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

We are given $$Z = \frac{A^2 B^3}{C^4}$$ and need to find the relative error in $$Z$$.

Apply the error propagation rule: For a quantity expressed as a product or quotient of powers, the relative error is found by taking the sum of the absolute values of (power multiplied by relative error) for each variable.

If $$Z = \frac{A^a \cdot B^b}{C^c}$$, then the relative error in $$Z$$ is:

$$\frac{\Delta Z}{Z} = a \cdot \frac{\Delta A}{A} + b \cdot \frac{\Delta B}{B} + c \cdot \frac{\Delta C}{C}$$

Substitute the powers from the given expression: Here $$a = 2$$, $$b = 3$$, and $$c = 4$$. Substituting these values:

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

Note that in error propagation, all terms are added with positive signs because errors always accumulate in the worst case, regardless of whether the variable appears in the numerator or denominator.

The correct answer is Option A.