Problem 61 - College Algebra with Corequisite Support

Question

For the following exercises, determine the function described and then use it to answer the question. The period \(T,\) in seconds, of a simple pendulum as a function of its length \(l\), in feet, is given by \(T(l)=2 \pi \sqrt{\frac{l}{32.2}}\). Express \(l\) as a function of \(T\) and determine the length of a pendulum with period of 2 seconds.

Step-by-Step Solution

Answer

The length of the pendulum is approximately 3.262 feet.

Step 1: Understand the Given Function

We are given the function for the period of a pendulum, \(T(l) = 2 \pi \sqrt{\frac{l}{32.2}}\). This describes how the period \(T\) depends on the length \(l\) of the pendulum. Our task is to express \(l\) as a function of \(T\).

Step 2: Solve for Length (l) in terms of Period (T)

To express \(l\) as a function of \(T\), we need to isolate \(l\) in the equation:1. Start with the given equation: \(T = 2 \pi \sqrt{\frac{l}{32.2}}\).2. Divide both sides by \(2 \pi\): \(\frac{T}{2 \pi} = \sqrt{\frac{l}{32.2}}\).3. Square both sides to eliminate the square root: \(\left(\frac{T}{2 \pi}\right)^2 = \frac{l}{32.2}\).4. Multiply both sides by 32.2 to solve for \(l\): \(l = 32.2 \cdot \left(\frac{T}{2 \pi}\right)^2\).Now, \(l(T) = 32.2 \left(\frac{T}{2 \pi}\right)^2\).

Step 3: Calculate Length for Given Period

Substitute the given period \(T = 2\) seconds into the function \(l(T)\):1. \(l = 32.2 \left(\frac{2}{2 \pi}\right)^2\).2. Simplify the expression: \(l = 32.2 \left(\frac{1}{\pi}\right)^2\).3. Simplify further: \(l = 32.2 \cdot \frac{1}{\pi^2}\).4. Calculate the numerical value: \(\pi^2 \approx 9.8696\), so \(l \approx 32.2 \cdot \frac{1}{9.8696}\).5. Finally, \(l \approx 3.262\) feet.

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a powerful tool that enables the solving of equations by systematically rearranging and simplifying expressions.
In the context of the given exercise, we needed to express the length of a pendulum, represented by \(l\), in terms of its period \(T\). To do this, we employed several algebraic techniques.
Here’s a simplified breakdown of the steps:

• **Division**: We started with the equation \(T = 2\pi \sqrt{\frac{l}{32.2}}\). By dividing both sides by \(2\pi\), we aimed to isolate the square root term: \(\frac{T}{2\pi} = \sqrt{\frac{l}{32.2}}\).
• **Squaring**: To eliminate the square root, we squared both sides of the equation, resulting in \(\left(\frac{T}{2\pi}\right)^2 = \frac{l}{32.2}\).
• **Multiplication**: Finally, to fully isolate \(l\), we multiplied both sides by 32.2, giving us \(l = 32.2 \left(\frac{T}{2\pi}\right)^2\).

By following these sequential manipulations, we successfully expressed the pendulum's length as a function of its period.

Function Inversion

Function inversion is the process of rearranging a function to solve for a different variable, essentially reversing the roles of dependent and independent variables.
In our scenario, we began with a function \(T(l) = 2\pi \sqrt{\frac{l}{32.2}}\), which dictates how the period \(T\) depends on the length \(l\).
To determine the length as a function of the period, we used these steps:

• First, we addressed the placement of \(T\) and \(l\) in the equation, aiming to make \(l\) the focus.
• Throughout the process, prevailing emphasis was placed on isolating \(l\) using algebraic manipulation steps, such as dividing, squaring, and multiplying.
• Upon completing these transformations, the function became \(l(T) = 32.2 \left(\frac{T}{2\pi}\right)^2\), indicating a successful inversion from period-based expression to length-oriented one.

This approach enhanced our ability to determine the length by directly plugging in the period, showcasing the versatility and utility of function inversion.

Square Root Elimination

When dealing with equations containing square roots, one common challenge arises: removing the square root to isolate variables. In this exercise, we encountered this while transitioning the period-based function to a length-based one.
To eliminate the square root, one key step was taken:

• **Square Both Sides**: From \(\frac{T}{2\pi} = \sqrt{\frac{l}{32.2}}\), we squared each side, transforming it into \(\left(\frac{T}{2\pi}\right)^2 = \frac{l}{32.2}\).
• This action effectively neutralized the square root by moving the subject out of the radical form.

By squaring both sides, the simplified expression allowed further manipulation, leading ultimately to the solution of \(l\) as a function of \(T\).
Understanding square root elimination helps us in simplification and aids us in accurate derivations of unknown quantities.