Water enters a tank at 3t + 4t^2 m^3/s and leaves at t - 16 m^3/s. At what t is there no change in the level of water in the tank.
The rate at which water flows into the tank is dependent on the time t and given by `F_ i = 3t + 4t^2` . The rate at which water flows out of the tank is given by `F_o = t - 16` . At the moment when there is no change in the water level, the rate of water flowing in is equal to the rate at which water flows out. This can be arrived at by solving: 3t + 4t^2 = t - 16
=> 4t^2 + 2t + 16 = 0
In the equation above 2^2 = 4 < 4*16*4 = 256. As the value of 4 - 256 is negative the equation only has imaginary roots.
Therefore the level of water in the tank is never constant, it is always changing.
check Approved by eNotes Editorial