Integrated Concentration-Time Profiles After an IV Bolus

From a Saturable Elimination Rate to a Concentration-Time Profile

After an intravenous bolus, concentration may decline according to an elimination process whose rate depends on the current concentration. Under Michaelis-Menten elimination, the rate approaches a maximum at high concentration and becomes approximately proportional to concentration at low concentration. The differential equation for that process is straightforward to state, but its integrated concentration-time profile is implicit: concentration appears both as a linear term and inside a logarithm.

This tutorial derives that relation and shows how to use it. The calculations assume one well-mixed compartment with constant volume and a single Michaelis-Menten elimination pathway. They therefore describe concentration in that compartment; they do not by themselves describe tissue distribution or changes in enzyme activity.

Definition: Let $C(t)$ be concentration at time $t$ after an IV bolus, $C_0$ the concentration immediately after the bolus, $V_{\max}$ the maximum elimination rate expressed in concentration per unit time, and $K_m$ the Michaelis-Menten concentration constant. The model is $$ \frac{dC}{dt}=-\frac{V_{\max}C}{K_m+C}, \qquad C(0)=C_0. $$ Here $C$, $C_0$, and $K_m$ have concentration units; $V_{\max}$ has concentration/time units. If the bolus dose is $D$ and the compartment volume $V_d$ is constant, $C_0=D/V_d$.

The negative sign indicates declining concentration. The model assumes $C_0>0$, $K_m>0$, and $V_{\max}>0$; a fixed compartment volume; no further input after the bolus; and no elimination route other than the stated pathway. It also assumes the parameters remain constant over the time interval considered. These assumptions are important: a profile influenced by continuing absorption, distribution between compartments, or changing enzyme capacity need not satisfy this equation.

Deriving the Implicit Integrated Equation

Separate variables while $C>0$. Dividing by the elimination term and multiplying by $dt$ gives

\left(1+\frac{K_m}{C}\right)dC=-V_{\max}\,dt.

Integrate from the initial condition $C_0$ at time zero to concentration $C(t)$ at time $t$. The logarithm is written as a ratio so its argument is dimensionless.

\int_{C_0}^{C(t)}\left(1+\frac{K_m}{C}\right)dC =-V_{\max}\int_0^t dt

C(t)-C_0+K_m\ln\left(\frac{C(t)}{C_0}\right)=-V_{\max}t.

Rearranging produces the useful form

C_0-C(t)+K_m\ln\left(\frac{C_0}{C(t)}\right)=V_{\max}t.

Key result: For an IV bolus followed by constant-parameter Michaelis-Menten elimination, concentration at time $t$ satisfies $$ C_0-C+K_m\ln\left(\frac{C_0}{C}\right)=V_{\max}t. $$ The equation gives time directly when $C$ is specified. To find $C$ at a specified time, solve the equation numerically; in general, it is not an explicit formula for $C(t)$.

The equation is dimensionally consistent: each term on the left has concentration units, and $V_{\max}t$ does as well. The logarithm itself is dimensionless. At $t=0$, setting $C=C_0$ makes both sides zero. For $t>0$, the solution must satisfy $0<C<C_0$, as expected for elimination without input.

Existence, Uniqueness, and Numerical Solution

For a specified time, define $g(C)=C_0-C+K_m\ln(C_0/C)$ for $C>0$. Its derivative is

g'(C)=-1-\frac{K_m}{C}<0.

Thus $g(C)$ is strictly decreasing. At $C=C_0$, $g(C_0)=0$; as $C$ approaches zero from above, the logarithmic term grows without bound. Consequently, for every $V_{\max}t>0$, there is exactly one concentration between zero and $C_0$ satisfying $g(C)=V_{\max}t$. This establishes that the implicit equation has one physically relevant solution for each positive time.

A numerical root finder can locate that solution. For example, Newton's method applied to $g(C)-V_{\max}t=0$ gives the update

C_{\text{new}}=C_{\text{old}}+ \frac{g(C_{\text{old}})-V_{\max}t} {1+K_m/C_{\text{old}}}.

A bracketed method, such as bisection, is often preferable when robustness matters: choose lower and upper concentrations for which the residual has opposite signs, then repeatedly narrow the bracket. In either case, verify the resulting concentration by substituting it back into the original integrated equation. Rounding should be done after the check, not during the iterations.

Worked Example: Concentration Five Hours After a Bolus

Suppose a hypothetical one-compartment model has $C_0=30.0\ \mathrm{mg/L}$, $K_m=10.0\ \mathrm{mg/L}$, and $V_{\max}=2.00\ \mathrm{mg/L/h}$. Find $C$ at $t=5.00\ \mathrm{h}$. The assumed conditions are a fixed compartment volume and constant parameters, with no input after the bolus.

The integrated equation has right-hand side $V_{\max}t=(2.00)(5.00)=10.0\ \mathrm{mg/L}$. Therefore solve

30.0-C+10.0\ln\left(\frac{30.0}{C}\right)=10.0.

Try $C=22.75\ \mathrm{mg/L}$. Its left-hand side is $30.0-22.75+10.0\ln(30.0/22.75)\approx7.25+2.7663=10.0163\ \mathrm{mg/L}$, slightly above the target. One Newton update gives $22.75+(10.0163-10.0)/(1+10.0/22.75)\approx22.7613\ \mathrm{mg/L}$.

Substitution gives $30.0-22.7613+10.0\ln(30.0/22.7613)\approx7.2387+2.7613=10.0000\ \mathrm{mg/L}$, within rounding. Thus the predicted concentration is $C(5.00\ \mathrm{h})\approx22.76\ \mathrm{mg/L}$, below the initial $30.0\ \mathrm{mg/L}$, as required.

Calculating a Concentration from a Different Parameter Set

The numerical solution depends on all three quantities $C_0$, $K_m$, and $V_{\max}$, as well as elapsed time. A useful calculation checks both the arithmetic and whether the solution remains within the physically meaningful interval $(0,C_0)$.

Worked Example: Solving at a Specified Time

Consider a second hypothetical bolus with $C_0=8.00\ \mathrm{mg/L}$, $K_m=4.00\ \mathrm{mg/L}$, and $V_{\max}=1.20\ \mathrm{mg/L/h}$. Find the concentration at $6.00\ \mathrm{h}$, assuming the same one-compartment and constant-parameter conditions.

Here $V_{\max}t=(1.20)(6.00)=7.20\ \mathrm{mg/L}$, so

8.00-C+4.00\ln\left(\frac{8.00}{C}\right)=7.20.

At $C=3.79\ \mathrm{mg/L}$, the left-hand side is $8.00-3.79+4.00\ln(8.00/3.79)\approx4.21+2.9883=7.1983\ \mathrm{mg/L}$. The derivative of the left-hand side with respect to $C$ is $-1-4.00/C$; using the unrounded left-hand side, a Newton correction is approximately $(7.1983017-7.20)/(1+4.00/3.79)=-0.0008262\ \mathrm{mg/L}$. The refined concentration is about $3.78917\ \mathrm{mg/L}$, which rounds to $3.7892\ \mathrm{mg/L}$.

Substitution at $3.7892\ \mathrm{mg/L}$ gives $8.00-3.7892+4.00\ln(8.00/3.7892)\approx4.2108+2.98915=7.19995\ \mathrm{mg/L}$. Thus $C(6.00\ \mathrm{h})\approx3.789\ \mathrm{mg/L}$. The result is positive and less than $C_0$, and its substitution reproduces the required concentration-time term.

Calculating Time Between Two Concentrations

When the starting and ending concentrations are known, no root-finding step is needed. Rearranging the integrated equation gives

t=\frac{C_0-C+K_m\ln(C_0/C)}{V_{\max}}.

This expression requires $C>0$. For elimination after a bolus, the ending concentration should be below the starting concentration, and the calculated elapsed time should be positive.

Worked Example: Time to Decline to a Specified Concentration

For a third hypothetical parameter set, let $C_0=40.0\ \mathrm{mg/L}$, $K_m=5.00\ \mathrm{mg/L}$, and $V_{\max}=3.00\ \mathrm{mg/L/h}$. Calculate the time for concentration to decline to $10.0\ \mathrm{mg/L}$.

Substitute the concentrations into the time equation:

t=\frac{40.0-10.0+5.00\ln(40.0/10.0)} {3.00\ \mathrm{mg/L/h}}.

Since $\ln(4)=1.38629$, the numerator is $30.0+5.00(1.38629)=36.9315\ \mathrm{mg/L}$. Dividing by $3.00\ \mathrm{mg/L/h}$ gives $t=12.3105\ \mathrm{h}$, or about $12.31\ \mathrm{h}$. The positive time is consistent with a decline from $40.0$ to $10.0\ \mathrm{mg/L}$, and the units reduce to hours.

Checks, Alternatives, and Limitations

The integrated equation is valuable because it converts the differential model into a direct relation between elapsed time and concentration. If the goal is a concentration at a specified time, solving this scalar equation is often simpler than simulating the differential equation over many time steps. Numerical integration of the differential equation is a reasonable alternative and can be useful when the model includes additional processes, but it should agree with the implicit relation when it uses precisely the same single-compartment assumptions and parameters.

Several checks help detect mistakes. First, concentration must remain positive and cannot exceed $C_0$ after a bolus with no further input. Second, the left-hand side $C_0-C+K_m\ln(C_0/C)$ must equal $V_{\max}t$ within numerical tolerance. Third, both sides must have concentration units. Finally, the predicted concentration should decline as time increases. A concentration greater than $C_0$, a negative elapsed time for a lower target concentration, or a large residual after substitution signals an error in the equation, units, or numerical calculation.

A common mistake is to treat $V_{\max}$ as a first-order clearance and write $C(t)=C_0e^{-V_{\max}t}$. That expression does not follow from the Michaelis-Menten rate law: $V_{\max}$ has concentration/time units, and the elimination rate depends on $C/(K_m+C)$. The integrated logarithmic relation is required under this model. Another error is to take the logarithm of a dimensional concentration by itself; use the dimensionless ratio $C_0/C$.

The model is limited to a well-mixed compartment with constant parameters and a single elimination process. A PBPK model with changing tissue concentrations or multiple processes may require a system of equations rather than this scalar relation. The integrated equation should therefore be used only when its assumptions match the situation being represented; a good numerical fit alone does not establish that they do.

Key takeaway: Integrating Michaelis-Menten elimination after an IV bolus yields $C_0-C+K_m\ln(C_0/C)=V_{\max}t$. Use it directly for time calculations, or solve its unique positive root below $C_0$ for concentration at a specified time, then verify the result by substitution.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

For $C_0=16.0\ \mathrm{mg/L}$, $K_m=4.00\ \mathrm{mg/L}$, and $V_{\max}=2.00\ \mathrm{mg/L/h}$, write the equation that determines concentration at $t=3.00\ \mathrm{h}$.
Why is there exactly one positive concentration below $C_0$ for any specified $t>0$ under this model?
For $C_0=20.0\ \mathrm{mg/L}$, $K_m=5.00\ \mathrm{mg/L}$, and $V_{\max}=2.00\ \mathrm{mg/L/h}$, calculate the time to reach $C=10.0\ \mathrm{mg/L}$.
A numerical solution at positive time gives $C=22.0\ \mathrm{mg/L}$ when $C_0=18.0\ \mathrm{mg/L}$. Name a necessary check that fails and give one possible explanation.

Solutions. (1) The right-hand side is $(2.00)(3.00)=6.00\ \mathrm{mg/L}$, so $16.0-C+4.00\ln(16.0/C)=6.00$, with concentrations in $\mathrm{mg/L}$. (2) The left-hand side $g(C)$ is strictly decreasing because $g'(C)=-1-K_m/C<0$; it is zero at $C_0$ and tends to infinity as $C$ tends to zero, so it crosses any positive $V_{\max}t$ exactly once. (3) $t=[20.0-10.0+5.00\ln(20.0/10.0)]/2.00=[10.0+3.46574]/2.00=6.73287\ \mathrm{h}$, or $6.733\ \mathrm{h}$. (4) The concentration must be positive and below $C_0$, so this result violates the decline-after-bolus bound. Possible causes include a mistaken parameter or initial condition, a sign error, or solving the wrong equation.