If Z = frac{A^{2} B^{3}}{C^{4}},then the rela | vedclass

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(C) Given the expression $Z = \frac{A^{2} B^{3}}{C^{4}}$.According to the rules of propagation of errors,for a quantity $Z = \frac{A^p B^q}{C^r}$,the

Vedclass · Accepted Answer

$\frac{2 \Delta A}{A} + \frac{3 \Delta B}{B} + \frac{4 \Delta C}{C}$

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(C) Given the expression $Z = \frac{A^{2} B^{3}}{C^{4}}$.According to the rules of propagation of errors,for a quantity $Z = \frac{A^p B^q}{C^r}$,the relative error is given by $\frac{\Delta Z}{Z} = p \frac{\Delta A}{A} + q \frac{\Delta B}{B} + r \frac{\Delta C}{C}$.Applying this rule to the given expression,where $p=2$,$q=3$,and $r=4$:$\frac{\Delta Z}{Z} = 2 \frac{\Delta A}{A} + 3 \frac{\Delta B}{B} + 4 \frac{\Delta C}{C}$.Therefore,the relative error in $Z$ is $\frac{2 \Delta A}{A} + \frac{3 \Delta B}{B} + \frac{4 \Delta C}{C}$.

If $Z = \frac{A^{2} B^{3}}{C^{4}}$,then the relative error in $Z$ will be

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