# Question 302c8

Jun 5, 2016

13%?

#### Explanation:

An experiment measures quantities a, b, c and x is calculated from the following relation.

$x = \frac{a {b}^{2}}{c} ^ 3$

To deal with this type of problem we are to take 'log' of the expression. Here we get

$\log x = \log a + 2 \log b - 3 \log c$

On differentiation this becomes

$\frac{\Delta x}{x} = \frac{\Delta a}{a} + 2 \frac{\Delta b}{b} - 3 \frac{\Delta c}{c}$

Now the maximum % of error is calculated on the basis of fluctuation obtained in this relation.
Only thing we are to remember is that only absolute value of error is to be taken for calculation of maximum error.

So maximum % of error will be

((Deltax)/x)_"max"%=(Deltaa)/a%+2(Deltab)/b%+3(Deltac)/c%X

Now given
(Deltaa)/a%=1%;(Deltab)/b%=3%;(Deltac)/c%=2%

So

((Deltax)/x)_"max"%=(1+2*3+3*2)%=13%#