Applications of polynomial functions of degree three

Basic functions: Polynomial functions

Applications of polynomial functions of degree three

You might think that in practice you will not encounter polynomials of degree three or higher. But this is a misconception: Three examples illustrate this.

A simple mathematical model of phenomena of epilepsy describes the dynamics of a neural network with \[P(t+1)=4C\cdot P(t)^3-6C\cdot P(t)^2+(1+2C)\cdot P(t),\] where \(C\) is a positive constant and \(P(t)\) is the fraction of neurons in the network is that fires at time \(t\).

The measles infection in the human body can be modeled with a third-degree function. Research indicates that the maximum number of infected cells is reached about two weeks after infection, when also the symptoms of the disease becvome visible, and that the infection persists, about three weeks. Check that the following third-degree function, whith which one can calculate the number of infected cells per microliter of plasma on day \(t\) after infection, \[f(t)=t\cdot(21-t)\cdot (t+1)\] has the properties mentioned. You can also use the below graph of \(f\).

graph of a cubic function of a measles infection

The third example is taken from chemistry: The solubility of an electrolyte in water in the presence of other substances.

Compute the solubility of \(\mathrm{PbCl}_2\) in a \(1.5\,\mathrm{M}\) solution of \(\mathrm{Pb}(\mathrm{NO}_3)_2\).
You can take \(K_s=1.2\times 10^{-5}\) for the solubility product.
Round to 2 significant digits and use the scientific E-notation because the outcome is often very small (so use 3.4E-12 instead \(3.4\times 10^{-12}\)).

Use the below calculator for computations.

For the heterogeneous equilibrium of a solid \(\mathrm{PbCl}_2\) saturated aqueous solution containing in addition \(1.5\,\mathrm{M}\) \(\mathrm{Pb}(\mathrm{NO}_3)_2\) \[\mathrm{PbCl}_2\rightleftharpoons \mathrm{Pb}^{2+}+2\mathrm{Cl}^{-}\] holds: \[K_s= \bigl[\mathrm{Pb}^{2+}]\cdot\bigl[\mathrm{Cl}^{-}\bigr]^2\] Now suppose that \(x\) mole of the electrolyte \(\mathrm{PbCl}_2\) dissolves in 1 liter of the aqueous solution. Then it follows from the stoichiometry of the equilibrium and from the presence of completely soluble \(\mathrm{Pb}(\mathrm{NO}_3)_2\) that \[\bigl[\mathrm{Pb}^{2+}\bigr]=x+1.5,\quad \bigl[\mathrm{Cl}^{-}\bigr]=2x\] and thus \[K_s= (x+1.5)\cdot (2x)^2\] Here, a cubic polynomial appears!
Since the solubility product is known \(\bigl(K_s=1.2\times 10^{-5}\bigr)\), you can find \(x\) numerically.
The solution of the equation can also be approximated by using the following neglect:
Because \(x\) is small, we may also use \(1.5\) instead of \(x+1.5\). It follows: \[K_s=6 x^2\] What we have achieved is that the equation equation in \(x\) to be solved is simple.
We have: \(x={}\) 1.4E-3

The solubility of \(\mathrm{PbCl}_2\) in a \(1.5\,\mathrm{M}\) solution of \(\mathrm{Pb}(\mathrm{NO}_3)_2\) is therefore 1.4E-3 mol/L.

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