Infusion Model: Constants, Plateau and Inverse
A pharmacology team models the plasma concentration of a drug during a controlled infusion by C(t)=At/(t+B), where t is hours from the start, C is in mg/L, and A and B are positive constants. Two measurements are treated as reliable: at t=2 h the concentration is 8 mg/L, and at t=6 h it is 12 mg/L. The team must decide whether the concentration reaches 5 mg/L within the first hour; if it does not, the infusion rate will be adjusted. They also want the theoretical plateau for planning, and a colleague proposes using the inverse function to answer questions of the form 'when does the concentration reach a given level?' instead of solving a new equation each time. The model is intended only for t≥0 during the infusion and for concentrations below saturation. Before the guideline is issued, the constants, the horizontal asymptote, and the domain of the inverse must all be checked for consistency.
1. Determine A and B from the two measurements, then state and interpret the horizontal asymptote of C(t).
Read the worked answer
Substitute t=2: 2A/(2+B)=8, so A=4(2+B)=8+4B. Substitute t=6: 6A/(6+B)=12, so A=2(6+B)=12+2B. Equating gives 8+4B=12+2B, hence 2B=4 and B=2. Then A=8+8=16. Thus C(t)=16t/(t+2). For t≥0, divide numerator and denominator by t: C(t)=16/(1+2/t), so as t→∞, C(t)→16. The plateau is 16 mg/L. Under the fitted model, no finite infusion at these settings can produce a steady concentration above 16 mg/L; targets at or above 16 are therefore unattainable. The asymptote is a limiting value, not a concentration reached in finite time.
2. Find the inverse function giving time as a function of concentration, state its contextual domain, and decide whether the 5 mg/L level is reached within the first hour.
Read the worked answer
Start from y=16t/(t+2). Then y(t+2)=16t, so yt+2y=16t, hence t(16−y)=2y and t=2y/(16−y). Because t≥0 and C(t) is increasing from 0 towards 16, the inverse has domain 0≤y<16. At y=5, t=2·5/(16−5)=10/11≈0.909 h. This is less than 1 h, so the 5 mg/L level is reached within the first hour, with about 5.5 minutes to spare. For comparison, y=10 gives t=20/6≈3.33 h. The inverse is useful only inside its domain; it cannot be evaluated at y=16.
3. A colleague tries to use the inverse at y=16 to find the time when the plateau is reached. Explain why this is invalid and what limiting statement is correct.
Read the worked answer
Substituting y=16 into t=2y/(16−y) gives 32/0, which is undefined. The reason is that C(t)=16t/(t+2)=16−32/(t+2), so C(t)<16 for every finite t≥0. The value 16 is approached only as t→∞. Therefore there is no finite time at which the concentration equals the plateau; the correct statement is lim(t→∞)C(t)=16. The inverse's domain excludes y=16, and any dosing guideline that requires a finite time to reach exactly 16 mg/L is asking for something the model does not allow.