Problem 57 - Precalculus: Functions and Graphs

Question

The formula occurs in the indicated application. Solve for the specified variable. \(\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\) for \(q\) (lens equation)

Step-by-Step Solution

Answer

The solution for \(q\) is \(q = \frac{pf}{p - f}\).

Step 1: Understand the formula and identify the target

The lens equation is given by \(\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\). Our goal is to solve for \(q\), which means that we need to isolate \(q\) on one side of the equation.

Step 2: Subtract \(\frac{1}{p}\) from both sides

To isolate \(\frac{1}{q}\) on one side, subtract \(\frac{1}{p}\) from both sides of the equation: \(\frac{1}{f} - \frac{1}{p} = \frac{1}{q}\).

Step 3: Simplify the left side of the equation

The left side, \(\frac{1}{f} - \frac{1}{p}\), can be simplified by finding a common denominator, which is \(pf\). Therefore, the expression becomes \(\frac{p - f}{pf} = \frac{1}{q}\).

Step 4: Take the reciprocal of both sides

To solve for \(q\), take the reciprocal of both sides of the equation: \(q = \frac{pf}{p - f}\).

Key Concepts

Variable Isolation

In mathematics, one of the common tasks is to solve equations for a specific variable. This is referred to as "variable isolation." In our exercise, the lens equation \( \frac{1}{f}=\frac{1}{p}+\frac{1}{q} \) needs to be rearranged to solve for the variable \( q \). To isolate a variable, you must manipulate the equation so that the variable stands alone on one side. Here's how you do it:

• Identify which variable you need to solve for. Here, it's \( q \).
• Perform operations such as addition, subtraction, multiplication, or division to both sides of the equation to "remove" other terms from the side with your variable.

In our example, we subtracted \(\frac{1}{p}\) from both sides to isolate the term \(\frac{1}{q}\) on one side of the equation, resulting in \( \frac{1}{f} - \frac{1}{p} = \frac{1}{q} \). This step is vital because it sets up the equation for simpler manipulation and understanding.

Rational Expressions

Rational expressions consist of fractions with polynomials in the numerator and the denominator. In our lens equation, terms like \( \frac{1}{f} \), \( \frac{1}{p} \), and \( \frac{1}{q} \) are examples of rational expressions. When working with rational expressions, you frequently need to find a common denominator to combine or simplify terms. In our step-by-step solution, after subtracting \(\frac{1}{p}\) from each side, we needed to simplify \( \frac{1}{f} - \frac{1}{p} \). The common denominator here is \( pf \), which allows us to rewrite the expression as \( \frac{p - f}{pf} \). This technique of finding a common denominator is crucial because:

• It allows the combining of fraction terms effectively.
• It simplifies the expression to make calculations more straightforward.

Mastering rational expressions can significantly impact your problem-solving skills, especially in algebra and calculus.

Reciprocal

The concept of a reciprocal is fundamental in fractions and equations involving inverses. A reciprocal of a number \( x \), is simply \( \frac{1}{x} \). For example, the reciprocal of \( 5 \) is \( \frac{1}{5} \), and the reciprocal of \(\frac{1}{7}\) is \( 7 \).In our original step-by-step solution for the lens equation, we reached a point where \( \frac{1}{q} = \frac{p - f}{pf} \). To solve for \( q \), it was necessary to take the reciprocal of both sides, thus flipping the fraction. The concept looks like this:

• If \( \frac{1}{q} = \frac{p-f}{pf} \), then \( q = \frac{pf}{p-f} \).

Taking reciprocals is a standard technique used to isolate a variable in rational expressions, particularly when the variable appears in the denominator. Understanding this concept will greatly aid you in rearranging various equations and expressions throughout your study of math.