# Thread: Dif EQ Word Prob

1. ## Dif EQ Word Prob

A tank contains 5000L of water. 20kg of salt is dissolved in the water. Brine, containing 0.03kg of salt per liter of water, makes its way in the tank at a rate of 25L per minute. Note that the solution is thoroughly mixed and the solution will drain at the same rate from the tank.

1.) Give a Dif EQ for the amount of salt, denoted A(t), in the tank at time t.

2.) What's the equilibrium solution of the Dif EQ. Describe what this means in relation to the problem.

2. Originally Posted by Ideasman
A tank contains 5000L of water. 20kg of salt is dissolved in the water. Brine, containing 0.03kg of salt per liter of water, makes its way in the tank at a rate of 25L per minute. Note that the solution is thoroughly mixed and the solution will drain at the same rate from the tank.
Let $\displaystyle A(t)$ be the amount in kilograms of salt in water at time $\displaystyle t$. Thus, we know that $\displaystyle A(0)=20$.
Thus,
$\displaystyle \frac{dA}{dt} = \mbox{ rate in }-\mbox{ rate out}$
The rate is is $\displaystyle (.03)(25) = .75 \mbox{kg}/\mbox{min}$.
The rate out is $\displaystyle \mbox{concentration} \times 25 \mbox{L}/\mbox{min}$. The concentration is given by $\displaystyle \mbox{amount}/\mbox{volume}$ the amount is $\displaystyle A(t)$ and volume is $\displaystyle 5000$. Thus we have $\displaystyle A(t)/200$.
The proper differencial equation is,
$\displaystyle \frac{dA}{dt} = .75 - \frac{A(t)}{200}$.

3. Originally Posted by ThePerfectHacker
Let $\displaystyle A(t)$ be the amount in kilograms of salt in water at time $\displaystyle t$. Thus, we know that $\displaystyle A(0)=20$.
Thus,
$\displaystyle \frac{dA}{dt} = \mbox{ rate in }-\mbox{ rate out}$
The rate is is $\displaystyle (.03)(25) = .75 \mbox{kg}/\mbox{min}$.
The rate out is $\displaystyle \mbox{concentration} \times 25 \mbox{L}/\mbox{min}$. The concentration is given by $\displaystyle \mbox{amount}/\mbox{volume}$ the amount is $\displaystyle A(t)$ and volume is $\displaystyle 5000$. Thus we have $\displaystyle A(t)/200$.
The proper differencial equation is,
$\displaystyle \frac{dA}{dt} = .75 - \frac{A(t)}{200}$.
Sorry I don't quite understand $\displaystyle A(t)/200$. How'd you get that the volume was 200? Also I don't understand what b is asking for the "equilibrium" solution

4. Originally Posted by Ideasman
Sorry I don't quite understand $\displaystyle A(t)/200$. How'd you get that the volume was 200?
Becuase the rate out of salt is the amount of water that comes out times the concentration of salt in water. The concetration of salt is the amount of salt $\displaystyle A(t)$ divided by the volume which is $\displaystyle A(t)/5000$ (think of this as the precentage of salt in water). And now multiply this by $\displaystyle 25$ because the precentage of salt is $\displaystyle A(t)/5000$. Thus, the rate out of salt is $\displaystyle \frac{25A(t)}{5000} = \frac{A(t)}{200}$.