Consider a spherical bubble whose diameter is expanding at the constant rate of 4 cm/s. What is approximately the rate of change of the volume of the bubble when its diameter is 10 cm? 628.3 cm³/s 2513.3 cm³/s 3 O 5026.6 cm³/s 1047.2 cm³/s 8377.6 cm³/s
Consider a spherical bubble whose diameter is expanding at the constant rate of 4 cm/s. What is approximately the rate of change of the volume of the bubble when its diameter is 10 cm? 628.3 cm³/s 2513.3 cm³/s 3 O 5026.6 cm³/s 1047.2 cm³/s 8377.6 cm³/s
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
Consider a spherical bubble whose diameter is expanding at the constant rate of 4 cm/s. What is approximately the rate of change of the volume of the bubble when its diameter is 10 cm?
### Multiple Choice Options
1. \[ \text{628.3 } \text{cm}^3/\text{s} \]
2. \[ \text{2513.3 } \text{cm}^3/\text{s} \]
3. \[ \text{5026.6 } \text{cm}^3/\text{s} \] (Selected Option)
4. \[ \text{1047.2 } \text{cm}^3/\text{s} \]
5. \[ \text{8377.6 } \text{cm}^3/\text{s} \]
### Explanation
To determine the rate of change of the volume of the spherical bubble, we need to use the formula relating the volume of a sphere to its radius and the chain rule for differentiation.
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
where:
- \( V \) is the volume,
- \( r \) is the radius of the sphere,
- \( \pi \) is the mathematical constant Pi (approximately 3.14159).
Since the diameter \( D \) is expanding at a constant rate, we have the relationship:
\[ D = 2r \]
Given that \( \frac{dD}{dt} = 4 \text{ cm/s} \), the rate of change of the diameter, we can find the rate of change of the radius \( r \) since \( r = \frac{D}{2} \):
\[ \frac{dD}{dt} = 2 \frac{dr}{dt} \]
\[ 4 \text{ cm/s} = 2 \frac{dr}{dt} \]
\[ \frac{dr}{dt} = 2 \text{ cm/s} \]
We now need to find the rate of change of the volume with respect to time, \( \frac{dV}{dt} \), when the diameter is 10 cm (thus, the radius \( r = 5 \text{ cm} \)):
\[ \frac{dV}{dt} = \frac{dV](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4c3310c5-cf1b-4860-895e-b0d1f5af2b55%2Fc879f77a-135b-42f0-9ac6-f2e120846010%2Fdyepbz_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Consider a spherical bubble whose diameter is expanding at the constant rate of 4 cm/s. What is approximately the rate of change of the volume of the bubble when its diameter is 10 cm?
### Multiple Choice Options
1. \[ \text{628.3 } \text{cm}^3/\text{s} \]
2. \[ \text{2513.3 } \text{cm}^3/\text{s} \]
3. \[ \text{5026.6 } \text{cm}^3/\text{s} \] (Selected Option)
4. \[ \text{1047.2 } \text{cm}^3/\text{s} \]
5. \[ \text{8377.6 } \text{cm}^3/\text{s} \]
### Explanation
To determine the rate of change of the volume of the spherical bubble, we need to use the formula relating the volume of a sphere to its radius and the chain rule for differentiation.
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
where:
- \( V \) is the volume,
- \( r \) is the radius of the sphere,
- \( \pi \) is the mathematical constant Pi (approximately 3.14159).
Since the diameter \( D \) is expanding at a constant rate, we have the relationship:
\[ D = 2r \]
Given that \( \frac{dD}{dt} = 4 \text{ cm/s} \), the rate of change of the diameter, we can find the rate of change of the radius \( r \) since \( r = \frac{D}{2} \):
\[ \frac{dD}{dt} = 2 \frac{dr}{dt} \]
\[ 4 \text{ cm/s} = 2 \frac{dr}{dt} \]
\[ \frac{dr}{dt} = 2 \text{ cm/s} \]
We now need to find the rate of change of the volume with respect to time, \( \frac{dV}{dt} \), when the diameter is 10 cm (thus, the radius \( r = 5 \text{ cm} \)):
\[ \frac{dV}{dt} = \frac{dV
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