Brine containing one pound of salt per gallon is added to a tank containing 30 g

Question

Brine containing one pound of salt per gallon is added to a tank containing 30 gallons of pure water at a rate of three gallons per minute. The solution is constantly stirred so its concentration remains uniform. At the same time, solution from the tank is drained off and discarded at the rate of three gallons per minute.

(a) If y = y(t) is the number of pounds of salt in the tank t minutes after starting, what initial value problem does y(t) satisfy?

(b) Solve the initial value problem showing all work.

(c) How much salt is in the tank 10 minutes after starting? After a very long time, approximately how much salt is in the tank?

Explanation / Answer

1) rate of salt inflow =3*1 =3pounds per minute

rate of salt out flow =3*y/((3-3)t +30)=3y/30 =y/10 pounds per minute

rate of amount of change of salt in tank dy/dt = rate of salt inflow-rate of salt outflow

dy/dt =3-(y/10)

initially tank contain pure water,so amount of salt is zero => y(0)=0

b)dy/dt =3-(y/10)

dy +(y/10)dt =3dt

integrating fator =e(1/10)dt

integrating fator =e(1/10)t

multiply on both sides by e(1/10)t

dye(1/10)t +(y/10)e(1/10)tdt =3e(1/10)tdt

(ye(1/10)t)' =3e(1/10)tdt

integrate on both sides

(ye(1/10)t)' =3e(1/10)tdt

(ye(1/10)t) =3*10e(1/10)t+C

(ye(1/10)t) =30e(1/10)t+C

y =30+(C/e(1/10)t)

y(0)=0

0 =30+(C/e0)

0=30+C

C=-30

y =30-(30/e(1/10)t) is the solution to initial value problem

c)after 10min, t=10

y =30-(30/e(1/10)10)

y =30-(30/e)

y =18.9636

18.96 pounds of  salt is in the tank 10 minutes after starting

long time means t->

y =limt->30-(30/e(1/10)t)

y=30-0

y=30 pounds

After a very long time, approximately 30 pounds salt is in the tank

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.