Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Complete the squares to find the center and radius of the sphere whose equation is given (see Example 2). \(x^{2}+y^{2}+z^{2}+8 x-4 y-22 z+77=0\)

Step-by-Step Solution

Answer

The center is (-4, 2, 11) and the radius is 8.

Step 1: Rearrange the Equation

First, rearrange the equation to group the terms involving the same variable. Arrange them in the form:\[ x^2 + 8x + y^2 - 4y + z^2 - 22z = -77 \] This sets up the equation for completing the square.

Step 2: Complete the Square for x

To complete the square for the terms involving \(x\), take half of the coefficient of \(x\), which is 8, divide it by 2 to get 4, and then square it to get 16. Add and subtract 16 inside the equation:\[ (x^2 + 8x + 16) + y^2 - 4y + z^2 - 22z = -77 + 16 \]This allows us to write \(x^2 + 8x + 16\) as \((x + 4)^2\).

Step 3: Complete the Square for y

For the \(y\)-terms, take half of -4, which is -2, and square it to get 4. Add and subtract 4:\[ (x+4)^2 + (y^2 - 4y + 4) + z^2 - 22z = -61 + 4 \]Now, \(y^2 - 4y+4\) can be simplified to \((y-2)^2\).

Step 4: Complete the Square for z

For the \(z\)-terms, take half of -22, which is -11, and square it to get 121. Add and subtract 121:\[ (x+4)^2 + (y-2)^2 + (z^2 - 22z + 121) = -57 + 121 \]This simplifies \(z^2 - 22z + 121\) to \((z-11)^2\).

Step 5: Simplify and Identify the Center and Radius

Now our equation is completely squared:\[ (x+4)^2 + (y-2)^2 + (z-11)^2 = 64 \]This represents a sphere with the center \((-4, 2, 11)\) and radius equal to the square root of 64, which is 8.

Key Concepts

Completing the Square

Completing the square is a technique used in algebra to simplify quadratic equations. It is essential when finding the equation of a circle or a sphere. To "complete the square," you take a quadratic expression and transform it into an equivalent expression that includes a perfect square trinomial. This is achieved by:

• Identifying the linear term's coefficient (e.g., for the term with "x" or "y").
• Dividing this coefficient by 2.
• Squaring the result and adding it inside the equation.

This process facilitates factoring the trinomial into a perfect square. For instance:

• If you have a term involving "x" like \(x^2 + 8x\), you take half of 8 to get 4, then square it to add 16.
• The expression becomes \((x+4)^2\).

Repeating this method for "y" and "z" terms helps to reformat the equation into a neat ('center-radius') form.

Center of a Sphere

The center of a sphere in a quadratic equation form can be determined once the equation is rewritten in a completed square format. The center is composed of the values that zero out each square term in the equation. When an equation is expressed as:\[(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \]The center of the sphere is given by the coordinates

• \((h, k, l)\)

of the equation — crucial for visual understanding. In our example, the equation becomes:\[(x+4)^2 + (y-2)^2 + (z-11)^2 = 64 \]Thus, the center of the sphere is

• \((-4, 2, 11)\)

Remember, the signs are reversed when identifying the center from the squared terms due to the equation's format from completing the square.

Radius of a Sphere

The radius of a sphere is derived from the resulting equation once you have completed the square on each variable term. The term that represents the sphere's size is on the equation's right side. After you have pressed each variable term into square terms, you end up with an equation in the form:\[(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \]Here, the radius squared is the constant \(r^2\). To find the actual radius:

• Take the square root of this constant.

In the example equation:\[(x+4)^2 + (y-2)^2 + (z-11)^2 = 64 \]The square of the radius is 64, so:

• The radius is the square root of 64, which is 8.

This provides a clear visual picture of the sphere's size in 3-dimensional space.