Question

Psychologists sometimes use the function$$L(t)=A\left(1-e^{-k t}\right)$$ to measure the amount \(L\) learned at time \(t .\) Here \(A\) represents the amount to be learned, and the number \(k\) measures the rate of learning. Suppose that a student has an amount \(A\) of 200 vocabulary words to learn. A psychologist determines that the student has learned 20 vocabulary words after 5 minutes. (a) Determine the rate of learning \(k\). (b) Approximately how many words will the student have learned after 10 minutes? (c) After 15 minutes? (d) How long does it take for the student to learn 180 words?

Short answer

a) \(k \approx 0.0212\) b) \(38.4\) words in 10 mins c) \(54.6\) words in 15 mins d) \(109.67\) minutes

Step-by-step solution

Setting Up the Equation

We are given the function \[ L(t) = A(1 - e^{-kt}) \]In this case, \(A = 200\) words. Therefore, the function becomes \[ L(t) = 200(1 - e^{-kt}) \]

Solving for k

We know that \(L(5) = 20\) words have been learned in 5 minutes. Substitute these values into the equation to find \(k\):\[ 20 = 200(1 - e^{-5k}) \]Simplify to solve for \(k\):\[ 0.1 = 1 - e^{-5k} \]\[ -0.9 = -e^{-5k} \]\[ 0.9 = e^{-5k} \]Take the natural logarithm of both sides:\[ \text{ln}(0.9) = -5k \]\[ k = -\frac{\text{ln}(0.9)}{5} \]Calculate this value to get \(k\approx 0.0212\)

Amount Learned After 10 Minutes

Substitute \(t = 10\) and \(k = 0.0212\) into the equation:\[ L(10) = 200(1 - e^{-0.0212 \times 10}) \]\[ L(10) = 200(1 - e^{-0.212}) \]Using a calculator, we find \[ e^{-0.212} \approx 0.808 \]Therefore, \[ L(10) \approx 200(1 - 0.808) \]\[ L(10) \approx 200(0.192) \]\[ L(10) \approx 38.4 \] words

Amount Learned After 15 Minutes

Substitute \(t = 15\) and \(k = 0.0212\) into the equation:\[ L(15) = 200(1 - e^{-0.0212 \times 15}) \]\[ L(15) = 200(1 - e^{-0.318}) \]Using a calculator, we find \[ e^{-0.318} \approx 0.727 \]Therefore, \[ L(15) \approx 200(1 - 0.727) \]\[ L(15) \approx 200(0.273) \]\[ L(15) \approx 54.6 \] words

Time to Learn 180 Words

Set \(L(t) = 180\) and solve for \(t\):\[ 180 = 200(1 - e^{-0.0212t}) \]\[ 0.9 = 1 - e^{-0.0212t} \]\[ -0.1 = -e^{-0.0212t} \]\[ 0.1 = e^{-0.0212t} \]Take the natural logarithm of both sides:\[ \text{ln}(0.1) = -0.0212t \]\[ t = -\frac{\text{ln}(0.1)}{0.0212} \]Calculate this value to get \(t\approx 109.67\) minutes

Key concepts

Exponential Functions

Exponential functions are mathematical expressions in which a variable appears in the exponent. They are commonly represented in the form y = a * e^(bx), where e is the base of natural logarithms (approximately equal to 2.71828), and a and b are constants. These functions are widely used in various fields like biology, economics, and psychology to model growth and decay processes.
For example, in the original problem, the exponential function L(t) = 200(1 - e^{-0.0212t}) is used to measure the amount of words learned over time. Here, the term e^{-0.0212t} represents the decayed part of the learning curve.
Key characteristics of exponential functions include:

• Rapid growth or decay: Depending on the sign of the exponent, the function can describe quick increases or decreases over time.
• Asymptotic behavior: The function tends to approach a certain value as time moves towards infinity, but never actually reaches it.
• Continuous and smooth curve: The graphical representation is a smooth curve that continues infinitely in both directions.

Understanding exponential functions is crucial for interpreting data that changes exponentially, like the learning rate in our example.

Natural Logarithms

Natural logarithms are logarithms to the base e, where e is approximately 2.71828. Represented as ln(x), the natural logarithm of a number x is the exponent to which e must be raised to produce that number. In mathematical terms, if e^y = x, then ln(x) = y.
Natural logarithms are particularly useful in solving equations involving exponential functions, as shown in the original solution. For example, to find the rate of learning k, we took the natural logarithm of both sides of the equation e^{-5k} = 0.9, resulting in ln(0.9) = -5k.
Here are some essential properties of natural logarithms:

• ln(1) = 0 because e^0 = 1
• ln(e) = 1 because e^1 = e
• ln(x*y) = ln(x) + ln(y) (Product Rule)
• ln(x/y) = ln(x) - ln(y) (Quotient Rule)
• ln(x^n) = n*ln(x) (Power Rule)

Natural logarithms simplify the process of working with exponential equations, making it easier to solve for variables in complex algebraic expressions.

Solving for a Variable

Solving for a variable involves isolating the variable on one side of the equation to find its value. This process often requires a combination of algebraic techniques including addition, subtraction, multiplication, division, and taking logarithms.
For instance, in the given exercise, we needed to solve for the variable k to determine the learning rate. This required several steps:

• Start with the given equation: 20 = 200(1 - e^{-5k})
• Isolate the exponential term: 0.1 = 1 - e^{-5k}
• Simplify: 0.9 = e^{-5k}
• Take the natural logarithm of both sides: ln(0.9) = -5k
• Solve for k: k = -ln(0.9)/5

These steps illustrate the importance of breaking down a problem and applying the appropriate algebraic operations to isolate and solve for the unknown variable.

Applications in Psychology

Mathematical functions, especially exponential functions, have vital applications in psychology for modeling various processes like learning, memory retention, and behavior prediction. The exercise demonstrates how an exponential function can describe the learning curve of a student.
In this scenario, psychologists may use the function L(t) = 200(1 - e^{-0.0212t}) to gauge how many words a student will remember over time. The parameters A and k represent the total words to be learned and the learning rate, respectively. Such functions help in:

• Understanding learning rates: By determining the rate constant k, psychologists can identify how quickly someone acquires new information.
• Predicting learning outcomes: The function allows for prediction of the number of words learned at any given time, which helps in optimizing teaching strategies.
• Evaluating interventions: Changes in the learning rate can indicate the effectiveness of different educational interventions.

Understanding these applications helps bring mathematics to life, showing students the real-world relevance of what they learn in their textbooks.