Question

A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate.Required:a. How much sugar is in the tank at the begining?b. Find the amount of sugar after t minutes.c. As t becomes large, what value is y(t) approaching ?

Answer 1

(a) Let $A(t)$ denote the amount of sugar in the tank at time $t$. The tank starts with only pure water, so $\boxed{A(0)=0}$.

(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 $A(t)$ is governed by the ODE,

$\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}$

or

$A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}$

Multiply both sides by the integrating factor $e^{7t/1080}$ to condense the left side into the derivative of a product:

$e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}$

$\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}$

Integrate both sides:

$e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt$

$e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C$

Solve for $A(t)$:

$A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}$

Given that $A(0)=0$, we find

$0=\dfrac{378}5+C\implies C=-\dfrac{378}5$

so that the amount of sugar at any time $t$ is

$\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}$

(c) As $t\to\infty$, the exponential term converges to 0 and we're left with

$\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5$

or 75.6 kg of sugar.