Problem 80 - College Algebra with Corequisite Support

Question

For the following exercises, find the dimensions of the right circular cylinder described. The radius is \(\frac{1}{3}\) meter greater than the height. The volume is \(\frac{98}{9} \pi\) cubic meters.

Step-by-Step Solution

Answer

The radius is \(\frac{7}{3}\) meters and the height is 2 meters.

Step 1: Understand the problem

We need to find the dimensions (radius and height) of a right circular cylinder. We're given that the radius \( r \) is \( \frac{1}{3} \) meter greater than the height \( h \), and the volume of the cylinder is \( \frac{98}{9} \pi \) cubic meters.

Step 2: Apply the volume formula

The volume of a cylinder can be calculated using the formula \( V = \pi r^2 h \). We know the volume \( V = \frac{98}{9} \pi \). We'll substitute and solve for \( r \) and \( h \) while keeping this in mind.

Step 3: Create an equation using known relationships

Since \( r = h + \frac{1}{3} \), we substitute \( r \) in the volume formula: \( \pi (h + \frac{1}{3})^2 h = \frac{98}{9} \pi \). We can cancel \( \pi \) from both sides.

Step 4: Simplify the equation

Simplifying the equation \( (h + \frac{1}{3})^2 h = \frac{98}{9} \), expand the squared term: \( (h^2 + \frac{2}{3}h + \frac{1}{9})h \). Now expand: \( h^3 + \frac{2}{3} h^2 + \frac{h}{9} = \frac{98}{9} \).

Step 5: Clear the fractions by multiplying through

Multiply every term by \( 9 \) to eliminate the fractions: \( 9h^3 + 6h^2 + h = 98 \). Rearrange to form a cubic equation: \( 9h^3 + 6h^2 + h - 98 = 0 \).

Step 6: Solve the cubic equation

Use trial and error, factoring, or numerical methods to find \( h \). Trying \( h = 2 \), we substitute: \( 9(2)^3 + 6(2)^2 + 2 = 98 \). This simplifies to \( 72 + 24 + 2 = 98 \), which holds true.

Step 7: Determine the radius

Since \( h = 2 \), the radius \( r = h + \frac{1}{3} = 2 + \frac{1}{3} = \frac{7}{3} \).

Step 8: Validate the solution

Substitute back into the volume formula to ensure correctness: \( V = \pi (\frac{7}{3})^2 (2) = \frac{98}{9} \pi \), validating that our solution is correct.

Key Concepts

Volume Calculation

To find the dimensions of a right circular cylinder, it's essential to understand how to calculate its volume. The formula for the volume of a cylinder is given by:\[ V = \pi r^2 h \]Where:

• \( V \) is the volume
• \( r \) is the radius
• \( h \) is the height

In this particular problem, we are provided with the total volume \( V = \frac{98}{9} \pi \) cubic meters. Knowing this volume is crucial, as it allows us to set up an equation that involves the radius and height of the cylinder. By substituting the known values and relationships into the volume formula, we can solve for their dimensions.

Cubic Equation Solving

After establishing the relationship and plugging values into the formula, a cubic equation often emerges when determining cylinder dimensions. The cubic equation formed in this problem is:\[ 9h^3 + 6h^2 + h - 98 = 0 \]Solving a cubic equation involves finding the value(s) of \( h \) that satisfy the equation. Common methods include:

• Factoring, which is not always straightforward for cubic equations.
• Trial and error, checking potential values by substituting back into the equation.
• Numerical methods or graphing for more complex cubics if factoring or simple substitution doesn't work.

In our exercise, using trial and error with potential height values, \( h = 2 \) emerged as a solution that satisfied the equation, as substituting it back confirms the equality. Solving cubic equations is pivotal in such problems as it directly leads us to one of the cylinder’s dimensions.

Radius and Height Relationship

In cylinder problems, understanding the relationship between the radius and the height can simplify the solving process. For our exercise, it is given that the radius \( r \) is \( \frac{1}{3} \) meter more than the height \( h \). This relationship is crucial:\[ r = h + \frac{1}{3} \]Knowing this relationship helps in expressing one dimension in terms of the other. In this exercise, we express the radius in terms of the height and substitute it into the volume equation:\[ \pi (h + \frac{1}{3})^2 h = \frac{98}{9} \pi \]Substitution is often necessary as it allows reducing the number of variables to solve for. Once \( h \) is found, calculating \( r \) becomes straightforward by adding \( \frac{1}{3} \) to \( h \). This type of relational understanding is a key strategy in tackling geometry problems efficiently.