# Newton's Law of Cooling and Further Growth

> Things that grow or decay towards a level that is not zero. Model cooling coffee and warming drinks with N = P + Ae^(kt).

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/further-growth-and-decay/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The cup of tea
A cup of tea is poured at $90^\circ$C in a room at $20^\circ$C. At first it cools fast. As it gets closer to room temperature it cools more and more slowly. It never drops below $20^\circ$C, no matter how long you wait.

Isaac Newton noticed the rule behind this: the tea cools at a rate that depends on the **gap** between its temperature and the room. A big gap means fast cooling. A small gap means slow cooling. That one idea predicts the temperature at any time.

**In short:** When a rate of change depends on the gap to a fixed value, $\dfrac{dN}{dt} = k(N - P)$, the solution is $N = P + Ae^{kt}$. Newton's law of cooling is an example: a hot drink's temperature $T = A + Be^{-kt}$ falls towards the room temperature $A$.

## Starting point: what you already know

- **Exponential growth and decay** (Mathematics Advanced): if $\dfrac{dN}{dt} = kN$ then $N = N_0e^{kt}$. Positive $k$ grows, negative $k$ decays.
- **Derivative of an exponential:** $\dfrac{d}{dt}e^{kt} = ke^{kt}$.
- **Logarithms undo exponentials:** if $e^{x} = a$ then $x = \ln a$. Also $\ln\dfrac{1}{a} = -\ln a$.
- **As $t$ gets very large,** $e^{-kt} \to 0$ when $k > 0$.

## The new rule

In plain growth the rate depends on $N$. Here it depends on the gap between $N$ and a fixed level $P$:

### Key idea: Growth or decay towards a level
$$\frac{dN}{dt} = k(N - P) \quad\text{has solution}\quad N = P + Ae^{kt}$$
For cooling, $k$ is negative. Newton's law of cooling is often written $\dfrac{dT}{dt} = -k(T - A)$ with $k > 0$, giving $T = A + Be^{-kt}$, where $A$ is the room temperature.

**Check it works.** If $N = P + Ae^{kt}$ then $\dfrac{dN}{dt} = kAe^{kt}$. Also $N - P = Ae^{kt}$. So $\dfrac{dN}{dt} = k(N - P)$. It fits.

What the letters mean:

- $P$ is the level it settles at (room temperature, for example).
- $A$ is the starting gap, found from $N$ when $t = 0$.
- $k$ sets how fast it happens, found from a second reading.

## Solving a cooling problem

### Worked example: The cup of tea
Tea at $90^\circ$C in a $20^\circ$C room is $60^\circ$C after 5 minutes. When does it reach $40^\circ$C?

**Set up:** $T = 20 + Be^{-kt}$.

**Find B:** at $t = 0$, $90 = 20 + B$, so $B = 70$.

**Find k:** $60 = 20 + 70e^{-5k}$, so $e^{-5k} = \dfrac{4}{7}$, giving $k = \dfrac{1}{5}\ln\dfrac{7}{4} \approx 0.112$.

**Answer:** $40 = 20 + 70e^{-kt}$, so $e^{-kt} = \dfrac{2}{7}$ and $t = \dfrac{\ln 3.5}{k} \approx 11.2$ minutes.

*Diagram: Cooling curve. Temperature starts at 90 and falls quickly, then more slowly, levelling off towards the dashed line at 20 degrees.*
*The tea cools towards room temperature but never goes below it.*

### Watch out: Do not forget P
$T = Be^{-kt}$ would cool the tea all the way to $0^\circ$C. The room level $P$ must be added on.

### Calculator tip: Finding k with ln
From $60 = 20 + 80e^{-10k}$, get $e^{-10k} = 0.5$, then type `ln(0.5) ÷ -10` to find $k \approx 0.0693$. Store $k$ with `STO` so later answers are not rounded early.

**Mind map**

```text
Growth towards a level
  Rule
    Rate depends on the gap
    dN/dt = k(N - P)
  Solution
    N = P + Ae^(kt)
    Check by differentiating
  Constants
    P: level it settles at
    A: starting gap
    k: from a second reading
  Long run
    e^(-kt) goes to 0
    N goes to P
  Uses
    Cooling
    Warming
    Drug levels
```

### Year 11: Warming up
A drink at $4^\circ$C is put in a $25^\circ$C room. Now the drink starts **below** the level, so $T = 25 + Be^{-kt}$ with $B = 4 - 25 = -21$.

$T = 25 - 21e^{-kt}$. The curve rises towards 25 instead of falling. The same formula covers both cases: the sign of the starting gap decides the direction.

### Year 11: What happens in the long run?
As $t \to \infty$, $e^{-kt} \to 0$, so $T \to A$. Exam questions often ask for this "limiting" value, or for a sketch with the horizontal asymptote $T = A$.

**The rate at a moment.** Put the temperature into the rule itself. If $\dfrac{dT}{dt} = -0.1(T - 20)$, then at $T = 50$ it cools at $0.1 \times 30 = 3^\circ$C per minute.

**Halving gaps.** If the gap from room temperature halves every 5 minutes, then a gap of 80 becomes 40, 20, 10 after 5, 10 and 15 minutes. Spotting this saves solving with logs. In general, after each equal time the gap is multiplied by the same factor: if 275 becomes 175 in 10 minutes, after 20 minutes the gap is $275 \times \left(\frac{175}{275}\right)^2$.

### Year 12: Solving the differential equation directly
Start from $\dfrac{dT}{dt} = -k(T - A)$ and flip both sides: $\dfrac{dt}{dT} = \dfrac{-1}{k(T - A)}$.

Integrate with respect to $T$: $t = -\dfrac{1}{k}\ln|T - A| + c$.

Rearrange: $\ln|T - A| = -k(t - c)$, so $T - A = Be^{-kt}$ for some constant $B$.

This is how the formula is found, not just checked. The chapter on differential equations does the same for other rates.

**Equilibrium.** A rule like $\dfrac{dP}{dt} = 0.04P - 200$ is $0.04(P - 5000)$, growth away from the level 5000. The **equilibrium** is where the rate is zero: $P = 5000$. Above it the population grows; below it, it falls.

**Mixing tanks.** A 100 L tank has brine (10 g of salt per litre) flowing in at 3 L/min and mixed liquid flowing out at 3 L/min. Salt comes in at 30 g/min and leaves at $\dfrac{S}{100} \times 3$ g/min, so $\dfrac{dS}{dt} = 30 - 0.03S = -0.03(S - 1000)$. In the long run $S \to 1000$ g, the same concentration as the incoming brine.

## Quick summary

- When the rate depends on a gap, $\dfrac{dN}{dt} = k(N - P)$, the solution is $N = P + Ae^{kt}$.
- Newton's law of cooling: $T = A + Be^{-kt}$, with room temperature $A$.
- Find $B$ from the start, then $k$ from a second reading, using logs.
- In the long run $T \to A$. The graph has a horizontal asymptote there.

## Quiz

Try these questions from the chapter quiz.

1. In Newton's law of cooling, the rate of cooling depends on:
   - A. the time only
   - B. the gap between the object and the room
   - C. the room temperature only
   - D. nothing, it is constant
   - Answer: B. the gap between the object and the room. $\dfrac{dT}{dt} = -k(T - A)$.

2. $T = 20 + 70e^{-kt}$. What is $T$ when $t = 0$?
   - A. 20
   - B. 90
   - C. 70
   - D. 50
   - Answer: B. 90. $e^0 = 1$, so $20 + 70 = 90$.

3. $T = 20 + 70e^{-kt}$ with $k > 0$. What does $T$ approach in the long run?
   - A. 20
   - B. 0
   - C. 70
   - D. 90
   - Answer: A. 20. $e^{-kt} \to 0$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/further-growth-and-decay/#quiz