Explain the propagation of errors in subtraction, quotient and power of a quantity.

Explanation

(i) Error in the difference of two qunatities:
Let `DeltaA and DeltaB` be the absolute errors in the two quantities, A and B, respectively Then,
Measured vlaue of A
`=Apm Delta`
Measured value of B
`=Bpm DeltaB`
Difference is, `Z=A-B`
The error `DeltaZ` in Z is then given by `Z pm DeltaZ`
`=(Apm DeltaA)-(B pm DeltaB)`
`=(A0B)pm(DeltaA+DeltaB)`
`=Zpm (DeltaA+DeltaB)`
(or) `DeltaZ=DeltaA+DeltaB`
(ii) Error in thedivision or quotient of two quantities : Let `DeltaA and Delta B` be the absolute error in the two quantities A and B respectively.
Consider the quotient,
`Z=(A)/(B)`
The error `DeltaZ` in Z is given by
`Zpm DeltaZ=(Apm DeltaA)/(Bpm DeltaB)=(A(1pm (DeltaA)/(A)))/(B(1pm(DeltaB)/(B)))`
`=(A)/(B)(1pm(Delta)/(A))(1pm (DeltaB)/(B))^(-1)`
`or Z pm DeltaZ=Z(1pm(DeltaA)/(A))(1pm(DeltaB)/(B))`
`["using "(1+x)^(n)~~1+nx," when "x lt lt1]`
Dividing both sides by Z, we get,
`1pm (DeltaZ)/(Z)=(1pm (DeltaA)/(A))(1pm (DeltaB)/(B))`
`=1pm (DeltaA)/(A)pm (DeltaB)/(B)pm(DeltaA)/(A).(DeltaB)/(B)`
As the terms `DeltaA//A and DeltaB//B` are samll, their product term can be neglected.
The maximum fractional error in Z is given by `(DeltaZ)/(Z)=((DeltaA)/(A)+(DeltaB)/(B))`
(iii) Error in the power of a quantity : Consider the `n^("th")` powero of `A, Z=A^(n)` The error `DeltaZ` in Z is given by
`Zpm DeltaZ=(Apm DeltaA)^(n)=A^(n)(1pm (DeltaA)/(A))^(n)`
`=Z(1pm n(DeltaA)/(A))`
We get `[(1+x)^(n)~~1+nx,` when `x lt lt 1]` neglecting remaining terms, Dividing both sides by Z.
`1pm (DeltaZ)/(Z)=1pm n(DeltaA)/(A)or (DeltaZ)/(Z)=n(DeltaA)/(A)`