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 . Suppose that the half-life of hydrocodone bitartrate in the body is 3.7 hours, and that the oral dose taken is 12 mg.
(a) Write a differential equation for the quantity, , of hydrocodone bitartrate in the body at time , in hours, since the drug was fully absorbed.
amount = 12 (1/2)^(t/3.7) where t is in hours.
To write a differential equation for the quantity of hydrocodone bitartrate in the body at time t, we can use the concept of exponential decay.
Let Q(t) be the quantity of hydrocodone bitartrate in the body at time t.
Since the quantity of the drug decreases at a rate proportional to the amount left in the body, the rate of change of Q with respect to time (dQ/dt) is proportional to Q itself. The constant of proportionality is given as -k, where k is a positive constant.
Therefore, we can represent this relationship using a differential equation as follows:
dQ/dt = -k * Q
Now, to find the value of the constant k, we can use the half-life information provided. The half-life is the time it takes for the quantity of the drug to reduce by half. In this case, the half-life is given as 3.7 hours.
Using the concept of half-life, we know that when t = 3.7 hours, Q(t) = 0.5 * Q(0), where Q(0) is the initial quantity of the drug.
Since we are given the initial dose as 12 mg, Q(0) = 12 mg.
Therefore, at t = 3.7 hours, Q(3.7) = 0.5 * 12 mg = 6 mg.
Now, let's use this information to find the value of k.
Q(t) = Q(0) * e^(-kt)
At t = 3.7 hours and Q(3.7) = 6 mg:
6 mg = 12 mg * e^(-k * 3.7)
Dividing both sides by 12 mg:
0.5 = e^(-k * 3.7)
To solve for k, take the natural logarithm (ln) of both sides:
ln(0.5) = -k * 3.7
Now, solve for k:
k = -ln(0.5) / 3.7
Substituting this value of k back into the differential equation, we get the final form of the differential equation:
dQ/dt = -(-ln(0.5) / 3.7) * Q
Simplifying:
dQ/dt = (ln(0.5) / 3.7) * Q
Therefore, the differential equation for the quantity of hydrocodone bitartrate in the body at time t is:
dQ/dt = (ln(0.5) / 3.7) * Q
Oops forgot about the 15. Toss it in as needed.
However, to answer the question,
da/dt = ka
The boundary conditions will reveal that
k = ln(1/2)/3.7