related rates

Question

Water is leaking out of an inverted conical tank at a rate of 0.008 m^3/min. At the same time water is being pumped into the tank at a constant rate. The tank has height 14 meters and the diameter at the top is 6.5 meters. If the water level is rising at a rate of 0.28 m/min when the height of the water is 2.5 meters, find the rate at which water is being pumped into the tank.

Answer 1

So, V = (1/3)πr²h,  and h = 14 m., when d = 6.5 m. = 13/2 m.,  which means r = ½d = 13/4 m.

 

dV/dt = (-0.008 + k) m³/min  (leaking 0.008 and increasing by a constant,m k) at the time when h = 2.5 m. and dh/dt = +.28 m/min.

 

I would stat by finding r in terms of h.

 

r/h = (13/4)/14.  So r = 13h/56.

 

V = (1/3)π(13h/56)²h = (1/3)(196h³/175616)     dV/dt = (196h²/175616)(dh/dt) = -0.008 + k.

 

If h = 2.5 and dh/dt = +.28  find k.

 

It looks complicated, but ...

 

I hope this helps.