Problem 43 - Precalculus with Calculus Previews

Question

Complete the square in \(x, y,\) and \(z\) to find the center and radius of the given sphere. $$ x^{2}+y^{2}+z^{2}+8 x-6 y-4 z-7=0 $$

Step-by-Step Solution

Answer

The center is \[ (-4, 3, 2) \] and the radius is 6.

Step 1: Group Terms

Start by rearranging the equation to group terms by variable: \[ x^{2} + 8x + y^{2} - 6y + z^{2} - 4z = 7 \]

Step 2: Complete the Square for x

Take the x terms \( x^2 + 8x \) and complete the square. - Take the coefficient of x, which is 8, divide it by 2 to get 4, then square it to get 16.- Add and subtract 16 to the x terms: \[ (x^{2} + 8x + 16) - 16 \]- This can be rewritten as: \[ (x + 4)^2 - 16 \]

Step 3: Complete the Square for y

Take the y terms \( y^2 - 6y \) and complete the square.- Take the coefficient of y, which is -6, divide it by 2 to get -3, then square it to get 9.- Add and subtract 9 to the y terms:\[ (y^{2} - 6y + 9) - 9 \]- This can be rewritten as:\[ (y - 3)^2 - 9 \]

Step 4: Complete the Square for z

Take the z terms \( z^2 - 4z \) and complete the square.- Take the coefficient of z, which is -4, divide it by 2 to get -2, then square it to get 4.- Add and subtract 4 to the z terms:\[ (z^{2} - 4z + 4) - 4 \]- This can be rewritten as:\[ (z - 2)^2 - 4 \]

Step 5: Rewrite and Simplify

Plug these completed squares back into the equation:\[ (x+4)^2 - 16 + (y-3)^2 - 9 + (z-2)^2 - 4 = 7 \]Simplify the equation by combining constants:\[ (x+4)^2 + (y-3)^2 + (z-2)^2 = 36 \]

Step 6: Identify the Center and Radius

The equation is now in the standard form of a sphere: \[ (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \]where the center of the sphere is \[ (h, k, l) = (-4, 3, 2) \]and the radius is \[ r = \sqrt{36} = 6 \].

Key Concepts

Completing the Square

Completing the square is a useful algebraic technique often employed to simplify quadratic expressions, especially when working with equations of conic sections like circles and spheres. It involves manipulating the expression to make perfect squares, which are much simpler to interpret.To complete the square:

• Focus on each variable term squared separately, like \(x^2 + bx\).
• Take the linear term's coefficient, \(b\), divide it by 2, and then square it. For example, \(b = 8\); \((8/2)^2 = 16\).
• Add and subtract this squared value. This transforms the expression into a perfect square trinomial.

This approach can make quadratic expressions easier to work with, especially for rewriting them in vertex form. This is exactly what we do when deriving the equation of a sphere in a neat format.

Equation of a Sphere

The equation of a sphere is typically expressed in the standard form: \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\]Here, \((h, k, l)\) represents the center of the sphere, and \(r\) is the radius. Each squared term corresponds to one dimension of space, relating to xyz coordinates.To derive this equation from a complex polynomial, as shown in the example, it is vital to:

• Group all the terms by variables.
• Complete the square for each variable, moving constants independently.
• Simplify the equation by consolidating these completed squares.

This transforms what may initially seem like a complicated polynomial into a much clearer and approachable formula. With practice, this becomes a straightforward process, resulting in a formula that immediately communicates key properties of the sphere.

Center and Radius of a Sphere

Identifying the center and radius of a sphere from its equation requires recognizing its standard form. Once the equation is neatly organized into the form \[(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\]you can easily read the center of the sphere as the coordinates \((h, k, l)\) and the radius as \(r\).Given the equation from the solution, \((x+4)^2 + (y-3)^2 + (z-2)^2 = 36\):

• The center is found by noting the opposite sign in the standard form transformation, thus \((-4, 3, 2)\).
• The radius is the square root of 36, which is \(6\).

Recognizing this form allows you to break down and visualize spheres in geometric problems with ease, and navigate from equation to spherical parameters intuitively.