Question

A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt per gallon of water is pumped into the tank at the rate of 5 gallons per minute, and the well-stirred mixture is pumped out at the same rate. Let LaTeX: A\left(t\right)A(t)[Math Processing Error] represent the amount of salt (measured in kilograms) in the tank at time LaTeX: t[Math Processing Error].

Answer 1

Therefore, after long period of time 80kg of salt will remain in tank

Step-by-step explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

`dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt`

integrating factor

`=e^(\int\limits (1/40) \, dt)`

integrating factor
`=e^((1/40)t)`

multiply on both sides by
`=e^((1/40)t)`

`dAe^((1/40)t)+(A/40)e^((1/40)t) dt =2e^((1/40)t)t\\\\(Ae^((1/40)t))=2e^((1/40)t)t`

integrate on both sides

`\int\limits(Ae^((1/40)t))=\int\limits2e^((1/40)t)dt\\\\(Ae^((1/40)t))=2*40e^((1/40)t)+C\\\\A=80+(C/e^((1/40)t))\\\\A(0)=0.3\\\\0.3=80+(C/e^((1/40)t)^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^((1/40)t))`

b)

after long period of time means t - > ∞

`{t \to \infty}\\\\ \lim_(t \to \infty) A_t \\\\ \lim_(t \to \infty) (80)-(79/{e^((1/40)t)}\\\\=80-(0)\\\\=80`

Therefore, after long period of time 80kg of salt will remain in tank