Problem 8
Question
Give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. $$y^{2}+z^{2}=1, \quad x=0$$
Step-by-Step Solution
Verified Answer
The solution describes a circle of radius 1 on the yz-plane at x = 0.
1Step 1: Identify the First Equation
The equation \( y^2 + z^2 = 1 \) represents a circular cross-section with radius 1 in the yz-plane. This means for any fixed value of \( x \), if you look at the slice in the yz-plane, the set of points will form a circle centered at the origin (0,0) with radius 1.
2Step 2: Identify the Second Equation
The equation \( x = 0 \) specifies all points where the x-coordinate is zero. This restricts all points to the yz-plane.
3Step 3: Combine the Equations
Combining \( y^2 + z^2 = 1 \) and \( x = 0 \), we see that the set of points that satisfy both equations are those lying on a circle of radius 1 in the yz-plane. Hence, these equations describe a circle on the yz-plane at \( x = 0 \).
4Step 4: Geometric Description
Thus, the set of points satisfying both equations is a circle of radius 1 centered at the origin (0,0) in the yz-plane.
Key Concepts
Geometric DescriptionEquations in SpaceCircle in Plane
Geometric Description
To describe something geometrically involves using shapes, sizes, and relative positions of figures in space. We're tasked with finding the coordinates of points that fulfill specific criteria in space.
This exercise focuses on identifying the collective location of points, or what shape they create, given certain equations.
The first equation, \( y^2 + z^2 = 1 \), forms a circle when viewed in the yz-plane. The circle is centered at the origin (0,0) and has a radius of 1.
This exercise focuses on identifying the collective location of points, or what shape they create, given certain equations.
The first equation, \( y^2 + z^2 = 1 \), forms a circle when viewed in the yz-plane. The circle is centered at the origin (0,0) and has a radius of 1.
- The term \( y^2 + z^2 = 1 \) closely aligns with the standard form of a circle's equation \( x^2 + y^2 = r^2 \).
- This means points that satisfy this equation form a circle around the origin.
Equations in Space
Equations in space are used to define the location and shape of geometric figures in three dimensions. They serve as mathematical rules, outlining where particular points can exist.
For this exercise:
For this exercise:
- \( y^2 + z^2 = 1 \) forms a circle in the yz-plane. It's a common geometric figure where only the y and z-coordinates are needed.
- \( x = 0 \) indicates the plane where all points with a zero x-coordinate reside, making it essentially an overlay of a shape on the yz-plane.
Circle in Plane
The concept of a circle in a plane simplifies the spatial interaction by focusing on what happens on a specific flat surface. In this context, a plane can encompass shapes using just two axes, reminiscent of drawing on paper.
Circles are defined by equations \( x^2 + y^2 = r^2 \), showcasing that they're perfect round shapes centered on a point (0,0) with a fixed radius.
For this situation:
Circles are defined by equations \( x^2 + y^2 = r^2 \), showcasing that they're perfect round shapes centered on a point (0,0) with a fixed radius.
For this situation:
- The circle is represented by \( y^2 + z^2 = 1 \) and is located strictly in the yz-plane due to the \( x = 0 \) constraint.
- This means the circle maintains a consistent radius of 1, perfectly balancing out from the origin in the yz-plane.
Other exercises in this chapter
Problem 8
Find the length and direction (when defined) of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u}\). $$\mathbf{u}=\frac{3}{2} \mathbf{i}-\frac
View solution Problem 8
Let \(\mathbf{u}=\langle 3,-2\rangle\) and \(\mathbf{v}=\langle-2,5\rangle .\) Find the (a) component form and (b) magnitude (length) of the vector. $$-\frac{5}
View solution Problem 9
Find parametric equations for the lines. The line through (0,-7,0) perpendicular to the plane \(x+2 y\) \(+2 z=13\)
View solution Problem 9
Find the angles between the vectors in Exercises \(9-12\) to the nearest hundredth of a radian. $$\mathbf{u}=2 \mathbf{i}+\mathbf{j}, \quad \mathbf{v}=\mathbf{i
View solution