Problem 43 - Calculus an Applied Approach

Question

Find the sphere's center and radius. $$ x^{2}+y^{2}+z^{2}-2 x+6 y+8 z+1=0 $$

Step-by-Step Solution

Answer

The sphere's center is at the point (1, -3, -4) and its radius is 5 units.

Step 1: Identify the coefficients

From the given equation, we see that we have coefficients \(g = -1\), \(f = 3\) and \(h = 4\) and \(c = 1\).

Step 2: Calculate the sphere's center

Based on the formula for the center of a sphere, the center is \(-g, -f, -h\). Plugging in the values yields \((1, -3, -4)\).

Step 3: Calculate the sphere's radius

The radius of the sphere is \(\sqrt{g^2 + f^2 + h^2 - c}\). Hence, substituting the values will have \(\sqrt{(1)^2 + (-3)^2 + (-4)^2 - 1} = 5\).

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This technique is vital in geometry, especially when dealing with sphere equations, as it allows you to rearrange the terms to expose the standard form of a sphere. Follow these steps to complete the square:

• First, group the terms by variable. For the given sphere equation: \[ x^2 - 2x + y^2 + 6y + z^2 + 8z + 1 = 0 \]
• Adjust each group (terms in \(x\), \(y\), and \(z\)) by adding and subtracting the necessary constants to make each a perfect square trinomial: - For \(x\), the terms are \(x^2 - 2x\). Add and subtract \( (\frac{-2}{2})^2 = 1 \). - For \(y\), the terms are \(y^2 + 6y\). Add and subtract \( (\frac{6}{2})^2 = 9 \). - For \(z\), the terms are \(z^2 + 8z\). Add and subtract \( (\frac{8}{2})^2 = 16 \).

By completing the square, the equation becomes:\[ (x - 1)^2 + (y + 3)^2 + (z + 4)^2 = 25 \]This makes it straightforward to identify both the center and the radius of the sphere.

Center of a Sphere

The center of a sphere in the 3-dimensional space is represented as a point \((x_0, y_0, z_0)\). It is directly derived from the equation of the sphere in its completed squared form:\[ (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2 \]Once you complete the square for the equation, the values \(x_0 = 1, y_0 = -3,\) and \(z_0 = -4\) can be read directly from the equation. In our example, the sphere's center is located at \((1, -3, -4)\). Always remember:

• A positive value of \(x_0\) comes from a term \((x - x_0)^2\).
• A negative value of \(y_0\) or \(z_0\) results from terms like \((y + 3)^2\) or \((z + 4)^2\).

Knowing how to find a sphere's center is essential for geometric interpretations and solving spatial problems.

Radius of a Sphere

The radius of a sphere indicates its size and is the distance from the center of the sphere to any point on the surface. The radius can be extracted easily from the completed square form of the equation.Let's break it down:- From the equation, \((x - 1)^2 + (y + 3)^2 + (z + 4)^2 = 25\), it is clear that the radius squared is 25.- Therefore, the radius \(r\) is the square root of 25, which is 5.Understanding the formula:- The general form \(r^2 = (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2\) reveals that \(r = \sqrt{25} = 5\), as deduced from our calculations.In any sphere equation, once the equation is properly squared and simplified, you will find \(r^2\) on one side of the equation. This makes it very simple to determine the radius. Recognizing how the radius is derived is crucial. It allows for practical applications such as calculating volume or determining circumference.