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JH
Numerade Educator

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Problem 72 Hard Difficulty

After injection of a does $ D $ of insulin, the concentration of insulin in patient's system decays exponentially and so it can be written as $ De^{-at}, $ where $ t $ represents time in hours and $ a $ is a positive constant.
(a) If a dose $ D $ is injected every $ T $ hours, write an expression for the sum of the residual concentrations just before the $ (n + 1) $st injection.
(b) Determine the limiting pre-injection concentration.
(c) If the concentration of insulin must always remain at or above a critical value $ C, $ determine a minimal dosage $ D $ in terms of $ C, a, $ and $ T. $

Answer

(A). $D e^{-aT} \frac{\left(1-e^{-a n T)}\right.}{1-e^{-a-T}}$
(B). $\frac{D }{e^{a T}-1}$
(C). $D=C\left(e^{a \top}-1\right)$

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Video Transcript

so for party here, let's go ahead and answer this. Notice that just before injection number two, the residual concentration is given by T E to the minus eighty. This is given, and this is just by the fact that were given an injection every t ours do it to the fact that injection is given, given type of there every t ours now before the third, immediately before the third. So let's just say just before injection number three, the residual this time is T E to the minus two eighty. So following this pattern, we see that residual just before the end plus first injection. It's just the minus a n t. Now let's go to the next page and add these up some of residuals, residual concentrations. So we're still in party here Now. This is just and these are look, and it turns out here we could see that this is a geometric series and it's fine. It's a fine, a geometric Siri's. Now we have a formula for this. We could see that the first term a one is C E to the minus eighty. We also see that ours just e to the minus eight see the common ratio there. So we're using. This is a fine eight. Let me spell it out over here. Fine. A geometric Siri's, which has a different form. Your slightly different formula than the infinite series one minus E minus eighty. And that's our answer for party, eh? Let's move on to Part B. So for Part B, we just take the limit from party, eh? Now, let's go ahead and do some algebra here. So go ahead and divide top and bottom by the same quantity. So also on the denominator that's eight times t down there over E to the minus eighty. Now go ahead and clean that up after. Let's do that division D one minus one over E to the anti over each of the eighty minus one. This's just have to do it. Some self replication not going to take that limit because there's n goes to infinity. The circle part goes to identity so that the fraction goes zero and go ahead and just simplify this and we're basically done with party here. This is our answer for party. Let's go on to party. So here we need cedes to satisfy this So this is due to the fact that the concentration must remain above or at the critical level C. So this is why we have this inequality over here going and saw that for D. So the minimal dosage me is just He will see either the eighty minus one. And that's our final answer for party, and that completes the problem.