Problem 38 - Beginning and Intermediate Algebra

Question

The volume of a cylinder varies jointly as its height and the square of its radius. The volume of a cylindrical can is \(108 \pi \mathrm{cm}^{3}\) when its radius is \(3 \mathrm{cm}\) and it is \(12 \mathrm{cm}\) high. Find the volume of a cylindrical can with a radius of \(4 \mathrm{cm}\) and a height of \(3 \mathrm{cm}\)

Step-by-Step Solution

Answer

The volume of the second cylindrical can with a radius of 4 cm and a height of 3 cm is \(144\pi \mathrm{cm}^3\).

Step 1: Create an equation relating the volume of a cylinder with its radius and height

We are told that the volume of a cylinder (V) varies jointly as its height (h) and the square of its radius (r). Mathematically, this is represented as follows: \(V = k \cdot h \cdot r^2\) where k is the constant of variation.

Step 2: Plug in the given values and find the constant of variation

We know the volume, radius, and height of a specific cylindrical can. To find the constant of variation k, plug these values into the equation: \(108 \pi = k \cdot 12 \cdot 3^2\) Now, solve for k: \(k = \dfrac{108 \pi}{(12)(3^2)} = 3\)

Step 3: Plug the values of the second cylindrical can to find its volume

We now know the constant of variation k, and we have the dimensions of the second cylindrical can (radius: 4 cm, height: 3 cm). Plug these values into the equation to find the volume of this cylindrical can: V = 3 * 3 * 4^2

Step 4: Calculate the volume of the second cylindrical can

Multiply the values to obtain the volume of the second cylindrical can: V = 3 * 3 * 16 = 144 cm³

Step 5: Multiply the volume by π to obtain the actual volume

Finally, to obtain the actual volume, multiply the calculated volume by π: V_actual = 144 * π = 144π cm³ So, the volume of the second cylindrical can with a radius of 4 cm and a height of 3 cm is \(144\pi \mathrm{cm}^3\).

Key Concepts

Cylinder Volume

The volume of a cylinder is a measure of how much space is inside the cylinder. Cylinders are common in everyday objects like cans or barrels, and understanding how to calculate their volume can be very practical.
For a cylinder, the volume formula is derived from its height and the area of its circular base. The formula is:

• \( V = \pi r^2 h \)

Where:

• \( V \) is the volume,
• \( r \) is the radius of the base,
• \( h \) is the height of the cylinder,
• \( \pi \) (pi) is approximately 3.14159, but usually denoted symbolically for exact calculations.

This formula combines both the area of the circle (\( \pi r^2 \)) and the height (\( h \)), ensuring you multiply these properties to find out how much space the cylinder encloses. Whether solving a math problem or filling a tank, knowing how to calculate a cylinder's volume is incredibly useful.

Algebraic Expressions

Algebraic expressions play a crucial role in representing and solving mathematical problems. In the context of variations and word problems, they often simplify the data we need to work with. Let's break down how algebraic expressions can be used to solve problems concerning joint variations.
An algebraic expression is a combination of numbers, operations, and variables that represents a specific value or relationship. When tackling a cylinder volume problem with joint variations, we typically set it up as:

• \( V = k \cdot h \cdot r^2 \)

Here, \( V \) stands for volume, \( k \) is a constant that relates the specific cylinder dimensions, \( h \) signifies height, and \( r \) is the radius. By manipulating this equation, we can find unknown values provided certain data, like volume and radius. By substituting known values into the equation, you can solve for the unknowns efficiently. This algebraic approach allows us to create equations that can be adjusted and solved accordingly, showing why they're foundational in solving math problems.

Constant of Variation

The constant of variation is essential in understanding problems involving variations, whether direct, inverse, or joint. In joint variation, like with our cylinder problem, the constant of variation (\( k \)) ties together the variables involved, showing how one set of dimensions or conditions changes with another.
In our example, we establish the equation \( V = k \cdot h \cdot r^2 \) to explore how volume varies with height and the square of the radius. The constant \( k \) is found by rearranging the variables and solving for it with known cylinder dimensions.

• \( k = \frac{V}{h \cdot r^2} \)

By re-substituting \( k \) back into the equation with different values, such as a new height or radius, we can determine how these variations affect the volume. The constant thus provides a bridge that helps us understand the proportional relationships between dimensions, enabling us to solve for unknown quantities in practical scenarios. This application of joint variation is seen in fields ranging from geometry to physics, where relationships between variables must be clearly understood.