Problem 45 - Precalculus Student Solutions Manual 5th

Question

Solve each formula for the indicated variable. Assume that the denominator is not 0 if variables appear in the denominator. See Examples \(4(a)\) and \((b)\). \(S=2 \pi r h+2 \pi r^{2}, \quad\) for \(h \quad\) (surface area of a right circular cylinder)

Step-by-Step Solution

Answer

h = \frac{S - 2 \pi r^2}{2 \pi r}

Step 1 - Identify the given equation and target variable

The given equation is the surface area formula for a right circular cylinder: \[ S = 2 \pi r h + 2 \pi r^2 \] The target variable to solve for is \( h \).

Step 2 - Isolate the term containing the variable h

Subtract \(2 \pi r^2\) from each side of the equation to isolate the term with \(h\).\[ S - 2 \pi r^2 = 2 \pi r h \]

Step 3 - Solve for h

Divide each side of the equation by \(2 \pi r\) to solve for \( h \):\[ h = \frac{S - 2 \pi r^2}{2 \pi r} \]

Key Concepts

Solving Equations

Solving equations is a critical skill in mathematics, which allows you to find the value of unknown variables. In the given exercise, we’re dealing with the formula for the surface area of a right circular cylinder. To solve this equation for the variable h (height), we use a series of algebraic steps. Here are some key points to remember about solving equations:

• Identify the equation and the variable you need to solve for.
• Isolate the term that contains the variable.
• Perform algebraic operations to get the variable by itself.

Always ensure each step you take is balanced by performing the same operation on both sides of the equation. This keeps the equation valid and helps you arrive at the correct answer.

Cylinder Surface Area

The surface area of a right circular cylinder is represented by the formula \( S = 2 \pi rh + 2 \pi r^{2} \). This equation combines the lateral (side) surface area with the areas of the two circular bases.

• 2 \pi rh represents the lateral surface area of the cylinder.
• 2 \pi r^{2} accounts for the area of both circular bases.

Understanding this formula is crucial for problems involving cylinders. It helps you comprehend how changes in radius or height affect the total surface area.

Variable Isolation

Variable isolation involves rearranging an equation to solve for a specific variable. In the example given, we needed to solve for h, the height of the cylinder. Here’s the process:

• Start with the equation: \( S = 2 \pi r h + 2 \pi r^{2} \).
• Subtract \( 2 \pi r^{2} \) from both sides to isolate the term with h: \( S - 2 \pi r^{2} = 2 \pi rh \).
• Divide both sides by \( 2 \pi r \): \( h = \frac{S - 2 \pi r^{2}}{2 \pi r} \).

This isolates h, giving you its value in terms of other variables. Practicing this skill helps with a wide range of mathematical problems.

Algebraic Manipulation

Algebraic manipulation involves the use of operations and properties of equality to solve equations. In our surface area formula problem, we used several manipulation techniques:

• Subtraction: We subtracted \(2 \pi r^{2} \) to isolate the term containing h.
• Division: We divided both sides by \( 2 \pi r \) to solve for h.

Other common techniques include addition and multiplication. Mastering these techniques is essential for efficiently solving complex equations and understanding their underlying relationships. Make sure to consistently apply these operations to both sides of an equation to maintain its balance and correctness.