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?
Water is pumped out of a holding tank at a rate of 4-4e^(-.14t) liters/minute. Lets call that function w(t). So if we integrate w(t) over a time period that will give us the total water removed over that time period. So lets integrate it from t=0 (time when pump started) to t=60 (one hour later): ∫ w(t) from t=0 to t=60 ∫ w(t)=∫ 4-4e^(-.14t)=4t+(4/.14)e^(-.14t) evaluated from 0 to 60 ∫ w(t)=4(60)+(4/.14)e^(-.14*60)-4(0)-(4/.1... ∫ w(t)=211.435 liters removed If the holding tank began at 1000 liters and 211.435 were removed... Remaining water=1000-211.435 ***Remaining water=788.565 liters***
