Question

In your pottery class, you are given a lump of clay with a volume of 200 cubic centimeters and are asked to make a cylindrical pencil holder. The pencil holder should be 9 centimeters high and have an inner radius of 3 centimeters. What thickness \(x\) should your pencil holder have if you want to use all of the clay?

Short answer

The exact thickness \(x\) of the pencil holder depending on the calculated value from the equation

Step-by-step solution

Determine the volume of clay

The total volume of the clay is stated in the problem as 200 cubic centimeters.

Formulate the equation

We need to formulate an equation using the volume of the clay and the volume of the cylinder, considering the thickness. The equation becomes \(200 = \pi (r_{2}^{2} - r_{1}^{2}) h\), where \(r_{1}\) is the inner radius, \(r_{2}\) is the outer radius (yet to be discovered), and \(h\) is the height of the cylinder.

Insert the known values

Substitute the given values of \(r_{1} = 3 cm\) and \(h = 9 cm\) into the equation, giving us \(200 = \pi (r_{2}^{2} - 9) * 9\).

Solve for \(r_{2}\)

Solve the equation for \(r_{2}\) to find the value of the outer radius. Re-arranging the equation and taking square root gives us \(r_{2} = \sqrt{\frac{200}{{9\pi}} + 9}\). Calculate the value to get the outer radius \(r_{2}\).

Find the thickness \(x\)

Now, remember that the thickness \(x\) of the pencil holder is defined as the difference between the outer and inner radii, \(x = r_{2} - r_{1}\). Substitute the values of \(r_{2}\) and \(r_{1}\) into the equation to find the thickness of the pencil holder.

Key concepts

Cylinder Volume

Understanding the volume of a cylinder is crucial in tackling exercises that involve such shapes. The volume of a cylinder can be calculated using the formula: \[ V = \pi r^2 h \]where:

• \(V\) represents the total volume of the cylinder.
• \(r\) is the radius of the base.
• \(h\) stands for the height of the cylinder.

This formula indicates that the volume is directly proportional to the square of the radius and the height. This means that even a small change in the radius or height can significantly affect the volume. It's important to understand this basic property when modeling objects, like a cylindrical pencil holder in a pottery class.

Radius and Diameter

In geometry, knowing the relationship between radius and diameter is indispensable, particularly in cylinder-related problems. The radius, denoted as \(r\), is the distance from the center of the circle to any point on its edge. In contrast, the diameter is twice the length of the radius, expressed as \(d = 2r\). Understand that changing the radius alters both the diameter and the cylinder's overall dimensions. For instance, the exercise involves an inner radius \(r_1 = 3\) cm, which helps in calculating the outer radius \(r_2\), essential for determining the thickness of the pencil holder. Always remember, radius and diameter are core measurements that influence the cylinder's volume and thickness in problem-solving.

Problem-Solving

Approaching geometry problems can initially seem complex but breaking them down into simple steps can simplify the process. In our exercise, the goal is to determine the thickness of a pencil holder from a fixed amount of clay. Start by identifying what you know: the volume of clay, the height of the cylinder, and the inner radius. Then, express what you need: the outer radius and eventually the thickness. Formulate an equation that accounts for these variables. In this case, subtract the volume of the inner cylinder from the total clay volume to determine the outer volume. This serves as a reliable framework in problem-solving, allowing you to manage what's often seen as overwhelming information systematically.

Volume Calculation

Calculating the volume of more complex forms involves understanding the concept thoroughly and applying the correct formulas. Here, the problem extends beyond calculating a simple cylinder's volume by considering the thickness, which involves an outer and an inner radius. The formula \( V = \pi (r_2^2 - r_1^2)h \) accounts for this difference, where \(r_2\) is the outer and \(r_1\) is the inner radius. By substituting known values, you can resolve the equation to find the missing dimensions like the outer radius. This calculation shows how volume properties apply to real-world objects beyond simple theoretical exercises, emphasizing the practical uses of geometry in everyday tasks like crafting or designing.