Problem 52 - Thomas Calculus

Question

Find equations for the spheres whose centers and radii are given in Exercises \(51-54 .\) $$\frac{\text { Center }}{(0,-1,5)} \frac{\text { Radius }}{2}$$

Step-by-Step Solution

Answer

The sphere equation is: \(x^2 + y^2 + z^2 + 2y - 10z + 22 = 0\).

Step 1: Understanding the Problem

We need to form the equation of a sphere based on the given center and radius. The standard equation for a sphere with center at \((h, k, l)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\).

Step 2: Identify Parameters

Identify the center coordinates \((h, k, l)\) and radius \(r\) from the given problem. Here, the center is \((0, -1, 5)\) and the radius is 2.

Step 3: Substitute Into the Formula

Substitute \(h=0\), \(k=-1\), \(l=5\), and \(r=2\) into the sphere equation: \[(x - 0)^2 + (y - (-1))^2 + (z - 5)^2 = 2^2\]

Step 4: Simplification

Simplify the equation by calculating each component:- \((x - 0)^2 = x^2\)- \( (y + 1)^2 = y^2 + 2y + 1 \)- \((z - 5)^2 = z^2 - 10z + 25\)- \(2^2 = 4\)

Step 5: Final Equation

Combine all the terms to form the final equation of the sphere: \[x^2 + y^2 + 2y + 1 + z^2 - 10z + 25 = 4\] Simplified further: \[x^2 + y^2 + z^2 + 2y - 10z + 26 = 4\] Rearranging, obtain:\[x^2 + y^2 + z^2 + 2y - 10z + 22 = 0\]

Key Concepts

Center of a Sphere

In geometry, understanding the location of the center of a sphere is crucial for solving problems. The center of a sphere is defined by a set of coordinates in a 3-dimensional space. These coordinates are usually denoted by \(h, k, l\)\, where:

• \(h\) represents the x-coordinate
• \(k\) represents the y-coordinate
• \(l\) represents the z-coordinate

In our problem, the center of the sphere is given as \(0, -1, 5\)\. This means:

• The sphere's center is located at the origin along the x-axis \(h = 0\)
• The y-coordinate is at \(-1\)
• The z-coordinate is at \(5\)

Knowing the center helps form the basis for the sphere's equation.

Radius of a Sphere

The radius is a vital part of a sphere's definition. It is the distance from the center to any point on the surface of the sphere. It remains constant no matter which point on the surface you choose. In mathematical problems, it is symbolized by \(r\).
In this example, the radius of the sphere is given as 2. This simple number tells us:

• Every point on the sphere's surface is 2 units away from the center point \(0, -1, 5\)

The radius plays a key role in the sphere's equation, as it determines the sphere's size.

Standard Equation of a Sphere

The standard equation for a sphere in three-dimensional space is critical for understanding its geometric properties. This equation is: \[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \]
To derive this formula:

• Subtract the center coordinates from the respective x, y, and z variables
• Square each of these terms
• Add them together
• Set them equal to the square of the radius

For our specific problem, we substitute the center \(0, -1, 5\)\ and radius \(r = 2\)\ into the standard equation:

\[ (x - 0)^2 + (y - (-1))^2 + (z - 5)^2 = 2^2 \]
After simplification, we get:
\[ x^2 + y^2 + z^2 + 2y - 10z + 22 = 0 \]
This equation encapsulates all the geometrical information of our sphere. It shows how the sphere is situated in the 3D space, centered and with a specific radius.