A tank contains 1640 L of pure water. Solution that contains 0.05 kg of sugar pe

Question

A tank contains 1640 L of pure water. Solution that contains 0.05 kg of sugar per liter enters the tank at the rate 9 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate.
(a) How much sugar is in the tank at the begining (in kg)?
y(0)= ?

(b) Find the amount of sugar after t minutes (kg).
y(t)= ?

(c) As t becomes large, what value is y(t) approaching ? In other words, calculate the following limit (in kg).

lim t?? y(t)= ?

please explain your answer bc it is most vital to me that i know how to do it myself. best explanation will be awarded points. thank you

Explanation / Answer

Rate going in : 0.05 kg/L * 9L / min = 0.45 kg per minute

Rate out : (s(t) / 1640) * (9 L / min) = 9s(t) / 1640 kg per min

s'(t) = rate in - rate out

s'(t) = 0.45 - 9s(t) / 1640

s'(t) + (9/1640)s(t) = 0.45

Integrating factor = e^(integral of 9/1640) = e^(9t/1640)

Multiply the above DE by the integrating factor, we get :

e^(9t/1640) * [s'(t) + (9/1640)s(t) = 0.45]

d/dt(s * e^(9t/1640)) = 0.45 * e^(9t/1640)

Integrating both sides :

s*e^(9t/1640) = 0.45 * 1640/9 * e^(9t/1640) + C

s*e^(9t/1640) = 82* e^(9t/1640) + C

s(t) = 82 + Ce^(-9t/1640)

a) How much sugar is in the tank at the begining (in kg)?
(recall that the tank is filled with pure water at the start)
So, the mass of sugar at t = 0 is : 0 --> ANSWER
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b) Find the amount of sugar after t minutes (kg).
We got
y(t) = 82 + Ce^(-9t/1640)
We know y(0) = 0, which we use to find C....

0 = 82 + Ce^0)

C = -82

So, y(t) = 82 - 82e^(-9t/1640) ----> ANSWER
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c)
When t ---> large value
9t/1640 ---> also a large value
-9t/1640 ---> large negative value
e^(large negative value) ----> equal to 0

y(t) = 82 - 82 * (0)

y(t) = 82

So, the limit is : 82 kgs ----> ANSWER

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