# Find the amount of salt in a tank at any time 𝑡. ?

A tank initially holds 60𝑔𝑎𝑙𝑠 of brine solution containing 1 𝑙𝑏 of salt 6

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

Relevance

Vo = 60 gallons

a = 1 lb

b = 2 lb

e =  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

mixture is leaving the tank

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

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

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

=> Q(t) = 10(60-2t) + C1(60-2t)²

Applying the initial condition

at t= 0 , Q = a = 10 we get : 10(60) + C1(60)²

=> C1 = -59/360

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

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

when Q(t) = 8

=> t = 29.59 minutes

• dQ/dt= 2*5 – 7Q/(60 – 2t)

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

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

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

Q(t)= 2(60 – 2t) + c(60 – 2t)^(7/2)

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

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

Q(28)≈ 8 lb salt