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

MATH102 December 2014

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Question C 02 (d)

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 ${\frac {dc}{dt}}=P(c)-D(c).$ Suppose that $P_{m}$ and $k$ are as given above. For what value of $r$ does the differential equation for $c(t)$ have only two steady states?

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
How to solve the steady state? When does a quadratic equation have only one solution?

Hint 2
In part a), we determine the value of the steady state by setting $c(rc^{2}-P_{m}c+k^{2}r)$ equals 0.

Hint 3
The quadratic equation $ax^{2}+bx+c$ has one solution when $b^{2}-4ac=0$

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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. Same as in C2(a), to solve the steady states, we just need to solve the following equation $c(rc^{2}-P_{m}c+k^{2}r)=0,$ Obviously $c=0$ is one of the steady states and other steady states are roots of the quadratic equation $rc^{2}-P_{m}c+k^{2}r=0.$ To get only two steady state, from the quadratic formula we need to have $b^{2}-4ac=0$ and in this case $(P_{m})^{2}-4\cdot r\cdot k^{2}r=0.$ Thus with the given values of $P_{m}$ and $k$ we have $r^{2}={\frac {P_{m}^{2}}{4k^{2}}}={\frac {1}{16}},$ To obtain positive steady state $c$, we get $r={\frac {1}{4}}.$

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