Establish and solve differential equations of the form $\frac{dN}{dt} = k N$ and $\frac{dT}{dt} = k(T - T_a)$ and apply them to growth, decay and Newton's law of cooling

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.

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 $T_d = \frac{\ln 2}{k}$.

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 $\frac{d}{dt}((T_0 - T_a) e^{k t}) = k (T - T_a)$.

Worked example: radioactive decay

A radioactive sample has a half-life of $20$ years and an initial mass of $80$ grams. How much remains after $50$ years?

Use $N(t) = N_0 \cdot (1/2)^{t / \tau}$ with $N_0 = 80$, $\tau = 20$, $t = 50$.

$N(50) = 80 \cdot (1/2)^{50/20} = 80 \cdot (1/2)^{2.5}$.

$(1/2)^{2.5} = 2^{-2.5} = \frac{1}{2^{2.5}} = \frac{1}{4 \sqrt{2}} \approx 0.1768$.

$N(50) \approx 80 \cdot 0.1768 \approx 14.1$ grams.

Equivalently, find $k$ from $\tau = \frac{\ln 2}{|k|}$, giving $k = -\frac{\ln 2}{20}$, then use $N = N_0 e^{k t}$.

Worked example: limited growth via Newton's form

A cold drink at $5$°C is placed in a $25$°C room. After $10$ minutes it has warmed to $15$°C. When will it reach $20$°C?

$T(t) = 25 + (5 - 25) e^{k t} = 25 - 20 e^{k t}$.

Use $T(10) = 15$: $15 = 25 - 20 e^{10 k}$, so $e^{10 k} = \frac{1}{2}$, giving $k = -\frac{\ln 2}{10}$.

Set $T(t) = 20$: $20 = 25 - 20 e^{k t}$, so $e^{k t} = \frac{1}{4}$.

$k t = \ln \frac{1}{4} = -2 \ln 2$, so $t = \frac{-2 \ln 2}{k} = \frac{-2 \ln 2}{-\ln 2 / 10} = 20$ minutes.

Common traps

Confusing the two equations. Pure exponential change uses $\frac{dN}{dt} = k N$. Cooling involves a constant ambient term: $\frac{dT}{dt} = k(T - T_a)$. Do not forget the $-T_a$.

Wrong sign of $k$. For decay and cooling, $k$ is negative. The exponent $k t$ should drive the function towards its limit, not away from it.

Linear versus exponential time variable. $N(t) = N_0 \cdot 2^{-t/\tau}$ and $N(t) = N_0 e^{k t}$ describe the same decay only if $k = -\frac{\ln 2}{\tau}$.

Dropping $T_a$ at the end. In a cooling problem the final temperature approaches $T_a$, not zero. The solution always includes a constant term equal to the ambient temperature.

Using $\log$ instead of $\ln$. In Maths Advanced, the natural logarithm $\ln$ (base $e$) is the inverse of $e^x$. Switching to $\log_{10}$ adds an unnecessary factor.

In one sentence

When a rate of change is proportional to a quantity (with or without an ambient offset), the result is an exponential function of time, fitted to the data by the initial value and one further data point.