1) The document discusses the determination of the time to reach maximum drug concentration (tmax) after oral administration of a drug. 2) It presents equations that model the absorption and elimination rates of drugs in the body over time and how these rates impact the drug concentration in plasma. 3) It derives an equation that shows tmax is independent of dose and depends only on the absorption and elimination rate constants (ka and k). The tmax occurs when the rates of drug absorption and elimination are equal, making the rate of concentration change equal zero.

1. Determination
of 𝒕 𝒎𝒂𝒙
Prepared by:
Imran Ahammad chowdhury

2. The rate of change of the amount of drug in the body after oral
administration is a fraction of both the absorption rate 𝑑AA
𝑑𝑡
and
elimination rate
𝑑AE
𝑑𝑡
.
That is-

𝑑𝐴
𝑑𝑡
=
𝑑𝐴 𝐴
𝑑𝑡
−
𝑑𝐴 𝐸
𝑑𝑡
or,
𝑑𝐶𝑝
𝑑𝑡
=
1
𝑉𝐷
𝑑𝐴 𝐴
𝑑𝑡
−
𝑑𝐴 𝐸
𝑑𝑡
…………..1
Here, 𝑉𝐷 =
𝐴
𝐶 𝑝
∴ 𝐴 = 𝑉𝐷 𝐶 𝑝
A=Amount of drug in the body

3. Now How much drug is absorbed from the GIT, that means rate
of absorption from the GIT is given by the following expression:

𝑑𝐷 𝐺𝐼
𝑑𝑡
= −𝐾 𝑎 𝐷 𝐺𝐼 =
𝑑𝐴 𝐴
𝑑𝑡
……. 2
So, from 1.

𝑑𝐶𝑝
𝑑𝑡
=
1
𝑉 𝐷
−𝐾 𝑎. 𝐷 𝐺𝐼 −𝐾𝐷 𝐵 … … . . 3
Similarly, rate of elimination
𝑑𝐴 𝐸
𝑑𝑡
= −k𝐷 𝐵

4. From equation 2 in case of multiple oral dosage regimen, we can
write,
𝑑𝐷 𝐺𝐼
𝑑𝑡
=-𝐾 𝑎 𝑛.τ. 𝐷 𝐺𝐼
𝑑𝐷 𝐺𝐼
𝑑𝑡
=-𝐾 𝑎 𝑛.τ.dt
𝑑𝐷 𝐺𝐼
𝑑𝑡
= 𝐾𝑎 𝑛.τ.dt
𝐷 𝐺𝐼 = −𝐷 𝑜 𝑒−𝐾 𝑎 𝑛.τ.𝑡
𝐷 𝐺𝐼 = −𝐹𝐷 𝑜 𝑒−𝐾 𝑎 𝑛.τ.𝑡
if fraction of dose is absorbed.
Where,
n=number of dose
τ =dosing interval

5. So, putting this value in equation 3 we get-

𝑑𝐶𝑝
𝑑𝑡
=
1
𝑉 𝐷
𝐹𝐾 𝑎. 𝐷 𝑜 𝑒−𝑛.𝐾 𝑎.τ.𝑡
− 𝐾𝐷 𝐵 … … … .4
Now, integration this equation we get, the general oral absorption
equation for calculating the drug concentration (𝐶 𝑝) in the
plasma, at any time t,
 𝐶 𝑝=
𝐹𝐾 𝑎 𝐷 𝑜
𝑉 𝐷(𝑘−𝑘 𝑎)
1−𝑒−𝑛.𝐾 𝑎.τ.𝑡
1−𝑒−𝐾 𝑎.τ 𝑒−𝑘 𝑎 𝑡 −
1−𝑒−𝑛𝑘τ
1−𝑒−𝑘τ 𝑒−𝑘𝑡 ….5

6. For a single oral dose, the equation can be written as follows-
 𝐶 𝑝=
𝐹𝐾 𝑎 𝐷 𝑜
𝑉 𝐷(𝑘−𝑘 𝑎)
𝑒−𝑘 𝑎 𝑡 − 𝑒−𝑘𝑡 …. ……6
The time needed to reach maximum concentration is independent
of dose & is dependent on the rate constants for absorption (𝑘 𝑎)
& elimination (k). At the maximum concentration, the rate of
drug absorbed is equal to the rate of drug eliminated. Therefore,
the rate of concentration change is equal to zero.

7. The rate of concentration change can be obtained by differentiating the
equation 6.
𝑑𝐶 𝑝
𝑑𝑡
=
𝐹𝑘 𝑎 𝐷 𝑜
𝑉 𝐷(𝑘−𝑘 𝑎)
−𝑘 𝑎. 𝑒−𝑘 𝑎 𝑡
+ 𝑘. 𝑒−𝑘𝑡
The rate of concentration change can be obtained by differentiating the
equation 6.
𝑑𝐶 𝑝
𝑑𝑡
=
𝐹𝑘 𝑎 𝐷 𝑜
𝑉 𝐷(𝑘−𝑘 𝑎)
−𝑘 𝑎. 𝑒−𝑘 𝑎 𝑡
+ 𝑘. 𝑒−𝑘𝑡
But at 𝐶∞
𝑚𝑎𝑥
𝑑𝐶 𝑝
𝑑𝑡
= 0
0=
𝐹𝑘 𝑎 𝐷 𝑜
𝑉 𝐷(𝑘−𝑘 𝑎)
−𝑘 𝑎. 𝑒−𝑘 𝑎 𝑡 + 𝑘. 𝑒−𝑘𝑡
𝐹𝑘 𝑎 𝐷 𝑜
𝑉𝐷(𝑘 − 𝑘 𝑎)
≠ 0

8. −𝑘 𝑎. 𝑒−𝑘 𝑎 𝑡 + 𝑘. 𝑒−𝑘𝑡 = 0
or, 𝑘 𝑎. 𝑒−𝑘 𝑎 𝑡 = 𝑘. 𝑒−𝑘𝑡
or, ln 𝑘 − 𝑘𝑡 = ln 𝑘 𝑎 − 𝑘 𝑎t
or, 𝑘 𝑎t − 𝑘𝑡 = ln 𝑘 𝑎 − ln 𝑘
or, t(𝑘 𝑎 − 𝑘)= ln 𝑘 𝑎 − ln 𝑘
or, 𝑡 𝑚𝑎𝑥 =
ln 𝑘 𝑎−𝑘
𝑘 𝑎−𝑘
=
2.303 log
𝑘 𝑎
𝑘
𝑘 𝑎−𝑘

9. Thank you