Problem 52 - University Calculus: Early Transcendentals

Question

Find the center \(C\) and the radius \(a\) for the spheres. $$(x-1)^{2}+\left(y+\frac{1}{2}\right)^{2}+(z+3)^{2}=25$$

Step-by-Step Solution

Answer

The center is \(C(1, -\frac{1}{2}, -3)\) and the radius is 5.

Step 1: Understand the general equation of a sphere

The general equation of a sphere is given by \[(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\]where \(C(h, k, l)\) is the center of the sphere, and \(r\) is the radius.

Step 2: Identify the center from the given equation

In the given equation \[(x-1)^{2}+ig(y+\frac{1}{2}\big)^{2}+(z+3)^{2}=25\]compare it with the general form to find the center coordinates \((h, k, l)\).Here, \(h = 1\), \(k = -\frac{1}{2}\), and \(l = -3\). Thus, the center \(C\) is at \((1, -\frac{1}{2}, -3)\).

Step 3: Identify the radius

The given equation \[(x-1)^{2}+ig(y+\frac{1}{2}\big)^{2}+(z+3)^{2}=25\]has \(25\) on the right side, which is \(r^2\). Take the square root of \(25\) to find the radius \(r\).Therefore, the radius \(a\) is \(5\).

Key Concepts

Center of a Sphere

To understand the center of a sphere, let's start with its position in the three-dimensional space. A sphere is a perfectly round geometric object in 3D space, like a ball. It has a specific point known as the 'center,' which is the same distance from any point on the sphere's surface.
In the equation of a sphere \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), the center is represented by the coordinates \((h, k, l)\). This center helps determine the sphere's position in space.

• **h**: The x-coordinate of the center.
• **k**: The y-coordinate of the center.
• **l**: The z-coordinate of the center.

In our exercise, by matching the equation \((x-1)^2 + (y+\frac{1}{2})^2 + (z+3)^2 = 25\) with the general equation, we find the center at \((1, -\frac{1}{2}, -3)\). By identifying these coordinates, you can easily set the sphere's spatial position.

Radius of a Sphere

The radius of a sphere is the distance from the center point of the sphere to any point on its surface. It is a fixed length that characterizes the size of the sphere. Mathematically, the radius is denoted by **r** and it is a key component in the general equation of the sphere: \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). This equation shows that every point \((x, y, z)\) on the sphere is at a distance **r** from the center \((h, k, l)\).

When solving the exercise, you identify **r** by taking the square root of the number on the right side of the given equation. For instance, if the equation is \((x-1)^2 + (y+\frac{1}{2})^2 + (z+3)^2 = 25\), then:

• **r² = 25**
• Take the square root: **r = \sqrt{25} = 5**

Thus, in this example, the radius of the sphere is 5, indicating how far each point on its surface is from the center.

General Equation of a Sphere

The general equation of a sphere is fundamental in geometry. It describes all the points that form the exterior surface of the sphere in 3D space. Written as \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), it combines the sphere's geometric properties, where:

• **h, k, l**: Define the sphere's center.
• **r**: Represents the sphere's radius.

You can derive specific characteristics of the sphere, such as its position and size, by manipulating this equation.

In our given problem, you are tasked with finding both the center and the radius by comparing the equation \((x-1)^2 + (y+\frac{1}{2})^2 + (z+3)^2 = 25\) against this general equation.

• The terms \((x-1)^2\), \((y+\frac{1}{2})^2\), \((z+3)^2\) help identify the center \((1, -\frac{1}{2}, -3)\).
• The value on the right side, **25**, indicates the square of the radius \(r^2\), so the radius is \(r = 5\).

By using the conceptual framework of the general equation, you can handle similar problems with ease.