Problem 56 - University Calculus: Early Transcendentals

Question

Find the center \(C\) and the radius \(a\) for the spheres. $$x^{2}+y^{2}+z^{2}-6 y+8 z=0$$

Step-by-Step Solution

Answer

The center is \(C(0, 3, -4)\) and the radius is \(5\).

Step 1: Identify the standard form

The standard form of a sphere's equation is \((x-h)^2 + (y-k)^2 + (z-l)^2 = a^2\), where \((h, k, l)\) is the center and \(a\) is the radius. We must first rewrite the given equation to match this form.

Step 2: Rewrite the equation by completing the square for y

Rewrite the \(y\) terms as a completed square: Given equation: \(x^{2}+y^{2}+z^{2}-6y+8z=0\).Focus on completing the square for \(y\): \((y^2 - 6y) = (y-3)^2 - 9\).

Step 3: Rewrite the equation by completing the square for z

Complete the square for \(z\):\(z^2 + 8z = (z+4)^2 - 16\).

Step 4: Substitute completed squares back into the equation

Substitute the completed squares for \(y\) and \(z\) into the equation and simplify:\(x^2 + (y-3)^2 - 9 + (z+4)^2 - 16 = 0\).This becomes:\(x^2 + (y-3)^2 + (z+4)^2 = 25\).

Step 5: Identify the sphere's center and radius

From the final equation \((x-0)^2 + (y-3)^2 + (z+4)^2 = 5^2\), we can see the center is \(C(0, 3, -4)\) and the radius \(a = 5\).

Key Concepts

Completing the Square

Completing the square is a valuable algebraic method used to transform a quadratic expression into a perfect square trinomial. This is essential for rewriting equations, such as the equation of a sphere, into a recognizable standard form. Here's how you can complete the square for different variables:

• Start by looking at the squared terms of the variable you want to complete. For instance, in the equation given, you will notice the terms for both \(y\) and \(z\).
• For \(y\), you start with \(y^2 - 6y\). You would take half of the linear coefficient (\(-6\)), which is \(-3\), and then square it to get \(9\). Add and subtract this \(9\) in the expression to keep it balanced: \((y-3)^2 - 9\).
• Apply the same method for \(z\). With \(z^2 + 8z\), take half of \(8\) to get \(4\), square it to get \(16\). Adjust it like \((z+4)^2 - 16\).
• These newly formed perfect squares can now be substituted back into the sphere equation.

By completing the square, the messy equation is transformed into a neater form that is much easier to interpret.

Center of a Sphere

The center of a sphere is the point that is equidistant from any point on the surface of the sphere. It's a crucial part of understanding the geometry of spheres. When you transform the general equation of a sphere using the completing the square technique, you reveal the center's coordinates.

The standard form of a sphere's equation is:\[(x-h)^2 + (y-k)^2 + (z-l)^2 = a^2\]

• In this equation, \((h, k, l)\) represents the coordinates of the center of the sphere.
• From our problem, we rewrote the equation to \((x-0)^2 + (y-3)^2 + (z+4)^2 = 5^2\), showing the center to be \((0, 3, -4)\).
• This indicates that every point on the sphere is an equal distance away from \( (0, 3, -4) \).

Being able to find the center of a sphere can facilitate solving geometrical problems, such as finding intersections with planes or other spheres.

Radius of a Sphere

The radius of a sphere is one of its most defining features - the constant distance from the center to any point on the sphere's surface. Once you complete the square and rewrite an equation into the sphere's standard form, the radius becomes easy to identify.

• From the standard form \((x-h)^2 + (y-k)^2 + (z-l)^2 = a^2\), the term \(a\) is the radius.
• In our example, the sphere's equation \((x-0)^2 + (y-3)^2 + (z+4)^2 = 5^2\) shows us that \(a^2 = 25\).
• Thus, calculating \(a = \sqrt{25} = 5\) gives us the sphere's radius.
• The radius, being always positive, defines the size of the sphere and impacts its volume and surface area.

Understanding how to derive and interpret the radius from the equation ensures a clear grasp of the sphere's properties.