Water is leaking out of an inverted conical tank at a rate of 11500.000 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 9.000 meters and the diameter at the top is 4.500 meters. If the water level is rising at a rate of 21.000 centimeters per minute when the height of the water is 2.000 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
Water is leaking out of an inverted conical tank at a rate of 11500.000 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 9.000 meters and the diameter at the top is 4.500 meters. If the water level is rising at a rate of 21.000 centimeters per minute when the height of the water is 2.000 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem Description
Water is leaking out of an inverted conical tank at a rate of 11,500.000 cubic centimeters per minute at the same time that water is being pumped into the tank at a constant rate. The tank has a height of 9.000 meters and the diameter at the top is 4.500 meters. If the water level is rising at a rate of 21.000 centimeters per minute when the height of the water is 2.000 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
### Explanation
To solve this problem, we need to use the concept of related rates in calculus. Let's break this down step-by-step:
1. **Convert Units:**
- Height of the tank: \(9.000\) meters = \(900.000\) centimeters
- Diameter of the tank: \(4.500\) meters = \(450.000\) centimeters
- Height of the water: \(2.000\) meters = \(200.000\) centimeters
- Rate at which the water level is rising: \(21.000\) centimeters per minute
2. **Formulas and Relationships:**
- Volume of a cone:
\[ V = \frac{1}{3} \pi r^2 h \]
- Given the linear relationships due to similar triangles in the tank:
\[ \frac{r}{h} = \frac{225}{900} = \frac{1}{4} \]
- Simplifying, \( r = \frac{h}{4} \)
3. **Substitute r into the Volume Formula:**
\[ V = \frac{1}{3} \pi \left(\frac{h}{4}\right)^2 h = \frac{\pi h^3}{48} \]
4. **Differentiate with respect to time:**
\[ \frac{dV}{dt} = \frac{\pi}{48} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{16} \cdot \frac{dh}{dt} \]
5. **Substitute Given Values:**
- At the height of \( 200.000 \) cm:
\[ h = 200 \]
- Rate of water level](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F276d7283-01d9-45ed-a762-7e45846115eb%2Ff6cca19f-e816-4ec3-8ace-d4f4ff5e5172%2Fjbksix_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Description
Water is leaking out of an inverted conical tank at a rate of 11,500.000 cubic centimeters per minute at the same time that water is being pumped into the tank at a constant rate. The tank has a height of 9.000 meters and the diameter at the top is 4.500 meters. If the water level is rising at a rate of 21.000 centimeters per minute when the height of the water is 2.000 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
### Explanation
To solve this problem, we need to use the concept of related rates in calculus. Let's break this down step-by-step:
1. **Convert Units:**
- Height of the tank: \(9.000\) meters = \(900.000\) centimeters
- Diameter of the tank: \(4.500\) meters = \(450.000\) centimeters
- Height of the water: \(2.000\) meters = \(200.000\) centimeters
- Rate at which the water level is rising: \(21.000\) centimeters per minute
2. **Formulas and Relationships:**
- Volume of a cone:
\[ V = \frac{1}{3} \pi r^2 h \]
- Given the linear relationships due to similar triangles in the tank:
\[ \frac{r}{h} = \frac{225}{900} = \frac{1}{4} \]
- Simplifying, \( r = \frac{h}{4} \)
3. **Substitute r into the Volume Formula:**
\[ V = \frac{1}{3} \pi \left(\frac{h}{4}\right)^2 h = \frac{\pi h^3}{48} \]
4. **Differentiate with respect to time:**
\[ \frac{dV}{dt} = \frac{\pi}{48} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{16} \cdot \frac{dh}{dt} \]
5. **Substitute Given Values:**
- At the height of \( 200.000 \) cm:
\[ h = 200 \]
- Rate of water level
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