Question

Water flows into a tank according to the rate F(t) = t+6/1+t, and at the same time empties out at the rate E(t) = ln(t+2)/t+1, with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest gallon, is in the tank at time t = 10 minutes. You must show your setup but can use your calculator for all evaluations.

Answer 1

Answer:

12.29minutes

Step-by-step explanation:

The net flow rate = flow in - flow out = F(t) - E(t) = t+6/1+t -  ln(t+2)/t+1 = [(t + 6) - ln(t+2)]/(t+1)

So, the amount of water present in the tank at time, t is A = net flow rate × time = [(t + 6) - ln(t+2)]/(t+1) × t = t[(t + 6) - ln(t+2)]/(t+1)

At t = 10 minutes,

A = t[(t + 6) - ln(t+2)]/(t+1)

A = 10[(10 + 6) - ln(10+2)]/(10+1)

A = 10[16 - ln12]/11

A = 10[16 - 2.485]/11

A = 10[13.515]/11

A = 135.15/11

A = 12.29 minutes