In Specialist Mathematics, modelling involves translating real-world scenarios—where the rate of change of a variable depends on the variable itself or on time—into mathematical equations. These are known as Ordinary Differential Equations (ODEs).
A differential equation is an equation involving an unknown function and its derivatives.
* Order: The order of the DE is the highest derivative present (VCE focuses primarily on first-order DEs).
* Solution: A function that satisfies the DE for all values in its domain.
* General Solution: A solution containing an arbitrary constant $C$, representing a family of curves.
* Particular Solution: A specific solution where $C$ is determined using given conditions.
KEY TAKEAWAY: In modelling, the derivative $\frac{dy}{dt}$ represents the rate of change of the quantity $y$ with respect to time $t$. If the rate is increasing, $\frac{dy}{dt} > 0$; if decreasing, $\frac{dy}{dt} < 0$.
The first step in modelling is converting a written description into a derivative statement.
| Verbal Description | Mathematical Equation |
|---|---|
| The rate of change of $N$ is proportional to $N$. | $\frac{dN}{dt} = kN$ |
| The rate of change of $V$ is inversely proportional to $t$. | $\frac{dV}{dt} = \frac{k}{t}$ |
| The rate of change of $x$ is proportional to the square root of $x$. | $\frac{dx}{dt} = k\sqrt{x}$ |
| The rate of change of $T$ is proportional to the difference between $T$ and $A$. | $\frac{dT}{dt} = k(T - A)$ |
COMMON MISTAKE: Be careful with the sign of the proportionality constant $k$. If a quantity is “decaying” or “decreasing,” ensure you account for the negative sign, either by stating $k < 0$ or by writing the equation as $\frac{dy}{dt} = -ky$ (where $k > 0$).
To find a unique solution to a differential equation, extra information is required to solve for the constant of integration.
An initial value is a known data point at the start of the process, usually at $t = 0$.
* Example: If a population starts at 500, then $P(0) = 500$.
A boundary value provides information at a specific point or “boundary” of the system, not necessarily at the start.
* Example: A chemical reaction ends when the concentration $C = 0$ at $t = 10$.
EXAM TIP: VCAA often awards marks for the correct separation of variables. Even if you cannot complete the integration, ensure you show the variables separated with their respective differentials ($dy$ and $dx$).
Used for population growth, radioactive decay, and compound interest.
* Model: $\frac{dy}{dt} = ky$
* Solution: $y = Ae^{kt}$ (where $A$ is the initial value $y_0$)
The rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature.
* Model: $\frac{dT}{dt} = -k(T - T_a)$
* Solution: $T = T_a + (T_0 - T_a)e^{-kt}$
The rate of change of a substance $Q$ in a tank is the difference between the rate at which it enters and the rate at which it leaves.
$$\frac{dQ}{dt} = \text{Rate In} - \text{Rate Out}$$
$$\frac{dQ}{dt} = (c_{in} \times r_{in}) - \left(\frac{Q(t)}{V(t)} \times r_{out}\right)$$
Where $c$ is concentration, $r$ is flow rate, and $V$ is volume.
In Unit 4, acceleration is often given as a function of velocity $v$ (e.g., air resistance).
* To find $v$ in terms of $t$: Use $a = \frac{dv}{dt}$
* To find $v$ in terms of $x$: Use $a = v\frac{dv}{dx}$
VCAA FOCUS: Mixing problems and kinematics involving $v\frac{dv}{dx}$ are frequent high-mark questions in Exam 2. Practice setting up these equations from worded descriptions carefully.
Once a solution is found, students must be able to:
1. Predict future values: Substitute a value of $t$ to find $y$.
2. Find limits: Determine the “steady state” or “carrying capacity” by calculating $\lim_{t \to \infty} y(t)$.
3. Sketch the solution: Use the DE to identify stationary points (where $\frac{dy}{dt} = 0$) and the behavior of the gradient.
STUDY HINT: When solving $\frac{dy}{dt} = k(y - A)$, the solution involves a logarithm: $\ln|y - A| = kt + C$. Always remember the modulus signs and consider the physical context to determine if the expression inside the log is positive or negative.
