Calculus I · The Function That Is Its Own Slope

What all of this is for

Time to collect on the three real-world situations that started the chapter. They were all the same equation, $\dfrac{dN}{dt} = kN$ — “the rate of change is proportional to the amount.” We can now write down its solution and check it ourselves:

$$N(t) = N_{0}\,e^{kt}.$$

Here $N_0$ is the amount at the start (set $t=0$: $e^0 = 1$, so $N(0) = N_0$ — good). Does it actually solve the equation? Differentiate, using the chain rule:

$$\frac{dN}{dt} = N_{0}\cdot k\,e^{kt} = k\,(N_{0}e^{kt}) = k\,N.$$

Yes. The derivative came back equal to $k$ times the function itself, which is exactly the equation we needed to satisfy. We have not just been handed the answer — we have verified it.

Growth: money and rabbits

When $k > 0$, $N(t) = N_0 e^{kt}$ grows, and it grows faster the bigger it gets. This is continuous compounding: a bank account growing at rate $r$, compounded not yearly or monthly but every instant, reaches $N_0 e^{rt}$ after time $t$. (This is also where $\left(1+\frac1n\right)^n$ secretly lives — compounding $n$ times a year and letting $n \to \infty$ is what produces $e$ in the first place. The promise from earlier, peeking through again.)

Decay and half-life: atoms and medicine

When $k < 0$, write it as $k = -\lambda$ with $\lambda > 0$, and the same function now decays:

$$N(t) = N_{0}\,e^{-\lambda t}.$$

This describes a radioactive sample losing atoms, and — the example to keep in your pocket if you care about drug design — a medicine being cleared from the bloodstream. Most drugs leave the body by first-order kinetics: the liver and kidneys remove an amount proportional to how much is currently circulating, which is exactly $\frac{dN}{dt} = -\lambda N$.

The number everyone quotes for such a process is its half-life: the time $t_{1/2}$ for the amount to fall to half of whatever it was. We can derive it instead of memorizing it. We want the time when $N$ has dropped to $N_0/2$:

$$N_{0}\,e^{-\lambda t_{1/2}} = \tfrac{1}{2}N_{0}.$$

Cancel $N_0$, then take $\ln$ of both sides to pull the exponent down (this is the moment logarithms were built for — undoing an exponential):

$$e^{-\lambda t_{1/2}} = \tfrac12 \;\Longrightarrow\; -\lambda t_{1/2} = \ln\tfrac12 = -\ln 2 \;\Longrightarrow\; \boxed{\,t_{1/2} = \dfrac{\ln 2}{\lambda}\,}.$$

Worked example. A painkiller has an elimination rate constant $\lambda = 0.231$ per hour. What is its half-life, and how much of a $400\text{ mg}$ dose remains after $9$ hours?

Half-life: $t_{1/2} = \dfrac{\ln 2}{0.231} = \dfrac{0.693}{0.231} = 3.0$ hours. So the amount halves every $3$ hours.

After $9$ hours: that is exactly three half-lives, so we expect three halvings: $400 \to 200 \to 100 \to 50\text{ mg}$. Check it the precise way too: $N(9) = 400\,e^{-0.231\times 9} = 400\,e^{-2.079} = 400 \times 0.125 = 50\text{ mg}$. The two methods agree, which is how you know you did it right. (Figure 5.4: the decay curve with the dose halving at $t = 3, 6, 9$ hours marked.)

Why log scales exist

Because $\ln$ turns multiplication into addition, it squashes enormous multiplicative ranges into walkable additive ones. That is why the loud, scientific scales of the world are logarithmic: earthquake energy (Richter), sound (decibels), acidity (pH). A jump of “one unit” on these scales means times ten, not plus ten. When someone tells you a magnitude-$7$ quake is “a bit worse” than a magnitude-$5$, the logarithm is the reason they are off by a factor of a hundred.

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