Problem 372 - Precalculus

Question

Use the given volume and radius of a cylinder to express the height of the cylinder algebraically. Volume is \(\pi\left(25 x^{3}-65 x^{2}-29 x-3\right),\) radius is \(5 x+1.\)

Step-by-Step Solution

Answer

The height is \( h = \frac{25x^3 - 65x^2 - 29x - 3}{25x^2 + 10x + 1} \).

Step 1: Recall the Formula for the Volume of a Cylinder

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

Step 2: Identify the Given Values

In this problem, the given volume of the cylinder is \( \pi(25x^3 - 65x^2 - 29x - 3) \) and the radius is \( 5x + 1 \). We'll use these values in our formula to solve for the height.

Step 3: Substitute into the Volume Formula

Plug the given values into the volume formula: \[ \pi(25x^3 - 65x^2 - 29x - 3) = \pi (5x + 1)^2 h \] We have substituted \( r = 5x + 1 \) and \( V = \pi(25x^3 - 65x^2 - 29x - 3) \).

Step 4: Simplify and Solve for the Height

First, cancel \( \pi \) from both sides of the equation:\[ 25x^3 - 65x^2 - 29x - 3 = (5x + 1)^2 h \]Next, simplify \((5x + 1)^2\) to get \(25x^2 + 10x + 1\). Substitute into the equation:\[ 25x^3 - 65x^2 - 29x - 3 = (25x^2 + 10x + 1)h \]Now, solve for \( h \):\[ h = \frac{25x^3 - 65x^2 - 29x - 3}{25x^2 + 10x + 1} \]

Step 5: Finalize the Expression for the Height

The height \( h \) of the cylinder is expressed in terms of \( x \) as:\[ h = \frac{25x^3 - 65x^2 - 29x - 3}{25x^2 + 10x + 1} \] This is the algebraic expression for the height using the given volume and radius.

Key Concepts

Algebraic Expression

An algebraic expression is a mathematical phrase that can contain numbers, variables, and arithmetic operations.
In our cylinder problem, we are tasked with expressing the height of a cylinder using an algebraic expression.
To find this expression, we need to manipulate formula components algebraically.
Specifically, the formula for the volume of a cylinder is key in deriving the expression.
In our problem, the volume was given as a complex polynomial \( \pi(25x^3 - 65x^2 - 29x - 3) \).
Solving for the height means rearranging the formula to express the height as:

• Identify and cancel like terms.
• Rearrange the equation to isolate the desired variable, in this case, the height.

These operations yield the algebraic expression for the height:
\[ h = \frac{25x^3 - 65x^2 - 29x - 3}{25x^2 + 10x + 1} \] Each term in this expression serves a specific purpose, allowing the height to flexibly adjust based on the variable \(x\).

Cylinder Height

The height of a cylinder is a major component in determining its volume.
The volume formula is defined as \( V = \pi r^2 h \).
This means that manipulating either the radius or height impacts the volume. To determine the height algebraically, we must first express the volume equation in terms of height \( h \):

• Insert given values into the volume formula.
• Isolate the variable \( h \) to determine the height expression.

In our example, after substituting the values, we followed through cancelling and rearranging the equation:
From \( \pi (25x^3 - 65x^2 - 29x - 3) = \pi (5x + 1)^2 h \) to reach:
\[ h = \frac{25x^3 - 65x^2 - 29x - 3}{25x^2 + 10x + 1} \] This form shows how the height directly correlates to the cylinder's variable aspects, specifically \(x\).

Radius of a Cylinder

The radius of a cylinder is another fundamental element related to its dimensions and volume.
For our problem, the radius has been defined as the algebraic term \( r = 5x + 1 \).
The role of the radius in the volume calculation is significant.
It is squared in the formula \( V = \pi r^2 h \), thus heavily influencing the cylinder's overall volume.
We use this radius to help express the height algebraically.
After substituting the radius into the volume formula, we needed to simplify:

• Calculate the square of the radius, resulting in \((5x + 1)^2\).
• Use this squared value to help solve for the height.

Since the radius is squared, any change in \( x \) directly affects both the height and the resulting volume.
This manipulation of the radius within the expression showcases its importance in algebraic problem-solving.