Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Problem Statement:
**Objective:** Find the standard equation of the sphere with the given characteristics.
**Given Data:**
- Endpoints of a diameter of the sphere: \( (0, 0, 8) \) and \( (4, 6, 0) \)
### Solution:
To find the equation of the sphere, follow these steps:
1. **Determine the center of the sphere (C):**
- The center of the sphere is the midpoint of the diameter. To find the midpoint, use the midpoint formula for coordinates:
\[
\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)
\]
- Plugging in the given endpoints \((0, 0, 8)\) and \((4, 6, 0)\):
\[
C = \left( \frac{0 + 4}{2}, \frac{0 + 6}{2}, \frac{8 + 0}{2} \right) = (2, 3, 4)
\]
2. **Determine the radius (r) of the sphere:**
- The radius is half the length of the diameter. First, calculate the distance between the two endpoints using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]
- For endpoints \((0, 0, 8)\) and \((4, 6, 0)\):
\[
d = \sqrt{(4 - 0)^2 + (6 - 0)^2 + (0 - 8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29}
\]
- Therefore, the radius \( r \) is half of the diameter \( d \):
\[
r = \frac{d}{2} = \frac{2\sqrt{29}}{2} = \sqrt{29}
\]
3. **Formulate the standard equation](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd09d3726-e6d4-40fb-b56b-e32001dd8288%2F165d3d31-2303-425c-bb1e-73794d5f28d5%2Fvxcl8r3_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
**Objective:** Find the standard equation of the sphere with the given characteristics.
**Given Data:**
- Endpoints of a diameter of the sphere: \( (0, 0, 8) \) and \( (4, 6, 0) \)
### Solution:
To find the equation of the sphere, follow these steps:
1. **Determine the center of the sphere (C):**
- The center of the sphere is the midpoint of the diameter. To find the midpoint, use the midpoint formula for coordinates:
\[
\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)
\]
- Plugging in the given endpoints \((0, 0, 8)\) and \((4, 6, 0)\):
\[
C = \left( \frac{0 + 4}{2}, \frac{0 + 6}{2}, \frac{8 + 0}{2} \right) = (2, 3, 4)
\]
2. **Determine the radius (r) of the sphere:**
- The radius is half the length of the diameter. First, calculate the distance between the two endpoints using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]
- For endpoints \((0, 0, 8)\) and \((4, 6, 0)\):
\[
d = \sqrt{(4 - 0)^2 + (6 - 0)^2 + (0 - 8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29}
\]
- Therefore, the radius \( r \) is half of the diameter \( d \):
\[
r = \frac{d}{2} = \frac{2\sqrt{29}}{2} = \sqrt{29}
\]
3. **Formulate the standard equation
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