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Integration Question

by Shaybay92
Tags: absorption, integration, rate
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Shaybay92
#1
Nov9-09, 12:24 AM
P: 122
1. The problem statement, all variables and given/known data
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.

3. 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 [tex]\alpha[/tex] 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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clamtrox
#2
Nov9-09, 12:39 AM
P: 939
Quote Quote by Shaybay92 View Post
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 [tex]\alpha[/tex] D
This is correct. I'd only write dD/dt though.


Quote Quote by Shaybay92 View Post
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).
Shaybay92
#3
Nov9-09, 12:40 AM
P: 122
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

clamtrox
#4
Nov9-09, 01:09 AM
P: 939
Integration Question

That's right, but you can certainly show that d(50-D)/dt = -dD/dt. What was your question again? :p
Shaybay92
#5
Nov9-09, 01:11 AM
P: 122
How does d(50-D)/dt turn into -dD/dt???
clamtrox
#6
Nov9-09, 01:18 AM
P: 939
Differentiation is linear, so that d(50-D)/dt = d 50 /dt - dD/dt. Differentiation of a constant gives zero.
Shaybay92
#7
Nov9-09, 01:28 AM
P: 122
Thankyou!
lanedance
#8
Nov9-09, 01:34 AM
HW Helper
P: 3,307
i don't think you need the 50 at all to formulate the differntial 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
[tex]\frac{dD(t)}{dt} = -kD(t) [/tex]
for some as yet undetermined constant k

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


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