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andriy [413]
3 years ago
12

g Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another

brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min. If the concentration of the solution entering is 2 lb/gal, determine a differential equation (in lb/min) for the amount of salt A(t) (in lb) in the tank at time t > 0. (Use A for A(t).)
Mathematics
1 answer:
stellarik [79]3 years ago
6 0

Answer:

the required differential equation is dA/dt - \frac{2A}{t+500} = 6  

A(0) = 50lb

Step-by-step explanation:

Given the data in the question;

A(t) is the mass of the salt in in the tank at time t

the new volume of the solution in the tank increases by one gallon each minute

so

dA/dt = ( Rate in) - ( Rate out)

dA/dt = ( 3 gal/min × 2 lb/gal ) - ( 2 gal/min × \frac{A}{t+500} lb/gal )

dA/dt = 6 - \frac{2A}{t+500}

dA/dt - \frac{2A}{t+500} = 6  

A(0) = 50lb

Therefore,  the required differential equation is dA/dt - \frac{2A}{t+500} = 6  

A(0) = 50lb

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