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ranger

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A tank initially contains 60 gal of pure water. Brine (saltwater) containing 1lb of salt per gallon enters the tank at 2gal/min. and the (perfectly mixed) solution leaves the tank at 3 gal/min; thus the tank is empty after exactly 1 hr.

a)Find the amount of salt in the tank after t minutes.

b)What is the maximum amount of salt ever in the tank?

Let Q be the amount of salt.

I'll work with part a) first:

[tex]\frac{dQ}{dt} = rate_{entering} - rate_{leaving}[/tex]

Rate of entering: 1lb/gal * 2gal/min = 2lb/min

Rate of leaving: [tex]\frac{Q}{(60-t)} \cdot \frac{3gal}{min} = \frac{3Q}{60-t}[/tex]

This gives me the first order linear differential to be:

[tex]\frac{dQ}{dt} = 2 - \frac{3Q}{60-t}[/tex]

[tex]\frac{dQ}{dt} + \frac{3Q}{60-t} = 2[/tex]

Well P(x) or P(t) in this case would be 3/(60-t), which when worked out would give me an integration factor of (60-t)^3.

I would eventually get:

[tex]Q\cdot (60-t)^3 = \int 2(60-t)^3[/tex]

[tex]Q\cdot (60-t)^3 = \frac{(60-t)^4}{2} + C[/tex]

[tex]Q = \frac{60-t}{2} + \frac{C}{(60-t)^3}[/tex]

Before I go ahead and find a specific solution, is my general solution correct?

a)Find the amount of salt in the tank after t minutes.

b)What is the maximum amount of salt ever in the tank?

Let Q be the amount of salt.

I'll work with part a) first:

[tex]\frac{dQ}{dt} = rate_{entering} - rate_{leaving}[/tex]

Rate of entering: 1lb/gal * 2gal/min = 2lb/min

Rate of leaving: [tex]\frac{Q}{(60-t)} \cdot \frac{3gal}{min} = \frac{3Q}{60-t}[/tex]

This gives me the first order linear differential to be:

[tex]\frac{dQ}{dt} = 2 - \frac{3Q}{60-t}[/tex]

[tex]\frac{dQ}{dt} + \frac{3Q}{60-t} = 2[/tex]

Well P(x) or P(t) in this case would be 3/(60-t), which when worked out would give me an integration factor of (60-t)^3.

I would eventually get:

[tex]Q\cdot (60-t)^3 = \int 2(60-t)^3[/tex]

[tex]Q\cdot (60-t)^3 = \frac{(60-t)^4}{2} + C[/tex]

[tex]Q = \frac{60-t}{2} + \frac{C}{(60-t)^3}[/tex]

Before I go ahead and find a specific solution, is my general solution correct?

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