# Thread: A tank of water.

1. ## A tank of water.

I am getting perplexed with this problem. I have my method below following the question but the more I look at it the more it doesn't seem right.

A tank contains a perfectly mixed solution of 5kg of salt and 500 litres of water. Starting at t=0, fresh water is poured into the tank at a rate of 4 litres/min. A mixing device mainntains homogeneity. The solution leaves the tank at a rate of 4 litres/min.

a). What is the differential equation governing the amount of salt in the tank at any time?
There is a part b to this question but I can figure that part out if I can do this part.

I need to find $\frac{dm}{dt}$ where m is the mass of salt in kg and t is the time in minutes. Since we have 5kg and 500 litres to start with the concentration is $\frac{5}{500}=\frac{1}{100}$kg/litre.

Using the relationship $\frac{dm}{dt}=\frac{dm}{dv} \frac{dv}{dt}$ (from the chain rule) I get:

$\frac{dm}{dt}=\frac{1}{100}-4t\frac{1}{100}$.

Here $\frac{dm}{dv}=\frac{1}{100}-4t\frac{1}{100}$ and $\frac{dv}{dt}=1$ since the volume of the tank is staying the same.

To me it makes sense because the mass of the salt with respect to time is getting smaller, but as time increases the concentration would get further from $\frac{1}{100}$ so it wouldn't be $4t\frac{1}{100}$ at the end of the expression. I can't see a way past this unless I could somehow incorporate natural logarthms (ie. exponential decay).

Am I approaching this from the right angle?

2. A tank contains a perfectly mixed solution of 5kg of salt and 500 litres of water. Starting at t=0, fresh water is poured into the tank at a rate of 4 litres/min. A mixing device mainntains homogeneity. The solution leaves the tank at a rate of 4 litres/min.

a). What is the differential equation governing the amount of salt in the tank at any time?
Let $y(t)$ be the amount of salt at time $t$.

This means,
$\frac{dy}{dt} = \text{ rate in } - \text{ rate out }$

The rate (of salt) in is 0 since only clean water is coming in.

The rate (of salt) out is $\frac{y}{500}\cdot 5 = \frac{y}{100}$.

Thus, $\frac{dy}{dt} = - \frac{y}{100} \text{ and }y(0)=5$

3. awesome!

This allowed me to do part B (i figured i'd post this because something I predicted happens!)

b). In how many minutes will the concentration of salt reach a 0.1% level (ie. initial concentration is 1%)?
$\frac{dy}{dt}=\frac{-y}{100}$

Solving this gives:

$y=Ae^{\frac{-t}{100}}$ <---- The natural logarthm I mentioned! =O

$t=0, \ y=5$

$5=A$

$y=5e^{\frac{-t}{100}}$

$\frac{0.5}{5}=e^{\frac{-t}{100}}$

$ln\frac{0.5}{5}=\frac{-t}{100}$

$t=-100ln\frac{0.5}{5}$

$t=230 \ minutes \ and \ 16 \ seconds$

EDIT: I was talking to one of my friends today and she pointed out that the step $
\frac{y}{500}\cdot 5 = \frac{y}{100}
$
was a little strange. Wouldn't it make more sense if it was $
\frac{y}{500}\cdot 4 = \frac{y}{125}
$
since 4 litres of water are leaving the tank?