Answer:
a. dQ/dt = -kQ
b. [tex]Q = 9e^{-kt}[/tex]
c. k = 0.178
d. Q = 1.063 mg
Step-by-step explanation:
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.
Let Q be the quantity of drug left in the body.
Since the rate of decrease of the quantity of drug -dQ/dt is directly proportional to the quantity of drug left, Q then
-dQ/dt ∝ Q
-dQ/dt = kQ
dQ/dt = -kQ
This is the required differential equation.
b) Solve your differential equation, assuming that at the patient has just absorbed the full 9 mg dose of the drug.
with t = 0, Q(0) = 9 mg
dQ/dt = -kQ
separating the variables, we have
dQ/Q = -kdt
Integrating we have
∫dQ/Q = ∫-kdt
㏑Q = -kt + c
[tex]Q = e^{-kt + c}\\Q = e^{-kt}e^{c}\\Q = Ae^{-kt} (A = e^{c})[/tex]
when t = 0, Q = 9
[tex]Q = Ae^{-kt} \\9 = Ae^{-k0}\\9 = Ae^{0}\\9 = A\\A = 9[/tex]
So, [tex]Q = 9e^{-kt}[/tex]
c) Use the half-life to find the constant of proportionality k.
At half-life, Q = 9/2 = 4.5 mg and t = 3.9 hours
So,
[tex]Q = 9e^{-kt}\\4.5 = 9e^{-kX3.9}\\\frac{4.5}{9} = e^{-kX3.9}\\\frac{1}{2} = e^{-3.9k}\\[/tex]
taking natural logarithm of both sides, we have
[tex]ln\frac{1}{2} = ln(e^{-3.9k})\\\\-ln2 = -3.9k\\k = -ln2/-3,9 \\k = -0.693/-3.9\\k = 0.178[/tex]
d) How much of the 9 mg dose is still in the body after 12 hours?
Since k = 0.178,
[tex]Q = 9e^{-0.178t}[/tex]
when t = 12 hours,
[tex]Q = 9e^{-0.178t}\\Q = 9e^{-0.178X12}\\Q = 9e^{-2.136}\\Q = 9 X 0.1181\\Q = 1.063 mg[/tex]
