Problem 62 - College Algebra

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 is approximately 3.99 meters.

Step 1: Express radius as a function of volume

We are given the equation for the volume of a cylinder: \( V = \pi r^2 h \). Substitute \( h = 6 \) meters into the equation to get \( V = \pi r^2 \times 6 \). Simplify this to \( V = 6\pi r^2 \). From this equation, solve for \( r \) by isolating it: \( r^2 = \frac{V}{6\pi} \). Now, take the square root of both sides to express \( r \) in terms of \( V \): \( r = \sqrt{\frac{V}{6\pi}} \).

Step 2: Find the radius for a given volume

Now that we have the function \( r = \sqrt{\frac{V}{6\pi}} \), let's find the radius when the volume \( V \) is 300 cubic meters. Substitute \( V = 300 \) into the equation: \( r = \sqrt{\frac{300}{6\pi}} \). Simplify the fraction: \( r = \sqrt{\frac{50}{\pi}} \). Now calculate the square root: \( r = \sqrt{\frac{50}{3.14159}} \approx \sqrt{15.92} \approx 3.99 \) meters.

Key Concepts

Function of Radius

When discussing cylinders, the **function of the radius** refers to how the radius depends on other variables like volume and height. Here, we need the radius expressed in terms of volume.Given the formula for the volume of a cylinder, \( V = \pi r^2 h \), if the height \( h \) is known or constant, we can express the radius \( r \) as:

• Rearranging the equation to solve for \( r \), starting with: \( V = \pi r^2 h \)
• If \( h = 6 \), the equation simplifies to \( V = 6\pi r^2 \)
• To find \( r \), solve for \( r^2 \), giving us \( r^2 = \frac{V}{6\pi} \)

Finally, apply the square root to find \( r \). This function shows us the relationship "radius as a function of volume." It helps calculate the radius when volume and height are given.

Algebraic Manipulation

**Algebraic manipulation** is a powerful tool in mathematics, allowing us to reshape equations to isolate and solve for specific variables. In the cylinder volume problem, we applied algebraic manipulation to express the radius.Here are the steps we used:

• Start with the volume equation, \( V = \pi r^2 h \)
• Substitute the known value for height, \( h = 6 \), resulting in \( V = 6\pi r^2 \)
• Rearrange the equation to solve for \( r^2 \), obtaining \( r^2 = \frac{V}{6\pi} \)
• Conclude with solving for \( r \) using the square root operation

These steps highlight manipulating algebraic expressions such as distributing, isolating variables, and performing operations like division, are crucial for solving complex geometric problems like finding the radius of a cylinder.

Geometry

In **geometry**, the properties of shapes and figures are explored. For a cylinder, understanding these properties is necessary to apply geometric formulas correctly.A few important geometric characteristics of a cylinder include:

• The cylinder has circular bases, with radius \( r \), which stack on top of each other to form its 3D shape.
• The height \( h \,\) represents the distance between the bases.
• Volume calculation considers space inside the cylinder, using \( V = \pi r^2 h \), reflecting its circular and length dimensions.

In our problem, we used these geometric properties combined with given algebraic equations to work out the radius of the cylinder when its volume is given, reinforcing how geometric understanding informs problem-solving.

Square Root Operation

The **square root operation** is a mathematical function that finds a value, which, when multiplied by itself, equals the original number. It's essential for solving equations where the variable is squared, like finding the radius in cylinder volume problems.To extract \( r \) from \( r^2 = \frac{V}{6\pi} \) in our exercise:

• Take the square root of both sides: \( r = \sqrt{\frac{V}{6\pi}} \)
• Substitute known values, such as \( V = 300 \), giving \( r = \sqrt{\frac{300}{6\pi}} \)
• Simplify the value inside the square root, then compute \( r = \sqrt{\frac{50}{\pi}} \)

This operation ensures that the radius value is logically derived for any given volume. Practicing this step of taking square roots is crucial since it frequently appears in solving quadratic terms.