Problem 74 - Algebra and Trigonometry

Question

Radius of a Sphere \(\quad\) A jeweler has three small solid spheres made of gold, of radius \(2 \mathrm{mm}, 3 \mathrm{mm},\) and 4 \(\mathrm{mm}\) . He decides to melt these down and make just one sphere out of them. What will the radius of this larger sphere be?

Step-by-Step Solution

Answer

The radius of the larger sphere is approximately 4.64 mm.

Step 1: Calculate the Volume of Each Sphere

To find the total volume of the gold, we must first calculate the volume of each smaller sphere. The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius. Substitute \( r = 2 \), \( 3 \), and \( 4 \) mm into the formula:- For the sphere with \( r = 2 \) mm: \( V_1 = \frac{4}{3} \pi (2)^3 = \frac{32}{3} \pi \) cubic mm.- For the sphere with \( r = 3 \) mm: \( V_2 = \frac{4}{3} \pi (3)^3 = 36 \pi \) cubic mm.- For the sphere with \( r = 4 \) mm: \( V_3 = \frac{4}{3} \pi (4)^3 = \frac{256}{3} \pi \) cubic mm.

Step 2: Calculate the Total Volume

Add the volumes of all three spheres to get the total volume of gold before melting. This sum will give us the volume of the larger sphere:\( V_{total} = V_1 + V_2 + V_3 = \frac{32}{3} \pi + 36\pi + \frac{256}{3} \pi = \frac{32 + 256}{3} \pi + 36\pi = 96\pi + 36\pi = 132\pi \) cubic mm.

Step 3: Find the Radius of the Larger Sphere

Now, use the total volume to find the radius of the new, larger sphere. Setting the total volume equal to the volume formula for a sphere, solve for \( r \):\( \frac{4}{3} \pi r^3 = 132\pi \).First, divide both sides by \( \pi \):\( \frac{4}{3} r^3 = 132 \).Then, multiply both sides by \( \frac{3}{4} \) to solve for \( r^3 \):\( r^3 = \frac{3 \times 132}{4} = 99 \).Finally, take the cube root of both sides to find \( r \):\( r = \sqrt[3]{99} \approx 4.64 \) mm.

Key Concepts

Radius

The radius of a sphere is a crucial concept in geometry, and it helps us understand the size of a sphere. The radius is the distance from the center of the sphere to any point on its surface. In simple terms, think of it as the spoke of a wheel that stretches from the middle to the edge. The radius is half the diameter, which is the longest distance passing through the center of the sphere.

To calculate the volume of a sphere, the radius is used within various formulas. Knowing the radius helps determine other properties, like the surface area and volume. In our example, the initial spheres have radii of 2 mm, 3 mm, and 4 mm. This establishes the groundwork for volume calculations, as each contributes differently depending on its size. An understanding of the radius enables more complex operations, such as melting these three spheres together to form a new one.

For tasks involving spheres, measuring or being given the radius is often the first step in solving intricate problems.

Sphere

A sphere is a perfectly symmetrical 3D shape, often described as a round ball. Every point on its surface is equidistant from its center, defining its radius. This shape is common in nature and used extensively in various fields of science and engineering.

Spheres have several key characteristics:

• They have no edges or corners.
• The radius remains constant throughout.
• They boast symmetrical properties, meaning one side of a sphere mirrors the other.

In our exercise, we consider three small solid spheres made of gold. By melting them together, we can create a larger sphere. To understand how these transformations occur, one must grasp the inherent symmetry and uniformity of spheres.

The sphere's unique properties make it an ideal shape for minimizing surface area for a given volume, which explains why bubbles and planets often take this form. Understanding spheres can pave the way for comprehending a variety of physical phenomena.

Volume Calculation

Calculating the volume of a sphere involves a specific formula. To do this, the radius is crucial. The formula is given by \[ V = \frac{4}{3} \pi r^3 \] where \( V \) represents the volume, and \( r \) is the radius. This equation allows us to determine how much space is contained within a sphere.

In the given exercise, we first compute the volume of each of the smaller gold spheres using their respective radii. By substituting the radii into the formula, we find:

• The sphere with radius 2 mm has a volume of \( \frac{32}{3} \pi \) cubic mm.
• The sphere with radius 3 mm has a volume of \( 36\pi \) cubic mm.
• The sphere with radius 4 mm has a volume of \( \frac{256}{3} \pi \) cubic mm.

Summing these volumes gives the total gold volume before melting, which in turn helps calculate the radius of the new larger sphere. Solving for the radius stems from rearranging the volume formula to find \( r \), allowing us to work backward from knowing the total volume.

Understanding this method is essential in various contexts, from creating jewelry to advanced scientific applications, where the distribution of matter or energy within a spherical shape needs quantification.