Homework Help: What happens to uncertainty when I divide values by a constant and squ

1. Apr 11, 2014

Akbar123

The values below are different wire thicknesses measured with a micrometer screw gauge. The uncertainty of it is +/- 0.01mm.

Diameter/thickness of wire/mm
0.02
0.10
0.14
0.30
0.42

I need to halve (or divide by two) these values to calculate radius and hence calculate cross sectional area of the wires. What happens to the uncertainty of +/- 0.01mm.

In addition what happens to the uncertainty of the radius when I square these values using the equation area of circle = πr2 ?

This is for a physics investigation.

2. Apr 11, 2014

BvU

You calculate cross section with $a = {\pi\over 4} d^2$.
Propagation of independent uncertainties works as follows:
$$\bigl ( \Delta f(x_1, x_2, x_3 ... x_N) \bigr )^2 = \Sigma_{i=1}^N \ ({\partial f \over \partial x_i})^2 (\Delta x_i)^2$$
Examples:
$f = a\,x \quad \Rightarrow \quad \Delta f = |a|\,\Delta x \quad$ or: $\quad {\Delta f \over f} = |a|\,{\Delta x\over x}$
$f = x_1 + x_2 \quad \Rightarrow \quad (\Delta f)^2 = (\Delta x_1)^2 + (\Delta x_2)^2$
$f = x_1 \times x_2 \quad \Rightarrow \quad (\Delta f)^2 = x_2^2 (\Delta x_1)^2 + x_1^2(\Delta x_2)^2 \quad \Rightarrow \quad ({\Delta f\over f})^2 = ({\Delta x_1 \over x_1})^2 + ({\Delta x_2 \over x_2})^2$
but $f = x_1^2 \quad \Rightarrow \quad (\Delta f)^2 = 4 x_1^2 (\Delta x_1)^2 \quad \Rightarrow \quad ({\Delta f\over f}) = 2 ({\Delta x_1 \over x_1})$

In short:
squares: double relative error, square root: halve relative error

In your case ${\Delta a \over a }= 2 {\Delta d \over d}$

3. Apr 11, 2014

patrickmoloney

You would also divide the uncertainty (or error) by 2. If you make the measurement smaller, you also make the associated uncertainty with that measurement smaller, in this case x2 smaller. Squaring the r value will result in the uncertainty being doubled. So your $r^2$ will actually just be +/- 0.01 mm uncertainty. If it was the diameter squared, the uncertainty would be doubled, so +/- 0.02 mm .

4. Apr 11, 2014

BvU

Demonstrably incorrect: $(10 \pm 1)^2 = 100 \pm 20 \quad$(*) and not $100 \pm 2$

What he means is that the relative error is doubled.

(*) here you see the effect of taking only the first derivative and ignoring the higher orders. It's a small effect, given that this shows up marginally (namely 1%) even at a 10% relative error.