Problem 63 - Precalculus: Functions and Graphs

Question

The formula occurs in the indicated application. Solve for the specified variable. \(d=\frac{1}{2} \sqrt{4 R^{2}-C^{2}}\) for \(C\) (segments of circles)

Step-by-Step Solution

Answer

\(C = 2\sqrt{R^{2} - d^2}\)

Step 1: Understand the given formula

We begin with the equation \(d=\frac{1}{2} \sqrt{4 R^{2}-C^{2}}\). We need to solve for \(C\), which involves isolating \(C\) on one side of the equation.

Step 2: Eliminate the Fraction

Multiply both sides of the equation by 2 to remove the fraction: \(2d = \sqrt{4R^{2} - C^{2}}\).

Step 3: Remove the Square Root

Square both sides of the equation to remove the square root: \((2d)^2 = 4R^{2} - C^{2}\).

Step 4: Expand the Square

Expand \((2d)^2\): \(4d^2 = 4R^{2} - C^{2}\).

Step 5: Isolate \(C^2\)

Rearrange the equation to solve for \(C^2\): \(C^{2} = 4R^{2} - 4d^2\).

Step 6: Solve for \(C\)

Take the square root of both sides to solve for \(C\): \(C = \sqrt{4R^{2} - 4d^2}\). Simplify further if possible: \(C = 2\sqrt{R^{2} - d^2}\).

Key Concepts

Circle Geometry

Circle geometry is a branch of mathematics that deals with properties and relations of points, lines, angles, and figures on a circle. It's an important part of geometry because many shapes and configurations involve circles or parts of circles. In this exercise, the formula given relates to a segment of a circle. A circle segment is the region between a chord and the corresponding arc of the circle.

• A circle has several key parts like the radius (R), diameter (d), arc length, and chords.
• Understanding the role of each is crucial for solving geometry problems.
• In our exercise, we focus on the radius and its relationship to other segments.
• The radius is the distance from the center of the circle to any point on its circumference.

Understanding these elements helps in applying formulas related to circle segments as seen in our given problem: solving for the chord length (C) when given d and R.

Isolation of Variables

Isolation of variables is a fundamental technique used to solve equations in algebra. The goal is to get the variable you are solving for by itself on one side of the equation. In the given exercise, we needed to isolate \( C \) from the equation.
To isolate a variable, follow these steps:

• Identify the variable you want to solve for.
• Use inverse operations to "undo" the operations surrounding your variable.
• Rearrange the equation step-by-step until the variable is on its own.

In our example, we systematically eliminated the fraction, squared both sides to remove the square root, and rearranged terms to isolate \( C^2 \), eventually leading us to \( C \). This clear step-by-step approach ensures that the variable is isolated correctly.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations using algebraic rules. It is essential for solving complex problems by breaking them into simpler steps.
In the given exercise, multiple algebraic manipulations were applied:

• Multiplication to eliminate fractions: \(2d = \sqrt{4R^2 - C^2}\).
• Squaring both sides to remove square roots: \((2d)^2 = 4R^2 - C^2\).
• Rearranging terms to isolate \(C^2\): \(C^2 = 4R^2 - 4d^2\).
• Finally, solving for \(C\): \(C = 2\sqrt{R^2 - d^2}\).

Each step uses basic algebra principles like addition, subtraction, multiplication, division, and handling exponents to simplify and solve the equation. It highlights the importance of careful and systematic manipulation for accuracy in solutions.