/*! 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 78 Suppose that you are given the t... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that you are given the task of learning \(100 \%\) of a block of knowledge. Human nature tells us that we would retain only a percentage \(P\) of the knowledge \(t\) weeks after we have learned it. The Ebbinghaus learning model asserts that \(P\) is given by $$ P(t)=Q+(100 \%-Q) e^{-h t} $$ where \(Q\) is the percentage that we would never forget and \(h\) is a constant that depends on the knowledge learned. Suppose that \(Q=40 \%\) and \(k=0.7\) a) Find the percentage retained after \(0 \mathrm{wk}\); 1 wk; 2 wk; 6 wk; 10 wk. b) Find \(\lim _{\ell \rightarrow \infty} P(\iota)\) c) Sketch a graph of \(P\). d) Find the rate of change of \(P\) with respect to time \(t, P^{\prime}(t)\). e) Interpret the meaning of the derivative.

Short Answer

Expert verified
Retained knowledge after 0, 1, 2, 6, 10 weeks is 100%, 69.8%, 54.8%, 40.9%, 40% respectively. As \(t\) approaches infinity, retained knowledge approaches 40%. The rate of change of retained knowledge is \(P'(t) = -42e^{-0.7t}\).

Step by step solution

01

- Identify Given Parameters

Given parameters in the problem are:\(Q = 40 \%\) (percentage never forgotten)\(h = 0.7\) (constant)The provided formula is:\[P(t) = Q + (100 \%- Q) e^{-ht}\].Use these parameters to solve each part of the problem.
02

- Calculate Percentage Retained After Specified Weeks

Substitute \(Q\), \(h\), and various values of \(t\) into the formula \[P(t) = 40 + (100 - 40) e^{-0.7t}\]- For \(t = 0\) weeks:\[P(0) = 40 + 60 e^{0} = 100 \%\]- For \(t = 1\) week:\[P(1) = 40 + 60 e^{-0.7} \approx 40 + 60 (0.4966) \approx 69.8 \%\]- For \(t = 2\) weeks:\[P(2) = 40 + 60 e^{-1.4} \approx 40 + 60 (0.2466) \approx 54.8 \%\]- For \(t = 6\) weeks:\[P(6) = 40 + 60 e^{-4.2} \approx 40 + 60 (0.0149) \approx 40.9 \%\]- For \(t = 10\) weeks:\[P(10) = 40 + 60 e^{-7} \approx 40 + 60 (0.0009) \approx 40 \%\]
03

- Calculate the Limit as \(t\) Approaches Infinity

To find the limit as \(t\) approaches infinity:\[\lim_{t \rightarrow \infty} P(t) = \lim_{t \rightarrow \infty} \left( 40 + 60 e^{-0.7t} \right) = 40 + 60 \left( \lim_{t \rightarrow \infty} e^{-0.7t} \right) = 40 \%\]
04

- Sketch the Graph of \(\mathbf{P(t)}\)

Create a graph with the x-axis representing time \(t\) in weeks and the y-axis representing the percentage retained \(P(t)\). The starting point is \(100 \%\), and the curve decreases exponentially, approaching \(40 \%\) asymptotically as \(t\) increases.
05

- Find the Derivative \(\mathbf{P'(t)}\)

Differentiate \(P(t)\) with respect to time \(t\):\[P(t) = 40 + 60 e^{-0.7t}\]\[P'(t) = 60 \left( -0.7 \right) e^{-0.7t} = -42 e^{-0.7t}\]
06

- Interpret the Meaning of the Derivative

The derivative \(P'(t) = -42 e^{-0.7t}\) indicates the rate of change of the retained knowledge over time. It shows that the rate of forgetting is fastest at \(t = 0\) and decreases exponentially over time.

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Key Concepts

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

Knowledge Retention
Knowledge retention is crucial in understanding how much of what we learn stays with us over time. The Ebbinghaus learning model helps predict this retention. When we first learn something, we retain almost all of it. However, over time, we tend to forget a portion. This model assumes that we will always retain a minimum percentage of the knowledge, represented by the variable \( Q \). This is referred to as the percentage of knowledge that is never forgotten, and it varies depending on the individual and the type of knowledge learned.
In our example, \( Q \) is 40%, meaning no matter how much time passes, we will always retain at least 40% of the learned material. The retention percentage at any given time \( t \) can be calculated using the formula:
\[P(t) = Q + (100 \% - Q) e^{-ht}\]
where \( h \) is a constant that affects the rate of forgetting.
Decay Constant
In the formula for knowledge retention, \( h \) represents the decay constant. This constant indicates how quickly we forget the information over time. A larger \( h \) means faster forgetting, and a smaller \( h \) means slower forgetting.
In our example, \( h = 0.7 \), which means we have a moderate rate of forgetting. The decay constant affects the exponent in the formula, which dictates how quickly the memory retention percentage drops over time.
Understanding the decay constant can help in designing better learning schedules and revisions. By knowing how quickly information is forgotten, one can plan revisions to boost retention just as it starts to decline.
Exponential Function
The Ebbinghaus learning model uses the exponential function to describe how knowledge retention decreases over time. The exponential function is represented as \( e^{-ht} \). It is a mathematical function in which the decrease happens in proportion to the value of the function itself.
This function characterizes the rapid drop in retention initially, which then slows down as time progresses. For example, after one week, the retention might drop significantly, but the rate of forgetting slows down in subsequent weeks.
The formula \( P(t) = 40 + 60 e^{-0.7t} \) shows this exponential decay, where \( e \) is the base of natural logarithms, and \( -0.7t \) represents the exponent that includes the decay constant and time.
Rate of Change
The rate of change in the retention percentage \( P(t) \) shows how fast the retention level is decreasing over time. It is found by taking the derivative of \( P(t) \), symbolized as \( P'(t) \).
In our example, the rate of change is given by:
\[P'(t) = -42 e^{-0.7t}\]
This formula indicates that the rate of forgetting is initially faster and then slows down over time. When \( t = 0 \), the rate of change is at its maximum because \( e^0 = 1 \). As \( t \) increases, the value of the exponential term \( e^{-0.7t} \) becomes smaller, reducing the rate at which knowledge is forgotten.
Derivative Interpretation
Interpreting the derivative, \( P'(t) \), and understanding its significance helps us grasp the nature of knowledge decay. The derivative indicates how quickly the retention percentage is changing at any moment in time.
In our case, \( P'(t) = -42 e^{-0.7t} \) shows the rate at which knowledge is forgotten. Initially, at \( t = 0 \), the value is \( -42 \), meaning the retention percentage is decreasing quickly. As time progresses, the negative value becomes less significant due to the exponential decay, indicating a slowdown in forgetting.
This understanding is essential for educators and learners alike to optimize study schedules and improve long-term retention. It provides insights into when revisions are most effective, helping to combat the natural tendency to forget learned material over time.

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Most popular questions from this chapter

In an art class, students were tested at the end of the course on a final exam. Then they were retested with an equivalent test at subsequent time intervals. Their scores after time \(t\), in months, are given in the following table. $$ \begin{array}{|c|c|} \hline \text { Time, } t \text { (in months) } & \text { Score, } y \\ \hline 1 & 84.9 \% \\ 2 & 84.6 \% \\ 3 & 84.4 \% \\ 4 & 84.2 \% \\ 5 & 84.1 \% \\ 6 & 83.9 \% \\ \hline \end{array} $$ a) Use the REGRESSION feature on a grapher to fit a logarithmic function \(y=a+b \ln x\) to the data. b) Use the function to predict test scores after \(8 \mathrm{mo} ; 10 \mathrm{mo} ; 24 \mathrm{mo} ; 36 \mathrm{mo}\) c) After how long will the test scores fall below \(82 \% ?\) d) Find the rate of change of the scores and interpret its meaning.

Suppose that \(P_{0}\) is invested in a savings account in which interest is compounded continuously at \(6.5 \%\) per year. That is, the balance P grows at the rate given by $$ \frac{d P}{d t}=0.065 P $$ a) Find the function that satisfies the equation. List it in terms of \(P_{0}\) and \(0.065\). b) Suppose that $$\$ 1000$$ is invested. What is the balance after 1 yr? after 2 yr? c) When will an investment of $$\$ 1000$$ double itself?

Medication Conccntration. The concentration \(C\) in parts per million, of a medication in the body \(t\) hours after ingestion is given by the function $$ C(t)=10 t^{2} e^{-t} $$ a) Find the concentration after \(0 \mathrm{hr} ; \mathrm{l} \mathrm{hr} ; 2 \mathrm{hr} ;\) \(3 \mathrm{hr} ; 10 \mathrm{hr}\). b) Sketch a graph of the function for \(0 \leq t \leq 10\) c) Find the rate of change of the concentration \(C^{\prime}(t) .\) d) Find the maximum value of the concentration and where it occurs. e) Interpret the meaning of the derivative.

A bank advertises that it compounds interest continuously and that it will double your money in 10 yr. What is its exponential growth rate?

Graph each of the following and find the relative extrema. $$ f(x)=e^{-x^{2}} $$

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