Water is pumped out of a holding tank at a rate of 6-6e^(-.09t) liters/minute, where t is in minutes since the pump is started. If the holding tank contains 1000 liters of water when the pump is started, how much water does it hold one hour later??Round your answer to one decimal place.
Let V(t) be the volume of water (in Liters) in the tank after t minutes. We are given that V(0) = 1000, and we wish to find V(60). We are also given the rate of change of the volume: V'(t) = -[6 - 6e^(-0.09t)]. Note that a negative sign needs to be placed because the rate represents water leaving the tank, i.e., being pumped out of the tank. By the Fundamental Theorem of Calculus, Integral (0 to 60) V'(t) dt = V(60) - V(0), so that V(60) = V(0) + Integral (0 to 60) V'(t) dt V(60) = V(0) + Integral (0 to 60) -[6 - 6e^(-0.09t)] dt V(60) = V(0) + Integral (0 to 60) [-6 + 6e^(-0.09t)] dt V(60) = V(0) + [-6t + (6/-0.09)e^(-0.09t)] (Evaluated at t = 60 - Evaluated at t = 0] V(60) = V(0) + [-6(60) + (6/-0.09)e^(-0.09(60))] - [-6(0) + (6/-0.09)e^(-0.09(0))] V(60) = 1000 + [-360 - (200/3)e^(-5.4)] - [0 - (200/3)] V(60) = 1000 - 360 + 200/3[1 - e^(-5.4)] V(60) = 640 + 200/3[1 - e^(-5.4)] V(60) = approx. 706.4 Therefore, approximately 706.4 L of water occupy the holding tank 1 hour after the pump is started.
