Visual walkthrough — Exponential growth and decay models — half-life, doubling time

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The characters (defined before we use them)

Before any equation, meet the four letters we will need. Each is a thing you can point to.

Why do we need $\frac{dN}{dt}$ at all? Because the only fact we start with is a fact about speed — "the more you have, the faster it changes." Speed of a changing quantity is exactly what a derivative measures, so that is the tool the problem hands us. We could not phrase the starting fact without it.

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Step 1 — Draw the starting law: speed is proportional to amount

WHAT. We picture jars holding different amounts and ask how fast each is changing.

WHY. Every exponential story begins with one sentence: "twice as much stuff → twice as fast a change." We must turn that sentence into a picture before we trust any symbol.

PICTURE. Below, three jars. The first has a few dots, the second twice as many, the third twice again. The arrows above each jar show the change-per-second — and each arrow is twice as long as the one before, exactly matching the dot count.

Reading the arrows off the picture and writing "is proportional to" as $\propto$:

$\underset{\text{speed of change}}{\underbrace{\frac{dN}{dt}}}\propto \underset{\text{amount now}}{\underbrace{N}}$

Proportional means "a constant times", and that constant is our $k$:

$\frac{dN}{dt}=\underset{\text{fixed number}}{\underbrace{k}}\underset{\text{amount now}}{\underbrace{N}}$

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Step 2 — Separate the two worlds: $N$ on the left, $t$ on the right

WHAT. We rearrange so that everything about amount sits on one side and everything about time on the other.

WHY. We have two changing things tangled in one equation. We cannot integrate a mix. Splitting them lets us handle each on its own turf — this is the whole trick of separable equations. Think of it as putting the $N$-tools in one drawer and the $t$-tools in another.

PICTURE. Two shelves: the left shelf collects the $N$-piece $\frac{1}{N}dN$, the right shelf collects the $t$-piece $kdt$.

Divide both sides by $N$ and multiply both by $dt$:

$\underset{\text{only }N\text{ lives here}}{\underbrace{\frac{1}{N}dN}}=\underset{\text{only }t\text{ lives here}}{\underbrace{kdt}}$

Each term earns its place: $\frac{1}{N}$ is large when $N$ is tiny (change matters most relatively when little is left), and $dN$, $dt$ are the tiny steps in amount and time.

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Step 3 — Add up all the tiny steps (integrate)

WHAT. We sum every microscopic step across the whole journey — this summing is what the $\int$ sign means.

WHY. The equation in Step 2 is only true for an instant. To know $N$ after real time passes, we glue infinitely many instants together. Integration is exactly "glue the slivers".

PICTURE. The left side sums slivers of width $\frac{1}{N}$; the area under $\frac{1}{N}$ from a start amount up to $N$ is the natural logarithm. The right side sums flat slivers of height $k$, giving the straight line $kt$.

$\int \frac{1}{N}dN=\int kdt⟹\underset{\text{area under }1/N}{\underbrace{\ln |N|}}=\underset{\text{area under flat }k}{\underbrace{kt}}+\underset{\text{leftover constant}}{\underbrace{C}}$

Why does $\int \frac{1}{N}dN$ give $\ln$ and not a power? Because $\ln$ is defined as that running area — see Natural logarithm and e. No other function has slope exactly $\frac{1}{N}$, so it is forced on us, not chosen.

The lone $C$ is a constant of integration: integrating tells us the shape but not the height it starts at. $C$ is the missing height, and Step 5 will pin it down.

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Step 4 — Undo the log to set $N$ free

WHAT. We apply $e^{(\cdot )}$ to both sides to release $N$ from inside the logarithm.

WHY. $N$ is trapped: it sits inside $\ln$. The exponential $e^x$ is the exact inverse of $\ln$ — feed a log into $e$ and you get the original number back. So $e$ is the key cut for this specific lock.

PICTURE. A lock-and-key: $\ln$ locked $N$ away; $e$ is the one key that opens it. The split $e^{kt+C}=e^C{e}^{kt}$ is drawn as splitting one block into a fixed block $e^C$ times a growing block $e^{kt}$.

$|N|=e^{kt+C}=\underset{\text{fixed number, call it }A}{\underbrace{e^C}}\underset{\text{the growing part}}{\underbrace{e^{kt}}}$

The split uses a law of indices: $e^{a+b}=e^a{e}^b$. Since $e^C$ is just some fixed positive number, we rename it $A$. Dropping the now-unneeded bars ($N>0$):

$N=A{e}^{kt}$

We have the shape — but $A$ is still an unknown height.

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Step 5 — Pin the height using the start ($t=0$)

WHAT. We plug in the one moment we actually know — the beginning, $t=0$, where the amount is $N_0$.

WHY. The general solution $A{e}^{kt}$ fits infinitely many curves stacked at different heights. Reality has one starting amount. Substituting $t=0$ reads off which curve is ours.

PICTURE. A family of curves of the same shape at different heights; the horizontal line $N=N_0$ selects exactly one — the one crossing the axis at our starting count.

Set $t=0$. Since $e^0=1$:

$N(0)=A{e}^{k\cdot 0}=A\cdot 1=A⟹A=N_0$

So the mystery constant was the starting amount all along. Substituting back:

$\boxed{N(t)=\underset{\text{start height}}{\underbrace{N_0}}\underset{\begin{array}{c}\text{multiply-factor}\\ \text{after time }t\end{array}}{\underbrace{e^{kt}}}}$

Every letter now has a job: $N_0$ sets where we begin, $k$ sets how fierce the growth/decay is, $t$ is how long we wait, and $e$ is the natural base forced on us by the $\ln$ in Step 3.

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Step 6 — The decay case: what changes when $k<0$

WHAT. We re-run the same formula but with a negative $k$, and watch the curve flip from climbing to sinking.

WHY. The parent note warns: never drop the sign of $k$. The algebra is identical; only the geometry inverts. Covering this case means a decay problem never surprises you.

PICTURE. Two curves sharing the same $N_0$: orange with $k>0$ shooting up, teal with $k<0$ sliding toward — but never touching — the $t$-axis.

\begin{cases} k>0 \Rightarrow e^{kt}\text{ grows} & \text{(curve rises)}\\[4pt] k<0 \Rightarrow e^{kt}\text{ shrinks} & \text{(curve falls toward }0)\\[4pt] k=0 \Rightarrow e^{0}=1 & \text{(flat: nothing changes)} \end{cases}$$ The $k=0$ **degenerate case** is worth naming: no rate means $\frac{dN}{dt}=0$, a horizontal line stuck at $N_0$ forever. And note the decay curve *approaches* zero but never arrives — an [[Graphs of exponential functions|asymptote]] — which is exactly why "half of what remains" never empties the jar. --- ## Step 7 — Reading the doubling time and half-life off the curve **WHAT.** We ask "how long to reach a fixed multiple?" and read it as a fixed horizontal step on the curve. **WHY.** These are the two headline numbers of the whole topic. The picture shows *why they are constant*: the same vertical multiplying-factor always needs the same horizontal time, no matter where you stand on the curve. **PICTURE.** On the growth curve, three equal-width horizontal bands each double the height ($N_0 \to 2N_0 \to 4N_0 \to 8N_0$). On the decay curve, equal steps each halve it. The equal widths are the point. ![[deepdives/dd-maths-3.2.05-d2-s07.png]] **Doubling** — set the amount to $2N_0$ and solve for the time $T_d$: $$2N_0=N_0 e^{kT_d}\;\Rightarrow\; \underbrace{2=e^{kT_d}}_{N_0\text{ cancels}}\;\Rightarrow\;\ln 2 = kT_d \;\Rightarrow\; \boxed{T_d=\frac{\ln 2}{k}}$$ **Halving** — set the amount to $\tfrac12 N_0$ and solve for $T_{1/2}$: $$\tfrac12 N_0=N_0 e^{kT_{1/2}}\;\Rightarrow\;\tfrac12=e^{kT_{1/2}}\;\Rightarrow\;\ln\tfrac12=kT_{1/2}\;\Rightarrow\;\boxed{T_{1/2}=\frac{\ln 2}{|k|}}$$ The crucial visual fact: $N_0$ **cancels in the very first step**, so the answer cannot depend on where you started. That is why the bands in the figure are equal-width regardless of starting height — the ==signature of exponential behaviour==. > [!example] Reading it back — one quick check > If a drug halves in $\tau=2.71$ h, then $k=-\dfrac{\ln 2}{2.71}\approx -0.2557$ /h. Plug into $N_0 e^{kt}$ at $t=2.71$: $e^{-0.2557\times 2.71}=e^{-0.693}=\tfrac12$. ✓ The formula and the picture agree. --- ## The one-picture summary Everything above, compressed into a single flow: the starting law becomes a separated equation, becomes two areas, becomes a freed $N$, becomes the pinned curve — with growth, decay, and the constant doubling/half-life bands all shown at once. ![[deepdives/dd-maths-3.2.05-d2-s08.png]] > [!recall]- Feynman: the whole walkthrough in plain words > Picture a jar of splitting bacteria. Rule one: **the more there are, the faster new ones appear** — I drew this as arrows over jars, each arrow twice as long when the jar holds twice as much. That single rule is $\frac{dN}{dt}=kN$. > > To solve it, I sorted the tools into two drawers — all the "amount" stuff on the left, all the "time" stuff on the right — so I could deal with each alone. Then I glued together infinitely many tiny time-slivers (that's integrating); the amount side happened to add up to the natural log, and the time side to a plain straight line. The log had $N$ trapped inside it, so I used its one and only key, $e$, to set $N$ free. That left a mystery starting-height, which I found by looking at the very first moment $t=0$ — it was just the amount I began with, $N_0$. Result: $N(t)=N_0 e^{kt}$. > > Flip the sign of $k$ and the same curve sinks toward zero instead of climbing — but never quite lands, which is why "half of what's left" goes on forever. Finally, asking "how long to double?" or "how long to halve?", the starting amount cancels out immediately, so the answer is always the same length of time. Same ratio, same time. That's the whole magic. > [!mnemonic] Five words for five steps > **Split · Sum · Free · Pin · Read.** (Separate variables · integrate · undo the log with $e$ · pin the constant at $t=0$ · read off doubling/half-life.) --- ## Connections - [[Exponential growth and decay models — half-life, doubling time (index 3.2.5)]] — the parent this page draws out. - [[Solving first-order separable ODEs]] — the split-and-sum machinery of Steps 2–3. - [[Natural logarithm and e]] — why the area under $1/N$ is $\ln$, and why $e$ is the freeing key. - [[Laws of logarithms]] — the index law $e^{a+b}=e^a e^b$ used in Step 4. - [[Graphs of exponential functions]] — the rising/falling shapes and the zero-asymptote. - [[Radioactive decay (Physics)]] — half-life in the wild.