Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Write the equation of the sphere with the given center and radius. (a) \((1,2,3) ; 5\) (b) \((-2,-3,-6) ; \sqrt{5}\) (c) \((\pi, e, \sqrt{2}) ; \sqrt{\pi}\)

Step-by-Step Solution

Answer

(a) \((x-1)^2+(y-2)^2+(z-3)^2=25\); (b) \((x+2)^2+(y+3)^2+(z+6)^2=5\); (c) \((x-\pi)^2+(y-e)^2+(z-\sqrt{2})^2=\pi\)

Step 1: Recall the Equation of a Sphere

The general equation of a sphere in three-dimensional space with center at \((h, k, l)\) and radius \(r\) is given by: \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\]

Step 2: Substitute Values for Part (a)

For part (a), the center is \((1, 2, 3)\) and the radius is \(5\). Substitute these values into the sphere equation:\[(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 5^2\]Simplifying gives:\[(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 25\]

Step 3: Substitute Values for Part (b)

For part (b), the center is \((-2, -3, -6)\) and the radius is \(\sqrt{5}\). Substitute these into the sphere equation:\[(x + 2)^2 + (y + 3)^2 + (z + 6)^2 = (\sqrt{5})^2\]Simplifying gives:\[(x + 2)^2 + (y + 3)^2 + (z + 6)^2 = 5\]

Step 4: Substitute Values for Part (c)

For part (c), the center is \((\pi, e, \sqrt{2})\) and the radius is \(\sqrt{\pi}\). Substitute these values into the sphere equation:\[(x - \pi)^2 + (y - e)^2 + (z - \sqrt{2})^2 = (\sqrt{\pi})^2\]Simplifying gives:\[(x - \pi)^2 + (y - e)^2 + (z - \sqrt{2})^2 = \pi\]

Key Concepts

Understanding 3D Geometry

3D geometry deals with objects that have length, width, and height. This is different from 2D geometry, which only considers length and width. In 3D space, points are determined by coordinates

• These coordinates are typically written as \((x, y, z)\), where are the positions along the three axes: x-axis (left to right), y-axis (front to back), and z-axis (up and down).
• Many shapes exist in 3D space, such as spheres, cubes, and cylinders.
• Understanding the location of points in relation to these shapes is crucial in 3D geometry.

A sphere is a perfect example of a 3D shape. It is a space object where every point on the surface is equidistant from its center. This equal distance is called the radius, and it varies depending on the size of the sphere.

Radius Calculation

The radius is a key component when dealing with spheres. It is the distance from the center of the sphere to any point on its surface. To find the radius in sphere problems, you need to understand the formula of the sphere's equation:

• To calculate the radius, we often deal with an equation in the format:
• \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\)

This equation has a center at \((h, k, l)\) and its radius noted as \(r\). For example, if the center of the sphere is \((1, 2, 3)\) and the radius is 5, the equation simplifies to \((x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 25\). Radius calculation is often about making sure we substitute correctly and simplify the given equation to find the values.

Equation Solving

Equation solving for spheres involves substituting the center coordinates and the given radius into the general sphere equation. It's essential to carefully perform each arithmetic operation to ensure accuracy.

• The formula \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\) requires substituting the known values and simplifying.
• For example, in part (b) of our problem, we leverage the center \((-2, -3, -6)\) and radius \(\sqrt{5}\). We insert these values as:
• \((x + 2)^2 + (y + 3)^2 + (z + 6)^2 = 5\)
• Notice that positive signs are used when substituting negative values for the center coordinates. It's crucial to apply these changes correctly to avoid errors.

With practice, solving such equations becomes straightforward, allowing students to locate spheres in 3D space without difficulty.