Science:Math Exam Resources/Courses/MATH101/April 2009/Question 08 (a)
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Question 08 (a)
Consider the chemical reaction
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
where k is a positive constant whole value can be determined by experiment.
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!
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.
Cross multiplying our equation, we have that
Now we integrate. The right hand side integrates to
As for the left hand side, we integrate by partial fractions.
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
when , we have that
and so . When , we have
and so . Hence, we have
(technically the above should add a constant but we can include this in a minute.) Hence
Using the initial condition of , we see that our constant is
and simplifying, this becomes
Our equation simplifies to
Exponentiating the far left and the far right hand sides of the above gives
Taking the tenth root of both sides yields
Removing the absolute values, we have
Plugging in the initial condition of , we see that only the positive value above works. Hence
Now, we cross multiply to see that
Bringing the values to one side gives
Lastly, isolating for gives
completing the question.