Problem 28 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve the formula for the volume of a cone for \(h\)

Step-by-Step Solution

Answer

So, solving the formula for the volume of a cone for \(h\), we get \(h = \frac{3V}{\pi r^2 }\)

Step 1: Identify the formula

The formula for the volume of a cone is given as \(V=\frac{1}{3}\pi r^2 h\). The goal is to isolate \(h\) in this equation.

Step 2: Remove \( \frac{1}{3}\pi r^2 \) from the right hand side

Since \(h\) is being multiplied by \( \frac{1}{3}\pi r^2 \), divide each side of the equal sign by \( \frac{1}{3}\pi r^2 \), resulting in the equation \( \frac{V}{\frac{1}{3}\pi r^2 } = h \)

Step 3: Simplify the new equation for \( h \)

Now simplify the left-hand side of the equation. The equation simplifies to \( h = \frac{3V}{\pi r^2 }\)

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a critical skill in mathematics that involves rearranging equations and expressions to solve for particular variables. When approaching a problem like solving the formula for the volume of a cone to find the height \(h\), strong algebraic manipulation skills allow you to isolate the variable of interest. In this process, understanding and correctly applying the rules of algebra—such as the order of operations, and the properties of equality—is essential for success.

For instance, if you have an equation like \(V = \frac{1}{3}\pi r^2 h\), where you want to solve for \(h\), you must perform a series of operations to manipulate the equation to get \(h\) by itself on one side of the equal sign. This process typically involves operations such as adding, subtracting, multiplying, dividing, factoring, and canceling out terms. In the given example, by dividing both sides by \(\frac{1}{3}\pi r^2\), you're using division to undo the multiplication that's applying to the variable \(h\).

This manipulation leads to the new equation \(h = \frac{3V}{\pi r^2}\), thus making the value of \(h\) explicit.

Volume Formulas

Volume formulas are essential tools for calculating the capacity of three-dimensional objects. In geometry, different shapes have specific volume formulas associated with them. To solve volume problems effectively, you need to know the correct formula to apply to the given situation.

For a cone, the volume formula is \(V = \frac{1}{3}\pi r^2 h\), where \(V\) represents volume, \(r\) is the radius of the base, and \(h\) is the height of the cone. The \(\frac{1}{3}\) in the formula accounts for the fact that a cone can be thought of as one-third of a cylinder that has the same base and height. The \(\pi r^2\) portion represents the area of the base circle.

This formula is derived by integrating the area of circles of varying sizes to form a cone. Remembering that volume measurement is in cubic units (such as cubic centimeters or cubic meters) is important, as it represents the three-dimensional space occupied within a shape.

Geometric Problem Solving

Geometric problem-solving combines the understanding of shapes and their properties with mathematical strategies to solve for unknowns. When tackling a problem that requires finding a missing dimension like the height of a cone based on its volume, geometric intuition plays a vital role.

Understanding the relationship between different dimensions of a cone and visualizing the shape can aid in formulating a strategy for solving the problem. Identifying that the cone's volume depends on both its height and the area of its base is the first step. Subsequently, applying the volume formula for the cone and performing algebraic manipulation can yield the height.

The ability to disassemble tricky geometric problems into individual components that can be tackled with algebraic manipulation and volume formulas ensures a systematic and successful approach to finding solutions. Moreover, practicing such problems enhances your capability to view complex three-dimensional figures in simpler terms and solve for any unknown variable easily.