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Answer the following · Q10

Q.If the measured values of two quantities are A ± ΔA and B ± ΔB, ΔA and ΔB being the mean absolute errors. What is the maximum possible error in A ± B? Show that if Z=A/BZ = A/B, then ΔZZ=ΔAA+ΔBB\dfrac{\Delta Z}{Z} = \dfrac{\Delta A}{A} + \dfrac{\Delta B}{B}.

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Step 1. Maximum error in A ± B. Let Z=A+BZ = A + B (or A−BA - B). Writing Z±ΔZ=(A±ΔA)±(B±ΔB)Z \pm \Delta Z = (A \pm \Delta A) \pm (B \pm \Delta B) and expanding, ±ΔZ=±ΔA±ΔB\pm\Delta Z = \pm\Delta A \pm \Delta B in every case. The four sign combinations give at most ΔZ=ΔA+ΔB\Delta Z = \Delta A + \Delta B, so the maximum possible absolute error in A±BA \pm B is ΔA+ΔB\Delta A + \Delta B — the individual absolute errors simply add, regardless of whether A and B are added or subtracted.

Step 2. Proof for Z = A/B. Let Z=A/BZ = A/B, i.e. A=ZBA = ZB. Since AA is now a product of ZZ and BB, the product-error rule (Equation 1.7) applies directly: ΔAA=ΔZZ+ΔBB\dfrac{\Delta A}{A} = \dfrac{\Delta Z}{Z} + \dfrac{\Delta B}{B}. …

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