Absolute Relative Error and Graphs

[Procrastinate]

I'm calculating absolute relative error at the moment for a practical I did. However, I've hit a brick when attempting to find a general relative error I can use because I need to gradient to find something else which also requires a relative error calculation.

I have frequency (+/- 0.0005) graphed against stopping potential (+/- 0..000005) and obviously because both of them differ, the relative percentages are very different. How do I find the general absolute relative error from a gradient.

I'm sorry if I didn't explain it that well but I can't think of a better way to convey what I'm trying to say.

 

[Feldoh]

Absolute relative error? Typical we talk about either

relative error: Error = \frac{|(Exp - Act)|}{Act}

OR

absolute error: Error = |Exp - Actual|

However I think what you're trying to do is something similar the derivative method of uncertainty. Say you have a function you're trying to find experimentally g(f,v) where there is error in both f (\delta_{f}) and v (\delta_{v}) from measurements. Then the error in g is:

\sqrt{\delta_{gf}^2 + \delta_{gv}^2}

Where:

\delta_{gf} = |\frac{\partial g}{\partial f}| * \delta_{f}

\delta_{gv} = |\frac{\partial g}{\partial v}| * \delta_{v}

 

[Procrastinate]

I'm a bit new to derivatives, could you please show me a quick example? Thanks.

 

[diazona]

For example, say you have measured distance x and time t (each with its own uncertainty) and you're trying to compute the average velocity v. The formula for average velocity is
v = \frac{x}{t}
So to figure out the uncertainty in average velocity, you would first calculate the contribution from the uncertainty in distance:
\delta_{vx} = \left|\frac{\partial v}{\partial x}\right| \delta_x = \frac{1}{t} \delta_x
and the contribution from the uncertainty in time:
\delta_{vt} = \left|\frac{\partial v}{\partial t}\right| \delta_t = \frac{x}{t^2} \delta_t
(As for how exactly you figure out those derivatives: when taking the derivative with respect to distance, you treat time as a constant, so you are basically taking the derivative of a constant times x. That's just the constant. When taking the derivative with respect to time, you treat distance as a constant, so you are taking the derivative of the function constant/t, and that's -(constant)/t²... but the absolute value removes the negative sign.)

Having done that, just square those two quantities, add them up, and take the square root to get the overall uncertainty in velocity,
\delta_v = \sqrt{\delta_{vx}^2 + \delta_{vt}^2} = \sqrt{\left(\frac{1}{t} \delta_x\right)^2 + \left(\frac{x}{t^2} \delta_t\right)^2}
And of course plug in whatever numbers you actually measured.