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

MATH103 April 2015

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Question 02 (e)
The concentration of a certain hormone in the blood changes periodically over 24h. The graph of the production rate p(t) and removal rate r(t) of the hormone are shown below. graph In the figure, clearly mark the times (i) $t_{1}$ at which the total concentration of the hormone is highest, and (ii) $t_{2}$ at which the hormone concentration is increasing at the fastest rate.

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Hint
Find the relation between the total concentration $c(t)$ of the hormone in $t$ hours and the two given functions $p$ and $r$.

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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. 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.