1 Modelling with first-order equations

1.1 Applying Newton’s law of cooling

In Section 19.1 we introduced Newton’s law of cooling. The model equation is

$\frac{d\theta }{dt}=-k\left(\theta -{\theta }_{\text{s}}\right)\theta ={\theta }_0\text{at}t=0.$ $\left(5\right)$

where $\theta =\theta \left(t\right)$ is the temperature of the cooling object at time $t$ , ${\theta }_{\text{s}}$ the temperature of the environment (assumed constant) and $k$ is a thermal constant related to the object, ${\theta }_0$ is the initial temperature of the liquid.

Task!

Solve this initial value problem:

$\frac{d\theta }{dt}=-k\left(\theta -{\theta }_s\right),\theta ={\theta }_0$ at    $t=0$

Separate the variables to obtain an equation connecting two integrals:

Answer

Now integrate both sides of this equation:

Answer

Answer

The graph of $\theta$ against $t$ for $\theta ={\theta }_s+\left({\theta }_0-{\theta }_s\right){\text{e}}^{-kt}$ is shown in Figure 4 below.

Figure 4

We see that as time increases ( $t\to \infty$ ), then the temperature of the object cools down to that of the environment, that is: $\theta \to {\theta }_{\text{s}}$ .

We could have solved (5) by the integrating factor method, which you are now asked to do.

Task!

We can write the equation for Newton’s law of cooling (5) as

$\frac{d\theta }{dt}+k\theta =k{\theta }_{\text{s}},\theta ={\theta }_0\text{at}t=0$ (6)

State the integrating factor for this equation:

Answer

Multiplying (6) by this factor we find that

${\text{e}}^{kt}\frac{d\theta }{dt}+k{\text{e}}^{kt}\theta =k{\theta }_{\text{s}}{\text{e}}^{kt}\text{or, rearranging,}\frac{d}{dt}\left({\text{e}}^{kt}\theta \right)=k{\theta }_{\text{s}}{\text{e}}^{kt}$

Now integrate this equation and apply the initial condition:

Answer

Another application of first-order differential equations arises in the modelling of electrical circuits.

In Section 19.1 the differential equation for the RL circuit in Figure 5 below was shown to be

$L\frac{di}{dt}+Ri=E$

in which the initial condition is $i=0$ at $t=0$ .

Figure 5

First we write this equation in standard form { $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ } and obtain the integrating factor.

Dividing the differential equation through by $L$ gives

$\frac{di}{dt}+\frac{R}{L}i=\frac{E}{L}$

which is now in standard form. The integrating factor is ${\text{e}}^{\int \; <!--nolimits\; -->\frac{R}{L}dt}={\text{e}}^{Rt∕L}.$

Multiplying the equation in standard form by the integrating factor gives

${\text{e}}^{Rt∕L}\frac{di}{dt}+{\text{e}}^{Rt∕L}\frac{R}{L}i=\frac{E}{L}{\text{e}}^{Rt∕L}$

or, rearranging,

$\frac{d}{dt}\left({\text{e}}^{Rt∕L}i\right)=\frac{E}{L}{\text{e}}^{Rt∕L}.$

Now we integrate both sides and apply the initial condition to obtain the solution.

Integrating the differential equation gives:

${\text{e}}^{Rt∕L}i=\frac{E}{R}{\text{e}}^{Rt∕L}+C$

where $C$ is a constant so

$i=\frac{E}{R}+C{\text{e}}^{-Rt∕L}$

Applying the initial condition $i=0$ when $t=0$ gives

$0=\frac{E}{R}+C$

so that $C=-\frac{E}{R}$ .

Finally, $i=\frac{E}{R}\left(1-{\text{e}}^{-Rt∕L}\right)$ .

Note that as $t\to \infty ,$ $i\to \frac{E}{R}$ so as $t$ increases the effect of the inductor diminishes to zero.

Task!

A spherical pill with volume $V$ and surface area $S$ is swallowed and slowly dissolves in the stomach, releasing an active component. In one model it is assumed that the capsule dissolves in the stomach acids such that the rate of change in volume, $\frac{dV}{dt}$ , is directly proportional to the pill’s surface area.

1. Show that $\frac{dV}{dt}=-k{V}^{2∕3}$ where $k$ is a positive real constant and solve this given that $V=V_0$ at $t=0$ .
2. Experimental measurements indicate that for a 4 mm pill, half of the volume has dissolved after 3 hours. Find the rate constant $\text{k}$ ( $\text{m}{\text{s}}^{-1}$ ).
3. Estimate the time required for 95% of the pill to dissolve.

1. First write down the formulae for volume of a sphere ( $V$ ) and surface area of a sphere $\left(S\right)$ and so express $S$ in terms of $V$ by eliminating $r$ :

   Answer

   $V=\frac{4}{3}\pi r^{3}S=4\pi r^2$

   From the $V$ equation $r={\left(\frac{3V}{4\pi }\right)}^{1∕3}$ so $S={\left(36\pi \right)}^{1∕3}{V}^{2∕3}=k{V}^{2∕3}$ for constant $k$ .

   Now write down the differential equation modelling the solution:

   Answer

   $\frac{dV}{dt}=-k{V}^{2∕3}$ (negative to represent a decrease with time)

   Using the condition $V=V_0$ when $t=0$ , solve the differential equation:

   Answer

   Solving by separation of variables gives

   $V={\left\{\frac{1}{3}\left(C-kt\right)\right\}}^{1∕3}$

   and setting $V=V_0$ when $t=0$ means

   $V_0={\left(\frac{1}{3}C\right)}^3$ so $C=3V_0^{1∕3}$ and the solution is

   $V={\left\{V_0^{1∕3}-\frac{kt}{3}\right\}}^3$

2. Impose the condition that half the volume has dissolved after 3 hours to find $k$ :

   Answer

   $V={\left\{V_0^{1∕3}-\frac{kt}{3}\right\}}^3$

   and when $t=3$ , $V=\frac{V_0}{2}$ so

   ${\left(\frac{V_0}{2}\right)}^{1∕3}=V_0^{1∕3}-k$ and so $k=V_0^{1∕3}\left(1-{\left(0.5\right)}^{1∕3}\right)$

3. First write down the solution to the differential equation inserting the value of $k$ obtained in (b) and then use it to estimate the time to 95% dissolving:

   Answer

   $V={\left\{V_0^{1∕3}-V_0^{1∕3}\left(1-{\left(0.5\right)}^{1∕3}\right)\frac{t}{3}\right\}}^3$ i.e. $V=V_0{\left\{1-\left(1-{\left(0.5\right)}^{1∕3}\right)\frac{t}{3}\right\}}^3$

   When 95% dissolved $V=0.05V_0$ so

   $0.05V_0=V_0{\left\{1-\left(1-{\left(0.5\right)}^{1∕3}\right)\frac{t}{3}\right\}}^3$ so ${\left(0.05\right)}^{1∕3}=1-\left(1-{\left(0.5\right)}^{1∕3}\right)\frac{t}{3}$

   so

   $t=3\left\{\frac{1-{\left(0.05\right)}^{1∕3}}{1-{\left(0.5\right)}^{1∕3}}\right\}\approx 9.185\approx 9$  hr  $11$ min