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

MATH102 December 2014

• QA 1 • QA 2 • QA 3 • QA 4 • QA 5 • QA 6 • QA 7 • QA 8 • QB 1 • QB 2 • QB 3 • QB 4 • QB 5 • QB 6 • QB 7 • QC 1 • QC 2(a) • QC 2(b) • QC 2(c) • QC 2(d) • QC 3 •

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[hide] Question C 02 (b)
The graphs to the right show two functions, Graphs $P(c)=P_{m}{\frac {c^{2}}{k^{2}+c^{2}}},\quad D(c)=rc$ 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 ${\frac {dc}{dt}}=P(c)-D(c).$ Determine the stability of each steady state. Explain how you arrived at your conclusions either in words or using a diagram.

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[show] Hint
A steady state is stable if states that are initially close enough to that steady state will get closer to it with time. A steady state is unstable, if states that are initially very close to it eventually move away from that steady state. For example, if we start above(below) a steady state, and the derivative $dc/dt$ is positive(negative), then the state will increase(decrease) and eventually move away from the steady state so it is an unstable steady state.

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[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. To determine the stability of each steady state, we just need to find the increasing and decreasing regions. As $c$ represents the concentration of a substance, we only need to consider the positive half of the real axis. The three steady states divide it into three intervals ${\begin{aligned}c\in (0,5),\quad P(c)-D(c)<0\quad {\text{decreasing}},\\c\in (5,20),\quad P(c)-D(c)>0\quad {\text{increasing}},\\c\in (20,\infty ),\quad P(c)-D(c)<0\quad {\text{decreasing}},\end{aligned}}$ If we choose a initial condition in the interval $c_{0}\in (0,5)$ , as $P(c)-D(c)$ is negative, $c$ would decrease and approach the steady state $c=0$ as time goes on. So $c=0$ is stable. Similarly, we find $c=5$ is unstable and $c=20$ is stable.