### 5 Growth and decay to a limit

Consider a function intended to represent the speed of a parachutist after the opening of the parachute where $v\text{m}{\text{s}}^{-1}$ is the instantanous speed at time $t$ s. An appropriate function is

$\phantom{\rule{2em}{0ex}}v=12-8{e}^{-1.25t}\phantom{\rule{2em}{0ex}}\left(t\ge 0\right),$

We will look at some of the properties and modelling implications of this function. Consider first the value of $v$ when $t=0$ :

$\phantom{\rule{2em}{0ex}}v=12-8{e}^{0}=12-8=4$

This means the function predicts that the parachutist is moving at $4\phantom{\rule{1em}{0ex}}\text{m}{\text{s}}^{-1}$ when the parachute opens. Consider next the value of $v$ when $t$ is arbitrarily large. For such a value of $t,8{e}^{-1.25t}$ would be arbitrarily small, so $v$ would be very close to the value 12. The modelling interpretation of this is that eventually the speed becomes very close to a constant value, $12\phantom{\rule{1em}{0ex}}\text{m}{\text{s}}^{-1}$ which will be maintained until the parachutist lands.

The steady speed which is approached by the parachutist (or anything else falling against air resistance) is called the terminal velocity . The parachute, of course, is designed to ensure that the terminal velocity is sufficiently low ( $12\phantom{\rule{1em}{0ex}}\text{m}{\text{s}}^{-1}$ in the specific case we have looked at here) to give a reasonably gentle landing and avoid injury.

Now consider what happens as $t$ increases from near zero. When $t$ is near zero, the speed will be near $4\phantom{\rule{1em}{0ex}}\text{m}{\text{s}}^{-1}$ . The amount being subtracted from 12, through the term 8 ${\text{e}}^{-1.25t}$ , is close to 8 because ${e}^{0}=1$ . As $t$ increases the value of 8 ${\text{e}}^{-1.25t}$ decreases fairly rapidly at first and then more gradually until $v$ is very nearly 12. This is sketched in Figure 10. In fact $v$ is never equal to 12 but gets imperceptibly close as anyone would like as $t$ increases. The value shown as a horizontal broken line in Figure 10 is called an asymptotic limit for $v$ .

Figure 10 : The model concerned the approach of a parachutist’s velocity to terminal velocity but the kind of behaviour portrayed by the resulting function is useful generally in modelling any growth to a limit .

A general form of this type of growth-to-a-limit function is

$\phantom{\rule{2em}{0ex}}y=a-b{e}^{-kx}\phantom{\rule{2em}{0ex}}\left(C\le x\le D\right)$

where $a,b$ and $k$ are positive constants (parameters) and $C$ and $D$ represent values of the independent variable between which the function is valid. We will now check on the properties of this general function. When $x=0,y=a-b{e}^{0}=a-b$ . As $x$ increases the exponential factor ${e}^{-kx}$ gets smaller, so $y$ will increase from the value $a-b$ but at an ever-decreasing rate. As $b{e}^{-kx}$ becomes very small, $y$ , approaches the value $a$ . This value represents the limit, towards which $y$ grows. If a function of this general form was being used to create a model of population growth to a limit, then $a$ would represent the limiting population, and $a-b$ would represent the starting population.

There are three parameters, $a,b$ , and $k$ in the general form. Knowledge of the initial and limiting population only gives two pieces of information. A value for the population at some non-zero time is needed also to evaluate the third parameter $k$ .

As an example we will obtain a function to describe a food-limited bacterial culture that has 300 cells when first counted, has 600 cells after 30 minutes but seems to have approached a limit of 4000 cells after 18 hours.

We start by assuming the general form of growth-to-a-limit function for the bacteria population, with time measured in hours

$\phantom{\rule{2em}{0ex}}P=a-b{e}^{-kt}\phantom{\rule{2em}{0ex}}\left(0\le t\le 18\right).$

When $t$ = 0 (the start of counting), $P$ = 300. Since the general form gives $P=a-b$ when $t$ = 0, this means that

$\phantom{\rule{2em}{0ex}}a-b=300.$

The limit of $P$ as $t$ gets large, according to the general form $P=a-{b}^{-kt}$ , is $a$ , so $a$ = 4000. From this and the value of $a-b$ , we deduce that $b$ = 3700. Finally, we use the information that $P$ = 600 when $t$ (measuring time in hours) = 0.5. Substitution in the general form gives

$\phantom{\rule{2em}{0ex}}600=4000-3700{e}^{-0.5k}$

$\phantom{\rule{2em}{0ex}}3400=3700{e}^{-0.5k}$

$\phantom{\rule{2em}{0ex}}\frac{3400}{3700}={e}^{-0.5k}$

Taking natural logs of both sides:

$\phantom{\rule{2em}{0ex}}ln\left(\frac{3400}{3700}\right)=-0.5k$ so $k=-2ln\left(\frac{34}{37}\right)=0.1691$

Note, as a check, that $k$ turns out to be positive as required for a growth-to-a-limit behaviour. Finally the required function may be written

$\phantom{\rule{2em}{0ex}}P=4000-3700{e}^{-0.1691t}\phantom{\rule{2em}{0ex}}\left(0\le t\le 18\right).$

As a check we should substitute $t$ = 18 in this equation. The result is $P$ = 3824 which is close to the required value of 4000.

Find a function that could be used to model the growth of a population that has a value of 3000 when counts start, reaches a value of 6000 after 1 year but approaches a limit of 12000 after a period of 10 years.

1. First find the modelling equation:

$\phantom{\rule{2em}{0ex}}P=a-b{e}^{-kt}\phantom{\rule{2em}{0ex}}\left(0\le t\le 10\right).$

where $P$ is the number of members of the population at time $t$ years. The given data requires that $a$ is 12000 and that $a-b=3000$ , so $b=9000.$

The corresponding curve must pass through ( $t=1,\phantom{\rule{1em}{0ex}}P=6000$ ) so

$\phantom{\rule{2em}{0ex}}6000=12000-9000{e}^{-k}$

$\phantom{\rule{2em}{0ex}}{\text{e}}^{-k}=\frac{12000-6000}{9000}=\frac{2}{3}\phantom{\rule{2em}{0ex}}\text{so}\phantom{\rule{1em}{0ex}}{\text{e}}^{-kt}={\left({\text{e}}^{-k}\right)}^{t}={\left(\frac{2}{3}\right)}^{t}$ (using Rule 3b, Table 1, page 42)

So the population function is

$\phantom{\rule{2em}{0ex}}P=12000-9000{\left(\frac{2}{3}\right)}^{t}\phantom{\rule{1em}{0ex}}\left(0\le t\le 10\right)$ .

Note that $P$ (10) according to this formula is approximately 11840, which is reasonably close to the required value of 12000.

2. Now sketch this function: 