Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully a

Question

Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully absorbed, the quantity of drug in the body decreases at a rate proportional to the amount left in the body, with constant of proportionality k. Suppose that the half-life of hydrocodone bitartrate in the body is 3.8 hours, and that the oral dose taken is 10 mg. Write a differential equation for the quantity, Q. of hydrocodone bitartrate in the body at time t. in hours, since the drug was fully absorbed. (Write your answer in terms of Q and t, e.g., Q' = 3t (1 - kQ).) Solve your differential equation, assuming that at t = Othe patient has just absorbed the full 10 mg dose of the drug.

Explanation / Answer

y = Ce^(kt)

Where C is the original amount, k is the constant of proportionality, and t is the time.

Here, the question tells you that "the oral does taken is 10 mg", so C is 10
y = 10e^(kt)
This will be the answer to part A

(a) Q = 10e^(kt)

For part b, it looks like your supposed to set t=0

Q = 10e^(0)
Q = 10

I'm actually really not sure about that one, so let's move.

For part c

It tells you that the half life is 3.8 hours. So, t = 3.8

Q = 10e^(3.8k)

At t= 3.8, Q will be half of 10 (hence the name half-life)

5 = 10e^(3.8k)

5/10 = e^(3.8k)

0.5 = e^(3.8k)

Take the natural log of both sides

ln(0.5) = 3.8k

k = ln(0.5) / 3.8

And that's the constant of proportionality.

For d, just plug everything into the equation

Q = 10e^(kt)

Q = 1.121 mg

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