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Differentials

ayahouyee

New member
Nov 4, 2013
12
Suppose the concentration of a drug in the blood stream exponentially decays and is
given by c(t) = a0e^-kt (0 is a subscript) where t is the time elapsed and a0 is the initial concentration after one dose. Further assume that doses of the drug are administered at time intervals of T.

(a) After the first dosage the concentration of the drug is a0. Assuming each dosage
is also going to be a0 of the drug, what is the concentration immediately after the second dosage? After the third dosage? After n dosages? What does the concentration approach as n approaches infinity (this is the equilibrium value)?
You may use the sum 1 + r + r^2 + ... + r^(n-1) = (1-r^n)/(1-r)

(b) Let a1 denote the equilibrium value found in (a). Now suppose that the first dosage
is a1 of the drug and the following doses continue to be a0 of the drug. Following
this model what is the maximum concentration of the drug in the system? Show
why. (hint: consider the concentration after the subsequent dosages)

Thanks again in advance! :D
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,506
Re: Help with differentials question please?! :)

Suppose the concentration of a drug in the blood stream exponentially decays and is
given by c(t) = a0e^-kt (0 is a subscript) where t is the time elapsed and a0 is the initial concentration after one dose. Further assume that doses of the drug are administered at time intervals of T.

(a) After the first dosage the concentration of the drug is a0. Assuming each dosage
is also going to be a0 of the drug, what is the concentration immediately after the second dosage?
Surely you can answer that.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Suppose the concentration of a drug in the blood stream exponentially decays and is
given by c(t) = a0e^-kt (0 is a subscript) where t is the time elapsed and a0 is the initial concentration after one dose. Further assume that doses of the drug are administered at time intervals of T.

(a) After the first dosage the concentration of the drug is a0. Assuming each dosage
is also going to be a0 of the drug, what is the concentration immediately after the second dosage? After the third dosage? After n dosages? What does the concentration approach as n approaches infinity (this is the equilibrium value)?
You may use the sum 1 + r + r^2 + ... + r^(n-1) = (1-r^n)/(1-r)

(b) Let a1 denote the equilibrium value found in (a). Now suppose that the first dosage
is a1 of the drug and the following doses continue to be a0 of the drug. Following
this model what is the maximum concentration of the drug in the system? Show
why. (hint: consider the concentration after the subsequent dosages)

Thanks again in advance! :D
I should have mentioned to you earlier that when I and others here bring questions from other sites, we give full solutions in order to increase our knowledge base of worked problems. For our members posting questions, we ask that work be given so our helpers can see where you are stuck and how best to help. I apologize for not having made this clear.

So, if you can show what you have tried, we will be happy to offer guidance based on that. :D
 

ayahouyee

New member
Nov 4, 2013
12
can you please give me a hint on how to start because honestly i dont know where to start? :((
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Let's look at the second dose. What is the concentration immediately before the second dose? And then immediately after?
 

ayahouyee

New member
Nov 4, 2013
12
before a0e^-k

after a0e^-k2

is that right?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
before a0e^-k

after a0e^-k2

is that right?
No, how much time has elapsed in between the first and second doses? Each time a dose is administered, how is the concentration affected?
 

Nabouabo

New member
Nov 5, 2013
2
So I worked through this and got (a0/1-e^-k) being the limit as n approached infinity, however I am unsure of how to proceed with part b.

I got an equation for the concentration at point n being: a0(1+e^-k...e^(n-2)k + (e^(n-1)k/1-e^-k)), but I don't know what this is supposed to to tell me.
 

Nabouabo

New member
Nov 5, 2013
2
So I worked through this and got (a0/1-e^-k) being the limit as n approached infinity, however I am unsure of how to proceed with part b.

I got an equation for the concentration at point n being: a0(1+e^-k...e^(n-2)k + (e^(n-1)k/1-e^-k)), but I don't know what this is supposed to to tell me.
Nevermind, I didn't think to do literally what I did in part a...
I got the answer.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Since there are two people working this problem, I am going to go ahead and offer my thoughts on how I feel this problem should be worked. Since the doses occur at time intervals of $T$, we have a concentration immediately before the second dose of:

\(\displaystyle \lim_{t\to T^{-}}c(t)=a_0e^{-kT}\)

We assume immediate absorption of the drug, so immediately after the second dose, we have a concentration of:

\(\displaystyle c(T)=a_0+a_0e^{-kT}=a_0\sum_{j=0}^1e^{j(-kT)}\)

And thus, after the third dose, we find:

\(\displaystyle c(2T)=a_0\sum_{j=0}^2e^{j(-kT)}\)

And after the $n$th dose:

\(\displaystyle c\left((n-1)T \right)=a_0\sum_{j=0}^{n-1}e^{j(-kT)}=a_0\frac{1-e^{n(-kT)}}{1-e^{-kT}}\)

We find the equilibrium value:

\(\displaystyle \lim_{n\to\infty}c\left((n-1)T \right)=\frac{a_0}{1-e^{-kT}}\)

This should provide enough to tackle part b).