Problem 66 - Algebra and Trigonometry with Analytic Geometry

Question

The formula occurs in the indicated application. Solve for the specified variable. \(V=\frac{1}{3} \pi r^{2} h\) for \(h\)

Step-by-Step Solution

Answer

The solution for height \( h \) is \( h = \frac{3V}{\pi r^{2}} \).

Step 1: Review the Formula

The given formula is the volume of a cone: \[ V = \frac{1}{3} \pi r^{2} h \] where \( V \) represents the volume, \( r \) is the radius of the base of the cone, and \( h \) is the height. We need to solve this formula for the height \( h \).

Step 2: Isolate the Term Containing h

To solve for \( h \), we should first isolate the term that contains \( h \) on one side of the equation. The formula can be rewritten by multiplying both sides by \( 3 \) to eliminate the fraction:\[ 3V = \pi r^{2} h \]

Step 3: Solve for h

Next, we need \( h \) by itself, so we divide both sides of the equation by \( \pi r^{2} \):\[ h = \frac{3V}{\pi r^{2}} \] This formula now represents height in terms of volume and radius.

Key Concepts

Solving Equations

Solving equations involves finding the value of the unknown variable that makes the equation true. In this exercise, we aim to solve for the height \( h \) of a cone given its volume formula \( V = \frac{1}{3} \pi r^2 h \). The process generally starts with ensuring that the variable of interest is isolated on one side of the equation. This is done through a series of operations that effectively "undo" any relationships bundling the variable with numerical or literal constants.
The key steps include applying inverse operations. For example:

• If the variable is being multiplied by a number, you divide both sides by that number.
• If the variable is a part of a fraction, you might multiply through by the denominator to clear the fraction.

By rigorously following these steps, we systematically rearrange the given formula to make \( h \) the subject.

Geometry

Geometry offers us many tools to understand the shapes and forms that exist in our world, with cones being one such fascinating figure. A cone is a three-dimensional shape with a circular base that tapers smoothly to a point called the apex or vertex. The one discussed here is a right circular cone, meaning that if you were to draw a line from the center of the base to the apex, it would align perfectly with the axis running through the center of the circular base.
Discerning the volume of such a cone relies on understanding the relationship between its base radius \( r \), height \( h \), and the volume \( V \) itself. The formula \( V = \frac{1}{3} \pi r^2 h \) reflects the fact that the volume of a cone is one-third of the volume of a cylinder with the same base and height. This intuitive reduction is key to solving and rearranging the equation for different variables.

Algebraic Manipulation

Algebraic manipulation refers to the transformation of an equation or formula to achieve a specific arrangement or to solve for a particular variable. This skill is especially useful when working with formulas like the volume of a cone. In the exercise, the original formula \( V = \frac{1}{3} \pi r^2 h \) needed adjustment to solve for \( h \).
The primary steps of algebraic manipulation are:

• Rewriting the formula by clearing fractions, such as multiplying both sides by 3 to get \( 3V = \pi r^2 h \).
• Isolating the desired variable \( h \) by dividing both sides by the known quantities, \( \pi r^2 \) in this case, resulting in \( h = \frac{3V}{\pi r^2} \).

Through careful application of these methods, particularly focusing on maintaining equal balancing equations, we achieve a formula where \( h \) is now expressed in terms of calculable quantities, ensuring we can determine this dimension based on the cone's volume and radius.