(a) Let [tex]A(t)[/tex] denote the amount of sugar in the tank at time [tex]t[/tex]. The tank starts with only pure water, so [tex]\boxed{A(0)=0}[/tex].
(b) Sugar flows in at a rate of
(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min
and flows out at a rate of
(A(t)/1080 kg/L) * (7 L/min) = 7A(t)/1080 kg/min
so that the net rate of change of [tex]A(t)[/tex] is governed by the ODE,
[tex]\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}[/tex]
or
[tex]A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}[/tex]
Multiply both sides by the integrating factor [tex]e^{7t/1080}[/tex] to condense the left side into the derivative of a product:
[tex]e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}[/tex]
[tex]\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}[/tex]
Integrate both sides:
[tex]e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt[/tex]
[tex]e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C[/tex]
Solve for [tex]A(t)[/tex]:
[tex]A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}[/tex]
Given that [tex]A(0)=0[/tex], we find
[tex]0=\dfrac{378}5+C\implies C=-\dfrac{378}5[/tex]
so that the amount of sugar at any time [tex]t[/tex] is
[tex]\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}[/tex]
(c) As [tex]t\to\infty[/tex], the exponential term converges to 0 and we're left with
[tex]\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5[/tex]
or 75.6 kg of sugar.
