Answer the following question.
Show that if Z = `"A"/"B"`, `(triangle"Z")/"Z" = (triangle"A")/"A" + (triangle "B")/"B"`
Solution
Errors in divisions:
Suppose, Z = `"A"/"B"` and measured values of A and B are (A ± ΔA) and (B ± ΔB) then,
Z ± ΔZ = `("A" +- triangle"A")/("B"+-triangle"B")`
∴ Z`(1 +- (triangle"Z")/"Z") = ("A"[1 +-(Δ"A"//"A")])/("B"[1 +-(Δ"B"//"B")])`
`= "A"/"B" xx [1 +-(Δ"A"//"A")]/[1 +-(Δ"B"//"B")]`
As, `(triangle"B")/"B" ≪ 1,` expanding using Binomial theorem,
`"Z"(1 +- (triangle"Z")/"Z") = "Z" (1 +- (triangle"A")/"A") xx (1 bar+ (Δ"B")/"B")` ....`(because "A"/"B" = "Z")`
∴ `1 +- (Δ"Z")/"Z" = 1 +- (Δ"A")/"A" bar+ (Δ"B")/"B" +- (Δ"A")/"A" xx (Δ"B")/"B"`
Ignoring term `(Δ"A")/"A" xx (Δ"B")/"B", (Δ"Z")/"Z" = +- (Δ"A")/"A" bar+ (Δ"B")/"B"`
This gives four possible values of `(Δ"Z")/"Z"` as
`(+ (Δ"A")/"A" - (Δ"B")/"B"), (+(Δ"A")/"A" + (Δ"B")/"B"), (-(Δ"A")/"A" - (Δ"B")/"B") and (-(Δ"A")/"A" + (Δ"B")/"B")`
∴ Maximum relative error of `(Δ"Z")/"Z" +- ((Δ"A")/"A" + (Δ"B")/"B")`
Thus, when two quantities are divided, the maximum relative error in the result is the sum of relative errors in each quantity.
