Question

If $Z=\frac{A^{4}\,B^{1/3}}{CD^{3/2}}$. and $\Delta A$, $\Delta B$, $\Delta C$, and $\Delta D$ are their absolute errors in $A$, $B$, $C$ and $D$ respectively. The relative error in $Z$ is

• A. $\frac{\Delta Z}{Z}=4 \frac{\Delta A}{A}+\frac{1}{3} \frac{\Delta B}{B}+\frac{\Delta C}{C}+\frac{3}{2} \frac{\Delta D}{D}$
• B. $\frac{\Delta Z}{Z}=4 \frac{\Delta A}{A}+\frac{1}{3} \frac{\Delta B}{B}-\frac{\Delta C}{C}-\frac{3}{2} \frac{\Delta D}{D}$
• C. $\frac{\Delta Z}{Z}=4 \frac{\Delta A}{A}+\frac{1}{3} \frac{\Delta B}{B}+\frac{\Delta C}{C}-\frac{3}{2} \frac{\Delta D}{D}$
• D. $\frac{\Delta Z}{Z}=4 \frac{\Delta A}{A}+\frac{1}{3} \frac{\Delta B}{B}-\frac{\Delta C}{C}+\frac{3}{2} \frac{\Delta D}{D}$

Answer 1

Answer: (A)

$Z=\frac{A^{4}\,B^{1/3}}{CD^{3/2}}$. The relative error in $Z$ is given by
$\frac{\Delta Z}{Z}=4 \frac{\Delta A}{A}+\frac{1}{3} \frac{\Delta B}{B}+\frac{\Delta C}{C}+\frac{3}{2} \frac{\Delta D}{D}$