Problem 18 - Precalculus Mathematics for Calculus

Question

Show that the equation represents a sphere, and find its center and radius. $$x^{2}+y^{2}+z^{2}=14 y-6 z$$

Step-by-Step Solution

Answer

The sphere has center \((0, 7, 3)\) and radius \(\sqrt{58}\).

Step 1: Rearrange the equation

First, we need to rearrange the given equation to make it resemble the standard form of a sphere. Starting with the equation \( x^2 + y^2 + z^2 = 14y - 6z \), move all terms to one side to get \( x^2 + y^2 - 14y + z^2 - 6z = 0 \).

Step 2: Complete the square for y-term

To complete the square for the \( y \) term, focus on \( y^2 - 14y \). Rewrite this in a completed square form: \( (y - 7)^2 - 49 \). Therefore, substitute \( y^2 - 14y \) with \( (y - 7)^2 - 49 \) in the equation.

Step 3: Complete the square for z-term

Next, complete the square for the \( z \) term, \( z^2 - 6z \). Rewrite this as \( (z - 3)^2 - 9 \). Substitute \( z^2 - 6z \) with \( (z - 3)^2 - 9 \) in the equation.

Step 4: Simplify the equation

Now, substitute these completed squares back into the equation: \( x^2 + (y - 7)^2 - 49 + (z - 3)^2 - 9 = 0 \). Simplify it to get: \( x^2 + (y - 7)^2 + (z - 3)^2 = 58 \).

Step 5: Identify the center and radius

The simplified equation \( x^2 + (y - 7)^2 + (z - 3)^2 = 58 \) corresponds to the standard form of a sphere \( (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \), where the center is \((h, k, l) = (0, 7, 3)\) and the radius \( r = \sqrt{58} \).

Key Concepts

Completing the Square

Completing the square is a technique used to help transform quadratic expressions into a perfect square trinomial, making them easier to manage and solve. This is crucial when dealing with the equation of a sphere because it helps us simplify and rearrange terms into a recognizable format. To complete the square, you follow these steps:

• Take the quadratic term and linear term you want to simplify, for instance, the terms involving \( y \) are \( y^2 - 14y \).
• Identify the coefficient of the linear term, which is \( -14 \) in this case, divide it by 2, and then square the result. Specifically, \( \left(\frac{-14}{2}\right)^2 = 49 \).
• Add and subtract this square (49 in this case) to transform the expression into a perfect square trinomial. This gives us \( (y-7)^2 - 49 \).

Completing the square reshapes the expression, making such problems more straightforward and preparing them to be inserted into a standard equation format.

Center of a Sphere

In the equation of a sphere, the standard form is \[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \] where \((h, k, l)\) are coordinates of the sphere's center. Recognizing the center of a sphere involves identifying these values from the equation.After completing the square in the given problem, the equation \[ x^2 + (y - 7)^2 + (z - 3)^2 = 58 \] clearly reveals the center's coordinates:

• \( h = 0 \), as there is no \( (x - )^2 \) term, meaning the \( x \)-coordinate is 0.
• \( k = 7 \), derived from \((y-7)^2\).
• \( l = 3 \), derived from \((z-3)^2\).

Thus, the center of the sphere is at the point \((0, 7, 3)\). Understanding how to extract these from the equation is key to solving sphere-related problems.

Radius of a Sphere

A sphere's radius in its equation determines how far each point on the surface is from its center. In the standard form for a sphere's equation, \[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \] \( r \) represents the radius.From the equation we derived through completing the square, \[ x^2 + (y - 7)^2 + (z - 3)^2 = 58 \], we identify that \( r^2 = 58 \).To find the radius \( r \), simply take the square root: \( r = \sqrt{58} \).The radius is a measure of size for the sphere, and knowing how to retrieve it will help you in various applications, from geometry problems to practical real-world measurements.