Find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Question

Water is leaking out of an inverted conical tank at a rate of 8300.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 15.0 meters and the diameter at the top is 6.5 meters. If the water level is rising at a rate of 18.0 centimeters per minute when the height of the water is 1.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. 

Note: Let "R" be the unknown rate at which water is being pumped in. Then you know that if V is volume of water, dV/dt = R - 8300.0. Use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time. Recall that the volume of a cone with base radius r and height h is given by 1/3pir²h.

Answer 1

The size of this tank is bh/3=3.25m^2pi15m/3= 165.915cu/m

To get to 18cm the tank will hold 3.25m^2pi18m/3=199.1cu/m;

so 33.183 is needed.

Now we're losing 8300.0 cubic centimeters per min or -.0083c/m/min.

=33.145cu/m/min is poured in.

Water level at 1.5m the new volume is?

I need more imfo on this 1.5 height; either an angle or the new radius.

I realize the tank is 15m tall.

Answer 2

dV/dt = R - 80000, V = 1/3πr²h. Now h/2r = 15/6.5, 6.5 h = 30r, r = 6.5/30h = 13/60h.

Then V = 1/3π(13/60)²h³ = 0.04916h^3.

dV/dt = 0.14748h^2dh/dt and so R = 80000 + 0.14748h^2dh/dt = 80000 + 0.14748·(150)²·18 = 139730 cm³/min