Calculus Question Help! (1 Viewer)

[Jmc-Automatic]

Brine containing 2 grams of salt per litre runs into a vessel initially containing 100 litres of water and 25 grams of salt. If the brine enters at 5 litres/minute, the contents of the vessel are kept uniform by stirring and the mixture flows out at the same rate, show that if Q grams of salt are in the vessel at any time t:

dQ/dt = 0.05(200-Q)

I just can't get the required answer, if any one could help that would be great!
Thanks in advance.

 

[Michaelmoo]

Trick would have been to recognise it as an exponential growth. If you think about it, the original salt concentration in the vessel is 25/100 = 0.25 grams per litre. As you add a 2g/L and remove the equivalent, the concentration in the vessel will approach 2g/L. So the amount of salt at any time is in the form:



now as t approaches infinity, the concentration will approach 2g/L (from above). And, since the volume input is equal to the volume output, the volume is always 100L. so as t approaches infinity, approaches 0, and Q approaches 100 x 2 = 200grams. So C is equal to 200:



Differentiating with respect to t:







Now INITIALLY, salt enters the vessel at 5 x 2 = 10g/min [since the concentration is 2g/L and volume is dispensed at 5L/min]

Also, salt leaves the vessel at 25/20 = 1.25 g/min [since one 20th of the volume of water in the vessel leaves at any time, and initially there is 25g of salt --> 5g/L is one 20th of 100mL]

So the net change in volume of salt INITIALLY is 10-1.25 = 8.75g/min. In other words when t = 0 and Q = 25.

Sub that in above and you get k = 0.05


So the final equation is :



Sorry bit long. Might be a shorter way =\

 

[Jmc-Automatic]

Brilliant! I was trying to get there by using dQ/dt = [rate of inflow - rate of outflow] but couldn't quite figure out the right values for each in terms of Q. Kudos on your detailed method and thanks for your help.