Question

A tank contains 350 liters of fluid in which 50 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate. Let the A(t) be the number of grams of salt in the tank at time t. Find the number A(t).)

Answer 1

Answer:

$A(t)=50e^{-\frac{t}{70}}$

Explanation:

The volume of fluid in the tank =350 liters

Initial Amount of Salt, A(0)=50 grams

Amount of Salt in the Tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

Pure water is then pumped into the tank at a rate of 5 L/min. Therefore:

$R_{in}$=(concentration of salt in inflow)(input rate of brine)

$=(0\frac{g}{L})( 5\frac{L}{min})\\\\R_{in}=0\frac{g}{min}$

$R_{out}$=(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{350})( 5\frac{L}{min})\\R_{out}=\frac{A(t)}{70}$

Therefore:

$\dfrac{dA}{dt}=0-\dfrac{A}{70}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{70}=0\\ The integrating factor: e^{\int \frac{1}{70}dt} =e^{\frac{t}{70}}\\ Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{70}}+\dfrac{A}{70}e^{\frac{t}{70}}=0\\(Ae^{\frac{t}{70}})'=0$

Taking the integral of both sides

$\int(Ae^{\frac{t}{70}})'=\int 0 dt\\Ae^{\frac{t}{70}}=C\\ Divide both sides by e^{\frac{t}{70}}\\A(t)=Ce^{-\frac{t}{70}}$

Recall that when t=0, A(0)=50 grams (our initial condition)

$A(t)=Ce^{-\frac{t}{70}}\\50=Ce^{-\frac{0}{70}}\\C=50$

Therefore, the number of grams of salt in the tank at time t is:

$A(t)=50e^{-\frac{t}{70}}$