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Chapter 12: Problem 6

give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations \(x^{2}+y^{2}=4, \quad z=-2\)

Short Answer

Expert verified
A circle of radius 2 in the plane z = -2, centered at (0, 0, -2).

Step by step solution

01

Understand Each Equation

The first equation is \( x^2 + y^2 = 4 \). This represents a circle in the xy-plane with a radius of 2, centered at the origin (0,0).The second equation is \( z = -2 \). This represents a plane parallel to the xy-plane, located at z = -2.
02

Combine the Equations

To find the geometric description, combine both equations. The circle \( x^2 + y^2 = 4 \) lies in a specific plane given by \( z = -2 \). Therefore, every point \( (x, y, z) \) that satisfies both equations lies on this circle in the plane \( z = -2 \).
03

Describe the Geometry

The intersection of the circle and the plane results in a circular path. Therefore, the set of points that satisfy both equations form a circle of radius 2 centered at \( (0, 0, -2) \) in the three-dimensional coordinate space.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Coordinates
Coordinates are numerical values representing a point's position in space. They provide a unique identifier for any location in a given space. In three-dimensional space, we use three numbers, typically denoted as \((x, y, z)\), as coordinates.
  • The first number, \(x\), indicates the position along the x-axis, which typically runs horizontally.
  • The second number, \(y\), specifies the position along the y-axis, usually running vertically.
  • The third number, \(z\), represents the height or depth, indicating the position along the z-axis.
In our exercise, we see equations involving these coordinates to describe a particular geometry. Each coordinate plays a crucial role in defining where precisely in space a point lies.
Exploring Three-dimensional Space
Three-dimensional (3D) space extends in three directions, typically marked by x, y, and z axes. This space can be visualized as a box, with each point within it having three coordinates. The concept of dimensionality is key:
  • A point has no dimension, only location.
  • A line or curve is one-dimensional, extending along one axis.
  • A plane is two-dimensional, having length and width but no height.
  • Three-dimensional space combines all these: length, width, and height.
In the exercise, the equations given are plotted in this 3D space. The circle from the first equation lies in a specific plane dictated by the second equation. This combination highlights how shapes can be fixed within different dimensions of space.
The Concept of a Circle in Geometry
A circle is a fundamental geometric shape consisting of all points in a plane that are a fixed distance (the radius) from a center point. It is a two-dimensional shape when considered by itself, but it can be placed in a three-dimensional context. Key properties of a circle include:
  • The center, denoted here as \((0, 0)\) in the xy-plane.
  • The radius, the constant distance from the center to any point on the circle. In the exercise, this is 2 units.
  • The circumference, the total distance around the circle.
In our exercise, the circle defined by \(x^2 + y^2 = 4\) lies in the xy-plane, implying it is two-dimensional within that plane. However, with the introduction of the equation \(z = -2\), this circle takes a fixed position in three-dimensional space, specifically in the plane where \(z = -2\).
Understanding Planes in Three-dimensional Space
Planes are flat, two-dimensional surfaces extending infinitely within their dimensions. In three-dimensional space, a plane can be imagined as a sheet of paper extending without end.
  • A plane can be defined by a point and a normal (perpendicular) vector, or by an equation involving coordinates.
  • In our case, the equation \(z = -2\) describes a plane parallel to the xy-plane but shifted down by 2 units along the z-axis.
  • This plane holds all points in space where the z-coordinate is constantly -2.
When another geometric shape lies within this plane, it is confined to two dimensions within the plane's expanse, but still considered part of the three-dimensional space. Here, the circle determined by \(x^2 + y^2 = 4\) is entirely within the plane \(z = -2\), introducing a blend of these geometric concepts.

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Most popular questions from this chapter

Use a CAS to plot the surfaces in Exercises \(53-58 .\) Identify the type of quadric surface from your graph. $$\frac{x^{2}}{9}+\frac{y^{2}}{36}=1-\frac{z^{2}}{25}$$

$$ \begin{array}{c}{\text { a. Find the volume of the solid bounded by the hyperboloid }} \\\ {\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1} \\ {\text { and the planes } z=0 \text { and } z=h, h>0}\end{array} $$ $$ \begin{array}{c}{\text { b. Express your answer in part (a) in terms of } h \text { and the areas } A_{0}} \\ {\text { and } A_{h} \text { of the regions cut by the hyperboloid from the planes }} \\ {z=0 \text { and } z=h .} \\\ {\text { c. Show that the volume in part (a) is also given by the formula }} \\\ {\text { where } A_{m} \text { is the area of the region cut by the hyperboloid }} \\ {\text { from the plane } z=h / 2}\end{array} $$

Write inequalities to describe the sets in Exercises \(45-50 .\) \begin{equation} \begin{array}{l}{\text { The solid cube in the first octant bounded by the coordinate planes }} \\ {\text { and the planes } x=2, y=2, \text { and } z=2}\end{array} \end{equation}

Show that the point \(P(3,1,2)\) is equidistant from the points \(A(2,-1,3)\) and \(B(4,3,1)\) .

Show that the volume of the segment cut from the paraboloid $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\frac{z}{c}$$ by the plane \(z=h\) equals half the segment's base times its altitude.
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