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a large tank is filled to capacity with 300 gallons of pure water. brine containing 2 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. find the number a(t) of pounds of salt in the tank at time t. a(t)

Sagot :

The number A(t) of pounds of salt in the tank at time t is, [tex]A(t) = 1200(1-e^-^0^.^0^1^t)[/tex].

What is pounds and gallons?

The name of a unit of currency is pound. Both presently and in the past, it was employed in many different nations.

A gallon, which is equal to eight pints or 4.564 liters, is a unit of measurement for liquids.

Let A(t) be the number of pounds of salts at time t.

Input rate = (4 Ib/gal)(3 gal/min) = 12 Ib/min

Output rate = ((a/300) Ib/gal )(3 gal/min) = 0.01a Ib/min

[tex]\frac{dA}{dt} = 12 - 0.01A\\ \frac{dA}{dt} + 0.01A = 12\\[/tex]

[tex]I.F = e^\int\limits^{0.01}\, ^d^t = e^0^.^0^1^t[/tex]

Solution:

[tex]A(e^0^.^0^1^t)=\int\limits{12(e^0^.^0^1^tdt+c}\\= 12(\frac{e^0^.^0^1^t}{0.01} +c\\Ae^0^.^0^1^t = 1200 e^0^.^0^1^t + c\\A = 1200 + ce^-^0^.^0^1^t\\\\A(0) = 0\\So,\\0 = 1200 + c\\c = -1200\\A(t) = 1200 - 1200e^-^0^.^0^1^t\\A(t) = 1200(1-e^-^0^.^0^1^t)[/tex]

This is the number of pounds of salt in the tank at time t.

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