A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/min, and the mixture is allowed to to flow out of the tank at a rate of 2 gal/min. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Find the concentration (in pounds per gallon) of salt in the tank when it is at the point of overflow. Compare this concentration with the theoretical concentration if the tank had infinite capacity.

A) 484 lb of salt is the amount of salt at anytime prior to the moment when the solution begins to overflow.

B) The concentration of salt at that same time above is 0.968 lb/gal

C)The theoretical concentration if the tank had infinite capacity is 1 lb/gal

Step-by-step explanation:

Let the amount of salt at anytime (t) be y(t).

Now from Balance laws, we know that time rate of change (y') is the difference between the salt inflow and outflow rates.

From the question, 3 gallons of salt water runs in the tank per minute and contains 1 lb of salt.

Therefore, the salt inflow rate ;

3 x 1 = 3 gal.lb/min

Now, the amount of water in the tank at any moment 't' will be written as:

200 + (3-2)t which gives 200 + t.

From the question, the tank initially contained 200gal of water and the rate at which salt water is pumped into the top of the tank is 1gal/min.

This means that we have one new gallon of water in the tank at every minute.

Now, the outflow is 2 gal/min. Which can be written as a fraction of the total solution content in the tank as; 2/(200+t). But when we include the salt content y(t) , we have;2y(t)/(200 + t)

Since the tank initially contains 100lb of salt, we therefore get the initial condition as;

y'(t) = 3 - {(2y(t))/(200+t)} and y(0) = 100lb

If we rearrange the equation of y'(t) to make 3 the subject, we get;

y'(t) + {(2y(t))/(200+t)} = 3

This is a linear ordinary differential equation. From our initial derivation, we know P(t) = 2/(200+t) and Q(t) = 3.

Thus, h = ∫ P dt = 2 ∫ dt/(200 + t)

=In (t + 200)^2

Therefore, an integrating factor is;

e^h = (200 + t)^2

Thus, general solution is;

y(t) = (e^-h) (c + ∫ Q(e^h) dt

= 1/(200 + t)^2 [c + ∫3(200 + t)^2dt]

= {c/(200 + t) ^2} + 200 + t

We can now use the initial condition to determine the numeric value of c.

Substitute 0 for t and 100 for y in the last equation to get ;

100 = y(0) - c/(200+0)^2 + 200 + 0 to give; - 100 = c/200^2

Therefore, c = - 4 x 10^6

So, y(t) = {(- 4 x 10^6) / (200+t)^2} + 200 + t. This is the amount of salt in the tank at any moment t.

Now, we know that the capacity of the tank is 500 gallons and it contained 300 gallons initially. The remaining space in the tank at the beginning will be = 500 - 200 = 300 gallons.

We now need to calculate y(300) to find the amount of salt at anytime prior to the moment when the solution begins to overflow;

y(300) = {(- 4 x 10^6) / (200+300)^2} + 200 + 300 = 484lb of salt

To get the concentration of salt at this same moment of time, we divide the amount of salt by the amount of water in the tank at that moment.

So concentration at t=300;

= 484/500 = 0.968 lb/gal

If the tank had infinite capacity, then;

Lim, t →∞ (y(t) /(200+t) = Lim, t →∞ [{(4x10^6)/(200+t)^2} + (200 + t)} /(200 +t)]

= Lim, t →∞ [{(4x10^6)/(200 + t)^3} + 1] = 1lb/gal