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 m s 1 is the instantanous speed at time t s. An appropriate function is

v = 12 8 e 1.25 t ( t 0 ) ,

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

v = 12 8 e 0 = 12 8 = 4

This means the function predicts that the parachutist is moving at 4 m 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 1.25 t 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 m 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 m 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 m s 1 . The amount being subtracted from 12, through the term 8 e 1.25 t , is close to 8 because e 0 = 1 . As t increases the value of 8 e 1.25 t 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 :

{ Graph of a parachutist's speed against time}

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

y = a b e k x ( C x D )

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 k x gets smaller, so y will increase from the value a b but at an ever-decreasing rate. As b e k x 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

P = a b e k t ( 0 t 18 ) .

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

a b = 300.

The limit of P as t gets large, according to the general form P = a b k t , 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

600 = 4000 3700 e 0.5 k

3400 = 3700 e 0.5 k

3400 3700 = e 0.5 k

Taking natural logs of both sides:

ln 3400 3700 = 0.5 k so k = 2 ln ( 34 37 ) = 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

P = 4000 3700 e 0.1691 t ( 0 t 18 ) .

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:

    Start with

    P = a b e k t ( 0 t 10 ) .

    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 , P = 6000 ) so

    6000 = 12000 9000 e k

    e k = 12000 6000 9000 = 2 3 so e k t = ( e k ) t = 2 3 t (using Rule 3b, Table 1, page 42)

    So the population function is

    P = 12000 9000 2 3 t ( 0 t 10 ) .

    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:

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