Problem 44 - Thomas Calculus

Question

Find the centers and radii of the spheres in Exercises \(41-44\) $$ x^{2}+\left(y+\frac{1}{3}\right)^{2}+\left(z-\frac{1}{3}\right)^{2}=\frac{29}{9} $$

Step-by-Step Solution

Answer

Center: \((0, -\frac{1}{3}, \frac{1}{3})\); Radius: \(\frac{\sqrt{29}}{3}\).

Step 1: Identify the Sphere Equation

The given equation is \( x^2 + \left( y + \frac{1}{3} \right)^2 + \left( z - \frac{1}{3} \right)^2 = \frac{29}{9} \). This is in the standard form \( (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \) for a sphere.

Step 2: Extract the Center Coordinates

Compare the given equation with the standard sphere equation to find the center. Here, \( h = 0 \), \( k = -\frac{1}{3} \), and \( l = \frac{1}{3} \). Hence, the center of the sphere is \( (0, -\frac{1}{3}, \frac{1}{3}) \).

Step 3: Determine the Radius

The right side of the equation \( \frac{29}{9} \) represents \( r^2 \), where \( r \) is the radius of the sphere. To find \( r \), take the square root: \( r = \sqrt{\frac{29}{9}} = \frac{\sqrt{29}}{3} \).

Step 4: State the Center and Radius

From our calculations, the center of the sphere is \( (0, -\frac{1}{3}, \frac{1}{3}) \) and the radius of the sphere is \( \frac{\sqrt{29}}{3} \).

Key Concepts

Center of a Sphere

Understanding the center of a sphere is a fundamental concept in geometry. In a three-dimensional space, a sphere's center is the point that is equidistant from all points on its surface. To find the center from a sphere equation, we need to identify the values of \(h, k, l\) in the standard form equation: \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\). In this exercise, the given equation is \(x^{2} + \left(y + \frac{1}{3}\right)^{2} + \left(z - \frac{1}{3}\right)^{2} = \frac{29}{9}\). When we compare it with the standard form, we can see:

• \(h = 0\)
• \(k = -\frac{1}{3}\)
• \(l = \frac{1}{3}\)

Thus, the coordinates of the center of the sphere are \( (0, -\frac{1}{3}, \frac{1}{3}) \). This is crucial because it tells us exactly where the center point of our sphere is located in a three-dimensional coordinate system.

Radius of a Sphere

The radius of a sphere is the distance from its center to any point on its surface. It plays a vital role in defining the sphere's size. In the standard sphere equation \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\), the term \(r^2\) represents the square of the radius. To find the radius, we simply take the square root of this term.
For our given equation \(x^{2} + \left(y + \frac{1}{3}\right)^{2} + \left(z - \frac{1}{3}\right)^{2} = \frac{29}{9}\), \(r^2\) is \(\frac{29}{9}\). Hence, the radius \(r\) is calculated as follows:

• \(r = \sqrt{\frac{29}{9}}\)
• \(r = \frac{\sqrt{29}}{3}\)

This means that the radius is \(\frac{\sqrt{29}}{3}\), which provides us with a numerical value for the sphere's size.

Standard Form of Sphere Equation

The standard form of the sphere equation is a crucial tool for understanding the geometry of spheres. It is expressed as \( (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \). This equation gives us complete information about the sphere:

• \(h, k, l\): These values represent the coordinates of the center of the sphere.
• \(r\): This is the radius of the sphere, derived from \( r^2 \).

Knowing how to use the standard form helps in identifying and extracting the sphere's geometric properties. For example, from the equation \(x^{2} + \left(y + \frac{1}{3}\right)^{2} + \left(z - \frac{1}{3}\right)^{2} = \frac{29}{9}\), by comparing it to the standard form, we are able to find the center \( (0, -\frac{1}{3}, \frac{1}{3}) \) and the radius \(\frac{\sqrt{29}}{3}\) easily. Understanding standard form makes it quicker to interpret any sphere equation accurately.