Drug Concentration After Multiple Dosages

Click For Summary
SUMMARY

The concentration of a drug in the bloodstream, modeled by the equation c(t) = a0e^-kt, exhibits exponential decay over time. After multiple dosages administered at intervals of T, the concentration immediately after the nth dosage can be expressed as c((n-1)T) = a0(1 - e^(-n*k*T))/(1 - e^(-k*T)). As n approaches infinity, the equilibrium concentration is a1 = a0/(1 - e^(-kT)). If the first dosage is set to this equilibrium value, the maximum concentration in the system can be derived from subsequent doses, confirming the importance of understanding dosage timing and decay rates.

PREREQUISITES
  • Understanding of exponential decay functions
  • Familiarity with dosage interval calculations
  • Knowledge of limits and series summation
  • Basic pharmacokinetics principles
NEXT STEPS
  • Study the pharmacokinetics of drug absorption and elimination
  • Learn about the application of differential equations in pharmacology
  • Explore the concept of steady-state concentration in drug dosing
  • Investigate the effects of varying dosage intervals on drug concentration
USEFUL FOR

Pharmacologists, medical professionals, students in pharmacokinetics, and anyone involved in drug dosage planning and analysis.

ayahouyee
Messages
12
Reaction score
0
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
 
Physics news on Phys.org
Re: Help with differentials question please?! :)

ayahouyee said:
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.
 
ayahouyee said:
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
 
can you please give me a hint on how to start because honestly i don't know where to start? :((
 
Let's look at the second dose. What is the concentration immediately before the second dose? And then immediately after?
 
before a0e^-k

after a0e^-k2

is that right?
 
ayahouyee said:
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?
 
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 said:
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.
 
  • #10
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:

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

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

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

And thus, after the third dose, we find:

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

And after the $n$th dose:

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

We find the equilibrium value:

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

This should provide enough to tackle part b).
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 131 ·
5
Replies
131
Views
10K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 2 ·
Replies
2
Views
9K