Science:Math Exam Resources/Courses/MATH101/April 2013/Question 08

MATH101 April 2013

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Question 08

Full-Solution Problems. In questions 4–12, justify your answers and show all your work. If a box is provided, write your final answer there. Unless otherwise indicated, simplification of numerical answers is required in these questions. A tank contains 1000 L (litres) of pure water. A solution that contains 0.01 kg/L of sugar is poured into the tank at a rate of 20 L/min. The tank's contents are throughly mixed and drain out of the tank at the same rate. How much sugar is in the tank after one hour?

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
Set up what you know and what you want in variables. What is the relevant equation which relates the variables and quantities?

Hint 2
Don't forget the absolute value in $\ln |x|$. How does this affect isolating for your desired variable?

Hint 3
What are the initial conditions, and how does that affect your solution?

Hint 4
Be careful about your units!

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. Let $\displaystyle y(t)$ be the mass of sugar in the tank at time $\displaystyle t$ (in minutes), with units of kg. The rate of change of $\displaystyle y(t)$ is the difference between the incoming rate of salt and the outgoing rate of salt. The rate is simply the concentration multiplied by the flow rate; for incoming, we know the concentration is 0.01 kg/L and the flow rate is 20 L/min; for the outgoing, the concentration is $\displaystyle y/1000$ and the flow rate is still 20 L/min. $\displaystyle {\begin{aligned}{\frac {dy}{dt}}&=({\text{rate in}})-({\text{rate out}})\\&=0.01\cdot 20-{\frac {y}{1000}}\cdot 20\\&={\frac {10-y}{50}}.\end{aligned}}$ If $\displaystyle y\neq 10$, then $\displaystyle {\begin{aligned}{\frac {1}{10-y}}{\frac {dy}{dt}}&={\frac {1}{50}}\\\int {\frac {1}{10-y}}{\frac {dy}{dt}}\ dt&=\int {\frac {1}{50}}\ dt\\-\ln |10-y|&={\frac {1}{50}}t+C\end{aligned}}$ We have an initial condition of $y(0)=0$, so $\displaystyle -\ln 10=C.$ (Incidentally, this initial condition implies that $\displaystyle y\neq 10$, since $\displaystyle y=10$ is a steady-state solution, where the mass never changes from 10.) Now we solve $\displaystyle {\begin{aligned}-\ln |10-y|&={\frac {1}{50}}t-\ln 10\\|10-y|&=10e^{-t/50}\end{aligned}}$ Since $\displaystyle y(t)$ is continuous, $\displaystyle 10-y(0)>0$, and the right-hand side of the equation above is always positive, we conclude $\displaystyle 10-y(t)>0$ for all $\displaystyle t$. Then $\displaystyle {\begin{aligned}10-y&=10e^{-t/50}\\y&=10-10e^{-t/50}\end{aligned}}$ The amount of sugar in the tank after 1 hour, or 60 minutes, is $\displaystyle y(60)=10-10e^{-6/5}$

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