Problem 58 - College Algebra with Corequisite Support

Question

For the following exercises, determine the function described and then use it to answer the question. The volume, \(V\), of a sphere in terms of its radius, \(r\), is given by \(V(r)=\frac{4}{3} \pi r^{3}\). Express \(r\) as a function of \(V\), and find the radius of a sphere with volume of 200 cubic feet.

Step-by-Step Solution

Answer

The radius of the sphere is approximately 3.63 feet.

Step 1: Understanding the Problem

We are given the volume of a sphere as a function of its radius: \( V(r) = \frac{4}{3} \pi r^3 \). We need to express the radius \( r \) as a function of volume \( V \), then find \( r \) when \( V = 200 \) cubic feet.

Step 2: Rearranging the Equation

Start with the equation for the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \). To find \( r \) as a function of \( V \), solve for \( r \) by isolating \( r^3 \). Multiply both sides by \( \frac{3}{4} \) to get \( r^3 = \frac{3V}{4\pi} \).

Step 3: Solving for the Radius

Take the cube root of both sides of the equation \( r^3 = \frac{3V}{4\pi} \) to solve for \( r \). This gives \( r = \left(\frac{3V}{4\pi}\right)^{1/3} \).

Step 4: Calculating the Radius for Given Volume

Substitute \( V = 200 \) into the expression for \( r \): \( r = \left(\frac{3 \times 200}{4\pi}\right)^{1/3} \). Simplify the expression and compute the cube root: \( r = \left(\frac{600}{4\pi}\right)^{1/3} = \left(\frac{150}{\pi}\right)^{1/3} \). Use a calculator to approximate \( r \approx 3.63 \) feet.

Key Concepts

Radius as a Function of Volume

In the realm of geometry, understanding how to express one variable in terms of another is crucial. When dealing with spheres, we often know the volume and want to find the radius. The volume of a sphere, given by the formula \( V = \frac{4}{3} \pi r^3 \), shows that volume is a function of the radius. However, sometimes we need the reverse: expressing radius \( r \) as a function of volume \( V \). This requires us to manipulate our formula. To do this, we rearrange it to solve for \( r \).
This is done by isolating \( r^3 \) first, then taking the cube root to get \( r \) in terms of \( V \). Understanding and using this radius function is essential for solving sphere related problems when only the volume is provided.

Algebraic Manipulation

Algebraic manipulation is a skill that greatly aids in transforming equations into useful forms. In our sphere volume formula, we start with \( V = \frac{4}{3} \pi r^3 \) and seek to find \( r \). To do this, several algebraic steps are employed:

• Multiply both sides by \( \frac{3}{4} \) to cancel the \( \frac{4}{3} \) fraction on the right-hand side.
• This gives \( r^3 = \frac{3V}{4\pi} \), neatly isolating \( r^3 \).

Each step requires careful execution to maintain the equality.
This ability to manipulate equations is fundamental in solving for variables within any mathematical context, providing the tools needed to find unknowns based on given information.

Taking the Cube Root

A pivotal step in solving for the radius \( r \) is taking the cube root of the expression \( r^3 = \frac{3V}{4\pi} \). The cube root \( \sqrt[3]{x} \) finds the number which, when multiplied by itself twice more, yields \( x \). This is necessary when solving for a variable that is raised to the third power.
To find \( r \), we write:

• \( r = \left(\frac{3V}{4\pi}\right)^{1/3} \)

This notation simply means that we are taking the cube root of the entire expression inside the parentheses. Understanding cube roots is important in manipulating and interpreting such expressions, especially when dealing with geometric shapes where volume is a cube function of linear dimensions.

Understanding and Using Geometric Formulas

Geometric formulas like the volume of a sphere are fundamental tools in mathematics. They allow us to describe and calculate properties of geometric shapes. The formula \( V = \frac{4}{3} \pi r^3 \) is used to calculate the volume based on a given radius.
However, by manipulating these formulas, we can also solve for other variables, showcasing their flexibility.

• Spheres, being perfectly symmetrical, offer simpler computation models with their formulas.
• They allow transformations and flexibility through substitution and algebraic manipulation methods.

Understanding these formulas not only aids in solving specific problems but also enhances overall mathematical literacy by providing insights into how these shapes work and relate to each other.