Homework Help: Propagation of uncertainties

1. Oct 15, 2008

Oerg

1. The problem statement, all variables and given/known data
Given that $$f=\frac{\bar u \bar v}{\bar u +\bar v}$$

show that

$$e_f=f^2({\frac{e_u}{\bar u^2} + \frac{e_v}{\bar v^2})$$

where $$e$$ refers to the error. ok so I added up the fractional uncertainties

2. Relevant equations

3. The attempt at a solution
and I got this

$$\frac{e_f}{f}=\frac{e_u}{u}+\frac{e_v}{v}+\frac{e_u+e_v}{u+v}$$

after some simplifying, I got to this,

$$e_f=f^2(\frac{e_u(u+v)}{u^2v}+\frac{e_v(u+v)}{v^2u}+\frac{e_u+e_v}{uv})$$

and then I realized that I could never get the answer, however, if this term was negative,
$$\frac{e_u+e_v}{uv}$$, i would get the answer perfectly, but how can it be negative??? Problem is even in division, shouldn't the fractional uncertianties add up??

2. Oct 15, 2008

Oerg

any help?

3. Oct 16, 2008

Oerg

e_f is the maximum error, I have no idea why the term should be negative

4. Oct 16, 2008

atyy

f(u,v)

df=(df/du)du+(df/dv)dv

I didn't notate it properly, but the derivatives are partial derivatives. Treat df as ef, du as eu, dv as ev

5. Oct 16, 2008

Oerg

Im sorry, I dont understand your post. how is my working affected by this equation?

6. Oct 16, 2008

atyy

I don't follow your working at all, so I'm not sure if it's a right way that I don't know about.

Are you familiar with differentiation?

7. Oct 16, 2008

Oerg

yeh but not with partial derivatives,

What I did was to add up the fractional uncertainties.

I tried evaluating your equation, but I arrived at an ugly equation that doesnt fit the final bill.

8. Oct 16, 2008

atyy

OK, that's good. Partial differentiation is exactly like ordinary differentiation. So for examle df(u,v)/du just means normal differentiation of f with respect to u, holding v constant. It may look ugly, but it should work out.

So first evaluate df(u,v)/du.

9. Oct 16, 2008

Oerg

ok, I tried working out the coefficient of e_u according to your equation. Which is the partial derivative of f wrt u.

$$\frac{df}{du}=\frac{-uv}{(u+v)^2}+\frac{v}{u+v}$$

unfortunately, it evaluaes out to be

$$f \times (\frac{v}{u(u+v)})$$

which isn't what the question asked for. Have I differentiated wrongly?

10. Oct 16, 2008

atyy

Try a different rearrangement, maybe take the common denominator to be (u+v)2

11. Oct 16, 2008

Oerg

I have already combined the two terms in my previous post, but i got $$f \times (\frac{v}{u(u+v)})$$ for the coefficient of $$e_u$$ when it should have been
$$\frac{f}{u^2}$$

12. Oct 16, 2008

Oerg

ok sorry, i got the answer, i have to factorise another f out again. THANKS FOR YOUR TIME, IM VERY VERY GRATEFUL!!!!

13. Oct 16, 2008

atyy

Wait - you have to do it for df(u,v)/dv also!

14. Oct 16, 2008

Oerg

yeh it's the same for the other term, it should evaluate out to be the same.

15. Oct 16, 2008

atyy

And you understand what I meant in #4?

16. Oct 16, 2008

Oerg

yeh i do,

df=partial derivative of f wrt u times du + partial derivative of f wrt v times dv

\

from df du and dv to ef eu and ev is an approximation for a small change.

17. Oct 16, 2008

atyy

OK, great (assuming what I said is correct )