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Nonamiya [84]
3 years ago
7

A large tank is filled to capacity with 600 gallons of pure water. brine containing 4 pounds of salt per gallon is pumped into t

he tank at a rate of 6 gal/min. the well-mixed solution is pumped out at the same rate. find the number a(t) of pounds of salt in the tank at time t.
Mathematics
1 answer:
nignag [31]3 years ago
4 0
If A(t) is the amount of salt in the tank at time t, then the rate at which the amount of salt in the tank changes is given by

\dfrac{\mathrm dA(t)}{\mathrm dt}=\dfrac{4\text{ lbs}}{1\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}-\dfrac{A(t)\text{ lbs}}{600\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}
\dfrac{\mathrm dA}{\mathrm dt}=24\dfrac{\text{lb}}{\text{min}}-\dfrac{A(t)}{100}\dfrac{\text{lb}}{\text{min}}

Let's drop the units for now. We have

\dfrac{\mathrm dA(t)}{\mathrm dt}+\dfrac{A(t)}{100}=24
e^{t/100}\dfrac{\mathrm dA(t)}{\mathrm dt}+e^{t/100}\dfrac{A(t)}{100}=24e^{t/100}
\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/100}A(t)\right]=24e^{t/100}
e^{t/100}A(t)=\displaystyle24\int e^{t/100}\,\mathrm dt
e^{t/100}A(t)=2400e^{t/100}+C
A(t)=2400+Ce^{-t/100}

We're given that the water is pure at the start, so A(0)=0, giving

A(0)=0=2400+Ce^{-0/100}\implies C=-2400

So the amount of salt in the tank (in lbs) at time t is

A(t)=2400\left(1-e^{-t/100}\right)
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