Question

Need help in solving this problem. Section is under limits at infinity, horizontal asymptotes...

(a) A tank contains 5000L of pure water. Brine that contains 30g of salt per liter of water is pumped into the tank at a rate of 25L/min. Show that the concentration of salt after t minutes (in grams per liter) is
C(t)+ 30t/200+t
(b) what happens to the concentration as t--> infinity? @Mathematics

Answer 1

Do you have any ideas on what might happen as t gets larger?

Answer 2

Where us the equal sign? (in part (a))

Answer 3

Is it "C(t)= 30t/200 +t" or is it C(t)=30t/(200+t)?

Answer 4

a) The tank starts with 5000 L of water and 25 L of brine pours in each minute, the number of liters of water in the tank is

5000+25t

On the other hand, the tank starts with just pure water and no salt, and for each liter of brine that pours in, 30 g of salt pour in. So the number of grams of salt in the tank is

25(30t) (since 25 L/min x 30 g/L x t minutes gets the number of grams!)

So the salt concentration is given by

\[C(t)=25(30t)/(5000+25t)=25(30t)/(25)(200+t)=30t/(200+t)\]

b) A t approaches infinity, the concentration of salt is given by

\[\lim_{t \rightarrow \infty }C(t)=\lim_{t \rightarrow \infty}30t/(200+t)\]

If we divide both top and bottom by t, we get

\[\lim_{t \rightarrow \infty}(1/t)(30t)/(1/t)(200+t)=\lim_{t \rightarrow \infty}30/(200/t+1)=30\]

So eventually the concentration of salt in the tank approaches 30 g/L, which is the same as the concentration of salt in the brine. The point is that the brine will eventually overpower the pure water. Of course, this is blatantly false, since brine is largely water, so the amount of water in the tank will also increase over time, and the brine should never overpower the water in the real world. But we'll roll with it.

Answer 5

Very good GoBlue, you get a medal