Question

Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 12 meters and the diameter at the top is 3 meters. If the water level is rising at a rate of 21 centimeters per minute when the height of the water is 3.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Answer 1

Answer:

1508527.582 cm³/min

Step-by-step explanation:

The net rate of flow dV/dt = flow rate in - flow rate out

Let flow rate in = k. Since flow rate out = 6800 cm³/min,

dV/dt = k - 6800

Now, the volume of a cone V = πr²h/3 where r = radius of cone and h = height of cone

dV/dt = d(πr²h/3)/dt = (πr²dh/dt)/3 + 2πrhdr/dt (since dr/dt is not given we assume it is zero)

So, dV/dt = (πr²dh/dt)/3

Let h = height of tank = 12 m, r = radius of tank = diameter/2 = 3/2 = 1.5 m, h' = height when water level is rising at a rate of 21 cm/min = 3.5 m and r' = radius when water level is rising at a rate of 21 cm/min

Now, by similar triangles, h/r = h'/r'

r' = h'r/h = 3.5 m × 1.5 m/12 m = 5.25 m²/12 m = 2.625 m = 262.5 cm

Since the rate at which the water level is rising is dh/dt = 21 cm/min, and the radius at that point is r' = 262.5 cm.

The net rate of increase of water is dV/dt = (πr'²dh/dt)/3

dV/dt = (π(262.5 cm)² × 21 cm/min)/3

dV/dt = (π(68906.25 cm²) × 21 cm/min)/3

dV/dt = 1447031.25π/3 cm³

dV/dt = 4545982.745/3 cm³

dV/dt = 1515327.582 cm³/min

Since dV/dt = k - 6800 cm³/min

k = dV/dt - 6800 cm³/min

k = 1515327.582 cm³/min - 6800 cm³/min

k = 1508527.582 cm³/min

So, the rate at which water is pumped in is 1508527.582 cm³/min