My Math Forum Infinite geometric series application

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 May 12th, 2011, 10:55 AM #1 Newbie   Joined: Dec 2010 Posts: 9 Thanks: 1 Infinite geometric series application A doctor prescribes a 240 milligram (mg), pain-reducing drug to a patient who has chronic pain. The medical instructions read that this drug should be taken every 4 hours. After 4 hours, 60% of the original dose leaves the body. Under these conditions, the amount of drug remaining in the body, at 4-hour intervals, forms a geometric series. 1. Supposing that the patient takes just one dose of the medicine write an equation for the amount of the drug in the patient's blood stream t hours after taking the medicine Qe^-(ct) where c is a positive constant? 2. How many mgs of the drug are present in the body after 4 hours? (just after second dose?) 3. Graph the amount of medicine in the blood stream for the first 24 hour period. 4. Show that the amount of medicine in the patient's bloodstream after the Nth dose can be expressed by a geometric series. Use sigma notation to express the series. 5. The minimum lethal dosage of this pain-reducing drug is 600mg. if the patient is continuously (infinitely) taking the drug as prescribed, will the patient ever have this much of the drug in their body? 6. If the patient decides to abuse the drug and take it every 2 hours, against the doctor’s orders, then will the patient meet reach the minimal lethal dosage? Thanks from Elizabeth Thomas Last edited by skipjack; December 12th, 2018 at 08:41 PM.
 May 12th, 2011, 11:05 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Infinite geometric series application 1. Since we are told the concentration of the drug decays exponentially, we know the time rate of change of the amount of the drug in the patient's bloodstream is proportional to the amount A(t). We may assume initially at time t = 0, A(t) = 240 mg. This gives rise to the IVP: $\displaystyle \frac{dA}{dt}=-kA$, (k > 0), $\displaystyle A(0)=A_0$ Separate variables: $\displaystyle \frac{1}{A}\,dA=-k\,dt$ Integrate, using boundaries as limits of integration: $\displaystyle \int_{A_0}^A\frac{1}{A}\,dA=-k\int_0^t\,dt$ $\displaystyle \ln\left(\frac{A}{A_0}\right)=-kt$ Convert from logarithmic to exponential form: $\displaystyle \frac{A}{A_0}=e^{-kt}$ $\displaystyle A=A_0e^{-kt}$ We are told it takes 4 hours for 60% of the dose to leave the body. This means A(4) = 0.4A(0)...this gives: $\displaystyle \ln\left(\frac{\frac{2}{5}A_0}{A_0}\right)=-k(4)$ $\displaystyle \ln\left(\frac{2}{5}\right)=-4k$ $\displaystyle \frac{\ln\left(\frac{2}{5}\right)}{4}=-k$ $\displaystyle -k=\frac{1}{4}\ln\left(\frac{2}{5}\right)=\ln\left( \left(\frac{2}{5}\right)^{\frac{1}{4}}\right)$ Now, substituting for -k into the expression for A gives: $\displaystyle A = A_0e^{\ln\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)t}$ $\displaystyle A = A_0\left(e^{\ln\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)}\right)^t$ $\displaystyle A = A_0\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)^t$ $\displaystyle A = A_0\left(\frac{2}{5}\right)^{\frac{t}{4}}$ So, with $\displaystyle A_0=240\text{ mg}$ we have: $\displaystyle A(t) = 240\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ 2. Just after the second dose, we would then have: $\displaystyle A(t)=\left(240\left(\frac{2}{5}\right)^{\frac{4}{4 }} + 240\right)\text{ mg} = \frac{7}{5}240\text{ mg} = 336\text{ mg}$ 3.) I don't own any graphing software, so we'll have to piece this together. As we found in part 1, the first 4 hours (up until the 2nd dose is taken) is given by: $\displaystyle A(t) = 240\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ Then, for the 2nd four hour period, using the result from part 2, we have: $\displaystyle A(t) = 336\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ For the 3rd four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}336\right) \left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 374.4\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text {mg}$ For the 4th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}374.4\right)\left(\frac{2}{5} \right)^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 389.76\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ For the 5th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}389.76\right)\left(\frac{2}{5 }\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 395.904\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ For the 6th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}395.904\right)\left(\frac{2}{ 5}\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 398.3616\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ Graph these six pieces for each successive 4 hour period. 4. As we saw in part 3, the amount A after the doses is: dose...amount (mg) 1..........240 2..........240 + 0.4�240 = 1.4�240 = 336 3..........240 + 0.4(240 + 0.4�240) = 240(1+0.4+0.4�) = 374.4 We see then that the amount after the nth dose will be given by: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\frac{2}{5}\right)^k$ 5. Using the formula for the first n terms of a geometric series, we get: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\frac{2}{5}\right)^k=240\frac{1-\left(\frac{2}{5}\right)^n}{1-\frac{2}{5}} = 400\left(1-\left(\frac{2}{5}\right)^n\right)$ Thus, we find: $\displaystyle \lim_{n\to\infty}A_n = 400\text{ mg}$ So we see the patient will never reach the lethal dose of 600 mg. 6. To answer this, we need to know what portion of the drug remains in the body after 2 hours. Using the formula from part 1: $\displaystyle A = A_0\left(\frac{2}{5}\right)^{\frac{2}{4}} = A_0\sqrt{\frac{2}{5}}$ So now, we have: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\sqrt{\frac{2}{5}}\right)^k = 240\frac{1-\left(\sqrt{\frac{2}{5}}\right)^n}{1-\sqrt{\frac{2}{5}}} = \left(400+80\sqrt{10}\right)\left(1-\left(\sqrt{\frac{2}{5}}\right)^n\right)$ $\displaystyle \lim_{n\to\infty}A_n = 400+80\sqrt{10}\text{ mg}\approx652.98\text{ mg}$ So we see the patient will eventually reach the lethal dosage. To find after which dosage, we may use: $\displaystyle \left(400+80\sqrt{10}\right)\left(1-\left(\sqrt{\frac{2}{5}}\right)^n\right) = 600$ Now we solve for n: $\displaystyle 1-\left(\sqrt{\frac{2}{5}}\right)^n = \frac{600}{400+80\sqrt{10}}=\frac{5-\sqrt{10}}{2}$ $\displaystyle \left(\frac{2}{5}\right)^{\frac{n}{2}} = 1-\frac{5-\sqrt{10}}{2}=\frac{\sqrt{10}-3}{2}$ Take the natural logarithm of both sides: $\displaystyle \ln\left(\left(\frac{2}{5}\right) ^{\frac{n}{2}}\right) = \ln\left(\frac{\sqrt{10}-3}{2}\right)$ $\displaystyle \frac{n}{2}\ln\left(\frac{2}{5}\right) = \ln\left(\frac{\sqrt{10}-3}{2}\right)$ $\displaystyle n = 2\frac{\ln\left(\frac{\sqrt{10}-3}{2}\right)}{\ln\left(\frac{2}{5}\right)}\approx5 .48$ Thus, the patient will reach a lethal amount after the 6th dose. Last edited by skipjack; December 12th, 2018 at 08:34 PM.
 May 12th, 2011, 04:53 PM #3 Newbie   Joined: Dec 2010 Posts: 9 Thanks: 1 Re: Infinite geometric series application Thanks a ton for the solution! I'll post back if I come up with any questions. Last edited by skipjack; December 12th, 2018 at 08:36 PM.
 May 12th, 2011, 04:57 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Infinite geometric series application You're welcome!
 September 16th, 2015, 10:37 AM #5 Newbie   Joined: Sep 2015 From: malaysia Posts: 1 Thanks: 0 I would like to ask... where did 4c/t come from (the solution of fourth four hour period)? The second question is why I did not get the value of lethal dose of paracetamol when I use the infinite limit geometric series? Last edited by skipjack; December 12th, 2018 at 08:37 PM.
December 11th, 2018, 12:45 AM   #6
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 Originally Posted by MarkFL 1. Since we are told the concentration of the drug decays exponentially, we know the time rate of change of the amount of the drug in the patient's bloodstream is proportional to the amount A(t). We may assume initially at time t = 0, A(t) = 240 mg. This gives rise to the IVP: $\displaystyle \frac{dA}{dt}=-kA$, (k > 0), $\displaystyle A(0)=A_0$ Separate variables: $\displaystyle \frac{1}{A}\,dA=-k\,dt$ Integrate, using boundaries as limits of integration: $\displaystyle \int_{A_0}^A\frac{1}{A}\,dA=-k\int_0^t\,dt$ $\displaystyle \ln\left(\frac{A}{A_0}\right)=-kt$ Convert from logarithmic to exponential form: $\displaystyle \frac{A}{A_0}=e^{-kt}$ $\displaystyle A=A_0e^{-kt}$ We are told it takes 4 hours for 60% of the dose to leave the body. This means A(4) = 0.4A(0)...this gives: $\displaystyle \ln\left(\frac{\frac{2}{5}A_0}{A_0}\right)=-k(4)$ $\displaystyle \ln\left(\frac{2}{5}\right)=-4k$ $\displaystyle \frac{\ln\left(\frac{2}{5}\right)}{4}=-k$ $\displaystyle -k=\frac{1}{4}\ln\left(\frac{2}{5}\right)=\ln\left( \left(\frac{2}{5}\right)^{\frac{1}{4}}\right)$ Now, substituting for -k into the expression for A gives: $\displaystyle A = A_0e^{\ln\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)t}$ $\displaystyle A = A_0\left(e^{\ln\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)}\right)^t$ $\displaystyle A = A_0\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)^t$ $\displaystyle A = A_0\left(\frac{2}{5}\right)^{\frac{t}{4}}$ So, with $\displaystyle A_0=240\text{ mg}$ we have: $\displaystyle A(t) = 240\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ 2. Just after the second dose, we would then have: $\displaystyle A(t)=\left(240\left(\frac{2}{5}\right)^{\frac{4}{4 }} + 240\right)\text{ mg} = \frac{7}{5}240\text{ mg} = 336\text{ mg}$ 3.) I don't own any graphing software, so we'll have to piece this together. As we found in part 1, the first 4 hours (up until the 2nd dose is taken) is given by: $\displaystyle A(t) = 240\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ Then, for the 2nd four hour period, using the result from part 2, we have: $\displaystyle A(t) = 336\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ For the 3rd four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}336\right) \left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 374.4\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text {mg}$ For the 4th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}374.4\right)\left(\frac{2}{5} \right)^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 389.76\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ For the 5th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}389.76\right)\left(\frac{2}{5 }\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 395.904\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ For the 6th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}395.904\right)\left(\frac{2}{ 5}\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 398.3616\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ Graph these six pieces for each successive 4 hour period. 4. As we saw in part 3, the amount A after the doses is: dose...amount (mg) 1..........240 2..........240 + 0.4�240 = 1.4�240 = 336 3..........240 + 0.4(240 + 0.4�240) = 240(1+0.4+0.4�) = 374.4 We see then that the amount after the nth dose will be given by: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\frac{2}{5}\right)^k$ 5. Using the formula for the first n terms of a geometric series, we get: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\frac{2}{5}\right)^k=240\frac{1-\left(\frac{2}{5}\right)^n}{1-\frac{2}{5}} = 400\left(1-\left(\frac{2}{5}\right)^n\right)$ Thus, we find: $\displaystyle \lim_{n\to\infty}A_n = 400\text{ mg}$ So we see the patient will never reach the lethal dose of 600 mg. 6. To answer this, we need to know what portion of the drug remains in the body after 2 hours. Using the formula from part 1: $\displaystyle A = A_0\left(\frac{2}{5}\right)^{\frac{2}{4}} = A_0\sqrt{\frac{2}{5}}$ So now, we have: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\sqrt{\frac{2}{5}}\right)^k = 240\frac{1-\left(\sqrt{\frac{2}{5}}\right)^n}{1-\sqrt{\frac{2}{5}}} = \left(400+80\sqrt{10}\right)\left(1-\left(\sqrt{\frac{2}{5}}\right)^n\right)$ $\displaystyle \lim_{n\to\infty}A_n = 400+80\sqrt{10}\text{ mg}\approx652.98\text{ mg}$ So we see the patient will eventually reach the lethal dosage. To find after which dosage, we may use: $\displaystyle \left(400+80\sqrt{10}\right)\left(1-\left(\sqrt{\frac{2}{5}}\right)^n\right) = 600$ Now we solve for n: $\displaystyle 1-\left(\sqrt{\frac{2}{5}}\right)^n = \frac{600}{400+80\sqrt{10}}=\frac{5-\sqrt{10}}{2}$ $\displaystyle \left(\frac{2}{5}\right)^{\frac{n}{2}} = 1-\frac{5-\sqrt{10}}{2}=\frac{\sqrt{10}-3}{2}$ Take the natural logarithm of both sides: $\displaystyle \ln\left(\left(\frac{2}{5}\right) ^{\frac{n}{2}}\right) = \ln\left(\frac{\sqrt{10}-3}{2}\right)$ $\displaystyle \frac{n}{2}\ln\left(\frac{2}{5}\right) = \ln\left(\frac{\sqrt{10}-3}{2}\right)$ $\displaystyle n = 2\frac{\ln\left(\frac{\sqrt{10}-3}{2}\right)}{\ln\left(\frac{2}{5}\right)}\approx5 .48$ Thus, the patient will reach a lethal amount after the 6th dose.
Was wondering whether you still had the photos of the work you did for this problem. Unfortunately, I am not seeing them and they would help me immensely. Thank you!

Last edited by skipjack; December 12th, 2018 at 08:38 PM.

 December 12th, 2018, 08:42 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,927 Thanks: 2205 Is it displayed properly now?

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