Question

A tank contains 2100 L of pure water. Solution that contains 0.05 kg of sugar per liter enters the tank at the rate 44 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?

y(0)=

(b) Find the amount of sugar after t minutes.

y(t)=

(c) As t becomes large, what value is y(t) approaching ? In other words, calculate the following limit. lim t→∞ y(t)=

Answer 1

Answer :

(a)

Let y(t) be the amount of sugar in the tank at time in seconds.

Then y' = ( rate in ) - ( rate out )

y' = (0.05)(44) - (y/(2100)(44) with y(0) = 0

Where y(0) = 0 represents the initial amount of sugar.

y' = 2.2 - 11* y/525

y' +(11/525)y = 2.2

The integrating factor is

The general solution is

Use y(0) = 0

c = - 105

Which gives the amount of sugar in the tank after t minutes.

(C) As t becomes very very large