What is the error analysis for a function of Newton's cooling law?

[dziech]

Hi guys,

I have a silly question, but I seem to be confused about it. Let's say I have a function of Newton's cooling law. I measured the exponential drop of the temperature of some system and now I want to make an error analysis. Do I treat this function as a two variable function of tau (time constant in exponent) and time ? If yes, according to the error analysis I need to take partial derivatives over time and tau. This results in having time as a product in df/dtau result. Shall this be a measured time of the temperature falling to the constant level ?

In equations :

$\frac{df(\tau,t)}{d\tau} = \tau e^{-\tau t}$
$\frac{df(\tau,t)}{dt} = t e^{-\tau t }$

 

[Mark44]

Hi guys,

I have a silly question, but I seem to be confused about it. Let's say I have a function of Newton's cooling law. I measured the exponential drop of the temperature of some system and now I want to make an error analysis. Do I treat this function as a two variable function of tau (time constant in exponent) and time ? If yes, according to the error analysis I need to take partial derivatives over time and tau. This results in having time as a product in df/dtau result. Shall this be a measured time of the temperature falling to the constant level ?

In equations :

$\frac{df(\tau,t)}{d\tau} = \tau e^{-\tau t}$
$\frac{df(\tau,t)}{dt} = t e^{-\tau t }$

Since $\tau$ is a time constant (your words), it's not a variable, so don't differentiate with respect to it. It could be that here $\tau$ is a parameter, a value that can have different values for different scenarios, but isn't considered to be a variable.

 

[dziech]

I asked my question in a wrong way. The uncertainty i want to measure is the one if the time constant. Then it's a function of tine and temperature. The question remains, is the t value that stays in derivative a value over i made my calculations? It makes sense if i take the uncertainty of the arguments and multiply them by time so kind of number of samples.

 

[Mark44]

I asked my question in a wrong way. The uncertainty i want to measure is the one if the time constant.

If the time constant is what? This isn't a complete thought.

Then it's a function of tine and temperature. The question remains, is the t value that stays in derivative a value over i made my calculations?

I don't understand what you're asking.

It makes sense if i take the uncertainty of the arguments and multiply them by time so kind of number of samples.

I don't understand this either.

 

[dziech]

Ok, maybe more context will help : )

I have samples of a solid cooling down. I do data fitting with Levenberg Marquardt algorithm to find $\tau$ - time constant of the exponential function. Now I want to calculate the uncertainty of the time constant $\tau$. Let's put it simple - what are the possibilities to do that ?

 

[Mark44]

OK, that's clearer. The bit about a time constant threw me off, since it's not actually known.

I guess I would approach this as f being a function of both t and $\tau$.

The total differential, df, would be
$$df \approx \frac{\partial f}{\partial \tau}\cdot \Delta \tau + \frac{\partial f}{\partial t} \cdot \Delta t$$

 

[dziech]

Yes, sorry, sometimes I forget to keep things straight forward : )

Ok, great - this is exactly where I got stuck: the derivative with respect to $\tau$ will in the end include time wouldn't it ? So :
$f = T_{ambient} + (T(0)-T_{ambient})e^{-kt}$
Now I keep the t as a constant (if I understand correctly) and :
$\frac{df}{d\tau} \Delta\tau= -t(T(0)-T_{ambient})\Delta\tau$

I am confused what t is in my example ?

But the question still remains how to calculate the $\Delta\tau$. I got this parameter from data fitting. I just wonder if I can turn the function around so that the $\tau(T,t)$ and then calculate $\Delta\tau$