Problem 12 - Essential Calculus Early Transcendentals

Question

Find an equation of the sphere that passes through the origin and whose center is \((1,2,3) .\)

Step-by-Step Solution

Answer

The sphere's equation is \( (x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 14 \).

Step 1: Write the General Equation of a Sphere

The general equation of a sphere with center \( (h, k, l) \) and radius \( r \) is given by \[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2. \] For this problem, the center is given as \( (1, 2, 3) \), so the equation becomes: \[ (x - 1)^2 + (y - 2)^2 + (z - 3)^2 = r^2. \]

Step 2: Use the Condition that the Sphere Passes Through the Origin

Since the sphere passes through the origin \( (0, 0, 0) \), we can substitute these coordinates into the equation of the sphere to find \( r^2 \). Substitute \( x = 0, y = 0, z = 0 \) into the sphere's equation: \[ (0 - 1)^2 + (0 - 2)^2 + (0 - 3)^2 = r^2. \]

Step 3: Calculate the Radius Squared, r^2

Calculate each term on the left side of the equation: \[ (0 - 1)^2 + (0 - 2)^2 + (0 - 3)^2 = 1 + 4 + 9 = 14. \] Thus, \( r^2 = 14 \).

Step 4: Formulate the Final Equation of the Sphere

Rewrite the equation of the sphere by substituting the value of \( r^2 \) obtained in the previous step back into the general equation. The equation of the sphere is: \[ (x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 14. \]

Key Concepts

Center of a Sphere

The center of a sphere is a vital concept in geometry, defining the fixed point in 3D space equidistant from every point on the sphere's surface. Think of it as the sphere's heart. It is represented by a point \((h, k, l)\). The parameters \(h\), \(k\), and \(l\) are the coordinates in the x, y, and z dimensions respectively.

In our example, the sphere's center is given as the point \((1, 2, 3)\). This means:

• 1 unit along the x-axis,
• 2 units along the y-axis, and
• 3 units along the z-axis.

Identifying the center is crucial for forming the sphere's equation, as it involves integrating these values directly into the sphere's standard equation format. This center acts as a reference, making it easy to measure distances within the 3D space and identify relationships to other points, such as the sphere’s surface or external points like the origin.

Radius Calculation

The radius of a sphere is the distance from its center to any point on its surface. In mathematical terms, it measures how far the sphere extends in all directions around its center.

The radius \(r\) can be found using the sphere's equation: \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2.\] Here, \(r^2\) is simply the square of the radius, making \(r\) the square root of \(r^2\).

In problems where the sphere passes through a known point, like the origin \((0, 0, 0)\), you substitute the point into the sphere's equation to find \(r^2\). For our sphere with center \((1, 2, 3)\), if you place the origin's coordinates into the equation, you deduce:

• \( (0-1)^2 = 1 \)
• \( (0-2)^2 = 4 \)
• \( (0-3)^2 = 9 \)

Adding these gives \( r^2 = 14 \), so the radius \( r \) is the square root of 14, illustrating a straightforward method to find the radius whenever specific points are involved.

3D Geometry

Three-dimensional geometry deals with shapes located in 3D space, which includes width, height, and depth, unlike the flat surfaces of 2D geometry. Understanding 3D geometry is fundamental, especially when studying shapes like spheres, which exist in all three dimensions simultaneously.

Key aspects of 3D geometry include:

• Knowing how to represent points in a 3D coordinate system (e.g. \((x, y, z)\)).
• Interpreting distances and relationships between these points.
• Visualizing how objects like spheres occupy space.

In the context of spheres, 3D geometry helps visualize and solve for how a sphere passes through points, such as how any point, like the origin, is equidistant from the sphere's center at \((1, 2, 3)\). By understanding these relationships, we can create and manipulate mathematical equations that define and describe spheres in a spatial context. This knowledge not only solves textbook problems but also enriches our understanding of the physical world, where such geometric concepts are constantly at play.