A tank contains 2380 L of pure water Solution that contains

A tank contains 2380 L of pure water. Solution that contains 0.03 kg of sugar per liter enters the tank at the rate 3 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate. How much sugar is in the tank at the begining? y(0) = (kg) Find the amount of sugar after t minutes. y(t) = (kg) As t becomes large, what value is y(t) approaching ? In other words, calculate the following limit, lim_t rightarrow infinity y(t) = (kg)

Solution

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)
(b)
Rate of sugar flowing into tank:
0.03 kg/L * 3 L/min = 0.09 kg/min

Rate of sugar flowing out of tank:
concentration of sugar in tank * 3 L/min
x kg / 2380L * 3L/min = 3x/2380 kg/min

dx/dt = rate in rate out
dx/dt = 0.03 3x/2380
dx/dt = 3/2380 (23.8x)
dx/(23.8x) = 3/2380 dt

Integrate both sides
dx/(23.8x) = 3/2380 dt
ln|23.8x| = 3/2380 t + C
ln|23.8x| = 3t/2380 + C
23.8 x = Ce^(3t/2380)
x = 23.8 Ce^(9t/2380)

x(0) = 0
23.8 Ce^0 = 0
23.8 C = 0
C = 23.8

x(t) = 23.8-23.8e^(3t/2380)
(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] 23.8 23.8e^(3t/2380)
= 23.8 23.8e^()
= 23.8 23.8(0)
= 23.8 kg
Note that concentration of sugar entering tank is 0.3 kg/L, so you would expect concentration in tank to reach this same concentration over time

Since tank contains 2380L, then 0.3 kg/L * 2380L = 714 kg