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The following message was posted to: PharmPK
Hello all
I'm stuck... I set a exercise to derive the equation for X(or Cp)
versus time using a general infusion equation and ended up not being
able to complete the Laplace based derivation myself ;-(
From http://www.boomer.org/c/p4/c02/c07/c0707.html
or Benet's paper reference in the footer
IV infusion, drug amount in the central compartment, one compartment
gives the Laplace
-a x s -z x s
k0 x (e - e ) 1
X(bar) = ------------------------ x ---------
s (s + kel)
Roots = 0, and -kel
Finger print method seems to give
k0 [ z x kel ] -kel x t
X = --- [e - 1] x e
kel [ ]
which doesn't seem to be the same as the expected
k0 -kel x z -kel x t
X = --- [1 - e ] x e
kel
Help!
where did I do wrong? :-)
Thanks, David
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The following message was posted to: PharmPK
David,
Your derivation is correct (assuming a=0) but your expectation is
wrong :)
To see is, set t=z when z is very large. You should get k/kel (as
follows from the steady-state equation and from your Laplace
derivation) but your expectation would give incorrect value (zero).
Leonid
--
Leonid Gibiansky, Ph.D.
President, QuantPharm LLC
web: www.quantpharm.com
e-mail: LGibiansky at quantpharm.com
--
Thanks Leonid
I did assume a = 0
So my derivation
> k0 [ kel x z ] -kel x t
> X = --- [e - 1] x e
> kel [ ]
is correct. If z = t (continuous infusion) this transforms into
k0 [ kel x t ] -kel x t
X = --- [e - 1] x e
kel [ ]
k0 [ (t x kel - t x kel) -kel x t]
or X = --- [e - e ]
kel [ ]
k0 [ -kel x t]
or X = --- [1 - e ]
kel [ ]
And when t > z my derivation is another form of the equation for X
after the infusion has stopped.
k0 [ -kel*z] -kel x (t-z)
X = --- [1 - e ] x e
kel [ ]
(Checked numerically :)
Great, thanks again.
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The following message was posted to: PharmPK
Hello,
you can use the method described here:
http://www.uef.sav.sk/CXT-Main
to complete the Laplace based derivation.
Regards,
Maria Durisova
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