Exponential and logistic growth models (WACE Year 12 Mathematics Specialist Unit 4)

What this dot point is asking

SCSA wants you to set up these two models from a description, solve them (logistic via separation and partial fractions), and interpret the parameters and long-term behaviour.

The exponential model

When the rate of change is proportional to the current amount, $\tfrac{dy}{dt} = ky$. Separating variables and integrating gives

where $y_0$ is the initial value and $k$ is the constant of proportionality. Positive $k$ models unrestricted growth; negative $k$ models decay such as radioactive half-life. This model has no upper bound, which is unrealistic over long times.

Solving the exponential equation in full

It is worth seeing the separation of variables that produces the exponential solution, because SCSA can ask for the derivation, not just the formula. Starting from $\dfrac{dy}{dt} = ky$, separate the variables to $\dfrac{1}{y}\,dy = k\,dt$ and integrate both sides: $\ln|y| = kt + c$. Exponentiating gives $|y| = e^{kt + c} = e^c e^{kt}$, so $y = A e^{kt}$ where $A = \pm e^c$ absorbs the constant. Applying the initial condition $y(0) = y_0$ fixes $A = y_0$, giving $y = y_0 e^{kt}$. The same steps, with a sign, handle decay, where the half-life satisfies $T = \dfrac{\ln 2}{|k|}$.

The logistic model

Real populations are limited by resources. The logistic equation adds a braking factor:

where $M$ is the carrying capacity. When $P$ is small the bracket is near $1$ and growth is nearly exponential; as $P$ approaches $M$ the bracket approaches $0$ and growth slows to a halt.

Equilibria and the S-curve

Solving the logistic equation uses separation of variables with a partial-fraction decomposition of $\tfrac{1}{P(1 - P/M)}$, leading to a solution of the form $P = \tfrac{M}{1 + A e^{-kt}}$.

Interpreting the parameters

Each symbol in the logistic solution has a clear meaning that examiners ask students to read off. The carrying capacity $M$ is the horizontal asymptote the curve rises toward, the long-run population the environment can sustain. The growth-rate constant $k$ controls how fast the S-curve climbs: a larger $k$ gives a steeper rise and a faster approach to $M$. The constant $A$ is fixed by the initial condition through $A = \dfrac{M - P_0}{P_0}$, so a small starting population gives a large $A$ and a long slow phase before the rapid middle growth. Sketching the solution means drawing the asymptote at $P = M$, the starting value $P_0$, and the S-shape with its inflection at $P = \dfrac{M}{2}$.

Choosing between the two models

Deciding which model a context calls for is itself examinable. Use the exponential model when nothing limits the growth, for example a short-term investment at compound interest or radioactive decay, where the rate stays proportional to the current amount forever. Use the logistic model when the context names or implies a ceiling, such as the maximum population an ecosystem can support or the saturation level of a market. A useful diagnostic is the early behaviour: both models look exponential while the quantity is small, but only the logistic model bends over and levels off, so any mention of a limit, saturation or carrying capacity points to logistic growth.