Exponential growth and decay: dN/dt = kN, Newton's law of cooling and applications

What this dot point is asking

NESA wants you to recognise that a rate of change proportional to the quantity itself produces exponential growth or decay, set up the appropriate differential equation, solve it, and apply it to populations, radioactive material and cooling bodies.

The answer

The growth and decay equation

If a quantity $N(t)$ changes at a rate proportional to its current value,

the solution is

where $N_0 = N(0)$ is the initial amount. If $k > 0$ the quantity grows; if $k < 0$ it decays.

The deep idea, and the one NESA is really testing, is that "rate of change proportional to the amount present" is the verbal cue for this whole model. Whenever a population grows faster the bigger it is, or a radioactive sample decays in proportion to how much is left, the rate is $kN$ and the answer is an exponential. Nothing else has the property that its derivative is a constant multiple of itself, which is exactly why $e^{kt}$ appears.

Half-life. For decay with rate constant $k < 0$, the half-life is $\tau = \frac{\ln 2}{|k|}$. Equivalently $N(t) = N_0 \cdot (1/2)^{t / \tau}$.

Doubling time. For growth, the doubling time is $T_d = \frac{\ln 2}{k}$. Both half-life and doubling time are independent of where you start: it takes the same time to fall from $80$ to $40$ as from $40$ to $20$, which is the hallmark of exponential change.

Newton's law of cooling

If $T(t)$ is the temperature of a body and $T_a$ is the ambient temperature (assumed constant), then

with $k < 0$. The solution is

where $T_0 = T(0)$. As $t \to \infty$, $T \to T_a$.

Why these solutions work

Differentiate $N(t) = N_0 e^{k t}$ to get $\frac{dN}{dt} = k N_0 e^{k t} = k N$, which matches the equation. The same check works for the cooling solution, since $\frac{d}{dt}(T_a) = 0$ and $\frac{d}{dt}((T_0 - T_a) e^{k t}) = k (T - T_a)$.

The shapes behind dN/dt = kN, stage by stage

The differential equation $\frac{dN}{dt} = kN$ says one thing: the rate of change is always proportional to how much you currently have. The four stages below show what that forces the graph to look like, and how the sign of $k$ flips growth into decay and cooling.

Stage 1, growth when $k > 0$. The solution is $N = N_0 e^{kt}$. Starting from the initial amount $N_0$ on the vertical axis, it climbs and gets ever steeper, because the bigger $N$ becomes, the faster it grows.

Stage 2, why the curve bends the way it does. Draw the tangent at several points: the gradient is $\frac{dN}{dt} = kN$, so it is small where the curve is low and large where the curve is high. The tangents visibly steepen as the height grows, which is the geometric meaning of "rate proportional to value".

Stage 3, decay when $k < 0$. Flip the sign of $k$ and the same equation gives decay: $N = N_0 e^{kt}$ falls towards zero. Every fixed time step multiplies $N$ by the same factor, so after one half-life $\tau = \frac{\ln 2}{|k|}$ the amount halves, after $2\tau$ it quarters, and so on.

Stage 4, cooling towards an ambient value. Newton's law $\frac{dT}{dt} = k(T - T_a)$ is the same idea applied to the gap above the ambient temperature $T_a$. The solution $T = T_a + (T_0 - T_a)e^{kt}$ decays, but towards $T_a$ rather than zero, so the curve levels off at the ambient line as its horizontal asymptote.

Fitting a model to data

A typical exam question gives you the initial value and one further data point, and asks you to fit the model. The recipe is fixed:

1. Write the general solution, $N(t) = N_0 e^{k t}$ (or the cooling form).
2. Use the initial condition $N(0) = N_0$ to fix the front constant.
3. Substitute the second data point and solve for $k$, which almost always requires taking a natural logarithm.
4. Substitute the now-complete model to answer the actual question (a value at a time, or a time for a value).

The verb "establish" in a NESA question means show that the proposed solution satisfies the differential equation, not just state it. To establish $N = N_0 e^{kt}$, differentiate it and substitute into $\frac{dN}{dt} = kN$ to confirm both sides agree, then note the initial value.

Reading the rate constant

The sign and size of $k$ carry meaning. A positive $k$ is growth and a negative $k$ is decay or cooling. The magnitude of $k$ sets the speed: a larger $|k|$ means a shorter half-life or doubling time. Because $\tau = \frac{\ln 2}{|k|}$, halving $|k|$ doubles the half-life.

How exam questions ask about exponential models

The wording tells you which step of the recipe is being tested:

• "Show that / establish that $N = N_0 e^{kt}$ satisfies $\frac{dN}{dt} = kN$". Differentiate the given solution and substitute; do not just assert it. "Establish" and "show" both demand the verification.
• "A quantity changes at a rate proportional to ..." then "set up a differential equation". Write $\frac{dN}{dt} = kN$ (or the cooling form with the ambient term). Recognising the proportional-rate phrasing is the whole task.
• "Find $k$ / the growth rate / the decay constant". Substitute the second data point into the solution and solve, which needs a natural logarithm. Watch the sign: decay and cooling give $k < 0$.
• "Find the value at time $t$" or "how much remains after ...". Substitute $t$ into the fully fitted model.
• "Find the time for the quantity to reach / halve / double". Set the model equal to the target value and solve for $t$ with a logarithm.
• "Find the half-life / doubling time". Use $\tau = \frac{\ln 2}{|k|}$ or $T_d = \frac{\ln 2}{k}$, or set $N = \tfrac{1}{2}N_0$ (or $2N_0$) and solve.
• "What temperature does the body approach?" The limit as $t \to \infty$, which for cooling is the ambient $T_a$, not zero (stage 4 above).

Almost every question reduces to the four-step recipe: write the solution, fix $N_0$, find $k$ from a second point, then answer. The logarithm appears at the "solve for $k$" or "solve for $t$" step, and it is the natural log $\ln$, never $\log_{10}$.