Problem 21 - Algebra for College Students

Question

Solve each of the following for the indicated variable. \(C=2 \pi r\) for \(r \quad\) (Circumference of a circle)

Step-by-Step Solution

Answer

The solution for \( r \) is \( r = \frac{C}{2\pi} \).

Step 1: Identify the formula

The given formula is the expression for the circumference of a circle: \[ C = 2 \pi r \] where \( C \) is the circumference and \( r \) is the radius.

Step 2: Isolate the variable

To solve for \( r \), we need to isolate it on one side of the equation. Start by dividing both sides of the equation by \( 2\pi \): \[ r = \frac{C}{2\pi} \]

Step 3: Confirm the isolated form

Check that the expression \( r = \frac{C}{2\pi} \) properly isolates \( r \) on one side, and it expresses \( r \) in terms of \( C \).

Key Concepts

Circumference of a Circle

The circumference of a circle is the total distance around the circle's edge or boundary. It's like measuring the length of a fence that goes all the way around a circular garden. This concept is essential in geometry because it helps in calculating properties and measurements related to circles. The formula for the circumference of a circle is given by:

• \( C = 2\pi r \), where \( C \) is the circumference and \( r \) is the radius.

The number \( \pi \) (pi) is a constant approximately equal to 3.14159. This constant represents the ratio of a circle's circumference to its diameter, making it a vital part of circle-related calculations. By understanding this formula, you can quickly determine or verify how long the path is around any circle, provided you know the radius.
Whether you're wrapping a string around a jar or designing a circular racetrack, knowing how to compute circumference is invaluable.

Isolating a Variable

Isolating a variable is a crucial skill in algebra, and it refers to manipulating an equation in order to express a specific variable on its own, typically on one side of the equation. This step is foundational in solving equations and can be used across various branches of mathematics. In the exercise, the goal is to solve for the radius \( r \) in terms of the circumference \( C \) using the formula \( C = 2\pi r \).
To isolate \( r \), we need to get it by itself on one side of the equation. We can achieve this by performing inverse operations. Here, you have the multiplication of \( r \) by \( 2\pi \). To undo this multiplication, divide both sides by \( 2\pi \).

• Divide both sides by \( 2\pi \): \( r = \frac{C}{2\pi} \)

This process of isolating \( r \) is essential for solving for unknowns and understanding the relationships between different variables in an equation. It's a systematic way of reversing operations to get the variable you are interested in by itself.

Algebraic Manipulation

Algebraic manipulation involves rearranging and modifying equations to arrive at a desired form. This often includes performing operations such as addition, subtraction, multiplication, division, or even factoring and expanding. Proper handling of these operations is necessary to solve equations and inequalities efficiently. In the provided exercise, we needed to use algebraic manipulation to solve for \( r \).
Starting with the equation for the circumference, the task was straightforward because it required basic operations. However, understanding what each step means is crucial:

• Recognizing that \( 2\pi \) is a coefficient: Here, \( 2\pi \) multiplies \( r \), so we reverse this by dividing.
• Performing division: Division by a number such as \( 2\pi \) involves distributing this division across the equation, essentially reversing the multiplication process.

These steps are relatively simple yet highlight the importance of algebraic manipulation. Once you become comfortable with these techniques, they allow you to solve much more complex equations, manipulating them into simpler forms to reveal unknown values.