Science:Math Exam Resources/Courses/MATH101/April 2009/Question 08 (a)

MATH101 April 2009

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q2  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q8 (a)  •  Q8 (b)  •  

   Other MATH101 Exams

  •  April 2011  •  April 2010  •  April 2009  •  April 2008  •  April 2007  •  April 2005  •  April 2006  •  April 2012  •  April 2013  •  April 2014  •  April 2015  •  April 2016  •  April 2018  •  April 2017  •  

[hide] Question 08 (a)

Consider the chemical reaction $\displaystyle A+B\rightarrow C+D.$ Suppose at time t = 0 sec the concentration of chemical A is 0.1 mol/L, the concentration of chemical B is 0.2 mol/L, and the concentrations of chemicals C and D are both 0. For t ≥ 0, let x(t) be the concentration of chemical D in mol/L. It can be shown that x(t) is the solution to the initial-value problem ${\frac {dx}{dt}}=k(0.1-x)(0.2-x),\quad x(0)=0,$ where k is a positive constant whole value can be determined by experiment. (a) Solve the initial-value problem to find x(t) explicitly.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint
Start off by cross multiplying. Then integrate (you may want to use partial fractions to integrate). I hope you ate your Wheaties because this is a long one...

Checking a solution serves two purposes: helping you if, after having used the hint, 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.

[show] 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. Cross multiplying our equation, we have that ${\frac {1}{(0.1-x)(0.2-x)}}\,dx=k\,dt$ Now we integrate. The right hand side integrates to $\int k\,dt=kt$ As for the left hand side, we integrate by partial fractions. ${\begin{aligned}{\frac {1}{(0.1-x)(0.2-x)}}&={\frac {A}{(0.1-x)}}+{\frac {B}{(0.2-x)}}\\&={\frac {A(0.2-x)+B(0.1-x)}{(0.1-x)(0.2-x)}}\\&={\frac {(-A-B)x+A(0.2)+B(0.1)}{(0.1-x)(0.2-x)}}\end{aligned}}$ Now we can either compare coefficients or plug in values for x. I will choose to plug in values for x. The above tells us that $\displaystyle 1=A(0.2-x)+B(0.1-x)$ when $x=0.2$, we have that $\displaystyle 1=A(0.2-0.2)+B(0.1-0.2)=B(-0.1)$ and so $B=-10$. When $x=0.1$, we have $\displaystyle 1=A(0.2-0.1)+B(0.1-0.1)=A(0.1)$ and so $A=10$. Hence, we have ${\begin{aligned}\int {\frac {1}{(0.1-x)(0.2-x)}}\,dx&=\int {\frac {10}{(0.1-x)}}\,dx-\int {\frac {10}{(0.2-x)}}\,dx\\&=-10\ln |0.1-x|+10\ln |0.2-x|.\end{aligned}}$ (technically the above should add a constant but we can include this in a minute.) Hence $\int {\frac {1}{(0.1-x)(0.2-x)}}\,dx=\int k\,dt$ becomes $\displaystyle 10\ln |0.2-x|-10\ln |0.1-x|=kt+C$ Using the initial condition of $x(0)=0$, we see that our constant is $\displaystyle 10\ln |0.2|-10\ln |0.1|=C$ and simplifying, this becomes $\displaystyle C=10(\ln {\tfrac {0.2}{0.1}})=10\ln(2)$ Our equation simplifies to $\displaystyle 10\ln \left|{\frac {0.2-x}{0.1-x}}\right|=10\ln |0.2-x|-10\ln |0.1-x|=kt+10\ln(2)$ Exponentiating the far left and the far right hand sides of the above gives $\left|{\frac {0.2-x}{0.1-x}}\right|^{10}=2^{10}e^{kt}$ Taking the tenth root of both sides yields $\left|{\frac {0.2-x}{0.1-x}}\right|=2e^{\tfrac {kt}{10}}$ Removing the absolute values, we have ${\frac {0.2-x}{0.1-x}}=\pm 2e^{\tfrac {kt}{10}}$ Plugging in the initial condition of $x(0)=0$, we see that only the positive value above works. Hence ${\frac {0.2-x}{0.1-x}}=2e^{\tfrac {kt}{10}}$ Now, we cross multiply to see that $\displaystyle 0.2-x=2e^{\tfrac {kt}{10}}(0.1-x)=0.2e^{\tfrac {kt}{10}}-x(2e^{\tfrac {kt}{10}})$ Bringing the $x$ values to one side gives $\displaystyle x(2e^{\tfrac {kt}{10}})-x=0.2e^{\tfrac {kt}{10}}-0.2$ Lastly, isolating for $x$ gives $\displaystyle x={\frac {0.2e^{\tfrac {kt}{10}}-0.2}{2e^{\tfrac {kt}{10}}-1}}$ completing the question.

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Initial value problem, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag