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

MATH102 December 2014

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Question C 02 (c)
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 ${\frac {dc}{dt}}=P(c)-D(c).$ If the concentration is initially $c(0)=8$, what concentration does $c(t)$ eventually approach?

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Hint
Is the derivative $dc/dt$ positive or negative at $c(0)=8$? Will the function increase or decrease after $t=0$?

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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. Observe that $P(c(0))-D(c(0))=P(8)-D(8)>0,$according to the graph. Therefore, ${\frac {dc}{dt}}(0)=P(c(0))-D(c(0))$ is positive initially, so that $c$ would increase with time. On the other hand, the value $c(0)=8$ falls in the interval $(5,20)$, i.e., $5<8<20$. Remember that when the graphs of $P(c)$ and $D(c)$ intersect, we have ${\frac {dc}{dt}}=P(c)-D(c)=0$. In other words, the steady states $c$ are obtained at the intersection points. In the graph, when we start from $c(0)=8$ and move to the right (because it increases with time), the first steady state that we meet is $c=20$. Therefore, $c$ approaches the steady state $c=20$. Answer: $\color {blue}c=20$

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