Looking for the Formula for Uncertainty for Velocity In a lab I'm working on, we used a formula for uncertainty of area:
$$\sqrt{\left(\frac{δl}{l}\right)^2+\left(\frac{δw}{w}\right)^2}$$
We are asked to now find uncertainty for velocity (m/s) and the hint was to use the same formula above, but I'm not sure how.  This is my only guess, but it seems odd and incorrect to me:
$$\sqrt{\left(\frac{δm}{m}\right)^2+\left(\frac{δs}{s}\right)^2}$$
Any help would be greatly appreciated.  Thanks.
 A: Here's what you do:
Start with the area formula (no explanation needed):
$$ A = lw $$
Add the derivatives in quadrature:
$$ (dA)^2 = (l\cdot dw)^2 + (w \cdot dl)^2 $$
Use the area formula's inverses:
$$ (dA)^2 = \left(\frac A w dw\right)^2 + \left(\frac A l \cdot dl\right)^2 $$
Divide by $A^2$ and take the square root:
$$ \frac{dA}A = \sqrt{\left(\frac{dw}w\right)^2 + \left(\frac{dl}l\right)^2 }$$
So that checks out. This is now in a "relative error form"--that is, e.g., a 10% error in length gives a 10% error in area--makes sense.
So if I do it in meter-Hertz:
$$ v = xf $$
(distance $x$ times frequency $f$). I can safely say:
$$ dv = \sqrt{\left(\frac{dx}x\right)^2 + \left(\frac{df}f\right)^2 }$$
Then I just need to convert $df/f$ to a seconds based result--since it's a relative error, I suspect it should work--a 10% error in frequency should have the same (up to sign and to 1st order) effect as a 10% error in time--and we don't care about sign because we're adding in quadrature:
$$ f = \frac 1 T $$
$$ df = \frac{-dT}{T^2} $$
so that
$$ \frac{df}{f} = \frac{\frac{-dT}{T^2}}{\frac 1 T} = -\frac{dT} T$$
Your professor was right.
A: The error formula for a product and for a ratio is the same if you write it in terms of relative errors. 
The general combination of uncorrelated errors formula for $f(x,y)$ is
$\sigma_f^2=\left({\partial f \over \partial x}\right)^2 \sigma_x^2+\left({\partial f \over \partial y}\right)^2 \sigma_y^2$
For the special case $f=xy$ this is $\sigma_f^2 = y^2 \sigma_x^2+x^2 \sigma_y^2$, which can be made to look prettier by dividing throughout by $f$ (using $f=xy$ on the RHS)
$\left({\sigma_f\over f}\right)^2 =\left({\sigma_x\over x}\right)^2 +\left({\sigma_y\over y}\right)^2 $
For the special case $f=x/y$ the formula gives
$\sigma_f^2=\left({1 \over y}\right)^2 \sigma_x^2+\left(-{x\over y^2}\right)^2 \sigma_y^2$
which looks horrible, but that minus sign goes away as it is squared, and if you again divide by $f$ on the $LHS$ and $x/y$ on the RHS what emerges is
$\left({\sigma_f\over f}\right)^2 =\left({\sigma_x\over x}\right)^2 +\left({\sigma_y\over y}\right)^2 $
the same as for the product. Which is nice - but is (I think) just a coincidence with no deep meaning, though it is handy. Other forms (sums, differences, powers) do not fit this pattern.
