(1 pt) Hydrocodone bitartrate is used as a cough suppressant. After the drug is
ID: 2866501 • Letter: #
Question
(1 pt) 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 3.7 hours, and that the oral dose taken is 11 mg. (a) 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. Q' = (Write your answer in terms of Q and t, e.g., Q' = 3t(1 - kQ).) (b) Solve your differential equation, assuming that at t = 0 the patient has just absorbed the full 11 mg dose of the drug. Q = _________ (Your solution may include the constant of proporationality k, but should not include any other unspecified constant.) (c) Use the half-life to find the constant of proportionality, k. k = (d) How much of the 11 mg dose is still in the body after 12 hours? Amount = (include units)Explanation / Answer
You'll start with an equation in the form of:
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 11 mg", so C is 11
y = 11e^(kt)
This will be the answer to part A
(a) Q = 11e^(kt)
For part b, it looks like your supposed to set t=0
Q = 11e^(0)
Q = 11
I'm actually really not sure about that one, so let's move.
For part c
It tells you that the half life is 3.7 hours. So, t = 3.7
Q = 11e^(3.7k)
At t= 3.7, Q will be half of 11 (hence the name half-life)
5.5 = 11e^(3.7k)
5.5/11 = e^(3.7k)
0.5 = e^(3.7k)
Take the natural log of both sides
ln(0.5) = 3.7k
k = ln(0.5) / 3.7
And that's the constant of proportionality.
For d, just plug everything into the equation
Q = 11e^(kt)
Q = 1.23mg
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.