Problem 50 - Calculus an Applied Approach

Question

Sketch the \(x y\) -trace of the sphere. $$ x^{2}+y^{2}+z^{2}-4 y+2 z-60=0 $$

Step-by-Step Solution

Answer

The x-y trace of the given sphere is a circle centered at (0, 2) with a radius of \(\sqrt{65}\).

Step 1: Simplify Given Equation

Rearrange the terms in the given equation \(x^{2}+y^{2}+z^{2}-4 y+2 z-60=0\) to represent the equation of a sphere. That is, complete the squares for each variable to obtain an equation of the form \((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\), which is the standard form for the equation of a sphere.

Step 2: Find the Center and the Radius

By simplifying the equation, the equation is obtained as \((x - 0)^2 + (y -2)^2 + (z - 1)^2 = 65\). Thus, we find that the center \((a, b, c) = (0, 2, 1)\) and the radius \(r = \sqrt{65}\).

Step 3: Obtain the x-y Trace

To sketch the x-y trace of the sphere, set \(z = 0\). Substituting \(z = 0\) in the equation of the sphere we get \((x - 0)^2 + (y -2)^2 = 65\), which is the equation of a circle in the x-y plane, with center at (0, 2) and radius of \(\sqrt{65}\).

Key Concepts

Completing the Square

If you've come across a quadratic equation in two variables, you may wonder how to solve or visualize it. One method to make sense of such equations is 'completing the square'. This technique transforms quadratic expressions into a perfect square trinomial, leading to an equation that is much easier to solve or graph.

For example, in the exercise where we are given the equation of a sphere, completing the square can hypothetically turn each variable's expression, such as \( x^2 \) or \( y^2 \), into \( (x-a)^2 \) or \( (y-b)^2 \), by adding and subtracting terms inside the bracket. The values \( a \) and \( b \) are determined based on the coefficients of the quadratic term and the linear term. To actually complete the square for \( y^2 - 4y \), add and subtract \( (4/2)^2 \), which is \( 4 \) in this case. The expression becomes \( (y^2 - 4y + 4) - 4 \), and finally \( (y-2)^2 \), which is a perfect square.

Equation of a Sphere

An equation for a sphere in three-dimensional space resembles that of a circle, but with an extra dimension added. The standard form of the equation of a sphere is \( (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2 \), where \( (a, b, c) \) is the center and \( r \) is the radius of the sphere.

Upon analyzing our given equation \( x^2 + y^2 + z^2 - 4y + 2z - 60 = 0 \), it's not immediately clear if it represents a sphere because it's not in standard form. But, after completing the square for the \( x \), \( y \), and \( z \) terms and rearranging, we derive an equation that matches the standard sphere form. This process clarifies the otherwise hidden spherical nature of the geometric body it represents.

Center of a Sphere

The center of a sphere is the fixed point in space from which every point on the surface is equidistant. In the equation \( (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2 \), the coordinates \( (a, b, c) \) define the sphere's center. In our exercise, after completing the square and simplifying the equation, we find that the center is at \( (0, 2, 1) \).

This tells us that all points on the surface of the sphere are exactly \( r \) units away from this central point. Identifying the center is crucial for understanding the position of the sphere in space and serves as a primary step towards sketching and representing the sphere geometrically.

Radius of a Sphere

The radius of a sphere is the distance from its center to any point on its surface. It's a constant length that's crucial for defining the size of the sphere. In the standard equation of a sphere, the value of \( r \) after the term \( (x-a)^2 + (y-b)^2 + (z-c)^2 \) indicates the radius, which is squared to match the squared terms on the left.

From our problem, when the simplified equation is \( (x-0)^2 + (y-2)^2 + (z-1)^2 = 65 \), taking the square root of the constant term on the right-hand side, \( 65 \), gives us the radius of the sphere, \( r = \sqrt{65} \). This radius helps not only in calculating the volume and surface area of the sphere but also in graphing its two-dimensional traces, like the \( x-y \) trace.

Analytic Geometry

Traces in Analytic Geometry

Analytic geometry combines algebra and geometry to describe geometric figures using equations. In this realm, the concept of a 'trace' is used to represent cross sections of three-dimensional objects by intersecting them with planes. The x-y trace, for example, is the intersection of the object with the x-y plane, which in our context, is obtained by setting \( z = 0 \).

For the equation \( x^2 + y^2 + z^2 - 4y + 2z - 60 = 0 \), setting \( z = 0 \) simplifies it to a two-dimensional form, revealing a circle on the x-y plane. Analytic geometry allows us to visualize and analyze spheres and other complex figures through these simpler slices, making the study of their properties more accessible and practical.