Problem 53 - Thomas Calculus

Question

Sketch the surfaces in Exercises \(13-76\) $$ x^{2}+y^{2}-z^{2}=4 $$

Step-by-Step Solution

Answer

The surface is a hyperboloid of one sheet with circular symmetry in the xy-plane.

Step 1: Identify the Surface Type

The equation \(x^2 + y^2 - z^2 = 4\) is a type of quadratic surface known as a hyperboloid. Specifically, it is a hyperboloid of one sheet because it can be rewritten as \(\frac{x^2}{4} + \frac{y^2}{4} - \frac{z^2}{4} = 1\).

Step 2: Compare with Standard Form

The standard form of a hyperboloid of one sheet is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\). Comparing \(\frac{x^2}{4} + \frac{y^2}{4} - \frac{z^2}{4} = 1\) with the standard form, we have \(a^2 = 4\), \(b^2 = 4\), and \(c^2 = 4\), implying \(a = 2\), \(b = 2\), \(c = 2\).

Step 3: Sketch Axes and Surface Shape

Begin by sketching the coordinate axes. Since \(a = b = 2\), the surface has circular symmetry in the \(xy\)-plane and opens along the z-axis. Draw the vertex circle in the \(xy\)-plane with radius \(2\). The hyperboloid extends infinitely along the \(z\)-axis, forming two tubular sections which narrow towards the \(xy\)-plane.

Step 4: Hyperboloid Structure Details

At \(z = 0\), the cross-section is a circle with radius \(2\). For \(z > 0\) or \(z < 0\), the cross-sections are ellipses, expanding as \(|z|\) increases. This indicates the surface's breadth grows along \(z\)-axis.

Key Concepts

Quadratic Surfaces

Quadratic surfaces are intriguing three-dimensional shapes defined by second-degree polynomial equations in three variables. These surfaces include parabolas, ellipsoids, and hyperboloids, among others. The general form of a quadratic surface is given by:\[ Ax^2 + By^2 + Cz^2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0 \]Here, the coefficients \(A, B, C, D, E, F, G, H, I, J\) determine the specific type and orientation of the surface.

In our exercise, we have a simplified surface known as a hyperboloid of one sheet. This particular quadratic surface exhibits unique structural features that differentiate it from other types, such as its open, tubular shape that is symmetrical in one plane but extends infinitely in another direction. Understanding the nature of these surfaces can vividly illustrate complex mathematical concepts and enhance spatial visualization skills.

Coordinate Axes

Understanding coordinate axes is crucial for visualizing three-dimensional surfaces. The coordinate axes, typically labeled as the \(x\)-axis, \(y\)-axis, and \(z\)-axis, form a basis to describe any point in a 3D space. These axes intersect at the origin \( (0, 0, 0) \) and are mutually perpendicular to each other.

When sketching a surface, identifying significant features like symmetry or symmetry planes with respect to these axes is key. For a hyperboloid of one sheet, the symmetry around the \(xy\)-plane is crucial for visualizing its structure. The elliptical and circular cross-sections are easily plotted by understanding how they intercept with these axes, which helps in sketching an accurate representation of the surface.

Standard Form

The standard form of a hyperboloid of one sheet helps simplify its equation to identify basic properties and to assist in sketching. The standard equation is as follows:\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \]Here, \(a\), \(b\), and \(c\) are constants that determine the size and shape of the hyperboloid's cross-sections.

In our problem, the equation \( \frac{x^2}{4} + \frac{y^2}{4} - \frac{z^2}{4} = 1 \) represents a hyperboloid of one sheet. Calculating \(a = 2\), \(b = 2\), and \(c = 2\) gives the radii of cross-sections parallel to the coordinate planes. This form makes it apparent that the hyperboloid is symmetric around the \(xy\)-plane and stretches along the \(z\)-axis without bound.

Circular Symmetry

Circular symmetry in a surface means that its shape remains consistent when rotated around a particular axis. For the hyperboloid of one sheet, this circular symmetry exists primarily in the \(xy\)-plane.

This characteristic is evident because the coefficients of \(x^2\) and \(y^2\) are equal, resulting in circular cross-sections in the \(xy\)-plane. At \(z = 0\), the section of the surface is a circle, signifying perfect symmetry in this plane.

As we move away from the \(z = 0\) plane, the circular symmetry gives way to elliptical cross-sections, indicating the structure's expansion along the \(z\)-axis. This unique symmetry helps in visualizing and sketching the hyperboloid's continuous structure, aiding in the understanding of its geometric properties.