Problem 37 - Calculus an Applied Approach

Question

Find the standard equation of the sphere. $$ \text { Endpoints of a diameter: }(2,0,0),(0,6,0) $$

Step-by-Step Solution

Answer

The standard equation of the sphere is \((x-1)^2 + (y-3)^2 + z^2 = 10 \)

Step 1: Compute Midpoint

The midpoint, M, of two points in three-dimensional space, (x1, y1, z1) and (x2, y2, z2), is given as \( M = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2}, \frac{z1 + z2}{2}\right) \). Here, the given points are (2, 0, 0) and (0, 6, 0), so the midpoint is \( M = \left( \frac{2 + 0}{2}, \frac{0 + 6}{2}, \frac{0 + 0}{2}\right) = (1, 3, 0) \). Therefore, the center of the sphere is at (1, 3, 0)

Step 2: Compute Radius

The distance between two points in 3D space, (x1, y1, z1) and (x2, y2, z2), is calculated as \(d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2}\). Applying this formula to the first endpoint (2, 0, 0) and the center (1, 3, 0), the radius is computed as \( r = \sqrt{(1 - 2)^2 + (3 - 0)^2 + (0 - 0)^2} = \sqrt{1^2 + 3^2 + 0^2} = \sqrt{10}\)

Step 3: Write Standard Equation

We now substitute the center (a,b,c) = (1,3,0) and the radius r = \(\sqrt{10}\) into the standard equation for a sphere which is \( (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2\). Substitutions will render the standard equation of the sphere to be \((x-1)^2 + (y-3)^2 + z^2 = 10 \)

Key Concepts

Understanding Midpoint Calculation

Calculating the midpoint is crucial when dealing with geometry in three-dimensional space. The midpoint of two points lies exactly halfway between them and is also the center of a sphere when those points form a diameter. To find the midpoint between two points, \(x_1, y_1, z_1\) and \(x_2, y_2, z_2\), we use the formula: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)\] This formula tells us to add the respective coordinates of the two points and divide each sum by 2.

• This gives us the average of each coordinate, hence pinpointing the middle.

In the given exercise, the endpoints of the diameter are \(2, 0, 0\) and \(0, 6, 0\). Using the midpoint formula, we compute: \[ M = \left( \frac{2 + 0}{2}, \frac{0 + 6}{2}, \frac{0 + 0}{2} \right) = (1, 3, 0) \] So in this case, the sphere's center is at \(1, 3, 0\). Finding the midpoint is foundational for solving many geometric problems, especially those involving spheres and circles.

Applying the Distance Formula in 3D

The distance formula in three-dimensional space helps us calculate the straight-line distance between two points. Knowing this distance assists in determining the radius of a sphere when given endpoints of a diameter. The formula is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

• This formula expands the Pythagorean theorem into three dimensions.
• By squaring the differences and summing them, it effectively measures a straight line between the points in space.

In our example, we have one endpoint of the diameter \(2, 0, 0\) and the sphere's center \(1, 3, 0\). Replacing these into the formula, we find: \[ r = \sqrt{(1 - 2)^2 + (3 - 0)^2 + (0 - 0)^2} = \sqrt{1^2 + 3^2 + 0^2} = \sqrt{10} \] This distance, \(\sqrt{10}\), becomes our radius. Knowing how to compute this accurately ensures clarity in geometric modeling and visualization.

Expressing the Standard Equation of a Sphere

Once the center and the radius have been determined, writing the standard equation of a sphere is quite straightforward. The standard equation is derived from the general structure for a sphere's equation in space, which is: \[ (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2 \] In this formula:

• \(a, b, c\) represents the coordinates of the center of the sphere.
• \(r\) is the sphere's radius.

Replacing the values from our exercise, where the center \(a, b, c\) is \(1, 3, 0\) and the radius \(r\) is \(\sqrt{10}\), into the equation provides: \[ (x - 1)^2 + (y - 3)^2 + z^2 = 10 \] This equation now represents the sphere in a mathematically precise way, reflecting its size, position in space, and symmetry. Mastery of this concept is essential for advanced studies in geometry and helps in understanding shapes in a three-dimensional environment.