Answer :
According to implicit differentiation, the depth of water varies at a rate of -0.0764 feet per minute.
The volume of a cone with radius r and height h may be calculated as follows:
V = (πr²h) ÷ 3
Using implicit differentiation, determine that it is the rate of change, as follows:
dV ÷ dt = ((2πrh ÷ 3) × (dr ÷ dt)) + ((πr² ÷ 3) × (dh ÷ dt))
In this case, the height is 12 feet and the radius is 10 feet, therefore h = 12 and r = 10
Because the radius does not change, dr ÷ dt = 0
Water seeps at an 8 cubic foot per minute pace, therefore dV ÷ dt = -8
Then:
dV ÷ dt = ((2πrh ÷ 3) × (dr ÷ dt)) + ((πr² ÷ 3) × (dh ÷ dt))
-8 = (π(10)² ÷ 3) × (dh ÷ dt)
dh ÷ dt = -24 ÷ 100π
dh ÷ dt = -0.0764
Water depth varies at a rate of -0.0764 feet per minute.
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The question is -
Consider a tank in the shape of an inverted right circular cone that is leaking water. The dimensions of the conical tank are a height of 12 ft and a radius of 10 ft. How fast does the depth of the water change when the water is 10 ft high if the cone leaks at a rate of 8 cubic feet per minute?
