Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of
(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min
and flows out at a rate of
(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min
Then the net flow rate is governed by the differential equation
[tex]\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}[/tex]
Solve for S(t):
[tex]\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}[/tex]
[tex]e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}[/tex]
The left side is the derivative of a product:
[tex]\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}[/tex]
Integrate both sides:
[tex]e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt[/tex]
[tex]e^{t/800}S(t)=64e^{t/800}+C[/tex]
[tex]S(t)=64+Ce^{-t/800}[/tex]
There's no sugar in the water at the start, so (a) S(0) = 0, which gives
[tex]0=64+C\impleis C=-64[/tex]
and so (b) the amount of sugar in the tank at time t is
[tex]S(t)=64\left(1-e^{-t/800}\right)[/tex]
As [tex]t\to\infty[/tex], the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.
