Question

If $Q = \dfrac{{{X^n}}}{{{Y^m}}}$ and $\Delta X$ is absolute error in the measurement of $X$, $\Delta Y$ is absolute error in the measurement of $Y$, then the absolute error $\Delta Q$ in $Q$ is
A) $\Delta Q = \pm \left( {n\dfrac{{\Delta X}}{X} + m\dfrac{{\Delta Y}}{Y}} \right)$
B) $\Delta Q = \pm \left( {n\dfrac{{\Delta X}}{X} + m\dfrac{{\Delta Y}}{Y}} \right)Q$
C) $\Delta Q = \pm \left( {n\dfrac{{\Delta X}}{X} - m\dfrac{{\Delta Y}}{Y}} \right)$
D) $\Delta Q = \pm \left( {n\dfrac{{\Delta X}}{X} - m\dfrac{{\Delta Y}}{Y}} \right)Q$

Answer 1

In this solution, we will use the principles of error propagation. The absolute error in Q will depend on its dependence on X and Y as well as their absolute errors.

Complete step by step answer:
We’ve been given the relation of Q as $Q = \dfrac{{{X^n}}}{{{Y^m}}}$. This relation can be broken down into two parts to calculate the error propagation and determine the absolute error $\Delta Q$.
In the formula of $Q = \dfrac{{{X^n}}}{{{Y^m}}}$, let us take the natural log on both sides of the equation, which gives us
$\ln Q = n\ln X - m\ln Y$
Differentiating this equation, we get
$\dfrac{{dq}}{Q} = \dfrac{n}{X}dX - \dfrac{m}{Y}dY$
We can write the small elements $dX$ and $dY$ as $\Delta X$ and $\Delta Y$. These correspond to the errors of the measurement of X and Y. However errors always add in any case and they cannot be subtracted. So, we can write
$\dfrac{{\Delta Q}}{Q} = \dfrac{n}{X}\Delta X + \dfrac{m}{Y}\Delta Y$
Since we want to find the absolute error of Q i.e. $\Delta Q$, we can rearrange the above equation as
$\Delta Q = \left( {n\dfrac{{\Delta X}}{X} + m\dfrac{{\Delta Y}}{Y}} \right)Q$
Which is the absolute error in Q. Hence the correct choice is option (B).

Note: The general steps in calculating the errors for different formulae involving ratios and powers involves taking the natural log of the equation and then differentiating it. Here we have assumed that the quantities X and Y are independent of each other. In general, it can be helpful to know the propagation of errors for such formulae. For terms in ratios, the relative errors add up so if Q was equal to $Q = \dfrac{X}{Y}$ the relation of the relative errors would be $\dfrac{{\Delta Q}}{Q} = \left( {\dfrac{{\Delta X}}{X} + \dfrac{{\Delta Y}}{Y}} \right)$. Then if the X and Y terms had corresponding powers, they would be multiplied as constant to their individual variables so
$\dfrac{{\Delta Q}}{Q} = \left( {n\dfrac{{\Delta X}}{X} + m\dfrac{{\Delta Y}}{Y}} \right)$