 0 \). Solving for \( A \), we get \( A = \frac{k_1 M}{k_1 + k_2} \). This is the value \( A(t) \) approaches as \( t \to \infty \).
03

Interpret the Limiting Behavior

The equilibrium point \( A = \frac{k_1 M}{k_1 + k_2} \) represents the amount that will be permanently memorized as time goes to infinity. It indicates how \( A(t) \) stabilizes due to the competing effects of memorization \( k_1 \) and forgetfulness \( k_2 \).
04

Solve the Differential Equation

The differential equation can be solved using an integrating factor or separation of variables. For this linear differential equation, an integrating factor approach leads to the solution \( A(t) = \frac{k_1 M}{k_1 + k_2} + C e^{-(k_1 + k_2)t} \). Using the initial condition \( A(0) = 0 \), we find \( C = -\frac{k_1 M}{k_1 + k_2} \). Thus, \( A(t) = \frac{k_1 M}{k_1 + k_2}(1 - e^{-(k_1 + k_2)t}) \).
05

Sketch the Graph of A(t)

Sketching \( A(t) = \frac{k_1 M}{k_1 + k_2}(1 - e^{-(k_1 + k_2)t}) \), you observe that \( A(t) \) starts at 0, increases monotonically, and asymptotes to \( \frac{k_1 M}{k_1 + k_2} \). This verifies the prediction that \( A(t) \to \frac{k_1 M}{k_1 + k_2} \) as \( t \to \infty \).

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Key Concepts

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

Autonomous Equation
An autonomous equation is a type of differential equation that does not depend explicitly on the independent variable, often time \( t \). In the context of our problem, the differential equation \( \frac{d A}{d t} = k_1(M-A) - k_2 A \) is autonomous. This means that the changes in the amount memorized, \( A \), depend solely on \( A \) itself and not directly on time. Autonomous equations are particularly useful in modeling real-world phenomena, where the rate of change is determined by the current state rather than the passage of time. Key advantages include:
  • Simpler mathematical handling since functions only depend on state variables.
  • Ability to analyze stability and dynamics without specifying an initial time.
Recognizing that an equation is autonomous allows us to utilize phase portraits efficiently to study the system's behavior over time without solving the equation in a time-dependent manner.
Equilibrium Point
In differential equations, an equilibrium point is where the rate of change of the system is zero, meaning the system is in a steady state. For the equation \( \frac{d A}{d t} = k_1M - (k_1 + k_2)A \), the equilibrium point is found by setting the derivative equal to zero.Solving \( k_1M - (k_1 + k_2)A = 0 \), we find the equilibrium point to be \( A = \frac{k_1 M}{k_1 + k_2} \). This point represents a balance between the forces of memorization and forgetfulness. As \( t \to \infty \), \( A(t) \) approaches this equilibrium, indicating a stable memorized amount.Understanding equilibrium points helps predict long-term behavior without having to extend calculations to infinity. They highlight the stable outcomes of a process, such as in our memorization model, suggesting how much information will be retained over time.
Phase Portrait
A phase portrait is a graphical representation that depicts the trajectories of a dynamical system in phase space. For autonomous equations like ours, phase portraits help visualize how solutions evolve over time.In the memorization context described by our equation, the phase portrait would show how \( A(t) \) moves from its initial state towards the equilibrium point \( A = \frac{k_1 M}{k_1 + k_2} \).Key features of a phase portrait include:
  • Stability: Shows whether trajectories remain bounded and approach equilibrium points.
  • Direction: Arrows indicate the flow towards stability, providing insights into system behavior.
Phase portraits are valuable in identifying the stability of equilibrium points, offering a visual summary of all possible behaviors of the system.
Memorization Model
The memorization model in this exercise describes how information is retained and forgotten over time. Governed by the differential equation \( \frac{d A}{d t} = k_1(M-A) - k_2 A \), it models memory as a dynamic process with two competing effects: learning and forgetting. In this model:
  • \( k_1(M-A) \): Represents the rate of new memorization based on the amount left to learn.
  • \( k_2 A \): Represents the rate of forgetting the information already memorized.
The solution to this problem tells us how much information, \( A(t) \), is retained after some time \( t \). With the initial condition \( A(0) = 0 \), solving gives us \( A(t) = \frac{k_1 M}{k_1 + k_2}(1 - e^{-(k_1 + k_2)t}) \), showing an exponential approach towards the equilibrium amount \( \frac{k_1 M}{k_1 + k_2} \).This simple yet insightful model emphasizes the natural interplay between learning new information and losing previously acquired knowledge, with applications in educational strategies and cognitive modeling.

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