Problem 46 - Calculus an Applied Approach

Question

Find the sphere's center and radius. $$ 4 x^{2}+4 y^{2}+4 z^{2}-8 x+16 y+11=0 $$

Step-by-Step Solution

Answer

The center of the sphere is at point \((1, -2, 0)\) and the radius of the sphere is \(\frac{\sqrt{5}}{2}\)

Step 1: Write the standard form of a sphere

The standard form of a sphere equation is \( (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \) where \((h,k,l)\) are the coordinates of the center and \( r \) is the radius.

Step 2: Rephrase the given equation

Rewrite the equation given, perhaps by completing the square, to match it to the standard form of the equation of a sphere. The given equation is \(4 x^{2}+4 y^{2}+4 z^{2}-8 x+16 y+11=0 \). Divide all terms by 4 to simplify: \(x^{2}+y^{2}+z^{2}-2 x+4 y+\frac{11}{4}=0 \). Now, complete the square for x and y. The equation becomes \( (x-1)^2 + (y+2)^2 + z^2 = r^2 \) where r is yet to be found.

Step 3: Compute the center and the radius

The center of the sphere is the point \((h, k, l) = (1, -2, 0)\) which are the constants subtracted from x, y, and z in the standard equation. To compute the radius, we set \( (x-1)^2 + (y+2)^2 + z^2 = r^2 \) to the constant on the right side of the original rephrased equation, which was \( \frac{11}{4} \). Solving for r gives \( r^2 = 4-11/4 = \frac{5}{4} \), so \( r = \sqrt{\frac{5}{4}} \). So, the radius is \( r = \frac{\sqrt{5}}{2} \)

Key Concepts

Center of a Sphere

Understanding the center of a sphere is essential when dealing with sphere equations. For any sphere, the center is represented by coordinates

• \((h, k, l)\), where \(h\), \(k\), and \(l\) are constants subtracted from \(x\), \(y\), and \(z\) respectively in the equation.
• It indicates the point from which all points on the surface of the sphere are equidistant.

In the standard form equation of a sphere \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), identify these constants to find the center. So, in our specific case, given the equations \((x-1)^2 + (y+2)^2 + z^2\), the center becomes \((1, -2, 0)\). This tells us exactly where the center of the sphere is located in three-dimensional space.

Radius of a Sphere

The radius of a sphere is a measurement of how far any point on the surface is from the center. It's essentially the distance from the sphere's center to any point on its surface.

• In the standard sphere equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), \(r\) represents the radius.
• To find the radius, look for the square root of \(r^2\).

From our equation, once rearranged and completed into \((x-1)^2 + (y+2)^2 + z^2 = \frac{5}{4}\), the radius \(r\) is calculated as \(\sqrt{\frac{5}{4}}\).
This simplifies to \(\frac{\sqrt{5}}{2}\). Now you know how large the sphere is from its center to any point on its surface.

Completing the Square

Completing the square is a helpful algebraic technique used to form a perfect square trinomial. This method helps convert the general quadratic forms into more workable forms such as the standard form of a sphere equation. Here's how you can do it:

• Group the like terms. For instance, in the equation \(x^2 - 2x\), first handle the \(x\) terms.
• Take half of the coefficient of \(x\), square it, and add and subtract it inside the equation. So for \(-2x\), half is \(-1\), and squaring gives \(1\).
• Rewrite the equation as \((x-1)^2\).

Applying this technique to both \(x\) and \(y\) terms helped transform our original equation into a cleaner form \((x-1)^2 + (y+2)^2 + z^2 = \frac{5}{4}\). This process simplifies finding sphere attributes like the center and radius.

Standard Form of a Sphere

The standard form of a sphere is a compact way to express the sphere equation, making it easier to identify important characteristics like the center and radius.

• It is expressed as \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\).
• This form clearly shows both the center \((h, k, l)\) and the square of the radius \(r^2\).

In our problem, the process of reducing then rearranging the given equation into this standard form involved completing the square and factoring for simplicity. The transformation to \((x-1)^2 + (y+2)^2 + z^2 = \frac{5}{4}\) made it clear where the center and radius lie. Using this format is crucial for quickly understanding a sphere's properties in geometric problems.