Problem 12
Question
Give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. $$x^{2}+(y-1)^{2}+z^{2}=4, \quad y=0$$
Step-by-Step Solution
Verified Answer
The set of points forms a circle in the xz-plane with radius \(\sqrt{3}\).
1Step 1: Understand the First Equation
The first equation is \( x^2 + (y-1)^2 + z^2 = 4 \). This represents a sphere in three-dimensional space. The general form of a sphere's equation is \( (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2 \), where \((a, b, c)\) is the center of the sphere and \(r\) is the radius. For our equation, the center is \((0, 1, 0)\) and the radius is \(2\).
2Step 2: Interpret the Second Equation
The second equation is \( y = 0 \). This represents a plane in three-dimensional space where \(y\) is always \(0\). This plane is parallel to the xz-plane and intersects the y-axis at \(0\).
3Step 3: Combine Both Equations
We need to find the intersection of the sphere and the plane. Substitute \(y = 0\) into the sphere's equation: \( x^2 + (0-1)^2 + z^2 = 4 \), which simplifies to \( x^2 + 1 + z^2 = 4 \). Simplifying further, we get \( x^2 + z^2 = 3 \).
4Step 4: Analyze the Resulting Equation
The equation \( x^2 + z^2 = 3 \) describes a circle in the xz-plane. The center of this circle is at \((0, 0)\) and it has a radius of \(\sqrt{3}\). This circle is the intersection shadow of the sphere on the plane \(y = 0\).
Key Concepts
Geometric DescriptionSphere and Plane Intersection3D Geometry
Geometric Description
When we think of geometric description in the context of algebraic equations, we're talking about visualizing shapes and forms these equations represent in space. Here, the equation \(x^2 + (y-1)^2 + z^2 = 4\) characterizes a sphere in a 3D coordinate system. To break this down, the geometric description involves:
- The form of the sphere, determined by comparing with the standard equation \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\).
- Identifying its center, \((a, b, c)\), based on the constants in the equation. Here, it is \((0, 1, 0)\).
- Determining the radius, \(r\), which in this case equates to \(2\) as observed from \(r^2 = 4\).
Sphere and Plane Intersection
Understanding the intersection of a sphere and a plane is like observing how a knife cuts through a fruit. When the plane \(y = 0\) slices through the sphere defined by \(x^2 + (y-1)^2 + z^2 = 4\), it reveals a circle.
The step-by-step analysis includes:
The step-by-step analysis includes:
- Substitute the constraint \(y = 0\) into the sphere’s equation which gives us \(x^2 + (0-1)^2 + z^2 = 4\).
- This simplifies to \(x^2 + 1 + z^2 = 4\), then further to \(x^2 + z^2 = 3\).
- The resultant equation \(x^2 + z^2 = 3\) describes a circle on the y=0 plane, centered at \((0, 0)\) with a radius of \(\sqrt{3}\).
3D Geometry
Three-dimensional geometry is all about visualizing figures and their interactions within a space with height, width, and depth. This exercise dives into aspects of 3D geometry by examining a sphere and a plane:
- A sphere, a perfect symmetrical object defined by all points equidistant from its center, occurs frequently in real-world scenarios, providing meaningful geometrical insights.
- The plane, a flat, two-dimensional surface extending infinitely within that context, intersects other objects, including spheres, to create different geometries, like circles here.
- Key concepts include the ability to manipulate and understand dimensions, recognizing how altering one dimension impacts the overall geometry.
Other exercises in this chapter
Problem 12
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Sketch the surfaces CYLINDERS $$x^{2}+y^{2}=4$$
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