A tank contains 80 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drains from the tank at the rate 7 L/min.
Let y be the number of kg of salt in the tank after t minutes.
The differential equation for this situation would be ??
I don't see how this question is even possible to answer. How does one know the change in salt when I don't know anything about what happening to salt? The question doesn't specify. For all I know the salt never leaves the tank right?
The answer given is
(-7*y)/(1000-1*t)
But this answer doesn't make any sense. It doesn't take into consideration the salt or the 6 liters entering the tank?
Is this the correct answer?
Can someone explain this to me ? Thank you!
Best Answer
Let $x$ be the amount of salt left in the tank after $t$ minutes.
Each minute $6L$ of water is pumped into the tank, and $7L$ leaves the tank, so overall $1L$ of water is drained from the tank. Then after $t$ minutes water left in the tank would be $(1000-t)L$.
Now at $t$ minutes, new $6L$ of water is pumped into the tank, so there are $1000-t+6$ water in the tank. The solution is mixed up, so that the concentration of salt is uniform throughout the solution. Concentration is now
$$ \frac{x}{1006-t} $$
(unit is kg/L, but doesn't really matter).
Now exactly $7L$ of this solution is drained, meaning the salt drained out is exactly
$$ \frac{x}{1006-t}*7 (\text{kg}) $$
edit: above is slightly wrong because I didn't consider the fact that amount of pure water is decreased a bit more each time solution is drained from the tank. But now you have the idea, so you might prove that amount of pure water in the tank is constantly $1000L$ which proves the answer given is correct