Problem 11 - Single Variable Calculus

Question

The change from religious to lay teachers at Roman Catholic schools has been partly attributed to the decline in the number of women and men entering religious orders. The percentage of teachers who are lay teachers is given by $$ f(t)=\frac{98}{1+2.77 e^{-t}} \quad 0 \leq t \leq 4 $$ where \(t\) is measured in decades, with \(t=0\) corresponding to the beginning of 1960 . a. What percentage of teachers were lay teachers at the beginning of 1990 ? b. Find the year when the percentage of lay teachers was increasing most rapidly.

Step-by-Step Solution

Answer

a. At the beginning of 1990, approximately 85.27% of teachers were lay teachers. b. The percentage of lay teachers was increasing most rapidly at around the year 1970.

Step 1: Find the value of t for 1990

To find the percentage of lay teachers at the beginning of 1990, first, we need to determine the value of t. Since t=0 corresponds to the beginning of 1960, we can calculate the value of t for 1990 as: t = Years since 1960 / 10 = (1990 - 1960) / 10 = 30 / 10 = 3

Step 2: Calculate f(t) for t=3

Now, plug t=3 into the function f(t) to find the percentage of lay teachers in 1990: f(3) = \(\frac{98}{1 + 2.77e^{-3}}\) Calculate the value of f(3) to get the percentage of lay teachers in 1990: f(3) ≈ 85.27% b. Finding the year when the percentage of lay teachers was increasing most rapidly:

Step 3: Compute the derivative of f(t)

We will find the derivative of f(t) with respect to t to understand how it is changing with time: f'(t) = \(\frac{d}{dt} \; (\frac{98}{1 + 2.77e^{-t}})\) Using the quotient rule: f'(t) = \(\frac{(1 + 2.77e^{-t})(98 * 0) - 98(2.77 * -1 * e^{-t})}{(1 + 2.77e^{-t})^{2}}\) f'(t) = \(\frac{98 * 2.77 * e^{-t}}{(1 + 2.77e^{-t})^{2}}\)

Step 4: Set f'(t) to 0 and Solve for t

To find the point of maximum increase, we will find the critical points by setting f'(t) = 0. 0 = \(\frac{98 * 2.77 * e^{-t}}{(1 + 2.77e^{-t})^{2}}\) Rewriting the equation, to solve for t, initially, we calculate the value of e^t: \(e^t = 2.77\) Take the natural logarithm of both sides: ln(\(e^t\)) = ln(2.77) t = ln(2.77) Calculate the value of t: t ≈ 1.02

Step 5: Determine the corresponding year for t ≈ 1.02

Convert the value of t back to the corresponding year: Years since 1960 = t * 10 = 1.02 * 10 = 10.2 Year = 1960 + 10.2 = 1970.2 So, the percentage of lay teachers was increasing most rapidly at 1970, approximately.

Key Concepts

Rate of Change

Understanding the rate of change is crucial in mathematics, especially when analyzing how a function behaves over time. Essentially, the rate of change tells us how quickly or slowly a variable is changing relative to another variable. In the context of the exercise, we consider how the percentage of lay teachers changes over time.

A function's rate of change can be constant or variable. When it's constant, the function is linear, indicating a steady increase or decrease. When it's variable, like with most real-world scenarios, the function is usually non-linear.

For example, in this problem, the rate of change is determined by the function's derivative. By examining the derivative, you can see how fast the percentage of lay teachers is increasing or decreasing at any given moment. When you calculate this on the interval from 0 to 4 decades (1960 to 2000), it helps identify periods of significant change.

Derivative

The derivative is a fundamental tool in calculus used to measure how a function changes at any point. It's like a mathematical magnifying glass, giving detailed insights into a function's behavior.

To calculate a derivative of a function, such as the given percentage function of lay teachers, we consider the rate of change of the function's output with respect to the input variable, in this case, time.

• The derivative of a function provides the slope of the tangent line at any point on the curve of the function.
• A positive derivative indicates that the function is increasing at that point, while a negative derivative means it's decreasing.
• The more detailed we examine the derivative, the better we grasp rate and nature of changes in the function's trend.

In the exercise, by determining the derivative \( f'(t) \), we can find out where this rate of change (i.e., the percentage of lay teachers) is most rapid. This is done by setting the derivative equal to zero, a method used to find critical points where maxima or minima occur.

Exponential Functions

Exponential functions model situations where growth or decay accelerates rapidly, rather than occurring at a constant pace. These functions are a key part of calculus and help describe phenomena in fields like finance, biology, and in this case, education.

The function in the exercise, \( f(t) = \frac{98}{1 + 2.77 e^{-t}} \), is an example of an exponential function. Here's why exponential functions matter in this context:

• They can model non-linear growth, which means that instead of adding the same amount each period, the value grows by a certain factor. Here, the decrease of religious teachers is modeled exponentially.
• The component \( e^{-t} \) represents an exponential decay factor, which suggests as time goes on (from 1960 to 1990), the value approaches a limiting value or a maximum threshold, here likely around 98%.
• Using exponential models helps identify long-term trends, such as how teacher demographics shift over decades.

Understanding how exponential functions work is critical for grasping longer-term trends that occur in many natural and social phenomena, including the shift from religious to lay teaching staff in the exercise.