Problem 54 - University Calculus: Early Transcendentals

Question

Find the center \(C\) and the radius \(a\) for the spheres. $$x^{2}+\left(y+\frac{1}{3}\right)^{2}+\left(z-\frac{1}{3}\right)^{2}=\frac{16}{9}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Form of the Sphere Equation

The given equation is \(x^2 + (y + \frac{1}{3})^2 + (z - \frac{1}{3})^2 = \frac{16}{9}\). It is in the standard form \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\), where \((h, k, l)\) is the center and \(r\) is the radius.

Step 2: Determine the Center Coordinates

From the equation \(x^2 + (y + \frac{1}{3})^2 + (z - \frac{1}{3})^2 = \frac{16}{9}\), we can see that there is no term involving \(x\), so \(h = 0\). The term \((y + \frac{1}{3})^2\) suggests \(k = -\frac{1}{3}\). The term \((z - \frac{1}{3})^2\) suggests \(l = \frac{1}{3}\). Thus, the center \(C\) is \((0, -\frac{1}{3}, \frac{1}{3})\).

Step 3: Calculate the Radius

The equation is \(x^2 + (y + \frac{1}{3})^2 + (z - \frac{1}{3})^2 = \frac{16}{9}\). By comparing with the form \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\), we get \(r^2 = \frac{16}{9}\). Solving for \(r\), we find that \(r = \sqrt{\frac{16}{9}} = \frac{4}{3}\).

Key Concepts

Sphere Equations

The equation of a sphere in three-dimensional space often comes in the form \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\] Here, each variable represents a component of the sphere's attributes. The

• \((h, k, l)\) are the coordinates of the center of the sphere.
• \(r\) stands for the radius of the sphere.

The expression on the left of the equation shows the squared distances from each point on the sphere to its center. These distances are balanced by the radius squared on the right.
This indicates that every point \((x, y, z)\) on the sphere is exactly \(r\) units away from the center. This geometric property defines the sphere entirely. Understanding its equation helps in identifying the sphere's characteristics, such as its size and position.
In the exercise, the equation provided is \[x^2 + (y + \frac{1}{3})^2 + (z - \frac{1}{3})^2 = \frac{16}{9}\] By comparing this with the standard form, you can determine key features like center and radius, which are essential in sphere calculations and applications.

Center of a Sphere

To determine the center of a sphere described by its equation, we find the values of \(h, k,\) and \(l\) from the standard equation form: \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\] Each of these values represents one of the coordinate axes in a 3D space.

• The value \(h\) determines the horizontal shift along the \(x\)-axis.
• \(k\) moves along the \(y\)-axis and affects the vertical position.
• \(l\) aligns the center along the \(z\)-axis.

For our specific equation \[x^2 + (y + \frac{1}{3})^2 + (z - \frac{1}{3})^2 = \frac{16}{9}\], let's see:- The absence of transformation in \(x\) implies \(h = 0\).- The term \((y + \frac{1}{3})^2\) suggests a downward \(y\)-axis shift, resulting in \(k = -\frac{1}{3}\).- The \(z\) transformation \((z - \frac{1}{3})^2\) indicates an upward shift, deriving \(l = \frac{1}{3}\).
Putting it all together, behind these operations lies the center point \((0, -\frac{1}{3}, \frac{1}{3})\). This point is crucial for understanding the sphere's placement in a 3D environment.

Radius of a Sphere

The radius of a sphere is a critical measure indicating how far any point on the sphere is from its center. It is the defining measurement of the sphere's size.
For any sphere equation of the form \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\], \(r^2\) is positioned on the right of the equation. To find the radius \(r\), take the square root of \(r^2\).
Let's look at this in the context of the exercise:The given sphere's equation is \[x^2 + (y + \frac{1}{3})^2 + (z - \frac{1}{3})^2 = \frac{16}{9}\]. Here, \(r^2 = \frac{16}{9}\). Taking the square root of both sides gives:\[r = \sqrt{\frac{16}{9}}\] which simplifies to \(\frac{4}{3}\). Thus, the radius of the sphere is \(\frac{4}{3}\). Understanding the radius helps in visualizing the sphere's scale and applying it in practical geometric problems, ranging from simple calculations to complex spatial configurations.