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A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into the tank at a rate of 3 gal/min. the well-mixed solution is pumped out at the same rate.
a.set up an initial value problem (differential equation and initial condition) to find ????(????). (6pts)
b.find the number ????(????) of pounds of salt in the tank at time ????. (6pts)
c.how much salt will be in the tank in 10 minutes? (2pts)

Respuesta :

LammettHash
If [tex]A(t)[/tex] is the amount of salt in the tank at time [tex]t[/tex], then the rate at which this amount changes over time is given by the ODE

[tex]A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}[/tex]

[tex]A'(t)+\dfrac1{100}A(t)=15[/tex]

We're told that the tank initially starts with no salt in the water, so [tex]A(0)=0[/tex].

Multiply both sides by an integrating factor, [tex]e^{t/100}[/tex]:

[tex]e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}[/tex]
[tex]\left(e^{t/100}A(t)\right)'=15e^{t/100}[/tex]
[tex]e^{t/100}A(t)=1500e^{t/100}+C[/tex]
[tex]A(t)=1500+Ce^{-t/100}[/tex]

Since [tex]A(0)=0[/tex], we have

[tex]0=1500+C\implies C=-1500[/tex]

so that the amount of salt in the tank over time is given by

[tex]A(t)=1500(1-e^{-t/100})[/tex]

After 10 minutes, the amount of salt in the tank is

[tex]A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}[/tex]

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