# Find the amount of salt in a tank at any time π‘.Β ?

Β A tank initially holds 60ππππ of brine solution containing 1 ππ of salt 6

per gallon. At π‘ = 0, another brine solution containing 2 ππ of salt per gallon is poured into the tank at the rate of 5πππ/πππ, while the well- stirred mixture leaves the tank at the rate of 7πππ/πππ. Find the amount of salt in the tank at any time π‘. Also the time at which the mixture in the tank contains 8 ππ of salt.

### 2 Answers

- jacob sLv 72 months agoFavourite answer
Vo = 60 gallons

a = 1 lb

b = 2 lb

e = Β 5gal/min , f = 7gal/min

Vo is the initial volume in the tank

a = brine solution in the tank initially

b = amount of salt present in the other brine solution

e and f = rate at which new solution is being poured in the tank = rate at which well stirred

Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β mixture is leaving the tank

=> dQ/dt + f/[Q/(Vo + et - ft)]=be

dQ/dt +7*[Q/(100 + 5t - 7t)]=2*5Β

dQ/dt = 2*5 - 7*[Q/(100 - 2t)]

=> Q(t) = 10(60-2t) + C1(60-2t)Β²

Applying the initial condition

at t= 0 , Q = a = 10 we get : 10(60) + C1(60)Β²

=> C1 = -59/360

Therefore, the amount of salt in the tank at any time t

=> Q(t) Β = 10(60-2t) + -(59/360)(60-2t)Β²

when Q(t) = 8Β

=> t = 29.59 minutes

Your comments is greatly appreciated . Thank you.

- BryceLv 71 month ago
dQ/dt= 2*5 β 7Q/(60 β 2t)

integrating factor: e^β« 7/(60 β 2t) dt= (-7/2)ln(60 β 2t)= (60 β 2t)^(-7/2)

d/dt[Q(60 β 2t)^(-7/2)]= 10(60 β 2t)^(-7/2)

Q(60 β 2t)^(-7/2)= 10*-2/5*-1/2(60 β 2t)^(-5/2) + c

Q(t)= 2(60 β 2t) + c(60 β 2t)^(7/2)

Q(0)= 60, c= -1/60^(5/2)

Q(t)= 2(60 β 2t) β 1/60^(5/2)(60 β 2t)^(7/2)

Q(28)β 8 lb salt