Case Study 2 — Why Your Medication Reaches a Plateau: Drug Accumulation as a Geometric Series
Field: Medicine and pharmacology (with a bouncing-ball coda in physics) Calculus used: Geometric series — partial sums and infinite sums (§21.3, §21.10)
A Question from the Clinic
A pharmacist hands you a prescription: take 100 mg of a drug every eight hours. You swallow the first tablet, and over the next eight hours your body clears most — but not all — of it. Then you take the second tablet on top of the leftover. Then a third on top of that residue. A natural worry surfaces: if every dose leaves something behind, doesn't the drug keep piling up without limit? Will the concentration climb forever until it becomes toxic?
The answer is no, and the reason is a geometric series. The body does not accumulate the drug without bound; instead the concentration climbs to a stable plateau — a steady-state level it hovers around for the rest of the course — and stays there. Understanding why the plateau is finite, and computing where it sits, is one of the most direct medical applications of this chapter: the bouncing ball of §21.10 wearing a lab coat.
The Model
The key biological fact is that the body eliminates drugs by first-order kinetics: in any fixed time interval, a constant fraction of the drug present is removed, regardless of how much is there. This is exactly the multiplicative structure that produces a geometric series. Let
- $D$ = the dose taken each interval (here $D = 100$ mg),
- $r$ = the fraction of drug remaining in the body at the moment the next dose is taken, with $0 < r < 1$.
The fraction $r$ is set by the drug's half-life and the spacing of doses. If the dosing interval equals exactly one half-life, then half the drug remains and $r = \tfrac12$. For our drug, suppose the dosing interval is a bit shorter than a half-life, so that $r = 0.7$ — meaning 70% of each dose is still present eight hours later, when the next tablet goes down.
Now track the body load — the amount of drug present just after each dose. Just after dose 1, the load is simply $D$. By the time dose 2 is taken, the first dose has decayed to $rD$, and the new tablet adds $D$ on top, giving $D + rD$. By dose 3, both earlier amounts have decayed one more step and a fresh $D$ arrives. After the $k$-th dose, the load is
$$L_k = D + rD + r^2 D + \cdots + r^{k-1}D = D\sum_{j=0}^{k-1} r^{j}.$$
That sum is the partial sum of a geometric series — the very object §21.3 taught us to evaluate in closed form. Using $\sum_{j=0}^{k-1} r^j = \dfrac{1 - r^k}{1 - r}$,
$$\boxed{\;L_k = D\,\frac{1 - r^k}{1 - r}\;}.$$
This single formula contains the whole story of accumulation. Each successive dose adds less net increase than the last, because the older doses are simultaneously decaying. The amounts being stacked form a geometric sequence, and a geometric sequence with ratio below one has a finite total.
The Plateau
What happens in the long run, as $k \to \infty$? The only piece of $L_k$ that depends on $k$ is $r^k$, and since $0 < r < 1$, the powers $r^k \to 0$ — precisely the behavior from §21.3 that makes a geometric series converge. Therefore the body load climbs toward a finite limit:
$$L_\infty = \lim_{k \to \infty} D\,\frac{1 - r^k}{1 - r} = \frac{D}{1 - r}.$$
The drug does not accumulate without bound. It plateaus at the steady-state peak $L_\infty = \dfrac{D}{1-r}$. For our numbers,
$$L_\infty = \frac{100\text{ mg}}{1 - 0.7} = \frac{100}{0.3} \approx 333.3\text{ mg}.$$
So although each tablet is only 100 mg, at steady state the body carries about 333 mg just after each dose — more than three times a single dose, but a fixed, predictable ceiling rather than a runaway climb. This is the accumulation factor $\dfrac{1}{1-r}$ at work: with $r = 0.7$ it equals $\dfrac{1}{0.3} \approx 3.33$, telling us the steady-state peak is 3.33 single doses.
The intuition matches the perpetuity of Case Study 1: infinitely many contributions, each a fixed fraction of the previous, summing to a finite total because geometric decay outruns accumulation. The body reaching a drug plateau is the formula $a/(1-r)$ made physiological.
Why clinicians care about the trough too. Just before the next dose, the load has decayed once more, to $r \cdot L_\infty = \dfrac{rD}{1-r}$. For us that is $0.7 \times 333.3 \approx 233.3$ mg. So at steady state the body load oscillates between a trough of about 233 mg and a peak of about 333 mg. Effective therapy requires keeping the trough above the minimum effective concentration and the peak below the toxic threshold — and both endpoints fall straight out of the geometric series.
How Fast Do You Reach the Plateau?
A patient (and a physician) wants to know not only where the plateau is but when the drug becomes fully effective. The gap between the current load $L_k$ and the plateau $L_\infty$ is
$$L_\infty - L_k = \frac{D}{1-r} - \frac{D(1 - r^k)}{1-r} = \frac{D}{1-r}\,r^k = L_\infty\, r^k.$$
So the fractional shortfall is exactly $r^k$ — the load reaches $(1 - r^k)$ of its plateau after $k$ doses. To get within 90% of plateau we need $1 - r^k \ge 0.9$, i.e. $r^k \le 0.1$:
$$0.7^k \le 0.1 \quad\Longrightarrow\quad k \ge \frac{\ln 0.1}{\ln 0.7} = \frac{-2.303}{-0.357} \approx 6.46.$$
Since $0.7^6 \approx 0.1176 > 0.1$ while $0.7^7 \approx 0.0824 \le 0.1$, it takes 7 doses to reach 90% of the steady-state level. This is why some medications take a couple of days of regular dosing to "kick in," and why physicians sometimes prescribe a larger loading dose at the start — a deliberate jump straight to $L_\infty$ to skip the slow geometric climb. The loading dose is nothing more than the partial-sum formula solved for the shortcut: give $\dfrac{D}{1-r}$ up front, and you begin at the plateau instead of crawling toward it.
The Same Series, Now Bouncing
It is worth seeing that this is literally the same calculation as the bouncing ball of §21.10, because that equivalence is the recurring lesson of the textbook: one mathematical structure, many fields. A ball dropped from height $h$ rebounds to a fraction $r$ of its height each bounce. It falls $h$; then rises and falls $rh$; then rises and falls $r^2h$; and so on. The total vertical distance is
$$d = h + 2rh + 2r^2h + \cdots = h + 2rh\sum_{n=0}^{\infty} r^n = h + 2rh\cdot\frac{1}{1-r} = h\,\frac{1+r}{1-r}.$$
For $h = 10$ m and $r = 0.6$, the total path length is $10 \cdot \dfrac{1.6}{0.4} = 40$ m — a finite distance covered in infinitely many bounces. Drug molecules decaying by a fixed fraction between doses, and a ball losing a fixed fraction of its height between bounces, are governed by the identical geometric series. The factor $\dfrac{1}{1-r}$ that sets the drug plateau is the same factor that makes the ball's path finite. When you have internalized one, you have internalized both.
A Caution the Chapter Demands
The whole analysis rests on $0 < r < 1$. If a patient's elimination were impaired — say liver or kidney disease so severe that more than the eliminated amount were somehow retained, pushing $r$ toward or past $1$ — the geometric series would no longer converge, and the model would predict unbounded accumulation. That is precisely the divergence boundary of §21.6: a geometric series converges if and only if $|r| < 1$. In the clinic, $r$ creeping toward $1$ is the mathematics of a drug that builds to toxicity, which is why dosing must be adjusted for patients who clear drugs slowly. The same inequality that decides a series' fate decides a patient's safety.
Discussion Questions
- Explain in terms of partial sums why a drug taken repeatedly reaches a finite plateau rather than accumulating without bound. Which property of $r$ is essential, and what would happen if $r \ge 1$ (connect to §21.6)?
- Our drug plateaued at about 333 mg with $r = 0.7$. Recompute the steady-state peak if the dosing interval is lengthened so that only $r = 0.5$ remains between doses. Why does a longer interval lower the plateau?
- Derive the steady-state trough (the load just before a dose) and explain why both peak and trough matter for safe, effective dosing.
- A loading dose lets a patient start at the plateau immediately. Using $L_\infty = D/(1-r)$, what single first dose would put our patient at steady state at once? How does this relate to the partial-sum formula?
- Show explicitly that the drug-accumulation series and the bouncing-ball series share the same common ratio $r$ and the same convergence factor $1/(1-r)$. What does this say about the textbook's theme that one structure recurs across fields?
A Short Annotated Reading
- Rowland & Tozer, Clinical Pharmacokinetics and Pharmacodynamics (4th ed.). The standard reference; its multiple-dose chapter derives the accumulation factor $1/(1-r)$ and loading doses exactly as above, in clinical units.
- Shargel & Yu, Applied Biopharmaceutics and Pharmacokinetics. A more introductory treatment connecting half-life, dosing interval, and the steady-state peak and trough.
- Stewart, Calculus: Early Transcendentals (9th ed.), §11.2. The drug-concentration accumulation problem appears among Stewart's geometric-series applications, alongside the bouncing ball — a chance to see the same series in a second textbook's voice.
A repeated dose, a bouncing ball, a perpetuity: three faces of one geometric series. The factor $1/(1-r)$ that turns infinitely many contributions into a finite total is among the most quietly useful results in all of calculus.