Problem 39 - Thomas Calculus

Question

Write inequalities to describe the sets in Exercises \(35-40\) The (a) interior and (b) exterior of the sphere of radius 1 centered at the point \((1,1,1)\)

Step-by-Step Solution

Answer

Interior: \((x-1)^2 + (y-1)^2 + (z-1)^2 < 1\); Exterior: \((x-1)^2 + (y-1)^2 + (z-1)^2 > 1\).

Step 1: Determine the Equation of the Sphere

The equation of a sphere with a radius of 1 centered at the point \((1, 1, 1)\) is given by \((x - 1)^2 + (y - 1)^2 + (z - 1)^2 = 1\). This equation represents all points on the surface of the sphere.

Step 2: Define the Interior of the Sphere

The interior of the sphere includes all points such that the sum of squared differences from \((1, 1, 1)\) is less than 1. Therefore, the inequality for the interior is:\[(x - 1)^2 + (y - 1)^2 + (z - 1)^2 < 1\]

Step 3: Define the Exterior of the Sphere

The exterior of the sphere includes all points such that the sum of squared differences from \((1, 1, 1)\) is greater than 1. Therefore, the inequality for the exterior is:\[(x - 1)^2 + (y - 1)^2 + (z - 1)^2 > 1\]

Key Concepts

Equation of a Sphere

A sphere in three-dimensional geometry is a perfectly symmetrical object. The equation of a sphere describes all the points around a central point, forming its surface. For instance, if you have a sphere with a radius of 1 that is centered at the point \((1, 1, 1)\), the equation is \((x - 1)^2 + (y - 1)^2 + (z - 1)^2 = 1\). This equation represents all the points that lie exactly on the surface of the sphere.

Let's break this down. The formula is derived from the Pythagorean theorem in three dimensions. It calculates the distance from any point \((x, y, z)\) to the center of the sphere \((1, 1, 1)\). When this distance equals the radius, the point is on the surface. Thus, if a point satisfies the equation, it's on the sphere's surface.

Interior and Exterior of a Sphere

Understanding the interior and exterior of a sphere involves inequalities. These inequalities help determine whether a point in space is inside or outside the sphere.

• **Interior of the Sphere**: The interior includes all points where the sum of the squared differences from \((1, 1, 1)\) is less than 1. This is represented by the inequality \((x - 1)^2 + (y - 1)^2 + (z - 1)^2 < 1\). This means that any point inside would have a radius (distance from the center) smaller than 1.
• **Exterior of the Sphere**: The exterior consists of all points where the sum of the squared differences from \((1, 1, 1)\) is more than 1. The inequality for this is \((x - 1)^2 + (y - 1)^2 + (z - 1)^2 > 1\). Here, a point outside would have a radius greater than 1, meaning it's beyond the boundary.

These inequalities give a clear distinction between the interior and exterior of a sphere, using the center and radius as references.

Three-Dimensional Geometry

Three-dimensional geometry involves the study of figures in space that have width, height, and depth. It is a crucial part of mathematics that extends beyond the flat surfaces of two dimensions. For a sphere, three-dimensional geometry helps us visualize how it occupies space.

In three-dimensional space, every point is uniquely defined by three coordinates \((x, y, z)\). A sphere, being a set of points equidistant from a center point, uses these coordinates in its equation to define its shape.

• Visualization: Visualizing in 3D can be challenging but understanding the relationship between the coordinates and distances helps. Imagine slicing the sphere with different planes; the cross-sections would form circles.
• Applications: Spheres are seen not only in mathematics but in real-world applications like modeling planets or atoms. They provide a simplified yet powerful way to compute spatial relationships.

Overall, mastering these concepts gives one the ability to navigate complex geometrical problems with ease.