Answer:
dQ/dt = 3 - 2Q/(t+200), Q(0) = 100
Step-by-step explanation:
The rate of change dQ/dt describes the amount of salt in the tank at a given time, this rate of change can be express as rate-in minus the rate-out and those individual rates can be expressed as the rate at which liquid enters or leaves the tank (gal/min) by the amount of salt entering or leaving the tank at a given time (lb/gal).
- Rate-in: the amount of liquid entering the tank per minute remains constant at 3 gal/min, and the amount of salt per gallon remains constant as well at 1 lb/gal, therefore, the rate-in is r1 = 3 gal/min * 1 lb/gal = 3 lb/min
- Rate-out: the amount of liquid leaving the tank per minute remains constant at 2 gal/min but the amount of salt per gallon does not because the concentration of the salt is constantly changing due to the entering and leaving liquid, to calculate the variable amount of salt per gallon we divide the amount of salt in the tank Q at a given time by the volume of liquid at given time, this volume can be calculated as the initial content of liquid (200 gal) plus the net amount of liquid entering the tank (liquid entering the tank minus liquid leaving it) multiplied by time, the volume at a given time is ((3 gal/min - 2 gal/min)*t + 200 gal), therefore, the rate-out is r2 = 2 gal/min * Q/(t*(1 gal/min)+200)
Collecting everything said:
dQ/dt = r1 - r2 = 3 lb/min - 2 gal/min * Q/(t*(1 gal/min)+200)
Note:
- The units are really important when solving the equation
- As you can see from our result the amount of solution in the tank depends on the time
- The initial condition Q(0) = 100 describe the amount of salt at t = 0 and is fundamental to solve the differential equation
