Water is leaking out of an inverted conical tank at a rate of 8200.08200.0 cm3/min cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 11.0 m11.0 m and the the diameter at the top is 4.5 m4.5 m. If the water level is rising at a rate of 16.0 cm/min16.0 cm/min when the height of the water is 3.0 m3.0 m, find the rate at which water is being pumped into the tank in cubic centimeters per minute. Answer: cm3/min

I' = 197,474.47 cm^3/min

the rate at which water is being pumped into the tank in cubic centimeters per minute is 197,474.47 cm^3/min

Step-by-step explanation:

Given;

Tank radius r = d/2 = 4.5/2 = 2.25 m = 225 cm

height = 11 m

Change in height dh/dt = 16 cm/min

The volume of a conical tank is;

V = (1/3) πr^2 h ... 1

The ratio of radius to height for the cone is

r/h = 2.25/11

r = 2.25/11 * h

Substituting into equation 1.

V = (1/3 * (2.25/11) ^2) πh^3

the change in volume in tank is

dV/dt = dV/dh. dh/dt

dV/dt = ((2.25/11) ^2) πh^2. dh/dt ... 2

And change in volume dV/dt is the aggregate rate at which water is pumped into the tank.

dV/dt = inlet - outlet rate

Let I' represent the rate of water inlet and O' represent the rate of water outlet.

dV/dt = I' - O'

Water outlet O' is given as 8200 cm^3/min

dV/dt = I' - 8200

Substituting into equation 2;

I' - 8200 = ((2.25/11) ^2) πh^2. dh/dt

I' = ((2.25/11) ^2) πh^2. dh/dt + 8200

h = 3.0 m = 300 cm (water height)

Substituting the given values;

I' = ((2.25/11) ^2) * π*300^2 * 16 + 8200

I' = 197,474.47 cm^3/min

the rate at which water is being pumped into the tank in cubic centimeters per minute is 197,474.47 cm^3/min