Question 1179182: A drug is administered to a patient every 4 hours. The patient's metabolism cause the concentration of the drug in the patient's blood to decrease by 75% between doses. The dose causes an instantaneous rise in blood concentration of 0.2 units. Then what will be the concentration of drug in patient's blood immediately after the nth dose
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's analyze the drug concentration over time:
* **Initial Dose:** After the first dose, the concentration is 0.2 units.
* **Time Between Doses:** The drug concentration decreases by 75%, meaning 25% remains. This is equivalent to multiplying the concentration by 0.25.
* **Instantaneous Rise:** Each dose causes an instantaneous rise of 0.2 units.
Here's how the concentration evolves:
* **After 1st dose:** 0.2
* **Before 2nd dose:** 0.2 * 0.25
* **After 2nd dose:** (0.2 * 0.25) + 0.2
* **Before 3rd dose:** ((0.2 * 0.25) + 0.2) * 0.25
* **After 3rd dose:** ((0.2 * 0.25) + 0.2) * 0.25 + 0.2
* **After nth dose:** 0.2 + 0.2(0.25) + 0.2(0.25)^2 + ... + 0.2(0.25)^(n-1)
This is a geometric series with:
* First term (a): 0.2
* Common ratio (r): 0.25
* Number of terms (n): n
The sum of a geometric series is given by:
* S_n = a * (1 - r^n) / (1 - r)
Applying this to our problem:
* Concentration after nth dose = 0.2 * (1 - 0.25^n) / (1 - 0.25)
* Concentration after nth dose = 0.2 * (1 - 0.25^n) / 0.75
* Concentration after nth dose = (0.2 / 0.75) * (1 - 0.25^n)
* Concentration after nth dose = (2/7.5) * (1 - 0.25^n)
* Concentration after nth dose = (4/15) * (1 - 0.25^n)
Therefore, the concentration of the drug in the patient's blood immediately after the nth dose is:
* (4/15) * (1 - (1/4)^n)
This formula allows you to calculate the drug concentration after any number of doses.
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