My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree1Thanks
  • 1 Post By blueapples
Reply
 
LinkBack Thread Tools Display Modes
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.
blueapples is offline  
 
May 12th, 2011, 11:05 AM   #2
Senior Member
 
MarkFL's Avatar
 
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.
MarkFL is offline  
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.
blueapples is offline  
May 12th, 2011, 04:57 PM   #4
Senior Member
 
MarkFL's Avatar
 
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!
MarkFL is offline  
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.
eunina is offline  
December 11th, 2018, 12:45 AM   #6
Newbie
 
Joined: Dec 2018
From: United States

Posts: 1
Thanks: 0

Quote:
Originally Posted by MarkFL View Post
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.
chrisbeezyy is offline  
December 12th, 2018, 08:42 PM   #7
Global Moderator
 
Joined: Dec 2006

Posts: 20,652
Thanks: 2085

Is it displayed properly now?
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
application, geometric, infinite, series



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Fermat Hyperbolic Geometric Infinite Sum petroljose Calculus 3 August 8th, 2013 09:06 AM
sum of infinite geometric series commonteal Calculus 2 January 18th, 2013 02:21 PM
Infinite geometric series fe phi fo Algebra 5 January 6th, 2012 04:17 PM
Convergent series -> series of geometric means converges The Chaz Real Analysis 11 February 7th, 2011 04:52 AM
Infinite geometric series fe phi fo Advanced Statistics 1 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.