Problem 45 - Calculus an Applied Approach

Question

Find the sphere's center and radius. $$ 2 x^{2}+2 y^{2}+2 z^{2}-4 x-12 y-8 z+3=0 $$

Step-by-Step Solution

Answer

The center of the sphere is at (1, 3, 2) and the radius is \( \sqrt{\frac{21}{2}} \).

Step 1: Rewrite the equation

Start off by grouping the x, y, and z terms together and dividing through by 2: \( x^{2}-2x + y^{2}-6y + z^{2}-4z = -\frac{3}{2} \)

Step 2: Complete the squares

To bring the equation into standard form and find the center of sphere, complete the square on the x, y, and z terms:\( (x-1)^{2} + (y-3)^{2} + (z-2)^{2} = -\frac{3}{2} + 1 + 9 + 4 = \frac{21}{2} \)

Step 3: Find the center and radius

Now the equation is in standard form. The center (h, k, l) of the sphere is the point (1, 3, 2) and the radius r of the sphere is \( \sqrt{\frac{21}{2}} \).

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic expression, making it easier to solve or graph. When dealing with equations that describe spheres, this method helps rearrange expressions into a form that highlights the sphere's properties.

In our problem, we start by grouping the x, y, and z terms. Each variable's terms need adjustment by adding and subtracting the same constant. For instance, the expression involving x,\( x^2 - 2x \), is transformed into \((x-1)^2\) by adding and subtracting 1 (this constant comes from \((\frac{-2}{2})^2\)).

Similarly, for y and z, the same process applies. For y, \(y^2 - 6y\) becomes \((y-3)^2\) after adding and subtracting 9, and for z, \(z^2 - 4z\) turns into \((z-2)^2\) after adding and subtracting 4.

• The objective is to form perfect square trinomials like \((a-b)^2\).
• This makes the general form: \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\).
• In our example, these manipulations get us closer to understanding the sphere's properties.

Standard Form of a Sphere

The standard form of a sphere's equation is a neat way to express the position and size of a sphere in 3D space, and it is given by the equation \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\).

Here, \( (h, k, l) \) represents the coordinates of the sphere's center, while \( r \) is the sphere's radius. In our problem, we reorganized the equation by completing the square for each variable.

Once the square completion is finished for x, y, and z, sum all adjustments you've made to balance the equation. This reformulation to standard form clarifies the geometrical information encoded in the equation. In our case, the equation transforms into \((x-1)^2 + (y-3)^2 + (z-2)^2 = \frac{21}{2}\), revealing the sphere's properties directly.

• This form underscores the geometric meaning of each part of the equation.
• From this form, identifying the center and radius becomes straightforward.

Center of a Sphere

The center of a sphere is an important concept because it defines the sphere's position in space. In the standard form of the sphere equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), the point \((h, k, l)\) is the center.

Finding the center involves looking at the completed square terms. Each of these terms has a center coordinate: \(x-1\) becomes \(x = 1\), \(y-3\) turns into \(y = 3\), and \(z-2\) gives \(z = 2\). Hence, the sphere's center is at point (1, 3, 2).

The center plays a crucial role, as it helps you visualize and understand how the sphere is oriented in relation to other objects in space. Knowing the center lets you predict distances and interactions with the surrounding environment, effectively enabling calculations and understanding of spatial relationships in various applications.

• The center is a reference point for understanding the sphere's dimensions and location.
• It's a key aspect when addressing problems involving fields like physics or geometry.