We need to describe and sketch the graphs of the following equations in three-space: Equation 18: $y = \cos x$ Equation 20: $z = \sqrt{x^2 + y^2 + 1}$

Geometry3D GeometryGraphingSurfacesCylindersHyperboloids
2025/4/15

1. Problem Description

We need to describe and sketch the graphs of the following equations in three-space:
Equation 18: y=cosxy = \cos x
Equation 20: z=x2+y2+1z = \sqrt{x^2 + y^2 + 1}

2. Solution Steps

Equation 18: y=cosxy = \cos x
In the xyxy-plane, y=cosxy = \cos x is the standard cosine curve. Since there is no restriction on zz, zz can take on any value. Thus, in three-space, the graph is a surface formed by translating the cosine curve along the zz-axis. The surface is a cosine cylinder.
Equation 20: z=x2+y2+1z = \sqrt{x^2 + y^2 + 1}
Square both sides to get z2=x2+y2+1z^2 = x^2 + y^2 + 1.
Rearranging, we have z2x2y2=1z^2 - x^2 - y^2 = 1.
Since z=x2+y2+1z = \sqrt{x^2 + y^2 + 1}, zz is always non-negative (z0z \ge 0).
This is the equation of a hyperboloid of two sheets, but because of the square root, we only consider the upper sheet where z0z \ge 0. The equation can also be interpreted as a hyperboloid of one sheet, with its axis aligned along z axis, with its lower part truncated at z =
1.

3. Final Answer

Equation 18: y=cosxy = \cos x is a cosine cylinder.
Equation 20: z=x2+y2+1z = \sqrt{x^2 + y^2 + 1} is the upper sheet of a hyperboloid of two sheets (or a hyperboloid of one sheet truncated at z=1z=1).

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