Question

In a group project on learning theory, a mathematical model for the percent \(P\) (in decimal form) of correct responses after \(n\) trials was found to be \(P=\frac{0.98}{1+e^{-0.3 n}}, \quad n \geq 0\) (a) After how many trials will \(80 \%\) of the responses be correct? (That is, for what value of \(n\) will \(P=0.8\) ?) (b) Use a graphing utility to graph the memory model and confirm the result found in part (a). (c) Write a paragraph describing the memory model.

Short answer

Around 6.24 trials are needed for \(80 \%\) of the responses to be correct according to the given model. The model represents a logistic growth in learning, with learning being slow initially, then accelerating, and finally slowing down as the learner approaches their maximum potential.

Step-by-step solution

Set Up the Equation

In order to find the number of trials needed for \(80 \%\) of the responses to be correct, we need to set \(P= 0.8\) in the equation. The equation we'll use then is \[0.8 = \frac{0.98}{1 + e^{-0.3n}}\]

Solve the Equation

Now, we can solve this equation for \(n\). First, cross-multiply to get rid of the denominator on the right-hand side, giving us \[0.8(1 + e^{-0.3n}) = 0.98\]. Simplify to obtain \[0.8 + 0.8e^{-0.3n} = 0.98\]. Next, subtract 0.8 from both sides to get \[0.8e^{-0.3n} = 0.18\]. Dividing both sides by 0.8 gives us \[e^{-0.3n} = 0.225\]. Taking natural logarithm (ln) on both sides to get the exponent on its own yields \[-0.3n = ln(0.225)\]. Finally, divide by -0.3 on both sides to solve for \(n\), giving us \[n = \frac{ln(0.225)}{-0.3}\]

Calculate the Exact Value

Calculate the value of \(n\) using a calculator, providing an approximation of \(n \approx 6.24\) trials.

Using Graphing Utility

By using a graphing calculator or any graphing software, plot the equation \(P = \frac{0.98}{1 + e^{-0.3n}}\). Identify the point at which \(P=0.8\). This value should align with the previous calculation, providing a visual confirmation. Please note, this step requires a graphing tool and won't be shown here.

Describing the Model

The function \(P=\frac{0.98}{1+e^{-0.3n}}\) is a logistic function, which models the learning process. It predicts that the percentage of correct responses, \(P\), will increase as the number of trials, \(n\), increases. This model posits that learning is slow initially and then accelerates before eventually plateauing as maximum potential is approached.

Key concepts

Logistic Function

The logistic function is a complex yet insightful representation often used in the mathematical modeling of growth processes. At its core, the logistic equation describes how a population or quantity grows rapidly at first, then slows down as it approaches a maximum limit, known as the carrying capacity.

For instance, in the problem provided, the learning process is captured by the logistic function \( P = \frac{0.98}{1 + e^{-0.3n}} \). Initially, the learning rate is slow, which can be seen from the flat start of the curve. As trials, denoted as \( n \), increase, learning accelerates rapidly until eventually slowing as it gets closer to a perfect score, or a \( P \) value of 0.98, indicating a plateau effect.

The equation \( P = 0.8 \) is used to find the specific point at which 80% of the possible knowledge has been acquired. The logistic model here is particularly great at predicting how learning doesn't increase indefinitely but instead reaches a saturation point - reflecting a more realistic learning curve than what a linear or exponential model would predict.

Exponential Growth and Decay

Exponential growth and decay phenomena are ubiquitous in nature and can be modeled by functions that have a constant ratio of change in the quantity per unit time. This leads to either an explosive increase or decrease over time, depending on the context.

In the setting of our learning model example, the rate of learning can initially appear to demonstrate exponential growth, represented by the function \( e^{-0.3n} \) within the logistic function. However, the logistic function ultimately modifies this growth with its structure to account for the plateauing effect.

The 'decay' aspect comes into play when we look at the inverse proportionality; as trials increase, the exponential term creates a decaying effect on the denominator, signifying a diminishing marginal learning rate with each additional trial. This interplay between growth and decay intricately models the complex process of learning over time.

Graphing Utility

A graphing utility is an indispensable tool in visualizing mathematical functions, particularly when it comes to comprehending behavioral trends that are otherwise difficult to discern. When dealing with the logistic function like the one presented in our learning model example, a graph provides a clear illustration of the saturation effect as trials increase.

In the solution steps, graphing the function \( P = \frac{0.98}{1 + e^{-0.3n}} \) allows students to see how the percentage of correct responses (\( P \)) changes as a function of trials (\( n \)). By entering the equation into the utility, one can find the exact point where \( P \) reaches 0.8 or 80%, thus visually confirming the findings from our algebraic calculations.

Using a graphing utility is also a form of exercise improvement for students, as it supports the process of verification and reinforces understanding of the concepts at play within a graphical context.