Problem 22 - Algebra and Trigonometry

Question

A Spherical Buoy A spherical buoy of radius 2 feet floats in a calm lake. Six inches of the buoy are submerged. Place a coordinate system with the origin at the center of the sphere. (a) Find an equation of the sphere. (b) Find an equation of the circle formed at the waterline of the buoy.

Step-by-Step Solution

Answer

(a) Equation of sphere: \(x^2 + y^2 + z^2 = 4\). (b) Circle at waterline: \(x^2 + y^2 = 1.75\), at \(z = 1.5\).

Step 1: Understand the Geometry

The buoy is a sphere with a radius of 2 feet, meaning its diameter is 4 feet. We need to place a coordinate system at the sphere's center (origin). The sphere equation will use this coordinate system.

Step 2: Write the Equation of the Sphere

The general equation for a sphere centered at the origin is \(x^2 + y^2 + z^2 = r^2\).Here, the radius \(r = 2\) feet, so the equation is: \[x^2 + y^2 + z^2 = 4\]

Step 3: Identify Submersion Details

The buoy is submerged by 6 inches, which is \(\frac{1}{2}\) foot. This means the circle formed at the waterline is just \(2 - \frac{1}{2} = 1.5\) feet from the sphere's center on the z-axis.

Step 4: Find the Equation of the Circle

At the waterline, the circle is parallel to the xy-plane and has a center at (0, 0, 1.5). The radius of the circle is found using Pythagorean Theorem; in this case, it's \(\sqrt{2^2 - 1.5^2} = \sqrt{4 - 2.25} = \sqrt{1.75}\). The circle equation is:\[x^2 + y^2 = 1.75\],\,\text{ at } z = 1.5.

Key Concepts

Coordinate System

Understanding a coordinate system is essential when dealing with geometric shapes like spheres. In this context, we are working with a three-dimensional coordinate system where the origin is placed at the center of the sphere. This is a common choice because it simplifies the formulation and understanding of the sphere's geometry.

• The x, y, and z coordinates all have the origin point at (0, 0, 0), which aligns with the sphere's center.
• Each coordinate axis is perpendicular to the others, and all intersect at the origin.
• This setup is especially convenient for symmetrical objects like spheres.

By positioning the coordinate system in this manner, calculations become more straightforward, allowing us to easily express the sphere's equation and analyze other geometric features, such as the waterline circle.

Submersion Calculation

Submersion calculation helps us understand how much of the buoy is beneath the water surface. For this spherical buoy, we know that 6 inches, or 0.5 feet, of the buoy is submerged beneath the water. Thus, we need to find the distance from the center of the sphere to this waterline.

• Subtract the submersion depth from the sphere's radius: 2 feet - 0.5 feet = 1.5 feet.
• This distance of 1.5 feet is along the z-axis in our coordinate system, so the waterline is at z = 1.5.

This submersion aspect is crucial for determining where the circle formed by the waterline intersects the sphere, which leads us to further calculations involving the circle at the waterline.

Circle Equation at Waterline

The circle equation at the waterline is a two-dimensional representation derived from the three-dimensional sphere. Since the waterline circle lies in a plane parallel to the xy-plane, its equation must reflect this.

• The center of the circle at the waterline is at (0, 0, 1.5), aligned with the z-axis submersion level.
• The radius of the circle is not the same as the sphere's radius. Instead, it is determined by the relationship between the sphere's radius and the submerged depth.

The mathematical expression for this circle is derived using Pythagorean theorem in our given plane, resulting in:\[x^2 + y^2 = 1.75, \quad z = 1.5\]This equation emphasizes how much radius the circle at the waterline actually has, distinct from the sphere's radius.

Pythagorean Theorem Application

The Pythagorean theorem is a powerful tool in finding unknown lengths in right triangles, and it becomes essential in determining the radius of the circle at the waterline.

• In this instance, the circle's radius is derived from the difference between the sphere's radius and the vertical distance from the center to the waterline.
• By applying the theorem: \(c^2 = a^2 + b^2\), where \(c\) is the sphere’s radius, \(a\) is the unknown radius of the circle, and \(b\) is the submersion depth.

Using the numbers:\[c = 2, \quad b = 1.5\]Thus:\[a^2 = c^2 - b^2\]Plug in the known values:\[a^2 = 4 - 2.25 = 1.75\]\[a = \sqrt{1.75}\]This calculation completes our understanding of the waterline circle's geometry based on the sphere's properties.