Problem 88 - Algebra and Trigonometry

Question

A learning curve is a graph of a function \(P(t)\) that measures the performance of someone learning a skill as a function of the training time \(t\) . At first, the rate of learning is rapid. Then, as performance increases and approaches a maximal value \(M\) , the rate of learning decreases. It has been found that the function $$P(t)=M-C e^{-k t}$$ where \(k\) and \(C\) are positive constants and \(C< M\) is a reasonable model for learning. (a) Express the learning time \(t\) as a function of the performance level \(P .\) (b) For a pole-vaulter in training, the learning curve is given by $$P(t)=20-14 e^{-0.024 t}$$ where \(P(t)\) is the height he is able to pole-vault after \(t\) months. After how many months of training is he able to vault 12 \(\mathrm{ft}\) ? (c) Draw a graph of the learning curve in part (b).

Step-by-Step Solution

Answer

(a) \( t = -\frac{1}{k} \ln \left( \frac{M - P}{C} \right) \); (b) 50.11 months; (c) Graph the curve, showing approach to maximum height 20 ft.

Step 1: Solving for Learning Time t as a Function of Performance Level P

Given the function \( P(t) = M - C e^{-kt} \), we need to express time \( t \) in terms of performance level \( P \). To do this, rearrange the equation to isolate the exponential term:\[ P = M - C e^{-kt} \]Subtract \( M \) from both sides to get:\[ P - M = -C e^{-kt} \]Divide both sides by \( -C \) to isolate \( e^{-kt} \):\[ \frac{M - P}{C} = e^{-kt} \]Taking the natural logarithm of both sides gives:\[ -kt = \ln \left( \frac{M - P}{C} \right) \]Solve for \( t \):\[ t = -\frac{1}{k} \ln \left( \frac{M - P}{C} \right) \]

Step 2: Finding the Training Time for a Given Performance Level

For a pole-vaulter in training, the learning curve is given by \( P(t) = 20 - 14 e^{-0.024t} \). We need to find \( t \) when \( P = 12 \). Substitute \( P = 12 \) into the equation:\[ 12 = 20 - 14 e^{-0.024t} \]Rearrange to solve for \( e^{-0.024t} \):\[ 14 e^{-0.024t} = 20 - 12 \]\[ e^{-0.024t} = \frac{8}{14} = \frac{4}{7} \]Take the natural logarithm of both sides:\[ -0.024t = \ln \left( \frac{4}{7} \right) \]Solve for \( t \):\[ t = \frac{-\ln \left( \frac{4}{7} \right)}{0.024} \approx 50.11 \text{ months} \]

Step 3: Graphing the Learning Curve

To graph the learning curve described by the function \( P(t) = 20 - 14 e^{-0.024t} \), note that as \( t \) increases, \( P(t) \) approaches the maximal value \( M = 20 \).Plot \( P(t) \) against \( t \) for a range of values to visualize the function. The graph starts at a point where \( P(t) \) rapidly rises and then flattens as it approaches 20.- Vertical axis: Performance level \( P(t) \)- Horizontal axis: Training time \( t \) in monthsThe curve will have its steepest slope at the beginning (rapid learning) and will level off as it approaches the maximum value of 20.

Key Concepts

Exponential Function

The concept of an exponential function is pivotal to understanding learning curves. In the given learning model, the function is expressed as \( P(t) = M - C e^{-kt} \). Here, \( e^{-kt} \) signifies the exponential function, where \( e \) is the base of natural logarithms, approximately equal to 2.718. This function shows how performance becomes a function of time.

• When \( t \) is very small, \( e^{-kt} \) is close to 1, so the learning starts at a rapid rate.
• As \( t \) increases, \( e^{-kt} \) approaches zero, causing the learning to slow down as it nears the maximum value \( M \).
• This creates a curve where learning is rapid initially but slows as it reaches the skill's optimal performance level.

Understanding how exponential decay works helps in predicting learning outcomes over time. It is important in performance-based activities, as it captures the essence of initial enthusiasm slowing as mastery is achieved.

Performance Measurement

Performance measurement through the learning curve highlights how potential capabilities or skills are achieved over time. In the model, \( P(t) \) represents the performance level at time \( t \), and its measurement is dependent on various constants.

• \( M \) is the maximum performance level achievable over an indefinite period of practice.
• \( C \) is the extent to which initial performance can improve, given that \( C < M \).
• Performance measurement in this model is not linear, it adapts as learning progresses, slowing down as optimal performance is approached.

The exponential nature of \( P(t) = M - C e^{-kt} \) offers a realistic portrayal of skill acquisition. It provides insights into how different factors, such as time devoted to practice and the nature of training, influence the learning rate. By analyzing such curves, educators and trainers can optimize training conditions to ensure consistent progress.

Training Time

Training time is a crucial element in the learning curve model, reflecting the duration needed to achieve a certain performance level. Expressing \( t \) as a function of \( P \) allows one to precisely calculate the time required to reach a desired skill level.

• The formula \( t = -\frac{1}{k} \ln \left( \frac{M - P}{C} \right) \) is derived from manipulating the original performance function \( P(t) = M - C e^{-kt} \).
• This equation calculates how long it will take for performance \( P \) to reach a specific level.
• For instance, in a pole-vaulting context, to find when the athlete reaches 12 feet, substitute \( P = 12 \) and solve for \( t \).

Understanding training time's dependency on other variables like the learning rate constant \( k \) assists in better training design. It clarifies how incremental improvements occur and adjusts expectations based on realistic timeframes for achieving proficiency.