Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 9 cubic feet per minute. If the pool has radius 5 feet and height 5 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 3 feet? (Include help (units) with your answer.)
Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 9 cubic feet per minute. If the pool has radius 5 feet and height 5 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 3 feet? (Include help (units) with your answer.)
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:**
Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 9 cubic feet per minute. If the pool has a radius of 5 feet and height of 5 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 3 feet?
*[Include help (units) with your answer.]*
**Analysis Steps:**
1. **Understand the Shape and Given Values:**
- The pool is a right circular cylinder.
- Radius (\(r\)) = 5 feet.
- Height (\(h\)) of the water is changing and depends on the volume.
- Volume flow rate (\(\frac{dV}{dt}\)) = 9 cubic feet per minute.
2. **Formula for Volume of a Cylinder:**
- \( V = \pi r^2 h \).
3. **Differentiate the Volume with Respect to Time:**
- \(\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}\).
4. **Substitute Known Values:**
- \(\frac{dV}{dt} = 9 \) cubic feet per minute.
- \(r = 5\) feet.
- \(h = 3\) feet (when determining the rate of change of height).
- Solve for \(\frac{dh}{dt}\).
5. **Calculate:**
- Use \(\frac{dV}{dt} = \pi (5)^2 \frac{dh}{dt}\).
- \(9 = 25\pi \frac{dh}{dt}\).
- \(\frac{dh}{dt} = \frac{9}{25\pi}\).
6. **Conclusion:**
- The rate of change of the height of the water in the pool when the depth is 3 feet is \(\frac{9}{25\pi}\) feet per minute.
7. **Units:**
- Ensure all units are consistent (cubic feet, feet, minutes).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8256360-52ce-4e6c-9144-e8e1be652f97%2F95b2d153-40a9-4a35-950d-67ef061f2414%2F0fa2pjb_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 9 cubic feet per minute. If the pool has a radius of 5 feet and height of 5 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 3 feet?
*[Include help (units) with your answer.]*
**Analysis Steps:**
1. **Understand the Shape and Given Values:**
- The pool is a right circular cylinder.
- Radius (\(r\)) = 5 feet.
- Height (\(h\)) of the water is changing and depends on the volume.
- Volume flow rate (\(\frac{dV}{dt}\)) = 9 cubic feet per minute.
2. **Formula for Volume of a Cylinder:**
- \( V = \pi r^2 h \).
3. **Differentiate the Volume with Respect to Time:**
- \(\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}\).
4. **Substitute Known Values:**
- \(\frac{dV}{dt} = 9 \) cubic feet per minute.
- \(r = 5\) feet.
- \(h = 3\) feet (when determining the rate of change of height).
- Solve for \(\frac{dh}{dt}\).
5. **Calculate:**
- Use \(\frac{dV}{dt} = \pi (5)^2 \frac{dh}{dt}\).
- \(9 = 25\pi \frac{dh}{dt}\).
- \(\frac{dh}{dt} = \frac{9}{25\pi}\).
6. **Conclusion:**
- The rate of change of the height of the water in the pool when the depth is 3 feet is \(\frac{9}{25\pi}\) feet per minute.
7. **Units:**
- Ensure all units are consistent (cubic feet, feet, minutes).
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