Water is leaking out of an inverted conical tank at a rate of 6400 cubic centimeters per min at the same time

Water is leaking out of an inverted conical tank at a rate of 6400 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 11 meters and the diameter at the top is 7 meters. If the water level is rising at a rate of 27 centimeters per minute when the height of the water is 5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

First, we need everything in common terms (centimeters)

H = 11 m = 1100 cm
D = 7 m = 700 cm

h = 5m = 500 cm


Now, we need to set up a formula for volume:

V = (1/3) * pi * r^2 * h

Now, we need to find r when h = 500

r = 0 when h = 0
r = 350 when h = 1100

(1100 - 0) / (350 - 0) = 110 / 35 = 22 / 7

h = (22/7) * r
r = (7/22) * h

V = (1/3) * pi * r^2 * h
V = (1/3) * pi * (7/22)^2 * h^2 * h
V = (1/3) * pi * (49/484) * h^3
dV/dt = 3 * (1/3) * pi * (49/484) * h^2 * dh/dt
dV/dt = (49 * pi / 484) * h^2 * dh/dt

Now, we know that 6400 cm^3 of water is draining out every minute, so we need to modify our volume derivative:

dV/dt = (49 * pi / 484) * h^2 * dh/dt - 6400

h = 500
dh/dt = 27
dV/dt = ?

dV/dt = (49 * pi / 484) * 500^2 * 27 - 6400
dV/dt = (250000 * 27 * 49 * pi - 484 * 6400) / 484
dV/dt = 2140463.1615182316397138828492142

So, 2,140,463.16 cm^3 is being pumped in every minute

V = 1/3 pi r^2 h

11r = 3.5h
r = 7/22h

V = (1/3)(pi)(7/22 h)^2 (h)
dV/dt = (21/22)(pi)(h^2)(dh/dt)
dV/dt = (21/22)(pi)(5^2)(27)

Now, dV/dt = -6400 + the amount that is being added in

figure it out from there :D