An inverted conical tank (with vertex down) is 14 feet across the top and 24 feet deep. If water is flowing in at a rate of 12 ft3/min, find the rate of change of the depth of the water when the water is 10 feet deep.
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To find the rate of change of the depth of the water, we need to take the derivative of the depth with respect to time, given that the water is flowing in at a constant rate.
Let's denote the depth of the water as h(t), where t represents time.
From the problem statement, we know that the rate of change of the volume of water is given by dV/dt = 12 ft^3/min.
Now, we need to express the volume of the water in terms of the depth. Since the tank has the shape of an inverted cone, we can use the volume formula for a cone.
The volume V of a cone with radius r and height h is given by V = (1/3)πr^2h.
In this case, the radius r is half the diameter of the top of the tank, so r = 14 ft / 2 = 7 ft.
Plugging in the values, we have V = (1/3)π(7^2)h = (49/3)πh.
Now, we can differentiate both sides of the equation with respect to time t to get dV/dt:
dV/dt = d/dt[(49/3)πh]
= (49/3)π(dh/dt).
Since dV/dt = 12 ft^3/min, we can substitute this value into the equation:
12 ft^3/min = (49/3)π(dh/dt).
Now, we can solve for dh/dt, which represents the rate of change of the depth of the water:
dh/dt = (12 ft^3/min) * (3/(49π))
= 36/(49π) ft/min.
Therefore, the rate of change of the depth of the water when the water is 10 feet deep is 36/(49π) ft/min.