Growth, decay, inflation and Newton's law of cooling

The contexts named in the standard

• The standard lists the applications explicitly: growth and decay, inflation, Newton's Law of Cooling and similar situations.
• All of them are the same differential equation dressed in different words, so recognising the structure is worth more than memorising four separate models.

Exponential growth and decay

• The words to look for: "the rate of change is proportional to the amount present".

$$\frac{dP}{dt} = kP \quad \Longrightarrow \quad P = P_0e^{kt}$$

• Each part:
  • $P_0$ — the initial amount, the value at $t = 0$
  • $k$ — the rate constant; positive for growth, negative for decay
• Note that a decay problem can be written either as $\frac{dP}{dt} = -kP$ with $k$ positive, or as $\frac{dP}{dt} = kP$ with $k$ negative. Read the question to see which convention it uses.

Half-life and doubling time

• Half-life — the time for a quantity to fall to half its value:

$$\frac{P_0}{2} = P_0e^{kt} \quad \Rightarrow \quad t_{1/2} = \frac{\ln\frac{1}{2}}{k} = \frac{-\ln 2}{k}$$

• Doubling time — the time to double:

$$t_2 = \frac{\ln 2}{k}$$

• Both are independent of the starting amount, because $P_0$ cancels. That is the defining feature of exponential change.

Inflation

• Inflation is compound growth applied to prices, and it uses the same model:

$$\frac{dC}{dt} = rC \quad \Longrightarrow \quad C = C_0e^{rt}$$

• $r$ is the inflation rate as a decimal — 3% per year is $r = 0.03$.
• A New Zealand context: if the Reserve Bank of New Zealand holds inflation at 2% per year, an item costing $40 today is modelled as costing $40e^{0.02t}$ dollars in $t$ years — about $48.86 after 10 years.
• Note the distinction between this continuous model and the discrete compound-interest formula $C = C_0(1 + r)^t$. They give slightly different answers; use whichever the question sets up.

Newton's law of cooling

• The statement: the rate at which an object cools is proportional to the difference between its temperature and the surrounding temperature.

$$\frac{dT}{dt} = -k\left(T - T_s\right)$$

• Each part:

  • $T$ — the object's temperature at time $t$
  • $T_s$ — the surrounding (ambient) temperature, a constant
  • $k$ — a positive cooling constant

• The key structural point: it is the difference $T - T_s$ that decays exponentially, not $T$ itself.

• Solving it by separation:

$$\int \frac{1}{T - T_s}\,dT = \int -k\,dt$$

$$\ln\left|T - T_s\right| = -kt + c$$

$$T - T_s = Ae^{-kt}$$

$$T = T_s + Ae^{-kt}$$

• $A$ is the initial temperature difference, $T_0 - T_s$.
• As $t \to \infty$, $e^{-kt} \to 0$, so $T \to T_s$ — the object approaches room temperature but never quite reaches it. That limiting value is the horizontal asymptote of the cooling curve.

The general method for an applied question

• Identify the model from the wording — proportional to the amount, or to a difference.
• Write the differential equation in symbols.
• Separate and integrate to get the general solution.
• Use the first condition (usually at $t = 0$) to find the constant $A$.
• Use the second condition to find $k$, which needs logarithms.
• Answer the question, with units and in context.

Interpreting $k$

• A larger $|k|$ means faster change. A hot drink in a cold room has a larger $k$ than one in a warm room.
• $k$ has units of per unit time — per second, per year — and this is worth stating.
• The sign carries meaning. A positive $k$ in a growth model and a negative one in a decay model are not interchangeable, and a sign error inverts the entire prediction.