Problem 42 - Thomas Calculus

Question

Find the centers and radii of the spheres in Exercises \(41-44\) $$ \left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}+\left(z+\frac{1}{2}\right)^{2}=\frac{21}{4} $$

Step-by-Step Solution

Answer

The center is \((-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2})\) and the radius is \(\frac{\sqrt{21}}{2}\).

Step 1: Identify the Sphere Equation Format

The given equation \(\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}+\left(z+\frac{1}{2}\right)^{2}=\frac{21}{4}\) is in the standard form for a sphere's equation. The general equation of a sphere is \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), where \((h, k, l)\) are the coordinates of the center and \(r\) is the radius.

Step 2: Determine the Sphere's Center

Compare the given equation with the general equation to identify the center. From \((x-h)^2\), \((y-k)^2\), and \((z-l)^2\), we equate values: \(h = -\frac{1}{2}\), \(k = -\frac{1}{2}\), and \(l = -\frac{1}{2}\). Thus, the center of the sphere is \((-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2})\).

Step 3: Calculate the Sphere's Radius

Find the radius by comparing \(r^2\) in the sphere equation. Given \(r^2 = \frac{21}{4}\), take the square root of both sides: \(r = \sqrt{\frac{21}{4}} = \frac{\sqrt{21}}{2}\). Thus, the radius of the sphere is \(\frac{\sqrt{21}}{2}\).

Key Concepts

Sphere Center

To find the center of a sphere, it is essential to understand how the standard equation of a sphere is structured. The standard form can be written as \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), where \((h, k, l)\) represents the center coordinates of the sphere. By examining the given equation:
\[ \left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}+\left(z+\frac{1}{2}\right)^{2}=\frac{21}{4} \]
It matches this standard form. When you compare it directly with the standard form, you can identify:

• \(h = -\frac{1}{2}\)
• \(k = -\frac{1}{2}\)
• \(l = -\frac{1}{2}\)

So, the center of the sphere is at the point \((-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2})\). This means that if you were to position the center in a coordinate system, you'd move \(-\frac{1}{2}\) units along each of the x, y, and z axes.

Sphere Radius

The radius of the sphere is another crucial piece of information that can be derived from the standard form equation. Let’s look again at the standard form:\((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). The term \(r^2\) on the right side represents the square of the sphere's radius.
In the equation provided:
\[ \left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}+\left(z+\frac{1}{2}\right)^{2}=\frac{21}{4} \]
We identify the right side, \(\frac{21}{4}\), as \(r^2\). To find \(r\), the actual radius, we take the square root of \(r^2\):

• \(r = \sqrt{\frac{21}{4}}\)
• \(r = \frac{\sqrt{21}}{2}\)

Thus, the radius of the sphere is \(\frac{\sqrt{21}}{2}\), providing the length from the sphere's center to any point on its surface.

Standard Form of a Sphere Equation

Understanding the standard form of a sphere's equation is the key to analyzing many geometric problems involving spheres. This form is expressed as \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). Here, each part of the equation plays a unique role:

• The expressions \((x-h)^2\), \((y-k)^2\), and \((z-l)^2\) shift the sphere from the origin to its center at \((h, k, l)\).
• The \(r^2\) on the right side accounts for the sphere's size, as it is the radius squared.

By placing specific values of \(h\), \(k\), \(l\), and \(r^2\) into the equation, you can represent any sphere. For example, the equation:
\[\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}+\left(z+\frac{1}{2}\right)^{2}=\frac{21}{4}\]
Is already in the standard form, aiding us to quickly derive the center and radius, thus understanding the sphere's location and size in three-dimensional space.