Science:Math Exam Resources/Courses/MATH102/December 2014/Question C 02 (a)

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MATH102 December 2014

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Question C 02 (a)
Graphs The graphs to the right show two functions, $P(c)=P_{m}{\frac {c^{2}}{k^{2}+c^{2}}},\quad D(c)=rc$ where $P_{m}=5$, $k=10$ and $r=1/5$. Suppose that $c(t)$ is the concentration of a substance involved in a chemical reaction and satisfies the equation ${\frac {dc}{dt}}=P(c)-D(c).$ Calculate the steady states of the differential equation (show your work).

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!

Hint
A Steady state is a state in which a system is not changing. So it is calculated by setting ${\frac {dc}{dt}}=0$ and find the solution $c$.

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Solution

By the definition of steady state, we look for the value of $c$ such that ${\frac {dc}{dt}}=0,\quad {\text{or }}P(c)-D(c)=0.$ Thus we need to solve ${\begin{aligned}P_{m}{\frac {c^{2}}{k^{2}+c^{2}}}&=rc\\P_{m}\cdot c^{2}&=rc(k^{2}+c^{2})\\P_{m}\cdot c^{2}&=k^{2}rc+rc^{3}\\P_{m}\cdot c^{2}-k^{2}rc-rc^{3}&=0\\c(P_{m}c-k^{2}r-rc^{2})&=0.\end{aligned}}$ Substitute values of $P_{m}=5$, $k=10$ and $r=1/5$ and simplify the expression, we obtain ${\begin{aligned}c(P_{m}c-k^{2}r-rc^{2})&=0\\c(5c-20-c^{2}/5)&=0\\{\frac {-c}{5}}(-25c+100+c^{2})&=0\\{\frac {1}{5}}c(c-5)(c-20)=0\end{aligned}}$ which gives $c=0,\ 5,\ 20$ are the steady states.

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