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Write rule for error in result due to multiplication and division |
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Answer» Solution :Suppose two physical quantities A and B have measured values `A+-DELTAA,B +-DeltaB` respectively. Where `DeltaA and DeltaB` are their absolute errors. For a PRODUCT : Suppose the product of A and B is Z and the absolute error in Z is `DeltaZ`. `:.Z= AB` `:.Z+-DeltaZ=(A+-DeltaA)(B+-DeltaB)` `:.Z+DeltaZ=AB+-DeltaB+-BDeltaA+_(DeltaA DeltaB)` Dividing the left side by 2 and right side by AB, `1+-(DeltaZ)/(Z)=1+(DeltaA)/(A)+-(DeltaB)/(B)+-((DeltaA)/(A))((DeltaB)/(B))` `(DeltaA)/(A) and (DeltaB)/(B)` are very small hence their product will be very small and can be neglected. Neglecting negative sign for maximum error `(DeltaZ)/(Z)=(DeltaA)/(A)+(DeltaB)/(B)` For a Quotient : Suppose the quotient of A and B is Z and the absolute error of Z is `DeltaZ` `:.Z=(A)/(B)` `:.Z+DeltaZ=(A+-DeltaA)/(B+-DeltaB)` `Z(1+-(DeltaZ)/(Z))==(A(1+-(DeltaA)/(A)))/(B(1+-(DeltaB)/(B)))` `(1+(DeltaB))/(B))^(-1)` writing first two terms of binomial expansion, `(1+-(DeltaB)/(B))^(-1)=(1-+-(DeltaB)/(B))` From EQUATION (1), `:.1+-(DeltaZ)/(Z)=(1+(DeltaA)/(A)) (1+-(DeltaB)/(B))` `:.1+-(DeltaZ)/(Z)=1+(DeltaA)/(A)+-(DeltaB)/(B)+-((DeltaA)/(A))((DeltaB)/(B))` But `((DeltaA)/(A)) and ((DeltaB)/(B))` are smaller, so their product will be zero by ignoring them, `+(DeltaZ)/(Z)=+-(DeltaA)/(A)+-(DeltaB)/(B)` For relaive errros in Z, ignoring, regative signs `(DeltaZ)/(Z)=(DeltaA)/(A)+(DeltaB)/(B)` Hence the rule : When two quantities are multiplied or divided, the relative errorin the RESULT is the sm of the relative errors in the multipliers. |
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