Question

Show that when two quantities are divided the relative error on the result is the sum of relative error in the quotient

Answer 1

Explanation:

let A and B be the true value and ∆A and ∆ B are the errors in them

let R=A/B

R±∆R=(A±∆A)÷(B±∆B)

R±∆R=(A±∆A)(B±∆B)-¹

R±∆R=A(1±∆A/A)B-¹(1±∆B/B)-¹

since ∆B/B is very small so using binomial theorem we write (1±∆B/B)-¹ as

1±(-1)(∆B/B)

R±∆R=A(1±∆A/A)B-¹(1±∆B/B)

R±∆R=(A/B)(1±∆A/A)(1±∆B/B)

R±∆R=R(1±∆A/A)(1±∆B/B)

1±∆R/R=1±(∆B/B)±(∆A/A)±(∆A∆B/AB)

since ∆A∆B/AB is very small so we can neglect it

±∆R/R=±∆A/A±∆B/B

∆R/R=∆A/A+∆B/B