Water is Pumped into an underground tank at a constant rate of 8 gallons per minute.Water leaks out of the tank at the rate of (t+1)^½ gallons per minute, for 0<t<120 minutes. At time t=0, the tank contains 30 gallons of water.

Question

How many gallons of water leak out of the tank from time t=0 to t=3 minutes?

Answer 1

the effective rate of change of the water is dV/dt = 8 - (t+1)^½ + c
so integrating we get
V = 8t - (2/3)(t+1)^(3/2) + c , where c is a constant
but when t=0 , V = 30
30 = 0 - 2/3(1)^(3/2) + c
c = 92/3

so V = 8t - (2/3)(t+1)^(3/2) + 92/3

when t = 3 minutes
V = 24 - (2/3)(4)^(3/2) + 92/3
= 24 - 16/3 + 92/3
= 148/3

so the increase is 19.3333333
but we poured in 8 gallons/min for 3 minutes which is 24 gallons.
so the leakage must be 24 - 19.3333
or 4.66666 gallons

Answer 2

You work is extremely confusing. You started off with dv/dt and then the next line you have V = 8t - (2/3)(t+1)^(3/2) + c? wouldn't the derivative of sqrt(t+1) be (1/2)(t+1)^(-3/2)?