A tank contains 1000 gal of fresh water. At time t = 0 min, brine containing 5 o

Question

A tank contains 1000 gal of fresh water. At time t = 0 min, brine containing 5 oz of salt per gallon of brine is poured into the tank at a rate of 10 gal/min, and the mixed solution is drained from the tank at the same rate. After 15 min that process is stopped and fresh water is poured into the tank at the rate of 5 gal/min, and the mixed solution is drained from the tank at the same rate. a) Find an equation for the amount of salt at time t, when 0 < t < 15. b) Find an equation for the amount of salt at time t, when 15 < t < 30. c) Find the amount of salt in the tank at time t = 30 min. A tank contains 1000 gal of fresh water. At time t = 0 min, brine containing 5 oz of salt per gallon of brine is poured into the tank at a rate of 10 gal/min, and the mixed solution is drained from the tank at the same rate. After 15 min that process is stopped and fresh water is poured into the tank at the rate of 5 gal/min, and the mixed solution is drained from the tank at the same rate. a) Find an equation for the amount of salt at time t, when 0 < t < 15. b) Find an equation for the amount of salt at time t, when 15 < t < 30. c) Find the amount of salt in the tank at time t = 30 min. A tank contains 1000 gal of fresh water. At time t = 0 min, brine containing 5 oz of salt per gallon of brine is poured into the tank at a rate of 10 gal/min, and the mixed solution is drained from the tank at the same rate. After 15 min that process is stopped and fresh water is poured into the tank at the rate of 5 gal/min, and the mixed solution is drained from the tank at the same rate. a) Find an equation for the amount of salt at time t, when 0 < t < 15. b) Find an equation for the amount of salt at time t, when 15 < t < 30. c) Find the amount of salt in the tank at time t = 30 min.

Explanation / Answer

let amount of salt in tank at t min is y(t) or simply y

y =0 at t=0

a)

net rate = rate of input -rate of output

dy/dt = 10*5 - 10*(y/1000)

=>dy/dt = 50 - (y/100)

=>dy + (y/100)dt =50 dt

multiply by integration factor e(1/100)t

=>dye(1/100)t + (y/100)e(1/100)t dt =50e(1/100)t dt

=>(ye(1/100)t)' =50e(1/100)t  dt

=>ye(1/100)t =50e(1/100)t  dt

=>ye(1/100)t =5000e(1/100)t +c

=>y =5000   +ce(-t/100)

y =0 at t =0

=>0 =5000   +ce(-0/100)

=>0=5000+ c

=>c =-5000

=>y =5000 -5000e(-t/100)

=>y =5000(1 -e(-t/100))

equation for the amount of salt at time t, when 0 < t < 15 is y =5000(1 -e(-t/100))

at t=15,y =5000(1 -e(-15/100))

b)

net rate = rate of input -rate of output

dy/dt = 5*0 - 5*(y/1000)

=>dy/dt = - (y/200)

=>dy/y = - (1/200)dt

=> dy/y = - (1/200)dt

=>ln(y) =(-t/200)+c

=>y =e(-t/200)+c

=>y =Ce(-t/200)

at t=15,y =5000(1 -e(-15/100))

=>Ce(-15/200) =5000(1 -e(-15/100))

=>C =5000e(15/200)(1 -e(-15/100))

=>y =5000e(15/200)(1 -e(-15/100))e(-t/200)

=>y =5000(1 -e(-15/100))e((15-t)/200)

equation for the amount of salt at time t, when 15 < t < 30 is y =5000(1 -e(-15/100))e((15-t)/200)

c)

y =5000(1 -e(-15/100))e((15-30)/200)

=> y =5000(1 -e(-15/100))e(-15/200)

=> y =646.1363378458792

the amount of salt in the tank at time t = 30 min is 646 oz approximately

please rate if helpful. please comment if you have any doubt

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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