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Suppose that a large mixing tank initially holds 400 gallons of water in which 70 pounds of salt have been dissolved. Pure water is pumped into the tank at a rate of 4 gal/min, and when the solution is well stirred, it is then pumped out at the same rate. Determine a differential equation for the amount of salt A(t) in the tank at time t > 0. What is A(0)? (Use A for A(t).)
dA dt =
A(0) =

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Answer:

Step-by-step explanation:

Given that a large mixing tank initially holds 400 gallons of water in which 70 pounds of salt have been dissolved. Pure water is pumped into the tank at a rate of 4 gal/min, and when the solution is well stirred, it is then pumped out at the same rate.

Let Q(t) be the quantity of salt at time t.

Q(0) = 70 pounds

Inflow of water = outflow = 4 gal/min

So resulting volume of tank at time t = 400 gallons

Rate of change of salt = inflow - outflow

= 0 - [tex]\frac{Q(t)}{400}[/tex]

i.e. [tex]\frac{dQ}{dt} =\frac{-Q(t)}{400}[/tex]

This is a variable separable equation

[tex]\frac{dQ}{Q}=-\frac{dt}{400} \\ln Q = {-\frac{t}{400}} +C\\Q = A e^{-\frac{t}{400}}[/tex]

Use when t =0 Q = 70

A =70

So A(t) = Q(t) = [tex]70 e^{-\frac{t}{400}}[/tex] and

A(0) = Q(0) = 70 lbs.

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