water is being poured into a cylindrical tank at 9 ft/sec. If the tank has a base radius of 5ft and a height of 10ft, how fast is water being poured in when the height is at 6ft?

Question

anonymous · Answer

V = pi * r^2 * h
dV = pi * r^2 * dh
dV/dt = pi * r^2 * dh/dt
since dh/dt=9 , r=5, we subsititute these into that eqauation
dV/dt = pi * 25 *9 = 225pi
if you meant 9 ft^3 / sec then that would be dV/dt and you would be trying to find dh/dt so 9/25pi

anonymous · Answer

I'm so sorry!! I meant tank shaped like a cone.

anonymous · Answer

V = pi * r^2 * h/3
dV = pi *(2r * dr * h + r^2 *dh)
also 5r = 10h so r = 2h and dr = 2dh
dV = pi *(2(2h) * (2dh) * h + (2h)^2 *dh)
divide both sides by dt  and plug in 9for dh/dt, and 6 for h