Question

Show that the maximum fractional error in the product of two quantities is equal to the sum of the
fractional errors in the individual quantities.

​

Answer 1

Answer:

Explanation:

Here is your answer

Answer 2

Answer:

Maximum % error in $\mathrm{x}$ = maximum % error in $\mathrm{a}$+ maximum % error in b.

                                                   $\\&\frac{\Delta x}{x}=\frac{\Delta a}{a}+\frac{\Delta b}{b}\end{aligned}$

Explanation:

Suppose a result $\mathrm{x}$ is obtained by the product of two quantities say $\mathrm{a}$ and $\mathrm{b}$

i.e. $\mathbf{x}=\mathrm{a} \times \mathrm{b}$  (1)

Let $\Delta \mathrm{a} and \Delta \mathrm{b}$ are absolute errors in the measurement of a and b. $\Delta \mathrm{x}$ be the corresponding absolute error in $\mathrm{x} .$

$\therefore \mathrm{x} \pm \Delta \mathrm{x}=(\mathrm{a} \pm \Delta \mathrm{a}) \mathrm{x}(\mathrm{b} \pm \Delta \mathrm{b})$

$\therefore \mathrm{x} \pm \Delta \mathrm{x}=\mathrm{ab} \pm \mathrm{a} \Delta \mathrm{b} \pm \mathrm{b} \Delta \mathrm{a} \pm \Delta \mathrm{a} \Delta \mathrm{b} \\\therefore \mathrm{x} \pm \Delta \mathrm{x}=\mathrm{x} \pm \mathrm{a} \Delta \mathrm{b} \pm \mathrm{b} \Delta \mathrm{a} \pm \Delta \mathrm{a} \Delta \mathrm{b} \\\therefore \pm \Delta \mathrm{x}=\pm \mathrm{a} \Delta \mathrm{b} \pm \mathrm{b} \Delta \mathrm{a} \pm \Delta \mathrm{a} \Delta \mathrm{b} \ .$(2)

Dividing equation (2) by (1) we have

$\begin{aligned}\pm \frac{\Delta x}{x} &=\pm \frac{b \Delta a}{a b} \pm \frac{a \Delta b}{a b} \pm \frac{\Delta a \Delta b}{a b} \\\therefore \pm \frac{\Delta x}{x} &=\pm \frac{\Delta a}{a} \pm \frac{\Delta b}{b} \pm \frac{\Delta a}{a} \times \frac{\Delta b}{b}\end{aligned}$

$The quantities \Delta \mathrm{a} / \mathrm{a}, \Delta \mathrm{b} / \mathrm{b} and \Delta \mathrm{x} / \mathrm{x} are called relative errors in the values of \mathrm{a}, \mathrm{b} and \mathrm{x} respectively. The product of relative errors in \mathrm{a} and b i.e. \Delta \mathrm{a} \times \Delta \mathrm{b} is very small hence is neglected.$$\begin{aligned}&\therefore \pm \frac{\Delta x}{x}=\pm \frac{\Delta a}{a} \pm \frac{\Delta b}{b} \\&\therefore \frac{\Delta x}{x}=\frac{\Delta a}{a}+\frac{\Delta b}{b}\end{aligned}$

$Hence maximum relative error in \mathrm{x}= maximum relative error in \mathrm{a}+ maximum relative error in b$

Hence maximum % error in $\mathrm{x}$ = maximum % error in $\mathrm{a}$+ maximum % error in b