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Question: If \(Z = \frac{A^{4}B^{1/3}}{CD^{3/2}}\) and \(\Delta A,\Delta B,and\Delta D\)are their absolute err...

If Z=A4B1/3CD3/2Z = \frac{A^{4}B^{1/3}}{CD^{3/2}} and ΔA,ΔB,andΔD\Delta A,\Delta B,and\Delta Dare their absolute errors in A, B, C and D respectively. The relative error in Z is

A

ΔZZ=4ΔAA+13ΔBB+ΔCC+32ΔDD\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

ΔZZ=4ΔAA+13ΔBB−ΔCC−32ΔDD\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

ΔZZ=4ΔAA+13ΔBB+ΔCC−32ΔDD\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

ΔZZ=4ΔAA+13ΔBB−ΔCC+32ΔDD\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

ΔZZ=4ΔAA+13ΔBB+ΔCC+32ΔDD\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}

Explanation

Solution

Z=A4B1/3CD3/2Z = \frac{A^{4}B^{1/3}}{CD^{3/2}}

The relative error in Z is given by

ΔZZ=4ΔAA+13ΔBB+ΔCC+32ΔDD\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}