Problem 63 - University Calculus: Early Transcendentals

Question

Find equations for the spheres whose centers and radii are given. $$\begin{array}{l}\text { (Radius) } \\\\\sqrt{14} \\\2 \\\\\frac{4}{9} \\\7\end{array}$$ (Center) $$\left(-1, \frac{1}{2},-\frac{2}{3}\right)$$

Step-by-Step Solution

Answer

The sphere equations are: 1) \((x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = 14\); 2) \((x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = 4\); 3) \((x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = \frac{16}{81}\); 4) \((x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = 49\)."

Step 1: Understanding the Sphere Equation

The equation of a sphere with center \( (h, k, l) \) and radius \( r \) is given by \[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \] In this context, \( h = -1 \, k = \frac{1}{2} \, l = -\frac{2}{3}.\)

Step 2: Finding Equation for Sphere with Radius \\( \sqrt{14} \\\\)

Given the center \( (-1, \frac{1}{2}, -\frac{2}{3}) \) and radius \( r = \sqrt{14} \), we substitute these values into the sphere equation formula:\( (x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = (\sqrt{14})^2 \).Thus, the equation becomes: \[ (x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = 14 \]

Step 3: Finding Equation for Sphere with Radius \\( 2 \\\\)

For a sphere with center \( (-1, \frac{1}{2}, -\frac{2}{3}) \) and radius \( 2 \), substitute the values into the formula:\( (x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = 2^2 \).The equation simplifies to:\[ (x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = 4 \]

Step 4: Finding Equation for Sphere with Radius \\( \frac{4}{9} \\\\)

For a sphere with the same center and radius \( \frac{4}{9} \), substitute as follows:\( (x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = \left(\frac{4}{9}\right)^2 \).This results in:\[ (x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = \frac{16}{81} \]

Step 5: Finding Equation for Sphere with Radius \\( 7 \\\\)

Given the radius \( 7 \), substitute into the standard equation:\( (x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = 7^2 \).This simplifies to:\[ (x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = 49 \]

Key Concepts

Center of a Sphere

The center of a sphere is a key point that defines the position of the sphere in a three-dimensional space. It is like the heart of the sphere—the point from which every point on the surface of the sphere is equidistant. This center is represented by coordinates in the form \( h, k, l \).

In our exercise, the center of the sphere is given as \( (-1, \frac{1}{2}, -\frac{2}{3})\). This means that in a 3D coordinate system:

• The sphere is placed 1 unit left on the x-axis (since we use \( h = -1\)).
• The center lies halfway up along the y-axis (because \( k = \frac{1}{2}\)).
• It is also located two-thirds down on the z-axis ( ext{as} \( l = -\frac{2}{3}\)).

Remembering the center's placement helps comprehend how the sphere is situated spatially. Every equation of a sphere must have its center properly defined by these three coordinates.

Radius

The radius of a sphere is the distance from the center of the sphere to any point on its surface. It determines the size of the sphere and is always positive.

For each sphere in the exercise, we are given different radii:

• The first example uses a radius of \( \sqrt{14} \). This translates to a sphere with a distance of \( \sqrt{14} \) units from the center to any point on its surface.
• For the second sphere, the radius is 2. This means the sphere extends 2 units outwards from its center.
• The third example involves a radius of \( \frac{4}{9} \), which indicates a smaller sphere compared to the first two.
• The fourth sphere has a radius of 7, making it larger than the others discussed here.

Understanding the radius is vital for comprehending how each sphere varies in size. The larger the radius, the bigger the sphere.

Equation Simplification

Simplifying the equation of a sphere involves substituting the center's coordinates and the radius into the standard sphere equation and then performing basic arithmetic operations. The formula for the equation of a sphere with center \((h, k, l)\) and radius \(r\) is: \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\]

Let's see how these substitutions work using our solution steps:
If the center is \( (-1, \frac{1}{2}, -\frac{2}{3}) \) and the radius is defined, you substitute these into the equation:

• Replace \( h, k, l \) with \( -1, \frac{1}{2}, -\frac{2}{3}\) respectively.
• Use your given radius, say \( \sqrt{14}\), square it to get 14.
• Insert into the equation, like \( (x + 1)^2 + (y - \frac{1}{2})^2 + (z + \frac{2}{3})^2 = 14\).

Now your equation is simplified and ready for interpretation or graphing. Simplification is fundamental in making the equations neat and understandable.

Coordinate Geometry

Coordinate geometry is a branch of geometry where the position of points on a plane is described using an ordered pair of numbers. In three dimensions, like in our exercise, coordinate geometry helps plot points as \( (x, y, z) \) coordinates.

Understanding this is crucial when working with spheres since:

• The center of each sphere is defined using three coordinates (x, y, z).
• The radius, measured as distance from the center, assists in visualizing the sphere in the coordinate space.
• The equation \( (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \) interprets the spatial existence of the sphere in coordinate geometry.

Consider how each sphere's equation places it within the 3D coordinate system. Coordinate geometry thus acts as a bridge that translates numbers and formulas into tangible shapes and forms in space.