25. Newton's Law of Cooling and Further Growth

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

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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