Problem 65 - College Algebra with Corequisite Support

Question

For the following exercises, determine the function described and then use it to answer the question. Consider a cone with height of 30 feet. Express the radius, \(r,\) in terms of the volume, \(V,\) and find the radius of a cone with volume of 1000 cubic feet.

Step-by-Step Solution

Answer

The radius of the cone is approximately 5.64 feet.

Step 1: Understand the Formula for the Volume of a Cone

The volume of a cone is given by the formula \( V = \frac{1}{3} \pi r^2 h \), where \( V \) is the volume, \( r \) is the radius of the base, and \( h \) is the height. We are given the height \( h = 30 \) feet.

Step 2: Rearrange the Formula to Express Radius in Terms of Volume

Starting with the volume formula \( V = \frac{1}{3} \pi r^2 h \), substitute \( h = 30 \) to get \( V = \frac{1}{3} \pi r^2 \cdot 30 \). Simplify to \( V = 10 \pi r^2 \). Now solve for \( r^2 \): \( r^2 = \frac{V}{10\pi} \). Then, solve for \( r \) by taking the square root: \( r = \sqrt{\frac{V}{10\pi}} \).

Step 3: Substitute the Volume into the Radius Formula

Substitute \( V = 1000 \) cubic feet into the formula \( r = \sqrt{\frac{V}{10\pi}} \) to find \( r = \sqrt{\frac{1000}{10\pi}} \).

Step 4: Calculate the Radius

Calculate \( \frac{1000}{10\pi} = \frac{100}{\pi} \), and then take the square root to find \( r = \sqrt{\frac{100}{\pi}} \). Using a calculator, find the approximate value: \( r \approx \frac{10}{\sqrt{\pi}} \approx 5.64 \) feet.

Key Concepts

Mathematical Formulas

When dealing with the volume of a cone, it's crucial to understand the mathematical formula used to determine it. The volume formula is: \[ V = \frac{1}{3} \pi r^2 h \]where:

• \( V \) represents the volume of the cone,
• \( r \) is the radius of the cone's base, and
• \( h \) is the height of the cone.

This formula shows that the volume is one-third of the product of the base's area \( (\pi r^2) \) and the height \( h \). Understanding this relationship helps in rearranging the formula to solve for different variables, depending on which parameters are known or unknown.
For our specific problem, we are provided with a height of 30 feet and need to determine the radius when the volume is 1000 cubic feet.

Geometry Concepts

In geometry, a cone is a three-dimensional shape with a circular base and a pointed top, known as the apex. The line segment from the apex to the center of the base is the height, and the radius extends from the center of the base to its perimeter.
Visualizing the cone as a combination of its 2D circular base and 3D properties can clarify how the volume formula comes to be. The cone's base area is calculated as \( \pi r^2 \). The height \( h \), when combined with this base area, helps determine the cone's capacity or volume.
This concept of integrating a base area with a height is central to other geometric shapes like cylinders, where the volume formula \( \pi r^2 h \) shows a full base-area multiplication unlike the cone's one-third factor.
Such geometric insights are key to making sense of the transformations in the cone's volume formula.

Problem Solving Steps

Solving the problem of finding the radius of a cone with an unknown volume involves clear, methodical steps.First, return to the general volume formula \( V = \frac{1}{3} \pi r^2 h \) and recognize what needs to be isolated. Here, our task is to express \( r \) in terms of \( V \).
Start by incorporating the given height, which simplifies our formula to \( V = 10 \pi r^2 \). This step condenses the equation to focus on \( r^2 \), allowing you to isolate \( r \) by forming the equation \( r^2 = \frac{V}{10\pi} \).
Finally, solve for \( r \) by taking the square root of the entire expression:\[ r = \sqrt{\frac{V}{10\pi}} \]Next, substitute the specific volume value, \( V = 1000 \) cubic feet. Calculate\[ r = \sqrt{\frac{1000}{10\pi}} = \sqrt{\frac{100}{\pi}} \]This gives an approximate radius of 5.64 feet when calculated.
Understanding each manipulation and substitution step helps in not just solving similar problems, but also broadening your comprehension of the relationships within geometry equations.