[EXAMPLES] Modelling with first-order equations

Newton's law of cooling

If $A$ is a constant ambient temperature, the rate of change of the temperature difference is proportional to that difference:

\frac{d}{dt}(T-A)=-k(T-A), \qquad k>0.

Equivalently,

T'=-k(T-A).

If $t$ is measured in hours, then $k$ has units of $\mathrm{h}^{-1}$ ; the same inverse-time units apply to any first-order rate constant.

Separation or the linear method gives

T(t)=A+(T_0-A)e^{-kt}.

The difference $T-A$ keeps its sign and decays exponentially. A hotter object cools monotonically, a colder object warms monotonically and

\lim_{t\to\infty}T(t)=A.

As a forensic application, a body's temperature is measured at 11:30 am and again an hour later. Let $t=0$ be 11:30 am. The data are

T(0)=24.5,\qquad T(1)=24.0,\qquad A=20.

Using the first measurement,

T(t)=20+4.5e^{-kt}.

The second measurement gives

24=20+4.5e^{-k} \quad\Longrightarrow\quad e^{-k}=\frac89 \quad\Longrightarrow\quad k=\ln\frac98.

If death occurred $\tau$ hours before 11:30 am, the body was at normal temperature $36.5^\circ\mathrm{C}$ at $t=-\tau$ :

36.5=20+4.5e^{k\tau}.

Therefore

\tau =\frac{\ln(16.5/4.5)}{k} =\frac{\ln(11/3)}{\ln(9/8)} \approx11.03\text{ hours}.

The estimated time of death is about 12:28 am. This estimate assumes a constant ambient temperature and cooling coefficient over the whole period; the ODE alone cannot verify those assumptions.

[exam]

[1]:Translate a rate law

A salt solution

Let $S(t)$ be the amount of salt in grams and $V(t)$ the volume in litres. For a well-stirred tank,

\text{salt rate in} =(\text{inflow rate})(\text{inflow concentration}),

and

\text{salt rate out} =(\text{outflow rate})\frac{S(t)}{V(t)}.

Suppose $S(0)=50$ , $V(0)=300$ , the inflow is $3$ litres per minute at $2$ grams per litre and the outflow is $2$ litres per minute. Assuming the tank remains well mixed and no salt is created or removed except by the streams, the volume is

V(t)=300+t,

and conservation gives

S'=6-\frac{2S}{300+t}.

The equation is linear:

S'+\frac{2}{300+t}S=6.

An integrating factor is $(300+t)^2$ , so

\bigl((300+t)^2S\bigr)'=6(300+t)^2.

Integrating gives $(300+t)^2S=2(300+t)^3+C$ . At $t=0$ , $C=300^2(50)-2(300)^3=-49\,500\,000$ . Thus

S(t)=2(300+t)-\frac{49\,500\,000}{(300+t)^2} \quad\text{grams}.

The concentration tends to $2$ grams per litre as $t$ increases, which matches the inflow concentration.

The volume grows because inflow exceeds outflow. The physical domain is $t\geq0$ (so $V(t)=300+t>0$ ); extending the formula to earlier times would need a separate physical justification.