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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![**Mathematics - Calculus Problem**
**Problem Statement:**
The radius of a sphere is increasing at a rate of 2 mm/s. How fast is the volume increasing (in mm³/s) when the diameter is 40 mm?
**Solution Explanation:**
1. **Identifying given values:**
- The rate of change of the radius, \( dr/dt \), is \( 2 \, \text{mm/s} \).
- The diameter of the sphere is 40 mm, which means the radius, \( r \), is \( \frac{40}{2} = 20 \, \text{mm} \).
2. **Formula for volume of a sphere:**
\[
V = \frac{4}{3} \pi r^3
\]
3. **Differentiate the volume with respect to time \( t \):**
\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right)
\]
Using the chain rule, we get:
\[
\frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt}
\]
4. **Substitute the known values into the differentiated formula:**
- \( r = 20 \, \text{mm} \)
- \( \frac{dr}{dt} = 2 \, \text{mm/s} \)
\[
\frac{dV}{dt} = 4 \pi (20 \, \text{mm})^2 \cdot (2 \, \text{mm/s})
\]
5. **Calculate the rate of change of the volume:**
\[
\frac{dV}{dt} = 4 \pi (400 \, \text{mm}^2) \cdot (2 \, \text{mm/s})
\]
\[
\frac{dV}{dt} = 3200 \pi \, \text{mm}^3/\text{s}
\]
Hence, the volume of the sphere is increasing at a rate of \( 3200 \pi \, \text{mm}^3/\text{s} \).
**Concepts Covered:**
- Differential calculus](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc1c90340-a748-4160-8298-b9e689632892%2F2bf3c869-e65c-4e94-ba1b-e6f8efcacf47%2Fnhmycq6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Mathematics - Calculus Problem**
**Problem Statement:**
The radius of a sphere is increasing at a rate of 2 mm/s. How fast is the volume increasing (in mm³/s) when the diameter is 40 mm?
**Solution Explanation:**
1. **Identifying given values:**
- The rate of change of the radius, \( dr/dt \), is \( 2 \, \text{mm/s} \).
- The diameter of the sphere is 40 mm, which means the radius, \( r \), is \( \frac{40}{2} = 20 \, \text{mm} \).
2. **Formula for volume of a sphere:**
\[
V = \frac{4}{3} \pi r^3
\]
3. **Differentiate the volume with respect to time \( t \):**
\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right)
\]
Using the chain rule, we get:
\[
\frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt}
\]
4. **Substitute the known values into the differentiated formula:**
- \( r = 20 \, \text{mm} \)
- \( \frac{dr}{dt} = 2 \, \text{mm/s} \)
\[
\frac{dV}{dt} = 4 \pi (20 \, \text{mm})^2 \cdot (2 \, \text{mm/s})
\]
5. **Calculate the rate of change of the volume:**
\[
\frac{dV}{dt} = 4 \pi (400 \, \text{mm}^2) \cdot (2 \, \text{mm/s})
\]
\[
\frac{dV}{dt} = 3200 \pi \, \text{mm}^3/\text{s}
\]
Hence, the volume of the sphere is increasing at a rate of \( 3200 \pi \, \text{mm}^3/\text{s} \).
**Concepts Covered:**
- Differential calculus
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