Problem 87 - College Algebra and Calculus: An Applied Approach

Question

Students in a mathematics class were given an exam and then retested monthly with equivalent exams. The average scores \(S\) (on a 100 -point scale) for the class can be modeled by \(S=80-14 \ln (t+1)\) \(0 \leq t \leq 12\), where \(t\) is the time in months. (a) What was the average score on the original exam? (b) What was the average score after 4 months? (c) After how many months was the average score 46 ?

Step-by-Step Solution

Answer

a) The average score for the original exam was 80. b) After 4 months, the average score of the class was approximately 58.51. c) It took about 5.77 months for the average score to drop to 46.

Step 1: Determine the average score on the original exam

For the original exam, the time \(t\) is 0 since we're considering the score of the initial exam before any retests. Substitute \(t = 0\) into the model \(S = 80 - 14 \ln (t+1)\). So, \(S = 80 - 14 \ln (0+1)\), which simplifies to \(S = 80\).

Step 2: Calculate the average score after 4 months

Here, \(t = 4\) months. Substitute \(t = 4\) into the model \(S = 80 - 14 \ln (t+1)\). Perform the calculations to get: \(S = 80 - 14 \ln (4+1)\), which simplifies to \(S \approx 58.51\) when rounded to 2 decimal places.

Step 3: Determine after how many months the average score was 46

Since the average score \(S\) was 46, substitute \(S = 46\) into the model and solve for \(t\). Doing this, we get: \(46 = 80 - 14 \ln (t+1)\). Rearranging and solving, we obtain \(\ln (t+1) = \frac{80 - 46}{14}\) and then from this \(t+1 = e^{\frac{80 - 46}{14}}\). Subtracting 1 from both sides finally gives \(t = e^{\frac{80 - 46}{14}} - 1 \approx 5.77\) months when rounded to 2 decimal places.

Key Concepts

average score

Understanding average scores in mathematical contexts like exams is crucial. In this exercise, the average score refers to the arithmetic mean of the tests taken at different intervals. Initially, the students have a full score of 80 points on the original exam when no time has passed, so the formula simplifies to an average score of 80.

The average score decreases over time, modeled by an exponential decay pattern. This means as time progresses, students on average tend to score less on re-examinations. This model allows us to calculate how the class's performance changes over months. After more specific durations, such as four months, the score can be calculated through substitution in the provided formula. The importance of noting the score trends is vital as it provides insights into learning retention and the effectiveness of teaching methods.

• Original score: 80 (at time zero)
• Score trend: decreases over months
• Score periodically reassessed

Analyzing average scores allows educators to adjust their instructions depending on how the students are scoring as time goes by.

mathematical modeling

Mathematical modeling is a fascinating and powerful tool used in this exercise to represent real-world phenomena. In this scenario, the average test scores of students over twelve months are modeled using a logarithmic function. The purpose of such modeling is to predict outcomes and analyze changes over time with a mathematical equation.

By creating a model like this, the complex phenomenon of how scores might naturally degrade over time is simplified into a manageable and predictable pattern. The model here is expressed as \( S = 80 - 14 \ln (t+1) \). This equation factors in time \(t\) in months and a logarithmic term which explains the decay in average scores post-assessment.

• Models make sense of complex data
• Predict future outcomes
• Used for decision-making and analysis

The strength of mathematical modeling lies in its ability to give educators a clear picture of academic retention and effectiveness of learning over time.

logarithmic function

The logarithmic function plays an integral role in this exercise by representing the decrease of average scores over time. A logarithm, generally speaking, is a mathematical operation that is the inverse of exponentiation, providing insight into rates of decay, growth, or change.

In the context of this exercise, the function \( S = 80 - 14 \ln(t+1) \) implies a slower decay as time progresses rather than a sharp drop-off. The inclusion of the logarithmic term \( \ln(t+1) \) reveals how small time (\(t\)) changes impact average scores significantly at first, but less so as more time passes. The constant 14 signifies the rate at which scores decrease under a natural logarithmic decay.

• Logarithms transform multiplicative changes into additive ones
• Provides a more nuanced view of decay or growth patterns
• Highlights initial high-impact periods of change

This feature of the logarithm is particularly useful in capturing the essence of natural phenomena such as memory retention, as seen in student test scores. It allows us to visualize and anticipate changes with precision and clarity.