Problem 78 - College Algebra with Corequisite Support

Question

For the following exercises, write the polynomial function that models the given situation. A cylinder has a radius of \(x+2\) units and a height of 3 units greater. Express the volume of the cylinder as a polynomial function.

Step-by-Step Solution

Answer

The polynomial function for the cylinder's volume is \( V(x) = \pi (x^3 + 9x^2 + 24x + 20) \).

Step 1: Recall the Formula for Cylinder Volume

The volume of a cylinder is given by the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. We'll use this formula to express the volume.

Step 2: Identify Radius and Height

For this problem, the radius \( r \) of the cylinder is \( x+2 \) units. The height \( h \) is 3 units greater than the radius, so \( h = (x+2) + 3 = x+5 \) units.

Step 3: Substitute Radius and Height into Volume Formula

Substitute the expressions for \( r \) and \( h \) into the volume formula: \[ V = \pi (x+2)^2 (x+5) \]

Step 4: Expand the Polynomial

First, expand \((x+2)^2\): \((x+2)^2 = (x+2)(x+2) = x^2 + 4x + 4\).Next, substitute back into the equation: \[ V = \pi (x^2 + 4x + 4)(x+5) \]

Step 5: Multiply to Get a Polynomial

Expand the expression \((x^2 + 4x + 4)(x+5)\): 1. Multiply each term in \(x^2 + 4x + 4\) by \(x\) to get \(x^3 + 4x^2 + 4x\)2. Multiply each term in \(x^2 + 4x + 4\) by \(5\) to get \(5x^2 + 20x + 20\)3. Add the results: \[ x^3 + 4x^2 + 4x + 5x^2 + 20x + 20 = x^3 + 9x^2 + 24x + 20 \]Thus, \[ V = \pi (x^3 + 9x^2 + 24x + 20) \]

Step 6: Final Polynomial Function

The polynomial function expressing the volume of the cylinder is \[ V(x) = \pi (x^3 + 9x^2 + 24x + 20) \].

Key Concepts

Volume of a Cylinder

Calculating the volume of a cylinder involves using a specific formula that relates the dimensions of the cylinder. Cylinders are a common shape in geometry and are characterized by having a circular base and straight sides.

The formula to determine the volume is as follows:

• \[ V = \pi r^2 h \]

Here, \( V \) represents the volume, \( r \) is the radius of the circular base, and \( h \) is the height of the cylinder. This formula applies to any cylinder, regardless of its size.

In this specific example, the radius is given as \( x + 2 \) units, which means the height, also specified as 3 units greater than the radius, evaluates to \( x + 5 \) units. Substituting these values into our formula gives the volume as:

• \[ V = \pi (x+2)^2 (x+5) \]

Understanding this setup is crucial before proceeding to expand the polynomial expression. It’s essential to know how each component of the formula relates to the dimensions of the cylinder to correctly calculate the volume.

Expanding Polynomials

Expanding polynomials is a fundamental skill in algebra, especially when dealing with expressions that have been factored. To expand means to multiply out the brackets to present the expression as a sum of terms.

Let's explore this using our expression for the cylinder volume:
First, focus on expanding \((x+2)^2\):

• This is equivalent to \((x+2)(x+2)\), which upon expansion results in \(x^2 + 4x + 4\).

This part is derived by applying the distributive property (also known as "FOIL" for binomials). It ensures every term is multiplied by each other, covering all combinations.

Next, substitute the expanded \(x^2 + 4x + 4\) back into the volume expression:

• \[ V = \pi (x^2 + 4x + 4)(x+5) \]

Then, expand further by multiplying each term in the polynomial \((x^2 + 4x + 4)\) by each term in \((x+5)\):

• This requires using distributive property again, and results in multiple terms, which when simplified yield \(x^3 + 9x^2 + 24x + 20\).

Expanding polynomials allows you to simplify expressions into a manageable form, providing clarity on the algebraic relationships within the expression.

Algebraic Expressions

Algebraic expressions are an essential part of algebra, representing numbers and operations symbolically. Understanding how to manipulate these expressions is crucial for solving many mathematical problems.

They often involve variables (such as \(x\) in our example), constants (numbers like 2, 5), and operations (addition, multiplication). When tasked with expressing the volume of our cylinder, we use these concepts:

• Identify each component of the cylinder as an algebraic expression: here, \((x+2)\) as the radius, and \((x+5)\) as the height.
• Substitute these expressions into the volume formula and expand as needed.

This process of combining and rearranging symbols is what turns a real-world problem like determining volume into a solvable equation.

It's vital to master algebraic expressions because they form the backbone of solving equations, calculating with functions, and even modeling real-world situations in mathematics and sciences.