In Newton's law of cooling, the rate of cooling depends on:

the gap between the object and the room

Newton's Law of Cooling and Further Growth

In short: When a rate of change depends on the gap to a fixed value, $\frac{d N}{d t} = k (N - P)$ , the solution is $N = P + A e^{k t}$ . Newton's law of cooling is an example: a hot drink's temperature $T = A + B e^{- k t}$ falls towards the room temperature $A$ .

Starting point: what you already know

Exponential growth and decay (Mathematics Advanced): if $\frac{d N}{d t} = k N$ then $N = N_{0} e^{k t}$ . Positive $k$ grows, negative $k$ decays.
Derivative of an exponential: $\frac{d}{d t} e^{k t} = k e^{k t}$ .
Logarithms undo exponentials: if $e^{x} = a$ then $x = ln a$ . Also $ln \frac{1}{a} = - ln a$ .
As $t$ gets very large, $e^{- k t} \to 0$ 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 $P$ :

Check it works. If $N = P + A e^{k t}$ then $\frac{d N}{d t} = k A e^{k t}$ . Also $N - P = A e^{k t}$ . So $\frac{d N}{d t} = k (N - P)$ . 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

The tea cools towards room temperature but never goes below it.

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 12Solving the differential equation directly

Start from $\frac{d T}{d t} = - k (T - A)$ and flip both sides: $\frac{d t}{d T} = \frac{- 1}{k ( T - A )}$ .

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

Rearrange: $ln ∣ T - A ∣ = - k (t - c)$ , so $T - A = B e^{- k t}$ 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 $\frac{d P}{d t} = 0.04 P - 200$ 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 $\frac{S}{100} \times 3$ g/min, so $\frac{d S}{d t} = 30 - 0.03 S = - 0.03 (S - 1000)$ . In the long run $S \to 1000$ g, the same concentration as the incoming brine.

Quick summary

When the rate depends on a gap, $\frac{d N}{d t} = k (N - P)$ , the solution is $N = P + A e^{k t}$ .
Newton's law of cooling: $T = A + B e^{- k t}$ , 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

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Try 3 questions now

In Newton's law of cooling, the rate of cooling depends on:
1. A the time only
2. B the gap between the object and the room
3. C the room temperature only
4. D nothing, it is constant
Show answer
B: the gap between the object and the room
$\frac{d T}{d t} = - k (T - A)$ .
$T = 20 + 70 e^{- k t}$ . What is $T$ when $t = 0$ ?
1. A 20
2. B 90
3. C 70
4. D 50
Show answer
B: 90
$e^{0} = 1$ , so $20 + 70 = 90$ .
$T = 20 + 70 e^{- k t}$ with $k > 0$ . What does $T$ approach in the long run?
1. A 20
2. B 0
3. C 70
4. D 90
Show answer
A: 20
$e^{- k t} \to 0$ .