Question

If $Z=\frac{A^4{B}^{1/3}}{C{D}^{3/2}}$ 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 then

• 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

The correct option is A $\frac{\Delta Z}{Z}=4\frac{\Delta A}{A}+\frac{1}{3}\frac{\Delta B}{B}+\frac{\Delta C}{C}+\frac{3}{2}\frac{\Delta D}{D}$

Given:

The value of Z is expressed as: $Z=\frac{A^4{B}^{1/3}}{C{D}^{3/2}}$

The relative error in Z is given as the sum of all the relative errors.
$\frac{\Delta Z}{Z}=4\frac{\Delta A}{A}+\frac{1}{3}\frac{\Delta B}{B}+\frac{\Delta C}{C}+\frac{3}{2}\frac{\Delta D}{D}$

Option $A$ is correct.