Drug Clearance and Dosing Interval on the Ward
A pharmacology student models the blood concentration $C(t)$ of a drug, in mg/L, $t$ hours after a single intravenous dose. She assumes elimination is proportional to the concentration present, so $dC/dt=-kC$. Ward records show that 2 hours after a 20 mg/L dose the concentration has fallen to 16 mg/L. The therapeutic window requires $C$ to stay at or above 5 mg/L, and the ward currently redoses every 8 hours.
The student claims that after 10 hours less than 30% of the original dose remains, and argues from this that a 10-hour interval is safe. The clinical pharmacist wants the elimination constant and half-life, the time at which concentration falls to 5 mg/L, and a check of the 30% claim before the interval is changed. She is also considering a reduced 15 mg/L dose on the same 8-hour schedule and wants to know whether the 5 mg/L floor still holds.
1. Solve the differential equation, find k and the half-life, and find the time at which the concentration reaches 5 mg/L.
Read the worked answer
Separating variables, $\dfrac{dC}{C}=-k,dt$, so $\ln C=-kt+\ln C_0$ and $C=C_0e^{-kt}$ with $C_0=20$.
Using the 2-hour reading: $16=20e^{-2k}$, so $e^{-2k}=0.8$ and
$$k=-\tfrac{1}{2}\ln 0.8=\tfrac{0.22314}{2}\approx 0.1116\text{ h}^{-1}$$
Half-life: $t_{1/2}=\dfrac{\ln 2}{k}=\dfrac{0.6931}{0.1116}\approx 6.21$ h.
Time to reach 5 mg/L: $5=20e^{-kt}$, so $e^{-kt}=0.25$ and
$$t=\frac{\ln 4}{k}=\frac{1.3863}{0.1116}\approx 12.4\text{ h}$$
So the model predicts the floor is crossed only after about 12.4 hours, not at 5 mg/L after 8 hours.
2. Check the student's claim about 10 hours, and assess whether a 10-hour dosing interval would actually be safe under the stated therapeutic rule.
Read the worked answer
At $t=10$: $C=20e^{-0.1116\times 10}=20e^{-1.116}\approx 20(0.3277)\approx 6.55$ mg/L.
The remaining fraction is about 32.8%, which is more than 30%, so the numerical claim is false.
More importantly, the reasoning is wrong in principle. The safety rule is a threshold of 5 mg/L, not '30% of the original dose'. The 5 mg/L level corresponds to 25% of 20 mg/L, reached at 12.4 h, so the 30% figure has no clinical meaning here.
On the actual criterion, $C(8)=20e^{-0.8926}\approx 8.19$ mg/L and $C(10)\approx 6.55$ mg/L, both above 5 mg/L. A 10-hour interval would therefore pass the rule under this single-dose model, but the student's justification for it does not.
3. With a reduced 15 mg/L dose on the existing 8-hour schedule, does the concentration stay at or above 5 mg/L? Compute the margin and the time at which the floor would be reached.
Read the worked answer
Scale the model: $C(t)=15e^{-0.1116t}$.
At $t=8$: $C=15e^{-0.8926}\approx 15(0.4096)\approx 6.14$ mg/L, which is above 5 mg/L, with a margin of about 1.14 mg/L.
The floor is reached when $15e^{-kt}=5$, i.e. $e^{-kt}=\tfrac13$, so
$$t=\frac{\ln 3}{k}=\frac{1.0986}{0.1116}\approx 9.85\text{ h}$$
The floor holds through the 8-hour point but only for about 1.85 hours beyond it. That margin is thin: a delayed dose, a missed reading, or faster-than-assumed elimination in a patient would breach 5 mg/L, so the reduced dose should not be combined with any lengthening of the interval.