Problem 36 - Calculus an Applied Approach

Question

Find the standard equation of the sphere. $$ \text { Center: }(4,-1,1) ; \text { radius: } 5 $$

Step-by-Step Solution

Answer

The standard equation of the sphere is \((x-4)^2 + (y+1)^2 + (z-1)^2 = 25\).

Step 1: Identify the given values

In this problem, the center of the sphere is given as (4,-1,1) which corresponds to h=4, k=-1, and l=1. The radius of the sphere is given as 5, which corresponds to r=5.

Step 2: Plug in the values into the standard equation

Substitute the values for h, k, l and r into the standard equation of a sphere. This gives \((x-4)^2 + (y+1)^2 + (z-1)^2 = 5^2\).

Step 3: Simplify the Equation

5^2 is 25. So, the equation simplifies to \((x-4)^2 + (y+1)^2 + (z-1)^2 = 25\). This is the standard equation of a sphere with the given center and radius.

Key Concepts

Center of a Sphere

The center of a sphere is crucial because it is the point from which all points on the surface of the sphere are equidistant. In our exercise, the center is provided as the coordinates \((4, -1, 1)\).

• The 'x'-coordinate is denoted by \(h\), which is 4 in this case.
• The 'y'-coordinate is denoted by \(k\), which is -1.
• The 'z'-coordinate is denoted by \(l\), which is 1.

In 3D space, the center plays a similar role to the center in a circle, but now we are working in a volume rather than a flat plane.
Understanding the center helps define the position of the sphere in space, and it directly influences how we write the sphere's equation.

Radius of a Sphere

The radius is the distance from the center of the sphere to any point on its surface. It's a fixed value that shapes the sphere. In our example, the radius is given as 5.
The radius affects the size of the sphere but does not alter its position. This is because the radius is uniform across all directions from the center.

• The radius can be visualized as a straight line from the center point \((4, -1, 1)\) to any surface point.
• In mathematical terms, it is denoted by \(r\) and here \(r = 5\).

The calculation of the standard equation will use this radius value, squaring it to determine how far the sphere extends in 3D space.

3D Geometry

3D Geometry involves studying spaces, shapes, and surfaces in three dimensions—length, width, and height. It's fundamental for understanding spheres since they exist in a 3-dimensional space.
A sphere is a perfect example of a 3D shape; unlike a circle, which is 2D with just length and height, a sphere adds depth, forming a complete volume.

• 3D geometry helps to visualize where the center and edges of the sphere lie in space.
• It involves understanding coordinates and how they correspond to positions in 3D space.
• For the sphere centered at \((4, -1, 1)\), it involves considering all possible x, y, z combinations that satisfy the sphere's equation.

Understanding these spatial dimensions enables us to build mental models of where and how the sphere exists in space.

Standard Equation

The standard equation of a sphere provides a way to describe any sphere's position and size in 3D space with mathematical precision. The overall format is:
\[(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\]This equation encapsulates:

• \((h, k, l)\), the center of the sphere, which determines its location.
• \(r\), the radius, squaring it provides the sphere's reach in space.

For the sphere in our exercise, substituting the given center \((4, -1, 1)\) and radius \(5\), we form:
\[(x-4)^2 + (y+1)^2 + (z-1)^2 = 25\]This is the standard equation tailored to the specific sphere, succinctly summarizing its dimensions and placement.