Drug Absorption: Proportional to Amount Present at Time t

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Homework Help Overview

The discussion revolves around a problem in pharmacokinetics, specifically focusing on the absorption rate of a drug in relation to its concentration over time. The original poster presents a scenario where a drug is administered, and the amount remaining after a certain period is given, prompting the need to derive a function representing the amount of drug present at any time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the amount of drug absorbed and the amount remaining, questioning how to correctly formulate the differential equation. There is discussion about whether to express the rate of absorption in terms of the amount remaining or the amount absorbed.

Discussion Status

The discussion is active, with participants providing insights into the formulation of the differential equation and clarifying the relationships between the variables involved. Some participants suggest that the initial conditions and constants can be determined from the problem's constraints, while others seek clarification on the mathematical expressions used.

Contextual Notes

Participants note that the problem involves specific initial conditions (50 mL administered and 50% remaining after 4 hours) that are crucial for determining the constants in the differential equation. There is also mention of the need to understand the implications of the proportionality statement in the context of the problem.

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Homework Statement


The rate at which a particular drug is absorbed by the body is proportional to the amount of the drug present (D) at any time t. If 50mL is initially administered to a patient and 50% remains after 4 hours, find

A) D as a function of t.

The Attempt at a Solution



I'm confused as we have to make D=the amount present (not yet absorbed)

This means that it would be:

d(50-D)/dt \alpha D

Is this correct, in that the rate of absorption is Amount absorbed/time, so 50mL-D = amount absorbed?

By doing this, I integrate and just get Ae^kt = D
 
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Shaybay92 said:
I'm confused as we have to make D=the amount present (not yet absorbed)

This means that it would be:

d(50-D)/dt \alpha D

This is correct. I'd only write dD/dt though.


Shaybay92 said:
By doing this, I integrate and just get Ae^kt = D

Correct as well, now you just have two constants and two constraints on them: D(0) and D(4 hours).
 
But the rate of absorption is the amount absorbed over time? D represents the amount left, so it would have to be d(50-D)/dt? Because 50-amount left = amount absorbed
 
That's right, but you can certainly show that d(50-D)/dt = -dD/dt. What was your question again? :p
 
How does d(50-D)/dt turn into -dD/dt?
 
Differentiation is linear, so that d(50-D)/dt = d 50 /dt - dD/dt. Differentiation of a constant gives zero.
 
Thankyou!
 
i don't think you need the 50 at all to formulate the differential equation

the statement
The rate at which a particular drug is absorbed by the body is proportional to the amount of the drug present (D)

is enough to give you the differential equation
\frac{dD(t)}{dt} = -kD(t)
for some as yet undetermined constant k

integrate, then use the given values to find the constants
assume an initial value
D(0) = d
value at time = 4hrs
D(4) = d/2
 

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