Problem 30
Question
Describe in words the region of \(\mathbb{R}^{3}\) represented by theequations or inequalities. \(x^{2}+y^{2}+z^{2}>2 z\)
Step-by-Step Solution
Verified Answer
The region is the exterior of a sphere, centered at (0, 0, 1) with radius 1.
1Step 1: Simplify the Inequality
Start by simplifying the inequality: \(x^2 + y^2 + z^2 > 2z\).Move \(2z\) to the left side to obtain \(x^2 + y^2 + z^2 - 2z > 0\).
2Step 2: Complete the Square
For the term \(z^2 - 2z\), complete the square to make it easier to interpret geometrically. Write \(z^2 - 2z\) as \((z-1)^2 - 1\). This transforms the inequality into: \(x^2 + y^2 + (z-1)^2 - 1 > 0\).
3Step 3: Rearrange the Terms
Rearrange the terms to make the inequality more understandable:\(x^2 + y^2 + (z-1)^2 > 1\).
4Step 4: Interpret the Geometric Meaning
Recognize that \(x^2 + y^2 + (z-1)^2 = 1\) describes a sphere of radius 1 centered at the point \((0, 0, 1)\) in \(\mathbb{R}^3\).Therefore, the inequality \(x^2 + y^2 + (z-1)^2 > 1\) represents the set of all points outside this sphere.
Key Concepts
Completing the SquareGeometric InterpretationSphere in Mathematics
Completing the Square
Completing the square is a method used in algebra to transform a quadratic expression into a perfect square trinomial, making the equation or inequality easier to analyze and solve. Let's take a closer look at how this technique works in the context of geometry. In the given exercise, we start with the inequality
x^2 + y^2 + z^2 > 2z
. By rearranging terms as
x^2 + y^2 + z^2 - 2z > 0
, we focus on the
z^2 - 2z
portion. To complete the square, we add and subtract the same constant, turning
z^2 - 2z
into
(z - 1)^2 - 1
.
- Find a constant to add and subtract that makes a perfect square: Here, 1 completes the square for z^2 - 2z .
- Rewrite z^2 - 2z as (z-1)^2 - 1 to make it manageable in geometric analysis.
Geometric Interpretation
The concept of geometric interpretation involves visualizing algebraic equations and inequalities as shapes or regions in space. Once we completed the square in the inequality, we arrived at a more familiar form:
x^2 + y^2 + (z-1)^2 > 1
. This new form is the key to understanding the spatial relationship in three-dimensional space.
Visualizing the inequality,
x^2 + y^2 + (z-1)^2 = 1
represents a sphere centered at the point (0, 0, 1) with a radius of 1. The inequality
x^2 + y^2 + (z-1)^2 > 1
describes the region outside this sphere, because it includes all points beyond the surface of the sphere. By rearranging mathematical expressions through completing the square, we can map these into intuitive geometric concepts.
- The center of the sphere derives from shifting the z-term as z' = z - 1 .
- The radius, 1, comes from the equation's equality part, denoting points exactly on the sphere.
Sphere in Mathematics
A sphere in mathematics is a perfectly round geometric object in three-dimensional space, analogous to a circle in two dimensions. The equation of a sphere centered at a point
(h, k, l)
with radius
r
is given by
(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2
. In the exercise example, this sphere is centered at (0, 0, 1) and has a radius of 1. The inequality
x^2 + y^2 + (z-1)^2 > 1
denotes all points lying outside this sphere.
- The center (0, 0, 1) indicates a vertical shift along the z-axis.
- The inequality implies an open set outside of the sphere surface.
Other exercises in this chapter
Problem 30
(a) Find a nonzero vector orthogonal to the plane through the points \(P, Q,\) and \(R,\) and (b) find the area of triangle \(P Q R .\) $$P(0,0,-3), \quad Q(4,2
View solution Problem 30
Sketch the region bounded by the paraboloids \(z=x^{2}+y^{2}\) and \(z=2-x^{2}-y^{2}\)
View solution Problem 30
\(29-32\) Find the scalar and vector projections of b onto a. $$\mathbf{a}=\langle 1,4\rangle, \quad \mathbf{b}=\langle 2,3\rangle$$
View solution Problem 31
Find the tangential and normal components of the acceleration vector. $$\mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+t \mathbf{k}$$
View solution