1. Water is flowing into a tank at a rate of r(t)=(t^2)((ft^3)/(min)). At the same time, water is leaking out of the tank at a constant rate of .25((ft^3)/(min)). How much will the amount of water in the tank change from t=1 minute to t=5 minutes?

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1 Answer

dv/dt = water in - water out

dv/dt = t^2 - 0.25

Amount of water changed between t= 1 to 5 minutes

 V = int(1,5) t^2 - 0.25 = int(1,5) t^3/3 - 0.25t = 40.333 ft^3

by Level 7 User (27.4k points)

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