Problem 104 - Calculus and its applications

Question

As part of a study, students in a psychology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After \(t\) months, the average score \(S(t),\) as a percentage, was found to be given by $$ S(t)=78-15 \ln (t+1), \quad t \geq 0 $$. a) What was the average score when they initially took the test, \(t=0 ?\) b) What was the average score after 4 months? c) What was the average score after 24 months? d) What percentage of their original answers did the students retain after 2 years ( 24 months)? e) Find \(S^{\prime}(t)\). f) Find the maximum and minimum values, if they exist. g) Find \(\lim _{t \rightarrow \infty} S(t)\) and discuss its meaning.

Step-by-Step Solution

Answer

a) 78%; b) 53.9%; c) 29.7%; d) 38.1%; e) \( S'(t) = -\frac{15}{t+1} \); f) Max: 78, No Min; g) \( -\infty \), indicates continuous forgetting.

Step 1: Initial Score Calculation

To find the average score when students initially took the test, plug \( t = 0 \) into the function \( S(t) = 78 - 15 \ln(t + 1) \). When \( t = 0 \), the equation becomes \( S(0) = 78 - 15 \ln(0 + 1) = 78 - 15(0) = 78 \). Therefore, the initial score is 78%.

Step 2: Score After 4 Months

To determine the average score after 4 months, substitute \( t = 4 \) into the function. Calculate \( S(4) = 78 - 15 \ln(4 + 1) = 78 - 15 \ln(5) \). Calculate \( \ln(5) \approx 1.609 \), thus \( S(4) = 78 - 15 \times 1.609 \approx 78 - 24.135 = 53.865 \). The average score after 4 months is approximately 53.9%.

Step 3: Score After 24 Months

For the score after 24 months, substitute \( t = 24 \) into \( S(t) \). Thus, \( S(24) = 78 - 15 \ln(24 + 1) = 78 - 15 \ln(25) \). Since \( \ln(25) \approx 3.218 \), then \( S(24) = 78 - 15 \times 3.218 \approx 78 - 48.27 = 29.73 \). So, the average score after 24 months is approximately 29.7%.

Step 4: Percentage of Retained Answers After 2 Years

The original score was 78% (when \( t = 0 \)). After 24 months, the score is approximately 29.7%. To find the percentage of the original score retained after 2 years: \( \frac{29.7}{78} \times 100 \approx 38.08\% \). Thus, students retained about 38.1% of their original answers after 2 years.

Step 5: Finding the Derivative of S(t)

To find the derivative \( S'(t) \) of \( S(t) = 78 - 15 \ln(t + 1) \), apply the derivative \((d/dt)\). The derivative of \(-15 \ln(t + 1)\) is \(-15 \times \frac{1}{t+1} \times 1 \) using the chain rule. So, \( S'(t) = -\frac{15}{t+1} \).

Step 6: Evaluating Maximum and Minimum

Since \( S'(t) = -\frac{15}{t+1} \), it is clear that \( S'(t) < 0 \) for all \( t > 0 \). This implies that \( S(t) \) is decreasing as \( t \) increases. Therefore, the maximum value occurs at \( t=0 \) with \( S(0)=78 \). As \( t \) tends to infinity, the function decreases without a lower bound, so no minimum value exists.

Step 7: Limit as t Approaches Infinity

The limit of \( S(t) \) as \( t \rightarrow \infty \) is found by considering the term \( 15 \ln(t+1) \). As \( t \rightarrow \infty \), \( \ln(t+1) \rightarrow \infty \) resulting in \( S(t) \rightarrow -\infty \). The meaning here is that over a long period, the average score can decrease indefinitely, suggesting significant forgetting over an extended time.

Key Concepts

Exponential Decay

Exponential decay is a crucial concept in calculus, often used in real-life scenarios such as radioactive decay or population decrease. In the context of the given exercise, it reflects how students gradually forget what they've learned over time. Although the formula provided, \( S(t) = 78 - 15 \ln(t+1) \), is not a pure exponential function, it shares the characteristic of exponential decay, where the score diminishes as time progresses.

One hallmark of exponential decay is its steady percentage rate of reduction over equal time intervals. For example, with each passing month, the amount of knowledge retained diminishes by a constant proportion, similar to how radioactive substances decay by a fixed rate each year.

While the logarithmic function here controls the decay rate, the effect is nonetheless a decrease that resembles exponential behavior. By recognizing the exponential trend in such problems, students can better predict outcomes over longer periods following the structure of results found in the initial analysis.

Logarithmic Functions

Logarithmic functions are essential tools in understanding data that involves exponential relationships. In this scenario, the function \( S(t) = 78 - 15 \ln(t+1) \) includes a natural logarithm \( \ln(t+1) \) which plays a crucial role in mapping the decay of test scores over time.

The natural logarithm is the inverse function of the exponential function \( e^x \), making it perfect for modeling situations where data grows or declines rapidly over time and then slows down. It effectively helps to linearize multiplicative processes, making them more manageable for calculations and predictions.

In our case, as the variable \( t \) increases in value, \( \ln(t+1) \) grows, meaning the term \( -15 \ln(t+1) \) causes a larger reduction in the score \( S(t) \). Therefore, students can visualize how logarithms allow them to gauge the slowly accumulating effect of time and aid in calculating the rate of knowledge depreciation accurately.

Derivative Concepts

Calculus allows us to understand the behavior of functions through derivatives, revealing changes over time. Finding the derivative of a function provides insight into its rate of change. In the exercise, the derivative \( S'(t) = -\frac{15}{t + 1} \) represents the rate at which the students' scores decrease over time.

The negative sign indicates a decreasing trend. However, as time \( t \) increases, the magnitude of \( S'(t) \) decreases. This decline means students are forgetting more slowly as time continues, aligning with real-world observations that retention initially drops rapidly but stabilizes gradually.

Understanding derivatives in this context helps students grasp how the rate of forgetting changes and becomes a powerful tool for anticipating future outcomes. It can also guide educators in tailoring review periods or interventions in order to optimize long-term learning retention.

Limit Behavior

The concept of limits is foundational in calculus, mainly when evaluating the behavior of functions over extended periods. With the score function given as \( S(t) = 78 - 15 \ln(t+1) \), evaluating its behavior as \( t \) approaches infinity is critical.

The limit \( \lim_{t \to \infty} S(t) \) focuses on what happens to the function as time goes on forever. In this study's context, as \( t \) gets extremely large, \( \ln(t+1) \) also becomes very large. Hence, \( 15 \ln(t+1) \) could exceed the initial value of 78, resulting in \( S(t) \to -\infty \). This situation implies that test scores theoretically continue to decrease indefinitely, showcasing a long-term projection of knowledge fading over time.

Recognizing this kind of limit behavior is crucial for students as it helps them comprehend scenarios beyond immediate calculations, offering insights into potential outcomes of extensive time-scale processes.