Problem 91 - Algebra and Trigonometry

Question

Solve the equation for the indicated variable. \(V=\frac{1}{3} \pi r^{2} h, \quad\) for \(r\)

Step-by-Step Solution

Answer

\( r = \sqrt{\frac{3V}{\pi h}} \)

Step 1: Substitute the expression

The formula given is the volume of a cone, \( V = \frac{1}{3} \pi r^2 h \). We need to solve for \( r \). First, we'll eliminate the fractional coefficient by multiplying both sides by 3, resulting in: \( 3V = \pi r^2 h \).

Step 2: Isolate the term containing r

Next, we want to isolate the term \( r^2 \). We can do this by dividing both sides by \( \pi h \). Thus, we have: \( r^2 = \frac{3V}{\pi h} \).

Step 3: Solve for r

To solve for \( r \), take the square root of both sides. This gives us \( r = \sqrt{\frac{3V}{\pi h}} \).

Key Concepts

Volume of a Cone

The volume of a cone is a topic that pops up frequently in geometry and algebra. It's important to understand the shape and dimensions we're dealing with.
A cone is a 3-dimensional shape with a circular base and a point called the apex. To find the volume of a cone, we use the formula: \[ V = \frac{1}{3} \pi r^2 h \] Where:

• \( V \) is the volume
• \( r \) is the radius of the base
• \( h \) is the height of the cone

This formula allows us to calculate the space inside the cone, measured in cubic units. The key here is to visualize it as one-third of the volume of a cylinder with the same base and height.
By understanding this formula, you can plug in known measurements of a cone to find its volume. This is essential for many practical applications, such as determining how much liquid can fit into a conical container.

Solving for a Variable

Solving for a variable is often necessary when you need to express one variable in terms of others. In our original problem, we are tasked with solving for the radius \( r \) in the volume formula of a cone.
The process involves rearranging the formula to isolate the desired variable. Here’s how it’s done:

• The equation we started with is \( V = \frac{1}{3} \pi r^2 h \).
• We wanted \( r \), so first, we multiplied both sides by 3 to remove the fraction: \( 3V = \pi r^2 h \).
• Next, to isolate \( r^2 \), we divided both sides by \( \pi h \): \( r^2 = \frac{3V}{\pi h} \).
• Finally, we took the square root of both sides to solve for \( r \): \( r = \sqrt{\frac{3V}{\pi h}} \).

This sequence of steps, known as "isolating the variable," is crucial in algebra. Understanding how to manipulate equations allows you to solve not just geometric problems, but a wide range of algebraic problems.

Mathematical Operations

Mathematical operations are the building blocks of solving equations. Breaking down each step into smaller operations makes complex problems simpler.
In our exercise, there were several operations used to arrive at the final answer.

• Multiplication: We started by multiplying both sides of the initial equation by 3. This dealt with the fraction \( \frac{1}{3} \).
• Division: Next, we divided both sides by \( \pi h \) to isolate \( r^2 \). Division helps to "move" terms and adjust an equation's balance.
• Square Root: Finally, we employed the square root to transition from \( r^2 \) to \( r \). Taking square roots is a common method for solving squared variables.

Embracing each mathematical operation individually, while maintaining the integrity of the equation, is key to effective problem solving. Understanding how and why each operation is performed allows students to tackle a variety of algebraic challenges with confidence.