Science:Math Exam Resources/Courses/MATH102/December 2014/Question C 02 (c)/Solution 1

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: $\color {blue}c=20$