Question

If the relative errors of three quantities to be multiplied together are 0.001, 0.008 and 0.007, what is the relative error of the resulting quantity? Give your answer to three decimal places Answer:

Answer 1

Let the quantity be A which is given by

............(1)

Let the quantity be A which is given by A = P times q times r (1) Let the absolute errors in measurement of A, p q, and r be delta A, delta P, delta q, delta r respectively. Hence A plusminus delta A = (p plusminus delta p) times (q plusminus delta q) times (r plusminus delta r) rightdoublearrow A plusminus delta A = pqr plusminus p delta qr plusminus p delta q delta r plusminus pq delta r plusminus delta pqr plusminus delta p delta qr plusminus de;ta p delta q delta r plusminus delta pq delta r plusminus delta A = plusminus p delta qr plusminus p delta q delta r plusminus pq delta r plusminus delta pqr plusminus delta p delta qr plusminus delta p delta q delta r plusminus delta pq delta r (2) dividing equation (2) by equation (1), we get plusminus delta A/A =

Let the absolute errors in measurement of A, p q, and r be $\Delta A, \Delta p , \Delta q, \Delta r$ respectively.

hence,

${A}\pm \Delta A = (p\pm \Delta p) \times (q\pm\Delta q) \times (r\pm \Delta r) \\ \Rightarrow \cancel{A}\pm \Delta A = \cancel{pqr} \pm p \Delta q r \pm p\Delta q \Delta r \pm pq\Delta r \pm \Delta p qr\pm \Delta p \Delta q r \pm \Delta p \Delta q \Delta r \pm \Delta p q \Delta r \\ \Rightarrow \pm \Delta A = \pm p \Delta q r \pm p\Delta q \Delta r \pm pq\Delta r \pm \Delta p qr\pm \Delta p \Delta q r \pm \Delta p \Delta q \Delta r \pm \Delta p q \Delta r$ ........(2)

Dividing equation (2) by equation (1), we get

$\pm \frac{\Delta A}{A} = \pm \frac{\Delta q}{q} \pm \frac{\Delta q \Delta r}{qr} \pm \frac{\Delta r}{r} \pm \frac{\Delta p}{p} \pm \frac{\Delta p \Delta q}{pq} \pm \frac{\Delta p \Delta q \Delta r}{pqr} \pm \frac{\Delta p \Delta r}{pr} \\$

Now, we can write this properly as

$\pm \frac{\Delta A}{A} = \pm \frac{\Delta p}{p} \pm \frac{\Delta q}{q} \pm \frac{\Delta r}{r} \pm\left ( \frac{\Delta p}{p} \times \frac{\Delta q}{q} \right ) \pm \left ( \frac{\Delta q}{q} \times \frac{\Delta r}{r} \right ) \pm \left ( \frac{\Delta p}{p} \times \frac{\Delta r}{r} \right ) \pm \left ( \frac{\Delta p}{p} \times \frac{\Delta q}{q} \times \frac{\Delta r}{r} \right )$ .......(3)

We know that the relative errors are given by

$Relative \ error \ in \ A = \pm \frac{\Delta A}{A}$

$Relative \ error \ in \ p = \pm \frac{\Delta p}{p} = 0.001$

$Relative \ error \ in \ q = \pm \frac{\Delta q}{q} = 0.008$

$Relative \ error \ in \ r = \pm \frac{\Delta r}{r} = 0.007$

hence, we can calculat the relative error in the measurement of the quantity using equation (3)

$\frac{\Delta A}{A} = 0.001 + 0.008+ 0.007 +(0.001\times 0.008) + (0.008\times 0.007) +( 0.001\times 0.007) + (0.001\times 0.008\times 0.007)\\ \Rightarrow \frac{\Delta A}{A} = 0.016 + 0.000008 + 0.000056+ 0.000007+ 0.000000056 = 0.016071056 \\ \Rightarrow \frac{\Delta A}{A} \approx 0.016$

Hence, the relative error in the measurement of the quantity is 0.016.

Note that, all other terms that appear in the relative error calculation dimish because of the product. Only the sum of individual relative errors of 3 quantities is good enough estimate the relative error of product.