A tank contains 2960 L of pure water Solution that contains

A tank contains 2960 L of pure water. Solution that contains 0.06 kg of sugar per liter enters the tank at the rate 4 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate.

1 pt) A tank contains 2960l L of pure water. Solution that contains 0.06lkg of sugar per liter enters the tank at the rate 4l/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) (kg) (b) Find the amount of sugar after t minutes. y(t) =1 (kg) (c) As t becomes large, what value is y()l approaching? In other words, calculate the following limit. lim y(t) (kg) 1+00

Solution

Ans-

Let x(t) = amount of sugar in tank (in kg) at time t (in min)

(a) x(0) = 0 (since tank initially contains pure water)

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(b)

Rate of sugar flowing into tank:
0.1 kg/L * 9 L/min = 0.9 kg/min

Rate of sugar flowing out of tank:
concentration of sugar in tank * 9 L/min
x kg / 1060L * 9L/min = 9x/1060 kg/min

dx/dt = rate in rate out
dx/dt = 0.9 9x/1060
dx/dt = 9/1060 (106x)
dx/(106x) = 9/1060 dt

Integrate both sides
dx/(106x) = 9/1060 dt
ln|106x| = 9/1060 t + C
ln|106x| = 9t/1060 + C
106 x = Ce^(9t/1060)
x = 106 Ce^(9t/1060)

x(0) = 0
106 Ce^0 = 0
106 C = 0
C = 106

x(t) = 106 106e^(9t/1060)

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(c)

As t becomes large, the value that is approaching infinity is t itself

I think you want to find limit of amount of sugar (x(t)) as t approaches infinity.

lim[t] 106 106e^(9t/1060)
= 106 106e^()
= 106 106(0)
= 106 kg