IV Infusion: Reaching and Holding a Therapeutic Level
A patient is started on a continuous intravenous infusion of a drug at rate R mg/h. The drug distributes in an apparent volume V = 10 L and is eliminated at a rate proportional to its concentration, with elimination constant k = 0.20 h⁻¹. The therapeutic window is 2–4 mg/L; below 2 mg/L the drug is ineffective, above 4 mg/L it is toxic. The clinical team first sets R = 4 mg/h. At that rate the steady-state concentration is R/(kV). The patient's concentration is 0 mg/L at the start of infusion. A nurse asks when the concentration will reach 1.5 mg/L, and what happens if the infusion is stopped once steady state is reached. A second proposal is to start at R = 4 mg/h while the patient already has 1.0 mg/L from a previous dose. The team must model the approach to steady state, check whether the chosen rate can ever enter the therapeutic window, and decide whether a loading dose or a different rate is needed.
1. Solve the infusion model for R = 4 mg/h and C(0) = 0. Find the time to reach 1.5 mg/L and comment on whether that is therapeutically adequate.
Read the worked answer
The model is
dC/dt = R/V - kC = 4/10 - 0.20C = 0.4 - 0.2C.
The steady state is Css = R/(kV) = 4/(0.20 × 10) = 2 mg/L. Solving the linear differential equation with C(0) = 0:
C(t) = 2(1 - e^(-0.2t)).
Set C = 1.5: 2(1 - e^(-0.2t)) = 1.5 ⇒ e^(-0.2t) = 0.25 ⇒ -0.2t = ln(0.25) ⇒ t = ln(4)/0.2 ≈ 6.93 h.
So it takes about 6.93 h to reach 1.5 mg/L. But 1.5 mg/L is below the therapeutic lower limit of 2 mg/L. The concentration approaches 2 mg/L only asymptotically and never exceeds it, so R = 4 mg/h has no margin above the ineffective threshold. It is not an adequate maintenance rate for a 2–4 mg/L target.
2. If the infusion is stopped at steady state, find the concentration decay and the time to fall below 0.5 mg/L. Then solve the second proposal: R = 4 mg/h with C(0) = 1.0 mg/L.
Read the worked answer
At steady state C = 2 mg/L. With infusion stopped, dC/dt = -0.2C, so
C(t) = 2e^(-0.2t).
Set C = 0.5: 2e^(-0.2t) = 0.5 ⇒ e^(-0.2t) = 0.25 ⇒ t = ln(4)/0.2 ≈ 6.93 h.
For the second proposal, R = 4 mg/h and C(0) = 1:
C(t) = Css + (C0 - Css)e^(-0.2t) = 2 + (1 - 2)e^(-0.2t) = 2 - e^(-0.2t).
Set C = 1.5: 2 - e^(-0.2t) = 1.5 ⇒ e^(-0.2t) = 0.5 ⇒ t = ln(2)/0.2 ≈ 3.47 h.
Starting from 1.0 mg/L reaches 1.5 mg/L in about 3.47 h, but it still approaches only 2 mg/L, so the maintenance rate remains at the lower boundary.
3. Recommend a maintenance rate that targets 3 mg/L, and compute the time to reach 2 mg/L from zero. Explain why simply raising the rate without checking steady state is unsafe.
Read the worked answer
For a target steady state Css = 3 mg/L:
R = kVCss = (0.20)(10)(3) = 6 mg/h.
The solution from C(0) = 0 is
C(t) = 3(1 - e^(-0.2t)).
Set C = 2: 3(1 - e^(-0.2t)) = 2 ⇒ 1 - e^(-0.2t) = 2/3 ⇒ e^(-0.2t) = 1/3 ⇒ t = ln(3)/0.2 ≈ 5.49 h.
So R = 6 mg/h gives a steady state of 3 mg/L, comfortably inside 2–4 mg/L, and reaches 2 mg/L in about 5.49 h. If faster loading is needed, a loading dose can raise C0, but the maintenance rate should still be 6 mg/h.
Simply raising R without checking R/(kV) is unsafe: a rate above 8 mg/h would give Css above 4 mg/L and risk toxicity. The correct check is the steady-state formula plus the approach curve, not the infusion rate alone.