Science:Math Exam Resources/Courses/MATH103/April 2015/Question 02 (e)/Solution 1

Let $c(t)$ be the total concentration of the hormone in $t$ hours.

Since $p(t)$ and $r(t)$ represent the production and the removal rate of the concentration of the hormone, respectively, the rate of change for the total concentration of the hormone can be written as

$c'(t)=p(t)-r(t).$

This implies that $c(t)=\int _{0}^{t}p(s)-r(s)ds+c(0)$.

Using this, we can find the time $t_{1}$ at which $c(t)$ achieves its maximum as in the graph shown below.

graphs

The reason is as follows. In the region $A$ and $C$, we have $c'(t)=p(t)-r(t)<0$, while in the region $B$, we obtain $c'(t)=p(t)-r(t)>0$. Therefore, $c$ is decreasing at the beginning, increasing until $t=t_{1}$, then decreasing after that. Therefore either at $t=0$ or at $t=t_{1}$ the maximum of $c$ is achieved. But if we compare the area of the region $a$ and $b$, obviously the area of $b$ is greater than that of $A$. Therefore, we have

$c(t_{1})=\int _{0}^{t_{1}}p(s)-r(s)ds+c(0)=-{\text{Area }}a+{\text{Area }}b+c(0)>c(0)$.

Now, we claim that $t_{2}$ is as in the graph. As we explained above, only in the interval $B$, the concentration is increasing. Since $c'(t)=p(t)-r(t)$ is the greatest at $t_{2}$, at $t_{2}$ we have the fastest increasing rate.