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

[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.

 

[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:
addition and subtraction: add absolute errors quadratically
multiplication and division: add relative errors quadratically
squares: double relative error, square root: halve relative error

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

 

[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 .

 

[BvU]

Squaring the r value will result in the uncertainty being doubled

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.