MATH103 April 2015
• Q1 (a) (i) • Q1 (a) (ii) • Q1 (a) (iii) • Q1 (b) (i) • Q1 (b) (ii) • Q1 (b) (iii) • Q1 (c) (i) • Q1 (c) (ii) • Q1 (c) (iii) • Q1 (d) (i) • Q1 (d) (ii) • Q1 (e) (i) • Q1 (e) (ii) • Q1 (e) (iii) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q2 (e) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (b) • Q6 (c) • Q7 (a) • Q7 (b) • Q7 (c) • Q7 (d) • Q8 • Q9 • Q10 (a) • Q10 (b) •
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
In the figure, clearly mark the times
(i) at which the total concentration of the hormone is highest, and
(ii) at which the hormone concentration is increasing at the fastest rate.
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Find the relation between the total concentration of the hormone in hours and the two given functions and .
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Let be the total concentration of the hormone in hours.
Since and 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
This implies that .
Using this, we can find the time at which achieves its maximum as in the graph shown below.
The reason is as follows. In the region and , we have , while in the region , we obtain . Therefore, is decreasing at the beginning, increasing until , then decreasing after that. Therefore either at or at the maximum of is achieved. But if we compare the area of the region and , obviously the area of is greater than that of . Therefore, we have
Now, we claim that is as in the graph. As we explained above, only in the interval , the concentration is increasing. Since is the greatest at , at we have the fastest increasing rate.