Problem 66 - Elementary and Intermediate Algebra

Question

Solve each equation for the specified variable or expression. $$ d=\sqrt[3]{\frac{12 V}{\pi}} \text { for } V $$

Step-by-Step Solution

Answer

The solution for \( V \) is \( V = \frac{\pi d^3}{12} \).

Step 1: Understand the Equation

We start with the equation given: \[ d = \sqrt[3]{\frac{12 V}{\pi}} \]We are tasked with solving for \( V \). The equation involves a cube root, which suggests we'll need to raise both sides to the power of three to eliminate the root.

Step 2: Remove the Cube Root

To eliminate the cube root, we'll cube both sides of the equation:\[ d^3 = \left(\sqrt[3]{\frac{12 V}{\pi}}\right)^3 \]Simplifying the right side, we have:\[ d^3 = \frac{12 V}{\pi} \]

Step 3: Isolate the Variable V

Next, to solve for \( V \), we need to isolate \( V \) on one side of the equation. We do this by multiplying both sides by \( \pi \) to remove the fraction:\[ \pi d^3 = 12 V \]Then, divide both sides by 12:\[ V = \frac{\pi d^3}{12} \]

Step 4: Solution Verification

With \( V = \frac{\pi d^3}{12} \), plug this back into the original formula to verify. The solution holds as it satisfies the original equation when re-substituting.

Key Concepts

Cube Root

The term *cube root* refers to a special mathematical function. Instead of squaring a number, you 'cube' it by raising it to the power of three. When you take the cube root, you are essentially finding a number which, when multiplied by itself twice, results in the original number. In the context of equations, dealing with cube roots often involves raising both sides to the power of three to simplify the operation.
This operation is crucial for solving equations where the variable is inside a cube root. It allows us to remove the cube root, making it easier to work with. For example, if you have an equation like \(d = \sqrt[3]{x}\), cubing both sides gives you \(d^3 = x\). This step is often the first logical move when solving for variables within cube roots in algebraic expressions.

Isolate Variables

In algebra, *isolating variables* means manipulating an equation so that a single variable is by itself on one side of the equation. This process is fundamental in solving equations, especially when working with complex mathematical expressions or solving for specific variables.
To efficiently isolate a variable, you'll generally perform inverse operations, such as:

• Adding or subtracting terms from both sides to eliminate constant terms surrounding the variable.
• Multiplying or dividing both sides to eliminate coefficients or fractions.

These steps require a strong understanding of the properties of equality, which ensure that both sides of an equation remain balanced. For example, if you start with an expression \(\pi d^3 = 12 V\) and you need \( V \) isolated, you would divide both sides by 12 to isolate \( V \) and get \( V = \frac{\pi d^3}{12} \).
Keeping the equation balanced while isolating variables is crucial and a core technique in algebra.

Algebra

*Algebra* is a branch of mathematics that uses symbols and letters to represent numbers and quantities in equations and formulas. It's a powerful tool that allows us to solve problems by generalizing arithmetic operations and understanding relationships between variables.
In algebra, we often work with equations, which are statements that assert the equality of two expressions. Solving an equation involves transforming it into simpler forms while looking for solutions of the variable that satisfy the equation.
Key algebraic techniques include:

• Combining like terms to simplify expressions.
• Applying the distributive property to expand expressions.
• Using inverse operations to solve equations.

Understanding algebra involves a lot of pattern recognition and logical deduction to solve problems efficiently. In our example, algebra helps us manipulate an expression like \(d = \sqrt[3]{\frac{12 V}{\pi}}\) to eventually isolate and solve for \( V \). This ability to transform and solve expressions is what makes algebra a valuable tool in mathematics.