Question

Find relative error in $Z$, given that $Z={\displaystyle \frac{1}{A^2}}$ is:

Answer 1

Hint: If two or more quantities are multiplied or divided, the sum of the relative errors of the quantities is the relative error of the result.The absolute error is inadequate due to the fact that it does not give any details regarding the importance of the error.

Complete step by step answer:
Absolute error is defined as the difference between the actual value of a quantity and the measured value . If $x$ is the actual value of a quantity and $x_0$ is the measured value, then absolute error represented by $\mathrm{\Delta }x$ can be calculated as $\mathrm{\Delta }x=x_0-x$. Relative error is the absolute error divided by the actual value of the quantity.

It is used to express how close the measured quantity is to the actual value. If $x$ is the actual value of a quantity , $x_0$is the measured value and the absolute error is $\mathrm{\Delta }x$ then, relative error can be calculated as: ${\displaystyle \frac{\mathrm{\Delta }x}{x}}={\displaystyle \frac{x_0-x}{x}}$
Let us assume $Z=P^A\times Q^B$
Let us apply log on both sides:
$\ln (Z)=\ln (P^A\times Q^B)$
$\Rightarrow \ln (Z)=\ln (P^A)+\ln (Q^B)$
Differentiating both sides with Z, we get:
${\displaystyle \frac{1}{Z}}\times {\displaystyle \frac{dZ}{dZ}}={\displaystyle \frac{d(\ln (P^A)+\ln (Q^B))}{dZ}}$
$\Rightarrow {\displaystyle \frac{1}{Z}}\times {\displaystyle \frac{dZ}{dZ}}={\displaystyle \frac{d\{\ln (P^A)\}}{dZ}}+{\displaystyle \frac{d\{\ln (Q^B)\}}{dZ}}$
$\Rightarrow {\displaystyle \frac{1}{Z}}=A\times {\displaystyle \frac{1}{P}}\times {\displaystyle \frac{dP}{dZ}}+B\times {\displaystyle \frac{1}{Q}}\times {\displaystyle \frac{dQ}{dZ}}$

Multiplying by $dZ$ we get:
${\displaystyle \frac{dZ}{Z}}=\pm \{A\times {\displaystyle \frac{dP}{P}}+B\times {\displaystyle \frac{dQ}{Q}}\}$
We are given that $Z={\displaystyle \frac{1}{A^2}}$
Taking log on both sides:
$\ln (Z)=\ln ({\displaystyle \frac{1}{A^2}})$
Differentiating both sides with respect to $dZ$
${\displaystyle \frac{1}{Z}}\times {\displaystyle \frac{dZ}{dZ}}={\displaystyle \frac{d(A^{-2})}{dZ}}$
$\Rightarrow {\displaystyle \frac{1}{Z}}=\pm \{-2\times {\displaystyle \frac{1}{A}}\times {\displaystyle \frac{dA}{dZ}}\}$
$\Rightarrow {\displaystyle \frac{dZ}{Z}}=\pm (-2\times {\displaystyle \frac{dA}{A}})$
$\therefore {\displaystyle \frac{dZ}{Z}}=2\times {\displaystyle \frac{dA}{A}}$
Therefore the relative error in $Z$ is ${\displaystyle \frac{dZ}{Z}}$ is two times or twice the relative error in $A$ that is ${\displaystyle \frac{dA}{A}}$. Therefore the relative error of the result is the sum of relative errors of the multipliers where the power of the quantity is multiplied by its relative error.

Note:This can also be directly calculated by summing up the relative errors of the multiplied/divided quantities .Also keep in mind to multiply the power/exponent to which the quantity is raised with its relative error.