Science:Math Exam Resources/Courses/MATH101/April 2011/Question 05 (b)

MATH101 April 2011

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Question 05 (b)

Bird-Bath and Beyond Incorporated is famous for its Quetzal attracting bird-feeder solution made from water, honey and cane-sugar. To make their solution, both honey and a cane-sugar solution are poured into a 200 L mixing tank. The honey is poured in at a rate of 1 L per minute while the sugar solution is poured in at 9 L per minute. Note that 1 L of honey contains 1 kg of sugar, while the cane-sugar solution contains 100g of sugar per L. Unfortunately today there is a problem with the mixing tank. It was thoroughly cleaned and is initially filled with pure water, but main valve was broken and the water cannot be drained. When the mixing process is started, the honey and sugar-solutions are poured into the tank, and the excess fluid flows out of an emergency valve at 10 L per minute and onto the floor. You should assume that the solutions mix immediately and thoroughly in the tank. (b) How much sugar is in the tank after time t?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Solve the differential equation found in part (a).

Hint 2
To solve the differential equation, separate the variables.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that the differential equation is given by ${\frac {dS}{dt}}=1.9-{\frac {1}{20}}S$ We use separation of variables and bring all terms with S to the left hand side, and all terms involving t to the right hand side. So we multiply both sides with $\displaystyle dt$ and divide by $\displaystyle 1.9-{\frac {1}{20}}S$: ${\frac {dS}{1.9-{\frac {1}{20}}S}}=dt$ Let's now integrate both sides: $\int {\frac {dS}{1.9-{\frac {1}{20}}S}}=\int dt$ The integral on the right hand side is simply $\displaystyle t+C_{1}.$ To solve the left hand side, use the substitution $u=1.9-{\frac {1}{20}}S,\ dS=-20\,du$ and obtain $\int {\frac {dS}{1.9-{\frac {1}{20}}S}}=\int {\frac {-20\,du}{u}}=-20\ln |u|=-20\ln \left|1.9-{\frac {S}{20}}\right|=t+C_{1}.$ Solving the above for S yields ${\begin{aligned}-20\ln \left|1.9-{\frac {S}{20}}\right|&=t+C_{1}\\\ln \left|1.9-{\frac {S}{20}}\right|&=-{\frac {t}{20}}-{\frac {C_{1}}{20}}\\\left|1.9-{\frac {S}{20}}\right|&=e^{-{\frac {t}{20}}-{\frac {C_{1}}{20}}}\\1.9-{\frac {S}{20}}&=\pm e^{-{\frac {C_{1}}{20}}}e^{-{\frac {t}{20}}}\\S&=20\left(1.9-C_{2}e^{-t/20}\right),\end{aligned}}$ with a different constant $C_{2}=\pm e^{-C_{1}/20}$. Now we use the initial condition S(0) = 0 to solve for C2: $0=20\left(1.9-C_{2}e^{0}\right)=20(1.9-C_{2}),$ hence C2 = 1.9 and therefore the final answer is $S(t)=20\left(1.9-1.9e^{-t/20}\right).$

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