Problem 50 - Thomas Calculus

Question

Find the centers and radii of the spheres in Exercises \(47-50\) $$x^{2}+\left(y+\frac{1}{3}\right)^{2}+\left(z-\frac{1}{3}\right)^{2}=\frac{16}{9}$$

Step-by-Step Solution

Answer

Center: \((0, -\frac{1}{3}, \frac{1}{3})\); Radius: \(\frac{4}{3}\).

Step 1: Identify the Sphere Equation

A sphere's equation generally looks like \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), where \((h, k, l)\) is the center and \(r\) is the radius. In this problem, the given equation is \(x^{2} + \left(y + \frac{1}{3}\right)^{2} + \left(z - \frac{1}{3}\right)^{2} = \frac{16}{9}\).

Step 2: Extract the Center

Compare the given equation \(x^{2} + \left(y + \frac{1}{3}\right)^{2} + \left(z - \frac{1}{3}\right)^{2} = \frac{16}{9}\) with the standard form \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). Here, \(h=0\), \(k=-\frac{1}{3}\), and \(l=\frac{1}{3}\). This gives the center of the sphere as \((0, -\frac{1}{3}, \frac{1}{3})\).

Step 3: Calculate the Radius

In the standard sphere equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), \(r^2\) is given as \(\frac{16}{9}\). To find the radius \(r\), take the square root: \(r = \sqrt{\frac{16}{9}} = \frac{4}{3}\).

Step 4: Concluding Results

The center of the sphere is \((0, -\frac{1}{3}, \frac{1}{3})\) and the radius is \(\frac{4}{3}\).

Key Concepts

Center of Sphere

The center of a sphere in a three-dimensional space is a key concept when working with the sphere's equation. A sphere's equation typically follows the format \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), where \((h, k, l)\) represents the coordinates of the center of the sphere.
To determine the center, compare the sphere's equation side-by-side with the standard form. For the equation given in the problem, \(x^2 + (y+\frac{1}{3})^2 + (z-\frac{1}{3})^2 = \frac{16}{9}\), the terms inside each squared expression indicate the center coordinates:

• For \((x-h)^2\), since there's no \(h\) term in the \(x\) component, this implies \(h=0\).
• The term \((y+\frac{1}{3})\) implies \(k=-\frac{1}{3}\).
• The term \((z-\frac{1}{3})\) suggests \(l=\frac{1}{3}\).

Therefore, the center of the sphere is located at the point \((0, -\frac{1}{3}, \frac{1}{3})\), which acts as a pivotal point from which all radii of the sphere extend outwards.

Radius of Sphere

Understanding the radius of a sphere is crucial in geometry since it defines the size of the sphere. The radius is the distance from the center of the sphere to any point on its surface.
In the equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), the right side, \(r^2\), represents the square of the radius. To find the actual radius \(r\), you need to take the square root of this value.
For the example given, the equation provided is \(x^2 + (y+\frac{1}{3})^2 + (z-\frac{1}{3})^2 = \frac{16}{9}\). Hence, we have:

• \(r^2 = \frac{16}{9}\)
• Taking the square root gives \(r = \sqrt{\frac{16}{9}}\)
• Simplifying further, \(r = \frac{4}{3}\)

Thus, the radius of the sphere is \(\frac{4}{3}\), which indicates how far each point on the surface is from its center.

Standard Form of Sphere Equation

The standard form of a sphere's equation is a fundamental concept that simplifies identifying a sphere's geometric properties, such as its center and radius. This standard form is given by the equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\).
Here, \((h, k, l)\) are the coordinates of the center of the sphere, and \(r\) is the radius.
To put any sphere equation into its standard form, carefully adjust the equation by comparing it with \((x-h)^2, (y-k)^2, (z-l)^2\) terms.
This structure provides a clear path to quickly determine:

• The center: by identifying \(h\), \(k\), and \(l\) values.
• The radius: by solving \(r = \sqrt{r^2}\).

The equation directly illustrates how each variable contributes to the position and size of the sphere, making calculations intuitive and straightforward.