Antibiotic concentration after a single dose
A veterinary clinic treats a sheep with a single intravenous dose of an antibiotic. A simplified two-compartment model predicts plasma concentration C(t)=a t e^(-kt) mg/L, where t is hours after injection and a,k>0. Measurements from a similar animal show that the concentration peaks at t=2.5 h and reaches 18 mg/L at that moment. The therapeutic window is 8 to 30 mg/L: below 8 mg/L the treatment is ineffective, above 30 mg/L there is a risk of toxicity. The clinic must decide whether one dose keeps the animal inside the window for at least 6 hours. The veterinarian also wants to report the mean concentration over the first 10 hours and the time when the drug falls below the therapeutic threshold on the descending side. Assume the model applies continuously and that no second dose is given during the observation period.
1. Determine a and k from the peak information, and justify that the critical point is a maximum rather than a minimum.
Read the worked answer
Differentiate C(t)=a t e^(-kt):
C'(t)=a e^(-kt) - a k t e^(-kt)=a e^(-kt)(1-kt).
Since a e^(-kt)>0, the only interior critical point satisfies 1-kt=0, so t=1/k. The measured peak occurs at t=2.5, hence k=1/2.5=0.4 h^-1.
At the peak, C(2.5)=a(2.5)e^(-1)=18, so a=18e/2.5=7.2e≈19.57 mg/L per hour. The sign of C' is controlled by 1-0.4t: positive for t<2.5 and negative for t>2.5. Therefore C increases up to t=2.5 and decreases afterwards; together with C(0)=0 and C(t)→0 as t→∞, this proves the critical point is the global maximum. Its value is 18 mg/L, below the toxicity threshold of 30 mg/L, so toxicity is not predicted by this model.
2. Compute the mean concentration over the first 10 hours. Then find, numerically if necessary, the interval during which C(t)≥8 mg/L and decide whether one dose keeps the animal therapeutic for at least 6 hours.
Read the worked answer
The mean value is
(1/10)∫_0^10 a t e^(-kt) dt.
An antiderivative of t e^(-kt) is -(t/k+1/k^2)e^(-kt). With a=19.57 and k=0.4,
∫_0^10 C(t)dt = [-48.93(t+2.5)e^(-0.4t)]_0^10 = -11.20 + 122.31 ≈111.11 mg·h/L.
Dividing by 10 gives a mean concentration of about 11.1 mg/L.
For the therapeutic threshold, solve C(t)=8:
19.57 t e^(-0.4t)=8, so t e^(-0.4t)=0.4088.
The function t e^(-0.4t) rises to its maximum at t=2.5 and then falls, so there are two solutions. Numerically, t1≈0.50 h and t2≈7.16 h. Thus the concentration is at least 8 mg/L from about 0.50 h to 7.16 h, a duration of about 6.66 h. Since 6.66>6, one dose does keep the animal in the therapeutic window for at least 6 hours under the model. The maximum concentration is only 18 mg/L, so the upper bound of 30 mg/L is not reached.