Problem 19 - Precalculus Mathematics for Calculus

Question

Describe the trace of the sphere $$(x+1)^{2}+(y-2)^{2}+(z+10)^{2}=100$$ in (a) the \(y z\) -plane and (b) the plane \(x=4\)

Step-by-Step Solution

Answer

YZ-plane: Circle centered at (2, -10) with radius √99. Plane x=4: Circle centered at (2, -10) with radius √75.

Step 1: Understanding the Sphere Equation

The given equation \((x+1)^2 + (y-2)^2 + (z+10)^2 = 100\) represents a sphere with center at \((-1, 2, -10)\) and radius \(10\), because the equation is of the form \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\).

Step 2: Trace in the YZ-Plane

In the \(yz\)-plane, set \(x=0\). Substituting it into the sphere equation, we have: \((0+1)^2 + (y-2)^2 + (z+10)^2 = 100\) which simplifies to \(1 + (y-2)^2 + (z+10)^2 = 100\). Rearranging gives \((y-2)^2 + (z+10)^2 = 99\). This is the equation of a circle centered at \((2, -10)\) with radius \(\sqrt{99}\).

Step 3: Trace in the Plane x=4

In the plane \(x=4\), substitute \(x=4\) into the sphere equation. This gives \((4+1)^2 + (y-2)^2 + (z+10)^2 = 100\) which simplifies to \(25 + (y-2)^2 + (z+10)^2 = 100\) or \((y-2)^2 + (z+10)^2 = 75\). This corresponds to a circle centered at \((2, -10)\) with radius \(\sqrt{75}\).

Key Concepts

Equation of a Sphere

The equation of a sphere in 3D geometry is represented in the standard form as \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). Here, \((h,k,l)\) is the center of the sphere, while \(r\) denotes the radius. Breaking it down:

• \(h, k,\) and \(l\) are the coordinates for the center.
• \(r^2\) is the square of the radius.

This form allows us to easily identify both the position of the sphere in space and its size. In the given exercise, the equation \((x+1)^2 + (y-2)^2 + (z+10)^2 = 100\) implies:

• The center of the sphere is at \((-1, 2, -10)\).
• The radius is \(\sqrt{100} = 10\).

Understanding this equation is vital as it gives us a foundation for analyzing spans and projections in various coordinate planes.

Coordinate Planes

In three-dimensional geometry, we often analyze objects concerning the coordinate planes: the XY-plane, YZ-plane, and XZ-plane. Each of these planes allows us to examine how a sphere interacts in a different 2D view. For instance:

• The XY-plane assumes \(z = 0\) and focuses on the \(x\) and \(y\) coordinates.
• The YZ-plane considers \(x = 0\), focusing on \(y\) and \(z\).
• The XZ-plane takes \(y = 0\), focusing on \(x\) and \(z\).

In our problem, we examined traces in the YZ-plane by setting \(x = 0\) and plane \(x = 4\). This allows for simplification into 2D circle equations within these specific slices of the 3D space.

Circle in 3D Geometry

Even in 3D geometry, we can find instances of familiar 2D shapes such as circles. When a sphere is intersected by a plane, the result is often a circle. This occurs because the plane cuts a symmetric cross-section through the sphere. Let's look into the equations derived in our solution:

• For the YZ-plane, the equation \((y-2)^2 + (z+10)^2 = 99\) represents a circle.
• In the plane where \(x = 4\), the equation becomes \((y-2)^2 + (z+10)^2 = 75\).

Both scenarios describe circles centered at \((2, -10)\) but with different radii, \(\sqrt{99}\) and \(\sqrt{75}\) respectively. This demonstrates the elegance of 3D geometry, where such intersections convert high-dimensional shapes into familiar, more manageable 2D objects.