A tank contains 1360 L of pure water. Solution that contains 0.04 kg of sugar pe

Question

A tank contains 1360 L of pure water. Solution that contains 0.04 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 beginning? y(0)= __________ (kg) (b) Find the amount of sugar after t minutes. y(t)= _____________(kg) (c) As t becomes large, what value is y(t) approaching ? In other words, calculate the following limit. limit??y(t)= ___________ (kg) Please show some work.

Explanation / Answer

Let s(t) be a function of the total amount of sugar as a function of time.

It starts as pure water, so
A)

s(0) = 0 kg

B)

Amount of sugar being added per minute:
( Rate of Volume Addition )×( Concentration )
= ( V / time )×( sugar / V )
= ( 9 L/min )·( 0.04 kg/L ) = 0.36 kg/min

Thus the initial rate is:
s'(0) = 0.36 kg/min

Amount being removed (which depends on s of the solution):
= ( 9 L/min )· [ s /( 1360L ) ] = s/( 151.11 min )

ds/dt = (added) - (removed) = 0.36 kg/min - s/( 151.11 min )
s' = 0.36 - s/151.11
s' + s/151.11 = 0.36

The derived solution is :
s(t) = c1*e^( 0*t ) + c2*e^( -t/151.11 )
s(t) = c1 + c2*e^( -t/151.11 )

Solve for the constants using two initial values.

s(0) = 0 = c1 + c2*e^( -0/151.11 )
0 = c1 + c2
c1 = -c2

s'(t) = (-c2/151.11)*e^( -t/151.11 )
s'(0) = 0.36 = (-c2/151.11)e^( -0/151.11 )
0.36 = -c2/151.11
-54 = c2

hence
c1 = -c2 = 54

amount of sugar after t minutes
s(t) = 54 - 54·e^( -t/151.11 )

c)

{ lim t} s(t) = 54 - 54·e^( -/151.11 )
= 54 - 54·e^( - )
= 54 - 54*( 0 )
= 54

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