By the end of this week you should be able to:
describe the modelling cycle and apply it to a simple real-world situation;
identify the dimensions of physical quantities and check dimensional homogeneity;
derive and solve a simple first-order ODE arising from a rate law;
interpret the parameters of an exponential model in context.
Mathematical modelling is the process of translating a real-world phenomenon into the language of mathematics, analysing the result, and then translating back to make predictions or gain understanding.
A mathematical model is a set of equations, together with assumptions and parameter values, that describes or approximates the behaviour of a real-world system.
No model is a perfect description of reality — and that is fine. A model is a tool, judged by whether it is useful for the question at hand. As the statistician George Box put it: “All models are wrong, but some are useful.”
Building and using a model typically follows a cycle:
Observe: identify the phenomenon and the question you want to answer.
Simplify: make assumptions to reduce the problem to something tractable.
Formulate: express the assumptions as equations.
Solve: find solutions analytically or numerically.
Interpret: translate the mathematical output back into the real-world context.
Validate: compare predictions with data or known behaviour.
Refine: relax assumptions, add complexity, and repeat.
We will follow this cycle throughout the course, starting with very simple models and building up to richer ones.
Every physical quantity has dimensions — the type of measurement it represents — and is expressed in particular units. The most common dimensions are:
| Dimension | Symbol | SI unit |
|---|---|---|
| Length | \(L\) | metre (m) |
| Mass | \(M\) | kilogram (kg) |
| Time | \(T\) | second (s) |
| Temperature | \(\Theta\) | kelvin (K) |
We write \([Q]\) for the dimensions of a quantity \(Q\). For example, \([\text{speed}] = L\,T^{-1}\), i.e. metres per second.
Dimensional homogeneity: every term in a valid equation must have the same dimensions. You can use this as a quick sanity check on any formula.
Newton’s second law states \(F = ma\). Check dimensional homogeneity.
\([F] = M\,L\,T^{-2}\) (force, in newtons), \([m] = M\), \([a] = L\,T^{-2}\). So \([ma] = M \cdot L\,T^{-2} = M\,L\,T^{-2}\).
The period of a simple pendulum of length \(\ell\) in gravitational field \(g\) is claimed to be \[T = 2\pi\sqrt{\frac{\ell}{g}}.\] Check dimensional homogeneity. What are the dimensions of \(g\)?
Suppose a population of bacteria doubles every hour under ideal conditions. Let \(N(t)\) denote the population size at time \(t\) (measured in hours).
The simplest assumption is that the rate of change of the population is proportional to the current population: \[\ddt{N} = r\,N,\] where \(r > 0\) is the growth rate (units: per hour, i.e. \([r] = T^{-1}\)). This is a first-order ordinary differential equation (ODE).
Separate and integrate: \[\frac{\dd N}{N} = r\,\dd t \implies \ln N = r\,t + C \implies N(t) = N_0\,e^{rt},\] where \(N_0 = N(0)\) is the initial population.
Suppose \(N_0 = 100\) and the population doubles every hour, so \(N(1) = 200\). Find \(r\).
From \(200 = 100\,e^{r \cdot 1}\) we get \(r = \ln 2 \approx 0.693\) per hour. The population at time \(t\) is \(N(t) = 100\,e^{0.693\,t}\).
\(N_0\): the initial value — set the scale of the population.
\(r > 0\): exponential growth; \(r < 0\): exponential decay; \(r = 0\): constant.
Doubling time: the time \(t_d\) such that \(N(t_d) = 2N_0\) is \(t_d = \ln 2 / r\).
Half-life: for decay (\(r < 0\)), the time to halve is \(t_{1/2} = \ln 2 / |r|\).
The same equation \(\ddt{N} = rN\) models radioactive decay (with \(r < 0\)), continuously-compounded interest, and Newton’s law of cooling (with a shift). This universality is one reason ODEs are so powerful.
A radioactive isotope has a half-life of 5730 years (carbon-14).
Find the decay rate \(r\).
What fraction of the original material remains after 1000 years?
Write a sentence explaining what the sign of \(r\) means physically.
A cup of coffee is at 90°C in a room at 20°C. Assume Newton’s law of cooling: \[\ddt{T} = -k(T - T_{\text{room}}),\quad k > 0.\] Let \(\theta = T - T_{\text{room}}\). Show that \(\theta(t) = \theta_0\,e^{-kt}\) and hence \(T(t) = T_{\text{room}} + (T_0 - T_{\text{room}})\,e^{-kt}\).