Asked by Mindy
my circular swimming pool has a 29 foot diameter and is 8 feet tall. at the beginning of the summer i need to fill the pool with water. the flow rate from my spigot into the pool is about 0.75 cubic feet per minute. how fast is the water level, h, rising at the moment when the water level is at a height is 3.7 feet. Plus find the equation relating the two changing quantities.
Answers
Answered by
Steve
The pool has Volume = π * 29^2 * 8/4 = 1682π ft^3
However, the water h feet deep has volume π * 29^2 * 1/4 * h ft^3 = 210.25πh ft^3
V = 841πh/4
dV = 841π/4 dh
0.75 = 841π/4 dh
dh = 0.00113544 ft/min
This rate does not change, since the pool has a circular cross-section, and the radius is constant.
Not only that, it will take 8/0.00028386 = 7045 min to fill the pool. That's 4.9 days!
However, the water h feet deep has volume π * 29^2 * 1/4 * h ft^3 = 210.25πh ft^3
V = 841πh/4
dV = 841π/4 dh
0.75 = 841π/4 dh
dh = 0.00113544 ft/min
This rate does not change, since the pool has a circular cross-section, and the radius is constant.
Not only that, it will take 8/0.00028386 = 7045 min to fill the pool. That's 4.9 days!
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