Problem 10 - Thomas Calculus

Question

In Exercises \(1-16,\) give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. $$x^{2}+y^{2}+z^{2}=25, \quad y=-4$$

Step-by-Step Solution

Answer

The solution is a circle centered at (0, -4, 0) with radius 3.

Step 1: Identify the Equation of a Sphere

The first equation \(x^2 + y^2 + z^2 = 25\) represents a sphere in 3D space. The center of this sphere is located at the origin (0,0,0) and its radius is \(\sqrt{25} = 5\).

Step 2: Analyze the Plane Equation

The second equation \(y = -4\) represents a plane parallel to the xz-plane. This plane is located at \(y = -4\) and does not depend on the values of \(x\) and \(z\).

Step 3: Intersection of the Sphere and the Plane

To find the geometric description of the intersection, substitute \(y = -4\) into the sphere's equation. This gives \(x^2 + (-4)^2 + z^2 = 25\), simplifying to \(x^2 + z^2 = 9\).

Step 4: Describe the Geometric Shape of the Intersection

The equation \(x^2 + z^2 = 9\) represents a circle in the xz-plane at \(y = -4\). The circle is centered at (0, -4, 0) and has a radius of 3, which is \(\sqrt{9}\).

Step 5: Deliver the Final Geometric Description

The set of points is described geometrically as a circle with its center at (0, -4, 0) and radius 3 in the xz-plane at \(y = -4\).

Key Concepts

Sphere Equation

A sphere in three-dimensional space is a perfectly round geometrical object, similar to a circle in two dimensions. The equation of a sphere with its center at the origin (0, 0, 0) is expressed as:\[x^2 + y^2 + z^2 = r^2\]
where \(r\) is the radius of the sphere. This equation essentially states that any point \((x, y, z)\) on the sphere is at a constant distance \(r\) from the center.

• In our specific exercise, the sphere equation is \(x^2 + y^2 + z^2 = 25\).
• The center of this sphere is at the origin (0, 0, 0).
• The radius is given by the square root of 25, which is 5.

Understanding the sphere equation helps us visualize which points in space are included within it.

Plane Equation

A plane equation represents a flat, two-dimensional surface extending infinitely in 3D space. The simplest form of a plane equation with one fixed variable can be something like \(y = c\), where \(c\) is a constant.
This creates a plane parallel to the xz-plane and shifts vertically by the amount \(c\). It does not vary with \(x\) and \(z\).

• In our exercise, the plane equation is \(y = -4\).
• This represents a plane parallel to and 4 units below the xz-plane.
• Every point on this plane has a y-coordinate of -4.

This simple representation will help us understand how this plane interacts with other geometrical shapes like a sphere.

Intersection of Sphere and Plane

When a plane intersects with a sphere, the intersection is often a shape like a circle, or it can be empty (no intersection) or a single point, depending on the plane's position.
To find this intersection, we substitute the plane equation into the sphere equation. In our exercise, where the plane is \(y = -4\), substitute \(-4\) for \(y\) in the sphere's equation, resulting in the formula:
\[x^2 + (-4)^2 + z^2 = 25\]
Simplifying gives:
\[x^2 + z^2 = 9\]
This equation represents all points where the sphere and plane intersect. Importantly, it tells us that this intersection is not a full sphere nor a simple flat plane but forms another recognizable shape, which we detail next.

Circle in 3D Space

In three dimensions, a circle is the result of the intersection between a plane and a sphere. For the circle to exist, they must intersect in a way that creates a round edge.
From our previous calculations, the equation \(x^2 + z^2 = 9\) describes a circle. This equation is typical for a circle in the 2D xy-plane, but here it applies to the xz-plane at a fixed \(y\) value.

• The circle is centered at the point (0, -4, 0) in 3D space.
• The radius of the circle is 3, derived from \(\sqrt{9}\).
• Every point on the circle shares the \(y\)-coordinate of -4, remaining consistent with the plane equation.

Visualizing circles in three-dimensional space can be challenging, but understanding its formation from sphere-plane intersections aids comprehension of such spatial geometry.