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Related Rates: A Tank of Two Saltwater Concentrations

Date: 08/05/97 at 03:17:15
From: G Ferguson
Subject: Calculus: Related rates

A tank contains 150 litres of brine solution. The concentration is 
0.7 kg of salt per litre. Brine containing 0.5 kg of salt per litre 
runs into the tank at 7 litres per minute. The mixture, kept uniform 
by stirring, is running out at the same rate. Find the time when the 
salt content will be 90 kg.

I tried Concentration = (Initial conc)*e^kt where k = -0.2 * 7/150.

I calculated  a time of 16.5 min., which is 2 min. out!

Can you please find my error?

Date: 08/05/97 at 08:07:53
From: Doctor Anthony
Subject: Re: Calculus: Related rates

With this type of problem it is better to write down the differential 
equation and solve it, rather than assuming a solution.

Initial salt content is 150 (.7) = 105 kg. We require to reduce this 
to 90 kg., i.e., with concentration 90/150 = 0.6

Let the concentration at time t be s kg/litre. Then if 7 litres per 
min are running out you are losing 7s kg of salt per minute. At the 
same time you are gaining salt at a rate 7(.5) = 3.5 kg of salt per 

The change in salt content  150.ds = (3.5 - 7s)dt

  So  150.ds/(7s-3.5) = -dt   and integrating both sides

   150(1/7)ln(7s-3.5) = -t + const.

           ln(7s-3.5) = -7/150 t + const.

             7s - 3.5 = Ae^(-7/150 t)

                    s = 1/2 + A/7 e^(-7/150 t)

Now at t = 0  s = 0.7  so  0.7 = 0.5 + A/7  so A/7 = 0.2

   Our equation becomes      s = 0.5 + 0.2e^(-7/150 t)

Now find t for s = 0.6     0.6 = 0.5 + 0.2e^(-7/150 t)

                           0.1 = 0.2e^(-7/150 t)

   and so         e^(-7/150 t) = 0.5

            -7/150 t = ln(0.5) =  -.6933

   and therefore             t = (150 x .6933)/7 = 14.856

                               = 14 mins 51 secs
-Doctor Anthony,  The Math Forum
 Check out our web site!   

Date: 08/05/97 at 08:28:40
From: Doctor Jerry
Subject: Re: Calculus: Related rates

Hi G Ferguson,

I think you should formulate the proper differential equation and 
solve it, rather than to try to short-circuit this process.

If S(t) is the amount of salt present at time t, then

S(t) = original amount + amount in up to t - amount out up to t
S(t) = 150*0.7 + 7*0.5*t - integral from T=0 to T=t of (S/150)*7*dt.

You may prefer the differential equation form, which you can obtain by 
differentiating the above or directly.  Rate of change of salt equals 
rate in minus rate out:

dS/dt = 7*0.5 - S*7/150.

The initial condition is S(0) = 150*0.7 = 105.

Just solve this differential equation (variables separable), use the 
initial condition, and then set S = 90.

I did all of that and got t = 14.85min

-Doctor Jerry,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Calculus

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