Applications

Basic functions: Rational functions

Applications

Rational functions are often used in mathematical models of biological processes of change. But there are also simple applications; we give three examples.

The thin lens formula for a thin lens with a focal length $f$ is

$\frac{1}{b}+\frac{1}{v}=\frac{1}{f},$ where $v$ is the distance from the object to the lens and $b$ is the distance of the image to the lens. Given an object distance $v$ and focal length $f$ you can compute the image distance $b$ .

Verify that the image distance as a function of the object distance $v$ is equivalent to the following rational function:

$b(v)=\frac{f\cdot v}{v-f}$

$\begin{aligned} \frac{1}{b}+\frac{1}{v}=\frac{1}{f} &\implies \frac{1}{b}=\frac{1}{f}-\frac{1}{v}\\ \\ &\implies \frac{1}{b}=\frac{v}{f\cdot v}-\frac{f}{f\cdot v} \\ \\ &\implies b=\frac{f\cdot v}{v-f}\end{aligned}$

In an enzyme-catalysed biochemical reaction, the reaction rate $v$ depends on the substrate concentration $s.$
The Michaelis-Menten equation for this relation is

$v=\frac{V\cdot s}{K+s}$ where $V$ is the maximum reaction rate and $K$ the Michaelis-Menten constant.
Verify that $v\approx V$ for large $s$ and $\displaystyle v\approx \frac{V}{K}\cdot s$ for small $s$ .

$\begin{aligned} v=\frac{V\cdot s}{K+s} &\implies v=\frac{V}{\displaystyle\frac{K}{s}+1}\\ \\ &\implies v \approx V \text{ for large }s \text{ because then }\frac{K}{s}\approx 0 \\ \\ v=\frac{V\cdot s}{K+s} &\implies v=\frac{\displaystyle\frac{V}{K}\cdot s}{\displaystyle1+\frac{s}{K}}\\ \\ &\implies v \approx \frac{V}{K}\cdot s \text{ for small }s\text{ because then }\frac{s}{K}\approx 0\end{aligned}$

Degree of protolysis of a weak acid In the calculation of the degree of protolysis of a weak acid, that is, the fraction of the original molecules (or ions) that donate a proton, rational functions also play a role.

Suppose that we dissolve $n$ moles of an acid $A$ in pure water up to $V$ liter, then the reaction

$A+\mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}_3\mathrm{O}^{+} +\mathrm{B^-}$ where $A$ and $B$ is an acid-base pair (for example acetic acid $\mathrm{HAc}$ with $\mathrm{Ac}^{-}$ as short notation for the acetate ion $(\mathrm{CH}_3)\!\mathrm{COO}^{-}$ . Under equilibrium conditions, the following holds:

$K_a=\frac{\bigl[\mathrm{H}_3\mathrm{O}^{+}\bigr]\cdot\bigl[B\bigr]}{\bigl[A\bigr]},$ where $K_a$ is the acidity dissociation constant (for example, $K_{\mathrm{HAc}}=1.7\times 10^{-5}$ at room temperature).

Suppose that the degree of protolysis is equal to $\alpha$ , the volume of the solution is equal to $V$ , and the original number of acid molecules is equal to $n$ , then:

$\bigl[\mathrm{H}_3\mathrm{O}^{+}\bigr]=\bigl[B^-\bigr]=\frac{n\alpha}{V}, \quad \bigl[A\bigr]=\frac{n(1-\alpha)}{V}$ It follows that:

$K_a=\frac{n}{V}\cdot \frac{\alpha^2}{1-\alpha}$ The acidity dissociation constant $K_a$ is a rational function in the degree of protolysis $\alpha$ . When $K_a$ is known, one can calculate the degree of protolysis $\alpha$ from the data.

If the degree of protolysis $\alpha$ is small, then you may use in the denominator of the above formula instead of $1-\alpha$ also the number 1. It follows then (check yourself):

$\alpha\approx \sqrt{\frac{V}{n}\cdot K_{a}}\;,\quad \text{for small }\alpha\text.$