Case Study 1 — Drug Pharmacokinetics: When Is a Drug Most Effective?

Field: Pharmacology, medicine Calculus used: Critical points, the first and second derivative tests, concavity, and L'Hôpital's rule — all from §9.3, §9.5, §9.6, and §9.8.

The clinical question

A patient swallows a tablet of an antibiotic. Within minutes the drug begins crossing the gut wall into the bloodstream; meanwhile the liver and kidneys begin clearing it out. The concentration in the blood does not jump to a peak and stay there — it rises, crests, and falls. A clinician needs to answer three questions that sound like medicine but are entirely calculus:

  1. When is the blood concentration highest, and how high does it get? (Too low and the drug fails; too high and it is toxic.)
  2. How fast is the concentration changing at any moment — and when does the rise turn into a fall?
  3. How long until the drug has cleared enough to give the next dose safely?

This is the classic one-compartment oral absorption model, and it is one of the most-used functions in clinical pharmacology. We will analyze it with nothing more than the derivative tools of this chapter.

The model

After a single oral dose $D$, the standard model (the Bateman function) gives the blood concentration at time $t \ge 0$ as

$$C(t) = \frac{D\,k_a}{V\,(k_a - k_e)}\left(e^{-k_e t} - e^{-k_a t}\right),$$

where $V$ is the volume of distribution, $k_a > 0$ is the absorption rate constant (how fast the drug enters the blood), and $k_e > 0$ is the elimination rate constant (how fast it leaves). For an oral drug, absorption is faster than elimination, so $k_a > k_e$. To keep the arithmetic readable, fix realistic numbers: $D = 500$ mg, $V = 10$ L, $k_a = 1.0\ \text{h}^{-1}$, $k_e = 0.2\ \text{h}^{-1}$. The leading constant becomes

$$\frac{D\,k_a}{V\,(k_a - k_e)} = \frac{500 \cdot 1.0}{10 \cdot 0.8} = 62.5\ \text{mg/L},$$

so

$$C(t) = 62.5\left(e^{-0.2 t} - e^{-1.0 t}\right).$$

At $t = 0$, $C(0) = 62.5(1 - 1) = 0$ — no drug in the blood yet, exactly right. As $t \to \infty$, both exponentials vanish and $C \to 0$ — the drug fully clears. Between these, the function rises to a peak and falls. We find the peak with the first derivative test.

Finding the peak concentration (the first derivative test)

The clinically critical moment is $t_{\max}$, the time of peak concentration, and $C_{\max}$, its value. These are a textbook critical-point problem (§9.3, §9.5).

Differentiate:

$$C'(t) = 62.5\left(-0.2\,e^{-0.2 t} + 1.0\,e^{-1.0 t}\right).$$

Set $C'(t) = 0$:

$$1.0\,e^{-1.0 t} = 0.2\,e^{-0.2 t} \quad\Longrightarrow\quad \frac{e^{-1.0 t}}{e^{-0.2 t}} = \frac{0.2}{1.0} = 0.2.$$

The left side is $e^{-0.8 t}$, so $e^{-0.8 t} = 0.2$. Take logarithms:

$$-0.8\,t = \ln(0.2) = -\ln 5 \quad\Longrightarrow\quad t_{\max} = \frac{\ln 5}{0.8} \approx \frac{1.609}{0.8} \approx 2.01\ \text{hours}.$$

Now apply the first derivative test to confirm this is a maximum, not a minimum. For small $t$ (say $t = 0$), $C'(0) = 62.5(-0.2 + 1.0) = 62.5(0.8) > 0$ — concentration is rising. For large $t$, the slower-decaying $e^{-0.2t}$ term dominates and its coefficient $-0.2$ is negative, so $C'(t) < 0$ — concentration is falling. Thus $C'$ changes from $+$ to $-$ at $t_{\max}$: a local (and, on $[0,\infty)$, absolute) maximum. The peak is real.

Evaluate the peak height. At $t_{\max} \approx 2.01$:

$$e^{-0.2(2.01)} = e^{-0.402} \approx 0.669, \qquad e^{-1.0(2.01)} = e^{-2.01} \approx 0.134,$$ $$C_{\max} = 62.5(0.669 - 0.134) \approx 62.5(0.535) \approx 33.4\ \text{mg/L}.$$

So the model predicts a peak of about 33 mg/L roughly two hours after the dose. If the drug's toxic threshold is, say, 45 mg/L and its minimum effective concentration is 10 mg/L, this single dose sits comfortably in the therapeutic window at its peak. That is exactly the judgment a dosing decision rests on — and it came straight out of a first-derivative-test calculation.

Reading the curvature (the second derivative)

The clinician also cares about the shape of the rise and fall, because it governs how long the drug stays in the effective range. That is a concavity question (§9.6). The second derivative is

$$C''(t) = 62.5\left(0.04\,e^{-0.2 t} - 1.0\,e^{-1.0 t}\right).$$

Setting $C''(t) = 0$: $1.0\,e^{-1.0 t} = 0.04\,e^{-0.2 t}$, i.e. $e^{-0.8 t} = 0.04 = \tfrac{1}{25}$, so $-0.8 t = -\ln 25$ and

$$t_{\text{inf}} = \frac{\ln 25}{0.8} = \frac{2\ln 5}{0.8} \approx \frac{3.219}{0.8} \approx 4.02\ \text{hours}.$$

For $t < t_{\text{inf}}$ the fast term $e^{-1.0t}$ still dominates and $C'' < 0$ (the curve is concave down — it rises steeply, then flattens as it approaches the peak, then begins falling). For $t > t_{\text{inf}}$, the slow term wins and $C'' > 0$ (the curve is concave up — the decline decelerates as the concentration eases toward zero). The inflection point at about 4 hours marks the transition from the "still falling fast" phase to the "long slow tail" phase — pharmacologically, the moment the drug shifts from active clearance dynamics into its terminal elimination phase.

Notice the elegant consistency check: $t_{\text{inf}} \approx 4.02$ h is exactly twice $t_{\max} \approx 2.01$ h. That is no accident — both come from $e^{-0.8t} = (\text{ratio})$, and $0.04 = 0.2^2$, so $\ln 25 = 2\ln 5$. The structure of the model forces the inflection to fall at $2\,t_{\max}$ whenever $k_a = 5 k_e$.

A limit that explains the terminal slope (L'Hôpital)

How fast does the drug clear in the long run? Pharmacologists summarize this with the terminal half-life — but there is a subtler question that L'Hôpital's rule (§9.8) answers cleanly. Suppose a researcher models a near-equal pair of rate constants, $k_a$ very close to $k_e$. The leading constant $\frac{D k_a}{V(k_a - k_e)}$ blows up as $k_a \to k_e$, while the bracket $(e^{-k_e t} - e^{-k_a t}) \to 0$. The concentration is a genuine $\frac{0}{0} \times \infty$ indeterminate form in the parameter $k_a$.

Fix $t$ and let $k_a \to k_e$. Write the concentration as a function of $k_a$:

$$C(t) = \frac{D\,k_a}{V}\cdot\frac{e^{-k_e t} - e^{-k_a t}}{k_a - k_e}.$$

The fraction is $\frac{0}{0}$ as $k_a \to k_e$. Apply L'Hôpital, differentiating numerator and denominator with respect to $k_a$ (treating $t$ and $k_e$ as constants):

$$\lim_{k_a \to k_e} \frac{e^{-k_e t} - e^{-k_a t}}{k_a - k_e} = \lim_{k_a \to k_e} \frac{t\,e^{-k_a t}}{1} = t\,e^{-k_e t}.$$

So in the limit $k_a \to k_e$ the model collapses gracefully to

$$C(t) = \frac{D\,k_e}{V}\,t\,e^{-k_e t},$$

a perfectly finite function with a single peak at $t = 1/k_e$. The apparent singularity was an artifact of the algebra, not the biology — L'Hôpital's rule reveals the well-behaved function hiding inside the $\frac{0}{0}$. This is precisely the §9.8 lesson that an indeterminate form does not mean "undefined"; it means "look closer."

What the calculus delivered

Three derivative tools answered three clinical questions. The first derivative test located the peak — when the drug is most concentrated (≈ 2 h) and how high it climbs (≈ 33 mg/L), the inputs to a safety judgment. The second derivative located the inflection (≈ 4 h), distinguishing the rapid-decline phase from the long terminal tail. And L'Hôpital's rule showed that the model degrades smoothly when absorption and elimination rates coincide, rescuing a formula that naively appears to divide by zero. None of this required a single experimental measurement beyond the four rate constants — the shape of the entire concentration curve was readable from $C'$ and $C''$ alone, the central promise of this chapter.

Discussion Questions

  1. The minimum effective concentration is 10 mg/L. Using the falling branch of $C(t) = 62.5(e^{-0.2t} - e^{-1.0t})$, estimate when a second dose is needed (when does $C$ drop back to 10?). Why is the rising branch irrelevant to this question?
  2. If a pediatric patient has a smaller volume of distribution $V = 5$ L (half the adult value), how does $C_{\max}$ change, and does $t_{\max}$ change? Explain using the structure of the formula.
  3. The inflection point fell at exactly $2\,t_{\max}$ because $k_a = 5k_e$. For a drug with $k_a = 4k_e$, would the inflection still be at $2t_{\max}$? Set up (but do not fully solve) the equation that decides this.
  4. A colleague applies L'Hôpital's rule to $\lim_{t\to\infty} C(t)/e^{-0.2t}$ to study the tail. Is this an indeterminate form? Check before differentiating (recall the §9.8.3 pitfall).

Annotated Reading

  • Stewart, Calculus: Early Transcendentals (9th ed.), §4.3 and §4.4. The increasing/decreasing test, concavity, and the derivative tests applied to applied functions — the direct analog of the analysis above.
  • OpenStax Calculus Vol. 1, §4.5 ("Derivatives and the Shape of a Graph"). Free, with worked extrema-and-concavity examples in an applied style.
  • Rowland & Tozer, Clinical Pharmacokinetics and Pharmacodynamics (4th ed.). The Bateman function and $t_{\max}$/$C_{\max}$ derivations in their native clinical setting — the same calculus, written for pharmacologists.