Problem 62 - College Algebra with Corequisite Support

Question

For the following exercises, determine the function described and then use it to answer the question. The volume of a cylinder, \(V\), in terms of radius, \(r\), and height, \(h,\) is given by \(V=\pi r^{2} h .\) If a cylinder has a height of 6 meters, express the radius as a function of \(V\) and find the radius of a cylinder with volume of 300 cubic meters.

Step-by-Step Solution

Answer

The radius of the cylinder is approximately 3.99 meters for a volume of 300 cubic meters.

Step 1: Express the Function

Start with the formula for the volume of a cylinder: \[ V = \pi r^2 h. \] Given that the height \( h = 6 \) meters, we can substitute \( h \) into the equation: \[ V = \pi r^2 \times 6. \] Simplify this to get: \[ V = 6\pi r^2. \] Our goal is to express the radius \( r \) in terms of \( V \).

Step 2: Isolate the Radius

To express \( r \) as a function of \( V \), divide both sides by \( 6\pi \): \[ \frac{V}{6\pi} = r^2. \] Next, take the square root of both sides to solve for \( r \): \[ r = \sqrt{\frac{V}{6\pi}}. \] This equation expresses the radius as a function of the volume \( V \).

Step 3: Calculate the Radius for Given Volume

Now that we have the formula \( r = \sqrt{\frac{V}{6\pi}} \), plug in \( V = 300 \) to find \( r \): \[ r = \sqrt{\frac{300}{6\pi}}. \] Simplifying inside the square root: \[ \frac{300}{6} = 50, \] so the equation becomes: \[ r = \sqrt{\frac{50}{\pi}}. \] Calculate this to find \( r \).

Step 4: Final Calculation

Using a calculator, approximate \( \sqrt{\frac{50}{\pi}} \). The value of \( \pi \) is approximately 3.14159, so: \[ \frac{50}{\pi} \approx 15.92. \] Then, calculate the square root: \[ \sqrt{15.92} \approx 3.99. \] Thus, the radius \( r \) is approximately 3.99 meters.

Key Concepts

Volume of a Cylinder

The volume of a cylinder is a fundamental concept in geometry and is expressed in cubic units. It describes how much space is enclosed within the cylinder's surface.
Understanding the formula for the volume of a cylinder is essential. The formula is given by:

• \[ V = \pi r^2 h \]

in which:

• \( V \) represents the volume,
• \( r \) symbolizes the radius of the circular base,
• \( h \) denotes the height of the cylinder.

By plugging known measurements of the radius and height into this equation, you can determine the volume of a cylinder. In exercises, sometimes you need to manipulate this formula to find other unknown variables, such as the radius or height when the volume is given.

Radius as a Function

In many problems, you may need to express one quantity as a function of another. Here, we are asked to express the radius as a function of the volume, which means we want a rule that tells us how to compute the radius given a specific volume.
We begin with the volume formula for a cylinder:

• \[ V = \pi r^2 h \]

Since the height \( h \) is known (e.g., 6 meters), we substitute \( h \) in:

• \[ V = 6\pi r^2 \]

To isolate \( r \), divide both sides by \( 6\pi \):

• \[ \frac{V}{6\pi} = r^2 \]

Finally, take the square root of both sides to solve for \( r \):

• \[ r = \sqrt{\frac{V}{6\pi}} \]

This expression allows you to calculate the radius when the volume of the cylinder is known.

Algebraic Expressions

An algebraic expression is a combination of numbers, variables, and arithmetic operations, such as addition, subtraction, multiplication, and division. In this context, the formula for the volume of a cylinder is an algebraic expression with variables \( r \) and \( h \).Algebraic manipulations often involve the following steps:

• Substitution: Replace variables with given numbers or expressions. For example, substituting \( h = 6 \) into \( V = \pi r^2 h \).
• Simplification: Combine like terms and perform arithmetic operations. In our example, \( V = 6\pi r^2 \) is a simplified form of the expression with \( h \) substituted.
• Rewriting: Rearrange the expression to solve for a particular variable, which involved isolating \( r \) to express it in terms of \( V \).

Understanding these basic manipulations is crucial for solving equations and finding unknowns.

Solving Equations

Solving equations is a process where we find the value of unknown variables that make the equation true. It is a fundamental skill in algebra.When solving equations like the one in our exercise, we:

• Identify what we need to solve for (e.g., \( r \) as a function of \( V \)).
• Manipulate the equation using arithmetic operations, such as division and extraction of square roots, to isolate the desired variable.
• Utilize approximation techniques with known constants like \( \pi \). For instance, in our step-by-step solution, after finding \( \frac{50}{\pi} \), we calculated its square root to find \( r \).

This ability to solve for unknown values is key in many areas of science and mathematics, enabling us to adapt formulas to meet specific conditions or solve real-world problems.