I also need help with this problem. I started by creating equations to describe the rate of flow of salt for both tanks. I found:
A'(t) = 5 gal/s * 20 lb/100 gal *A(t) ==>> A'(t) = -A(t). ==>> A(t) = 20exp(-t)
This is wrong. Since the salt is uniformly mixed into the water in A, as the water drains out, the amount of salt in A declines but
not the density in lbs per gallon. there is 20/100= 1/5 lb of salt in every gallon of water so every second 5(1/5)= 1 lb of salt flows out, irrespective of how much salt is left in A. The differential equation, with A(t) the amount of salt, in pounds, left in A after t seconds, is A'(t)= -1, not A'(t)= -A, with initial condition A(0)= 20.
B'(t) = A'(t) - rate out ==>> B'(t) = A'(t) - 1/2 (B(t))
If this is, indeed, the same problem as in the first post, the amount of water coming
in to B every second is 5 gallons while the amount of water flowing out of B is 2.5 gallons. Assuming that B was originally empty, the amount of
water in B after t seconds is 2.5t gallons. Taking B(t) to be the amount of salt, in B after t seconds, the density of salt, in pounds per gallon, in B is B(t)/(2.5t) and the amount of salt flowing out each second it (2.5)B(t)/(2.5t)= B(t)/t lb. Since the amount of salt flowing out of A every second is -1 lb,, the amount of salt flowing into B every second is +1 lb. The differential equation for B(t) is B'(t)= 1- B(t)/t.
From here, I used the Variation of Parameters method to solve for B(t). I Came up with:
Bh = Aexp(-t/2)
v = (40/A) * exp(-t/2)
B(t) = V* Bh = 40exp(-t)
This answer is obviously not correct because it implies that the amount of salt in tank B gets exponentially smaller with time, when we know by inspection that it should decrease slowly.
Can anyone tell what I did wrong?
