Question

Mixture problem: A tank contains 1000 L of solution consisting of 100kg of salt dissolved in water. Pure water is pumped into the tank at a rate of 5 L/s, and the mixture, kept uniform by stirring, is pumped out at the same rate. How long until only 10 kg of salt remains in the tank?

Answer 1

So, I think I got this, but not really sure cus it seemed like it didn't really involve any differential equation...but, we would be looking
time t when x(t) = 10 right?

Then, \[\frac{dy}{dx} = (rate in) - (rate out)\]
where the rate in would be \[\frac{0kg}{second} = 0\]and rate out would be \[\frac{5L}{second}*\frac{x(t)}{100-t} = \frac{5L}{second}*\frac{10}{100-t}=\frac{50}{100-t}\]

Answer 2

so, the equation would become \[\frac{dx}{dt} = \frac{50}{100-t}\] and you would just integrate both sides and solve for t right?

Answer 3

(sorry, bottom should be 100+t not minus...)

Answer 4

\[\begin{align*}\frac{dy}{dt}&=\text{(rate in)}-\text{(rate out)}\\
&=\text{(volume in)}\times\text{(concentration in)}-\text{(volume out)}\times\text{(conc. out)}\\
&=5\times0-5\times\frac{y}{1000+(5-5)t}\\
&=-\frac{5y}{1000+t}
\end{align*}\]
This is definitely a differential equation... A DE is any equation that contains a derivative. Do you need help solving it?