Problem 72 - College Algebra with Corequisite Support

Question

For the following exercises, use the given volume and radius of a cylinder to express the height of the cylinder algebraically. Volume is \(\pi\left(4 x^{3}+12 x^{2}-15 x-50\right)\), radius is \(2 x+5\)

Step-by-Step Solution

Answer

The height of the cylinder is \( h = \frac{4x^3 + 12x^2 - 15x - 50}{4x^2 + 20x + 25} \).

Step 1: Identify the Cylinder Volume Formula

The formula to find the volume of a cylinder is given by \( V = \pi r^2 h \) where \( V \) is volume, \( r \) is radius, and \( h \) is the height of the cylinder.

Step 2: Substitute Given Values into the Formula

We are given \( V = \pi(4x^3 + 12x^2 - 15x - 50) \) and \( r = 2x + 5 \). Substitute these values into the volume formula: \( \pi(4x^3 + 12x^2 - 15x - 50) = \pi (2x+5)^2 h \).

Step 3: Simplify the Equation

Remove the \( \pi \) factor from both sides, as it is common: \( 4x^3 + 12x^2 - 15x - 50 = (2x+5)^2 h \).

Step 4: Expand the Radius Factor

Calculate \((2x+5)^2\) by expanding it: \((2x+5)^2 = (2x+5)(2x+5) = 4x^2 + 20x + 25\).

Step 5: Express Height in Terms of Volume and Radius

Now replace \((2x+5)^2\) in the equation: \( 4x^3 + 12x^2 - 15x - 50 = (4x^2 + 20x + 25)h \). Solve for \( h \) by dividing both sides by \(4x^2 + 20x + 25\): \[ h = \frac{4x^3 + 12x^2 - 15x - 50}{4x^2 + 20x + 25} \].

Key Concepts

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and arithmetic operations like addition, subtraction, multiplication, and division. In this exercise, we are using algebraic expressions to handle the volume of a cylinder and manipulate it to find the height. The given volume of our cylinder is represented by the expression \( \pi(4x^3 + 12x^2 - 15x - 50) \). This expression includes:

• Coefficients: Numbers like 4, 12, -15, and -50.
• Variables: The variable \( x \), which is raised to different powers, shown as exponents.
• Arithmetic operations: Included are the addition and subtraction signs in the expression.

To work with these algebraic expressions efficiently, it is crucial to understand how to simplify and manipulate them. In mathematics, simplifying expressions is essential to make problem-solving easier and ensure a clearer representation of the mathematical problem at hand.

Polynomial Division

Polynomial division is a technique used to divide one polynomial by another, similar to long division with numbers. In this exercise, we use polynomial division to find the height of the cylinder.
When given the volume, \(4x^3 + 12x^2 - 15x - 50\), we are required to divide it by the expanded radius squared, \(4x^2 + 20x + 25\), to solve for the cylinder's height. The formula:

• \[ h = \frac{4x^3 + 12x^2 - 15x - 50}{4x^2 + 20x + 25} \]

is simplified using polynomial division. This process involves:

• Arranging the polynomial in descending order of the variable powers.
• Identifying the leading terms of the numerator and denominator.
• Using these leading terms to perform iterative divisions, similar to steps in long division.

Successfully carrying out these steps will lead to an expression that represents the height of the cylinder in terms of \( x \).

Cylinder Height Calculation

Calculating the height of a cylinder given its volume and radius involves substituting known values into the volume formula and solving for the unknown height. The standard formula for the volume of a cylinder is:

• \( V = \pi r^2 h \)

In this scenario, we know the volume \( \pi(4x^3 + 12x^2 - 15x - 50) \) and the radius \( 2x + 5 \). To find the height, we:

• Substitute these into the formula: \( \pi (2x+5)^2 h = \pi(4x^3 + 12x^2 - 15x - 50) \).
• Cancel out the common factor \( \pi \) from both sides, simplifying our equation to \( 4x^3 + 12x^2 - 15x - 50 = (2x+5)^2 h \).
• Calculate the expanded form of \( (2x+5)^2 \), which is \( 4x^2 + 20x + 25 \).
• Divide the volume expression by the expanded squared radius to find \( h \).

Thus, the height \( h \) is found as a simplified algebraic expression: \[ h = \frac{4x^3 + 12x^2 - 15x - 50}{4x^2 + 20x + 25} \]. This operation connects various algebraic techniques to find a practical solution for the cylinder's dimensions.