Problem 73 - College Algebra with Modeling and Visualization

Question

Solve for the specified variable. $$ V=\frac{1}{3} \pi r^{2} h \text { for } r $$

Step-by-Step Solution

Answer

The formula for \( r \) is \( r = \sqrt{\frac{3V}{\pi h}} \).

Step 1: Isolate the term with the variable

The given formula is for the volume of a cone: \( V = \frac{1}{3} \pi r^2 h \). We want to solve for \( r \). Start by isolating the term with \( r^2 \) by eliminating other factors on the right-hand side. First, multiply both sides by 3 to get \( 3V = \pi r^2 h \).

Step 2: Eliminate the coefficient of the squared variable

To further isolate \( r^2 \), divide both sides of the equation by \( \pi h \). This gives \( r^2 = \frac{3V}{\pi h} \).

Step 3: Solve for the variable

Now that you have \( r^2 = \frac{3V}{\pi h} \), take the square root of both sides to solve for \( r \). This results in \( r = \sqrt{\frac{3V}{\pi h}} \).

Key Concepts

Understanding Geometry

Geometry is an essential branch of mathematics focused on the study of shapes, sizes, and the properties of space. It helps us in understanding and describing the physical world around us. When it comes to solving problems related to geometric figures, it often involves calculating dimensions, areas, and volumes.
A few key elements in geometry include angles, lines, shapes (like triangles, circles, and polygons), and more complex structures such as spheres and cones.

• Lines and angles help us understand the basics of shape construction.
• Triangles form the foundation for understanding more complex shapes.
• Circles and conical shapes bring us into the world of curved surfaces.

Understanding geometry allows us to solve problems related to spatial understanding and is invaluable when dealing with real-world applications ranging from architecture to physics.

Calculating the Volume of a Cone

The volume of a cone is a measure of the space it occupies in a three-dimensional space. It's essential in various applications, such as determining the amount of a substance that a conical container can hold. The fundamental formula to calculate the volume of a cone is: \[ V = \frac{1}{3} \pi r^2 h \]This formula dictates that the volume \( V \) is based on three dimensions:

• \( r \) is the radius of the circular base – the distance from the center to the edge of the base.
• \( h \) is the height of the cone – the perpendicular distance from the base to the tip of the cone.
• \( \pi \), a mathematical constant, is approximately 3.14159.

The multiplication by \( \frac{1}{3} \) accounts for the conical shape, being exactly a third of the volume of a cylinder with the same base and height. This formula helps to understand how volume scales with changes in radius and height.

Applying Algebraic Manipulation

Algebraic manipulation is a crucial mathematical technique enabling us to solve equations and express variables in terms of others. It involves rearranging equations to isolate variables and finding their values, as seen in solving for \( r \) in the volume formula for a cone.
Here’s a simplified breakdown of the process of rearranging formulas:

• Isolating the Variable: Start by moving terms to either side of the equation to leave the term with the desired variable by itself. In our example, multiply both sides of the equation by 3 to clear the fraction, resulting in \( 3V = \pi r^2 h \).
• Removing Coefficients: Divide or multiply as required to remove coefficients around the variable term. Example: divide by \( \pi h \) to obtain \( r^2 = \frac{3V}{\pi h} \).
• Solving for the Variable: Once the main term is isolated, employ root or other operations to solve for the variable itself. We take the square root, giving us \( r = \sqrt{\frac{3V}{\pi h}} \).

Mastering algebraic manipulation provides a powerful tool for solving equations and understanding real-world relationships between different quantities.