Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 13-16, complete the squares to find the center and \(\mathrm{ra}\) dius of the sphere whose equation is given (see Example 2). x^{2}+y^{2}+z^{2}-12 x+14 y-8 z+1=0

Step-by-Step Solution

Answer

Center: (6, -7, 4); Radius: 10.

Step 1: Rearrange the Equation

Start with the given equation and move the constant term to the right side: x^{2} + y^{2} + z^{2} - 12x + 14y - 8z = -1.

Step 2: Complete the Square for x

To complete the square for the terms involving x, take half of the coefficient of x (-12), square it, and add it inside the equation for x. We have: (x - 6)^2 = x^2 - 12x + 36. Add 36 to the right side: (x - 6)^2 + ... = 35.

Step 3: Complete the Square for y

Next, complete the square for the y terms by taking half the coefficient of y (14), squaring it, and adding it inside the equation for y. We have: (y + 7)^2 = y^2 + 14y + 49. Add 49 to the right side: (x - 6)^2 + (y + 7)^2 + ... = 84.

Step 4: Complete the Square for z

For the z terms, take half of the coefficient of z (-8), square it, and add it inside the equation. We have: (z - 4)^2 = z^2 - 8z + 16. Add 16 to the right side: (x - 6)^2 + (y + 7)^2 + (z - 4)^2 = 100.

Step 5: Identify Center and Radius

The equation of the sphere is now in the form (x - 6)^2 + (y + 7)^2 + (z - 4)^2 = 10^2. Here, the center is (6, -7, 4) and the radius is 10.

Key Concepts

Completing the Square

Completing the square is a powerful method to express quadratic equations in a more convenient form. It enables you to transform an expanded quadratic equation into a compact square form. This technique is essential for solving equations and understanding geometric figures like spheres.

Here's how completing the square works:

• Take the coefficient of the linear term (the number next to the variable), divide it by two, and then square the result.
• Add and subtract this squared number to transform your quadratic expression into a perfect square trinomial.
• Adjust the equation by moving constant terms across to maintain equality.

By completing the square for each variable in a three-dimensional object like a sphere, we rewrite its equation into a neat form, making it easier to identify characteristics like the center and radius.

Equation of a Sphere

The equation of a sphere in three-dimensional space is an essential concept in geometry. A generic equation of a sphere is written as:\[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \]Here,

• \((h, k, l)\) represents the center of the sphere.
• \(r\) signifies the radius.

This form emerges after rewriting the original expanded quadratic equation by completing the square, as shown in our step-by-step solution. This specific equation succinctly describes all points \((x, y, z)\) that lie on the surface of the sphere of a given radius and center location, offering a clear and precise geometrical representation.

Center of a Sphere

The center of a sphere is a crucial point that helps in defining the sphere’s location in space. It is represented in the sphere's equation as the point \((h, k, l)\). When you transform the general quadratic equation of a sphere by completing the square, the terms \((x - h)^2, (y - k)^2,\) and \((z - l)^2\) naturally unveil the center.

In our solved example:

• We found \((h, k, l) = (6, -7, 4)\).
• This indicates the sphere is centered at \((6, -7, 4)\) in the coordinate space.

Identifying the center is important in understanding the position of the sphere, especially when it interacts with other geometric entities in space.

Radius of a Sphere

The radius of a sphere is the distance from the center of the sphere to any point on its surface. In the equation \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\), \(r\) is the radius.

To find the radius after transforming the equation into its complete square form:

• Look at the number on the right side of the equation, which is \(r^2\).
• Take the square root of this number to determine the value of the radius.

In our solution example, the right side was 100, indicating a radius of 10. Understanding the radius is vital for calculating various properties of spheres, such as volume and surface area.