Infusion Rate and Therapeutic Window
A patient receives a drug by constant intravenous infusion at R = 6 mg/h. The drug distributes in an apparent volume V = 30 L and is eliminated at a rate proportional to concentration, with elimination constant k = 0.20 h⁻¹. Let C(t) be plasma concentration in mg/L; before infusion C(0)=0. The therapeutic window is 0.60 to 1.40 mg/L. The clinical team wants to know when the drug first reaches the window, what steady-state concentration the current rate would eventually produce, and how long the concentration stays above 0.60 mg/L after the infusion is stopped at t = 12 h. A nurse suggests doubling the infusion rate to reach the window faster. The pharmacist warns that this may overshoot the upper limit. The team needs a quantitative answer, assuming the one-compartment model and no loading dose.
1. Set up and solve the differential equation for C(t) during infusion. Find the time to reach 0.60 mg/L and the steady-state concentration.
Read the worked answer
During infusion, dC/dt = R/V − kC = 6/30 − 0.20C = 0.20(1 − C).
Separate: dC/(1 − C) = 0.20 dt, with C(0)=0. This gives −ln(1 − C) = 0.20t, so C(t) = 1 − e^(−0.20t).
Set C = 0.60: e^(−0.20t) = 0.40, so t = −ln(0.40)/0.20 ≈ 4.58 h. The steady state is C∞ = R/(kV) = 6/(0.20 × 30) = 1.00 mg/L. The current rate can therefore enter the window and stay below 1.40 mg/L.
2. At t=12 h the infusion is stopped. Write C(t) for t>12 and find how long after stopping the concentration remains at or above 0.60 mg/L.
Read the worked answer
At t = 12 h, C(12) = 1 − e^(−2.4) = 1 − 0.0907 = 0.9093 mg/L.
After stopping, dC/dt = −0.20C, so for s = t − 12 ≥ 0, C = 0.9093e^(−0.20s). Set C = 0.60: e^(−0.20s) = 0.60/0.9093 = 0.660, so s = −ln(0.660)/0.20 ≈ 2.08 h.
The concentration remains at or above 0.60 mg/L for about 2.1 h after the infusion is stopped, then falls below the therapeutic window.
3. Evaluate doubling the rate to 12 mg/h. What steady state would be reached, when would the concentration cross 1.40 mg/L, and is a loading dose a better option?
Read the worked answer
Doubling to 12 mg/h gives R/V = 0.40, so C(t) = 2(1 − e^(−0.20t)) and the steady state is 2.00 mg/L, above the 1.40 mg/L upper limit.
The 1.40 level is crossed when 2(1 − e^(−0.20t)) = 1.40, i.e. e^(−0.20t) = 0.30, t ≈ 6.02 h. The 0.60 level is crossed at t ≈ 1.78 h, so the patient would be inside the window only for about 4.2 h before overshooting.
Doubling is not a safe maintenance strategy. If a faster rise is needed, a loading dose followed by the 6 mg/h maintenance rate is better; it raises C quickly without a 2.0 mg/L steady state.