1. ## rates

A tank contains kg of salt and L of water. A solution of a concentration kg of salt per liter enters a tank at the rate L/min. The solution is mixed and drains from the tank at the same rate.

Find the amount of salt in the tank after hours.

2. Originally Posted by amiv4
A tank contains kg of salt and L of water. A solution of a concentration kg of salt per liter enters a tank at the rate L/min. The solution is mixed and drains from the tank at the same rate.

Find the amount of salt in the tank after hours.
You should know how to set up the differential equation that models this situation:

$\frac{dx}{dt} = (0.04)(8) - \left(\frac{x}{1000}\right) (8)$

where x is the amount (kg) of salt in the tank at time t. That is,

$\frac{dx}{dt} = \frac{40 - x}{125}$, subject to the boundary condition x = 80 when t = 0.

Solve the DE and hence solve for the value of x when t = 5.

3. k i solved it and got
-ln(40-x) = t/125
is that right cuz its not giving me the right answer

4. Originally Posted by amiv4
k i solved it and got
-ln(40-x) = t/125
is that right cuz its not giving me the right answer
If you remember to include the arbitrary constant of integration and then use the given boundary condition to find its value then perhaps you might get the right answer.