Problem 17 - Precalculus Mathematics for Calculus

Question

Show that the equation represents a sphere, and find its center and radius. $$x^{2}+y^{2}+z^{2}=12 x+2 y$$

Step-by-Step Solution

Answer

The equation represents a sphere with center (6, 1, 0) and radius \(\sqrt{37}\).

Step 1: Identify the Equation Form

The given equation is \(x^{2}+y^{2}+z^{2}=12x+2y\). This resembles the general form of a sphere equation: \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\). We need to rewrite the given equation in this form.

Step 2: Rearrange the Equation

Move all terms to one side of the equation to set it equal to zero: \(x^2 - 12x + y^2 - 2y + z^2 = 0\).

Step 3: Complete the Square for x-terms

Focus on \(x^2 - 12x\). To complete the square, take half of \(-12\), which is \(-6\), and square it, adding \(36\). This transforms to \((x-6)^2 - 36\).

Step 4: Complete the Square for y-terms

For \(y^2 - 2y\), take half of \(-2\), which is \(-1\), and square it, adding \(1\). This gives \((y-1)^2 - 1\).

Step 5: Rewrite Equation with Completed Squares

Substitute the completed squares back into the equation: \((x-6)^2 - 36 + (y-1)^2 - 1 + z^2 = 0\).

Step 6: Simplify the Equation

Combine the constant terms: \((x-6)^2 + (y-1)^2 + z^2 = 37\). Now the equation is in the standard form of a sphere.

Step 7: Identify Center and Radius

The equation \((x-6)^2 + (y-1)^2 + z^2 = 37\) reveals the center of the sphere at \((6, 1, 0)\) and the radius as \(\sqrt{37}\).

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial. This technique is very useful for rewriting the equation of a sphere into its standard form. When we apply completing the square in the context of a sphere, the goal is to arrange the terms of each variable into a squared binomial.
To complete the square, follow these steps:

• Identify and isolate the terms of the variable you are focusing on. For example, if focusing on the x-terms in the equation, they might be grouped as something like \(x^2 - 12x\).
• Take the coefficient of the linear term (the term without a square), divide it by 2, and then square it. This squared value is what you add and subtract within the equation to complete the square. For \(-12x\), you take half of \(-12\) to get \(-6\), squaring it gives \(36\).
• Rewrite the original terms as a binomial squared, like \((x-6)^2\), while remembering to also subtract the squared value you added to balance the equation. In this way, \(x^2 - 12x\) becomes \((x-6)^2 - 36\).

This technique is applied to all variable terms in the equation until each variable is part of a perfect square.

Standard Form of a Sphere

The standard form of a sphere equation is a key concept, as it clearly defines the parameters of the sphere such as its center and radius. This form looks like \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\). Each of these binomial terms squared represents the respective x, y, and z coordinates of the sphere's center.

• The variables \(a\), \(b\), and \(c\) represent the x, y, and z coordinates of the sphere's center in 3-dimensional space.
• The right side of the equation, \(r^2\), represents the radius squared of the sphere.

Transforming a given equation into this standard form allows one to easily identify both the center and the radius. By using completing the square, you typically rearrange and simplify to this format. For example, equation terms \((x-6)^2 + (y-1)^2 + z^2 = 37\), clearly showcase this standard form where the sphere's parameters can be easily read off.

Center and Radius of a Sphere

Identifying the center and the radius of a sphere from its equation in standard form is straightforward. Once the equation is in the standard form, \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\), you can directly read the center and radius from it.

• The center is located at the point \((a, b, c)\), which you determine from the terms inside the parentheses next to each variable. For example, \((x-6)^2 + (y-1)^2 + z^2\) has a center at \((6, 1, 0)\).
• The radius \(r\) is found by taking the square root of the number on the right side of the equation, which is \(r^2\). Thus, with the equation being equal to 37, the radius \(r\) is \(\sqrt{37}\).

Understanding these components is crucial because they give you the geometric properties of the sphere, notably defining its exact size and location in a three-dimensional space. Reading them correctly from the equation allows for better visualization and understanding of the sphere's attributes.