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 minute. 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! http://mathforum.org/dr.math/
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! http://mathforum.org/dr.math/
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