Question

A tank contains 1000 liters (L) of a solution consisting of 100 kg of salt dissolved in water. Pure water is pumped into the tank at the rate of 5 L/s, and the mixture - kept uniform by stirring - is pumped out at the same rate. How long will it be until only 10 kg of salt remains in the tank?
I need help with this problem plz.

Answer 1

\[\begin{align*}\frac{dA(t)}{dt}&=\text{(rate in)}-\text{(rate out)}\\
&=\left(0\frac{\text{kg}}{\text{L}}\right)\left(5\frac{\text{L}}{\text{s}}\right)-\left(\frac{A(t)}{1000}\frac{\text{kg}}{\text{L}}\right)\left(5\frac{\text{L}}{\text{s}}\right)\\\\
&=-\frac{1}{200}A(t)
\end{align*}\]
Do you see how we get this setup?

The equation is separable, so finding the solution should be fairly straightforward.

Answer 2

could you plz finish it so i can get the idea

Answer 3

Are you telling me you don't know how to solve this equation:
\[y'=-\frac{1}{200}y~~?\]
It's separable, consider the differential notation for the derivative:
\[\begin{align*}
\frac{dy}{dx}&=-\frac{1}{200}y\\\\
\frac{dy}{y}&=-\frac{dx}{200}\\\\
\int\frac{dy}{y}&=-\frac{1}{200}\int dx
\end{align*}\]

Answer 4

so i will get y=Ce^(-x/200)
and C=100
y(x)=100 e^(-x/200)

Answer 5

Correct