Problem 25 - Basic Technical Mathematics with Calculus

Question

\(\text {Solve the given problems.}\) The radius of a cylinder is twice as long as the radius of a cone, and the height of the cylinder is half as long as the height of the cone. What is the ratio of the volume of the cylinder to that of the cone?

Step-by-Step Solution

Answer

The ratio of the volume of the cylinder to that of the cone is 6:1.

Step 1: Identify Given Relationships

Let's identify the relationship between the radii and heights of the cylinder and cone. Let the radius of the cone be \( r_c \) and the height of the cone be \( h_c \). Then, the radius of the cylinder is \( r_c' = 2r_c \) and the height of the cylinder is \( h_c' = \frac{h_c}{2} \).

Step 2: Write Volume Formulas

The volume of the cone is given by the formula \( V_{cone} = \frac{1}{3} \pi r_c^2 h_c \). The volume of the cylinder is \( V_{cyl} = \pi (r_c')^2 h_c' \).

Step 3: Substitute Relationships

Substitute the relationships from Step 1 into the volume formulas: \( V_{cone} = \frac{1}{3} \pi r_c^2 h_c \) and \( V_{cyl} = \pi (2r_c)^2 \frac{h_c}{2} \).

Step 4: Simplify Volume of Cylinder

Calculate the volume of the cylinder: \( V_{cyl} = \pi (4r_c^2) \frac{h_c}{2} = \pi \cdot 4r_c^2 \cdot \frac{h_c}{2} = 2\pi r_c^2 h_c \).

Step 5: Calculate Volume Ratio

Now that we have the expressions for the volumes, calculate the ratio \( \frac{V_{cyl}}{V_{cone}} = \frac{2\pi r_c^2 h_c}{\frac{1}{3} \pi r_c^2 h_c} = \frac{2}{\frac{1}{3}} \).

Step 6: Final Ratio

Divide to find the ratio: \( \frac{2}{1/3} = 2 \times 3 = 6 \). Therefore, the ratio of the volume of the cylinder to the volume of the cone is 6:1.

Key Concepts

Cylinder Volume Calculation

The volume of a cylinder can be thought of as the amount of space inside the 3-dimensional shape. To find the volume of a cylinder, you can use the formula:

• \( V_{cyl} = \pi r^2 h \)

Where \( r \) is the radius of the base of the cylinder and \( h \) is the height.
To visualize, imagine filling the cylinder with water or any liquid and figuring out how much it can hold.

In our volume ratio problem, the cylinder's radius is double that of the cone, and its height is half that of the cone. By substituting these into the formula, we consider how these changes affect the volume.
This change in dimensions influences the overall volume, so when you calculate with these variables, you will effectively cover the variations and correctly solve the problem.

Cone Volume Calculation

The volume of a cone, much like the cylinder, relies on its base and height, but it also accounts for the fact that a cone comes to a point (a vertex) unlike a cylinder. To calculate the volume of a cone, use the formula:

• \( V_{cone} = \frac{1}{3} \pi r^2 h \)

Here, \( r \) is the radius of the circular base, and \( h \) is the height from the base to the tip, or vertex, of the cone.
Notice the factor of \( \frac{1}{3} \) in the formula. This reflects that a cone's volume is a third of the volume of a cylinder that has the same base and height, due to its tapering shape.
In our problem, it's key to track how adjustments in the radius and height affect the volume. By substituting these relationships into the formula, you can observe how these properties interplay to define the cone's volume in relation to the cylinder.

Volume Ratio Problem-Solving

Investigating volume ratios is about understanding how changes in dimensions of geometric shapes affect their volume in relation to each other. Here, we explore how the cylinder's dimensions being twice and half that of the cone lead to an impactful difference in volume.
By calculating each shape's volume with their respective formulas, and substituting the relationships we discovered (double radius, half height), you can measure the effects mathematically through the calculated ratio.

• The ratio of volumes: \( \frac{V_{cyl}}{V_{cone}} = \frac{2\pi r_c^2 h_c}{\frac{1}{3} \pi r_c^2 h_c} \)
• Simplifying gives \( \frac{2}{1/3} = 6 \).

This is a useful problem-solving strategy for comparing volumes and seeing how even small changes in dimensions can lead to large changes in spatial capacity.