91Ó°ÊÓ

When dealing with surfaces in three dimensions, a tangent plane is a flat plane that just touches a surface at a single point. This plane acts as a local approximation to the surface around that point. Let's consider the surface described by the equation \(xyz = k\). At any given point \((x_0, y_0, z_0)\) on this hyperbolic surface, the gradient vector, \((yz, xz, xy)\), provides crucial information to form the tangent plane.

Substituting \((x_0, y_0, z_0)\) into the gradient vector, we derive the tangent plane equation:

• \(yz_0(x-x_0) + xz_0(y-y_0) + xy_0(z-z_0) = 0\).

This equation maintains the slope characteristics of the original surface in the immediate vicinity of \((x_0, y_0, z_0)\).

For this particular problem, the equation simplifies to \(yz_0x + xz_0y + xy_0z = k\), ensuring that for any valid point \((x_0, y_0, z_0)\) on the surface, the tangent plane only grazes the surface at that unique point.

In three-dimensional space, the coordinate planes serve as the boundaries and are defined by setting each variable equal to zero in turn:

• \(x = 0\) defines the yz-plane.
• \(y = 0\) defines the xz-plane.
• \(z = 0\) defines the xy-plane.

These coordinate planes intersect to form axes of the 3D coordinate system.

The problem involves finding the volume of a region in the first octant (where all coordinates are positive) bounded by these coordinate planes and a particular tangent plane. By considering these intersections, we derive the intercepts of the tangent plane on each coordinate plane:

- **x-Intercept:** Set \(y = 0\) and \(z = 0\), yielding \(x = \frac{k}{y_0z_0}\).
- **y-Intercept:** Set \(x = 0\) and \(z = 0\), yielding \(y = \frac{k}{x_0z_0}\).
- **z-Intercept:** Set \(x = 0\) and \(y = 0\), yielding \(z = \frac{k}{x_0y_0}\).

These intercepts are crucial for determining the shape and size of the tetrahedron formed within the first octant.

A tetrahedron is a polyhedron with four triangular faces, and the volume of such a shape can be calculated from the intercepts found on the coordinate planes. In this exercise, the region enclosed by the tangent plane and the coordinate planes forms a tetrahedron in the first octant.

The formula for the volume \(V\) of a tetrahedron given its vertex coordinates is:

• \(V = \frac{1}{6} \times \text{(Base Area)} \times \text{Height}\)

However, with intercepts at \(x = \frac{k}{y_0z_0}\), \(y = \frac{k}{x_0z_0}\), and \(z = \frac{k}{x_0y_0}\), the volume is more readily calculated with:

• \[V = \frac{1}{6} \left( \frac{k}{y_0z_0} \right) \left( \frac{k}{x_0z_0} \right) \left( \frac{k}{x_0y_0} \right)\]

By substituting \(x_0y_0z_0 = k\), it simplifies to:

• \[V = \frac{1}{6} \cdot \frac{k^3}{k^2} = \frac{k}{6}\]

Thus, the volume remains constant and is solely determined by \(k\), showing independence from the specific point of tangency on the original surface. This insight beautifully illustrates the elegance and simplicity often found within geometry.