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.
