Problem 70 - University Calculus: Early Transcendentals

Question

Find an equation for the set of all points equidistant from the point (0,0,2) and the \(x y\) -plane.

Step-by-Step Solution

Answer

The equation is \( z = \frac{x^2 + y^2 + 4}{4} \).

Step 1: Understanding the Problem

We need to find the equation of the locus of points that are equidistant from a given point (0,0,2) and the xy-plane. This means that for any point (x,y,z) on this locus, the distance to the point (0,0,2) is equal to the perpendicular distance to the xy-plane.

Step 2: Distance Formula from a Point

The distance from any point \(x, y, z\) to the point \(0, 0, 2\) is calculated using the Euclidean distance formula: \\[ D_1 = \sqrt{(x - 0)^2 + (y - 0)^2 + (z - 2)^2} = \sqrt{x^2 + y^2 + (z-2)^2} \]

Step 3: Distance to the xy-plane

The distance from any point \(x, y, z\) to the xy-plane is simply the absolute value of its z-coordinate: \[ D_2 = |z| \]

Step 4: Equating the Distances

We set the two distances equal to each other as the problem states the locus is for equidistant points: \[ \sqrt{x^2 + y^2 + (z-2)^2} = |z| \]

Step 5: Simplifying the Equation

Square both sides to eliminate the square root: \[ (x^2 + y^2 + (z-2)^2) = z^2 \] Then, simplify by expanding and rearranging:\[ x^2 + y^2 + z^2 - 4z + 4 = z^2 \]Cancel out \z^2 from both sides: \[ x^2 + y^2 - 4z + 4 = 0 \] Rearrange to express it in a standard form: \[ x^2 + y^2 = 4z - 4 \]

Step 6: Final Equation Form

Finally, solve for z to present the equation in terms of z:\[ z = \frac{x^2 + y^2 + 4}{4} \] This is the equation of the set of points equidistant from the point \(0,0,2\) and the xy-plane.

Key Concepts

Distance Formula

The distance formula is a fundamental tool in analytic geometry used to determine the distance between two points in a coordinate system. Originating from the Pythagorean theorem, it helps us calculate the shortest path between any two points in space.
To find the distance between two points \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\), we use the three-dimensional version of the formula:

• The square root of the sum of the squares of the differences in each coordinate.\[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

This formula is versatile, applicable in many mathematical problems that involve points, distances, and measurements.
In the original exercise, we used the distance formula to calculate how far a point \(x, y, z\) lays from a fixed point \(0, 0, 2\). This kind of calculation is central to understanding spatial relationships in three dimensions.

Equidistant Points

Equidistant points are locations in space that share the same distance from one or more specified points or lines.
This concept involves finding points that lie at the same distance from given reference points or geometric structures, like planes.
In our exercise, we're exploring points equidistant from a specific point and the xy-plane. Such points form what is known as a locus, a collection of points that fulfill a particular condition.

• This condition is the same distance from both a point and a plane, which here results in a specific geometric configuration.
• To find these points, we equated their distances from both the point (0,0,2) and the xy-plane using the distance formula.

By solving this equation, we derived a formula representing the surface where this equidistance holds true, further illustrating the power of geometry in describing space and order.

Plane Geometry

Plane geometry is an area of mathematics that deals with shapes such as points, lines, circles, and polygons on a two-dimensional surface. However, when we introduce another dimension, these concepts become part of three-dimensional analytic geometry, allowing us to work with volumes and surfaces.
In the context of the xy-plane, it is defined by the equation \(z = 0\), representing a flat, two-dimensional surface stretching infinitely along the x and y axes, with no variation in the z-coordinate.

• When considering the distance from a point like \(x, y, z\) to the xy-plane, the distance is simply the absolute value of the z-coordinate \(|z|\).
• This simple calculation provides essential insights into the point's position relative to the plane.

By understanding plane geometry, we gain an appreciation for how simple equations can describe complex spatial situations, such as the one in our exercise where the locus is described in part by its relationship to the xy-plane.