16 The population density \(P\), in suitable units, of a certain bacterium at time \(t\) hours is to be modelled by a differential equation. Initially, the population density is zero, and its long-term value is \(A\).
- One simple model is to assume that the rate of change of population density is directly proportional to \(A - P\).
- Formulate a differential equation for this model.
- Verify that \(P = A \left( 1 - \mathrm { e } ^ { - k t } \right)\), where \(k\) is a positive constant, satisfies
- this differential equation,
- the initial condition,
- the long-term condition.
An alternative model uses the differential equation
$$\frac { \mathrm { d } P } { \mathrm {~d} t } - \frac { P } { t \left( 1 + t ^ { 2 } \right) } = \mathrm { Q } ( t )$$
where \(\mathrm { Q } ( t )\) is a function of \(t\).- Find the integrating factor for this differential equation, showing that it can be written in the
$$\text { form } \frac { \sqrt { 1 + t ^ { 2 } } } { t } \text {. }$$
- Suppose that \(\mathrm { Q } ( t ) = 0\).
$$\text { (i) Show that } P = \frac { A t } { \sqrt { 1 + t ^ { 2 } } } \text {. }$$
(ii) Find the time predicted by this model for the population density to reach half its longterm value. Give your answer correct to the nearest minute.
- Now suppose that \(\mathrm { Q } ( t ) = \frac { t \mathrm { e } ^ { - t } } { \sqrt { 1 + t ^ { 2 } } }\).
$$\text { Show that } \left. P = \frac { A t - t e ^ { - t } } { \sqrt { 1 + t ^ { 2 } } } \text {. [You may assume that } \lim _ { t \rightarrow \infty } t e ^ { - t } = 0 . \right]$$
It is found that the long-term value of \(P\) is 10, and \(P\) reaches half this value after 37 minutes.
- Determine which of the models proposed in parts (c) and (d) is more consistent with these data.