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
$\phantom{\rule{2em}{0ex}}v=128{e}^{1.25t}\phantom{\rule{2em}{0ex}}\left(t\ge 0\right),$
We will look at some of the properties and modelling implications of this function. Consider first the value of $v$ when $t=0$ :
$\phantom{\rule{2em}{0ex}}v=128{e}^{0}=128=4$
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 growthtoalimit function is
$\phantom{\rule{2em}{0ex}}y=ab{e}^{kx}\phantom{\rule{2em}{0ex}}\left(C\le x\le D\right)$
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 $x=0,y=ab{e}^{0}=ab$ . As $x$ increases the exponential factor ${e}^{kx}$ gets smaller, so $y$ will increase from the value $ab$ but at an everdecreasing 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 $ab$ 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 nonzero time is needed also to evaluate the third parameter $k$ .
As an example we will obtain a function to describe a foodlimited 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 growthtoalimit function for the bacteria population, with time measured in hours
$\phantom{\rule{2em}{0ex}}P=ab{e}^{kt}\phantom{\rule{2em}{0ex}}\left(0\le t\le 18\right).$
When $t$ = 0 (the start of counting), $P$ = 300. Since the general form gives $P=ab$ when $t$ = 0, this means that
$\phantom{\rule{2em}{0ex}}ab=300.$
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 $ab$ , 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
$\phantom{\rule{2em}{0ex}}600=40003700{e}^{0.5k}$
$\phantom{\rule{2em}{0ex}}3400=3700{e}^{0.5k}$
$\phantom{\rule{2em}{0ex}}\frac{3400}{3700}={e}^{0.5k}$
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 growthtoalimit behaviour. Finally the required function may be written
$\phantom{\rule{2em}{0ex}}P=40003700{e}^{0.1691t}\phantom{\rule{2em}{0ex}}\left(0\le t\le 18\right).$
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.
Task!
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.

First find the modelling equation:
Start with
$\phantom{\rule{2em}{0ex}}P=ab{e}^{kt}\phantom{\rule{2em}{0ex}}\left(0\le t\le 10\right).$
where $P$ is the number of members of the population at time $t$ years. The given data requires that $a$ is 12000 and that $ab=3000$ , so $b=9000.$
The corresponding curve must pass through ( $t=1,\phantom{\rule{1em}{0ex}}P=6000$ ) so
$\phantom{\rule{2em}{0ex}}6000=120009000{e}^{k}$
$\phantom{\rule{2em}{0ex}}{\text{e}}^{k}=\frac{120006000}{9000}=\frac{2}{3}\phantom{\rule{2em}{0ex}}\text{so}\phantom{\rule{1em}{0ex}}{\text{e}}^{kt}={\left({\text{e}}^{k}\right)}^{t}={\left(\frac{2}{3}\right)}^{t}$ (using Rule 3b, Table 1, page 42)
So the population function is
$\phantom{\rule{2em}{0ex}}P=120009000{\left(\frac{2}{3}\right)}^{t}\phantom{\rule{1em}{0ex}}\left(0\le t\le 10\right)$ .
Note that $P$ (10) according to this formula is approximately 11840, which is reasonably close to the required value of 12000.

Now sketch this function: